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Let $(M,\nabla)$ be a Riemannian manifold with the Levi-Cevita connection. Let $c:I \rightarrow M$ be a differentiable curve in $M$, and let $\frac{D}{dt}$ be the covariant derivative. Then

$\frac{D}{dt}(\frac{dc}{dt})=\nabla_{c'(t)}c'(t)$

I am confused about how we get this equality. I know it follows somehow from the following fact:

If $V$ is a vector field along $c$ that is induced by a vector field $Y \in \mathfrak{X}(M)$, i.e. $V(t)=Y(c(t))$ then $\frac{DV}{dt}=\nabla_{c'(t)}Y$.

This arises in do Carmo's textbook and he says it follows by "extending $c'(t)$ to a neighborhood of $c(t)$ in $M$". But this is confusing to me because on page 43 do Carmo says that $c$ cannot necessarily be extended to a vector field on an open set of $M$.

I know from the Smooth Extension Lemma that we can necessarily extend a vector field along only closed subsets to any open subset of the manifold.

So, I would appreciate it if someone could explain clearly how this works.

Tuo
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1 Answers1

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To define $DV/dt$ along a curve, you only need the vector field $V$ defined along the curve. I don't find the particular equality you have highlighted in doCarmo; where is it?

Note that you run into trouble with extending a vector field $V$ along $c$ if the differentiable curve $c$ is not embedded — if the curve is not one-to-one, there's not even a well-defined vector field on $M$ defined on the image of $c$ that gives $V$.

EDIT: However, if $\gamma$ is an embedding, there is no problem extending $V$ (use a tubular neighborhood and then a partition of unity to extend to the rest of $M$). Thus, we can always do a local extension of the vector field on $\gamma|_J$ so long as $\gamma|_J$ is an immersion. That is the context of the proof you referred to in Lemma 2.6.

Ted Shifrin
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