### 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 $X_1(N)$, which result from quotienting the hyperbolic plane (or complex upper half-plane) by the action of the subgroup $\Gamma_1(N)$ of $PSL(2, \mathbb{Z})$. We do this by first quotienting by $\Gamma(N)$, giving a “mod $N$” 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 $X(N)$ into cones, pictorially representing the second quotient process. Today, I want to take advantage of the fact that several of the $X_1(N)$ for small $N$ have genus $0$, 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!

**$X_1(8)$**

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 $0$, that is $\frac{1}{0}$ and $\frac{3}{0}$. 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:

Next, there is one face with denominator $2$. This results from the identification of the fractions $\frac{1}{2}$, $\frac{3}{2}$, $\frac{5}{2}$ and $\frac{7}{2}$. I’m using $\frac{1}{2}$ 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:

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

This leaves two octagons, corresponding to denominators $1$ and $3$. But since I’ve tried to make the other faces look pretty, these have been distorted into weird, irregular shapes to fit in:

See the holes where the denominator $0$ 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:

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

**Other $N$**

$N=7$

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 $3$—one on each side.

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

$N=9$

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:

And opaque:

We can see clearly the three-fold symmetry as with $X_1(7)$, but this time the triangular panes on each side, containing $\frac{1}{3}$ and $\frac{2}{3}$, 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 $0$ or a unit:

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

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

Because we have a possibly even bigger star coming out …

$N=10$

Observe that, since $10$ is squarefree, we have the simple phenomenon that the faces with denominator $5=\frac{10}{2}$ have $2$ sides—they’re bigons—and faces with denominator $2$ or $4$ (i.e. $gcd(N, d)=2=\frac{10}{5}$) have $5$ sides—they are (heart-shaped) pentagons. Faces with denominator $1$ or $3$ (i.e. $gcd(N, d)=1$) have $10$ sides.

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

The largest $N$ for which $X_1(N)$ has genus $0$ is $12$:

And opaque for clarity:

And more from the side:

And opaque:

Those are all the more complicated genus-$0$ cases, which we haven’t seen before. I’ll also briefly add the cases with $N\le 6$ for the sake of completeness. We have seen these before, but in conical rather than spherical form.

First, $N=1$ and $N=2$, 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):

For $X_1(1)$ (which is the same as $X(1)$), 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 $2$ at one end of the line, and an elliptic point of period $3$ at the other end, the period-$3$ elliptic point being at the vertex between two of the edges of the triangle, and the period-$2$ elliptic point being at the midpoint of the third edge, where it is folded in half.

For $X_1(2)$, 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 $2$ at the end of the stalk (or seam) you can see, where one edge of the bigon is folded back on itself.

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

$X_1(3)$, like $X_1(2)$, 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 $3$, and occurs at the vertex where two edges of the blue triangle meet.

$X_1(4)$ has three faces: a red monogon (denominator $0$), a green monogon (denominator $2$) and a blue square that takes up the rest of the sphere (denominator $1$).

Finally, $X_1(5)$ has four faces: red monogons with corresponding to $\frac{1}{0}$ and $\frac{2}{0}$, and a blue and a green pentagon with denominators $1$ and $2$; and $X_1(6)$ also has four faces: a monogon with denominator $0$, a blue hexagon with denominator $1$ (which takes up most of the sphere), a green triangle with denominator $2$, and an orange bigon with denominator $3$.

And that is all the $N$ for which $X_1(N)$ has genus $0$.

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

$N=14$ and $N=15$ are also genus $1$.

Here is $N=14$.

And here is $N=15$.

Aren’t they pretty!

And now you’ve seen all the genus $0$ and genus $1$ 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 $X_1(N)$ folks! Next time, we’ll be looking at $X_0(N)$.

## Re: Pictures of Modular Curves (IX)

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