2

What is an example of an orientable compact $2n$ dimensional manifold $M$ whose all even dimensional De Rham cohomology groups $H_{\mathrm{DeR}}^{2i}(M)$ are nonzero, but $M$ does not admit any symplectic structure?

Added: As it is indicated in the comments, this post is not a duplicated post.

Ali Taghavi
  • 312
  • 8
    $S^2\times S^4$? $3\mathbb{CP}^2$? – Marco Golla Oct 13 '18 at 21:42
  • 1
    Just to mention that examples can't be surfaces, as mentioned in https://math.stackexchange.com/questions/2116869/does-torus-admit-symplectic-structure/2954354#2954354 – YCor Oct 13 '18 at 21:43
  • 4
    The linked question (claimed duplicate) does not refer to the nonvanishing of the even De Rham cohomology, and precisely the answer to this question is this nonvanishing. So the answer to the linked question does not answer the current question, so it's not a duplicate. – YCor Oct 13 '18 at 22:18
  • 4
    p12 of https://www.math.tecnico.ulisboa.pt/~xvi-iwgp/talks/ACannas.pdf, it is mentioned that the 4-manifold $(S^2\times S^2)# (S^2\times S^2)$ admits no almost complex (and hence no symplectic) structure. Its $b_2$ is clearly nonzero, and hence it satisfies the condition on even Betti numbers. – YCor Oct 13 '18 at 22:25
  • @YCor Thank you very much for your 3 comments and your edit. – Ali Taghavi Oct 15 '18 at 21:12
  • Maybe a good answer would provide an example for each even $n\ge 2$. – YCor Oct 15 '18 at 21:22
  • 1
    @ChrisGerig This gives one example in each dimension multiple of 4, right? You probably know examples in dimension $4n+2\ge 6$ as well? – YCor Oct 15 '18 at 21:52
  • @YCor $S^2 \times S^4 \times \cdots \times S^4$ – mme Oct 15 '18 at 22:12
  • 1
    Similarly $S^2 \times S^2$ times enough factors of $S^4$ gets every doubly even dimension past $4$. – mme Oct 15 '18 at 22:17
  • 1
    @MikeMiller thanks; what's its obstruction to admit a symplectic stucture (and, if it has no almost complex structure, what's the obstruction?) – YCor Oct 15 '18 at 22:28
  • 1
    @YCor In the first every closed 2-form has $\omega^2 = 0$ in cohomology. In the second, $\omega^3$. But the top power of $\omega$ should be a volume form, by definition of symplectic form. ACS is out of my pay grade for a quick comment. – mme Oct 15 '18 at 22:29

1 Answers1

6

For your $n=2k$, $\mathbb{C} P^n\#\mathbb{C} P^n$ does not even admit an almost complex structure, so it cannot be symplectic.

See also: 1) Goertsches-Konstantis' paper "Almost complex structures on connected sums of complex projective spaces" which answers the following MO question
2) Does $\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}$ ever support an almost complex structure?

Including Miller's comment for the remaining (your $n=2k+1$) dimensions: $S^2\times(S^4\times\cdots\times S^4)$

Chris Gerig
  • 17,130