# Flat torus embedding in euclidean space?

#### Highwind

hi,

for most of you this might be a simple question:

Is it possible to embed the flat torus in Euclidean space?

If we, for example, take a rectangle and identify the left and the right sides we get a cylinder shell, that can be embedded easily in R^3. If we construct the flat torus by identifying opposite sides, can we find an embedding in higher dimensions? Why is it possible or not?

Thanx for help.

Related Differential Geometry News on Phys.org

#### mathwonk

Homework Helper
take a circle in R^2, and another circle in another R^2. then the product of those two circles embedded in the product of those two R^2's is my guess at an embedding of a flat torus in R^4.

at least the curvature seems uniform all over, whence it must be zero everywhere, since the integral of the gaussian curvature should equal the euler characteristic which is zero.

#### Highwind

Your argumentation is not completeley clear:

Why exactly do you think that an embedding in R^4 can be found with constant gauss curvature? When I take the product of the two S^1 (circles) I can easily embedd the result (namely the torus) in R^3 (as a doughnut). This is of course also a valid embedding in R^4 but does not have constant curvature. Actually the topological product you talked about has nothing to do with the embedding you choose, therefore it does not help at all.

The question is rather:
Is it possible to find an embedding of the flat torus in R^4 or higher with constant gauss curvature? Because then I agree: The curvature has to be zero because of Gauss-Bonnet.

#### Hurkyl

Staff Emeritus
Gold Member
There's a canonical isomorphism R2xR2 &rarr; R4.

#### mathwonk

Homework Helper
have you read my argument again, more carefully, high wind? still puzzled?

#### Highwind

I understand what you're at. I also think that it is true, but
the only gap in the argumentation is this:

You take a circle (having constant curvatrue) and another circle (also having constant curvature) in another R^2. Why exactly has the product of those circles (in the product of those R^2) also constant curvature?

As I mentioned, products of constantly curved objects do not necessarily have constant curvature (as can be seen in the case of the doughnut torus in R^3).

#### Hurkyl

Staff Emeritus
Gold Member
Look at the symmetries of the embedding!

#### mathwonk

Homework Helper
you took a product of circles and then gave it a completely different embedding. i took the "product embedding".

#### Doodle Bob

Highwind said:
I understand what you're at. I also think that it is true, but
the only gap in the argumentation is this:

You take a circle (having constant curvatrue) and another circle (also having constant curvature) in another R^2. Why exactly has the product of those circles (in the product of those R^2) also constant curvature?

As I mentioned, products of constantly curved objects do not necessarily have constant curvature (as can be seen in the case of the doughnut torus in R^3).
I don't think that MW is arguing that. As hinted by Hurkyl, the argument is that the embedding in R^2xR^2 is homogeneous under a subgroup of isometries in R^4, i.e. rotations about the origin in either copy of R^2, and compositions thereof. Thus, they are isometries of the induced metric on T^2, i.e. they preserve the induced metric. So, they preserve the curvature (take your pick on which kind: Riemannian, sectional, Gaussian, etc.). In particular, they preserve the sectional curvature of the embedded torus, which on a Riemann surface is the same as the Gaussian curvature, i.e. the Gaussian curvature is constant.

As argued by MW before, we know by Gauss-Bonnet that the integral of the Gaussian curvature over the embedded torus is zero, and it is the same number at each point. So, the Gaussian curvature is zero everywhere.

Another way to see it is that with the R^4 embedding of the torus, you've got an basis of its tangent bundle, orthonormal with respect to the induced metric and whose Lie bracket of each other is zero, namely,

X1= -(sin u)e1 + (cos u) e2,
X2= -(sin v) e3 + (cos v) e4,

at the point (cos u, sin u, cos v, sin v) of the embedding. The existence of this guarantees that the embedding is flat, since it means that we locally can put the induced metric in the form: ds^2= dx^2+dy^2, for a local coordinate system (x,y) on the embedded torus .

Last edited:

#### mathwonk

Homework Helper
wonderful post doodlebob. do you follow highwind? he is saying that there is a distance preserving map, namely a rotation, or composition of rotations, taking any point of the product embedded torus in R^4 to any other point. hence the local metric struture must be the same everywhere.

then he goes on a little over my head, with what looks like a nice alternate approach....

#### Highwind

Yes, thanks.
Now it's clear. If you actually take the parametrization f(u,v) = (cos u, sin u, cos v, sin v) you can easily calculate the first and second fundamental tensor (being the unity matrix). Therefore the surface is flat.
Now only the question if the universe is a 3 dimensional torus remains to be answered. I think we should leave this for a different thread.

#### mathwonk

Homework Helper
einsteins theory says that massive bodies contribute non zero curvature, if what i read in the funny papers is correct, so this is inconsistent with saying the universe is flat.

#### Highwind

yes, einstein says that space-time is curved. But still the universe might be of torodial shape (not necessarily flat).

#### mathwonk

Homework Helper
now you are back to talking only about the topology which has nothing to do with the (local) "shape", only the global connectivity properties.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving