Cohomology & De Rham Theory
Cohomology & De Rham Theory
Cohomology $H^n(X)$ is the dual of homology, assigning groups via cochains $C^n=\text{Hom}(C_n,\mathbb{Z})$ and coboundary maps $\delta^n$. De Rham cohomology $H^n_{\text{dR}}(M)$ uses differential forms on smooth manifolds: $H^n_{\text{dR}}=\ker d^n/\text{Im}\,d^{n-1}$ where $d$ is the exterior derivative. De Rham's theorem equates $H^n_{\text{dR}}(M;\mathbb{R})\cong H^n(M;\mathbb{R})$, linking differential geometry to topology.
Differential Forms & De Rham Cohomology
A $k$-form $\omega$ on a smooth manifold $M$ is a smooth section of $\Lambda^k T^*M$. The exterior derivative $d:\Omega^k\to\Omega^{k+1}$ satisfies $d^2=0$: $\omega$ is closed if $d\omega=0$; exact if $\omega=d\eta$. $H^k_{\text{dR}}=\{$closed$\}/\{$exact$\}$. Examples: $H^0_{\text{dR}}(M)=\mathbb{R}^c$ (one $\mathbb{R}$ per component); $H^1_{\text{dR}}(S^1)=\mathbb{R}$ (1-form $d\theta$ is closed, not exact on $S^1$); $H^n_{\text{dR}}(S^n)=\mathbb{R}$.
Stokes' Theorem (Generalized)
For a compact oriented $n$-manifold-with-boundary $M$ and $(n-1)$-form $\omega$: $$\int_M d\omega=\int_{\partial M}\omega.$$ This unifies: Green's theorem ($n=2$); Stokes' curl theorem ($n=3$, $M$ a surface); Gauss' divergence theorem ($n=3$, $M$ a region). Topological consequence: if $\omega$ is exact ($\omega=d\eta$) then $\int_{\partial M}\omega=\int_M d^2\eta=0$ (since $d^2=0$). This is the origin of the 'closed implies exact on contractible domains' (Poincaré lemma).
Example 1
Show $d\theta=(-y\,dx+x\,dy)/(x^2+y^2)$ is closed but not exact on $\mathbb{R}^2\setminus\{0\}$.
Solution: $d(d\theta)=0$ by $d^2=0$: direct computation also gives 0 (or note $d\theta$ is a pullback of $d\theta$ on $S^1$, which is closed). Not exact: if $d\theta=df$ for some $f:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}$, then $\int_{S^1}d\theta=\int_{S^1}df=0$ (exact form integrates to 0 on closed curve). But $\int_{S^1}d\theta=2\pi\neq 0$. Hence $d\theta$ is not exact, representing a non-trivial class in $H^1_{\text{dR}}(\mathbb{R}^2\setminus\{0\})\cong\mathbb{R}$.
Example 2
Compute $H^*_{\text{dR}}(T^2)$ using the Künneth formula.
Solution: De Rham Künneth: $H^*_{\text{dR}}(X\times Y)\cong H^*_{\text{dR}}(X)\otimes H^*_{\text{dR}}(Y)$ (for compact manifolds). $T^2=S^1\times S^1$: $H^0=\mathbb{R}$ (connected); $H^1=\mathbb{R}\otimes\mathbb{R}\oplus\mathbb{R}\otimes\mathbb{R}=\mathbb{R}^2$ (two 1-forms $d\theta_1$ and $d\theta_2$); $H^2=\mathbb{R}^1$ (the area form $d\theta_1\wedge d\theta_2$). $\chi=1-2+1=0$. ✓
Practice
- Prove the Poincaré lemma: every closed form on a contractible manifold is exact.
- Use Stokes' theorem to derive Gauss's divergence theorem.
- Show $H^n_{\text{dR}}(S^n)\cong\mathbb{R}$ using Mayer-Vietoris.
- State Poincaré duality: $H^k_{\text{dR}}(M)\cong H^{n-k}_{\text{dR}}(M)$ for a compact oriented $n$-manifold, and verify it for $S^n$ and $T^2$.
Show Answer Key
1. A closed form $\omega$ ($d\omega=0$) on a contractible manifold: the contraction $H:M\times[0,1]\to M$ with $H(x,0)=x_0$, $H(x,1)=x$ gives a homotopy operator $K$ (via integration over the fiber) such that $\omega=d(K\omega)$. So $\omega$ is exact.
2. Stokes: $\int_M d\omega=\int_{\partial M}\omega$. For a region $V\subset\mathbb{R}^3$ with $\omega=F_1\,dy\wedge dz+F_2\,dz\wedge dx+F_3\,dx\wedge dy$: $d\omega=(\nabla\cdot\mathbf{F})\,dx\wedge dy\wedge dz$. So $\int_V\nabla\cdot\mathbf{F}\,dV=\int_{\partial V}\mathbf{F}\cdot d\mathbf{S}$.
3. Mayer-Vietoris with $U,V$ covering $S^n$ (open hemispheres): $U\cap V\simeq S^{n-1}$, $U\simeq V\simeq\mathbb{R}^n$. The sequence gives $H^k_{dR}(S^n)\cong H^{k-1}_{dR}(S^{n-1})$ for $k\ge2$. By induction from $H^0_{dR}(S^0)=\mathbb{R}^2$: $H^n_{dR}(S^n)\cong\mathbb{R}$.
4. Poincaré duality: for a compact oriented $n$-manifold $M$, $H^k_{dR}(M)\cong H^{n-k}_{dR}(M)$ via the wedge product and integration. For $S^n$: $H^0\cong H^n\cong\mathbb{R}$, all others 0 — consistent. For $T^2$: $H^0\cong H^2\cong\mathbb{R}$, $H^1\cong\mathbb{R}^2$ — self-dual at $k=1$. ✓