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November 22, 2010

Pictures of Modular Curves (IV)

Posted by Guest

guest post by Tim Silverman

Well, here we are again, trying to understand modular curves in the simplest possible way. The last couple of times, I displayed lots of pictures, and talked about tilings. This time, I’m going to display better pictures—with colour!—and I’m going to talk about denominators. Denominators? Yes, denominators: those things on the bottom of fractions.

Why? Well, we want to understand a little bit more about how the NN-gons are patched together—not so much topologically as from as a number-theoretical point of view (though that’s a rather grandiose phrase for some mostly simple arithmetic). I also want to look at some larger groups of transformations than just Γ(N)\Gamma(N). And I’m going to explore this by considering particularly the denominators of the mod-NN-reduced fractions.

Striping by Denominator

The crude reason that this seems as though it might be worth considering runs roughly like this: we can layer the fractions in Farey sequences into rows by denominator, like this:

... 10 01 11 21 31 12 32 52 13 23 43 53 73 83 14 34 54 74 94 114 15 25 35 45 65 75 85 95 115 125 135 145\array{ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&...&\frac{1}{0}\\ \frac{0}{1}&&&&&&&&&&\frac{1}{1}&&&&&&&&&&\frac{2}{1}&&&&&&&&&&\frac{3}{1}\\ &&&&&\frac{1}{2}&&&&&&&&&&\frac{3}{2}&&&&&&&&&&\frac{5}{2}\\ &&&\frac{1}{3}&&&&\frac{2}{3}&&&&&&\frac{4}{3}&&&&\frac{5}{3}&&&&&&\frac{7}{3}&&&&\frac{8}{3}\\ &&\frac{1}{4}&&&&&&\frac{3}{4}&&&&\frac{5}{4}&&&&&&\frac{7}{4}&&&&\frac{9}{4}&&&&&&\frac{11}{4}\\ &\frac{1}{5}&&&\frac{2}{5}&&\frac{3}{5}&&&\frac{4}{5}&&\frac{6}{5}&&&\frac{7}{5} &&\frac{8}{5}&&&\frac{9}{5}&&\frac{11}{5}&&&\frac{12}{5}&&\frac{13}{5}&&&\frac{14}{5}}

We might very crudely represent this as a series of stripes labelled by denominator.

0 1 2 3 4 5 ... 10 01 11 21 31 12 32 52 13 23 43 53 73 83 14 34 54 74 94 114 15 25 35 45 65 75 85 95 115 125 135 145\array{ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&...&\frac{1}{0}\\ \frac{0}{1}&&&&&&&&&&\frac{1}{1}&&&&&&&&&&\frac{2}{1}&&&&&&&&&&\frac{3}{1}\\ &&&&&\frac{1}{2}&&&&&&&&&&\frac{3}{2}&&&&&&&&&&\frac{5}{2}\\ &&&\frac{1}{3}&&&&\frac{2}{3}&&&&&&\frac{4}{3}&&&&\frac{5}{3}&&&&&&\frac{7}{3}&&&&\frac{8}{3}\\ &&\frac{1}{4}&&&&&&\frac{3}{4}&&&&\frac{5}{4}&&&&&&\frac{7}{4}&&&&\frac{9}{4}&&&&&&\frac{11}{4}\\ &\frac{1}{5}&&&\frac{2}{5}&&\frac{3}{5}&&&\frac{4}{5}&&\frac{6}{5}&&&\frac{7}{5} &&\frac{8}{5}&&&\frac{9}{5}&&\frac{11}{5}&&&\frac{12}{5}&&\frac{13}{5}&&&\frac{14}{5}}
Denominator bands

Adding 11 to all fractions is a symmetry of the triangulation which shifts everything to the right within its own band (so the bands are preserved by a non-trivial subgroup of the symmetry group, helping us partially break down the full complexity).

Moving over to the situation mod NN, the numbers wrap around, so the bands become circles:

0 1 2 3 4 5
Denominator bands mod N

Now obviously this picture is an oversimplification—the denominators will repeat mod NN as you move out, the bands are sometimes broken up and partly interleaved, etc—but some pictures will show how it works in practice (SVG makes overlapping translucent colours look a bit strange to my eye, but that’s what’s happening in the pictures below, if you were wondering. The matches of colours to denominators are the same as in the simplified bands above.)

2|1,0|0,0|0|false|false|false|15.000000|10.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000|false|false|1.000000|false|1|false|1|1,0| e | h | 0:red,, 1:blue,, N = 2: coloured by denominator|||| 10 11 01
N = 2: coloured by denominator
3|1,0|0,0|0|false|false|false|5.000000|20.000000|5.000000|1.000000|false|1.200000|1.000000|false|false|false|true|1.100000|false|false|1.000000|false|1|false|1|1,0| e | h | 0:red,,, 1:blue,,, N = 3: coloured by denominator|||| 11 21 10 01
N = 3: coloured by denominator
4|1,0|0,0|1|false|false|false|35.000000|5.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000|false|false|1.000000|false|1|false|1|1,0| e | h | 0:red,,,, 1:blue,,,, 2:green,,,, N = 4: coloured by denominator|||| 10 21 31 12 11 01
N = 4: coloured by denominator
5|1,0|0,0|2|false|false|false|35.000000|5.000000|5.000000|1.000000|false|1.200000|1.000000|false|false|false|true|1.100000|false|false|1.000000|false|1|false|1|1,0| e | h | 0:red,,,,, 1:blue,,,,, 2:green,,,,, N = 5: coloured by denominator|||| 10 21 31 41 32 02 22 20 11 01 12 42
N = 5: coloured by denominator

Well, perhaps that’s not perfectly clear, because of the way SVG handles translucent colours—here’s a version of the last one with a front face removed to show that faces with denominator 22 are really all green.

5|1,0|0,0|2|false|false|false|32.000000|15.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000|false| e h 5-5; 0:red,,,,, 1:blue,,,,, 2:green,,,,, 10 21 31 41 32 02 22 20 11 12 42
N = 5: coloured, 0/1 removed

A couple more.

N = 6: coloured by denominator
N = 6: coloured by denominator
N = 7: coloured by denominator
N = 7: coloured by denominator

Supposing this is not completely unhelpful, I want to talk in more detail about organising the tiling by denominator, this time looking at arithmetic aspects a little more carefully.


First of all, what symmetries of the tiling will preserve denominators? A matrix (a b c d)\left(\array{a&b\\c&d}\right) in PSL(2, N)PSL(2,\mathbb{Z}_N) will act on a fraction pq\frac{p}{q} to give ap+bqcp+dq\frac{a p+b q}{c p+d q}. Hence if we want the denominator to still be qq after the transformation (regardless of what pp and qq are), we require c=0c=0 and d=1d=1. The requirement that the determinant must also be 11 then forces a=1a=1 also. So we are dealing with matrices of the form (1 * 0 1)\left(\array{1&*\\0&1}\right), where ** can be anything. Since the product (1 b 1 0 1)(1 b 2 0 1)\left(\array{1&b_1\\0&1}\right)\left(\array{1&b_2\\0&1}\right) is (1 b 1+b 2 0 1)\left(\array{1&b_1+b_2\\0&1}\right), this is a subgroup isomorphic to the additive group in N\mathbb{Z}_N. We can think of these as the translations (mod NN) of the projective line over N\mathbb{Z}_N—the matrix (1 b 0 1)\left(\array{1&b\\0&1}\right) adds bb to everything (mod NN).

Of course, translations over N\mathbb{Z}_N are cyclic— a kind of rotation, about the centre at 10\frac{1}{0}. But notice that, setting q=0q=0, we get p0p0\frac{p}{0}\rightarrow\frac{p}{0}, whatever pp is. Not only do these transformations act as rotations around 10\frac{1}{0}, they also act as rotations around p0\frac{p}{0} for any other unit pp!

This fact makes it look as though the different fractions with 00 denominator might be equivalent to each other under some interesting group of symmetries. And this is so, although they are not quite as equivalent as we might think at first, since although rotations about one of them are the same as rotations about the others, the rotations about different centres generally occur at different speeds!

Affine Transformations

So, in pursuit of this idea of going beyond organising the surface into concentric rings by denominator, let us investigate the relationship between different centres of rotation.

We can start by considering a more general group of transformations that respect denominators, namely those that, though they may send a fraction with one denominator to a fraction with a different denominator, nevertheless ensure that two fractions which share the same denominator get sent to fractions which still share a denominator. For this, we only require c=0c=0 (so a denominator qq gets consistently sent to dqd q), so this subgroup consists of the matrices of the form (* * 0 *)\left(\array{*&*\\0&*}\right), or (to give a more specific notation) (a b 0 d)\left(\array{a&b\\0&d}\right) for general aa, bb and dd. However, the requirement of determinant 11 forces ad=1a d=1, so these latter elements must actually be units of N\mathbb{Z}_N, and reciprocal to each other. In particular, we have a smaller subgroup within this one, consisting of matrices of the form (a 0 0 d)\left(\array{a&0\\0&d}\right). Since we have (a 1 0 0 d 1)(a 2 0 0 d 2)=(a 1a 2 0 0 d 1d 2)\left(\array{a_1&0\\0&d_1}\right)\left(\array{a_2&0\\0&d_2}\right)=\left(\array{a_1 a_2&0\\0&d_1 d_2}\right), this subgroup is isomorphic to the multiplicative group (of units) in N\mathbb{Z}_N, mod {1,1}\{-1,1\}, and we might consider it the “rescalings” mod NN. The full subgroup (* * 0 *)\left(\array{*&*\\0&*}\right) might then be considered the “affine transformations” mod NN. Note also the following: if the denominator q=0q=0, then we have (a b 0 d)(p 0)=(ap 0)\left(\array{a&b\\0&d}\right)\left(\array{p\\0}\right)=\left(\array{a p\\0}\right). This is not only the condition we originally wanted (these points are the “centres of rotation”) but also agrees with our description of these transformations as “affine”, since fractions with denominator 00 might be considered to be points “at infinity”, and these transformations keep these points at infinity, as we expect from affine transformations.

So to summarise: the translations, (1 b 0 1)\left(\array{1&b\\0&1}\right), which preserve denominators, are a subgroup of the group of affine transformations, (a b 0 d)\left(\array{a&b\\0&d}\right), which preserve the relation of having the same denominator. The latter is generated by the translations (1 b 0 1)\left(\array{1&b\\0&1}\right) together with the scalings, (a 0 0 d)\left(\array{a&0\\0&d}\right).

Extending Γ(N)\Gamma(N)

Now the usual procedure when looking at these groups is to start with Γ(N)\Gamma(N) (i.e. matrices of the form (1 0 0 1)\left(\array{1&0\\0&1}\right) mod NN), and then first to extend Γ(N)\Gamma(N) by translations, giving Γ 1(N)\Gamma_1(N) (i.e. matrices of the form (1 b 0 1)\left(\array{1&b\\0&1}\right) mod NN), and then to extend to the full group Γ 0(N)\Gamma_0(N) (i.e. matrices of the form (a b 0 d)\left(\array{a&b\\0&d}\right) mod NN). And indeed, in due course, I will be following this path. But before I start extending Γ(N)\Gamma(N) by translations, I want to have a preliminary discussion of its extension by rescalings (i.e. matrices of the form (a 0 0 d)\left(\array{a&0\\0&d}\right) mod NN), and talk about the action of rescalings on X(N)X(N); the reason is that that I think the action is easier to understand if we haven’t first quotiented out by translations.

So at this point I want to look into the action of the group of rescalings.

However, there is a limit to how much text the blogging software (not to mention my readers) can swallow at one go, so I think I will leave that for next time.

Posted at November 22, 2010 6:30 PM UTC

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Re: Pictures of Modular Curves (IV)

Man, I never thought I’d be the only person posting to the nCafé for five days at a stretch. Is the entire rest of the world at a conference in a remote underground cavern with no phones, or something? Should I be stocking up on canned food?

Posted by: Tim Silverman on November 23, 2010 9:49 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

I can’t speak for anyone else, but I’m wiped out with various teaching tasks at the moment: I should be back with it in a couple of weeks. Keep minding the shop Tim! You’re doing a grand job.

Posted by: Simon Willerton on November 23, 2010 11:05 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Thanks for the encouragement, Simon! And good luck with the teaching load …

Posted by: Tim Silverman on November 24, 2010 12:39 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Yes, it has been awfully quiet. Often it happens that when one puts up a long post, people take a couple of days to absorb it and start commenting. But it hasn’t only been this post that people have been quiet about.

Anyway, let’s not turn this into a thread about the Café. Let me ask a slightly contentful question instead. Are the pictures you’ve drawn standard? I mean, are they standard ways of portraying things, or are you being innovative?

I’m not particularly interested in the technology of it — though of course it does sometimes happen that technology enables the drawing of diagrams that couldn’t have been drawn before (e.g. in dynamical systems). I really want to know how originally you’re thinking.

Posted by: Tom Leinster on November 23, 2010 11:27 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Tom asked:

Are the pictures you’ve drawn standard?

Ah, that is indeed an interesting question.

At one level, the most naive level, I think the answer is basically, “No. At least, I don’t think so.” That is, I’m not copying these pictures out of textbooks or papers or off people’s websites—well, obviously, I’d attribute them properly if I did that, but I mean I’m not implementing prettier or more colourful versions of things that I’ve seen depicted or even described. In fact, the reason I did this series is that I was trying to understand this stuff, and thought it ought to be possible to visualise it very prettily, so I looked around desperately for actual illustrations, couldn’t find them, started drawing them myself, and then though I should do them properly and systematically for the world to see. Of course, it’s perfectly possible that I’ve missed something out there—I have very limited access to information.

But at another level, this stuff is lurking right below the surface of what people do show, and people do know about it.

For instance, one of the strange paradoxes of conventions in mathematical illustrations is that, on the one hand, people who like making pictures of tesselations in the hyperbolic plane usually like to use the Poincaré Disk, which shows the whole plane in a compact space, and shows off the symmetry nicely; but when people draw of tilings of modular curves, they tend to show two or three tiles on the complex upper half-plane (or equivalently, the “Poincaré Half-plane model” of the hyperbolic plane), which seems (to me at least) much less enlightening. So in a sense, all I’m doing is bringing these two drawing conventions together.

Likewise, people have known for a very long time that Platonic solids were involved, as well as the Klein Quartic with its many amazing properties (which I’ve ignored), etc, etc. It’s curiously hard to find overt statements to that effect but they are out there lurking in odd places. I think McKean and Moll talk about this in their book Elliptic Curves. In fact, they must do, because they have a whole chapter named after Klein’s Ikosaeder! Given that, PSL(2,5)PSL(2, 5) is isomorphic to the symmetry group of the dodecahedron, etc, this would be quite difficult to miss, and must have been known to Klein (maybe not in exactly the modern form).

On the Farey sequences: there is lots and lots of stuff out there about Farey sequences including many excellent pictures, some of them resembling my own. The reduction of Farey sequences mod NN is something I haven’t come across in that overt form, and I sort of stumbled across it by accident, but it’s implicit at a quite shallow level in various places (for instance in Diamond and Shurman’s book A First Course in Modular Forms).

I did a bunch of calculations with matrices, fractions and so on, but they are not hard, just a bit dull, and I am sure they have been done many times before.

So it’s like that thing John sometimes complains of—there are things that are known to the experts but it’s devilishly difficult to pull together a coherent story out of the scattered hints spread across dozens of places, and it’s sometimes easier to sit down and work it out for yourself from scratch. And you find yourself exclaiming, “You mean it’s just <whatever>? Why didn’t they just &@!*ing say so!”

Posted by: Tim Silverman on November 24, 2010 12:36 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Thanks, Tim. That’s a very interesting answer.

I think this kind of work (if you think of it as work) is often very underrated.

I’m kind of curious about your background. Of course, don’t say anything if you don’t want to—maybe you want to keep your mystique—but… from various things you’ve said over the years (e.g. just now, about having limited access to information) I take it you’re not in an academic job. On the other hand, you obviously haven’t got this from reading the kind of books you find in WHSmith. So…?

Posted by: Tom Leinster on November 24, 2010 3:03 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

I’m somewhat entertained by the idea that I might have a mystique, and having discovered it, I’m almost tempted to hang on to it. But I’m not really that mysterious.

As you say, I don’t have an academic job (and don’t want one—that would turn a hobby into a chore). However, WHSmith is not the only bookshop. And I’m insatiably curious, and learning and studying is one of the main ways I like to relax and enjoy myself. And so, with quite a moderate allocation of spare time and book budget, I’ve built up an acquaintance with a fair number of subjects over several decades, simply by dabbling in whatever interested me at the time.

So I guess my background is … whatever took my fancy. I haven’t taken a formal maths course for a very long time.

Posted by: Tim Silverman on November 24, 2010 10:48 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

In my case it is too: “my background is … whatever took my fancy”. For me too, modular forms etc. has been a “strange attractor” into looking at fascinating mathematics. In the moment I am wondering if the connections between noncommutative geometry and modular forms belongs to my fancy.

Posted by: Thomas on November 25, 2010 10:30 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Tim wrote:

So I guess my background is … whatever took my fancy.

Okay, I think you’re hanging on to your mystique just fine, while modestly denying you have one.

It’s actually very important to have a bit of mystique when you’re blogging: it makes people want to keep on reading. You could fill your blog posts with photos of your apartment, and your pet hamster, and yourself… but don’t. It’s much better to remain a shadowy figure.

I really like the pictures this time! Up to now I think this was stuff I’d already seen in my head. But I never visualized the effect of Γ 1(N)\Gamma_1(N). And certainly not in color!

Number theorists everywhere must be cursing you: you are making this subject easier to understand, which makes them seem less like geniuses.

Posted by: John Baez on November 25, 2010 2:10 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Understanding the substructures in the modular group could be fascinating. All finite simple groups (including even the Monster!) are lurking in the modular group as some subquotient. I asked a queston abbout this in sci.math.research here.

As you can read there, I was already mixing up subgroups, quotients, and group homomorphisms then. Here too, I havn’t yet straightened out what the coloring means in terms of subgroups/quotients.

I know that Gamma(N) is a normal subgroup of Gamma, so the quotient is a group. (Leading to the Platonics, as you showed). But I am not sure about Gamma_1(N).


Posted by: Gerard Westendorp on November 24, 2010 8:59 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

Understanding the structure of PSL(2,)PSL(2, \mathbb{Z}) is indeed fascinating. It’s a vast and remarkable terrain. It’s great that you’re thinking about this stuff!

To answer a specific question: Γ 1(N)\Gamma_1(N) is not a normal subgroup of Γ\Gamma. It fixes a point (\infty) and is conjugate to other subgroups that fix a point.

Posted by: Tim Silverman on November 24, 2010 12:32 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IV)

A lot of Tim’s svg pictures no longer work in these modular curve articles. Let’s see about this one:
0 1 2 3 4 5 ... 10 01 11 21 31 12 32 52 13 23 43 53 73 83 14 34 54 74 94 114 15 25 35 45 65 75 85 95 115 125 135 145\array{ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&...&\frac{1}{0}\\ \frac{0}{1}&&&&&&&&&&\frac{1}{1}&&&&&&&&&&\frac{2}{1}&&&&&&&&&&\frac{3}{1}\\ &&&&&\frac{1}{2}&&&&&&&&&&\frac{3}{2}&&&&&&&&&&\frac{5}{2}\\ &&&\frac{1}{3}&&&&\frac{2}{3}&&&&&&\frac{4}{3}&&&&\frac{5}{3}&&&&&&\frac{7}{3}&&&&\frac{8}{3}\\ &&\frac{1}{4}&&&&&&\frac{3}{4}&&&&\frac{5}{4}&&&&&&\frac{7}{4}&&&&\frac{9}{4}&&&&&&\frac{11}{4}\\ &\frac{1}{5}&&&\frac{2}{5}&&\frac{3}{5}&&&\frac{4}{5}&&\frac{6}{5}&&&\frac{7}{5} &&\frac{8}{5}&&&\frac{9}{5}&&\frac{11}{5}&&&\frac{12}{5}&&\frac{13}{5}&&&\frac{14}{5}}
Denominator bands

Hmm, it sort of works as a comment, though it seems a bit wonky.

Posted by: John Baez on March 4, 2024 10:48 PM | Permalink | Reply to this

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