No symplectic structure on $S^{2n},\ n>1$

Question

I am trying to show that there is no symplectic structure on the $2n$-dimensional sphere $S^{2n}$, where $n>1$. I've tried following these steps:

(a) Given a compact $2n$-dimensional symplectic manifold $M$ with symplectic structure $\omega$, $[\omega^n]\in H^{2n}(M)$ is nontrivial, or in other words $\omega^{n}=\mathrm{d}\alpha$ is never true.

(b) $[\omega]$ itself is nontrivial.

(c) If $n>1$ there is no symplectic structure on the $2n$-dimensional sphere $S^{2n}$ .

My try: (a) Let $\omega^{n}=\mathrm{d}\alpha$, then by Stokes Theorem. $$ \int_M\omega^{n}=\int_M\mathrm{d}\alpha=\int_{\partial M}\alpha=\int_{0}\alpha=0, $$ were $\partial M=0$ because $M$ is compact. Then, since for compact manifolds Poincaré Duality Theorem reads: $H^{k}(M) = H^{2n-k}(M)^\ast$, we get $H^{2n}(M) = H^{0}(M)^\ast$ that is the dual to the one-dimensional space of constants ( $f$ such that $\mathrm{d}f=0$).Therefore $dim\ H^{2n}(M) = dim\ H^{0}(M)^\ast=1$. Finally (NOT sure of this passage) $$ \int_M\omega^{2n}=0 \iff \omega^{2n}=0, $$ which is absurd, since a volume form must yield locally nonzero volumes.

(b)$[\omega]$ is nonzero too, thanks to the annulus structure given by wedge product.

(c)We have a general result $$ dim\ H^{2n}(S^{2n}) = dim\ H^{0}(S^{2n})=1$$ $$ dim\ H^{k}(S^{2n})=0 $$ if $k\not=0,\ 2n$. Thus, by comparison with the previous results, the only sphere compatible with the existence of a symplectic form is $S^2$.

First of all did I go wrong somewhere? Could someone explain the point in (a) I'm not sure of? Finally, is there a simple way to grasp that $ dim\ H^{k}(S^{2n})=0 $?

Answer 1

See if the following makes sense.

1. By definition of a symplectic form, for any $2n$-dimensional symplectic manifold, $\omega^n$ is a nowhere vanishing volume form.

2. For any compact manifold, the deRham cohomology class of a nowhere vanishing volume form is non-zero. This you can prove by integration, using the Stokes theorem (no need to use Poincare duality).

3. Conclude from 1+2 that for a compact symplectic manifold, the class $[\omega^n]$ is non-zero. The latter is $[\omega]^n$, by definition of multiplication in deRham cohomology, hence $[\omega]$ is a degree 2 non-zero class.

4. To show that $H^2(S^n)=0$ for $n>2$, you can use for example the Mayer-Vietoris argument (see any standard textbook on deRham cohomology, e.g. Bott-Tu).

Answer 2

Here is my favorite way to compute $H^*(S^k)$ for $k > 1$.

We can realize $S^k$ as a CW-complex with one cell in dimension 0 and one cell in dimension $k$. That is, glue the boundary of the $k$-disk $D^k$ to a point $e^0$ and you get $S^k$.

The cellular chain groups $C^n, n \geq 0$ for this CW-structure on $S^k$ are nonzero only in dimension $0$ and $k$, and in these dimensions $C^0 \cong C^k \cong \mathbb{Z}$ because $C^0$ is generated by $e^0$ and $C^k$ is generated by $D^k$. So since we assumed $k > 1$ we now see that all boundary maps in the cellular chain complex are trivial, hence the nontrivial cohomology of $S^k$ is $H^n(S^k) \cong C^n$ for all $n$. That is, $H^n(S^k) = \mathbb{Z}$ if $n = 0,k$ and $H^n(S^k) = 0$ otherwise.