-1
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.