- Commit
- 9bbab87c16919043a2db660c996645b7bbfd745e
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- 7abeda5bc9d129a3946bb2cb668dcf4cae777625
- Author
- Pablo <pablo-escobar@riseup.net>
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Renamed some files
lie-algebras-and-their-representations
Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
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5 files changed, 2657 insertions, 2657 deletions
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| Added | sections/fin-dim-simple.tex | 1 file changed | 1044 | 0 |
| Deleted | sections/mathieu.tex | 1 file changed | 0 | 1611 |
| Deleted | sections/semisimple-algebras.tex | 1 file changed | 0 | 1044 |
| Added | sections/simple-weight.tex | 1 file changed | 1611 | 0 |
| Modified | tcc.tex | 2 files changed | 2 | 2 |
diff --git /dev/null b/sections/fin-dim-simple.tex @@ -0,0 +1,1044 @@ +\chapter{Finite-Dimensional Simple Modules} + +In this chapter we classify the finite-dimensional simple +\(\mathfrak{g}\)-modules for a finite-dimensional semisimple Lie algebra +\(\mathfrak{g}\) over \(K\). At the heart of our analysis of +\(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) was the decision to consider +the eigenspace decomposition +\begin{equation}\label{sym-diag} + M = \bigoplus_\lambda M_\lambda +\end{equation} + +This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the +rational behind it and the reason why equation (\ref{sym-diag}) holds are +harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace +decomposition associated with an operator \(M \to M\) is a very well-known +tool, and readers familiarized with basic concepts of linear algebra should be +used to this type of argument. On the other hand, the eigenspace decomposition +of \(M\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h} +\subset \mathfrak{gl}(M)\) is neither well-known nor does it hold in general: +as indicated in the previous chapter, it may very well be that +\[ + \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda \subsetneq M +\] + +We should note, however, that these two cases are not as different as they may +sound at first glance. Specifically, we can regard the eigenspace decomposition +of a \(\mathfrak{sl}_2(K)\)-module \(M\) with respect to the eigenvalues of the +action of \(h\) as the eigenvalue decomposition of \(M\) with respect to the +action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\). +Furthermore, in both cases \(\mathfrak{h} \subset \mathfrak{sl}_n(K)\) is the +subalgebra of diagonal matrices, which is Abelian. The fundamental difference +between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) for +\(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim \mathfrak{h} > 1\) for +\(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The question then is: why did we +choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for +\(\mathfrak{sl}_3(K)\)? + +The rational behind fixing an Abelian subalgebra \(\mathfrak{h}\) is a simple +one: we have seen in the previous chapter that representations of Abelian +algebras are generally much simpler to understand than the general case. Thus +it make sense to decompose a given \(\mathfrak{g}\)-module \(M\) of into +subspaces invariant under the action of \(\mathfrak{h}\), and then analyze how +the remaining elements of \(\mathfrak{g}\) act on these subspaces. The bigger +\(\mathfrak{h}\) is, the simpler our problem gets, because there are fewer +elements outside of \(\mathfrak{h}\) left to analyze. + +Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} +\subset \mathfrak{g}\), which leads us to the following definition. + +\begin{definition}\index{Cartan subalgebra} + A subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan + subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in + \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in + \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\), + \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if + it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in + \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former + two properties. +\end{definition} + +\begin{proposition} + There exists a Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{g}\). +\end{proposition} + +\begin{proof} + Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose + elements act as diagonal operators via the adjoint \(\mathfrak{g}\)-module. + Indeed, \(0\), the only element of \(0 \subset \mathfrak{g}\), is such that + \(\operatorname{ad}(0) = 0\) is a diagonalizable operator. Furthermore, given + a chain of Abelian subalgebras + \[ + 0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots + \] + such that \(\operatorname{ad}(H)\) is a diagonal operator for each \(H \in + \mathfrak{h}_i\), the subalgebra \(\bigcup_i \mathfrak{h}_i \subset + \mathfrak{g}\) is Abelian, and its elements also act diagonally in + \(\mathfrak{g}\). It then follows from Zorn's Lemma that there exists a + subalgebra \(\mathfrak{h}\) which is maximal with respect to both these + properties, also known as a Cartan subalgebra. +\end{proof} + +We have already seen some concrete examples. Namely\dots + +\begin{example}\label{ex:cartan-of-gl} + The Lie subalgebra + \[ + \mathfrak{h} = + \begin{pmatrix} + K & 0 & \cdots & 0 \\ + 0 & K & \cdots & 0 \\ + \vdots & \vdots & \ddots & \vdots \\ + 0 & 0 & \cdots & K + \end{pmatrix} + \subset \mathfrak{gl}_n(K) + \] + of diagonal matrices is a Cartan subalgebra. + Indeed, every pair of diagonal matrices commutes, so that \(\mathfrak{h}\) + is an Abelian -- and hence nilpotent -- subalgebra. A + simple calculation also shows that if \(i \ne j\) then the coefficient of + \(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\) + in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i}, + X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is + self-normalizing. +\end{example} + +\begin{example} + Let \(\mathfrak{h}\) be as in Example~\ref{ex:cartan-of-gl}. Then the + subalgebra \(\mathfrak{h} \cap \mathfrak{sl}_n(K)\) of traceless diagonal + matrices is a Cartan subalgebra of \(\mathfrak{sl}_n(K)\). +\end{example} + +\begin{example}\label{ex:cartan-direct-sum} + Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras and + \(\mathfrak{h}_i \subset \mathfrak{g}_i\) be Cartan subalgebras. Then + \(\mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan subalgebra of + \(\mathfrak{g}_1 \oplus \mathfrak{g}_2\). +\end{example} + +\index{Cartan subalgebra!simultaneous diagonalization} +The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the +subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of +\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the +subalgebras described the previous chapter. The remaining question then is: if +\(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(M\) is a +\(\mathfrak{g}\)-module, does the eigenspace decomposition +\[ + M = \bigoplus_\lambda M_\lambda +\] +of \(M\) hold? The answer to this question turns out to be yes. This is a +consequence of something known as \emph{simultaneous diagonalization}, which is +the primary tool we will use to generalize the results of the previous section. +What is simultaneous diagonalization all about then? + +\begin{definition}\label{def:sim-diag} + Given a \(K\)-vector space \(V\), a set of operators \(\{T_j : V \to V\}_j\) + is called \emph{simultaneously diagonalizable} if there is a basis \(\{v_1, + \ldots, v_n\}\) for \(V\) such that \(T_j v_i\) is a scalar multiple of + \(v_i\), for all \(i, j\). +\end{definition} + +\begin{proposition} + Given a \emph{finite-dimensional} vector space \(V\), a set of diagonalizable + operators \(V \to V\) is simultaneously diagonalizable if, and only if all of + its elements commute with one another. +\end{proposition} + +We should point out that simultaneous diagonalization \emph{only works in the +finite-dimensional setting}. In fact, simultaneous diagonalization is usually +framed as an equivalent statement about diagonalizable \(n \times n\) matrices. +Simultaneous diagonalization implies that to show \(M = \bigoplus_\lambda +M_\lambda\) it suffices to show that \(H\!\restriction_M : M \to M\) is a +diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we +introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract +Jordan decomposition of a semisimple Lie algebra}. + +\begin{proposition}[Jordan] + Given a finite-dimensional vector space \(V\) and an operator \(T : V \to + V\), there are unique commuting operators \(T_{\operatorname{ss}}, + T_{\operatorname{nil}} : V \to V\), with \(T_{\operatorname{ss}}\) + diagonalizable and \(T_{\operatorname{nil}}\) nilpotent, such that \(T = + T_{\operatorname{ss}} + T_{\operatorname{nil}}\). The pair + \((T_{\operatorname{ss}}, T_{\operatorname{nil}})\) is known as \emph{the Jordan + decomposition of \(T\)}. +\end{proposition} + +\begin{proposition}\index{abstract Jordan decomposition} + Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are + \(X_{\operatorname{ss}}, X_{\operatorname{nil}} \in \mathfrak{g}\) such that \(X + = X_{\operatorname{ss}} + X_{\operatorname{nil}}\), \([X_{\operatorname{ss}}, + X_{\operatorname{nil}}] = 0\), \(\operatorname{ad}(X_{\operatorname{ss}})\) is a + diagonalizable operator and \(\operatorname{ad}(X_{\operatorname{nil}})\) is a + nilpotent operator. The pair \((X_{\operatorname{ss}}, X_{\operatorname{nil}})\) + is known as \emph{the Jordan decomposition of \(X\)}. +\end{proposition} + +It should be clear from the uniqueness of +\(\operatorname{ad}(X)_{\operatorname{ss}}\) and +\(\operatorname{ad}(X)_{\operatorname{nil}}\) that the Jordan decomposition of +\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = +\operatorname{ad}(X_{\operatorname{ss}}) + +\operatorname{ad}(X_{\operatorname{nil}})\). What is perhaps more remarkable is +the fact this holds for \emph{any} finite-dimensional \(\mathfrak{g}\)-module. +In other words\dots + +\begin{proposition}\label{thm:preservation-jordan-form} + Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(X + \in \mathfrak{g}\). Denote by \(X\!\restriction_M\) the action of \(X\) on + \(M\). Then \(X_{\operatorname{ss}}\!\restriction_M = + (X\!\restriction_M)_{\operatorname{ss}}\) and + \(X_{\operatorname{nil}}\!\restriction_M = + (X\!\restriction_M)_{\operatorname{nil}}\). +\end{proposition} + +This last result is known as \emph{the preservation of the Jordan form}, and a +proof can be found in appendix C of \cite{fulton-harris}. As promised this +implies\dots + +\begin{corollary}\label{thm:finite-dim-is-weight-mod} + Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset + \mathfrak{g}\) be a Cartan subalgebra and \(M\) be any finite-dimensional + \(\mathfrak{g}\)-module. Then there is a basis \(\{m_1, \ldots, + m_r\}\) of \(M\) so that each \(m_i\) is simultaneously an eigenvector of all + elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as + a diagonal matrix in this basis. In other words, there are linear functionals + \(\lambda_i \in \mathfrak{h}^*\) so that + \( + H \cdot m_i = \lambda_i(H) m_i + \) + for all \(H \in \mathfrak{h}\). In particular, + \[ + M = \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda + \] +\end{corollary} + +\begin{proof} + Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_M + : M \to M\) is a diagonalizable operator. + + If we write \(H = H_{\operatorname{ss}} + H_{\operatorname{nil}}\) for the + abstract Jordan decomposition of \(H\), we know + \(\operatorname{ad}(H_{\operatorname{ss}}) = + \operatorname{ad}(H)_{\operatorname{ss}}\). But \(\operatorname{ad}(H)\) is a + diagonalizable operator, so that \(\operatorname{ad}(H)_{\operatorname{ss}} = + \operatorname{ad}(H)\). This implies + \(\operatorname{ad}(H_{\operatorname{nil}}) = + \operatorname{ad}(H)_{\operatorname{nil}} = 0\), so that + \(H_{\operatorname{nil}}\) is a central element of \(\mathfrak{g}\). Since + \(\mathfrak{g}\) is semisimple, \(H_{\operatorname{nil}} = 0\). + Proposition~\ref{thm:preservation-jordan-form} then implies + \((H\!\restriction_M)_{\operatorname{nil}} = + H_{\operatorname{nil}}\!\restriction_M = 0\), so \(H\!\restriction_M = + (H\!\restriction_M)_{\operatorname{ss}}\) is a diagonalizable operator. +\end{proof} + +We should point out that this last proof only works for semisimple Lie +algebras. This is because we rely heavily on +Proposition~\ref{thm:preservation-jordan-form}, as well in the fact that +semisimple Lie algebras are centerless. In fact, +Corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie +algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the +Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a +\(\mathfrak{g}\)-module is simply a vector space \(M\) endowed with an operator +\(M \to M\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on +\(M\). In particular, if we choose an operator \(M \to M\) which is \emph{not} +diagonalizable we find \(M \ne \bigoplus_{\lambda \in \mathfrak{h}^*} +M_\lambda\). + +However, Corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive +\(\mathfrak{g}\) if we assume that the \(\mathfrak{g}\)-module \(M\) in +question is simple, since central elements of \(\mathfrak{g}\) act on simple +\(\mathfrak{g}\)-modules as scalar operators. The hypothesis of +finite-dimensionality is also of huge importance. For instance, consider\dots + +\begin{example}\label{ex:regular-mod-is-not-weight-mod} + Let \(\mathcal{U}(\mathfrak{g})\) denote the regular \(\mathfrak{g}\)-module. + Notice that \(\mathcal{U}(\mathfrak{g})_\lambda = 0\) for all \(\lambda \in + \mathfrak{h}^*\). Indeed, since \(\mathcal{U}(\mathfrak{g})\) is a domain, if + \((H - \lambda(H)) u = 0\) for some nonzero \(H \in \mathfrak{h}\) then \(u = + 0\). In particular, + \[ + \bigoplus_{\lambda \in \mathfrak{h}^*} \mathcal{U}(\mathfrak{g})_\lambda + = 0 \neq \mathcal{U}(\mathfrak{g}) + \] +\end{example} + +As a first consequence of Corollary~\ref{thm:finite-dim-is-weight-mod} we +show\dots + +\begin{corollary} + The restriction of the Killing form \(\kappa\) to \(\mathfrak{h}\) is + non-degenerate. +\end{corollary} + +\begin{proof} + Consider the root space decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus + \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint + \(\mathfrak{g}\)-module, where \(\alpha\) ranges over all nonzero eigenvalues + of the adjoint action of \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = + \mathfrak{h}\). + + Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h}) + \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the + other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in + \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) -- + i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigenspace decomposition + becomes + \[ + \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha + \] + + We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to + \(\mathfrak{g}_\alpha\) with respect to \(\kappa\) for any \(\alpha \ne 0\). + Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\) + with \(\alpha(H_1) \ne 0\) we have + \[ + \alpha(H_1) \cdot \kappa(X, H_2) + = \kappa([H_1, X], H_2) + = - \kappa([X, H_1], H_2) + = - \kappa(X, [H_1, H_2]) + = 0 + \] + + Hence the non-degeneracy of \(\kappa\) implies the non-degeneracy of its + restriction. +\end{proof} + +We should point out that the restriction of \(\kappa\) to \(\mathfrak{h}\) is +\emph{not} the Killing form of \(\mathfrak{h}\). In fact, since +\(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is +hardly ever a non-degenerate form. + +\begin{note} + Since \(\kappa\) induces an isomorphism \(\mathfrak{h} \isoto + \mathfrak{h}^*\), it induces a bilinear form \((\kappa(X, \cdot), \kappa(Y, + \cdot)) \mapsto \kappa(X, Y)\) in \(\mathfrak{h}^*\). As in + section~\ref{sec:sl3-reps}, we denote this form by \(\kappa\) as well. +\end{note} + +We now have most of the necessary tools to reproduce the results of the +previous chapter in a general setting. Let \(\mathfrak{g}\) be a +finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\) +and let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. We will +proceed, as we did before, by generalizing the results of the previous two +sections in order. By now the pattern should be starting to become clear, so we +will mostly omit technical details and proofs analogous to the ones on the +previous sections. Further details can be found in appendix D of +\cite{fulton-harris} and in \cite{humphreys}. + +\section{The Geometry of Roots and Weights} + +We begin our analysis, as we did for \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\), by investigating the locus of roots of and weights of +\(\mathfrak{g}\). Throughout chapter~\ref{ch:sl3} we have seen that the weights +of any given finite-dimensional module of \(\mathfrak{sl}_2(K)\) or +\(\mathfrak{sl}_3(K)\) can only assume very rigid configurations. For instance, +we have seen that the roots of \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\) are symmetric with respect to the origin. In this +chapter we will generalize most results from chapter~\ref{ch:sl3} regarding the +rigidity of the geometry of the set of weights of a given module. + +As for the aforementioned result on the symmetry of roots, this turns out to be +a general fact, which is a consequence of the non-degeneracy of the restriction +of the Killing form to the Cartan subalgebra. + +\begin{proposition}\label{thm:weights-symmetric-span} + The roots \(\alpha\) of \(\mathfrak{g}\) are symmetrical about the origin -- + i.e. \(- \alpha\) is also a root -- and they span all of \(\mathfrak{h}^*\). +\end{proposition} + +\begin{proof} + We will start with the first claim. Let \(\alpha\) and \(\beta\) be two + roots. Notice \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset + \mathfrak{g}_{\alpha + \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and + \(Y \in \mathfrak{g}_\beta\) then + \[ + [H [X, Y]] + = [X, [H, Y]] - [Y, [H, X]] + = (\alpha + \beta)(H) \cdot [X, Y] + \] + for all \(H \in \mathfrak{h}\). + + This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) + \operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then + \[ + (\operatorname{ad}(X) \operatorname{ad}(Y))^r Z + = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ] + \in \mathfrak{g}_{r \alpha + r \beta + \gamma} + = 0 + \] + for \(r\) large enough. In particular, \(\kappa(X, Y) = + \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if + \(- \alpha\) is not an eigenvalue we find \(\kappa(X, \mathfrak{g}_\beta) = + 0\) for all roots \(\beta\), which contradicts the non-degeneracy of + \(\kappa\). Hence \(- \alpha\) must be an eigenvalue of the adjoint action of + \(\mathfrak{h}\). + + For the second statement, note that if the roots of \(\mathfrak{g}\) do not + span all of \(\mathfrak{h}^*\) then there is some nonzero \(H \in + \mathfrak{h}\) such that \(\alpha(H) = 0\) for all roots \(\alpha\), which is + to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in + \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of + the center \(\mathfrak{z} = 0\) of \(\mathfrak{g}\), a contradiction. +\end{proof} + +Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and +\(\mathfrak{sl}_3(K)\) one can show\dots + +\begin{proposition}\label{thm:root-space-dim-1} + The root spaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional. +\end{proposition} + +The proof of the first statement of +Proposition~\ref{thm:weights-symmetric-span} highlights something interesting: +if we fix some eigenvalue \(\alpha\) of the adjoint action of \(\mathfrak{h}\) +on \(\mathfrak{g}\) and a eigenvector \(X \in \mathfrak{g}_\alpha\), then for +each \(H \in \mathfrak{h}\) and \(m \in M_\lambda\) we find +\[ + H \cdot (X \cdot m) + = X H \cdot m + [H, X] \cdot m + = (\lambda + \alpha)(H) X \cdot m +\] + +Thus \(X\) sends \(m\) to \(M_{\lambda + \alpha}\). We have encountered this +formula twice in these notes: again, we find \(\mathfrak{g}_\alpha\) \emph{acts +on \(M\) by translating vectors between eigenspaces}. In particular, if we +denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) then\dots + +\begin{theorem}\label{thm:weights-congruent-mod-root}\index{weights!root lattice} + The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) are + all congruent modulo the root lattice \(Q = \mathbb{Z} \Delta\) of + \(\mathfrak{g}\). In other words, all weights of \(M\) lie in the same + \(Q\)-coset \(\xi \in \mfrac{\mathfrak{h}^*}{Q}\). +\end{theorem} + +Again, we may leverage our knowledge of \(\mathfrak{sl}_2(K)\) to obtain +further restrictions on the geometry of the locus of weights of \(M\). Namely, +as in the case of \(\mathfrak{sl}_3(K)\) we show\dots + +\begin{proposition}\label{thm:distinguished-subalgebra} + Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace + \(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha} + \oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra + isomorphic to \(\mathfrak{sl}_2(K)\). +\end{proposition} + +\begin{corollary}\label{thm:distinguished-subalg-rep} + For all weights \(\mu\), the subspace + \[ + \bigoplus_k M_{\mu - k \alpha} + \] + is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\) + and the weight spaces in this string match the eigenspaces of \(h\). +\end{corollary} + +The proof of Proposition~\ref{thm:distinguished-subalgebra} is very technical +in nature and we won't include it here, but the idea behind it is simple: +recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both +\(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- +\alpha}]\) is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- +\alpha}] \ne 0\) and that no generator of \([\mathfrak{g}_\alpha, +\mathfrak{g}_{- \alpha}]\) is annihilated by \(\alpha\), so that by adjusting +scalars we can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in +\mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\) +satisfies +\begin{align*} + [H_\alpha, F_\alpha] & = -2 F_\alpha & + [H_\alpha, E_\alpha] & = 2 E_\alpha +\end{align*} + +The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely +determined by this condition, but \(H_\alpha\) is. As promised, the second +statement of Corollary~\ref{thm:distinguished-subalg-rep} imposes strong +restrictions on the weights of \(M\). Namely, if \(\lambda\) is a weight, +\(\lambda(H_\alpha)\) is an eigenvalue of \(h\) on some +\(\mathfrak{sl}_2(K)\)-module, so it must be an integer. In other words\dots + +\begin{definition}\label{def:weight-lattice}\index{weights!weight lattice} + The lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) \in + \mathbb{Z} \, \forall \alpha \in \Delta \} \subset \mathfrak{h}^*\) is called + \emph{the weight lattice of \(\mathfrak{g}\)}. We call the elements of \(P\) + \emph{integral}. +\end{definition} + +\begin{proposition}\label{thm:weights-fit-in-weight-lattice} + The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) of + all lie in the weight lattice \(P\). +\end{proposition} + +Proposition~\ref{thm:weights-fit-in-weight-lattice} is clearly analogous to +Corollary~\ref{thm:sl3-weights-fit-in-weight-lattice}. In fact, the weight +lattice of \(\mathfrak{sl}_3(K)\) -- as in Definition~\ref{def:weight-lattice} +-- is precisely \(\mathbb{Z} \langle \alpha_1, \alpha_2, \alpha_3 \rangle\). To +proceed further, we would like to take \emph{the highest weight of \(M\)} as in +section~\ref{sec:sl3-reps}, but the meaning of \emph{highest} is again unclear +in this situation. We could simply fix a linear function \(\mathbb{Q} P \to +\mathbb{Q}\) -- as we did in section~\ref{sec:sl3-reps} -- and choose a weight +\(\lambda\) of \(M\) that maximizes this functional, but at this point it is +convenient to introduce some additional tools to our arsenal. These tools are +called \emph{basis}. + +\begin{definition}\label{def:basis-of-root}\index{weights!basis} + A subset \(\Sigma = \{\beta_1, \ldots, \beta_r\} \subset \Delta\) of linearly + independent roots is called \emph{a basis for \(\Delta\)} if, given \(\alpha + \in \Delta\), there are \(k_1, \ldots, k_r \in \mathbb{N}\) such that + \(\alpha = \pm(k_1 \beta_1 + \cdots + k_r \beta_r)\). +\end{definition} + +The interesting thing about basis for \(\Delta\) is that they allow us to +compare weights of a given \(\mathfrak{g}\)-module. At this point the reader +should be asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly +all that intuitive. Well, the thing is that any choice of basis induces a +partial order in \(Q\), where elements are ordered by their \emph{heights}. + +\begin{definition}\index{weights!orderings of roots} + Let \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) be a basis for \(\Delta\). + Given \(\alpha = k_1 \beta_1 + \cdots + k_r \beta_r \in Q\) with \(k_1, + \ldots, k_r \in \mathbb{Z}\), we call the number \(\operatorname{ht}(\alpha) + = k_1 + \cdots + k_r \in \mathbb{Z}\) \emph{the height of \(\alpha\)}. We say + that \(\alpha \preceq \beta\) if \(\operatorname{ht}(\alpha) \le + \operatorname{ht}(\beta)\). +\end{definition} + +\begin{definition} + Given a basis \(\Sigma\) for \(\Delta\), there is a canonical + partition\footnote{Notice that $\operatorname{ht}(\alpha) = 0$ if, and only + if $\alpha = 0$. Since $0$ is, by definition, not a root, the sets $\Delta^+$ + and $\Delta^-$ account for all roots.} \(\Delta^+ \cup \Delta^- = \Delta\), + where \(\Delta^+ = \{ \alpha \in \Delta : \alpha \succ 0 \}\) and \(\Delta^- + = \{ \alpha \in \Delta : \alpha \prec 0 \}\). The elements of \(\Delta^+\) + and \(\Delta^-\) are called \emph{positive} and \emph{negative roots}, + respectively. +\end{definition} + +\begin{definition} + Let \(\Sigma\) be a basis for \(\Delta\). The subalgebra \(\mathfrak{b} = + \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is + called \emph{the Borel subalgebra associated with \(\mathfrak{h}\) and + \(\Sigma\)}. +\end{definition} + +It should be obvious that the binary relation \(\preceq\) in \(Q\) is a partial +order. In addition, we may compare the elements of a given \(Q\)-coset +\(\lambda + Q\) by comparing their difference with \(0 \in Q\). In other words, +given \(\lambda \in \mu + Q\), we say \(\lambda \preceq \mu\) if \(\lambda - +\mu \preceq 0\). In particular, since the weights of \(M\) all lie in a single +\(Q\)-coset, we may compare them in this way. Given a basis \(\Sigma\) for +\(\Delta\) we may take ``the highest weight of \(M\)'' as a maximal weight +\(\lambda\) of \(M\). The obvious question then is: can we always find a basis +for \(\Delta\)? + +\begin{proposition} + There is a basis \(\Sigma\) for \(\Delta\). +\end{proposition} + +The intuition behind the proof of this proposition is similar to our original +idea of fixing a direction in \(\mathfrak{h}^*\) in the case of +\(\mathfrak{sl}_3(K)\). Namely, one can show that \(\kappa(\alpha, \beta) \in +\mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form +\(\kappa\) restricts to a nondegenerate \(\mathbb{Q}\)-bilinear form +\(\mathbb{Q} \Delta \times \mathbb{Q} \Delta \to \mathbb{Q}\). We can then fix +a nonzero vector \(\gamma \in \mathbb{Q} \Delta\) and consider the orthogonal +projection \(f : \mathbb{Q} \Delta \to \mathbb{Q} \gamma \cong \mathbb{Q}\). We +say a root \(\alpha \in \Delta\) is \emph{positive} if \(f(\alpha) > 0\), and +we call a positive root \(\alpha\) \emph{simple} if it cannot be written as the +sum two other positive roots. The subset \(\Sigma \subset \Delta\) of all +simple roots is a basis for \(\Delta\), and all other basis can be shown to +arise in this way. + +Fix some basis \(\Sigma\) for \(\Delta\), with corresponding decomposition +\(\Delta^+ \cup \Delta^- = \Delta\). Let \(\lambda\) be a maximal weight of +\(M\). We call \(\lambda\) \emph{the highest weight of \(M\)}, and we call any +nonzero \(m \in M_\lambda\) \emph{a highest weight vector}. The strategy then +is to describe all weight spaces of \(M\) in terms of \(\lambda\) and \(m\), as +in Theorem~\ref{thm:sl3-irr-weights-class}. Unsurprisingly we do so by +reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). + +First, we note that any highest weight vector \(m \in M_\lambda\) is +annihilated by all positive root spaces, for if \(\alpha \in \Delta^+\) then +\(E_\alpha \cdot m \in M_{\lambda + \alpha}\) must be zero -- or otherwise we +would have that \(\lambda + \alpha\) is a weight with \(\lambda \prec \lambda + +\alpha\). In particular, +\[ + \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k \alpha} + = \bigoplus_{k \in \mathbb{N}} M_{\lambda - k \alpha} +\] +and \(\lambda(H_\alpha)\) is the right-most eigenvalue of the action of \(h\) +on the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k M_{\lambda - k \alpha}\). + +This has a number of important consequences. For instance\dots + +\begin{corollary} + If \(\alpha \in \Delta^+\) and \(\sigma_\alpha : \mathfrak{h}^* \to + \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to + \(\alpha\) with respect to the Killing form, the weights of \(M\) occurring + in the line joining \(\lambda\) and \(\sigma_\alpha\) are precisely the \(\mu + \in P\) lying between \(\lambda\) and \(\sigma_\alpha(\lambda)\). +\end{corollary} + +\begin{proof} + Notice that any \(\mu \in P\) in the line joining \(\lambda\) and + \(\sigma_\alpha(\lambda)\) has the form \(\mu = \lambda - k \alpha\) for some + \(k\), so that \(M_\mu\) corresponds the eigenspace associated with the + eigenvalue \(\lambda(H_\alpha) - 2k\) of the action of \(h\) on \(\bigoplus_k + M_{\lambda - k \alpha}\). If \(\mu\) lies between \(\lambda\) and + \(\sigma_\alpha(\lambda)\) then \(k\) lies between \(0\) and + \(\lambda(H_\alpha)\), in which case \(M_\mu \neq 0\) and therefore \(\mu\) + is a weight. + + On the other hand, if \(\mu\) does not lie between \(\lambda\) and + \(\sigma_\alpha(\lambda)\) then either \(k < 0\) or \(k > + \lambda(H_\alpha)\). Suppose \(\mu\) is a weight. In the first case \(\mu + \succ \lambda\), a contradiction. On the second case the fact that \(M_\mu + \ne 0\) implies \(M_{\lambda + (k - \lambda(H_\alpha)) \alpha} = + M_{\sigma_\alpha(\mu)} \ne 0\), which contradicts the fact that \(M_{\lambda + + \ell \alpha} = 0\) for all \(\ell \ge 0\). +\end{proof} + +This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we +found that the weights of the simple \(\mathfrak{sl}_3(K)\)-modules formed +continuous strings symmetric with respect to the lines \(K \alpha\) with +\(\kappa(\alpha_i - \alpha_j, \alpha) = 0\). As in the case of +\(\mathfrak{sl}_3(K)\), the same class of arguments leads us to the +conclusion\dots + +\begin{definition}\index{Weyl group} + We refer to the group \(W = \langle \sigma_\alpha : \alpha \in + \Delta^+ \rangle \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl + group of \(\mathfrak{g}\)}. +\end{definition} + +\begin{theorem}\label{thm:irr-weight-class} + The weights of a simple \(\mathfrak{g}\)-module \(M\) with highest weight + \(\lambda\) are precisely the elements of the weight lattice \(P\) congruent + to \(\lambda\) modulo the root lattice \(Q\) lying inside the convex hull of + the orbit of \(\lambda\) under the action of the Weyl group \(W\). +\end{theorem} + +\index{Weyl group!actions} +Aside from showing up in the previous theorem, the Weyl group will also play an +important role in chapter~\ref{ch:mathieu} by virtue of the existence of a +canonical action of \(W\) on \(\mathfrak{h}\). By its very nature, +\(W\) acts in \(\mathfrak{h}^*\). If we conjugate the action +\(\sigma\!\restriction_{\mathfrak{h}^*} : \mathfrak{h}^* \isoto +\mathfrak{h}^*\) of some \(\sigma \in W\) by the isomorphism +\(\mathfrak{h}^* \isoto \mathfrak{h}\) afforded by the restriction of the +Killing for to \(\mathfrak{h}\) we get a linear automorphism +\(\sigma\!\restriction_{\mathfrak{h}} : \mathfrak{h} \isoto \mathfrak{h}\). As +it turns out, the \(\sigma\!\restriction_{\mathfrak{h}}\) can be extended to an +automorphism of Lie algebras \(\mathfrak{g} \isoto \mathfrak{g}\). This +translates into the following results, which we do not prove -- but see +\cite[sec.~14.3]{humphreys}. + +\begin{proposition}\label{thm:weyl-group-action} + Given \(\alpha \in \Delta^+\), let\footnote{Notice that since $\mathfrak{g}$ + is finite-dimensional, $\operatorname{ad}(X)$ is nilpotent for each root + vector $X \in \mathfrak{g}$, so that the linear automorphism + $e^{\operatorname{ad}(X)} = \operatorname{Id} + \operatorname{ad}(X) + + \frac{\operatorname{ad}(X)^2}{2!} + \cdots : \mathfrak{g} \isoto + \mathfrak{g}$ is well defined.} \(\tilde{\sigma}_\alpha = + e^{\operatorname{ad}(E_\alpha)} e^{- \operatorname{ad}(F_\alpha)} + e^{\operatorname{ad}(E_\alpha)} : \mathfrak{g} \isoto \mathfrak{g}\). Then + \(\tilde\sigma_\alpha\) is an automorphism of Lie algebras, and this defines + an action of \(W\) on \(\mathfrak{g}\) which is compatible with the + canonical action of \(W\) on \(\mathfrak{h}\) -- i.e. + \(\tilde\sigma\!\restriction_{\mathfrak{h}} = + \sigma\!\restriction_{\mathfrak{h}}\) for all \(\sigma \in W\). +\end{proposition} + +\begin{note} + Notice that the action of \(W\) on \(\mathfrak{g}\) from + Proposition~\ref{thm:weyl-group-action} is not canonical, since it depends on + the choice of \(E_\alpha\) and \(F_\alpha\). Nevertheless, \(\mathfrak{h}\) + is stable under the action of \(W\) -- i.e. \(W \cdot + \mathfrak{h} \subset \mathfrak{h}\) -- and the restriction of this action to + \(\mathfrak{h}\) is independent of any choices. +\end{note} + +We should point out that the results in this section regarding the geometry +roots and weights are only the beginning of a well develop axiomatic theory of +the so called \emph{root systems}, which was used by Cartan in the early 20th +century to classify all finite-dimensional simple complex Lie algebras in terms +of Dynking diagrams. This and much more can be found in \cite[III]{humphreys} +and \cite[ch.~21]{fulton-harris}. Having found all of the weights of \(M\), the +only thing we are missing for a complete classification is an existence and +uniqueness theorem analogous to Theorem~\ref{thm:sl2-exist-unique} and +Theorem~\ref{thm:sl3-existence-uniqueness}. This will be the focus of the next +section. + +\section{Verma Modules \& the Highest Weight Theorem} + +It is already clear from the previous discussion that if \(\lambda\) is the +highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots +\(\alpha\). In other words, having \(\lambda(H_\alpha) \ge 0\), for all +\(\alpha \in \Delta^+\), is a necessary condition for the existence of a simple +\(\mathfrak{g}\)-module with highest weight given by \(\lambda\). Surprisingly, +this condition is also sufficient. In other words\dots + +\begin{definition}\index{weights!dominant weight}\index{weights!integral weight} + An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all + \(\alpha \in \Delta^+\) is referred to as an \emph{dominant integral weight + of \(\mathfrak{g}\)}. +\end{definition} + +\begin{theorem}\label{thm:dominant-weight-theo} + For each dominant integral \(\lambda \in P\) there exists precisely one + finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest weight + is \(\lambda\). +\end{theorem} + +\index{weights!Highest Weight Theorem} This is known as \emph{the Highest +Weight Theorem}, and its proof is the focus of this section. The ``uniqueness'' +part of the theorem follows at once from the arguments used for +\(\mathfrak{sl}_3(K)\). However, the ``existence'' part of the theorem is more +nuanced. + +Our first instinct is, of course, to try to generalize the proof used for +\(\mathfrak{sl}_3(K)\). The issue is that our proof relied heavily on our +knowledge of the roots of \(\mathfrak{sl}_3(K)\). It is thus clear that we need +a more systematic approach for the general setting. We begin by asking a +simpler question: how can we construct \emph{any} -- potentially +infinite-dimensional -- \(\mathfrak{g}\)-module \(M\) of highest weight +\(\lambda\)? In the process of answering this question we will come across a +surprisingly elegant solution to our problem. + +If \(M\) is a module with highest weight vector \(m^+ \in M_\lambda\), we +already know \(H \cdot m^+ = \lambda(H) m^+\) for all \(\mathfrak{h}\) and \(X +\cdot m^+ = 0\) for \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). If +\(M\) is simple we find \(M = \mathcal{U}(\mathfrak{g}) \cdot m^+\), which +implies the restriction of \(M\) to the Borel subalgebra \(\mathfrak{b} \subset +\mathfrak{g}\) has a prescribed action. On the other hand, we have essentially +no information about the action of the rest of \(\mathfrak{g}\) on \(M\). +Nevertheless, given a \(\mathfrak{b}\)-module we may obtain a +\(\mathfrak{g}\)-module by formally extending the action of \(\mathfrak{b}\) +via induction. This leads us to the following definition. + +\begin{definition}\label{def:verma}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules} + The \(\mathfrak{g}\)-module \(M(\lambda) = + \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\), where the action of + \(\mathfrak{b}\) on \(K m^+\) is given by \(H \cdot m^+ = \lambda(H) m^+\) + for all \(H \in \mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in + \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma + module of weight \(\lambda\)}. +\end{definition} + +It turns out that \(M(\lambda)\) enjoys many of the features we've grown used +to in the past chapters. Explicitly\dots + +\begin{proposition}\label{thm:verma-is-weight-mod} + The Verma module \(M(\lambda)\) is generated \(m^+ = 1 \otimes m^+ \in + M(\lambda)\) as in Definition~\ref{def:verma}. The weight spaces + decomposition + \[ + M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu + \] + holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in + \mathfrak{h}^*\) and \(\dim M(\lambda)_\lambda = 1\). Finally, \(\lambda\) is + the highest weight of \(M(\lambda)\), with highest weight vector given by + \(m^+ \in M(\lambda)\). +\end{proposition} + +\begin{proof} + The PBW Theorem implies that \(M(\lambda)\) is spanned by the vectors + \(F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+\) for + \(\Delta^+ = \{\alpha_1, \ldots, \alpha_r\}\) and \(F_{\alpha_i} \in + \mathfrak{g}_{- \alpha_i}\) as in the proof of + Proposition~\ref{thm:distinguished-subalgebra}. But + \[ + \begin{split} + H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+) + & = ([H, F_{\alpha_{i_1}}] + F_{\alpha_{i_1}} H) + F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ + & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ + + F_{\alpha_{i_1}} ([H, F_{\alpha_{i_2}}] + F_{\alpha_{i_2}} H) + F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ + & \;\; \vdots \\ + & = (- \alpha_{i_1} - \cdots - \alpha_{i_s})(H) + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ + + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m^+ \\ + & = (\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s})(H) + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ + & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ + \in M(\lambda)_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}} + \end{split} + \] + + Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*} + M(\lambda)_\mu\), as desired. In fact we have established + \[ + M(\lambda) + \subset + \bigoplus_{k_i \in \mathbb{N}} + M(\lambda)_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r} + \] + where \(\{\alpha_1, \ldots, \alpha_r\} = \Delta^+\), so that all weights of + \(M(\lambda)\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots - + k_r \cdot \alpha_r\). + + This already gives us that the weights of \(M(\lambda)\) are bounded by + \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that + \(m^+\) is nonzero weight vector. Clearly \(m^+ \in M_\lambda\). The + Poincaré-Birkhoff-Witt Theorem implies \(\mathcal{U}(\mathfrak{g})\) is a + free \(\mathfrak{b}\)-module, so that + \[ + M(\lambda) + \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right) + \otimes_{\mathcal{U}(\mathfrak{b})} K m^+ + \cong \bigoplus_i \mathcal{U}(\mathfrak{b}) + \otimes_{\mathcal{U}(\mathfrak{b})} K m^+ + \cong \bigoplus_i K m^+ + \ne 0 + \] + as \(\mathfrak{b}\)-modules. We then conclude \(m^+ \ne 0\) in + \(M(\lambda)\), for if this was not the case we would find \(M(\lambda) = + \mathcal{U}(\mathfrak{g}) \cdot m^+ = 0\). Hence \(M_\lambda \ne 0\) and + therefore \(\lambda\) is the highest weight of \(M(\lambda)\), with highest + weight vector \(m^+\). + + To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only + finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots + F_{\alpha_s}^{k_s}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots + + k_s \cdot \alpha_s\). Since \(M(\lambda)_\mu\) is spanned by the images of + \(m^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In + particular, there is a single monomials \(F_{\alpha_1}^{k_1} + F_{\alpha_2}^{k_2} \cdots F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda + + k_1 \cdot \alpha_1 + \cdots + k_s \cdot \alpha_s\) -- which is, of course, + the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\). +\end{proof} + +\begin{example}\label{ex:sl2-verma} + If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K + h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in + \mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) = + \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action of + \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by the formulas in + (\ref{eq:sl2-verma-formulas}). Visually, + \begin{center} + \begin{tikzcd} + \cdots \rar[bend left=60]{-10} + & M(\lambda)_{-6} \rar[bend left=60]{-4} \lar[bend left=60]{1} + & M(\lambda)_{-4} \rar[bend left=60]{0} \lar[bend left=60]{1} + & M(\lambda)_{-2} \rar[bend left=60]{2} \lar[bend left=60]{1} + & M(\lambda)_0 \rar[bend left=60]{2} \lar[bend left=60]{1} + & M(\lambda)_2 \lar[bend left=60]{1} + \end{tikzcd} + \end{center} + where \(M(\lambda)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows + represent the action of \(e\) and the bottom arrows represent the action of + \(f\). The scalars labeling each arrow indicate to which multiple of \(f^{k + \pm 1} \cdot m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The + string of weight spaces to the left of the diagram is infinite. + \begin{equation}\label{eq:sl2-verma-formulas} + \begin{aligned} + f^k \cdot m^+ & \overset{e}{\mapsto} (2 - k(k + 1)) f^{k - 1} \cdot m^+ & + f^k \cdot m^+ & \overset{f}{\mapsto} f^{k + 1} \cdot m^+ & + f^k \cdot m^+ & \overset{h}{\mapsto} (2 - 2k) f^k \cdot m^+ & + \end{aligned} + \end{equation} +\end{example} + +The Verma module \(M(\lambda)\) should really be though-of as ``the freest +highest weight \(\mathfrak{g}\)-module of weight \(\lambda\)''. Unfortunately +for us, this is not a proof of Theorem~\ref{thm:dominant-weight-theo}, since in +general \(M(\lambda)\) is neither simple nor finite-dimensional. Indeed, the +dimension of \(M(\lambda)\) is the same as the codimension of +\(\mathcal{U}(\mathfrak{b})\) in \(\mathcal{U}(\mathfrak{g})\), which is always +infinite. Nevertheless, we may use \(M(\lambda)\) to prove +Theorem~\ref{thm:dominant-weight-theo} as follows. + +Given a \(\mathfrak{g}\)-module \(M\), any \(\mathfrak{g}\)-homomorphism \(f : +M(\lambda) \to M\) is determined by the image of \(m^+\). Indeed, \(f(u \cdot +m^+) = u \cdot f(m^+)\) for all \(u \in \mathcal{U}(\mathfrak{g})\). In +addition, it is clear that +\[ + H \cdot f(m^+) = f(H \cdot m^+) = f(\lambda(H) m^+) = \lambda(H) f(m^+) +\] +for all \(H \in \mathfrak{h}\) and, similarly, \(X \cdot f(m^+) = 0\) for all +\(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). This leads us to the +universal property of \(M(\lambda)\). + +\begin{definition} + Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M\). If \(X \cdot m = 0\) + for all \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\), then \(m\) is + called \emph{a singular vector of \(M\)}. +\end{definition} + +\begin{proposition} + Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M_\lambda\) be a singular + vector. Then there exists a unique \(\mathfrak{g}\)-homomorphism \(f : + M(\lambda) \to M\) such that \(f(m^+) = m\). Furthermore, all homomorphisms + \(M(\lambda) \to M\) are given in this fashion. + \[ + \operatorname{Hom}_{\mathfrak{g}}(M(\lambda), M) + \cong \{ m \in M_\lambda : m \ \text{is singular}\} + \] +\end{proposition} + +\begin{proof} + The result follows directly from Proposition~\ref{thm:frobenius-reciprocity}. + Indeed, by the Frobenius Reciprocity Theorem, a \(\mathfrak{g}\)-homomorphism + \(f : M(\lambda) \to M\) is the same as a \(\mathfrak{b}\)-homomorphism \(g : + K m^+ \to M = \operatorname{Res}_{\mathfrak{b}}^{\mathfrak{g}} M\). More + specifically, given a \(\mathfrak{b}\)-homomorphism \(g : K m^+ \to M\), + there exists a unique \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) + such that \(f(u \otimes m) = u \cdot g(m)\) for all \(m \in K m^+\), and all + \(\mathfrak{g}\)-homomorphism \(M(\lambda) \to M\) arise in this fashion. + + Any \(K\)-linear map \(g : K m^+ \to M\) is determined by the image + \(g(m^+)\) of \(m^+\) and such an image is a singular vector if, and only if + \(g\) is a \(\mathfrak{b}\)-homomorphism. +\end{proof} + +Notice that any highest weight vector is a singular vector. Now suppose \(M\) +is a simple finite-dimensional \(\mathfrak{g}\)-module of highest weight vector +\(m \in M_\lambda\). By the last proposition, there is a +\(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) = m\). +Since \(M\) is simple, \(M = \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore +\(M \cong \mfrac{M(\lambda)}{\ker f}\). It then follows from the simplicity of +\(M\) that \(\ker f \subset M(\lambda)\) is a maximal +\(\mathfrak{g}\)-submodule. Maximal submodules of Verma modules are thus of +primary interest to us. As it turns out, these can be easily classified. + +\begin{proposition}\label{thm:max-verma-submod-is-weight} + Every submodule \(N \subset M(\lambda)\) is the direct sum of its weight + spaces. In particular, \(M(\lambda)\) has a unique maximal submodule + \(N(\lambda)\) and a unique simple quotient \(L(\lambda) = + \sfrac{M(\lambda)}{N(\lambda)}\). +\end{proposition} + +\begin{proof} + Let \(N \subset M(\lambda)\) be a submodule and take any nonzero \(n \in N\). + Because of Proposition~\ref{thm:verma-is-weight-mod}, we know there are + \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in + M(\lambda)_{\mu_i}\) such that \(n = m_1 + \cdots + m_r\). We want to show + \(m_i \in N\) for all \(i\). + + Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\). + Then + \[ + m_1 + - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_3 + - \cdots + - \frac{(\mu_r - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_r + = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n + \in N + \] + + Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By + applying the same procedure again we get + \begin{multline*} + m_1 + - + \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)} + {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_4 + - \cdots - + \frac{(\mu_r - \mu_3)(H_3) \cdot (\mu_r - \mu_1)(H_2)} + {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_r \\ + = + \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right) + \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n + \in N + \end{multline*} + + By applying the same procedure over and over again we can see that \(m_1 = u + \cdot n \in N\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if + we reproduce all this for \(m_2 + \cdots + m_r = n - m_1 \in N\) we get that + \(m_2 \in N\). All in all we find \(m_1, \ldots, m_r \in N\). Hence + \[ + N = \bigoplus_\mu N_\mu = \bigoplus_\mu M(\lambda)_\mu \cap N + \] + + Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\), if \(N\) is a + proper submodule then \(m^+ \notin N\). Hence any proper submodule lies in + the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum + \(N(\lambda)\) of all such submodules is still proper. This implies + \(N(\lambda)\) is the unique maximal submodule of \(M(\lambda)\) and + \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique simple + quotient. +\end{proof} + +\begin{example}\label{ex:sl2-verma-quotient} + If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we + can see from Example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge + 3} K f^k \cdot m^+\), so that \(L(\lambda)\) is the \(3\)-dimensional simple + \(\mathfrak{sl}_2(K)\)-module -- i.e. the finite-dimensional simple module + with highest weight \(\lambda\) constructed in chapter~\ref{ch:sl3}. +\end{example} + +All its left to prove the Highest Weight Theorem is verifying that the +situation encountered in Example~\ref{ex:sl2-verma-quotient} holds for any +\(\lambda \in P\). In other words, we need to show\dots + +\begin{proposition}\label{thm:verma-is-finite-dim} + If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then + the unique simple quotient \(L(\lambda)\) of \(M(\lambda)\) is + finite-dimensional. +\end{proposition} + +The proof of Proposition~\ref{thm:verma-is-finite-dim} is very technical and we +won't include it here, but the idea behind it is to show that the set of +weights of \(L(\lambda)\) is stable under the natural action of the Weyl group +\(W\) on \(\mathfrak{h}^*\). One can then show that the every weight +of \(L(\lambda)\) is conjugate to a single dominant integral weight of +\(L(\lambda)\), and that the set of dominant integral weights of \(L(\lambda)\) +is finite. Since \(W\) is finitely generated, this implies the set of +weights of the unique simple quotient of \(M(\lambda)\) is finite. But +each weight space is finite-dimensional. Hence so is the simple quotient +\(L(\lambda)\). + +We refer the reader to \cite[ch. 21]{humphreys} for further details. We are now +ready to prove the Highest Weight Theorem. + +\begin{proof}[Proof of Theorem~\ref{thm:dominant-weight-theo}] + We begin with the ``existence'' part of the theorem by showing that + \(L(\lambda)\) is indeed a finite-dimensional simple module whose + highest-weight is \(\lambda\). It suffices to show that the highest weight of + \(L(\lambda)\) is \(\lambda\). We have already seen that \(m^+ \in + M(\lambda)_\lambda\) is a highest weight vector. Now since \(m^+\) lies + outside of the maximal submodule of \(M(\lambda)\), the projection \(m^+ + + N(\lambda) \in L(\lambda)\) is nonzero. + + We now claim that \(m^+ + N(\lambda) \in L(\lambda)_\lambda\). Indeed, + \[ + H \cdot (m^+ + N(\lambda)) + = H \cdot m^+ + N(\lambda) + = \lambda(H) (m^+ + N(\lambda)) + \] + for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of + \(L(\lambda)\), with weight vector \(m^+ + N(\lambda)\). Finally, we remark + that \(\lambda\) is the highest weight of \(L(\lambda)\), for if this was not + the case we could find a weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ + \lambda\). + + Now suppose \(M\) is some other finite-dimensional simple + \(\mathfrak{g}\)-module with highest weight vector \(m \in M_\lambda\). By + the universal property of the Verma module, there is a + \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) = + m\). As indicated before, since \(M\) is simple, \(M = + \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore \(f\) is surjective. It + then follows \(M \cong \mfrac{M(\lambda)}{\ker f}\). + + Since \(M\) is simple, \(\ker f \subset M(\lambda)\) is maximal and therefore + \(\ker f = N(\lambda)\). In other words, \(M \cong \mfrac{M(\lambda)}{\ker f} + = L(\lambda)\). We are done. +\end{proof} + +We should point out that Proposition~\ref{thm:verma-is-finite-dim} fails for +non-dominant \(\lambda \in P\). While \(\lambda\) is always a maximal weight of +\(M(\lambda)\), one can show that if \(\lambda \in P\) is not dominant then +\(N(\lambda) = 0\) and \(M(\lambda)\) is simple. For instance, if +\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto -2\) then the +action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by +\begin{center} + \begin{tikzcd} + \cdots \rar[bend left=60]{-20} + & M(\lambda)_{-8} \rar[bend left=60]{-12} \lar[bend left=60]{1} + & M(\lambda)_{-6} \rar[bend left=60]{-6} \lar[bend left=60]{1} + & M(\lambda)_{-4} \rar[bend left=60]{-2} \lar[bend left=60]{1} + & M(\lambda)_{-2} \lar[bend left=60]{1} + \end{tikzcd}, +\end{center} +so we can see that \(M(\lambda)\) has no proper submodules. Verma modules can +thus serve as examples of infinite-dimensional simple modules. Our next +question is: what are \emph{all} the infinite-dimensional simple +\(\mathfrak{g}\)-modules?
diff --git a/sections/mathieu.tex /dev/null @@ -1,1611 +0,0 @@ -\chapter{Simple Weight Modules}\label{ch:mathieu} - -In this chapter we will expand our results on finite-dimensional simple modules -of semisimple Lie algebras by considering \emph{infinite-dimensional} -\(\mathfrak{g}\)-modules, which introduces numerous complications to our -analysis. - -For instance, in the infinite-dimensional setting we can no longer take -complete-reducibility for granted. Indeed, we have seen that even if -\(\mathfrak{g}\) is a semisimple Lie algebra, there are infinite-dimensional -\(\mathfrak{g}\)-modules which are not semisimple. For a counterexample look no -further than Example~\ref{ex:regular-mod-is-not-semisimple}: the regular -\(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) is never semisimple. -Nevertheless, for simplicity -- or shall we say \emph{semisimplicity} -- we -will focus exclusively on \emph{semisimple} \(\mathfrak{g}\)-modules. Our -strategy is, once again, that of classifying simple modules. The regular -\(\mathfrak{g}\)-module hides further unpleasant surprises, however: recall -from Example~\ref{ex:regular-mod-is-not-weight-mod} that -\[ - \bigoplus_\lambda \mathcal{U}(\mathfrak{g})_\lambda - = 0 - \subsetneq \mathcal{U}(\mathfrak{g}) -\] -and the weight space decomposition fails for \(\mathcal{U}(\mathfrak{g})\). - -Indeed, our proof of the weight space decomposition in the finite-dimensional -case relied heavily in the simultaneous diagonalization of commuting operators -in a finite-dimensional space. Even if we restrict ourselves to simple modules, -there is still a diverse spectrum of counterexamples to -Corollary~\ref{thm:finite-dim-is-weight-mod} in the infinite-dimensional -setting. For instance, any \(\mathfrak{g}\)-module \(M\) whose restriction to -\(\mathfrak{h}\) is a free module satisfies \(M_\lambda = 0\) for all -\(\lambda\) as in Example~\ref{ex:regular-mod-is-not-weight-mod}. These are -called \emph{\(\mathfrak{h}\)-free \(\mathfrak{g}\)-modules}, and rank \(1\) -simple \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules where first -classified by Nilsson in \cite{nilsson}. Dimitar's construction of the so -called \emph{exponential tensor \(\mathfrak{sl}_n(K)\)-modules} in -\cite{dimitar-exp} is also an interesting source of counterexamples. - -Since the weight space decomposition was perhaps the single most instrumental -ingredient of our previous analysis, it is only natural to restrict ourselves -to the case it holds. This brings us to the following definition. - -\begin{definition}\label{def:weight-mod}\index{\(\mathfrak{g}\)-module!weight modules}\index{weights!weight modules}\index{\(\mathfrak{g}\)-module!(essential) support} - A \(\mathfrak{g}\)-module \(M\) is called a \emph{weight - \(\mathfrak{g}\)-module} if \(M = \bigoplus_{\lambda \in \mathfrak{h}^*} - M_\lambda\) and \(\dim M_\lambda < \infty\) for all \(\lambda \in - \mathfrak{h}^*\). The \emph{support of \(M\)} is the set - \(\operatorname{supp} M = \{\lambda \in \mathfrak{h}^* : M_\lambda \ne 0\}\). -\end{definition} - -\begin{example} - Corollary~\ref{thm:finite-dim-is-weight-mod} is equivalent to the fact that - every finite-dimensional module of a semisimple Lie algebra is a weight - module. More generally, every finite-dimensional simple module of a reductive - Lie algebra is a weight module. -\end{example} - -\begin{example}\label{ex:reductive-alg-equivalence} - We have seen that every finite-dimensional \(\mathfrak{g}\)-module is a - weight module for semisimple \(\mathfrak{g}\). In particular, if - \(\mathfrak{g}\) is finite-dimensional then the adjoint - \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module. More generally, - a finite-dimensional Lie algebra \(\mathfrak{g}\) is reductive if, and only - if the adjoint \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module, - in which case its weight spaces are given by the root spaces of - \(\mathfrak{g}\) -\end{example} - -\begin{example}\label{ex:submod-is-weight-mod} - Proposition~\ref{thm:verma-is-weight-mod} and - Proposition~\ref{thm:max-verma-submod-is-weight} imply that the Verma module - \(M(\lambda)\) and its maximal submodule are both weight modules. In - fact, the proof of Proposition~\ref{thm:max-verma-submod-is-weight} is - actually a proof of the fact that every submodule \(N \subset M\) of - a weight module \(M\) is a weight module, and \(N_\lambda = M_\lambda \cap - N\) for all \(\lambda \in \mathfrak{h}^*\). -\end{example} - -\begin{example}\label{ex:quotient-is-weight-mod} - Given a weight module \(M\), a submodule \(N \subset M\) and \(\lambda \in - \mathfrak{h}^*\), it is clear that \(\mfrac{M_\lambda}{N} \subset - \left(\mfrac{M}{N}\right)_\lambda\). In addition, \(\mfrac{M}{N} = - \bigoplus_{\lambda \in \mathfrak{h}^*} \mfrac{M_\lambda}{N}\). Hence - \(\mfrac{M}{N}\) is weight \(\mathfrak{g}\)-module with - \(\left(\mfrac{M}{N}\right)_\lambda = \mfrac{M_\lambda}{N} \cong - \mfrac{M_\lambda}{N_\lambda}\). -\end{example} - -\begin{example}\label{ex:tensor-prod-of-weight-is-weight} - Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras, \(M_1\) be a - weight \(\mathfrak{g}_1\)-module and \(M_2\) a weight - \(\mathfrak{g}_2\)-module. Recall from Example~\ref{ex:cartan-direct-sum} - that if \(\mathfrak{h}_i \subset \mathfrak{g}_i\) are Cartan subalgebras then - \(\mathfrak{h} = \mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan - subalgebra of \(\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2\) with - \(\mathfrak{h}^* = \mathfrak{h}_1^* \oplus \mathfrak{h}_2^*\). In this - setting, one can readily check that \(M_1 \otimes M_2\) is a weight - \(\mathfrak{g}\)-module with - \[ - (M_1 \otimes M_2)_{\lambda_1 + \lambda_2} - = (M_1)_{\lambda_1} \otimes (M_2)_{\lambda_2} - \] - for all \(\lambda_i \in \mathfrak{h}_i^*\) and \(\operatorname{supp} M_1 - \otimes M_2 = \operatorname{supp} M_1 \oplus \operatorname{supp} M_2 = \{ - \lambda_1 + \lambda_2 : \lambda_i \in \operatorname{supp} M_i \subset - \mathfrak{h}_i^*\}\). -\end{example} - -\begin{example}\label{thm:simple-weight-mod-is-tensor-prod} - Let \(\mathfrak{g} = \mathfrak{z} \oplus \mathfrak{s}_1 \oplus \cdots \oplus - \mathfrak{s}_r\) be a reductive Lie algebra, where \(\mathfrak{z}\) is the - center of \(\mathfrak{g}\) and \(\mathfrak{s}_1, \ldots, \mathfrak{s}_r\) are - its simple components. As in - Example~\ref{ex:all-simple-reps-are-tensor-prod}, any simple weight - \(\mathfrak{g}\)-module \(M\) can be decomposed as - \[ - M \cong Z \otimes M_1 \otimes \cdots \otimes M_r - \] - where \(Z\) is a \(1\)-dimensional representation of \(\mathfrak{z}\) and - \(M_i\) is a simple weight \(\mathfrak{s}_i\)-module. The modules \(Z\) and - \(M_i\) are uniquely determined up to isomorphism. -\end{example} - -\begin{example}\label{ex:adjoint-action-in-universal-enveloping-is-weight} - We would like to show that the requirement of finite-dimensionality in - Definition~\ref{def:weight-mod} is not redundant. Let \(\mathfrak{g}\) be a - finite-dimensional reductive Lie algebra and consider the adjoint - \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) -- where \(X \in - \mathfrak{g}\) acts by taking commutators. Given \(\alpha \in Q\), a simple - computation shows \(K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in - \mathfrak{g}_{\alpha_i}, H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha = - \alpha_1 + \cdots + \alpha_n \rangle \subset - \mathcal{U}(\mathfrak{g})_\alpha\). The PBW Theorem and - Example~\ref{ex:reductive-alg-equivalence} thus imply that - \(\mathcal{U}(\mathfrak{g}) = \bigoplus_{\alpha \in Q} - \mathcal{U}(\mathfrak{g})_\alpha\) where \(\mathcal{U}(\mathfrak{g})_\alpha = - K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in \mathfrak{g}_{\alpha_i}, - H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha = \alpha_1 + \cdots + - \alpha_n \rangle\). However, \(\dim \mathcal{U}(\mathfrak{g})_\alpha = - \infty\). For instance, \(\mathcal{U}(\mathfrak{g})_0\) is \emph{precisely} - the commutator of \(\mathfrak{h}\) in \(\mathcal{U}(\mathfrak{g})\), which - contains \(\mathcal{U}(\mathfrak{h})\) and is therefore infinite-dimensional. -\end{example} - -\begin{note} - We should stress that the weight spaces \(M_\lambda \subset M\) of a given - weight \(\mathfrak{g}\)-module \(M\) are \emph{not} - \(\mathfrak{g}\)-submodules. Nevertheless, \(M_\lambda\) is a - \(\mathfrak{h}\)-submodule. More generally, \(M_\lambda\) is a - \(\mathcal{U}(\mathfrak{g})_0\)-submodule, where - \(\mathcal{U}(\mathfrak{g})_0\) is the centralizer of \(\mathfrak{h}\) in - \(\mathcal{U}(\mathfrak{g})\) -- which coincides with the weight space of \(0 - \in \mathfrak{h}^*\) in the adjoint \(\mathfrak{g}\)-module - \(\mathcal{U}(\mathfrak{g})\), as seen in - Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}. -\end{note} - -A particularly well behaved class of examples are the so called -\emph{bounded} modules. - -\begin{definition}\index{\(\mathfrak{g}\)-module!bounded modules}\index{\(\mathfrak{g}\)-module!(essential) support} - A weight \(\mathfrak{g}\)-module \(M\) is called \emph{bounded} if \(\dim - M_\lambda\) is bounded. The lowest upper bound \(\deg M\) for \(\dim - M_\lambda\) is called \emph{the degree of \(M\)}. The \emph{essential - support} of \(M\) is the set \(\operatorname{supp}_{\operatorname{ess}} M = - \{ \lambda \in \mathfrak{h}^* : \dim M_\lambda = \deg M \}\). -\end{definition} - -\begin{example}\label{ex:supp-ess-of-tensor-is-product} - Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras with Cartan - subalgebras \(\mathfrak{h}_i \subset \mathfrak{g}_i\) and take \(\mathfrak{g} - = \mathfrak{g}_1 \oplus \mathfrak{g}_2\). Given bounded - \(\mathfrak{g}_i\)-modules \(M_i\), it follows from - Example~\ref{ex:tensor-prod-of-weight-is-weight} that \(M_1 \otimes M_2\) is - a bounded \(\mathfrak{g}\)-module with \(\deg M_1 \otimes M_2 = \deg M_1 - \cdot \deg M_2\) and - \[ - \operatorname{supp}_{\operatorname{ess}} M_1 \otimes M_2 - = \operatorname{supp}_{\operatorname{ess}} M_1 \oplus - \operatorname{supp}_{\operatorname{ess}} M_2 - = \{ - \lambda_1 + \lambda_2 : \lambda_i \in - \operatorname{supp}_{\operatorname{ess}} M_i \subset \mathfrak{h}_i^* - \} - \] -\end{example} - -\begin{example}\label{ex:laurent-polynomial-mod} - There is a natural action of \(\mathfrak{sl}_2(K)\) on the space \(K[x, - x^{-1}]\) of Laurent polynomials, given by the formulas in - (\ref{eq:laurent-polynomials-cusp-mod}). One can quickly verify \(K[x, - x^{-1}]_{2 k} = K x^k\) and \(K[x, x^{-1}]_\lambda = 0\) for any \(\lambda - \notin 2 \mathbb{Z}\), so that \(K[x, x^{-1}] = \bigoplus_{k \in \mathbb{Z}} - K x^k\) is a degree \(1\) bounded weight \(\mathfrak{sl}_2(K)\)-module. It - follows from the remark at the end of Example~\ref{ex:submod-is-weight-mod} - that any nonzero submodule \(N \subset K[x, x^{-1}]\) must contain a - monomial \(x^k\). But since the operators \(-\frac{\mathrm{d}}{\mathrm{d}x} + - \frac{x^{-1}}{2}, x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} : K[x, - x^{-1}] \to K[x, x^{-1}]\) are both injective, this implies all other - monomials can be found in \(N\) by successively applying \(f\) and \(e\). - Hence \(N = K[x, x^{-1}]\) and \(K[x, x^{-1}]\) is a simple module. - \begin{align}\label{eq:laurent-polynomials-cusp-mod} - e \cdot p - & = \left( x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} \right) p & - f \cdot p - & = \left(- \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x^{-1}}{2} \right) p & - h \cdot p - & = 2 x \frac{\mathrm{d}}{\mathrm{d}x} p - \end{align} -\end{example} - -Notice that the support of \(K[x, x^{-1}]\) is the trivial \(2 -\mathbb{Z}\)-coset \(0 + 2 \mathbb{Z}\). This is representative of the general -behavior in the following sense: if \(M\) is a simple weight -\(\mathfrak{g}\)-module, since \(M[\lambda] = \bigoplus_{\alpha \in Q} -M_{\lambda + \alpha}\) is stable under the action of \(\mathfrak{g}\) for all -\(\lambda \in \mathfrak{h}^*\), \(\bigoplus_{\alpha \in Q} M_{\lambda + -\alpha}\) is either \(0\) or all of \(M\). In other words, the support of a -simple weight module is always contained in a single \(Q\)-coset. - -However, the behavior of \(K[x, x^{-1}]\) deviates from that of an arbitrary -bounded \(\mathfrak{g}\)-module in the sense its essential support is -precisely the entire \(Q\)-coset it inhabits -- i.e. -\(\operatorname{supp}_{\operatorname{ess}} K[x, x^{-1}] = 2 \mathbb{Z}\). This -isn't always the case. Nevertheless, in general we find\dots - -\begin{proposition}\label{thm:ess-supp-is-zariski-dense} - Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra and \(M\) - be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module. The - essential support \(\operatorname{supp}_{\operatorname{ess}} M\) is - Zariski-dense\footnote{Any choice of basis for $\mathfrak{h}^*$ induces a - $K$-linear isomorphism $\mathfrak{h}^* \isoto K^n$. In particular, a choice - of basis induces a unique topology in $\mathfrak{h}^*$ such that the map - $\mathfrak{h}^* \to K^n$ is a homeomorphism onto $K^n$ with the Zariski - topology. Any two basis induce the same topology in $\mathfrak{h}^*$, which - we call \emph{the Zariski topology of $\mathfrak{h}^*$}.} in - \(\mathfrak{h}^*\). -\end{proposition} - -This proof was deemed too technical to be included in here, but see Proposition -3.5 of \cite{mathieu} for the case where \(\mathfrak{g} = \mathfrak{s}\) is a -simple Lie algebra. The general case then follows from -Example~\ref{thm:simple-weight-mod-is-tensor-prod}, -Example~\ref{ex:supp-ess-of-tensor-is-product} and the asserting that the -product of Zariski-dense subsets in \(K^n\) and \(K^m\) is Zariski-dense in -\(K^{n + m} = K^n \times K^m\). - -We now begin a systematic investigation of the problem of classifying the -infinite-dimensional simple weight modules of a given Lie algebra -\(\mathfrak{g}\). As in the previous chapter, let \(\mathfrak{g}\) be a -finite-dimensional semisimple Lie algebra. As a first approximation of a -solution to our problem, we consider the induction functors -\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} : -\mathfrak{p}\text{-}\mathbf{Mod} \to \mathfrak{g}\text{-}\mathbf{Mod}\), where -\(\mathfrak{p} \subset \mathfrak{g}\) is some subalgebra. - -% TODOO: Are you sure that these are indeed the weight spaces of the induced -% module? Check this out? -These functors have already proved themselves a powerful tool for constructing -modules in the previous chapters. Our first observation is that if -\(\mathfrak{p} \subset \mathfrak{g}\) contains the Borel subalgebra -\(\mathfrak{b}\) then \(\mathfrak{h}\) is a Cartan subalgebra of -\(\mathfrak{p}\) and \((\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} -M)_\lambda = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{p})} -M_\lambda\) for all \(\lambda \in \mathfrak{h}^*\). In particular, -\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}}\) takes weight -\(\mathfrak{p}\)-modules to weight \(\mathfrak{g}\)-modules. This leads us to -the following definition. - -\begin{definition}\index{Lie subalgebra!parabolic subalgebra} - A subalgebra \(\mathfrak{p} \subset \mathfrak{g}\) is called \emph{parabolic} - if \(\mathfrak{b} \subset \mathfrak{p}\). -\end{definition} - -% TODOO: Why is the fact that p is not reductive relevant?? Why do we need to -% look at the quotient by nil(p)?? -Parabolic subalgebras thus give us a process for constructing weight -\(\mathfrak{g}\)-modules from modules of smaller (parabolic) subalgebras. Our -hope is that by iterating this process again and again we can get a large class -of simple weight \(\mathfrak{g}\)-modules. However, there is a small catch: a -parabolic subalgebra \(\mathfrak{p} \subset \mathfrak{g}\) needs not to be -reductive. We can get around this limitation by modding out by -\(\mathfrak{nil}(\mathfrak{p})\) and noticing that -\(\mathfrak{nil}(\mathfrak{p})\) acts trivially in any weight -\(\mathfrak{p}\)-module \(M\). By applying the universal property of quotients -we can see that \(M\) has the natural structure of a -\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module, which is always -a reductive algebra. -\begin{center} - \begin{tikzcd} - \mathfrak{p} \rar \dar & - \mathfrak{gl}(M) \\ - \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})} \arrow[dotted]{ur} & - \end{tikzcd} -\end{center} - -Let \(\mathfrak{p}\) be a parabolic subalgebra and \(M\) be a simple -weight \(\mathfrak{p}\)-module. We should point out that while -\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} M\) is a weight -\(\mathfrak{g}\)-module, it isn't necessarily simple. Nevertheless, we can -use it to produce a simple weight \(\mathfrak{g}\)-module via a -construction very similar to that of Verma modules. - -\begin{definition}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules} - Given any \(\mathfrak{p}\)-module \(M\), the module \(M_{\mathfrak{p}}(M) = - \operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} M\) is called \emph{a - generalized Verma module}. -\end{definition} - -\begin{proposition}\label{thm:generalized-verma-has-simple-quotient} - Given a simple \(\mathfrak{p}\)-module \(M\), the generalized Verma module - \(M_{\mathfrak{p}}(M)\) has a unique maximal \(\mathfrak{p}\)-submodule - \(N_{\mathfrak{p}}(M)\) and a unique irreducible quotient - \(L_{\mathfrak{p}}(M) = \mfrac{M_{\mathfrak{p}}(M)}{N_{\mathfrak{p}}(M)}\). - The irreducible quotient \(L_{\mathfrak{p}}(M)\) is a weight module. -\end{proposition} - -The proof of Proposition~\ref{thm:generalized-verma-has-simple-quotient} is -entirely analogous to that of Proposition~\ref{thm:max-verma-submod-is-weight}. -This leads us to the following definitions. - -\begin{definition}\index{\(\mathfrak{g}\)-module!parabolic induced modules}\index{\(\mathfrak{g}\)-module!cuspidal modules} - A \(\mathfrak{g}\)-module is called \emph{parabolic induced} if it is - isomorphic to \(L_{\mathfrak{p}}(M)\) for some proper parabolic subalgebra - \(\mathfrak{p} \subsetneq \mathfrak{g}\) and some \(\mathfrak{p}\)-module - \(M\). An \emph{simple cuspidal \(\mathfrak{g}\)-module} is a simple - \(\mathfrak{g}\)-module which is \emph{not} parabolic induced. -\end{definition} - -Since every weight \(\mathfrak{p}\)-module \(M\) is an -\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module, it makes sense -to call \(M\) \emph{cuspidal} if it is a cuspidal -\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module. The first -breakthrough regarding our classification problem was given by Fernando in his -now infamous paper \citetitle{fernando} \cite{fernando}, where he proved that -every simple weight \(\mathfrak{g}\)-module is parabolic induced. In other -words\dots - -\begin{theorem}[Fernando] - Any simple weight \(\mathfrak{g}\)-module is isomorphic to - \(L_{\mathfrak{p}}(M)\) for some parabolic subalgebra \(\mathfrak{p} \subset - \mathfrak{g}\) and some simple cuspidal \(\mathfrak{p}\)-module \(M\). -\end{theorem} - -We should point out that the relationship between simple weight -\(\mathfrak{g}\)-modules and pairs \((\mathfrak{p}, M)\) -- where -\(\mathfrak{p}\) is some parabolic subalgebra and \(M\) is a simple cuspidal -\(\mathfrak{p}\)-module -- is not one-to-one. Nevertheless, this relationship -is well understood. Namely, Fernando himself established\dots - -\begin{proposition}[Fernando] - Given a parabolic subalgebra \(\mathfrak{p} \subset \mathfrak{g}\), there - exists a basis \(\Sigma\) for \(\Delta\) such that \(\Sigma \subset - \Delta_{\mathfrak{p}} \subset \Delta\), where \(\Delta_{\mathfrak{p}}\) - denotes the set of roots of \(\mathfrak{p}\). Furthermore, if \(\mathfrak{p}' - \subset \mathfrak{g}\) is another parabolic subalgebra, \(M\) is a simple - cuspidal \(\mathfrak{p}\)-module and \(N\) is a simple cuspidal - \(\mathfrak{p}'\)-module then \(L_{\mathfrak{p}}(M) \cong - L_{\mathfrak{p}'}(N)\) if, and only if \(\mathfrak{p}' = - \twisted{\mathfrak{p}}{\sigma}\) and \(M \cong \twisted{N}{\sigma}\) for - some\footnote{Here $\twisted{\mathfrak{p}}{\sigma}$ denotes the image of - $\mathfrak{p}$ under the automorphism of $\sigma : \mathfrak{g} \to - \mathfrak{g}$ given by the canonical action of $W$ on $\mathfrak{g}$ and - $\twisted{N}{\sigma}$ is the $\mathfrak{p}$-module given by composing the map - $\mathfrak{p}' \to \mathfrak{gl}(N)$ with the restriction - $\sigma\!\restriction_{\mathfrak{p}} : \mathfrak{p} \to \mathfrak{p}'$.} - \(\sigma \in W_M\), where - \[ - W_M - = \langle - \sigma_\beta : \beta \in \Sigma, H_\beta + \mathfrak{nil}(\mathfrak{p}) - \ \text{is central in}\ \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})} - \ \text{and}\ H_\beta\ \text{acts on \(M\) as a positive integer} - \rangle - \subset W - \] -\end{proposition} - -\begin{note} - The definition of the subgroup \(W_M \subset W\) is independent of the choice - of basis \(\Sigma\). -\end{note} - -As a first consequence of Fernando's Theorem, we provide two alternative -characterizations of cuspidal modules. - -\begin{corollary}[Fernando]\label{thm:cuspidal-mod-equivs} - Let \(M\) be a simple weight \(\mathfrak{g}\)-module. The following - conditions are equivalent. - \begin{enumerate} - \item \(M\) is cuspidal. - \item \(F_\alpha\) acts injectively on \(M\) for all - \(\alpha \in \Delta\) -- this is what is usually referred - to as a \emph{dense} module in the literature. - \item The support of \(M\) is precisely one \(Q\)-coset -- this is - what is usually referred to as a \emph{torsion-free} module in the - literature. - \end{enumerate} -\end{corollary} - -\begin{example} - As noted in Example~\ref{ex:laurent-polynomial-mod}, the element \(f \in - \mathfrak{sl}_2(K)\) acts injectively on the space of Laurent polynomials. - Hence \(K[x, x^{-1}]\) is a cuspidal \(\mathfrak{sl}_2(K)\)-module. -\end{example} - -Having reduced our classification problem to that of classifying simple -cuspidal modules, we are now faced the daunting task of actually classifying -them. Historically, this was first achieved by Olivier Mathieu in the early -2000's in his paper \citetitle{mathieu} \cite{mathieu}. To do so, Mathieu -introduced new tools which have since proved themselves remarkably useful -throughout the field, known as\dots - -\section{Coherent Families} - -We begin our analysis with a simple question: how to do we go about -constructing cuspidal modules? Specifically, given a cuspidal -\(\mathfrak{g}\)-module, how can we use it to produce new cuspidal modules? To -answer this question, we look back at the single example of a cuspidal module -we have encountered so far: the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\) -of Laurent polynomials -- i.e. Example~\ref{ex:laurent-polynomial-mod}. - -Our first observation is that \(\mathfrak{sl}_2(K)\) acts on \(K[x, x^{-1}]\) -via differential operators. In other words, the action map -\(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{End}(K[x, x^{-1}])\) -factors through the inclusion of the algebra \(\operatorname{Diff}(K[x, -x^{-1}]) = K\left[x, x^{-1}, \frac{\mathrm{d}}{\mathrm{d}x}\right]\) of -differential operators in \(K[x, x^{-1}]\). -\begin{center} - \begin{tikzcd} - \mathcal{U}(\mathfrak{sl}_2(K)) \rar & - \operatorname{Diff}(K[x, x^{-1}]) \rar & - \operatorname{End}(K[x, x^{-1}]) - \end{tikzcd} -\end{center} - -The space \(K[x, x^{-1}]\) can be regarded as a \(\operatorname{Diff}(K[x, -x^{-1}])\)-module in the natural way, and we can produce new -\(\operatorname{Diff}(K[x, x^{-1}])\)-modules by twisting \(K[x, x^{-1}]\) by -automorphisms of \(\operatorname{Diff}(K[x, x^{-1}])\). For example, given -\(\lambda \in K\) we may take the automorphism -\begin{align*} - \varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) & - \to \operatorname{Diff}(K[x, x^{-1}]) \\ - x & \mapsto x \\ - x^{-1} & \mapsto x^{-1} \\ - \frac{\mathrm{d}}{\mathrm{d} x} & \mapsto \frac{\mathrm{d}}{\mathrm{d} x} + - \frac{\lambda}{2} x^{-1} -\end{align*} -and consider the twisted module \(\twisted{K[x, x^{-1}]}{\varphi_\lambda} = -K[x, x^{-1}]\), where some operator \(P \in \operatorname{Diff}(K[x, x^{-1}])\) -acts as \(\varphi_\lambda(P)\). - -By composing the action map \(\operatorname{Diff}(K[x, x^{-1}]) \to -\operatorname{End}(\twisted{K[x, x^{-1}]}{\varphi_\lambda})\) with the -homomorphism of algebras \(\mathcal{U}(\mathfrak{sl}_2(K)) \to -\operatorname{Diff}(K[x, x^{-1}])\) we can give \(\twisted{K[x, -x^{-1}]}{\varphi_\lambda}\) the structure of an \(\mathfrak{sl}_2(K)\)-module. -Diagrammatically, we have -\begin{center} - \begin{tikzcd} - \mathcal{U}(\mathfrak{sl}_2(K)) \rar & - \operatorname{Diff}(K[x, x^{-1}]) \rar{\varphi_\lambda} & - \operatorname{Diff}(K[x, x^{-1}]) \rar & - \operatorname{End}(K[x, x^{-1}]) - \end{tikzcd}, -\end{center} -where the maps \(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x, -x^{-1}])\) and \(\operatorname{Diff}(K[x, x^{1}]) \to \operatorname{End}(K[x, -x^{-1}])\) are the ones from the previous diagram. - -Explicitly, we find that the action of \(\mathfrak{sl}_2(K)\) on -\(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is given by -\begin{align*} - p & \overset{e}{\mapsto} - \left( - x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 + \lambda}{2} x - \right) p & - p & \overset{f}{\mapsto} - \left( - - \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 - \lambda}{2} x^{-1} - \right) p & - p & \overset{h}{\mapsto} - \left( 2 x \frac{\mathrm{d}}{\mathrm{d}x} + \lambda \right) p, -\end{align*} -so we can see \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}_{2 k + -\frac{\lambda}{2}} = K x^k\) for all \(k \in \mathbb{Z}\) and \(\twisted{K[x, -x^{-1}]}{\varphi_\lambda}_\mu = 0\) for all other \(\mu \in \mathfrak{h}^*\). - -Hence \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a degree \(1\) bounded -\(\mathfrak{sl}_2(K)\)-module with \(\operatorname{supp} \twisted{K[x, -x^{-1}]}{\varphi_\lambda} = \frac{\lambda}{2} + 2 \mathbb{Z}\). One can also -quickly check that if \(\lambda \notin 1 + 2 \mathbb{Z}\) then \(e\) and \(f\) -act injectively in \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\), so that -\(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is simple. In particular, if -\(\lambda, \mu \notin 1 + 2 \mathbb{Z}\) with \(\lambda \notin \mu + 2 -\mathbb{Z}\) then \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) and -\(\twisted{K[x, x^{-1}]}{\varphi_\mu}\) are non-isomorphic simple cuspidal -\(\mathfrak{sl}_2(K)\)-modules, since their supports differ. These cuspidal -modules can be ``glued together'' in a \emph{monstrous concoction} by summing -over \(\lambda \in K\), as in -\[ - \mathcal{M} - = \bigoplus_{\lambda + 2 \mathbb{Z} \in \mfrac{K}{2 \mathbb{Z}}} - \twisted{K[x, x^{-1}]}{\varphi_\lambda}, -\] - -To a distracted spectator, \(\mathcal{M}\) may look like just another, -innocent, \(\mathfrak{sl}_2(K)\)-module. However, the attentive reader may have -already noticed some of the its bizarre features, most noticeable of which is -the fact that \(\mathcal{M}\) is very big. In fact, \(\mathcal{M}\) is as big a -degree \(1\) bounded module gets: \(\operatorname{supp} \mathcal{M} -= \operatorname{supp}_{\operatorname{ess}} \mathcal{M}\) is the entirety of -\(\mathfrak{h}^*\). This may look very alien the reader familiarized with the -finite-dimensional setting, where the configuration of weights is very rigid. -For this reason, \(\mathcal{M}\) deserves to be called ``a monstrous -concoction''. - -On a perhaps less derogatory note, \(\mathcal{M}\) also deserves to be called -\emph{a family}. This is because \(\mathcal{M}\) consists of lots of smaller -cuspidal modules which fit together inside of it in a \emph{coherent} fashion. -Mathieu's ingenious breakthrough was the realization that \(\mathcal{M}\) is a -particular example of a more general pattern, which he named \emph{coherent -families}. - -\begin{definition}\index{coherent family} - A \emph{coherent family \(\mathcal{M}\) of degree \(d\)} is a weight - \(\mathfrak{g}\)-module \(\mathcal{M}\) such that - \begin{enumerate} - \item \(\dim \mathcal{M}_\lambda = d\) for \emph{all} \(\lambda \in - \mathfrak{h}^*\) -- i.e. \(\operatorname{supp}_{\operatorname{ess}} - \mathcal{M} = \mathfrak{h}^*\). - \item For any \(u \in \mathcal{U}(\mathfrak{g})\) in the centralizer - \(\mathcal{U}(\mathfrak{g})_0\) of \(\mathfrak{h}\) in - \(\mathcal{U}(\mathfrak{g})\), the map - \begin{align*} - \mathfrak{h}^* & \to K \\ - \lambda & \mapsto - \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\lambda}) - \end{align*} - is polynomial in \(\lambda\). - \end{enumerate} -\end{definition} - -\begin{example}\label{ex:sl-laurent-family} - The module \(\mathcal{M} = \bigoplus_{\lambda + 2 \mathbb{Z} \in - \mfrac{K}{2 \mathbb{Z}}} \twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a - degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family. -\end{example} - -\begin{example} - Given \(\lambda \in K\), \(\mathcal{M}(\lambda) = \bigoplus_{\mu \in K} K - x^\mu\) with - \begin{align*} - p & \overset{e}{\mapsto} - \left(x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \lambda x\right) p & - p & \overset{f}{\mapsto} - \left(-\frac{\mathrm{d}}{\mathrm{d}x} + \lambda x^{-1}\right) p & - p & \overset{h}{\mapsto} 2 x \frac{\mathrm{d}}{\mathrm{d}x} p, - \end{align*} - is a degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family -- where \(x^{\pm - 1}, \sfrac{\mathrm{d}}{\mathrm{d}x} : \mathcal{M}(\lambda) \to - \mathcal{M}(\lambda)\) are given by \(x^{\pm 1} x^\mu = x^{\mu \pm 1}\) and - \(\sfrac{\mathrm{d}}{\mathrm{d}x} x^\mu = \mu x^{\mu - 1}\). It is easy to - check \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family} is isomorphic - to \(\mathcal{M}(\sfrac{1}{2})\) and \((\mathcal{M}(\sfrac{1}{2}))[0] \cong - K[x, x^{-1}]\). -\end{example} - -\begin{note} - We would like to stress that coherent families have proven themselves useful - for problems other than the classification of cuspidal - \(\mathfrak{g}\)-modules. For instance, Nilsson's classification of rank 1 - \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules is based on the - notion of coherent families and the so called \emph{weighting functor}. -\end{note} - -Our hope is that given a simple cuspidal module \(M\), we can somehow fit \(M\) -inside of a coherent \(\mathfrak{g}\)-family, such as in the case of \(K[x, -x^{-1}]\) and \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family}. In -addition, we hope that such coherent families are somehow \emph{uniquely -determined} by \(M\). This leads us to the following definition. - -\begin{definition}\index{coherent family!coherent extension} - Given a bounded \(\mathfrak{g}\)-module \(M\) of degree \(d\), a - \emph{coherent extension \(\mathcal{M}\) of \(M\)} is a coherent family - \(\mathcal{M}\) of degree \(d\) that contains \(M\) as a subquotient. -\end{definition} - -Our goal is now showing that every simple bounded module has a coherent -extension. The idea then is to classify coherent families, and classify which -submodules of a given coherent family are actually simple cuspidal modules. If -every simple bounded \(\mathfrak{g}\)-module fits inside a coherent extension, -this would lead to classification of all simple cuspidal -\(\mathfrak{g}\)-modules, which we now know is the key for the solution of our -classification problem. However, there are some complications to this scheme. - -Leaving aside the question of existence for a second, we should point out that -coherent families turn out to be rather complicated on their own. In fact they -are too complicated to classify in general. Ideally, we would like to find -\emph{nice} coherent extensions -- ones we can actually classify. For instance, -we may search for \emph{irreducible} coherent extensions, which are defined as -follows. - -\begin{definition}\index{coherent family!irreducible coherent family} - A coherent family \(\mathcal{M}\) is called \emph{irreducible} if it contains - no proper coherent subfamilies -- i.e. \(\mathcal{M}\) is a simple object in - the full subcategory of \(\mathfrak{g}\text{-}\mathbf{Mod}\) consisting of - coherent families. Equivalently, we call \(\mathcal{M}\) irreducible if - \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module - for some \(\lambda \in \mathfrak{h}^*\). -\end{definition} - -Another natural candidate for the role of ``nice extensions'' are the -semisimple coherent families -- i.e. families which are semisimple as -\(\mathfrak{g}\)-modules. These turn out to be very easy to produce. Namely, -there is a construction, known as \emph{the semisimplification of a coherent -family}, which takes a coherent extension of \(M\) to a semisimple coherent -extension of \(M\). - -% Mathieu's proof of this is somewhat profane, I don't think it's worth -% including it in here -% TODO: Move this somewhere else? This holds in general for weight modules -% whose suppert is contained in a single Q-coset -\begin{lemma}\label{thm:component-coh-family-has-finite-length} - Given a coherent family \(\mathcal{M}\) and \(\lambda \in \mathfrak{h}^*\), - \(\mathcal{M}[\lambda]\) has finite length as a \(\mathfrak{g}\)-module. -\end{lemma} - -\begin{proposition}\index{coherent family!semisimplification} - Let \(\mathcal{M}\) be a coherent family of degree \(d\). There exists a - unique semisimple coherent family \(\mathcal{M}^{\operatorname{ss}}\) of - degree \(d\) such that the composition series of - \(\mathcal{M}^{\operatorname{ss}}[\lambda]\) is the same as that of - \(\mathcal{M}[\lambda]\) for all \(\lambda \in \mathfrak{h}^*\), called - \emph{the semisimplification of \(\mathcal{M}\)}. - - Namely, if \(\lambda \in \mathfrak{h}^*\) and \(0 = \mathcal{M}_{\lambda 0} - \subset \mathcal{M}_{\lambda 1} \subset \cdots \subset \mathcal{M}_{\lambda - r_\lambda} = \mathcal{M}[\lambda]\) is a composition series\footnote{Notice - that $\mathcal{M}[\lambda] = \mathcal{M}[\mu]$ for any $\mu \in \lambda + Q$. - Hence the sum $\bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}} - \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}$ is - independent of the choice of representative for $\lambda + Q$ -- provided we - choose $\mathcal{M}_{\mu i} = \mathcal{M}_{\lambda i}$ for all $\mu \in - \lambda + Q$ and $i$.}, - \[ - \mathcal{M}^{\operatorname{ss}} - \cong \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} - \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} - \] -\end{proposition} - -\begin{proof} - The uniqueness of \(\mathcal{M}^{\operatorname{ss}}\) should be clear: - since \(\mathcal{M}^{\operatorname{ss}}\) is semisimple, so is - \(\mathcal{M}^{\operatorname{ss}}[\lambda]\). Hence by the Jordan-Hölder - Theorem - \[ - \mathcal{M}^{\operatorname{ss}}[\lambda] - \cong - \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} - \] - - As for the existence of the semisimplification, it suffices to show - \[ - \mathcal{M}^{\operatorname{ss}} - = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} - \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} - \] - is indeed a semisimple coherent family of degree \(d\). - - We know from Examples~\ref{ex:submod-is-weight-mod} and - \ref{ex:quotient-is-weight-mod} that each quotient - \(\mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}\) is a weight - module. Hence \(\mathcal{M}^{\operatorname{ss}}\) is a weight module. - Furthermore, given \(\mu \in \mathfrak{h}^*\) - \[ - \mathcal{M}_\mu^{\operatorname{ss}} - = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} - \left( - \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} - \right)_\mu - = \bigoplus_i - \left( - \mfrac{\mathcal{M}_{\mu i + 1}}{\mathcal{M}_{\mu i}} - \right)_\mu - \cong \bigoplus_i - \mfrac{(\mathcal{M}_{\mu i + 1})_\mu} - {(\mathcal{M}_{\mu i})_\mu} - \] - - In particular, - \[ - \dim \mathcal{M}_\mu^{\operatorname{ss}} - = \sum_i - \dim (\mathcal{M}_{\mu i + 1})_\mu - \dim (\mathcal{M}_{\mu i})_\mu - = \dim \mathcal{M}[\mu]_\mu - = \dim \mathcal{M}_\mu - = d - \] - - Likewise, given \(u \in \mathcal{U}(\mathfrak{g})_0\) the value - \[ - \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu^{\operatorname{ss}}}) - = \sum_i - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i + 1})_\mu}) - - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i})_\mu}) - = \operatorname{Tr}(u\!\restriction_{\mathcal{M}[\mu]_\mu}) - = \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu}) - \] - is polynomial in \(\mu \in \mathfrak{h}^*\). -\end{proof} - -\begin{note} - Although we have provided an explicit construction of - \(\mathcal{M}^{\operatorname{ss}}\) in terms of \(\mathcal{M}\), we should - point out this construction is not functorial. First, given a - \(\mathfrak{g}\)-homomorphism \(f : \mathcal{M} \to \mathcal{N}\) between - coherent families, it is unclear what \(f^{\operatorname{ss}} : - \mathcal{M}^{\operatorname{ss}} \to \mathcal{N}^{\operatorname{ss}}\) is - supposed to be. Secondly, and this is more relevant, our construction depends - on the choice of composition series \(0 = \mathcal{M}_{\lambda 0} \subset - \cdots \subset \mathcal{M}_{\lambda r_\lambda} = \mathcal{M}[\lambda]\). - While different choices of composition series yield isomorphic results, there - is no canonical isomorphism. In addition, there is no canonical choice of - composition series. -\end{note} - -The proof of Lemma~\ref{thm:component-coh-family-has-finite-length} is -extremely technical and will not be included in here. It suffices to note that, -as in Proposition~\ref{thm:ess-supp-is-zariski-dense}, the general case follows -from the case where \(\mathfrak{g}\) is simple, which may be found in -\cite{mathieu} -- see Lemma 3.3. As promised, if \(\mathcal{M}\) is a coherent -extension of \(M\) then so is \(\mathcal{M}^{\operatorname{ss}}\). - -\begin{proposition} - Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\) - be a coherent extension of \(M\). Then \(\mathcal{M}^{\operatorname{ss}}\) is - a coherent extension of \(M\), and \(M\) is in fact a submodule of - \(\mathcal{M}^{\operatorname{ss}}\). -\end{proposition} - -\begin{proof} - Since \(M\) is simple, its support is contained in a single \(Q\)-coset. - This implies that \(M\) is a subquotient of \(\mathcal{M}[\lambda]\) for any - \(\lambda \in \operatorname{supp} M\). If we fix some composition series \(0 - = \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_r = - \mathcal{M}[\lambda]\) of \(\mathcal{M}[\lambda]\) with \(M \cong - \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}\), there is a natural inclusion - \[ - M - \isoto \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i} - \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j} - \cong \mathcal{M}^{\operatorname{ss}}[\lambda] - \] -\end{proof} - -Given the uniqueness of the semisimplification, the semisimplification of any -semisimple coherent extension \(\mathcal{M}\) is \(\mathcal{M}\) -itself and therefore\dots - -\begin{corollary}\label{thm:bounded-is-submod-of-extension} - Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\) - be a semisimple coherent extension of \(M\). Then \(M\) is - contained in \(\mathcal{M}\). -\end{corollary} - -These last results provide a partial answer to the question of existence of -well behaved coherent extensions. As for the uniqueness \(\mathcal{M}\) in -Corollary~\ref{thm:bounded-is-submod-of-extension}, it suffices to show that -the multiplicities of the simple weight \(\mathfrak{g}\)-modules in -\(\mathcal{M}\) are uniquely determined by \(M\). These multiplicities may be -computed via the following lemma. - -\begin{lemma}\label{thm:centralizer-multiplicity} - Let \(M\) be a semisimple weight \(\mathfrak{g}\)-module. Then \(M_\lambda\) - is a semisimple \(\mathcal{U}(\mathfrak{g})_0\)-module for any \(\lambda \in - \operatorname{supp} M\). Moreover, if \(L\) is a simple weight - \(\mathfrak{g}\)-module such that \(\lambda \in \operatorname{supp} L\) then - \(L_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and the - multiplicity \(L\) in \(M\) coincides with the multiplicity of \(L_\lambda\) - in \(M_\lambda\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module. -\end{lemma} - -\begin{proof} - We begin by showing that \(L_\lambda\) is simple. Let \(N \subset L_\lambda\) - be a nontrivial \(\mathcal{U}(\mathfrak{g})_0\)-submodule. We want to - establish that \(N = L_\lambda\). - - If \(\mathcal{U}(\mathfrak{g})_\alpha\) denotes the root space of \(\alpha\) - in \(\mathcal{U}(\mathfrak{g})\) under the adjoint action of \(\mathfrak{g}\) - as in Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}, - \(\alpha \in Q\), a simple calculation shows - \(\mathcal{U}(\mathfrak{g})_\alpha \cdot N \subset L_{\lambda + \alpha}\). - Since \(L\) is simple and \(N\) is nonzero, it follows from - Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight} that - \[ - L - = \mathcal{U}(\mathfrak{g}) \cdot N - = \bigoplus_{\alpha \in Q} \mathcal{U}(\mathfrak{g})_\alpha \cdot N - \] - and thus \(L_{\lambda + \alpha} = \mathcal{U}(\mathfrak{g})_\alpha \cdot N\). - In particular, \(L_\lambda = \mathcal{U}(\mathfrak{g})_0 \cdot N \subset N\) - and \(N = L_\lambda\). - - Now given a semisimple weight \(\mathfrak{g}\)-module \(M = \bigoplus_i M_i\) - with \(M_i\) simple, it is clear \(M_\lambda = \bigoplus_i (M_i)_\lambda\). - Each \((M_i)_\lambda\) is either \(0\) or a simple - \(\mathcal{U}(\mathfrak{g})_0\)-module, so that \(M_\lambda\) is a semisimple - \(\mathcal{U}(\mathfrak{g})_0\)-module. In addition, to see that the - multiplicity of \(L\) in \(M\) coincides with the multiplicity of - \(L_\lambda\) in \(M_\lambda\) it suffices to show that if \((M_i)_\lambda - \cong (M_j)_\lambda\) are both nonzero then \(M_i \cong M_j\). - - If \(I(M_i) = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0} - (M_i)_\lambda\), the inclusion of \(\mathcal{U}(\mathfrak{g})_0\)-modules - \((M_i)_\lambda \to M_i\) induces a \(\mathfrak{g}\)-homomorphism - \begin{align*} - I(M_i) & \to M_i \\ - u \otimes m & \mapsto u \cdot m - \end{align*} - - Since \(M_i\) is simple and \(\lambda \in \operatorname{supp} M_i\), \(M_i = - \mathcal{U}(\mathfrak{g}) \cdot (M_i)_\lambda\). The homomorphism \(I(M_i) - \to M_i\) is thus surjective. Similarly, if \(I(M_j) = - \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0} - (M_j)_\lambda\) then there is a natural surjective - \(\mathfrak{g}\)-homomorphism \(I(M_j) \to M_j\). Now suppose there is an - isomorphism of \(\mathcal{U}(\mathfrak{g})_0\)-modules \(f: (M_i)_\lambda - \isoto (M_j)_\lambda\). Such an isomorphism induces an isomorphism of - \(\mathfrak{g}\)-modules - \begin{align*} - \tilde f : I(M_i) & \isoto I(M_j) \\ - u \otimes m & \mapsto u \otimes f(m) - \end{align*} - - By composing \(\tilde f\) with the projection \(I(M_j) \to M_j\) we get a - surjective homomorphism \(I(M_i) \to M_j\). We claim \(\ker (I(M_i) \to M_i) - = \ker (I(M_i) \to M_j)\). To see this, notice that \(\ker(I(M_i) \to M_i)\) - coincides with the largest submodule \(Z(M_i) \subset I(M_i)\) contained in - \(\bigoplus_{\alpha \ne 0} \mathcal{U}(\mathfrak{g})_\alpha - \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda\). Indeed, a simple - computation shows \(\ker (I(M_i) \to M_i) \cap (\mathcal{U}(\mathfrak{g})_0 - \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda) = 0\), which implies - \(\ker(I(M_i) \to M_i) \subset Z(M_i)\). Since \(M_i\) is simple, \(\ker - (I(M_i) \to M_i)\) is maximal and thus \(\ker(I(M_i) \to M_i) = Z(M_i)\). By - the same token, \(\ker (I(M_j) \to M_j)\) is the largest submodule of - \(I(M_j)\) contained in \(\bigoplus_{\alpha \ne 0} - \mathcal{U}(\mathfrak{g})_\alpha \otimes_{\mathcal{U}(\mathfrak{g})_0} - (M_j)_\lambda\) and therefore \(\ker(I(M_i) \to M_i) = - \tilde{f}^{-1}(\ker(I(M_j) \to M_j)) = \ker(I(M_i) \to M_j)\). - - Hence there is an isomorphism \(\mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \isoto - M_j\) satisfying - \begin{center} - \begin{tikzcd} - I(M_i) \rar{\tilde f} \dar & I(M_j) \dar \\ - \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \rar{\sim} & M_j - \end{tikzcd} - \end{center} - and finally \(M_i \cong \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \cong M_j\). -\end{proof} - -A complementary question now is: which submodules of a \emph{nice} coherent -family are cuspidal representations? - -\begin{proposition}[Mathieu] - Let \(\mathcal{M}\) be an irreducible coherent family of degree \(d\) and - \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent. - \begin{enumerate} - \item \(\mathcal{M}[\lambda]\) is simple. - \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for - all \(\alpha \in \Delta\). - \item \(\mathcal{M}[\lambda]\) is cuspidal. - \end{enumerate} -\end{proposition} - -\begin{proof} - The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly - from Corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the - corollary that \strong{(iii)} implies \strong{(ii)}. All it is left is to - show \strong{(ii)} implies \strong{(iii)}. This isn't already clear from - Corollary~\ref{thm:cuspidal-mod-equivs} because, at first glance, - $\mathcal{M}[\lambda]$ may not be simple for some $\lambda$ satisfying - \strong{(ii)}. We will show this is never the case. - - Suppose \(F_\alpha\) acts injectively on the submodule - \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since - \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains - an infinite-dimensional simple \(\mathfrak{g}\)-submodule \(M\). Moreover, - again by Corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(M\) is a - cuspidal module, and its degree is bounded by \(d\). We want to show - \(\mathcal{M}[\lambda] = M\). - - We claim the set \(U = \{\mu \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is - a simple $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. If we - suppose this is the case for a moment or two, it follows from the fact that - \(M\) is simple and \(\operatorname{supp}_{\operatorname{ess}} M\) is - Zariski-dense that \(U \cap \operatorname{supp}_{\operatorname{ess}} M\) is - non-empty. In other words, there is some \(\mu \in \mathfrak{h}^*\) such that - \(\mathcal{M}_\mu\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and - \(\dim M_\mu = \deg M\). - - In particular, \(M_\mu \ne 0\), so \(M_\mu = \mathcal{M}_\mu\). Now given any - simple \(\mathfrak{g}\)-module \(L\), it follows from - Lemma~\ref{thm:centralizer-multiplicity} that the multiplicity of \(L\) - in \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(L_\mu\) in - \(\mathcal{M}_\mu\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module -- which is, - of course, \(1\) if \(L \cong M\) and \(0\) otherwise. Hence - \(\mathcal{M}[\lambda] = M\) and \(\mathcal{M}[\lambda]\) is cuspidal. -\end{proof} - -To finish the proof, we now show\dots - -\begin{lemma} - Let \(\mathcal{M}\) be a coherent family. The set \(U = \{\lambda \in - \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple - $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. -\end{lemma} - -\begin{proof} - For each \(\lambda \in \mathfrak{h}^*\) we introduce the bilinear form - \begin{align*} - B_\lambda : \mathcal{U}(\mathfrak{g})_0 \times \mathcal{U}(\mathfrak{g})_0 - & \to K \\ - (u, v) - & \mapsto \operatorname{Tr}(u v \!\restriction_{\mathcal{M}_\lambda}) - \end{align*} - and consider its rank -- i.e. the dimension of the image of the induced - operator - \begin{align*} - \mathcal{U}(\mathfrak{g})_0 & \to \mathcal{U}(\mathfrak{g})_0^* \\ - u & \mapsto B_\lambda(u, \cdot) - \end{align*} - - Our first observation is that \(\operatorname{rank} B_\lambda \le d^2\). This - follows from the commutativity of - \begin{center} - \begin{tikzcd} - \mathcal{U}(\mathfrak{g})_0 \rar \dar & - \mathcal{U}(\mathfrak{g})_0^* \\ - \operatorname{End}(\mathcal{M}_\lambda) \rar{\sim} & - \operatorname{End}(\mathcal{M}_\lambda)^* \uar - \end{tikzcd}, - \end{center} - where the map \(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)\) is given by the action of - \(\mathcal{U}(\mathfrak{g})_0\), the map - \(\operatorname{End}(\mathcal{M}_\lambda)^* \to - \mathcal{U}(\mathfrak{g})_0^*\) is its dual, and the isomorphism - \(\operatorname{End}(\mathcal{M}_\lambda) \isoto - \operatorname{End}(\mathcal{M}_\lambda)^*\) is induced by the trace form - \begin{align*} - \operatorname{End}(\mathcal{M}_\lambda) \times - \operatorname{End}(\mathcal{M}_\lambda) & \to K \\ - (T, S) & \mapsto \operatorname{Tr}(T S) - \end{align*} - - Indeed, \(\operatorname{rank} B_\lambda \le - \operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)) \le \dim - \operatorname{End}(\mathcal{M}_\lambda) = d^2\). Furthermore, if - \(\operatorname{rank} B_\lambda = d^2\) then we must have - \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)) = d^2\) -- i.e. the map - \(\mathcal{U}(\mathfrak{g})_0 \to \operatorname{End}(\mathcal{M}_\lambda)\) - is surjective. In particular, if \(\operatorname{rank} B_\lambda = d^2\) then - \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module, - for if \(M \subset \mathcal{M}_\lambda\) is invariant under the action of - \(\mathcal{U}(\mathfrak{g})_0\) then \(M\) is invariant under any - \(K\)-linear operator \(\mathcal{M}_\lambda \to \mathcal{M}_\lambda\), so - that \(M = 0\) or \(M = \mathcal{M}_\lambda\). - - On the other hand, if \(\mathcal{M}_\lambda\) is simple then by Burnside's - Theorem on matrix algebras the map \(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. Hence the - commutativity of the previously drawn diagram, as well as the fact that - \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)) = - \operatorname{rank}(\operatorname{End}(\mathcal{M}_\lambda)^* \to - \mathcal{U}(\mathfrak{g})_0^*)\), imply that \(\operatorname{rank} B_\lambda - = d^2\). This goes to show that \(U\) is precisely the set of all \(\lambda\) - such that \(B_\lambda\) has maximal rank \(d^2\). We now show that \(U\) is - Zariski-open. First, notice that - \[ - U = - \bigcup_{\substack{V \subset \mathcal{U}(\mathfrak{g})_0 \\ \dim V = d}} - U_V, - \] - where \(U_V = \{\lambda \in \mathfrak{h}^* : \operatorname{rank} - B_\lambda\!\restriction_V = d^2 \}\). Here \(V\) ranges over all - \(d\)-dimensional subspaces of \(\mathcal{U}(\mathfrak{g})_0\) -- \(V\) is - not necessarily a \(\mathcal{U}(\mathfrak{g})_0\)-submodule. - - Indeed, if \(\operatorname{rank} B_\lambda = d^2\) it follows from the - subjectivity of the map \(\mathcal{U}(\mathfrak{g})_0 \to - \operatorname{End}(\mathcal{M}_\lambda)\) that there is some \(V \subset - \mathcal{U}(\mathfrak{g})_0\) with \(\dim V = d\) such that the restriction - \(V \to \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. The - commutativity of - \begin{center} - \begin{tikzcd} - V \rar \dar & V^* \\ - \operatorname{End}(\mathcal{M}_\lambda) \rar{\sim} & - \operatorname{End}(\mathcal{M}_\lambda)^* \uar - \end{tikzcd} - \end{center} - then implies \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\). In - other words, \(U \subset \bigcup_V U_V\). - - Likewise, if \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\) for some - \(V\), then the commutativity of - \begin{center} - \begin{tikzcd} - V \rar \dar & V^* \\ - \mathcal{U}(\mathfrak{g})_0 \rar & - \mathcal{U}(\mathfrak{g})_0^* \uar - \end{tikzcd} - \end{center} - implies \(\operatorname{rank} B_\lambda \ge d^2\), which goes to show - \(\bigcup_V U_V \subset U\). - - Given \(\lambda \in U_V\), the surjectivity of \(V \to - \operatorname{End}(\mathcal{M}_\lambda)\) and the fact that \(\dim V < - \infty\) imply \(V \to V^*\) is invertible. Since \(\mathcal{M}\) is a - coherent family, \(B_\lambda\) depends polynomially in \(\lambda\). Hence so - does the induced maps \(V \to V^*\). In particular, there is some Zariski - neighborhood \(U'\) of \(\lambda\) such that the map \(V \to V^*\) induced by - \(B_\mu\!\restriction_V\) is invertible for all \(\mu \in U'\). - - But the surjectivity of the map induced by \(B_\mu\!\restriction_V\) implies - \(\operatorname{rank} B_\mu = d^2\), so \(\mu \in U_V\) and therefore \(U' - \subset U_V\). This implies \(U_V\) is open for all \(V\). Finally, \(U\) is - the union of Zariski-open subsets and is therefore open. We are done. -\end{proof} - -The major remaining question for us to tackle is that of the existence of -coherent extensions, which will be the focus of our next section. - -\section{Localizations \& the Existence of Coherent Extensions} - -Let \(M\) be a simple bounded \(\mathfrak{g}\)-module of degree \(d\). Our -goal is to prove that \(M\) has a (unique) irreducible semisimple coherent -extension \(\mathcal{M}\). Since \(M\) is simple, we know \(M \subset -\mathcal{M}[\lambda]\) for any \(\lambda \in \operatorname{supp} M\). Our first -task is constructing \(\mathcal{M}[\lambda]\). The issue here is that -\(\operatorname{supp}_{\operatorname{ess}} M\) may not be all of \(\lambda + Q -= \operatorname{supp}_{\operatorname{ess}} \mathcal{M}[\lambda]\), so we may -find \(M \subsetneq \mathcal{M}[\lambda]\). In fact, we may find -\(\operatorname{supp} M \subsetneq \lambda + Q\). - -This wasn't an issue an Example~\ref{ex:laurent-polynomial-mod} because we -verified that the action of \(f \in \mathfrak{sl}_2(K)\) on \(K[x, x^{-1}]\) is -injective. Since all weight spaces of \(K[x, x^{-1}]\) are \(1\)-dimensional, -this implies the action of \(f\) is actually bijective, so we can obtain a -nonzero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\) -by translating between weight spaced using \(f\) and \(f^{-1}\) -- here -\(f^{-1}\) denotes the \(K\)-linear operator \((- -\sfrac{\mathrm{d}}{\mathrm{d}x} + \sfrac{x^{-1}}{2})^{-1}\), which is the -inverse of the action of \(f\) on \(K[x, x^{-1}]\). -\begin{center} - \begin{tikzcd} - \cdots \rar[bend left=60]{f^{-1}} - & K x^{-2} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} - & K x^{-1} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} - & K \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} - & K x \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} - & K x^2 \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} - & \cdots \lar[bend left=60]{f} - \end{tikzcd} -\end{center} - -In the general case, the action of some \(F_\alpha \in \mathfrak{g}\) with -\(\alpha \in \Delta\) in \(M\) may not be injective. In fact, we have seen that -the action of \(F_\alpha\) is injective for all \(\alpha \in \Delta^+\) if, and -only if \(M\) is cuspidal. Nevertheless, we could intuitively \emph{make it -injective} by formally inverting the elements \(F_\alpha \in -\mathcal{U}(\mathfrak{g})\). This would allow us to obtain nonzero vectors in -\(M_\mu\) for all \(\mu \in \lambda + Q\) by successively applying elements of -\(\{F_\alpha^{\pm 1}\}_{\alpha \in \Delta}\) to a nonzero weight vector \(m \in -M_\lambda\). Moreover, if the actions of the \(F_\alpha\) were to be -invertible, we would find that all \(M_\mu\) are \(d\)-dimensional for \(\mu -\in \lambda + Q\). - -In a commutative domain, this can be achieved by tensoring our module by the -field of fractions. However, \(\mathcal{U}(\mathfrak{g})\) is hardly ever -commutative -- \(\mathcal{U}(\mathfrak{g})\) is commutative if, and only if -\(\mathfrak{g}\) is Abelian -- and the situation is more delicate in the -non-commutative case. For starters, a non-commutative \(K\)-algebra \(A\) may -not even have a ``field of fractions'' -- i.e. an over-ring where all elements -of \(A\) have inverses. Nevertheless, it is possible to formally invert -elements of certain subsets of \(A\) via a process known as -\emph{localization}, which we now describe. - -\begin{definition}\index{localization!multiplicative subsets}\index{localization!Ore's condition} - Let \(A\) be a \(K\)-algebra. A subset \(S \subset A\) is called - \emph{multiplicative} if \(s \cdot t \in S\) for all \(s, t \in S\) and \(0 - \notin S\). A multiplicative subset \(S\) is said to satisfy \emph{Ore's - localization condition} if for each \(a \in A\) and \(s \in S\) there exists - \(b, c \in A\) and \(t, t' \in S\) such that \(s a = b t\) and \(a s = t' - c\). -\end{definition} - -\begin{theorem}[Ore-Asano]\index{localization!Ore-Asano Theorem} - Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization - condition. Then there exists a (unique) \(K\)-algebra \(S^{-1} A\), with a - canonical algebra homomorphism \(A \to S^{-1} A\), enjoying the universal - property that each algebra homomorphism \(f : A \to B\) such that \(f(s)\) is - invertible for all \(s \in S\) can be uniquely extended to an algebra - homomorphism \(S^{-1} A \to B\). \(S^{-1} A\) is called \emph{the - localization of \(A\) by \(S\)}, and the map \(A \to S^{-1} A\) is called - \emph{the localization map}. - \begin{center} - \begin{tikzcd} - A \dar \rar{f} & B \\ - S^{-1} A \urar[swap, dotted] & - \end{tikzcd} - \end{center} -\end{theorem} - -If we identify an element with its image under the localization map, it follows -directly from Ore's construction that every element of \(S^{-1} A\) has the -form \(s^{-1} a\) for some \(s \in S\) and \(a \in A\). Likewise, any element -of \(S^{-1} A\) can also be written as \(b t^{-1}\) for some \(t \in S\), \(b -\in A\). - -Ore's localization condition may seem a bit arbitrary at first, but a more -thorough investigation reveals the intuition behind it. The issue in question -here is that in the non-commutative case we can no longer take the existence of -common denominators for granted. However, the existence of common denominators -is fundamental to the proof of the fact the field of fractions is a ring -- it -is used, for example, to define the sum of two elements in the field of -fractions. We thus need to impose their existence for us to have any hope of -defining consistent arithmetics in the localization of an algebra, and Ore's -condition is actually equivalent to the existence of common denominators -- -see the discussion in the introduction of \cite[ch.~6]{goodearl-warfield} for -further details. - -We should also point out that there are numerous other conditions -- which may -be easier to check than Ore's -- known to imply Ore's condition. For -instance\dots - -\begin{lemma} - Let \(S \subset A\) be a multiplicative subset generated by finitely many - locally \(\operatorname{ad}\)-nilpotent elements -- i.e. elements \(s \in S\) - such that for each \(a \in A\) there exists \(r > 0\) such that - \(\operatorname{ad}(s)^r a = [s, [s, \cdots [s, a]]\cdots] = 0\). Then \(S\) - satisfies Ore's localization condition. -\end{lemma} - -In our case, we are more interested in formally inverting the action of -\(F_\alpha\) on \(M\) than in inverting \(F_\alpha\) itself. To that end, we -introduce one further construction, known as \emph{the localization of a -module}. - -\begin{definition}\index{localization!localization of modules} - Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization - condition and \(M\) be an \(A\)-module. The \(S^{-1} A\)-module \(S^{-1} M = - S^{-1} A \otimes_A M\) is called \emph{the localization of \(M\) by \(S\)}, - and the homomorphism of \(A\)-modules - \begin{align*} - M & \to S^{-1} M \\ - m & \mapsto 1 \otimes m - \end{align*} - is called \emph{the localization map of \(M\)}. -\end{definition} - -Notice that the \(S^{-1} A\)-module \(S^{-1} M\) has the natural structure of -an \(A\)-module, where the action of \(A\) is given by the localization map \(A -\to S^{-1} A\). - -It is interesting to observe that, unlike in the case of the field of fractions -of a commutative domain, in general the localization map \(A \to S^{-1} A\) -- -i.e. the map \(a \mapsto \frac{a}{1}\) -- may not be injective. For instance, -if \(S\) contains a divisor of zero \(s\), its image under the localization map -is invertible and therefore cannot be a divisor of zero in \(S^{-1} A\). In -particular, if \(a \in A\) is nonzero and such that \(s a = 0\) or \(a s = 0\) -then its image under the localization map has to be \(0\). However, the -existence of divisors of zero in \(S\) turns out to be the only obstruction to -the injectivity of the localization map, as shown in\dots - -\begin{lemma} - Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization - condition and \(M\) be an \(A\)-module. If \(S\) acts injectively on \(M\) - then the localization map \(M \to S^{-1} M\) is injective. In particular, if - \(S\) has no zero divisors then \(A\) is a subalgebra of \(S^{-1} A\). -\end{lemma} - -Again, in our case we are interested in inverting the actions of the -\(F_\alpha\) on \(M\). However, for us to be able to translate between all -weight spaces associated with elements of \(\lambda + Q\), \(\lambda \in -\operatorname{supp} M\), we only need to invert the \(F_\alpha\)'s for -\(\alpha\) in some subset of \(\Delta\) which spans all of \(Q = \mathbb{Z} -\Delta\). In other words, it suffices to invert \(F_\beta\) for all \(\beta\) -in some basis \(\Sigma\) for \(\Delta\). We can choose such a basis to be -well-behaved. For example, we can show\dots - -\begin{lemma}\label{thm:nice-basis-for-inversion} - Let \(M\) be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module. - There is a basis \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) for \(\Delta\) - such that the elements \(F_{\beta_i}\) all act injectively on \(M\) and - satisfy \([F_{\beta_i}, F_{\beta_j}] = 0\). -\end{lemma} - -\begin{note} - The basis \(\Sigma\) in Lemma~\ref{thm:nice-basis-for-inversion} may very - well depend on the representation \(M\)! This is another obstruction to the - functoriality of our constructions. -\end{note} - -The proof of the previous Lemma is quite technical and was deemed too tedious -to be included in here. See Lemma 4.4 of \cite{mathieu} for a full proof. Since -\(F_\alpha\) is locally \(\operatorname{ad}\)-nilpotent for all \(\alpha \in -\Delta\), we can see\dots - -\begin{corollary} - Let \(\Sigma\) be as in Lemma~\ref{thm:nice-basis-for-inversion} and - \((F_\beta)_{\beta \in \Sigma} \subset \mathcal{U}(\mathfrak{g})\) be the - multiplicative subset generated by the \(F_\beta\)'s. The \(K\)-algebra - \(\Sigma^{-1} \mathcal{U}(\mathfrak{g}) = (F_\beta)_{\beta \in \Sigma}^{-1} - \mathcal{U}(\mathfrak{g})\) is well defined. Moreover, if we denote by - \(\Sigma^{-1} M\) the localization of \(M\) by \((F_\beta)_{\beta \in - \Sigma}\), the localization map \(M \to \Sigma^{-1} M\) is injective. -\end{corollary} - -From now on let \(\Sigma\) be some fixed basis for \(\Delta\) satisfying the -hypothesis of Lemma~\ref{thm:nice-basis-for-inversion}. We now show that -\(\Sigma^{-1} M\) is a weight \(\mathfrak{g}\)-module whose support is an -entire \(Q\)-coset. - -\begin{proposition}\label{thm:irr-bounded-is-contained-in-nice-mod} - The restriction of the localization \(\Sigma^{-1} M\) is a bounded - \(\mathfrak{g}\)-module of degree \(d\) with \(\operatorname{supp} - \Sigma^{-1} M = Q + \operatorname{supp} M\) and \(\dim \Sigma^{-1} M_\lambda - = d\) for all \(\lambda \in \operatorname{supp} \Sigma^{-1} M\). -\end{proposition} - -\begin{proof} - Fix some \(\beta \in \Sigma\). We begin by showing that \(F_\beta\) and - \(F_\beta^{-1}\) map the weight space \(\Sigma^{-1} M_\lambda\) to - \(\Sigma^{-1} M_{\lambda - \beta}\) and \(\Sigma^{-1} M_{\lambda + \beta}\), - respectively. Indeed, given \(m \in M_\lambda\) and \(H \in \mathfrak{h}\) we - have - \[ - H \cdot (F_\beta \cdot m) - = ([H, F_\beta] + F_\beta H) \cdot m - = F_\beta (-\beta(H) + H) \cdot m - = (\lambda - \beta)(H) F_\beta \cdot m - \] - - On the other hand, - \[ - 0 - = [H, 1] - = [H, F_\beta F_\beta^{-1}] - = F_\beta [H, F_\beta^{-1}] + [H, F_\beta] F_\beta^{-1} - = F_\beta [H, F_\beta^{-1}] - \beta(H) F_\beta F_\beta^{-1}, - \] - so that \([H, F_\beta^{-1}] = \beta(H) \cdot F_\beta^{-1}\) and therefore - \[ - H \cdot (F_\beta^{-1} \cdot m) - = ([H, F_\beta^{-1}] + F_\beta^{-1} H) \cdot m - = F_\beta^{-1} (\beta(H) + H) \cdot m - = (\lambda + \beta)(H) F_\beta^{-1} \cdot m - \] - - From the fact that \(F_\beta^{\pm 1}\) maps \(M_\lambda\) to \(\Sigma^{-1} - M_{\lambda \pm \beta}\) follows our first conclusion: since \(M\) is a weight - module and every element of \(\Sigma^{-1} M\) has the form \(s^{-1} \cdot m = - s^{-1} \otimes m\) for \(s \in (F_\beta)_{\beta \in \Sigma}\) and \(m \in - M\), we can see that \(\Sigma^{-1} M = \bigoplus_\lambda \Sigma^{-1} - M_\lambda\). Furthermore, since the action of each \(F_\beta\) on - \(\Sigma^{-1} M\) is bijective and \(\Sigma\) is a basis for \(Q\) we obtain - \(\operatorname{supp} \Sigma^{-1} M = Q + \operatorname{supp} M\). - - Again, because of the bijectivity of the \(F_\beta\)'s, to see that \(\dim - \Sigma^{-1} M_\lambda = d\) for all \(\lambda \in \operatorname{supp} - \Sigma^{-1} M\) it suffices to show that \(\dim \Sigma^{-1} M_\lambda = d\) - for some \(\lambda \in \operatorname{supp} \Sigma^{-1} M\). We may take - \(\lambda \in \operatorname{supp} M\) with \(\dim M_\lambda = d\). For any - finite-dimensional subspace \(V \subset \Sigma^{-1} M_\lambda\) we can find - \(s \in (F_\beta)_{\beta \in \Sigma}\) such that \(s \cdot V \subset M\). If - \(s = F_{\beta_{i_1}} \cdots F_{\beta_{i_r}}\), it is clear \(s \cdot V - \subset M_{\lambda - \beta_{i_1} - \cdots - \beta_{i_r}}\), so \(\dim V = - \dim s \cdot V \le d\). This holds for all finite-dimensional \(V \subset - \Sigma^{-1} M_\lambda\), so \(\dim \Sigma^{-1} M_\lambda \le d\). It then - follows from the fact that \(M_\lambda \subset \Sigma^{-1} M_\lambda\) that - \(M_\lambda = \Sigma^{-1} M_\lambda\) and therefore \(\dim \Sigma^{-1} - M_\lambda = d\). -\end{proof} - -We now have a good candidate for a coherent extension of \(M\), but -\(\Sigma^{-1} M\) is still not a coherent extension since its support is -contained in a single \(Q\)-coset. In particular, \(\operatorname{supp} -\Sigma^{-1} M \ne \mathfrak{h}^*\) and \(\Sigma^{-1} M\) is not a coherent -family. To obtain a coherent family we thus need somehow extend \(\Sigma^{-1} -M\). To that end, we will attempt to replicate the construction of the coherent -extension of the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\). Specifically, -the idea is that if twist \(\Sigma^{-1} M\) by an automorphism which shifts its -support by some \(\lambda \in \mathfrak{h}^*\), we can construct a coherent -family by summing these modules over \(\lambda\) as in -Example~\ref{ex:sl-laurent-family}. - -For \(K[x, x^{-1}]\) this was achieved by twisting the -\(\operatorname{Diff}(K[x, x^{-1}])\)-module \(K[x, x^{-1}]\) by the -automorphisms \(\varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) \to -\operatorname{Diff}(K[x, x^{-1}])\) and restricting the results to -\(\mathcal{U}(\mathfrak{sl}_2(K))\) via the map -\(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x, x^{-1}])\), but -this approach is inflexible since not every \(\mathfrak{sl}_2(K)\)-module -factors through \(\operatorname{Diff}(K[x, x^{-1}])\). Nevertheless, we could -just as well twist \(K[x, x^{-1}]\) by automorphisms of -\(\mathcal{U}(\mathfrak{sl}_2(K))_f\) directly -- where -\(\mathcal{U}(\mathfrak{sl}_2(K))_f = (f)^{-1} \mathcal{U}(\mathfrak{g})\) is -the localization of \(\mathcal{U}(\mathfrak{sl}_2(K))\) by the multiplicative -subset generated by \(f\). - -In general, we may twist the \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module -\(\Sigma^{-1} M\) by automorphisms of \(\Sigma^{-1} -\mathcal{U}(\mathfrak{g})\). For \(\lambda = \beta \in \Sigma\) the map -\begin{align*} - \theta_\beta : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & \to - \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ - u & \mapsto F_\beta u F_\beta^{-1} -\end{align*} -is a natural candidate for such a twisting automorphism. Indeed, we will soon -see that \(\twisted{(\Sigma^{-1} M)}{\theta_\beta}_\lambda = \Sigma^{-1} -M_{\lambda + \beta}\). However, this is hardly useful to us, since \(\beta \in -Q\) and therefore \(\beta + \operatorname{supp} \Sigma^{-1} M = -\operatorname{supp} \Sigma^{-1} M\). If we want to expand the support of -\(\Sigma^{-1} M\) we will have to twist by automorphisms that shift its support -by \(\lambda \in \mathfrak{h}^*\) lying \emph{outside} of \(Q\). - -The situation is much less obvious in this case. Nevertheless, it turns out we -can extend the family \(\{\theta_\beta\}_{\beta \in \Sigma}\) to a family of -automorphisms \(\{\theta_\lambda\}_{\lambda \in \mathfrak{h}^*}\). -Explicitly\dots - -\begin{proposition}\label{thm:nice-automorphisms-exist} - There is a family of automorphisms \(\{\theta_\lambda : \Sigma^{-1} - \mathcal{U}(\mathfrak{g}) \to \Sigma^{-1} - \mathcal{U}(\mathfrak{g})\}_{\lambda \in \mathfrak{h}^*}\) such that - \begin{enumerate} - \item \(\theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u) = F_{\beta_1}^{k_1} - \cdots F_{\beta_r}^{k_r} u F_{\beta_r}^{- k_r} \cdots F_{\beta_1}^{- - k_1}\) for all \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and \(k_1, - \ldots, k_r \in \mathbb{Z}\). - - \item For each \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) the map - \begin{align*} - \mathfrak{h}^* & \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ - \lambda & \mapsto \theta_\lambda(u) - \end{align*} - is polynomial. - - \item If \(\lambda, \mu \in \mathfrak{h}^*\), \(N\) is a \(\Sigma^{-1} - \mathcal{U}(\mathfrak{g})\)-module whose restriction to - \(\mathcal{U}(\mathfrak{g})\) is a weight \(\mathfrak{g}\)-module and - \(\twisted{N}{\theta_\lambda}\) is the \(\Sigma^{-1} - \mathcal{U}(\mathfrak{g})\)-module \(N\) twisted by the automorphism - \(\theta_\lambda\) then \(N_\mu = \twisted{N}{\theta_\lambda}_{\mu + - \lambda}\). In particular, \(\operatorname{supp} - \twisted{N}{\theta_\lambda} = \lambda + \operatorname{supp} N\). - \end{enumerate} -\end{proposition} - -\begin{proof} - Since the elements \(F_\beta\), \(\beta \in \Sigma\) commute with one - another, the endomorphisms - \begin{align*} - \theta_{k_1 \beta_1 + \cdots + k_r \beta_r} - : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & - \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ - u & \mapsto - F_{\beta_1}^{k_1} \cdots F_{\beta_r}^{k_r} - u - F_{\beta_1}^{- k_r} \cdots F_{\beta_r}^{- k_1} - \end{align*} - are well defined for all \(k_1, \ldots, k_r \in \mathbb{Z}\). - - Fix some \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\). For any \(s \in - (F_\beta)_{\beta \in \Sigma}\) and \(k > 0\) we have \(s^k u = \binom{k}{0} - \operatorname{ad}(s)^0 u s^{k - 0} + \cdots + \binom{k}{k} - \operatorname{ad}(s)^k u s^{k - k}\). Now if we take \(\ell\) such - \(\operatorname{ad}(F_\beta)^{\ell + 1} u = 0\) for all \(\beta \in \Sigma\) - we find - \[ - \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u) - = \sum_{i_1, \ldots, i_r = 1, \ldots, \ell} - \binom{k_1}{i_1} \cdots \binom{k_r}{i_r} - \operatorname{ad}(F_{\beta_1})^{i_1} \cdots - \operatorname{ad}(F_{\beta_r})^{i_r} - u - F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r} - \] - for all \(k_1, \ldots, k_r \in \mathbb{N}\). - - Since the binomial coefficients \(\binom{x}{k} = \frac{x (x-1) \cdots (x - k - + 1)}{k!}\) can be uniquely extended to polynomial functions in \(x \in K\), - we may in general define - \[ - \theta_\lambda(u) - = \sum_{i_1, \ldots, i_r \ge 0} - \binom{\lambda_1}{i_1} \cdots \binom{\lambda_r}{i_r} - \operatorname{ad}(F_{\beta_1})^{i_1} \cdots - \operatorname{ad}(F_{\beta_r})^{i_r} - r - F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r} - \] - for \(\lambda_1, \ldots, \lambda_r \in K\), \(\lambda = \lambda_1 \beta_1 + - \cdots + \lambda_r \beta_r \in \mathfrak{h}^*\). - - It is clear that the \(\theta_\lambda\) are endomorphisms. To see that the - \(\theta_\lambda\) are indeed automorphisms, notice \(\theta_{- k_1 \beta_1 - - \cdots - k_r \beta_r} = \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}^{-1}\). - The uniqueness of the polynomial extensions then implies \(\theta_{- \lambda} - = \theta_\lambda^{-1}\) in general: given \(u \in \Sigma^{-1} - \mathcal{U}(\mathfrak{g})\), the map - \begin{align*} - \mathfrak{h}^* & \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ - \lambda & \mapsto \theta_\lambda(\theta_{-\lambda}(u)) - u - \end{align*} - is a polynomial extension of the zero map \(\mathbb{Z} \beta_1 \oplus \cdots - \oplus \mathbb{Z} \beta_r \to \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and is - therefore identically zero. - - Finally, let \(N\) be a \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module - whose restriction is a weight module. If \(n \in N\) then - \[ - n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda} - \iff \theta_\lambda(H) \cdot n = (\mu + \lambda)(H) n - \, \forall H \in \mathfrak{h} - \] - - But - \[ - \theta_\beta(H) - = F_\beta H F_\beta^{-1} - = ([F_\beta, H] + H F_\beta) F_\beta^{-1} - = (\beta(H) + H) F_\beta F_\beta^{-1} - = \beta(H) + H - \] - for all \(H \in \mathfrak{h}\) and \(\beta \in \Sigma\). In general, - \(\theta_\lambda(H) = \lambda(H) + H\) for all \(\lambda \in \mathfrak{h}^*\) - and hence - \[ - \begin{split} - n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda} - & \iff (\lambda(H) + H) \cdot n = (\mu + \lambda)(H) n - \; \forall H \in \mathfrak{h} \\ - & \iff H \cdot n = \mu(H) n \; \forall H \in \mathfrak{h} \\ - & \iff n \in N_\mu - \end{split}, - \] - so that \(\twisted{N}{\theta_\lambda}_{\mu + \lambda} = N_\mu\). -\end{proof} - -It should now be obvious\dots - -\begin{proposition}[Mathieu] - There exists a coherent extension \(\mathcal{M}\) of \(M\). -\end{proposition} - -\begin{proof} - Take\footnote{Here we fix some $\lambda_\xi \in \xi$ for each $Q$-coset $\xi - \in \mfrac{\mathfrak{h}^*}{Q}$. While there is a natural isomorphism - $\twisted{(\Sigma^{-1} M)}{\theta_\lambda} \isoto \twisted{(\Sigma^{-1} - M)}{\theta_\mu}$ for each $\mu \in \lambda + Q$, they are not the same - \(\mathfrak{g}\)-modules strictly speaking. This is yet another obstruction - to the functoriality of our constructions.} - \[ - \mathcal{M} - = \bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}} - \twisted{(\Sigma^{-1} M)}{\theta_\lambda} - \] - - It is clear \(M\) lies in \(\Sigma^{-1} M = \twisted{(\Sigma^{-1} - M)}{\theta_0}\) and therefore \(M \subset \mathcal{M}\). On the other hand, - \(\dim \mathcal{M}_\mu = \dim \twisted{(\Sigma^{-1} M)}{\theta_\lambda}_\mu = - \dim \Sigma^{-1} M_{\mu - \lambda} = d\) for all \(\mu \in \lambda + Q\) -- - \(\lambda\) standing for some fixed representative of its \(Q\)-coset. - Furthermore, given \(u \in \mathcal{U}(\mathfrak{g})_0\) and \(\mu \in - \lambda + Q\), - \[ - \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu}) - = \operatorname{Tr} - (\theta_\lambda(u)\!\restriction_{\Sigma^{-1} M_{\mu - \lambda}}) - \] - is polynomial in \(\mu\) because of the second item of - Proposition~\ref{thm:nice-automorphisms-exist}. -\end{proof} - -Lo and behold\dots - -\begin{theorem}[Mathieu]\index{coherent family!Mathieu's \(\mExt\) coherent extension} - There exists a unique semisimple coherent extension \(\mExt(M)\) of \(M\). - More precisely, if \(\mathcal{M}\) is any coherent extension of \(M\), then - \(\mathcal{M}^{\operatorname{ss}} \cong \mExt(M)\). Furthermore, \(\mExt(M)\) - is a irreducible coherent family. -\end{theorem} - -\begin{proof} - The existence part should be clear from the previous discussion: it suffices - to fix some coherent extension \(\mathcal{M}\) of \(M\) and take - \(\mExt(M) = \mathcal{M}^{\operatorname{ss}}\). - - To see that \(\mExt(M)\) is irreducible, recall from - Corollary~\ref{thm:bounded-is-submod-of-extension} that \(M\) is a - \(\mathfrak{g}\)-submodule of \(\mExt(M)\). Since the degree of \(M\) is the - same as the degree of \(\mExt(M)\), some of its weight spaces have maximal - dimension inside of \(\mExt(M)\). In particular, it follows from - Lemma~\ref{thm:centralizer-multiplicity} that \(\mExt(M)_\lambda = - M_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module for some - \(\lambda \in \operatorname{supp} M\). - - As for the uniqueness of \(\mExt(M)\), fix some other semisimple coherent - extension \(\mathcal{N}\) of \(M\). We claim that the multiplicity of a given - simple \(\mathfrak{g}\)-module \(L\) in \(\mathcal{N}\) is determined by its - \emph{trace function} - \begin{align*} - \mathfrak{h}^* \times \mathcal{U}(\mathfrak{g})_0 & - \to K \\ - (\lambda, u) & - \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda}) - \end{align*} - - It is a well known fact of the theory of modules that, given an associative - \(K\)-algebra \(A\), a finite-dimensional semisimple \(A\)-module \(L\) is - determined, up to isomorphism, by its \emph{character} - \begin{align*} - \chi_L : A & \to K \\ - a & \mapsto \operatorname{Tr}(a\!\restriction_L) - \end{align*} - - In particular, the multiplicity of \(L\) in \(\mathcal{N}\), which is the - same as the multiplicity of \(L_\lambda\) in \(\mathcal{N}_\lambda\), is - determined by the character \(\chi_{\mathcal{N}_\lambda} : - \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all simple weight - \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is determined by - its trace function. Of course, the same holds for \(\mExt(M)\). We now claim - that the trace function of \(\mathcal{N}\) is the same as that of - \(\mExt(M)\). Clearly, - \(\operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda}) = - \operatorname{Tr}(u\!\restriction_{M_\lambda}) = - \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda - \in \operatorname{supp}_{\operatorname{ess}} M\), \(u \in - \mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(M\) is - Zariski-dense and the maps \(\lambda \mapsto - \operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda})\) and \(\lambda \mapsto - \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) are polynomial in - \(\lambda \in \mathfrak{h}^*\), it follows that these maps coincide for all - \(u\). - - In conclusion, \(\mathcal{N} \cong \mExt(M)\) and \(\mExt(M)\) is unique. -\end{proof} - -% This is a very important theorem, but since we won't classify the coherent -% extensions in here we don't need it, and there is no other motivation behind -% it. Including this would also require me to explain what central characters -% are, which is a bit of a pain -%\begin{proposition}[Mathieu] -% The central characters of the irreducible submodules of -% \(\operatorname{Ext}(M)\) are all the same. -%\end{proposition} - -We have thus concluded our classification of cuspidal modules in terms of -coherent families. Of course, to get an explicit construction of all simple -\(\mathfrak{g}\)-modules we would have to classify the irreducible semisimple -coherent \(\mathfrak{g}\)-families themselves, which is the subject of sections -7, 8 and 9 of \cite{mathieu}. In addition, in sections 11 and 12 of -\cite{mathieu} Mathieu provides an explicit construction of coherent families. -We unfortunately do not have the necessary space to discuss these results in -detail, but we will now provide a brief overview. - -First and foremost, notice that because of -Example~\ref{thm:simple-weight-mod-is-tensor-prod} the problem of classifying -the simple weight \(\mathfrak{g}\)-modules can be reduced to that of -classifying the simple weight modules of its simple components. In addition, it -turns out that very few simple Lie algebras admit cuspidal modules at all. -Specifically\dots - -\begin{proposition}[Fernando]\label{thm:only-sl-n-sp-have-cuspidal} - Let \(\mathfrak{s}\) be a finite-dimensional simple Lie algebra. Suppose - there exists a simple cuspidal \(\mathfrak{s}\)-module. Then \(\mathfrak{s} - \cong \mathfrak{sl}_n(K)\) or \(\mathfrak{s} \cong \mathfrak{sp}_{2 n}(K)\). -\end{proposition} - -Hence it suffices to classify the irreducible semisimple coherent families of -\(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\). These can be described -either algebraically, using combinatorial invariants -- which Mathieu does in -sections 7, 8 and 9 of his paper -- or geometrically, using algebraic varieties -and differential forms -- which is done in sections 11 and 12. While rather -complicated on its own, the geometric construction is more concrete than its -combinatorial counterpart. - -This construction also brings us full circle to the beginning of these notes, -where we saw in Proposition~\ref{thm:geometric-realization-of-uni-env} that -\(\mathfrak{g}\)-modules may be understood as geometric objects. In fact, -throughout the previous four chapters we have seen a tremendous number of -geometrically motivated examples, which further emphasizes the connection -between representation theory and geometry. I would personally go as far as -saying that the beautiful interplay between the algebraic and the geometric is -precisely what makes representation theory such a fascinating and charming -subject. - -Alas, our journey has come to an end. All it is left is to wonder at the beauty -of Lie algebras and their representations. - -\label{end-47}
diff --git a/sections/semisimple-algebras.tex /dev/null @@ -1,1044 +0,0 @@ -\chapter{Finite-Dimensional Simple Modules} - -In this chapter we classify the finite-dimensional simple -\(\mathfrak{g}\)-modules for a finite-dimensional semisimple Lie algebra -\(\mathfrak{g}\) over \(K\). At the heart of our analysis of -\(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) was the decision to consider -the eigenspace decomposition -\begin{equation}\label{sym-diag} - M = \bigoplus_\lambda M_\lambda -\end{equation} - -This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the -rational behind it and the reason why equation (\ref{sym-diag}) holds are -harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace -decomposition associated with an operator \(M \to M\) is a very well-known -tool, and readers familiarized with basic concepts of linear algebra should be -used to this type of argument. On the other hand, the eigenspace decomposition -of \(M\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h} -\subset \mathfrak{gl}(M)\) is neither well-known nor does it hold in general: -as indicated in the previous chapter, it may very well be that -\[ - \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda \subsetneq M -\] - -We should note, however, that these two cases are not as different as they may -sound at first glance. Specifically, we can regard the eigenspace decomposition -of a \(\mathfrak{sl}_2(K)\)-module \(M\) with respect to the eigenvalues of the -action of \(h\) as the eigenvalue decomposition of \(M\) with respect to the -action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\). -Furthermore, in both cases \(\mathfrak{h} \subset \mathfrak{sl}_n(K)\) is the -subalgebra of diagonal matrices, which is Abelian. The fundamental difference -between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) for -\(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim \mathfrak{h} > 1\) for -\(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The question then is: why did we -choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} > 1\) for -\(\mathfrak{sl}_3(K)\)? - -The rational behind fixing an Abelian subalgebra \(\mathfrak{h}\) is a simple -one: we have seen in the previous chapter that representations of Abelian -algebras are generally much simpler to understand than the general case. Thus -it make sense to decompose a given \(\mathfrak{g}\)-module \(M\) of into -subspaces invariant under the action of \(\mathfrak{h}\), and then analyze how -the remaining elements of \(\mathfrak{g}\) act on these subspaces. The bigger -\(\mathfrak{h}\) is, the simpler our problem gets, because there are fewer -elements outside of \(\mathfrak{h}\) left to analyze. - -Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h} -\subset \mathfrak{g}\), which leads us to the following definition. - -\begin{definition}\index{Cartan subalgebra} - A subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan - subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in - \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in - \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\), - \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if - it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in - \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former - two properties. -\end{definition} - -\begin{proposition} - There exists a Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{g}\). -\end{proposition} - -\begin{proof} - Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose - elements act as diagonal operators via the adjoint \(\mathfrak{g}\)-module. - Indeed, \(0\), the only element of \(0 \subset \mathfrak{g}\), is such that - \(\operatorname{ad}(0) = 0\) is a diagonalizable operator. Furthermore, given - a chain of Abelian subalgebras - \[ - 0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots - \] - such that \(\operatorname{ad}(H)\) is a diagonal operator for each \(H \in - \mathfrak{h}_i\), the subalgebra \(\bigcup_i \mathfrak{h}_i \subset - \mathfrak{g}\) is Abelian, and its elements also act diagonally in - \(\mathfrak{g}\). It then follows from Zorn's Lemma that there exists a - subalgebra \(\mathfrak{h}\) which is maximal with respect to both these - properties, also known as a Cartan subalgebra. -\end{proof} - -We have already seen some concrete examples. Namely\dots - -\begin{example}\label{ex:cartan-of-gl} - The Lie subalgebra - \[ - \mathfrak{h} = - \begin{pmatrix} - K & 0 & \cdots & 0 \\ - 0 & K & \cdots & 0 \\ - \vdots & \vdots & \ddots & \vdots \\ - 0 & 0 & \cdots & K - \end{pmatrix} - \subset \mathfrak{gl}_n(K) - \] - of diagonal matrices is a Cartan subalgebra. - Indeed, every pair of diagonal matrices commutes, so that \(\mathfrak{h}\) - is an Abelian -- and hence nilpotent -- subalgebra. A - simple calculation also shows that if \(i \ne j\) then the coefficient of - \(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\) - in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i}, - X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is - self-normalizing. -\end{example} - -\begin{example} - Let \(\mathfrak{h}\) be as in Example~\ref{ex:cartan-of-gl}. Then the - subalgebra \(\mathfrak{h} \cap \mathfrak{sl}_n(K)\) of traceless diagonal - matrices is a Cartan subalgebra of \(\mathfrak{sl}_n(K)\). -\end{example} - -\begin{example}\label{ex:cartan-direct-sum} - Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras and - \(\mathfrak{h}_i \subset \mathfrak{g}_i\) be Cartan subalgebras. Then - \(\mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan subalgebra of - \(\mathfrak{g}_1 \oplus \mathfrak{g}_2\). -\end{example} - -\index{Cartan subalgebra!simultaneous diagonalization} -The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the -subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of -\(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the -subalgebras described the previous chapter. The remaining question then is: if -\(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(M\) is a -\(\mathfrak{g}\)-module, does the eigenspace decomposition -\[ - M = \bigoplus_\lambda M_\lambda -\] -of \(M\) hold? The answer to this question turns out to be yes. This is a -consequence of something known as \emph{simultaneous diagonalization}, which is -the primary tool we will use to generalize the results of the previous section. -What is simultaneous diagonalization all about then? - -\begin{definition}\label{def:sim-diag} - Given a \(K\)-vector space \(V\), a set of operators \(\{T_j : V \to V\}_j\) - is called \emph{simultaneously diagonalizable} if there is a basis \(\{v_1, - \ldots, v_n\}\) for \(V\) such that \(T_j v_i\) is a scalar multiple of - \(v_i\), for all \(i, j\). -\end{definition} - -\begin{proposition} - Given a \emph{finite-dimensional} vector space \(V\), a set of diagonalizable - operators \(V \to V\) is simultaneously diagonalizable if, and only if all of - its elements commute with one another. -\end{proposition} - -We should point out that simultaneous diagonalization \emph{only works in the -finite-dimensional setting}. In fact, simultaneous diagonalization is usually -framed as an equivalent statement about diagonalizable \(n \times n\) matrices. -Simultaneous diagonalization implies that to show \(M = \bigoplus_\lambda -M_\lambda\) it suffices to show that \(H\!\restriction_M : M \to M\) is a -diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we -introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract -Jordan decomposition of a semisimple Lie algebra}. - -\begin{proposition}[Jordan] - Given a finite-dimensional vector space \(V\) and an operator \(T : V \to - V\), there are unique commuting operators \(T_{\operatorname{ss}}, - T_{\operatorname{nil}} : V \to V\), with \(T_{\operatorname{ss}}\) - diagonalizable and \(T_{\operatorname{nil}}\) nilpotent, such that \(T = - T_{\operatorname{ss}} + T_{\operatorname{nil}}\). The pair - \((T_{\operatorname{ss}}, T_{\operatorname{nil}})\) is known as \emph{the Jordan - decomposition of \(T\)}. -\end{proposition} - -\begin{proposition}\index{abstract Jordan decomposition} - Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are - \(X_{\operatorname{ss}}, X_{\operatorname{nil}} \in \mathfrak{g}\) such that \(X - = X_{\operatorname{ss}} + X_{\operatorname{nil}}\), \([X_{\operatorname{ss}}, - X_{\operatorname{nil}}] = 0\), \(\operatorname{ad}(X_{\operatorname{ss}})\) is a - diagonalizable operator and \(\operatorname{ad}(X_{\operatorname{nil}})\) is a - nilpotent operator. The pair \((X_{\operatorname{ss}}, X_{\operatorname{nil}})\) - is known as \emph{the Jordan decomposition of \(X\)}. -\end{proposition} - -It should be clear from the uniqueness of -\(\operatorname{ad}(X)_{\operatorname{ss}}\) and -\(\operatorname{ad}(X)_{\operatorname{nil}}\) that the Jordan decomposition of -\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = -\operatorname{ad}(X_{\operatorname{ss}}) + -\operatorname{ad}(X_{\operatorname{nil}})\). What is perhaps more remarkable is -the fact this holds for \emph{any} finite-dimensional \(\mathfrak{g}\)-module. -In other words\dots - -\begin{proposition}\label{thm:preservation-jordan-form} - Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(X - \in \mathfrak{g}\). Denote by \(X\!\restriction_M\) the action of \(X\) on - \(M\). Then \(X_{\operatorname{ss}}\!\restriction_M = - (X\!\restriction_M)_{\operatorname{ss}}\) and - \(X_{\operatorname{nil}}\!\restriction_M = - (X\!\restriction_M)_{\operatorname{nil}}\). -\end{proposition} - -This last result is known as \emph{the preservation of the Jordan form}, and a -proof can be found in appendix C of \cite{fulton-harris}. As promised this -implies\dots - -\begin{corollary}\label{thm:finite-dim-is-weight-mod} - Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset - \mathfrak{g}\) be a Cartan subalgebra and \(M\) be any finite-dimensional - \(\mathfrak{g}\)-module. Then there is a basis \(\{m_1, \ldots, - m_r\}\) of \(M\) so that each \(m_i\) is simultaneously an eigenvector of all - elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as - a diagonal matrix in this basis. In other words, there are linear functionals - \(\lambda_i \in \mathfrak{h}^*\) so that - \( - H \cdot m_i = \lambda_i(H) m_i - \) - for all \(H \in \mathfrak{h}\). In particular, - \[ - M = \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda - \] -\end{corollary} - -\begin{proof} - Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_M - : M \to M\) is a diagonalizable operator. - - If we write \(H = H_{\operatorname{ss}} + H_{\operatorname{nil}}\) for the - abstract Jordan decomposition of \(H\), we know - \(\operatorname{ad}(H_{\operatorname{ss}}) = - \operatorname{ad}(H)_{\operatorname{ss}}\). But \(\operatorname{ad}(H)\) is a - diagonalizable operator, so that \(\operatorname{ad}(H)_{\operatorname{ss}} = - \operatorname{ad}(H)\). This implies - \(\operatorname{ad}(H_{\operatorname{nil}}) = - \operatorname{ad}(H)_{\operatorname{nil}} = 0\), so that - \(H_{\operatorname{nil}}\) is a central element of \(\mathfrak{g}\). Since - \(\mathfrak{g}\) is semisimple, \(H_{\operatorname{nil}} = 0\). - Proposition~\ref{thm:preservation-jordan-form} then implies - \((H\!\restriction_M)_{\operatorname{nil}} = - H_{\operatorname{nil}}\!\restriction_M = 0\), so \(H\!\restriction_M = - (H\!\restriction_M)_{\operatorname{ss}}\) is a diagonalizable operator. -\end{proof} - -We should point out that this last proof only works for semisimple Lie -algebras. This is because we rely heavily on -Proposition~\ref{thm:preservation-jordan-form}, as well in the fact that -semisimple Lie algebras are centerless. In fact, -Corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie -algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the -Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a -\(\mathfrak{g}\)-module is simply a vector space \(M\) endowed with an operator -\(M \to M\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on -\(M\). In particular, if we choose an operator \(M \to M\) which is \emph{not} -diagonalizable we find \(M \ne \bigoplus_{\lambda \in \mathfrak{h}^*} -M_\lambda\). - -However, Corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive -\(\mathfrak{g}\) if we assume that the \(\mathfrak{g}\)-module \(M\) in -question is simple, since central elements of \(\mathfrak{g}\) act on simple -\(\mathfrak{g}\)-modules as scalar operators. The hypothesis of -finite-dimensionality is also of huge importance. For instance, consider\dots - -\begin{example}\label{ex:regular-mod-is-not-weight-mod} - Let \(\mathcal{U}(\mathfrak{g})\) denote the regular \(\mathfrak{g}\)-module. - Notice that \(\mathcal{U}(\mathfrak{g})_\lambda = 0\) for all \(\lambda \in - \mathfrak{h}^*\). Indeed, since \(\mathcal{U}(\mathfrak{g})\) is a domain, if - \((H - \lambda(H)) u = 0\) for some nonzero \(H \in \mathfrak{h}\) then \(u = - 0\). In particular, - \[ - \bigoplus_{\lambda \in \mathfrak{h}^*} \mathcal{U}(\mathfrak{g})_\lambda - = 0 \neq \mathcal{U}(\mathfrak{g}) - \] -\end{example} - -As a first consequence of Corollary~\ref{thm:finite-dim-is-weight-mod} we -show\dots - -\begin{corollary} - The restriction of the Killing form \(\kappa\) to \(\mathfrak{h}\) is - non-degenerate. -\end{corollary} - -\begin{proof} - Consider the root space decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus - \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint - \(\mathfrak{g}\)-module, where \(\alpha\) ranges over all nonzero eigenvalues - of the adjoint action of \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = - \mathfrak{h}\). - - Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h}) - \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the - other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in - \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) -- - i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigenspace decomposition - becomes - \[ - \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha - \] - - We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to - \(\mathfrak{g}_\alpha\) with respect to \(\kappa\) for any \(\alpha \ne 0\). - Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\) - with \(\alpha(H_1) \ne 0\) we have - \[ - \alpha(H_1) \cdot \kappa(X, H_2) - = \kappa([H_1, X], H_2) - = - \kappa([X, H_1], H_2) - = - \kappa(X, [H_1, H_2]) - = 0 - \] - - Hence the non-degeneracy of \(\kappa\) implies the non-degeneracy of its - restriction. -\end{proof} - -We should point out that the restriction of \(\kappa\) to \(\mathfrak{h}\) is -\emph{not} the Killing form of \(\mathfrak{h}\). In fact, since -\(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is -hardly ever a non-degenerate form. - -\begin{note} - Since \(\kappa\) induces an isomorphism \(\mathfrak{h} \isoto - \mathfrak{h}^*\), it induces a bilinear form \((\kappa(X, \cdot), \kappa(Y, - \cdot)) \mapsto \kappa(X, Y)\) in \(\mathfrak{h}^*\). As in - section~\ref{sec:sl3-reps}, we denote this form by \(\kappa\) as well. -\end{note} - -We now have most of the necessary tools to reproduce the results of the -previous chapter in a general setting. Let \(\mathfrak{g}\) be a -finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\) -and let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. We will -proceed, as we did before, by generalizing the results of the previous two -sections in order. By now the pattern should be starting to become clear, so we -will mostly omit technical details and proofs analogous to the ones on the -previous sections. Further details can be found in appendix D of -\cite{fulton-harris} and in \cite{humphreys}. - -\section{The Geometry of Roots and Weights} - -We begin our analysis, as we did for \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\), by investigating the locus of roots of and weights of -\(\mathfrak{g}\). Throughout chapter~\ref{ch:sl3} we have seen that the weights -of any given finite-dimensional module of \(\mathfrak{sl}_2(K)\) or -\(\mathfrak{sl}_3(K)\) can only assume very rigid configurations. For instance, -we have seen that the roots of \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\) are symmetric with respect to the origin. In this -chapter we will generalize most results from chapter~\ref{ch:sl3} regarding the -rigidity of the geometry of the set of weights of a given module. - -As for the aforementioned result on the symmetry of roots, this turns out to be -a general fact, which is a consequence of the non-degeneracy of the restriction -of the Killing form to the Cartan subalgebra. - -\begin{proposition}\label{thm:weights-symmetric-span} - The roots \(\alpha\) of \(\mathfrak{g}\) are symmetrical about the origin -- - i.e. \(- \alpha\) is also a root -- and they span all of \(\mathfrak{h}^*\). -\end{proposition} - -\begin{proof} - We will start with the first claim. Let \(\alpha\) and \(\beta\) be two - roots. Notice \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset - \mathfrak{g}_{\alpha + \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and - \(Y \in \mathfrak{g}_\beta\) then - \[ - [H [X, Y]] - = [X, [H, Y]] - [Y, [H, X]] - = (\alpha + \beta)(H) \cdot [X, Y] - \] - for all \(H \in \mathfrak{h}\). - - This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) - \operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then - \[ - (\operatorname{ad}(X) \operatorname{ad}(Y))^r Z - = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ] - \in \mathfrak{g}_{r \alpha + r \beta + \gamma} - = 0 - \] - for \(r\) large enough. In particular, \(\kappa(X, Y) = - \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if - \(- \alpha\) is not an eigenvalue we find \(\kappa(X, \mathfrak{g}_\beta) = - 0\) for all roots \(\beta\), which contradicts the non-degeneracy of - \(\kappa\). Hence \(- \alpha\) must be an eigenvalue of the adjoint action of - \(\mathfrak{h}\). - - For the second statement, note that if the roots of \(\mathfrak{g}\) do not - span all of \(\mathfrak{h}^*\) then there is some nonzero \(H \in - \mathfrak{h}\) such that \(\alpha(H) = 0\) for all roots \(\alpha\), which is - to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in - \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of - the center \(\mathfrak{z} = 0\) of \(\mathfrak{g}\), a contradiction. -\end{proof} - -Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and -\(\mathfrak{sl}_3(K)\) one can show\dots - -\begin{proposition}\label{thm:root-space-dim-1} - The root spaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional. -\end{proposition} - -The proof of the first statement of -Proposition~\ref{thm:weights-symmetric-span} highlights something interesting: -if we fix some eigenvalue \(\alpha\) of the adjoint action of \(\mathfrak{h}\) -on \(\mathfrak{g}\) and a eigenvector \(X \in \mathfrak{g}_\alpha\), then for -each \(H \in \mathfrak{h}\) and \(m \in M_\lambda\) we find -\[ - H \cdot (X \cdot m) - = X H \cdot m + [H, X] \cdot m - = (\lambda + \alpha)(H) X \cdot m -\] - -Thus \(X\) sends \(m\) to \(M_{\lambda + \alpha}\). We have encountered this -formula twice in these notes: again, we find \(\mathfrak{g}_\alpha\) \emph{acts -on \(M\) by translating vectors between eigenspaces}. In particular, if we -denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) then\dots - -\begin{theorem}\label{thm:weights-congruent-mod-root}\index{weights!root lattice} - The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) are - all congruent modulo the root lattice \(Q = \mathbb{Z} \Delta\) of - \(\mathfrak{g}\). In other words, all weights of \(M\) lie in the same - \(Q\)-coset \(\xi \in \mfrac{\mathfrak{h}^*}{Q}\). -\end{theorem} - -Again, we may leverage our knowledge of \(\mathfrak{sl}_2(K)\) to obtain -further restrictions on the geometry of the locus of weights of \(M\). Namely, -as in the case of \(\mathfrak{sl}_3(K)\) we show\dots - -\begin{proposition}\label{thm:distinguished-subalgebra} - Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace - \(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha} - \oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra - isomorphic to \(\mathfrak{sl}_2(K)\). -\end{proposition} - -\begin{corollary}\label{thm:distinguished-subalg-rep} - For all weights \(\mu\), the subspace - \[ - \bigoplus_k M_{\mu - k \alpha} - \] - is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\) - and the weight spaces in this string match the eigenspaces of \(h\). -\end{corollary} - -The proof of Proposition~\ref{thm:distinguished-subalgebra} is very technical -in nature and we won't include it here, but the idea behind it is simple: -recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both -\(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{- -\alpha}]\) is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{- -\alpha}] \ne 0\) and that no generator of \([\mathfrak{g}_\alpha, -\mathfrak{g}_{- \alpha}]\) is annihilated by \(\alpha\), so that by adjusting -scalars we can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in -\mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\) -satisfies -\begin{align*} - [H_\alpha, F_\alpha] & = -2 F_\alpha & - [H_\alpha, E_\alpha] & = 2 E_\alpha -\end{align*} - -The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely -determined by this condition, but \(H_\alpha\) is. As promised, the second -statement of Corollary~\ref{thm:distinguished-subalg-rep} imposes strong -restrictions on the weights of \(M\). Namely, if \(\lambda\) is a weight, -\(\lambda(H_\alpha)\) is an eigenvalue of \(h\) on some -\(\mathfrak{sl}_2(K)\)-module, so it must be an integer. In other words\dots - -\begin{definition}\label{def:weight-lattice}\index{weights!weight lattice} - The lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) \in - \mathbb{Z} \, \forall \alpha \in \Delta \} \subset \mathfrak{h}^*\) is called - \emph{the weight lattice of \(\mathfrak{g}\)}. We call the elements of \(P\) - \emph{integral}. -\end{definition} - -\begin{proposition}\label{thm:weights-fit-in-weight-lattice} - The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) of - all lie in the weight lattice \(P\). -\end{proposition} - -Proposition~\ref{thm:weights-fit-in-weight-lattice} is clearly analogous to -Corollary~\ref{thm:sl3-weights-fit-in-weight-lattice}. In fact, the weight -lattice of \(\mathfrak{sl}_3(K)\) -- as in Definition~\ref{def:weight-lattice} --- is precisely \(\mathbb{Z} \langle \alpha_1, \alpha_2, \alpha_3 \rangle\). To -proceed further, we would like to take \emph{the highest weight of \(M\)} as in -section~\ref{sec:sl3-reps}, but the meaning of \emph{highest} is again unclear -in this situation. We could simply fix a linear function \(\mathbb{Q} P \to -\mathbb{Q}\) -- as we did in section~\ref{sec:sl3-reps} -- and choose a weight -\(\lambda\) of \(M\) that maximizes this functional, but at this point it is -convenient to introduce some additional tools to our arsenal. These tools are -called \emph{basis}. - -\begin{definition}\label{def:basis-of-root}\index{weights!basis} - A subset \(\Sigma = \{\beta_1, \ldots, \beta_r\} \subset \Delta\) of linearly - independent roots is called \emph{a basis for \(\Delta\)} if, given \(\alpha - \in \Delta\), there are \(k_1, \ldots, k_r \in \mathbb{N}\) such that - \(\alpha = \pm(k_1 \beta_1 + \cdots + k_r \beta_r)\). -\end{definition} - -The interesting thing about basis for \(\Delta\) is that they allow us to -compare weights of a given \(\mathfrak{g}\)-module. At this point the reader -should be asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly -all that intuitive. Well, the thing is that any choice of basis induces a -partial order in \(Q\), where elements are ordered by their \emph{heights}. - -\begin{definition}\index{weights!orderings of roots} - Let \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) be a basis for \(\Delta\). - Given \(\alpha = k_1 \beta_1 + \cdots + k_r \beta_r \in Q\) with \(k_1, - \ldots, k_r \in \mathbb{Z}\), we call the number \(\operatorname{ht}(\alpha) - = k_1 + \cdots + k_r \in \mathbb{Z}\) \emph{the height of \(\alpha\)}. We say - that \(\alpha \preceq \beta\) if \(\operatorname{ht}(\alpha) \le - \operatorname{ht}(\beta)\). -\end{definition} - -\begin{definition} - Given a basis \(\Sigma\) for \(\Delta\), there is a canonical - partition\footnote{Notice that $\operatorname{ht}(\alpha) = 0$ if, and only - if $\alpha = 0$. Since $0$ is, by definition, not a root, the sets $\Delta^+$ - and $\Delta^-$ account for all roots.} \(\Delta^+ \cup \Delta^- = \Delta\), - where \(\Delta^+ = \{ \alpha \in \Delta : \alpha \succ 0 \}\) and \(\Delta^- - = \{ \alpha \in \Delta : \alpha \prec 0 \}\). The elements of \(\Delta^+\) - and \(\Delta^-\) are called \emph{positive} and \emph{negative roots}, - respectively. -\end{definition} - -\begin{definition} - Let \(\Sigma\) be a basis for \(\Delta\). The subalgebra \(\mathfrak{b} = - \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is - called \emph{the Borel subalgebra associated with \(\mathfrak{h}\) and - \(\Sigma\)}. -\end{definition} - -It should be obvious that the binary relation \(\preceq\) in \(Q\) is a partial -order. In addition, we may compare the elements of a given \(Q\)-coset -\(\lambda + Q\) by comparing their difference with \(0 \in Q\). In other words, -given \(\lambda \in \mu + Q\), we say \(\lambda \preceq \mu\) if \(\lambda - -\mu \preceq 0\). In particular, since the weights of \(M\) all lie in a single -\(Q\)-coset, we may compare them in this way. Given a basis \(\Sigma\) for -\(\Delta\) we may take ``the highest weight of \(M\)'' as a maximal weight -\(\lambda\) of \(M\). The obvious question then is: can we always find a basis -for \(\Delta\)? - -\begin{proposition} - There is a basis \(\Sigma\) for \(\Delta\). -\end{proposition} - -The intuition behind the proof of this proposition is similar to our original -idea of fixing a direction in \(\mathfrak{h}^*\) in the case of -\(\mathfrak{sl}_3(K)\). Namely, one can show that \(\kappa(\alpha, \beta) \in -\mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form -\(\kappa\) restricts to a nondegenerate \(\mathbb{Q}\)-bilinear form -\(\mathbb{Q} \Delta \times \mathbb{Q} \Delta \to \mathbb{Q}\). We can then fix -a nonzero vector \(\gamma \in \mathbb{Q} \Delta\) and consider the orthogonal -projection \(f : \mathbb{Q} \Delta \to \mathbb{Q} \gamma \cong \mathbb{Q}\). We -say a root \(\alpha \in \Delta\) is \emph{positive} if \(f(\alpha) > 0\), and -we call a positive root \(\alpha\) \emph{simple} if it cannot be written as the -sum two other positive roots. The subset \(\Sigma \subset \Delta\) of all -simple roots is a basis for \(\Delta\), and all other basis can be shown to -arise in this way. - -Fix some basis \(\Sigma\) for \(\Delta\), with corresponding decomposition -\(\Delta^+ \cup \Delta^- = \Delta\). Let \(\lambda\) be a maximal weight of -\(M\). We call \(\lambda\) \emph{the highest weight of \(M\)}, and we call any -nonzero \(m \in M_\lambda\) \emph{a highest weight vector}. The strategy then -is to describe all weight spaces of \(M\) in terms of \(\lambda\) and \(m\), as -in Theorem~\ref{thm:sl3-irr-weights-class}. Unsurprisingly we do so by -reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). - -First, we note that any highest weight vector \(m \in M_\lambda\) is -annihilated by all positive root spaces, for if \(\alpha \in \Delta^+\) then -\(E_\alpha \cdot m \in M_{\lambda + \alpha}\) must be zero -- or otherwise we -would have that \(\lambda + \alpha\) is a weight with \(\lambda \prec \lambda + -\alpha\). In particular, -\[ - \bigoplus_{k \in \mathbb{Z}} M_{\lambda - k \alpha} - = \bigoplus_{k \in \mathbb{N}} M_{\lambda - k \alpha} -\] -and \(\lambda(H_\alpha)\) is the right-most eigenvalue of the action of \(h\) -on the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k M_{\lambda - k \alpha}\). - -This has a number of important consequences. For instance\dots - -\begin{corollary} - If \(\alpha \in \Delta^+\) and \(\sigma_\alpha : \mathfrak{h}^* \to - \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to - \(\alpha\) with respect to the Killing form, the weights of \(M\) occurring - in the line joining \(\lambda\) and \(\sigma_\alpha\) are precisely the \(\mu - \in P\) lying between \(\lambda\) and \(\sigma_\alpha(\lambda)\). -\end{corollary} - -\begin{proof} - Notice that any \(\mu \in P\) in the line joining \(\lambda\) and - \(\sigma_\alpha(\lambda)\) has the form \(\mu = \lambda - k \alpha\) for some - \(k\), so that \(M_\mu\) corresponds the eigenspace associated with the - eigenvalue \(\lambda(H_\alpha) - 2k\) of the action of \(h\) on \(\bigoplus_k - M_{\lambda - k \alpha}\). If \(\mu\) lies between \(\lambda\) and - \(\sigma_\alpha(\lambda)\) then \(k\) lies between \(0\) and - \(\lambda(H_\alpha)\), in which case \(M_\mu \neq 0\) and therefore \(\mu\) - is a weight. - - On the other hand, if \(\mu\) does not lie between \(\lambda\) and - \(\sigma_\alpha(\lambda)\) then either \(k < 0\) or \(k > - \lambda(H_\alpha)\). Suppose \(\mu\) is a weight. In the first case \(\mu - \succ \lambda\), a contradiction. On the second case the fact that \(M_\mu - \ne 0\) implies \(M_{\lambda + (k - \lambda(H_\alpha)) \alpha} = - M_{\sigma_\alpha(\mu)} \ne 0\), which contradicts the fact that \(M_{\lambda - + \ell \alpha} = 0\) for all \(\ell \ge 0\). -\end{proof} - -This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we -found that the weights of the simple \(\mathfrak{sl}_3(K)\)-modules formed -continuous strings symmetric with respect to the lines \(K \alpha\) with -\(\kappa(\alpha_i - \alpha_j, \alpha) = 0\). As in the case of -\(\mathfrak{sl}_3(K)\), the same class of arguments leads us to the -conclusion\dots - -\begin{definition}\index{Weyl group} - We refer to the group \(W = \langle \sigma_\alpha : \alpha \in - \Delta^+ \rangle \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl - group of \(\mathfrak{g}\)}. -\end{definition} - -\begin{theorem}\label{thm:irr-weight-class} - The weights of a simple \(\mathfrak{g}\)-module \(M\) with highest weight - \(\lambda\) are precisely the elements of the weight lattice \(P\) congruent - to \(\lambda\) modulo the root lattice \(Q\) lying inside the convex hull of - the orbit of \(\lambda\) under the action of the Weyl group \(W\). -\end{theorem} - -\index{Weyl group!actions} -Aside from showing up in the previous theorem, the Weyl group will also play an -important role in chapter~\ref{ch:mathieu} by virtue of the existence of a -canonical action of \(W\) on \(\mathfrak{h}\). By its very nature, -\(W\) acts in \(\mathfrak{h}^*\). If we conjugate the action -\(\sigma\!\restriction_{\mathfrak{h}^*} : \mathfrak{h}^* \isoto -\mathfrak{h}^*\) of some \(\sigma \in W\) by the isomorphism -\(\mathfrak{h}^* \isoto \mathfrak{h}\) afforded by the restriction of the -Killing for to \(\mathfrak{h}\) we get a linear automorphism -\(\sigma\!\restriction_{\mathfrak{h}} : \mathfrak{h} \isoto \mathfrak{h}\). As -it turns out, the \(\sigma\!\restriction_{\mathfrak{h}}\) can be extended to an -automorphism of Lie algebras \(\mathfrak{g} \isoto \mathfrak{g}\). This -translates into the following results, which we do not prove -- but see -\cite[sec.~14.3]{humphreys}. - -\begin{proposition}\label{thm:weyl-group-action} - Given \(\alpha \in \Delta^+\), let\footnote{Notice that since $\mathfrak{g}$ - is finite-dimensional, $\operatorname{ad}(X)$ is nilpotent for each root - vector $X \in \mathfrak{g}$, so that the linear automorphism - $e^{\operatorname{ad}(X)} = \operatorname{Id} + \operatorname{ad}(X) + - \frac{\operatorname{ad}(X)^2}{2!} + \cdots : \mathfrak{g} \isoto - \mathfrak{g}$ is well defined.} \(\tilde{\sigma}_\alpha = - e^{\operatorname{ad}(E_\alpha)} e^{- \operatorname{ad}(F_\alpha)} - e^{\operatorname{ad}(E_\alpha)} : \mathfrak{g} \isoto \mathfrak{g}\). Then - \(\tilde\sigma_\alpha\) is an automorphism of Lie algebras, and this defines - an action of \(W\) on \(\mathfrak{g}\) which is compatible with the - canonical action of \(W\) on \(\mathfrak{h}\) -- i.e. - \(\tilde\sigma\!\restriction_{\mathfrak{h}} = - \sigma\!\restriction_{\mathfrak{h}}\) for all \(\sigma \in W\). -\end{proposition} - -\begin{note} - Notice that the action of \(W\) on \(\mathfrak{g}\) from - Proposition~\ref{thm:weyl-group-action} is not canonical, since it depends on - the choice of \(E_\alpha\) and \(F_\alpha\). Nevertheless, \(\mathfrak{h}\) - is stable under the action of \(W\) -- i.e. \(W \cdot - \mathfrak{h} \subset \mathfrak{h}\) -- and the restriction of this action to - \(\mathfrak{h}\) is independent of any choices. -\end{note} - -We should point out that the results in this section regarding the geometry -roots and weights are only the beginning of a well develop axiomatic theory of -the so called \emph{root systems}, which was used by Cartan in the early 20th -century to classify all finite-dimensional simple complex Lie algebras in terms -of Dynking diagrams. This and much more can be found in \cite[III]{humphreys} -and \cite[ch.~21]{fulton-harris}. Having found all of the weights of \(M\), the -only thing we are missing for a complete classification is an existence and -uniqueness theorem analogous to Theorem~\ref{thm:sl2-exist-unique} and -Theorem~\ref{thm:sl3-existence-uniqueness}. This will be the focus of the next -section. - -\section{Verma Modules \& the Highest Weight Theorem} - -It is already clear from the previous discussion that if \(\lambda\) is the -highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots -\(\alpha\). In other words, having \(\lambda(H_\alpha) \ge 0\), for all -\(\alpha \in \Delta^+\), is a necessary condition for the existence of a simple -\(\mathfrak{g}\)-module with highest weight given by \(\lambda\). Surprisingly, -this condition is also sufficient. In other words\dots - -\begin{definition}\index{weights!dominant weight}\index{weights!integral weight} - An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all - \(\alpha \in \Delta^+\) is referred to as an \emph{dominant integral weight - of \(\mathfrak{g}\)}. -\end{definition} - -\begin{theorem}\label{thm:dominant-weight-theo} - For each dominant integral \(\lambda \in P\) there exists precisely one - finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest weight - is \(\lambda\). -\end{theorem} - -\index{weights!Highest Weight Theorem} This is known as \emph{the Highest -Weight Theorem}, and its proof is the focus of this section. The ``uniqueness'' -part of the theorem follows at once from the arguments used for -\(\mathfrak{sl}_3(K)\). However, the ``existence'' part of the theorem is more -nuanced. - -Our first instinct is, of course, to try to generalize the proof used for -\(\mathfrak{sl}_3(K)\). The issue is that our proof relied heavily on our -knowledge of the roots of \(\mathfrak{sl}_3(K)\). It is thus clear that we need -a more systematic approach for the general setting. We begin by asking a -simpler question: how can we construct \emph{any} -- potentially -infinite-dimensional -- \(\mathfrak{g}\)-module \(M\) of highest weight -\(\lambda\)? In the process of answering this question we will come across a -surprisingly elegant solution to our problem. - -If \(M\) is a module with highest weight vector \(m^+ \in M_\lambda\), we -already know \(H \cdot m^+ = \lambda(H) m^+\) for all \(\mathfrak{h}\) and \(X -\cdot m^+ = 0\) for \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). If -\(M\) is simple we find \(M = \mathcal{U}(\mathfrak{g}) \cdot m^+\), which -implies the restriction of \(M\) to the Borel subalgebra \(\mathfrak{b} \subset -\mathfrak{g}\) has a prescribed action. On the other hand, we have essentially -no information about the action of the rest of \(\mathfrak{g}\) on \(M\). -Nevertheless, given a \(\mathfrak{b}\)-module we may obtain a -\(\mathfrak{g}\)-module by formally extending the action of \(\mathfrak{b}\) -via induction. This leads us to the following definition. - -\begin{definition}\label{def:verma}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules} - The \(\mathfrak{g}\)-module \(M(\lambda) = - \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\), where the action of - \(\mathfrak{b}\) on \(K m^+\) is given by \(H \cdot m^+ = \lambda(H) m^+\) - for all \(H \in \mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in - \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma - module of weight \(\lambda\)}. -\end{definition} - -It turns out that \(M(\lambda)\) enjoys many of the features we've grown used -to in the past chapters. Explicitly\dots - -\begin{proposition}\label{thm:verma-is-weight-mod} - The Verma module \(M(\lambda)\) is generated \(m^+ = 1 \otimes m^+ \in - M(\lambda)\) as in Definition~\ref{def:verma}. The weight spaces - decomposition - \[ - M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu - \] - holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in - \mathfrak{h}^*\) and \(\dim M(\lambda)_\lambda = 1\). Finally, \(\lambda\) is - the highest weight of \(M(\lambda)\), with highest weight vector given by - \(m^+ \in M(\lambda)\). -\end{proposition} - -\begin{proof} - The PBW Theorem implies that \(M(\lambda)\) is spanned by the vectors - \(F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+\) for - \(\Delta^+ = \{\alpha_1, \ldots, \alpha_r\}\) and \(F_{\alpha_i} \in - \mathfrak{g}_{- \alpha_i}\) as in the proof of - Proposition~\ref{thm:distinguished-subalgebra}. But - \[ - \begin{split} - H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+) - & = ([H, F_{\alpha_{i_1}}] + F_{\alpha_{i_1}} H) - F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ - & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ - + F_{\alpha_{i_1}} ([H, F_{\alpha_{i_2}}] + F_{\alpha_{i_2}} H) - F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ - & \;\; \vdots \\ - & = (- \alpha_{i_1} - \cdots - \alpha_{i_s})(H) - F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ - + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m^+ \\ - & = (\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s})(H) - F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\ - & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ - \in M(\lambda)_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}} - \end{split} - \] - - Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*} - M(\lambda)_\mu\), as desired. In fact we have established - \[ - M(\lambda) - \subset - \bigoplus_{k_i \in \mathbb{N}} - M(\lambda)_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r} - \] - where \(\{\alpha_1, \ldots, \alpha_r\} = \Delta^+\), so that all weights of - \(M(\lambda)\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots - - k_r \cdot \alpha_r\). - - This already gives us that the weights of \(M(\lambda)\) are bounded by - \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that - \(m^+\) is nonzero weight vector. Clearly \(m^+ \in M_\lambda\). The - Poincaré-Birkhoff-Witt Theorem implies \(\mathcal{U}(\mathfrak{g})\) is a - free \(\mathfrak{b}\)-module, so that - \[ - M(\lambda) - \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right) - \otimes_{\mathcal{U}(\mathfrak{b})} K m^+ - \cong \bigoplus_i \mathcal{U}(\mathfrak{b}) - \otimes_{\mathcal{U}(\mathfrak{b})} K m^+ - \cong \bigoplus_i K m^+ - \ne 0 - \] - as \(\mathfrak{b}\)-modules. We then conclude \(m^+ \ne 0\) in - \(M(\lambda)\), for if this was not the case we would find \(M(\lambda) = - \mathcal{U}(\mathfrak{g}) \cdot m^+ = 0\). Hence \(M_\lambda \ne 0\) and - therefore \(\lambda\) is the highest weight of \(M(\lambda)\), with highest - weight vector \(m^+\). - - To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only - finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots - F_{\alpha_s}^{k_s}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots - + k_s \cdot \alpha_s\). Since \(M(\lambda)_\mu\) is spanned by the images of - \(m^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In - particular, there is a single monomials \(F_{\alpha_1}^{k_1} - F_{\alpha_2}^{k_2} \cdots F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda - + k_1 \cdot \alpha_1 + \cdots + k_s \cdot \alpha_s\) -- which is, of course, - the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\). -\end{proof} - -\begin{example}\label{ex:sl2-verma} - If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K - h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in - \mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) = - \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action of - \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by the formulas in - (\ref{eq:sl2-verma-formulas}). Visually, - \begin{center} - \begin{tikzcd} - \cdots \rar[bend left=60]{-10} - & M(\lambda)_{-6} \rar[bend left=60]{-4} \lar[bend left=60]{1} - & M(\lambda)_{-4} \rar[bend left=60]{0} \lar[bend left=60]{1} - & M(\lambda)_{-2} \rar[bend left=60]{2} \lar[bend left=60]{1} - & M(\lambda)_0 \rar[bend left=60]{2} \lar[bend left=60]{1} - & M(\lambda)_2 \lar[bend left=60]{1} - \end{tikzcd} - \end{center} - where \(M(\lambda)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows - represent the action of \(e\) and the bottom arrows represent the action of - \(f\). The scalars labeling each arrow indicate to which multiple of \(f^{k - \pm 1} \cdot m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The - string of weight spaces to the left of the diagram is infinite. - \begin{equation}\label{eq:sl2-verma-formulas} - \begin{aligned} - f^k \cdot m^+ & \overset{e}{\mapsto} (2 - k(k + 1)) f^{k - 1} \cdot m^+ & - f^k \cdot m^+ & \overset{f}{\mapsto} f^{k + 1} \cdot m^+ & - f^k \cdot m^+ & \overset{h}{\mapsto} (2 - 2k) f^k \cdot m^+ & - \end{aligned} - \end{equation} -\end{example} - -The Verma module \(M(\lambda)\) should really be though-of as ``the freest -highest weight \(\mathfrak{g}\)-module of weight \(\lambda\)''. Unfortunately -for us, this is not a proof of Theorem~\ref{thm:dominant-weight-theo}, since in -general \(M(\lambda)\) is neither simple nor finite-dimensional. Indeed, the -dimension of \(M(\lambda)\) is the same as the codimension of -\(\mathcal{U}(\mathfrak{b})\) in \(\mathcal{U}(\mathfrak{g})\), which is always -infinite. Nevertheless, we may use \(M(\lambda)\) to prove -Theorem~\ref{thm:dominant-weight-theo} as follows. - -Given a \(\mathfrak{g}\)-module \(M\), any \(\mathfrak{g}\)-homomorphism \(f : -M(\lambda) \to M\) is determined by the image of \(m^+\). Indeed, \(f(u \cdot -m^+) = u \cdot f(m^+)\) for all \(u \in \mathcal{U}(\mathfrak{g})\). In -addition, it is clear that -\[ - H \cdot f(m^+) = f(H \cdot m^+) = f(\lambda(H) m^+) = \lambda(H) f(m^+) -\] -for all \(H \in \mathfrak{h}\) and, similarly, \(X \cdot f(m^+) = 0\) for all -\(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). This leads us to the -universal property of \(M(\lambda)\). - -\begin{definition} - Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M\). If \(X \cdot m = 0\) - for all \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\), then \(m\) is - called \emph{a singular vector of \(M\)}. -\end{definition} - -\begin{proposition} - Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M_\lambda\) be a singular - vector. Then there exists a unique \(\mathfrak{g}\)-homomorphism \(f : - M(\lambda) \to M\) such that \(f(m^+) = m\). Furthermore, all homomorphisms - \(M(\lambda) \to M\) are given in this fashion. - \[ - \operatorname{Hom}_{\mathfrak{g}}(M(\lambda), M) - \cong \{ m \in M_\lambda : m \ \text{is singular}\} - \] -\end{proposition} - -\begin{proof} - The result follows directly from Proposition~\ref{thm:frobenius-reciprocity}. - Indeed, by the Frobenius Reciprocity Theorem, a \(\mathfrak{g}\)-homomorphism - \(f : M(\lambda) \to M\) is the same as a \(\mathfrak{b}\)-homomorphism \(g : - K m^+ \to M = \operatorname{Res}_{\mathfrak{b}}^{\mathfrak{g}} M\). More - specifically, given a \(\mathfrak{b}\)-homomorphism \(g : K m^+ \to M\), - there exists a unique \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) - such that \(f(u \otimes m) = u \cdot g(m)\) for all \(m \in K m^+\), and all - \(\mathfrak{g}\)-homomorphism \(M(\lambda) \to M\) arise in this fashion. - - Any \(K\)-linear map \(g : K m^+ \to M\) is determined by the image - \(g(m^+)\) of \(m^+\) and such an image is a singular vector if, and only if - \(g\) is a \(\mathfrak{b}\)-homomorphism. -\end{proof} - -Notice that any highest weight vector is a singular vector. Now suppose \(M\) -is a simple finite-dimensional \(\mathfrak{g}\)-module of highest weight vector -\(m \in M_\lambda\). By the last proposition, there is a -\(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) = m\). -Since \(M\) is simple, \(M = \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore -\(M \cong \mfrac{M(\lambda)}{\ker f}\). It then follows from the simplicity of -\(M\) that \(\ker f \subset M(\lambda)\) is a maximal -\(\mathfrak{g}\)-submodule. Maximal submodules of Verma modules are thus of -primary interest to us. As it turns out, these can be easily classified. - -\begin{proposition}\label{thm:max-verma-submod-is-weight} - Every submodule \(N \subset M(\lambda)\) is the direct sum of its weight - spaces. In particular, \(M(\lambda)\) has a unique maximal submodule - \(N(\lambda)\) and a unique simple quotient \(L(\lambda) = - \sfrac{M(\lambda)}{N(\lambda)}\). -\end{proposition} - -\begin{proof} - Let \(N \subset M(\lambda)\) be a submodule and take any nonzero \(n \in N\). - Because of Proposition~\ref{thm:verma-is-weight-mod}, we know there are - \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in - M(\lambda)_{\mu_i}\) such that \(n = m_1 + \cdots + m_r\). We want to show - \(m_i \in N\) for all \(i\). - - Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\). - Then - \[ - m_1 - - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_3 - - \cdots - - \frac{(\mu_r - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_r - = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n - \in N - \] - - Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By - applying the same procedure again we get - \begin{multline*} - m_1 - - - \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)} - {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_4 - - \cdots - - \frac{(\mu_r - \mu_3)(H_3) \cdot (\mu_r - \mu_1)(H_2)} - {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_r \\ - = - \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right) - \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n - \in N - \end{multline*} - - By applying the same procedure over and over again we can see that \(m_1 = u - \cdot n \in N\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if - we reproduce all this for \(m_2 + \cdots + m_r = n - m_1 \in N\) we get that - \(m_2 \in N\). All in all we find \(m_1, \ldots, m_r \in N\). Hence - \[ - N = \bigoplus_\mu N_\mu = \bigoplus_\mu M(\lambda)_\mu \cap N - \] - - Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\), if \(N\) is a - proper submodule then \(m^+ \notin N\). Hence any proper submodule lies in - the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum - \(N(\lambda)\) of all such submodules is still proper. This implies - \(N(\lambda)\) is the unique maximal submodule of \(M(\lambda)\) and - \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique simple - quotient. -\end{proof} - -\begin{example}\label{ex:sl2-verma-quotient} - If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we - can see from Example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge - 3} K f^k \cdot m^+\), so that \(L(\lambda)\) is the \(3\)-dimensional simple - \(\mathfrak{sl}_2(K)\)-module -- i.e. the finite-dimensional simple module - with highest weight \(\lambda\) constructed in chapter~\ref{ch:sl3}. -\end{example} - -All its left to prove the Highest Weight Theorem is verifying that the -situation encountered in Example~\ref{ex:sl2-verma-quotient} holds for any -\(\lambda \in P\). In other words, we need to show\dots - -\begin{proposition}\label{thm:verma-is-finite-dim} - If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then - the unique simple quotient \(L(\lambda)\) of \(M(\lambda)\) is - finite-dimensional. -\end{proposition} - -The proof of Proposition~\ref{thm:verma-is-finite-dim} is very technical and we -won't include it here, but the idea behind it is to show that the set of -weights of \(L(\lambda)\) is stable under the natural action of the Weyl group -\(W\) on \(\mathfrak{h}^*\). One can then show that the every weight -of \(L(\lambda)\) is conjugate to a single dominant integral weight of -\(L(\lambda)\), and that the set of dominant integral weights of \(L(\lambda)\) -is finite. Since \(W\) is finitely generated, this implies the set of -weights of the unique simple quotient of \(M(\lambda)\) is finite. But -each weight space is finite-dimensional. Hence so is the simple quotient -\(L(\lambda)\). - -We refer the reader to \cite[ch. 21]{humphreys} for further details. We are now -ready to prove the Highest Weight Theorem. - -\begin{proof}[Proof of Theorem~\ref{thm:dominant-weight-theo}] - We begin with the ``existence'' part of the theorem by showing that - \(L(\lambda)\) is indeed a finite-dimensional simple module whose - highest-weight is \(\lambda\). It suffices to show that the highest weight of - \(L(\lambda)\) is \(\lambda\). We have already seen that \(m^+ \in - M(\lambda)_\lambda\) is a highest weight vector. Now since \(m^+\) lies - outside of the maximal submodule of \(M(\lambda)\), the projection \(m^+ + - N(\lambda) \in L(\lambda)\) is nonzero. - - We now claim that \(m^+ + N(\lambda) \in L(\lambda)_\lambda\). Indeed, - \[ - H \cdot (m^+ + N(\lambda)) - = H \cdot m^+ + N(\lambda) - = \lambda(H) (m^+ + N(\lambda)) - \] - for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of - \(L(\lambda)\), with weight vector \(m^+ + N(\lambda)\). Finally, we remark - that \(\lambda\) is the highest weight of \(L(\lambda)\), for if this was not - the case we could find a weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ - \lambda\). - - Now suppose \(M\) is some other finite-dimensional simple - \(\mathfrak{g}\)-module with highest weight vector \(m \in M_\lambda\). By - the universal property of the Verma module, there is a - \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) = - m\). As indicated before, since \(M\) is simple, \(M = - \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore \(f\) is surjective. It - then follows \(M \cong \mfrac{M(\lambda)}{\ker f}\). - - Since \(M\) is simple, \(\ker f \subset M(\lambda)\) is maximal and therefore - \(\ker f = N(\lambda)\). In other words, \(M \cong \mfrac{M(\lambda)}{\ker f} - = L(\lambda)\). We are done. -\end{proof} - -We should point out that Proposition~\ref{thm:verma-is-finite-dim} fails for -non-dominant \(\lambda \in P\). While \(\lambda\) is always a maximal weight of -\(M(\lambda)\), one can show that if \(\lambda \in P\) is not dominant then -\(N(\lambda) = 0\) and \(M(\lambda)\) is simple. For instance, if -\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto -2\) then the -action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by -\begin{center} - \begin{tikzcd} - \cdots \rar[bend left=60]{-20} - & M(\lambda)_{-8} \rar[bend left=60]{-12} \lar[bend left=60]{1} - & M(\lambda)_{-6} \rar[bend left=60]{-6} \lar[bend left=60]{1} - & M(\lambda)_{-4} \rar[bend left=60]{-2} \lar[bend left=60]{1} - & M(\lambda)_{-2} \lar[bend left=60]{1} - \end{tikzcd}, -\end{center} -so we can see that \(M(\lambda)\) has no proper submodules. Verma modules can -thus serve as examples of infinite-dimensional simple modules. Our next -question is: what are \emph{all} the infinite-dimensional simple -\(\mathfrak{g}\)-modules?
diff --git /dev/null b/sections/simple-weight.tex @@ -0,0 +1,1611 @@ +\chapter{Simple Weight Modules}\label{ch:mathieu} + +In this chapter we will expand our results on finite-dimensional simple modules +of semisimple Lie algebras by considering \emph{infinite-dimensional} +\(\mathfrak{g}\)-modules, which introduces numerous complications to our +analysis. + +For instance, in the infinite-dimensional setting we can no longer take +complete-reducibility for granted. Indeed, we have seen that even if +\(\mathfrak{g}\) is a semisimple Lie algebra, there are infinite-dimensional +\(\mathfrak{g}\)-modules which are not semisimple. For a counterexample look no +further than Example~\ref{ex:regular-mod-is-not-semisimple}: the regular +\(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) is never semisimple. +Nevertheless, for simplicity -- or shall we say \emph{semisimplicity} -- we +will focus exclusively on \emph{semisimple} \(\mathfrak{g}\)-modules. Our +strategy is, once again, that of classifying simple modules. The regular +\(\mathfrak{g}\)-module hides further unpleasant surprises, however: recall +from Example~\ref{ex:regular-mod-is-not-weight-mod} that +\[ + \bigoplus_\lambda \mathcal{U}(\mathfrak{g})_\lambda + = 0 + \subsetneq \mathcal{U}(\mathfrak{g}) +\] +and the weight space decomposition fails for \(\mathcal{U}(\mathfrak{g})\). + +Indeed, our proof of the weight space decomposition in the finite-dimensional +case relied heavily in the simultaneous diagonalization of commuting operators +in a finite-dimensional space. Even if we restrict ourselves to simple modules, +there is still a diverse spectrum of counterexamples to +Corollary~\ref{thm:finite-dim-is-weight-mod} in the infinite-dimensional +setting. For instance, any \(\mathfrak{g}\)-module \(M\) whose restriction to +\(\mathfrak{h}\) is a free module satisfies \(M_\lambda = 0\) for all +\(\lambda\) as in Example~\ref{ex:regular-mod-is-not-weight-mod}. These are +called \emph{\(\mathfrak{h}\)-free \(\mathfrak{g}\)-modules}, and rank \(1\) +simple \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules where first +classified by Nilsson in \cite{nilsson}. Dimitar's construction of the so +called \emph{exponential tensor \(\mathfrak{sl}_n(K)\)-modules} in +\cite{dimitar-exp} is also an interesting source of counterexamples. + +Since the weight space decomposition was perhaps the single most instrumental +ingredient of our previous analysis, it is only natural to restrict ourselves +to the case it holds. This brings us to the following definition. + +\begin{definition}\label{def:weight-mod}\index{\(\mathfrak{g}\)-module!weight modules}\index{weights!weight modules}\index{\(\mathfrak{g}\)-module!(essential) support} + A \(\mathfrak{g}\)-module \(M\) is called a \emph{weight + \(\mathfrak{g}\)-module} if \(M = \bigoplus_{\lambda \in \mathfrak{h}^*} + M_\lambda\) and \(\dim M_\lambda < \infty\) for all \(\lambda \in + \mathfrak{h}^*\). The \emph{support of \(M\)} is the set + \(\operatorname{supp} M = \{\lambda \in \mathfrak{h}^* : M_\lambda \ne 0\}\). +\end{definition} + +\begin{example} + Corollary~\ref{thm:finite-dim-is-weight-mod} is equivalent to the fact that + every finite-dimensional module of a semisimple Lie algebra is a weight + module. More generally, every finite-dimensional simple module of a reductive + Lie algebra is a weight module. +\end{example} + +\begin{example}\label{ex:reductive-alg-equivalence} + We have seen that every finite-dimensional \(\mathfrak{g}\)-module is a + weight module for semisimple \(\mathfrak{g}\). In particular, if + \(\mathfrak{g}\) is finite-dimensional then the adjoint + \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module. More generally, + a finite-dimensional Lie algebra \(\mathfrak{g}\) is reductive if, and only + if the adjoint \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module, + in which case its weight spaces are given by the root spaces of + \(\mathfrak{g}\) +\end{example} + +\begin{example}\label{ex:submod-is-weight-mod} + Proposition~\ref{thm:verma-is-weight-mod} and + Proposition~\ref{thm:max-verma-submod-is-weight} imply that the Verma module + \(M(\lambda)\) and its maximal submodule are both weight modules. In + fact, the proof of Proposition~\ref{thm:max-verma-submod-is-weight} is + actually a proof of the fact that every submodule \(N \subset M\) of + a weight module \(M\) is a weight module, and \(N_\lambda = M_\lambda \cap + N\) for all \(\lambda \in \mathfrak{h}^*\). +\end{example} + +\begin{example}\label{ex:quotient-is-weight-mod} + Given a weight module \(M\), a submodule \(N \subset M\) and \(\lambda \in + \mathfrak{h}^*\), it is clear that \(\mfrac{M_\lambda}{N} \subset + \left(\mfrac{M}{N}\right)_\lambda\). In addition, \(\mfrac{M}{N} = + \bigoplus_{\lambda \in \mathfrak{h}^*} \mfrac{M_\lambda}{N}\). Hence + \(\mfrac{M}{N}\) is weight \(\mathfrak{g}\)-module with + \(\left(\mfrac{M}{N}\right)_\lambda = \mfrac{M_\lambda}{N} \cong + \mfrac{M_\lambda}{N_\lambda}\). +\end{example} + +\begin{example}\label{ex:tensor-prod-of-weight-is-weight} + Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras, \(M_1\) be a + weight \(\mathfrak{g}_1\)-module and \(M_2\) a weight + \(\mathfrak{g}_2\)-module. Recall from Example~\ref{ex:cartan-direct-sum} + that if \(\mathfrak{h}_i \subset \mathfrak{g}_i\) are Cartan subalgebras then + \(\mathfrak{h} = \mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan + subalgebra of \(\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2\) with + \(\mathfrak{h}^* = \mathfrak{h}_1^* \oplus \mathfrak{h}_2^*\). In this + setting, one can readily check that \(M_1 \otimes M_2\) is a weight + \(\mathfrak{g}\)-module with + \[ + (M_1 \otimes M_2)_{\lambda_1 + \lambda_2} + = (M_1)_{\lambda_1} \otimes (M_2)_{\lambda_2} + \] + for all \(\lambda_i \in \mathfrak{h}_i^*\) and \(\operatorname{supp} M_1 + \otimes M_2 = \operatorname{supp} M_1 \oplus \operatorname{supp} M_2 = \{ + \lambda_1 + \lambda_2 : \lambda_i \in \operatorname{supp} M_i \subset + \mathfrak{h}_i^*\}\). +\end{example} + +\begin{example}\label{thm:simple-weight-mod-is-tensor-prod} + Let \(\mathfrak{g} = \mathfrak{z} \oplus \mathfrak{s}_1 \oplus \cdots \oplus + \mathfrak{s}_r\) be a reductive Lie algebra, where \(\mathfrak{z}\) is the + center of \(\mathfrak{g}\) and \(\mathfrak{s}_1, \ldots, \mathfrak{s}_r\) are + its simple components. As in + Example~\ref{ex:all-simple-reps-are-tensor-prod}, any simple weight + \(\mathfrak{g}\)-module \(M\) can be decomposed as + \[ + M \cong Z \otimes M_1 \otimes \cdots \otimes M_r + \] + where \(Z\) is a \(1\)-dimensional representation of \(\mathfrak{z}\) and + \(M_i\) is a simple weight \(\mathfrak{s}_i\)-module. The modules \(Z\) and + \(M_i\) are uniquely determined up to isomorphism. +\end{example} + +\begin{example}\label{ex:adjoint-action-in-universal-enveloping-is-weight} + We would like to show that the requirement of finite-dimensionality in + Definition~\ref{def:weight-mod} is not redundant. Let \(\mathfrak{g}\) be a + finite-dimensional reductive Lie algebra and consider the adjoint + \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) -- where \(X \in + \mathfrak{g}\) acts by taking commutators. Given \(\alpha \in Q\), a simple + computation shows \(K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in + \mathfrak{g}_{\alpha_i}, H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha = + \alpha_1 + \cdots + \alpha_n \rangle \subset + \mathcal{U}(\mathfrak{g})_\alpha\). The PBW Theorem and + Example~\ref{ex:reductive-alg-equivalence} thus imply that + \(\mathcal{U}(\mathfrak{g}) = \bigoplus_{\alpha \in Q} + \mathcal{U}(\mathfrak{g})_\alpha\) where \(\mathcal{U}(\mathfrak{g})_\alpha = + K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in \mathfrak{g}_{\alpha_i}, + H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha = \alpha_1 + \cdots + + \alpha_n \rangle\). However, \(\dim \mathcal{U}(\mathfrak{g})_\alpha = + \infty\). For instance, \(\mathcal{U}(\mathfrak{g})_0\) is \emph{precisely} + the commutator of \(\mathfrak{h}\) in \(\mathcal{U}(\mathfrak{g})\), which + contains \(\mathcal{U}(\mathfrak{h})\) and is therefore infinite-dimensional. +\end{example} + +\begin{note} + We should stress that the weight spaces \(M_\lambda \subset M\) of a given + weight \(\mathfrak{g}\)-module \(M\) are \emph{not} + \(\mathfrak{g}\)-submodules. Nevertheless, \(M_\lambda\) is a + \(\mathfrak{h}\)-submodule. More generally, \(M_\lambda\) is a + \(\mathcal{U}(\mathfrak{g})_0\)-submodule, where + \(\mathcal{U}(\mathfrak{g})_0\) is the centralizer of \(\mathfrak{h}\) in + \(\mathcal{U}(\mathfrak{g})\) -- which coincides with the weight space of \(0 + \in \mathfrak{h}^*\) in the adjoint \(\mathfrak{g}\)-module + \(\mathcal{U}(\mathfrak{g})\), as seen in + Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}. +\end{note} + +A particularly well behaved class of examples are the so called +\emph{bounded} modules. + +\begin{definition}\index{\(\mathfrak{g}\)-module!bounded modules}\index{\(\mathfrak{g}\)-module!(essential) support} + A weight \(\mathfrak{g}\)-module \(M\) is called \emph{bounded} if \(\dim + M_\lambda\) is bounded. The lowest upper bound \(\deg M\) for \(\dim + M_\lambda\) is called \emph{the degree of \(M\)}. The \emph{essential + support} of \(M\) is the set \(\operatorname{supp}_{\operatorname{ess}} M = + \{ \lambda \in \mathfrak{h}^* : \dim M_\lambda = \deg M \}\). +\end{definition} + +\begin{example}\label{ex:supp-ess-of-tensor-is-product} + Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras with Cartan + subalgebras \(\mathfrak{h}_i \subset \mathfrak{g}_i\) and take \(\mathfrak{g} + = \mathfrak{g}_1 \oplus \mathfrak{g}_2\). Given bounded + \(\mathfrak{g}_i\)-modules \(M_i\), it follows from + Example~\ref{ex:tensor-prod-of-weight-is-weight} that \(M_1 \otimes M_2\) is + a bounded \(\mathfrak{g}\)-module with \(\deg M_1 \otimes M_2 = \deg M_1 + \cdot \deg M_2\) and + \[ + \operatorname{supp}_{\operatorname{ess}} M_1 \otimes M_2 + = \operatorname{supp}_{\operatorname{ess}} M_1 \oplus + \operatorname{supp}_{\operatorname{ess}} M_2 + = \{ + \lambda_1 + \lambda_2 : \lambda_i \in + \operatorname{supp}_{\operatorname{ess}} M_i \subset \mathfrak{h}_i^* + \} + \] +\end{example} + +\begin{example}\label{ex:laurent-polynomial-mod} + There is a natural action of \(\mathfrak{sl}_2(K)\) on the space \(K[x, + x^{-1}]\) of Laurent polynomials, given by the formulas in + (\ref{eq:laurent-polynomials-cusp-mod}). One can quickly verify \(K[x, + x^{-1}]_{2 k} = K x^k\) and \(K[x, x^{-1}]_\lambda = 0\) for any \(\lambda + \notin 2 \mathbb{Z}\), so that \(K[x, x^{-1}] = \bigoplus_{k \in \mathbb{Z}} + K x^k\) is a degree \(1\) bounded weight \(\mathfrak{sl}_2(K)\)-module. It + follows from the remark at the end of Example~\ref{ex:submod-is-weight-mod} + that any nonzero submodule \(N \subset K[x, x^{-1}]\) must contain a + monomial \(x^k\). But since the operators \(-\frac{\mathrm{d}}{\mathrm{d}x} + + \frac{x^{-1}}{2}, x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} : K[x, + x^{-1}] \to K[x, x^{-1}]\) are both injective, this implies all other + monomials can be found in \(N\) by successively applying \(f\) and \(e\). + Hence \(N = K[x, x^{-1}]\) and \(K[x, x^{-1}]\) is a simple module. + \begin{align}\label{eq:laurent-polynomials-cusp-mod} + e \cdot p + & = \left( x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} \right) p & + f \cdot p + & = \left(- \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x^{-1}}{2} \right) p & + h \cdot p + & = 2 x \frac{\mathrm{d}}{\mathrm{d}x} p + \end{align} +\end{example} + +Notice that the support of \(K[x, x^{-1}]\) is the trivial \(2 +\mathbb{Z}\)-coset \(0 + 2 \mathbb{Z}\). This is representative of the general +behavior in the following sense: if \(M\) is a simple weight +\(\mathfrak{g}\)-module, since \(M[\lambda] = \bigoplus_{\alpha \in Q} +M_{\lambda + \alpha}\) is stable under the action of \(\mathfrak{g}\) for all +\(\lambda \in \mathfrak{h}^*\), \(\bigoplus_{\alpha \in Q} M_{\lambda + +\alpha}\) is either \(0\) or all of \(M\). In other words, the support of a +simple weight module is always contained in a single \(Q\)-coset. + +However, the behavior of \(K[x, x^{-1}]\) deviates from that of an arbitrary +bounded \(\mathfrak{g}\)-module in the sense its essential support is +precisely the entire \(Q\)-coset it inhabits -- i.e. +\(\operatorname{supp}_{\operatorname{ess}} K[x, x^{-1}] = 2 \mathbb{Z}\). This +isn't always the case. Nevertheless, in general we find\dots + +\begin{proposition}\label{thm:ess-supp-is-zariski-dense} + Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra and \(M\) + be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module. The + essential support \(\operatorname{supp}_{\operatorname{ess}} M\) is + Zariski-dense\footnote{Any choice of basis for $\mathfrak{h}^*$ induces a + $K$-linear isomorphism $\mathfrak{h}^* \isoto K^n$. In particular, a choice + of basis induces a unique topology in $\mathfrak{h}^*$ such that the map + $\mathfrak{h}^* \to K^n$ is a homeomorphism onto $K^n$ with the Zariski + topology. Any two basis induce the same topology in $\mathfrak{h}^*$, which + we call \emph{the Zariski topology of $\mathfrak{h}^*$}.} in + \(\mathfrak{h}^*\). +\end{proposition} + +This proof was deemed too technical to be included in here, but see Proposition +3.5 of \cite{mathieu} for the case where \(\mathfrak{g} = \mathfrak{s}\) is a +simple Lie algebra. The general case then follows from +Example~\ref{thm:simple-weight-mod-is-tensor-prod}, +Example~\ref{ex:supp-ess-of-tensor-is-product} and the asserting that the +product of Zariski-dense subsets in \(K^n\) and \(K^m\) is Zariski-dense in +\(K^{n + m} = K^n \times K^m\). + +We now begin a systematic investigation of the problem of classifying the +infinite-dimensional simple weight modules of a given Lie algebra +\(\mathfrak{g}\). As in the previous chapter, let \(\mathfrak{g}\) be a +finite-dimensional semisimple Lie algebra. As a first approximation of a +solution to our problem, we consider the induction functors +\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} : +\mathfrak{p}\text{-}\mathbf{Mod} \to \mathfrak{g}\text{-}\mathbf{Mod}\), where +\(\mathfrak{p} \subset \mathfrak{g}\) is some subalgebra. + +% TODOO: Are you sure that these are indeed the weight spaces of the induced +% module? Check this out? +These functors have already proved themselves a powerful tool for constructing +modules in the previous chapters. Our first observation is that if +\(\mathfrak{p} \subset \mathfrak{g}\) contains the Borel subalgebra +\(\mathfrak{b}\) then \(\mathfrak{h}\) is a Cartan subalgebra of +\(\mathfrak{p}\) and \((\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} +M)_\lambda = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{p})} +M_\lambda\) for all \(\lambda \in \mathfrak{h}^*\). In particular, +\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}}\) takes weight +\(\mathfrak{p}\)-modules to weight \(\mathfrak{g}\)-modules. This leads us to +the following definition. + +\begin{definition}\index{Lie subalgebra!parabolic subalgebra} + A subalgebra \(\mathfrak{p} \subset \mathfrak{g}\) is called \emph{parabolic} + if \(\mathfrak{b} \subset \mathfrak{p}\). +\end{definition} + +% TODOO: Why is the fact that p is not reductive relevant?? Why do we need to +% look at the quotient by nil(p)?? +Parabolic subalgebras thus give us a process for constructing weight +\(\mathfrak{g}\)-modules from modules of smaller (parabolic) subalgebras. Our +hope is that by iterating this process again and again we can get a large class +of simple weight \(\mathfrak{g}\)-modules. However, there is a small catch: a +parabolic subalgebra \(\mathfrak{p} \subset \mathfrak{g}\) needs not to be +reductive. We can get around this limitation by modding out by +\(\mathfrak{nil}(\mathfrak{p})\) and noticing that +\(\mathfrak{nil}(\mathfrak{p})\) acts trivially in any weight +\(\mathfrak{p}\)-module \(M\). By applying the universal property of quotients +we can see that \(M\) has the natural structure of a +\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module, which is always +a reductive algebra. +\begin{center} + \begin{tikzcd} + \mathfrak{p} \rar \dar & + \mathfrak{gl}(M) \\ + \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})} \arrow[dotted]{ur} & + \end{tikzcd} +\end{center} + +Let \(\mathfrak{p}\) be a parabolic subalgebra and \(M\) be a simple +weight \(\mathfrak{p}\)-module. We should point out that while +\(\operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} M\) is a weight +\(\mathfrak{g}\)-module, it isn't necessarily simple. Nevertheless, we can +use it to produce a simple weight \(\mathfrak{g}\)-module via a +construction very similar to that of Verma modules. + +\begin{definition}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules} + Given any \(\mathfrak{p}\)-module \(M\), the module \(M_{\mathfrak{p}}(M) = + \operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} M\) is called \emph{a + generalized Verma module}. +\end{definition} + +\begin{proposition}\label{thm:generalized-verma-has-simple-quotient} + Given a simple \(\mathfrak{p}\)-module \(M\), the generalized Verma module + \(M_{\mathfrak{p}}(M)\) has a unique maximal \(\mathfrak{p}\)-submodule + \(N_{\mathfrak{p}}(M)\) and a unique irreducible quotient + \(L_{\mathfrak{p}}(M) = \mfrac{M_{\mathfrak{p}}(M)}{N_{\mathfrak{p}}(M)}\). + The irreducible quotient \(L_{\mathfrak{p}}(M)\) is a weight module. +\end{proposition} + +The proof of Proposition~\ref{thm:generalized-verma-has-simple-quotient} is +entirely analogous to that of Proposition~\ref{thm:max-verma-submod-is-weight}. +This leads us to the following definitions. + +\begin{definition}\index{\(\mathfrak{g}\)-module!parabolic induced modules}\index{\(\mathfrak{g}\)-module!cuspidal modules} + A \(\mathfrak{g}\)-module is called \emph{parabolic induced} if it is + isomorphic to \(L_{\mathfrak{p}}(M)\) for some proper parabolic subalgebra + \(\mathfrak{p} \subsetneq \mathfrak{g}\) and some \(\mathfrak{p}\)-module + \(M\). An \emph{simple cuspidal \(\mathfrak{g}\)-module} is a simple + \(\mathfrak{g}\)-module which is \emph{not} parabolic induced. +\end{definition} + +Since every weight \(\mathfrak{p}\)-module \(M\) is an +\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module, it makes sense +to call \(M\) \emph{cuspidal} if it is a cuspidal +\(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module. The first +breakthrough regarding our classification problem was given by Fernando in his +now infamous paper \citetitle{fernando} \cite{fernando}, where he proved that +every simple weight \(\mathfrak{g}\)-module is parabolic induced. In other +words\dots + +\begin{theorem}[Fernando] + Any simple weight \(\mathfrak{g}\)-module is isomorphic to + \(L_{\mathfrak{p}}(M)\) for some parabolic subalgebra \(\mathfrak{p} \subset + \mathfrak{g}\) and some simple cuspidal \(\mathfrak{p}\)-module \(M\). +\end{theorem} + +We should point out that the relationship between simple weight +\(\mathfrak{g}\)-modules and pairs \((\mathfrak{p}, M)\) -- where +\(\mathfrak{p}\) is some parabolic subalgebra and \(M\) is a simple cuspidal +\(\mathfrak{p}\)-module -- is not one-to-one. Nevertheless, this relationship +is well understood. Namely, Fernando himself established\dots + +\begin{proposition}[Fernando] + Given a parabolic subalgebra \(\mathfrak{p} \subset \mathfrak{g}\), there + exists a basis \(\Sigma\) for \(\Delta\) such that \(\Sigma \subset + \Delta_{\mathfrak{p}} \subset \Delta\), where \(\Delta_{\mathfrak{p}}\) + denotes the set of roots of \(\mathfrak{p}\). Furthermore, if \(\mathfrak{p}' + \subset \mathfrak{g}\) is another parabolic subalgebra, \(M\) is a simple + cuspidal \(\mathfrak{p}\)-module and \(N\) is a simple cuspidal + \(\mathfrak{p}'\)-module then \(L_{\mathfrak{p}}(M) \cong + L_{\mathfrak{p}'}(N)\) if, and only if \(\mathfrak{p}' = + \twisted{\mathfrak{p}}{\sigma}\) and \(M \cong \twisted{N}{\sigma}\) for + some\footnote{Here $\twisted{\mathfrak{p}}{\sigma}$ denotes the image of + $\mathfrak{p}$ under the automorphism of $\sigma : \mathfrak{g} \to + \mathfrak{g}$ given by the canonical action of $W$ on $\mathfrak{g}$ and + $\twisted{N}{\sigma}$ is the $\mathfrak{p}$-module given by composing the map + $\mathfrak{p}' \to \mathfrak{gl}(N)$ with the restriction + $\sigma\!\restriction_{\mathfrak{p}} : \mathfrak{p} \to \mathfrak{p}'$.} + \(\sigma \in W_M\), where + \[ + W_M + = \langle + \sigma_\beta : \beta \in \Sigma, H_\beta + \mathfrak{nil}(\mathfrak{p}) + \ \text{is central in}\ \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})} + \ \text{and}\ H_\beta\ \text{acts on \(M\) as a positive integer} + \rangle + \subset W + \] +\end{proposition} + +\begin{note} + The definition of the subgroup \(W_M \subset W\) is independent of the choice + of basis \(\Sigma\). +\end{note} + +As a first consequence of Fernando's Theorem, we provide two alternative +characterizations of cuspidal modules. + +\begin{corollary}[Fernando]\label{thm:cuspidal-mod-equivs} + Let \(M\) be a simple weight \(\mathfrak{g}\)-module. The following + conditions are equivalent. + \begin{enumerate} + \item \(M\) is cuspidal. + \item \(F_\alpha\) acts injectively on \(M\) for all + \(\alpha \in \Delta\) -- this is what is usually referred + to as a \emph{dense} module in the literature. + \item The support of \(M\) is precisely one \(Q\)-coset -- this is + what is usually referred to as a \emph{torsion-free} module in the + literature. + \end{enumerate} +\end{corollary} + +\begin{example} + As noted in Example~\ref{ex:laurent-polynomial-mod}, the element \(f \in + \mathfrak{sl}_2(K)\) acts injectively on the space of Laurent polynomials. + Hence \(K[x, x^{-1}]\) is a cuspidal \(\mathfrak{sl}_2(K)\)-module. +\end{example} + +Having reduced our classification problem to that of classifying simple +cuspidal modules, we are now faced the daunting task of actually classifying +them. Historically, this was first achieved by Olivier Mathieu in the early +2000's in his paper \citetitle{mathieu} \cite{mathieu}. To do so, Mathieu +introduced new tools which have since proved themselves remarkably useful +throughout the field, known as\dots + +\section{Coherent Families} + +We begin our analysis with a simple question: how to do we go about +constructing cuspidal modules? Specifically, given a cuspidal +\(\mathfrak{g}\)-module, how can we use it to produce new cuspidal modules? To +answer this question, we look back at the single example of a cuspidal module +we have encountered so far: the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\) +of Laurent polynomials -- i.e. Example~\ref{ex:laurent-polynomial-mod}. + +Our first observation is that \(\mathfrak{sl}_2(K)\) acts on \(K[x, x^{-1}]\) +via differential operators. In other words, the action map +\(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{End}(K[x, x^{-1}])\) +factors through the inclusion of the algebra \(\operatorname{Diff}(K[x, +x^{-1}]) = K\left[x, x^{-1}, \frac{\mathrm{d}}{\mathrm{d}x}\right]\) of +differential operators in \(K[x, x^{-1}]\). +\begin{center} + \begin{tikzcd} + \mathcal{U}(\mathfrak{sl}_2(K)) \rar & + \operatorname{Diff}(K[x, x^{-1}]) \rar & + \operatorname{End}(K[x, x^{-1}]) + \end{tikzcd} +\end{center} + +The space \(K[x, x^{-1}]\) can be regarded as a \(\operatorname{Diff}(K[x, +x^{-1}])\)-module in the natural way, and we can produce new +\(\operatorname{Diff}(K[x, x^{-1}])\)-modules by twisting \(K[x, x^{-1}]\) by +automorphisms of \(\operatorname{Diff}(K[x, x^{-1}])\). For example, given +\(\lambda \in K\) we may take the automorphism +\begin{align*} + \varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) & + \to \operatorname{Diff}(K[x, x^{-1}]) \\ + x & \mapsto x \\ + x^{-1} & \mapsto x^{-1} \\ + \frac{\mathrm{d}}{\mathrm{d} x} & \mapsto \frac{\mathrm{d}}{\mathrm{d} x} + + \frac{\lambda}{2} x^{-1} +\end{align*} +and consider the twisted module \(\twisted{K[x, x^{-1}]}{\varphi_\lambda} = +K[x, x^{-1}]\), where some operator \(P \in \operatorname{Diff}(K[x, x^{-1}])\) +acts as \(\varphi_\lambda(P)\). + +By composing the action map \(\operatorname{Diff}(K[x, x^{-1}]) \to +\operatorname{End}(\twisted{K[x, x^{-1}]}{\varphi_\lambda})\) with the +homomorphism of algebras \(\mathcal{U}(\mathfrak{sl}_2(K)) \to +\operatorname{Diff}(K[x, x^{-1}])\) we can give \(\twisted{K[x, +x^{-1}]}{\varphi_\lambda}\) the structure of an \(\mathfrak{sl}_2(K)\)-module. +Diagrammatically, we have +\begin{center} + \begin{tikzcd} + \mathcal{U}(\mathfrak{sl}_2(K)) \rar & + \operatorname{Diff}(K[x, x^{-1}]) \rar{\varphi_\lambda} & + \operatorname{Diff}(K[x, x^{-1}]) \rar & + \operatorname{End}(K[x, x^{-1}]) + \end{tikzcd}, +\end{center} +where the maps \(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x, +x^{-1}])\) and \(\operatorname{Diff}(K[x, x^{1}]) \to \operatorname{End}(K[x, +x^{-1}])\) are the ones from the previous diagram. + +Explicitly, we find that the action of \(\mathfrak{sl}_2(K)\) on +\(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is given by +\begin{align*} + p & \overset{e}{\mapsto} + \left( + x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 + \lambda}{2} x + \right) p & + p & \overset{f}{\mapsto} + \left( + - \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 - \lambda}{2} x^{-1} + \right) p & + p & \overset{h}{\mapsto} + \left( 2 x \frac{\mathrm{d}}{\mathrm{d}x} + \lambda \right) p, +\end{align*} +so we can see \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}_{2 k + +\frac{\lambda}{2}} = K x^k\) for all \(k \in \mathbb{Z}\) and \(\twisted{K[x, +x^{-1}]}{\varphi_\lambda}_\mu = 0\) for all other \(\mu \in \mathfrak{h}^*\). + +Hence \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a degree \(1\) bounded +\(\mathfrak{sl}_2(K)\)-module with \(\operatorname{supp} \twisted{K[x, +x^{-1}]}{\varphi_\lambda} = \frac{\lambda}{2} + 2 \mathbb{Z}\). One can also +quickly check that if \(\lambda \notin 1 + 2 \mathbb{Z}\) then \(e\) and \(f\) +act injectively in \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\), so that +\(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is simple. In particular, if +\(\lambda, \mu \notin 1 + 2 \mathbb{Z}\) with \(\lambda \notin \mu + 2 +\mathbb{Z}\) then \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) and +\(\twisted{K[x, x^{-1}]}{\varphi_\mu}\) are non-isomorphic simple cuspidal +\(\mathfrak{sl}_2(K)\)-modules, since their supports differ. These cuspidal +modules can be ``glued together'' in a \emph{monstrous concoction} by summing +over \(\lambda \in K\), as in +\[ + \mathcal{M} + = \bigoplus_{\lambda + 2 \mathbb{Z} \in \mfrac{K}{2 \mathbb{Z}}} + \twisted{K[x, x^{-1}]}{\varphi_\lambda}, +\] + +To a distracted spectator, \(\mathcal{M}\) may look like just another, +innocent, \(\mathfrak{sl}_2(K)\)-module. However, the attentive reader may have +already noticed some of the its bizarre features, most noticeable of which is +the fact that \(\mathcal{M}\) is very big. In fact, \(\mathcal{M}\) is as big a +degree \(1\) bounded module gets: \(\operatorname{supp} \mathcal{M} += \operatorname{supp}_{\operatorname{ess}} \mathcal{M}\) is the entirety of +\(\mathfrak{h}^*\). This may look very alien the reader familiarized with the +finite-dimensional setting, where the configuration of weights is very rigid. +For this reason, \(\mathcal{M}\) deserves to be called ``a monstrous +concoction''. + +On a perhaps less derogatory note, \(\mathcal{M}\) also deserves to be called +\emph{a family}. This is because \(\mathcal{M}\) consists of lots of smaller +cuspidal modules which fit together inside of it in a \emph{coherent} fashion. +Mathieu's ingenious breakthrough was the realization that \(\mathcal{M}\) is a +particular example of a more general pattern, which he named \emph{coherent +families}. + +\begin{definition}\index{coherent family} + A \emph{coherent family \(\mathcal{M}\) of degree \(d\)} is a weight + \(\mathfrak{g}\)-module \(\mathcal{M}\) such that + \begin{enumerate} + \item \(\dim \mathcal{M}_\lambda = d\) for \emph{all} \(\lambda \in + \mathfrak{h}^*\) -- i.e. \(\operatorname{supp}_{\operatorname{ess}} + \mathcal{M} = \mathfrak{h}^*\). + \item For any \(u \in \mathcal{U}(\mathfrak{g})\) in the centralizer + \(\mathcal{U}(\mathfrak{g})_0\) of \(\mathfrak{h}\) in + \(\mathcal{U}(\mathfrak{g})\), the map + \begin{align*} + \mathfrak{h}^* & \to K \\ + \lambda & \mapsto + \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\lambda}) + \end{align*} + is polynomial in \(\lambda\). + \end{enumerate} +\end{definition} + +\begin{example}\label{ex:sl-laurent-family} + The module \(\mathcal{M} = \bigoplus_{\lambda + 2 \mathbb{Z} \in + \mfrac{K}{2 \mathbb{Z}}} \twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a + degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family. +\end{example} + +\begin{example} + Given \(\lambda \in K\), \(\mathcal{M}(\lambda) = \bigoplus_{\mu \in K} K + x^\mu\) with + \begin{align*} + p & \overset{e}{\mapsto} + \left(x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \lambda x\right) p & + p & \overset{f}{\mapsto} + \left(-\frac{\mathrm{d}}{\mathrm{d}x} + \lambda x^{-1}\right) p & + p & \overset{h}{\mapsto} 2 x \frac{\mathrm{d}}{\mathrm{d}x} p, + \end{align*} + is a degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family -- where \(x^{\pm + 1}, \sfrac{\mathrm{d}}{\mathrm{d}x} : \mathcal{M}(\lambda) \to + \mathcal{M}(\lambda)\) are given by \(x^{\pm 1} x^\mu = x^{\mu \pm 1}\) and + \(\sfrac{\mathrm{d}}{\mathrm{d}x} x^\mu = \mu x^{\mu - 1}\). It is easy to + check \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family} is isomorphic + to \(\mathcal{M}(\sfrac{1}{2})\) and \((\mathcal{M}(\sfrac{1}{2}))[0] \cong + K[x, x^{-1}]\). +\end{example} + +\begin{note} + We would like to stress that coherent families have proven themselves useful + for problems other than the classification of cuspidal + \(\mathfrak{g}\)-modules. For instance, Nilsson's classification of rank 1 + \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules is based on the + notion of coherent families and the so called \emph{weighting functor}. +\end{note} + +Our hope is that given a simple cuspidal module \(M\), we can somehow fit \(M\) +inside of a coherent \(\mathfrak{g}\)-family, such as in the case of \(K[x, +x^{-1}]\) and \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family}. In +addition, we hope that such coherent families are somehow \emph{uniquely +determined} by \(M\). This leads us to the following definition. + +\begin{definition}\index{coherent family!coherent extension} + Given a bounded \(\mathfrak{g}\)-module \(M\) of degree \(d\), a + \emph{coherent extension \(\mathcal{M}\) of \(M\)} is a coherent family + \(\mathcal{M}\) of degree \(d\) that contains \(M\) as a subquotient. +\end{definition} + +Our goal is now showing that every simple bounded module has a coherent +extension. The idea then is to classify coherent families, and classify which +submodules of a given coherent family are actually simple cuspidal modules. If +every simple bounded \(\mathfrak{g}\)-module fits inside a coherent extension, +this would lead to classification of all simple cuspidal +\(\mathfrak{g}\)-modules, which we now know is the key for the solution of our +classification problem. However, there are some complications to this scheme. + +Leaving aside the question of existence for a second, we should point out that +coherent families turn out to be rather complicated on their own. In fact they +are too complicated to classify in general. Ideally, we would like to find +\emph{nice} coherent extensions -- ones we can actually classify. For instance, +we may search for \emph{irreducible} coherent extensions, which are defined as +follows. + +\begin{definition}\index{coherent family!irreducible coherent family} + A coherent family \(\mathcal{M}\) is called \emph{irreducible} if it contains + no proper coherent subfamilies -- i.e. \(\mathcal{M}\) is a simple object in + the full subcategory of \(\mathfrak{g}\text{-}\mathbf{Mod}\) consisting of + coherent families. Equivalently, we call \(\mathcal{M}\) irreducible if + \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module + for some \(\lambda \in \mathfrak{h}^*\). +\end{definition} + +Another natural candidate for the role of ``nice extensions'' are the +semisimple coherent families -- i.e. families which are semisimple as +\(\mathfrak{g}\)-modules. These turn out to be very easy to produce. Namely, +there is a construction, known as \emph{the semisimplification of a coherent +family}, which takes a coherent extension of \(M\) to a semisimple coherent +extension of \(M\). + +% Mathieu's proof of this is somewhat profane, I don't think it's worth +% including it in here +% TODO: Move this somewhere else? This holds in general for weight modules +% whose suppert is contained in a single Q-coset +\begin{lemma}\label{thm:component-coh-family-has-finite-length} + Given a coherent family \(\mathcal{M}\) and \(\lambda \in \mathfrak{h}^*\), + \(\mathcal{M}[\lambda]\) has finite length as a \(\mathfrak{g}\)-module. +\end{lemma} + +\begin{proposition}\index{coherent family!semisimplification} + Let \(\mathcal{M}\) be a coherent family of degree \(d\). There exists a + unique semisimple coherent family \(\mathcal{M}^{\operatorname{ss}}\) of + degree \(d\) such that the composition series of + \(\mathcal{M}^{\operatorname{ss}}[\lambda]\) is the same as that of + \(\mathcal{M}[\lambda]\) for all \(\lambda \in \mathfrak{h}^*\), called + \emph{the semisimplification of \(\mathcal{M}\)}. + + Namely, if \(\lambda \in \mathfrak{h}^*\) and \(0 = \mathcal{M}_{\lambda 0} + \subset \mathcal{M}_{\lambda 1} \subset \cdots \subset \mathcal{M}_{\lambda + r_\lambda} = \mathcal{M}[\lambda]\) is a composition series\footnote{Notice + that $\mathcal{M}[\lambda] = \mathcal{M}[\mu]$ for any $\mu \in \lambda + Q$. + Hence the sum $\bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}} + \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}$ is + independent of the choice of representative for $\lambda + Q$ -- provided we + choose $\mathcal{M}_{\mu i} = \mathcal{M}_{\lambda i}$ for all $\mu \in + \lambda + Q$ and $i$.}, + \[ + \mathcal{M}^{\operatorname{ss}} + \cong \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} + \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} + \] +\end{proposition} + +\begin{proof} + The uniqueness of \(\mathcal{M}^{\operatorname{ss}}\) should be clear: + since \(\mathcal{M}^{\operatorname{ss}}\) is semisimple, so is + \(\mathcal{M}^{\operatorname{ss}}[\lambda]\). Hence by the Jordan-Hölder + Theorem + \[ + \mathcal{M}^{\operatorname{ss}}[\lambda] + \cong + \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} + \] + + As for the existence of the semisimplification, it suffices to show + \[ + \mathcal{M}^{\operatorname{ss}} + = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} + \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} + \] + is indeed a semisimple coherent family of degree \(d\). + + We know from Examples~\ref{ex:submod-is-weight-mod} and + \ref{ex:quotient-is-weight-mod} that each quotient + \(\mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}\) is a weight + module. Hence \(\mathcal{M}^{\operatorname{ss}}\) is a weight module. + Furthermore, given \(\mu \in \mathfrak{h}^*\) + \[ + \mathcal{M}_\mu^{\operatorname{ss}} + = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}} + \left( + \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}} + \right)_\mu + = \bigoplus_i + \left( + \mfrac{\mathcal{M}_{\mu i + 1}}{\mathcal{M}_{\mu i}} + \right)_\mu + \cong \bigoplus_i + \mfrac{(\mathcal{M}_{\mu i + 1})_\mu} + {(\mathcal{M}_{\mu i})_\mu} + \] + + In particular, + \[ + \dim \mathcal{M}_\mu^{\operatorname{ss}} + = \sum_i + \dim (\mathcal{M}_{\mu i + 1})_\mu - \dim (\mathcal{M}_{\mu i})_\mu + = \dim \mathcal{M}[\mu]_\mu + = \dim \mathcal{M}_\mu + = d + \] + + Likewise, given \(u \in \mathcal{U}(\mathfrak{g})_0\) the value + \[ + \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu^{\operatorname{ss}}}) + = \sum_i + \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i + 1})_\mu}) + - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i})_\mu}) + = \operatorname{Tr}(u\!\restriction_{\mathcal{M}[\mu]_\mu}) + = \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu}) + \] + is polynomial in \(\mu \in \mathfrak{h}^*\). +\end{proof} + +\begin{note} + Although we have provided an explicit construction of + \(\mathcal{M}^{\operatorname{ss}}\) in terms of \(\mathcal{M}\), we should + point out this construction is not functorial. First, given a + \(\mathfrak{g}\)-homomorphism \(f : \mathcal{M} \to \mathcal{N}\) between + coherent families, it is unclear what \(f^{\operatorname{ss}} : + \mathcal{M}^{\operatorname{ss}} \to \mathcal{N}^{\operatorname{ss}}\) is + supposed to be. Secondly, and this is more relevant, our construction depends + on the choice of composition series \(0 = \mathcal{M}_{\lambda 0} \subset + \cdots \subset \mathcal{M}_{\lambda r_\lambda} = \mathcal{M}[\lambda]\). + While different choices of composition series yield isomorphic results, there + is no canonical isomorphism. In addition, there is no canonical choice of + composition series. +\end{note} + +The proof of Lemma~\ref{thm:component-coh-family-has-finite-length} is +extremely technical and will not be included in here. It suffices to note that, +as in Proposition~\ref{thm:ess-supp-is-zariski-dense}, the general case follows +from the case where \(\mathfrak{g}\) is simple, which may be found in +\cite{mathieu} -- see Lemma 3.3. As promised, if \(\mathcal{M}\) is a coherent +extension of \(M\) then so is \(\mathcal{M}^{\operatorname{ss}}\). + +\begin{proposition} + Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\) + be a coherent extension of \(M\). Then \(\mathcal{M}^{\operatorname{ss}}\) is + a coherent extension of \(M\), and \(M\) is in fact a submodule of + \(\mathcal{M}^{\operatorname{ss}}\). +\end{proposition} + +\begin{proof} + Since \(M\) is simple, its support is contained in a single \(Q\)-coset. + This implies that \(M\) is a subquotient of \(\mathcal{M}[\lambda]\) for any + \(\lambda \in \operatorname{supp} M\). If we fix some composition series \(0 + = \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_r = + \mathcal{M}[\lambda]\) of \(\mathcal{M}[\lambda]\) with \(M \cong + \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}\), there is a natural inclusion + \[ + M + \isoto \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i} + \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j} + \cong \mathcal{M}^{\operatorname{ss}}[\lambda] + \] +\end{proof} + +Given the uniqueness of the semisimplification, the semisimplification of any +semisimple coherent extension \(\mathcal{M}\) is \(\mathcal{M}\) +itself and therefore\dots + +\begin{corollary}\label{thm:bounded-is-submod-of-extension} + Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\) + be a semisimple coherent extension of \(M\). Then \(M\) is + contained in \(\mathcal{M}\). +\end{corollary} + +These last results provide a partial answer to the question of existence of +well behaved coherent extensions. As for the uniqueness \(\mathcal{M}\) in +Corollary~\ref{thm:bounded-is-submod-of-extension}, it suffices to show that +the multiplicities of the simple weight \(\mathfrak{g}\)-modules in +\(\mathcal{M}\) are uniquely determined by \(M\). These multiplicities may be +computed via the following lemma. + +\begin{lemma}\label{thm:centralizer-multiplicity} + Let \(M\) be a semisimple weight \(\mathfrak{g}\)-module. Then \(M_\lambda\) + is a semisimple \(\mathcal{U}(\mathfrak{g})_0\)-module for any \(\lambda \in + \operatorname{supp} M\). Moreover, if \(L\) is a simple weight + \(\mathfrak{g}\)-module such that \(\lambda \in \operatorname{supp} L\) then + \(L_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and the + multiplicity \(L\) in \(M\) coincides with the multiplicity of \(L_\lambda\) + in \(M_\lambda\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module. +\end{lemma} + +\begin{proof} + We begin by showing that \(L_\lambda\) is simple. Let \(N \subset L_\lambda\) + be a nontrivial \(\mathcal{U}(\mathfrak{g})_0\)-submodule. We want to + establish that \(N = L_\lambda\). + + If \(\mathcal{U}(\mathfrak{g})_\alpha\) denotes the root space of \(\alpha\) + in \(\mathcal{U}(\mathfrak{g})\) under the adjoint action of \(\mathfrak{g}\) + as in Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}, + \(\alpha \in Q\), a simple calculation shows + \(\mathcal{U}(\mathfrak{g})_\alpha \cdot N \subset L_{\lambda + \alpha}\). + Since \(L\) is simple and \(N\) is nonzero, it follows from + Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight} that + \[ + L + = \mathcal{U}(\mathfrak{g}) \cdot N + = \bigoplus_{\alpha \in Q} \mathcal{U}(\mathfrak{g})_\alpha \cdot N + \] + and thus \(L_{\lambda + \alpha} = \mathcal{U}(\mathfrak{g})_\alpha \cdot N\). + In particular, \(L_\lambda = \mathcal{U}(\mathfrak{g})_0 \cdot N \subset N\) + and \(N = L_\lambda\). + + Now given a semisimple weight \(\mathfrak{g}\)-module \(M = \bigoplus_i M_i\) + with \(M_i\) simple, it is clear \(M_\lambda = \bigoplus_i (M_i)_\lambda\). + Each \((M_i)_\lambda\) is either \(0\) or a simple + \(\mathcal{U}(\mathfrak{g})_0\)-module, so that \(M_\lambda\) is a semisimple + \(\mathcal{U}(\mathfrak{g})_0\)-module. In addition, to see that the + multiplicity of \(L\) in \(M\) coincides with the multiplicity of + \(L_\lambda\) in \(M_\lambda\) it suffices to show that if \((M_i)_\lambda + \cong (M_j)_\lambda\) are both nonzero then \(M_i \cong M_j\). + + If \(I(M_i) = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0} + (M_i)_\lambda\), the inclusion of \(\mathcal{U}(\mathfrak{g})_0\)-modules + \((M_i)_\lambda \to M_i\) induces a \(\mathfrak{g}\)-homomorphism + \begin{align*} + I(M_i) & \to M_i \\ + u \otimes m & \mapsto u \cdot m + \end{align*} + + Since \(M_i\) is simple and \(\lambda \in \operatorname{supp} M_i\), \(M_i = + \mathcal{U}(\mathfrak{g}) \cdot (M_i)_\lambda\). The homomorphism \(I(M_i) + \to M_i\) is thus surjective. Similarly, if \(I(M_j) = + \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0} + (M_j)_\lambda\) then there is a natural surjective + \(\mathfrak{g}\)-homomorphism \(I(M_j) \to M_j\). Now suppose there is an + isomorphism of \(\mathcal{U}(\mathfrak{g})_0\)-modules \(f: (M_i)_\lambda + \isoto (M_j)_\lambda\). Such an isomorphism induces an isomorphism of + \(\mathfrak{g}\)-modules + \begin{align*} + \tilde f : I(M_i) & \isoto I(M_j) \\ + u \otimes m & \mapsto u \otimes f(m) + \end{align*} + + By composing \(\tilde f\) with the projection \(I(M_j) \to M_j\) we get a + surjective homomorphism \(I(M_i) \to M_j\). We claim \(\ker (I(M_i) \to M_i) + = \ker (I(M_i) \to M_j)\). To see this, notice that \(\ker(I(M_i) \to M_i)\) + coincides with the largest submodule \(Z(M_i) \subset I(M_i)\) contained in + \(\bigoplus_{\alpha \ne 0} \mathcal{U}(\mathfrak{g})_\alpha + \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda\). Indeed, a simple + computation shows \(\ker (I(M_i) \to M_i) \cap (\mathcal{U}(\mathfrak{g})_0 + \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda) = 0\), which implies + \(\ker(I(M_i) \to M_i) \subset Z(M_i)\). Since \(M_i\) is simple, \(\ker + (I(M_i) \to M_i)\) is maximal and thus \(\ker(I(M_i) \to M_i) = Z(M_i)\). By + the same token, \(\ker (I(M_j) \to M_j)\) is the largest submodule of + \(I(M_j)\) contained in \(\bigoplus_{\alpha \ne 0} + \mathcal{U}(\mathfrak{g})_\alpha \otimes_{\mathcal{U}(\mathfrak{g})_0} + (M_j)_\lambda\) and therefore \(\ker(I(M_i) \to M_i) = + \tilde{f}^{-1}(\ker(I(M_j) \to M_j)) = \ker(I(M_i) \to M_j)\). + + Hence there is an isomorphism \(\mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \isoto + M_j\) satisfying + \begin{center} + \begin{tikzcd} + I(M_i) \rar{\tilde f} \dar & I(M_j) \dar \\ + \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \rar{\sim} & M_j + \end{tikzcd} + \end{center} + and finally \(M_i \cong \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \cong M_j\). +\end{proof} + +A complementary question now is: which submodules of a \emph{nice} coherent +family are cuspidal representations? + +\begin{proposition}[Mathieu] + Let \(\mathcal{M}\) be an irreducible coherent family of degree \(d\) and + \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent. + \begin{enumerate} + \item \(\mathcal{M}[\lambda]\) is simple. + \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for + all \(\alpha \in \Delta\). + \item \(\mathcal{M}[\lambda]\) is cuspidal. + \end{enumerate} +\end{proposition} + +\begin{proof} + The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly + from Corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the + corollary that \strong{(iii)} implies \strong{(ii)}. All it is left is to + show \strong{(ii)} implies \strong{(iii)}. This isn't already clear from + Corollary~\ref{thm:cuspidal-mod-equivs} because, at first glance, + $\mathcal{M}[\lambda]$ may not be simple for some $\lambda$ satisfying + \strong{(ii)}. We will show this is never the case. + + Suppose \(F_\alpha\) acts injectively on the submodule + \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since + \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains + an infinite-dimensional simple \(\mathfrak{g}\)-submodule \(M\). Moreover, + again by Corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(M\) is a + cuspidal module, and its degree is bounded by \(d\). We want to show + \(\mathcal{M}[\lambda] = M\). + + We claim the set \(U = \{\mu \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is + a simple $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. If we + suppose this is the case for a moment or two, it follows from the fact that + \(M\) is simple and \(\operatorname{supp}_{\operatorname{ess}} M\) is + Zariski-dense that \(U \cap \operatorname{supp}_{\operatorname{ess}} M\) is + non-empty. In other words, there is some \(\mu \in \mathfrak{h}^*\) such that + \(\mathcal{M}_\mu\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and + \(\dim M_\mu = \deg M\). + + In particular, \(M_\mu \ne 0\), so \(M_\mu = \mathcal{M}_\mu\). Now given any + simple \(\mathfrak{g}\)-module \(L\), it follows from + Lemma~\ref{thm:centralizer-multiplicity} that the multiplicity of \(L\) + in \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(L_\mu\) in + \(\mathcal{M}_\mu\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module -- which is, + of course, \(1\) if \(L \cong M\) and \(0\) otherwise. Hence + \(\mathcal{M}[\lambda] = M\) and \(\mathcal{M}[\lambda]\) is cuspidal. +\end{proof} + +To finish the proof, we now show\dots + +\begin{lemma} + Let \(\mathcal{M}\) be a coherent family. The set \(U = \{\lambda \in + \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple + $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. +\end{lemma} + +\begin{proof} + For each \(\lambda \in \mathfrak{h}^*\) we introduce the bilinear form + \begin{align*} + B_\lambda : \mathcal{U}(\mathfrak{g})_0 \times \mathcal{U}(\mathfrak{g})_0 + & \to K \\ + (u, v) + & \mapsto \operatorname{Tr}(u v \!\restriction_{\mathcal{M}_\lambda}) + \end{align*} + and consider its rank -- i.e. the dimension of the image of the induced + operator + \begin{align*} + \mathcal{U}(\mathfrak{g})_0 & \to \mathcal{U}(\mathfrak{g})_0^* \\ + u & \mapsto B_\lambda(u, \cdot) + \end{align*} + + Our first observation is that \(\operatorname{rank} B_\lambda \le d^2\). This + follows from the commutativity of + \begin{center} + \begin{tikzcd} + \mathcal{U}(\mathfrak{g})_0 \rar \dar & + \mathcal{U}(\mathfrak{g})_0^* \\ + \operatorname{End}(\mathcal{M}_\lambda) \rar{\sim} & + \operatorname{End}(\mathcal{M}_\lambda)^* \uar + \end{tikzcd}, + \end{center} + where the map \(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)\) is given by the action of + \(\mathcal{U}(\mathfrak{g})_0\), the map + \(\operatorname{End}(\mathcal{M}_\lambda)^* \to + \mathcal{U}(\mathfrak{g})_0^*\) is its dual, and the isomorphism + \(\operatorname{End}(\mathcal{M}_\lambda) \isoto + \operatorname{End}(\mathcal{M}_\lambda)^*\) is induced by the trace form + \begin{align*} + \operatorname{End}(\mathcal{M}_\lambda) \times + \operatorname{End}(\mathcal{M}_\lambda) & \to K \\ + (T, S) & \mapsto \operatorname{Tr}(T S) + \end{align*} + + Indeed, \(\operatorname{rank} B_\lambda \le + \operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)) \le \dim + \operatorname{End}(\mathcal{M}_\lambda) = d^2\). Furthermore, if + \(\operatorname{rank} B_\lambda = d^2\) then we must have + \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)) = d^2\) -- i.e. the map + \(\mathcal{U}(\mathfrak{g})_0 \to \operatorname{End}(\mathcal{M}_\lambda)\) + is surjective. In particular, if \(\operatorname{rank} B_\lambda = d^2\) then + \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module, + for if \(M \subset \mathcal{M}_\lambda\) is invariant under the action of + \(\mathcal{U}(\mathfrak{g})_0\) then \(M\) is invariant under any + \(K\)-linear operator \(\mathcal{M}_\lambda \to \mathcal{M}_\lambda\), so + that \(M = 0\) or \(M = \mathcal{M}_\lambda\). + + On the other hand, if \(\mathcal{M}_\lambda\) is simple then by Burnside's + Theorem on matrix algebras the map \(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. Hence the + commutativity of the previously drawn diagram, as well as the fact that + \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)) = + \operatorname{rank}(\operatorname{End}(\mathcal{M}_\lambda)^* \to + \mathcal{U}(\mathfrak{g})_0^*)\), imply that \(\operatorname{rank} B_\lambda + = d^2\). This goes to show that \(U\) is precisely the set of all \(\lambda\) + such that \(B_\lambda\) has maximal rank \(d^2\). We now show that \(U\) is + Zariski-open. First, notice that + \[ + U = + \bigcup_{\substack{V \subset \mathcal{U}(\mathfrak{g})_0 \\ \dim V = d}} + U_V, + \] + where \(U_V = \{\lambda \in \mathfrak{h}^* : \operatorname{rank} + B_\lambda\!\restriction_V = d^2 \}\). Here \(V\) ranges over all + \(d\)-dimensional subspaces of \(\mathcal{U}(\mathfrak{g})_0\) -- \(V\) is + not necessarily a \(\mathcal{U}(\mathfrak{g})_0\)-submodule. + + Indeed, if \(\operatorname{rank} B_\lambda = d^2\) it follows from the + subjectivity of the map \(\mathcal{U}(\mathfrak{g})_0 \to + \operatorname{End}(\mathcal{M}_\lambda)\) that there is some \(V \subset + \mathcal{U}(\mathfrak{g})_0\) with \(\dim V = d\) such that the restriction + \(V \to \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. The + commutativity of + \begin{center} + \begin{tikzcd} + V \rar \dar & V^* \\ + \operatorname{End}(\mathcal{M}_\lambda) \rar{\sim} & + \operatorname{End}(\mathcal{M}_\lambda)^* \uar + \end{tikzcd} + \end{center} + then implies \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\). In + other words, \(U \subset \bigcup_V U_V\). + + Likewise, if \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\) for some + \(V\), then the commutativity of + \begin{center} + \begin{tikzcd} + V \rar \dar & V^* \\ + \mathcal{U}(\mathfrak{g})_0 \rar & + \mathcal{U}(\mathfrak{g})_0^* \uar + \end{tikzcd} + \end{center} + implies \(\operatorname{rank} B_\lambda \ge d^2\), which goes to show + \(\bigcup_V U_V \subset U\). + + Given \(\lambda \in U_V\), the surjectivity of \(V \to + \operatorname{End}(\mathcal{M}_\lambda)\) and the fact that \(\dim V < + \infty\) imply \(V \to V^*\) is invertible. Since \(\mathcal{M}\) is a + coherent family, \(B_\lambda\) depends polynomially in \(\lambda\). Hence so + does the induced maps \(V \to V^*\). In particular, there is some Zariski + neighborhood \(U'\) of \(\lambda\) such that the map \(V \to V^*\) induced by + \(B_\mu\!\restriction_V\) is invertible for all \(\mu \in U'\). + + But the surjectivity of the map induced by \(B_\mu\!\restriction_V\) implies + \(\operatorname{rank} B_\mu = d^2\), so \(\mu \in U_V\) and therefore \(U' + \subset U_V\). This implies \(U_V\) is open for all \(V\). Finally, \(U\) is + the union of Zariski-open subsets and is therefore open. We are done. +\end{proof} + +The major remaining question for us to tackle is that of the existence of +coherent extensions, which will be the focus of our next section. + +\section{Localizations \& the Existence of Coherent Extensions} + +Let \(M\) be a simple bounded \(\mathfrak{g}\)-module of degree \(d\). Our +goal is to prove that \(M\) has a (unique) irreducible semisimple coherent +extension \(\mathcal{M}\). Since \(M\) is simple, we know \(M \subset +\mathcal{M}[\lambda]\) for any \(\lambda \in \operatorname{supp} M\). Our first +task is constructing \(\mathcal{M}[\lambda]\). The issue here is that +\(\operatorname{supp}_{\operatorname{ess}} M\) may not be all of \(\lambda + Q += \operatorname{supp}_{\operatorname{ess}} \mathcal{M}[\lambda]\), so we may +find \(M \subsetneq \mathcal{M}[\lambda]\). In fact, we may find +\(\operatorname{supp} M \subsetneq \lambda + Q\). + +This wasn't an issue an Example~\ref{ex:laurent-polynomial-mod} because we +verified that the action of \(f \in \mathfrak{sl}_2(K)\) on \(K[x, x^{-1}]\) is +injective. Since all weight spaces of \(K[x, x^{-1}]\) are \(1\)-dimensional, +this implies the action of \(f\) is actually bijective, so we can obtain a +nonzero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\) +by translating between weight spaced using \(f\) and \(f^{-1}\) -- here +\(f^{-1}\) denotes the \(K\)-linear operator \((- +\sfrac{\mathrm{d}}{\mathrm{d}x} + \sfrac{x^{-1}}{2})^{-1}\), which is the +inverse of the action of \(f\) on \(K[x, x^{-1}]\). +\begin{center} + \begin{tikzcd} + \cdots \rar[bend left=60]{f^{-1}} + & K x^{-2} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} + & K x^{-1} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} + & K \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} + & K x \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} + & K x^2 \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f} + & \cdots \lar[bend left=60]{f} + \end{tikzcd} +\end{center} + +In the general case, the action of some \(F_\alpha \in \mathfrak{g}\) with +\(\alpha \in \Delta\) in \(M\) may not be injective. In fact, we have seen that +the action of \(F_\alpha\) is injective for all \(\alpha \in \Delta^+\) if, and +only if \(M\) is cuspidal. Nevertheless, we could intuitively \emph{make it +injective} by formally inverting the elements \(F_\alpha \in +\mathcal{U}(\mathfrak{g})\). This would allow us to obtain nonzero vectors in +\(M_\mu\) for all \(\mu \in \lambda + Q\) by successively applying elements of +\(\{F_\alpha^{\pm 1}\}_{\alpha \in \Delta}\) to a nonzero weight vector \(m \in +M_\lambda\). Moreover, if the actions of the \(F_\alpha\) were to be +invertible, we would find that all \(M_\mu\) are \(d\)-dimensional for \(\mu +\in \lambda + Q\). + +In a commutative domain, this can be achieved by tensoring our module by the +field of fractions. However, \(\mathcal{U}(\mathfrak{g})\) is hardly ever +commutative -- \(\mathcal{U}(\mathfrak{g})\) is commutative if, and only if +\(\mathfrak{g}\) is Abelian -- and the situation is more delicate in the +non-commutative case. For starters, a non-commutative \(K\)-algebra \(A\) may +not even have a ``field of fractions'' -- i.e. an over-ring where all elements +of \(A\) have inverses. Nevertheless, it is possible to formally invert +elements of certain subsets of \(A\) via a process known as +\emph{localization}, which we now describe. + +\begin{definition}\index{localization!multiplicative subsets}\index{localization!Ore's condition} + Let \(A\) be a \(K\)-algebra. A subset \(S \subset A\) is called + \emph{multiplicative} if \(s \cdot t \in S\) for all \(s, t \in S\) and \(0 + \notin S\). A multiplicative subset \(S\) is said to satisfy \emph{Ore's + localization condition} if for each \(a \in A\) and \(s \in S\) there exists + \(b, c \in A\) and \(t, t' \in S\) such that \(s a = b t\) and \(a s = t' + c\). +\end{definition} + +\begin{theorem}[Ore-Asano]\index{localization!Ore-Asano Theorem} + Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization + condition. Then there exists a (unique) \(K\)-algebra \(S^{-1} A\), with a + canonical algebra homomorphism \(A \to S^{-1} A\), enjoying the universal + property that each algebra homomorphism \(f : A \to B\) such that \(f(s)\) is + invertible for all \(s \in S\) can be uniquely extended to an algebra + homomorphism \(S^{-1} A \to B\). \(S^{-1} A\) is called \emph{the + localization of \(A\) by \(S\)}, and the map \(A \to S^{-1} A\) is called + \emph{the localization map}. + \begin{center} + \begin{tikzcd} + A \dar \rar{f} & B \\ + S^{-1} A \urar[swap, dotted] & + \end{tikzcd} + \end{center} +\end{theorem} + +If we identify an element with its image under the localization map, it follows +directly from Ore's construction that every element of \(S^{-1} A\) has the +form \(s^{-1} a\) for some \(s \in S\) and \(a \in A\). Likewise, any element +of \(S^{-1} A\) can also be written as \(b t^{-1}\) for some \(t \in S\), \(b +\in A\). + +Ore's localization condition may seem a bit arbitrary at first, but a more +thorough investigation reveals the intuition behind it. The issue in question +here is that in the non-commutative case we can no longer take the existence of +common denominators for granted. However, the existence of common denominators +is fundamental to the proof of the fact the field of fractions is a ring -- it +is used, for example, to define the sum of two elements in the field of +fractions. We thus need to impose their existence for us to have any hope of +defining consistent arithmetics in the localization of an algebra, and Ore's +condition is actually equivalent to the existence of common denominators -- +see the discussion in the introduction of \cite[ch.~6]{goodearl-warfield} for +further details. + +We should also point out that there are numerous other conditions -- which may +be easier to check than Ore's -- known to imply Ore's condition. For +instance\dots + +\begin{lemma} + Let \(S \subset A\) be a multiplicative subset generated by finitely many + locally \(\operatorname{ad}\)-nilpotent elements -- i.e. elements \(s \in S\) + such that for each \(a \in A\) there exists \(r > 0\) such that + \(\operatorname{ad}(s)^r a = [s, [s, \cdots [s, a]]\cdots] = 0\). Then \(S\) + satisfies Ore's localization condition. +\end{lemma} + +In our case, we are more interested in formally inverting the action of +\(F_\alpha\) on \(M\) than in inverting \(F_\alpha\) itself. To that end, we +introduce one further construction, known as \emph{the localization of a +module}. + +\begin{definition}\index{localization!localization of modules} + Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization + condition and \(M\) be an \(A\)-module. The \(S^{-1} A\)-module \(S^{-1} M = + S^{-1} A \otimes_A M\) is called \emph{the localization of \(M\) by \(S\)}, + and the homomorphism of \(A\)-modules + \begin{align*} + M & \to S^{-1} M \\ + m & \mapsto 1 \otimes m + \end{align*} + is called \emph{the localization map of \(M\)}. +\end{definition} + +Notice that the \(S^{-1} A\)-module \(S^{-1} M\) has the natural structure of +an \(A\)-module, where the action of \(A\) is given by the localization map \(A +\to S^{-1} A\). + +It is interesting to observe that, unlike in the case of the field of fractions +of a commutative domain, in general the localization map \(A \to S^{-1} A\) -- +i.e. the map \(a \mapsto \frac{a}{1}\) -- may not be injective. For instance, +if \(S\) contains a divisor of zero \(s\), its image under the localization map +is invertible and therefore cannot be a divisor of zero in \(S^{-1} A\). In +particular, if \(a \in A\) is nonzero and such that \(s a = 0\) or \(a s = 0\) +then its image under the localization map has to be \(0\). However, the +existence of divisors of zero in \(S\) turns out to be the only obstruction to +the injectivity of the localization map, as shown in\dots + +\begin{lemma} + Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization + condition and \(M\) be an \(A\)-module. If \(S\) acts injectively on \(M\) + then the localization map \(M \to S^{-1} M\) is injective. In particular, if + \(S\) has no zero divisors then \(A\) is a subalgebra of \(S^{-1} A\). +\end{lemma} + +Again, in our case we are interested in inverting the actions of the +\(F_\alpha\) on \(M\). However, for us to be able to translate between all +weight spaces associated with elements of \(\lambda + Q\), \(\lambda \in +\operatorname{supp} M\), we only need to invert the \(F_\alpha\)'s for +\(\alpha\) in some subset of \(\Delta\) which spans all of \(Q = \mathbb{Z} +\Delta\). In other words, it suffices to invert \(F_\beta\) for all \(\beta\) +in some basis \(\Sigma\) for \(\Delta\). We can choose such a basis to be +well-behaved. For example, we can show\dots + +\begin{lemma}\label{thm:nice-basis-for-inversion} + Let \(M\) be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module. + There is a basis \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) for \(\Delta\) + such that the elements \(F_{\beta_i}\) all act injectively on \(M\) and + satisfy \([F_{\beta_i}, F_{\beta_j}] = 0\). +\end{lemma} + +\begin{note} + The basis \(\Sigma\) in Lemma~\ref{thm:nice-basis-for-inversion} may very + well depend on the representation \(M\)! This is another obstruction to the + functoriality of our constructions. +\end{note} + +The proof of the previous Lemma is quite technical and was deemed too tedious +to be included in here. See Lemma 4.4 of \cite{mathieu} for a full proof. Since +\(F_\alpha\) is locally \(\operatorname{ad}\)-nilpotent for all \(\alpha \in +\Delta\), we can see\dots + +\begin{corollary} + Let \(\Sigma\) be as in Lemma~\ref{thm:nice-basis-for-inversion} and + \((F_\beta)_{\beta \in \Sigma} \subset \mathcal{U}(\mathfrak{g})\) be the + multiplicative subset generated by the \(F_\beta\)'s. The \(K\)-algebra + \(\Sigma^{-1} \mathcal{U}(\mathfrak{g}) = (F_\beta)_{\beta \in \Sigma}^{-1} + \mathcal{U}(\mathfrak{g})\) is well defined. Moreover, if we denote by + \(\Sigma^{-1} M\) the localization of \(M\) by \((F_\beta)_{\beta \in + \Sigma}\), the localization map \(M \to \Sigma^{-1} M\) is injective. +\end{corollary} + +From now on let \(\Sigma\) be some fixed basis for \(\Delta\) satisfying the +hypothesis of Lemma~\ref{thm:nice-basis-for-inversion}. We now show that +\(\Sigma^{-1} M\) is a weight \(\mathfrak{g}\)-module whose support is an +entire \(Q\)-coset. + +\begin{proposition}\label{thm:irr-bounded-is-contained-in-nice-mod} + The restriction of the localization \(\Sigma^{-1} M\) is a bounded + \(\mathfrak{g}\)-module of degree \(d\) with \(\operatorname{supp} + \Sigma^{-1} M = Q + \operatorname{supp} M\) and \(\dim \Sigma^{-1} M_\lambda + = d\) for all \(\lambda \in \operatorname{supp} \Sigma^{-1} M\). +\end{proposition} + +\begin{proof} + Fix some \(\beta \in \Sigma\). We begin by showing that \(F_\beta\) and + \(F_\beta^{-1}\) map the weight space \(\Sigma^{-1} M_\lambda\) to + \(\Sigma^{-1} M_{\lambda - \beta}\) and \(\Sigma^{-1} M_{\lambda + \beta}\), + respectively. Indeed, given \(m \in M_\lambda\) and \(H \in \mathfrak{h}\) we + have + \[ + H \cdot (F_\beta \cdot m) + = ([H, F_\beta] + F_\beta H) \cdot m + = F_\beta (-\beta(H) + H) \cdot m + = (\lambda - \beta)(H) F_\beta \cdot m + \] + + On the other hand, + \[ + 0 + = [H, 1] + = [H, F_\beta F_\beta^{-1}] + = F_\beta [H, F_\beta^{-1}] + [H, F_\beta] F_\beta^{-1} + = F_\beta [H, F_\beta^{-1}] - \beta(H) F_\beta F_\beta^{-1}, + \] + so that \([H, F_\beta^{-1}] = \beta(H) \cdot F_\beta^{-1}\) and therefore + \[ + H \cdot (F_\beta^{-1} \cdot m) + = ([H, F_\beta^{-1}] + F_\beta^{-1} H) \cdot m + = F_\beta^{-1} (\beta(H) + H) \cdot m + = (\lambda + \beta)(H) F_\beta^{-1} \cdot m + \] + + From the fact that \(F_\beta^{\pm 1}\) maps \(M_\lambda\) to \(\Sigma^{-1} + M_{\lambda \pm \beta}\) follows our first conclusion: since \(M\) is a weight + module and every element of \(\Sigma^{-1} M\) has the form \(s^{-1} \cdot m = + s^{-1} \otimes m\) for \(s \in (F_\beta)_{\beta \in \Sigma}\) and \(m \in + M\), we can see that \(\Sigma^{-1} M = \bigoplus_\lambda \Sigma^{-1} + M_\lambda\). Furthermore, since the action of each \(F_\beta\) on + \(\Sigma^{-1} M\) is bijective and \(\Sigma\) is a basis for \(Q\) we obtain + \(\operatorname{supp} \Sigma^{-1} M = Q + \operatorname{supp} M\). + + Again, because of the bijectivity of the \(F_\beta\)'s, to see that \(\dim + \Sigma^{-1} M_\lambda = d\) for all \(\lambda \in \operatorname{supp} + \Sigma^{-1} M\) it suffices to show that \(\dim \Sigma^{-1} M_\lambda = d\) + for some \(\lambda \in \operatorname{supp} \Sigma^{-1} M\). We may take + \(\lambda \in \operatorname{supp} M\) with \(\dim M_\lambda = d\). For any + finite-dimensional subspace \(V \subset \Sigma^{-1} M_\lambda\) we can find + \(s \in (F_\beta)_{\beta \in \Sigma}\) such that \(s \cdot V \subset M\). If + \(s = F_{\beta_{i_1}} \cdots F_{\beta_{i_r}}\), it is clear \(s \cdot V + \subset M_{\lambda - \beta_{i_1} - \cdots - \beta_{i_r}}\), so \(\dim V = + \dim s \cdot V \le d\). This holds for all finite-dimensional \(V \subset + \Sigma^{-1} M_\lambda\), so \(\dim \Sigma^{-1} M_\lambda \le d\). It then + follows from the fact that \(M_\lambda \subset \Sigma^{-1} M_\lambda\) that + \(M_\lambda = \Sigma^{-1} M_\lambda\) and therefore \(\dim \Sigma^{-1} + M_\lambda = d\). +\end{proof} + +We now have a good candidate for a coherent extension of \(M\), but +\(\Sigma^{-1} M\) is still not a coherent extension since its support is +contained in a single \(Q\)-coset. In particular, \(\operatorname{supp} +\Sigma^{-1} M \ne \mathfrak{h}^*\) and \(\Sigma^{-1} M\) is not a coherent +family. To obtain a coherent family we thus need somehow extend \(\Sigma^{-1} +M\). To that end, we will attempt to replicate the construction of the coherent +extension of the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\). Specifically, +the idea is that if twist \(\Sigma^{-1} M\) by an automorphism which shifts its +support by some \(\lambda \in \mathfrak{h}^*\), we can construct a coherent +family by summing these modules over \(\lambda\) as in +Example~\ref{ex:sl-laurent-family}. + +For \(K[x, x^{-1}]\) this was achieved by twisting the +\(\operatorname{Diff}(K[x, x^{-1}])\)-module \(K[x, x^{-1}]\) by the +automorphisms \(\varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) \to +\operatorname{Diff}(K[x, x^{-1}])\) and restricting the results to +\(\mathcal{U}(\mathfrak{sl}_2(K))\) via the map +\(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x, x^{-1}])\), but +this approach is inflexible since not every \(\mathfrak{sl}_2(K)\)-module +factors through \(\operatorname{Diff}(K[x, x^{-1}])\). Nevertheless, we could +just as well twist \(K[x, x^{-1}]\) by automorphisms of +\(\mathcal{U}(\mathfrak{sl}_2(K))_f\) directly -- where +\(\mathcal{U}(\mathfrak{sl}_2(K))_f = (f)^{-1} \mathcal{U}(\mathfrak{g})\) is +the localization of \(\mathcal{U}(\mathfrak{sl}_2(K))\) by the multiplicative +subset generated by \(f\). + +In general, we may twist the \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module +\(\Sigma^{-1} M\) by automorphisms of \(\Sigma^{-1} +\mathcal{U}(\mathfrak{g})\). For \(\lambda = \beta \in \Sigma\) the map +\begin{align*} + \theta_\beta : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & \to + \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ + u & \mapsto F_\beta u F_\beta^{-1} +\end{align*} +is a natural candidate for such a twisting automorphism. Indeed, we will soon +see that \(\twisted{(\Sigma^{-1} M)}{\theta_\beta}_\lambda = \Sigma^{-1} +M_{\lambda + \beta}\). However, this is hardly useful to us, since \(\beta \in +Q\) and therefore \(\beta + \operatorname{supp} \Sigma^{-1} M = +\operatorname{supp} \Sigma^{-1} M\). If we want to expand the support of +\(\Sigma^{-1} M\) we will have to twist by automorphisms that shift its support +by \(\lambda \in \mathfrak{h}^*\) lying \emph{outside} of \(Q\). + +The situation is much less obvious in this case. Nevertheless, it turns out we +can extend the family \(\{\theta_\beta\}_{\beta \in \Sigma}\) to a family of +automorphisms \(\{\theta_\lambda\}_{\lambda \in \mathfrak{h}^*}\). +Explicitly\dots + +\begin{proposition}\label{thm:nice-automorphisms-exist} + There is a family of automorphisms \(\{\theta_\lambda : \Sigma^{-1} + \mathcal{U}(\mathfrak{g}) \to \Sigma^{-1} + \mathcal{U}(\mathfrak{g})\}_{\lambda \in \mathfrak{h}^*}\) such that + \begin{enumerate} + \item \(\theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u) = F_{\beta_1}^{k_1} + \cdots F_{\beta_r}^{k_r} u F_{\beta_r}^{- k_r} \cdots F_{\beta_1}^{- + k_1}\) for all \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and \(k_1, + \ldots, k_r \in \mathbb{Z}\). + + \item For each \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) the map + \begin{align*} + \mathfrak{h}^* & \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ + \lambda & \mapsto \theta_\lambda(u) + \end{align*} + is polynomial. + + \item If \(\lambda, \mu \in \mathfrak{h}^*\), \(N\) is a \(\Sigma^{-1} + \mathcal{U}(\mathfrak{g})\)-module whose restriction to + \(\mathcal{U}(\mathfrak{g})\) is a weight \(\mathfrak{g}\)-module and + \(\twisted{N}{\theta_\lambda}\) is the \(\Sigma^{-1} + \mathcal{U}(\mathfrak{g})\)-module \(N\) twisted by the automorphism + \(\theta_\lambda\) then \(N_\mu = \twisted{N}{\theta_\lambda}_{\mu + + \lambda}\). In particular, \(\operatorname{supp} + \twisted{N}{\theta_\lambda} = \lambda + \operatorname{supp} N\). + \end{enumerate} +\end{proposition} + +\begin{proof} + Since the elements \(F_\beta\), \(\beta \in \Sigma\) commute with one + another, the endomorphisms + \begin{align*} + \theta_{k_1 \beta_1 + \cdots + k_r \beta_r} + : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & + \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ + u & \mapsto + F_{\beta_1}^{k_1} \cdots F_{\beta_r}^{k_r} + u + F_{\beta_1}^{- k_r} \cdots F_{\beta_r}^{- k_1} + \end{align*} + are well defined for all \(k_1, \ldots, k_r \in \mathbb{Z}\). + + Fix some \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\). For any \(s \in + (F_\beta)_{\beta \in \Sigma}\) and \(k > 0\) we have \(s^k u = \binom{k}{0} + \operatorname{ad}(s)^0 u s^{k - 0} + \cdots + \binom{k}{k} + \operatorname{ad}(s)^k u s^{k - k}\). Now if we take \(\ell\) such + \(\operatorname{ad}(F_\beta)^{\ell + 1} u = 0\) for all \(\beta \in \Sigma\) + we find + \[ + \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u) + = \sum_{i_1, \ldots, i_r = 1, \ldots, \ell} + \binom{k_1}{i_1} \cdots \binom{k_r}{i_r} + \operatorname{ad}(F_{\beta_1})^{i_1} \cdots + \operatorname{ad}(F_{\beta_r})^{i_r} + u + F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r} + \] + for all \(k_1, \ldots, k_r \in \mathbb{N}\). + + Since the binomial coefficients \(\binom{x}{k} = \frac{x (x-1) \cdots (x - k + + 1)}{k!}\) can be uniquely extended to polynomial functions in \(x \in K\), + we may in general define + \[ + \theta_\lambda(u) + = \sum_{i_1, \ldots, i_r \ge 0} + \binom{\lambda_1}{i_1} \cdots \binom{\lambda_r}{i_r} + \operatorname{ad}(F_{\beta_1})^{i_1} \cdots + \operatorname{ad}(F_{\beta_r})^{i_r} + r + F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r} + \] + for \(\lambda_1, \ldots, \lambda_r \in K\), \(\lambda = \lambda_1 \beta_1 + + \cdots + \lambda_r \beta_r \in \mathfrak{h}^*\). + + It is clear that the \(\theta_\lambda\) are endomorphisms. To see that the + \(\theta_\lambda\) are indeed automorphisms, notice \(\theta_{- k_1 \beta_1 - + \cdots - k_r \beta_r} = \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}^{-1}\). + The uniqueness of the polynomial extensions then implies \(\theta_{- \lambda} + = \theta_\lambda^{-1}\) in general: given \(u \in \Sigma^{-1} + \mathcal{U}(\mathfrak{g})\), the map + \begin{align*} + \mathfrak{h}^* & \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\ + \lambda & \mapsto \theta_\lambda(\theta_{-\lambda}(u)) - u + \end{align*} + is a polynomial extension of the zero map \(\mathbb{Z} \beta_1 \oplus \cdots + \oplus \mathbb{Z} \beta_r \to \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and is + therefore identically zero. + + Finally, let \(N\) be a \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module + whose restriction is a weight module. If \(n \in N\) then + \[ + n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda} + \iff \theta_\lambda(H) \cdot n = (\mu + \lambda)(H) n + \, \forall H \in \mathfrak{h} + \] + + But + \[ + \theta_\beta(H) + = F_\beta H F_\beta^{-1} + = ([F_\beta, H] + H F_\beta) F_\beta^{-1} + = (\beta(H) + H) F_\beta F_\beta^{-1} + = \beta(H) + H + \] + for all \(H \in \mathfrak{h}\) and \(\beta \in \Sigma\). In general, + \(\theta_\lambda(H) = \lambda(H) + H\) for all \(\lambda \in \mathfrak{h}^*\) + and hence + \[ + \begin{split} + n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda} + & \iff (\lambda(H) + H) \cdot n = (\mu + \lambda)(H) n + \; \forall H \in \mathfrak{h} \\ + & \iff H \cdot n = \mu(H) n \; \forall H \in \mathfrak{h} \\ + & \iff n \in N_\mu + \end{split}, + \] + so that \(\twisted{N}{\theta_\lambda}_{\mu + \lambda} = N_\mu\). +\end{proof} + +It should now be obvious\dots + +\begin{proposition}[Mathieu] + There exists a coherent extension \(\mathcal{M}\) of \(M\). +\end{proposition} + +\begin{proof} + Take\footnote{Here we fix some $\lambda_\xi \in \xi$ for each $Q$-coset $\xi + \in \mfrac{\mathfrak{h}^*}{Q}$. While there is a natural isomorphism + $\twisted{(\Sigma^{-1} M)}{\theta_\lambda} \isoto \twisted{(\Sigma^{-1} + M)}{\theta_\mu}$ for each $\mu \in \lambda + Q$, they are not the same + \(\mathfrak{g}\)-modules strictly speaking. This is yet another obstruction + to the functoriality of our constructions.} + \[ + \mathcal{M} + = \bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}} + \twisted{(\Sigma^{-1} M)}{\theta_\lambda} + \] + + It is clear \(M\) lies in \(\Sigma^{-1} M = \twisted{(\Sigma^{-1} + M)}{\theta_0}\) and therefore \(M \subset \mathcal{M}\). On the other hand, + \(\dim \mathcal{M}_\mu = \dim \twisted{(\Sigma^{-1} M)}{\theta_\lambda}_\mu = + \dim \Sigma^{-1} M_{\mu - \lambda} = d\) for all \(\mu \in \lambda + Q\) -- + \(\lambda\) standing for some fixed representative of its \(Q\)-coset. + Furthermore, given \(u \in \mathcal{U}(\mathfrak{g})_0\) and \(\mu \in + \lambda + Q\), + \[ + \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu}) + = \operatorname{Tr} + (\theta_\lambda(u)\!\restriction_{\Sigma^{-1} M_{\mu - \lambda}}) + \] + is polynomial in \(\mu\) because of the second item of + Proposition~\ref{thm:nice-automorphisms-exist}. +\end{proof} + +Lo and behold\dots + +\begin{theorem}[Mathieu]\index{coherent family!Mathieu's \(\mExt\) coherent extension} + There exists a unique semisimple coherent extension \(\mExt(M)\) of \(M\). + More precisely, if \(\mathcal{M}\) is any coherent extension of \(M\), then + \(\mathcal{M}^{\operatorname{ss}} \cong \mExt(M)\). Furthermore, \(\mExt(M)\) + is a irreducible coherent family. +\end{theorem} + +\begin{proof} + The existence part should be clear from the previous discussion: it suffices + to fix some coherent extension \(\mathcal{M}\) of \(M\) and take + \(\mExt(M) = \mathcal{M}^{\operatorname{ss}}\). + + To see that \(\mExt(M)\) is irreducible, recall from + Corollary~\ref{thm:bounded-is-submod-of-extension} that \(M\) is a + \(\mathfrak{g}\)-submodule of \(\mExt(M)\). Since the degree of \(M\) is the + same as the degree of \(\mExt(M)\), some of its weight spaces have maximal + dimension inside of \(\mExt(M)\). In particular, it follows from + Lemma~\ref{thm:centralizer-multiplicity} that \(\mExt(M)_\lambda = + M_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module for some + \(\lambda \in \operatorname{supp} M\). + + As for the uniqueness of \(\mExt(M)\), fix some other semisimple coherent + extension \(\mathcal{N}\) of \(M\). We claim that the multiplicity of a given + simple \(\mathfrak{g}\)-module \(L\) in \(\mathcal{N}\) is determined by its + \emph{trace function} + \begin{align*} + \mathfrak{h}^* \times \mathcal{U}(\mathfrak{g})_0 & + \to K \\ + (\lambda, u) & + \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda}) + \end{align*} + + It is a well known fact of the theory of modules that, given an associative + \(K\)-algebra \(A\), a finite-dimensional semisimple \(A\)-module \(L\) is + determined, up to isomorphism, by its \emph{character} + \begin{align*} + \chi_L : A & \to K \\ + a & \mapsto \operatorname{Tr}(a\!\restriction_L) + \end{align*} + + In particular, the multiplicity of \(L\) in \(\mathcal{N}\), which is the + same as the multiplicity of \(L_\lambda\) in \(\mathcal{N}_\lambda\), is + determined by the character \(\chi_{\mathcal{N}_\lambda} : + \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all simple weight + \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is determined by + its trace function. Of course, the same holds for \(\mExt(M)\). We now claim + that the trace function of \(\mathcal{N}\) is the same as that of + \(\mExt(M)\). Clearly, + \(\operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda}) = + \operatorname{Tr}(u\!\restriction_{M_\lambda}) = + \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda + \in \operatorname{supp}_{\operatorname{ess}} M\), \(u \in + \mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(M\) is + Zariski-dense and the maps \(\lambda \mapsto + \operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda})\) and \(\lambda \mapsto + \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) are polynomial in + \(\lambda \in \mathfrak{h}^*\), it follows that these maps coincide for all + \(u\). + + In conclusion, \(\mathcal{N} \cong \mExt(M)\) and \(\mExt(M)\) is unique. +\end{proof} + +% This is a very important theorem, but since we won't classify the coherent +% extensions in here we don't need it, and there is no other motivation behind +% it. Including this would also require me to explain what central characters +% are, which is a bit of a pain +%\begin{proposition}[Mathieu] +% The central characters of the irreducible submodules of +% \(\operatorname{Ext}(M)\) are all the same. +%\end{proposition} + +We have thus concluded our classification of cuspidal modules in terms of +coherent families. Of course, to get an explicit construction of all simple +\(\mathfrak{g}\)-modules we would have to classify the irreducible semisimple +coherent \(\mathfrak{g}\)-families themselves, which is the subject of sections +7, 8 and 9 of \cite{mathieu}. In addition, in sections 11 and 12 of +\cite{mathieu} Mathieu provides an explicit construction of coherent families. +We unfortunately do not have the necessary space to discuss these results in +detail, but we will now provide a brief overview. + +First and foremost, notice that because of +Example~\ref{thm:simple-weight-mod-is-tensor-prod} the problem of classifying +the simple weight \(\mathfrak{g}\)-modules can be reduced to that of +classifying the simple weight modules of its simple components. In addition, it +turns out that very few simple Lie algebras admit cuspidal modules at all. +Specifically\dots + +\begin{proposition}[Fernando]\label{thm:only-sl-n-sp-have-cuspidal} + Let \(\mathfrak{s}\) be a finite-dimensional simple Lie algebra. Suppose + there exists a simple cuspidal \(\mathfrak{s}\)-module. Then \(\mathfrak{s} + \cong \mathfrak{sl}_n(K)\) or \(\mathfrak{s} \cong \mathfrak{sp}_{2 n}(K)\). +\end{proposition} + +Hence it suffices to classify the irreducible semisimple coherent families of +\(\mathfrak{sl}_n(K)\) and \(\mathfrak{sp}_{2 n}(K)\). These can be described +either algebraically, using combinatorial invariants -- which Mathieu does in +sections 7, 8 and 9 of his paper -- or geometrically, using algebraic varieties +and differential forms -- which is done in sections 11 and 12. While rather +complicated on its own, the geometric construction is more concrete than its +combinatorial counterpart. + +This construction also brings us full circle to the beginning of these notes, +where we saw in Proposition~\ref{thm:geometric-realization-of-uni-env} that +\(\mathfrak{g}\)-modules may be understood as geometric objects. In fact, +throughout the previous four chapters we have seen a tremendous number of +geometrically motivated examples, which further emphasizes the connection +between representation theory and geometry. I would personally go as far as +saying that the beautiful interplay between the algebraic and the geometric is +precisely what makes representation theory such a fascinating and charming +subject. + +Alas, our journey has come to an end. All it is left is to wonder at the beauty +of Lie algebras and their representations. + +\label{end-47}
diff --git a/tcc.tex b/tcc.tex @@ -33,9 +33,9 @@ \input{sections/sl2-sl3} -\input{sections/semisimple-algebras} +\input{sections/fin-dim-simple} -\input{sections/mathieu} +\input{sections/simple-weight} \printbibliography \printindex