# Does a local isometry imply that the first fundamental form has F=0

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 $$F= 0$$?

• 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. – jgon May 22 at 2:21
• @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? – Link Morgan May 22 at 6:01
• 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 $\Bbb{R}^2$. – jgon May 22 at 20:25