# Compatible connection over a riemannian manifold

How do I prove the following assertion:

Let $\nabla$ be a connection on a riemannian manifold. $\nabla$ is compatible with the metric if and only if for all vector fields $X,Y,Z$ we must have:

$X\langle Y,Z \rangle = \langle\nabla_X^Y,Z\rangle+\langle Y,\nabla_X^Z\rangle$

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What's your definition of "compatibility"? –  Neal May 12 '12 at 15:24
For every $X,Y$ parallel vector fields along a curve we have: $\langle X,Y \rangle$ = constant, but suppose you know that the product rule can be applied to $\frac{d\langle X,Y \rangle}{dt}$ –  Jr. May 12 '12 at 16:35
$\frac{d\langle X,Y \rangle}{dt}=\langle \frac{DX}{dt},Y\rangle + \langle X,\frac{DY}{dt}\rangle$ –  Jr. May 12 '12 at 16:40