### Pictures of Modular Curves (X)

#### Posted by Guest

*guest post by Tim Silverman*

Welcome to the next part of our series The Big Colouring Book of Modular Curves. The last few times I was looking at $\Gamma_1(N)$ and $X_1(N)$. In this part, I want to move on to $\Gamma_0(N)$ and $X_0(N)$.

**Taking a Quotient by $\Gamma_0(N)$ in Three Easy Steps**

(For the sake of brevity, I shall be using the letter $\Gamma$ as a short name for $PSL(2, \mathbb{Z})$.)

Recall that, in order to get to $X(N)$, we quotient the complex upper half-plane by the action of $\Gamma(N)$, where $\Gamma(N)$ is the subgroup of $\Gamma$ consisting of matrices of the form $\left(\array{1&0\\0&1}\right)$ mod $N$. (To remind you where this definition comes from: if we reduce $\Gamma$ mod $N$ (by reducing the elements of the matrices in it mod $N$), then $\Gamma(N)$ is the kernel of this reduction.)

And to get to $X_1(N)$, we quotient the upper half-plane by the action of $\Gamma_1(N)$, which is the subgroup of $\Gamma$ consisting of matrices of the form $\left(\array{1&*\\0&1}\right)$ mod $N$—where $*$ may be anything—but we can get to it in two stages: first we quotient by $\Gamma(N)$, and then we take the quotient of the resulting curve by the action on it of the quotient group $\Gamma_1(N)/\Gamma(N)$. The latter can be thought of as a version of $\Gamma_1(N)$ which has been reduced mod $N$, and which acts on $X(N)$. Like $\Gamma_1(N)$, it consists of matrices of the form $\left(\array{1&*\\0&1}\right)$ (mod $\{1,-1\}$)—but they are matrices mod $N$.

Finally, to get from the upper half-plane to $X_0(N)$, we quotient by the action of $\Gamma_0(N)$. $\Gamma_0(N)$ is the subgroup of $\Gamma$ consisting of matrices of the form $\left(\array{*&*\\0&*}\right)$ (mod $N$), but we can do this in three steps: first doing the two-step quotient to $X_1(N)$ described above, and then appending an extra stage.

What does this extra stage look like? Well, we want to get from $\Gamma_1(N)$ to $\Gamma_0(N)$, or rather, from the former reduced mod $N$ to the latter reduced mod $N$.

Now, $\Gamma_0(N)$ reduced mod $N$—that is, the group of matrices with entries in $\mathbb{Z}_N$, determinant $1$, and of the form $\left(\array{*&*\\0&*}\right)$ (mod $\{1,-1\}$)—is generated by two subgroups: the “translations” (or “additions”), i.e. matrices of the form $\left(\array{1&*\\0&1}\right)$ and the “rescalings” (or “multiplications”), i.e. matrices of the form $\left(\array{*&0\\0&*}\right)$. So having quotiented by the action of the first, “translation”, subgroup to get down to $X_1(N)$, we can quotient by the action of the second, “rescaling”, subgroup to get right down to $X_0(N)$. And this second subgroup is isomorphic to the projective group of units of $\mathbb{Z}_N$—in fact, it acts on the denominators as multiplication by projective units.

I talked quite a bit about the action of the rescaling group on $X(N)$ a few articles ago. Its action on $X_1(N)$ is a simpler version of that, so I won’t go into it in as much detail, but will just describe it briefly.

Consider $X_1(5)$:

As we’ve seen before, with the full $X(5)$, the only nontrivial action of the rescaling group is to swap the blue and green halves, by rotating this shape $180^\circ$ about a horizontal axis through the midpoints of the edges half way down. Thus if we *quotient* by this action we get something looking like just *one* of the two identical pieces, with its lower lip folded back on itself.

Likewise, with $N=7$ (the next case with a non-trivial projective unit group). The photo and drawing below illustrate $X_1(7)$. The action of the projective group of units is just to rotate it by multiples of one third of a full turn, so we see that quotienting by the action of the group of units gives us just *one* of the three pieces, but with the edge-$2$s (the edges next to the seam) folded back so that they are glued to the edge-$3$s (the edges further along).

And this is generally the case for any $N$: $X_1(N)$ is made of several identical pieces, one for every member of the projective group of units (which permutes them), glued to each other. Quotienting by the action of that group precisely reduces us down to *one* piece, glued to itself.

Here is a prettier picture of $X_1(7)$, also showing the three-fold symmetry.

**Structure of $X_0(N)$: Elliptic Points**

At this point, we are in a position to say something about the elliptic points of $X_0(N)$.

In the case of $N=5$, as we mentioned, it’s clear that the rotation about $180^\circ$ rotates the edge-$2$s of the pentagons about their mid-points, so when quotienting by this action, we fold back these edges against themselves, resulting in two elliptic points of period $2$, one on each side, at the midpoints of the folded edges. (Being at the midpoints of edges, they lift the period-$2$ elliptic point at $i$ in the fundamental domain of $\Gamma$.) We might suspect that something similar would happen whenever the group of projective units is of even order, and for primes this is true: if $N$ is an odd prime, there are two elliptic points of period $2$ if $\frac{N-1}{2}$ is even, and none if it is odd. (With $N$ prime and $\frac{N-1}{2}$ even, there will be, in general, multiple edges passing through the centre of rotation—e.g. three on each side for $N=13$, and generally $\frac{N-1}{4}$ on each side for $X_1(N)$. However, these all get identified with each other under the action of the projective group of units, before being folded in half, and so between them give rise to just $2$ elliptic points.) If $N$ is an even prime—well, if you think back to the case $X_1(2)$, which we looked at a few weeks ago, there was a seam where one edge of a bigon folded back on itself, resulting in just one elliptic point of period $2$. And since the projective group of units in $\mathbb{Z}_2$ is trivial, $\X_0(2)$ is the same curve as $X_1(2)$, so also has one elliptic point of period $2$.

The more eagle-eyed among you will have noticed that the contents of the above paragraph can be expressed in terms of quadratic residues: for prime $N$, the number of elliptic points of period $2$ is $1+\left(\frac{-1}{N}\right)$ where $\left(\frac{-1}{N}\right)$ is the Legendre symbol.

For composite $N$, we proceed as follows. Take all the prime factors of $N$, counted only once each. Each of these contributes a factor of $1+\left(\frac{-1}{N}\right)$. (So if *any* of them is odd and also has a projective group of units of odd order, there are *no* elliptic points of period $2$). In addition, we need to apply the following rule: if there is a *single* factor of $2$ in the prime decomposition of $N$, that contributes a factor of $1$. But if there is *more* than one factor of $2$, we get a factor of $0$ instead, i.e. *no* elliptic points of period $2$. For instance, $X_0(4)$, which is the same as $X_1(4)$, has no elliptic points.

So if there is at most one factor of $2$, and no prime factors with an odd number of projective units, then each other prime factor (counted once) contributes a factor of $2$. So, for instance, $65=5\cdot 13$ has $4$ elliptic points of period $2$, as does $325=5^2\cdot 13$.

Something similar happens with elliptic points of period $3$. Consider $N=7$ as the primordial example. Obviously $X_1(7)$ has $3$-fold rotational symmetry, and this is about an axis passing through two vertices (where the edge-$2$s join the edge-$3$s of their own heptagon, one on each side). Quotienting out by the action of the projective group of units turns these vertices into elliptic points of order $3$, lifting the elliptic point at $\omega$, and there are two of them: one on each side. We might expect there to be something similar whenever the projective group of units has order $3$, and, as in the case of period $2$, something like this is true.

For prime $N$, if $N$ is not $3$, then it will have two elliptic points of period $3$ if $\frac{N-1}{2}$ is a multiple of $3$, and none otherwise, while if $N$ is $3$, it has a single elliptic point of order $3$, as we recall from when we looked at $X_1(3)$ a little while ago. Once again, we can express this in terms of quadratic residues: the number of elliptic points is $1+\left(\frac{-3}{N}\right)$. For composite $N$, the story is exactly the same as for period $2$ elliptic points, except with $3$ replacing $2$.

(This includes the fact that more than one factor of $3$ in $N$ means no elliptic points of period $3$, as we saw last time with $X_1(9)$.)

**Gluing Edges in $X_0(N)$**

Now I’d like to talk about the structure of $X_0(N)$ more generally. Of course, to understand how all these curves are constructed, we need not only to know what their tiles are, but also how they are glued together. So I’ll finish off this time by talking a bit more about how to find out which edges end up identified with which—at least in the case of prime $N$.

Recall from the time before last the way that we numbered the edges of the faces that have unit denominator (e.g. denominator $1$). We did this for $N=7$. To apply this numbering scheme, we start at the edge which has the face of denominator $0$ (e.g. $\frac{1}{0}$) on the other side, and label it $0$. Then we go out from that edge on both sides, labelling the edges in sequence, $1$, $2$, etc—one edge with each label on each side, left and right.

Now the key facts about the edge gluing in $X_1(N)$ are:

1) For the face with denominator $1$, edge-$i$ attaches it to a face with denominator $i$—edge-$0$ attaches to the denominator-$0$ face, edge-$1$ attaches to the denominator-$1$ face (i.e.—after gluing—the *other* edge-$1$ of the *same* face), edge-$2$ attaches to the denominator-$2$ face, etc. This convenient fact results from the mediant relationship between triples of faces, acting on the denominator $1$ (viz: one of the denominators must be the sum of the other two).

2) The action by rescaling maps the face with denominator $1$ onto other faces: in the case of prime $N$, onto all other non-$0$ faces.

But this enables us to work out what face each edge of any given face is adjacent to. For instance: edge-$2$ of face $1$ is adjacent to face $2$. Under the action of a unit $u$, the image of face $1$ is of course $u$, while the image of face $2$ is $2 u$. So of course edge-$2$ of face $u$ must be adjacent to face $2 u$. (This is not always immediately clear from pictures since we might be representing the denominator $2 u$ as $N-2 u$, if $2 u>\frac{N}{2}$.)

So for instance, for $N=13$, suppose we pick the element $2$ to generate the projective group of units:

$1\rightarrow 2\rightarrow 4\rightarrow 5\rightarrow 3\rightarrow 6\rightarrow 1$.

We know which denominator lies on the other side of any given edge of the denominator-$1$ face (viz. denominator $i$ lies on the other side of edge-$i$), and then, by applying the rescaling action to all denominators, we can determine the denominator lying on the other side of any given edge of any other face. We get the following table of faces on the other side of a given edge of a given face:

$\array{\arrayopts{\collines{solid none}\rowlines{solid none}}given face&edge\text{-}2&edge\text{-}3&edge\text{-}4&edge\text{-}5&edge\text{-}6\\ 1&2&3&4&5&6\\ 2&4&6&5&3&1\\ 4&5&1&3&6&2\\ 5&3&2&6&1&4\\ 3&6&4&1&2&5\\ 6&1&5&2&4&3}$

E.g. if we cross over edge-$5$ of face $3$, we find ourselves in face $2$.

Observe that the leftmost edge of face $1$ attaches to face $2$ and, going down to the second line, the rightmost edge of face $2$ connects back to face $1$ (that leftmost glues to rightmost is generally true—the nook next to the seam, made by the edge-$2$s, gets filled by the pointy tip made by the highest-numbered edges). So we can connect them by a line, indicating which faces are connected by which edges:

Let’s fill in connections corresponding to the other edges in the same way:

That is, of course, the table for $X_1(13)$. To get to the corresponding table for $X_0(13)$, we take the quotient of this table by the rescaling group. This means that the rows all get identified with each other, so the table collapses to a single row, giving the mutual identifications among the edges of the single remaining $13$-gon.

Now, it is obvious from the table above that when we do this, edge-$5$ will connect back to *itself*—this is why we have an elliptic point of period $2$ in the middle of it.

Actually, it’s not quite as obvious as that: there is another possibility, though it is not realised. There are two edge-$5$s, one on the left and one on the right, and the table is compatible with the possibility that these are identified with each other, rather than each of them being identified with itself. That is, after all, what happens with edge-$1$.

We can show geometrically why this doesn’t happen, looking at a part of $X_1(13)$. For suppose it *did* happen. Then we’d see something like this:

This shows the right-hand edge-$5$ of face $1$ identified with the left-hand edge-$5$ of face $5$. Of course, if we have this, then we must also identify the left-hand edge-$5$ of face $1$ with the right-hand edge-$5$ of edge-$5$:

Consider, though, the effect of multiplying by the projective unit $5$. Faces $1$ and $5$ swap places, which means that the pictures above swap places. But, comparing those pictures with each other, clearly face $4$ would have to be sent to itself, so we would have $5\cdot 4=\pm 4$. Obviously this isn’t true for $N=13$ (in fact, $5\cdot 4=-6$, confirming that edge-$5$ flips) but this argument can be used in a more general setting, if we replace $5$ by some other projective unit $u$, $13$ by $N$ and $4$ by some face $a$. If applying $u$ twice gives the identity, and the left-hand edge-$u$ is identified with the right-hand edge-$u$, we must have, by a generalisation of the pictures above, $u a=a$ (projectively). If $a$ is a unit, which will always be true if $N$ is prime, this can’t happen unless $u$ is $1$—and indeed, we do have precisely this situation for edge-$1$.

On the other hand, if $N$ is composite, we can and do get $u a=a$ for other values of $u$, when $a$ is not itself a unit, and so we can have other cases of a left edge-$u$ being identified with a right edge-$u$. For instance, last time we showed the case $N=15$, and we can see that the left edge-$4$ of one unit is identified with the right edge-$4$ of the neighbouring unit (except the picture is oriented so that they’re top and bottom rather than left and right). This is possible because $5$ and $3$, the neighbours of $4$, are not units in $\mathbb{Z}_15$:

So much for period $2$. From the gluing table, we can also see a cycle of order $3$:

edge-$3$ of face $1$ connects to face $3$

edge-$3$ of face $3$ connects to face $4$

edge-$3$ of face $4$ connects to face $1$

This is what gives rise to the period-$3$ elliptic points in $X_0(13)$. We have $3^2=4$ in the projective group of units of $\mathbb{Z}_13$, owing to the fact that $3^2=-4$ mod $13$. We see a situation like this:

Multiplying by $3$ cycles $1\rightarrow 3\rightarrow 4\rightarrow 1$.

I’ll talk about elliptic points in more generality next time, as I have skipped over some of the interesting algebra behind the calcuations I’ve been doing.

Anyway, I think that’s enough about $N=13$. I now want to repeat this exercise, with variations, for $N=11$, because, although there are no elliptic points, something else interesting shows up.

The table of edge connections looks like this:

Notice that there is now no edge that folds back on itself (no elliptic point of period $2$) and there is no $3$-cycle (no elliptic point of period $3$). However, we’re going to be looking at something else.

Recall that there are actually two of each kind of edge (other than edge-$0$)—one on the left side and one on the right side. First we’ll give each edge an orientation pointing towards larger-numbered edges:

Now, let’s construct a picture of how edges are glued together in $X_1(11)$ or $X_0(11)$. From the table, we see that edge-$2$ is identified with edge-$5$, and edge-$3$ is identified with edge-$4$, but the question is, *which* edge-$2$ is identified with *which* edge-$5$ and *which* edge-$3$ is identified with *which* edge-$4$—is the *left-hand* edge-$2$ identified with the *left-hand* edge-$5$, or with the *right-hand* one … etc? And what happens to the orientations of the edges, shown in the picture above—once two edges have been identified, do their corresponding orientations end up parallel or anti-parallel?

In pursuit of this, let us first observe that edge-$3$ of face $1$ is identified with edge-$4$ of face $3$, while edge-$4$ of face $1$ is identified with edge-$3$ of face $4$. Let’s illustrate this with the following diagram:

We know the orientation of the edges in face $1$ because we have two edges, so we know which direction the labels are increasing in. To complete the picture, we need the same information for faces $3$ and $4$, and for this, we can conveniently use the fact, from the table above, that edge-$2$ of face $4$ is edge-$5$ of face $3$, so we get the following:

We could go on to complete further edges, but this has told use the crucial information that we need: edge-$2$ and edge-$5$ are connected in opposite directions, implying that one is folded back to lie over the other. This happens if they are both on the *same* side of the $N$-gon (both left side or both right side). So, carrying out this first identification, of edge-$2$ with edge-$5$, on the single $11$-gon of $X_0(11)$, and leaving the identification of edge-$3$ with edge-$4$ aside for a moment, we get something like this on each side of the $11$-gon:

On the other hand, edge-$3$ and edge-$4$ are glued together in parallel. From the picture above, it can’t be the *adjacent* edge-$3$ and edge-$4$ on each side that are identified, since they run in opposite directions: it must be that the edge-$3$ from one side of the $N$-gon is joined to the edge-$4$ on the other side, and vice versa, forming a tunnel from one side to the other. (There has to be a $180^\circ$ twist in this tunnel as it crosses, in order to get edge-$3$ at the top of the left side to join the edge-$4$ from the bottom of the right side and vice versa as illustrated in the diagram above.) So the two sides of the $11$-gon are connected to each other, and $X_0(11)$ has the topology of a torus. Stretching the edges to make this clearer, and imagining that the grey area below is a hole in a sphere (comprising the tube formed by the $11$-gon with its edge-$1$s identified, capped by the cone formed by the single sector of denominator $0$):

First we roll up the bottom to meet the top, with the arrows on $2$ and $5$ meeting in opposite directions. This causes the concatenation of edge-$3$ and edge-$4$ on each side to roll into a circle. Then we identify those circles, with the arrows pointing in the same direction, making sure that edge-$3$ on the left joins to edge-$4$ on the right and vice versa. That’s $X_0(11)$.

**Calculating the Genus**

Since we’ve now worked out the construction of a curve of genus $\gt 0$, I think I’ll finish off by working out the genus formula. I’ll start with the Euler characteristic $\chi$, since it is related to the genus $g$ by $\chi=2-2g$.

To guide the discussion, here is a picture of an example that we can keep in mind:

Each pentagon is divided into little triangles. The total number of little triangles is the *degree*, $d$. They each correspond to a copy of the fundamental domain, so the number of them is the degree of the covering of $X(1)$ by the particular curve.

So we’ll calculate $\chi$ as $F-E+V$ where these are the numbers of faces, edges and vertices. We’ll use the little triangles as faces, so $F=d$, the degree. In general, each triangle has $3$ edges, each shared by $2$ faces, so $E=\frac{3d}{2}$. And generally, each triangle has $3$ vertices, but these are different from each other. The two vertices at the base (on the thick lines in the picture) are generally shared among six triangles, giving a contribution $\frac{2d}{6}=\frac{d}{3}$. However, the vertices at the apex are shared among various numbers of triangles ($N$ for $X(N)$, but we also want to handle $X_1(N)$, $X_0(N)$ and others). Luckily, these are precisely where the cusps are, so we can duck the issue by expressing the number of these vertices as the number of cusps, which we’ll represent by $e_\infty$. So, so far, we have $\chi=d-\frac{3d}{2}+\frac{d}{3}+e_\infty=-\frac{d}{6}+e_\infty$.

However, we have forgotten the elliptic points of period $2$ and $3$!

The elliptic points of period $2$ correspond to edges which project into a triangle and end in the middle of nowhere. These add nothing to the Euler characteristic. We need to place an extra vertex at the midpoint, so the net contribution to $V-E$ is $0$ (or we can add an extra, degenerate, face, bounded solely by the single edge). So for each elliptic point of period $2$, we have subtracted $\frac{1}{2}$ from the Euler characteristic when the actual contribution is $0$. So we need to add an extra $\frac{1}{2}$ for each elliptic point of period $2$. Denoting the total number of these by $e_2$, we have an extra term $\frac{e_2}{2}$.

Likewise, for each elliptic point of period $3$, we should count a whole vertex rather than a third of a vertex, so we need to add on an extra two thirds of a vertex for each one: $\frac{2 e_3}{3}$.

In conclusion, we have $2-2g=-\frac{d}{6}+\frac{e_2}{2}+\frac{2 e_3}{3}+e_\infty$, or, finally,

$g=1+\frac{d}{12}-\frac{e_2}{4}-\frac{e_3}{3}-\frac{e_\infty}{2}$

For example, for $X_0(N)$ with prime $N$, if $N$ is not $2$ or $3$, then there is some $k$ such that we have that $N=12 k+r$ with $r\in\{1,5,7,11\}$. We then have:

$\array{r&d&e_2&e_3&e_\infty&g\\ 1&12 k+2&2&2&2&k-1\\ 5&12 k+6&2&0&2&k\\ 7&12 k+8&0&2&2&k\\ 11&12 k+12&0&0&2&k+1}$

Next time, we’ll generalise our method for working out which edge is identified with which in $X_0(N)$, and use that to give a more pictorial way to illustrate the genus.

## Re: Pictures of Modular Curves (X)