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:

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

  • $\begingroup$ What's your definition of "compatibility"? $\endgroup$ – Neal May 12 '12 at 15:24
  • $\begingroup$ 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 $\frac{d\langle X,Y \rangle}{dt}$ $\endgroup$ – Jr. May 12 '12 at 16:35
  • $\begingroup$ $\frac{d\langle X,Y \rangle}{dt}=\langle \frac{DX}{dt},Y\rangle + \langle X,\frac{DY}{dt}\rangle$ $\endgroup$ – Jr. May 12 '12 at 16:40

Look at the Proposition 3.2 and Corollary 3.3

  • $\begingroup$ actualy I was trying to understand that corollary but stuck, but after some time I understand :) $\endgroup$ – Jr. May 12 '12 at 19:06
  • $\begingroup$ Dear Jr. Knowing what have tried out precisely, would be helpful to those who want to answer your questions. $\endgroup$ – Ehsan M. Kermani May 12 '12 at 19:09
  • $\begingroup$ 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.) $\endgroup$ – New day rising Jan 24 '17 at 8:31

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