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Let $G$ be a connected Lie group and $\Sigma$ a closed oriented surface. We know that principal $G$-bundles $P$ can be topologically classified by a characteristic class $c(P)\in H^2(\Sigma,\pi_1G)\cong\pi_1G$.

The following is my question:

Let $G$ be a semisimple connected (or even compact) Lie group, $\beta\colon\pi_1(\Sigma)\to G$ a group homomorphism, and $\Sigma$ a closed oriented surface. Consider the universal cover $\tilde{\Sigma}\to\Sigma$ which is a principal $\pi_1(\Sigma)$-bundle. Form the associated bundle $\tilde{\Sigma}\times_\beta G$ which is necessarily a principal $G$-bundle over $\Sigma$. Hence, it should have a characteristic class $c(\tilde{\Sigma}\times_\beta G)\in H^2(\Sigma,\pi_1G)\cong\pi_1G$. Is there a way to compute this characteristic class in terms of $\beta$?

The difficulty here is that $\pi_1(\Sigma)$ is not connected, and hence I cannot use the functoriality of the characteristic class. Another one is that I am not sure how to compute the induced map $B\beta\colon B\pi_1(\Sigma)\to BG$ concretely.

I've asked this question on MSE and haven't got any answers yet.

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Note sure if you can call it "calculating" but there is a rather straightforward construction. Let us fix some generators $A_i, B_i$ of $\pi_1(\Sigma)$ (i.e. a canonical system of curves such that slicing $\Sigma$ along them gives the fundamental polygon). Consider the central extension of $\pi_1$ $$ \mathbb{Z} \to \Gamma \to \pi_1(\Sigma),$$ where $\Gamma$ is generated by $A_i, B_i$ and the central element $J = \Pi_i [A_i, B_i]$ (with the commutator being $[A, B] = ABA^{-1}B^{-1}$). Since this extension is universal, every group homomorphism $\beta: \pi_1(\Sigma) \to G$ lifts to a group homomorphism $\hat{\beta}: \Gamma \to \tilde{G}$ where $\tilde{G}$ is the universal covering group. Then $\hat{\beta}(J) \in \pi_1(G)$ is the characteristic class of the bundle coming from $\beta$.

I think more details about this construction can be found in Atiyah and Bott's classical work on the Yang-Mills equation over surfaces.

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  • $\begingroup$ Actually, I asked this question exactly because I was reading Atiyah and Bott's paper. Therefore, could you explain why $\hat{\beta}(J)$ is the characteristic class of the bundle? They didn't offer many details actually. $\endgroup$ – YYF Nov 6 '17 at 15:05
  • $\begingroup$ Well, it is kind of like in Chern-Weil theory. There you integrate the curvature over the 2-dimensional fundamental polygon. However, now that everything is discrete, this integral has to be replaced with the holonomy of the boundary curve. Now the "holonomy" is $\hat{\beta}$ and the boundary curve $J$. Admittedly, this reasoning is a bit heuristic but should give you a general idea (sorry, I don't have the time right now for a more detailed answer). $\endgroup$ – Tobias Diez Nov 7 '17 at 12:41

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