Question

I know that all compact Riemann surfaces with the same genus are topologically equivalent. Moreover they are diffeomorphic. But are they biholomorphic, too? In other words, is the complex structure conserved?

Answer 1

Some magic words for this question are "moduli space" or "moduli stack". In the early days, one was interested in a variety or variety-like object which would classify projective complex curves (compact Riemann surfaces) of given genus $g$, i.e., whose points correspond to isomorphism classes of curves (or biholomorphism classes of compact Riemann surfaces). This is nowadays called a "coarse moduli space". As GH and François commented, there is a whole continuum of points in the coarse moduli space of genus 1; the same is true for any genus $g > 1$.

Over time, it became apparent that the coarse moduli space is not a very pleasant thing the most fundamental object of study. Some information that is desirable to have that the coarse moduli space misses is: what are the possible automorphisms on a fixed compact Riemann surface? For example, in the case of an elliptic curve (genus 1), the automorphism group is infinite and acts transitively on the curve. (Edit: this remark may be slightly misleading because it is more usual to consider elliptic curves with a chosen origin, and this cuts way down on the automorphism group. Thanks to Donu Arapura for pointing this out in comments.) Not so in higher genus; curves of higher genus are much more rigid, and in fact have only finite automorphism groups.

(I think to me this was a bigger shock than finding out about the plenitude of complex manifold structures on a given curve. In ordinary smooth manifold theory, all the points are pretty much alike, in that one can construct a diffeomorphism that takes one point to another. But in complex curve theory, points can have different "personalities"; for example, cf. Weierstrass points.)

Anyway, the better object of study in these questions, which parametrizes not only isomorphism classes of curves but also isomorphisms between them, is called a moduli stack. You can begin reading about them here.

Answer 2

The answer is no. For example, if $\Lambda_1$ and $\Lambda_2$ are two lattices in $\mathbb{C}$, then the surfaces $\mathbb{C}/\Lambda_1$ and $\mathbb{C}/\Lambda_2$ are conformally equivalent if and only if $\Lambda_1$ and $\Lambda_2$ are similar. This follows from the theory of elliptic functions (or elliptic curves).

Answer 3

Identify the opposite sides of the unit square to get a torus $A$. Identify the opposite sides of a rectangle of side lengths $\pi$ and $\frac{1}{\pi}$ to get a torus $B$.

The extremal length of every closed curve in $A$ is an algebraic integer, which is not true of $B$. Since the set of extremal lengths of curves is a conformal invariant, $A$ and $B$ are not biholomorphic.