3
$\begingroup$

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 \begin{equation}\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) \end{equation} 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 \begin{equation} (\nabla^A)^{0,1}:\Omega^0(\Sigma,\mathfrak{g}_P)\rightarrow\Omega^{0,1}(\Sigma,\mathfrak{g}_P) \end{equation} 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 \begin{equation} \textrm{dim }G(1-g)=\frac{\chi}{2}\textrm{dim }G \end{equation} 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)

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Sebastian Goette
    Commented Apr 8, 2017 at 5:07
  • $\begingroup$ May I know if the answer above and your additional comment hold in the case where the Riemann surface has boundaries? $\endgroup$
    – Mtheorist
    Commented Apr 8, 2017 at 6:36
  • $\begingroup$ The Euler operator admits local boundary conditions, and you can choose either absolute or relative ones (as for cohomology). The Dolbeault operator does not, so now you should stick to the Euler operator. Because the boundary is odd-dimensional, it has Euler number 0, and the index is still given by the formula above both for absolute and for relative boundary conditions. If you are interested, read the intro to Atiyah, Patodi, Singer, Spectral asymmetry and Riemannian geometry I, Math. Proc. Camb. Phil. Soc. 77 (1975), 43-69. $\endgroup$
    – Sebastian Goette
    Commented Apr 8, 2017 at 13:13
  • $\begingroup$ Thank you for your answer and the reference. Am I right to say that since the Dolbeault operator does not admit local boundary conditions, then we cannot compute the index of the linearized operator in reference [1] for an open Riemann surface, since the other part of the linearized operator is a Cauchy-Riemann operator, as shown on page 28? (I was hoping to use the local boundary conditions which you discussed in mathoverflow.net/questions/266520/…) $\endgroup$
    – Mtheorist
    Commented Apr 8, 2017 at 17:07
  • $\begingroup$ So, you are studying the operator (23)? For the Cauchy-Riemann operator, there are typically no local boundary conditions. Also, I don't see anything else you could pair this with to get local boundary conditions as in the other question. Is there a reason that you need local boundary conditions? If you like, we can continue the discussion by email. $\endgroup$
    – Sebastian Goette
    Commented Apr 9, 2017 at 3:56

You must log in to answer this question.

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