2
$\begingroup$

I want to know what is the symplectic reduction of a symplectic linear space.

Suppose $(V,\omega)$ is vector space with a nondegenerate bilinear form $\omega$. We can assume it is $(\mathbb R^{2n},\omega)$ such that \begin{equation} \omega(u,v)=(Ju)^tv \end{equation} where \begin{equation} J=\begin{pmatrix} 0&I\\-I&0 \end{pmatrix} \end{equation}

The symplectic group is \begin{equation} Sp(2n)=\{A\in\mathbb R^{2n\times 2n}\colon A^tJA=J \} \end{equation}

The moment map for the $Sp(2n)$ action on $(\mathbb R^{2n},\omega)$ is \begin{equation} \mu\colon \mathbb R^{2n}\to \mathfrak{sp}(2n),\quad v\mapsto \frac{-1}{2}Jvv^t \end{equation} with $\mathfrak{sp}(2n)\cong\mathfrak{sp}(2n)^*$ by the inner product $\langle A,B\rangle=\textbf{tr}AB^t$.

Let $H\le Sp(2n)$ be a Lie subgroup, and let \begin{equation} \phi\colon \mathbb R^{2n}\to\mathfrak h \end{equation} be the composition of $\mu$ and $\text{pr}\colon \mathfrak{sp}(2n)\to\mathfrak h$, the orthogonal projection.

Then $(\mathbb R^{2n},\omega,H,\phi)$ is a Hamiltonian $H$-space.

My question is: do we have some theorems about the symplectic reduction $\mathbb R^{2n}/H$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.