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Let $S$ be a compact R.S of genus $\geq 2$. In the paper "Stable and unitary vector bundles on compact Riemann surfaces" (by Narasimhan and Seshadri), they claim that there is a branched covering map from the upper half plane to $S$ which is ramified at exactly one point (with index $N$) (i.e. $S$ is the quotient of $\mathbb{H}$ by the group $\langle A_i, B_i, C \vert \Pi [A_i, B_i] = C, [A_i,C] = [B_i, C] = I, C^N=I \rangle$). I am not able to access the reference (Grothendieck) pointed out in that paper. Is there any other reference (or any easy proof of this?) ? Also, does a similar fact hold for non-compact Riemann surfaces (compact minus a finite collection of points).

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I am not sure about the group you wrote down there. It looks like a central extension of the fundamental group of a Riemann surface by $\mathbb{Z}$. –  Charlie Frohman Jul 9 '11 at 22:38
    
Indeed it is a central extension of the fundamental group. –  Vamsi Jul 10 '11 at 0:13
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2 Answers

up vote 10 down vote accepted

From your group-theoretic description, it seems to me that you are asking for a covering $S' \to S$ which is only ramified over one point of $S$, and the ramification index of each point of the inverse image divides $N$. So, Felipe's construction should work.

Here is a more direct construction. Suppose that $S$ is compact. Pick a point $p\in S$; the fundamental group of $S\smallsetminus \{p\}$ is the free group on $2N$ generators $A_1, \dots, A_g$, $B_1, \dots, B_g$; the product $C = \prod_{i=1}^g[A_i, B_i]$ represents a small loop around $p$. Consider the homomorphism $\pi_1(S\smallsetminus \{p\})\to \mathrm{S}_{2N}$ that sends $A_1$ into $(1 \dots N)$, $B_1$ into $(1 \dots 2N)$, and all the other $A_i$ and $B_i$ into the identity. It is easy to see that $C$ goes into $(1 \dots N)(N+1 \dots 2N)$, and that this homomorphism has transitive image. Hence the corresponding ramified cover $S' \to S$ has two ramification points over $p$, each of index $N$.

When $S$ is non-compact, then the small loop $C$ is part of a free set of generators of $\pi_1(S\smallsetminus \{p\}$, and the construction is even easier (for example, you can easily construct a cyclic cover $S' \to S$ of degree $N$ which is totally ramified over $p$, and nowhere else; this is impossible in the compact case).

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Take a finite cover $S'$ of $S$ ramified at only one point $P$. Then $\mathbb{H} \to S' \to S$ is the map you want, where the first map just exhibits the universal cover of $S'$ and is unramified. To construct $S'$, first take the unramified cover $S_1$ of $S$ obtained by pulling back the map multiplication-by-2 on the Jacobian. The pre-image of $P$ in $S_1$ can then be split in two subsets of same cardinality such that there is a function on $S_1$ with simple poles on the first subset and simple zeros on the second and no other. Now take the $N$-th root of this function to get $S'$.

This construction is part of the Kodaira-Parshin trick. I have no idea if that is what Narashiman and Seshadri had in mind or if there is an easier construction.

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Thanks! I apologise for my question, but this won't work for a compact R.S minus a finite number of points would it? If it doesn't, then is there an analogue? Also, would the branching index of $S^{'}$ not be $N$ times the number of sheets in $S_1$? –  Vamsi Jul 10 '11 at 0:11
    
I didn't think about the non-compact case, I don't know. Also, re-reading my answer and your follow-up question, there is a problem. Indeed, the ramification is on $N$ times the number of sheets in $S_1$ points. If we could somehow reverse the order of the arrows $S' \to S_1 \to S$ it would be OK, but I don't think that's right, the cover is Galois but not necessarily abelian. –  Felipe Voloch Jul 10 '11 at 0:22
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