MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Short answer : no, for example complex tori are not all biholomorphic. – François Brunault Sep 29 '11 at 11:43
Miranda's book on algebraic curves is a very nice introductory book to RSs – Tommaso Centeleghe Sep 29 '11 at 11:55
In general, there are uncountably many. More precisely, the set of biholomorphism classes of genus $g=1$ surfaces is parameterized by the line $\mathbb{C}$ via the so called $j$-invariant. For $g>1$, the set is a complex space of dimension $3g-3$. – Donu Arapura Sep 29 '11 at 12:28
(better suited for SE? btw I didn't vote to close) – Qfwfq Sep 29 '11 at 14:16
@unknowngoogle: Why didn't you vote to close? Why did I not vote to close? Do either one of us believe this is a research-level question? Is it 1850 yet? – Igor Rivin Sep 29 '11 at 20:40
up vote 9 down vote accepted

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.

share|cite|improve this answer
While appreciating the benefits of the fine moduli space, I (mildly) object to the statement that the coarse moduli space is "not a very pleasant thing to study". – Richard Kent Sep 29 '11 at 12:49
Thanks, Richard -- I think you're right. I've edited. – Todd Trimble Sep 29 '11 at 13:26
Todd, your answer is good, and I don't mean to criticize, but if I may suggest another correction: Elliptic curves are generally understood as genus one curves with a chosen origin, so the automorphism groups are finite. Consequently the moduli stack is not as wild as it would be otherwise. Also those of us who tend to avoid stacks, when possible, can kill the difficulties by passing to finite level if necessary. – Donu Arapura Sep 29 '11 at 14:36
@Donu: I freely confess that I'm very far from expert here; expert correction or criticism is definitely welcomed. I incorporated part of your comment in an edit which I hope is correct. – Todd Trimble Sep 29 '11 at 15:10
I think it would be best to simply remove the word "elliptic" from that sentence. In the context of moduli problems, elliptic curves are defined as genus one curves equipped with a rational point (more precisely, a smooth morphism with one dimensional genus one geometric fibers, equipped with a section). There is a forgetful functor to the category of genus one curves, and it is not the identity. There seems to be near-universal agreement on this definition, with the exception of Hartshorne's text. – S. Carnahan Sep 29 '11 at 16:58

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).

share|cite|improve this answer

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.

share|cite|improve this answer

The answers here are great, but I think anyone drawn to this question should find the keyword "Teichmüller space" somewhere and maybe some references. Here is the wikipedia page as a start.

And Chapters 10-12 in A primer on mapping class groups serves as a great introduction. See also here for more suggestions.

Note: with the top of page 10 (ibid) in mind, in general complex smooth varieties that are biholomorphic need not be biregular (see here).

share|cite|improve this answer

Your Answer


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.