lie-algebras-and-their-representations

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -748,12 +748,11 @@ Moreover, we find\dots
   \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots +
   k_n \cdot \alpha_n\).
 
-  % TODO: Note that PBW implies U(g) is a free b-module
   This already gives us that the weights of \(M(\lambda)\) are bounded by
-  \(\lambda\) -- in the sense that no weight of \(M(\lambda)\) is ``higher''
-  than \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that
+  \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that
   \(v^+\) is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The
-  Poincaré-Birkhoff-Witt theorem implies
+  Poincaré-Birkhoff-Witt theorem implies \(\mathcal{U}(\mathfrak{g})\) is a
+  free \(\mathcal{U}(\mathfrak{b})\)-module, so that
   \[
     M(\lambda)
     \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
@@ -763,10 +762,11 @@ Moreover, we find\dots
     \cong \bigoplus_i K v^+
     \ne 0
   \]
-  as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v^+ \ne 0\) -- for if this was
-  not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+
-  = 0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest
-  weight of \(M(\lambda)\), with highest weight vector \(v^+\).
+  as \(\mathcal{U}(\mathfrak{b})\)-modules. We then conclude \(v^+ \ne 0\) in
+  \(M(\lambda)\), for if this was not the case we would find \(M(\lambda) =
+  \mathcal{U}(\mathfrak{g}) \cdot v^+ = 0\). Hence \(V_\lambda \ne 0\) and
+  therefore \(\lambda\) is the highest weight of \(M(\lambda)\), with highest
+  weight vector \(v^+\).
 
   To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only
   finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots