# Proof of a tensor identity involved in the derivation of the Einstein field equations?

On the wikipedia page for the einstein-hilbert action, the section for the derivation of the einstein field equations cites this identity: $$\sqrt{g} \nabla_\mu A^\mu = \partial_\mu (\sqrt{g} A^\mu)$$ Where $g$ is the determinant of the metric tensor, $\nabla_\mu$ is the covariant derivative, $\partial_\mu$ is the ordinary derivative with respect to the $x^\mu$ coordinate, $A^\mu$ is an arbitrary rank 1 tensor, Einstein summation notation is implied. So I tryed to prove this and started by first expanding the right side: $$\partial_\mu (\sqrt{g} A^\mu) = \frac{\partial_\mu g}{2\sqrt{g}} + \sqrt{g}\partial_\mu A^\mu = \sqrt{g}(\frac{\partial_\mu g}{2g}A^\mu + \partial_\mu A^\mu)$$ From here one must simply show that the expression in parentheses is equal to the definition of $\nabla_\mu A^\mu$ which is $$\partial_\mu A^\mu + \Gamma^\mu_{\mu \nu} A^\nu$$

Since we see both terms have the partial derivatives we just have to prove that $$\frac{\partial_\mu g}{2g}A^\mu = \Gamma^\mu_{\mu \nu}A^\nu$$

or $$\frac{\partial_\nu g}{g} = g^{\mu \alpha}(\partial_\mu g_{\nu \alpha} + \partial_\nu g_{\mu \alpha} - \partial_\alpha g_{\mu \nu})$$

I've tried this by plugging in the definition of the metric tensor's determinant both in tensor notation and fully written out. Is there some formula in tensor notation for the components of the inverse of the metric tensor? I feel like that would help a lot. Otherwise, could you direct me to a proof of the identity itself? Thank you!

The first thing to notice is in your last equation, the symmetry of $g^{\mu\alpha}$ means that $$g^{\mu\alpha} \partial_{\mu} g_{\nu \alpha} = g^{\mu\alpha} \partial_{\alpha} g_{\mu\nu},$$ so those two terms cancel and we are left with proving that $$(\partial_{\nu} \log{g} =) \, \frac{\partial_{\nu} g}{g} = g^{\mu\alpha}\partial_{\nu} g_{\mu\alpha} \tag{1}$$ We have two options here: we can use some identities involving matrix logarithms, but easier is to remember how we derive the inverse of a matrix.

Recall the expansion of the determinant in minors, $$\det{g} = \sum_{\nu \text{ only}} g_{\mu\nu} M^{\mu\nu}, \tag{2}$$ where $M^{\mu\nu}$ is $(-1)^{\mu+\nu}$ times the subdeterminant formed by deleting the $\mu$th row and $\nu$th column of the matrix of $g$, (there's also a transpose in here somewhere that we can ignore since the metric is symmetric) and most importantly for us, $M^{\mu\nu}$ does not depend on $g_{\mu\nu}$. Then the inverse matrix is $$g^{\mu\nu} = \frac{1}{g}M^{\mu\nu}$$

Now, we can get to (1): differentiating (2) with respect to $x^{\alpha}$, we find $$\partial_{\alpha} g = \frac{\partial g}{\partial g_{\mu\nu}} g_{\mu\nu,\alpha} = M^{\mu\nu}g_{\mu\nu,\alpha} = g \, g^{\mu\nu}g_{\mu\nu,\alpha},$$ which is (1) after we change the indices and divide by $g$.

Use metric compatibility on the left hand side. First, note that the traditional expression for metric compatibility is

$$\nabla \sqrt{g} = 0$$

Note also that this implies, for a unit pseudoscalar $\hat \epsilon$,

$$\nabla [\sqrt{g} \hat \epsilon] = \nabla \epsilon = 0$$

where $\epsilon$ is the coordinate pseudoscalar (volume form). This is, perhaps, the more fundamental equation in the first place.

Now, clifford multiply the left-hand side by $\hat \epsilon$, and we get

$$(\nabla \cdot A) \sqrt{g}\hat \epsilon = (\nabla \cdot A) \epsilon = \nabla \wedge (A \epsilon)$$

Again, the last equality follows by metric compatibility.

Because the connection is torsion-free, $\nabla \wedge X = \partial \wedge X$, so we get

$$\nabla \wedge (A \epsilon) = \partial \wedge (A \epsilon)$$

Note that $\hat \epsilon$ can be moved out of this expression, leaving $\sqrt{g}$ behind, and we get

$$(\nabla \cdot A) \sqrt{g} \hat \epsilon = (\partial \cdot [A \sqrt{g}]) \hat \epsilon$$

Cancel out the $\hat \epsilon$ from each side, and we're done.