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I started learning symplectic geometry, but apparently, i forgot some of my differential geometry. I am studying from the book by Aebischer et al.

In it, it is claimed that $S^4$ is not a symplectic manifold, and the proof goes as follows: suppose $\omega$ is a symplectic form on $S^4$. Because of vanishing of De-Rham cohomology, $\omega$ must be of the form $\omega = d(\alpha)$ for some 1-form $\alpha$.

But then, the volume form $\Omega = \omega \wedge \omega$ is also exact, with $d(\alpha \wedge \omega) = \Omega$.

Now, by Stokes theorem,

$\int_{S^4} \Omega = \int_{\partial S^4} \alpha \wedge \omega = 0$.

And the book claims that this is impossible for a volume form.

My two (probably very naive) questions: 1. Why is this intgeral zero? 2. Why can't the integral of a volume form vanish?

guest
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2 Answers2

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More generally, if $(X^{2n}, \omega)$ is a compact symplectic manifold, then $\omega^n$, as a volume form, represents a generator of top cohomology $H^{2n}(X, \mathbb{R})$, from which it follows that the image of $\omega$ in $H^2(X, \mathbb{R})$ must have the property that its $n^{th}$ cup power can't vanish; in particular, all of the even-dimensional cohomology groups $H^{2k}(X, \mathbb{R})$ contain at least one nontrivial element $\omega^k$ and hence also can't vanish. Conversely, if $X$ is a compact manifold such that any of the even-dimensional cohomology groups $H^{2k}(X, \mathbb{R})$ vanish, $0 \le k \le n$, then $X$ cannot be a symplectic manifold.

This implies, for example, that $S^2$ is the only sphere that can be a symplectic manifold, and also that, for example, $S^1 \times S^3$ cannot be a symplectic manifold. The stronger condition that there is some element of $H^2(X, \mathbb{R})$ whose $n^{th}$ power is nontrivial also shows that, for example, $S^2 \times S^4$ cannot be a symplectic manifold, despite having nontrivial even-dimensional cohomology groups.

From this perspective, arguably the "minimal" examples of compact symplectic manifolds are the complex projective spaces $\mathbb{CP}^n$, which are symplectic when equipped with the imaginary part of their Fubini-Study forms, and whose cohomologies are in fact generated by the powers of their symplectic forms.

Qiaochu Yuan
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  • Qiaochu, I don't understand the argument for why $S^2\times S^4$ is not symplectic. In this case, I only know $H^2,H^4, H^6$ are isomorphic to $\mathbb{R}$, but how can we conclude it is not symplectic? – rmdmc89 May 09 '18 at 18:47
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    @rmdmc89: the argument in the first paragraph establishes several facts; the weaker fact is that if any of the even cohomology vanishes then the closed manifold can't be symplectic, and the stronger fact is that if no class in $H^2$ has nonvanishing $n^{th}$ cup power then the closed manifold can't be symplectic, and I'm using that stronger fact for $S^2 \times S^4$. – Qiaochu Yuan May 10 '18 at 04:39
  • Oh, I see, so in order to use that fact, you need to verify that if $[\omega]\in H^2(S^2\times S^4)$, then $[\omega^3]=[0]$, right? How do I prove that? – rmdmc89 May 10 '18 at 16:40
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    @mdmc89: every nonzero class in $H^2(S^2 \times S^4)$ is the pullback of a class in $H^2(S^2)$, and the third cup power of a class in $H^2(S^2)$ is a class in $H^6(S^2)$, which vanishes. Alternatively you can use the Kunneth theorem to compute the cohomology ring of $S^2 \times S^4$ as the tensor product of the cohomology rings of $S^2$ and $S^4$. – Qiaochu Yuan May 10 '18 at 18:07
  • one last question, I just want tô make sure I get everything straight : in deed, using Kunneth I can see that $H^2(S^2\times S^4)\simeq H^2(S^2)$, which matches what you said. But how did you know every nonzero class in $H^2(S^2\times S^4)$ is the pullback of a class in $H^2(S^2)$? – rmdmc89 May 11 '18 at 00:12
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    The map you get coming from the Kunneth theorem is taking pullbacks along the product projections. – Qiaochu Yuan May 11 '18 at 00:16
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  1. The boundary of $S^4$ is empty.

  2. By definition, the integral of a volume form is greater than zero.

Najib Idrissi
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  • Can you elaborate on the second issue? why is the integral of a non-vanishing top form greater then zero? – guest Nov 05 '14 at 14:27
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    You can find the answer here. What I said was a bit abusive -- it's always nonzero, and by convention we generally look at volume forms with positive integral (but in this case $-\Omega$ is also a volume form with negative integral). – Najib Idrissi Nov 05 '14 at 14:30
  • Got it, thanks! – guest Nov 05 '14 at 14:33