## December 14, 2010

### Pictures of Modular Curves (VII)

#### Posted by Guest

guest post by Tim Silverman

Welcome once more to our exploration of the structure of modular curves. So far, we’ve looked at the curves $X(N)$—that is, quotients of the complex upper half plane by the groups $\Gamma(N)$ for various $N$. We’ve seen that we can tile them with regular $N$-gons, three $N$-gons meeting at each vertex, and each labelled with a fraction (reduced mod $N$) at its centre; and that we can break this tiling up into several identical pieces, corresponding with the units in $\mathbb{Z}_N$ mod $\{1, -1\}$, which I’ve called the “projective” group of units. A unit $u$ corresponds to a matrix $\left(\array{v&0\\0&u}\right)$ (mod $N$) where $u v=1$, which acts on the fractions mod $N$ that lie at the centres of the $N$-gons, thus permuting the $N$-gons while preserving the tiling, and, in particular, carrying the piece centred on $\frac{1}{0}$ to the piece centred on $\frac{v}{0}$. These units, acting by multiplication, may be considered as the rescalings of the “projective line” over $\mathbb{Z}_N$.

The Action by Translations

This time, I’m going to consider instead the additive or translation action of the group of matrices which have the form $\left(\array{1&b\\0&1}\right)$ (with entries in $\mathbb{Z}_N$) for arbitrary $b$, sometimes written $\left(\array{1&*\\0&1}\right)$. If we take the lift of this group to the integers (that is, the integer-valued matrices of the form $\left(\array{1&b\\0&1}\right)$ mod $N$ lying in $PSL(2,\mathbb{Z})$) then we get the subgroup known as $\Gamma_1(N)$, and taking the quotient of the upper halfplane by this group gives a space known as $X_1(N)$. So one way of looking at what we are about to do is that, in order to get to $X_1(N)$, we are taking a quotient by $\Gamma_1(N)$ in two stages: first taking a quotient by $\Gamma(N)$ (which does the “mod $N$” part), then taking a quotient by the mod-$N$ image of $\Gamma_1(N)$ (which does the remaining “additive” part of the operation).

Now, over the integers, $\left(\array{1&b\\0&1}\right)$ just adds $b$, and, not surprisingly, the mod $N$ version adds $b$ mod $N$. And this, in turn, just amounts to a rotation, by $\frac{1}{N}$ of a full turn, about the central face $\frac{1}{0}$.

For example, we can slice the spherical dodecahedron into five sectors:

N = 5: sectors

The division is perhaps easier to see in this view:

N = 5: sectors, from above

Under the action of $\left(\array{1&1\\0&1}\right)$, which rotates by $\frac{1}{5}$ of a circle, we identify these sectors with each other, so roll up one sector into a sort of cone.

To be clearer: one fifth (one sector) of the top pentagon ($\frac{1}{0}$) rolls up into a cone, like this:

Spherical Segment: N = 5, top cone

The rather funny-shaped curve that makes up the bottom of the cone is a single edge of the original pentagon, wrapped round on itself, with the pointy tip corresponding to the vertices at the two ends of the original edge, identified together.

And then the $5$ pentagons with denominator $1$ get identified with each other, and roll up like this:

Spherical Segment: N = 5, denominator 1

The curve at the top is one edge of the pentagon, curled back on itself; the vertical line is the seam where the edges on either side of that get sewn together; and then the lip at the bottom is made up of the remaining two edges. Putting these together, we get the following:

Spherical Segment: N = 5, denominator 0 and 1

Then the bottom half of the sphere—the denominator $2$ pentagon, and the segment coming from the pentagon with fraction $\frac{2}{0}$—makes up another piece with the same shape as the above, but upside down; so that finally, putting them all together, we get this:

Spherical Segment: N = 5

And that’s $X_1(5)$.

We can do the same thing for $N=4$:

Spherical Segment: N = 4

$Next, N=3$:

Spherical Segment: N = 3

and $N=2$:

Spherical Segment: N = 2

And those are the curves $X_1(N)$! (At least, for $N<6$.)

Elliptic Points on $X_1(N)$

Since these conical figures have pointy bits at the ends, which one might suspect are points that are special in some way, I should, before moving on, say a little about elliptic points.

Elliptic points are points where, roughly, going around the point by some fraction of $360^\circ$ brings you back to where you started. Since we’re working on surfaces with conformal structures, with a good notion of angle, this is all well-defined and a big deal. Consider, for instance, the standard fundamental domain of $\Gamma$, i.e. of $PSL(2,\mathbb{Z})$. (Below, $\omega$ and $\omega^2$ are the complex cube roots of $1$):

Fundamental domain $\omega$ $i$ $-\omega^2$ $\array{&\uparrow\\to&\infty}$
Fundamental domain of Γ

The left and right edges are identified with each other, and in addition, the circular bottom segment is folded in half about its midpoint at $i$, so that the left half is identified with the right half.

The consequences are three:

Firstly, going halfway around $i$ (along the semicircle shown) brings you back to where you started: that is, after going around an angle of only $180^{\circ}$. Hence there is an elliptic point at $i$ of period $2$.

Secondly, starting vertically above the point at $\omega$ and going clockwise $60^\circ$ takes you to a point on the left part of the circular arc, which is identical to the mirror-opposite point on the right of the circular arc, and continuing clockwise from there a further $60^\circ$ takes you to a point above $-\omega^2$ which is identical to where you started: in short, a complete circuit of the point in only $120^\circ$ degrees. So there is an elliptic point at $\omega$ of period $3$.

Finally, the two vertical edges form a long, tapering tube going off to $\infty$, and tending toward meeting at an angle of $0^\circ$, so going around $\infty$ by $0^\circ$ brings you back to where you started, which makes it kind of like an elliptic point of period $\infty$. However, when we compactify by adding points (“cusps”) at the ends of these tubes (a standard procedure), the cusps end up as quite ordinary points in $X(N)$—not elliptic points. So we can forget about them. We don’t do this for other elliptic points, so we can’t forget about them: we need to think about them. And we are thinking about them—right now!

Now, in the surfaces $X(N)$ for $N>1$, multiple copies of the fundamental domain get packed together ($N$ per $N$-gon). There is no longer any folding over, so (once the cusps have been added) there are no elliptic points at all on these surfaces.

We can take a look at these fundamental domains on the sphere for, e.g. $N=5$:

Dodecahedron, triangulated: N = 5

Or, making it opaque for clarity:

Dodecahedron, triangulated (front only): N = 5

Each edge of an $N$-gon (the thick lines) is the base of a fundamental triangle, with a lift of $i$ at its midpoint and lifts of $\omega$ at its endpoints. The centres of the $N$-gons are the cusps. We can see that, since two triangles are based on each edge of an $N$-gon, and six triangles around each vertex, both the period-$2$ and the period-$3$ elliptic points have been unfolded and do not lift to elliptic points on these surfaces.

What about the surfaces $X_1(N)$? These have pointy bits at their tops and bottoms, which look as though they might be elliptic points. However, with two exceptions, these are actually cusps (they come from the centre of an $N$-gon), and since the compactification by cusps occurs after we take the quotient, they go. The pointy appearance is actually a bit misleading.

The exceptions are the cases $N=2$ and $N=3$. There, the top points come from cusps, but the bottom points don’t.

Let’s take a look at the triangulation of these cases by lifts of the fundamental domain:

Tetrahedron, triangulated: N = 3
Trihedron, triangulated: N = 2

Or opaque (and viewed at a different angle) for clarity:

Tetrahedron, triangulated, opaque: N = 3
Trihedron, triangulated, opaque: N = 2

When forming the bottom part of $X_1(3)$:

Spherical Segment (bottom piece): N = 3

we take one of the triangular faces of the tetrahedron and fold it so that two of its edges are sewn together: that is the seam you see running up from the bottom. That means that the point at the bottom occurs at a vertex of the triangle, and only covers $120^\circ$. So it is an elliptic point of period $3$. (It is at a vertex, so it is a lift of the elliptic point of period $3$ at $\omega$ in the fundamental domain.)

Similarly, for $N=2$:

Spherical Segment (bottom piece): N = 2

The bottom part is formed of a single bigon. The bigon only has two edges: one of them forms the loop at the top, and the other is actually folded back on itself to form the seam you see. Thus the conical tip at the bottom is at the midpoint of an edge, so is an elliptic point of period $2$, lifting the elliptic point of period $2$ at $i$ in the fundamental domain.

These are the only elliptic points in the curves $X_1(N)$ (except for the trivial $N=1$ case, which is just another name for the same curve as $X(1)$).

Well, there hasn’t been much content this time, but there’ve been a lot of pictures, so I’ll stop here. Next time, I’ll talk about $X_1(N)$ for $N>5$, and in the posts after that I’ll go on to talk about $\Gamma_0(N)$ and $X_0(N)$. That will complete this set of posts about what modular curves look like. After that, I’ll move on to Hecke operators, but I’m not sure how I’m going to organise that, so it may be a long time before it appears. Though hopefully not as long as it took to get this series prepared.

Posted at December 14, 2010 12:21 PM UTC

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

Tim, I’m afraid I again haven’t got anything intelligent to say; unfortunately I haven’t kept up. But I wanted to ask whether you’d given any thought to collecting the whole story together, when you’re finished, and publishing it somehow.

Of course it’s great that you’re publishing it here. But it seems pretty clear that your story in eight parts belongs together, and I can see some advantages to it being available in print.

Or maybe you don’t want to think about this, because you’re enjoying coasting along, writing posts whenever you feel like it, not feeling the pressure to produce a grand masterpiece. In which case, just ignore this.

Posted by: Tom Leinster on December 14, 2010 2:52 PM | Permalink | Reply to this

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

John suggested converting it into a paper that I could put on the arxiv. This seems reasonable, though I haven’t yet given much thought to how exactly I’m going to do this. I have given no thought to anything beyond that.

By the way, that last paragraph is a bit misleading. When I wrote it, I thought I was close to the limit of what I could display in pretty pictures, but I subsequently realised that there was a lot more that I could show. So although it’s correct about what I’m going to do next, it’s going to take more than one post to do it! I guess I should have changed the paragraph, but when I read through it again, I ended up “reinterpreting” it in my own mind in the light of my new knowledge, so it seemed to describe the new situation perfectly …

Posted by: Tim Silverman on December 14, 2010 4:26 PM | Permalink | Reply to this

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

If you want to change the post and don’t have a login, just drop me a mail and I’ll do it.

Posted by: Tom Leinster on December 14, 2010 4:53 PM | Permalink | Reply to this

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

Thanks, Tom. I tweaked it through the guest login.

Posted by: Tim Silverman on December 14, 2010 6:23 PM | Permalink | Reply to this

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

Please don’t ignore that suggestion completely, take you time, but hopefully it will eventually become a book on paper in living color.

Btw, there’s a lot of content in pictures like those.

Posted by: jim stasheff on December 15, 2010 12:45 PM | Permalink | Reply to this

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

Oh, I’m not going to ignore the suggestion. The thing is, I planned this as a series of posts on the n-Category Café, i.e. with a certain format, and certain assumptions about the background of the audience (not necessarily correct assumptions, but it gave me something clear to aim at). And if I was to try and publish it in a different medium, I’d have to rethink all that. Which can be done, of course, but takes time. Also, it was a massive effort just to get it to this stage, so I wanted to get this finished (and see what people thought of it) before worrying about what came next.

But I’m not ignoring the suggestion.

Posted by: Tim Silverman on December 15, 2010 2:58 PM | Permalink | Reply to this

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

Just one naive question: for n= 3 and n=2, why is the bottom `half’ bigger than the top?

Posted by: jim stasheff on December 15, 2010 12:47 PM | Permalink | Reply to this

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

for $n=3$ and $n=2$, why is the bottom ‘half’ bigger than the top?

OK. Well, for $N=3$, we start with $X(3)$, a tetrahedron:

Tetrahedron: N = 3

(I’d normally draw this spherical, but it’s probably clearer what’s going on when it’s drawn like this.)

Then we mod out by rotation through $\frac{1}{3}$ turn about the centre of the top face.

So as far as the top face (representing $\frac{1}{0}$) is concerned, effectively we take one third of it, like this:

As a result of the quotienting, we end up taking just the grey sector and identifying the dotted edges with each other.

On the other hand, the other three faces of the tetrahedron (representing $\frac{0}{1}$, $\frac{1}{1}$, and $\frac{2}{1}$) get identified with each other. So in that case, we boil down from three whole equilateral triangles to one whole equilateral triangle, like this:

And we identify the two dotted edges of that with each other.

So the “top” bit of the quotient surface is $\frac{1}{3}$ of one triangle, whereas the “bottom” bit is a whole triangle. And that’s why it’s bigger.

For $N=2$, the story is the same, except we have bigons and half a turn, instead of triangles and a third of a turn.

I hope that makes things clearer.

Posted by: Tim Silverman on December 15, 2010 1:50 PM | Permalink | Reply to this

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

Very clearly! thanks

Does anyone know why I (just lately) have to input my name and e-address each time I try to reply?

Posted by: jim stasheff on December 16, 2010 1:25 PM | Permalink | Reply to this

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

No, I don’t know, but I’ve had the same problem. I’ll post a query on the TeXnical issues thread.

Posted by: Tom Leinster on December 16, 2010 1:46 PM | Permalink | Reply to this
Read the post Pictures of Modular Curves (VIII)
Weblog: The n-Category Café
Excerpt: We continue modding out by translations, for N greater than 5
Tracked: December 23, 2010 4:58 PM

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