lie-algebras-and-their-representations

diff --git a/sections/coherent-families.tex b/sections/coherent-families.tex
@@ -168,13 +168,34 @@ combinatorial counterpart.
 % TODO: Note we will prove that central characters are also invariants of
 % coherent families
 
-% TODOO: Prove this
-\begin{proposition}[Mathieu]\label{thm:coherent-family-has-uniq-central-char}
+\begin{proposition}\label{thm:coherent-family-has-uniq-central-char}
   Suppose \(\lambda, \mu \in \mathfrak{h}^*\) are such that \(L(\lambda)\) and
   \(L(\mu)\) are both bounded and \(\mExt(L(\lambda)) \cong \mExt(L(\mu))\).
-  Then \(\chi_\lambda = \chi_\mu\).
+  Then \(\chi_\lambda = \chi_\mu\). In particular, \(\mu \in W \bullet
+  \lambda\).
 \end{proposition}
 
+\begin{proof}
+  Fix \(u \in \mathcal{U}(\mathfrak{g})_0\). It is clear that
+  \(\operatorname{Tr}(u\!\restriction_{\mExt(L(\lambda))_\nu}) =
+  \operatorname{Tr}(u\!\restriction_{L(\lambda)_\nu}) = d \chi_\lambda(u)\) for
+  all \(\nu \in \operatorname{supp}_{\operatorname{ess}} L(\lambda)\). Since
+  \(\operatorname{supp}_{\operatorname{ess}} L(\lambda)\) is Zariski-dense and
+  the map
+  \(\nu \mapsto \operatorname{Tr}(u\!\restriction_{\mExt(L(\lambda))_\nu})\)
+  is polynomial, it follows that
+  \(\operatorname{Tr}(u\!\restriction_{\mExt(L(\lambda))_\nu}) = d
+  \chi_\lambda(u)\) for all \(\nu \in \mathfrak{h}^*\). But by the same token
+  \[
+    d \chi_\lambda(u)
+    = \operatorname{Tr}(u\!\restriction_{\mExt(L(\lambda))_\nu})
+    = \operatorname{Tr}(u\!\restriction_{\mExt(L(\mu))_\nu})
+    = d \chi_\mu(u)
+  \]
+  for any \(\nu \in \operatorname{supp}_{\operatorname{ess}} L(\mu)\) and thus
+  \(\chi_\lambda(u) = \chi_\mu(u)\).
+\end{proof}
+
 % TODO: Remark that the probability of σ_β ∙ λ ∈ P+ is slight: there precisely
 % one element in the orbit of λ which is dominant integral, so the odds are
 % 1/|W ∙ λ|