On 13/01/2021 00:56, Dan Asimov wrote:
But something different happens when considering Riemann surfaces that are topologically a torus, T^2. In this case it can be shown that the set of conformal equivalence classes of tori are in bijective correspondence with the points of the set
X = {z ∊ ℂ | |z| > 1 and -1/2 ≤ Re(z) < 1/2}.
(Note that its boundary consists of two semi-infinite lines and a 60º arc.)
I think you need part of that arc to be in the set.
A finer tuning of this fact gives a geometric bijection between the set of conformal equivalence classes of tori with the *quotient* of the set
X = {z ∊ ℂ | |z| > 1 and -1/2 ≤ Re(z) ≤ 1/2}
(note that a less-than has become a less-than-or-equal-to) after any boundary point z ∊ ∂X and its reflection z' about the y-axis are identified with each other.
Again, I think you want the part-of-a-circle arc included there. (The two special points you mention are both on that arc.) I think it's worth saying a bit more about this business of equivalence classes of complex tori, because there's some quite math-fun-shaped stuff in this vicinity. It relates to the famous fact that exp(pi sqrt(163)) is scarily close to being an exact integer. So, the bijection Dan mentions between tori and points in his fundamental region (let's call it R) comes about because any complex torus can be had by taking the complex plane C and identifying points that differ by m+nz where m,n are integers and z is some (fixed) complex number: we're making a torus by "rolling up" the complex plane in two different directions. We need z not to be real (else the thing collapses and we have a cylinder instead). And we might as well take z to be in the upper half-plane -- if it isn't, then instead of using {1,z} as the basis for the lattice we quotient by we can use {1/z,1} and now 1/z _is_ in the upper half-plane. But there are some less obvious equivalences, and these are what reduce the whole upper half-plane to Dan's region R. For instance, z+1 produces the exact same lattice as z, hence the same quotient. And -1/z produces, not the same lattice, but an equivalent one, for basically the same reason as we gave above for being able to restrict attention to the upper half-plane. These two operations z -> z+1 and z -> -1/z generate a group called PSL(2,Z) acting on the upper half-plane. It turns out that these are all the equivalences; if z1 and z2 are _not_ related by a transformation in PSL(2,z) then their tori are different. Now. It happens that there is a function called j, the "elliptic modular function", which maps H to the complex plane, in such a way that z1 and z2 are equivalent iff j(z1) = j(z2). That is, it exactly distinguishes inequivalent tori. (Its values at the two special points mentioned by Dan are 1728 and 0 respectively.) Clearly j is periodic with period 1 because z -> z+1 is in PSL(2,Z). So j(z) is a function of exp(2 pi i z), which is traditionally denoted q. We can write it as a power series in q, and it happens to begin 1/q + 744 + 196884 q + ... Now, here's the cool thing. If z is a non-real quadratic irrational number -- i.e., the solution to az^2+bz+c=0 where a,b,c are integers and b^2-4ac<0 -- then j(z) is an algebraic integer (i.e., the solution to a polynomial equation with integer coefficients and leading coefficient 1). And its degree (i.e., the degree of that polynomial, if we take it to have degree as small as possible) is the so-called "class number" of the quadratic field generated by z. In particular, if the class number is 1 then j(z) is an (ordinary rational) integer. In some sense the class number measures how _untrue_ it is that the integers in that quadratic field have unique factorization. In particular, the class number is 1 iff they do have unique factorization. This turns out to happen only for a finite number of those quadratic fields, those generated by the square roots of -1, -2, -3, -7, -11, -19, -43, -67, -163. OK. Now we can put the pieces together and get something startling. Consider z = [1 + sqrt(-163)]/2. This is a complex quadratic number in the upper half-plane, so j(z) is an algebraic integer. The corresponding quadratic field has class number 1, so j(z) is an ordinary integer. The value q = exp(2 pi i z) equals - exp(-pi sqrt 163); in particular, it is extremely small. And j(z) = 1/q + 744 + 196884q + higher-order terms. Since q is very small (and the series behaves reasonably well) all the terms in q, q^2, etc., are very small. So we have integer = j(z) = 1/q + 744 + very small, so exp pi sqrt 163 = -1/q = j(z) - 744 - very small = integer - 744 - very small, so exp pi sqrt 163 is just very slightly smaller than a (large) integer. The large integer in question turns out to be 640320^3, a nicely round number. Similarly, exp pi sqrt 67 and exp pi sqrt 43 and the others are slightly smaller than integers. Once we move to smaller exponents than those, though, q isn't small enough to outweigh the sizeable coefficients in the expansion of the j-function. E.g., exp pi sqrt 19 is 96^3 + 744 - 0.2223ish, which isn't close enough to an integer to be so startling. A related curiosity: exp pi sqrt 163 is _also_ almost exactly x^24-24, where x is a root of x^3 - 6x^2 + 4x - 2 = 0. This approximation is even closer than the integer one. -- g