Isometry (Riemannian geometry) on a simple example

Question

I'm trying to understand the concept of isometry (Isometry (Riemannian geometry)) on a specific problem. I think I have to interpret $f^*$ as $$g(v,w) = f^* g'(v,w) = (df_{v,w})^* g'(f(v),f(w)),$$ where $(df_{v,w})^*$ is the transpose matrix of $df_{v,w}$. Could anyone be able to explain to me that concept on a simple example?

Answer 1

One example is translation in $\mathbb{R}^{d}$.

Perhaps a slightly more interesting case is polar coordinates. Take $M=M'=\mathbb{R}^{2}$, $f=$ identity. On $\left(M,g\right)$, take polar coordinates, and on $\left(M',g'\right)$ the Euclidean chart.

Fix a point $p\in M\backslash\left\{ 0\right\} $, $p=\left(r,\theta\right)=\left(x,y\right)$, with $x=r\cos\theta$, $y=r\sin\theta$. Then $g'=dx^{2}+dy^{2}$, and \begin{align*} f_{*}\left(\partial_{r}\right) & =\cos\theta\partial_{x}+\sin\theta\partial_{y}\\ f_{*}\left(\partial_{\theta}\right) & =-r\sin\theta\partial_{x}+r\cos\theta\partial_{y}\\ g\left(\partial_{r},\partial_{r}\right) & =g'\left(f_{*}\left(\partial_{r}\right),f_{*}\left(\partial_{r}\right)\right)=1\\ g\left(\partial_{r},\partial_{\theta}\right) & =g'\left(f_{*}\left(\partial_{r}\right),f_{*}\left(\partial_{\theta}\right)\right)=0\\ g\left(\partial_{\theta},\partial_{\theta}\right) & =g'\left(f_{*}\left(\partial_{\theta}\right),f_{*}\left(\partial_{\theta}\right)\right)=r^{2}. \end{align*} That is, $g=dr^{2}+r^{2}d\theta^{2}$.