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De Rham cohomology

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Closed and exact forms vs topology

Let's review what we have learned about this relation.

The main idea is that the "difference" between the sets of closed and exact forms reveals the topology of the domain (and vice versa):

  • path-connectedness via $0$-forms;
  • simple-connectedness via $1$-forms.

Path-connectedness

Closed forms are constants, if $R$ is path-connected. But what if it's not? What if $R$ has two path components, etc?

In general, closed forms are the ones piecewise constant, i.e., constant on each path-component:

PathComponentDiagram.png

However, these values may differ from one component to the next.

Recall:

Theorem: For path-connected $R$:

  • $\ker d_0 \simeq {\bf R}$.

Further, for $2$ path components:

  • $\ker d_0 \simeq {\bf R} \times {\bf R}$

If $R$ has $p$ path components:

  • $\ker d_0 \simeq {\bf R}^p$.

To generalize:

Definition. $\dim \ker d$ is the $0$th Betti number.

Why no mention of exact $0$-forms? Because $\Omega^{-1}(R)=0$, so $im \hspace{3pt} d=0$.

Simple connectedness

Recall:

  • The exterior derivative is a linear operator $d_1 \colon \Omega^1(R) \rightarrow \Omega^2(R)$;
  • Closed $1$-forms = $\ker d_1$, subspace of $\Omega^1(R)$;
  • Exact $1$-forms = $im d_0$, subspace of $\Omega^1(R)$.

Theorem: If $R \subset {\bf R}^n$ is simply connected, then every closed $1$-form is exact.

In other words, $${\rm im} \hspace{3pt} d_0 = \ker d_1,$$ which together imply that there is no "hole".

More generally, the "difference" between ${\rm im} \hspace{3pt} d$ and $\ker d$ is in the dimensions of these two spaces. If we subtract the dimensions, we "find" the topology of $R$, i.e., the number of holes.

And the difference of dimensions corresponds to the dimension of the quotient of the spaces, see below.

de Rham cohomology

As we have seen, the interaction between closed and exact forms reveals the topology of the domain.

In particular for $0$-forms,

  • ${\rm dim} \hspace{3pt} {\rm ker} \hspace{3pt} d - {\rm dim} \hspace{3pt} {\rm im} \hspace{3pt} d$ = # of path-components.

Another way to see that is to consider ${\bf R}^n / {\bf R}^m = {\bf R}^{n-m}$ where "/" means "modulo" in the quotient set.

Since $d_{k+1}d_k=0$, we have $${\rm im} \hspace{3pt} d_k \subset \ker d_{k+1}.$$ So it makes sense to define the equivalence class $$[\psi] \in \ker d_{k+1} / {\rm im} \hspace{3pt} d_k,$$ for $\psi\in\ker d_{k+1}.$ Then each class is given by: $$[\psi] = \psi + {\rm im} \hspace{3pt} d_k.$$

So we have introduced an equivalence relation on closed forms: $$\varphi \sim \psi \hspace{3pt} \mathrm{ if } \hspace{3pt} \varphi - \psi \in {\rm im} \hspace{3pt} d,$$ i.e.,

two forms are equivalent -- cohomologous -- if their difference is exact.

Example: For $0$-forms, if $f$ and $g$ are functions and $f \sim g$, then $f-g = {\rm constant}$. So $f'=g'$. That's Calc 1!

Definition: $R \subset {\bf R}^n$ is a region. The $k^{\rm th}$ de Rham cohomology of $R$ is the vector space $$H_{dR}^k(R) = \ker d_k / {\rm im} \hspace{3pt} d_{k-1}.$$

Here's where this comes from: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \Omega^{k-1} & \ra{d_{k-1}} & \Omega^k & \ra{d_k} & \Omega^{k+1} \\ & & \cup & & \\ & & {\rm ker \hspace{3pt}} d_k & & \\ & & \cup & & \\ & & {\rm im \hspace{3pt}}d_{k-1} & & \end{array} $$

Computing the de Rham cohomology

Let's see what we can actually compute with the data that we have.

We do know a lot about $0$-forms.

For a path-connected $R$, we have found everything about closed and exact forms. So, we can compute the cohomology: $$\begin{align*} {\rm closed} \hspace{3pt} 0{\rm -forms} / {\rm exact} \hspace{3pt} 0{\rm -forms} &= {\rm constants} / 0 \\ &= {\rm constants} \\ &= {\bf R}. \end{align*}$$ Similarly:

  • $H_{dR}^0(B(0,1) \cup B(3,1)) = {\bf R}^2,$
  • $H_{dR}^0({\bf R}^n) = {\bf R}$;
  • $H_{dR}^0(B((0,1)) = {\bf R}$;

etc.

This is not so simple with the $1$-forms.

Frequently, we can prove that the domain is simply connected, then we have trivial $1$-cohomology: $$H_{dR}^1({\bf R}^3 \backslash \{(0,0,0)\})=0.$$

If it is not simply connected, we have less to say. For example, we know that the $1$-form $$\varphi = \frac{1}{x^2+y^2} (-ydx + xdy)$$ is closed but not exact. Therefore

  • $H_{dR}^1({\bf R}^1 \backslash \{(0,0)\})\ne 0;$

We know that it is actually ${\bf R}$.

More examples of results that we can compute with some effort:

  • $H_{dR}^1({\bf R}^3 \backslash \{(0,0,0)\}) = 0$;
  • $H_{dR}^1({\bf R}^3 \backslash {\rm z-axis})={\bf R}$.

For higher degrees:

  • $H_{dR}^k({\bf R}^n)=0, k=1,2,\ldots.$

But what about the torus, cone, pretzel, etc?

If we can't compute anything more complicated, what is that we have accomplished?

We are in about the same place we are close to the end of Calc 1. You first learn that "you can compute areas under graphs by means of definite integrals" but later run into functions like $e^{x^2}$. You then realize that the examples have been carefully selected and you can't always compute what you want or need. Then you wonder what's been accomplished?

And the answer is, you didn't used to understand areas and now you do. Areas under graphs are definite integrals. So, holes are what is captured by $\Omega ^1$, whether we can compute it or not.

To appreciate the challenge see Tunnels.

Meanwhile, the solution of always being able to compute cohomology lies with cubical complexes and discrete forms.