I have a donut. Its boundary is a two-dimensional surface embedded in three-dimensional space, and surely is homeomorphic to a torus. If we fix a Riemannian metric on the space, it induces a two-dimensional Riemannian metric on the donut. Forgetting the scaling, we get a conformal structure on the surface. If we fix an orientation, this defines a complex structure.

Complex surfaces homeomorphic to a torus are classified by their $j$-invariant. Thus, I would like to calculate the $j$-invariant of my donut. There are several variants of this question:

  1. Suppose my donut is modeled in cylindrical coordinates $(r,z,\theta)$ by the equation $(r-1)^2 + z^2=\rho^2$. What is the $j$-invariant as a function of $\rho$?

  2. Find an actual, real donut, and approximate it by a surface of this form. Estimate $\rho$ and thereby estimate its $j$-invariant.

  3. Determine how your donut differs from this model, and estimate in which direction that might change its $j$-invariant. Or solve a different model that better fits your donut.

I would be happy to see progress on any of these vitally important problems, which you will no doubt agree have many fruitful applications to mathematics and our daily life.

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    $\begingroup$ I tried 2, but ended up eating the doughnut! $\endgroup$ – Lord Shark the Unknown May 1 '17 at 16:30
  • $\begingroup$ @LordSharktheUnknown is woke af for eating the 🍩 $\endgroup$ – Dave huff May 1 '17 at 17:19
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    $\begingroup$ If your donut is a jelly donut then it will not be homeomorphic to a torus and will thus not be able to be classified by a j-invariant. You will have to use a related invariant called a jelly-invariant. $\endgroup$ – dxdydz May 1 '17 at 17:20

For $\rho\in (0,1)$, write $D_{\rho}$ for your donut $(r-1)^2 + z^2 = \rho^2$. For $x\in\mathbb{R}$, define $$ q_\rho(x):=2\sqrt{1-\rho^2}\tan\left(\frac{x\sqrt{1-\rho^2}}{2\rho}\right)-\rho. $$ We define a parametrization of $D_\rho$: $$ \begin{align*} f_{\rho}:\mathbb{C}&\to D_{\rho}\\ x+iy&\mapsto \left(1+\rho\frac{2 q_\rho(x)}{1+q_\rho(x)^2},\rho\frac{1-q_\rho(x)^2}{1+q_\rho(x)^2},y\right). \end{align*} $$ Here the image is written in $(r,z,\theta)$ coordinates. Note that this function is defined even where $q_\rho(y)$ has poles. It is straightforward but tedious to verify that $f_\rho$ is conformal. If we write $r(x)$, $z(x)$ for the $r$- and $z$-coordinates, this amounts to the identity $$ r'(x)^2+z'(x)^2=r(x)^2. $$ I used a CAS to find the formula for $f_\rho$.

From the formula, we see that $f_{\rho}$ is doubly periodic, with minimal periods $2\pi i$ and $2\pi\rho/\sqrt{1-\rho^2}$. Thus $D_\rho$ is conformally equivalent to the torus $$ \frac{\mathbb{C}}{\mathbb{Z}+i\frac{\sqrt{1-\rho^2}}{\rho}\mathbb{Z}}. $$ Write $j(\tau)$ for the $j$-function, which takes $\tau$ in the upper half plane to the $j$-invariant of $\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})$. The $j$-invariant of $D_{\rho}$ is then $$ j\left(i\frac{\sqrt{1-\rho^2}}{\rho}\right). $$

I don't have a donut in front of me, but the internet suggests that a Krispy Kream original glazed has outter diameter $3.5''$ and width from edge to hole $1.4''$. Scaling, this gives the value $\rho=\frac{2}{3}$. Sage then gives the estimate $$ j\approx 2060.94946856905. $$


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