6.2 de Rham cohomology groups

6.2.1 Definitions

Definition 6.1

Let $M$ be an $m$ -dimensional differentiable manifold. The set of closed $r$ -forms is called the $r$ th cocycle group, denoted $Z^r(M)$ . The set of exact $r$ -forms is called the $r$ th coboundary group, denoted $B^r(M)$ . These are vector spaces with $\mathbb{R}$ -coefficients. It follows from ${\mathrm{d}}^2=0$ that $Z^r(M)\supset B^r(M)$ .

(a) if $\omega \in Z^r(M)$ and $\psi \in Z^s(M)$ , then $\omega \wedge \psi \in Z^{r+s}(M)$ ;
(b) if $\omega \in Z^r(M)$ and $\psi \in B^s(M)$ , then $\omega \wedge \psi \in B^{r+s}(M)$ ; and

(c) if $\omega \in B^r(M)$ and $\psi \in B^s(M)$ , then $\omega \wedge \psi \in B^{r+s}(M)$ .

Definition 6.2

The $r$ th de Rham cohomology group is defined by

$$H^r(M;\mathbb{R})\equiv Z^r(M)/B^r(M).\text{\hspace{1em}(6.11)}$$

If $r\le -1$ or $r\ge m+1$ , $H^r(M;\mathbb{R})$ may be defined to be trivial. In the following, we omit the explicit quotation of $\mathbb{R}$ -coefficients.

Let $\omega \in Z^r(M)$ . Then $[\omega ]\in H^r(M)$ is the equivalence class $\{{\omega }^{\prime }\in Z^r(M)|{\omega }^{\prime }=\omega +\mathrm{d}\psi ,\psi \in {\Omega }^{r-1}(M)\}$ . Two forms which differ by an exact form are called cohomologous. We will see later that $H^r(M)$ is isomorphic to $H_r(M)$ . The following examples will clarify the idea of de Rham cohomology groups.

Example 6.1

When $r=0$ , $B^0(M)$ has no meaning since there is no $(-1)$ -form. We define ${\Omega }^{-1}(M)$ to be empty, hence $B^0(M)=0$ . Then $H^0(M)=Z^0(M)=\{f\in {\Omega }^0(M)=\mathcal{F}(M)|\mathrm{d}f=0\}$ . If $M$ is connected, the condition $\mathrm{d}f=0$ is satisfied if and only if $f$ is constant over $M$ . Hence, $H^0(M)$ is isomorphic to the vector space $\mathbb{R}$ ,

$$H^0(M)\cong \mathbb{R}.\text{\hspace{1em}(6.12)}$$

If $M$ has $n$ connected components, $\mathrm{df}=0$ is satisfied if and only if $f$ is constant on each connected component, hence it is specified by $n$ real numbers,

$$H^0(M)\cong \underset{n}{\underbrace{\mathbb{R}\oplus \mathbb{R}\oplus \cdots \oplus \mathbb{R}}}.\text{\hspace{1em}(6.13)}$$

Example 6.2

Let $M=\mathbb{R}$ . From example 6.1, we have $H^0(\mathbb{R})=\mathbb{R}$ . Let us find $H^1(\mathbb{R})$ next. Let $x$ be a coordinate of $\mathbb{R}$ . Since $\dim \mathbb{R}=1$ , any one-form $\omega \in {\Omega }^1(\mathbb{R})$ is closed, $\mathrm{d}\omega =0$ . Let $\omega =f\mathrm{d}x$ , where $f\in \mathcal{F}(\mathbb{R})$ . Define a function $F(x)$ by

$$F(x)={\int }_0^{x}f(s)\mathrm{d}s\in \mathcal{F}(\mathbb{R})={\Omega }^0(\mathbb{R}).$$

Since $\mathrm{d}F(x)/\mathrm{d}x=f(x)$ , $\omega$ is an exact form,

$$\omega =f\mathrm{d}x=\frac{\mathrm{d}F(x)}{\mathrm{d}x}\mathrm{d}x=\mathrm{d}F.$$

Thus, any one-form is closed as well as exact. We have established

$$H^1(\mathbb{R})=\{0\}.\text{\hspace{1em}(6.14)}$$

Example 6.3

Let $S^1=\{{\mathrm{e}}^{\mathrm{i}\theta }|0\le \theta <2\pi \}$ . Since $S^1$ is connected, we have $H^0(S^1)=\mathbb{R}$ . We compute $H^1(S^1)$ next. Let $\omega =f(\theta )\mathrm{d}\theta \in {\Omega }^1(S^1)$ . Is it

possible to write $\omega =\mathrm{d}F$ for some $F\in \mathcal{F}(S^1)$ ? Let us repeat the analysis of the previous example. If $\omega =\mathrm{d}F$ , then $F\in \mathcal{F}(S^1)$ must be given by

$$F(\theta )={\int }_0^{\theta }f\left({\theta }^{\prime }\right)d{\theta }^{\prime }.$$

For $F$ to be defined uniquely on $S^1$ , $F$ must satisfy the periodicity $F(2\pi )=F(0)(=0)$ . Namely $F$ must satisfy

$$F(2\pi )={\int }_0^{2\pi }f({\theta }^{\prime })\mathrm{d}{\theta }^{\prime }=0.$$

If we define a map $\lambda :{\Omega }^1(S^1)\to \mathbb{R}$ by

$$\lambda :\omega =fd\theta \mapsto {\int }_0^{2\pi }f\left({\theta }^{\prime }\right)d{\theta }^{\prime }\text{\hspace{1em}(6.15)}$$

then $B^1(S^1)$ is identified with $\ker \lambda$ . Now we have (theorem 3.1)

$$H^1\left(S^1\right)={\Omega }^1\left(S^1\right)/\ker \lambda =\mathrm{im}\lambda =\mathbb{R}.\text{\hspace{1em}(6.16)}$$

This is also obtained from the following consideration. Let $\omega$ and ${\omega }^{\prime }$ be closed forms that are not exact. Although $\omega -{\omega }^{\prime }$ is not exact in general, we can show that there exists a number $a\in \mathbb{R}$ such that ${\omega }^{\prime }-a\omega$ is exact. In fact, if we put

$$a={\int }_0^{2\pi }{\omega }^{\prime }/{\int }_0^{2\pi }\omega$$

we have

$${\int }_0^{2\pi }({\omega }^{\prime }-a\omega )=0.$$

This shows that, given a closed form $\omega$ which is not exact, any closed form ${\omega }^{\prime }$ is cohomologous to $a\omega$ for some $a\in \mathbb{R}$ . Thus, each cohomology class is specified by a real number $a$ , hence $H^1(S^1)=\mathbb{R}$ .

Exercise 6.3

Let $M={\mathbb{R}}^2-\{0\}$ . Define a one-form $\omega$ by

$$\omega =\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy.\text{\hspace{1em}(6.17)}$$

(a) Show that $\omega$ is closed.
(b) Define a ‘function’ $F(x,y)={\tan }^{-1}(y/x)$ . Show that $\omega =\mathrm{d}F$ . Is $\omega$ exact?

6.2.2 Duality of $H_r(M)$ and $H^r(M)$ ; de Rham’s theorem

As the name itself suggests, the cohomology group is a dual space of the homology group. The duality is provided by Stokes’ theorem. We first define the inner product of an $r$ -form and an $r$ -chain in $M$ . Let $M$ be an $m$ -dimensional manifold and let $C_r(M)$ be the chain group of $M$ . Take $c\in C_r(M)$ and $\omega \in {\Omega }^r(M)$ where $1\le r\le m$ . Define an inner product $(\cdot ,\cdot ):C_r(M)\times {\Omega }^r(M)\to \mathbb{R}$ by

$$c,\omega \mapsto (c,\omega )\equiv {\int }_c\omega .\text{\hspace{1em}(6.18)}$$

Clearly, $(c,\omega )$ is linear in both $c$ and $\omega$ and $(\omega ,\omega )$ may be regarded as a linear map acting on $c$ and vice versa,

$$\left(c_1+c_2,\omega \right)={\int }_{c_1+c_2}\omega ={\int }_{c_1}\omega +{\int }_{c_2}\omega \text{\hspace{1em}(6.19a)}$$

$$(c,{\omega }_1+{\omega }_2)={\int }_c({\omega }_1+{\omega }_2)={\int }_c{\omega }_1+{\int }_c{\omega }_2.\text{\hspace{1em}(6.19b)}$$

Now Stokes’ theorem takes a compact form:

$$(c,\mathrm{d}\omega )=(\partial c,\omega ).\text{\hspace{1em}(6.20)}$$

In this sense, the exterior derivative operator $\mathfrak{d}$ is the adjoint of the boundary operator $\partial$ and vice versa.

Exercise 6.4

Let (i) $c\in B_r(M)$ , $\omega \in Z^r(M)$ or (ii) $c\in Z_r(M)$ , $\omega \in B^r(M)$ . Show, in both cases, that $(c,\omega )=0$ .

The inner product $(\cdot ,\cdot )$ naturally induces an inner product $\lambda$ between the elements of $H_r(M)$ and $H^r(M)$ . We now show that $H_r(M)$ is the dual of $H^r(M)$ . Let $[c]\in H_r(M)$ and $[\omega ]\in H^r(M)$ and define an inner product $\Lambda :H_r(M)\times H^r(M)\to \mathbb{R}$ by

$$\Lambda ([c],[\omega ])\equiv (c,\omega )={\int }_c\omega .\text{\hspace{1em}(6.21)}$$

This is well defined since (6.21) is independent of the choice of the representatives. In fact, if we take $c+\partial c^{\prime }$ , $c^{\prime }\in C_{r+1}(M)$ , we have, from Stokes’ theorem,

$$(c+\partial c^{\prime },\omega )=(c,\omega )+\left(c^{\prime },\mathrm{d}\omega \right)=(c,\omega )$$

where $\mathrm{d}\omega =0$ has been used. Similarly, for $\omega +\mathrm{d}\psi ,\psi \in {\Omega }^{r-1}(M)$

$$(c,\omega +d\psi )=(c,\omega )+(\partial c,\psi )=(c,\omega )$$

since $\partial c=0$ . Note that $\Lambda (\omega ,[c])$ is a linear map $H_r(M)\to \mathbb{R}$ , and $\Lambda ([c],\omega )$ is a linear map $H^r(M)\to \mathbb{R}$ . To prove the duality of $H_r(M)$ and $H^r(M)$ , we have

to show that $\Lambda (\mathbf{\Omega },[\omega ])$ has the maximal rank, that is, $\dim H_r(M)=\dim H^r(M)$ . We accept the following theorem due to de Rham without the proof which is highly non-trivial.

Theorem 6.2 · de Rham's theorem

If $M$ is a compact manifold, $H_r(M)$ and $H^r(M)$ are finite dimensional. Moreover the map

$$\Lambda :H_r(M)\times H^r(M)\to \mathbb{R}$$

is bilinear and non-degenerate. Thus, $H^r(M)$ is the dual vector space of $H_r(M)$ .

A period of a closed $r$ -form $\omega$ over a cycle $c$ is defined by $(c,\omega )={\int }_c\omega$ . Exercise 6.4 shows that the period vanishes if $\omega$ is exact or if $c$ is a boundary. The following corollary is easily derived from de Rham’s theorem.

Corollary 6.1

Let $M$ be a compact manifold and let $k$ be the $r$ th Betti number (see section 3.4). Let $c_1,c_2,\dots ,c_k$ be properly chosen elements of $Z_r(M)$ such that $[c_i]\ne [c_j]$ .

(a) A closed $r$ -form $\psi$ is exact if and only if

$${\int }_{c_i}\psi =0(1\le i\le k).\text{\hspace{1em}(6.22)}$$

(b) For any set of real numbers $b_1,b_2,\dots ,b_k$ there exists a closed $r$ -form $\omega$ such that

$${\int }_{c_i}\omega =b_i(1\le i\le k).\text{\hspace{1em}(6.23)}$$

Proof

(a) de Rham’s theorem states that the bilinear form $\Lambda ([c],[\omega ])$ is nondegenerate. Hence, if $\Lambda ([c_i],\dots )$ is regarded as a linear map acting on $H^r(M)$ , the kernel consists of the trivial element, the cohomology class of exact forms. Accordingly, $\psi$ is an exact form.

(b) de Rham’s theorem ensures that corresponding to the homology basis $\{[c_i]\}$ , we may choose the dual basis $\{[{\omega }_i]\}$ of $H^r(M)$ such that

$$\Lambda \left(\left[c_i\right],\left[{\omega }_j\right]\right)={\int }_{c_i}{\omega }_j={\delta }_{ij}.\text{\hspace{1em}(6.24)}$$

If we define $\omega \equiv {\sum }_{i=1}^k{b}_i{\omega }_i$ , the closed $r$ -form $\omega$ satisfies

$${\int }_{c_i}\omega =b_i$$

as claimed.

For example, we observe the duality of the following groups.

(a) $H^0(M)\cong H_0(M)\cong \underset{n}{\underbrace{\mathbb{R}\oplus \cdots \oplus \mathbb{R}}}$ if $M$ has $n$ connected components.
(b) $H^1(S^1)\cong H_1(S^1)\cong \mathbb{R}.$

Since $H^r(M)$ is isomorphic to $H_r(M)$ , we find that

$$b^r(M)\equiv \dim H^r(M)=\dim H_r(M)=b_r(M)\text{\hspace{1em}(6.25)}$$

where $b_r(M)$ is the Betti number of $M$ . The Euler characteristic is now written as

$$\chi (M)=\sum _{r=1}^m(-1{)}^r{b}^r(M).\text{\hspace{1em}(6.26)}$$

This is quite an interesting formula; the LHS is purely topological while the RHS is given by an analytic condition (note that $\mathrm{d}\omega =0$ is a set of partial differential equations). We will frequently encounter this interplay between topology and analysis.

In summary, we have the chain complex $C(M)$ and the de Rham complex ${\Omega }^{\ast }(M)$ ,

$$\begin{array}{rl}& \leftarrow C_{r-1}(M)\stackrel{{\partial }_r}{\leftarrow }{C}_r(M)\stackrel{{\partial }_{r+1}}{\leftarrow }{C}_{r+1}(M)\leftarrow \\ & ⟶{\Omega }^{r-1}(M)\stackrel{{\mathrm{d}}_r}{\to }{\Omega }^r(M)\stackrel{{\mathrm{d}}_{r+1}}{\to }{\Omega }^{r+1}(M)⟵\end{array}\text{\hspace{1em}(6.27)}$$

for which the $r$ th homology group is defined by

$$H_r(M)=Z_r(M)/B_r(M)=\ker {\partial }_r/\mathrm{im}{\partial }_{r+1}$$

and the $r$ th de Rham cohomology group is defined by

$$H^r(M)=Z^r(M)/B^r(M)=\ker {\mathrm{d}}_{r+1}/\mathrm{im}{\mathrm{d}}_r.$$

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