lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -263,37 +263,35 @@
   (\operatorname{Ext}(V))[\lambda]\).
 \end{proposition}
 
+% TODO: Do we need to use the fact that U(g) is netherian to conclude that V is
+% the quotient of subsequent terms of a composition series?
 \begin{proof}
   Fix some coherent extension \(\mathcal{M}\) of \(V\), so that \(V\) is a
-  subquotient of \(\mathcal{M}\). Since \(V\) is irreducible, it can be
-  realized as the quotient of consecutive terms of a composition series \(0 =
-  \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_n =
-  \mathcal{M}\). But
+  subquotient of \(\mathcal{M}\). More precisely, since \(V\) is irreducible 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
   \[
-    \operatorname{Ext}(V)
-    \cong \mathcal{M}^{\operatorname{ss}}
-    = \bigoplus_i \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i},
+    (\operatorname{Ext}(V))[\lambda]
+    \cong \mathcal{M}^{\operatorname{ss}}[\lambda]
+    = \bigoplus_i \mfrac{\mathcal{M}_{i + 1}[\lambda]}{\mathcal{M}_i[\lambda]},
   \]
-  so that \(V\) is contained in \(\operatorname{Ext}(V)\).
+  so that \(V\) is contained in \((\operatorname{Ext}(V))[\lambda]\).
 
-  Furtheremore, the irreducibility of \(V\) implies \(V \subset
-  (\operatorname{Ext}(V))[\lambda]\), and \(V_\mu \subset
-  \operatorname{Ext}(V)_\mu\) for any \(\mu \in \mathfrak{h}^*\). Hence it
-  suffices to show that \(V_\mu = \operatorname{Ext}(V)_\mu\) for any \(\mu \in
-  \lambda + Q\). But this is already clear from the fact that
+  Hence it suffices to show that \(V_\mu = \operatorname{Ext}(V)_\mu\) 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 \(v \in
-  V_\mu\), \(H \in \mathfrak{h}\) and \(X \in
+  V_\mu\), \(H \in \mathfrak{h}\) and \(u \in
   C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\) we find
   \[
-    H X v = X H v = \mu(H) \cdot X v,
+    H u v = u H v = \mu(H) \cdot u v,
   \]
   so that \(V_\mu\) is a
   \(C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\)-submodule of
   \(\operatorname{Ext}(V)_\mu\).
-
-  Since \(V\) is cuspidal and \(\mu \in \lambda + Q\), \(V_\mu \ne 0\) and
-  hence \(V_\mu = \operatorname{Ext}(V)_\mu\) -- because
-  \(\operatorname{Ext}(V)_\mu\) is an irreducible
+  Since \(V\) is cuspidal, \(\mu \in \lambda + Q = \operatorname{supp} V\)
+  implies \(V_\mu \ne 0\), and hence \(V_\mu = \operatorname{Ext}(V)_\mu\) --
+  because \(\operatorname{Ext}(V)_\mu\) is an irreducible
   \(C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\)-module.
 \end{proof}