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January 7, 2011

Pictures of Modular Curves (IX)

Posted by Guest

guest post by Tim Silverman

Welcome back to our series A Glorious Technicolor Panorama Documentary of Modular Form Country.

We’ve been looking at the curves X 1(N)X_1(N), which result from quotienting the hyperbolic plane (or complex upper half-plane) by the action of the subgroup Γ 1(N)\Gamma_1(N) of PSL(2,)PSL(2, \mathbb{Z}). We do this by first quotienting by Γ(N)\Gamma(N), giving a “mod NN” version of our surface, and then further quotienting out by the additive or “translation” subgroup of the remaining action.

Last time, and the time before, we looked at rolling up sectors of X(N)X(N) into cones, pictorially representing the second quotient process. Today, I want to take advantage of the fact that several of the X 1(N)X_1(N) for small NN have genus 00, to try to represent everything smoothly on the surface of a sphere.

I should warn you that the pictures you’re about to see are to some extent “hand-sketched”. I didn’t really have guides to how to lay out the sizes and proportions of the variously-shaped tiles, unlike in the case of the Platonic solids or other regular tilings, or the conical rolled-up sectors. So there’s no guarantee that these images are even conformally faithful, let alone in some sense canonically “correct”. I’ve done my best to make them look nice and at least be combinatorially correct, but … well … you have been warned. Take them with a pinch of salt. Not actual size. Contents may have settled in transit. Batteries not included. Cape does not really confer ability to fly.

With that out of the way, let’s plunge right in!

X 1(8)X_1(8)

X1(8)

Whoa! That’s a lot of pieces! And a lot of colours! And a lot of shapes!

Let’s go through it bit by bit.

First, there are the the faces corresponding to fractions with denominator 00, that is 10\frac{1}{0} and 30\frac{3}{0}. When we take conical sectors, these each consist of just one sector of a single octagon, made into a conical cap with a single edge looped round to join itself around the base of the cone. Since these have a single edge, and the supposed cones aren’t genuinely conical, I’ve flattened them out into circles. I’m not convinced this is really the right thing to do, but they look pretty:

X1(8): Denominator 0

Next, there is one face with denominator 22. This results from the identification of the fractions 12\frac{1}{2}, 32\frac{3}{2}, 52\frac{5}{2} and 72\frac{7}{2}. I’m using 12\frac{1}{2} to stand for this class. Since there were originally four faces, the single face that results from this identification consists of four copies of the fundamental domain, or, to put matters more plainly, it’s a square:

X1(8): Denominator 2

Next, there is one face with denominator 44. There were originally two fractions with denominator 44, back in X(8)X(8), namely 14\frac{1}{4} and 34\frac{3}{4}, but 14+1=54=54=34\frac{1}{4}+1=\frac{5}{4}=\frac{-5}{-4}=\frac{3}{4}, so these get identified. I use the fraction 14\frac{1}{4} to represent the class. Since there were originally two faces, we get a big yellow bigon (I’m afraid I’ve changed the colour\rightarrownumber mapping from the one I’ve been using up to now, mostly just to make it look prettier):

X1(8): Denominator 4

This leaves two octagons, corresponding to denominators 11 and 33. But since I’ve tried to make the other faces look pretty, these have been distorted into weird, irregular shapes to fit in:

X1(8): Denominator 1
X1(8): Denominator 3

See the holes where the denominator 00 faces snugly fit, and the seam where two edges have been identified.

Now we can put all these faces together to get the shocking concoction we saw at the beginning:

X1(8)

Now lets give a few different values of NN the same treatment.

Other NN

N=7N=7

X1(7)

The threefold symmetry should hopefully be very clear here, as well as the fact that, if we quotient the surface out by it, it gives rise to two elliptic points of period 33—one on each side.

And here’s a view from another angle (and further away).

X1(7) (Front View)

N=9N=9

This one is a little more complex, so I’ll try to get it to pose for the paparazzi from a few angles.

From the side:

X1(9)

And opaque:

X1(9) opaque

We can see clearly the three-fold symmetry as with X 1(7)X_1(7), but this time the triangular panes on each side, containing 13\frac{1}{3} and 23\frac{2}{3}, ensure that quotienting by the threefold symmetry does not produce any elliptic points.

And now from the front, so we can get a better look at the faces with denominator 00 or a unit:

X1(9) (Front View)

And with the front panels removed so we can see inside:

X1(9) (Front View, Panel Removed)

OK, that’s enough from you, X 1(9)X_1(9). You can go inside now.

Because we have a possibly even bigger star coming out …

N=10N=10

X1(10)

Observe that, since 1010 is squarefree, we have the simple phenomenon that the faces with denominator 5=1025=\frac{10}{2} have 22 sides—they’re bigons—and faces with denominator 22 or 44 (i.e. gcd(N,d)=2=105gcd(N, d)=2=\frac{10}{5}) have 55 sides—they are (heart-shaped) pentagons. Faces with denominator 11 or 33 (i.e. gcd(N,d)=1gcd(N, d)=1) have 1010 sides.

Observe also that the relation |adbc|=1\vert a d-b c\vert = 1 is still visible between adjacent faces (though not always perfectly clearly).

The largest NN for which X 1(N)X_1(N) has genus 00 is 1212:

X1(12)

And opaque for clarity:

X1(12) opaque

And more from the side:

X1(12) side

And opaque:

X1(12) side opaque

Those are all the more complicated genus-00 cases, which we haven’t seen before. I’ll also briefly add the cases with N6N\le 6 for the sake of completeness. We have seen these before, but in conical rather than spherical form.

First, N=1N=1 and N=2N=2, side by side (I’ve refrained from painting the blue on the back of the sphere in order to bring out the colours on the front more clearly):

X1(1) and X1(2)

For X 1(1)X_1(1) (which is the same as X(1)X(1)), the whole sphere is a single triangular face, sewn up with a seam which you can see as a line on the surface of the sphere. There is an elliptic point of period 22 at one end of the line, and an elliptic point of period 33 at the other end, the period-33 elliptic point being at the vertex between two of the edges of the triangle, and the period-22 elliptic point being at the midpoint of the third edge, where it is folded in half.

For X 1(2)X_1(2), there are two faces, one a red monogon, and the other a blue bigon which takes up the rest of the sphere. There is an elliptic point of period 22 at the end of the stalk (or seam) you can see, where one edge of the bigon is folded back on itself.

Here are X 1(3)X_1(3) and X 1(4)X_1(4) (again, I’ve left the blue colour off the back of the sphere):

X1(3) and X1(4)

X 1(3)X_1(3), like X 1(2)X_1(2), has a two faces—one a red monogon, and the other a blue triangle which takes up the rest of the sphere. It also has an elliptic point at the end of the stalk, but it is of period 33, and occurs at the vertex where two edges of the blue triangle meet.

X 1(4)X_1(4) has three faces: a red monogon (denominator 00), a green monogon (denominator 22) and a blue square that takes up the rest of the sphere (denominator 11).

X1(5) and X1(5)

Finally, X 1(5)X_1(5) has four faces: red monogons with corresponding to 10\frac{1}{0} and 20\frac{2}{0}, and a blue and a green pentagon with denominators 11 and 22; and X 1(6)X_1(6) also has four faces: a monogon with denominator 00, a blue hexagon with denominator 11 (which takes up most of the sphere), a green triangle with denominator 22, and an orange bigon with denominator 33.

And that is all the NN for which X 1(N)X_1(N) has genus 00.

However, in going all the way from N=1N=1 to N=12N=12—albeit in a funny order—we have skipped over a genus-11 case: N=11N=11. This is also easy to represent. So here is a tiling of the plane corresponding to this torus:


N=14N=14 and N=15N=15 are also genus 11.

Here is N=14N=14.


And here is N=15N=15.


Aren’t they pretty!

And now you’ve seen all the genus 00 and genus 11 cases. I’m not going to show anything of higher genus—I’m too lazy, and I’ve run out of time. So that’s all from X 1(N)X_1(N) folks! Next time, we’ll be looking at X 0(N)X_0(N).

Posted at January 7, 2011 6:28 PM UTC

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3 Comments & 0 Trackbacks

Re: Pictures of Modular Curves (IX)

Tim, if you’re “too lazy”, I don’t know what that makes the rest of us…

Posted by: Tom Leinster on January 7, 2011 10:59 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IX)

Heh! Bear in mind that it took a lot longer to prepare this material than it has done to post it. And you write actual papers containing new results …

Posted by: Tim Silverman on January 8, 2011 11:39 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (IX)

Truly a labor of love!

Posted by: jim stasheff on January 10, 2011 1:52 PM | Permalink | Reply to this

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