diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -81,9 +81,22 @@ eigenvalues. To do so, we suppose without any loss in generality that
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 \(V\).
+ The set \(\{v, f v, f^2, \ldots\}\) 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}
+ f^k v & \overset{e}{\mapsto} k(n + 1 - k) f^{k - 1} v
+ & f^k v & \overset{f}{\mapsto} f^{k + 1} v
+ & f^k v & \overset{h}{\mapsto} (n - 2 k) f^k v
+ \end{aligned}
+ \end{equation}
\end{proposition}
+\begin{note}
+ For these formulas to work we fix the convention that \(f^{-1} v = 0\) --
+ which reflects the fact that \(e v = 0\).
+\end{note}
+
\begin{proof}
First of all, notice \(f^k v\) lies in \(V_{\lambda - 2 k}\), so that \(\{v,
f v, f^2 v, \ldots\}\) is a set of linearly independent vectors. Hence it
@@ -95,8 +108,9 @@ 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, \(h f^k v = (\lambda - 2 k) f^k v \in K \langle v, f v, f^2, v,
- \ldots \rangle\). Seeing \(e f^k v \in K \langle v, f v, f^2 v, \ldots
- \rangle\) is a bit more complex. Clearly,
+ \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,
\[
\begin{split}
e f v
@@ -120,19 +134,14 @@ V_\lambda\) and consider the set \(\{v, f v, f^2, v, \ldots\}\).
The pattern is starting to become clear: \(e\) sends \(f^k v\) to a multiple
of \(f^{k - 1} v\). Explicitly, it's not hard to check by induction that
\[
- e f^k v = k (\lambda + 1 - k) f^{k - 1} v
+ e f^k v = k (\lambda + 1 - k) f^{k - 1} v,
\]
+ which which is the first formula of (\ref{eq:irr-rep-of-sl2}).
\end{proof}
-\begin{note}
- For this last formula to work we fix the convention that \(f^{-1} v = 0\) --
- which is to say \(e v = 0\).
-\end{note}
-
-Proposition~\ref{thm:basis-of-irr-rep} may seem unrelated to our problem at first,
-but its significance lies in the fact that we have just provided a complete
-description of the action of \(\mathfrak{sl}_2(K)\) on \(V\). In other
-words\dots
+The significance of proposition~\ref{thm:basis-of-irr-rep} should be
+self-evident: we have just provided a complete description of the action of
+\(\mathfrak{sl}_2(K)\) on \(V\). In particular, this goes to show\dots
\begin{corollary}
\(V\) is completely determined by the right-most eigenvalue \(\lambda\) of
@@ -175,8 +184,7 @@ Other important consequences of proposition~\ref{thm:basis-of-irr-rep} are\dots
\begin{proof}
If \(f^m\) is the lowest power of \(f\) that annihilates \(v\), it follows
- from the formula for \(e f^k v\) obtained in the proof of
- proposition~\ref{thm:basis-of-irr-rep} that
+ from the formulas in (\ref{eq:irr-rep-of-sl2}) that
\[
0 = e 0 = e f^m v = m (\lambda + 1 - m) f^{m - 1} v
\]
@@ -208,39 +216,12 @@ the direct sums of the eigenspaces of such irreducible components.
In particular, if the eigenvalues of \(V\) all have the same parity -- i.e.
they are either all even integers or all odd integers -- and the dimension of
-each eigenspace is no greater than \(1\) then \(V\) must be irreducible, for
-if \(U, W \subset V\) are subrepresentations with \(V = W \oplus U\) then
-either \(W_\lambda = 0\) for all \(\lambda\) or \(U_\lambda = 0\) for all
-\(\lambda\).
-
-We now know every irreducible representation \(V\) of \(\mathfrak{sl}_2(K)\)
-has the form
-\begin{center}
- \begin{tikzcd}
- \cdots \arrow[bend left=60]{r}
- & V_{n - 6} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}
- & V_{n - 4} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
- & V_{n - 2} \arrow[bend left=60]{r}{e} \arrow[bend left=60]{l}{f}
- & V_n \arrow[bend left=60]{l}{f}
- \end{tikzcd}
-\end{center}
-where \(V_{n - 2 k}\) is the 1-dimensional eigenspace of \(h\) associated to
-\(n - 2 k\) and \(n = \dim V - 1\). Even more so, we explicitly know
-\[
- V = \bigoplus_{k = 0}^n K f^k v
-\]
-and
-\begin{equation}\label{eq:irr-rep-of-sl2}
- \begin{aligned}
- f^k v & \overset{e}{\mapsto} k(n + 1 - k) f^{k - 1} v
- & f^k v & \overset{f}{\mapsto} f^{k + 1} v
- & f^k v & \overset{h}{\mapsto} (n - 2 k) f^k v
- \end{aligned}
-\end{equation}
-
-To conclude our analysis all it's left is to show that for each \(n\) such
-irreducible \(V\) does indeed exist. Surprisingly, we have already encountered
-such a \(V\).
+each eigenspace is no greater than \(1\) then \(V\) must be irreducible, for if
+\(U, W \subset V\) are subrepresentations with \(V = W \oplus U\) then either
+\(W_\lambda = 0\) for all \(\lambda\) or \(U_\lambda = 0\) for all \(\lambda\).
+To conclude our analysis all it's left is to show that for each \(n\) there is
+some finite-dimensional irreducible \(V\) whose highest weight is \(\lambda\).
+Surprisingly, we have already encountered such a \(V\).
\begin{theorem}\label{thm:sl2-exist-unique}
For each \(n \ge 0\) there exists a unique irreducible representation of