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This is a crosspost from math.stackexchange

Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then we have to ways to map $T_pM$ to $T_{c(t_0)} M$ isomorphically. One is the parallel transport along $c$, let's call it $P_{c, 0, t_0}$ and the other is given by $$ d \exp_p|_{tX}: T_{tX}T_pM \cong T_p M \longrightarrow T_{\exp_p(tX)}M = T_{c(t_0)}.$$

My question is: What is the relation between those two? Are there formulas which relate the two concepts with curvature terms?

The parallel transport is a linear isometry, and the derivative of the exponential map is a radial isometry by the Gauss lemma, meaning $$ \langle d \exp_p|_{tX} \cdot Y, \dot{c}(t) \rangle = \langle Y, X \rangle $$ for all $Y \in T_pM$.

In two dimensions, this means that the two mappings coincide up to scaling, as there is only on orthogonal direction to the radial one. However, in higher dimensions, this is not true, i guess. However, on $S^3$, I computed that the derivative of the exponential map conincides with the parallel transport except that vectors orthogonal to the direction of parallel transport are multiplied by $\frac{\sin r}{r}$.

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    $\begingroup$ If you crosspost, make sure to make a link in both directions so that people don't end up wasting their time... $\endgroup$ – Anthony Quas Oct 22 '13 at 6:36
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To understand the realationship between $Q$ and $P$, it suffices to study how they act on an orthornormal(o.n.) basis. Here is the detailed argument:

Let $J_i(t)$ be the Jacobi field along $c$, with $J_i(0)=0$ and $J_i'(0)=e_i$, where $e_1, \cdots, e_{n-1}, X$ is an o.n basis of $T_p(M)$. Then one can show that $J_i(t)=Q_t(e_i)$, since $$ Q_t(e_i)=dexp_{tX} (te_i). $$

Now extend $e_i$ to $E_i(t)$ along $c$ via parallel transport $P$. One can write the $J_i$ (i.e. $Q_t$) in terms of the o.n. basis $E_i(t)$: $$ Q_t(e_i)=J_i(t)=\sum_j a_{ij}E_j(t)=\sum_j a_{ij}P_t(e_j). $$

Let $A=(a_{ij})$. Using Jacobi equation, one can show that $$ A''(t)+R(t)A(t)=0, $$ where $R(t)=(R_{ij}(t))$ is the curvature term defined by $R_{ij}(t)=\langle R(E_j, \dot{c}) \dot{c}, E_i \rangle$.

This is essentially Jacobi equation. So roughly speaking, $Q$ and $P$ satisfy the Jacobi equation.

remark: It seems my definition of $Q$ differs by a rescaling fact $t$. So if we use $N$ to denote your map, then $Q_t(e_i)= N(t e_i)$.

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  • $\begingroup$ Shouldn't the curvature term be $R_{ij}(t) = \langle R(E_j, \dot{c}) \dot{c}, E_i \rangle$? I submitted an edit. $\endgroup$ – seub Aug 30 '18 at 20:13
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Let $c$ be a geodesic with $c'\ne 0$ in a Riemann manifold $(M,g)$ and let $Y$ be a Jacobi field along $c$ with $Y(0)=0$. Then we have $$ Y(t) = T_{t.\dot c(0)}(\exp_{c(0)}) {vl}\big(t.\dot c(0),t.(\nabla_{\partial_t}Y)(0)\big). $$ Here $vl: TM\times_M TM \to T^2M$ is the vertical lift. See page 350 of (here). The ODE for Jacobi fields is second order and involves the curvature operator. The ODE for parallel transport is first order constant.

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This is sort of a non-answer answer, but possibly illustrative.

Note that your identification $T_{t_0X}(T_pM)\cong T_p M$ is precisely parallel translation in the Euclidean space $T_p M$. In some sense, by wanting to compare your revised $d\exp_p|_{t_0X}$ to $P_{c,0,t_0}$, you're just "undoing" that identification. The "true" $d\exp_p|_{t_0X}$ maps $T_{t_0X}(T_pM)$ to $T_{c(t_0)}M$, and it is just measuring distortion of normal coordinates at $t_0X$.

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Let $P_t:T_p\to T_{c(t)}$ be the parallel transport and $Q_t:T_p\to T_{c(t)}$ be your map. Given $J\in T_p$, the field $J(t)=Q_t(J)$ is a Jacoby field and $J'(0)=0$. Set $$W_t=P_t-Q_t.$$ It follows that

  • $W_t=O(t^2)$;
  • the value $W_t''$ can be expressed from the curvature tensor.
  • As you noticed $W_t(\tau)\equiv 0$ where $\tau=c'(0)\in T_p$.

I guess that as much as you can get.

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  • $\begingroup$ If I'm not mistaken, is $J'(0)=0$ or rather $J(0)=0, J'(0)=Y$ where $J(t)=(Dexp_p)_{tX}(tY)$? If this is the case, why is $W(t)=O(t^2)$, as at $0$, the difference is Y? $\endgroup$ – Let's talk math Apr 27 '15 at 22:47
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A detailed answer to the original question is Sections 3.1/3.3 in arXiv/1012.3980, which expand on the earlier answers.

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Cartan's Theorem (see for example Do Carmo, "Riemannian Geometry", p. 157) gives an answer to the question. I give here some detail:

Let $M,\widetilde{M}$ be two Riemannian manifolds of same dimension and let $i:T_pM\rightarrow T_{\widetilde{p}}\widetilde{M}$ be a linear isomorphism. Then we have the local diffeomorphism $$f(q)={\rm exp}_{\widetilde{p}}\circ i\circ {\rm exp}_p^{-1}(q)$$ Also, using parallel transport we can define a map (set $q=\gamma(t)$, $\gamma$ normalized radial geodesic starting from $p$) $$\phi_t=\widetilde{P}_t\circ i\circ P_t^{-1}:T_qM\rightarrow T_{f(q)}\widetilde{M}.$$ Cartan's Theorem states that if $\phi_t$ is compatible with the curvature tensors of the two manifolds, then it coincides with the Jacobian of $f$. In such case, $f$ is a local isometry.

You can find the answer to your question by setting $M:=T_{\widetilde{p}}\widetilde{M}$, $i=$ identity, $p=0\in T_{\widetilde{p}}\widetilde{M}$.

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