The n-Category Café

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.

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.)

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).

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 §3.8 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 $PSL(2,\mathbb{Z}/7)$.

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.

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.

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.

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.

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.

Here a link to interesting course notes.

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.

See this page for more pictures.

Gerard

Here is a modular cabinet.

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.

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.