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.

Definition

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

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:

Examples

1. 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 $k>0$.

2. 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.

3. 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 $0<k<n$.