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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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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](http://en.wikipedia.org/wiki/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.2012-09-19
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    I guess in the last line you mean $\nabla_{v}X$ instead of $\Delta_{v}X$.2012-09-19
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    @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)$2012-09-19
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    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 $\partial_{\gamma'(t)}a =\gamma_1'(t)\cdot\frac{\partial a}{\partial x} + \gamma_2'(t)\cdot\frac{\partial a}{\partial y}$.2012-09-19
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    Ah so it is just the directional derivative along the gradient of $a$2012-09-19
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    @BerciPecsi One final stupid question. Is it correct that we are profiting now from the fact that the Christoffel symbols are identically zero for Euclidean co-ordinates? If I had written down the problem in polar co-ordinates you would have to do more work.2012-09-19