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I am having a hard time understanding covariant derivatives. My main problem is working with concrete example. So I was wondering if anybody could help me with explaining it by using simple example.

Let us say we have the curve

\begin{equation*} \gamma(x):=(x,x^2) \end{equation*}

in $\mathbb{R}^2$ with $\mathbb{x} \in \mathbb{R}$. Now suppose I have a vector field $X$ in $\mathbb{R}^2$ how do I write down the covariant derivate of $X$ along $\gamma$.

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1 Answer 1

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In $\mathbb R^2$ it is easy, if say $X=a(x,y)\mathrm dx+ b(x,y)\mathrm dy$, then you have to consider the derivatives of both $a$ and $b$ in the $\gamma'(t)$ direction (and speed) at each $t\in \mathbb R$ moment, i.e. $$\nabla_\gamma X(t) = (\partial_{\gamma'(t)}a)\mathrm dx + (\partial_{\gamma'(t)}b)\mathrm dy$$ For more sophisticated spaces, a Riemmann metric (local scalar products) is needed on the manifold.

The point is, that for a manifold $M$, and $z\in M$, for any individual vector $v\in T_zM$ one can define $\nabla_v X$ for any vector field defined in a neighborhood of $z$.

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@user29751: It is maybe worth adding that on an arbitrary pseudo-Riemannian manifold (i.e one does not require the scalar products on the tangent spaces to be positive-definite) there are in general many different choices of a covariant derivative, but one which is very "natural": The Levi-Civita connection (where in this context, connection is synonymous for covariant derivative). Note that if one endows $\mathbb{R}^{2}$ with the standard metric, then the Levi-Civita connection is just the usual directional derivative of smooth functions. –  Nils Matthes Sep 19 '12 at 12:49
    
I guess in the last line you mean $\nabla_{v}X$ instead of $\Delta_{v}X$. –  Nils Matthes Sep 19 '12 at 12:53
    
@BerciPecsi Perhaps a stupid question. I am not sure what you mean by the direction in $\gamma'(t)$ direction. For $\gamma=(\gamma_1, \gamma_2)$ can the term $\partial_{\gamma'(t)} a dx $ be written as $(d\gamma_1/dt) \cdot (\partial a / \partial x)$ –  Novo Sep 19 '12 at 14:09
    
Errhh.. not exactly, but you're close: in your notation $\gamma'(t)=d\gamma/dt$, it has first coordinate $\gamma_1'(t) = d\gamma_1/dt$, and I meant $\partial_{\gamma'(t)}a =\gamma_1'(t)\cdot\frac{\partial a}{\partial x} + \gamma_2'(t)\cdot\frac{\partial a}{\partial y}$. –  Berci Sep 19 '12 at 17:39
    
Ah so it is just the directional derivative along the gradient of $a$ –  Novo Sep 19 '12 at 17:41

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