If I'm given $\xi^b$, how do I calculate $\omega^{ab}$ given $\xi^b = \nabla_a \omega ^{ab}$?

Question

I'm given a one-form, $\xi^b$. Then, I'm given a relation for a two-form, $\omega^{ab}$, which says that

$$ \xi^b = \nabla_a \omega^{ab}$$

How do I calculate $\omega^{ab}$ given this relation?

Answer 1

You can calculate it in terms of the Christoffel symbols $$\nabla_a \omega^{ab} = \delta_{c}^a \nabla_a \omega^{cb} = \delta_{c}^a (\partial_a\omega^{cb} + \Gamma_{da}^{c} \omega^{db}+\Gamma_{ad}^{b} \omega^{cd}) = \partial_a\omega^{ab} + \Gamma_{da}^{a} \omega^{db}+\Gamma_{ad}^{b} \omega^{ad},$$

since $$\nabla_a \omega^{cb}=\partial_a\omega^{cb} + \Gamma_{da}^{c} \omega^{db}+\Gamma_{ad}^{b} \omega^{cd}.$$

If you suppose the connection is Levi-Civita, then you can simplify it more.

Notice $\Gamma_{ad}^{b} \omega^{ad} = 0$ because $$\sum_{a,d}\Gamma_{ad}^{b} \omega^{ad} = \sum_{a<d}\Gamma_{ad}^{b} \omega^{ad}+ \sum_{a>d}\Gamma_{ad}^{b} \omega^{ad} = \sum_{a<d}\Gamma_{ad}^{b} \omega^{ad} + \sum_{d>a}\Gamma_{da}^{b} \omega^{da} =0,$$ since $\omega^{ab}$ is an antisymetric tensor.

So we have $\nabla_a \omega^{ab} = \partial_a\omega^{ab} + \Gamma_{da}^{a} \omega^{db}$.

You can also write down $\nabla_a \omega^{ab}$ in terms of the metric using the following formula $$\Gamma_{ab}^c = \frac{g^{cl}}{2}(\partial_a g_{lb} + \partial_b g_{la} - \partial_{l}g_{ab}).$$

Answer 2

You can't "calculate" $\omega$ in terms of $\xi$. This is an underdetermined partial differential equation, and sometimes it has a solution and sometimes not; when it does have a solution, it often has an infinite-dimensional space of them.

Your equation can be written invariantly as $\delta\omega = \xi$, where $\delta$ is the codifferential. Since $\delta\circ\delta = 0$, a necessary condition for the existence of a solution, even locally, is $\delta\xi = 0$.

If $\xi$ satisfies this compatibility condition, then there will be solutions, at least locally in a neighborhood of each point. But if $\omega_0$ is any solution at all, then $\omega_0 + \eta$ is also a solution whenever $\eta$ is any $2$-form satisfying $\delta\eta=0$. For example, in dimensions greater than $2$, you can take $\alpha$ to be any $3$-form at all, and take $\eta = \delta\alpha$. This gives an infinite-dimensional space of solutions to your problem. (In dimension $2$, the only possibilities for $\eta$ are locally constant multiples of the Riemannian volume form.)

If you're just interested in local solutions, then $\delta\xi=0$ is the only obstruction to the existence of solutions.

Theorem. Suppose $\xi$ is a smooth $1$-form on a Riemannian manifold satisfying $\delta\xi=0$. Then in a neighborhood of each point, there is a smooth $2$-form $\omega$ such that $\delta\omega=-\xi$.

Proof: The equation we want to solve can be written in terms of the Hodge star operator as $$ (-1)^{n+1} * d * \omega = - \xi,\tag{1} $$ where $n$ is the dimension of the manifold. Since $d*\omega$ is an $(n-1)$-form, $\ast\ast(d\ast\omega) = (-1)^{n+1}d\ast\omega$, so $(1)$ is equivalent to $$ d(*\omega) = -*\xi. \tag{2} $$ The assumption $\delta\xi=0$ implies that $*\xi$ is closed, so $(2)$ has a solution in a neighborhood of each point by the Poincaré lemma. $\square$

Most proofs of the Poincaré lemma give some sort of explicit formula that you might use to construct a local solution, at least in low dimensions.

If you want a global solution, the theory is most satisfactory on a compact smooth manifold (without boundary). In that case, one can prove the following.

Theorem. Suppose $(M,g)$ is a smooth compact Riemannian manifold and $\xi$ is a smooth $1$-form on $M$. Then there exists a smooth $2$-form $\omega$ satisfying $\delta\omega=-\xi$ if and only if $\delta\xi=0$ and $\int_M \langle \xi,\beta\rangle\,dV_g = 0$ for every harmonic $1$-form $\beta$.

I don't have a handy reference for the proof, but it's a fairly standard application of the Hodge theorem for de Rham cohomology.