Divergence of commutator

Question

Let $M$ be a Riemannian manifold and $X,Y$ two vector fields. I want to derive some closed expression for $\mbox{div }[X,Y]$ where $[X,Y]$ is the commutator. Does anyone know some sugestion or reference?

Answer 1

div([X,Y])=X(div(Y))-Y(div(X))

Answer 2

I like zhanghtam's answer, as the fundamental challenge here is that the divergence operation must after all be applied to vector fields; $XY$ and $YX$ are not such as discussed on this related post.

Otherwise, This pdf very thoroughly disusses how to form the divergence of a vector field in a Riemannian geometry setting. What it ends up deriving is

$$\mathrm{div} \vec{F} = \frac{\partial F^i}{\partial x^i} + F^k \Gamma^{i}_{ki}$$

where their is implied summation over the $k$ index and $\Gamma^{i}_{ki}$ are the usual Christoffel symbols {of the second kind).

Answer 3

Let me show you a calculation.

With the conventions from this article, $$ \mathrm{div}X := \nabla_a X^a $$ for a vector field $X$ and the Levi-Civita connection $\nabla$ of a Riemannian metric $g$, chosen in the manifold $M$ under consideration. Here I am using the abstract index notation.

Since the Levi-Civita connection $\nabla$ is torsion-free, we can write $$ [X, Y] = \nabla_X Y - \nabla_Y X $$ or, adding the abstract indices, $$ [X, Y]^b = X^a \nabla_a Y^b - Y^a \nabla_a X^b $$ With these identities at hand, we can compute: $$ \mathrm{div} [X, Y] = \nabla_b (X^a \nabla_a Y^b - Y^a \nabla_a X^b) \\ = (\nabla_b X^a) \nabla_a Y^b + X^a \nabla_b \nabla_a Y^b - (\nabla_b Y^a) \nabla_a X^b - Y^a \nabla_b \nabla_a X^b \\ = X^a \nabla_b \nabla_a Y^b - Y^a \nabla_b \nabla_a X^b \\ = X^a \nabla_a \nabla_b Y^b - X^a R_{a b}{}^b{}_c Y^c - Y^a \nabla_a \nabla_b X^b + Y^a R_{a b}{}^b{}_c X^c \\ = X^a \nabla_a \nabla_b Y^b- Y^a \nabla_a \nabla_b X^b \\ = X \mathrm{div} Y - Y \mathrm{div} X $$ as claimed by zhanghtam.

Of course, one needs to make use of the Ricci identity $$ \nabla_a \nabla_b X^c - \nabla_b \nabla_a X^c = R_{a b}{}^c{}_d X^d $$ and the swap symmetry of the Riemannian curvature $R_{a b}{}^c{}_d$: $$ R_{abcd} = R_{cdab} $$ to be precise.