diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -1021,770 +1021,7 @@ irreducible representations of \(\mathfrak{sl}_2(K)\) and
motivation behind our methods. In particular, we did not explain why we chose
\(h\) and \(\mathfrak{h}\), and neither why we chose to look at their
eigenvalues. Apart from the obvious fact we already knew it would work a
-priory, why did we do all that? In the following section we will attempt to
+priory, why did we do all that? In the following chapter we will attempt to
answer this question by looking at what we did in the last chapter through more
abstract lenses and studying the representations of an arbitrary
finite-dimensional semisimple Lie algebra \(\mathfrak{g}\).
-
-\section{Simultaneous Diagonalization \& the General Case}
-
-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}
- V = \bigoplus_\lambda V_\lambda
-\end{equation}
-
-This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the
-reasoning behind it, as well as the mere fact equation (\ref{sym-diag}) holds,
-are harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace
-decomposition associated with an operator \(V \to V\) is a very well-known
-tool, and this type of argument should be familiar to anyone familiar with
-basic concepts of linear algebra. On the other hand, the eigenspace
-decomposition of \(V\) with respect to the action of an arbitrary subalgebra
-\(\mathfrak{h} \subset \mathfrak{gl}(V)\) is neither well-known nor does it
-hold in general: as previously stated, it may very well be that
-\[
- \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V
-\]
-
-We should note, however, that this two cases are not as different as they may
-sound at first glance. Specifically, we can regard the eigenspace decomposition
-of a representation \(V\) of \(\mathfrak{sl}_2(K)\) with respect to the
-eigenvalues of the action of \(h\) as the eigenvalue decomposition of \(V\)
-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 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 representation \(V\) of
-\(\mathfrak{g}\) into subspaces invariant under the action of \(\mathfrak{h}\),
-and then analyze how the remaining elements of \(\mathfrak{g}\) act on this
-subspaces. The bigger \(\mathfrak{h}\) 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}
- An 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 representation. Indeed,
- \(0\) -- the only element of \(0 \subset \mathfrak{g}\) -- is such that
- \(\operatorname{ad}(0) = 0\). 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 -- i.e. a Cartan subalgebra.
-\end{proof}
-
-We have already seen some concrete examples. For instance, one can readily
-check that every pair of diagonal matrices commutes, so that
-\[
- \mathfrak{h} =
- \begin{pmatrix}
- K & 0 & \cdots & 0 \\
- 0 & K & \cdots & 0 \\
- \vdots & \vdots & \ddots & \vdots \\
- 0 & 0 & \cdots & K
- \end{pmatrix}
-\]
-is an Abelian -- and hence nilpotent -- subalgebra of \(\mathfrak{gl}_n(K)\). 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. Hence \(\mathfrak{h}\) is a Cartan subalgebra of
-\(\mathfrak{gl}_n(K)\).
-
-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 two sections. The remaining question then
-is: if \(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\)
-is a representation of \(\mathfrak{g}\), does the eigenspace decomposition
-\[
- V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda
-\]
-of \(V\) 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'll 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
--- where \(n\) is, of course, finite.
-
-Simultaneous diagonalization implies that to show \(V = \bigoplus_\lambda
-V_\lambda\) it suffices to show that \(H\!\restriction_V : V \to V\) 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 \(T_s, T_n : V \to V\), with
- \(T_s\) diagonalizable and \(T_n\) nilpotent, such that \(T = T_s + T_n\).
- The pair \((T_s, T_n)\) is known as \emph{the Jordan decomposition of \(T\)}.
-\end{proposition}
-
-\begin{proposition}
- Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are
- \(X_s, X_n \in \mathfrak{g}\) such that \(X = X_s + X_n\), \([X_s, X_n] =
- 0\), \(\operatorname{ad}(X_s)\) is a diagonalizable operator and
- \(\operatorname{ad}(X_n)\) is a nilpotent operator. The pair \((X_s, X_n)\)
- is known as \emph{the Jordan decomposition of \(X\)}.
-\end{proposition}
-
-It should be clear from the uniqueness of \(\operatorname{ad}(X)_s\) and
-\(\operatorname{ad}(X)_n\) that the Jordan decomposition of
-\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = \operatorname{ad}(X_s) +
-\operatorname{ad}(X_n)\). What's perhaps more remarkable is the fact this holds
-for \emph{any} finite-dimensional representation of \(\mathfrak{g}\). In other
-words\dots
-
-\begin{proposition}\label{thm:preservation-jordan-form}
- Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(X
- \in \mathfrak{g}\). Denote by \(X\!\restriction_V\) the action of \(X\) in
- \(V\). Then \(X_s\!\restriction_V = (X\!\restriction)_s\) and
- \(X_n\!\restriction_V = (X\!\restriction)_n\).
-\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}. We should point out
-this fails spectacularly in positive characteristic. Furthermore, the statement
-of proposition~\ref{thm:preservation-jordan-form} only makes sense for
-\emph{semisimple} Lie algebras -- i.e. the algebras \(\mathfrak{g}\) for which
-the abstract Jordan decomposition of \(\mathfrak{g}\) is defined. Nevertheless,
-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 \(V\) be any finite-dimensional
- representation of \(\mathfrak{g}\). Then there is a basis \(\{v_1, \ldots,
- v_n\}\) of \(V\) so that each \(v_i\) 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
- \(
- H v_i = \lambda_i(H) \cdot v_i
- \)
- for all \(H \in \mathfrak{h}\). In particular,
- \[
- V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda
- \]
-\end{corollary}
-
-\begin{proof}
- Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_V
- : V \to V\) is a diagonalizable operator.
-
- If we write \(H = H_s + H_n\) for the abstract Jordan decomposition of \(H\),
- we know \(\operatorname{ad}(H_s) = \operatorname{ad}(H)_s\). But
- \(\operatorname{ad}(H)\) is a diagonalizable operator, so that
- \(\operatorname{ad}(H)_s = \operatorname{ad}(H)\). This implies
- \(\operatorname{ad}(H_n) = \operatorname{ad}(H)_n = 0\), so that \(H_n\) is a
- central element of \(\mathfrak{g}\). Since \(\mathfrak{g}\) is semisimple,
- \(H_n = 0\). Proposition~\ref{thm:preservation-jordan-form} then implies
- \((H\!\restriction_V)_n = (H_n)\!\restriction_V = 0\), so \(H\!\restriction_V
- = (H\!\restriction_V)_s\) 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 \(V\) endowed with an operator
-\(V \to V\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) in
-\(V\). In particular, if we choose an operator \(V \to V\) which is \emph{not}
-diagonalizable we find \(V \ne \bigoplus_{\lambda \in \mathfrak{h}^*}
-V_\lambda\).
-
-However, corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive
-\(\mathfrak{g}\) if we assume that the representation in question is
-irreducible, since central elements of \(\mathfrak{g}\) act on irreducible
-representations as scalar operators. The hypothesis of finite-dimensionality is
-also of huge importance. In the next chapter we will encounter
-infinite-dimensional \(\mathfrak{g}\)-modules for which the eigenspace
-decomposition \(V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails.
-As a first consequence of corollary~\ref{thm:finite-dim-is-weight-mod}
-
-\begin{corollary}
- The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate.
-\end{corollary}
-
-\begin{proof}
- Consider the eigenspace decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
- \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, 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 \(B\) 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
- \[
- \alpha(H_1) \cdot B(X, H_2)
- = B([H_1, X], H_2)
- = - B([X, H_1], H_2)
- = - B(X, [H_1, H_2])
- = 0
- \]
-
- Hence the non-degeneracy of \(B\) implies the non-degeneracy of its
- restriction.
-\end{proof}
-
-We should point out that the restriction of \(B\) 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 \(B\) induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\), it
- induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in
- \(\mathfrak{h}^*\). We denote this form by \(B\).
-\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 \(V\) be a finite-dimensional irreducible representation of
-\(\mathfrak{g}\). We will proceed, as we did before, by generalizing the
-results about of the previous two sections in order. By now the pattern should
-be starting 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}.
-
-We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) and
-\(\mathfrak{sl}_3(K)\), the roots were symmetric about the origin and spanned
-all of \(\mathfrak{h}^*\). 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 eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) in
- \(\mathfrak{g}\) are symmetrical about the origin -- i.e. \(- \alpha\) is
- also an eigenvalue -- and they span all of \(\mathfrak{h}^*\).
-\end{proposition}
-
-\begin{proof}
- We'll start with the first claim. Let \(\alpha\) and \(\beta\) be two
- eigenvalues of the adjoint action of \(\mathfrak{h}\). 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
- \[
- (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z
- = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ]
- \in \mathfrak{g}_{n \alpha + n \beta + \gamma}
- = 0
- \]
- for \(n\) large enough. In particular, \(B(X, Y) =
- \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if
- \(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\)
- for all eigenvalues \(\beta\), which contradicts the non-degeneracy of \(B\).
- Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
- \(\mathfrak{h}\).
-
- For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do
- not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\)
- non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), 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 \(\mathfrak{z}\) of \(\mathfrak{g}\), which is zero by the
- semisimplicity -- 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 eigenspaces \(\mathfrak{g}_\alpha\) 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 some eigenvalue \(\alpha\) of the adjoint action of
-\(\mathfrak{h}\) in \(\mathfrak{g}\) and a eigenvector \(X \in
-\mathfrak{g}_\alpha\), then for each \(H \in \mathfrak{h}\) and \(v \in
-V_\lambda\) we find
-\[
- H (X v)
- = X (H v) + [H, X] v
- = (\lambda + \alpha)(H) \cdot X v
-\]
-so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). We have encountered
-this formula twice in this chapter: again, we find \(\mathfrak{g}_\alpha\)
-\emph{acts on \(V\) by translating vectors between eigenspaces}. In other
-words, if we denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\)
-then\dots
-
-\begin{theorem}\label{thm:weights-congruent-mod-root}
- The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) are
- all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\).
-\end{theorem}
-
-% TODOO: Turn this into a proper discussion of basis and give the idea of the
-% proof of existance of basis?
-To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a
-direction in \(\mathfrak{h}^*\) -- i.e. we fix a linear function
-\(\mathfrak{h}^* \to \QQ\) such that \(Q\) lies outside of its kernel. This
-choice induces a partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of
-roots of \(\mathfrak{g}\) and once more we find\dots
-
-\begin{definition}
- The elements of \(\Delta^+\) and \(\Delta^-\) are called \emph{positive} and
- \emph{negative roots}, respectively. The subalgebra \(\mathfrak{b} =
- \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is
- called \emph{the Borel subalgebra associated with \(\mathfrak{h}\)}.
-\end{definition}
-
-\begin{theorem}
- There is a weight vector \(v \in V\) that is killed by all positive root
- spaces of \(\mathfrak{g}\).
-\end{theorem}
-
-% TODO: Here we may take a weight of maximal height, but why is it unique?
-% TODO: We don't really need to talk about height tho, we may simply take a
-% weight that maximizes B(gamma, lambda) in QQ
-% TODOO: Either way, we need to move this to after the discussion on the
-% integrality of weights
-\begin{proof}
- It suffices to note that if \(\lambda\) is the weight of \(V\) lying the
- furthest along the direction we chose and \(V_{\lambda + \alpha} \ne 0\) for
- some \(\alpha \in \Delta^+\) then \(\lambda + \alpha\) is a weight that is
- furthest along the direction we chose than \(\lambda\), which contradicts the
- definition of \(\lambda\).
-\end{proof}
-
-Accordingly, we call \(\lambda\) \emph{the highest weight of \(V\)}, and we
-call any \(v \in V_\lambda\) \emph{a highest weight vector}. The strategy then
-is to describe all weight spaces of \(V\) in terms of \(\lambda\) and \(v\), as
-in theorem~\ref{thm:sl3-irr-weights-class}, and unsurprisingly we do so by
-reproducing the proof of the case of \(\mathfrak{sl}_3(K)\). Namely, 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
- \[
- V_\mu[\alpha] = \bigoplus_k V_{\mu + k \alpha}
- \]
- 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 \([\mathfrak{g}_\alpha, \mathfrak{g}_{-
-\alpha}] \ne 0\) 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. The second statement of
-corollary~\ref{thm:distinguished-subalg-rep} imposes a restriction on the
-weights of \(V\). Namely, if \(\mu\) is a weight, \(\mu(H_\alpha)\) is an
-eigenvalue of \(h\) in some representation of \(\mathfrak{sl}_2(K)\), so
-that\dots
-
-\begin{proposition}
- The weights \(\mu\) of an irreducible representation \(V\) of
- \(\mathfrak{g}\) are so that \(\mu(H_\alpha) \in \ZZ\) for each \(\alpha \in
- \Delta\).
-\end{proposition}
-
-Once more, the lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha)
-\in \ZZ, \forall \alpha \in \Delta \}\) is called \emph{the weight lattice of
-\(\mathfrak{g}\)}, and we call the elements of \(P\) \emph{integral}. Finally,
-another important consequence of theorem~\ref{thm:distinguished-subalgebra}
-is\dots
-
-\begin{corollary}
- If \(\alpha \in \Delta^+\) and \(T_\alpha : \mathfrak{h}^* \to
- \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to
- \(\alpha\) with respect to the Killing form,
- corollary~\ref{thm:distinguished-subalg-rep} implies that all \(\nu \in P\)
- lying inside the line connecting \(\mu\) and \(T_\alpha \mu\) are weights --
- i.e. \(V_\nu \ne 0\).
-\end{corollary}
-
-\begin{proof}
- It suffices to note that \(\nu \in V_\mu[\alpha]\) -- see appendix D of
- \cite{fulton-harris} for further details.
-\end{proof}
-
-\begin{definition}
- We refer to the group \(\mathcal{W} = \langle T_\alpha : \alpha \in \Delta^+
- \rangle \subset \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of
- \(\mathfrak{g}\)}.
-\end{definition}
-
-This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we
-found that the weights of the irreducible representations were symmetric with
-respect to the lines \(K \alpha\) with \(B(\alpha_i - \alpha_j, \alpha) = 0\).
-Indeed, the same argument leads us to the conclusion\dots
-
-\begin{theorem}\label{thm:irr-weight-class}
- The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) 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 image of \(\lambda\) under the action of the Weyl group
- \(\mathcal{W}\).
-\end{theorem}
-
-Now 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}. Lo and behold\dots
-
-\begin{definition}
- An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all
- \(\alpha \in \Delta^+\) is referred to as an \emph{integral dominant weight
- of \(\mathfrak{g}\)}.
-\end{definition}
-
-\begin{theorem}\label{thm:dominant-weight-theo}
- For each dominant integral \(\lambda \in P\) there exists precisely one
- irreducible finite-dimensional representation \(V\) of \(\mathfrak{g}\) whose
- highest weight is \(\lambda\).
-\end{theorem}
-
-Fix some dominant integral \(\lambda \in P\). The ``uniqueness'' part of the
-theorem follows at once from the argument used for \(\mathfrak{sl}_3(K)\). The
-``existence'' part 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 \(\mathfrak{sl}_3(K)\).
-Instead, we need a new strategy for the general setting. To that end, we
-introduce a special class of \(\mathfrak{g}\)-modules, known as \emph{Verma
-modules}.
-
-\begin{definition}\label{def:verma}
- The \(\mathfrak{g}\)-module \(M(\lambda) =
- \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v^+\), where the action of
- \(\mathfrak{b}\) in \(K v^+\) is given by \(H v^+ = \lambda(H) \cdot v^+\)
- for all \(H \in \mathfrak{h}\) and \(X v^+ = 0\) for \(X \in
- \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma
- module of weight \(\lambda\)}
-\end{definition}
-
-We should point out that, unlike most representations we've encountered so far,
-Verma modules are \emph{highly infinite-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,
-\(M(\lambda)\) turns out to be quite well behaved. For instance, by
-construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\) -- where
-\(v^+ = 1 \otimes v^+ \in M(\lambda)\) is as in definition~\ref{def:verma}.
-Moreover, we find\dots
-
-\begin{proposition}\label{thm:verma-is-weight-mod}
- 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 \(\dim M(\lambda) = 1\). Finally, \(\lambda\) is the
- highest weight of \(M(\lambda)\), with highest weight vector given by \(v^+ =
- 1 \otimes v^+ \in M(\lambda)\) as in definition~\ref{def:verma}.
-\end{proposition}
-
-\begin{proof}
- The Poincaré-Birkhoff-Witt theorem implies that \(M(\lambda)\) is spanned by
- the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+\) for
- \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as
- in the proof of proposition~\ref{thm:distinguished-subalgebra}. But
- \[
- \begin{split}
- H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+
- & = ([H, F_{\alpha_1}] + F_{\alpha_1} H)
- F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
- & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v^+
- + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H)
- F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
- & \;\; \vdots \\
- & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot
- F_{\alpha_1} \cdots F_{\alpha_n} v^+
- + F_{\alpha_1} \cdots F_{\alpha_n} H v^+ \\
- & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot
- F_{\alpha_1} \cdots F_{\alpha_n} v^+ \\
- & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v^+
- \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n}
- \end{split}
- \]
-
- Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*}
- M(\lambda)_\mu\), as desired. In fact we have established
- \[
- M(\lambda)
- \subset
- \bigoplus_{\substack{k_i \in \ZZ \\ k_i \ge 0}}
- M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n}
- \]
- where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of
- \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots +
- k_n \cdot \alpha_n\).
-
- This already gives us that the weights of \(M(\lambda)\) are bounded by
- \(\lambda\) -- in the sense that no weight of \(M(\lambda)\) is ``higher''
- than \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that
- \(v^+\) is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The
- Poincaré-Birkhoff-Witt theorem implies
- \[
- M(\lambda)
- \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
- \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
- \cong \bigoplus_i \mathcal{U}(\mathfrak{b})
- \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
- \cong \bigoplus_i K v^+
- \ne 0
- \]
- as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v^+ \ne 0\) -- for if this was
- not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+
- = 0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest
- weight of \(M(\lambda)\), with highest weight vector \(v^+\).
-
- To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only
- finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
- F_{\alpha_n}^{k_n}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
- + k_n \cdot \alpha_n\). Since \(M(\lambda)_\mu\) is spanned by the images of
- \(v^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In
- particular, there is a single monomials \(F_{\alpha_1}^{k_1}
- F_{\alpha_2}^{k_2} \cdots F_{\alpha_n}^{k_n}\) such that \(\lambda = \lambda
- + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n\) -- which is, of course,
- the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim V_\lambda = 1\).
-\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 \(M(\lambda) =
- \bigoplus_{k \ge 0} K f^k v^+\), and the action of \(\mathfrak{sl}_2(K)\) in
- \(M(\lambda)\) is given by
- \begin{align*}
- f^{k + 1} v^+ & \overset{e}{\mapsto} (2 - k (k - 1)) f^k v^+ &
- f^{k + 1} v^+ & \overset{f}{\mapsto} f^{k + 2} v^+ &
- f^{k + 1} v^+ & \overset{h}{\mapsto} - 2 k f^{k + 1} v^+ &
- \end{align*}
-
- In the language of the diagrams used in section~\ref{sec:sl2}, we write
- % TODO: Add a label to the righ of the diagram indicating that the top arrows
- % are the action of e and the bottom arrows are the action of f
- \begin{center}
- \begin{tikzcd}
- \cdots \arrow[bend left=60]{r}{-10}
- & M(\lambda)_{-6} \arrow[bend left=60]{r}{-4} \arrow[bend left=60]{l}{1}
- & M(\lambda)_{-4} \arrow[bend left=60]{r}{0} \arrow[bend left=60]{l}{1}
- & M(\lambda)_{-2} \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1}
- & M(\lambda)_0 \arrow[bend left=60]{r}{2} \arrow[bend left=60]{l}{1}
- & M(\lambda)_2 \arrow[bend left=60]{l}{1}
- \end{tikzcd}
- \end{center}
- where \(M(\lambda)_{2 - 2 k} = K f^k v\). In this case, unlike we have see in
- section~\ref{sec:sl2}, the string of weight spaces to left of the diagram is
- infinite.
-\end{example}
-
-What's interesting to us about all this is that we've just constructed a
-\(\mathfrak{g}\)-module whose highest weight is \(\lambda\). This is not a
-proof of theorem~\ref{thm:dominant-weight-theo}, however, since \(M(\lambda)\)
-is neither irreducible nor finite-dimensional. Nevertheless, we can use
-\(M(\lambda)\) to construct an irreducible representation of \(\mathfrak{g}\)
-whose highest weight is \(\lambda\).
-
-\begin{proposition}\label{thm:max-verma-submod-is-weight}
- Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its
- weight spaces. In particular, \(M(\lambda)\) has a unique maximal
- subrepresentation \(N(\lambda)\) and a unique irreducible quotient
- \(\sfrac{M(\lambda)}{N(\lambda)}\).
-\end{proposition}
-
-\begin{proof}
- Let \(V \subset M(\lambda)\) be a subrepresentation and take any nonzero \(v
- \in V\). Because of proposition~\ref{thm:verma-is-weight-mod}, we know there
- are \(\mu_1, \ldots, \mu_n \in \mathfrak{h}^*\) and nonzero \(v_i \in
- M(\lambda)_{\mu_i}\) such that \(v = v_1 + \cdots + v_n\). We want to show
- \(v_i \in V\) for all \(i\).
-
- Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\).
- Then
- \[
- v_1
- - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3
- - \cdots
- - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n
- = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v
- \in V
- \]
-
- 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*}
- v_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)} v_4
- - \cdots -
- \frac{(\mu_n - \mu_3)(H_3) \cdot (\mu_n - \mu_1)(H_2)}
- {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_n \\
- =
- \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) v
- \in V
- \end{multline*}
-
- By applying the same procedure over and over again we can see that \(v_1 = X
- v \in V\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we
- reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that
- \(v_2 \in V\). Now by applying the same procedure over and over we find
- \(v_1, \ldots, v_n \in V\). Hence
- \[
- V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V
- \]
-
- Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), \(V\) is a proper
- subrepresentation then \(v^+ \notin V\). 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. In fact, this implies
- \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and
- \(\sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible 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 \(N(\lambda) = \bigoplus_{k \ge
- 3} K f^k v^+\), so that \(\sfrac{M(\lambda)}{N(\lambda)}\) is the
- \(3\)-dimensional irreducible representation of \(\mathfrak{sl}_2(K)\) --
- i.e. the finite-dimensional irreducible representation with highest weight
- \(\lambda\) constructed in section~\ref{sec:sl2}.
-\end{example}
-
-This last example is particularly interesting to us, since it indicates that
-the finite-dimensional irreducible representations of \(\mathfrak{sl}_2(K)\) as
-quotients of Verma modules. This is because the quotient
-\(\sfrac{M(\lambda)}{N(\lambda)}\) in example~\ref{ex:sl2-verma-quotient}
-happened to be finite-dimensional. As it turns out, this is always the case for
-semisimple \(\mathfrak{g}\). Namely\dots
-
-\begin{proposition}\label{thm:verma-is-finite-dim}
- If \(\lambda\) is dominant integral then the unique irreducible quotient 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 \(\sfrac{M(\lambda)}{N(\lambda)}\) is stable under the natural
-action of the Weyl group \(\mathcal{W}\) in \(\mathfrak{h}^*\). One can then
-show that the every weight of \(V\) is conjugate to a single dominant integral
-weight of \(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant
-integral weights of such irreducible quotient is finite. Since \(W\) is
-finitely generated, this implies the set of weights of the unique irreducible
-quotient of \(M(\lambda)\) is finite. But each weight space is
-finite-dimensional. Hence so is the irreducible quotient.
-
-We refer the reader to \cite[ch. 21]{humphreys} for further details. What we
-are really interested in is\dots
-
-\begin{corollary}
- There is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose
- highest weight is \(\lambda\).
-\end{corollary}
-
-\begin{proof}
- Let \(V = \sfrac{M(\lambda)}{N(\lambda)}\). It suffices to show that its
- highest weight is \(\lambda\). We have already seen that \(v^+ \in
- M(\lambda)_\lambda\) is a highest weight vector. Now since \(v\) lies outside
- of the maximal subrepresentation of \(M(\lambda)\), the projection \(v^+ +
- N(\lambda) \in V\) is nonzero.
-
- % TODO: Why is V_mu = M(lambda)_mu + N(lambda)? Turn this into a proposition?
- We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed,
- \[
- H (v^+ + N(\lambda))
- = H v^+ + N(\lambda)
- = \lambda(H) \cdot (v^+ + N(\lambda))
- \]
- for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with
- weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
- the highest weight of \(V\), for if this was not the case we could find a
- weight \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\).
-\end{proof}
-
-% TODO: Write a conclusion and move this to the next chapter
-