# Confusion of definition of covariant derivative along curve

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?

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

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.

• 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 May 15 at 16:22
• 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. May 15 at 18:25
• 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 May 15 at 18:29

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?

• 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 $\gamma(I)$ May 15 at 15:25
• @Lorago: As this and the other answer explain, you do not need $\gamma'$ to be a (unique) vector field on anything. May 15 at 15:58
• @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 May 15 at 16:13
• @Lorago: I suggest you reread my comment. May 15 at 16:20
• @MoisheKohan it makes more sense now, thanks May 15 at 16:50