## November 14, 2010

### Pictures of Modular Curves (III)

#### Posted by Guest

guest post by Tim Silverman

Welcome back to this series of posts: Pretty Pictures of Modular Curves

The Story So Far

The last two times, we started looking at the curves $X(N)$, that is, the quotients of the upper half of the complex plane by the groups $\Gamma(N)$, the latter being defined as those subgroups of $PSL(2, \mathbb{Z})$ which consist of the matrices congruent to $\left(\array{1&0\\0&1}\right)$ mod $N$. We discovered that for $N=3$, $N=4$ and $N=5$, the resulting quotients (to be strictly accurate: once we have compactified by cusps) can basically be thought of as (the surfaces of) Platonic solids, respectively the tetrahedron, cube and dodecahedron, perhaps best thought of as the spherical versions of those solids. These are curved surfaces tiled regularly with a finite number of regular $N$-gons, with $3$ of the tiles meeting at each vertex. This view in terms of tiling also works for the $N=2$ case, which involves a tiling by $3$ bigons. The residual action of $PSL(2, \mathbb{Z})$ on these quotient surfaces is then precisely the symmetry of the tiling.

Non-Positive Curvature

Having done the cases $N=2, 3, 4$ and $5$, it’s time to move to $N=6$. Of course, we can’t fit together regular hexagons on a sphere; we need a (globally) flat surface. The $N=6$ case is a tiling of a torus by $12$ hexagons (or, dually, by $24$ triangles). However, rather than suffering the pain and confusion of trying to embed a tiled torus in $3$-space and then project it onto the plane for your viewing pleasure, I shall instead tile the whole plane with hexagons, and let the labelling by reduced fractions show how to identify different hexagons in order to take the quotient. I think, in any case, that this is more perspicuous.

Hexagons: N = 6

Actually, letting the numbers do all the work is a bit cruel, so to help make the quotient clearer, I’ll pick out a particular “unit cell” of $12$ hexagons so the repeat pattern is obvious.

12 distinct hexagons: N = 6

As you can see, by glancing at the picture, or by looking more carefully at where repeats occur, the repetition actually gives the $A_2$ lattice.

Even with these guidelines, it’s a lot harder to see what’s going on here than it is in the case of the spherical Platonic solids. You might enjoy trying to see the patterns in the fractions here. In any case, I’ll be discussing how they are organised in a lot more depth next time.

Now we can move on to the case $N=7$. By now, we need the curvature to be negative. As with the flat $N=6$ case, I’ll handle this by tiling the entire hyperbolic plane, and use the repetition of fractions to the indicate the quotient. I’m far too lazy to try to embed the multi-holed compact quotient surface in $3$-space (with its tiling!)—and besides, this has already been done, far better than I could manage, by Greg Egan, Mike Stay and Gerard Westendorp. (In fact, Gerard Westendorp’s page is full of beautiful and cool things relevant to this post, including many superb pictures, so do check it out!)

(I have to admit I’m slightly trepidatious about making this choice, because this surface is the famous Klein Quartic, and I have the impression that people have been expecting me to expound its wonderful properties. Well, maybe some time, but not here, because in this place I’m just slipping it into its place the parade of modular curves, and ignoring things that make it stand out. Sorry about that.)

I’ll again pick a “unit cell”, bordered by thickened edges, and unfortunately annoyingly large.

Heptagons: N = 7

Or this might be slightly clearer, putting $\frac{1}{1}$ in the centre instead of $\frac{1}{0}$.

Heptagons around 1/1: N = 7

Next time, we’ll be looking at more ways to make it clearer what’s going on in here.

Here’s $N=8$

Octagons: N = 8

$N=9$

Enniagons: N = 9

$N=15$

Another view of $N=15$

Pentadecagons, vertex centred: N = 15

And finally up to $N=\infty$, although I guess this could equally well be called $N=0$, since we are quotienting by the zero ideal in $\mathbb{Z}$.

Horocycles: N = infinity

We’ve now gone from regular $N$-gons to what we might call regular $\infty$-gons or regular $\mathbb{Z}$-gons. Their incircles are what are sometimes called horocycles.

And for good measure, I’ll finish up where I started in the last post, with the dual of that last picture:

Triangles: N = infinity

I’ll add a couple of notes about this picture and the previous one while I’m about it. Although we’ve been talking largely about the rational numbers and their reductions mod $N$, the interior of the Poincaré disk here is secretly intended to represent not just any old random hyperbolic plane but the upper half of the complex plane, with the circumference representing the real line. Having the metric is undoubtedly nice for drawing and tesselating, but ultimately we’re more concerned with the complex structure. Anyway, if we imagine the Poincaré disks in the last couple of pictures as sitting over the unit disk in the complex plane, then we can see them as images of the complex upper half-plane by taking the latter and applying the transformation $z\rightarrow\frac{z-\omega}{-\omega z+1}$, where $\omega$ is the complex cube root of unity which lies in the upper half plane. Then the rationals end up at the points indicated around the circumference. The incircles of the $\mathbb{Z}$-gons in the above picture are, I believe, the images of the famous Ford circles in the upper half of the complex plane.

Well, having shown you the full range of pictures, I suppose this would be as good a place as any to stop for a break. As usual, I’ve taken far more space than I expected to, and said much less than I intended to. I think I’m about half way through the article as originally planned. We haven’t even reached what I considered the most exciting part of it, where we get to take out our felt-tips and colour things in!—and where we attempt to herd the milling multitude of mod-$N$-reduced fractions into some sort of orderly system by staring very hard at their denominators.

Ah well. That’ll be something to talk about next time.

Posted at November 14, 2010 10:47 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2302

### Re: Pictures of Modular Curves (III)

Bravo!

I’m a bit confused about a phenomenon that seems to first show up (at least in this post) in the case N = 7:

How come you’re counting 0/2 and 0/3 as different from 0/1? Aren’t they all just zero, even if we work mod 7?

(With some work I could probably figure out what’s going on, but I figure some other people will have the same question but be too shy to ask it — so in my role as ‘class clown’, I’ll just blurt it out.)

Posted by: John Baez on November 15, 2010 1:39 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

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’…

Posted by: David Roberts on November 15, 2010 4:06 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

Now that Tim has explained why $0/1 \ne 0/2$, I’ll happily concede that $1/0 \ne 2/0$.

Posted by: John Baez on November 17, 2010 7:14 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

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.

Posted by: Tim Silverman on November 16, 2010 1:21 PM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

The fraction notation is great; I just need to think about what it means. And if your post didn’t confuse anyone, maybe nobody would post any comments, and you’d think that nobody had read it, or that everybody thought it was all obvious. So it’s good to be slightly confusing. And let’s face it: if math were never confusing at all, we wouldn’t like it.

Posted by: John Baez on November 17, 2010 5:40 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

… it’s good to be slightly confusing …

That is indeed true; but I’ve had to explain this particular bit three times now, so I rather suspect I didn’t do it well enough the first couple of times.

I’m glad you like the fraction notation, though, because I’m not going to redo all the pictures!

Posted by: Tim Silverman on November 18, 2010 11:37 PM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

One useful thing that the labeling by fractions does is tell you how to glue together the Riemann surface: You glue identical fractions on top of each other.

For example, you can print out the picture of the N=6 case, and try gluing it into a torus. The first part is easy, you can quickly see how to glue a cylinder. But then it is more tricky to attach the opposite ends of the cylinder. You need to introduce a 180 degree twist.

It can probably be done, if you are skilled in origami.

Likewise, it is not so hard to glue together Klein’s quartic into a kind of octagon. But then to glue the open ends together is much more difficult. (If you want to do it, better use identically sized heptagons rather than the Poincare disk ones)

Gerard

Posted by: Gerard Westendorp on November 24, 2010 8:46 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

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

Posted by: Gerard Westendorp on November 24, 2010 10:09 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

One more interesting observation:
The double cover also decomposes nicely into the inner ring and the outer ring. If you look on the outer side of the torus, you find each hexagon exactly once, although some are cut off and reappear at a place that is the iversion through the central point of the torus.

Gerard

Posted by: Gerard Westendorp on November 24, 2010 10:18 AM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

This is very cool!

I’m slightly embarrassed to admit that I was secretly hoping you’d see this and be inspired to make some models. I’m rather pleased to see that my wish has been granted.

Posted by: Tim Silverman on November 24, 2010 12:37 PM | Permalink | Reply to this

### Re: Pictures of Modular Curves (III)

Thanks.
I hope I get time to do some more stuff.
My compliments on your series `Pictures on Modular Curves´. I wish there was more of this on the web.

You could try painting the Klein quartic with fractions mod 7. A lot of work unfortunately.
One thing I want to do is make animations of ´warping´ tilings. I did one for Klein Quartic, but the 3D Platonic solids would be fun too.
Also I want to paint these figures with color coded plots of complex functions, like you can paint the J-invariant on the modular tiling. But I havn´t figured out how to do it yet.

Gerard

Posted by: Gerard Westendorp on November 29, 2010 10:25 PM | Permalink | Reply to this

Post a New Comment