Equivalent definition of metric compatibility for a connection: what does $\nabla g$ mean?

Question

So the definition I know for metric compatibility is:

$$Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ),$$

which make sense, as $g(Y,Z)$ is a smooth function from the manifold to reals and we think of $X$ as a derivation. So now I read this apparent equivalent definition that says $\nabla g=0$. Can someone explain what this means? How can I do $\nabla$ of $g$ I thought $g_p$ is an element of $T_p^*M\otimes T_p^*M$ at every point. Furthermore after you explain the meaning of this can you show me that these two definitions are indeed equivalent?

Answer 1

It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction.

Given a connection

\begin{align*} \nabla : \Gamma(TM) \times \Gamma(TM) &\to \Gamma(TM)\\ (X, Y) &\mapsto \nabla_XY \end{align*}

on $TM$, there is an associated connection (which I will also denote $\nabla$) on $T^*M$ given by

\begin{align*} \nabla : \Gamma(TM) \times \Gamma(T^*M) &\to \Gamma(T^*M)\\ (X, \alpha) &\mapsto \nabla_X\alpha \end{align*}

where $(\nabla_X\alpha)(Y) := X(\alpha(Y)) - \alpha(\nabla_XY)$. With this definition, together with the definition $\nabla_Xf = Xf$ for a smooth function $f$, we see that the following identity holds:

$$\nabla_X(\alpha(Y)) = (\nabla_X\alpha)(Y) + \alpha(\nabla_XY).$$

More generally, given a $(p, q)$-tensor $T$, we implicitly define the covariant derivative $\nabla_XT$, which is again a $(p, q)$-tensor, by the following equation:

\begin{align*} \nabla_X(T(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q)) =&\ (\nabla_XT)(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q)\\ &+ \sum_{i=1}^pT(Y_1, \dots, \nabla_XY_i, \dots, Y_p, \alpha_1, \dots, \alpha_q)\\ &+ \sum_{j=1}^qT(Y_1, \dots, Y_p, \alpha_1, \dots, \nabla_X\alpha_j, \dots, \alpha_q).\qquad (\ast) \end{align*}

One could instead consider the covariant derivative of $T$ as a $(p+1, q)$-tensor $\nabla T$ given by

$$(\nabla T)(X, Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q) := (\nabla_XT)(Y_1, \dots, Y_p, \alpha_1, \dots, \alpha_q).$$

Now, $g$ is a $(2, 0)$-tensor. So if $Y$ and $Z$ are vector fields, $g(Y, Z)$ is a smooth function and hence

$$\nabla_X(g(Y, Z)) = X(g(Y, Z)).$$

On the other hand, by $(\ast)$,

$$\nabla_X(g(Y, Z)) = (\nabla_Xg)(Y, Z) + g(\nabla_XY, Z) + g(Y, \nabla_XZ).$$

Using these two equations, we see that

$$(\nabla g)(X, Y, Z) = (\nabla_X g)(Y, Z) = X(g(Y, Z)) - g(\nabla_XY, Z) - g(Y, \nabla_XZ).$$

So $\nabla$ is compatible with the metric $g$ if and only if $(\nabla g)(X, Y, Z) = 0$ for all vector fields $X, Y, Z$ (i.e. $\nabla g = 0$).

Answer 2

It means the same thing, $\nabla g=0$ is equivalent to saying that $g$ is parallel relatively to the connection $\nabla$ which is equivalent to saying that $Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_X,Z)$.

It is the answer, the covariant derivative of the $n$ tensor $T$ is defined by $\nabla_XT(X_1,..,X_n)=X.T(X_1,..,X_n)-\sum_i(X_1,..,\nabla_XX_i,..,X_n)$