## Tuesday, October 31, 2017

### De Rham Cohomology

Guest post by Phd student  Enrique Alvarado

In the following, we will take a look at the motivation for considering $closed$ and $exact$ forms on manifolds. This will lead us to look for the closed forms which are $\it{not}$ exact -- which to put crudely, is what de Rham cohomology studies.

Let's first take an intuitive look at what differential forms are.

${\mathbf{1.}}$

$\color{purple}{\mathbf{Definition.}}$ A differential $k$-form on $\mathbb{R}^3$ is a differentiable mapping, $\varphi : \mathbb{R}^3 \to \Lambda^k$, that takes a point in 3-space to a $k$-covector.

So, what are $k$-covectors?

$\color{purple}{\mathbf{Definition.}}$ A $k$-$\it{covector}$ is a funciton, $\lambda : \Lambda_k \to \mathbb{R}$, that takes objects called $k$-$\it{vectors}$ to real numbers.

In other words, $\Lambda^k$ is the dual space of $\Lambda_k$.

Now, to understand the vector space of $k$-vectors, denoted $\Lambda_k$, let's take a little trip into Intuitionland by considering the cases for $k = 0, 1, 2$,  and $3$.

A $0$-vector in $\mathbb{R}^3$ can be thought to be a real number, a $1$-vector in $\mathbb{R}^3$ can be thought to be a vector in $\mathbb{R}^3$, and a $2$-vector in $\mathbb{R}^3$ can be pictured as the wedge of two linearly independent vectors, as shown below.

Similarly, a $3$-vector in $\mathbb{R}^3$ can be pictured as a wedge of three linearly independent vectors as shown below.

Now, although there is no geometric difference between $k$-vectors and $k$-covectors, there is an algebraic one. This reason can be intuitively explained by considering the difference between a $1$-vector and a $1$-covector. Notice that we are just saying that we are considering the difference between a vector, and a covector in 3-space.

If we think of 1-vectors as column vectors, $\left(\begin{array}{}x_1\\ y_1\\ z_1\\ \end{array}{} \right),$ we can then think about 1-covectors as $\it{row}$ vectors $\left(x_2, y_2, z_2\right)$ since we can then operate on the column vectors to get a real number as follows.

$\begin{array}{}\left(x_2, y_2, z_2\right) \end{array}{} \cdot \left(\begin{array}{} x_1\\ y_1\\ z_1\\ \end{array}{}\right) = x_1x_2 + y_1y_3 + z_1z_3$.

So what a differential $k$-form $\varphi$ does is that to every point $p$ in $\mathbb{R}^3$, we have an associated $k$-covector. The figure below is a mapping $p \mapsto \varphi \in \Lambda^3$.

Another way of defining a differential $k$-form on $\mathbb{R}^3$, is by saying that it is a $k$-covector field on $\mathbb{R}^3$. We will denote the space of all $k$-forms on a manifold $M$ as $\mathbf{C}^k(M)$.

As we have seen, $0$-forms can be identified to be scalar functions. In $\mathbb{R}^3$, 1-forms can be identified with vector fields, 2-forms can also be identified with vector fields via the right-hand rule, and 3-forms can be identified with scalar functions via a similar rule. There is a generalization of the gradient operator that is applied to forms.

\begin{align*}
d: \mathbf{C}^k(M) \to \mathbf{C}^{k+1}(M)
\end{align*}

Keeping in mind the ways we can identify 0-forms, 1-forms, and 2-forms, $\omega \mapsto d\omega$ is then identifiable to:

(1) The gradient operator $\omega \mapsto \nabla \omega$ when $\omega$ is a 0-form.

(2) The curl operation $\omega \mapsto \nabla \times \omega$ when $\omega$ is a 1-form.

(3) The divergence operation $\omega \mapsto \nabla \cdot \omega$ when $\omega$ is a 2-form.

${\mathbf{2.}}$

Now, differential forms may be used to give us global information about manifolds, rather than local. For example, let's consider the manifold $M := \mathbb{R}^2 - B$, where $B$ is some open ball centered about the origin. If we take any point in $M$, we can find a sufficiently small open ball that looks identical to some open ball in $\mathbb{R}^2$. Therefore, all local properties of $M$ are the same as those in $\mathbb{R}^2$. But the fact that the origin is missing is a global property.

Certain differential forms are interesting for the purpose of detecting these types of global properties. The interesting ones have their exterior derivative zero. Such differential forms are called closed. That is, a differential form $\varphi$ is $\it{closed}$ if $d\varphi = 0$.

So why are closed forms interesting when trying to investigate global properties?

Let $\omega$ be a closed $k$-form, and let's integrate it over a closed smooth $k$-chain $C$ (a chain $C$ is closed if it has no boundary) in a manifold $M$ that is at least $k$-dimensional. If $S$ is the boundary of an orientable, compact, smooth submanifold $S$ (i.e $\partial S = C$) of $M$, then Stokes' Theorem states

\begin{align*}
\int_C \omega &= \int_S d\omega \\
&= \int_S 0 \\
&= 0.
\end{align*}

Therefore, if we have a closed $k$-form $\omega$ on a submanifold $C$ of $M$ for which

\begin{align*}
\int_C\omega &\neq 0,
\end{align*}

we then know that $C$ must $\it{not}$ be $\it{the}$ boundary of any oriented, compact, smooth submanifold of $M$! The fact that there exists such a submanifold $C$ tells us about the global information of the manifold $M$.

If we want to be able to detect these global properties, we have to find reasonable forms to integrate over $S$. There might be forms which always integrate to 0, no matter what $S$ is!

Such forms are called $\it{exact}$.

$\color{purple}{\mathbf{Definition}}$ A $k$-form, $\omega$ is $\it{exact}$ if there exists a $(k-1)$-form such that $\omega = d\varphi$.

Note that $d:\mathbf{C}^{k}\to \mathbf{C}^{k+1}$ is an operator that takes k-forms and gives us (k+1)-forms, so the above definition makes sense.

Integrating exact forms over closed chains will always evaluate to 0. Let's prove this result for exact 1-forms and 1-chains.

Let's consider a 1-form $\omega$ and a 0-form $\varphi$ for which $\omega = d\varphi$, and let $C$ be any closed 1-chain. If we pick any two points $p, q \in C$, we may then say that $C = A + B$ where $A$ is the curve that goes from $p$ to $q$, and $B$ is the curve that goes from $q$ to $p$.

We can now compute the following,

\begin{align*}
\int_C\omega &= \int_{A + B}\omega\\
&= \int_{B}\omega + \int_{A}\omega\\
&= \int_{B}d\varphi + \int_{A}d\varphi\\
&= \int_{p - q}d\varphi + \int_{q - p} d\varphi\\
&= 0.
\end{align*}

So yes, integrating an exact 1-from over a closed 1-chain always gives us zero, and this result holds in general as well. You may however say that the only reason that we were able to find global properties of a manifold was by applying Stokes' Theorem to closed forms. So this would only be bad if all exact forms are closed. This is in fact true!

$\color{red}{\mathbf{Theorem.}}$ If $\omega \in \mathbf{C}^k$ is exact, then it is also closed. That is, for any differential form $\varphi$, $d\circ d\varphi = 0$.

What this means is that we cannot just integrate any closed form. We must choose closed forms which are not exact. Do there exists such closed forms? What does exactness depend on?

To investigate these questions a little further, let's take a look at a 1-form on the punctured plane, and then a 1-form on the half plane.

$\color{blue}{\mathbf{Example.}}$ Let $M := \mathbb{R}^2 - \{0\}$, and consider the 1-form on $M$

\begin{align*}
\omega &= \frac{xdy - ydx}{x^2 + y^2}.
\end{align*}

Let $\gamma : [0, 2\pi] \to M$ be the curve defined by $\gamma (t) = (\cos{t}, \sin{t})$, whose trace is the unit circle. By substituting $x = \cos{t}$ and $y = \sin{t}$ everywhere in the formula for $\omega$, we get that

\begin{align*}
\int_\gamma\omega &= \int_{[0,2\pi]}\frac{\cos{t}(\cos{t}\ dt) - \sin{t}(-\sin{t}\ dt)}{\sin^2{t} + \cos^2{t}} \\
&= \int_0^{2\pi}dt \\
&= 2\pi.
\end{align*}

This implies that $\omega$ is not exact; because if it were, then integrating it over any closed curve would give us $0$.

However, $\omega$ $\it{is}$ exact on some smaller domains such as the right half-plane $H := \{(x, y) \in \mathbb{R}^2 : x > 0\}$. In the right half-plane, we get that $\omega = d(\tan^{-1}({y/x}))$. In polar coordinates, we would get that $\omega = d\theta$.

This in fact is true in general, as the following theorem describes.

$\color{red}{\mathbf{Theorem.}}$ Let $M$ be a smooth manifold with or without boundary. Each point of $M$ has a neighborhood on which every closed form is exact.

What this tells us is that a form being exact is not a local property. So if our objective is to investigate global properties of manifolds, we must then find out which closed forms are $\it{not}$ exact. This is precisely what de Rham cohomology studies!

${\mathbf{3.}}$
The way we do this is by constructing the following equivalence relation among closed $k$-forms.

Two closed $k$-forms $\varphi$ and $\omega$ are $\it{equivalent}$ if their difference, $\omega - \varphi$ is an exact form. That is, if

\begin{align*}
\omega - \varphi &= d\phi
\end{align*}

for some $\phi \in C^{k-1}(M)$.

This will partition our space of closed $k$-forms into equivalence classes. So instead of talking about a specific form $\omega$, we will consider the equivalence class

$[\omega] := \{\varphi \in \mathbf{C}^k : \omega - \varphi = d\phi$ for some $\phi \in \mathbf{C}^{k-1}\}$.

We can then say that $[\omega] = [\varphi]$ for such forms $\varphi$.

Notice that constructing this equivalence relation is exactly what we get when we define the following quotient group.

Let's first define a couple of subspaces of $\mathbf{C}^k(M)$ when $M$ is a smooth manifold with or without boundary.

\begin{align*}
\mathcal{Z}^k(M) &:= \{{\rm closed } \ k{\rm -forms \ on } \ M\},\\
\mathcal{B}^k(M) &:= \{{\rm exact } \ k{\rm -forms \ on } \ M\}.
\end{align*}

Because $d: \mathbf{C}^k(M) \to \mathbf{C}^{k+1}(M)$ is linear, its kernel and image are linear subspaces. Together with the fact that every exact form is closed, we may define the de Rham cohomology group in degree $k$ as the following.

$\color{purple}{\mathbf{Definition.}\ (de\ Rham\ cohomology\ group)}$ We define the $pth$ de Rham group of $M$ to be the quotient vector space

\begin{align*}
H^k_{dR}(M) &= \mathcal{Z}^k(M)/\mathcal{B}^k(M).
\end{align*}

Notice that the diference between an exact form and the form which always returns the number $0$ is an exact form. Thus, if $\omega \in C^k$ is exact, then $\omega$ is equivalent to zero in $H_{dR}^k(M)$.

To get our hands a little dirty, let's try to reason out what we can get for $H^0(M)$ when $M$ is a smooth manifold. So let's begin by asking, when is a $0$-form closed? Recall that $0$-forms on $M$ are just functions $\varphi: M \to \mathbb{R}$ which assign to every point in $M$ a real number.

Thus, in local coordinates,

\begin{align*}
df &= \frac{\partial \varphi}{\partial x_1}dx_1 + ... + \frac{\partial\varphi}{\partial x_n}dx_n.
\end{align*}

Hence, a $0$-form $\varphi$ is closed if and only if its first partial derivatives vanish. That is, if it is locally constant. The only way for a locally constant function $M$ to not be constant on $M$ is for $M$ to have multiple connected components, say, $M_1, M_2, ..., M_m$. The most general such functions are $\varphi_i$ which take on the constant values $c_i$ on $M_i$ and $0$ elsewhere, for each $1\leq i \leq m$.

Now, if we are trying to find exact $0$-forms, we must be able to have $(-1)$-forms, which do not exist! Therefore we have $\mathcal{B}^0(M) = \{0\}$, the trivial group.

Hence,

\begin{align*}
H^0_{dR}(M) &= \mathcal{Z}^0(M)/\mathcal{B}^0(M)\\
&= \mathcal{Z}^0(M)/\{0\}\\
&\simeq \mathcal{Z}^0.
\end{align*}

This implies that $H^0_{dR}(M)$ is isomorphic (i.e., algebraically the same) to the space of locally constant functions on $M$. This space has dimension equal to the number of components $M$ has; in our case, dimension $m$.

Then, how do we know if a closed $k$-form is exact if it depends on the underlying space? We do have the following wonderful theorem!

$\color{red}{\mathbf{Theorem.}}$ Let $S$ be a $k$-dimensional manifold, $M$ a smooth manifold, and let $\omega$ be a differential $k$-form on $M$.  If

\begin{align*}
\int_S\phi^\ast\omega = 0
\end{align*}

for every map $\phi: S\to M$, then $\omega$ is exact.

Let's investigate what this theorem is saying by first looking at a slight variation of its statement for $k = 1$. Let $\varphi$ be a 1-form on $M$. If $\int_{\gamma}\varphi = 0$ for all closed curves $\gamma$, then $\varphi$ is exact.

If we consider some $k$-dimensional smooth manifold $S$ and a smooth manifold $M$, what the theorem is saying is that $\omega$ is exact only if its $\it{pullback}$ by all  maps $\phi:S\to M$ integrate to zero.

The pullback by $\phi$ is a $k$-form $\phi^\ast \omega$ on $S$. The intuition for considering different $\phi$'s is that they move $S$ around in $M$, and considering $\phi^\ast\omega$, let's us look at the form $\omega$ on these different images of $S$ in $M$. This is very much like considering all different k-dimensional manifolds $S$ in $M$, and looking at what $\omega$ integrates to over all these different manifolds.