Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Completed the proof that every finite-dimensional representation of a semisimple Lie algebra is a weight module
2 files changed, 99 insertions, 32 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/mathieu.tex | 5 | 4 | 1 |
| Modified | sections/semisimple-algebras.tex | 126 | 95 | 31 |
diff --git a/sections/mathieu.tex b/sections/mathieu.tex @@ -16,7 +16,9 @@ \begin{example} Corollary~\ref{thm:finite-dim-is-weight-mod} is equivalent to the fact that - every finite-dimensional \(\mathfrak{g}\)-module is a weight module. + every finite-dimensional representation of a semisimple Lie algebra is a + weight module. More generally, every finite-dimensional irreducible + representation of a reductive Lie algebra is a weight module. \end{example} % TODO: Is every quotient of a weight module a weight module too? @@ -60,6 +62,7 @@ which is \emph{not} parabolic induced. \end{definition} +% TODOO: w should be an element of the subgroup W(V) of W. Fix this! % TODO: Remark on the fact that any simple weight p-mod is a (p/u)-mod, so that % the notation of a cuspidal p-mod is well definited % TODO: Define the conjugation of a p-mod by an element of the Weil group
diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex @@ -1859,44 +1859,108 @@ What is simultaneous diagonalization all about then? 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 +Jorden 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} + +% TODOO: Prove this? It suffices to show that ad(X)_s, ad(X)_n in ad(g)! +\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 uniqueless 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 spetacularly in positive characteristic. Furtheremore, the statement +of proposition~\ref{thm:preservation-jordan-form} only makes sence for +\emph{semisimple} Lie algebras -- i.e. the algebras \(\mathfrak{g}\) for wich +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 Lie algebra, \(\mathfrak{h} \subset \mathfrak{g}\) - be an Abelian 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 + 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, + \[ + V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \] - for all \(H \in \mathfrak{h}\). \end{corollary} -% TODOO: Prove that the operators are diagonalizable \begin{proof} - It suffices to show that \(H : V \to V\) is a diagonalizable operator for - each \(H \in \mathfrak{h}\). + 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} -As promised, this implies\dots - -\begin{corollary} - Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra - and \(\mathfrak{h}\) be a Cartan subalgebra of \(\mathfrak{g}\). Given a - finite-dimensional representation \(V\) of \(\mathfrak{g}\), - \[ - V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda - \] -\end{corollary} - -% TODO: Add a reference to the next chapter when it is done -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. In the next chapter we will encounter -\(\mathfrak{g}\)-modules for which the eigenspace decomposition \(V = -\bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails. +We should point out that this last proof only works for semisimple Lie +algebras. This is because we rely heavely 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. @@ -1940,8 +2004,8 @@ We should point out that the restriction of \(B\) to \(\mathfrak{h}\) is hardly ever a non-degenerate form. \begin{note} - Since \(B\) is induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\), - it induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in + 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}