# Form of line element of a torus.

1. Jul 9, 2008

### arroy_0205

I noticed somewhere the line element of a two-dimensional torus is written in the form
$$ds^2=r^2(d\theta^2_1+d\theta^2_2)$$
The author only states that he assumes same radius parameter for simplicity and no further explanation is given. But I do not understand how that form is possible. I find, in such a case the line element should be
$$ds^2=r^2 d\theta^2_1+r^2(1+\cos\theta_1)^2d\theta^2_2)$$
I cannot reduce this form to the first form. Can anyone explain how the first form for 2D line element of a torus is possible?

2. Jul 9, 2008

### tiny-tim

ideal torus

Hi arroy_0205!

Your $$ds^2=r^2 d\theta^2_1+r^2(1+\cos\theta_1)^2d\theta^2_2$$ is for an ordinary common-or-garden torus, in which the "inside equator" is shorter than the "outside equator".

I think $$ds^2=r^2(d\theta^2_1+d\theta^2_2)$$ is for an "ideal torus", not embedded in any higher-dimensional space, in which all "equators" have the same length.

3. Jul 9, 2008

### arroy_0205

Hi tiny-tim,
Thanks for your response, but I do not completely get your point. You are trying to say something profound, that for a torus not embedded in a higher dimensional space... etc, but I cannot visualize a torus that way. Can you help? also I have taken the two radii same, that is why there is only one r in the line element that I wrote. My doubt is how embedding and the issue of two radii (same or different) are connected in this case. What is an "Ideal torus" by the way?

Also, do you mean that for a 2-torus which is $$S^1\times S^1$$, we can simply write
$$ds^2=r^2(d\theta^2_1+d\theta^2_2)$$
and as generalization, for a n-torus,
$$ds^2=r^2(d\theta^2_1+\cdots+d\theta^2_n)$$
will this be correct?

Last edited: Jul 9, 2008
4. Jul 9, 2008

### kahoomann

Is the geometry of torus flat or curved?
If it's flat, why it is flat?

5. Jul 9, 2008

### tiny-tim

Hi arroy_0205!
Yes for a 2-torus, not sure about an n-torus.
An "ideal torus" is S1 x S1.

It's like an Asteroids screen, or a piece of paper with opposite edges identified.

If the screen is square, there's only one r, it it's a rectangle, there are two rs.

The geometry of S1 x S1 is flat (locally identical to a Euclidean plane), so it can't be "joined up" in any higher dimensional Euclidean space.
Hi kahoomann!

The geometry of S1 x S1 is flat, because its geometry is locally identical to the Asteroids screen!

(Similarly, the geometry of the surface of a cone is flat.)

The geometry of a common-or-garden torus is curved.

6. Jul 10, 2008

### Doodle Bob

Re: ideal torus

I think this business about ideal versus garden-variety tori is a non-issue.

A common definition of an n-dimensional torus is simply a topological space that is diffeomorphic to the space $$S^1\times...\times S^1\subset R^{2n}.$$ When one says "a torus," usually (depending on the context) this is the 2-dimensional version, $$S^1\times S^1\subset C\times C$$, using the embedding $$S^1\subset C$$.

From the definition, one can see that there is no *one* Riemannian metric on a torus (what is being called a line element here), but many. This metric $$ds^2=r^2(d\theta^2_1+d\theta^2_2)$$ is the flat metric on the torus using as coordinates the angles in the embedding $$S^1\times S^1\subset C\times C$$. But the other one mentioned in the original post looks to me like the metric on the torus induced from the Euclidean metric when the torus is embedded in 3-space in the standard way. This is decidedly not a flat metric on the torus.

7. Jul 10, 2008

### tiny-tim

Re: ideal torus

Hi Doodle Bob!

But one could equally say that there is no *one* Riemannian metric on a sphere:

we can easily impose a metric on a sphere that "makes it an ellipsoid".

So when we define a sphere, we include the standard metric.

And when we define a torus, we should also include the metric.

An "ideal" torus and an "embedded" (common-or-garden) torus (with the second line element mentioned in the original post) are two different metric spaces, just as a sphere and an ellipsoid are.

8. Jul 10, 2008

### Doodle Bob

Re: ideal torus

But, your so-called "ideal" torus *is* "embedded" too; it's embeddable in R^4 as $$S^1\times S^1$$ and the metric that is induced by this embedding from the 4-dimensional Euclidean metric is, in fact, the standard flat metric on the torus, i.e., the Asteroids metric.

"Torus" is really reserved as topological term rather than a geometrical one. "Flat torus," e.g., is a better way of conveying a specific geometry rather than "ideal" since "ideal" conveys either a Platonistic point of view, which doesn't really to apply in this matter, or an algebraic point of view, which is equally as irrelevant here.

9. Jul 10, 2008

### tiny-tim

Re: ideal torus

Hi Doodle Bob!

So even though all its sides are curved, the geometry is flat, in exactly the same way as the surface of a cylinder in R3 is flat?

mmm … I never could think in four dimensions.

Yes, I suppose it does work in R4, because R4 has the direct product geometry R2 x R2, which S1 x S1 naturally slips into,

while S1 x S1 doesn't fit in with the direct product R3 = R2 x R1.

(I see you've had to put people right on this https://www.physicsforums.com/archive/index.php/t-80846.html".)

So if I say "embedded" in future … I'd better specify "in R3".

Thanks for the correction!

Last edited by a moderator: Apr 23, 2017