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By symplectic manifold I mean a pair $(M^{2n},\omega)$ consisting of a smooth, connected, even dimensional manifold and a non-degenerate $2$-form. I am interested in compact, boundarlyess examples where $\chi(M)=0$. If none such exist, can anyone provide a simple proof (understandable to a Topologist who knows a little geometry)?

In case the answer to the question in the title is a quick "yes", I have several follow up questions:

  1. What about if we restrict to the case $n=2$ (in which case $M$ would have to be non-simply-connected)?
  2. What about if we restrict to closed symplectically aspherical manifolds? [Recall that $(M^{2n},\omega)$ is called symplectically aspherical if the symplectic class $[\omega]\in H^2(M;\mathbb{R})\cong \mathrm{Hom}(H_2(M),\mathbb{R})$ vanishes on the image of the Hurewicz homomorphism $h\colon \pi_2(M)\to H_2(M)$.]

Thanks.

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3 Answers 3

up vote 9 down vote accepted

Yes. $T^2 \times T^2$ with the sum of the volume forms on each factor.

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Why not just $T^2$? –  Mariano Suárez-Alvarez Jan 3 '11 at 19:23
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Ah, right. @Mark: you can take $T^2\times\mathrm{your favorite closed symplectic manifold}$, in fact. –  Mariano Suárez-Alvarez Jan 3 '11 at 19:30
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In general if X->B is a bundle over an oriented surface whose fibers are closed oriented surfaces, the Thurston trick lets one construct a symplectic form on the total space of X as long as there is a class c in H^2(X) evaluating positively on the class of the fiber (start with a de Rham representative of c that restricts as a symplectic form to each of the fibers, and then add a large multiple of the pullback of a symplectic form on the base). If the fibers have any genus other than one then plus or minus the Euler class of the vertical tangent bundle is such a class c. ... –  Mike Usher Jan 3 '11 at 22:16
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Meanwhile Geiges (Duke Math J. 1992) showed that every T^2 bundle over T^2 admits a symplectic structure (in some cases these aren't like the ones from the previous comment in that the fibers are Lagrangian rather than symplectic). Combined with the previous comment this shows that every oriented surface bundle over T^2 admits a symplectic structure. –  Mike Usher Jan 3 '11 at 22:16
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And to state the obvious: if $\pi_1(M)$ is trivial then $\pi_2(M)=H_2(M)\ne 0$ so that there are no simply connected symplectically aspherical manifolds. Anybody know a simply connected $\chi=0$ 6-dimensional symplectic mfd? –  Paul Jan 4 '11 at 3:38
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You can find plenty of example of symplectic $4$-manifolds with $0$ Euler characteristic by taking $Y\times S^1$ where $Y$ is a fibered $3$-manifolds. See: http://www.zentralblatt-math.org/zmath/en/search/?q=an:0324.53031&format=complete for a proof (I couldn't find an on-line version of the paper, however the construction is outlined there: http://arxiv.org/abs/1001.0132).

For the vanishing of $\omega$ on $\pi_2$ just take the fiber to be something not a sphere (the fibers are symplectic so it couldn't work with the sphere as a fiber).

Those manifolds are non-trivial fibrations over the torus if you take the monodromy to be non trivial.

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OK Mike Usher was quicker and clearer I guess... –  Noz Jan 3 '11 at 22:18
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Also there are C.Y 3-folds with this property constructed via toric geometry (I think due to Batyrev)

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