Is $0 = \nabla_Xg(Y, -)(Z) = \nabla_X(g(Y, Z)) - g(Y, \nabla_XZ)$ if $\nabla g= 0$?

Question

Suppose a manifold $M$ with a metric $g_{\mu\nu}$.

I know that the covariant derivative (I'm assuming the connection induced by the metric) of a covector is given by:

$(\nabla_X \eta)(Y)=\nabla_X (\eta(Y))-\eta(\nabla_X Y)$

Also, $\nabla g=0$. So is it true that $0=\nabla_X g(Y,-)(Z)=\nabla_X(g(Y,Z))-g(Y,\nabla_X Z)$?

I think this is wrong since in abstract index notation we usually rise indices with the metric and then use the covariant derivative which is not always 0!

Could someone clarify this to me by using "index-free" point of view versus abstract index notation?

Answer 1

For any vector field $Y$, we obtain a covector field (i.e. one-form) $\eta$ given by $\eta(Z) = g(Y, Z)$. What you call $\nabla_Xg(Y, -)(Z)$ is precisely $(\nabla_X\eta)(Z)$. Using the identity in your question, we see that

\begin{align*} (\nabla_X\eta)(Z) &= \nabla_X(\eta(Z)) - \eta(\nabla_XZ)\\ &= \nabla_X(g(Y, Z)) - g(Y, \nabla_XZ)\\ &= g(\nabla_XY, Z) + g(Y, \nabla_XZ) - g(Y, \nabla_XZ)\\ &= g(\nabla_XY, Z) \end{align*}

where we have used implicitly the fact that $\nabla g = 0$ if and only if $\nabla_X(g(Y, Z)) = g(\nabla_XY, Z) + g(Y, \nabla_XZ)$ for all vector fields $X, Y,$ and $Z$ - see this question.

So we see that $(\nabla_X\eta)(Z) = 0$ if and only if $g(\nabla_XY, Z) = 0$. Furthermore, $\nabla_X\eta = 0$ if and only if $\nabla_XY = 0$.