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
- $$\frac{\mathrm{D}(\lambda X+\mu Y)}{\mathrm{d}t}=\lambda\frac{\mathrm{D}X}{\mathrm{d}t}+\mu\frac{\mathrm{D}Y}{\mathrm{d}t}$$
- $$\frac{\mathrm{D}(fX)}{\mathrm{d}t}=f'X+f\frac{\mathrm{D}X}{\mathrm{d}t}$$
- 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
- 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?
Best Answer
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.