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Closedness and exactness of 1-forms

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Recall:

Theorem: In $\Omega^0(R)$, with $R \subset {\bf R}^n$ path-connected, closed forms are constant, and vice versa.

Let's quickly review more of the related material as presented in calc 1 and use the ideas later.

This is how we solve the "exactness problem". Given a continuous function $f$, it is exact if it's the derivative of someone: $$f=F'.$$ In other words, there is an antiderivative of $f$. We can construct $F$ using nothing but continuity. Indeed $f$ is integrable on $[a,b]$ and its Riemann integral exists on all intervals within $[a,b]$. Therefore we can define $F=d^{-1}f$ via integration as follows: $$F(x)=\int _a ^x f(x)dx,x \in [a,b].$$ This is the idea, and the proof of, of a version of the Fundamental Theorem of Calculus. It will be used again for $1$-forms.

Next we consider $\Omega^1$.

We deal with closed and exact $1$-forms in ${\bf R}^1$ first.

All $1$-forms in ${\bf R}^1$ look like this: $\varphi = f(x) dx$, where $f$ is some function.

Question: When is such a $\varphi$ closed? When $d \varphi = 0$?

Answer: Always! There are no non-trivial $2$-forms, as we know.

Question: When is such a $\varphi$ exact? When $\varphi = d \psi$?

Well, if $d \psi = f(x) dx$, then $\psi$ is a $0$-form. And, in fact, $$\psi(x)=F(x) = \displaystyle\int f(x) dx,$$ i.e., $\psi$ is an antiderivative of $f$.

Answer: When $f$ is integrable.

Conclusion: $${\rm ker} d_1 = {\rm im} d_0$$ for all continuous ($C^0$) $1$-forms in ${\bf R}^1$. Remember, the index $i$ in $d_i$ stands for the dimension of the forms in the domain of this operator.

The issue isn't as simple in ${\bf R}^2$.

All $1$-forms in ${\bf R}^2$ look like this: $$\varphi = f dx + g dy,$$ where $f=f(x,y),g=g(x,y)$ are functions of two variables.

Question: When is such a $\varphi$ closed?

Let's compute! $$\begin{align*} d \varphi &= df \hspace{1pt} dx + dg \hspace{1pt} dy \\ &= (f_x dx + f_y dy)dx + (g_x dx + g_y dy)dy \\ &= f_y dy \hspace{1pt} dx + g_x dx \hspace{1pt} dy \\ &= (g_x - f_y) dx \hspace{1pt} dy \end{align*}$$

Observation: This expression is the rotation, $rot$, which is the integrand in the right hand side of Green's theorem.

The above computation implies the following theorem.

Theorem: $\varphi = fdx + gdy$ is closed if and only if $g_x=f_y$.

Note: Compare to the Cauchy-Riemann equation, a necessary condition for a conservative vector field $(f,g)$.

In other words, $$\ker d_1 =\{\varphi = fdx + gdy:g_x=f_y\}.$$

Recall that the problem of the number of path-components, the $0$-th Betti number, was solved by: $$b_0=\dim\ker d_0.$$ We are now after the number of holes, i.e., the $1$-st Betti number, but $\dim\ker d_1$ cannot be the answer! One simple reason is that this number is infinite (Exercise). We need to deal with exact forms as well.

Question: When is $\varphi= fdx + gdy$ exact, $\varphi = d \psi$?

Here $\psi$ is simply a function of two variables. So, we conclude that $$\psi _x=f,\psi _y =g.$$ And further $$\psi =\int f(x,y)dx, \psi =\int g(x,y)dy.$$

Example 1. Suppose $\varphi= xydx + x^2dy$. Then $$\psi =\int f(x,y)dx=\int xydx= \frac{1}{2}x^2y+C(y),$$ $$\psi =\int g(x,y)dy=\int x^2dy = x^2y+K(x).$$ If we set these two things equal to each other, we realize that this is impossible. So, this form isn't exact.

To be fair, it's not even closed! Indeed, $$f_y=x,g_x=2x.$$ What if it was closed?

Example 2. Consider now $$\varphi = (x^2y + x)dx + (\frac{1}{3}x^3 + y) dy.$$ Then if we set $f = x^2y+x$ and $g=\frac{1}{3}x^3 + y$, we have $f_y=x^2$ and $g_x=x^2$. Therefore $\varphi$ is closed.

Now exactness. It is exact when $\varphi = d \psi$ for some $\psi$, a $0$-form. Now, if $\psi = F(x,y)$, then $\varphi = d \psi = F_x dx + F_y dy$. Therefore $$F_x = x^2y + x, F_y= \frac{1}{3} x^3 + y$$ Now we integrate to get $$F= \frac{x^3}{3}y + \frac{x^2}{2} + C(y), F = \frac{1}{3}x^3y + \frac{y^2}{2} + K(x).$$ We conclude (calc 3) that $$F(x,y) = \frac{1}{3}x^3y + \frac{x^2}{2} + \frac{y^2}{2} + M,$$ for any constant $M$.

So this form is exact. Is it always the case?

Question. In $\Omega^1(R)$, what forms are exact?

Answer is below.

Theorem (Poincare Lemma). Every closed $1$-form in ${\bf R}^n$ is exact.

The meaning of the theorem is that our cochain complex is exact at dimension $1$. Below the idea is illustrated by a cochain complex which is exact at $\Omega ^k$ and non-exact at $\Omega ^{k+1}$:

Cochain exact.png

Proof. The proof is based on the same idea, integration with a variable upper limit, as the version of the Fundamental Theorem of Calculus discussed above. We will prove it for $C^1$-forms and $n=2$

Given a $1$-form $\varphi =fdx+gdy$ we construct a $0$-form $\psi$ with $d\psi =\varphi$ as a line integral. We fix a point $a \in {\bf R}^n$ and define $\psi$ as a function of $u=(x,y)$: $$\psi (u)=\int _C \varphi,$$ where $C$ is a path from $a$ to $u$.

Line integral for PL.png

Is this function well defined? To be well defined, $\psi$ should be independent of our choice of path $C$. There may be many paths from $a$ to a given $u$ and they all should produce the same value to be assigned to $\psi (u)$.

Line integrals for PL.png

Suppose we have another path $C'$ from $a$ to $u$ that doesn't intersect $C$ except at the end points. Then we need to prove that $$\int _C \varphi = \int _{C'} \varphi.$$

Recall that by paths we understand parametric curves (that we can integrate). The idea is to form a closed path from $a$ to $a$ from these two: $K=C \cup -C'$, where $-C'$ is $C$ parametrized in the opposite direction.

Line integral w closed path for PL.png

Then, on one hand, we compute $$\int _K \varphi =\int_C \varphi + \int _{-C'} \varphi = \int_C \varphi - \int _{C'} \varphi.$$ On the other hand $$\int _K \varphi = \int\int _D (g_x-f_y)dxdy ,$$ where $D$ is the region bounded by $K$, by Green's theorem. Therefore the integral is zero because the form $\varphi =fdx+gdy$ is closed. It follows then that $$\int_C \varphi - \int _{C'} \varphi =0.$$

Finally we differentiate the integral which is just a function: $$d\psi = \psi _x dx + \psi _y dy = fdx+gdy.$$ The result follows from the Fundamental Theorem of Calculus for parametric curves. $\blacksquare$

So, $$im\{d_0:\Omega ^0({\bf R}^n) \rightarrow \Omega ^1({\bf R}^n)\} = \ker\{d_1:\Omega ^1({\bf R}^n) \rightarrow \Omega ^2({\bf R}^n)\}.$$ However...

Example. Consider $$\theta = \frac{1}{x^2+y^2} (-ydx + xdy)$$ $$ = f dx + g dy.$$

This one is closed. Indeed, the numerators of the partial derivatives $\frac{\partial f}{\partial y}, \frac{\partial g}{\partial x}$ are: $-(x^2+y^2) + y \cdot 2y = -x^2 + y^2$ and $-y^2 + x^2$. They cancel.

But $\theta $ is not exact! (Proof long.)

How come? Since we are dividing by $x^2+y^2$ we must be careful about the domain, $R$. Here, $R = {\bf R}^2 - \{(0,0)\}.$

Conclusion: It is not true that every closed $1$-form in ${\bf R}^2 - \{(0,0)\}$ is exact.

What makes the difference is the topology of the region $R$.

Let's observe that in the example above he antiderivative of $\varphi$ is found to be $$F(x,y) = \frac{1}{3}x^3y + \frac{x^2}{2} + \frac{y^2}{2} + M,$$ for any constant $M$. If we think of $M$ as a function, it is the only closed $0$-form. So, if the derivative is the same then two forms differ by a closed form. More generally:

Theorem. If $d \psi_1 = d \psi_2$, then $\psi_1 - \psi_2$ is a closed form.

Proof. It follows that $d(\psi_1 - \psi_2) = 0$, so $\psi_1 - \psi_2$ is closed (we use linearity of $d$). $\blacksquare$

Example. If $\varphi = x dx + x dy$, is it exact? We differentiate, to get $F_x = x$ and $F_y = x$. Then, $F = \frac{x^2}{2} + C(y)$ and $F = xy + K(x)$, which is impossible! So $\varphi$ is not exact.