4
$\begingroup$

I've been reading Kamnitzer's survey Symplectic resolutions, symplectic duality, and Coulomb branches. Here the Higgs branch is defined as a projective GIT quotient, but I couldn't figure out how the construction gives you a symplectic structure on the variety.

Let me recall the construction given in Section 2.4 of the survey article. We begin with a reductive group $G$ and its representation $N$. Then the cotangent bundle $T^*N=N^*\oplus N$ is equipped with a Hamiltonian action of $G$. Let $\Phi:T^*N\rightarrow \mathfrak{g}^*$ be the moment map. Choose a character $\chi$ of $G$. Then the Higgs branch (which we denote by $Y$) is the projective GIT quotient $$Y=\text{Proj}\left( \bigoplus_{n\ge 0} \mathbb{C}[\Phi^{-1}(0)]^{G,n\chi}\right).$$ Let $X$ be the GIT quotient $X=\text{Spec}\left(\mathbb{C}[\Phi^{-1}(0)]^{G}\right)$. Then the canonical map $Y\rightarrow X$ is sometimes a symplectic resolution. I was wondering how this projective GIT quotient of Higgs branch tells you its symplectic structure (assuming the smoothness of Higgs branch).

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

This follows from Marsden-Weinstein symplectic reduction, but transported to the algebraic context. Essentially, taking the zero level of the moment map (taking $ \Phi^{-1}(0) $) gives you something coisotropic and then diving by the group action (the GIT quotient) restores the symplectic structure.

For a good exposition of this, specifically for quiver varieties, you could look at section 4 of these notes of Ginzburg (especially Lemma 4.1.7) https://arxiv.org/pdf/0905.0686.pdf

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thanks! For those who might take a look at Lemma 4.1.7 of Ginzburg's note, the vector space on the last line should be $T_{\alpha}(\mu^{-1}(\lambda)/G)=T_{\alpha}(\mu^{-1}(\lambda))/Lie(G)$. The vector space $T_{\alpha}(\mu^{-1}(\lambda))$ is something coisotropic mentioned in the answer and $\text{Lie}(G)$ (value of fundamental vector fields at $\alpha$) is its orthogonal subspace with respect to the symplectic form of $T^*X$. Therefore quotienting out by $\text{Lie}(G)$ gives nondegeneracy. $\endgroup$
    – Ji Woong Park
    Commented Mar 5 at 16:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .