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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.

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  • $\begingroup$ In (3), the second term vanishes due to antisymmetry, and the first and third terms give $\sqrt{g}\partial_bT^{ab}+\partial_b\sqrt{g}T^{ab}=\partial_b(\sqrt{g}T^{ab})$ by the Leibniz rule. In (4) there is a mistake since in the first term of (4) the derivative should act on $T$ only. $\endgroup$ Commented Oct 16, 2021 at 20:25
  • $\begingroup$ I meant to write the second term of (4) as the second term in (3), my mistake. I don't understand explicitly why the second term in (3) would cancel. Does that come from expanding the Christoffel symbol in terms of the metric? If I understand why it vanishes then my problem is solved. $\endgroup$
    – bsafaria
    Commented Oct 16, 2021 at 20:33

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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 $0$.

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