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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?

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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

1 Answer 1

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.

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