# How are geodesics and parallel transports actually defined?

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?

• What book are you reading? Most textbooks will discuss this issue. – Moishe Kohan Jan 22 at 18:52
• 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 – Cubi73 Jan 22 at 18:55
• 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 $\gamma(t)$. – Moishe Kohan Jan 22 at 20:32
• 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. – Neal Jan 22 at 21:05
• @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: $\dot\gamma\notin\Gamma(TM)$. – Cubi73 Jan 23 at 16:51

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.

• 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? – Cubi73 Jan 23 at 22:11
• $\nabla$ 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 $T_pM\times \Gamma(E)\to E_p$. – lulu Jan 23 at 22:26
• 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. – lulu Jan 23 at 22:27
• 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? – Cubi73 Jan 23 at 22:33
• 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. – lulu Jan 23 at 22:52