# Isometric Immersion of the Torus

1. Sep 25, 2011

### felper

Hello! I'm new here. I've been seeing this forum for a long time, but i never registered. I'd like to begin to contribute to this forum, i'll try it (even if my english doesn't help me).
So now, i ask you for help: i've to find an isometric immersion of the flat torus on $\mathbb{R}^4$. I know that the function $f:\mathbb{S}^1\times\mathbb{S}^1\rightarrow \mathbb{R}^4$ given by $f(\theta,\phi)=(\cos\theta,\sin\theta,\cos\phi, \sin \phi )$ is the indicated function, but i don't know how to demonstrate that it's differential is inyective.

2. Sep 26, 2011

### quasar987

Hi,

Notice that the matrix of the differential of f with respect to the (theta, phi) coordinates on TÂ² and the standard coordinates on R^4 is the 4x2 matrix

-sin(theta) 0
cos(theta) 0
0 -sin(phi)
0 cos(phi)

Since cos(x) and sin(x) never vanish simultaneously, it is easy to see that the linear map associated with this matrix has kernel={0} and so is injective.

3. Sep 27, 2011

### lavinia

just compute the Jacobian and prove that it has rank 2.

BTW: this is more than an immersion. It is an embedding. Further the length of every point considered as a vector in R^4 is constant so this torus lies on a 3 sphere. Try computing the equation of the torus in R^3 obtained from this one by stereographic projection. This new torus is not flat yet the mapping between it and the flat torus is conformal.

Your immersion is of the Euclidean plane into R^4.

Question: Is their an isometric embedding of the flat Klein bottle in R^4?

Last edited: Sep 27, 2011
4. Sep 27, 2011

### felper

Thanks, i could demonstrate it. I'll think about the question.