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