If $T^{ab}$ is an antisymmetric tensor, prove:
$$T^{ab}_{;b} = \frac{1}{\sqrt{g}}\partial_b(\sqrt{g}T^{ab}).$$
In this example, $g=|\text{det}g_{ab}|$. I already proved in a previous example that $\Gamma_{ab}^b = \partial_a\text{ln}\sqrt{g} = \frac{1}{\sqrt{g}}\partial_a\sqrt{g}$.
Attempt:
The covariant derivative can be expanded to
$$T^{ab}_{;b} = \partial_bT^{ab} + \Gamma_{bd}^aT^{bd} + \Gamma^{b}_{bd}T^{ad}. (1)$$
$\Gamma^{b}_{bd} = \Gamma^{b}_{db}$ can be written using the identity above, as
$$\Gamma^{b}_{bd} = \frac{1}{\sqrt{g}}\partial_b\sqrt{g}. (2)$$
Substituting this back into (1) and pulling out $1/\sqrt{g}$.
$$T^{ab}_{;b} = \frac{1}{\sqrt{g}}(\sqrt{g}\partial_bT^{ab} + \sqrt{g}\Gamma_{bd}^aT^{bd} + \partial_b\sqrt{g}T^{ad}). (3)$$
I then group the first and third terms.
$$T^{ab}_{;b} = \frac{1}{\sqrt{g}}\left[\partial_b(\sqrt{g}T^{ab}) + \sqrt{g}\Gamma_{bd}^aT^{bd}\right]. (4)$$
This is not right obviously. I'm not sure how to use the antisymmetric properties to arrive at the answer. I was able to prove some similar identity by taking the covariant derivative of a (1,0) tensor, so I tried the same thing here with this tensor. I'm new to tensor algebra/calculus so I'm not confident in my process.
Best Answer
Slight correction: since $\Gamma_{bd}^b=\frac{1}{\sqrt{|g|}}\partial_\color{red}{d}\sqrt{|g|}$,$$T_{;b}^{ab}=\partial_\color{blue}{d}T^{a\color{blue}{d}}+\color{limegreen}{\Gamma_{bd}^aT^{bd}}+\Gamma_{bd}^dT^{ad}=\frac{1}{\sqrt{|g|}}\partial_d\left(\sqrt{|g|}T^{ad}\right),$$where the blue indices have been renamed from $b$ to $d$ to make a subsequent use of the Leibniz rule easier to notice. The green term vanishes for the reason @BenceRacskó noticed, namely contracting a $bd$-symmetric Christoffel symbol with the antisymmetric tensor. For example, $\Gamma_{bd}^aT^{bd}$ includes both $\Gamma_{12}^0T^{12}$ and $\Gamma_{21}^0T^{21}=\Gamma_{12}^0\cdot-T^{12}=-\Gamma_{12}^0T^{12}$, but these sum to $0$.