Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!


share|improve this question
add comment

3 Answers 3

up vote 7 down vote accepted

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 $X$.

share|improve this answer
add comment

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.

share|improve this answer
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
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.