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I am currently studying Riemannian geometry, and have come across the following proposition:

Proposition. Let $(M,g)$ be a Riemannian manifold with Levi-Civita connection $\nabla$ and let $\gamma:I\to M$ be a $C^1$-curve in $M$. Then there exists a unique operator

$$\frac{\mathrm{D}}{\mathrm{d}t}:C^\infty_\gamma(\mathrm{T}M)\to C^\infty_\gamma(\mathrm{T}M),$$

such that for all $\lambda,\mu\in\mathbb{R}$, $f\in C^\infty(I)$, $X,Y\in C^\infty_\gamma(\mathrm{T}M)$, we have

  1. $$\frac{\mathrm{D}(\lambda X+\mu Y)}{\mathrm{d}t}=\lambda\frac{\mathrm{D}X}{\mathrm{d}t}+\mu\frac{\mathrm{D}Y}{\mathrm{d}t}$$
  2. $$\frac{\mathrm{D}(fX)}{\mathrm{d}t}=f'X+f\frac{\mathrm{D}X}{\mathrm{d}t}$$
  3. for each $t_0\in I$, there exists an open subinterval $J$ of $I$ such that $t_0\in J$ and if $\bar{X}\in C^\infty(\mathrm{T}M)$ is a vector field with $\bar{X}_{\gamma(t)}=X(t)$ for all $t\in J$, we have $$\frac{\mathrm{D}X}{\mathrm{d}t}(t_0)=(\nabla_{\dot{\gamma}}\bar{X})_{\gamma(t_0)}.$$

In the above, $C^\infty_\gamma(\mathrm{T}M)$ refers to all smooth vector fields along $\gamma$, i.e. smooth $X:I\to\mathrm{T}M$ with $\pi\circ X=\gamma$, and $C^\infty(\mathrm{T}M)$ refers to all smooth vector fields on $M$.

My problem with this proposition is that I don't quite think 3 makes sense, and I'm trying to figure out what it should be or how it should be interpreted. To see my problem, note first that the Levi-Civita connection on $(M,g)$ is a map

\begin{align*} \nabla:C^\infty(\mathrm{T}M)\times C^\infty(\mathrm{T}M)&\to C^\infty(\mathrm{T}M), \\ (X,Y)&\mapsto \nabla_XY. \end{align*}

Consequently the expression $\nabla_{\dot\gamma}\bar{X}$ does not make sense, as $\dot\gamma$ is not in $C^\infty(\mathrm{T}M)$. Even identifying $\dot\gamma$ with $t\mapsto(\gamma(t),\dot\gamma(t))$ we still only have a vector field along $\gamma$, i.e. an element of $C^\infty_\gamma(\mathrm{T}M)$. So the only way to try to make sense of this is to try to interpret $\dot\gamma$ as a vector field on $M$. If $\gamma$ is injective, this is not a problem. Simply note that $\gamma(I)$ defines a submanifold of $M$ locally, and so we can extend the map $\gamma(I)\to\mathrm{T}M$, $p\mapsto(p,\dot{\gamma}(\gamma^{-1}(p))$ to a vector field on $M$ locally around each point, and this vector field would then take the place of $\dot\gamma$ in $\nabla_{\dot\gamma}\bar{X}$, and everything would make sense. The problem is, however, that if we do not assume $\gamma$ to be injective, we may have self-intersections, and for a given point $p\in\gamma(I)$ with $t_1,t_2\in I$ such that $\gamma(t_1)=\gamma(t_2)=p$, we may have that $\dot{\gamma}(t_1)\neq\dot{\gamma}(t_2)$, meaning that the above construction would be ill-defined. So to get around this, I assume the point is to choose $J\subseteq I$ sufficiently small so that $\gamma\vert_{J}$ is injective, and then do the above construction locally. Is this what is meant by the above? So could be write 3 instead as

  1. for each $t_0\in I$, there exists an open subinterval $J$ of $I$ such that $t_0\in J$, $\gamma\vert_J$ is injective, and if $\bar{X}\in C^\infty(\mathrm{T}M)$ is a vector field with $\bar{X}_{\gamma(t)}=X(t)$ for all $t\in J$, we have $$\frac{\mathrm{D}X}{\mathrm{d}t}(t_0)=(\nabla_{\dot\gamma\circ(\gamma\vert_J)^{-1}}\bar{X})_{\gamma(t_0)}.$$

or something along those lines?

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  • $\begingroup$ There is the notion of extendibility of a curve, which sort of encompasses your issue. But it indeed seems the definition is somewhat clumsy. $\endgroup$
    – Pastudent
    May 15 at 15:05

2 Answers 2

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It is a well known fact (and I would assume that you know this - if not I would suggest you try to verify it) that, if $X, Y, Z$ are vector fields on $M$ with $X(p) = Y(p)$, then $\nabla_X Z|_p = \nabla_Y Z|_p$, i.e. the value of $X\mapsto \nabla_X Z$ in $p$ only depends on $X(p)$ and not on $X$ in a neighbourhood of $p$. With this observation $ \nabla_{\gamma^\prime(t_0)}\tilde{Z}|_{\gamma(t_0)}$ is well defined.

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  • $\begingroup$ This actually does make sense. So what we really want to do is take any vector field $Y$ on $M$ with $Y_{\gamma(t_0)}=\dot{\gamma}(t_0)$ and then by the equality we mean$$\frac{\mathrm{D}X}{\mathrm{d}t}(t_0)=(\nabla_Y\bar{X})_{\gamma(t_0)}$$? This corresponds roughly to what I did, and makes a lot of sense $\endgroup$
    – Lorago
    May 15 at 16:22
  • $\begingroup$ yes and no. The extension to a vector field you do only to verify that $X\mapsto \nabla_X Y$ is tensorial. It is actually equivalent to what A.J. has written (with different terms). Once you know that only the vector in $T_p M$ matters, you can ignore the construction with the extension. $\endgroup$
    – Thomas
    May 15 at 18:25
  • $\begingroup$ Yeah that makes perfect sense. I guess my problem was the proposition I posted wrote $\nabla_{\dot\gamma}X$ instead or $\nabla_{\dot\gamma(t_0)}X$. Now it makes perfect sense, thanks $\endgroup$
    – Lorago
    May 15 at 18:29
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You can avoid any problem by proving, prior to this proposition, that the covariant derivative is tensorial with respect to the first $C^\infty(\mathrm{T}M)$. That is, the covariant derivative only depends on the vector at the point, not on the vector field defined on a neighborhood of the point.

Are you sure this is not proven in the text, or left to the reader as an exercise?

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  • $\begingroup$ I know this, yes, hence why I continued by saying I think the solution is that we could choose a sufficiently small $J$ so that we don't run into problems there, but without doing so explicitly we don't have that $\dot\gamma$ defines a unique vector field on $\gamma(I)$ $\endgroup$
    – Lorago
    May 15 at 15:25
  • $\begingroup$ @Lorago: As this and the other answer explain, you do not need $\gamma'$ to be a (unique) vector field on anything. $\endgroup$ May 15 at 15:58
  • $\begingroup$ @MoisheKohan my problem really is that if $t_1,t_2\in I$ are two distinct points such that $p=\gamma(t_1)=\gamma(t_2)$ such that $\dot{\gamma}(t_1)\neq\dot{\gamma}(t_2)$, then how could I possibly make sense of the expression $(\nabla_{\dot{\gamma}}\bar{X})_p$? The given answers don't really address this $\endgroup$
    – Lorago
    May 15 at 16:13
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    $\begingroup$ @Lorago: I suggest you reread my comment. $\endgroup$ May 15 at 16:20
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    $\begingroup$ @MoisheKohan it makes more sense now, thanks $\endgroup$
    – Lorago
    May 15 at 16:50

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