diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -1333,32 +1333,30 @@ It should now be obvious\dots
Lo and behold\dots
\begin{theorem}[Mathieu]
- There exists a unique completely reducible coherent extension
- \(\mathcal{Ext}(M)\) of \(M\). More precisely, if \(\mathcal{M}\) is any
- coherent extension of \(M\), then \(\mathcal{M}^{\operatorname{ss}} \cong
- \mathcal{Ext}(M)\). Furthermore, \(\mathcal{Ext}(M)\) is
- a irreducible coherent family.
+ There exists a unique completely reducible coherent extension \(\mExt(M)\) of
+ \(M\). More precisely, if \(\mathcal{M}\) is any coherent extension of \(M\),
+ then \(\mathcal{M}^{\operatorname{ss}} \cong \mExt(M)\). Furthermore,
+ \(\mExt(M)\) is a irreducible coherent family.
\end{theorem}
\begin{proof}
The existence part should be clear from the previous discussion: it suffices
to fix some coherent extension \(\mathcal{M}\) of \(M\) and take
- \(\mathcal{Ext}(M) = \mathcal{M}^{\operatorname{ss}}\).
+ \(\mExt(M) = \mathcal{M}^{\operatorname{ss}}\).
- To see that \(\mathcal{Ext}(M)\) is irreducible, recall from
+ To see that \(\mExt(M)\) is irreducible, recall from
Corollary~\ref{thm:admissible-is-submod-of-extension} that \(M\) is a
- \(\mathfrak{g}\)-submodule of \(\mathcal{Ext}(M)\). Since the degree of \(M\)
- is the same as the degree of \(\mathcal{Ext}(M)\), some of its weight spaces
- have maximal dimension inside of \(\mathcal{Ext}(M)\). In particular, it
- follows from Proposition~\ref{thm:centralizer-multiplicity} that
- \(\mathcal{Ext}(M)_\lambda = M_\lambda\) is a simple
- \(\mathcal{U}(\mathfrak{g})_0\)-module for some \(\lambda \in
- \operatorname{supp} M\).
-
- As for the uniqueness of \(\mathcal{Ext}(M)\), fix some other completely
- reducible coherent extension \(\mathcal{N}\) of \(M\). We claim that the
- multiplicity of a given simple \(\mathfrak{g}\)-module \(L\) in
- \(\mathcal{N}\) is determined by its \emph{trace function}
+ \(\mathfrak{g}\)-submodule of \(\mExt(M)\). Since the degree of \(M\) is the
+ same as the degree of \(\mExt(M)\), some of its weight spaces have maximal
+ dimension inside of \(\mExt(M)\). In particular, it follows from
+ Proposition~\ref{thm:centralizer-multiplicity} that \(\mExt(M)_\lambda =
+ M_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module for some
+ \(\lambda \in \operatorname{supp} M\).
+
+ As for the uniqueness of \(\mExt(M)\), fix some other completely reducible
+ coherent extension \(\mathcal{N}\) of \(M\). We claim that the multiplicity
+ of a given simple \(\mathfrak{g}\)-module \(L\) in \(\mathcal{N}\) is
+ determined by its \emph{trace function}
\begin{align*}
\mathfrak{h}^* \times \mathcal{U}(\mathfrak{g})_0 &
\to K \\
@@ -1377,24 +1375,23 @@ Lo and behold\dots
In particular, the multiplicity of \(L\) in \(\mathcal{N}\), which is the
same as the multiplicity of \(L_\lambda\) in \(\mathcal{N}_\lambda\), is
determined by the character \(\chi_{\mathcal{N}_\lambda} :
- \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all simple
- weight \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is
- determined by its trace function. Of course, the same holds for
- \(\mathcal{Ext}(M)\). We now claim that the trace function of
- \(\mathcal{N}\) is the same as that of \(\mathcal{Ext}(M)\). Clearly,
- \(\operatorname{Tr}(u\!\restriction_{\mathcal{Ext}(M)_\lambda}) =
+ \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all simple weight
+ \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is determined by
+ its trace function. Of course, the same holds for \(\mExt(M)\). We now claim
+ that the trace function of \(\mathcal{N}\) is the same as that of
+ \(\mExt(M)\). Clearly,
+ \(\operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda}) =
\operatorname{Tr}(u\!\restriction_{M_\lambda}) =
\operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda
\in \operatorname{supp}_{\operatorname{ess}} M\), \(u \in
\mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(M\) is
Zariski-dense and the maps \(\lambda \mapsto
- \operatorname{Tr}(u\!\restriction_{\mathcal{Ext}(M)_\lambda})\) and
- \(\lambda \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\)
- are polynomial in \(\lambda \in \mathfrak{h}^*\), it follows that these maps
- coincide for all \(u\).
+ \operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda})\) and \(\lambda \mapsto
+ \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) are polynomial in
+ \(\lambda \in \mathfrak{h}^*\), it follows that these maps coincide for all
+ \(u\).
- In conclusion, \(\mathcal{N} \cong \mathcal{Ext}(M)\) and
- \(\mathcal{Ext}(M)\) is unique.
+ In conclusion, \(\mathcal{N} \cong \mExt(M)\) and \(\mExt(M)\) is unique.
\end{proof}
% This is a very important theorem, but since we won't classify the coherent