Covariant derivative

In my book the author makes the remark: If $X,Y$ are smooth vector fields, and $\nabla$ is a connection, then $\nabla_X Y(p)$ depends on the Value of $X(p)$ and the value of $Y$ along a curve, tangent to $X(p)$.

When I got it right, then we can consider a curve $c:I\rightarrow M$ with $c(0)=p$ and $c'(0)=X_p$.

I was wondering, why this is true. When I consider a coordinate representation around the point p, i.e. $X=\sum x^i\cdot \partial_i$ and $Y=\sum y^i \cdot \partial_i$, then we can calculate that $\nabla_X Y(p)$ depends on:

$x^i(p)$, $y^i(p)$ and $X_p(y^i)$.

This again is only depending on $X_p$ and the values of $y^i=Y(x^i)$ in a arbitrary small neighborhood of $p$. But I cannot see any curve ...

I hope you can help me!

Regards

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Write in coordinates $X = X^i \partial_i$ and $Y = Y^i \partial_i$ (using the Einstein summation convention).

Then $\nabla_X Y = X(Y^k)\partial_k + Y^j X^i \Gamma_{ij}^k \partial_k$.

The only part of this expression that depends on values of $Y$ other than at whatever fixed point $p$ you are looking at are the directional derivatives $X(Y^k)$. As always for directional derivatives of functions, these only depend on the values of the function along a curve tangent to $X$.

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You're right, the covariant derivative is the usual derivative along the coordinates with correction terms which tell how the coordinates change.

The covariant derivative $\nabla$ is a way of specifying a derivative along tangent vectors of a manifold.

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Yes, but why is $\nabla_X Y(p)$ depending just on the Values of Y along a curve? – Braten Apr 11 '12 at 12:30
Because the derivative is a "first order" difference, and so only depends on a "first order contact". Another way to say this is: if $\gamma:t\to M$ and $\eta:s\to M$ are two curves (parametrised by arc-length) tangent at $x = \gamma(0) = \eta(0)$. Then for any smooth function $f:M\to\mathbb{R}$: $$\lim_{t\to 0} \frac{1}{t}\left( f\circ\gamma(t) - f(x)\right) = \pm \lim_{s\to 0} \frac{1}{s}\left( f\circ\eta(s) - f(x) \right)$$ That is, while $\nabla_X Y(p)$ can be computed by just the value of $Y$ along a curve, there is considerable freedom in which curve you use. – Willie Wong Apr 11 '12 at 14:08

It follows as the connection is $C^\infty$ linear in that bottom slot, so that you can factor out the $x^i(p)$ terms out of the connection, then evaluate and at the point $p$. It then follows fairly simply.

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