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The problem statement is

Show that every local isometry is conformal.

I have seen this question posed several times, in particular here

Every diffeomorphism which is an isometry is also conformal

This is Problem 6.3.1 in Pressley's Elementary Differential Geometry book.

I used the fact that,

a local diffeomorphism $f: S_1 \rightarrow S_2$ is a local isometry if and only if, for any surface patch $\sigma_1$ of $S_1$, the patches $\sigma_1$ and $f \circ \sigma_1$ of $S_1$ and $S_2$, respectively have the same first fundamental form.

Then,

a local diffeomorphism $f: S_1 \rightarrow S_2$ is conformal if and only if, for any surface patch $\sigma$ of $S_1$, the first fundamental forms of the patches $\sigma$ of $S_1$and $f \circ \sigma$ of $S_2$ are proportional.

This totally makes sense. The constant of proportionality is 1. But then the author goes on to say that

In particular, a surface patch $\sigma(u,v)$ is conformal if and only if its first fundamental form is $\lambda(du^2+dv^2)$ for some smooth function $\lambda(u,v)$

This is my source of confusion. From the last statement, I am reading that the terms of the first fundamental form are $F_1=F_2=0$ and $E_1=E_2=G_1=G_2$, but the condition of an isometry is $E_1=E_2, F_1=F_2$, and $G_1=G_2$. The isometry condition is less restrictive. So how does having the condition of an isometry imply we are also satisfying the condition of conformal? How can I say that $E=G$ and $F= 0$?

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  • $\begingroup$ The metric on $\Bbb{R}^2$ the usual dot product, i.e., $du^2+dv^2$, so the surface patch will be conformal if and only if its first fundamental form is proportional to this. $\endgroup$ – jgon May 22 at 2:21
  • $\begingroup$ @jgon Then, since every local isometry is conformal, every isometry's first fundamental form must be proportional to the metric on $\mathbb{R}^2$? And this means that the coefficient of $dudv$ must be zero? $\endgroup$ – Link Morgan May 22 at 6:01
  • $\begingroup$ I think you're misunderstanding what they mean when they say that the surface patch is conformal. $\sigma$ is a diffeomorphism $U\subseteq \Bbb{R}^2\to S$ (I'm actually not sure which direction the diffeomorphism points in your book, but it's a diffeomorphism, so it doesn't matter). Thus for $\sigma$ itself to be conformal, it must be conformal as a map from $U\subseteq \Bbb{R}^2$ to $S$, which means its fundamental form must be a scalar multiple of the fundamental form on $\Bbb{R}^2$. $\endgroup$ – jgon May 22 at 20:25

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