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January 16, 2011

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 Γ 1(N)\Gamma_1(N) and X 1(N)X_1(N). In this part, I want to move on to Γ 0(N)\Gamma_0(N) and X 0(N)X_0(N).

Taking a Quotient by Γ 0(N)\Gamma_0(N) in Three Easy Steps

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

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

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

Finally, to get from the upper half-plane to X 0(N)X_0(N), we quotient by the action of Γ 0(N)\Gamma_0(N). Γ 0(N)\Gamma_0(N) is the subgroup of Γ\Gamma consisting of matrices of the form (* * 0 *)\left(\array{*&*\\0&*}\right) (mod NN), but we can do this in three steps: first doing the two-step quotient to X 1(N)X_1(N) described above, and then appending an extra stage.

What does this extra stage look like? Well, we want to get from Γ 1(N)\Gamma_1(N) to Γ 0(N)\Gamma_0(N), or rather, from the former reduced mod NN to the latter reduced mod NN.

Now, Γ 0(N)\Gamma_0(N) reduced mod NN—that is, the group of matrices with entries in N\mathbb{Z}_N, determinant 11, and of the form (* * 0 *)\left(\array{*&*\\0&*}\right) (mod {1,1}\{1,-1\})—is generated by two subgroups: the “translations” (or “additions”), i.e. matrices of the form (1 * 0 1)\left(\array{1&*\\0&1}\right) and the “rescalings” (or “multiplications”), i.e. matrices of the form (* 0 0 *)\left(\array{*&0\\0&*}\right). So having quotiented by the action of the first, “translation”, subgroup to get down to X 1(N)X_1(N), we can quotient by the action of the second, “rescaling”, subgroup to get right down to X 0(N)X_0(N). And this second subgroup is isomorphic to the projective group of units of N\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)X(N) a few articles ago. Its action on X 1(N)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)X_1(5):

Spherical Segment, showing 2-fold symmetry: N = 5

As we’ve seen before, with the full X(5)X(5), the only nontrivial action of the rescaling group is to swap the blue and green halves, by rotating this shape 180 180^\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=7N=7 (the next case with a non-trivial projective unit group). The photo and drawing below illustrate X 1(7)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-22s (the edges next to the seam) folded back so that they are glued to the edge-33s (the edges further along).

And this is generally the case for any NN: X 1(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)X_1(7), also showing the three-fold symmetry.

X1(7)

Structure of X 0(N)X_0(N): Elliptic Points

At this point, we are in a position to say something about the elliptic points of X 0(N)X_0(N).

In the case of N=5N=5, as we mentioned, it’s clear that the rotation about 180 180^\circ rotates the edge-22s 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 22, one on each side, at the midpoints of the folded edges. (Being at the midpoints of edges, they lift the period-22 elliptic point at ii in the fundamental domain of Γ\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 NN is an odd prime, there are two elliptic points of period 22 if N12\frac{N-1}{2} is even, and none if it is odd. (With NN prime and N12\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=13N=13, and generally N14\frac{N-1}{4} on each side for X 1(N)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 22 elliptic points.) If NN is an even prime—well, if you think back to the case X 1(2)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 22. And since the projective group of units in 2\mathbb{Z}_2 is trivial, X 0(2)\X_0(2) is the same curve as X 1(2)X_1(2), so also has one elliptic point of period 22.

X 0(2)X_0(2), showing elliptic point

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 NN, the number of elliptic points of period 22 is 1+(1N)1+\left(\frac{-1}{N}\right) where (1N)\left(\frac{-1}{N}\right) is the Legendre symbol.

For composite NN, we proceed as follows. Take all the prime factors of NN, counted only once each. Each of these contributes a factor of 1+(1N)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 22). In addition, we need to apply the following rule: if there is a single factor of 22 in the prime decomposition of NN, that contributes a factor of 11. But if there is more than one factor of 22, we get a factor of 00 instead, i.e. no elliptic points of period 22. For instance, X 0(4)X_0(4), which is the same as X 1(4)X_1(4), has no elliptic points.

So if there is at most one factor of 22, and no prime factors with an odd number of projective units, then each other prime factor (counted once) contributes a factor of 22. So, for instance, 65=51365=5\cdot 13 has 44 elliptic points of period 22, as does 325=5 213325=5^2\cdot 13.

Something similar happens with elliptic points of period 33. Consider N=7N=7 as the primordial example. Obviously X 1(7)X_1(7) has 33-fold rotational symmetry, and this is about an axis passing through two vertices (where the edge-22s join the edge-33s 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 33, lifting the elliptic point at ω\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 33, and, as in the case of period 22, something like this is true.

For prime NN, if NN is not 33, then it will have two elliptic points of period 33 if N12\frac{N-1}{2} is a multiple of 33, and none otherwise, while if NN is 33, it has a single elliptic point of order 33, as we recall from when we looked at X 1(3)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+(3N)1+\left(\frac{-3}{N}\right). For composite NN, the story is exactly the same as for period 22 elliptic points, except with 33 replacing 22.

(This includes the fact that more than one factor of 33 in NN means no elliptic points of period 33, as we saw last time with X 1(9)X_1(9).)

X 0(3)X_0(3), showing elliptic point

Gluing Edges in X 0(N)X_0(N)

Now I’d like to talk about the structure of X 0(N)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 NN.

Recall from the time before last the way that we numbered the edges of the faces that have unit denominator (e.g. denominator 11). We did this for N=7N=7. To apply this numbering scheme, we start at the edge which has the face of denominator 00 (e.g. 10\frac{1}{0}) on the other side, and label it 00. Then we go out from that edge on both sides, labelling the edges in sequence, 11, 22, etc—one edge with each label on each side, left and right.


Now the key facts about the edge gluing in X 1(N)X_1(N) are:

1) For the face with denominator 11, edge-ii attaches it to a face with denominator ii—edge-00 attaches to the denominator-00 face, edge-11 attaches to the denominator-11 face (i.e.—after gluing—the other edge-11 of the same face), edge-22 attaches to the denominator-22 face, etc. This convenient fact results from the mediant relationship between triples of faces, acting on the denominator 11 (viz: one of the denominators must be the sum of the other two).

2) The action by rescaling maps the face with denominator 11 onto other faces: in the case of prime NN, onto all other non-00 faces.

But this enables us to work out what face each edge of any given face is adjacent to. For instance: edge-22 of face 11 is adjacent to face 22. Under the action of a unit uu, the image of face 11 is of course uu, while the image of face 22 is 2u2 u. So of course edge-22 of face uu must be adjacent to face 2u2 u. (This is not always immediately clear from pictures since we might be representing the denominator 2u2 u as N2uN-2 u, if 2u>N22 u>\frac{N}{2}.)

So for instance, for N=13N=13, suppose we pick the element 22 to generate the projective group of units:

12453611\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-11 face (viz. denominator ii lies on the other side of edge-ii), 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:

givenface edge-2 edge-3 edge-4 edge-5 edge-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\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-55 of face 33, we find ourselves in face 22.

Observe that the leftmost edge of face 11 attaches to face 22 and, going down to the second line, the rightmost edge of face 22 connects back to face 11 (that leftmost glues to rightmost is generally true—the nook next to the seam, made by the edge-22s, 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:

givenface edge-2 edge-3 edge-4 edge-5 edge-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\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}

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

givenface edge-2 edge-3 edge-4 edge-5 edge-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\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}

That is, of course, the table for X 1(13)X_1(13). To get to the corresponding table for X 0(13)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 1313-gon.

Now, it is obvious from the table above that when we do this, edge-55 will connect back to itself—this is why we have an elliptic point of period 22 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-55s, 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-11.

We can show geometrically why this doesn’t happen, looking at a part of X 1(13)X_1(13). For suppose it did happen. Then we’d see something like this:

5 5 4 6 4 6 a a±1 4 6 1 5

This shows the right-hand edge-55 of face 11 identified with the left-hand edge-55 of face 55. Of course, if we have this, then we must also identify the left-hand edge-55 of face 11 with the right-hand edge-55 of edge-55:

5 5 4 6 4 6 a±1 a 4 6 5 1

Consider, though, the effect of multiplying by the projective unit 55. Faces 11 and 55 swap places, which means that the pictures above swap places. But, comparing those pictures with each other, clearly face 44 would have to be sent to itself, so we would have 54=±45\cdot 4=\pm 4. Obviously this isn’t true for N=13N=13 (in fact, 54=65\cdot 4=-6, confirming that edge-55 flips) but this argument can be used in a more general setting, if we replace 55 by some other projective unit uu, 1313 by NN and 44 by some face aa. If applying uu twice gives the identity, and the left-hand edge-uu is identified with the right-hand edge-uu, we must have, by a generalisation of the pictures above, ua=au a=a (projectively). If aa is a unit, which will always be true if NN is prime, this can’t happen unless uu is 11—and indeed, we do have precisely this situation for edge-11.

On the other hand, if NN is composite, we can and do get ua=au a=a for other values of uu, when aa is not itself a unit, and so we can have other cases of a left edge-uu being identified with a right edge-uu. For instance, last time we showed the case N=15N=15, and we can see that the left edge-44 of one unit is identified with the right edge-44 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 55 and 33, the neighbours of 44, are not units in 15\mathbb{Z}_15:


So much for period 22. From the gluing table, we can also see a cycle of order 33:

givenface edge-2 edge-3 edge-4 edge-5 edge-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\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}

edge-33 of face 11 connects to face 33
edge-33 of face 33 connects to face 44
edge-33 of face 44 connects to face 11

This is what gives rise to the period-33 elliptic points in X 0(13)X_0(13). We have 3 2=43^2=4 in the projective group of units of 13\mathbb{Z}_13, owing to the fact that 3 2=43^2=-4 mod 1313. We see a situation like this:

3 4 4 3 4 3 1 3 4

Multiplying by 33 cycles 13411\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=13N=13. I now want to repeat this exercise, with variations, for N=11N=11, because, although there are no elliptic points, something else interesting shows up.

The table of edge connections looks like this:

givenface edge-2 edge-3 edge-4 edge-5 edge-6 1 2 3 4 5 2 4 6 3 1 4 3 1 6 2 3 6 2 1 4 5 1 4 2 3\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\\ 2&4&6&3&1\\ 4&3&1&6&2\\ 3&6&2&1&4\\ 5&1&4&2&3}

Notice that there is now no edge that folds back on itself (no elliptic point of period 22) and there is no 33-cycle (no elliptic point of period 33). However, we’re going to be looking at something else.

Recall that there are actually two of each kind of edge (other than edge-00)—one on the left side and one on the right side. First we’ll give each edge an orientation pointing towards larger-numbered edges:

Directed edges for N = 11
Directed edges for N = 11

Now, let’s construct a picture of how edges are glued together in X 1(11)X_1(11) or X 0(11)X_0(11). From the table, we see that edge-22 is identified with edge-55, and edge-33 is identified with edge-44, but the question is, which edge-22 is identified with which edge-55 and which edge-33 is identified with which edge-44—is the left-hand edge-22 identified with the left-hand edge-55, 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-33 of face 11 is identified with edge-44 of face 33, while edge-44 of face 11 is identified with edge-33 of face 44. Let’s illustrate this with the following diagram:

1 3 4 3 4 4 3

We know the orientation of the edges in face 11 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 33 and 44, and for this, we can conveniently use the fact, from the table above, that edge-22 of face 44 is edge-55 of face 33, so we get the following:

1 3 4 3 4 4 3 5 2

We could go on to complete further edges, but this has told use the crucial information that we need: edge-22 and edge-55 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 NN-gon (both left side or both right side). So, carrying out this first identification, of edge-22 with edge-55, on the single 1111-gon of X 0(11)X_0(11), and leaving the identification of edge-33 with edge-44 aside for a moment, we get something like this on each side of the 1111-gon:

2 5 3 4

On the other hand, edge-33 and edge-44 are glued together in parallel. From the picture above, it can’t be the adjacent edge-33 and edge-44 on each side that are identified, since they run in opposite directions: it must be that the edge-33 from one side of the NN-gon is joined to the edge-44 on the other side, and vice versa, forming a tunnel from one side to the other. (There has to be a 180 180^\circ twist in this tunnel as it crosses, in order to get edge-33 at the top of the left side to join the edge-44 from the bottom of the right side and vice versa as illustrated in the diagram above.) So the two sides of the 1111-gon are connected to each other, and X 0(11)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 1111-gon with its edge-11s identified, capped by the cone formed by the single sector of denominator 00):

1 1 2 2 3 3 4 4 5 5

First we roll up the bottom to meet the top, with the arrows on 22 and 55 meeting in opposite directions. This causes the concatenation of edge-33 and edge-44 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-33 on the left joins to edge-44 on the right and vice versa. That’s X 0(11)X_0(11).

Calculating the Genus

Since we’ve now worked out the construction of a curve of genus >0\gt 0, I think I’ll finish off by working out the genus formula. I’ll start with the Euler characteristic χ\chi, since it is related to the genus gg by χ=22g\chi=2-2g.

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

Dodecahedron, triangulated (front only): N = 5

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

So we’ll calculate χ\chi as FE+VF-E+V where these are the numbers of faces, edges and vertices. We’ll use the little triangles as faces, so F=dF=d, the degree. In general, each triangle has 33 edges, each shared by 22 faces, so E=3d2E=\frac{3d}{2}. And generally, each triangle has 33 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 2d6=d3\frac{2d}{6}=\frac{d}{3}. However, the vertices at the apex are shared among various numbers of triangles (NN for X(N)X(N), but we also want to handle X 1(N)X_1(N), X 0(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 e_\infty. So, so far, we have χ=d3d2+d3+e =d6+e \chi=d-\frac{3d}{2}+\frac{d}{3}+e_\infty=-\frac{d}{6}+e_\infty.

However, we have forgotten the elliptic points of period 22 and 33!

The elliptic points of period 22 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 VEV-E is 00 (or we can add an extra, degenerate, face, bounded solely by the single edge). So for each elliptic point of period 22, we have subtracted 12\frac{1}{2} from the Euler characteristic when the actual contribution is 00. So we need to add an extra 12\frac{1}{2} for each elliptic point of period 22. Denoting the total number of these by e 2e_2, we have an extra term e 22\frac{e_2}{2}.

Likewise, for each elliptic point of period 33, 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: 2e 33\frac{2 e_3}{3}.

In conclusion, we have 22g=d6+e 22+2e 33+e 2-2g=-\frac{d}{6}+\frac{e_2}{2}+\frac{2 e_3}{3}+e_\infty, or, finally,

g=1+d12e 24e 33e 2g=1+\frac{d}{12}-\frac{e_2}{4}-\frac{e_3}{3}-\frac{e_\infty}{2}

For example, for X 0(N)X_0(N) with prime NN, if NN is not 22 or 33, then there is some kk such that we have that N=12k+rN=12 k+r with r{1,5,7,11}r\in\{1,5,7,11\}. We then have:

r d e 2 e 3 e g 1 12k+2 2 2 2 k1 5 12k+6 2 0 2 k 7 12k+8 0 2 2 k 11 12k+12 0 0 2 k+1\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)X_0(N), and use that to give a more pictorial way to illustrate the genus.

Posted at January 16, 2011 11:20 PM UTC

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3 Comments & 1 Trackback

Re: Pictures of Modular Curves (X)

Thank you very much for the beautiful pictures and explanations. I hope I can find more time in the future to read them thoroughly. That is I am badly lacking behind and thus I haven’t even yet understood how the mod(1,1)mod({1,-1}) part comes in when quotienting the upper half plane by the action with matrices ({1,b},{0,1}) (mod N) for X 1X_1. Does there exist a genus computation also for X 1X_1?
Posted by: student on January 18, 2011 11:44 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (X)

I’m glad you’ve liked it so far! I know what it’s like to never have enough time to learn the things one wants to.

There is a genus formula for X 1(N)X_1(N).

It is:

g=1+N 224 pp 21p 214 d|Nϕ(d)ϕ(Nd)g=1+\frac{N^2}{24}\prod_p{\frac{p^2-1}{p^2}}-\frac{1}{4}\sum_{d\vert N}{\phi(d)\phi(\frac{N}{d})}

where pp ranges over the distinct prime factors of NN, dd ranges over all factors, and ϕ\phi is the totient function: ϕ(n)\phi(n) is the number of natural numbers less than nn which are coprime to nn.

So, for instance, for N=20N=20, there are two distinct prime factors, 22 and 55, so the second term is 20 2242 212 25 215 2=40024342425=12\frac{20^2}{24}\cdot \frac{2^2-1}{2^2}\frac{5^2-1}{5^2}=\frac{400}{24}\frac{3}{4}\frac{24}{25}=12.

And the factors are 11, 22, 44, 55, 1010 and 2020, with totients of 11, 11, 22, 44, 44 and 88. So the third term is 14(18+14+24+42+41+81)=244=6\frac{1}{4}(1\cdot 8+1\cdot 4+2\cdot 4+4\cdot 2+4\cdot 1+8\cdot 1)=\frac{24}{4}=6.

So the genus is 1+126=71+12-6=7.

Posted by: Tim Silverman on January 19, 2011 1:26 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (X)

Perhaps I should have made clear: this uses the same underlying genus formula as X 0(N)X_0(N) and any other modular curves, viz. g=1+d12e 24e 33e 2g=1+\frac{d}{12}-\frac{e_2}{4}-\frac{e_3}{3}-\frac{e_\infty}{2}, since the underlying arithmetic for the vertices, edges and total number of tiles is the same. Also, the formula I gave above specificially for X 1(N)X_1(N) only works for N>4N>4, due to elliptic points and other funniness with small NN.

Posted by: Tim Silverman on January 19, 2011 1:32 PM | Permalink | Reply to this
Read the post Pictures of Modular Curves (XI)
Weblog: The n-Category Café
Excerpt: Working out how X0(N) is glued together
Tracked: January 27, 2011 5:59 PM

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