diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -253,40 +253,30 @@
is polynomial in \(\mu \in \mathfrak{h}^*\).
\end{proof}
-\begin{theorem}[Mathieu]
- Let \(V\) be an infinite-dimensional admissible irreducible
- \(\mathfrak{g}\)-module of degree \(d\). There exists a unique semisimple
- coherent extension \(\operatorname{Ext}(V)\) of \(V\). The central characters
- of the irreducible submodules of \(\operatorname{Ext}(V)\) are all the same.
- Furthermore, if \(\mathcal{M}\) is any coherent extension of \(V\), then
- \(\mathcal{M}^{\operatorname{ss}} \cong \operatorname{Ext}(V)\).
-\end{theorem}
-
-% TODO: Move this to before the proof of the existence of Ext?
\begin{proposition}
- Let \(V\) be a cuspidal representation of \(\mathfrak{g}\) and take any
- weight \(\lambda\) of \(V\). Then \(V \cong
- (\operatorname{Ext}(V))[\lambda]\).
+ Let \(V\) be a cuspidal representation of \(\mathfrak{g}\) and
+ \(\mathcal{M}\) be a semisimple coherent extension of \(V\) which is
+ irreducible as a coherent family. Then \(V = \mathcal{M}[\lambda]\) for any
+ \(\lambda \in \operatorname{supp} V\).
\end{proposition}
\begin{proof}
- Fix some coherent extension \(\mathcal{M}\) of \(V\), so that \(V\) is a
- subquotient of \(\mathcal{M}\). More precisely, since \(V\) is irreducible it
- is a subquotient of \(\mathcal{M}[\lambda]\) -- its support is contained in
- \(\lambda + Q\). Furthermore, once again it follows from the irreducibility
- of \(V\) that it can be realized as the quotient of consecutive terms of a
- composition series \(0 = \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots
- \subset \mathcal{M}_n = \mathcal{M}[\lambda]\). But
+ We know \(V\) is a subquotient of \(\mathcal{M}\). More precisely, since
+ \(V\) is irreducible it is a subquotient of \(\mathcal{M}[\lambda]\) -- its
+ support is contained in \(\lambda + Q\). Furthermore, once again it follows
+ from the irreducibility of \(V\) that it can be realized as the quotient of
+ consecutive terms of a composition series \(0 = \mathcal{M}_0 \subset
+ \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_n = \mathcal{M}[\lambda]\).
+ But since \(\mathcal{M}\) is semisimple
\[
- (\operatorname{Ext}(V))[\lambda]
- \cong \mathcal{M}^{\operatorname{ss}}[\lambda]
- = \bigoplus_i \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i},
+ \mathcal{M}[\lambda]
+ \cong \bigoplus_i \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i},
\]
- so that \(V\) is contained in \((\operatorname{Ext}(V))[\lambda]\).
+ so that \(V\) is contained in \(\mathcal{M}[\lambda]\).
- Hence it suffices to show that \(V_\mu = \operatorname{Ext}(V)_\mu\) for any
+ Hence it suffices to show that \(V_\mu = \mathcal{M}_\mu\) for any
\(\mu \in \lambda + Q\). But this is already clear from the fact that
- \(\operatorname{Ext}(V)\) is irreducible as a coherent family: given \(v \in
+ \(\mathcal{M}\) is irreducible as a coherent family: given \(v \in
V_\mu\), \(H \in \mathfrak{h}\) and \(u \in
C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\) we find
\[
@@ -294,10 +284,10 @@
\]
so that \(V_\mu\) is a
\(C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\)-submodule of
- \(\operatorname{Ext}(V)_\mu\). Since \(V\) is cuspidal, \(\mu \in \lambda + Q
+ \(\mathcal{M}_\mu\). Since \(V\) is cuspidal, \(\mu \in \lambda + Q
= \operatorname{supp} V\) -- i.e. the third equivalence of
corollary~\ref{thm:cuspidal-mod-equivs} -- implies \(V_\mu \ne 0\), and hence
- \(V_\mu = \operatorname{Ext}(V)_\mu\).
+ \(V_\mu = \mathcal{M}_\mu\).
\end{proof}
\begin{theorem}[Mathieu]
@@ -310,3 +300,15 @@
\item \(\mathcal{M}[\lambda]\) is cuspidal.
\end{enumerate}
\end{theorem}
+
+\section{Existance of Coherent Extensions}
+
+\begin{theorem}[Mathieu]
+ Let \(V\) be an infinite-dimensional admissible irreducible
+ \(\mathfrak{g}\)-module of degree \(d\). There exists a unique semisimple
+ coherent extension \(\operatorname{Ext}(V)\) of \(V\). The central characters
+ of the irreducible submodules of \(\operatorname{Ext}(V)\) are all the same.
+ Furthermore, if \(\mathcal{M}\) is any coherent extension of \(V\), then
+ \(\mathcal{M}^{\operatorname{ss}} \cong \operatorname{Ext}(V)\).
+\end{theorem}
+