diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -81,6 +81,31 @@
\mathfrak{h}^* : \dim V_\lambda = d \}\).
\end{definition}
+\begin{example}\label{ex:laurent-polynomial-mod}
+ There is a natural action of \(\mathfrak{sl}_2(K)\) in the space \(K[x,
+ x^{-1}]\) of Laurent polynomials given by the formulas in
+ (\ref{eq:laurent-polynomials-cusp-mod}). One can quickly verify \(K[x,
+ x^{-1}]_{2 k} = K x^k\) and \(K[x, x^{-1}]_\lambda = 0\) for any \(\lambda
+ \notin 2 \mathbb{Z}\), so that \(K[x, x^{-1}] = \bigoplus_{k \in \mathbb{Z}}
+ K x^k\) is a degree \(1\) admissible weight \(\mathfrak{sl}_2(K)\)-module. It
+ follows from example~\ref{ex:submod-is-weight-mod} that any non-zero
+ subrepresentation \(W \subset K[x, x^{-1}]\) must contain a monomial \(x^k\).
+ But since the operators \(-\frac{\mathrm{d}}{\mathrm{d}x} + \frac{x^{-1}}{2},
+ x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} : K[x, x^{-1}] \to K[x,
+ x^{-1}]\) are both injective, this implies all other monomials can be found
+ in \(W\) by successively applaying \(f\) and \(e\). Hence \(W = K[x,
+ x^{-1}]\) and \(K[x, x^{-1}]\) is an irreducible representation.
+ \begin{align}\label{eq:laurent-polynomials-cusp-mod}
+ f \cdot p
+ & = \left(- \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x^{-1}}{2} \right) p &
+ h \cdot p
+ & = 2 x \frac{\mathrm{d}}{\mathrm{d}x} p &
+ e \cdot p
+ & = \left( x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} \right) p
+ \end{align}
+\end{example}
+
+% TODO: Point out supp_ess K[x^+-1] is 2Z, which is zariski dense
% This proof is very technical, I don't think its worth including it
\begin{proposition}
Let \(V\) be an infinite-dimensional admissible representation of
@@ -171,6 +196,13 @@
\end{enumerate}
\end{corollary}
+\begin{example}
+ As noted in example~\ref{ex:laurent-polynomial-mod}, the elements \(e, f \in
+ \mathfrak{sl}_2(K)\) both act injectively in the space of Laurent
+ polynomials. Hence \(K[x, x^{-1}]\) is a cuspidal representation of
+ \(\mathfrak{sl}_2(K)\).
+\end{example}
+
\begin{proposition}
If \(\mathfrak{g} = \mathfrak{z} \oplus \mathfrak{s}_1 \oplus \cdots \oplus
\mathfrak{s}_n\), where \(\mathfrak{z}\) is the center of \(\mathfrak{g}\)