Covariant Derivative Clarification In my notes I have the following when taking the divergence, $\partial_\mu$ of $\partial_\alpha\varphi^\alpha g^{\mu\nu}$
$$
\partial_\mu \partial_\alpha \varphi^\alpha g^{\mu\nu}
= \partial_\nu \partial_\alpha \varphi^\alpha
$$
where $g^{\mu\nu}$ is the metric tensor.
I would have thought I should act the metric on the derivative to get obtain,
$$
\partial_\mu \partial_\alpha \varphi^\alpha g^{\mu\nu}
= \partial^\nu \partial^\alpha \varphi_\alpha
$$
where I just reversed the contracted indices to look pretty.
 A: Something fundamental to remember is that metric tensors are metrilinic with respect to covariant derivatives in the sense that:
$$\partial_\alpha g^{\mu\nu}=0,\text{ for all indices.}$$
So for this reason, it can be 'pulled through' any covariant derivatives. In addition, the metric tensor has the property that it can raise indices. So the expression is reduced to:
$$\begin{align}
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu}) \\
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=\partial_\mu(g^{\mu\nu}\partial_\alpha\phi^\alpha+\phi^\alpha\partial_\alpha g^{\mu\nu}) \\
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=\partial_\mu(g^{\mu\nu}\partial_\alpha\phi^\alpha) \\
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=\partial_\alpha\phi^\alpha\partial_\mu g^{\mu\nu}+g^{\mu\nu}\partial_\mu\partial_\alpha\phi^\alpha \\
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=g^{\mu\nu}\partial_\mu\partial_\alpha\phi^\alpha \\
\partial_\mu\partial_\alpha(\phi^\alpha g^{\mu\nu})&=\partial^\nu\partial_\alpha\phi^\alpha
\end{align}$$
This can be simplified in vector form as:
$$\nabla(\nabla\cdot\vec{\phi})$$
