Let $M$ denote a Riemannian manifold, $\gamma$ a geodesic of $M$ defined on $\mathbb{R}$. Let $t_{0} \in \mathbb{R}$ and $(\alpha,\beta) \in \mathbb{R}^{2}$. I define the reparametrized geodesic $\tilde{\gamma} \, : \, t \, \mapsto \, \gamma(\alpha t + \beta )$. Let $v \in T_{\tilde{\gamma}(t_{0})}M$. $\mathrm{P}_{t_{0},t,\tilde{\gamma}}(v)$ denotes the parallel transport of $v$ along the geodesic $\tilde{\gamma}$ from the point $\tilde{\gamma}(t_{0})$ to the point $\tilde{\gamma}(t)$.

I am wondering whether the following is true : $\boxed{\displaystyle \mathrm{P}_{t_{0},t,\tilde{\gamma}}(v) = \mathrm{P}_{\alpha t_{0}+\beta,\alpha t + \beta, \gamma}(v)}$

My idea is the following : let $V \, : \, t \, \mapsto \, \mathrm{P}_{\alpha t_{0}+\beta,\alpha t + \beta,\gamma}(v)$. $V$ is a vector field along the curve $\gamma$ and it satisfies : $V(t_{0})=v$. In order to prove that, for all $t$, $V(t) = \mathrm{P}_{t_{0},t,\tilde{\gamma}}(v)$, I would need to prove that :

$$ \frac{D_{\tilde{\gamma}}V}{dt}(t) = 0 $$

where $\frac{D_{\tilde{\gamma}}}{dt}$ denotes the covariant derivative of $V$ along the curve $\tilde{\gamma}$. After some calculations, I find that :

$$ \frac{D_{\tilde{\gamma}}V}{dt}(t)=\alpha \frac{D_{\gamma}V}{dt}(\alpha t + \beta) \tag{$\star$}$$

Since $V$ is parallel along $\gamma$, I get $\displaystyle \frac{D_{\tilde{\gamma}}V}{dt}=0$. Is that correct ?


This is correct. More is true: In your notation, for any continuous piecewise smooth curve $c$ in $M$ and reparameterization $f$ one has: $$ P_{t_0, f(t_1), c} = P_{t_0, t_1, c\circ f}\circ P_{t_0, f(t_0), c}. $$ See 24.1 of here, e.g. The proof is similar to yours.


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