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As an honest question (probably with some subjectivity), how many smooth oriented 4-manifolds are actually symplectic? Can I say half (perhaps under some mild assumptions)? I ask this question because every compact smooth oriented 4-manifold with $b^2_+\ge 1$ admits a near-symplectic form, i.e. a closed 2-form which is symplectic away from a finite set of circles.

Some results that might push the percentage one way or the other:
1) Gompf has shown that any finitely presented group can be realized as the fundamental group of a compact symplectic 4-manifold.
2) The Seiberg-Witten invariants are nonzero for symplectic 4-manifolds, and in a sense show that they are the "irreducible" basic forms of smooth 4-manifolds.
3) Every compact symplectic 4-manifold is a branched cover of $\mathbb{C}P^2$.

The responses/comments show that we can ask this question (on when can I expect my 4-manifold to be symplectic) in many different ways, each with different expectations. So I am interested in some further thoughts on Tim's and Dmitri's questions.

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I don't quite understand your sentence "every smooth oriented 4-manifold admits a near-symplectic form", since if $M$ is closed and near-symplectic, then $b^+(M)>0$, which is a nontrivial constraint. –  YangMills Oct 28 '12 at 0:39
Ah yes sorry for that gap. –  Chris Gerig Oct 28 '12 at 2:43
The BMY conjecture says that a symplectic 4-manifold satisfies $e(X)\ge 3\sigma(X)$, and it is hard to produce symplectic 4 manifolds with positive signature, so maybe a better question is "how many orientable 4-manifolds are symplectic (with some orientation)?" –  Paul Oct 28 '12 at 16:41
For 4-manifolds of the form $X=S^1\times M$, one could make the question precise using the Friedl-Vidussi theorem, that $X$ is symplectic iff $M$ fibers over $S^1$. One can make random 3-manifolds using the Dunfield-Thurston method: construct random Heegaard splittings of genus $g$ by gluing the handlebodies via a diffeomorphism obtained by taking a length $n$ random walk on a Cayley graph for the mapping class group. Maybe the large $n$ asymptotics are independent of the presentation? –  Tim Perutz Oct 29 '12 at 3:17

4 Answers 4

up vote 13 down vote accepted

I have to apologize, in fact the answer to the second question is still unknown. Namely, up to now all known symplectic manifolds of dimension 4 that have negative Euler characteristic are blow ups of ruled surfaces. However it is not known if there are no other examples. I have corrected the answer accordingly.

It is difficult to give a precise answer to this question, but personally I am pretty sure that a random four dimensional manifold has no reason at all to have a symplectic structure. What is true for sure is that symplectic manifolds are clustered among all four-manifolds, namely if you look in some places then probability to find a symplectic manifold equals zero. Let me give one example. Let us consider the following two questions:

Question 1. Is the probability for a 4-manifold to have negative Euler characteristic $>0$?

Question 2. Is the probability for a symplectic $4$-manifold to have negative Euler characteristic $>0$?

I am sure that one can give a relatively precise answer to Question 1 and my common sense tells me that since the probability for a number to be negative equals $\frac{1}{2}$, the answer to the question is morally yes.

What is very surprising is that the answer to the second question is probably NO. In 1995 Gompf asked if all such manifolds are blow ups of $S^2 \times \Sigma_g$ in at most $4g-5$ points where $\Sigma_g$ is a surface of genus $g$. This is still is an open question.

Added. It seems to me that it would not be unfair to say that the majority of constructions of compact symplectic 4-manifolds come for the moment from algebro-geometric analogues. In particular this is the case for Gompf sum, which is the most used construction. So in some sense the cloud of symplectic manifolds in dimension four that we see now is centred around algebraic surfaces. In particular this fact about symplectic 4-manifolds with negative Euler characteristic is inherited from the theory of algebraic surfaces (though one needs SW theory to prove it for symplectic manifolds). I think no one doubts that algebraic surfaces are not distributed among $4$-manifolds evenly.

I want to add that there is a conjecture saying that four manifolds of constant sectional curvature $-1$ are never symplectic. It is not clear (for me) how well this conjecture is justified but still there are no ideas of how to construct symplectic structures on these manifolds...

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This is a great perspective! What are other reasons to think that symplectic manifolds are found clustered in the `space' of 4-manifolds? –  Chris Gerig Oct 27 '12 at 20:12
It would be interesting to (make sense of the question and) figure out what the probability distribution of Euler characteristics is... –  Mariano Suárez-Alvarez Oct 28 '12 at 4:19
(Your argument for Q1 probably breaks when applied to «is the probability of a compact surface having positive Euler characteristic positive?») –  Mariano Suárez-Alvarez Oct 28 '12 at 5:23
Mariano, it seems to me that this depends on weather you consider connected or disconnected surfaces :) ... For connected ones you are right of course. There might indeed be some probabilistic models... –  Dmitri Oct 28 '12 at 15:08

Two results to "push the percentage down" (if the thing makes sense at all):

1) A symplectic $4$-manifold is in particular an almost complex manifold, so the Euler characteristic $e$ and the signature $\tau$ are such that $e+\tau\equiv 0$ $( \mathrm{mod}$ $4)$.

2) Fintushel and Stern have proved that there are infinitely many exotic $K3$ surfaces which do not admit any symplectic structure.

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In an attempt to get a sense for the answer to Dmitri's 1st question I checked the euler characteristics of the 4-manifolds in the census of triangulated closed 4-manifolds containing 6 or less 4-dimensional simplices. I like to call a 4-dimensional simplex a pentachoron.

Non-orientable 2-pentachoron triangulations: 2. All have Euler characteristic 0.

Orientable 2-pentachoron triangulations: 9. Three have euler characteristic equal to 0. 6 have Euler char equal to 2.

Non-orientable 4-pentachoron triangulations: 184. 119 have euler char 0, 65 have euler char 1.

Orientable 4-pentachoron triangulations: 785. 131 have euler char 0. 3 have euler char 1. 647 have euler char 2. 4 have euler char 3.

Non-orientable 6-pentachoron triangulations: 60229. 28831 have euler char 0. 20 have euler char -1. 30824 have euler char 1. 554 have euler char 2.

Orientable 6-pentachoron triangulations: 440496. 29294 have euler char 0. 1477 have euler char 1. 405282 have euler char 2. 4423 have euler char 3. 20 have euler char 4. There are no negative euler characteristics.

So from this point of view, positive euler characteristics tend to dominate, at least for small numbers of pentachora.

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Just curious, what did you use for this? –  Chris Gerig Nov 1 '12 at 17:47
Ben Burton and I generated the census using version 5.0 of Regina. That software isn't in general release yet but it will be soon. The core of the algorithm is the Rubinstein 3-sphere recognition algorithm, which Ben has done a lot of work on to make fairly efficient. –  Ryan Budney Nov 1 '12 at 18:26
Ryan, I just saw your answer, this is interesting :) I would not expect that there is such a strong tendency to have positive Euler characteristics but one can never know... :) –  Dmitri Nov 28 '12 at 10:42

By Donaldson, symplectic manifolds are the same as topological Lefschetz pencils. So you may ask how many 4-manifolds are topological Lefschetz pencils.

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