Something I learned this week: there is a relatively simple equation that defines a properly embedded punctured torus in R^3. Here it is: x(x^2+y^2+z^2-1)=2yz. I’ll explain where I got this.
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In the manifolds textbook we’re using for analysis, one of the exercises says that the set of 2x2 matrices with (Frobenius) norm 1 and determinant 0 is a torus. Since the Frobenius norm is just the standard Euclidean norm on R^4, this is a subset of the 3-sphere S^3.
We can parametrize S^3 using the inverse of stereographic projection: (x,y,z) -> (2x,2y,2z,x^2+y^2+z^2-1)/(x^2+y^2+z^2+1). Treating this vector as a 2x2 matrix, the det=0 condition becomes 2x(x^2+y^2+z^2-1)-4yz=0, which is equivalent to the equation I gave above.
The reason the torus is “punctured” is that the matrix [[0,0],[0,1]] is not in the image of our parametrization, although it has norm 1 and determinant 0. It is only reached by letting (x,y,z) go to infinity.
Fascinating! The defining equation is a polynomial of degree 3. Any punctured torus with a complex structure can be properly, holomorphically embedded in ℂ² where it is defined by a degree 3 polynomial complex equation (elliptic curves in Weierstrass form). Are the two embeddings related?
I noticed that too! But I haven’t pushed my thinking far enough in that direction. This surface came from intersecting the sphere |z|^2+|w|^2=1 in C^2 with the locus Im(z*conj(w))=0, then projecting stereographically to R^3. Does that make it obvious one way or the other?
It’s the lattice generated by (2,0) and (1,1), so I guess that’s the one from the B_2 root system.
I noticed the torus could be parametrized by
[cos(u)cos(v),cos(u)sin(v)]
[sin(u)cos(v),sin(u)sin(v)]
and that this map on the domain [0,2π)^2 is two-to-one.
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I noticed the torus could be parametrized by
[cos(u)cos(v),cos(u)sin(v)]
[sin(u)cos(v),sin(u)sin(v)]
and that this map on the domain [0,2π)^2 is two-to-one.