> 
> Out of all of my time learning General relativity, this is the one identity that I cannot get around.
>  $$ \Gamma\_{\alpha \beta}^{\alpha} = \partial\_{\beta}\ln\sqrt{-g} \tag{1}$$
>  where $g$ is the determinant of the metric tensor $g\_{\alpha \beta}$.
> 
> 
> 

With the Christoffel symbol, we start by contracting

$$ \begin{align}
 \Gamma\_{\alpha \beta}^{\alpha} &= \frac{1}{2} g^{\alpha\gamma} (\partial\_{\alpha} g\_{\beta\gamma} + \partial\_{\beta} g\_{\alpha\gamma} - \partial\_{\gamma} g\_{\alpha\beta} ) \\
&= \frac{1}{2} g^{\alpha\alpha} ( \partial\_{\beta} g\_{\alpha\alpha}) \\
&= \frac{1}{2g\_{\alpha\alpha}} ( \partial\_{\beta} g\_{\alpha\alpha}) 
\end{align}\tag{2}$$ 

where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g\_{\alpha\alpha}$.

The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,

$$\det A = \det||A^{\lambda}\_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}\_{\ \ \alpha}A^{1}\_{\ \ \beta}A^{2}\_{\ \ \gamma}A^{3}\_{\ \ \delta} $$

$$(A^{-1})^{\mu}\_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta\_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}\_{\ \ \nu} A^{\gamma}\_{\ \ \rho}A^{\delta}\_{\ \ \sigma} $$

$$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$
where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}\_{\ \ \mu}||$ whose entries are one-forms.

I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.