de Rham cohomology group

Let $M$ be a smooth manifold. The exterior derivative makes the graded vector space of differential forms into a cochain complex $(\Omega^\ast(M),d)$ . Its cohomology is the de Rham cohomology.

For $k\ge0$ , using $\Omega^{-1}(M)=0$ , define

$Z^k(M)$ as the space of closed $k$ -forms (those $\omega$ with $d\omega=0$ ),
$B^k(M)$ as the space of exact $k$ -forms (those $\omega$ with $\omega=d\eta$ ).

Because $d^2=0$ , every exact form is closed, so $B^k(M)\subseteq Z^k(M)$ . The $k$ th de Rham cohomology group is the quotient vector space

H^k_{\mathrm{dR}}(M)\coloneqq Z^k(M)\,/\,B^k(M).

An element $[\omega]\in H^k_{\mathrm{dR}}(M)$ is the equivalence class of a closed form $\omega$ , where $\omega\sim\omega'$ if $\omega-\omega'$ is exact.

Functoriality

If $F:M\to N$ is a smooth map, then the pullback of forms sends closed forms to closed forms and exact forms to exact forms (since $F^*$ commutes with $d$ ). Hence $F$ induces a linear map on cohomology:

F^*:H^k_{\mathrm{dR}}(N)\to H^k_{\mathrm{dR}}(M),\qquad [\omega]\mapsto [F^*\omega].

Examples

Euclidean space. For $M=\mathbb{R}^n$ , one has $H^0_{\mathrm{dR}}(\mathbb{R}^n)\cong \mathbb{R}$ and $H^k_{\mathrm{dR}}(\mathbb{R}^n)=0$ for all $k>0$ .

The circle. For $M=S^1$ , one has $H^0_{\mathrm{dR}}(S^1)\cong \mathbb{R}$ and $H^1_{\mathrm{dR}}(S^1)\cong \mathbb{R}$ . A generator of $H^1_{\mathrm{dR}}(S^1)$ can be represented by a closed 1-form whose integral around the circle is nonzero.

The sphere $S^n$ . For $M=S^n$ with $n\ge 1$ , one has $H^0_{\mathrm{dR}}(S^n)\cong \mathbb{R}$ , $H^n_{\mathrm{dR}}(S^n)\cong \mathbb{R}$ , and $H^k_{\mathrm{dR}}(S^n)=0$ for $0<k<n$ .