diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -374,13 +374,15 @@
\(K\)-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
+ On the other hand, if \(\mathcal{M}_\lambda\) is simple then by Jacobson's
+ density theorem the map \(\mathcal{U}(\mathfrak{g})_0 \to
+ \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. Hence the
+ commutativity of the previously drawn diagram, as well as the fact that
+ \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
+ \operatorname{End}(\mathcal{M}_\lambda)) =
+ \operatorname{rank}(\operatorname{End}(\mathcal{M}_\lambda)^* \to
+ \mathcal{U}(\mathfrak{g})_0^*)\), imply that \(\operatorname{rank} B_\lambda
+ = d^2\). 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
\[
@@ -404,10 +406,7 @@
\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
+ then implies \(\operatorname{rank} B_\lambda\!\restriction_W = d^2\). In
other words, \(U \subset \bigcup_W U_W\).
Likewise, if \(\operatorname{rank} B_\lambda\!\restriction_W = d^2\) for some