# transformation of vector fields under coordinate transformations

Consider two open subsets $\Omega, \Omega^{\prime}\subset \mathbb{R}^n$. Now consider a (volume preserving) diffeomorphism \begin{align} \varphi:\Omega^{\prime}\to\Omega; \alpha\to \varphi(\alpha) \end{align} If we have a (scalar) function $u\in C^1(\Omega)$, it will be transformed via $\tilde{u}=u\circ \varphi$. By differentiation, one can prove that the gradient of $u$ will be transformed according to \begin{align} (\nabla u)\circ \varphi=(\nabla \tilde{u})(D\varphi)^{-1} \end{align} where $D\varphi$ denotes the Jacobian matrix of $\varphi$ and $\nabla \tilde{u}$ is taken to be a row vector.

Now consider an arbitrary vector field $v$ in $\Omega$ (i.e. an electric field or a velocity field, etc.), which is not necessarily a gradient field of some function, how does this transform under $\varphi$? Does it transform the same way? And if so, how can I prove it (or where can I look it up)?

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On wikipedia: en.wikipedia.org/wiki/Pushforward_(differential). In general, a good place to start would be any of the older textbooks on tensor calculus. – Willie Wong Sep 9 '11 at 20:29
Thanks for pointing out where to start. Since I have absolutely no background in differential geometry yet, your advise helped me search for what I need. I'm having some problems with the notation, but I think I've got it. – Martin Sep 11 '11 at 19:12

Gradients are not vector fields; they are differential $1$-forms. Only in Riemannian geometry they have the option of becoming vector fields. The operation you described in the post is the pullback of a differential form.
For vector fields we have the pushforward operation. A vector field is a smooth selection of a tangent vector from every tangent space $T_pM$ (tangent at point $p$ of manifold $M$). The derivative of a smooth map $f:M\to N$ is a linear map from $T_pM$ to $T_{f(p)}N$. This linear map pushes the vector field from $M$ to $N$, in the forward direction. Compare this to the pullback, which moves differential forms from $N$ to $M$.