I am taking a differential geometry course and over 90% of the class doesn't understand the lecturer (the lecturer is a phd student and doesn't understand what we know already) , I was hoping someone can explain to me what you need to do to show that a one-form is or isn't a differential of a function, Thanks.
Given a 1-form such as $\alpha = x^2x^1 dx^1 - x^3x^2 dx^2 + x^3 x^1 dx^3$
How can you show $\alpha$ is or is not a differential of a function?
Note: the one-form is in 3-dimensional real space with coordinates $(x^1,x^2,x^3)$
Thanks for the help.
One can always proceed naively by supposing there is such a function, say, $f(x, y, z)$, differentiating to compute $df = \sum f_{x^a} \,dx^a$, comparing like coefficients, and solving the resulting p.d.e.
In the case of the example function, comparing the $dx^1$ coefficients gives $f_{x^1} = x^1 x^2$, so integrating gives $f(x^1, x^2, x^3) = \frac{1}{2} (x^1)^2 x^2 + g(x^2, x^3)$ for some $g$. Now, what does comparing the other coefficients give?
As Lord Shark the Known pointed out in a comment under the original question, if $\alpha = df$ for some $f$, then $d\alpha = d^2 f = 0$, giving a necessary condition for $\alpha$ to be a differential of a function. This has the advantage that the check requires only differentiation, not integration. In our case, we have $d\alpha = -x^1 dx^1 \wedge dx^2 + x^2 dx^2 \wedge dx^3 + x^3 dx^1 \wedge dx^3 \neq 0$, so we can conclude immediately that $\alpha$ is not exact. This necessary condition is also sufficient if the domain of $\alpha$ is simply connected, but otherwise it need not be.