Problem:
If $T^{ab}$ is an antisymmetric tensor, show $T^{ab}_{;ab} = 0$
Attempt:
Here are some identities I found earlier that might be useful:
$\Gamma^{a}_{ab} = \frac{1}{\sqrt{g}}\partial_b\sqrt{g}. (1)$
$T^{ab}_{;b} = \frac{1}{\sqrt{g}}\partial_b(\sqrt{g}T^{ab}). (2)$
$T^{ab} = -T^{ba}. (3)$
where $g=|det(g_{ab})|$. $g_{ab}$ is a metric tensor.
Now:
$$T^{ab}_{;a} = \partial_aT^{ab} + \Gamma^{a}_{ad}T^{db} + \Gamma^{b}_{ad}T^{ad}.(4)$$
The third term of $(4)$ is zero because of the contraction of the symmetric Christoffel with the antisymmetric tensor. Therefore we can express $T^{ab}_{;ab}$ as
$$T^{ab}_{;ab} = \nabla_b(\partial_aT^{ab}) + \nabla_b(\Gamma^{a}_{ad}T^{db}). (5)$$
So either both terms cancel, or both are zero. For the first term, it does not look like I could expand it to something useful. As for the second term
$$\nabla_b(\Gamma^{a}_{ad}T^{db}) = \nabla_b\Gamma^{a}_{ad}T^{db} + \Gamma^{a}_{ad}\nabla_bT^{db}. (6)$$
I cannot see anything immediately obvious. I tried expanding further and using antisymmetric properties to cancel terms, but got no luck with that.
We have: $$T^{ab}_{;ab} = - T^{ba}_{;ab} = -T^{ab}_{;ba} $$ The first comes from antisymetry, the second from relabeling (swap the names of $a$ and $b$. Because they are contracted the labels we use for them don't matter). We obtain: $$2T^{ab}_{;ab} = T^{ab}_{;ab} + T^{ab}_{;ab}= T^{ab}_{;ab} - T^{ab}_{;ba} = 2T^{ab}_{;[ab]}$$
The Ricci Identity allows us to calculate the anticommutator of a tensor with respect to the covariant derivative:
$$2T^{ab}_{;[cd]} = T^{ax}{R^{b}}_{xcd} + T^{xb}{R^{a}}_{xcd}\\ 2T^{ab}_{;[ab]} = T^{ax}{R^{b}}_{xab} + T^{xb}{R^{a}}_{xab} = \\= -T^{ax}{R^{b}}_{xba} + T^{xb}{R^{a}}_{xab} = \\ = -T^{ax} R_{ax} + T^{xb}R_{xb} $$ As the Ricci tensor is $R_{ab}$ is symmetric and $T^{ab}$ is antisymetric, we get: $$ T^{ab}_{;[ab]} = T^{ab}_{;ab} = 0$$