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How do I prove the following assertion:

Let $\nabla$ be a connection on a riemannian manifold, then $\nabla$ is compatible with the metric if and only if for all $X,Y,Z\in \mathfrak{X}(M)=\Gamma(TM)$ we have:

$X\langle Y,Z \rangle = \langle\nabla_XY,Z\rangle+\langle Y,\nabla_XZ\rangle$

Jr.
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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

1 Answers1

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Look at the Proposition 3.2 and Corollary 3.3

Ehsan M. Kermani
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  • actualy I was trying to understand that corollary but stuck, but after some time I understand :) – Jr. May 12 '12 at 19:06
  • Dear Jr. Knowing what have tried out precisely, would be helpful to those who want to answer your questions. – Ehsan M. Kermani May 12 '12 at 19:09
  • 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 08:31