de Rham cohomology

Remarks

Since $d \circ d = 0$, every exact form is closed, so the quotient makes sense. $H^0_{\mathrm{dR}}(M)$ is the space of locally constant functions, so when $M$ has finitely many connected components its dimension is their number. Example: on $\R^n$ every closed $k$-form with $k \ge 1$ is exact, by the Poincaré lemma, so $H^k_{\mathrm{dR}}(\R^n) = 0$ for $k \ge 1$ and $H^0_{\mathrm{dR}}(\R^n) = \R$. Non-example of exactness: on $M = \R^2 \setminus \set{0}$ the form $\omega = (x \, dy - y \, dx)/(x^2 + y^2)$ is closed but not exact. Its integral over the unit circle is $2\pi$, while the integral of an exact form $df$ over a closed loop is $0$; so $H^1_{\mathrm{dR}}(\R^2 \setminus \set{0}) \neq 0$, and in fact it is $\R$. A smooth map $f \colon M \to N$ induces linear maps $f^* \colon H^k_{\mathrm{dR}}(N) \to H^k_{\mathrm{dR}}(M)$ by pullback of forms, and the wedge product of forms makes $\bigoplus_k H^k_{\mathrm{dR}}(M)$ a graded ring.