Since it is possible to map a flat plane into 3d space through:
$$X=G_X(u,v)$$ $$Y=G_Y(u,v)$$ $$Z=G_Z(u,v)$$ Then you can find the tangental basis vectors through: $$\frac{\partial}{\partial u}=\frac{\partial X}{\partial u}\frac{\partial}{\partial X}+\frac{\partial Y}{\partial u}\frac{\partial}{\partial Y}+\frac{\partial Z}{\partial u}\frac{\partial}{\partial Z}$$
$$\frac{\partial}{\partial v}=\frac{\partial X}{\partial v}\frac{\partial}{\partial X}+\frac{\partial Y}{\partial v}\frac{\partial}{\partial Y}+\frac{\partial Z}{\partial v}\frac{\partial}{\partial Z}$$
$$\hat{e_u}=\frac{\partial X}{\partial u}\hat{X}+\frac{\partial Y}{\partial u}\hat{Y}+\frac{\partial Z}{\partial u}\hat{Z}$$
$$\hat{e_v}=\frac{\partial X}{\partial v}\hat{X}+\frac{\partial Y}{\partial v}\hat{Y}+\frac{\partial Z}{\partial v}\hat{Z}$$ And then the metric through:
$$g_{ij}=e_i \cdot e_j$$
Is there a way to find the metric for a transformation that maps the 2D plane to another 2D plane $$X = A(u,v)$$ $$Y = B(u,v)$$ Thanks
Too long for a comment.
Given a smooth coordinate mapping from $\mathbb{R}^n$ to $\mathbb{R}^n$, $(x_1,...,x_n)\to(\xi_1,...,\xi_n)$ which, say, could be given by some explicit transformation $\xi_i=f_i(x_1,...,x_n)$ the metric tensor of this transformation is defined as $$g_{i,j}=\frac{\partial \underline{x}}{\partial\xi_i}\boldsymbol{\cdot}\frac{\partial \underline{x}}{\partial\xi_j}$$ E.g, in the case of polar coordinates in $\mathbb{R}^2$ with transformation
$$(x,y)\to\left(r\cos\theta,r\sin\theta\right)$$ This is $$\mathbf{g} =\begin{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} & \begin{bmatrix} 0 & r \end{bmatrix} \end{bmatrix}$$ I've given it this flattened structure since $\mathbf{g}$ is a $(0,2)$ tensor, but it works just fine as a $2\times 2$ matrix as well.