Re: Homogeneous polynomials and SL(2,R)

From: Yves De Cornulier (decornul_at_clipper.ens.fr)
Date: 06/07/04 

Date: Mon, 7 Jun 2004 09:31:10 +0000 (UTC)

The following (latex source) is extracted from a paper of mine and should
answer your question. Note that the action of SL(2) on the homogeneous
polynomials of degree n is isomorphic to the only irreducible action on a
space of dimension n+1.

*************************************

Fix a field $K$ with $\textnormal{char}(K)=0$. Denote by
$\pi:\mathfrak{sl}(2,K)\to\textnormal{End}(V_m)$ the unique irreducible
$(m+1)$-dimensional representation. Take a basis $(e_0,\dots,e_m)$
of $V_m$, so that $\pi$ is in a standard form:
$$\pi(H)e_i=(m-2i)e_i,\qquad \pi(X)e_i=(m-i+1)e_{i-1}\;(i\ge 1),$$
$$\pi(Y)e_i=(i+1)e_{i+1}\quad(i\le m-1),\qquad \pi(X)e_0=\pi(Y)e_m=0.$$
We also denote by $\pi$ the corresponding representation of
$SL(2,K)$ on $V_m$.

\begin{lem}
Let $\mathcal{B}_n$ be the space of bilinear forms on $V_{2n-1}$
($n\ge 1$) preserved by $SL(2,K)$. Then $\mathcal{B}_n$ is
one-dimensional, and consists of symplectic forms. Namely,
$\mathcal{B}_n$ is generated by the form $\varphi$ defined by,
with $m=2n-1$
$$\varphi(e_i,e_{m-i})=(-1)^i\begin{pmatrix}
  i \\
  m \\
\end{pmatrix};\quad\varphi(e_i,e_j)=0 \textnormal{ if } i+j\neq
m.$$\label{inv}
\end{lem}

\textbf{Proof}: Let $\varphi$ be a bilinear form on $V_m$ preserved by
$\pi(SL(2,K))$, that is, making $\mathfrak{sl}(2,K)$ act by skew-adjoint
operators. Then, using that $\pi(H)$ is skew-adjoint, we obtain
$\varphi(e_i,e_j)=0$ if $i+j\neq m$. Write
$\varphi(e_i,e_{-i})=\lambda_i$. Then, using that $\pi(X)$ is
skew-adjoint, we obtain
$$\lambda_{i+1}=-\frac{m-i}{i+1}\lambda_i,\quad 0\le i\le n-1$$
hence $\lambda_i=(-1)^i\begin{pmatrix}
  i \\
  m \\
\end{pmatrix}\lambda_0$ for all $i$. It is straightforward that,
conversely, this defines a nondegenerate bilinear form that makes
$\pi(\mathfrak{sl}(2,K))$ skew-adjoint, which is symplectic (it would be
symmetric if $m$ were even).

Quantcast