Isometric Immersion of the Torus

[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.

Thanks for your help!

 

[quasar987]

Hi,

Notice that the matrix of the differential of f with respect to the (theta, phi) coordinates on T² 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.

 

[lavinia]

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.

Thanks for your help!

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?

 

[felper]

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

 

[blue_raver22]

Hello there! Welcome to the forum and thank you for your question. Isometric immersion of the torus is a fascinating topic in differential geometry. The function you mentioned, f(\theta,\phi)=(\cos\theta,\sin\theta,\cos\phi, \sin \phi ), is indeed a valid isometric immersion of the flat torus on \mathbb{R}^4. To demonstrate that its differential is injective, we can use the fact that an isometric immersion is a local isometry, meaning that the induced metric on the surface is equal to the metric on the ambient space.

In this case, the induced metric on the torus is given by ds^2=d\theta^2+d\phi^2, while the metric on \mathbb{R}^4 is given by ds^2=dx^2+dy^2+dz^2+dw^2. By taking the differential of f, we get df=(\sin\theta,-\cos\theta,\sin\phi,-\cos\phi)d\theta+(\cos\theta,\sin\theta,-\cos\phi,-\sin\phi)d\phi.

To show that df is injective, we need to show that the matrix of coefficients of df is invertible. In this case, the matrix is given by
\begin{pmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \\ \sin\phi & -\cos\phi \\ \cos\phi & \sin\phi \end{pmatrix}.

Using basic linear algebra, we can show that this matrix has full rank, meaning that its columns are linearly independent. Therefore, df is injective and the function f is an isometric immersion of the flat torus on \mathbb{R}^4.

I hope this helps and please don't hesitate to ask for further clarification. Happy contributing to the forum!