In $\mathbb R^3$ with Cartesian coordinates $(x,y,z),$ revolve the circle $(x-\sqrt 2)^2+z^2 =1,\ y=0$ about the $z$-axis.

This yields a torus embedded in $3$-space that is conformally equivalent to the flat torus $(\mathbb R/(2\pi))^2,$ via a mapping that takes $(x,y,z)=(1+\sqrt 2,0,0)$ to $(0,0)$ and $(x,z) = (-1-\sqrt 2,0, 0)$ to $(0,\pi),$ and $(-1+\sqrt 2,0,0)$ to $(\pi,0)$ and $(1-\sqrt 2,0,0)$ to $(\pi,\pi).$

One of Jacobi's elliptic functions gives us a $2$-to-$1$ mapping from the flat torus into $\mathbb C\cup\{\infty\}$ that takes $(0,0)$ to $0$ and $(0,\pi)$ to $\infty.$

Then the inverse of the stereographic projection takes $\mathbb C\cup\{\infty\}$ to the sphere $x^2+y^2+z^2=1$ while taking $0$ to $(1,0,0)$ and $\infty$ to $(-1,0,0).$

The composition of mappings takes the embedded torus to the sphere.

**Fact 1:** This composition of mappings takes the largest and the smallest of the four $x$-intercepts of the torus respectively to the largest and smallest of the two $x$-intercepts of the sphere.

**Fact 2:** This composition of mappings is conformal except at four points on the torus (corresponding to points at which the derivative of Jacobi's function is $0$).

(Unless I got some of the nitpicking details wrong about which points map to which) I believe I have reason to suspect the following really startling proposition:

The intersection of this sphere with each plane containing the $x$-axis is the image under this composition of mappings of the intersection of the torus with the

sameplane.

So I'm wondering:

- Is this true?
- Is it known?
- Can it be proved by using some known fact about Jacobi's function and maybe some elbow grease?

**APPENDIX:** The inverse of the conformal mapping from the embedded torus to the flat torus is this:
\begin{align}
\text{For }\alpha,\beta\in\mathbb R/(2\pi), \\[4pt]
x & = \cos\beta/(\sqrt2-\cos\alpha) \\[4pt]
y & = \sin\beta/(\sqrt2-\cos\alpha) \\[4pt]
z & = \sin\alpha/(\sqrt2-\cos\alpha) \\[4pt]
& {} \qquad \uparrow \\
& {} \qquad \text{same denominator in all three coordinates}
\end{align}

**POSTSCRIPT:** If I'm not mistaken, I've reduced this to the following.

You have a doubly periodic holomorphic function with periods $2\pi$ and $2\pi i$ that takes $0$ and $\pi + i\pi$ to $0$ and takes $\pi$ and $i\pi$ to $\infty.$

Now look at the inverse-images of straight lines through $0$ corresponding to arguments (i.e. angles from the real axis) between $0$ and $45^\circ$. Those inverse-images should be graphs of functions of the form $y=\arcsin(c\sin(x))$ where $x$ and $y$ are respectively the real and imaginary parts.

Is that a known proposition about Jacobi's functions?

couldsuspect you of sarcasm, but that would be petty. The people who make me wonder how I put up with them are a different group from those who don't answer things. $\endgroup$ – Michael Hardy Jan 29 '17 at 6:45