# Isometry between cone and plane

Question: Exercise 1, Section 4.2, Do Carmo Differential Geometry of Curves and Surfaces:
Let F be a parametrization $F: U\in \mathbb{R}^2\rightarrow \mathbb{R}^3$,

$F(u,v)=$(u sin$\alpha$ cos$v$, u sin$\alpha$ sin$v$, u cos$\alpha$),
$(u,v)\in U= \{(u,v)\in \mathbb{R}^2; u>0\}$, $\alpha=$const.

Is F a local isometry?

My attempt/ problem:
I believe the answer should be yes (since plane and cone are indeed locally isometric, and being isometric should not be dependent on parametrization.) To show that two surfaces are isometric, we need to show they have the same first fundamental forms. I do not know how to parametrize plane to get the same E, F, G as what we get for the above parametrization of cone, which is E= 1, F=0, G= $u^2sin^2\alpha$.

• I guess you are supposed to prove that $F$ is a local isometry by hand, not just saying "the plane and the cone are locally isometric, so it doesn't depend on the parametrization". – Javi Mar 21 '18 at 21:09
• Just because the plane and the cone are locally isometric, it does not mean that there are maps between them which are not local isometries. You have to carry the computation dutifully. – Ivo Terek Mar 21 '18 at 21:37
• Hint: Think about (slightly modified) polar coordinates on the plane. – Ted Shifrin Mar 22 '18 at 1:46
• Thank you. I see why you say polar coordinates ($E=1$, $F=0$, $G= \rho^2$), But I don't see the modification which doesn't change $E=1$, but makes $G=\rho^2 sin^2\alpha$. Would you please explain more? – Mathophile-Mathochist Mar 23 '18 at 15:16
• wow it was obvious. Sorry for bothering. take $\rho$ to be $usin\alpha$. My problem originated from the fact that I thought we need to look at the cone and the plane from the same coordinate system, with its origin on the vertex of the cone, and the plane be an arbitrary plane. – Mathophile-Mathochist Mar 23 '18 at 18:22