## September 23, 2023

### The Moduli Space of Acute Triangles

#### Posted by John Baez

I wrote a little article explaining the concept of ‘moduli space’ through an example. It’s due October 1st so I’d really appreciate it if you folks could take a look and see if it’s clear enough. It’s really short, and it’s written for people who know some math, but not necessarily anything about moduli spaces.

The cool part is the connection between the moduli space of acute triangles — that is, the space of all shapes an acute triangle can have — and the more famous moduli space of elliptic curves.

There’s a lot more one could do with this, e.g. describing the modular lambda function as the cross ratio of the 4 points on a sphere obtained by taking an acute triangle, dividing it into 4 similar acute triangles, and folding it up to form a tetrahedron, which is conformally equivalent to a sphere. But I didn’t have space for that here!

Okay, here goes.

### The moduli space of acute triangles

In mathematics we often like to classify objects up to isomorphism. Sometimes the classification is discrete, but sometimes we have a notion of when two objects are ‘close’. Then we can make the set of isomorphism classes into a topological space called a ‘moduli space’. A simple example is the moduli space of acute triangles. In simple terms, this is the space of all possible shapes that an acute triangle can have, where we count two triangles as having the same shape if they are similar.

As a first step, consider triangles with labeled vertices in the complex plane. Every triangle is similar to one with its first vertex at $0$, the second at $1$, and the third at some point in the upper half-plane. This triangle is acute precisely when its third vertex lies in this set:

$T = \left\{z \in \mathbb{C} \; \left\vert \; \mathrm{Im}(z) \gt 0, \; 0 \lt \mathrm{Re}(z) \lt 1, \; |z - \tfrac{1}{2}| \gt \tfrac{1}{2} \right. \right\}$

So, we say $T$ is the moduli space of acute triangles with labeled vertices. This set is colored yellow and purple above; the yellow and purple regions on top extend infinitely upward.

To get the moduli space of acute triangles with unlabeled vertices, we must mod out $T$ by the action of $S_3$ that permutes the three vertices. The 6 yellow and purple regions in $T$ are ‘fundamental domains’ for this $S_3$ action: that is, they each contain exactly one point from each orbit. If we reflect a labeled triangle corresponding to a point in a yellow region we get a triangle corresponding to a point in a purple region, and vice versa. Points on the boundary between two regions correspond to isosceles triangles. All 6 regions meet at the point that corresponds to an equilateral triangle.

The moduli space of acute triangles is closely related to a more famous moduli space: that of elliptic curves. The group $\mathrm{GL}(2,\mathbb{Z})$, consisting of invertible $2 \times 2$ integer matrices, acts on the upper half-plane

$\mathcal{H} = \left\{z \in \mathbb{C} \; \left\vert \; \mathrm{Im}(z) \gt 0 \right. \right\}$

as follows:

$\left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \colon z \mapsto \frac{a z + b}{c z + d} .$

The light and dark regions shown above are fundamental domains for this group action. Elements of $\mathrm{GL}(2,\mathbb{Z})$ with determinant $-1$ map light regions to dark ones and vice versa. Elements with determinant $1$ map light regions to light ones and dark ones to dark ones. People more often study the action of the subgroup $\mathrm{SL}(2,\mathbb{Z})$ consisting of $g \in \mathrm{GL}(2,\mathbb{Z})$ with determinant 1. The union of a light region and a dark one, touching each other, forms a fundamental domain of $\mathrm{SL}(2,\mathbb{Z})$.

For any point $z \in \mathcal{H}$ we can form a parallelogram with vertices $0, 1, z$ and $z+1$. If we identify the opposite edges of this parallelogram we get an elliptic curve: a torus equipped with the structure of a complex manifold. We can get every elliptic curve this way, at least up to isomorphism. Moreover, two points $z,z' \in \mathcal{H}$ in the upper half-plane give isomorphic elliptic curves iff $z' = g z$ for some $g \in \mathrm{SL}(2,\mathbb{Z})$. Thus the quotient space $\mathcal{H}/\mathrm{SL}(2,\mathbb{Z})$ is the moduli space of elliptic curves: points in this space correspond to isomorphism classes of elliptic curves.

Since $T$ is the union of three fundamental domains for $\mathrm{SL}(2,\mathbb{Z})$, there is a map

$p \colon T \to \mathcal{H}/\mathrm{SL}(2,\mathbb{Z})$

from the moduli space of acute triangles to the moduli space of elliptic curves, and generically this map is three-to-one. This map is not onto, but if we take the closure of $T$ inside $\mathcal{H}$ we get a larger set

$\overline{T} = \left\{z \in \mathbb{C} \; \vert \; \mathrm{Im}(z) \gt 0, \; 0 \le \mathrm{Re}(z) \le 1, \; |z - \tfrac{1}{2}| \ge \tfrac{1}{2} \right\}$

whose boundary consists of points corresponding to right triangles. Then $p$ extends to an onto map

$p \colon \overline{T} \to \mathcal{H}/\mathrm{SL}(2,\mathbb{Z}) .$

The existence of this map suggests that from any acute or right triangle in the plane we can construct an elliptic curve, in such a way that similar triangles give isomorphic elliptic curves. This is in fact true! How can we understand this more directly?

Take any acute or right triangle with labelled vertices in the complex plane. Rotating it 180° around the midpoint of any edge we get another triangle. The union of these two triangles is a parallelogram. Identifying opposite edges of this parallelogram we get a torus with a complex structure — and this is an elliptic curve! There are three choices of how to build this parallelogram, one for each edge of the original triangle, but they give isomorphic elliptic curves. Also, similar triangles give isomorphic elliptic curves. Even better, every elliptic curve is isomorphic to one arising from this construction. So this construction gives a map from $\overline{T}$ onto $\mathcal{H}/\mathrm{SL}(2,\mathbb{Z})$, and with a little thought one can see that this map is $p$.

I learned about the moduli space of acute triangles from James Dolan. There has also been interesting work on the moduli space of all triangles in the plane. Gaspar and Neto [2] noticed that this space is a triangle, and Ian Stewart later gave a more geometrical explanation [3]. In fact all the moduli spaces mentioned here are better thought of as moduli ‘stacks’: stacks give a way to understand the special role of more symmetrical objects, like isosceles and equilateral triangles. Kai Behrend [1] has written an introduction to stacks using various moduli stacks of triangles and the moduli space of elliptic curves as examples. Though he does not describe the map $p$ (or its stacky analogue), his work is a nice way to dig deeper into some of the material discussed here.

#### References

[1] K. Behrend, An introduction to algebraic stacks, in Moduli Spaces, eds. L. Brambila-Paz, P. Newstead, R. P. Thomas and O. García-Prada, Cambridge U. Press, Cambridge 2014, pp. 1–131. Also available at https://secure.math.ubc.ca/~behrend/math615A/stacksintro.pdf.

[2] J. Gaspar and O. Neto, All triangles at once, Amer. Math. Monthly 122 (2015), 982–982.

[3] I. Stewart, Why do all triangles form a triangle?, Amer. Math. Monthly 124 (2017), 70–73.

Posted at September 23, 2023 2:50 PM UTC

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### Re: The Moduli Space of Acute Triangles

In the first display, don’t you mean $|z - 1/2| \gt 1/2$?

Posted by: Tom Leinster on September 23, 2023 4:37 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Yes I do!

Posted by: John Baez on September 23, 2023 4:56 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

You asked about clarity. To me, the only point that sticks out is this: I wonder whether you should say what “fundamental domain” means, given that you use it so heavily. Personally, I know what it means only because I picked it up randomly at some point, but it’s not a term I ever learned in any course I took.

Posted by: Tom Leinster on September 23, 2023 7:38 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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.

Posted by: John Baez on September 23, 2023 11:02 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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”.

Posted by: Tom Leinster on September 24, 2023 11:38 AM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Another difference is that, in context, it’s probably easier for most mathematicians to guess the meaning of fundamental domain. (At least, I would guess so.) I also never learned that term in any course, and I’m pretty sure I also picked it up randomly by correctly guessing the definition after seeing it used.

Posted by: Mark Meckes on September 24, 2023 2:14 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Tom wrote:

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”.

Yes, that’s very short and possibly very helpful. I should do that.

I’d been put off by the fact that the full definition of ‘fundamental domain’ requires that the set be ‘almost open’ in some way that’s quite annoying to define. I know a fundamental domain when I see one, but don’t ask me to define it!

For example, if I had a non-measurable dense subset of the upper half-plane that contains exactly one point of each orbit of $GL(2,\mathbb{Z})$, I would never call that a fundamental domain.

But if I try to use an open set, that doesn’t really work. At best the triangles here can be open along some parts of their boundary and closed along the rest:

So Wikipedia winds up saying

Hints at a general definition

Given an action of a group $G$ on a topological space $X$ by homeomorphisms, a fundamental domain for this action is a set $D$ of representatives for the orbits. It is usually required to be a reasonably nice set topologically, in one of several precisely defined ways. One typical condition is that $D$ is almost an open set, in the sense that $D$ is the symmetric difference of an open set in $X$ with a set of measure zero, for a certain (quasi)invariant measure on $X$.

But I could make up ‘fundamental domains’ obeying this definition that still have horrible Cantorian pathologies.

Luckily I don’t need to dwell on this in my article!

I can just say what you said, without claiming I’m giving the full definition (whatever that is). And nobody is going to ask me what color are the points right on the edges of the yellow and purple regions! Right?

Posted by: John Baez on September 24, 2023 4:03 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

This is wonderful! I just want to note that these quotient objects can be interpreted on one hand as orbifolds (in which stackier points can have interesting isotropy groups) and on another other hand as categories [So for ex a group $G$ acting on itself by conjugation is an interesting 0-dim’l orbifold…]

Thanks also for the great references.

BTW re your mention of the modular function $\lambda$, doesn’t this construction associate a $j$-invariant to a triangle? I think we should be told (more…).

Posted by: jack morava on September 23, 2023 10:23 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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.

Posted by: John Baez on September 24, 2023 6:00 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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….]

Posted by: jack morava on September 26, 2023 3:39 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Maybe I buried the lede, which is perhaps that an equilateral triangle is an orbifold point of order three, while an isosceles right triangle is an orbifold point of order two. The point at $\infty$ isn’t a proper orbifold point somehow, because its isotropy group is infinite cyclic. All the other points have trivial isotropy, i.e. no nontrivial automorphisms.

This must be in Thurston’s works somehow but alas I don’t know much about that.

Posted by: jack morava on September 26, 2023 5:42 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Most of this is familiar to me in an intuitive way (but don’t ask me to prove it). But the cusp confuses me. Another thing that confuses me: the Picard group of the moduli stack of elliptic curves is $\mathbb{Z}/12$, but how exactly is that connected to the “almost periodic” behavior of the space of modular forms as a function of the weight?

I complained about my confusion on Mathstodon, as follows:

I’m confused! If

1) the Picard group of the moduli stack of elliptic curves is ℤ/12

and

2) a modular form is a section of a line bundle on the moduli stack of elliptic curves

then how come

3) the ring of modular forms is graded by an integer called the ‘weight’

instead of just an integer mod 12?

I must be making a mistake somewhere! Help!

Kyle Ormsby replied:

Is this just the difference between $\omega^{\otimes k}\cong \omega^{\otimes k+12}$ and $\omega^{\otimes k}=\omega^{\otimes k+12}$? The former is true but the latter isn’t. Then the ring of modular forms (over ℂ, say) is $\mathbb{C}[E_4,E_6,\Delta^{\pm 1}]$ where Δ implements the periodicity.

but this did not enlighten me, so I said:

That can’t be all there is to it. Since the Picard group of the moduli stack of elliptic curves is ℤ/12, there are just 12 isomorphism classes of line bundles on the moduli stack of elliptic curves, and the dimensions of their spaces of sections must be periodic mod 12. But the dimension of the space of modular forms of weight k does not depend on k in a way that’s periodic mod 12. It keeps getting bigger.

Now I think the issue is simply that the dimension of the space of sections I was describing is always infinite, and not all sections count as modular forms: only those obeying a certain growth condition at $\infty$, which depends on $k$. But Kyle Ormsby put it this way:

I stand by my answer if we are talking about 0-th cohomology of tensor powers of the canonical line bundle on the moduli of elliptic curves. But yes, if we pass to the compactified moduli space then the discriminant is no longer invertible!

which helped a little. Then someone named Lisanne said:

The mistake is in 2).

You should add in the cusp and take the compactification $\bar{M_{1,1}}$. Then modular forms are sections of powers of the log canonical sheaf. (The log structure sits exactly at the cusps)

The Picard group of $\bar{M_{1,1}}$ rigidified is $\mathbb{Z}[1/6]$

See The canonical ring of a stack curve, Lemma 6.2.3.

This intrigued me, but I didn’t know what a log structure or log canonical sheaf was, so I replied:

Thanks! I don’t know what a log canonical sheaf is, but I can learn about this stuff, and the first step was to figure out my error. Now I can try to read that paper you pointed me to!

Here’s another question, while I have your ear: what are sections of the 12 different line bundles on the moduli stack of elliptic curves?

Well, and another one: how are sections of line bundles on the stack of elliptic curves related to modular forms? They seem vaguely similar: both seem approximately like functions on the upper half-plane with some SL(2,ℤ) action on them, and when you work out what they are the number 12 shows up.

She replied:

The word “log” is just a strange word for something simple. The only thing we do is declare some points to be special (the cusps). The “log divisor” $\Delta$ is then the divisor consisting of these points. The log canoncial divisor is then $K_X + \Delta$ and the log canonical sheaf is $\omega_X(\Delta) := O(K_X + \Delta)$.

Now sections $H^0(\omega_X(\Delta)^{\otimes n})$ can be interpreted as $n$-differential forms on $X$ that are allowed to have a pole of order at most $n$ at $\Delta$ and nowhere else. (Here $n = k/2$, where $k$ is the weight of the modular form.) (Also $n$-differential forms are not differential $n$-forms. They look locally like $f(z)(d z)^n$ and transform appropriately.)

This formalism precisely captures the condition that modular forms should be “meromorphic at infinity”.

In our case (forgetting stacky structure) we just have $\mathbb{P}^1$ with $\Delta = \infty$. Now for $k = 1$ there are no differential forms without poles, but we do get the differential form $\frac{1}{z} dz = dlog(z)$, which has a simple pole at infinity. I have been told that this log is the reason log geometry is called log geometry.

I guess this description as differential forms still holds in the non-compactified setting, but the behaviour of the forms “at infinity” should be taken into account.

I was helped a lot reading her reference. Now I seem to have forgotten a lot of the details. If I were still writing This Week’s Finds I would figure it out and write it up in a way that people like me can understand.

Posted by: John Baez on September 28, 2023 3:02 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

I’ve grown to think that one of the coolest things mathematicians get to do, compared to most thinkers, is mix levels (and upend hierarchies). For example, function composition “o” goes from Hom(X,Y) x Hom(Y,Z) to Hom(X,Z), eats two functions to produce one function, but is itself a function. “Transpose” is a function Hom(V,W) -> Hom(W^ *,V^ *), and itself linear. And so on. No novelty to any reader here, of course, but very tricky to people first learning.

“Moduli space” is another place where the levels get mixed. What is a great big interesting object in one context (here, an acute triangle) is demoted to the dumbest possible one in another (a point; a point in the moduli space). Just the very idea that there isn’t a fundamental ontological level a mathematical object should be said to possess – rather, we can promote or demote it depending on context – is the difficult leap.

I was asked to give a talk to high school teachers once on something somehow related to the symplectic geometry Park City conference I was at. I decided to talk about this concept of moduli spaces, and used as example the space of not-necessarily-convex pentagons in the plane (up to translation and rotation but not reflection) with edge lengths 2,2,2,2,3. If I’ve got the lengths right, this is a genus two surface. I think they got it… or at least some of them…

Posted by: Allen Knutson on September 23, 2023 11:07 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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

but I can’t be more precise…

Posted by: jack morava on September 23, 2023 11:43 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

It might help to clarify the definition of isomorphisms of elliptic curves (as defined above). Are these just diffeomorphisms of the complex manifolds?

Posted by: unekdoud on September 24, 2023 7:43 AM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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.

Posted by: John Baez on September 24, 2023 3:44 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

There are often Mathematical Olympiad problems about triangle geometry. To draw a diagram for these problems it is useful if the triangle has no special properties, otherwise the diagram might give the misleading impression that something is true that doesn’t hold in the general case. So it shouldn’t be right angled or isoceles. This requirement is the same as saying that the triangle shouldn’t lie on any of the lines in your diagram! This can be surprisingly easy to mess up, accidentally making two sides too similar, or an angle too close to right.

To achieve this I draw two points, visualise the above diagram and aim to place the third point at the centre of the inscribed circle of the fundamental domain. This ensures that it’s as far away from any of the lines as possible.

But maybe the relation to the hyperbolic plane is telling us that the true most-scalene-triangle is actually at the ‘hyperbolic centre’ of this circle?

The other option is to get out the protractor and draw a 45°-60°-75°, triangle. This has its angles as far away as possible from being equal or 90°. Seen from the angle point of view, the Moduli Space of Triangles is just a ternary plot.

Posted by: Oscar Cunningham on September 24, 2023 9:50 AM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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.

Posted by: unekdoud on September 24, 2023 3:46 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

I like this construction!

Posted by: Oscar Cunningham on September 24, 2023 6:08 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

This is “a cute” post.

Posted by: Anon on September 24, 2023 3:28 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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.

Posted by: John Baez on September 26, 2023 9:55 AM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

Thanks for the great post!

Just wondering: is there a way to “see” that p is 3-to-1, using the direct description of putting two triangles together? Thanks!

Posted by: Anonymous on September 27, 2023 7:02 AM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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?

Posted by: John Baez on September 28, 2023 1:24 PM | Permalink | Reply to this

### Re: The Moduli Space of Acute Triangles

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

Posted by: Anonymous on September 29, 2023 2:37 PM | Permalink | Reply to this

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