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Let $\nabla\colon\Gamma(TM)\times\Gamma(E)\to\Gamma(E),(X,\sigma)\mapsto\nabla_X\sigma$ be a connection on a vector bundle $E$ over a smooth manifold $M$, and let $\gamma\in C^\infty(I,M),I:=[0,1]$ be a smooth curve in $M$. In our lecture we "defined" the parallel transport of $v_0\in E_{\gamma(0)}$ along $\gamma$ as follows.

The equation $\nabla_{\dot\gamma}\sigma\equiv0$ corresponds to an ODE which for any given initial value $\sigma_{\gamma(0)}=v_0\in E_{\gamma(0)}$ has a unique solution $\sigma\in\Gamma(E)$. We call $\sigma_{\gamma(1)}\in E_{\gamma(1)}$ the parallel transport of $v_0$ along $\gamma$.

I see one problem with this definition: The derivative $\dot\gamma=D\gamma(\partial_t)\colon I\to TM$ is not a vector field on $M$, i.e. it is not a section of $TM$. Even when viewed as a map $\gamma(I)\subset M\to M$, it isn't defined on all of $M$ making it unsuitable as an argument to $\nabla$. I am aware of the locality (or tensoriality) of the connection. But if there does not exist any vector field $X\in\Gamma(TM)$ which coincides with $\dot\gamma\colon\gamma(I)\to TM$ on $\gamma(I)$, then you can't use locality to argue that $\nabla_{\dot\gamma}\sigma$ is well-defined. If $\dot\gamma\colon\gamma(I)\to TM$ can actually be extended to a full section $\dot\gamma\in\Gamma(TM)$, then problem solved. But if $\gamma$ intersects itself with different velocities at the intersection, extending $\dot\gamma$ becomes impossible.

The only reasonable approach I could come up with is to define some pullback connection $\gamma^*\nabla\colon\Gamma(TI)\times\Gamma(\gamma^*E)\to\Gamma(\gamma^*E)$ on the pullback bundle $(\gamma^*E)_t=T_{\gamma(t)}M$ (over $I$) and define $\sigma\in\gamma^*E$ to be the solution of $(\gamma^*\nabla)_{\partial_t}\sigma\equiv0$ with initial value $\sigma_0=v_0\in T_{\gamma(0)}M$.

After a little bit of digging I found this Wikipedia article, where the author claims that there exists a unique connection $\gamma^*\nabla$ on $\gamma^*E$ satisfying $(\gamma^*\nabla)_X(\gamma^*\sigma)=\gamma^*(\nabla_{d\gamma(X)}\sigma)$ for all $X\in^?\Gamma(TI)$ and $\sigma\in\Gamma(E)$. But this seems to suffer the same problem as above, because $d\gamma(X)\colon I\to TM$ is not a section of $TM$ leaving $\nabla_{d\gamma(X)}\sigma$ undefined.

The same problem occurs in the definition of geodesics.

$\gamma$ is called a geodesic of $M$ with respect to an affine connection $\nabla\colon\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)$, if $\nabla_{\dot\gamma}\dot\gamma\equiv0$.

But again, $\dot\gamma\notin\Gamma(TM)$.

This brings me to my question: How are geodesics and the parallel transport of a vector $v_0\in E_{\gamma(0)}$ actually defined? Is there a rigorous way of defining these concepts?

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  • $\begingroup$ What book are you reading? Most textbooks will discuss this issue. $\endgroup$ Jan 22, 2020 at 18:52
  • $\begingroup$ None, I have lecture notes from the lecture. You can find them here (section C2 is about connections): www3.math.tu-berlin.de/geometrie/Lehre/WS19/DGII $\endgroup$
    – Cubi73
    Jan 22, 2020 at 18:55
  • $\begingroup$ My suggestion is to get a textbook, in addition to your class notes. This will clarify many issues. For instance, $\gamma'(t)$ should be regarded as a vector field along $\gamma$, i.e. a section of $\gamma^*(TM)$. Furthermore, the covariant derivative $\nabla_{\gamma'(t)}$ at $t_0$ depends only on the vector $\gamma'(t_0)$ and not on the vector field $\gamma'(t)$ along $\gamma(t)$. $\endgroup$ Jan 22, 2020 at 20:32
  • $\begingroup$ There always exists a vector field of $M$ that coincides with $\dot{\gamma}$. Extend $\dot{\gamma}$ to a section $\tilde{\gamma}$ of $\gamma^*TM$, use the exponential map to identify $\gamma^*TM$ with a tubular neighborhood of $\gamma(I)$ in $M$, push $\tilde{\gamma}$ forward with the exponential map, and multiply by your favorite cutoff function. Define $\nabla_{\dot{\gamma}}$ as $\nabla_{\mbox{exp}^*\tilde{\gamma}}$. Use tensoriality of the connection to argue that your choices don't matter. Conclude $\nabla_{\dot{\gamma}}$ is well-defined. $\endgroup$
    – Neal
    Jan 22, 2020 at 21:05
  • $\begingroup$ @Neal Why does using the exponential map make parallel transport well-defined? The exponential map is defined using geodesics, i.e. curves $\gamma\colon I\to M$ which are solution to the ODE $\nabla_{\dot\gamma}\dot\gamma\equiv0$. But as you can see, the term $\nabla_{\dot\gamma}\dot\gamma$ has the exact same problem as in my question above: $\dot\gamma\notin\Gamma(TM)$. $\endgroup$
    – Cubi73
    Jan 23, 2020 at 16:51

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The formally correct would be indeed to use the pullback connection, which exists by the following theorem:

If $f:S\to M$ is smooth map between smooth manifolds and $E$ is a smooth vectorbundle over $M$ with connection $\nabla$ then there is unique connection $f^*\nabla$ on $f^*E$ such that for all $p\in S$, $X_p\in T_pS$ and $\sigma\in\Gamma(E)$ $$(f^*\nabla)_{X_p}(f^*\sigma)=\nabla_{df(X_p)}\sigma$$

Note that since a connection is tensorial tensorial in the first slot it actually makes sense to plug in tangent vectors instead of vectorfields. The output will then be also just a tangent vector.

Now if $\nabla$ is a connection on $TM$ and $\gamma: I\to M$ is a smooth curve then $\dot\gamma\in\Gamma(\gamma^*TM)$, so if $\frac{\partial}{\partial s}$ is the standard vectorfield on $I$ (meaning the vectorfield coming from the curve $t\mapsto t$) then $\ddot \gamma(t)=(\gamma^*\nabla_{\frac{\partial}{\partial s}}\dot\gamma)(t)$ is well defined and may be also denoted by $(\nabla_{\dot\gamma}\dot\gamma)(t)$.

As a last word: If the curve $\gamma$ satisfies $\dot\gamma(t)\neq 0$ for all $t\in I$ then one can avoid using the pullback connection (even if $\gamma$ has self-intersections) by using the locality priciple and an extension argument, see for example here.

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  • $\begingroup$ Ahh, I think you just nailed down the problem I had with this. We introduced connections as maps $\Gamma(TM)\times\Gamma(E)\to\Gamma(E)$ whose properties make them tensorial, so I thought that talking about local properties nevertheless requires the existence of a full vector field. But from the comments and your answer it seems that one can define "pointwise connections" $T_pM\times\Gamma(E)\to\Gamma(E)$ which (under some extra conditions) can be "glued" together along $M$ to give a connection $\Gamma(TM)\times\Gamma(E)\to\Gamma(E)$. Is that right? $\endgroup$
    – Cubi73
    Jan 23, 2020 at 22:11
  • $\begingroup$ $\nabla$ does not need a vectorfield as input in the following sense: For $X_p\in T_pM$ and $\sigma\in\Gamma E$ one defines $\nabla_{X_p}\sigma=(\nabla_{\bar X}\sigma)(p)$ where $\bar X\in\Gamma (TM)$ is any vectorfield with $\bar X_p=X_p$. Such a vectorfield always exists and by tensoriality this definition does not depend on the choice of $\bar X$. So for fixed $p\in M$ $\nabla$ can be regarded as a map $T_pM\times \Gamma(E)\to E_p$. $\endgroup$
    – Claire
    Jan 23, 2020 at 22:26
  • $\begingroup$ I am not sure though if there are nice conditions under which "pointwise-connections" can be glued together to obtain a connection, but this is also not needed here. $\endgroup$
    – Claire
    Jan 23, 2020 at 22:27
  • $\begingroup$ For a single point that is correct, but $\dot\gamma$ cannot always be extended to a vector field. This is the reason why I asked about the definition of $\nabla_{\dot\gamma}$. Or maybe I misunderstood your comment and $(\nabla_{\dot\gamma}\sigma)_p$ really is just defined to be $\nabla_X\sigma$ for some vector field $X$ with $X_p=\dot\gamma(p)$ regardless of $\dot\gamma\notin\Gamma(TM)$. Is that the case? $\endgroup$
    – Cubi73
    Jan 23, 2020 at 22:33
  • $\begingroup$ An expression like $(\nabla_{\dot\gamma}\sigma)_p$ was not defined at all. All what was defined is the expression $(\nabla_{\dot\gamma}\dot\gamma)(t):=(\gamma^*\nabla_{\frac{\partial}{\partial s}}\dot\gamma)(t)$ which with the definition in the comment above can also be written as $\gamma^*\nabla_{\frac{\partial}{\partial s}(t)}\dot\gamma$, so you should regard the expression $\nabla_{\dot\gamma}\dot\gamma$. So you should regard the expression $\nabla_{\dot\gamma}\dot\gamma$ more like an abuse of notation. $\endgroup$
    – Claire
    Jan 23, 2020 at 22:52

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