Changes of coordinates in differential geometry I'm taking an undergraduate differential geometry class, and because either because the professor has chosen to forgo rigor (as a background in analysis isn't required and most of the class has negligible proof-writing experience) or else because I'm just not getting it, I'm having a lot of difficulty understanding the way he discusses changes of coordinates.
This is the general description he gives:
If we change local coordinates from $\{x^1, ..., x^i, ... x^n\} $ to $\{\hat{x}^1, ..., \hat{x}^{\alpha},..., \hat{x}^n\}$ by the map $F$, the vector field $U = u^iX_i$ to $\hat{U} = \hat{u}^{\alpha}\hat{X}_{\alpha}$
$\hat{U} = DF(U)$ and $\hat{u}^{\alpha} = \frac{\partial \hat{x}^{\alpha}}{\partial x^i}u^i$
I understand (I think) what all the symbols mean, and I'm comfortable with Einstein summation convention, and the idea of a derivative map (again, at an unrigorous, undergrad level). That said, I cannot for the life of me understand why if the coordinates transform under $F$, why the vector transforms under $DF$. 
I would like a) some intuition about this (if there's an intuitive way to view it) and b) a way to show that this is true (if there's a straightforward, fairly elementary way to show it).
Thanks!
 A: You haven't really specified enough to be able to answer this precisely. I will assume that your manifold $M$ is embedded in $\mathbf{R}^n$ for some $n$, since that is the typical assumption at this level. One has from calculus, or multivariable analysis that given a smooth map $F:\mathbf{R}^n\to \mathbf{R}^m$, there is an induced linear transformation $DF_p:\mathbf{R}^n\to \mathbf{R}^m$. This is usually called the total derivative of the map. (See Pugh's Real Mathematical Analysis, for instance.) $DF_p$ has matrix representation given by the so-called Jacobian matrix. If we think of $F=(F_1,\ldots, F_m)$, then
$$ J(F)_p=\begin{bmatrix}
\frac{\partial F_1}{\partial x_1}(p)&\cdots&\frac{\partial F_1}{\partial x_n}(p)\\
\vdots&\ddots&\vdots\\
\frac{\partial F_m}{\partial x_1}(p)&\cdots&\frac{\partial F_m}{\partial x_1}(p)
\end{bmatrix}.$$
The idea, is that this map is a linear transformation of tangent spaces. $DF_p: T_p\mathbf{R}^n\to T_{F(p)}\mathbf{R}^m$. In this case, since we are in Euclidean space, we have that the tangent space at $p$ is identified with $\mathbf{R}^n$ itself, so we view this transformation as simply going from $\mathbf{R}^n\to \mathbf{R}^m$ with no loss of precision. Now, if we have two coordinate charts on $M\subset \mathbf{R}^n$ given by $\mathbf{x}:U\to \mathbf{R}^n$ and $\mathbf{y}:V\to \mathbf{R}^n$, we have that $\mathbf{y}\circ \mathbf{x}^{-1}:\mathbf{x}(U\cap V)\to \mathbf{y}(U\cap V)$ is a smooth map of Euclidean spaces. As such, we can compute its Jacobian. Here, $\mathbf{x}=(x_1,\ldots, x_n)$ and $\mathbf{y}=(y_1,\ldots, y_n)$ can be thought of as your coordinate systems in your notation. There are some minor details missing. But, the idea is that this transformation also acts as a map $T_pM\to T_pM$ for $p\in U\cap V$.
Anyway, if you compute the Jacobian of the coordinate transformation of your change of coordinates, you will find that it is of the form
$$ J(\mathbf{y}\circ \mathbf{x}^{-1})_p=\begin{bmatrix}
\frac{\partial y_1}{\partial x_1}(p)&\cdots&\frac{\partial y_1}{\partial x_n}(p)\\
\vdots&\ddots&\vdots\\
\frac{\partial y_n}{\partial x_1}(p)&\cdots&\frac{\partial y_n}{\partial x_n}(p)
\end{bmatrix}.$$
Fine, so how does this work in vector field world? I'm uncertain as to what definition of vector field you are using. However, let's just think of a vector field $X$ on $M$ as a linear combination of the form
$$ X_p=\sum_{i=1}^na_i(p) \frac{\partial}{\partial x_i}\bigg|_p$$
where the $a_i\in C^{\infty}(M)$. This just means they are smooth real valued functions on $M$. Well, now we have that $X_p$ is acted upon naturally by $J(F)_p$ by
$$ \begin{bmatrix}
\frac{\partial y_1}{\partial x_1}(p)&\cdots&\frac{\partial y_1}{\partial x_n}(p)\\
\vdots&\ddots&\vdots\\
\frac{\partial y_n}{\partial x_1}(p)&\cdots&\frac{\partial y_n}{\partial x_n}(p)
\end{bmatrix}
\begin{bmatrix}
a_1(p)\\
\vdots\\
a_n(p)
\end{bmatrix}.$$
So, we have that the vector field $X$ re-expressed in $\mathbf{y}$ coordinates is of the form
$$ \sum_{j=1}^n\sum_{i=1}^na_i(p)\frac{\partial y_j}{\partial x_i}(p)\frac{\partial}{\partial y_j}\bigg|_p.$$
This is, I believe, everything you were trying to show.
A: I myself studied differential geometry and applications of it all without a formal education. As I understand it in the notation used by Pavel Grinfeld in his book "Introduction to Tensor Analysis and the Calculus of Moving Surfaces" The "map" you mentioned, $F$ is a vector function which goes from $\mathbb{R}^n$ to $\mathbb{R}^n$ with each component being described by $F$ so it can be represented in components as the following:
$$F^\alpha=\hat{x}^\alpha(x^i)=\hat{x}^\alpha(x)\\
\vec{F}=\hat{x}^\alpha(x)\hat{\textbf{Z}}_\alpha$$
where $\hat{\textbf{Z}}_\alpha$ is the basis in the $\hat{x}^\alpha$ coordinate system. Thus when we consider vector fields in the $\hat{x}^\alpha$ coordinate system, this is writ in the following format:
$$\textbf{P}'=\textbf{P}(\hat{x})$$
Since we are talking about a basis $\hat{\textbf{Z}}_\alpha$ which is defined globally all around the space of interest, we can safely assume that
$$\hat{\textbf{Z}}_\alpha=\frac{\partial \textbf{R}}{\partial \hat{x}^\alpha}$$
Where $\textbf{R}$ is an imaginary vector which can locate every point in the space. Expanding that definition of the basis vector, all the partial derivative means is that $\hat{\textbf{Z}}_\alpha$ is a method of directing distances between points along the coordinate directions. (Think about it, since the partial derivative is a method of observing differences in the directions of the coordinates). Now noticing how the basis changes from coordinate system to system, we see that using the chain rule:
$$\textbf{Z}_i=\frac{\partial \textbf{R}}{\partial x^i}=\frac{\partial \hat{x}^\alpha}{\partial x^i}\hat{\textbf{Z}}_\alpha$$
Now since we know that a tensor in the $\hat{x}^\alpha$ coordinate system will be represented as $\textbf{P}'=P^\alpha\hat{\textbf{Z}}_\alpha$ we can observe the way it compares to $\textbf{P}=P^i\textbf{Z}_i$ and by making the following correspondence $\textbf{P}'=\textbf{P}$ by geometrical intuition we see that:
$$P^\alpha=\left(\frac{\partial \hat{x}^\alpha}{\partial x^i}\right)P^i$$
To finish it off, noticing that $\vec{F}$ is a 1-form, we notice that its differential is given by:
$$D\vec{F}=\frac{\partial \hat{x}^\alpha}{\partial x^i}\textbf{Z}^i\otimes\hat{\textbf{Z}}_\alpha$$
And that indeed:
$$D\vec{F}\cdot\textbf{P}=D\vec{F}\cdot(P^j\textbf{Z}_j)=\frac{\partial \hat{x}^\alpha}{\partial x^i}P^i\hat{\textbf{Z}}_\alpha=P^\alpha\hat{\textbf{Z}}_\alpha=\textbf{P}' $$
This is the method by which contravariant vectors transform under your coordinate change :) I hope that helped...its slightly informal and a geometrical way of doing things but it really highlights the geometrical beauty of tensor calculus I think. Good Luck!
