# Index of linearized operator for symplectic vortex equations

In reference [1], the index of the linearized operator for the symplectic vortex equations is computed on page 27-28.

The first step of the proof says that the operator $$\tag{1} \Omega^1(\Sigma,\mathfrak{g}_P)\rightarrow\Omega^0(\Sigma,\mathfrak{g}_P)\oplus \Omega^0(\Sigma,\mathfrak{g}_P):\alpha\mapsto(-d^*_A\alpha,*d_A\alpha)$$ has index $-\chi\textrm{ dim }G$. How does one show that this is true?

My attempt to understand this is as follows. In reference [2], the index of the Cauchy-Riemann operator $$(\nabla^A)^{0,1}:\Omega^0(\Sigma,\mathfrak{g}_P)\rightarrow\Omega^{0,1}(\Sigma,\mathfrak{g}_P)$$ is shown to be equal to $c_1(\mathfrak{g}_P\rightarrow\Sigma)+(\textrm{dim }G)(1-g)$ via Hirzebruch-Riemann-Roch theorem, and is just $$\textrm{dim }G(1-g)=\frac{\chi}{2}\textrm{dim }G$$ for compact $G$. Hence, it seems to me that the operator in equation $(1)$ is equal to the adjoint of $2(\nabla^A)^{0,1}$. This seems to be supported by the statement below equation (24) of reference [2]. Is my understanding correct?

References

[1] "The symplectic vortex equations and invariants of Hamiltonian group actions" (https://arxiv.org/abs/math/0111176)

[2] "Twisting gauged nonlinear sigma models" (https://arxiv.org/abs/0707.2786)

The operator $D$ in (1) is the related to (odd part of) the de Rham operator $d+d^*$. If you omit the Hodge star in the last term, you end up in $\Omega^0\oplus\Omega^2$. The Atiyah-Singer index (or if you like, the twisted Gauß-Bonnet theorem) theorem gives \begin{align*} \operatorname{ind}(D)&=-\operatorname{ind}(d+d^*\colon\Omega^{\mathrm{even}}(\Sigma;\mathfrak g)\to\Omega^{\mathrm{odd}}(\Sigma;\mathfrak g))\\ &=-\int_\Sigma e(T\Sigma)\,\operatorname{ch}(\mathfrak g)\\ &=-\int_\Sigma e(T\Sigma)\,\dim\mathfrak g=-\chi(\Sigma)\,\dim\mathfrak g\;. \end{align*} The Euler class is homogeneous of degree $\dim\Sigma=2$, hence only the zero degree term $\dim\mathfrak g$ of the Chern character survives. In particular, this index does not see if the principal bundle $P$ is nontrivial.
• Maybe your explanation works, too, but for Riemann-Roch, you only consider the $\bar\partial$ part of $d$, so you have a different operator. You split $\Omega^1=\Omega^{0,1}\oplus\Omega^{1,0}$ and consider the sum of $\bar\partial^*$ and $\partial^*$, and then you would end up with the same formula. – Sebastian Goette Apr 8 '17 at 5:07