Connection and curve

43 Views Asked by At

Let $\nabla$ be a connection on a Riemannian manifold and let the differential of a curve be given by $$c'(t)=c_1'(t)\partial_1 + c_2'(t) \partial_2.$$

Now I was wondering how we define $\nabla_{\partial_2} c'$, since clearly $c'$ is a vector in the tangent space and $\partial_2$ another vector field along $c$? The problem is that we would normally get

$$\nabla_{\partial_2}c' = \partial_2(c_1') \partial_1+ \partial_2(c_2') \partial_2+ c_1' \nabla_{\partial_2} \partial_1 + c_2' \nabla_{\partial_2}\partial_2$$ if $c'$ was a normal vector field. But in this case, $c_1',c_2'$ are functions depending on time and not directly on space, so I don't know how to evaluate the first two terms in my equation above.

Hence I don't see how to define $\nabla_{\partial_2}c'$ explicitly?

Edit: I don't know if my expression for $\nabla_2{c'}$ makes any sense at all.

1

There are 1 best solutions below

0
On BEST ANSWER

It does not make sense to talk about $\nabla_{\partial_2}c'$.

$c'$ is a vector field on only a curve in the manifold $M$. $\nabla_{\partial_2}c'$ should give the derivative of the vector field $c'$ in the direction of $\partial_2$. This does not make sense, because $c'$ is not defined in that direction. We can only think about $\nabla_{c'}c'$.

This is all true in $\mathbb{R}^2$ as well, so maybe it's simpler to think about there. If you have a curve $c(t):[0,1] \to \mathbb{R}^2$, you have it's velocity $c'(t)$. It does not make sense to think about the derivative of $c'(t)$ in a direction that leaves the curve.