## October 6, 2010

### Pictures of Modular Curves (I)

#### Posted by John Baez

guest post by Tim Silverman

Introduction—and a Kind of Apology

It is with considerable relief that I finally find myself in a position to make a (hopefully) more substantive contribution here: namely, the second in the series I started here, an alarmingly long time ago. I took the long way around to generate the pretty pictures I wanted, and did a lot of reading and calculation—mostly for articles that I intend to appear after this one! Somehow, it all took a lot longer than I expected—even after taking into account that, as I know all too well, these things tend to take a lot longer than expected.

But we’re here at last! My aim in this series is to give the most elementary discussion of modular curves and modular forms that I can manage at each stage, partly for my own benefit, and partly for the benefit of, well, people like me, I guess. One of the strange (but nice) things about the theory of modular curves and modular forms is that, although it lies at the confluence of many areas of mathematics, and leads directly on to some subjects that are very hard indeed, nevertheless, there is a surprising amount that one can do in it using only quite basic geometry and arithmetic. This aspect of the subject is perhaps sometimes somewhat obscured by all the other aspects, but one of my aims here is to bring it out more clearly.

I’m a little embarrassed that this article, and the next few ones at least, are not really categorical, let alone n-categorical, but if things go according to plan, I will at least end up talking a lot about the sorts of things that people often talk about here, albeit at a more elementary level than usual. In any case, I hope that it will provide some interest and amusement to the n-Café patrons, and ideally also to the crowds of urchins in the street outside with their noses pressed against the café window, not to mention the more elegant ladies and gentlemen who pass by and look in but do not enter.

With that warning out of the way, let me start by summarising where we got to in the first article.

The Story so Far

A Farey sequence consists of those fractions, reduced to their lowest terms, which lie between $0$ and $1$ and have denominators less than some given natural number, the fractions being placed in order of increasing value (as rational numbers).

Here are the first few Farey sequences:

$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{5\phantom{\rule{thickmathspace}{0ex}}}\frac{2}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{4}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{6}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{4}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{5}{6}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$
$\frac{0}{1}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{6}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\phantom{\rule{thickmathspace}{0ex}}\frac{4}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{2}{3}\phantom{\rule{thickmathspace}{0ex}}\frac{5}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{3}{4}\phantom{\rule{thickmathspace}{0ex}}\frac{4}{5}\phantom{\rule{thickmathspace}{0ex}}\frac{5}{6}\phantom{\rule{thickmathspace}{0ex}}\frac{6}{7}\phantom{\rule{thickmathspace}{0ex}}\frac{1}{1}$

Given any two adjacent fractions $\frac{a}{b}$ and $\frac{c}{d}$ in any Farey sequence, we have $ad-bc=-1$. Given any three adjacent fractions $\frac{a}{b}$, $\frac{a\prime }{b\prime }$ and $\frac{a″}{b″}$ in any Farey sequence, where $\frac{a\prime }{b\prime }$ is in the middle, we have $\frac{a\prime }{b\prime }=\frac{a+a″}{b+b″}$. That is, the middle fraction is the mediant of the fractions on either side. We get later Farey sequences (with larger maximum denominator) by inserting mediants between pairs of adjacent fractions in earlier sequences.

Now lets make this more geometrical. Suppose, wherever two fractions are adjacent in some Farey sequence, we join them by an edge. From what we said above, it follows that when we have three such fractions of which one is the mediant of the other two, we find the three fractions give the vertices of a triangle. By taking the limit of increasing denominators, so that we eventually include all Farey sequences, we get to include all rational numbers between $0$ and $1$, and we can also extend the idea (in the obvious way!) to include all rational numbers—or (to express ourselves geometrically) to the entire affine line over $ℚ$. And from there, it’s just one short step to extend to the projective line over $ℚ$: we simply throw in the fraction $\frac{1}{0}$, which is just as capable of forming mediants as any other fraction. Having done this, we find that if we are given any pair of points on $ℚ{P}^{1}$ such that they both lie in some triangle (and hence are connected by an edge), then they also both lie in exactly one other triangle, on the “other side” of the edge, and indeed all the triangles together form a tiling of a (hyperbolic) surface. In the Poincaré Disc view it looks something like the picture below, where the edges appear as segments of circles and the elements of $ℚ{P}^{1}$ are distributed around the circumference of the disc. (Just to be clear: in the hyperbolic plane, all these triangles are (or can be made) congruent).

Triangles: N = infinity

Notice that each rational number is at the vertex of an infinite number of triangles. Note also that $\frac{1}{0}$ is adjacent to, precisely, the integers—and each integer is also adjacent to its predecessor and successor integers.

The Action of the Modular Group

Taking another, quite different, geometrical approach, we may consider points of $ℚ{P}^{1}$ to be lines in ${ℤ}^{2}$.

$\frac{1}{0}$ $\frac{0}{1}$ $\frac{2}{1}$ $\frac{3}{1}$ $\frac{3}{2}$ $\frac{-3}{1}$ $\frac{-3}{2}$ $\frac{-1}{1}$ $\frac{-2}{1}$ $\frac{1}{1}$ $\frac{1}{2}$ $\frac{1}{3}$ $\frac{2}{3}$ $\frac{-1}{2}$ $\frac{-1}{3}$ $\frac{-2}{3}$
Some lines in the ℤ plane

The group $\mathrm{PSL}\left(2,ℤ\right)$ acts on this plane, preserving lines (and preserving enclosed areas, if we consider ${ℤ}^{2}$ as a subset of the Euclidean plane). But more than that, $\mathrm{PSL}\left(2,ℤ\right)$ preserves crucial features of the structure of Farey sequences. It preserves the relationship $\mid ad-bc\mid =1$ between adjacent elements (since this just says that the points $\left(a,b\right)$ and $\left(c,d\right)$, corresponding to reduced fractions $\frac{a}{b}$ and $\frac{c}{d}$, form the vertices of a parallelogram of area $1$; and area is preserved). It also preserves mediancy, since, in the lattice ${ℤ}^{2}$, this just corresponds to the sum of vectors. So elements of $\mathrm{PSL}\left(2,ℤ\right)$ preserve the adjacency and mediancy relations, and hence preserve the triangles—at least combinatorially.

In fact, it’s much better than that. The metric structure on the hyperbolic plane gives it a natural conformal structure—which enables it to be considered as a region of $ℂ{P}^{1}$. Also, of course, $\mathrm{PSL}\left(2,ℤ\right)$ is a subgroup of $\mathrm{PSL}\left(2,ℂ\right)$, and the latter naturally acts on $ℂ{P}^{1}$, preserving its conformal structure. And the conformal structure implies a unique metric structure of constant curvature (up to an overall scale factor). So in this way $\mathrm{PSL}\left(2,ℤ\right)$ acts as a group of conformal transformations, hence as a group of isometries of the hyperbolic plane, and preserves the tiling by triangles. In fact, more even than that, this group is precisely the symmetry group of the tiling.

Reducing mod $N$

Now one thing we can do with $ℤ$ is reduce it mod $N$:

$ℤ\to {ℤ}_{N}$ $a↦a\mathrm{mod}N$

Since a lot of the things we have been talking about so far come from $ℤ$, it might be interesting to see to what extent reduction mod $N$ can be applied to these other things. The rings ${ℤ}_{N}$ provide, in a way, simpler microcosms of $ℤ$ itself, and we might achieve a similar simplification of related objects. So let’s see where we can go with this.

Obviously the surjective morphism of rings $ℤ\to {ℤ}_{N}$ will induce a morphism of the plane ${ℤ}^{2}\to {{ℤ}_{N}}^{2}$. But a line in the former plane is defined algebraically (as all points of the form $\left(cp,cq\right)$ for a certain point $\left(p,q\right)$). So the lines in ${ℤ}^{2}$ get carried down, in a well-defined way, to “lines” in the reduced plane ${{ℤ}_{N}}^{2}$. And so the projective line over $ℚ$—or perhaps it might almost be better here to say the “projective line” over $ℤ$, if that weren’t confusing for other reasons—gets sent to a sort of projective line over ${ℤ}_{N}$. If $N$ is prime, so that ${ℤ}_{N}$ is a field, then this is a completely ordinary projective line over a finite field. In any case, we can try to represent points of this projective line by fractions of some kind, but “reduced mod $N$”.

Here is how we reduce these fractions mod $N$: we identify two fractions $\frac{a}{b}$ and $\frac{a\prime }{b\prime }$ if $a\equiv a\prime \left(\mathrm{mod}N\right)$ and $b\equiv b\prime \left(\mathrm{mod}N\right)$. This will sometimes leave common factors between top and bottom (e.g. $\frac{13}{8}$ reduces to $\frac{3}{3}\mathrm{mod}\phantom{\rule{thickmathspace}{0ex}}5$) but we won’t cancel these factors unless they could also have been canceled back in fractions over $ℤ$—which means we only cancel factors of $-1$. This means that some lines may get associated with several different fractions—I’ll say a little more about this later.

What about Farey sequences? These can be reduced too! The fractions corresponding to points in the projective line over $ℚ$ go to reduced fractions corresponding to points in the reduced “projective line” over ${ℤ}_{N}$. But, in addition, the “adjacency” relation between successive members of a Farey sequence, viz. $\frac{a}{b}$ and $\frac{c}{d}$ are adjacent in some Farey sequence iff $ad-bc=-1$, also has a well-defined reduction mod $N$. So we get a reduced adjacency relation, and therefore a set of “reduced edges” between adjacent “reduced points” (i.e. adjacent reduced fractions). Not only that, but the mediancy relation, $\frac{a\prime }{b\prime }=\frac{a+a″}{b+b″}$ is also defined algebraically, so we can also reduce the triangles mod $N$! In fact, the whole tiled surface can be reduced mod $N$!

We can think of this whole thing alternatively in terms of quotients of the original surface by certain subgroups of $\mathrm{PSL}\left(2,ℤ\right)$.

Of course, $\mathrm{PSL}\left(2,ℤ\right)$ will act on the reduced plane ${{ℤ}_{N}}^{2}$ through its action on ${ℤ}^{2}$, hence on lines in the reduced plane, and so on the ‘projective line’ over ${ℤ}_{N}$. Now, the automorphisms of the latter are what we might call $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$. That is, they are given by the group of $2×2$ matrices whose entries are elements of ${ℤ}_{N}$ and whose determinants are equal to $1$ mod $N$, all modulo scalar multiplication by $\left\{1,-1\right\}$.

On the other hand, by reducing matrix elements mod $N$, we get the obvious surjection $\mathrm{PSL}\left(2,ℤ\right)↦\mathrm{PSL}\left(2,{ℤ}_{N}\right)$. And since the action of these matrices is defined in terms of addition and multiplication, the reduction mod $N$ commutes with the action: we can either act on ${ℤ}^{2}$ with $\mathrm{PSL}\left(2,ℤ\right)$, then reduce ${ℤ}^{2}$ to ${{ℤ}_{N}}^{2}$, or we can reduce $\mathrm{PSL}\left(2,ℤ\right)$ to $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$, and act with the latter on ${{ℤ}_{N}}^{2}$ and get the same result.

Now the kernel of the reduction $\mathrm{PSL}\left(2,ℤ\right)\to \mathrm{PSL}\left(2,{ℤ}_{N}\right)$ is the group of matrices congruent to the identity $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ mod $N$. This well-known and interesting subgroup of $\mathrm{PSL}\left(2,ℤ\right)$ is called $\Gamma \left(N\right)$.

$0\to \Gamma \left(N\right)\to \mathrm{PSL}\left(2,ℤ\right)\to \mathrm{PSL}\left(2,{ℤ}_{N}\right)\to 0$

So $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$ is the “residual” action of $\mathrm{PSL}\left(2,ℤ\right)$ left after we have taken the quotient of the latter by $\Gamma \left(N\right)$. And in the same way, the “projective line” over ${ℤ}_{N}$ is what is left after we have taken the quotient of the “projective line” over $ℤ$ by the action of $\Gamma \left(N\right)$ as matrices. And, finally, the reduced surface tiled with reduced triangles (made up of reduced edges joining reduced points) is the quotient of the full hyperbolic plane (or complex upper half-plane—which is what we are really after) by the action of $\Gamma \left(N\right)$ on the tiling. There is left a remaining group of symmetries of the reduced tiling by the residual (i.e. quotient) group $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$.

The special case of reduction mod $N=0$ obviously gives the identity map $ℤ\to ℤ$, and makes $\Gamma \left(N\right)$ the trivial group, so we get the full tiled hyperbolic plane. At the opposite extreme, reduction mod $N=1$ sends $ℤ$ to the trivial ring, and sets $\Gamma \left(N\right)$ to the entirety of $\mathrm{PSL}\left(ℤ,2\right)$, so reduces the surface down to one covered by a single tile.

Including $\Gamma \left(N\right)$ in larger subgroups enables us to quotient the reduced surfaces even further.

And this is basically what I’m going to be talking about for the next few weeks.

I’ll just say something about how I’m going to write the reduced fractions. I’ll represent the congruence classes of the numerator and denominator in the usual way by numbers lying in $\left[0,N\right)$. The fact that we can cancel a factor of $-1$ enables us to send $\frac{a}{b}$ to $\frac{-a}{-b}$, or in other words to $\frac{N-a}{N-b}$ (mod $N$), so we’ll use this to ensure that all denominators are $\le N/2$ when writing our reduced fractions down; also, by the same mechanism, we can ensure that if the denominator is $0$, or if $N$ is even and the denominator is $N/2$ (hence equal to minus itself), then the numerator is also $\le N/2$. So, for instance, (mod $10$) we’ll be writing $\frac{2}{3}$ rather than $\frac{8}{7}$, and $\frac{3}{5}$ rather than $\frac{7}{5}$.

The Plane over $ℤ/\left(4\right)$

Let’s look at an example of this. Here, in the picture below, are the six lines which lie in the plane over $ℤ/\left(4\right)$. Since $ℤ/\left(4\right)$ is cyclic, the plane over it is a kind of torus, and the lines wrap round! But since I’m forced to draw this on a flat surface, some of the lines end up broken into several pieces in the diagram below. I’ve given each line a different colour to make it clear what belongs with what.

Let’s look at an example of this. Here, in the picture below, are the six lines which lie in the plane over $ℤ/\left(4\right)$. Since $ℤ/\left(4\right)$ is cyclic, the plane over it is a kind of torus, and the lines wrap round! But since I’m forced to draw this on a flat surface, some of the lines end up broken into several pieces in the diagram below. I’ve given each line a different colour to make it clear what belongs with what.

$\frac{1}{0}$ $\frac{0}{1}$ $\frac{1}{1}$ $\frac{2}{1}$ $\frac{2}{1}$ $\frac{3}{1}$ $\frac{3}{1}$$\frac{3}{1}$ $\frac{3}{1}$ $\frac{1}{2}$ $\frac{1}{2}$
Lines in the plane over ℤ/(4)

Let’s be more explicit. Listing the lines by the points they contain, we have

$\frac{1}{0}:\left\{\frac{1}{0},\frac{2}{0},\frac{3}{0},\frac{0}{0}\right\}$
$\frac{0}{1}:\left\{\frac{0}{1},\frac{0}{2},\frac{0}{3},\frac{0}{0}\right\}$
$\frac{1}{1}:\left\{\frac{1}{1},\frac{2}{2},\frac{3}{3},\frac{0}{0}\right\}$
$\frac{2}{1}:\left\{\frac{2}{1},\frac{0}{2},\frac{2}{3},\frac{0}{0}\right\}$
$\frac{3}{1}:\left\{\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{0}{0}\right\}$
$\frac{1}{2}:\left\{\frac{1}{2},\frac{2}{0},\frac{3}{2},\frac{0}{0}\right\}$

Note that, e.g. $\frac{1}{3}=\frac{3}{1}$ but $\frac{1}{2}\ne \frac{2}{1}$.

Some, but not all, of the points “generate” the line they lie on: e.g. if we simultaneously multiply the numerator and denominator of $\frac{1}{3}$ by, successively $1$, $2$, $3$ and $0$, we get all the points on the line $\frac{3}{1}$ (viz. $\frac{1\cdot 1}{1\cdot 3}=\frac{1}{3}$, $\frac{2\cdot 1}{2\cdot 3}=\frac{2}{2}$, $\frac{3\cdot 1}{3\cdot 3}=\frac{3}{1}$, and $\frac{0\cdot 1}{0\cdot 3}=\frac{0}{0}$), but this does not work with $\frac{2}{2}$ (which, indeed, lies on two lines through the origin). We get a fraction which generates a line whenever the numerator and denominator do not have a factor in common which is also a factor of $N$. (Shared factors which are coprime to $N$ are fine.)

Because, in our rules for reducing mod $N$, we are not allowed to cancel common factors other than $-1$, the “fractions reduced mod $N$” correspond not to the lines in ${{ℤ}_{N}}^{2}$ themselves but to all the points which generate the lines—mod cancellation of factors of $-1$ as usual. Mod $4$, there are two generators for each line—but the ability to cancel factors of $-1$ means that they are secretly the same (e.g. $\frac{1}{3}=\frac{-1}{-3}=\frac{3}{1}$); so in the case of mod $4$, there is just one fraction per line. But this is not generally true. For instance, mod $5$ we have that $\frac{1}{1}$ is different from $\frac{2}{2}$, since they correspond to different generating points, even though they generate the same line.

We Want More!

So much for reduced points. But what about reduced edges? And reduced triangles? And the reduced tilings of reduced surfaces!

Since there are only a finite number of fractions mod $N$, we end up with only a finite number of mediant triangles in our tiling, which will tile a compact quotient surface of the hyperbolic plane. My real goal in this post is to show you some pictures of this happening.

Some Arrangements Of Triangles

Let’s start nice and small with $N=3$. Our rule for denominators (that they must be $\le \frac{N}{2}$) ensures that in this case the denominator must be either $0$ or $1$, and, in the case where the denominator is $0$, our rule for numerators ensures that the numerator must be either $0$ or $1$. However, $\frac{0}{0}$ is ruled out. The reason is that this would have to come from a fraction over $ℤ$ whose numerator and denominator were both multiples of $3$—but those original fractions over $ℤ$ were all reduced to their lowest form, and so wouldn’t have had this common factor in there in the first place. So the only fractions we are left with mod $3$ are $\frac{0}{1}$, $\frac{1}{1}$, $\frac{2}{1}$ and $\frac{1}{0}$. Any three of these form a mediant triangle (i.e. one is the mediant of the other two—mod $3$, of course!), so we get the following picture:

3|1,0|0,0|1|false|true|true|0.000000|0.000000|3.000000|1.000000|false|0.800000|1.000000|false|false|false|false|1.400000| e h 0:none,,, 1:none,,, 1/0 1/1 2/1 0/1
Triangles: N = 3

Lets go up to $N=4$.

Here, the allowed fractions are $\frac{1}{0}$, $\frac{0}{1}$, $\frac{1}{1}$, $\frac{2}{1}$, $\frac{3}{1}$ and $\frac{1}{2}$, as we saw above. Note that there is no $\frac{0}{2}$ or $\frac{2}{2}$ because $2$ is a factor of $N$ (viz. $4$) and so the parent fraction in $ℚ$ would have had to have had a common factor of $2$ on top and bottom—which it can’t have done because it was reduced to lowest form. Not all of these triplets form mediant triangles: the arrangement of triangles actually looks like this:

4|1,0|0,0|1|false|true|true|30.000000|10.000000|3.000000|1.000000|false|0.800000|1.000000|false|false|false|false|1.500000| e h 0:none,,,, 1:none,,,, 2:none,,,, 1/0 2/1 3/1 1/2 1/1 0/1
Triangles: N = 4

And now up to $N=5$. Here we get to see two different fractions with denominator $0$ (viz. $\frac{1}{0}$ and $\frac{2}{0}$)—symmetrically placed opposite one another, at top and bottom.

5|1,0|0,0|2|false|true|true|32.000000|15.000000|3.000000|1.000000|false|0.800000|1.000000|false|false|false|false|1.400000| e h 0:none,,,,, 1:none,,,,, 2:none,,,,, 1/0 2/1 3/1 4/1 3/2 0/2 2/2 2/0 1/1 0/1 1/2 4/2
Triangles: N = 5

Yes, if you’re not sure—that is an icosahedron.

Why would the Platonic solids fall out of modular arithmetic like that?

Well, actually it’s not really surprising. Consider the tiling of the hyperbolic plane induced by the full set of Farey sequences over $ℤ$. Consider, in particular, the $\infty$ vertex $\frac{1}{0}$ and its neighbours, which are just the integers. Adding $1$ to all vertex fractions is a symmetry of this tiling, which sends $\frac{1}{0}$ to itself and each integer to its neighbour. (The corresponding element of $\mathrm{PSL}\left(2,ℤ\right)$ is $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$). Obviously there are an infinite number of integers, so we can just keep adding $1$ forever.

But think now of how this reduces mod $N$. Now, $\frac{1}{0}$ reduces to $\frac{1}{0}\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{mod}N\right)$, $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ reduces to $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}\left(\mathrm{mod}N\right)$ and adding $1$ reduces to adding $1$ (mod $N$). And the integers reduce to integers mod $N$. So after adding $1$ $N$ times, we get back to where we started. So there are just $N$ triangles with a vertex at $\frac{1}{0}$, rather than an infinite number as with $ℚ{P}^{1}$. But all vertices are identical, so there are $N$ triangles meeting at every vertex. So we have a tiling of a surface by equilateral triangles, with $N$ of them meeting at each vertex. Which is pretty much all we need to get Platonic solids.

Some Arrangements Of $N$-gons

Actually, I don’t really like working with the triangular tiling, with $N$ triangles meeting at each vertex. I’d rather work with the dual tiling, where three $N$-gons meet at each vertex. As $N$ gets larger, I find the mass of triangles gets difficult to decipher and my eyes start to go funny. Also, my whole discussion obviously relies heavily on the importance of the fractions (reduced mod $N$) which in the triangle tiling get assigned to mere insubstantial vertices, whereas in the dual tiling they get placed into the centres of faces, which somehow I seem to feel is psychologically more appropriate to their important role—as well as making the numbers easier to read when you see them in a diagram. So here are the dual tilings. I’ve taken the liberty of pasting the fractions flat onto the faces instead of sticking them on invisible pins as in the previous pictures.

Here’s $N=3$.

3|1,0|0,0|0|false|true|false|-10.000000|20.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000| e h 0:white,,, 1:white,,, $\frac{1}{1}$ $\frac{2}{1}$ $\frac{1}{0}$ $\frac{0}{1}$
Tetrahedron: N = 3

Next, $N=4$.

4|1,0|0,0|1|false|true|false|30.000000|30.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000| e h 0:white,,,, 1:white,,,, 2:white,,,, $\frac{2}{1}$ $\frac{3}{1}$ $\frac{1}{2}$ $\frac{1}{0}$ $\frac{1}{1}$ $\frac{0}{1}$
Cube: N = 4

Observe how rotating around the top face ($\frac{1}{0}$) adds $1$: adding $1$ to the top face does nothing ($\infty +1=\infty$); adding $1$ to the integers on the side faces sends them to their anticlockwise (as seen from above) neighbours; and $\frac{1}{2}+1=\frac{3}{2}=\frac{-3}{-2}=\frac{1}{2}$, so the bottom face is also correctly sent to itself.

Observe also that rotating ${180}^{\circ }$ around the edge separating $\frac{1}{0}$ from $\frac{0}{1}$ sends each fraction $q\to \frac{-1}{q}$. E.g. $\frac{1}{1}\to \frac{-1}{1}=\frac{3}{1}$ and $\frac{1}{2}\to \frac{-2}{1}=\frac{2}{1}$.

Both of these facts relating arithmetic and geometry are generally true, not just for $N=4$. These two operations correspond to the matrices $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, which between them generate $\mathrm{PSL}\left(2,ℤ\right)$ (or $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$ as appropriate).

And here’s $N=5$.

5|1,0|0,0|2|false|true|false|32.000000|15.000000|3.000000|1.000000|false|1.400000|1.000000|false|false|false|true|1.100000| e h 0:white,,,,, 1:white,,,,, 2:white,,,,, $\frac{1}{0}$ $\frac{2}{1}$ $\frac{3}{1}$ $\frac{4}{1}$ $\frac{3}{2}$ $\frac{0}{2}$ $\frac{2}{2}$ $\frac{2}{0}$ $\frac{1}{1}$ $\frac{0}{1}$ $\frac{1}{2}$ $\frac{4}{2}$
Dodecahedron: N = 5
Posted at October 6, 2010 1:49 AM UTC

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

Great post, Tim! That’s gorgeous stuff.

Posted by: Tom Leinster on October 6, 2010 1:08 PM | Permalink | Reply to this

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

Tim’s gorgeous SVG graphics strained the capacities of this blog in some mysterious manner which required me to fiddle around a bit before they showed up — all except for one picture, a rotated view of the last one you see here! Presumably this shows the tantalizingly invisible top face of the dodecahedron.

Quiz question for people wanting to make sure they understand this stuff: how should the dodecahedron’s top face be labelled?

I’m sorry it took me so long to post this entry properly — there were some mistakes that I finally fixed just now, and there may be even more.

Posted by: John Baez on October 7, 2010 9:07 AM | Permalink | Reply to this

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

I wrote:

That’s gorgeous stuff.

John wrote:

Tim’s gorgeous SVG graphics

I don’t dispute the gorgeousness of the graphics, but actually my gorgeous referred to the mathematics. I’ve only skimmed it so far, but it seems an extraordinary story.

Posted by: Tom Leinster on October 7, 2010 12:33 PM | Permalink | Reply to this

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

Well, my fond hope is that the gorgeousness of the pictures (and thank you for the compliment) will reflect and illuminate the maths, and not be merely decorative! Not every article will have the same balance of ideas and eye-candy, but they should all have some of both, hopefully working together.

Posted by: Tim Silverman on October 8, 2010 5:30 PM | Permalink | Reply to this

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

It’s gorgeous in every way.

As Tim points out, the math here is pretty famous — and even its gorgeousness is famous! Modular curves and modular forms are one of those places where number theory and geometry and algebra and everything else meet in a kind of spire of perfection, leading all the way up to Fermat’s Last Theorem and beyond. But most books that explain these topics are so eager to get to the king’s palace that they rush through the surrounding fields of flowers and sort of trample on them. And they certainly don’t draw all the beautiful pictures that are implicit in this subject.

Posted by: John Baez on October 7, 2010 2:10 PM | Permalink | Reply to this

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

So is there visual gorgeousness to be had in the king’s palace? Or would we have to be beings who could visualise in many more dimensions to appreciate that?

Posted by: David Corfield on October 8, 2010 10:20 AM | Permalink | Reply to this

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

David wrote:

So is there visual gorgeousness to be had in the king’s palace?

I’m not sure yet — the guards didn’t let me in yet.

But I bet there is. The modularity theorem, which says that every elliptic curve over the rationals can be obtained from a modular curve, implies Fermat’s Last Theorem. Tim will show us how to visualize lots of modular curves. Elliptic curves are also very nice geometrical things. So if we were smart enough, it’s possible the statement of the modularity theorem would have some kind of visualizable content.

Grothendieck’s idea of children’s drawings get into the game here in a very nice way, too…

(Btw, the use of the phrase dessins d’enfant should be banned among English-speaking mathematicians. The phrase means “children’s drawings”, and Grothendieck used this term to emphasize the childlike simplicity of these cute little pictures. There’s no problem with this phrase if you’re speaking French! But when you’re speaking English and you say dessins d’enfant, you come across as a worldly sophisticate who summers in Paris and enjoys Château Lafite Rothschild — conveying exactly the opposite of the idea Grothendieck was trying to get at!)

Posted by: John Baez on October 10, 2010 2:58 PM | Permalink | Reply to this

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

I love how rotating around the top face adds one to the numbers labelling the faces, while rotating a half-turn around the edge separating $\frac{0}{1}$ and $\frac{1}{0}$ takes the negative reciprocal.

I’ve long been fascinated by these two generators of $\mathrm{PSL}\left(2,ℤ\right)$. In week228, I wrote:

There’s a wonderful game invented by John Conway called “rational tangles”. Here’s how it works. It involves two players and a referee.

The players, call them A and B, start by facing each other and holding ropes in each hand connecting them together like this:

  A   A
|   |
|   |
B   B


This is called “position 0”. The referee then cries out either add one! or take the negative reciprocal!. If the referee yells add one!, player B has to switch which hand he’s using to hold which rope, making sure to pass the right one over the left, like this:

  A   A
\ /
/
/ \
B   B


This is called “position 1”, since we started with “position 0” and then added one. But if the referee yells take the inverse reciprocal!, both players must cooperate to move all four ends of the ropes a quarter-turn clockwise, like this:

  A   A
\_/
_
/ \
B   B


This is called “position -1/0”, since we started with 0 and then took the negative reciprocal.

The referee keeps crying add one! or take the negative reciprocal! in whatever order she feels like, and players A and B keep doing the same sort of thing: either player B switches the ropes right over left, or both players rotate the whole tangle a quarter-turn clockwise. It’s actually best if the referee doesn’t start with take the negative reciprocal!, since some people refuse to divide by zero, for religious reasons. But, it’s perfectly legal in this game.

Anyway, after a while the ropes will be in “position p/q” for some complicated rational number p/q. The’ll be all tangled up - but in a special way, called a “rational tangle”.

Then the players have to undo the tangling and get back to “position 0”. They may not remember the exact sequence of moves that got them into the mess they are in. In fact the game is much more fun if they don’t remember. It’s best to do it at a party, possibly after a few drinks.

Luckily, any sequence of add one! and take the negative reciprocal! moves the players make that carry their number back to 0, will carry their tangle back to “position 0”. So they just need to figure out how to get their number back to 0, and the tangle will automatically untangle itself. That’s the cool part! It’s a highly nonobvious theorem due to Conway.

I’m vaguely aware of a few proofs of this fact. As far as I know, Conway’s original proof uses the Alexander-Conway polynomial:

16) John Horton Conway, An enumeration of knots and links and some of their algebraic properties, in Computational Problems of Abstract Algebra, ed. John Leech, Pergamon Press, Oxford, 1970, 329-358.

There’s also a proof by Goldman and Kauffman using the Jones polynomial:

17) Jay R. Goldman and Louis H. Kauffman, Rational tangles, Advances in Applied Mathematics 18 (1997), 300-332. Also available at http://www.math.uic.edu/~kauffman/RTang.pdf

There are also two proofs in here:

18) Louis H. Kauffman and Sofia Lambropoulou, On the classification of rational tangles, available as math.GT/0311499.

But here’s what I want to know: is there a proof that makes extensive use of the group PSL(2,Z) and its relation to topology?

After all, the basic operations on rational tangles are “adding one” and “negative reciprocal”, and these generate all the fractional linear transformations

         az + b
z |->   --------
cz + d


with a,b,c,d integer and ad-bc = 1. The group of these transformations is PSL(2,Z). It acts on rational tangles, and Conway’s theorem says this action is isomorphic to the obvious action of PSL(2,Z) as fractional linear transformations of the “rational projective line”, meaning the rational numbers together with a point at infinity. Since PSL(2,Z) has lots of relations to topology, there should be some proof of Conway’s theorem that uses these relations to get the job done.

Does anybody know one?

And then people supplied some nice proofs — see week229 for a discussion of those. If you want to really understand it all, make sure to read all the way to the Addendum.

By the time we’d gotten it all worked out, we’d understood a lot better how modular curves are related to the theory of tangles! But I bet there’s more left to discover…

Posted by: John Baez on October 7, 2010 9:13 AM | Permalink | Reply to this

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

So does n=6 give you a plane?

Also, there are various higher-dimensional generalizations of continued fractions (where, for instance, repetition implies that the number is the solution of a cubic equation); which of these are nicest from this modular point of view?

Posted by: Mike Stay on October 12, 2010 6:05 PM | Permalink | Reply to this

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

N=6 gives you not a whole plane but a quotient of a plane by a lattice—i.e. a torus. (I’ll be talking about the $N\ge 6$ cases in a future post.) Remember that $\mathrm{PSL}\left(2,{ℤ}_{N}\right)$, which is the symmetry group of the tiling, consists of (some of the) matrices whose elements are integers mod $N$; since there are only a finite number of the latter, the group itself is finite. So we end up with a finite tiling of a compact surface, whatever $N$ is.

Posted by: Tim Silverman on October 12, 2010 7:23 PM | Permalink | Reply to this

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

Those are some beautiful pictures - if I’m not mistaken, passing to N=7 yields a triangulation of the Klein quartic.

I have a very minor nitpick: The group SL_2(Z) acts on the plane Z^2 by area-preserving transformations, but the quotient PSL_2(Z) only acts projectively. In particular, -Id multiplies lattice vectors by -1 while representing the identity element in the quotient. You get an honest action of PSL_2(Z) when you pass to the set of lines in the lattice.

Posted by: Scott Carnahan on October 26, 2010 8:54 AM | Permalink | Reply to this

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

Thanks for the nicely illustrated post.

I didn’t understand how you are labeling your lines. That is for example if you are in Z_5 then the fraction which labels your line is secretely just a number in Z_5 that is you don’t know about “what the nominator and denominator is.” On the other hand if you would use the original labels from Z^2 then the lines labeled 1/1 and 12/7 for example would give the same line however the labels 1/1 and 2/2 (= (12 mod 5)/(7 mod 5) are different.

Are there references for the above ?

Posted by: student on October 28, 2010 7:54 PM | Permalink | Reply to this

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

Sorry for not being clear. The lines over (e.g.) ${ℤ}_{5}$ can be considered part of the story but, as you observe, they are not enough: one needs extra structure in order to get the labels for the vertices of the triangular tiling.

There are several ways to get the list of different labels.

We can start with a point $\left(m,n\right)$ in ${ℤ}^{2}$, cancel shared factors between the two components $m$ and $n$, reduce them each mod $N$, cancel factors of $-1$ (i.e. identify $\left(m,n\right)$ with $\left(-m,-n\right)$), and then put the reduced $n$ on top of the reduced $m$ to get something looking like a fraction (but allowing no further cancellation).

Or we can start with a point in $ℚ{P}^{1}$, take its expression as a fraction in lowest form, reduce numerator and denominator mod $N$, and cancel factors of $-1$.

Or we can start in ${{ℤ}_{N}}^{2}$, and look for points whose components are coprime relative to $N$ (i.e. share no prime factors which are also factors of $N$), cancel factors of $-1$, and put the second component over the first.

Or we can take the lines in ${ℤ}_{N}$, and search along each of them for points which generate the line they lie in (i.e. points $\left(m,n\right)$ such that the set of points $\left(km,kn\right)$ for $k\in {ℤ}_{N}$ consists of all the points in the line), and then select just these points, and then identify them in pairs by cancelling factors of $-1$.

So for instance, we can start with the point $\left(14,4\right)$ in ${ℤ}^{2}$, cancel the factor of $2$ to get $\left(7,2\right)$, reduce mod $5$ to get $\left(2,2\right)$, don’t cancel any factors of $-1$ because the pair is already in the form we like (with a second component less than $N/2$) and then represent it in fraction form as $\frac{2}{2}$.

Or we can start with $\frac{7}{2}$ in $ℚ{P}^{1}$, reduce numerator and denominator mod $5$ to give $\frac{2}{2}$, observe that the shared factor of $2$ is not also a factor of $5$, and therefore shouldn’t be cancelled, skip over cancellation of $-1$ as before, and so leave ourselves with $\frac{2}{2}$.

Or we could search through the points of ${{ℤ}_{5}}^{2}$, observe that the components of $\left(2,2\right)$ only share a factor of $2$, which is not, itself, a factor of $5$, skip cancellation by $-1$, and put the second component over the first to get $\frac{2}{2}$.

Or we could look at the line in ${{ℤ}_{5}}^{2}$ consisting of the five points $\left(0,0\right)$, $\left(1,1\right)$, $\left(2,2\right)$, $\left(3,3\right)$ and $\left(4,4\right)$, and observe that, among others, the point $\left(2,2\right)$ generates the line. (That is, multiplying both its components by, successively, $0$, $1$, $2$, $3$ and $4$ gives, successively, $\left(0,0\right)$, $\left(2,2\right)$, $\left(4,4\right)$, $\left(3,3\right)$ and $\left(1,1\right)$, which is all the points on the line. In fact, in this case, all of the points on the line, other than $\left(0,0\right)$, will generate the line.) So $\left(2,2\right)$ (identified with $\left(-2,-2\right)=\left(3,3\right)$) gives us one of the vertex labels.

What makes fractions look like a nice way to represent these special pairs of numbers, and which persuaded me to display them like that when labelling vertices (or, dually, faces) of the tilings, is that if we’re working with $ℚ{P}^{1}$, i.e. not doing any mod $N$ reduction, then we really do have a perfect match-up of points in $ℚ{P}^{1}$—represented by one fraction each—and vertices in the triangular tiling. So it’s nice to carry this over to the reduced-mod-$N$ version, even though we then end up with several different-looking fraction-labels corresponding to the same line in ${{ℤ}_{N}}^{2}$. Also, I think it looks cleaner on the pictures. I got used to working with these funny fractions quite quickly and forgot they were a bit strange.

References: I’ve never seen anything that lays all this out in the form I’ve been describing it, but here are some of the places I’ve looked at while trying to understand this stuff.

As so often, Wikipedia is a good place to start looking for basics and follow-up references: the articles on Modular group, Farey sequence, Modular curve, Stern-Brocot tree and Klein quartic contain lots of good stuff.

Gerard Westendorp’s site contains many excellent pictures. Lieven le Bruyn talks about related things in several places. I’ve made a lot of use of Diamond and Shurman’s rather heavy-duty book A First Course in Modular Forms, but you have to work hard to extract this pictorial sort of information from it. Oh yes, and I was much inspired by the hyperbolic tesselations in Don Hatch’s site. You’ll be seeing things like this in coming weeks.

Posted by: Tim Silverman on October 29, 2010 1:39 PM | Permalink | Reply to this

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

Thank you very much for the thorough reply. With it I think I was actually able to find all 24 reduced Farey numbers (labels of vertices) for Z mod 7, which were mentioned in the comment by John Baez. However since deriving them is a kind of painful procedure (in particular one is never really sure, whether one found them all) I wonder whether there exists a formula or table which says how many reduced Farey numbers there are for a given Z mod j?

Posted by: student on November 3, 2010 5:41 PM | Permalink | Reply to this

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

If ${p}_{i}$ are the distinct prime factors, the number of reduced fractions (i.e. faces of the tiling) is $\frac{{N}^{2}}{2}{\prod }_{i}\frac{{{p}_{i}}^{2}-1}{{{p}_{i}}^{2}}$.

E.g. $12={2}^{2}\cdot 3$, so for $N=12$ we have

$\frac{{12}^{2}}{2}{\prod }_{{p}_{i}=2,3}\frac{{{p}_{i}}^{2}-1}{{{p}_{i}}^{2}}=\frac{144}{2}\cdot \frac{{2}^{2}-1}{{2}^{2}}\cdot \frac{{3}^{2}-1}{{3}^{2}}=48$.

I can’t off the top of my head remember any way to make this obvious, but I do remember that the fraction of $\frac{1}{2}$ can be accounted for by the fact that we’re identifying $\frac{m}{n}$ with $\frac{-m}{-n}$, and I think the terms ${{p}_{i}}^{2}-1$ can be usefully treated as $\left({p}_{i}-1\right)\left({p}_{i}+1\right)$. Sorry I can’t be more helpful—I once convinced myself that this formula was “obvious”, but that was a long time ago.

Posted by: Tim Silverman on November 3, 2010 8:22 PM | Permalink | Reply to this

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

Thanks for the formula and patience. I find the formula sort of obvious for prime numbers but not for the general case, but probably I am just suffering of a numbertheoretical-dislexia.

Is there a similar formula for the edges?

I don’t understand the sentence:
“…while rotating a half-turn around the edge separating 0/1 and 1/0 takes the negative reciprocal.” which was written in John Baez comment at:
http://golem.ph.utexas.edu/category/2010/10/test.html#c035029

what do you mean by “edge separating 0/1 and 1/0”? If I didn’t miscompute then the reciprocal negatives for the N=5 case are e.g.
0/1 -> -1/0 = -1/-0 = 1/0
4/1 -> -4/1 = 1/1
2/2 -> 3/2
4/2 -> 2/1
3/1 -> 1/2
?
I don’t see how this goes together with a rotation around an edge.

Posted by: student on November 5, 2010 12:06 PM | Permalink | Reply to this

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

what do you mean by “edge separating 0/1 and 1/0”?

I’m working in the dual picture, where each fraction is associated with a face rather than a vertex. Then the faces labelled by $\frac{0}{1}$ and $\frac{1}{0}$ are separated by an edge, and flipping the polyhedron, so that this edge is sent to itself but pointing the opposite way, does the $q\to \frac{-1}{q}$ swap.

If you stare hard at the dodecahedron, you should be able to see this.

Posted by: Tim Silverman on November 5, 2010 12:51 PM | Permalink | Reply to this

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

Student wrote:

Thanks for the formula and patience. I find the formula sort of obvious for prime numbers but not for the general case, but probably I am just suffering from a numbertheoretical dyslexia.

I haven’t thought about it hard, but it probably helps to know that the ring $ℤ/N$ is a product of rings of the form $ℤ/{p}^{n}$ where the prime powers ${p}^{n}$ are the factors of $N$. This is called the Chinese remainder theorem. It should help reduce the general problem to the case of a prime power.

The Chinese remainder theorem is often attributed to a clever general who wanted to efficiently count his troops, but it first appeared in a book by Sun Tzu. No, not that Sun Tzu — not the author of the Art of War, but a mathematician by the same name.

By the way, are you the guy who invented the t-test?

Posted by: John Baez on November 5, 2010 12:26 PM | Permalink | Reply to this

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

By the way, are you the guy who invented the t-test?

You’re funny, John.

I guess the story behind “Student” is pretty well known, but for those who don’t know it, the etymology can be found here.

Posted by: Todd Trimble on November 5, 2010 1:45 PM | Permalink | Reply to this
Read the post Pictures of Modular Curves (II)
Weblog: The n-Category Café
Excerpt: Explaining some of the algebraic and geometrical background to the first article in the series.
Tracked: October 30, 2010 4:29 PM
Read the post Pictures of Modular Curves (III)
Weblog: The n-Category Café
Excerpt: Tilings of modular curves of level greater than 5, with non-positive curvature.
Tracked: November 14, 2010 10:57 PM

### Farey and the Golden Ratio

There is an interesting relation to the Golden Ratio, the “most irratrional number”. If you go down the Stern Brocot tree left-right-left-right-left-right-… you end up at the golden ratio. In the mean while you pay a visit to all fractions F_n+1/F_n, composed of consequetive Fibonacci numbers.
Here is the idea:

The idea of the golden ratio being the “most irratrional number” seems to be related, but I can’t prove it; it looks as if the circle density in the modular tiling near the Golden ratio is a minimum.

Gerard

Posted by: Gerard Westendorp on November 27, 2010 2:14 PM | Permalink | Reply to this

### Re: Farey and the Golden Ratio

This idea can be extended into the “Golden Stern Brocot Tree”:

The rationals are still ordered by their x-coordinate. But compared to a conventional Stern Brocot tree, there is more room for displaying the rationals, because we use 2 dimensions more efficiently.

Maybe even more efficient ways are possible. But it is late, so I’ll quit now.

Gerard

Posted by: Gerard Westendorp on November 29, 2010 12:20 AM | Permalink | Reply to this

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