# Showing f is an isometry

• May 7th 2013, 08:11 AM
Plato13
Showing f is an isometry
I have shown the second function is a isometry and if I can show the first function is as well, then that gurantess f is a local isometry right? The other thing is I'm not sure what showing f being well defined involves.

• May 7th 2013, 10:55 AM
xxp9
Re: Showing f is an isometry
1. To show that f is well-defined you need to show that $\displaystyle \phi$ is one to one, so that $\displaystyle f = \psi \circ \phi^{-1}$. This is easy since sinh is one to one.
2. To show that f is locally isometric, show that first fundamental forms of the two surfaces are the same: $\displaystyle I = \cosh^2v({du}^2+{dv}^2)$
3. To show that f is not globally isometric, choose the two points (u,v)=(0,0) and $\displaystyle (2\pi, 0)$
The two points on H is (0,0,0) and $\displaystyle (0,0,2\pi)$. The distance is $\displaystyle 2\pi$ since the z-axis lies on H.
The two points on C is (1,0,0) and (1,0,0), the same point. So the distance is 0.
• May 10th 2013, 07:45 AM
Plato13
Re: Showing f is an isometry
Thanks kindly.