lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -266,6 +266,31 @@
   is polynomial in \(\mu \in \mathfrak{h}^*\).
 \end{proof}
 
+\begin{corollary}\label{thm:admissible-is-submod-of-extension}
+  Let \(V\) be an irreducible admissible \(\mathfrak{g}\)-module and
+  \(\mathcal{M}\) be a completely reducible coherent extension of \(V\). Then
+  \(V\) is contained in \(\mathcal{M}\).
+\end{corollary}
+
+\begin{proof}
+  Since \(V\) is irreducible, its support is contained in a single \(Q\)-coset.
+  This implies that \(V\) is a subquotient of \(\mathcal{M}[\lambda]\) for any
+  \(\lambda \in \operatorname{supp} V\). If we fix some composition series \(0
+  = \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_n =
+  \mathcal{M}[\lambda]\) of \(\mathcal{M}[\lambda]\) with \(V \cong
+  \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}\), there is a natural inclusion
+  \[
+    V
+    \isoto \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}
+    \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j}
+    \cong \mathcal{M}^{\operatorname{ss}}[\lambda]
+  \]
+
+  It then follows from the uniqueness of the semisimplification of
+  \(\mathcal{M}\) that \(\mathcal{M} \cong \mathcal{M}^{\operatorname{ss}}\),
+  so we have an inclusion \(V \to \mathcal{M}\).
+\end{proof}
+
 \begin{theorem}[Mathieu]
   Let \(\mathcal{M}\) be an irreducible coherent family and \(\lambda \in
   \mathfrak{h}^*\). The following conditions are equivalent.
@@ -486,42 +511,27 @@
   There exists a unique completely reducible coherent extension
   \(\operatorname{Ext}(V)\) of \(V\). More precisely, if \(\mathcal{M}\) is any
   coherent extension of \(V\), then \(\mathcal{M}^{\operatorname{ss}} \cong
-  \operatorname{Ext}(V)\). Furthermore, \(V\) is itself contained in
-  \(\operatorname{Ext}(V)\) and \(\operatorname{Ext}(V)\) is irreducible as a
-  coherent family.
+  \operatorname{Ext}(V)\). Furthermore, \(\operatorname{Ext}(V)\) is
+  irreducible as a coherent family.
 \end{theorem}
 
-% TODOOO: Prove the uniqueness
+% TODOOO: Prove the uniqueness!
 \begin{proof}
-  The existence part should be clear from the previous discussion: let
-  \(\mathcal{M}\) be a coherent extension of \(V\) and take
-  \(\operatorname{Ext}(V) = \mathcal{M}^{\operatorname{ss}}\) of \(V\).
-
-  % TODOOOOOO: Extract this to a general result: any admissible module is
-  % contained in the semisimplification of any of its coherent extensions
-  We claim \(V\) is contained in \(\operatorname{Ext}(V)\). Indeed, if we fix
-  some composition series \(0 = \mathcal{M}_0 \subset \mathcal{M}_1 \subset
-  \cdots \subset \mathcal{M}_n = \mathcal{M}[\lambda]\) of
-  \(\mathcal{M}[\lambda]\) with \(V \cong \mfrac{\mathcal{M}_{i +
-  1}}{\mathcal{M}_i}\), \(\lambda \in \operatorname{supp} V\), there is a
-  natural inclusion
-  \[
-    V
-    \isoto \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}
-    \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j}
-    = \mathcal{M}^{\operatorname{ss}}[\lambda]
-    \subset \operatorname{Ext}(V)
-  \]
+  The existence part should be clear from the previous discussion: it suffices
+  to fix some coherent extension \(\mathcal{M}\) of \(V\) and take
+  \(\operatorname{Ext}(V) = \mathcal{M}^{\operatorname{ss}}\).
 
   % TODOOO: Prove that the weight spaces of any simple g-module are all simple
   % C(h)-modules
-  Since the degree of \(V\) is the same as the degree of
-  \(\operatorname{Ext}(V)\), some of its weight spaces have maximal dimension
-  inside of \(\operatorname{Ext}(V)\). In particular, it follows from the
-  irreducibility of \(V\) that \(\operatorname{Ext}(V)_\lambda = V_\lambda\) is
-  a simple \(C_{\mathcal{U}(\mathfrak{h})}(\mathfrak{h})\)-module for some
-  \(\lambda \in \operatorname{supp} V\), so that \(\operatorname{Ext}(V)\) is
-  irreducible as a coherent family.
+  To see that \(\operatorname{Ext}(V)\) is irreducible as a coherent family,
+  recall from corollary~\ref{thm:admissible-is-submod-of-extension} that \(V\)
+  is a subrepresentation of \(\operatorname{Ext}(V)\). Since the degree of
+  \(V\) is the same as the degree of \(\operatorname{Ext}(V)\), some of its
+  weight spaces have maximal dimension inside of \(\operatorname{Ext}(V)\). In
+  particular, it follows from the irreducibility of \(V\) that
+  \(\operatorname{Ext}(V)_\lambda = V_\lambda\) is a simple
+  \(C_{\mathcal{U}(\mathfrak{h})}(\mathfrak{h})\)-module for some \(\lambda \in
+  \operatorname{supp} V\).
 \end{proof}
 
 \begin{proposition}[Mathieu]