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

Solution 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).

Solution 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.