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I am teaching myself some elementary differential geometry and am stuck on the concept of the tangent vector field of a smooth curve. I have searched the web for an hour or so but cannot find anything pertaining specifically to my issue here.

For a smooth curve $\gamma:(0,1)\rightarrow M$, where $M$ is a manifold with a connection $\nabla$, we can define the notion of autoparallel transport. We say the curve is autoparallely transported if $\nabla_{\dot{\gamma}}\dot{\gamma} = 0$ on the curve, where $\dot{\gamma}$ is the tangent vector field to $\gamma$.

My question is, how do we define this guy in terms of an element in the section of the tangent bundle? This is my initial guess, sort of with an abuse of notation:

\begin{equation} \dot{\gamma}(\bullet) := \dot{\gamma}(\gamma(\bullet)) = \dot{\gamma}\circ \gamma : (0,1)\to TM ,\end{equation}

which I think would make the latter $\dot{\gamma}\in\Gamma(TM)$, i.e. a vector field sort of sending $M\lvert_{\gamma}\to TM$. But something feels off, is this enough to be able to continue?

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3 Answers 3

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The condition is indeed what you want: $\forall t\in (0,1),\gamma'(t)\in T_{\gamma(t)}M$. More formally, you can look at the definition of a pullback bundle: here you are actually defining a section of the bundle $\gamma^*TM$.

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The crucial point to understand why what you are attempting in your "abuse of notation" needs some more care is realizing that $\dot{\gamma}$ is essentially a section of $TTM$, not of $TM$.

To look "From the Point of View of the Tangent Bundle" (the name of a section, no pun intended, of Sakai's differential geometry book), the best way is arguably to realize a connection as a map $K: TTM \to TM$ (with some properties). This map splits $T_{(p,v)}TM$ into an horizontal subspace, which is the non-canonical part of the splitting, and a vertical subspace which is simply the kernel of $T\pi:TTM\to TM$ (i.e., essentially the fibers of $TTM$).

Then, given a vector field seen as a section $\gamma: M \to TM$, we have that at some point $p$, $$\nabla_{v} \gamma=(K \circ T \gamma)(p,v).$$ This is essentially projecting $T\gamma$ to the horizontal subspace, which is a nice way to think of a connection.

For an explicit definition of $K$ given the usual definition of covariant derivative, you can use the Christoffel symbols $\Gamma$ and define locally: given a chart $\varphi$, $$K:(x,\xi,y,\eta) \mapsto (x,\eta+\Gamma_{\varphi}(x)(y,\xi)).$$

The name "horizontal/vertical" may be confusing, given the usual way of thinking of $TM$ of submanifolds of Euclidean space. This construction of a connection works for any vector bundle $E$ (we have $K:TE \to E$ instead), and the name horizontal/vertical derives from the usual way we picture a fiber bundle as fibers vertically.

To adapt all this to your context of $\gamma$ being a smooth curve instead of a vector field, just consider the pull-back bundle $\gamma^*TM$ and the pull-back connection $K^*$. Then what we have is $\widetilde{\gamma}:(0,1) \to \gamma^*TM$ and $$\nabla_{\dot{\gamma}}\dot{\gamma}=(K^*\circ T\widetilde{\gamma})(t,\dot{\gamma}). $$

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Hopefully I've understood your question correctly. Let's add the assumption that $\gamma$ is such that its image is a regular submanifold of $M$. Technically the velocity field $V_{\gamma(t)}=\gamma'(t)$ is not a section of $TM$ because it is only defined on the submanifold $\text{im}(\gamma)$. With that being said, you can extend this to a (nonunique) vector field $\overline{V}$ defined on all of $M$. Then, $\overline{V}\in \Gamma(TM)=\mathfrak{X}(M)$. Conversely, you can restrict a section of $TM$ to a vector field on $\text{im}(\gamma)=N$. Given sufficient conditions on $\gamma$, its image is a submanifold $N\subseteq M$ and has a tangent bundle in its own right, $TN$. Then $\gamma'(t)=V_{\gamma(t)}$ is a section of $TN$.

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  • $\begingroup$ Ah, so when we take the covariant derivative (as in the def. of parallel transport), which must take a vector field as input, it is really with respect to some chosen extended vector field $\overline{V}$? $\endgroup$ Commented Feb 14, 2019 at 14:17
  • $\begingroup$ You can't always extend a vector field along a curve to a vector field on your manifold (think about the tangent vector of a "8" figure), but the operator that you are defining coincides with the covariant derivative when your vector field is extendible. $\endgroup$
    – Balloon
    Commented Feb 14, 2019 at 14:27
  • $\begingroup$ Editted, thanks for the comment. $\endgroup$ Commented Feb 14, 2019 at 15:32

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