# Symplectic structure of Higgs branch

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).

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.
• 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. Commented Mar 5 at 16:41