So I just learnt that $S^{2n}$ for n>1 doesn't have a symplectic structure. I saw the proof uses that the 2nd deRahm cohomology group of higher even dim spheres is 0.
But I have been asked to show that $S^{2n+1}$ has a contact structure.
Questions :
How do I provide a contact structure?
Why is it that the cohomology doesn't matter in the contact case and does this in a way say that symplectic geometry is more dependent on topology than contact geometry is?
Take $S^{2n+1}\subseteq\mathbb R^{2n+2}=\mathbb C^{n+1}$ as the unit sphere. For any $p\in S^{2n+1}$ consider the complex hyperplane $\pi_p$ orthogonal to $p$ and passing through $p$. It is a real $2n$ plane in the real tangent plane $T_p(S^{2n+1})$. The distribution of hyperplanes given by the $\pi_p$ provides a contract structure on $S^{2n+1}$.
As for question 2 I don't have a complete answer, but note that the non integrability condition on contact forms tells you that those forms in particular are not closed, hence do not represent cohomology classes.
(1) One standard construction uses the usual embedding $$i: S^{2 n + 1} \hookrightarrow \Bbb R^{2 n + 2} \cong \Bbb C^{n + 1}$$ of the $(2n + 1)$-sphere as the (real) unit sphere with respect to the usual inner product.
If we denote the standard complex structure on $\Bbb C$ by $$J : T \Bbb C^{n + 1} \to T \Bbb C^{n + 1} ,$$ then $$\boxed{H := T S^{2 n + 1} \cap J (TS^{2 n + 1})}$$ is a contact structure on $S^{2 n + 1}$.
To see this, notice that (for every $x \in S^{2 n + 1}$), $T_x S^{2 n + 1} = \operatorname{span}(x)^\perp$, so (since $J$ preserves the inner product) $$H_x := \operatorname{span}(x)^\perp \cap \operatorname{span}(J x)^\perp ,$$ or just as well, $$H_x = \{v \in T_x S^{2 n + 1} : v \in (\operatorname{span}(J x)^\perp)\} = \{v \in T_x S^{2 n + 1} : v \in \ker (Jx)^\flat\} = \ker i^* (Jx)^\flat. $$ So, it suffices to show that $\theta := i^* (Jx)^\flat$ is a contact form. But computing gives that in the standard coordinates on $\Bbb R^{2 n + 2}$, $$(J x)^\flat = -x^2 \,dx^1 + x^1 \,dx^2 - \cdots - x^{2 n + 2} \,dx^{2 n + 1} + x^{2 n + 1} \,dx^{2 n + 2}$$ and then that $$\theta \wedge d\theta \wedge \cdots \wedge d\theta$$ is a nonzero multiple of the standard volume form on $S^{2 n + 1}$; since volume forms vanish nowhere, $\theta$ is a contact form.
Remark Since $J\vert_H$ is thus a complex structure on $H$, $$(S^{2 n + 1}, H, J\vert_H)$$ is an example (in fact, the canonical example) of a CR structure.
(2) We shouldn't expect de Rham cohomology alone to say much about contact geometry in the way it does about symplectic geometry: (The proof you read was presumably that) a symplectic form $\omega$ on a $2m$-manifold $M$ is by definition closed and nondegenerate, that is, $\omega^m$ is a volume form, so $[\omega]^m = [\omega^m] \in H^{2m}(M)$ in nonzero and hence so is $[\omega] \in H^2(M)$. On the other hand, a contact form on a $(2 n + 1)$-manifold ($n \geq 1$) is not even closed (since for any local contact form $\eta$, $\eta \wedge (d\eta)^n$ vanishes nowhere), so it does not represent an element of $H^1(M)$.
This second part of the question is mostly outside my wheelhouse, but one can certainly say some interesting things about the interplay between topology and contact geometry: For example, Martinet proved that every compact, oriented $3$-manifold $M$ admits a contact structure (in particular irrespective of what $H^1(M)$ is), but the analogous statement is false in every dimension $2 n + 1 \geq 5$, and indeed, one can write down explicit topological obstructions to the existence of contact strutures. See this MathOverflow question for some more examples.
J. Martinet, "Formes de contact sur les variétés de dimension $3$". Proceedings of Liverpool Singularities Symposium, II (1969-1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.
