Do there exist closed symplectic manifolds with Euler characteristic zero? - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T21:05:58Zhttp://mathoverflow.net/feeds/question/51047http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/51047/do-there-exist-closed-symplectic-manifolds-with-euler-characteristic-zeroDo there exist closed symplectic manifolds with Euler characteristic zero?Mark Grant2011-01-03T19:11:48Z2011-01-09T19:14:20Z
<p>By <em>symplectic manifold</em> 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)?</p>
<p>In case the answer to the question in the title is a quick "yes", I have several follow up questions:</p>
<ol>
<li>What about if we restrict to the case $n=2$ (in which case $M$ would have to be non-simply-connected)? </li>
<li>What about if we restrict to closed symplectically aspherical manifolds? [Recall that $(M^{2n},\omega)$ is called <em>symplectically aspherical</em> if the symplectic class <code>$[\omega]\in H^2(M;\mathbb{R})\cong \mathrm{Hom}(H_2(M),\mathbb{R})$</code> vanishes on the image of the Hurewicz homomorphism <code>$h\colon \pi_2(M)\to H_2(M)$</code>.]</li>
</ol>
<p>Thanks.</p>
http://mathoverflow.net/questions/51047/do-there-exist-closed-symplectic-manifolds-with-euler-characteristic-zero/51048#51048Answer by Ben Webster for Do there exist closed symplectic manifolds with Euler characteristic zero?Ben Webster2011-01-03T19:18:29Z2011-01-03T19:18:29Z<p>Yes. $T^2 \times T^2$ with the sum of the volume forms on each factor.</p>
http://mathoverflow.net/questions/51047/do-there-exist-closed-symplectic-manifolds-with-euler-characteristic-zero/51070#51070Answer by Noz for Do there exist closed symplectic manifolds with Euler characteristic zero?Noz2011-01-03T22:15:36Z2011-01-03T22:15:36Z<p>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: <a href="http://www.zentralblatt-math.org/zmath/en/search/?q=an:0324.53031&format=complete" rel="nofollow">http://www.zentralblatt-math.org/zmath/en/search/?q=an:0324.53031&format=complete</a> for a proof (I couldn't find an on-line version of the paper, however the construction is outlined there: <a href="http://arxiv.org/abs/1001.0132" rel="nofollow">http://arxiv.org/abs/1001.0132</a>).</p>
<p>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).</p>
<p>Those manifolds are non-trivial fibrations over the torus if you take the monodromy to be non trivial.</p>
http://mathoverflow.net/questions/51047/do-there-exist-closed-symplectic-manifolds-with-euler-characteristic-zero/51574#51574Answer by Mohammad F.Tehrani for Do there exist closed symplectic manifolds with Euler characteristic zero?Mohammad F.Tehrani2011-01-09T19:14:20Z2011-01-09T19:14:20Z<p>Also there are C.Y 3-folds with this property constructed via toric geometry (I think due to Batyrev)</p>