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?
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4$\begingroup$ Short answer : no, for example complex tori are not all biholomorphic. $\endgroup$– François BrunaultCommented Sep 29, 2011 at 11:43
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$\begingroup$ Miranda's book on algebraic curves is a very nice introductory book to RSs $\endgroup$– Tommaso CentelegheCommented Sep 29, 2011 at 11:55
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1$\begingroup$ 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 $3g-3$. $\endgroup$– Donu ArapuraCommented Sep 29, 2011 at 12:28
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2$\begingroup$ (better suited for SE? btw I didn't vote to close) $\endgroup$– QfwfqCommented Sep 29, 2011 at 14:16
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$\begingroup$ @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? $\endgroup$– Igor RivinCommented Sep 29, 2011 at 20:40
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4 Answers
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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.
answered Sep 29, 2011 at 12:43
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2$\begingroup$ 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". $\endgroup$ Commented Sep 29, 2011 at 12:49
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2$\begingroup$ 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. $\endgroup$ Commented Sep 29, 2011 at 14:36
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$\begingroup$ @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. $\endgroup$ Commented Sep 29, 2011 at 15:10
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$\begingroup$ 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. $\endgroup$– S. Carnahan ♦Commented Sep 29, 2011 at 16:58
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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).
answered Sep 29, 2011 at 11:43
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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.
answered Sep 29, 2011 at 12:21
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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).
answered May 16, 2016 at 15:13