I am was reading the wikipedia page on metric tensors, when I saw something that was hard to grasp in the coordinate transformation section. This topic is a little bit uncomfortable to me, so maybe I have missed something, but there appears to be a factor of 4 that appears when working everything out by hand?
With r being a vector valued function $\vec{r}(u,\,v) = \bigl( x(u,\,v),\, y(u,\,v),\, z(u,\,v) \bigr)$, and with the following identity $$ \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} & \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} \\ \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} & \frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{bmatrix} = \begin{bmatrix} E & F \\ F & G \end{bmatrix} $$
the coordinate transformation is given by,
$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix}^\top \begin{bmatrix} E & F \\ F & G \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix} \end{aligned} $$
With the following substitution in the coordinate transformation matrix,
$$ \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} $$
The transformation then becomes,
$$ \begin{aligned} \begin{bmatrix} A & B \\ C & D \end{bmatrix}^\top \begin{bmatrix} E & F \\ F & G \end{bmatrix} \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} \underbrace{A^2 E + 2 ACF + C^2G}_{E^\prime} & \underbrace{ABE + BCF + AFD + CDG}_{F^\prime} \\ \underbrace{ABE + BCF + AFD + CDG}_{F^\prime} & \underbrace{B^2E + 2BFD + D^2G}_{G^\prime} \end{bmatrix} \end{aligned} $$
Plugging the values into the variables give the following expressions,
$$ \begin{aligned} E^\prime &= \frac{\partial u}{\partial u^\prime}\frac{\partial u}{\partial u^\prime} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + 2 \frac{\partial u}{\partial u^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v} + \frac{\partial v}{\partial u^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \\ F^\prime &= \frac{\partial u}{\partial u^\prime}\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + \frac{\partial u}{\partial v^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v} + \frac{\partial u}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v}\frac{\partial v}{\partial v^\prime} + \frac{\partial v}{\partial u^\prime}\frac{\partial v}{\partial v^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \\ G^\prime &= \frac{\partial u}{\partial v^\prime}\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + 2\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v}\frac{\partial v}{\partial v^\prime} + \frac{\partial v}{\partial v^\prime}\frac{\partial v}{\partial v^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{aligned} $$
after simplifying by cancelling out similar factors in the numerator and denominator and summing the result, everything comes out to,
$$ \require{cancel} \begin{aligned} E^\prime &= \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial u}}{\partial u^\prime} \frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + 2 \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}} + \frac{\cancel{\partial v}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \\ F^\prime &= \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + \frac{\cancel{\partial u}}{\partial v^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}} + \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}}\frac{\cancel{\partial v}}{\partial v^\prime} + \frac{\cancel{\partial v}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \\ G^\prime &= \frac{\cancel{\partial u}}{\partial v^\prime}\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + 2\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}}\frac{\cancel{\partial v}}{\partial v^\prime} + \frac{\cancel{\partial v}}{\partial v^\prime}\frac{\cancel{\partial v}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \end{aligned} $$
which then reduces to the following by adding the remaining terms,
$$ \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = 4\begin{bmatrix} \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} \\ \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} & \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} \end{bmatrix} $$
The Wikipedia article states (above equation 2') that the values of $E^\prime, F^\prime, G^\prime$ are in fact $ E^\prime = \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime}, \;\; F^\prime = \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime}, \;\; G^\prime = \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} $
which then means that I get the following expression after simplifying my manual transformation above, and comparing it with the definition of $E^\prime, F^\prime, G^\prime$ from Wikipedia.
$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} \neq 4\begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} \end{aligned} $$
Questions
How can I reconcile that this factor of 4 comes out? Is it just because the factor of 4 becomes irrelevant for an infinitesimal difference? Or have I made a terrible error somewhere?
Generally, how can I understand coordinate transformations when dealing with a Jacobian matrix, are they the same thing as a change of basis in linear algebra when we see the form $P^{-1}AP$? What is the significance here that this form is $P^\top AP$ with a transpose instead of an inverse? The Jacobian is highly unlikely to be orthonormal (right?), so the transpose is definitely not the inverse.
With r being a vector valued function $\vec{r}(u,\,v) = \bigl( x(u,\,v),\, y(u,\,v),\, z(u,\,v) \bigr)$, the partial derivatives $\frac{\partial r}{\partial u} \,\text{and}\, \frac{\partial r}{\partial v} \in \mathbb{R}^{1 \times 3}$, and with the following identity $$ \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} & \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} \\ \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} & \frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{bmatrix} = \begin{bmatrix} E & F \\ F & G \end{bmatrix} = \begin{bmatrix} \frac{\partial r}{\partial u} \\ \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial u} & \frac{\partial r}{\partial v} \end{bmatrix} $$
It is convenient to view the matrix as an outer product as we will see, the coordinate transformation is given by,
$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix}^\top \begin{bmatrix} \frac{\partial r}{\partial u} \\ \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial u} & \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix} \end{aligned} = P^\top P $$
$P$ then becomes the total derivative w.r.t. the new variables $u^\prime$ and $v^\prime$
$$ \begin{aligned} P = \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial u}{\partial u^\prime} + \frac{\partial r}{\partial v}\frac{\partial v}{\partial u^\prime} & \frac{\partial r}{\partial u}\frac{\partial u}{\partial v^\prime} + \frac{\partial r}{\partial v} \frac{\partial v}{\partial v^\prime} \end{bmatrix} = \begin{bmatrix} \frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial v^\prime} \end{bmatrix} \end{aligned} \implies P^\top P = \begin{bmatrix} \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} \\ \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} & \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} \end{bmatrix} $$
which gives the answer as specified in the Wikpedia page. The problem with the initial version is that I was accumulating terms of the total derivative which were not in fact like terms, but they were contributions to the total derivative from changes in the new variables $u^\prime$ and $v^\prime$.