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I have a question about the answer to this question

I don't know how to get the second equality in

$$\nabla_{E_i} E_j (\gamma(t)) = \nabla_{\gamma'} E_j (\gamma(t)) = 0$$

I tried to use the fact that the covariant derivative of the geodesic $\gamma$ is zero and to write the tangent vector along the curve in terms of the $E_j$'s but it doesn't work out.

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user136592
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  • That follows from the definition of $E_j$'s: For each $q$, let $\gamma$ be the (unique) geodesic so that $\gamma(0) = p, \gamma(1) = q$. Then define $E_j(q)$ to be the parallel transport of $E_j(0)$ along $\gamma$. Thus for each $t$, we have $t\mapsto E_j(\gamma(t))$ is the parallel transport along $\gamma$, thus $\nabla_{\gamma'} E_j(\gamma(t)) = 0$ for all $t$. –  Jun 02 '16 at 02:29

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