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I would like to know if the Hirzebruch-Riemann-Roch theorem exists for bundles over Riemann surfaces with a boundary. I am asking this because the Hirzebruch-Riemann-Roch theorem is used in the following paper (https://arxiv.org/pdf/0707.2786v2.pdf) on page 10 to compute the index of the following differential operators defined over fiber bundles on a Riemann surface $\Sigma$, \begin{align} (\nabla^A)^{0,1} &: \Omega^0 (\Sigma ; \mathfrak{g}_P) \longrightarrow \Omega^{0,1} (\Sigma ; \mathfrak{g}_P) \label{2.7} \\ (\phi^{\ast}\nabla^A)^{0,1} &: \Omega^0 (\Sigma ; \phi^\ast \ker \textrm{d} \pi_E) \longrightarrow \Omega^{0,1} (\Sigma ; \phi^\ast \ker \textrm{d} \pi_E) \ , \end{align} where the definitions of the fiber bundles $\mathfrak{g}_P$ and $\phi^\ast \ker \textrm{d} \pi_E$ are given on page 6 of the aforementioned paper.

The indices for these operators are said to be easily obtained from the Hirzebruch-Riemann-Roch theorem, and the result for a general compact $\Sigma$ with no boundary is \begin{align} \rm{index} (\nabla^A)^{0,1} &= c_1 (\mathfrak{g}_P \rightarrow \Sigma) + ({\rm dim}G)(1-g) \label{2.8} \\ \rm{index} (\phi^{\ast}\nabla^A)^{0,1} &= c_1 (\phi^\ast \ker \textrm{d} \pi_E \rightarrow \Sigma) + ({\rm dim}_\mathbb{C} X)(1-g) \ . \end{align}

I would like to know the generalizations of these formulae for general $\Sigma$ with boundary.

I suspect, based on the ordinary Riemann-Roch theorem, that the answer is replacing $(1-g)$ in the expressions above to $(1-g-\frac{b}{2})$, where $b$ is the number of boundaries. This is because the Euler characteristic for the Riemann surface is $\chi =2-2g-b$. Is this correct, and why? References would be highly appreciated.

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  • $\begingroup$ I think this is usually part of the Atiyah-Patodi-Singer theorem. $\endgroup$ – Jason Starr Oct 12 '16 at 8:16
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The Hodge-DeRham operator whose index on closed manifolds is the Euler characteristics admits local boundary conditions on manifolds with boundary, Dirichlet or Neumann type. That is not the case with the Dolbeault operator that does not admit local boundary conditions. (This is a rather nontrivial fact observed 50 years ago by Atiyah and Bott and involves some $K$-theory.)

In the Dolbeault case one has to use the Atiyah-Patodi-Singer boundary condition. This is a non-local condition. The resulting formula for the index of this boundary value problem is a bit more complicated than the one you suggested. For details see formula (2.8) in this paper.

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  • $\begingroup$ But it seems that a 'doubling trick' exists where one can glue two Riemann surfaces with boundaries to obtain the generalization I've mentioned, please see pages 811-813 of Mirror Symmetry by Hori et al., which are based on "Enumerative geometry of stable maps with Lagrangian boundary conditions and multiple covers of the disc" by Katz and Liu. Do these results agree with your paper? $\endgroup$ – Mtheorist Oct 14 '16 at 15:05
  • $\begingroup$ When you double the manifold you consider different boundary conditions on the two halves. The boundary contributions on the two sides the cancel each other. Let me also point out that the boundary contribution is metric dependent. It simplifies somewhat when the boundary is totally geodesic (say cylindrical metric near the boundary) but it still depends on the length of the boundary. $\endgroup$ – Liviu Nicolaescu Oct 14 '16 at 18:54
  • $\begingroup$ @LiviuNicolaescu Does simplification occur when instead of boundaries we consider a Riemann surface with punctures? $\endgroup$ – Wakabaloola Apr 4 at 11:27

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