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Suppose we have a vector bundle $E$ with connection $\nabla$ over a smooth manifold $M$. Let’s also say we have a sequence of smooth paths $\gamma_n\in C^\infty([0,1],M)$ starting at the same point $\gamma_n(0)=x$ and converging (in, say, the $C^1$ compact-open topology) to a smooth path $\gamma\in C^\infty([0,1],M)$.

The question, then, is this:

For $v\in E_x$, does the parallel transport $P_{\gamma_n}v$ of $v$ along $\gamma_n$ converge (in $E$) to the parallel transport $P_{\gamma}v$ of $v$ along $\gamma$?

To be honest, while I suspect this to be true, I have no clue how to go about showing it. It’s clear that we definitely need the paths to converge in a stronger sense than $C^0$-convergence, since otherwise any nontrivial curvature will mess things up. I’m also aware of the post here, which seems quite similar, though the accepted answer relies on a seemingly hand-wavy “convergence of equations implies convergence of solutions” claim that does not seem justified.

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    $\begingroup$ @WillieWong's answer below does not require the use of curvature and therefore works under low regularity assumptions. However, you can get a quantitative estimate of the convergence in terms of curvature using holonomy as described in the page below, as well as the references cited there. deaneyang.github.io/blog/blog/math/differential-geometry/… $\endgroup$
    – Deane Yang
    Commented Apr 1 at 16:18

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The following is a precise version of the "convergence of equations implies convergence of solutions" claim.

Theorem. Let $\Omega\subseteq \mathbb{R}^n$ be open, and $Q\subseteq \mathbb{R}$. Let $A:\Omega\times [0,1]\times Q\to \mathbb{R}^n$ be continuous. Consider the family of ordinary differential equations indexed by $q\in Q \subseteq \mathbb{R}$ given by $$ \dot{x}(t,q) = A(x(t),t,q) $$ Suppose we have:

  • $A(x,t,q)$ is uniformly Lipschitz in $x$; more precisely there exists $K > 0$ such that for all $x_1, x_2\in \Omega$ and $q\in Q$ and $t\in [0,1]$ we have $\|A(x_1,t,q) - A(x_2,t,q)\| \leq K\|x_1 - x_2\|$.
  • $A(x,t,q)$ is uniformly bounded in $\Omega\times Q$.

Let $q_k$ be a sequence converging to $q_\infty$ in $Q$, and suppose for each $k$, we have a solution $x(\cdot ,q_k):[0,1]\to \Omega$ to the ODE above. If furthermore $x(0,q_k)$ converges to $x(0,q_\infty)$, then $x(t,q_k)$ converges to $x(t,q_\infty)$ for every $t\in [0,1]$.


In the literature such results are usually called "continuous dependence on parameters" for ordinary differential equations. Henri Cartan devoted quite a few pages in his book Differential Calculus to this (I don't have my copy with me to track down a precise theorem number).


For parallel transport, the first condition is trivial since the relevant operator is linear, the second is also easy to check once you localize to a small neighborhood of the curve $\gamma$ and locally trivialize. The third condition is the only one that requires the curves to converge in $C^1$, since in local coordinates the connection coefficients depend continuously on the position of the curve and is linear on the velocity of the curve.

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    $\begingroup$ My colleague Huyi Hu's lecture notes contain a more general version of the Theorem stated above, covering the non-autonomous case too. $\endgroup$ Commented Mar 31 at 13:55
  • $\begingroup$ Thanks for the answer, but I’m having trouble understanding how it actually provides an answer to the question, since the ODE for parallel transport is definitely not autonomous. For example, paths and their parallel transports can loop back on themselves and have different derivatives at different times. Would you please clarify how you see the given result working in this context? $\endgroup$
    – user815293
    Commented Apr 1 at 11:38
  • $\begingroup$ @user815293: You are correct, I typed too fast (was thinking of geodesic spray). Let me correct it. $\endgroup$ Commented Apr 1 at 14:32
  • $\begingroup$ I've edited to use the version found in the linked lecture notes. $\endgroup$ Commented Apr 1 at 14:39

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