The n-Category Café

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):

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.

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:

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).

$1\cdot 10=10=-1$ 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 $N=17$:

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 $N=19$:

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 $N=23$:

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:

And $N=31$:

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 $1$.

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.