diff --git a/sections/fin-dim-simple.tex b/sections/fin-dim-simple.tex
@@ -1075,12 +1075,12 @@ We would now like to conclude this chapter by describing the situation where
out that Proposition~\ref{thm:verma-is-finite-dim} fails in the general
setting. For instance, consider\dots
-\begin{example}
+\begin{example}\label{ex:antidominant-verma}
The action of \(\mathfrak{sl}_2(K)\) on \(M(-4)\) is given by the following
diagram. In general, it is possible to check using formula
(\ref{eq:sl2-verma-formulas}) that \(e\) always maps \(f^{k + 1} \cdot m^+\)
to a nonzero multiple of \(f^k \cdot m^+\), so we can see that \(M(-4)\) has
- no proper submodules and \(M(-4) \cong L(-4)\).
+ no proper submodules, \(N(-4) = 0\) and thus \(L(-4) \cong M(-4)\).
\begin{center}
\begin{tikzcd}
\cdots \rar[bend left=60]{-28}
@@ -1094,12 +1094,12 @@ setting. For instance, consider\dots
While \(L(\lambda)\) is always a highest weight module of highest weight
\(\lambda\), one can show that if \(\lambda\) is not dominant integral then
-\(L(\lambda) \cong M(\lambda)\) is infinite-dimensional. Indeed, since the
-highest weight of a finite-dimensional simple \(\mathfrak{g}\)-module is always
-dominant integral, \(L(\lambda)\) is infinite-dimensional for any \(\lambda\)
-which is not dominant integral. Since the only \(\mathfrak{g}\)-submodules of
-\(M(\lambda)\) of infinite codimension is \(0\), it follows that \(N(\lambda) =
-0\) and \(L(\lambda) \cong M(\lambda)\).
+\(L(\lambda)\) is infinite-dimensional. Indeed, since the highest weight of a
+finite-dimensional simple \(\mathfrak{g}\)-module is always dominant integral,
+\(L(\lambda)\) is infinite-dimensional for any \(\lambda\) which is not
+dominant integral. If \(\lambda = k_1 \beta_1 + \cdots + k_r \beta_r \in P\) is
+integral and \(k_i < 0\) for all \(i\), then \(M(\lambda) \cong L(\lambda)\) as
+in Example~\ref{ex:antidominant-verma}.
Verma modules can thus serve as examples of infinite-dimensional simple
modules. In the next chapter we expand our previous results by exploring the