Timeline for Do there exist closed symplectic manifolds with Euler characteristic zero?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 9, 2011 at 19:21 | comment | added | Patrick I-Z | To Mike: it remains me a paper I wrote long time ago "Symplectisation des variétés de contact" math.huji.ac.il/~piz/Site/The%20Articles/… | |
| Jan 4, 2011 at 3:38 | comment | added | Paul | 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? | |
| Jan 3, 2011 at 22:16 | comment | added | Mike Usher | 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. | |
| Jan 3, 2011 at 22:16 | comment | added | Mike Usher | 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. ... | |
| Jan 3, 2011 at 22:04 | vote | accept | Mark Grant | ||
| Jan 3, 2011 at 21:57 | comment | added | Mark Grant | Thanks, guys. I forgot about tori...shows how often I think about these things. Do you know of examples which are non-trivial fibrations over $T^2$? | |
| Jan 3, 2011 at 19:30 | comment | added | Mariano Suárez-Álvarez | Ah, right. @Mark: you can take $T^2\times\mathrm{your favorite closed symplectic manifold}$, in fact. | |
| Jan 3, 2011 at 19:26 | comment | added | Ben Webster♦ | He asked for a 4-dimensional example. But otherwise T^2 is fine. | |
| Jan 3, 2011 at 19:23 | comment | added | Mariano Suárez-Álvarez | Why not just $T^2$? | |
| Jan 3, 2011 at 19:18 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |