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$(M,g)$ is a complete Riemannian manifold. $\gamma:[0,+\infty)\rightarrow M$ is geodesic. And $$ P_t: T_{\gamma(0)}M \rightarrow T_{\gamma(t)} M $$ is parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma (t)$. And $v(t)\subset T_{\gamma(0)}M$ is time-dependent. I feel there is $$ \frac{D}{dt}\Big |_{t=0}~ P_t(v(t)) = \frac{D}{dt}\Big |_{t=0}~ v(t) $$ where $\frac{D}{dt}$ is the covariant derivative along $\gamma$. Namely, the differential of parallel transport at $t=0$ is identity (in fact, I feel it always be identity). But I don't know how to prove it.

This problem is from the proof of 2.1 Theorem of chapter 8 of do Carmo's Riemannian Geometry.

Enhao Lan
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1 Answers1

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Here is a sketch: Let $X_i(t)$ be a parallel frame along $\gamma$ and write $v(t)=\sum_i\alpha_i(t)X_i(0)$. Then $P_t(v(t))=\sum_i\alpha_i(t)X_i(t)$ and hence by the product rule

$$ \frac{D}{dt}\Big |_{t=0}~ P_t(v(t))=\sum_i\alpha_i'(0)X_i(0)=\frac d{dt}\Big |_{t=0}v(t) $$ Note: $\gamma$ does not have to be a geodesic.

Claire
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