How to calculate $f(x, y, z)$ given $d f = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy +\frac{\partial f}{\partial z} dz$

Question

On a manifold with local coordinates $(x_1, \ldots, x_n)$ I have a closed 1-form $\omega$ for which $d \omega = 0$ holds. This means There must be a function $f(x_1, \ldots x_n)$ for which $d f = \omega$.

My question is, how to calculate this function $f$, only given the coefficients of $\omega$.

Example $\omega = dx_1 + \sin(x_3)dx_2 + x_2 \cos(x_3)dx_3$.

Of course, this example is solvable by a sharp look, but I look for an algorithm.

(In fact I already have one but it seems to complicated.)

Answer 1

I believe you want to calculate $f$ locally. So take a chart whose image is a ball which contains $0$ and identify it with a neighborhood of $x\in M$ you have:

$f(x)=\int_0^t\omega_{tx}(x)dt$.

Answer 2

Unfortunately, you really need to compute, either integrating (you know that integrals along any path with same initial and final points are equal), or solving: $$ \frac{\partial f}{\partial x_1}=1, \quad \frac{\partial f}{\partial x_2}=\sin(x_3),\quad \frac{\partial f}{\partial x_3}=x_2\cos{x_3},. $$ The computational difficulty is the same with both "methods". Certainly the same comment applies to any other example.

Answer 3

There is a general method for finding an antiderivative of a closed differential form defined on a star-shaped region (more generally, on some region $D$ that’s the image of a star-shaped region) that’s particularly simple in the case of a one-form. It’s basically the same as what’s described in Tsemo Aristide’s answer, elsewhere.

Step 1: Replace $x^i$ with $tx^i$ in the arguments of all the coefficient functions and $\mathrm{d}x^i$ with $x^i\,\mathrm{d}t$. (In the general case, make the replacement $\mathrm{d}x^i\to x^i\,\mathrm{d}t+t\,\mathrm{d}x^i$ instead.)

Step 2: Treat the $\mathrm{d}t$ as an ordinary integral, and integrate w/r $t$ from $0$ to $1$.

For a general differential form, there’s also an intermediate step, in which you discard all terms not involving $\mathrm{d}t$ and move $\mathrm{d}t$ to the left in the remaining terms, taking care to get the signs right.

In your case, you have $$\omega = \mathrm{d}x_1 + \sin x_3\,\mathrm{d}x_2 + x_2 \cos x_3\,\mathrm{d}x_3.$$ Step 1 produces $$x_1\,\mathrm{d}t + x_2\sin{tx_3}\,\mathrm{d}t + tx_2x_3\cos{tx_3}\,\mathrm{d}t,$$ and integrating with respect to $t$ yields $$\int_0^1x_1+x_2\sin{tx_3}+tx_2x_3\cos{tx_3}\,\mathrm{d}t = x_1+x_2\sin{x_3}.$$ As for why this method works in general, that takes more explaining than I care to do here.