MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question's not getting anywhere on, so let's see if someone here can say something:

In the $xyz$-space imagine a circle of radius $r>0$ in the $xz$-plane, whose center is on the $x$-axis at a distance $R>r$ from the $z$-axis, and revolve it about the $z$-axis, getting a torus embedded in $\mathbb R^3$.

Its intersection with planes parallel to the $xy$-planes let us call "parallel circles".

A curve winds $m$ times around the long way and $n$ times around the short way (and $m$ and $n$ are not both $0$), and returns to its starting point. It is situated so that it meets all parallel circles at the same angle.

As a function of $m$ and $n$ and $r$ and $R$ and the position on the curve, what are

  • the length of the curve;
  • the angle at which it meets the parallel circles;
  • the curvature;
  • the torsion?

When $m=0$ or $n=0$ the answers are obvious.

When $m=n=1$ the answer is surprising: the torsion is everywhere $0$ and the curvature is constant (and equal to $1/R$, so the arc length is $2\pi R$).

Are there other cases where the answer is surprising or elegant or of interest for other reasons?

share|cite|improve this question

I think those (1,1) curves are the Villarceau circles. Remarkably, from knowing that they have length $2\pi R$, you can calculate the length of the $(m,n)$ curves when $n\ne 0$.

There's a conformal map from the embedded torus to the flat torus formed by gluing the edges of a rectangle, and it takes all the parallel circles to lines parallel to one of the edges. (One way to see this is to draw circles on the torus orthogonal to the parallel circles. These form a grid, and if they're spaced right, the grid is made of approximate squares. The conformal map stretches and shrinks so that all the squares have the same size.) On the flat torus, it's easy to calculate all these curves -- they're just straight lines with slope is determined by $m$, $n$, and the side lengths of the rectangle.

Okay, so what are the side lengths? Consider the torus formed by gluing the edges of a $a \times b$ rectangle. There's some $a$ and $b$ and some metric $dg^2=f(x,y)(dx^2+dy^2)$ that makes it isometric to the embedded torus. In fact, by symmetry, $f(x,y)$ depends only on $y$, so $dg^2=h(y)^2(dx^2+dy^2)$. Take a closed curve of constant slope that winds $m$ times around the long way and $n$ times the short way. If $n\ne 0$, this curve spends the same amount of time at each $y$-coordinate, so its length wrt $g$ is its euclidean length times the average value of $h$. If we call that average value $M$, we get a length of $$\sqrt{(Mma)^2+(Mnb)^2}.$$

But we already know the lengths of two of these curves: when $m=1,n=1$, it has length $2\pi R$, and when $m=0,n=1$, it has length $2\pi r$. Therefore, $Ma=2\pi \sqrt{R^2-r^2}$ and $Mb=2\pi r$. So, the curve that goes around $m$ times the long way and $n$ times the short way has length $$2\pi \sqrt{m^2(R^2-r^2)+n^2r^2}$$ and intersects the parallel circles at slope $$\frac{nr}{m\sqrt{R^2-r^2}}.$$

share|cite|improve this answer
They are indeed Villarceau circles. A paper identified below says someone named Schoelcher proved that eacj Villarceau circle meets all of the parallel circles at the same angle. Villarceau apparently wrote about cutting the torus with certain planes, but the propsition I wrote above concludes, rather than assumes, that the curve lies in a plane.<br><br> Mannheim, M. A. (1903). "Sur le théorème de Schoelcher". Nouvelles Annales de Mathématiques (Paris: Carilian-Gœury et Vor. Dalmont) 4th series, volume 3: 105–107. – Michael Hardy Oct 17 '12 at 16:38
Someone on stackexchange a few months ago asserted in response to a question I raised, that not all embedded tori of the sort mentioned here are conformally equivalent to each other. And I'm not clear on whether any of them are conformally equivalent to a flat torus such as you describe. Might that depend on the shape of the rectangle? The interiors of different rectangles are conformally equivalent to each other, by the Riemann mapping theorem, but when you glue the edges together, are the equivalences still conformal at the edges? – Michael Hardy Oct 17 '12 at 17:40
That's neat; I'll have to look that paper up. Every torus with a Riemannian metric is conformally equivalent to a flat torus by the Uniformization Theorem, but there are lots of flat tori -- gluing the edges of any parallelogram gives a flat torus. Two flat tori are conformally equivalent if and only if they differ by a scaling, so there's a two-dimensional moduli space of conformal structures on a torus. In this case, the embedded torus has enough symmetry that the corresponding flat torus must come from a rectangle, but the aspect ratio of the rectangle is a conformal invariant. – Robert Young Oct 17 '12 at 21:15
I think in this case, an embedded torus of area 1 (to avoid the issue of different scalings) is only conformally equivalent to one other embedded torus of area 1. I haven't checked this, but I think there's an inversion of $\mathbb{R}^3$ which turns the torus inside out into a conformally equivalent torus, swapping the long and short circles. – Robert Young Oct 17 '12 at 21:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.