# 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:

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

• 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

• Do Carmo says at the beginning of the proof of Proposition 3.2: "It is obvious that the equation $$\frac d{dt} \langle X,Y \rangle = \langle \frac{DX}{dt},Y\rangle+\langle X,\frac{DY}{dt}\rangle, \quad t\in I$$ implies that $\nabla$ is compatible with $\langle \, , \, \rangle$." How exactly does this equation imply that $\frac d{dt}\langle X,Y\rangle=0$, so that the desired conclusion follows? (I'm self-studying this topic right now.) – New day rising Jan 24 '17 at 8:31