lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -9,9 +9,9 @@
 \end{definition}
 
 \begin{definition}
-  A weight \(\mathfrak{g}\)-module is called
-  \emph{an admissible} if \(\dim V_\lambda\) is bounded. The lowest upper bound
-  for \(\dim V_\lambda\) is called \emph{the degree of \(V\)}.
+  A weight \(\mathfrak{g}\)-module is called \emph{an admissible} if \(\dim
+  V_\lambda\) is bounded. The lowest upper bound for \(\dim V_\lambda\) is
+  called \emph{the degree of \(V\)}.
 \end{definition}
 
 \begin{example}
@@ -143,18 +143,49 @@
   \(\mathcal{M}^{\operatorname{ss}} \cong \operatorname{Ext}(V)\).
 \end{theorem}
 
-% TODO: Add a proof!
-% TODO: V is a submodule because of the definition of the semisimplification
-% TODO: All weights in the coset occur because V is cuspidal
-% TODO: The dimension of the weight spaces of V is maximal because Ext is
-% an irreducible coherent family
 % TODO: Define the notation for M[mu] somewhere else
+% TODO: Note somewhere else that the support of a cuspidal module is an entire
+% Q-coset
 \begin{proposition}
   Let \(V\) be a cuspidal representation of \(\mathfrak{g}\) and take any
   weight \(\lambda\) of \(V\). Then \(V \cong
   (\operatorname{Ext}(V))[\lambda]\).
 \end{proposition}
 
+\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 
+  \[
+    \operatorname{Ext}(V) 
+    \cong \mathcal{M}^{\operatorname{ss}}
+    = \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j},
+  \]
+  so that \(V\) is contained in \(\operatorname{Ext}(V)\).
+
+  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
+  \(\operatorname{Ext}(V)\) is irreducible as a coherent family: given \(v \in
+  V_\mu\), \(H \in \mathfrak{h}\) and \(X \in
+  C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\) we find
+  \[
+    H X v = X H v = \mu(H) \cdot X v,
+  \]
+  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
+  \(C_{\mathcal{U}(\mathfrak{g})}(\mathfrak{h})\)-module.
+\end{proof}
+
 \begin{theorem}[Mathieu]
   Let \(\mathcal{M}\) be an irreducible coherent family and \(\mu \in
   \mathfrak{h}^*\). The following conditions are equivalent.