lie-algebras-and-their-representations

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}
+