6.4 Structure of de Rham cohomology groups

de Rham cohomology groups exhibit quite an interesting structure that is very difficult or even impossible to appreciate with homology groups.

6.4.1 Poincaré duality

Let $M$ be a compact $m$ -dimensional manifold and let $\omega \in H^r(M)$ and $\eta \in H^{m-r}(M)$ . Noting that $\omega \wedge \eta$ is a volume element, we define an inner product $⟨,⟩:H^r(M)\times H^{m-r}(M)\to \mathbb{R}$ by

$$⟨\omega ,\eta ⟩\equiv {\int }_M\omega \wedge \eta .\text{\hspace{1em}(6.35)}$$

The inner product is bilinear. Moreover, it is non-singular, that is, if $\omega \ne 0$ or $\eta \ne 0$ , $⟨\omega ,\eta ⟩$ cannot vanish identically. Thus, (6.35) defines the duality of $H^r(M)$ and $H^{m-r}(M)$ ,

$$H^r(M)\cong H^{m-r}(M)\text{\hspace{1em}(6.36)}$$

called the Poincaré duality. Accordingly, the Betti numbers have a symmetry

$$b_r=b_{m-r}.\text{\hspace{1em}(6.37)}$$

It follows from (6.37) that the Euler characteristic of an odd-dimensional space vanishes,

$$\begin{array}{rl}\chi (M)& =\sum (-1{)}^r{b}_r=\frac{1}{2}\left\{\sum (-1{)}^r{b}_r+\sum (-1{)}^{m-r}{b}_{m-r}\right\}\\ & =\frac{1}{2}\left\{\sum (-1{)}^r{b}_r-\sum (-1{)}^{-r}{b}_r\right\}=0.\end{array}\text{\hspace{1em}(6.38)}$$

6.4.2 Cohomology rings

Let $[\omega ]\in H^q(M)$ and $[\eta ]\in H^r(M)$ . Define a product of $[\omega ]$ and $[\eta ]$ by

$$[\omega ]\wedge [\eta ]\equiv [\omega \wedge \eta ].\text{\hspace{1em}(6.39)}$$

It follows from exercise 6.2 that $\omega \wedge \eta$ is closed, hence $[\omega \wedge \eta ]$ is an element of $H^{q+r}(M)$ . Moreover, $[\omega \wedge \eta ]$ is independent of the choice of the representatives of $[\omega ]$ and $[\eta ]$ . For example, if we take ${\omega }^{\prime }=\omega +\mathrm{d}\psi$ instead of $\omega$ , we have

$$[{\omega }^{\prime }]\wedge [\eta ]\equiv [(\omega +d\psi )\wedge \eta ]=[\omega \wedge \eta +d(\psi \wedge \eta )]=[\omega \wedge \eta ].$$

Thus, the product $\wedge :H^q(M)\times H^r(M)\to H^{q+r}(M)$ is a well-defined map.

The cohomology ring $H^{\ast }(M)$ is defined by the direct sum,

$$H^{\ast }(M)\equiv \underset{r=1}{\overset{m}{⨁}}{H}^r(M).\text{\hspace{1em}(6.40)}$$

The product is provided by the exterior product defined earlier,

$$\wedge :H^{\ast }(M)\times H^{\ast }(M)\to H^{\ast }(M).\text{\hspace{1em}(6.41)}$$

The addition is the formal sum of two elements of $H^{\ast }(M)$ . One of the superiorities of cohomology groups over homology groups resides here. Products of chains are not well defined and homology groups cannot have a ring structure.

6.4.3 The Kunneth formula

Let $M$ be a product of two manifolds $M=M_1\times M_2$ . Let $\{{\omega }_i^p\}$ ( $1\le i\le b^p(M_1)$ ) be a basis of $H^p(M_1)$ and $\{{\eta }_i^p\}$ ( $1\le i\le b^p(M_2)$ ) be that of $H^p(M_2)$ . Clearly ${\omega }_i^p\wedge {\eta }_j^{r-p}$ ( $1\le p\le r$ ) is a closed $r$ -form in $M$ . We show that it is not exact. If it were exact, it would be written as

$${\omega }_i^p\wedge {\eta }_j^{r-p}=\mathrm{d}\left({\alpha }^{p-1}\wedge {\beta }^{r-p}+{\gamma }^p\wedge {\delta }^{r-p-1}\right)\text{\hspace{1em}(6.42)}$$

for some ${\alpha }^{p-1}\in {\Omega }^{p-1}(M_1)$ , ${\beta }^{r-p}\in {\Omega }^{r-p}(M_2)$ , ${\gamma }^p\in {\Omega }^p(M_1)$ and ${\delta }^{r-p-1}\in {\Omega }^{r-p-1}(M_2)$ . [If $p=0$ , we put ${\alpha }^{p-1}=0$ .] By executing the exterior derivative in (6.42), we have

$$\begin{array}{rl}& {\omega }_i^p\wedge {\eta }_j^{r-p}=d{\alpha }^{p-1}\wedge {\beta }^{r-p}+(-1{)}^{p-1}{\alpha }^{p-1}\wedge d{\beta }^{r-p}\\ & +\mathrm{d}{\gamma }^p\wedge {\delta }^{r-p-1}+(-1{)}^p{\gamma }^p\wedge \mathrm{d}{\delta }^{r-p-1}.\end{array}\text{\hspace{1em}(6.43)}$$

By comparing the LHS with the RHS, we find ${\alpha }^{p-1}={\delta }^{r-p-1}=0$ , hence ${\omega }_i^p\wedge {\eta }_j^{r-p}=0$ in contradiction to our assumption. Thus, ${\omega }_i^p\wedge {\eta }_j^{r-p}$ is a nontrivial element of $H^r(M)$ . Conversely, any element of $H^r(M)$ can be decomposed into a sum of a product of the elements of $H^p(M_1)$ and $H^{r-p}(M_2)$ for $0\le p\le r$ . Now we have obtained the Kunneth formula

$$H^r(M)=\underset{p+q=r}{⨁}[H^p(M_1)\otimes H^q(M_2)].\text{\hspace{1em}(6.44)}$$

This is rewritten in terms of the Betti numbers as

$$b^r(M)=\sum _{p+q=r}{b}^p\left(M_1\right)b^q\left(M_2\right).\text{\hspace{1em}(6.45)}$$

The Kunneth formula also gives a relation between the cohomology rings of the respective manifolds,

$$\begin{array}{rl}{H}^{\ast }(M)& =\sum _{r=1}^m{H}^r(M)=\sum _{r=1}^m\underset{p+q=r}{⨁}{H}^p\left(M_1\right)\otimes H^q\left(M_2\right)\\ & =\sum _p{H}^p\left(M_1\right)\otimes \sum _q{H}^q\left(M_2\right)=H^{\ast }\left(M_1\right)\otimes H^{\ast }\left(M_2\right).\end{array}\text{\hspace{1em}(6.46)}$$

Exercise 6.5

Let $M=M_1\times M_2$ . Show that

$$\chi (M)=\chi \left(M_1\right)\cdot \chi \left(M_2\right).\text{\hspace{1em}(6.47)}$$

Example 6.5

Let $T^2=S^1\times S^1$ be the torus. Since $H^0(S^1)=\mathbb{R}$ and $H^1(S^1)=\mathbb{R}$ , we have

$$H^0\left(T^2\right)=\mathbb{R}\otimes \mathbb{R}=\mathbb{R}\text{\hspace{1em}(6.48a)}$$

$$H^1\left(T^2\right)=(\mathbb{R}\otimes \mathbb{R})\oplus (\mathbb{R}\otimes \mathbb{R})=\mathbb{R}\oplus \mathbb{R}\text{\hspace{1em}(6.48b)}$$

$$H^2\left(T^2\right)=\mathbb{R}\otimes \mathbb{R}=\mathbb{R}.\text{\hspace{1em}(6.48c)}$$

Observe the Poincaré duality $H^0(T^2)=H^2(T^2)$ . [Remark: $\mathbb{R}\otimes \mathbb{R}$ is the tensor product and should not be confused with the direct product. Clearly the product of two real numbers is a real number.] Let us parametrize the coordinate of $T^2$

as $({\theta }_1,{\theta }_2)$ where ${\theta }_i$ is the coordinate of $S^1$ . The groups $H^r(T^2)$ are generated by the following forms:

$$r=0:{\omega }_0=c_0{c}_0\in \mathbb{R}$$

$$r=1:{\omega }_1=c_1\mathrm{d}{\theta }_1+c_1^{\prime }\mathrm{d}{\theta }_2{c}_1,c_1^{\prime }\in \mathbb{R}\text{\hspace{1em}(6.49a)}$$

$$r=2:{\omega }_2=c_2\mathrm{d}{\theta }_1\wedge \mathrm{d}{\theta }_2{c}_2\in \mathbb{R}.$$

Although the one-form $\mathrm{d}{\theta }_i$ looks like an exact form, there is no function ${\theta }_i$ which is defined uniquely on $S^1$ . Since $\chi (S^1)=0$ , we have $\chi (T^2)=0$ .

The de Rham cohomology groups of

$$T^n=\underset{n}{\underbrace{S^1\times \cdots \times S^1}}$$

are obtained similarly. $H^r(T^n)$ is generated by $r$ -forms of the form

$$\mathrm{d}{\theta }^{i_1}\wedge \mathrm{d}{\theta }^{i_2}\wedge \cdots \wedge \mathrm{d}{\theta }^{i_r}\text{\hspace{1em}(6.50)}$$

where $i_1<i_2<\cdots <i_r$ are chosen from $1,\dots ,n$ . Clearly

$$b^r=\dim H^r\left(T^n\right)=\left(\genfrac{}{}{0ex}{}{n}{r}\right).\text{\hspace{1em}(6.51)}$$

The Euler characteristic is directly obtained from (6.51) as

$$\chi \left(T^n\right)=\sum (-1{)}^r(\genfrac{}{}{0ex}{}{n}{r})=(1-1{)}^n=0.\text{\hspace{1em}(6.52)}$$

6.4.4 Pullback of de Rham cohomology groups

Let $f:M\to N$ be a smooth map. Equation (5.75) shows that the pullback $f^{\ast }$ maps closed forms to closed forms and exact forms to exact forms. Accordingly, we may define a pullback of the cohomology groups $f^{\ast }:H^r(N)\to H^r(M)$ by

$$f^{\ast }[\omega ]=\left[f^{\ast }\omega \right][\omega ]\in H^r(N).\text{\hspace{1em}(6.53)}$$

The pullback $f^{\ast }$ preserves the ring structure of $H^{\ast }(N)$ . In fact, if $[\omega ]\in H^p(N)$ and $[\eta ]\in H^q(N)$ , we find

$$\begin{array}{rl}{f}^{\ast }([\omega ]\wedge [\eta ])& =f^{\ast }[\omega \wedge \eta ]=[f^{\ast }(\omega \wedge \eta )]\\ & =\left[f^{\ast }\omega \wedge f^{\ast }\eta \right]=\left[f^{\ast }\omega \right]\wedge \left[f^{\ast }\eta \right].\end{array}\text{\hspace{1em}(6.54)}$$

6.4.5 Homotopy and $H^1(M)$

Let $f,g:M\to N$ be smooth maps. We assume $f$ and $g$ are homotopic to each other, that is, there exists a smooth map $F:M\times I\to N$ such that $F(p,0)=$

$f(p)$ and $F(p,1)=g(p)$ . We now prove that $f^{\ast }:H^r(N)\to H^r(M)$ is equal to $g^{\ast }:H^r(N)\to H^r(M)$ .

Lemma 6.1

Let $f^{\ast }$ and $g^{\ast }$ be defined as before. If $\omega \in {\Omega }^r(N)$ is a closed form, the difference of the pullback images is exact,

$$f^{\ast }\omega -g^{\ast }\omega =\mathrm{d}\psi \psi \in {\Omega }^{r-1}(M).\text{\hspace{1em}(6.55)}$$

Proof

We first note that

$$f=F\circ f_0,g=F\circ f_1$$

where $f_t:M\to M\times I$ ( $p\mapsto (p,t)$ ) has been defined in theorem 6.3. The LHS of (6.55) is

$$\begin{array}{rl}(F\circ f_0{)}^{\ast }\omega -(F\circ f_1{)}^{\ast }\omega & =f_0^{\ast }\circ F^{\ast }\omega -f_1^{\ast }\circ F^{\ast }\omega \\ & =-\left[\mathrm{d}P\left(F^{\ast }\omega \right)+P\mathrm{d}\left(F^{\ast }\omega \right)\right]=-\mathrm{d}P{F}^{\ast }\omega \end{array}$$

where (6.33) has been used. This shows that $f^{\ast }\omega -g^{\ast }\omega =\mathrm{d}(-P{F}^{\ast }\omega )$ .

Now it is easy to see that $f^{\ast }=g^{\ast }$ as the pullback maps $H^r(N)\to H^r(M)$ . In fact, from the previous lemma,

$$[f^{\ast }\omega -g^{\ast }\omega ]=[f^{\ast }\omega ]-[g^{\ast }\omega ]=[\mathrm{d}\psi ]=0.$$

We have established the following theorem.

Theorem 6.4

Let $f,g:M\to N$ be maps which are homotopic to each other. Then the pullback maps $f^{\ast }$ and $g^{\ast }$ of the de Rham cohomology groups $H^r(N)\to H^r(M)$ are identical.

Let $M$ be a simply connected manifold, namely ${\pi }_1(M)\cong \{0\}$ . Since $H_1(M)={\pi }_1(M)$ modulo the commutator subgroup (theorem 4.9), it follows that $H_1(M)$ is also trivial. In terms of the de Rham cohomology group this can be expressed as follows.

Theorem 6.5

Let $M$ be a simply connected manifold. Then its first de Rham cohomology group is trivial.

Proof

Let $\omega$ be a closed one-form on $M$ . It is clear that if $\omega =\mathrm{d}f$ , then a function $f$ must be of the form

$$f(p)={\int }_{p_0}^p\omega \text{\hspace{1em}(6.56)}$$

$p_0\in M$ being a fixed point.

We first prove that an integral of a closed form along a loop vanishes. Let $\alpha :I\to M$ be a loop at $p\in M$ and let $c_p:I\to M$ ( $t\mapsto p$ ) be a constant

loop. Since $M$ is simply connected, there exists a homotopy $F(s,t)$ such that $F(s,0)=\alpha (s)$ and $F(s,1)=c_p(s)$ . We assume $F:I\times I\to M$ is smooth. Define the integral of a one-form $\omega$ over $\alpha (I)$ by

$${\int }_{\alpha (I)}\omega ={\int }_{S^1}{\alpha }^{\ast }\omega \text{\hspace{1em}(6.57)}$$

where we have taken the integral domain in the RHS to be $S^1$ since $I=[0,1]$ in the LHS is compactified to $S^1$ . From lemma 6.1, we have, for a closed one-form $\omega$ ,

$${\alpha }^{\ast }\omega -c_p^{\ast }\omega =\mathrm{d}g\text{\hspace{1em}(6.58)}$$

where $g=-P{F}^{\ast }\omega$ . The pullback $c_p\omega$ vanishes since $c_p$ is a constant map. Then (6.57) vanishes since $\partial S^1$ is empty,

$${\int }_{S^1}{\alpha }^{\ast }\omega ={\int }_{S^1}\mathrm{d}g={\int }_{\partial S^1}g=0.\text{\hspace{1em}(6.59)}$$

Let $\beta$ and $\gamma$ be two paths connecting $p_0$ and $p$ . According to (6.59), integrals of $\omega$ along $\beta$ and along $\gamma$ are identical,

$${\int }_{\beta (I)}\omega ={\int }_{\gamma (I)}\omega .$$

This shows that (6.56) is indeed well defined, hence $\omega$ is exact.

$\square$

Example 6.6

The $n$ -sphere $S^n$ ( $n\ge 2$ ) is simply connected, hence

$$H^1\left(S^n\right)=0n\ge 2.\text{\hspace{1em}(6.60)}$$

From the Poincaré duality, we find

$$H^0\left(S^n\right)\cong H^n\left(S^n\right)=\mathbb{R}.\text{\hspace{1em}(6.61)}$$

It can be shown that

$$H^r\left(S^n\right)=01\le r\le n-1.\text{\hspace{1em}(6.62)}$$

$H^n(S^n)$ is generated by the volume element $\Omega$ . Since there are no $(n+1)$ -forms on $S^n$ , every $n$ -form is closed. $\Omega$ cannot be exact since if $\Omega =\mathrm{d}\psi$ , we would have

$${\int }_{S^n}\Omega ={\int }_{S^n}\mathrm{d}\psi ={\int }_{\partial S^n}\psi =0.$$

The Euler characteristic is

$$\chi \left(S^n\right)=1+(-1{)}^n=\{\begin{array}{ll}0& n\text{i s o d d},\\ 2& n\text{i s e v e n}.\end{array}\text{\hspace{1em}(6.63)}$$

Example 6.7

Take $S^2$ embedded in ${\mathbb{R}}^3$ and define

$$\Omega =\sin \theta d\theta \wedge d\varphi \text{\hspace{1em}(6.64)}$$

where $(\theta ,\varphi )$ is the usual polar coordinate. Verify that $\Omega$ is closed. We may formally write $\Omega$ as

$$\Omega =-\mathrm{d}(\cos \theta )\wedge \mathrm{d}\varphi =-\mathrm{d}(\cos \theta \mathrm{d}\varphi ).$$

Note, however, that $\Omega$ is not exact.

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