# Differential of parallel transport

$$(M,g)$$ 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.

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.