lie-algebras-and-their-representations

diff --git a/preamble.tex b/preamble.tex
@@ -130,6 +130,9 @@
 % the appropriate spacing
 \renewcommand{\chi}{\ensuremath \raisebox{\depth}{$\mathchar"11F$}}
 
+% Macro for Mathieu's coherent extension
+\newcommand{\mExt}{\mathcal{E\!x\!t}}
+
 % Isomorphism arrow
 \newcommand{\isoto}{\xlongrightarrow{\sim}}
 

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