lie-algebras-and-their-representations

diff --git a/sections/coherent-families.tex b/sections/coherent-families.tex
@@ -1,7 +1,6 @@
 \chapter{Classification of Coherent Families}
 
 % TODOOO: Is this decomposition unique??
-% TODOO: Prove this
 \begin{proposition}
   Suppose \(\mathfrak{g} = \mathfrak{s}_1 \oplus \cdots \oplus \mathfrak{s}_r\)
   and let \(\mathcal{M}\) be a semisimple irreducible coherent
@@ -12,6 +11,90 @@
   \]
 \end{proposition}
 
+\begin{proof}
+  Suppose \(\mathfrak{h}_i \subset \mathfrak{s}_i\) are Cartan subalgebras,
+  \(\mathfrak{h} = \mathfrak{h}_1 \oplus \cdots \oplus \mathfrak{h}_r\) and \(d
+  = \deg \mathcal{M}\). Let \(M \subset \mathcal{M}\) be any
+  infinite-dimensional simple submodule, so that \(\mathcal{M}\) is a
+  semisimple coherent extension of \(M\). By
+  Example~\ref{thm:simple-weight-mod-is-tensor-prod}, there exists (unique)
+  simple weight \(\mathfrak{s}_i\)-modules \(M_i\) such that \(M \cong M_1
+  \otimes \cdots \otimes M_r\). Take \(\mathcal{M}_i = \mExt(M_i)\). We will
+  show that \(\mathcal{M}_1 \otimes \cdots \mathcal{M}_r\) is a coherent
+  extension of \(M\).
+
+  It is clear that \(\mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r\) is a
+  degree \(d\) bounded \(\mathfrak{g}\)-module containing \(M\) as a submodule.
+  It thus suffices to show that \(\mathcal{M}\) is a coherent family. By
+  Example~\ref{ex:supp-ess-of-tensor-is-product},
+  \(\operatorname{supp}_{\operatorname{ess}} (\mathcal{M}_1 \otimes \cdots
+  \otimes \mathcal{M}_r) = \mathfrak{h}^*\). To see that the map
+  \begin{align*}
+    \mathfrak{h}^* & \to K \\
+    \lambda & \mapsto
+    \operatorname{Tr}
+    (
+      u\! \restriction_{(\mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r)_\lambda}
+    )
+  \end{align*}
+  is polynomial, notice that the natural isomorphism of algebras
+  \begin{align*}
+    f : \mathcal{U}(\mathfrak{s}_1) \otimes \cdots \mathcal{U}(\mathfrak{s}_1)
+    & \isoto \mathcal{U}(\mathfrak{g}) \\
+    u_1 \otimes \cdots \otimes u_r & \mapsto u_1 \cdots u_r
+  \end{align*}
+  described in Example~\ref{ex:univ-enveloping-of-sum-is-tensor} is a
+  \(\mathfrak{g}\)-homomorphism between the tensor product of the adjoint
+  \(\mathfrak{s}_i\)-modules \(\mathcal{U}(\mathfrak{s}_i)\) and the adjoint
+  \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\).
+
+  Indeed, given \(X = X_1 + \cdots + X_r \in \mathfrak{g}\) with \(X_i \in
+  \mathfrak{s}_i\) and \(u_i \in \mathcal{U}(\mathfrak{s}_i)\),
+  \[
+    \begin{split}
+      f(X \cdot (u_1 \otimes \cdots \otimes u_r))
+      & = f([X_1, u_1] \otimes u_2 \otimes \cdots \otimes u_r)
+        + \cdots
+        + f(u_1 \otimes \cdots \otimes u_{r-1} \otimes [X_r, u_r]) \\
+      & = [X_1, u_1] u_2 \cdots u_r + \cdots + u_1 \cdots u_{r-1} [X_r, u_r] \\
+      \text{(\([X_i, u_j] = 0\) for \(i \ne j\))}
+      & = [X_1, u_1u_2 \cdots u_r] + \cdots + [X_r, u_1 \cdots u_{r-1}u_r] \\
+      & = [X, f(u_1 \otimes \cdots \otimes u_r)]
+    \end{split}
+  \]
+
+  Hence by Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}
+  \(f\) restricts to an isomorphism of algebras \(\mathcal{U}(\mathfrak{s}_1)_0
+  \otimes \cdots \otimes \mathcal{U}(\mathfrak{s}_r)_0 \isoto
+  \mathcal{U}(\mathfrak{g})_0\) with image \(\mathcal{U}(\mathfrak{g})_0 =
+  \mathcal{U}(\mathfrak{s}_1)_0 \cdots \mathcal{U}(\mathfrak{s}_r)_0\). More
+  importantly, if we write \(\lambda = \lambda_1 + \cdots + \lambda_r\) for
+  \(\lambda_i \in \mathfrak{h}_i^*\) it is clear from
+  Example~\ref{ex:tensor-prod-of-weight-is-weight} that the
+  \(\mathcal{U}(\mathfrak{g})_0\)-module \((\mathcal{M}_1 \otimes \cdots
+  \otimes \mathcal{M}_r)_\lambda\) corresponds to exactly the
+  \(\mathcal{U}(\mathfrak{s}_1)_0 \otimes \cdots \otimes
+  \mathcal{U}(\mathfrak{s}_r)_0\)-module \((\mathcal{M}_1)_{\lambda_1} \otimes
+  \cdots \otimes (\mathcal{M}_r)_{\lambda_r}\), so we can see that the value
+  \[
+    \operatorname{Tr}
+    (
+      u_1 \cdots u_r \!\restriction_{(\mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r)_\lambda}
+    )
+    = \operatorname{Tr}(u_1\!\restriction_{(\mathcal{M}_1)_{\lambda_1}})
+      \cdots
+      \operatorname{Tr}(u_r\!\restriction_{(\mathcal{M}_r)_{\lambda_r}})
+  \]
+  varies polynomially with \(\lambda \in \mathfrak{h}^*\) for all \(u_i \in
+  \mathcal{U}(\mathfrak{s}_i)_0\).
+
+  Finally, \(\mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r\) is a coherent
+  extension of \(M\). Since the \(\mathcal{M}_i = \mExt(M_i)\) are semisimple,
+  so is \(\mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r\). It thus
+  follows from the uniqueness of semisimple coherent extensions that
+  \(\mathcal{M} \cong \mathcal{M}_1 \otimes \cdots \otimes \mathcal{M}_r\).
+\end{proof}
+
 % TODO: Rework this
 In addition, it turns out that very few simple Lie algebras admit cuspidal
 modules at all. Specifically\dots