Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite abelian group, and $C$ a compact Riemann surface (algebraic curve) of genus $g$. I am interested in topological Galois $G$-covers $X \to C$, aka \'etale $G$-principal bundles over $C$.

Let us write the abelian group as a direct sum of cyclic groups of order $d_1, \ldots, d_k$ with $d_1|d_2|\ldots|d_k$.

Is it true, and why, that if $k>2g$, then $X$ must be disconnected?

share|improve this question
    
Without condition on the $d_i$, they aren't uniquely defined (neither is $k$), so strictly, the answer is no. For example, you can take $N >> 1$ different primes $p_1, \dots, p_N$ and look at the (connected) Galois cover coming from a morphism $\pi_1 C \to \mathbb Z \to \mathbb Z/(p_1\cdots p_N)\mathbb Z \simeq \mathbb Z/p_1\mathbb Z \times \cdots \times \mathbb Z/p_N \mathbb Z$ as soon as $g\geq 1$. I guess you want the condition $d_i | d_{i+1}$. –  PseudoNeo Feb 24 '12 at 16:23
    
Yes sorry, thanks for your remark. I have edited the question. –  Calc Feb 24 '12 at 16:27
add comment

1 Answer

up vote 3 down vote accepted

By the Galois correspondance for covering spaces, you will have a connected topological $G$-cover $X \to C$ precisely if you can find a surjective morphism $$\pi_1 C = \langle a_1, b_1, \ldots, a_n, b_n \, | \, [a_1,b_1]\cdots[a_g,b_g]\rangle \to G.$$ (The complex/algebraic structure is irrelevant here).

As $\pi_1 C$ is generated by $2g$ elements, any such $G$ (abelian or not) must also be generated by $2g$ elements. If $G$ is abelian, it is then a quotient of $\mathbb Z^{2g}$, and the classification of finite type abelian groups tell you that this can only happen if its canonical decomposition $G = \oplus_{i=1}^r \mathbb Z/d_i$, (canonical means $d_i|d_{i+1}$ and we know that such a decomposition is unique) has at most $2g$ factors.

So yes, it is true.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.