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}