15

Reading some books on diferential geometry, a found that $S^{2n}$ (with $ n>1$) are not symplectic manifolds. They say it's because the de Rham cohomology of this spheres are R, but I do not understand this argument. It will be helpfull if anyone can explain me this fact (a silly explanation)...thx.

happymath
  • 6,297
Edgar Ortiz
  • 373
  • 2
    The second de Rham cohomology class of a closed symplectic manifold is non-trivial, and the only (even-dimensional) sphere that has this property is $ \mathbb{S}^{2} $. – Berrick Caleb Fillmore Apr 20 '15 at 14:50
  • I agree, in this case the second cohomology class is isomorphic to R; but how this fact implies s^2n (n>1) are not symplectic?? – Edgar Ortiz Apr 20 '15 at 14:53
  • 1
    $H^2(S^{2n}) = 0$ for $n > 1$. – anomaly Apr 20 '15 at 14:55
  • Use a Mayer-Vietoris sequence to prove that $ {H_{k}}(\mathbb{S}^{2 n}) = 0 $ for $ 0 < k < 2 n $. By Poincaré Duality, $ {H^{k}}(\mathbb{S}^{2 n}) = 0 $ for $ 0 < k < 2 n $. Hence, if $ n \geq 2 $, then $ {H^{2}}(\mathbb{S}^{2 n}) = 0 $. – Berrick Caleb Fillmore Apr 20 '15 at 15:00
  • @EdgarOrtiz: Hi Edgar. If you’re wondering why $ {H^{2}_{\text{dR}}}(\mathbb{S}^{2 n}) \cong {H^{2}}(\mathbb{S}^{2 n}) = 0 $ for $ n \geq 2 $, then please see the explanation above this comment. – Berrick Caleb Fillmore Apr 20 '15 at 15:11

2 Answers2

25

Compact symplectic manifolds must have nontrivial $H^2$.

A symplectic manifold $M^{2n}$, by definition, possesses a closed, nondegenerate 2-form $\omega$. Because $\omega$ is closed (i.e., $d\omega = 0$), $\omega$ represents a de Rham cohomology class $[\omega] \in H^2(M)$. The nondegeneracy condition implies $\omega^n$ (the wedge product of $\omega$ with itself $n$ times) is a volume form. This means that if $M$ is compact, $\int_M \omega^n$ is nonzero, and $[\omega]^n$ is nonzero in $H^{2n}(M)$, meaning $[\omega]$ must be nonzero in $H^2(M)$. In particular, $H^2(M)$ must be nonzero, which is not the case for $S^{2n}$ for $n>1$. ($H^\ast (S^k) = \mathbb{R} \text{ if } \ast = 0 \text{ or } k$ and is zero otherwise.)

Phillip Andreae
  • 4,230
  • Is there no other direct way except by going into de Rham cohomology? – Evangeline A. K. McDowell Jan 31 '17 at 04:48
  • 3
    @eri - Interesting question. I'm not an expert on symplectic geometry, so I don't know. Any "obstruction" to admitting a symplectic structure must be global as opposed to local, since local symplectic structures (say, in a coordinate patch in $\mathbb{R}^{2n}$) always exist. The only one of these global obstructions that I know about is the cohomological condition in my answer, but there are others. This MO question seems relevant: http://mathoverflow.net/questions/122119/what-prevents-a-manifold-to-be-symplectic – Phillip Andreae Jan 31 '17 at 13:24
  • "The nondegeneracy condition implies $\omega^n$ (the wedge product of $\omega$ with itself $n$ times) is a volume form." Why could you explain this fact, please? – Gordhob Brain Aug 24 '23 at 04:57
19

Suppose $(X, \omega)$ is a closed symplectic manifold of dimension $2n$. If $\omega = d\alpha$, then $$\int_X\omega^n = \int_X d(\alpha\wedge \omega^{n-1}) = \int_{\partial X}\alpha\wedge \omega^{n-1} = 0$$ but this is absurd as $\omega^n$ is a volume form. Therefore, a symplectic form on a closed manifold is not exact, so it defines a non-zero element of $H^2_{\text{dR}}(X)$. As $H^2_{\text{dR}}(S^{2n}) = 0$ for $n > 1$, $S^{2n}$ is not symplectic.

A similar argument can be used to show that $\omega^k$ is not exact for $k = 1, \dots, n$, so in fact $H^{2k}_{\text{dR}}(X) \neq 0$ for a closed symplectic manifold $X$.

Michael Albanese
  • 104,029