diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -78,10 +78,10 @@ around \(\lambda\).
Our main objective is to show \(V\) is determined by this string of
eigenvalues. To do so, we suppose without any loss in generality that
\(\lambda\) is the right-most eigenvalue of \(h\), fix some nonzero \(v \in
-V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
+V_\lambda\) and consider the set \(\{v, f v, f^2 v, \ldots\}\).
\begin{proposition}\label{thm:basis-of-irr-rep}
- The set \(\{v, f v, f^2, \ldots\}\) is a basis for \(V\). In addition, the
+ The set \(\{v, f v, f^2 v, \ldots\}\) is a basis for \(V\). In addition, the
action of \(\mathfrak{sl}_2(K)\) on \(V\) is given by the formulas
\begin{equation}\label{eq:irr-rep-of-sl2}
\begin{aligned}
@@ -102,7 +102,7 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
The fact that \(h f^k v \in K \langle v, f v, f^2 v, \ldots \rangle\) follows
immediately from our previous assertion that \(f^k v \in V_{\lambda - 2 k}\)
- -- indeed, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2, v,
+ -- indeed, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2 v,
\ldots \rangle\), which also goes to show one of the formulas in
(\ref{eq:irr-rep-of-sl2}). Seeing \(e f^k v \in K \langle v, f v, f^2 v,
\ldots \rangle\) is a bit more complex. Clearly,