The n-Category Café

Yes I do!

Hmm, interesting! I’ll think about that. In these articles — my column for the AMS Notices — I’m always trying to write in a way that ‘typical readers of the AMS Notices’ are likely to understand. So I imagine they know things like ‘fundamental domain’ and (earlier) ‘natural transformation’ and various other things. In reality not everyone will know all these things.

Ah, it’s for your Notices column.

One difference beteween “fundamental domain” and “natural transformation” is that the former can be defined in just a few words, e.g. “contains exactly one point of each orbit”.

Thanks, I’m glad you liked it! That means a lot.

Let’s see if I can figure some stuff out. If I understand correctly the modular lambda function is a function, not of elliptic curves per se, but of elliptic curves $E$ equipped with an ordered basis for their 2-torsion. For some reason this extra information is enough to define not just a branched double cover of a sphere with 4 branch points (which every elliptic curve gives, just taking $E \mapsto E/x \sim -x$) but a branched double cover with enough extra structure that the 4 points on the sphere have a well-defined cross ratio. Then the modular lambda function is this cross ratio!

Wikipedia then asserts:

By symmetrizing the lambda function under the canonical action of the symmetric group $S_3$ on $X(2)$, and then normalizing suitably, one obtains a function on the upper half-plane that is invariant under the full modular group $SL(2,\mathbb{Z})$, and it is in fact Klein’s modular $j$-invariant.

What’s $X(2)$? I guess it’s the moduli space of elliptic curves equipped with an ordered basis for their 2-torsion. What’s the canonical action of $S_3$ on $X(2)$? The 2-torsion of an elliptic curve consists of the identity together with 3 other points, which are all ‘equally good’. An ordered basis for the 2-torsion requires choosing 2 of these 3 points to be the first and second basis elements, leaving the third to be the odd man out. So I guess $S_3$ must act by permuting who counts as ‘first’, who counts as ‘second’, and who counts as ‘the odd man out’.

Why does an ordered basis for the 2-torsion put enough structure on the 4 branch points to make their cross-ratio well-defined? The cross ratio of 4 points on the sphere is actually invariant under some permutations of those points. According to Wikipedia, to define the cross ratio we just need to partition the 4 points into two 2-element subsets. So we just need the ordered basis for the 2-torsion to provide such a partition.

But it seems to me that an ordered basis of the 2-torsion of an elliptic curve does much more! An ordered basis $x, y$ actually gives us names for all four 2-torsion points: $0, x, y$ and $x+y$. So we can unambiguously compute their cross-ratio without even fussing about as I was in the previous paragraph. Seems too good to be true!

Okay, next: why does a ‘labeled’ acute triangle give an elliptic curve with an ordered basis for its 2-torsion? This will allow us to define the modular lambda function of a labeled acute triangle.

What I’m calling a ‘labeled’ acute triangle has vertices labeled 1, 2, 3. Given such a thing we can create a parallelogram by attaching to this triangle another triangle: a copy of the original triangle that’s been rotated $180^\circ$ rotated along the edge 2 3. This parallelogram then has names for all four of its vertices: 1, 2, 3 and ‘the other guy’.

With such a naming scheme in place, when we curl up our parallelogram to form an elliptic curve, we can choose an identity element for this elliptic curve (say the vertex 1) and also choose an ordered basis for its 2-torsion (say the points halfway along the edges 1 2 and 1 3). So, we have enough information to define its modular lambda function!

I’m hoping that when someone hands us an unlabeled acute triangle we can compute its modular $j$ function by labeling it in all possible ways, computing the modular lambda function as above for each way, and then averaging over labelings.

However, I’m a bit confused. How are the 3! labelings of the triangle related to the 3! choices of ordered basis of 2-torsion for the resulting elliptic curve? They seem different, e.g. we use the label of one vertex in our triangle just to decide what’s the identity element of our elliptic curve.

Perhaps we can only compute the modular $j$ function of an ‘oriented’ triangle.

I’ll quit here, but ultimately I hope the story will be quite nice.

The orbifold $\mathcal{H}$//$PSL_2(\mathbb{Z})$ can perhaps be visualized as a tricorn hat with a pigtail going off to $i \infty$ (the cusp ?), with two orbifold points with isotropy of order two and one of order three between them, as below (snatched from https://math.stackexchange.com/questions/3032808;there are many accounts of this out there):

The group $PSL_2(\mathbb{Z})$ is a free product $𝐶_2 ∗ 𝐶_3$, and hence its nontrivial torsion elements have order two or three. It has one conjugacy class of cyclic subgroups of order two (a representative is $z\mapsto z^{-1}$ with fixed point $i$), one conjugacy class of cyclic subgroup of order three (a representative being generated by $z \mapsto -(1 + z)^{-1}$ fixing $j$, the unique root of $1 + z +z^2$ in the upper half-plane). The elements of infinite order have no fixed point on $H$.

So the points with nontrivial stabilizer are those in the orbit of $i$ and the orbit of $j$. They are not in the same orbit since otherwise the stabilizer should contain an element of order six….

[I get confused because I keep trying to think of the cusp as having an infinite cyclic isotropy group….]

That’s reminiscent of Bring’s genus four Riemann surface

• Wikipedia, Bring’s curve.

but I can’t be more precise…

In informal presentations mathematicians often say “let me define a thing… now consider two isomorphic things”, and expect that the audience can guess the relevant concept of isomorphism from the definition of the particular kind of thing. It may be unfair but it’s how math works, and I need to take advantage of such tricks to keep this article no more than 2 pages long.

But I’m happy to explain it here:

I said that an elliptic curve is a torus (a topological thing) that’s been made into a complex manifold (which is a manifold with coordinate charts that are copies of $\mathbb{C}^n$, with transition functions that are analytic functions—but here $n = 1$). So an isomorphism between them must be a homeomorphism of tori that’s locally described by a complex-analytic function that has a complex-analytic inverse. We just go through the definition of the thing and read off the definition of isomorphism between those things. Sometimes there are difficult choices to be made, but not here.

Equivalently: imagine a clock face, then connect the points at 12, 9 and 4 o’clock.

If you don’t have a protractor, you can do this by eye starting with just a pair of roughly perpendicular lines, and the result will be good enough.

On Mathstodon someone asked why my picture chops the usual fundamental domains for $SL(2,\mathbb{Z})$ in two. The easy explanation is that the fundamental domains for $GL(2,\mathbb{Z})$ are half as big.

But another way to put it is this: we usually consider the moduli space of elliptic curves with holomorphic diffeomorphisms between them as isomorphisms; these are orientation-preserving. This is $\mathcal{H}/SL(2,\mathbb{Z})$. But we can also study the moduli space of elliptic curves with holomorphic or antiholomorphic diffeomorphisms as isomorphisms: antiholomorphic maps are orientation-reversing. This is $\mathcal{H}/GL(2,\mathbb{Z})$.

There’s a branched double cover

$$\mathcal{H}/SL(2,\mathbb{Z}) \to \mathcal{H}/GL(2,\mathbb{Z})$$

where the branch points come from reflection-invariant lattices.

Thanks!

Yes, it’s possible to see this directly.

First some review: remember that $p$ sends triangles in the complex plane with vertices labeled $1,2,3$ to elliptic curves as follows: we take the triangle, rotate it 180${}^\circ$ around the midpoint of any edge to get another triangle, notice that these two triangles form a parallelogram, and identify the opposite edges of this parallelogram to form a torus — which is an elliptic curve because our parallelogram was sitting in the complex plane.

Actually I was lying a bit here: $p$ sends isomorphism classes of triangles in the complex plane with vertices labeled $1,2,3$ to isomorphism classes of elliptic curves.

But anyway: if we cyclically permute the labels $1,2,3$ we typically get a nonisomorphic triangle in the complex plane with vertices labeled $1,2,3$, but this new triangle always gives an isomorphic elliptic curve. There are 3 cyclic permutations of $1,2,3$, so our map $p$ is generically 3-to-1.

The confusing part is why $p$ is not 6-to-1. That is: why are only cyclic permutations allowed here?

From one of view this seems obvious: if change the labeling with a non-cyclic permutation, you get a new labeled triangle that’s isomorphic to a reflected version of the original one, and this gives a reflected parallelogram. And when you reflect a parallelogram, you get a new parallelogram that typically gives a different elliptic curve, because an elliptic curve comes with an orientation.

But from another point of view it seems unobvious: why does the labeling matter at all in this game?

Ah right, thanks!

I guess labeling does not matter but orientation matters here, and would explain one needs to cut down 3! by a half