Notes on Differential Forms in Algebraic Topology: De Rham Cohomology with Compact Support

These notes are based on the course offered by Loring Tu at NCTS. The topic is Differential Forms in Algebraic Topology.

Support of Differential Forms

Definition. Support of differential form

Let $\omega$$\omega$ be a differential form on an open set $U\subseteq {\mathbb{R}}^n$$U\subseteq\mathbb{R}^n$, the support of $\omega$$\omega$, denoted as $\mathrm{supp}(\omega ):=\overline{\{p\in U|{\omega }_p\ne 0\}}$$\mathrm{supp}(\omega):=\overline{\{p\in U|\omega_p\neq 0\}}$.

A form $\omega$$\omega$ has compact support if $\mathrm{supp}(\omega )$$\mathrm{supp}(\omega)$ is compact. Let ${\mathrm{\Omega }}_c^k(U)=\{k\text{-forms with compact support in }U\}$$\Omega_c^k(U)=\{k \text{-forms with compact support in }U\}$.

The aim of this section is to build the De Rham cohomology of differential forms with compact support.

It is based on the following fact:

Proposition. $\mathrm{supp}(d\omega )\subset \mathrm{supp}(\omega )$$\mathrm{supp}(d\omega)\subset \mathrm{supp}(\omega)$. Also, if $\mathrm{supp}(\omega )$$\mathrm{supp}(\omega)$ is compact, then $\mathrm{supp}(d\omega )$$\mathrm{supp}({d}\omega)$ is compact as well since the closed subset of a compact set is compact.

Proof. We prove that $\mathrm{supp}(\omega {)}^c\subset \mathrm{supp}(d\omega {)}^c$$\mathrm{supp}(\omega)^c\subset\mathrm{supp}({d}\omega)^c$. Let $p\in \mathrm{supp}(\omega {)}^c$$p\in \mathrm{supp}(\omega)^c$. Since $\mathrm{supp}(\omega {)}^c$$\mathrm{supp}(\omega)^c$ is an open set, we can find an open neighborhood of $p$$p$, denoted as $V_p$$V_p$, such that $\mathrm{\forall }p\in V_p,{\omega }_p\equiv 0$$\forall p\in V_p,\omega_p\equiv 0$. Hence $\mathrm{\forall }p\in V_p,d{\omega }_p\equiv 0.$$\forall p\in V_p,d\omega_p\equiv 0.$ Thus $p\in \mathrm{supp}(d\omega {)}^c$$p\in \mathrm{supp}({d}\omega)^c$. $◻$$\square$

De Rham Cohomology with Compact Support

Definition. The complex of De Rham cohomology and the cohomology group.

By the proposition above, we can define the following complex:

$$0⟶{\mathrm{\Omega }}_c^0(U)\stackrel{d_0}{\to }{\mathrm{\Omega }}_c^1(U)\stackrel{d_1}{\to }{\mathrm{\Omega }}_c^2(U)\stackrel{d_2}{\to }...\stackrel{d_{n-1}}{\to }{\mathrm{\Omega }}_c^n(U)\stackrel{d_n}{\to }0$$

That is the complex of differential forms with compact support.

The $k$$k$-th cohomology group of differential forms with compact support is defined by

$$H_c^k(U):=\frac{Z^k(U)}{B^k(U)}$$

Where $Z^k(U)$$Z^k(U)$ is the kernel of $d_k$$d_k$, which is the closed $k$$k$-form with compact support and $B^k(U)$$B^k(U)$ is the image of $d_{k-1}$$d_{k-1}$, which is the exact $k$$k$-form with compact support.

Remark. Why ${\mathrm{\Omega }}_c^k(U)$$\Omega_c^k(U)$ forms a vector space?

You might ask how we know ${\mathrm{\Omega }}_c^k(U)$$\Omega_c^k(U)$ forms a vector space. Unfortunately, Loring Tu did not prove it. The details are as follows:

Firstly, $0\in {\mathrm{\Omega }}_c^k(U)$$0\in\Omega_c^k(U)$ since $\mathrm{\varnothing }$$\emptyset$ is compact. Obviously, ${\mathrm{\Omega }}_c^k(U)$$\Omega_c^k(U)$ is closed under scalar multiplication since $\mathrm{supp}(\omega )=\mathrm{supp}(\lambda \omega )$$\mathrm{supp}(\omega)=\mathrm{supp}(\lambda \omega)$.

Let $V(\omega )$$V(\omega)$ be the set where ${\omega }_p=0$$\omega_p= 0$ and $D(\omega )$$D(\omega)$ be its complement in $U$$U$.

To see that it is closed under addition, notice that $V(\omega )\cap V(\eta )\subseteq V(\omega +\eta )$$V(\omega)\cap V(\eta)\subseteq V(\omega +\eta)$, hence $D(\omega )\cup D(\eta )\supseteq D(\omega +\eta )$$D(\omega)\cup D(\eta)\supseteq D(\omega+\eta)$.

Therefore, taking the closure on both sides, we get:

$$\mathrm{supp}(\omega )\cup \mathrm{supp}(\eta )\supseteq \mathrm{supp}(\omega +\eta )$$

Since the finite union of compact sets is still compact, we get that $\mathrm{supp}(\omega +\eta )$$\mathrm{supp}(\omega+\eta)$ is compact as well.

In addition, notice that ${\mathrm{\Omega }}_c^0(U)$$\Omega_c^0(U)$ consists of smooth functions with compact support on $U$$U$, and it forms an ideal in $C^{\mathrm{\infty }}(U)$$C^{\infty}(U)$.

Let us prove it. We only need to prove that for $f\in C^{\mathrm{\infty }}(U),g\in {\mathrm{\Omega }}_c^0(U)$$f\in C^\infty(U),g\in\Omega _c^0(U)$, we have $f\cdot g\in {\mathrm{\Omega }}_c^0(U)$$f\cdot g\in\Omega_c^0(U)$.

Notice that $D(f\cdot g)=D(f)\cap D(g)\subset D(g)$$D(f\cdot g)=D(f)\cap D(g)\subset D(g)$, hence $\mathrm{supp}(f\cdot g)\subset \mathrm{supp}(g)$$\mathrm{supp}(f\cdot g)\subset\mathrm{supp}(g)$. $◻$$\square$

Example and Poincare duality

Let us compute $H_c^0(\mathbb{R})=\ker (d_0)$$H_c^0(\mathbb{R})=\mathrm{ker}(d_0)$. That is all the functions in ${\mathrm{\Omega }}_c^0$$\Omega_c^0$ that satisfy $df=0$$df=0$. It has to be a constant function, and the unique constant function with compact support is $0$$0$. Hence $H_c^0(\mathbb{R})\cong 0$$H_c^0(\mathbb{R})\cong 0$. But $H^0(\mathbb{R})\cong \mathbb{R}$$H^0(\mathbb{R})\cong\mathbb{R}$!

How about $H_c^1(\mathbb{R})$$H^1_c(\mathbb{R})$?

$$0⟶{\mathrm{\Omega }}_c^0(\mathbb{R})\stackrel{d}{\to }{\mathrm{\Omega }}_c^1(\mathbb{R})⟶0$$

Firstly, $H_c^1(\mathbb{R})\ne 0$$H^1_c(\mathbb{R})\neq 0$, we need to show that $d$$d$ is not surjective.

Let $gdx\in {\mathrm{\Omega }}_c^1(\mathbb{R})$$gdx\in\Omega_c^1(\mathbb{R})$, assume we have $df=gdx$$df=gdx$ for a $f\in {\mathrm{\Omega }}_c^0(U)$$f\in\Omega_c^0(U)$. Then $\mathrm{supp}(f)\subseteq [a,b]$$\mathrm{supp}(f)\subseteq [a,b]$ for some $a,b\in \mathbb{R}$$a,b\in\mathbb{R}$ since $\mathrm{supp}(f)$ is compact if and only if it is closed and bounded. Hence $\mathrm{\forall }x\in (-\mathrm{\infty },a)\cup (b,\mathrm{\infty }),f(x)=0$$\forall x\in(-\infty,a)\cup(b,\infty), f(x)=0$.

Hence ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}gdx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}^{\prime }dx=f{|}_{-\mathrm{\infty }}^{\mathrm{\infty }}=0-0=0}$$\displaystyle\int_{-\infty}^{\infty}gdx=\int_{-\infty}^\infty f'dx=f|_{-\infty}^\infty=0-0=0$. But not all the forms in ${\mathrm{\Omega }}_c^1(\mathbb{R})$$\Omega _c^1(\mathbb{R})$ have this property.

Therefore $d$$d$ is not surjective, and $H_c^1(\mathbb{R})\ne 0$$H^1_c(\mathbb{R})\neq 0$.

Now let us consider the integration map ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}:Z_c^1(\mathbb{R})={\mathrm{\Omega }}_c^1(\mathbb{R})\to \mathbb{R}}$$\displaystyle\int_{-\infty}^{\infty}:Z_c^1(\mathbb{R})=\Omega_c^1(\mathbb{R})\to\mathbb{R}$.

It is well defined since the form we are considering has compact support on $\mathbb{R}$$\mathbb{R}$, hence it is not equal to $0$$0$ only on a bounded set.

We have shown that $B_c^1(\mathbb{R})\subseteq \ker {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}$$B_c^1(\mathbb{R})\subseteq\mathrm{ker}\int_{-\infty}^{\infty}$. Now let us prove the converse, i.e.:

Proposition. ${\displaystyle B_c^1(\mathbb{R})=\ker {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}$$\displaystyle B^1_c(\mathbb{R})=\mathrm{ker}\int_{-\infty}^{\infty}$. This will imply that $H_c^1(\mathbb{R})\cong \mathbb{R}$$H_c^1(\mathbb{R})\cong\mathbb{R}$.

Proof.

We only need to prove that ${\displaystyle B_c^1(\mathbb{R})\supseteq \ker {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}$$\displaystyle B_c^1(\mathbb{R})\supseteq\mathrm{ker}\int_{-\infty}^\infty$, i.e., for $gdx\in {\mathrm{\Omega }}_c^1(\mathbb{R})$$gdx\in\Omega_c^1(\mathbb{R})$ such that ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}gdx=0}$$\displaystyle \int_{-\infty}^{\infty}gdx=0$, we need to find $f\in {\mathrm{\Omega }}_c^0(\mathbb{R})$$f\in\Omega_c^0(\mathbb{R})$ such that $df=gdx$$df=gdx$.

Let us define ${\displaystyle f(t)={\int }_{-\mathrm{\infty }}^{t}gdx}$$\displaystyle f(t)=\int_{-\infty}^t g dx$. It is well defined since $g(t)=0$$g(t)=0$ for all $t<a$$t<a$ as $g$$g$ has compact support.

Now we need to check that $f$$f$ has compact support. Let $\mathrm{supp}(g)\subseteq [a,b]$$\mathrm{supp}(g)\subseteq [a,b]$, then $\mathrm{supp}(f)\subseteq [a,b]$$\mathrm{supp}(f)\subseteq [a,b]$, hence it has compact support.

Since ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}$$\displaystyle \int_{-\infty}^\infty$ is not $0$$0$ and it is linear, it has to be surjective to $\mathbb{R}$$\mathbb{R}$. By the first isomorphism theorem, we have

$$H_c^1(\mathbb{R})\cong \mathbb{R}◻$$

Remark. Notice that $H^1(\mathbb{R})=0$$H^1(\mathbb{R})=0$ since for any $fdx\in {\mathrm{\Omega }}^1(\mathbb{R})$$fdx\in\Omega^1(\mathbb{R})$, define ${\displaystyle g(t)={\int }_0^{t}fdx}$$\displaystyle g(t)=\int_{0}^t fdx$. Then $g$$g$ is smooth and $dg=fdx$$dg=fdx$.

Now let us compare the results we have:

$$H^0(\mathbb{R})=\mathbb{R},H_c^0(\mathbb{R})=0,H^1(\mathbb{R})=0,H_c^1(\mathbb{R})=\mathbb{R}$$

In general, we have $H^k(U)\cong H_c^{n-k}(U)$$H^k(U)\cong H^{n-k}_c(U)$ for $U\subseteq M$$U\subseteq M$ where $M$$M$ is an orientable manifold. This is called Poincaré duality.

Then Loring Tu defines what is a manifold, tangent space, and $k$$k$-forms on manifolds.