lie-algebras-and-their-representations

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