## 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 $X_0(N)$ 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 $N$, 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 $0$ on the edge that separates it from a face with denominator $0$, and then increasing by $1$ 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 $N=13$.

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-$3$ of face $2$ connects it to face $6$.

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

The coloured lines connect faces on opposite sides of the same edge, e.g. edge-$6$ of face $1$ is the same as edge-$2$ of face $6$. Here’s the table for $N=11$:

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

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 $13$ and mod $11$ respectively—and always mod change of sign, of course. The missing column is for edge-$1$ and is omitted because it’s boring.)

For instance, look at the second row, for face $2$. The first column is for edge-$2$, we have $2\times 2=4$, and $4$ is what appears in the first column of the second row. The second column is for edge-$3$, we have $2\times 3=6$, but $-6=5$ mod $11$ and, since we’re modding out by sign changes and we’ve adopted the convention that the lesser of $+a$ and$-a$ will stand as the representative of both, $5$ is what appears in the second column of the second row. Similarly, $2\times 4=8$, $-8=3$ mod $11$, so $3$ 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 $1$ 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 $1$ 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-$2$ of face $1$ in the case $N=11$. This edge is the same as some edge of face $2$ (because edge-$u$ of face $1$ is always the same as some edge of face $u$). Which edge of face $2$ is it the same as? Well, we can look it up: the faces opposite each edge of face $2$ are given on row $2$, and we want to find the cell on row $2$ with $1$ in it. And this cell falls in the column belonging to edge-$5$. So edge-$5$ of face $2$ is the same as edge-$2$ of face $1$.

Yes, but why edge-$5$, particularly? Well, row $2$ is the same as row $1$ multiplied by $2$. So we want the edge whose number, when multiplied by $2$, gives $1$—in other words, the inverse of $2$ in the group of projective units mod $11$. Which is just $5$. And there’s nothing special about $2$ and $5$—this argument is quite general.

So, generally, edge-$u$ of one face is identified with edge-$v$ of another (or sometimes of the same) face precisely where $u v=1$ in the group of projective units. For edge-$u$ of face $1$, we can argue as in the paragraph above, looking at where $1$ falls in row $u$; and the pairings of edge labels on opposite sides of an edge are the same on all other faces as they are on face $1$, since the symmetry given by the action of projective units preserves edge labels (because it preserves denominators of $0$, from which the edge labellings derive).

Edge Directions

However, for any given non-zero edge number (such as $2$ or $5$) there are two edges with that number, one to the left and one to the right of edge-$0$. So, is the left edge-$2$ the same as the left edge-$5$, and the right edge-$2$ as the right edge-$5$—or is the left edge-$2$ identified with the right edge-$5$ and vice versa?

Also, we’ve given our edges an orientation (pointing away from edge-$0$). 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 $X_0(11)$, 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 $2$ and $5$ 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 $3$ and $4$ 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 $u$ and $v$, with $u v=1$ 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 $2\cdot 5=-1$, whereas $3\cdot 4=1$. It turns out that this is sufficient to tell us that edge-$2$ and edge-$5$ are joined same side/antiparallel, while edge-$3$ and edge-$4$ are joined opposite side/parallel.

We can remember this by imagining that the minus sign flips the direction of the edge, so if $u v=-1$, then edge-$u$ must be identified antiparallel with edge-$v$, and if $u v=1$, 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 $1$, and adjacent to that is face $a$. The edge between them must be numbered $a$ in the numbering of face $1$ (because on face $1$, edge labels match face numbers), while on the numbering of face $a$, we’ve determined in the discussion above that it must be the projective multiplicative inverse of $a$, which we denote by $a'$. So $a\cdot a'=\pm 1$.

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

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

Now we multiply faces numbers by $a'$.

We send $1\rightarrow a'$
We send $a\rightarrow 1$
We send $(a+1)\rightarrow (a+1)\cdot a'=(\pm 1+a')$

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

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

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

And if $a\cdot a'=-1$, then the right side of face $1$ must be identified with the right side of face $a$ (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 $11$ and see that, in $X_0(11)$, edge-$2$ is antiparallel to edge-$5$, while edge-$3$ is parallel to edge-$4$. 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 $0$, work our way anticlockwise (say) increasing the labels through all the numbers mod $N$. Orient the edges so they point in the direction of increasing numbers. Then edge-$u$ is identified with edge-$v$ precisely when $v=\frac{-1}{u}$, 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

$1\cdot 10=10=-1$ mod $11$.
$2\cdot 5=10=-1$ mod $11$.
$3\cdot 7=21=-1$ mod $11$.
$4\cdot 8=32=-1$ mod $11$.
$6\cdot 9=54=-1$ mod $11$.

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

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

This, again, has genus $1$. The cycle of edges $3\rightarrow 5\rightarrow 6\rightarrow 7$ on the left is joined to the same cycle on the right, but again with a $180^\circ$ twist. Since $4\cdot 4=-1$ mod $17$, this edge is folded back on itself and we have a pair of elliptic points of period 2.

Here’s $N=19$:

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

Since $7\cdot 8=-1$ mod $19$, these edges fold back against each other, and we have a pair of elliptic points of period $3$.

Here’s $N=23$:

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 $2$ case, and, not surprisingly, it’s a bit more complicated. Let’s take it step by step.

Step 1: identify edges $5$ and $9$ on the left ($5\cdot 9=45=-1$ mod $23$). This causes the sequence of edges $6\rightarrow 7\rightarrow 8$ to roll up into a loop. Do the same on the right.

Step 2: identify edges $2$ and $11$. With edges $5$ and $9$ already identified, this causes the sequence of edges $3\rightarrow 4\rightarrow 10$ to roll up into a loop.

Step 3: identify the $6\rightarrow 7\rightarrow 8$ loop on the left with the $3\rightarrow 4\rightarrow 10$ loop on the right, and vice versa. (To be precise, $6\rightarrow 7\rightarrow 8$ with $4\rightarrow 10\rightarrow 3$ in that order. E.g. $6\cdot 4=24=1$ mod $23$.) Each of these identifications produces a tube crossing from left to right, giving our genus $2$ curve.

$N=29$, another genus $2$ 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 $N=31$:

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 $N$, 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 $X_0(N)$ for non-prime $N$.

Elliptic Points in More Generality

Last time, we investigated how the elliptic points of $X_0(13)$ 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:

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

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

It is also not a coincidence that $x^2=-1$ is also true for $x=i$, $i$ being the location of the elliptic point of period $2$ in the complex upper half-plane under the action of $\Gamma$. 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 $i$, and that is the symmetry which sends $z\rightarrow\frac{-1}{z}$, and hence sends $i$ to itself.

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

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

(In fact, this is an example of a more general phenomenon. We have that $(x\pm 1)^p=\sum_{k=0}^p{\left(\array{p\\k}\right)(\pm 1)^k x^k}$, but $\left(\array{p\\k}\right)$ is a multiple of $p$ unless $k=0$ or $k=p$, so, mod $p$, $(x\pm 1)^p=x^p+(\pm 1)^p$. So, for instance, $x^2+1$ factorises as $(x-1)^2$, 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 $3$. As we said last time, we can see the following cycle in the table:

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 $1$

Note that the way the elliptic point arises is that edge-$3$ folds back against edge-$4$, 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 $3$) to send the relevant edge (here, edge-$3$) to an edge that shares a vertex with it (here, edge-$4$), that is, an edge whose number differs from its own by $1$. That it works in this case follows from the fact that $3^2=-4$ mod $13$, and $4$ differs from $3$ by $1$, i.e. $3^2=-(3+1)$ mod $13$—again, the minus sign ensures that edge-$3$ folds back against edge-$4$, rather than being parallel to it. Again, non-coincidentally, $x^2=-(x+1)$ is also true of $\omega$, the non-trivial cube root of $1$ in the upper complex half-plane, where the elliptic point of period $3$ lifts from.

And, still non-coincidentally, $3$ is also a non-trivial cube root of $1$—but mod $13$. Consider another way of looking at how the elliptic points of period $3$ arise. Multiplication by $3$ sends face $1$ to face $3$, face $3$ to face $4$, and face $4$ to face $1$.

3 4 4 3 4 3 1 3 4

When we quotient by this rotation, we get an elliptic point of period $3$ precisely because $3$ is a non-trivial cube root of $1$.

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

Now, these values are the solutions to $x^2+x+1=0$. Just as we saw with period $2$, there are normally two solutions to this equation, and there are normally two corresponding elliptic points of period $3$. For instance, mod $13$, there is one period-$3$ elliptic point between edge-$3$ and edge-$4$, and another between edge-$9$ and edge-$10$. But mod $3$, we have $(x-1)^3=x^3-1$, so the non-trivial cube roots of $1$ are the same as each other (and the same as the trivial root!)—namely $1$—and so $X_0(3)$ only has one elliptic point of period $3$.

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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### 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 $X_0(N)$ 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-$1$ piece sequentially from $0$ (where it contacts the denominator-$0$ piece) to $N-1$, then, for all edges, edge $n$ is glued to edge $\frac{-1}{n}$, 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 $N$ here. In that case, the only tiles are a denominator-$0$ piece with one edge, and a denominator-$1$ piece with $N$ edges (that’s because, if $N$ is prime, it will only have $2$ 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 $0$ in both directions, going up to $\frac{N-1}{2}$ 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 $n$ to either a) one of the edges labelled $\frac{1}{n}$ or b) one of the edges labelled $\frac{-1}{n}$. 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.

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 $N=11$ 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 $N \in \mathbb{N}$ is a prime (or maybe just if $N=13$ or $N= 11$?) then the curve called $X_0(N)$ may be built from gluing of its tiles (as defined in installment $x$) along their faces by the following rule: edge $k$ of face $n$ is glued to edge $k'$ of face $n'$ where $k' = ..$ and $n' = ...$ (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 $X_0(N)$ (as well as $X_1(N)$ and $X(N)$) can be tiled in the ways described in earlier episodes (which I’ll summarise a bit later in this comment).
• For prime $N$, $X_0(N)$ has just two tiles, one with a single edge (tile $0$) and one with with $N$ edges (tile $1$)
• One edge of tile $1$ is identified with the single edge of tile $0$
• The other $N-1$ edges of tile $1$ are identified with each other (or sometimes with themselves with reverse orientation) as follows
• If we number the edges of tile $1$ starting with $0$ on the edge shared with tile $0$, and continuing $1$, $2$, etc going around the tile clockwise (or anticlockwise) to edge $N-1$, then edge $u$ is identified with edge $\frac{-1}{u}$ mod $N$.
• If we give the edges of tile $1$ 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 $0$ to $N-1$. Rather, although I started at $0$ on the edge shared with tile $0$, I then labelled both of the edges adjacent to that one with a $1$, and then continued increasing the numbering down both sides of the tile $2$, $3$, etc to $\frac{N-1}{2}$. (An alternative description: I started with a labelling from $0$ to $N-1$ as described above, and then I worked my way half way around the tile replacing each label $u$ by $-u$.) There was a reason for this, but it makes the numbering ambiguous, since there are now generally two edges per label (exceptions being label $0$, and also label $\frac{N}{2}$ if $N$ 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 $u$ and $v$ are identified with each other, then we must have $u v=\pm 1$ (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 $360^\circ$. 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 $2$ or $3$ (in which case we say the elliptic points are of period $2$ or $3$ respectively). And there are a bunch of different but related ways to spot the elliptic points.

We start with the action of the group $PSL(2, \mathbb{Z})$ on the complex upper half plane, which works thus: $PSL(2, \mathbb{Z})$ is the group of $2\times 2$ matrices, of determinant $1$, with integer entries, quotiented by its subgroup of scalar matrices $\{I, -I\}$. These matrices act on the complex upper half plane by fractional linear transformations: $\left(\array{a&b\\c&d}\right):\tau\rightarrow\frac{a\tau+b}{c\tau+d}$.

In this case, there are two elliptic points. One is of period $2$ and lies at the image of $i$, the square root of $-1$. It’s the fixed point of the action of the matrix $\left(\array{0&1\\-1&0}\right)$ sending $\tau\rightarrow\frac{-1}{\tau}$. The second is of period $3$ and lies at the image of $\omega$, the complex cube root of $1$ lying in $\mathbf{H}$. It’s the fixed point of the action of the matrix $\left(\array{0&-1\\1&1}\right)$ sending $\tau\rightarrow\frac{-1}{\tau+1}$. I talked about this in episode VII.

All the other surfaces I talk about are quotients of the upper half plane $\mathbf{H}$ by subgroups of $PSL(2, \mathbb{Z})$; in other words, $\mathbf{H}/PSL(2, \mathbb{Z})$ is the smallest surfaces I deal with and all the others are coverings of it. The period $2$ and period $3$ elliptic points of those covering surfaces always lie over the period $2$ and period $3$ elliptic point (respectively) of the base surface.

We started, back in episodes I–III, by looking at quotients of $\mathbf{H}$ by the groups called $\Gamma(N)$, i.e. the subgroups of $PSL(2, \mathbb{Z})$ consisting of matrices which are element-wise congruent mod $N$ 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 $\mathbf{H}/\Gamma(N)$ by regular $N$-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 $N$-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 $N$, and so are their denominators. Hence we can express the congruence class by a “fraction reduced mod $N$”, with its numerator and denominator both drawn from the ring $\mathbb{Z}_N$ (integers mod $N$). (We also need to make the identification of reduced fractions $\frac{p}{q}\equiv\frac{-p}{-q}$.) Since the centre of each $N$-gon is the image of this congruence class, we can label the $N$-gons by the fractions reduced mod $N$. In fact, this is a bijection.

Also, the symmetry group of the tiling is isomorphic to the quotient group $PSL(2, \mathbb{Z})/\Gamma(N)$. That is, speaking somewhat metaphorically, after we have quotiented by the effect of the group $\Gamma(N)$, we have some residual action of $PSL(2, \mathbb{Z})$ left over, and this action can be treated as the group of symmetries of the tiling. The quotient group $PSL(2, \mathbb{Z})/\Gamma(N)$ is the group I have been calling $PSL(2, \mathbb{Z}_N)$. That is, it still consists of the $2\times 2$ matrices of determinant $1$, quotiented by the group of scalar matrices $\{I, -I\}$, but now its elements are drawn not from the integers $\mathbb{Z}$, but from the integers mod $N$, $\mathbb{Z}_N$. It acts by fractional linear transformations on the fractions reduced mod $N$, thus: $\left(\array{a&b\\c&d}\right):\frac{p}{q}\rightarrow\frac{a p+b q}{c p+d q}$. Since the fractions reduced mod $N$ are the labels on the $N$-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 $\mathbf{H}$ by the group $\Gamma(N)$ doesn’t give a surface with any elliptic points (except in the case $N=0$), but we now proceed to take quotients by larger groups. Each larger group contains $\Gamma(N)$ for some $N$, so we can proceed by first taking the quotient by $\Gamma(N)$, and then further taking a quotient by some subgroup of $PSL(2, \mathbb{Z}_N)$.

So first, we take the quotient by the group of matrices of the form $\left(\array{1&b\\0&1}\right)$, where $b$ is any element of $\mathbb{Z}_N$. This group fixes each reduced fraction whose denominator is $0$. 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 $0$. 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 $N$). In particular, if $N$ is prime, we get two classes of tile: $\frac{N-1}{2}$ $N$-gons, labelled by fractions of the form $\frac{1}{q}$, $q\neq 0$; and $\frac{N-1}{2}$ $1$-gons, labelled by fractions of the form $\frac{p}{0}$, $p\neq 0$. These surfaces are the surfaces called $X_1(N)$. They only have any elliptic points in the cases $N=2$—in which case we have one elliptic point of period $2$—and $N=3$—in which case we have one elliptic point of period $3$.

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

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 $N$) was that, not only do the square root of $-1$ and cube root of $1$ in the complex upper half plane give elliptic points in the quotient of $\mathbf{H}$ by $PSL(2, \mathbb{Z})$, but, also, square roots of $-1$ and cube roots of $1$ in the ring $\mathbb{Z}_N$ control elliptic points in the surface $X_0(N)$.

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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