The n-Category Café

I don’t have an answer, but the description of how to arrive at the list of reduced fractions is given here, for those interested. Also, the phenomenon John mentions turns up in the dodecahedron at the bottom of the first post in the series, with 0/1 and 0/2 appearing.

Mind you, would you complain, John, about the appearance of both 2/0 and 1/0 (e.g. at the dodecahedron at the link, or in your example) - they are both ‘infinity’…

Oh dear, I had no idea this would cause such confusion, or I’d have spent a lot more time on it at the start. Maybe I shouldn’t have written these labels as fractions, but just as ordered pairs of numbers … but that misses something too. And it seemed so neat and logical to me.

Anyway, yes, as elements of the projective line over $\mathbb{Z}_7$, $\frac{0}{1}$, $\frac{0}{2}$ and $\frac{0}{3}$ all represent the same thing. This doesn’t apply just to the zeros: since every non-zero element of $\mathbb{Z}_7 P^1$ is invertible, we can always cancel out any non-zero denominator. For example, $\frac{5}{3}=\frac{1}{2}=\frac{4}{1}$. Also, this phenomenon has already appeared with $N=5$. We get it whenever the group of units in $\mathbb{Z}_N$ has more than two elements.

However, the thing is, just using the points of the projective line over $\mathbb{Z}_N$ doesn’t give enough points to label the vertices (resp. faces) of the triangular (resp. $N$-gonal) tiling. If you want use fractions mod $N$ as labels, you have to include, as separate labels, certain different “fractions” which are equivalent to each other when evaluated as elements of $\mathbb{Z}_N P^1$, fractions which are equivalent because there are common factors in numerator and denominator which could be cancelled. When we are constructing labels, then out of all the possible pairs of numbers (mod $N$) that could make up the numerator and denominator of a fraction, we are allowed cancel out common factors if and only if they are also factors of $N$ itself. Shared factors between top and bottom which are coprime to $N$ can’t be cancelled. For instance, you can’t cancel a factor of $2$ or $3$ when $N=7$, although you can (and must) cancel shared factors of $2$ when $N=4$. (So these things are sort of like genuine fractions—you can cancel some factors. In fact, you cancel precisely the factors that you would have cancelled out of fractions representating elements of $\mathbb{Q}P^1$, before reducing mod $N$. So the mod-$N$ fractions used as labels sort of “remember” something about what they were when they were living back in $\mathbb{Q}P^1$.)

Maybe a good way to see why this is necessary is to think about $\Gamma(N)$. This is, if you recall, the subgroup $\left\{g\in PSL(2, \mathbb{Z})\vert g\equiv\left(\array{1&0\\0&1}\right) mod N\right\}$. We get to these modular curves by taking the quotient of the complex upper half plane by the action of $\Gamma(N)$, and we get the labels by taking the quotient of $\mathbb{Q}P^1$ by the action of $\Gamma(N)$.

Now the requirement that the elements of $\Gamma(N)$ should be equivalent to the identity mod $N$ is quite restrictive. From a fraction $\frac{m}{n}$, we can get fractions of the form $\frac{m+a m N+b n N}{n+c m N+d n N}$. That is, to the numerator we can add multiples of N times the numerator and multiples of N times the denominator, and likewise for the denominator. And this limits what we can get as a result.

For instance, suppose we have $\frac{11}{2}$ in $\mathbb{Q}$. Mod $7$, this is $\frac{4}{2}$. Is there some way to identify $\frac{11}{2}$ in $\mathbb{Q}$ with $\frac{4}{2}$ in $\mathbb{Q}$? The latter is just $\frac{2}{1}$ in $\mathbb{Q}$, which would reduce to $\frac{2}{1}$ mod $7$, enabling us to identify the fractions $\frac{4}{2}$ and $\frac{2}{1}$ mod $7$, the way my notation unfortunately makes us want to.

But no, we can’t make this identification using the action of $\Gamma(7)$. The basic trouble is that no element of $PSL(2, \mathbb{Z})$ can take a pair of numbers without a common factor and send them to a pair with a common factor, because this would violate the condition that the determinant is $1$. (It would take a parallelogram of unit area to one with a larger area).

To see how things go wrong, suppose we had $\left(\array{1+7a&7b\\7c&1+7d}\right)\left(\array{11\\2}\right)=\left(\array{4\\2}\right)$. Then $(1+7a)\cdot 11+7b\cdot2 = 4$, so $11a+2b=-1$, from which we must have $a$ odd, and hence $1+7a$ is even.

Also, $7c\cdot 11+(1+7d)\cdot 2=2$, so $11c+2d=0$, from which we must have $c$ even, and hence also $7c$ even.

So the alleged element of $\Gamma(7)$ will look like $\left(\array{even&?\\even&?}\right)$, which obviously can’t have determinant $1$.

On the other hand, we can’t send $\frac{11}{2}$ to $\frac{2}{1}$ directly, because elements of $\Gamma(7)$ preserve the mod-$7$ values of numerator and denominator separately.

I’m not sure if this makes up for the confusion caused by my notation, but I hope it makes it clearer why things have to be the way I’ve shown them.

Hmm…
As often, although I get really impatient when gluing, I did it anyway. An easier way to do the torus is to make a “double cover”. The 180 degree twist becomes 360 degree, much easier to glue. My new laptop has a webcam, which makes pictures easier to integrate, though perhaps not as clear. So here is the “Farey mod 6 torus double cover”:

It is a “C(0) isometric embedding”, ie it is isometrically embedded in 3D, but along 2 lines the extrinsic curvature is infinite (creased).
I did the 2 copies of the fundamental domain in red and blue respectively.

Gerard