lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -137,13 +137,15 @@
 
 % TODO: Remark that the support of a simple weight module is always contained
 % in a coset
-% TODO: Note that conditions (ii) and (iii) have special names
+% TODO: Note that conditions (ii), (iii) and (iv) have special names
+% TODO: Remove item (ii)? I don't think we use it anywhere
 \begin{corollary}[Fernando]\label{thm:cuspidal-mod-equivs}
   Let \(V\) be an irreducible weight \(\mathfrak{g}\)-module. The following
   conditions are equivalent.
   \begin{enumerate}
     \item \(V\) is cuspidal.
     \item \(E_\alpha\) acts injectively in \(V\) for all \(\alpha \in \Delta\).
+    \item \(F_\alpha\) acts injectively in \(V\) for all \(\alpha \in \Delta\).
     \item The support of \(V\) is precisely one \(Q\)-coset.
   \end{enumerate}
 \end{corollary}
@@ -313,18 +315,169 @@
   so we have an inclusion \(V \to \mathcal{M}\).
 \end{proof}
 
-% TODOOO: Prove this
+\begin{lemma}
+  Let \(\mathcal{M}\) be a coherent family. The set \(U = \{\lambda \in
+  \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple
+  $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open.
+\end{lemma}
+
+\begin{proof}
+  For each \(\lambda \in \mathfrak{h}^*\) we introduce the bilinear form
+  \begin{align*}
+    B_\lambda : \mathcal{U}(\mathfrak{g})_0 \times \mathcal{U}(\mathfrak{g})_0
+    & \to K \\
+    (u, v)
+    & \mapsto \operatorname{Tr}(u v \!\restriction_{\mathcal{M}_\lambda})
+  \end{align*}
+  and consider its rank -- i.e. the dimension of the image of the induced
+  operator
+  \begin{align*}
+    \mathcal{U}(\mathfrak{g})_0 & \to     \mathcal{U}(\mathfrak{g})_0^* \\
+                              u & \mapsto B_\lambda(u, \cdot)
+  \end{align*}
+
+  Our first observation is that \(\operatorname{rank} B_\lambda \le d^2\). This
+  follows from the commutativity of
+  \begin{center}
+    \begin{tikzcd}
+      \mathcal{U}(\mathfrak{g})_0 \arrow{r} \arrow{d} &
+      \mathcal{U}(\mathfrak{g})_0^* \\
+      \operatorname{End}(\mathcal{M}_\lambda)   \arrow{r}{\sim} &
+      \operatorname{End}(\mathcal{M}_\lambda)^* \arrow{u}
+    \end{tikzcd},
+  \end{center}
+  where the map \(\mathcal{U}(\mathfrak{g})_0 \to
+  \operatorname{End}(\mathcal{M}_\lambda)\) is given by the action of
+  \(\mathcal{U}(\mathfrak{g})_0\), the map
+  \(\operatorname{End}(\mathcal{M}_\lambda)^* \to
+  \mathcal{U}(\mathfrak{g})_0^*\) is its dual, and the isomorphism
+  \(\operatorname{End}(\mathcal{M}_\lambda) \isoto
+  \operatorname{End}(\mathcal{M}_\lambda)^*\) is induced by the trace form
+  \begin{align*}
+    \operatorname{End}(\mathcal{M}_\lambda) \times
+    \operatorname{End}(\mathcal{M}_\lambda) & \to K \\
+    (T, S) & \mapsto \operatorname{Tr}(T S)
+  \end{align*}
+
+  Indeed, \(\operatorname{rank} B_\lambda \le
+  \operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
+  \operatorname{End}(\mathcal{M}_\lambda)) \le \dim
+  \operatorname{End}(\mathcal{M}_\lambda) = d^2\). Furthermore, if
+  \(\operatorname{rank} B_\lambda = d^2\) then we must have
+  \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
+  \operatorname{End}(\mathcal{M}_\lambda)) = d^2\) -- i.e. the map
+  \(\mathcal{U}(\mathfrak{g})_0 \to \operatorname{End}(\mathcal{M}_\lambda)\)
+  is surjective. In particular, if \(\operatorname{rank} B_\lambda = d^2\) then
+  \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module,
+  for if \(V \subset \mathcal{M}_\lambda\) is invariant under the action of
+  \(\mathcal{U}(\mathfrak{g})_0\) then \(V\) is invariant under any
+  \(K\)-linear operator \(\mathcal{M}_\lambda \to \mathcal{M}_\lambda\), so
+  that \(W = 0\) or \(W = \mathcal{M}_\lambda\).
+
+  % TODOOO: Show this
+  On the other hand, if \(\mathcal{M}_\lambda\) is simple then we can find
+  \(u_1, \ldots, u_{d^2} \in \mathcal{U}(\mathfrak{g})_0\) with
+  \(B_\lambda(u_1, \cdot), \ldots, B_\lambda(u_{d^2}, \cdot) \in
+  \mathcal{U}(\mathfrak{g})_0^*\) linearly independent. In other words, if
+  \(\mathcal{M}_\lambda\) is simple then \(\operatorname{rank} B_\lambda \ge
+  d^2\). Hence \(U\) is precisely the set of \(\lambda\) such that
+  \(B_\lambda\) has maximal rank \(d^2\). We now show that \(U\) is
+  Zariski-open. First, notice that
+  \[
+    U = \bigcup_{
+          \substack{W \subset \mathcal{U}(\mathfrak{g})_0 \\ \dim W = d^2}
+        }
+      U_W,
+  \]
+  where \(U_W = \{\lambda \in \mathcal{U}(\mathfrak{g})_0 : \operatorname{rank}
+  B_\lambda\!\restriction_W = d^2 \}\).
+
+  Indeed, if \(\operatorname{rank} B_\lambda = d^2\) it follows from the
+  surjectivity of the map \(\mathcal{U}(\mathfrak{g})_0 \to
+  \operatorname{End}(\mathcal{M}_\lambda)\) that there is some \(W \subset
+  \mathcal{U}(\mathfrak{g})_0\) with \(\dim W = d^2\) such that the restriction
+  \(W \to \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. The
+  comutativity of
+  \begin{center}
+    \begin{tikzcd}
+      W \arrow{r} \arrow{d} & W^* \\
+      \operatorname{End}(\mathcal{M}_\lambda)   \arrow{r}{\sim} &
+      \operatorname{End}(\mathcal{M}_\lambda)^* \arrow{u}
+    \end{tikzcd}
+  \end{center}
+  and the fact that \(\operatorname{rank}(W \to
+  \operatorname{End}(\mathcal{M}_\lambda)) =
+  \operatorname{rank}(\operatorname{End}(\mathcal{M}_\lambda)^* \to W^*)\)
+  then imply \(\operatorname{rank} B_\lambda\!\restriction_W = d^2\). In
+  other words, \(U \subset \bigcup_W U_W\).
+
+  Likewise, if \(\operatorname{rank} B_\lambda\!\restriction_W = d^2\) for some
+  \(W\), then the commutativity of
+  \begin{center}
+    \begin{tikzcd}
+      W \arrow{r} \arrow{d} & W^* \\
+      \mathcal{U}(\mathfrak{g})_0   \arrow{r} &
+      \mathcal{U}(\mathfrak{g})_0^* \arrow{u}
+    \end{tikzcd}
+  \end{center}
+  implies \(\operatorname{rank} B_\lambda \ge d^2\), which goes to show
+  \(\bigcup_W U_W \subset U\).
+
+  Given \(\lambda \in U_W\), the surjectivity of \(W \to
+  \operatorname{End}(\mathcal{M}_\lambda)\) and the fact that \(\dim W <
+  \infty\) imply \(W \to W^*\) is invertible. Since \(\mathcal{M}\) is a
+  coherent family, \(B_\lambda\) depends polynomialy in \(\lambda\). Hence so
+  does the induced maps \(W \to W^*\). In particular, there is some Zariski
+  neighborhood \(V\) of \(\lambda\) such that the map \(W \to W^*\) induced by
+  \(B_\mu\!\restriction_W\) is invertible for all \(\mu \in V\).
+
+  But the surjectivity of the map induced by \(B_\mu\!\restriction_W\) implies
+  \(\operatorname{rank} B_\mu = d^2\), so \(\mu \in U_W\) and therefore \(V
+  \subset U_W\). This implies \(U_W\) is open for all \(W\). Finally, \(U\) is
+  the union of Zariski-open subsets and is therefore open. We are done.
+\end{proof}
+
 \begin{theorem}[Mathieu]
-  Let \(\mathcal{M}\) be an irreducible coherent family and \(\lambda \in
-  \mathfrak{h}^*\). The following conditions are equivalent.
+  Let \(\mathcal{M}\) be an irreducible coherent family of degree \(d\) and
+  \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent.
   \begin{enumerate}
     \item \(\mathcal{M}[\lambda]\) is irreducible.
-    \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for all
-      \(\alpha \in \Delta\).
+    \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for
+      all \(\alpha \in \Delta\).
     \item \(\mathcal{M}[\lambda]\) is cuspidal.
   \end{enumerate}
 \end{theorem}
 
+\begin{proof}
+  The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly
+  from corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the
+  corollary that \strong{(iii)} implies \strong{(ii)}. All it's left is to show
+  \strong{(ii)} implies \strong{(iii)}.
+
+  Suppose \(F_\alpha\) acts injectively in the subrepresentation
+  \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since
+  \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains
+  an infinite-dimensiona irreducible \(\mathfrak{g}\)-submodule \(V\).
+  Moreover, again by corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(V\)
+  is a cuspidal representation, and its degree is bounded by \(d\). We claim
+  \(\mathcal{M}[\lambda] = V\).
+
+  Since \(U = \{\mu \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is a
+  simple $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is a non-empty --
+  \(\mathcal{M}\) is irreducible -- open set and
+  \(\operatorname{supp}_{\operatorname{ess}} V\) is Zariski-dense, \(U \cap
+  \operatorname{supp}_{\operatorname{ess}} V\) is non-empty. In other words,
+  there is some \(\mu \in \mathfrak{h}^*\) such that \(\mathcal{M}_\mu\) is a
+  simple \(\mathcal{U}(\mathfrak{g})_0\)-module and \(\dim V_\mu = \deg V\).
+
+  In particular, \(V_\mu \ne 0\), so \(V_\mu = \mathcal{M}_\mu\). Now given any
+  irreducible \(\mathfrak{g}\)-module \(W\), the multiplicity of \(W\) in
+  \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(W_\mu\) in
+  \(\mathcal{M}\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module, which is, of
+  course, \(1\) if \(W \cong V\) and \(0\) otherwise. Hence
+  \(\mathcal{M}[\lambda] = V\) and \(\mathcal{M}[\lambda]\) is cuspidal.
+\end{proof}
+
 \section{Localizations \& the Existance of Coherent Extensions}
 
 % TODO: Comment on the intuition behind the proof: we can get vectors in a
@@ -655,7 +808,7 @@
   \begin{align*}
     \mathfrak{h}^* \times \mathcal{U}(\mathfrak{g})_0 &
     \to K \\
-    (\lambda, u) & 
+    (\lambda, u) &
     \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})
   \end{align*}