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January 27, 2011

Pictures of Modular Curves (XI)

Posted by Guest

guest post by Tim Silverman

Welcome, welcome to the antepenultimate part of our series illustrating modular curves and their tilings. Last time, we talked about how the curves X 0(N)X_0(N) are contructed out of their tiles, and illustrated how we could label the edges of the tiles, before going on to talk about how the labels of the edges could help understand how the edges glued together. But we only did this in a rather ad hoc way, and only for a couple of simple cases. We also talked about calculating the genus of modular curves.

This time, we’ll look at how, at least for prime NN, we can work out, systematically, how the edges are glued together in more complicated cases; and we’ll start to look at the genus in particular cases. We’ll also revisit the elliptic points of these curves in slightly greater generality than last time.

Edge Numbers

Remember that we number faces by their denominator; and we number the edges of a face by starting with 00 on the edge that separates it from a face with denominator 00, and then increasing by 11 as we go to successive edges away from that (so that there are left-hand and right-hand versions of each label).

Now, recall the gluing table for edges, reproduced below, for N=13N=13.

If we go to the face whose number appears in a given row on the left of the table, and pick the edge (of that face) whose label appears in a given column at the top of the table, and want to know which face we’ll end up in if cross over to the other side of that edge, we can find the answer in the number which appears in the cell at the intersection of the row and the column, e.g. edge-33 of face 22 connects it to face 66.

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}

The coloured lines connect faces on opposite sides of the same edge, e.g. edge-66 of face 11 is the same as edge-22 of face 66. Here’s the table for N=11N=11:

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

Closer examination reveals that these tables are simply multiplication tables, albeit with the rows rather oddly arranged and a column missing.

(They are, be precise, multiplication tables mod 1313 and mod 1111 respectively—and always mod change of sign, of course. The missing column is for edge-11 and is omitted because it’s boring.)

For instance, look at the second row, for face 22. The first column is for edge-22, we have 2×2=42\times 2=4, and 44 is what appears in the first column of the second row. The second column is for edge-33, we have 2×3=62\times 3=6, but 6=5-6=5 mod 1111 and, since we’re modding out by sign changes and we’ve adopted the convention that the lesser of +a+a anda-a will stand as the representative of both, 55 is what appears in the second column of the second row. Similarly, 2×4=82\times 4=8, 8=3-8=3 mod 1111, so 33 is what appears in the third column of the second row. And so on.

There is a good explanation for this.

The faces opposite given edges of face 11 have the same number (i.e. denominator) as the edge labels—that follows from the mediant relation between mutually adjacent triples of faces (whereby the denominator of one of them is the sum of the denominator of the other two). We get from face 11 to other faces by multiplying all denominators by a given projective unit (since this is a symmetry of the tiling). But if we do this, then obviously all face numbers—both the source faces listed on the left, and the target faces in the cells of the table—get multiplied by a given unit on a given row. So doubling the first row gives the second row, etc. (Mod change of sign, as always, of course.)

From this, we can easily work out which edge joins to which.

For instance, consider edge-22 of face 11 in the case N=11N=11. This edge is the same as some edge of face 22 (because edge-uu of face 11 is always the same as some edge of face uu). Which edge of face 22 is it the same as? Well, we can look it up: the faces opposite each edge of face 22 are given on row 22, and we want to find the cell on row 22 with 11 in it. And this cell falls in the column belonging to edge-55. So edge-55 of face 22 is the same as edge-22 of face 11.

Yes, but why edge-55, particularly? Well, row 22 is the same as row 11 multiplied by 22. So we want the edge whose number, when multiplied by 22, gives 11—in other words, the inverse of 22 in the group of projective units mod 1111. Which is just 55. And there’s nothing special about 22 and 55—this argument is quite general.

So, generally, edge-uu of one face is identified with edge-vv of another (or sometimes of the same) face precisely where uv=1u v=1 in the group of projective units. For edge-uu of face 11, we can argue as in the paragraph above, looking at where 11 falls in row uu; and the pairings of edge labels on opposite sides of an edge are the same on all other faces as they are on face 11, since the symmetry given by the action of projective units preserves edge labels (because it preserves denominators of 00, from which the edge labellings derive).

Edge Directions

However, for any given non-zero edge number (such as 22 or 55) there are two edges with that number, one to the left and one to the right of edge-00. So, is the left edge-22 the same as the left edge-55, and the right edge-22 as the right edge-55—or is the left edge-22 identified with the right edge-55 and vice versa?

Also, we’ve given our edges an orientation (pointing away from edge-00). So we want to know if a given pair of edges is glued with their orientations parallel or anti-parallel.

The left-right question and the parallel-antiparallel question are actually the same. Recall this picture of X 0(11)X_0(11), about to be sewn up, from last time (with coloured lines added to indicate what’s glued to what):

1 1 2 2 3 3 4 4 5 5

If the closed surface is to be oriented, edges on the same side must be joined antiparallel, by folding back the ones with higher numbers over those with lower numbers, while those on opposite sides must be joined parallel.

Alternatively, we can consider adjacent tiles, as in the picture below: if the arrows along shared edges are going anti-clockwise around both tiles—which happens with the shared edges 22 and 55 on the left hand side of the left hand picture below—then shared edges will have arrows pointing in opposite directions (and likewise for clockwise arrows and right sides). But if the arrows go clockwise around one tile and anticlockwise around another—as happens with the adjacent 33 and 44 arrows in the picture on the right below, where the right side edges are joined to left side edges—then shared edges have parallel arrows.

2 2 3 3 4 4 5 5 2 2 3 3 4 4 5 5 2 2 3 3 4 4 5 5 2 2 3 3 4 4 5 5

But we still don’t know whether any given pair of edges uu and vv, with uv=1u v=1 projectively, are joined same-side/antiparallel or other-side/parallel.

The answer is obtained by looking, not at the projective group of units, but at the full group of units. Here 25=12\cdot 5=-1, whereas 34=13\cdot 4=1. It turns out that this is sufficient to tell us that edge-22 and edge-55 are joined same side/antiparallel, while edge-33 and edge-44 are joined opposite side/parallel.

We can remember this by imagining that the minus sign flips the direction of the edge, so if uv=1u v=-1, then edge-uu must be identified antiparallel with edge-vv, and if uv=1u v=1, they must be identified parallel. However, here is a picture to show why this works:

a’ a a’±1 a+1 a+1 a 1 × a’ a’ a a’±1 a+1 a’±1 1 a’

On the left hand side, we show three adjacent faces, with edge numbers. One of the faces is face 11, and adjacent to that is face aa. The edge between them must be numbered aa in the numbering of face 11 (because on face 11, edge labels match face numbers), while on the numbering of face aa, we’ve determined in the discussion above that it must be the projective multiplicative inverse of aa, which we denote by aa'. So aa=±1a\cdot a'=\pm 1.

Let’s say we’ve chosen to show the right-hand side of face 11. That means that the edge numbering will increase clockwise, so the edge that is one step clockwise of edge-aa must be edge-(a+1)(a+1), as shown, which in turn means that the third face shown must be face (a+1)(a+1).

Now, depending on whether the right side of face 11 is identified with the left or the right side of face aa, the edge numbering of face aa will be either increasing or decreasing anticlockwise (respectively), so the edge one step anticlockwise of edge-aa' will be edge-a±1a'\pm1 (respectively).

Now we multiply faces numbers by aa'.

We send 1a1\rightarrow a'
We send a1a\rightarrow 1
We send (a+1)(a+1)a=(±1+a)(a+1)\rightarrow (a+1)\cdot a'=(\pm 1+a')

where the sign in the last equation depends on whether aa=1a\cdot a'=1 or aa=1a\cdot a'=-1.

But the edge numbering is unchanged, so now that face 11 is on the left, the third face must match the edge label, and hence must be a±1a'\pm 1 depending on whether the right side of face 11 was identified with the left or the right side of face aa (respectively). But the sign is also determined by the sign of aaa\cdot a'.

Hence if aa=+1a\cdot a'=+1, then the right side of face 11 must be identified with the left side of face aa (and hence they are joined parallel).

And if aa=1a\cdot a'=-1, then the right side of face 11 must be identified with the right side of face aa (and hence they are joined antiparallel).

QED

Now that we know that the minus sign flips directions, we can glance at the multiplication table mod 1111 and see that, in X 0(11)X_0(11), edge-22 is antiparallel to edge-55, while edge-33 is parallel to edge-44. And that in turn determines the sides they are on: if the joined edges are antiparallel, they are on the same side, folded back, while if they are parallel, they are on opposite sides, connected across the divide.

Another way to think of this rule is by labelling each edge uniquely: start at 00, work our way anticlockwise (say) increasing the labels through all the numbers mod NN. Orient the edges so they point in the direction of increasing numbers. Then edge-uu is identified with edge-vv precisely when v=1uv=\frac{-1}{u}, and they are always antiparallel (which is now a rather ordinary consequence of the fact that the numbering now gives the faces a consistent orientation).

1 10 2 9 3 8 4 7 5 6

110=10=11\cdot 10=10=-1 mod 1111.
25=10=12\cdot 5=10=-1 mod 1111.
37=21=13\cdot 7=21=-1 mod 1111.
48=32=14\cdot 8=32=-1 mod 1111.
69=54=16\cdot 9=54=-1 mod 1111.

Let’s illustrate this for a few more cases. Here’s N=17N=17:

1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1

This, again, has genus 11. The cycle of edges 35673\rightarrow 5\rightarrow 6\rightarrow 7 on the left is joined to the same cycle on the right, but again with a 180 180^\circ twist. Since 44=14\cdot 4=-1 mod 1717, this edge is folded back on itself and we have a pair of elliptic points of period 2.

Here’s N=19N=19:

1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1

Since 78=17\cdot 8=-1 mod 1919, these edges fold back against each other, and we have a pair of elliptic points of period 33.

Here’s N=23N=23:

1 2 3 4 5 6 7 8 9 10 11 11 10 9 8 7 6 5 4 3 2 1

Here is our first genus 22 case, and, not surprisingly, it’s a bit more complicated. Let’s take it step by step.

Step 1: identify edges 55 and 99 on the left (59=45=15\cdot 9=45=-1 mod 2323). This causes the sequence of edges 6786\rightarrow 7\rightarrow 8 to roll up into a loop. Do the same on the right.

Step 2: identify edges 22 and 1111. With edges 55 and 99 already identified, this causes the sequence of edges 34103\rightarrow 4\rightarrow 10 to roll up into a loop.

Step 3: identify the 6786\rightarrow 7\rightarrow 8 loop on the left with the 34103\rightarrow 4\rightarrow 10 loop on the right, and vice versa. (To be precise, 6786\rightarrow 7\rightarrow 8 with 41034\rightarrow 10\rightarrow 3 in that order. E.g. 64=24=16\cdot 4=24=1 mod 2323.) Each of these identifications produces a tube crossing from left to right, giving our genus 22 curve.

N=29N=29, another genus 22 curve, with a two-edge loop and a six-edge loop:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 13 12 11 10 9 8 7 6 5 4 3 2 1

And N=31N=31:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

The identification of edges, and the formation of tubes, gets considerably more complicated than this as NN, and the genus, increase. Next time, I’ll describe and illustrate a simpler kind of diagram that gives the same information but requires less space, thought and effort, and finish up with pictures of a few simple examples of X 0(N)X_0(N) for non-prime NN.

Elliptic Points in More Generality

Last time, we investigated how the elliptic points of X 0(13)X_0(13) can be spotted in its multiplication table. I’ll wrap up this time by fitting that discussion into a more general picture.

Let’s look at that table yet again:

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}

As we have seen, the existence of the elliptic points of period 22 is a result of the the fact that 5 2=15^2=-1 mod 1313. That there is a solution (namely x=5x=5) to the equation x 2=1x^2=-1 mod 1313 is also just a verbal restatement of the condition that (113)=1\left(\frac{-1}{13}\right)=1, the condition mentioned last time for Γ 0(13)\Gamma_0(13) to have elliptic points of period 22—so everything ties together.

It is also not a coincidence that x 2=1x^2=-1 is also true for x=ix=i, ii being the location of the elliptic point of period 22 in the complex upper half-plane under the action of Γ\Gamma. Taking the tiling by \mathbb{Z}-gons of the hyperbolic plane, there is a symmetry of the tiling which flips the edge whose midpoint is ii, and that is the symmetry which sends z1zz\rightarrow\frac{-1}{z}, and hence sends ii to itself.

Now, of course, x 2+1=0x^2+1=0 normally has two solutions, and that is indeed the case here: they are, of course, 55 and 5-5. The corresponding edges are just the left and right edge-55, and they both give an elliptic point, which is why X 0(13)X_0(13) has two elliptic points of period 22. Indeed, this is another way to see why, generally, X 0(p)X_0(p) for prime pp has two elliptic points of period 22, if it has any.

We can also see why X 0(2)X_0(2) has only one elliptic point: in 2\mathbb{Z}_2, x 2+1=0x^2+1=0 has only one solution, x=1x=1.

(In fact, this is an example of a more general phenomenon. We have that (x±1) p= k=0 p(p k)(±1) kx k(x\pm 1)^p=\sum_{k=0}^p{\left(\array{p\\k}\right)(\pm 1)^k x^k}, but (p k)\left(\array{p\\k}\right) is a multiple of pp unless k=0k=0 or k=pk=p, so, mod pp, (x±1) p=x p+(±1) p(x\pm 1)^p=x^p+(\pm 1)^p. So, for instance, x 2+1x^2+1 factorises as (x1) 2(x-1)^2, and both solutions are the same. We’ll see this sort of thing again in a moment.)

Now for some comments on the elliptic points of period 33. As we said last time, we can see the following cycle in the table:

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

Note that the way the elliptic point arises is that edge-33 folds back against edge-44, with the vertex between them being the elliptic point. In order for this to work, we need, of course, for the relevant projective unit (here, the unit 33) to send the relevant edge (here, edge-33) to an edge that shares a vertex with it (here, edge-44), that is, an edge whose number differs from its own by 11. That it works in this case follows from the fact that 3 2=43^2=-4 mod 1313, and 44 differs from 33 by 11, i.e. 3 2=(3+1)3^2=-(3+1) mod 1313—again, the minus sign ensures that edge-33 folds back against edge-44, rather than being parallel to it. Again, non-coincidentally, x 2=(x+1)x^2=-(x+1) is also true of ω\omega, the non-trivial cube root of 11 in the upper complex half-plane, where the elliptic point of period 33 lifts from.

And, still non-coincidentally, 33 is also a non-trivial cube root of 11—but mod 1313. Consider another way of looking at how the elliptic points of period 33 arise. Multiplication by 33 sends face 11 to face 33, face 33 to face 44, and face 44 to face 11.

3 4 4 3 4 3 1 3 4

When we quotient by this rotation, we get an elliptic point of period 33 precisely because 33 is a non-trivial cube root of 11.

Now, the non-trivial cube roots of 11 in the complex plane are of the form 12±32-\frac{1}{2}\pm\frac{\sqrt{-3}}{2}, and the same is true in N\mathbb{Z}_N. This is how the existence of elliptic points of period 33 gets to be equivalent to the condition that (3N)=1\left(\frac{-3}{N}\right)=1.

Now, these values are the solutions to x 2+x+1=0x^2+x+1=0. Just as we saw with period 22, there are normally two solutions to this equation, and there are normally two corresponding elliptic points of period 33. For instance, mod 1313, there is one period-33 elliptic point between edge-33 and edge-44, and another between edge-99 and edge-1010. But mod 33, we have (x1) 3=x 31(x-1)^3=x^3-1, so the non-trivial cube roots of 11 are the same as each other (and the same as the trivial root!)—namely 11—and so X 0(3)X_0(3) only has one elliptic point of period 33.

So again, everything ties together.

And I think that will do for this episode.

Next time, we’ll have more pictures and less alegbra.

Posted at January 27, 2011 5:51 PM UTC

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11 Comments & 0 Trackbacks

Re: Pictures of Modular Curves (XI)

I’d like to make an interesting comment on this post, but I suspect I’m like many of our readers here: I’m quite busy, and it looks like I’d have to go back and reread the last one before I could really understand this one… so I find myself staring at it goggle-eyed without much understanding. But I don’t want you to think I’m just ignoring it!

So you’re describing how some modular curves X 0(N)X_0(N) can be obtained by taking polygons and gluing their edges together…

Posted by: John Baez on February 1, 2011 3:25 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Oh dear, yes, I’ve overdone it again, haven’t I? It all seemed so clear the last time I read through it …

The key central fact is: if we number the edges of the denominator-11 piece sequentially from 00 (where it contacts the denominator-00 piece) to N1N-1, then, for all edges, edge nn is glued to edge 1n\frac{-1}{n}, and if the edges have a consistent orientation, then they are glued antiparallel. Everything else is an attempt to explain why this is true and follow through the consequences.

I’m sticking to prime NN here. In that case, the only tiles are a denominator-00 piece with one edge, and a denominator-11 piece with NN edges (that’s because, if NN is prime, it will only have 22 factors, and each of them gives rise to a tile). This makes the pictures simpler (!)

However, I’ve also decided not to number the edges sequentially all the way around, but to number increasing from 00 in both directions, going up to N12\frac{N-1}{2} on each side. This has the advantage of capturing a symmetry in the surfaces, but at the expense of having to keep track of signs and sides. That is, we might be connecting one of the two edges labelled nn to either a) one of the edges labelled 1n\frac{1}{n} or b) one of the edges labelled 1n\frac{-1}{n}. So now it becomes necessary to work out which.

After dealing with these surfaces a great deal, one gains a facility in making these calculations and they come to seem misleadingly obvious …

Posted by: Tim Silverman on February 1, 2011 12:21 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Oh dear, yes, I’ve overdone it again, haven’t I?

Not sure what you are thinking of. For whatever it’s worth my observation is: your exposition is maybe lacking clear statements or at least clear highlighting of statements that summarize the main insights gained.

You may not want to stick to the formal definition-theorem-proof yoga in a blog post, but I think a little bit of that kind of organizing structure would gain you a lot more active readership here among people who may have only 15 minutes per day to spend on this. What you currently have is in the style of “now consider this and then we do that” which may be very nice to read for readers who have followed what’s going on, but may make re-entry impossible for anyone who took a break after, say, the first half dozen of installments.

For instance above in the writeup I see that at some point you write

QED

I have to say that, looking briefly through the text, I cannot say easily what statement it is that you claim to have provided proof of here. WED? What was to be demonstrated?

I am sure it’s all clear and obvious to anyone who has read 11 posts first line to last line, but mabe few people did that. Maybe people would maybe appreciate a re-entering point. For istance a boldface Statement and then a crisp one-sentence “We now observe that …”. Followed by “Proof … QED”, to be skipped on first reading.

Posted by: Urs Schreiber on February 1, 2011 12:41 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Yes, thanks Urs. By “overdone”, I meant there was too much material all at once, but of course lack of a clear line through it exacerbates the difficulty.

Posted by: Tim Silverman on February 1, 2011 5:52 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Anyway, Tim, I wasn’t trying to say you were doing anything wrong — other than talking about math.

I was just wanting to apologize for not having time to carefully read it! I’m sure it’ll be really helpful to people who are struggling to understand modular curves.

Your summary explanation helps, though.

Posted by: John Baez on February 2, 2011 5:44 AM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

I wasn’t trying to say you were doing anything wrong

Well, it’s kind of you to say so, John … but if the whole audience is sitting in baffled silence after attempting to struggle through a post, then I don’t think I’ve achieved quite the outcome I was hoping for. I think (if it doesn’t strain everyone’s patience too much) I will add an episode in which I take one example—maybe N=11N=11 would be nice—briefly recapitulate the story so far as it pertains to that example, and then go though the key point above, step by step, specifically on that example. In trying to cut the whole series down to reasonable size(!), I left a lot of stuff on the cutting room floor, but some of that may have been a necessary part of the story I was trying to tell.

Posted by: Tim Silverman on February 2, 2011 1:53 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

I meant this as an honest question:

at the point where it says “QED” in the above, what is the statement that is meant to have been proven?

Could you just (re?)state this for me?

I believe I can deduce some parts of what it ought to be, but I am not sure if I know what the full statement is.

I am guessing it’s something like this:

If NN \in \mathbb{N} is a prime (or maybe just if N=13N=13 or N=11N= 11?) then the curve called X 0(N)X_0(N) may be built from gluing of its tiles (as defined in installment xx) along their faces by the following rule: edge kk of face nn is glued to edge kk' of face nn' where k=..k' = .. and n=...n' = ... (some algorithm applied to the gluing table) and where the gluing is either orientation preserving if the product of … with … is 1 or orientation reversing if it is -1.

Maybe that’s way off. Just a guess from looking at this installment. But I expect that you can make such a concise statement within one sentence or two, and that it would help (me, at least) get the message. As you say

So again, everything ties together.

but it would help me if you just said explicitly (again?) what “everything” is and maybe highlighted again what the conclusion induicated by “So” is explicitly.

(Sorry, that’s probably very frustrating for you to hear me ask this.)

I also have to admit that I don’t follow that description of the algorithm for the gluing tables, but now that I look at them again I am beginning to suspect that maybe the SVG is not rendering properly on my system (Firefox on Win), because on my system the table where it says

The coloured lines connect faces on opposite sides of the same edge, e.g. edge-6 of face 1

neither has a column labeled “edge-6” nor do the colored lines that I see seem to connect anything. They seem to start and end rather randomly. But as I said, now I am beginning to suspect that that’s a problem maybe with the SVG display on my end.

Posted by: Urs Schreiber on February 4, 2011 3:23 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Sorry—busy evening and I haven’t had a chance to reply. I’ll get back to this tomorrow. The lines on the SVG should line up though—they do for me (Firefox 3.6 on a Mac).

Posted by: Tim Silverman on February 4, 2011 10:22 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

I do apologise for the delay in answering. Last night I actually dreamt that I met you in person and you asked me about this! I’ve been trying to create the briefest possible summary of what’s happened so far in the series, in the hope of giving some context and plausibility to what I was trying to say in the post, but without success. So here is a largely context-free reply to the first part of your comment …

The answer to your first question is basically what I told John a little earlier, as follows:

  • The curves X 0(N)X_0(N) (as well as X 1(N)X_1(N) and X(N)X(N)) can be tiled in the ways described in earlier episodes (which I’ll summarise a bit later in this comment).
  • For prime NN, X 0(N)X_0(N) has just two tiles, one with a single edge (tile 00) and one with with NN edges (tile 11)
  • One edge of tile 11 is identified with the single edge of tile 00
  • The other N1N-1 edges of tile 11 are identified with each other (or sometimes with themselves with reverse orientation) as follows
  • If we number the edges of tile 11 starting with 00 on the edge shared with tile 00, and continuing 11, 22, etc going around the tile clockwise (or anticlockwise) to edge N1N-1, then edge uu is identified with edge 1u\frac{-1}{u} mod NN.
  • If we give the edges of tile 11 a consistent orientation, then edges that are identified with each other get glued together with opposite orientation.

However, when I introduced the edge numbering, I didn’t label the edges 00 to N1N-1. Rather, although I started at 00 on the edge shared with tile 00, I then labelled both of the edges adjacent to that one with a 11, and then continued increasing the numbering down both sides of the tile 22, 33, etc to N12\frac{N-1}{2}. (An alternative description: I started with a labelling from 00 to N1N-1 as described above, and then I worked my way half way around the tile replacing each label uu by u-u.) There was a reason for this, but it makes the numbering ambiguous, since there are now generally two edges per label (exceptions being label 00, and also label N2\frac{N}{2} if NN is even). So the last two bulletted statements above both get divided into two parts, one about how labels get paired up when we identify edges with each other, and then another one disambiguating between the two edges with the same label.

Thus I first argued, concerning the identifications of labels, that if edges with labels uu and vv are identified with each other, then we must have uv=±1u v=\pm 1 (that’s the first part). Then I tried to sort out the relationship among signs, sides and orientations (that’s the second part). And the second part is basically what I was trying to prove in the bit that ends “QED”, and which starts under the heading “Edge Directions”.

I’ll put a reply to the second half of your comment into a second comment.

Posted by: Tim Silverman on February 22, 2011 4:51 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

As for “everything ties together”, I confess that was very vague and probably shouldn’t have too much weight put on it, but I will try to indicate the sort of thing I was talking about:

I was talking about the elliptic points that arise on some of these tiled surfaces. In this context, elliptic points are points on a conformal surface where the angle around the point is less than 360 360^\circ. In the cases under discussion, we have taken the quotient of a surface by the action of a discrete group, and the elliptic points are the images (in the quotient) of points fixed by non-trivial elements of the group. The non-trivial stabiliser groups in this context are always of order 22 or 33 (in which case we say the elliptic points are of period 22 or 33 respectively). And there are a bunch of different but related ways to spot the elliptic points.

We start with the action of the group PSL(2,)PSL(2, \mathbb{Z}) on the complex upper half plane, which works thus: PSL(2,)PSL(2, \mathbb{Z}) is the group of 2×22\times 2 matrices, of determinant 11, with integer entries, quotiented by its subgroup of scalar matrices {I,I}\{I, -I\}. These matrices act on the complex upper half plane by fractional linear transformations: (a b c d):τaτ+bcτ+d\left(\array{a&b\\c&d}\right):\tau\rightarrow\frac{a\tau+b}{c\tau+d}.

In this case, there are two elliptic points. One is of period 22 and lies at the image of ii, the square root of 1-1. It’s the fixed point of the action of the matrix (0 1 1 0)\left(\array{0&1\\-1&0}\right) sending τ1τ\tau\rightarrow\frac{-1}{\tau}. The second is of period 33 and lies at the image of ω\omega, the complex cube root of 11 lying in H\mathbf{H}. It’s the fixed point of the action of the matrix (0 1 1 1)\left(\array{0&-1\\1&1}\right) sending τ1τ+1\tau\rightarrow\frac{-1}{\tau+1}. I talked about this in episode VII.

All the other surfaces I talk about are quotients of the upper half plane H\mathbf{H} by subgroups of PSL(2,)PSL(2, \mathbb{Z}); in other words, H/PSL(2,)\mathbf{H}/PSL(2, \mathbb{Z}) is the smallest surfaces I deal with and all the others are coverings of it. The period 22 and period 33 elliptic points of those covering surfaces always lie over the period 22 and period 33 elliptic point (respectively) of the base surface.

We started, back in episodes I–III, by looking at quotients of H\mathbf{H} by the groups called Γ(N)\Gamma(N), i.e. the subgroups of PSL(2,)PSL(2, \mathbb{Z}) consisting of matrices which are element-wise congruent mod NN to the identity matrix. After adding extra points to these quotient surfaces, corresponding to rational numbers and the point at \infty, we get compact surfaces. Importantly, we can tile the surface H/Γ(N)\mathbf{H}/\Gamma(N) by regular NN-gons in a way that corresponds to its algebraic and arithmetic structure.

To very briefly summarise the key arithmetico-geometric property I’ve been focusing on: the centre of each NN-gon is the image of a set of rationals sharing the property that, when they are expressed as fractions reduced to their lowest form (and adopting some conventions about how to cancel minus signs), their numerators are mutually congruent mod NN, and so are their denominators. Hence we can express the congruence class by a “fraction reduced mod NN”, with its numerator and denominator both drawn from the ring N\mathbb{Z}_N (integers mod NN). (We also need to make the identification of reduced fractions pqpq\frac{p}{q}\equiv\frac{-p}{-q}.) Since the centre of each NN-gon is the image of this congruence class, we can label the NN-gons by the fractions reduced mod NN. In fact, this is a bijection.

Also, the symmetry group of the tiling is isomorphic to the quotient group PSL(2,)/Γ(N)PSL(2, \mathbb{Z})/\Gamma(N). That is, speaking somewhat metaphorically, after we have quotiented by the effect of the group Γ(N)\Gamma(N), we have some residual action of PSL(2,)PSL(2, \mathbb{Z}) left over, and this action can be treated as the group of symmetries of the tiling. The quotient group PSL(2,)/Γ(N)PSL(2, \mathbb{Z})/\Gamma(N) is the group I have been calling PSL(2, N)PSL(2, \mathbb{Z}_N). That is, it still consists of the 2×22\times 2 matrices of determinant 11, quotiented by the group of scalar matrices {I,I}\{I, -I\}, but now its elements are drawn not from the integers \mathbb{Z}, but from the integers mod NN, N\mathbb{Z}_N. It acts by fractional linear transformations on the fractions reduced mod NN, thus: (a b c d):pqap+bqcp+dq\left(\array{a&b\\c&d}\right):\frac{p}{q}\rightarrow\frac{a p+b q}{c p+d q}. Since the fractions reduced mod NN are the labels on the NN-gons of the tiling, that tells us the action as symmetries of the tiling.

Anyway, back to elliptic points:

Now, taking the quotient of the complex upper half plane H\mathbf{H} by the group Γ(N)\Gamma(N) doesn’t give a surface with any elliptic points (except in the case N=0N=0), but we now proceed to take quotients by larger groups. Each larger group contains Γ(N)\Gamma(N) for some NN, so we can proceed by first taking the quotient by Γ(N)\Gamma(N), and then further taking a quotient by some subgroup of PSL(2, N)PSL(2, \mathbb{Z}_N).

So first, we take the quotient by the group of matrices of the form (1 b 0 1)\left(\array{1&b\\0&1}\right), where bb is any element of N\mathbb{Z}_N. This group fixes each reduced fraction whose denominator is 00. In terms of the tiling, it consists of the rotations of the tiling around the centre of any tile labeled with a fraction whose denominator is 00. From the tiled surfaces, we get a tiled quotient surface, tiled with an assortment of polygons; the polygons have various different numbers of edges (the numbers all being factors of NN). In particular, if NN is prime, we get two classes of tile: N12\frac{N-1}{2} NN-gons, labelled by fractions of the form 1q\frac{1}{q}, q0q\neq 0; and N12\frac{N-1}{2} 11-gons, labelled by fractions of the form p0\frac{p}{0}, p0p\neq 0. These surfaces are the surfaces called X 1(N)X_1(N). They only have any elliptic points in the cases N=2N=2—in which case we have one elliptic point of period 22—and N=3N=3—in which case we have one elliptic point of period 33.

Finally, we quotient even further, by the group of matrices of the form (a b 0 d)\left(\array{a&b\\0&d}\right), where aa, bb and dd are any element of N\mathbb{Z}_N, subject to the requirement that the matrix has determinant 11. The latter requirement forces aa and dd to be units in the ring N\mathbb{Z}_N, with ad=1a d=1. In particular, we can focus on the action of the group of matrices of the form (a 0 0 d)\left(\array{a&0\\0&d}\right), acting on the surface X 1(N)X_1(N) briefly described in the last paragraph. This group, which is isomorphic to the group of units in the ring N\mathbb{Z}_N, should help us to understand the elliptic points in the resulting quotient surface, which is called X 0(N)X_0(N).

And (to finally get to the point): what I was attempting unsuccessfully to explain in the second part of the post (at least for the case of prime NN) was that, not only do the square root of 1-1 and cube root of 11 in the complex upper half plane give elliptic points in the quotient of H\mathbf{H} by PSL(2,)PSL(2, \mathbb{Z}), but, also, square roots of 1-1 and cube roots of 11 in the ring N\mathbb{Z}_N control elliptic points in the surface X 0(N)X_0(N).

However, this comment has already run on far too long! Sometimes, the process of trying to get the whole story laid out clearly end to end for all to view feels like wresting a giant anaconda in a swamp. I’m sure there’s a way to make it lie down quietly but at the moment, when I get a grip on one part, another is sure to misbehave. I apologise again for letting this happen here.

I think I now have a better way to lay out this part of the story pictorially so I hope I will be able to be clearer in a future post.

Posted by: Tim Silverman on February 22, 2011 4:56 PM | Permalink | Reply to this

Re: Pictures of Modular Curves (XI)

Hi Tim,

sorry for appearing in your dreams, thanks for the reply, sorry for making you re-wrestle the anaconda but thanks for the summary. And finally sorry for my late reply.

I don’t have spent due time on this. But now I know a good entry point (your latest two comments) and may come back to this when I need to know more.

Posted by: Urs Schreiber on March 14, 2011 12:12 PM | Permalink | Reply to this

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