## February 1, 2008

### Modular Forms

#### Posted by John Baez Jim Dolan and I are trying to learn about modular forms and the Modularity Theorem, once known as the Taniyama–Shimura–Weil Conjecture. It’s an irresistible challenge. After all, this result implies Fermat’s Last Theorem, but it’s much more conceptual, and closely related to the Hecke operators we’re always talking about.

I think we’re making decent progress — Diamond and Shurman’s book is very helpful. But, I’m still lacking in intuition in many ways.

For example: I think I have a decent intuition for level-1 modular forms. It took me about 10 years. Now I want to understand higher levels equally well. I’m hoping it won’t take another decade.

Maybe I should say where I am.

I have a good picture of the moduli space of elliptic curves, $H/SL(2,\mathbb{Z})$. I’m friends with the Eisenstein series $g_2$ and $g_3$. I know they generate the ring of level-1 modular forms, and I know why the discriminant $\Delta = g_3^2 - 27 g_2^3$ generates the ideal of cusp forms.

Say we go up to level two, replacing $SL(2,\mathbb{Z})$ by the Hecke subgroup $\Gamma_0(2)$. I want an equally clear story in this case. What does the moduli space $H/\Gamma_0(2)$ look like? How many cusps does it have? What is the ring of level-2 modular forms like? What are some explicit generators and relations? What are the cusp forms like, in terms of these generators?

Presumably this information is buried in many places, e.g. in the chapter ‘Modular forms for Hecke subgroups’ of Knapp’s Elliptic Curves. These sources seem to tackle all levels at once. But I’m feeling lazy, so I’d like someone to just tell me what’s going on at level 2.

It would also be nice to see a kind of chart showing how things work for low levels.

Posted at February 1, 2008 7:29 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1590

### Re: Modular Forms

I’m not fully sure about this stuff but I think I understand it well enough at level 2. The moduli space looks like a Riemann sphere with three holes poked in it. (I think they’re actually called cusps, but I think of them as holes poked in the surface)

There are two level-one modular forms, which I know how to get in two ways:

The fourth powers of the Jacobi theta functions at z=0–there are three values but the fourth powers are not linearly independent, and the Jacobi identity brings it to two, or

The Weierstrass p functions at 1/2, $\tau/2$, and $(\tau+1)/2$(when you take 1 and $\tau$ to be the periods rather than the half-periods–I don’t know what the convention is), which always add up to 0. Also note that those three points are also precisely the points(modulo periods) where the derivative is zero.

Either basis can be described in terms of the other(and I think the Wikipedia pages describe how exactly to do that).

Higher-order(I can’t remember the exact term) modular forms are expressed in terms of these, and you can tell when it’s level-1 when you get one symmetric in the three generators. (That’s why you use three instead of two, despite only two being linearly independent)

There’s one fundamental modular function, the elliptic lambda, which takes values 0, 1, and infinity on the cusps.

I’m not fully sure about levels 3 and higher(most of this stuff I figured out myself/with the help of Wikipedia/MathWorld a few years ago, and I’m only seventeen so I probably shouldn’t know this at all, and take it with a grain of salt) but I do know that there’s a really nice link to Platonic solids on levels 3, 4, and 5. Can’t really tell you much more than what the moduli spaces look like, though.

On levels 3-5 the moduli space is a Riemann sphere with cusps.

On level 3, there are 4 cusps, the vertices of a tetrahedron.

On level 4, there are 6 cusps, the vertices of an octahedron.

On level 5, there are 12 cusps, the vertices of an icosahedron.

On level 6, you have a torus. I don’t remember how many cusps it has but if you fill in all the holes it’s isomorphic to the plane modulo the Eisenstein integers(or, in elliptic curve form, $x^{3}-1=y^{2}$)

On level 7 you have the Klein quartic. There are 24 cusps, related to tiling the quartic with heptagons and/or triangles.

And after that my knowledge is pretty much exhausted.

Posted by: William on February 1, 2008 11:33 PM | Permalink | Reply to this

### Re: Modular Forms

William said:

On level 6 you have a torus. I don’t remember how many cusps it has

12. $\mathbb{Z}_6$ only has 2 units, so each member of the “projective line” over $\mathbb{Z}_6$ gets one hexagon. It’s actually quite easy to build the correct labelling starting from a hexagonal tiling of the plane. Label two adjacent tiles with $0$ and $\infty$, and just keep applying the rules that rotating by 60 degrees about the $\infty$ hexagon adds $1$ to everything, while the $180$ degree rotation that switches $0$ and $\infty$ sends every $q\rightarrow\frac{-1}{q}$.

It’s helpful to build a table of equivalent fractions first, so that you can see at a glance that, for instance, $\frac{5}{2}=\frac{1}{4}$ (mod 6, of course).

It then also becomes obvious that the clumps of 12 hexagons give the tiling of the plane by equilateral triangles (that’s the $A_2$ lattice, isn’t it?)

Posted by: Tim Silverman on February 6, 2008 11:50 PM | Permalink | Reply to this

### Re: Modular Forms

William wrote:

I’m not fully sure about this stuff but I think I understand it well enough at level 2. The moduli space looks like a Riemann sphere with three holes poked in it. (I think they’re actually called cusps, but I think of them as holes poked in the surface.)

I just now happened to bump into a picture of the fundamental domain for level 2 – it’s on page 260 of Knapp’s Elliptic Curves. He says “A feature of this diagram is that $S(R)$ and $S T(R)$ contribute to the same cusp. Thus there are only two cusps, at $\infty$ and $0$, even though $[SL(2,\mathbb{Z}),\Gamma_0(2)] = 3$.”

So, it’s possible that the third ‘hole’ you’re talking about is not really what people call a cusp, but instead some other funny spot.

For example, at level 1, the moduli space looks like a Riemann sphere with one hole poked out of it — the cusp — but there are also two other funny spots, corresponding to the lattices with square symmetry and hexagonal symmetry — the Gaussian integers and the Eisenstein integers. Those other spots aren’t really cusps.

There are two level-one modular forms, which I know how to get in two ways:

I’m confused. Do you mean level-two?

By the way, my favorite level-2 modular form is the ‘cross ratio’. People usually think of that as an invariant of 4 points on the Riemann sphere, but you can reinterpret it as an invariant of elliptic curves with a little extra structure, as I explained near the end of week229. There’s a lot more about this in Lecture 9 of Dolgachev’s course. I guess I should re-read this now and see if I understand it better.

Higher-order (I can’t remember the exact term) modular forms are expressed in terms of these,

I think the term you want is ‘weight’.

I’m only seventeen so I probably shouldn’t know this at all…

Right — you should be out having sex and doing drugs, or whatever 17-year-olds are supposed to do these days.

On level 3, there are 4 cusps, the vertices of a tetrahedron.

On level 4, there are 6 cusps, the vertices of an octahedron.

On level 5, there are 12 cusps, the vertices of an icosahedron.

On level 7 you have the Klein quartic. There are 24 cusps, related to tiling the quartic with heptagons and/or triangles.

Cool! Incredible! I suspect these points aren’t all ‘cusps’ in the technical sense, but I think you’re on to something here… for example, I know that PSL(2,5) is the symmetry group of the dodecahedron, and PSL(2,7) is the symmetry group of the Klein quartic. Klein spent a lot of time studying both these things.

Speaking of cusps and the Klein quartic. I have a webpage on the Klein quartic, in which I claim this curve is the upper halfplane $H$ mod the group $\Gamma(7)$… and a while ago someone sent me an email saying that’s not quite right: you need to add on one point, a cusp, to get the Klein quartic. I can’t find that email now, but I think it sounds right. However, that suggests there’s just one cusp, and 23 other funny spots, for a total of 24 — the 24 vertices of this thing: Thanks to Greg Egan for this moving picture!

(I’m sort of confused about $\Gamma(7)$ versus $\Gamma_0(7)$, now.)

Posted by: John Baez on February 2, 2008 1:00 AM | Permalink | Reply to this

### Re: Modular Forms

So that William doesn’t feel too bad, I will add that his understanding corresponds to mine (right down to the holes poked in the surface). Of course, I’m just another amateur, so what I’m about to say may very well be completely idiotic, but the urge to babble has become irresistable, so here goes.

My understanding (if you can call it that) is that the inclusion of $\Gamma(N)$ in the modular group $\Gamma$ gives rise to a covering of $H/\Gamma$ by $H/\Gamma(N)$, and lifting the fundamental domain of $\Gamma$ to the covering gives a tiling of the surface by regular $N$-gons, 3 meeting at each vertex, each subdivided into $N$ triangles corresponding to the fundamental domain of the modular group, From this, it ought to be obvious that the cusps correspond to the centres of the faces. So if that’s not true, then something has going terribly wrong …

I thought that, except for the case of level 2, there are $\frac{N^2}{2}\prod_{p|N}(1-\frac{1}{p^2})$ N-gons (taking the product over all prime factors $p$ of $N$, each counted once. (In fact I’m sure I read this, or something like it, in Diamond and Shurman, only of course now I can’t find it.) This does agree with what William said about the Platonic solids. The faces can, I believe, be labelled by lines through the origin of the free 2-dimensional module over $\mathbb{Z}_N$, although in general, these labels will appear on multiple faces. (For prime $N$, these will be points in the projective line over the field with $N$ elements; otherwise they will be somewhat wackier points in a generalised “projective line” over a ring. E.g. for $N=6$ we get points $0$, $1$, $2$, $3$, $4$, $5$, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{1}{4}$, $\frac{3}{4}$ and $\infty$.)

However, the action of $\Gamma/\Gamma(N)$ will be consistent over where it sends the labels. So, for instance, the 24 faces of the Klein quartic can be labelled with the 8 points in the projective line over $\mathbb{F}_7$, each appearing three times, and if a member of $SL(2,7)$ sends, say one face labelled with 2 to a face labelled with $\infty$, then it will send all three faces labelled with 2 to faces labelled with $\infty$.

The modular group will then act as automorphisms of the tiling. The generator sending $z\rightarrow\z+1$ will rotate the faces labelled with $\infty$ by one step (ie one Nth of a full rotation), the generator sending $z\rightarrow-\frac{1}{z}$ will rotate 180 degrees about the edge(s) connecting the face(s) labelled with $\infty$ and the face(s) labelled with $0$, and likewise, order three elements rotate about vertices.

As William said, I thought at level 2, we get the tiling of the Riemann sphere with regular bigons, three meeting at each vertex, giving 3 faces (labelled with 1, 0 and $\infty$), 2 vertices and 3 edges.

Or perhaps this is all baloney.

Hmm, looking at what you said again, you’re talking not about $\Gamma(2)$, but about $\Gamma_0(2)$. That might account for the confusion. At least, if it did, I am less likely to be humiliated and equally likely to learn something when a genuine professional turns up to explain things. So I’m keeping my fingers crossed.

Posted by: Tim Silverman on February 2, 2008 4:33 PM | Permalink | Reply to this

### Re: Modular Forms

Ha! Diamond and Shurman, §3.8 (“More on cusps”) verifies at least the algebraic part of this, and goes into more detail.

On other elliptic points: the point at $i$ of order 2 lifts to the midpoints of edges, while the point of order 3 at $\omega$ (or $-\omega^2$ if we take the other boundary of the fundamental domain) lifts to the vertices of the tiling.

Posted by: Tim Silverman on February 2, 2008 9:31 PM | Permalink | Reply to this

### Re: Modular Forms

…we get the tiling of the Riemann sphere with regular bigons, three meeting at each vertex, giving 3 faces (labelled with 1, 0 and ∞), 2 vertices and 3 edges.

Yeah, the Grothendieck ribbon graph associated to the j-invariant!!! Cool!

Posted by: Kea on February 3, 2008 5:45 AM | Permalink | Reply to this

### Re: Modular Forms

Yes, that is certainly is one of the most fascinating themes in mathematics. Esp. the use of modular functions to build class fields in analogy to the use of the exponential function to produce roots of unity. Here a nice article. I found this book on that very thrilling. One of the IMHO most astonishing things is the existence of analoga over finite fields, e.g. here the “algebraist’s upper half plane”, related to other strange things. A beautifull introduction into such developments is the last chapter of Vladut “Kronecker’s Jugendtraum and Modular Functions”. Finally a related ppt and here and here articles about the strange conections between modular functions and continued fractions and Y.I. Manins “Alterstraum”.

Posted by: Thomas Riepe on February 2, 2008 4:35 PM | Permalink | Reply to this

### Re: Modular Forms

And yes, I meant level-two rather than level-one. I meant weight-one or weight-two there, anyways.

Posted by: William on February 2, 2008 7:18 PM | Permalink | Reply to this

### Re: Modular Forms

OK, got a little more.

The labels on the cusps of $\Gamma(N)$ are precisely those members of the group $\mathbb{Z}_N^2$ which are of order $N$, modulo negation. This means that we get a labelling by lines through the origin of the $\mathbb{Z}_N$-module $\mathbb{Z}_N^2$, each of which appears a number of times equal to half the order of the group of units in $\mathbb{Z}_N$. So twice in the case of $\Gamma(5)$, three times in the case of $\Gamma(7)$, once for $\Gamma(3)$, $\Gamma(4)$ and $\Gamma(6)$. ($\Gamma(2)$ is slightly different.)

Now, in the case of $\Gamma_1(N)$, we mod out by the action of the group generated by $z\rightarrow z+1$. So the $N$-gon(s) labelled by $\infty$ fold down to a single triangle made into a cone, with the cusp at its apex. (The triangles, I neglected to mention, have bases on the edges of the $N$-gon, and apices at its centre.) The faces labelled with integers all collapse into a single $N$-gon. Other faces (if any) may undergo various other forms of collapse.

In the case of $\Gamma_1(2)$, the faces labelled with $0$ and $1$ become identified, so we get left with 2 cusps, one at $0$ (or $1$) and one at $\infty$.

In the case of, say, $\Gamma_1(4)$, we get a cusp on the $\infty$ face, a cusp on the “integer” face (which results from identifying 0, 1, 2 and 3) and a cusp on the $\frac{1}{2}$ face.

Some of these cusps (not to mention the other elliptic points) get folded over in various ways which probably do things do their order that I haven’t worked out yet.

I still haven’t understood $\Gamma_0(N)$. It’s obviously somewhere in between the other two, but it seems to be more complicated than either.

Posted by: Tim Silverman on February 3, 2008 12:45 PM | Permalink | Reply to this

### Re: Modular Forms

I burbled

Some of these … elliptic points … get folded over …

I was very very confused here. It’s perfectly obvious from the tesselation picture that the lifts of the elliptic points get completely unfolded, so $\Gamma(N)$ doesn’t have any elliptic points for $N>1$. Larger congruence subgroups may leave some elliptic points still folded over, in which case they will retain their periods of 2 or 3. (However, $\Gamma_1(N)$ doesn’t in fact have any elliptic points for $N>3$. As yet, it’s only semi-clear to me intuitively why this is so. )

Posted by: Tim Silverman on February 5, 2008 8:33 PM | Permalink | Reply to this

### Re: Modular Forms

JB wrote:

PSL(2,5) is the symmetry group of the dodecahedron, and PSL(2,7) is the symmetry group of the Klein quartic

Yes, and PSL(2,3) is the symmetry group of the tetrahedron ($A_4$) and PSL(2,2) is the symmetry group of the tiling of the sphere by three bigons ($S_3$) and the group that one might call $PSL(2,\mathbb{Z}_4)$ (linear transformations of $\mathbb{Z}_N\times\mathbb{Z}_N$ of determinant 1, mod $\{-1,1\}$) is the symmetry group of the cube ($S_4$).

(Is there any deep reason for that progression of symmetric and alternating groups $S_3$, $A_4$, $S_4$, $A_5$, other than numerical coincidence?)

Posted by: Tim Silverman on February 3, 2008 12:56 PM | Permalink | Reply to this

### Re: Modular Forms

All these comments are great! I’m supposed to be writing a paper, so I’m trying not to blog too much. I will read these more carefully and reply when I feel like taking a break and having some fun.

Posted by: John Baez on February 4, 2008 3:22 AM | Permalink | Reply to this

### Re: Modular Forms

Here a book about the crime case behind this painting. The author Bernd Roeck is a well known historian and claims after studying old archives that Piero della Francesca’s “The Flagellation” is a painted testimony of a political murder case, where Federico da Montefeltro killed his stepbrother Oddantonio.

Conc. modular forms, Mumford’s article on compactifying the “universal elliptic curve” above a modular curve in “Smooth compactification of locally symmetric varieties” is very beautifull to read. By a “scissors and glue” construction instead of refering to general theorems he compactifies it in a very intuitive way with toric varieties, i.e. schemes over F1.

Posted by: Thomas Riepe on February 6, 2008 11:23 AM | Permalink | Reply to this

### Re: Modular Forms

Tim writes:

My understanding (if you can call it that) is that the inclusion of $\Gamma(N)$ in the modular group $\Gamma$ gives rise to a covering of $H/\Gamma$ by $H/\Gamma(N)$, and lifting the fundamental domain of $\Gamma$ to the covering gives a tiling of the surface by regular $N$-gons, 3 meeting at each vertex, each subdivided into $N$ triangles corresponding to the fundamental domain of the modular group.

I’m way behind you, so let’s see if you (or anyone!) can help me unnderstand this.

For our TV viewers out there, let’s first recall that $\Gamma$ is the group of $2 \times 2$ matrices with integer entries having determinant 1, better known as $SL(2,\mathbb{Z})$. Or, maybe it’s this group modulo $\pm 1$, better known as $PSL(2,\mathbb{Z})$. Which group we use doesn’t matter a lot for what we’re doing now, as long as we’re consistent about it. So, let me not mod out by $\pm 1$, at least in this comment.

Then $\Gamma(N)$ is the subgroup of $\Gamma = SL(2,\mathbb{Z})$ consisting of those matrices whose entries are equal, mod $N$, to those of the identity matrix.

This is a normal subgroup, and I guess

$\Gamma/\Gamma(N) = SL(2,\mathbb{Z}/N)$

at least when $N$ is prime. Is this right? Does this make sense when $N$ is not prime?

$\Gamma$ and thus $\Gamma(N)$ acts on the upper halfplane $H$, and we’re mainly interested in $H/\Gamma(N)$, which is a covering space of $H/\Gamma$.

I know and love $H/\Gamma$, and I want to know and love $H/\Gamma(N)$.

So, I’ll do what you say and think of $H/\Gamma(N)$ as a covering space of $H\Gamma$. Or, equivalently, I’ll think about a fundamental domain for the action of $\Gamma(N)$ on $H$ as made of a bunch of fundamental domains for the action of $\Gamma$ on $H$.

Here are a bunch of fundamental domains for the action of $\Gamma$ on $H$: The gray one is everybody’s favorite. It’s a ‘triangle’ with two corners having 60-degree angles and one corner infinitely far up the page with a 0-degree angle — that very pointy corner is what gives the ‘cusp’.

So, how do I create a fundamental domain for $\Gamma_0(N)$ by taking a bunch of fundamental domains for $\Gamma$?

For starters, how many do I need? It must be the cardinality of $\Gamma/\Gamma_0(N)$, no? That should be the cardinality of $SL(2,\mathbb{Z}/N)$, no? At least when that makes sense?

But wait — now I think the right answer is the cardinality of $PSL(2,\mathbb{Z}/N)$. We really need to think about groups that act faithfully on $H$, so we need to mod out by $\pm 1$ everywhere.

Which is it?

Let’s see: what’s the cardinality of $SL(2,\mathbb{Z}/p)$? First, what’s the cardinality of $GL(2,\mathbb{Z}/p)$? There are $p^2 - 1$ choices for the first row of the matrix, and $p^2 - p$ choices for the second row. So, $(p^2 - 1)(p^2 - p)$. Of these matrices, $1/(p-1)$ of them have determinant 1. So, there are

$(p+1)(p^2 - p) = p^3 - p$

matrices in $SL(2,\mathbb{Z}/p)$.

Let me check — I usually make mistakes in calculations like this. For $p = 2$, this formula gives 6, which fits the idea that $SL(2,\mathbb{Z}/2) = PSL(2,\mathbb{Z}/2) = S_3$.

For $p = 5$ this formula gives $120$, which matches the idea that $PSL(2,\mathbb{Z}/5)$, half as big, is $A_5$.

This also fits your formula which says that for prime $p$, there’s a tiling of $H/\Gamma_0(N)$ by

$\frac{p^2}{2}(1 - 1/p^2)$

$N$-gons, each of which consists of $2p$ triangles, for a total of

$p^3 - p$

triangles. Sort of like this for $p = 7$: I believe there are $7^3 - 7 = 336$ triangles in this picture: 14 triangles per heptagon, 24 heptagons in all. If I’m wrong, I must be making a serious mistake somewhere.

Hmm, so the number of triangles is the cardinality of $SL(2,\mathbb{Z}/N)$, not $PSL(2,\mathbb{Z}/N)$? At least when $N$ is prime?

Anyway, I eventually wish to get to more interesting questions, but for now let me quit while I’m ahead (or behind).

Posted by: John Baez on February 7, 2008 5:49 PM | Permalink | Reply to this

### Re: Modular Forms

The main next things I’d like to do are:

• Really understand the roles of $SL(2,\mathbb{Z}/N)$ versus $PSL(2,\mathbb{Z}/N)$ in this situation, at least when $N$ is prime. And do they make sense when $N$ isn’t prime?
• At least for $N = p$ an odd prime, see why the $p^3 - p$ triangles tiling $H/\Gamma(N)$ organize themselves into $p$-gons with $2p$ triangles each, for a total of $\frac{1}{2}(p^2 - 1)$ $p$-gons.
• Figure out how many of these $p$-gons meet at each vertex. It seems to be 3 a lot of the time. Is it always 3?
• If it’s always 3 $p$-gons meeting at a vertex, I can calculate the genus of $H/\Gamma(p)$, and see why we’re getting genus 0 (a sphere) for $p = 3$ and $5$, but genus 3 (a 3-holed torus, shown writhing above) for $p = 5$.
• It would then be fun to understand $p = 11$ — some sort of surface tiled by 120 hendecagons. • Of course, non-prime $N$ are also interesting.
Posted by: John Baez on February 7, 2008 7:12 PM | Permalink | Reply to this

### Re: Modular Forms

JB wrote

I’m way behind you

That’s a slightly surreal and very scary position for me to find myself in. Some of this is stuff I think I actually know, but lots of it is just strung together with connections that seem intuitively plausible, backed up by some numbers and groups I’ve calculated that turn out to match the answers in the books…

That said, let me make a couple of things clear which might have confused some of our readers. First, I’m only going to talk about $\Gamma(n)$, which I think I sort of understand, and not $\Gamma_0(N)$ or $\Gamma_1(N)$, which I’m much less clear about. Second, it’s as well to be explicit that there are two kinds of triangle that we are talking about. We can, for instance, take the fundamental domain of the modular group—the grey triangle in the first diagram above. Or we can slice it vertically up the middle to get two triangles—that’s also the kind of triangle in the second diagram, the one where the triangles are coloured. Obviously there are twice as many of the second type of triangle, so we need to be clear which kind we’re talking about. In my earlier posts, I was always talking about the first kind.

So that gives us the number of elements in $PSL(2,N)$, which is what I want. And yes, it does appear, from my not very rigorous or thorough investigations, that this can be made to work for non-prime $N$. We just have to think carefully about the points in $\mathbb{Z}_N^2$, and the effect on them of matrices taking values in $\mathbb{Z}_N$, and multiplication by scalars in $\mathbb{Z}_N$.

Darn. There’s plenty more I wanted to say, but it’s getting late, so maybe I’ll talk more tomorrow. In the meantime, I can refer you back to §3.8 of Diamond and Shurman, which is full of excellent stuff.

Posted by: Tim Silverman on February 7, 2008 10:32 PM | Permalink | Reply to this

### Re: Modular Forms

Concerning $PSL(2,\mathbb{Z}_N)$ for non-prime $N$: the problem you might expect is that, since there are non-zero elements of $\mathbb{Z}_N$ that are not invertible, there are matrices in $M_2(\mathbb{Z}_N)$ that have non-zero determinants but are not invertible. But we don’t care about these matrices: we only care about matrices of determinant $1$, which certainly are invertible. So at least we don’t have to worry about this particular problem.

Posted by: Tim Silverman on February 8, 2008 5:20 PM | Permalink | Reply to this

### Re: Modular Forms

Good point.

Posted by: John Baez on February 8, 2008 5:40 PM | Permalink | Reply to this

### Re: Modular Forms

Tim wrote:

That’s a slightly surreal and very scary position for me to find myself in.

I was trying to reassure you, not frighten you!

In the meantime, I can refer you back to ï¿½3.8 of Diamond and Shurman, which is full of excellent stuff.

Unfortunately Jim Dolan has that book checked out… but actually it’s not so bad: I’m more interested in chatting and doing calculations to build up my intuition, rather than looking up a bunch of facts.

First, I’m only going to talk about $\Gamma(n)$, which I think I sort of understand…

That’s fine. Knapp seems to focus on the groups $\Gamma_0(n)$. This, and the fact that they’re called ‘Hecke subgroups’, seems to suggest they’re especially important. But I don’t know why… and I agree with you that the groups $\Gamma(n)$ seem easier to start with. So, let’s think about those!

Second, it’s as well to be explicit that there are two kinds of triangle that we are talking about. We can, for instance, take the fundamental domain of the modular group — the grey triangle in the first diagram above. Or we can slice it vertically up the middle to get two triangles - that’s also the kind of triangle in the second diagram, the one where the triangles are coloured.

Okay, great! — that should clear up the annoying factor of 2 that’s been bugging me, which is related to my perpetual confusion about $SL(2)$ versus $PSL(2)$ versus $GL(2)$ versus $PGL(2)$.

For example, consider the dodecahedron. We can chop each of its 12 pentagons into 10 right triangles, for a total of 120. This is called the “Coxeter complex” for the group of all rotation and reflection symmetries of the dodecahedron — a 120-element group. The reason is that the set of these triangles forms a torsor for this group: given any two of these triangles, there exists a unique element of this group sending the first triangle to the second!

On the other hand, we can chop each of the dodecahedron’s 12 pentagons into 5 triangles, for a total of 60. Each of these triangles is built from two of the former triangles. The set of these triangles is a torsor for the group of rotation symmetries of the dodecahedron — a 60-element group.

This 60-element group is

$PSL(2,\mathbb{Z}/5) \cong \Gamma / \Gamma(5) \cong A_5,$

and indeed the whole story here is the $p = 5$ case of the general theory I’m trying to understand.

It looks like the 120-element group of all rotation and reflection symmetries of the dodecaheron is some other, not so relevant group. I guess it’s

$PSL(2,\mathbb{Z}/5) \times \mathbb{Z}/2 \cong A_5 \times \mathbb{Z}/2,$

because the extra symmetry $(x,y,z) \mapsto -(x,y,z)$ commutes with all the rotational symmetries of the icosahedron. And, I guess this group is not isomorphic to $PGL(2,\mathbb{Z}/5)$, because that 120-element group is isomorphic to $S_5$. Could this group be isomorphic to $SL(2,\mathbb{Z}/5)$? I guess so!

(I’m digressing. I just always get confused about this $SL(2)$ versus $PSL(2)$ versus $GL(2)$ versus $PGL(2)$ stuff.)

So, you’ve convinced me that the relevant tiling of $H/\Gamma(p)$ is the one with fewer triangles, namely $\frac{1}{2}(p^3 - p)$ triangles. So, for $p = 7$, not this tiling with $7^3 - 7 = 336$ triangles: but the one with 168 triangles, each twice as big — one for each element of $PSL(2,\mathbb{Z}/7)$.

Posted by: John Baez on February 8, 2008 6:35 AM | Permalink | Reply to this

### Re: Modular Forms

John wrote:

Knapp seems to focus on the groups $\Gamma_0(n)$. This, and the fact that they’re called “Hecke subgroups”, seems to suggest they’re especially important.

Jim Dolan replied via email:

is the structure on a lattice that the hecke subgroup stabilizes (say for prime n) essentially just an “n-local flag” or something like that? (in this dimension “flag” is just a bombastic way of saying “non-trivial subspace”, right?)

i didn’t actually try to calculate this or check any fine print or anything.

That sounds right, since this group consists of “upper triangular matrices mod n”, and in general upper triangular matrices preserve a flag.

Posted by: John Baez on February 8, 2008 4:46 PM | Permalink | Reply to this

### Re: Modular Forms

Jim wrote in reply to the above stuff:

whenever you have a “basic hecke operator” (aka “atomic geometric relationship”), there’s the “joint stabilizer” subgroup simultaneously stabilizing both of the features involved in the relationship; aka “intersection of the original stabilizers in the given relative position”. might this be what a “hecke subgroup” is, in general? so in this particular case, i’m imagining that maybe there’s a basic hecke operator something like “jump to an n-fold refinement of the lattice” (thought of as a random jump operator on lattices), and that the intersection of the two different sl(2,z)’s respectively stabilizing the original lattice and its n-fold refinement might be a “gamma_0(n)” or whatever it’s called; can you get something like that to work?

if this _is_ (more or less) what “hecke subgroups” are, then i’m a bit annoyed that i didn’t hear anyone say it outloud before. on the other hand the guess that i’m making here _is_ influenced by stuff gleaned from brown’s book, about the incidence geometry corresponding to the circular “augmented a_n series” dynkin diagrams.

Posted by: John Baez on February 8, 2008 6:10 PM | Permalink | Reply to this

### Re: Modular Forms

Conc. Hecke operators:

Yuri I. Manin found in the 1970’s a very beautifull, elementary and intuitive way to understand the 1st homology group of a modular curve and the action of Hecke operators on it through ‘modular symbols’. Here recent research on them, apparently many other related, fascinating things hide there to be found .

Posted by: Thomas Riepe on February 16, 2008 12:28 PM | Permalink | Reply to this

### Index of the Hecke algebra in its saturation; Re: Modular Forms

Excellent references, Thomas Riepe. Yuri I. Manin has discovered amazing things, and showed them to us clearly.

I have extracted and articulated an Integer Sequence which gives some of the flavor and data {do see the web page linked to, for its hotlinks and ability to show two graphs and the musical embodiment of the function}:

A135362 Index of the Hecke algebra in its saturation in End(J_0(n)).

n a(n)
40 1
41 1
42 1
43 1
44 2
45 1
46 2
47 1
48 1
49 1
50 1
51 1
52 1
53 1
54 3
55 1
56 2
57 1
58 1
59 1
60 2
61 1
62 2
63 1
64 2
65 1
66 1
67 1
68 2
69 1
70 1
71 1
72 2
73 1
74 1
75 1
76 2
77 1
78 2
79 1
80 4
81 1
82 1
83 1
84 2
85 1
86 1
87 1
88 8
89 1
90 1
91 1
92 16
93 1
94 4
95 1
96 8
97 1
98 1
99 9
100 1

OFFSET
40,5

COMMENT
“A quantity that controls the relation between the modular degree and congruences (the “congruence modulus”). This results in the following table, which suggests that if p | a(n) is a prime, then p^2 | 4 * n,a fact closely related to what Ken Ribet proved at the Raynaud birthday conference in Orsay a few years ago. Also, Mazur proved that a(p) = 1 when p is prime.”

Yuri I. Manin, Iterated Modular Symbols.

William Stein, Modular Symbols, Modular Forms and Modular Abelian Varieties in MAGMA. See table p. 54.

PROGRAM
(MAGMA) function f(N) J := JZero(N); T := HeckeAlgebra(J); return Index(Saturation(T), T); end function; for N in [1..120] do print N, f(N); end for;

KEYWORD
nonn,new

AUTHOR
Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 16 2008

Posted by: Jonathan Vos Post on February 19, 2008 5:20 PM | Permalink | Reply to this

### Re: Modular Forms

John wrote:

So, you’ve convinced me that the relevant tiling of $H/\Gamma(p)$ is the one with fewer triangles, namely $\frac{1}{2}(p^3−p)$ triangles.

So, next I want to see why these triangles are organized into a bunch of $p$-gons, with $p$ triangles per $p$-gon.

In other words: why does our tiling get a $\mathbb{Z}/p$ as a discrete rotational symmetry group?

For this, I guess we should use the fact that $PSL(2,\mathbb{Z}/p)$ acts on the set of triangles (in a free and transitive way). We should find a subgroup that looks like $\mathbb{Z}/p$. And, the obvious candidate is the group generated by

$T = \left( \array{ 1 & 1 \\ 0 & 1 } \right)$

This matrix generates a subgroup of infinite order in $SL(2,\mathbb{Z})$. In terms of the upper halfplane, this matrix acts by

$\tau \mapsto \tau + 1$

So, it slides this picture one notch to the right: In particular, it acts transitively on all the triangles like the gray one — the ones that share a vertex way up north, up at infinity… at the cusp.

So, when we work mod $p$, we get a similar picture, but with the cyclic group $\mathbb{Z}/p$ acting to cycle around $p$ triangles that share a vertex — a cusp.

So, I see why you guys like the picture of a Platonic solid with each face chopped into triangles sharing a vertex at the face’s center… a cusp.

I think someone told me that for the case $p = 7$, there’s just one cusp. But that doesn’t seem to make sense now.

Posted by: John Baez on February 8, 2008 5:11 PM | Permalink | Reply to this

### Re: Modular Forms

Yup. You’ve kind of followed the same reasoning that I used when working this stuff out, only backwards. I started out thinking, Well, mod $N$, that means the operation of adding $1$ should wrap round on itself after $N$ steps … and went on from there.

Maybe the guy who told you about the one cusp was thinking of the one cusp at $\infty$ which lifts to multiple cusps when we go to mod $N$.

Posted by: Tim Silverman on February 13, 2008 9:34 PM | Permalink | Reply to this

### Re: Modular Forms

Tim wrote:

Maybe the guy who told you about the one cusp was thinking of the one cusp at $\infty$ which lifts to multiple cusps when we go to mod $N$.

Maybe that’s it. Let’s say that’s it.

(This is the problem with ‘learning math on the streets’…. rumors spread and get a bit more distorted with each retelling.)

If I weren’t so busy right now, I’d post another entry on modular forms. I’ve made a lot of progress, mainly thanks to talking with Jim. I guess I’ll talk about it when I get time.

One cool fact: the Modularity Theorem says each rational elliptic curve is the moduli space for elliptic curves equipped with some sort of extra structure. Very conceptual and self-referential! Amazing that Fermat’s Last Theorem is a corollary.

Posted by: John Baez on February 15, 2008 12:31 AM | Permalink | Reply to this

### Re: Modular Forms

Even better, $SL(2,\mathbb{Z})$ is generated by that matrix $T$ above together with a matrix people call $S T$:

$S T = \left( \array{ 0 & -1 \\ 1 & 1 }\right)$

as discussed in week125. This matrix $ST$ has order 6 in $SL(2,\mathbb{Z})$ — in fact, it describes a 60 degree rotation in some weird basis. So, it has order 3 in $PSL(2,\mathbb{Z})$.

Why is this “even better”? Because it means our tiling of $H/\Gamma(n)$ will have 3-fold symmetry!

So, I’m feeling more confident that the tiling of $H/\Gamma(n)$ consists of a bunch of $p$-gons — $\frac{1}{2}(p^2 - 1)$ of them, to be precise — meeting three at at corner.

Check:

$p = 3$ gives us 4 triangles meeting 3 at a corner — the tetrahedron.

$p = 5$ gives us 12 pentagons meeting 3 at a corner — the dodecahedron.

$p = 7$ gives us 24 heptagons meeting 3 at a corner — Klein’s quartic.

$p = 11$ gives us 60 hendecagons meeting 3 at a corner — something I’d like to get to know.

I want to compute the genus of all such surfaces, which will be easy if this picture is correct. And, I should do some non-prime cases too! $N = 4$ should give us the cube, according to William. But I have to get to work… can’t have fun like this all day.

Posted by: John Baez on February 8, 2008 5:57 PM | Permalink | Reply to this

### p = 11 case is in the Math-Chemistry literature; Re: Modular Forms

The p = 11 case is in the literature after all – by Chemists!

“The undecakisicosahedral group and a 3-regular carbon network of genus 26”, by Erwin Lijnen, Arnout Ceulemans, Patrick W. Fowler, and Michel Deza

Abstract:

Three projective special linear groups PSL(2,p), those with p = 5, 7 and 11, can be seen as p-multiples of tetrahedral, octahedral and icosahedral rotational point groups, respectively. The first two have already found applications in carbon chemistry and physics, as PSL(2,5) ≡ I is the rotation group of the fullerene C_60 and dodecahedrane C_20H_20, and PSL(2,7) is the rotation group of the 56-vertex all-heptagon Klein map, an idealisation of the hypothetical genus-3 “plumber’s nightmare” allotrope of carbon.
Here, we present an analysis of PSL(2,11) as the rotation group of a 220-vertex, all 11-gon, 3-regular map, which provides the basis for a more exotic hypothetical sp^2 frame-work of genus 26. The group structure and character table of PSL(2,11) are developed in chemical notation and a three dimensional (3D) geometrical realisation of the 220-vertex map is derived in terms of a punctured polyhedron model where each of 12 pentagons of the truncated icosahedron is connected by a tunnel to an interior void and the 20 hexagons are connected tetrahedrally in sets of 4.
KEY WORDS: PSL(2,11), group theory, undecakisicosahedral group, topology, carbon allotrope
AMS: 05C10, 20B25, 57M20

1. Introduction

The three groups PSL(2,5), PSL(2,7) and PSL(2,11) form a special sub-set of the Projective Special Linear groups PSL(2,p) in view of their particular permutational structure. They can be viewed as multiples of the symmetry groups of the regular polyhedra in three dimensional (3D) space, and for this reason are also called the pollakispolyhedral groups . PSL(2,5) and PSL(2,7) correspond to the pentakistetrahedral,
5_T, and heptakisoctahedral group,
7_O, respectively. Both have found applications in chemistry and physics….
[truncated]

Bingo! Does that fit the bill?

Now, can Greg Egan animate this into a screensaver of dazzling brain-twisting gorgeousness?

Posted by: Jonathan Vos Post on February 13, 2008 10:14 PM | Permalink | Reply to this

### Re: Modular Forms

That’s interesting. On Tuesday, the mathematician Kostant visited UCR to talk about new connections between $E_8$ and the Standard Model — more about that later. At dinner, he reminded me that $PSL(2,11)$ has a deep relation to another molecule: the buckyball! He has explained this here:

The buckyball is has icosahedral symmetry, so you might expect a relation to $PSL(2,5)$, and there is — but it has a subtler relation to $PSL(2,11)$.

Posted by: John Baez on February 15, 2008 4:45 AM | Permalink | Reply to this

### Re: Modular Forms

Conc. “modular forms and $L$-series”:

When I thought about mentioning the connection between both as especially interesting, because in Hecke’s Theory the Mellin transform makes the functional equations of some L-series just the defining transformation rules of associated modular forms, I noticed that I know nothing about the p-adic analogon e.g. of the Mellin transform. Perhaps someone here knows where to look?

Concerning modular forms and critical $L$-values, here a descr. of a book by Shimura.

Posted by: Thomas Riepe on February 13, 2008 6:53 PM | Permalink | Reply to this

### Re: Modular Forms

Thomas Riepe wrote:

Conc. “modular forms and $L$-series”:

When I thought about mentioning the connection between both as esp. interesting, because in Hecke’s Theory the Mellin transform makes the functional eq.s of some $L$-series just the defining transformation rules of associated modular forms, I noticed that I know nothing about the $p$-adic analogon e.g. of the Mellin transform. Perhaps someone here knows where to look?

I’m no expert, but at my current stage of development I find it reassuring to regard the Mellin transform as a silly trick — part of a cute but distracting ‘explicit formula’ for the transform mapping the Taylor series

$\sum_{n \ge 1} a_n z^n$

to the Dirichlet series

$\sum_{n \ge 1} a_n n^z$

Here’s how I’d prefer to think about it:

$\mathbb{C}$-valued functions on a sufficiently well-behaved monoid $M$ form an algebra.

The algebra of Taylor series is the algebra of complex functions on the additive monoid of natural numbers.

The algebra of Dirichlet series is the algebra of complex functions on the multiplicative monoid of nonzero natural numbers.

Since these two monoids are both part of a single structure — the rig of natural numbers, the algebras of Taylor and Dirichlet series are part of a single algebraic structure: essentially, a ‘rig algebra’.

But, since most people have never thought about rig algebras, they jump back and forth between the two pieces of the rig algebra of $\mathbb{N}$ with the help of the Mellin transform and its inverse.

Unfortunately this brings analysis into what is at its core a simple algebraic idea which would make sense with any field (e.g. the $p$-adics) replacing $\mathbb{C}$.

I’m overstating my case here, because at certain points the analytic aspects of the relation between modular forms and $L$-functions do become important — e.g. when we gain information about algebraic gadgets from poles and zeroes of $L$-functions that we can cook up from them. At least, that’s what everyone keeps saying.

But, I still find it amusing that people seem loathe to simply define a transform by saying it sends

$\sum_{n \ge 1} a_n z^n$

to

$\sum_{n \ge 1} a_n n^z$

and worry later about whether there’s an integral formula for this transform. After all, as soon as you get integrals into the game you start having to worry about when the integrals converge. I’d prefer to worry about that only when it becomes crucial.

(By the way, in the stuff above, a monoid is ‘sufficiently nice’ if for every element $x$ there are finitely many pairs $y,z$ with $y z = x$. We could avoid this issue by working with the monoid algebra $\mathbb{C}[M]$, which you can think of as consisting of functions on $M$ that vanish except at finitely many points. However, the Dirichlet series appearing in number theory usually have infinitely many nonzero terms, so I’m not taking that approach.)

Posted by: John Baez on February 13, 2008 6:56 PM | Permalink | Reply to this

### Re: Modular Forms

Did’nt I see on this blog some days ago a now vanished question on a congruence of the Ramanujan tau numbers modulo 691 and how that relates to Galois representations? A cold prevented me to answer immediately, now here how I remember that and some links:

That congruence is a consequence of the Ramanujan conjecture, which was finally proved by Deligne. This conjecture is about estimating the coefficients of the Ramanujan function, a weight 12 modular form. If one Mellin-transforms that function into an L-series, these estimates of the coefficients look like similar estimates of coefficients in L-functions of the Weil conjecture. The later come from Galois representations on cohomology groups of varieties, so when one had no suitable verieties at hand, one looked for suitable else Galois representations. Eichler and Shimura were in part successfull with the Galois repr.s coming from the torsion points of the jacobians of modular curves. Then Deligne reduced the Ramanujan conjecture to the Weil conjecture by finding as suitable variety the “universal elliptic curve” above a modular curve. Like a usual ell. curve has torsion points, these universal curve have something like that (e.g. a representing scheme for the functor to the torsion points of each fiber). The first cohomology group of that thing with a sheaf of suitable weight finally solves the problem, and a Kuga-Sato variety (= a product of universal elliptic curves) connects it to the Weil conjecture. The Weil conjectures were proved by Deligne some years later.

Here and here Serres Bourbaki talks about the Ramanujan numbers and congruences coming from galois representations, here Delignes solution, here Brian Conrads lecture more detailed notes which use (and explain) the later and deeper “Weil II”. Here the very readable review article of Katz on Delignes proof of the Weil conjectures, here Katz lecture notes on “Weil-II” and finally here Katz general and simultaneous solution of congruence questions on coefficients of modular forms.

Posted by: Thomas Riepe on February 18, 2008 8:43 PM | Permalink | Reply to this

### Re: Modular Forms

” Here the very readable review article of Katz on Delignes proof of the Weil conjectures…”

“This paper makes accessible to non-specialists the principal ideas in P. Deligne’s proof of the following fundamental result:

The author begins with an historical and motivational survey of this Riemann hypothesis, explaining the connection with formulae and inequalities for the number of solutions of equations over finite fields, the ideas behind the early proofs of special cases, the historical role of the Riemann hypothesis and the other Weil conjectures in the development of algebraic geometry, and the heuristic argument for the conjectures based on classical (characteristic 0) phenomena.

Concerning Deligne’s proof itself, the author spotlights two ingredients—(1) the monodromy of Lefschetz pencils, and (2) modular forms and Rankin’s method for studying the Ramanujan function and describes their fundamental roles in the proof. Deligne’s proof is then outlined in more detail in the special case of odd-dimensional non-singular hypersurfaces. Here monodromy, l-adic cohomology, Rankin’s method and L-series occupy a central place. The author discusses applications to the Ramanujan-Petersson conjecture, estimation of exponential sums in several variables, and the “hard” Lefschetz theorem. He concludes with a discussion
of open questions. The paper includes a lengthy bibliography.”

Posted by: Thomas Riepe on February 19, 2008 5:37 PM | Permalink | Reply to this

### Re: Modular Forms

Aha, a discussion about the modular group!
See if I can join in.

Here are some of the things I learned lately.

First, about the relation between the quotients of gamma\n and other well known groups, such the cube, tetrahedron, Klein quartic. These groups are all automorphism groups of a
tessellation of triangles. I’ve displayed many of these tilings on my web page:
My page on geometry

If you compare at the triangles of let’s say the cube (or the octahedron) and that gamma\4, you see firstly that the number of triangles is the same. The corners are (pi/3, pi/2, pi/4) versus (pi/3, pi/2, pi/inf) respectively. So there is no conformal mapping between the tilings. However, there *is* a conformal mapping between gamma\4 and the “hyperbolic octahedron”, such as displayed here:
Hyperbolic Octahedron.

For each of the gamma\n tilings, you can make a “hyperbolic” figure that has a “cusp” at the centre of each p-gon. Now, you can warp this into an ordinary polyhedron by slowly
increasing the angles the cusp, (We are doing Ricci flow now!) until the intrinsic curvature is zero. Seems like a nice computer graphics project.

Second, a bit on modular forms. There is one easy trick to make an automorphic function of a tiling generated by Mobius transforms. Just take any function f(z), and apply all
elements of the automorphism group to it. Then, create a symmetric expression containing all of them, for example:

FF(z) = f(z1)+f(z2)+f(z3)+…

Now it is easy to see that that FF(z) is automorphic, because an element of the automorphism group will just create a rearrangement of the terms. But we need to be careful to avoid “trivial” cases: often, the terms cancel to produce a boring case, such as a constant. (Which is symmetric, but..)

For example, the 4 element group of Mobius transforms {z, -z, 1/z, -1/z}, with f(z) = …
But wait. There is an even nicer way.

Observe that these automorphic functions will have poles and zero’s on the vertices of the tiles. This is because in a region near a p-fold vertex, each value will exist p times. So there must be a zero of order p, or a pole of order p there. So just look at a tiling, the (2,2,2) case on my web page. Put zero’s and poles on the vertices in some systematic way. In this case, we could put 4-fold poles on z=1,-1,i,-i, and 2-fold zero’s on 0 and inf. We get : FF(z) = z^2/(z^4-1).

Now to this trick to the modular group. There is a nice picture of the j-invariant on Wikipedia:

j invariant

You can see by looking at the graphics that there are 3-fold zero’s on the pi/3 vertices, and inf-fold poles on the pi/inf vertices. These poles are all on the real axis. Knowing the poles and zero’s is sufficient to compute the function value at each point. So we have the j-invariant, a modular function.
Modular *forms* are not allowed to have infinite order poles on the real axis. So replace them by poles of order k, to get a modular form of weight k. This goes at the cost of full invariance. The invariance is replace by FF(z) = (cz+d)^k FF(z). (Something I need to understand better)
A picture for k=4 is at
Eisenstein weight 4.

Another point:
I read that a tessellation of the complex plane by (pi/p, pi/q, pi/r) triangles is related to the generalised Fermat curve u^p + v^q + w^r = 0. This I would like to understand.

Posted by: Gerard Westendorp on March 7, 2008 10:27 AM | Permalink | Reply to this

### Re: Modular Forms

Something I should mention:H. A. Verrill has a nice online Java applet that computes fundamental domains of Γ, Γ0 and Γ1. Using this applet, I made a table for p = 1 to p = 11. n is the order of the group, c is the number of cusps, and g is the genus.

$\begin{array}{cccccccccc}& \Gamma & & & {\Gamma }_{0}& & & {\Gamma }_{1}& & \\ p& n& c& g& n& c& g& n& c& g\\ 1& 1& 1& 0& 1& 1& 0& 1& 1& 0\\ 2& 6& 3& 0& 3& 2& 0& 3& 2& 0\\ 3& 12& 4& 0& 4& 2& 0& 4& 2& 0\\ 4& 24& 6& 0& 6& 3& 0& 6& 3& 0\\ 5& 60& 12& 0& 6& 2& 0& 12& 4& 0\\ 6& 72& 12& 1& 12& 4& 0& 12& 4& 0\\ 7& 168& 24& 3& 8& 2& 0& 24& 6& 0\\ 8& 192& 24& 5& 12& 4& 0& 24& 6& 0\\ 9& 324& 36& 10& 12& 4& 0& 36& 6& 0\\ 10& 360& 36& 13& 18& 4& 0& 36& 8& 0\\ 11& 660& 60& 26& 12& 2& 1& 60& 10& 1\end{array}$

The Gamma domains are (p,3) tilings with c tiles, so

n=c*p

and

g=1-n(1+p/3-p/2)/2

But for Gamma1 and Gamma0 these 2 observations are not valid, so they are not such nice tilings. I havn’t yet figured out what they are, but one clue seems to be that the

order on Gamma1 is just the number of cusps in Gamma.

Posted by: Gerard Westendorp on March 9, 2008 12:44 PM | Permalink | Reply to this

### Re: Modular Forms

Cool! I’m in the throes of painfully giving birth to a rather large paper, but when it’s out I’ll return to discussing this stuff.

Actually, one of my few forms of recreation the last few weeks has been talking about modular curves with Jim Dolan, and we’ve built up enough material for a fun This Week’s Finds. Verrill’s applet will fit in there nicely!

One of the neat things is that we can take a lot of the tilings you’ve drawn and convert them into child’s drawings, which express your Riemann surfaces as branched covers of the Riemann sphere. I bet you can see how it works. One nice thing about children’s drawings is that they convey the information purely topologically and quite efficiently. There are also important connections to Galois theory.

Posted by: John Baez on March 10, 2008 3:52 AM | Permalink | Reply to this
Read the post This Week's Finds in Mathematical Physics (Week 261)
Weblog: The n-Category Café
Excerpt: Learn about the Engraved Hourglass Nebula... and read an ode to the number 3.
Tracked: March 20, 2008 6:29 PM

### Re: Modular Forms

Here a link to interesting course notes.

Posted by: Thomas Riepe on April 5, 2008 5:00 PM | Permalink | Reply to this

### Re: Modular Forms I want to discuss in more detail the spaces $X(N)$, $X_1(N)$ and $X_0(N)$, that is, the quotients of the upper half plane $H$ by $\Gamma(N)$, $\Gamma_1(N)$ and $\Gamma_0(N)$ respectively (with cusps filled in). I also want to talk about the actions of the Hecke operators $<d>$ and $T_p$ on modular forms—or at least the corresponding actions on modular curves. Alas, I don’t think I’m going to get there in this post, but I can at least make a start.

But before I talk about all that, I want to talk about Farey sequences!

I was introduced to Farey sequences by a maths teacher at school, when I was about 13 or so, one afternoon when I’d done all my work and was feeling bored. They’re a simple and cute idea, and I found them kind of fascinating, but, not being much of a number theory person, I never looked into them all that deeply. But one stumbles across these things in the oddest places.

What is a Farey sequence? It’s very simple. Take all the reduced fractions from 0 to 1 (inclusive) with denominators no larger than some maximum $n$. Now arrange them in order of increasing value. And that’s a Farey sequence. Thus Farey sequences include

$\frac{0}{1},\frac{1}{1}$

$\frac{0}{1},\frac{1}{2},\frac{1}{1}$

$\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1}$

and so on. Here’s the ninth Farey sequence:

$\frac{0}{1},\frac{1}{9},\frac{1}{8},\frac{1}{7}, \frac{1}{6},\frac{1}{5},\frac{2}{9},\frac{1}{4}, \frac{2}{7},\frac{1}{3},\frac{3}{8},\frac{2}{5}, \frac{3}{7},\frac{4}{9},\frac{1}{2},\frac{5}{9}, \frac{4}{7},\frac{3}{5},\frac{5}{8},\frac{2}{3}, \frac{5}{7},\frac{3}{4},\frac{7}{9},\frac{4}{5}, \frac{5}{6},\frac{6}{7},\frac{7}{8},\frac{8}{9}, \frac{1}{1}$

There are some curious facts about Farey sequences. Here’s one: if $\frac{a}{b}$ and $\frac{c}{d}$ appear adjacent to one another in some Farey sequence, then $bc-ad=1$.

Here’s another: if $\frac{a}{b}$ and $\frac{c}{d}$ appear adjacent to one another in some Farey sequence, then the next fraction which gets inserted between them, as you work your way through the successive Farey sequences, is the fraction $\frac{a+c}{b+d}$. This funny combination is called the mediant of $\frac{a}{b}$ and $\frac{c}{d}$, since it always lies between them. By repeatedly taking mediants, we build up all Farey sequences, and hence get each rational number from $0$ to $1$ exactly once.

We can indicate the way that the mediant derives from two parent fractions by drawing lines from the parents to their mediant:

Farey Sequence as tree 0/1 1/1 1/2 1/3 2/3 1/4 3/4 1/5 4/5 2/5 3/5
A Farey sequence in tree form

Additionally, one of the parents must always derive from the other—it must have been inserted as a mediant itself at some earlier time. So if a fraction is connected to its parents by a pair of lines, the parents must also be connected to each other. So these lines actually make up a bunch of triangles, except at the top of the picture where everything starts. Taking the three numbers at the vertices of a triangle, any one of them can be derived from the other two either as their mediant, $\frac{a+b}{c+d}$, or as what we might call their ‘mediant difference’, $\left\vert\frac{a-b}{c-d}\right\vert$. In fact every pair of numbers joined by a line belongs to two triangles (one on either side of the line joining them), and one of the triangles will contain their mediant as its third point, while the other will contain their mediant difference instead!

So, can the Farey tree be extended to include fractions outside the interval $[0, 1]$? That’s something we might be wondering, and the answer is Yes! In fact, this would have happened already if we hadn’t foolishly forgotten something: we’ve included fractions with denominators of $1$, $2$, $3$, etc, but we’ve forgotten to include fractions with denominator $0$! How could we have been so careless? Let us remedy this oversight at once!

Including the fraction $\frac{1}{0}$ right at the top of the tree, it turns out that we can actually derive $\frac{1}{1}$ from this together with $\frac{0}{1}$, as their mediant, and then we can get $\frac{2}{1}$ from $\frac{1}{0}$ and $\frac{1}{1}$, and so forth, and by building up mediants, we can get all non-negative rational numbers that way. The whole tree is called the Stern-Brocot tree.

Stern-Brocot tree 1/0 0/1 1/1 2/1 1/2 3/2 1/3 2/3 4/3 5/3 1/4 3/4 5/4 7/4 1/5 4/5 2/5 3/5 6/5 9/5 7/5 8/5
A piece of Stern-Brocot Tree

Once we’ve made these extensions, and added an edge connecting $\frac{1}{0}$ and $\frac{0}{1}$, we see that every fraction in the tree, including the ones right at the top, belongs to a triangle—in fact, it belongs to an infinite number of triangles! We can just keep forming new mediants from it with its new neighbours forever. However, an infinite number of triangles hanging off every vertex is rather unmanageable. Is there some way we can get a toy version, with only a finite number of triangles instead? For instance, suppose we mod out the fractions by some natural number $N$? That is, we mod out both numerator and denominator by $N$, taking care not to forget that $\frac{a}{b}=\frac{-a}{-b}$? Then many different $\mathbb{Z}$-fractions will map to the same $\mathbb{Z}/(N)$-fraction, and if we identify the corresponding points, and also, of course, identify the corresponding edges connecting corresponding points, then we will get something finite.

For instance, suppose we take the above tree mod $3$ (reducing the fractions where necessary to their lowest form, and bearing in mind that $1=-2$, so $\frac{1}{2}=\frac{-1}{-2}=\frac{2}{1}$). We get this:

Stern-Brocot Mod 3 1/0 0/1 1/1 2/1 2/1 0/1 1/0 1/0 1/0 1/0 1/1 0/1 2/1 1/1 2/1 2/1 1/1 0/1 0/1 0/1 2/1 1/1
A piece of Stern-Brocot Tree Mod 3

Or reducing mod $4$ (and bearing in mind that $-1=3$ and $-2=2$, we get this:

Stern-Brocot Mod 4 1/0 0/1 1/1 2/1 1/2 1/2 3/1 2/1 0/1 2/1 1/0 1/0 1/0 1/0 1/1 0/1 2/1 3/1 2/1 3/1 3/1 0/1
A piece of Stern-Brocot Tree Mod 4

The rule about mediants and mediant differences still applies, but sometimes, with all the swapping of signs, we’ve switched around which fraction relates to which how!

Anyway, with these trees in hand, all we need to do is identify the vertices and edges correctly, and we’ll get something interesting and finite. In fact, while we’re about it, why not fill in those triangles? Then we’ll get a compact surface! It will be triangulated, and in fact each vertex will be the apex of just $N$ triangles (this is completely obvious in the case of the vertex $\frac{1}{0}$, but it applies to all vertices). Or, dually, every vertex will stand at the centre of an $N$-gon, while each of the triangular faces gives rise, dually, to a vertex at which $3$ $N$-gons meet.

Wait a minute …

That sounds familiar …

Could it be … ?

Yes it is! The compact surface we have just constructed is none other than $X(N)$, the quotient of $H$ by $\Gamma(N)$. Quel surprise! The fractions mod $N$ correspond to the cusps, and we can make the dual tiling completely regular if we pick the right constant-curvature metric (which will need positive curvature for $N<6$, negative curvature for $N>6$, and will have to be flat for $N=6$).

In fact, this also works for the full Stern-Brocot tree, without any modding out, with fractions over all integers. We can see this either arithmetically or geometrically.

Arithmetically, replace the fraction $\frac{a}{b}$ by the element $\left(\array{a\\b}\right)$ in $\mathbb{Z}^2$. Of course, $SL(2, \mathbb{Z})$ acts on this by ordinary matrix multiplication; indeed, $PSL(2, \mathbb{Z})$ acts on it too, more nicely, since switching the signs of both numerator and denominator doesn’t affect the value of the fraction.

In particular, if we pick an element of $PSL(2, \mathbb{Z})$ such as $\left(\array{a&b\\c&d}\right)$, then those two starting fractions at the top of the tree, $\frac{0}{1}$ and $\frac{1}{0}$, will get sent, under the action of that group element, to a pair of fractions $\frac{a}{c}$ and $\frac{b}{d}$ such that $ad-bc=-1$. (That minus sign is a result of listing the fractions the ‘wrong’ way around). This same relation holds true of any two adjacent elements in a Farey sequence, and since some member of $PSL(2, \mathbb{Z})$ is available to send $\frac{0}{1}$ and $\frac{1}{0}$ to any adjacent pair, we get all Farey sequences in this way—over all integers, not just in the interval $[0, 1]$.

Now, $\frac{0}{1}$ and $\frac{1}{0}$ are joined to each other by an edge, and, being adjacent elements of a Farey sequence, so will their images be. In addition, each of $\frac{0}{1}$ and $\frac{1}{0}$ is also joined by an edge to $\frac{1}{1}$, their mediant. But the mediant of fractions corresponds simply to the sum in $\mathbb{Z}^2$ (considered as a $\mathbb{Z}$-module). And obviously the sum relation is preserved by $PSL(2, \mathbb{Z})$. So if $\frac{0}{1}$ gets sent to $\frac{a}{c}$ and $\frac{1}{0}$ gets sent to $\frac{b}{d}$, then $\frac{1}{1}$ gets sent to the mediant of $\frac{a}{b}$ and $\frac{c}{d}$, meaning that the other two sides of that top triangle $\frac{0}{1}$-$\frac{1}{0}$-$\frac{1}{1}$ are also preserved—sent to sides of a triangle in the image.

So triangles are sent to triangles. Moreover, since the relation of being the mediant (or sum) of the other two vertices—as opposed to their mediant difference (or difference)—uniquely picks out one vertex of a given triangle, and since the sign of the determinant—or, more simply, which of the parent vertices is larger—uniquely picks an orientation of the triangle, each triangle is sent to another triangle in only one way. So $PSL(2, \mathbb{Z})$ acts freely on the set of triangles, and of course preserves their relations.

Geometrically, as we can see from Don Hatch’s nice Hyperbolic Tesselations web page, the tilings of the hyperbolic plane by regular $n$-gons, $3$ meeting at each vertex, have a limiting case with a tiling by what one might call regular $\mathbb{Z}$-gons, or regular $\infty$-gons, with an infinite number of sides! They appear in the website as {infinity, 3}. Dually, we get a tiling by triangles, an infinite number of which meet at the ‘centre’ of each $\mathbb{Z}$-gon. These triangles correspond precisely to the triangles in the Stern-Brocot tree. Rather confusingly, however, in the Poincaré disk model, the ‘centre’ of a $\mathbb{Z}$-gon appears at the edge of the disk—that’s the sort of thing that happens when you have an infinite number of sides and are trying to show them all in a bounded region.

If we compare the Poincaré disk model of the hyperbolic plane with the ‘upper half-plane’ model, we see that the real line of the latter goes to the boundary circle of the former. The $\mathbb{Z}$-gons, drawn in the Poincaré disk, are all tangent to the boundary circle, each with its own distinct point of tangency, and these points correspond precisely to the rational numbers (together with $\infty$). In turn, these rational numbers are the ones in the Stern-Brocot tree, i.e. at the vertices of the triangles, which can also be seen as tiling the hyperbolic plane, dually to the $\mathbb{Z}$-gons. In this way, we also get to label each $\mathbb{Z}$-gon by a unique rational number, and this is inherited when we go over to the version mod $N$—fractions also being reduced mod $N$.

Given all this, one way to try and understand the way that geometry and arithmetic interact in $X(N)$ is by working our way down the Stern-Brocot tree mod $N$, which will correspond to working our way outward from one of the $N$-gons in $X(N)$. What we’re going to see is a series of concentric circles of fractions with successively incrementing denominators.

But that, I think, is something for another post.

Posted by: Tim Silverman on May 29, 2008 9:33 PM | Permalink | Reply to this

### Re: Modular Forms

A nice way to display Farey sequences is on the “Dedekind tesselation”, such as here on Lieven le Bruyn’s weblog. The intersections of the tesselation with the real axis are precisely the rational numbers.
If you draw in the Stern-Brocot Tree, its edges will coincide with the perfect triangles inscribed in the tesselation. There is a nice section on this in the book “Indra’s Pearls”.

I like your idea of relating Farey sequences modulo N with tessellations.
I made one for the N=4 case: The arrows correspond to Farey additions: If you are in a triangle enclosed by blue arrows, then there is always one vertex (C) where 2 arrows point to, from (A) and (B) respecively. The corresponding Farey addition is C= Farey(A,B):

Gerard

Posted by: Gerard Westendorp on June 8, 2008 10:49 PM | Permalink | Reply to this

### Re: Modular Forms

I got a couple of new pictures.

First, the relation between a tile of the modular group and the rational number of its vertex on the real axis: This may help to understand how the stern brocot tree relates to the modular group.

Second, some “cusped polyhedra” like these: The tiles on these polyhedra are conformal images of tiles on the modular tiling. That is, they are (inf,3,2) hyperbolic triangles.
The shape is also a C1 embedding, except at the cusps.

Third, an animation showing how a slice of Gamma(7) warps into a slice of Klein’s quartic. This shows how the rational numbers modulo 7 travel to the centres of the heptagons as they warp into the Quartic.

Gerard

Posted by: Gerard Westendorp on December 7, 2008 8:52 PM | Permalink | Reply to this

### Re: Modular Forms

I interpreted the picture of the C1 embedding as having the cusps at infty.
That brought to mind the work of Igor Rivin

36. arXiv:math/0005234 [ps, pdf, other]
Title: Intrinsic geometry of convex ideal polyhedra in hyperbolic 3-space
Authors: Igor Rivin
Comments: 13 pages; appeared in proceedings of the 1992 nordic math congress; very difficult to find
Journal-ref: Analysis, algebra, and computers in mathematical research (Lule\aa, 1992), 275–291, Lecture Notes in Pure and Appl. Math., 156, Dekker, New York, 1994

and his more accessible paper in the Annals

whose classification turned out to be of the duals to the special polytopes Zwiebach used
in his development of the Linfty-algebra for closed string field theory.

arXiv:hep-th/9305026 [ps, pdf, other]
Title: Closed String Field Theory: An Introduction
Authors: B. Zwiebach
Subjects: High Energy Physics - Theory (hep-th)
14. arXiv:hep-th/9206084 [ps, pdf, other]
Title: Closed String Field Theory: Quantum Action and the BV Master Equation
Authors: Barton Zwiebach
Comments: 115 pages, 5 figures (not included). IASSNS-HEP-92/41
Journal-ref: Nucl.Phys. B390 (1993) 33-152
Subjects: High Energy Physics - Theory (hep-th)

Posted by: jim stasheff on December 9, 2008 6:34 PM | Permalink | Reply to this

### Re: Modular Forms

Here is a modular cabinet.

Posted by: Thomas on December 8, 2008 6:02 PM | Permalink | Reply to this

### Re: Modular Forms

Hi,

I’m just browsing a lecture by Katz on p-adic properties of modular schemes (Antwerpen summer school on modular functions, 1972, online on Katz’ website) and am startled about a theorem of Igusa on a surjectivity of a monodromy action around supersingular ell. curves on p. 149 -150. In case someone knows where to read more about that and the stories behind that, I’d be happy if links etc. were posted.

Posted by: Thomas on January 24, 2009 11:12 AM | Permalink | Reply to this

### Re: Modular Forms

e.g. what precisely means Katz with a “physical appearance” of a scheme? What are the analoga of “supersingularity” in other cases?

Posted by: Thomas on February 7, 2009 7:57 AM | Permalink | Reply to this

### Re: Modular Forms

Gunther Cornellisen’s interesting slides on Drinfeld modular curves compare classical modular forms with Drinfeld’s ones and relate the later to Mumford curves resp. orbifolds.

Posted by: 도마 on February 18, 2009 11:04 AM | Permalink | Reply to this
Read the post This Week's Finds in Mathematical Physics (Week 274)
Weblog: The n-Category Café
Excerpt: Learn about infinite-dimensional 2-vector spaces and representations of 2-groups on these.
Tracked: March 12, 2009 9:17 PM

Post a New Comment