diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -1334,28 +1334,28 @@ Lo and behold\dots
\begin{theorem}[Mathieu]
There exists a unique completely reducible coherent extension
- \(\operatorname{Ext}(V)\) of \(V\). More precisely, if \(\mathcal{M}\) is any
+ \(\mathcal{Ext}(V)\) of \(V\). More precisely, if \(\mathcal{M}\) is any
coherent extension of \(V\), then \(\mathcal{M}^{\operatorname{ss}} \cong
- \operatorname{Ext}(V)\). Furthermore, \(\operatorname{Ext}(V)\) is
+ \mathcal{Ext}(V)\). Furthermore, \(\mathcal{Ext}(V)\) is
a simple 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 \(V\) and take
- \(\operatorname{Ext}(V) = \mathcal{M}^{\operatorname{ss}}\).
+ \(\mathcal{Ext}(V) = \mathcal{M}^{\operatorname{ss}}\).
- To see that \(\operatorname{Ext}(V)\) is simple, recall from
+ To see that \(\mathcal{Ext}(V)\) is simple, 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
+ subrepresentation of \(\mathcal{Ext}(V)\). Since the degree of \(V\) is
+ the same as the degree of \(\mathcal{Ext}(V)\), some of its weight
+ spaces have maximal dimension inside of \(\mathcal{Ext}(V)\). In
particular, it follows from Proposition~\ref{thm:centralizer-multiplicity}
- that \(\operatorname{Ext}(V)_\lambda = V_\lambda\) is a simple
+ that \(\mathcal{Ext}(V)_\lambda = V_\lambda\) is a simple
\(\mathcal{U}(\mathfrak{g})_0\)-module for some \(\lambda \in
\operatorname{supp} V\).
- As for the uniqueness of \(\operatorname{Ext}(V)\), fix some other completely
+ As for the uniqueness of \(\mathcal{Ext}(V)\), fix some other completely
reducible coherent extension \(\mathcal{N}\) of \(V\). We claim that the
multiplicity of a given irreducible \(\mathfrak{g}\)-module \(W\) in
\(\mathcal{N}\) is determined by its \emph{trace function}
@@ -1380,21 +1380,21 @@ Lo and behold\dots
\mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all irreducible
weight \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is
determined by its trace function. Of course, the same holds for
- \(\operatorname{Ext}(V)\). We now claim that the trace function of
- \(\mathcal{N}\) is the same as that of \(\operatorname{Ext}(V)\). Clearly,
- \(\operatorname{Tr}(u\!\restriction_{\operatorname{Ext}(V)_\lambda}) =
+ \(\mathcal{Ext}(V)\). We now claim that the trace function of
+ \(\mathcal{N}\) is the same as that of \(\mathcal{Ext}(V)\). Clearly,
+ \(\operatorname{Tr}(u\!\restriction_{\mathcal{Ext}(V)_\lambda}) =
\operatorname{Tr}(u\!\restriction_{V_\lambda}) =
\operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda
\in \operatorname{supp}_{\operatorname{ess}} V\), \(u \in
\mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(V\) is
Zariski-dense and the maps \(\lambda \mapsto
- \operatorname{Tr}(u\!\restriction_{\operatorname{Ext}(V)_\lambda})\) and
+ \operatorname{Tr}(u\!\restriction_{\mathcal{Ext}(V)_\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, \(\mathcal{N} \cong \operatorname{Ext}(V)\) and
- \(\operatorname{Ext}(V)\) is unique.
+ In conclusion, \(\mathcal{N} \cong \mathcal{Ext}(V)\) and
+ \(\mathcal{Ext}(V)\) is unique.
\end{proof}
% This is a very important theorem, but since we won't classify the coherent