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

git clone: git://git.pablopie.xyz/lie-algebras-and-their-representations
Log
Files
Refs
README
LICENSE

Commit: aa6137d89836bdd213677a1638d5e7af8d3e6a33
Parent: f6e4bb679dc8ee9d56e58798289e6409486ba4fe
Author: Pablo <pablo-escobar@riseup.net>
Date: Sun, 25 Sep 2022 13:53:49 +0000

Completed the proof that every finite-dimensional representation of a semisimple Lie algebra is a weight module

Diffstat

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}