diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -25,28 +25,28 @@ form a basis for \(\mathfrak{sl}_2(K)\) and satisfy
\end{align*}
Let \(M\) be a finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module. We now
-turn our attention to the action of \(h\) on \(M\), in particular, to the
-eigenspace decomposition
-\[
- M = \bigoplus_{\lambda} M_\lambda
-\]
-of \(M\) -- where \(\lambda\) ranges over the eigenvalues of \(h\) and
-\(M_\lambda\) is the corresponding eigenspace. At this point, this is nothing
-short of a gamble: why look at the eigenvalues of \(h\)?
-
-The short answer is that, as we shall see, this will pay off. For now we will
-postpone the discussion about the real reason of why we chose \(h\). Let
-\(\lambda\) be any eigenvalue of \(h\). Notice \(M_\lambda\) is in general not
-a \(\mathfrak{sl}_3(K)\)-submodule of \(M\). Indeed, if \(m \in M_\lambda\)
-then
+turn our attention to the action of \(h\) on \(M\), in particular, we
+investigate the subspace \(\bigoplus_{\lambda} M_\lambda \subset M\) -- where
+\(\lambda\) ranges over the eigenvalues of \(h\!\restriction_M\) and
+\(M_\lambda\) is the corresponding eigenspace.
+
+At this point, this is nothing short of a gamble: why look at the eigenvalues
+of \(h\)? The short answer is that, as we shall see, this will pay off. We will
+postpone the discussion about the real reason of why we chose \(h\), but for
+now we may notice that, perhaps surprisingly, the action \(h\!\restriction_M\)
+of \(h\) on a finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module \(M\)
+is always a diagonalizable operator.
+
+Let \(\lambda\) be any eigenvalue of \(h\!\restriction_M\). Notice
+\(M_\lambda\) is in general not a \(\mathfrak{sl}_2(K)\)-submodule of \(M\).
+Indeed, if \(m \in M_\lambda\) then the identities
\begin{align*}
h \cdot (e \cdot m) &= 2e \cdot m + e h \cdot m = (\lambda + 2) e \cdot m \\
h \cdot (f \cdot m) &= -2f \cdot m + f h \cdot m = (\lambda - 2) f \cdot m
\end{align*}
-
-In other words, \(e\) sends an element of \(M_\lambda\) to an element of
-\(M_{\lambda + 2}\), while \(f\) sends it to an element of \(M_{\lambda - 2}\).
-Visually, we may draw
+follow. In other words, \(e\) sends an element of \(M_\lambda\) to an element
+of \(M_{\lambda + 2}\), while \(f\) sends it to an element of \(M_{\lambda -
+2}\). Visually, we may draw
\begin{center}
\begin{tikzcd}
\cdots \rar[bend left=60] &
@@ -57,27 +57,36 @@ Visually, we may draw
\end{tikzcd}
\end{center}
-This implies \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k}\) is a
+This implies \(\bigoplus_\lambda M_\lambda\) is a
+\(\mathfrak{sl}_2(K)\)-submodule, so that \(\bigoplus_\lambda M_\lambda\) is
+either \(0\) or the entirety of \(M\) -- recall that \(M\) is simple. Since
+\(M\) is finite dimensional, \(h\!\restriction_M\) has at least one eigenvalue
+and therefore
+\[
+ M = \bigoplus_\lambda M_\lambda
+\]
+
+Even more so, we have seen that for any eigenvalue \(\lambda \in K\) of
+\(h\!\restriction_M\), \(\bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k}\) is a
\(\mathfrak{sl}_2(K)\)-invariant subspace, which goes to show
\[
M = \bigoplus_{k \in \mathbb{Z}} M_{\lambda - 2 k},
\]
and the eigenvalues of \(h\) all have the form \(\lambda - 2 k\) for some
-\(k\). Even more so, if \(a = \max \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne
-0 \}\) and \(b = \min \{ k \in \mathbb{Z} : V_{\lambda - 2 k} \ne 0 \}\) we can
-see that
+\(k\). By the same token, if \(a\) is the greatest \(k \in \mathbb{Z}\) such
+that \(V_{\lambda - 2 k} \ne 0\) and, likewise, \(b\) is the smallest \(k \in
+\mathbb{Z}\) such that \(V_{\lambda - 2 k} \ne 0\) then
\[
- \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}} M_{\lambda - 2 k}
+ M = \bigoplus_{\substack{k \in \mathbb{Z} \\ a \le k \le b}}
+ M_{\lambda - 2 k}
\]
-is also a \(\mathfrak{sl}_2(K)\)-submodule, so that the eigenvalues of \(h\)
-form an unbroken string
+
+The eigenvalues of \(h\) thus form an unbroken string
\[
\ldots, \lambda - 4, \lambda - 2, \lambda, \lambda + 2, \lambda + 4, \ldots
\]
-around \(\lambda\).
-
-Our main objective is to show \(M\) is determined by this string of
-eigenvalues. To do so, we suppose without any loss in generality that
+around \(\lambda\). Our main objective is to show \(M\) 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 \(m \in
M_\lambda\) and consider the set \(\{m, f \cdot m, f^2 \cdot m, \ldots\}\).