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*}