Isogenies of Riemann surfaces

1. Fix $n$ natural. I want to characterize all compact Riemann surfaces $M$ such that $M$ is an unramified covering of degree $n$ over itself.

2. How do I construct this covering map?

This map is called an isogeny of $M$.

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Covering maps of compact spaces are multiplicative with respect to Euler characteristic, which implies that the only Riemann surfaces admitting nontrivial self-isogenies are tori. – Aaron Mazel-Gee Dec 17 '12 at 19:03
For 2., it depends on whether you want continuous or holomorphic covering maps . – Georges Elencwajg Dec 17 '12 at 23:35

Following the suggestion of the comment above we can applies Hurwitz's formula, since f is unramified covering we have,

$$X(M)=nX(M)$$

$n>1$ implies that $X(M)=0$, then $M$ must be the torus.

Now for the covering consider the application,

$$(z,w)\in \mathbb{T}\mapsto (z, w^k)\in \mathbb{T}$$

is a simple calculation to verify that the application is a covering application.

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Notice that your covering maps are not holomorphic (but it is not clear whether the OP wants holomorphic or just continuous maps). – Georges Elencwajg Dec 17 '12 at 23:37

If $f:M\to M$ is a ramified cover of degree $n$ (=non constant morphism= surjective morphism = finite morphism) , Riemann-Hurwitz's formula implies that for the canonical divisor class $K=K_M$ we have the relation
$$K= f^*K+R$$ where $R$ is the ramification divisor.
Taking degrees and remembering the expression $deg K=2g-2$ for the degree of a canonical divisor in terms of the genus of $g$ of $M$ yields $$2g-2=n(2g-2)+deg R$$ If the covering is known to be unramified (=étale), we have $deg (R)=0$ (actually even $R=0$) so that $$2g-2=n(2g-2)$$ which forces $n=1$ (duh!) or $g=1$.

Edit
Conversely, if $g=1$ we have an elliptic curve $M=\mathbb C/\Lambda$ and all of its holomorphic unramified covers are of the form $$M\to M:[z]\mapsto [az+b]$$ where $b\in \mathbb C$ is arbitrary and $a\in \mathbb C^*$ is a complex number satisfying $a\Lambda \subset \Lambda$.
They are called the isogenies of $M$.

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This response is very close to the answer given above, my impression is that you used bananas instead oranges. – user27456 Dec 17 '12 at 20:39
@kLEIN: My answer is one from algebraic geometry and I wrote it deliberately in order to give a different perspective on the problem. It is valid in characteristic $p$ as long as the morphism $f$ is separable, whereas the other answer depends on classical topology and relies,as should have been clear to you, on completely different techniques (divisors and canonical class versus Euler characteristic). Also, I find the tone of your comment quite unpleasant, especially coming from someone ignorant of such well-known results. – Georges Elencwajg Dec 17 '12 at 21:03