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my aim is to show that the $n$-Torus $T^n = S^1 \times \ldots \times S^1$ can be embedded into $\mathbb{R}^{n+1}$ by giving a function $f: \mathbb{R}^{n+1} \rightarrow \mathbb{R}$ such that the Torus is described by all points satisfying $\ f(x_1,\ldots,x_n)=0$. This is quite simple in the two dimensional case with the equation $(\sqrt{x^2+y^2}-R)^2+z^2=r^2$. Trying to find a similar equation for higher dimensions i was starting with a parametric description of $T^3$ and then tried to eliminate the parameters in order to get a implicit description. But already in this case it got extremely messy with a lot of case distinctions (since $\sin$ and $\cos$ can't be inverted at once for the whole interval of the parameters $(0,2\pi)$).

So does anybody know a more elegant way to achieve an implicit description of $T^3$ and eventually $T^n$ or is there just none?

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@Jason: Do any of the answers of the previous question give a function $f:\mathbb R^{n+1}\to\mathbb R$ whose zero level hypersurface is an $n$-torus? I don't see it, so this doesn't look like a duplicate to me. –  Rahul Feb 19 '13 at 20:23
@ℝⁿ: (It's hard to ping you with your new user name!). I see what you're saying, but it is a duplicate once the restriction on methods is lifted. Further, once one has an embedding, cooking up the desired $f$ can be done in a canonical fasion (as a signed distance to $T^n$ with larger values cut off). –  Jason DeVito Feb 19 '13 at 20:40

1 Answer 1

up vote 2 down vote accepted

Start with the equation $$x_1^2+x_2^2=r_1^2$$ which describes a centered circle in the place of radius $r_1$. Replace $x_2$ by $\sqrt{x_2^2+x_3^2}-r_2$, so get $$x_1^2+(\sqrt{x_2^2+x_3^2}-r_2)^2=r_1^2.$$ This is a $2$-torus with radii $r_1$ and $r_2$. Let's do this again: replace the last variable $x_3$ by $\sqrt{x_3^2+x_4^2}-r_3$, to get $$x_1^2+(\sqrt{x_2^2+(\sqrt{x_3^2+x_4^2}-r_3)^2}-r_2)^2=r_1^2.$$ This is a $3$-torus with radii $r_1$, $r_2$, $r_3$.

One can go on as long as one wants.

This is just using the procedure which goes from, say, a curve to the surface of revolution it generates, several times.

(Of course, the radii have to be such that the thing is an actual torus, without any self-intersections)

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You seem to give the formula for a 2-Torus in $\mathbb{R}^{n+1}$ and not for a n-Torus, or am I getting it wrong? –  user9784 Feb 18 '13 at 23:30
This gives a nice, round $n$-torus in $\mathbb R^{n+1}$. You should look up how to write down equations for hypersurfaces of revolution. –  Mariano Suárez-Alvarez Feb 18 '13 at 23:38
But if its a n-Torus = $S^1 \times\ldots \times S^1$ there should be n radii to specify it and not just two. –  user9784 Feb 19 '13 at 13:57
If you wanted that, it would have been useful to be explicit about it in the body of the question :-) –  Mariano Suárez-Alvarez Feb 19 '13 at 18:28
I'm sorry, i thought there was some kind of convention about what an n-Torus is. I clarified it in the question above. –  user9784 Feb 19 '13 at 19:21

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