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Is there a classification of symplectic surfaces, i.e. of surfaces equipped with an area form? Symplectic topology references like McDuff-Salamon seem to start their discussion of open questions with dimension four.

  • A surface admits a symplectic form iff it is orientable.
  • The Moser trick seems to show that on a compact orientable surface $M$, the unique invariant of a symplectic form is its total area? So the set of symplectic forms on $M$ (up to symplectomorphism) is parametrized by $\mathbb{R}^+$?
  • And, for non-compact orientable surfaces, ...?
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2 Answers

up vote 3 down vote accepted

This is a bit late answer to an old MO question, but Moser's theorem was generalized to open manifolds in

R. Greene and K. Shiohama, Diffeomorphisms and volume-preserving embeddings of noncompact manifolds, Trans. Amer. Math. Soc. 255 (1979), p. 403-414.


They gave a necessary and sufficient condition for existence of volume-preserving diffeomorphisms between open manifolds in terms of volumes of ends and overall volumes. This takes care of surfaces as a very special case.

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Thanks, nice reference! [Summary: there exists a symplectomorphism between surfaces $(M_1,\omega_1)$ and $(M_2, \omega_2)$, if and only if (1) they have the same area and (2) there exists a diffeomorphism between them sending finite- (respectively, infinite-) area ends to finite- (resp., infinite-) area ends.] This is the result that Dmitri effectively derived by hand in 2009, but I am going to accept Misha's and uncheck Dmitri's to highlight the journal reference. –  macbeth Mar 4 '12 at 19:10
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This is not a complete answer to the question, and I don't know if a complete answer is written down anywhere in the literature. In the first revison of the answer I tried to adress all the 10 comments that my previous answer recieved. The second revison contains a conjecture (that I am 99% sure of) describing the complete answer to this question.

The first point is that the classification of symplectic surfaces can not be simpler than the classification of surfaces up to a diffeo. And the classification up to a diffeo of non-compact surfaces is quite a delicate subject. It is desribed for example in the paper http://www.jstor.org/pss/1993768, 1963, ON THE CLASSIFICATION OF NONCOMPACT SURFACES, IAN RICHARDS. In particular one phenomena apears here -- a certain ideal boundry of the surface. This idel boundary is a totally disconnected, compact separable space. I guess a good illustration will be a disk from which we throw away a Cantor set on the x axis, Cantor set been the ideal boundary.

Let us give now some examples that illustrate the additional phenomena that happen for non-compact surfaces, if we take in account the symplectic form. First of all there is the simplest case when the surface has finite topological type. In this case we have two topological invariants, the fundamental group, a free group on $n$ generators plus the number of punctures (or boundary components) $m$. In this case a complete classification of symplectic forms can be given. Either the surface has a bouned are $A$, in this case this area is the only invariant. Or it has an infinite area. In this case there are $m$ types of surfaces. Namely for every boundary (or puncture) we can check if it has on open neighborhood that his finite area, or not. The number of components near which the area is unbounded can be any between $1$ and $m$.

Example. Consider $S^2$ with two delited points. Then either it has finite area, or it is symplectomorphic to an infinite cylinder $S^1\times R^1$ with the form $ds \wedge dt$, or it is symplectomorphic to $R^2\setminus 0$, with the form $dx\wedge dy$.

If the number of punctures is countable, and every puncture has a neighborhood that is diffeomorphic to a punctured disk, then the situation should be very similar to what I have described above. Namely 1) the area can be bounded. 2) The number of UNBOUNDED punctures for wich every neighborhood has infinite area is bounded, in such surfaces are enumerated by natural numbers. 3) The number of unbounded punctures is infinite, in this case we just need to cound the number of bounded punctures, that can be finite of inifinite.

CONJECTURE. Here is the conjecture telling what should be the complete answer to the question.

Take a non-compact surface. Then the set of symplectic structures of infinite area on it is in one to one correspondence with closed non-empty subsets of its ideal boundary. For every bounded A>0 there is a unique symplectic strucutre on the surface with given area.

In the case of a surface with puncutres, the ideal boundary is just the union of punctures. Below the statement of this conjecture is proven for some simplest examples of surfaces. I think, that the general case should not be much different.

For the simple examples that I have descirbed the proof should be ALMOST identical to Moser's argument. Indeed the simplest examples have the following property: these surfaces can be decomposed in a countable union of compact pieces, all of which apart from one piece are annuli, and one piece is a compact surface with a boundary. Now, on a compact surface with a boundary (as well as on the cylinder) the symplectic form is uniquely defined by its area -- this can be done by Moser argument (we need that in a neighbrohood of each boundary the symplectic form is strandard, which is automatic in our case). Now the symplectomorphism can be constructed inductively.

Consider for example the case of $R^2$ of infinite area. We can take any exostion of $R^2$ by cylinders. For example we indroduce some coordinates on $R^2$, and conisder cylinders of radiuce $n,n+1$. We don't care what is the exact expression of w. The crucial thing for us is that the sum of areas tends to infinity. Now we take the standard $R^2$ and take a decompositon in concetnric cylinders of needed area. Then define the symplectomorphism from the standrd $R^2$ cylinder by cylinder. For the class of surfaces, that I dealed with same thing should work. Though I am not sure that this is the best "proof" of the statement.

I think it is not hard to prove this conjecture as well in the case of the complement to a Cantor set in the unit disk.

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Thanks, that's really helpful! A couple of questions: 1. You appear to be using a (presumably standard) result which says that every non-compact surface is diffeomorphic to some sufficiently-many-times-punctured sphere. Is this true? Can you give me a reference? 2. Why is it true that - two symplectic forms on a non-compact surface, with the same finite area, are symplectomorphic? - two symplectic forms on $\mathbb{R}^2$ (say), both with infinite area (i.e. with no finite-area nhd of $\infty$), are symplectomorphic? etc. etc. -- is this again a variation on the Moser trick? –  macbeth Dec 27 '09 at 13:14
It is not true that every non-compact surface is homeomorphic to a multiply punctured sphere. For any nonnegative integers (g,k), let S(g,k) be a k-punctured orientable genus g surface. Then g and k are determined by the homeomorphism class of S(g,k), because k is the number of boundary components and the Euler characteristic is 2-2g-k. –  David Speyer Dec 27 '09 at 13:27
Okay, that was just some erroneous optimism from me. What exactly is the topological classification of non-compact surfaces that Dmitri is using, then? (Something like "every non-compact surface is diffeomorphic to a multiply-punctured compact surface"?) Presumably it's standard, but I'd be glad of a reference. –  macbeth Dec 27 '09 at 13:55
I think it is not true that every non-compact surface is homeomorphic to an open subset of a closed one. To see this, take an infinite chain of handles (start with a sphere, add a handle, add another handle to this handle, etc). This is an open surface that has, for every g, a subset homeomorphic to a genus g compact surface with a puncture. This implies it is not embeddable in a compact surface: for every g, we can find 2g cycles $a_i$, $b_i$ that intersect in the usual way: $(a_i,a_j)=(b_i,b_j)=0$, $(a_i,b_j=\delta_{ij}$. Such cycles prevent embedding into a surface of genus less than $g$. –  t3suji Dec 27 '09 at 14:20
I don't see how Moser's argument by itself is enough to give the claim that, for instance, any two infinite-area symplectic forms on the disc are symplectomorphic. Moser's argument entails taking the time-one map of the flow of a cleverly chosen vector field, and in noncompact cases there won't always be any reason to expect the flow to exist for the necessary amount of time. –  Mike Usher Dec 27 '09 at 14:39
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