The n-Category Café

The Action by Translations

This time, I’m going to consider instead the additive or translation action of the group of matrices which have the form $\left(\array{1&b\\0&1}\right)$ (with entries in $\mathbb{Z}_N$) for arbitrary $b$, sometimes written $\left(\array{1&*\\0&1}\right)$. If we take the lift of this group to the integers (that is, the integer-valued matrices of the form $\left(\array{1&b\\0&1}\right)$ mod $N$ lying in $PSL(2,\mathbb{Z})$) then we get the subgroup known as $\Gamma_1(N)$, and taking the quotient of the upper halfplane by this group gives a space known as $X_1(N)$. So one way of looking at what we are about to do is that, in order to get to $X_1(N)$, we are taking a quotient by $\Gamma_1(N)$ in two stages: first taking a quotient by $\Gamma(N)$ (which does the “mod $N$” part), then taking a quotient by the mod-$N$ image of $\Gamma_1(N)$ (which does the remaining “additive” part of the operation).

Now, over the integers, $\left(\array{1&b\\0&1}\right)$ just adds $b$, and, not surprisingly, the mod $N$ version adds $b$ mod $N$. And this, in turn, just amounts to a rotation, by $\frac{1}{N}$ of a full turn, about the central face $\frac{1}{0}$.

For example, we can slice the spherical dodecahedron into five sectors:

The division is perhaps easier to see in this view:

Under the action of $\left(\array{1&1\\0&1}\right)$, which rotates by $\frac{1}{5}$ of a circle, we identify these sectors with each other, so roll up one sector into a sort of cone.

To be clearer: one fifth (one sector) of the top pentagon ($\frac{1}{0}$) rolls up into a cone, like this:

The rather funny-shaped curve that makes up the bottom of the cone is a single edge of the original pentagon, wrapped round on itself, with the pointy tip corresponding to the vertices at the two ends of the original edge, identified together.

And then the $5$ pentagons with denominator $1$ get identified with each other, and roll up like this:

The curve at the top is one edge of the pentagon, curled back on itself; the vertical line is the seam where the edges on either side of that get sewn together; and then the lip at the bottom is made up of the remaining two edges. Putting these together, we get the following:

Then the bottom half of the sphere—the denominator $2$ pentagon, and the segment coming from the pentagon with fraction $\frac{2}{0}$—makes up another piece with the same shape as the above, but upside down; so that finally, putting them all together, we get this:

And that’s $X_1(5)$.

We can do the same thing for $N=4$:

$Next, N=3$:

and $N=2$:

And those are the curves $X_1(N)$! (At least, for $N<6$.)

Elliptic Points on $X_1(N)$

Since these conical figures have pointy bits at the ends, which one might suspect are points that are special in some way, I should, before moving on, say a little about elliptic points.

Elliptic points are points where, roughly, going around the point by some fraction of $360^\circ$ brings you back to where you started. Since we’re working on surfaces with conformal structures, with a good notion of angle, this is all well-defined and a big deal. Consider, for instance, the standard fundamental domain of $\Gamma$, i.e. of $PSL(2,\mathbb{Z})$. (Below, $\omega$ and $\omega^2$ are the complex cube roots of $1$):

The left and right edges are identified with each other, and in addition, the circular bottom segment is folded in half about its midpoint at $i$, so that the left half is identified with the right half.

The consequences are three:

Firstly, going halfway around $i$ (along the semicircle shown) brings you back to where you started: that is, after going around an angle of only $180^{\circ}$. Hence there is an elliptic point at $i$ of period $2$.

Secondly, starting vertically above the point at $\omega$ and going clockwise $60^\circ$ takes you to a point on the left part of the circular arc, which is identical to the mirror-opposite point on the right of the circular arc, and continuing clockwise from there a further $60^\circ$ takes you to a point above $-\omega^2$ which is identical to where you started: in short, a complete circuit of the point in only $120^\circ$ degrees. So there is an elliptic point at $\omega$ of period $3$.

Finally, the two vertical edges form a long, tapering tube going off to $\infty$, and tending toward meeting at an angle of $0^\circ$, so going around $\infty$ by $0^\circ$ brings you back to where you started, which makes it kind of like an elliptic point of period $\infty$. However, when we compactify by adding points (“cusps”) at the ends of these tubes (a standard procedure), the cusps end up as quite ordinary points in $X(N)$—not elliptic points. So we can forget about them. We don’t do this for other elliptic points, so we can’t forget about them: we need to think about them. And we are thinking about them—right now!

Now, in the surfaces $X(N)$ for $N>1$, multiple copies of the fundamental domain get packed together ($N$ per $N$-gon). There is no longer any folding over, so (once the cusps have been added) there are no elliptic points at all on these surfaces.

We can take a look at these fundamental domains on the sphere for, e.g. $N=5$:

Or, making it opaque for clarity:

Each edge of an $N$-gon (the thick lines) is the base of a fundamental triangle, with a lift of $i$ at its midpoint and lifts of $\omega$ at its endpoints. The centres of the $N$-gons are the cusps. We can see that, since two triangles are based on each edge of an $N$-gon, and six triangles around each vertex, both the period-$2$ and the period-$3$ elliptic points have been unfolded and do not lift to elliptic points on these surfaces.

What about the surfaces $X_1(N)$? These have pointy bits at their tops and bottoms, which look as though they might be elliptic points. However, with two exceptions, these are actually cusps (they come from the centre of an $N$-gon), and since the compactification by cusps occurs after we take the quotient, they go. The pointy appearance is actually a bit misleading.

The exceptions are the cases $N=2$ and $N=3$. There, the top points come from cusps, but the bottom points don’t.

Let’s take a look at the triangulation of these cases by lifts of the fundamental domain:

Or opaque (and viewed at a different angle) for clarity:

When forming the bottom part of $X_1(3)$:

we take one of the triangular faces of the tetrahedron and fold it so that two of its edges are sewn together: that is the seam you see running up from the bottom. That means that the point at the bottom occurs at a vertex of the triangle, and only covers $120^\circ$. So it is an elliptic point of period $3$. (It is at a vertex, so it is a lift of the elliptic point of period $3$ at $\omega$ in the fundamental domain.)

Similarly, for $N=2$:

The bottom part is formed of a single bigon. The bigon only has two edges: one of them forms the loop at the top, and the other is actually folded back on itself to form the seam you see. Thus the conical tip at the bottom is at the midpoint of an edge, so is an elliptic point of period $2$, lifting the elliptic point of period $2$ at $i$ in the fundamental domain.

These are the only elliptic points in the curves $X_1(N)$ (except for the trivial $N=1$ case, which is just another name for the same curve as $X(1)$).

Well, there hasn’t been much content this time, but there’ve been a lot of pictures, so I’ll stop here. Next time, I’ll talk about $X_1(N)$ for $N>5$, and in the posts after that I’ll go on to talk about $\Gamma_0(N)$ and $X_0(N)$. That will complete this set of posts about what modular curves look like. After that, I’ll move on to Hecke operators, but I’m not sure how I’m going to organise that, so it may be a long time before it appears. Though hopefully not as long as it took to get this series prepared.