Decomposition of cohomology group on $S^{n}$

Question

If we have decomposition of cohomology group on $S^{n}$ it looks like $H^{n}(S^{n})=H^{n}(S^{n})_{+}\oplus H^{n}(S^{n})_{-}$, where $H^{n}(S^{n})_{\pm}$ cohomology of invariant or anti-invariant $n$ form on $S^{n}$. Why one of $H^{n}(S^{n})_{\pm}$ are trival in dependence from $n$ odd or even?

Answer 1

As John points out, the answer to your first question has very little to do with cohomology.

As $H^n(S^n) = H^n_+(S^n) \oplus H^n_-(S^n)$, $\dim H^n(S^n) = \dim H^n_+(S^n) + \dim H^n_-(S^n)$ but $H^n(S^n) \cong \mathbb{R}$ as $S^n$ is a compact connected orientable $n$-dimensional manifold, so $\dim H^n(S^n) = 1$ so $\{\dim H^n_+(S^n), \dim H^n_-(S^n)\} = \{0, 1\}$.

As for the dependence on $n$, that does relate to cohomology.

What you need to know is that the antipodal map is orientation-preserving if and only if $n$ is odd (so it is orientation-reversing if and only if $n$ is even); this result is given as exercise $15$-$3$ in Lee's Introduction to Smooth Manifolds, second edition. Another way of stating this is $\dim H^n_+(S^n) = 1$ if and only if $n$ is odd (so $\dim H^n_-(S^n) = 1$ if and only if $n$ is even). The equivalence follows because $H^n(S^n)$ is generated by the cohomology class of an orientation form.