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Stream: deprecated: mathematics

Topic: moduli spaces

John Baez (Sep 23 2023 at 13:29):

I wrote a little article explaining the concept of 'moduli space' through an example:

• The moduli space of acute triangles.

This article is due October 1st so I'd really appreciate it if y'all could take a look and see if it's understandable. It's just 2 pages long, and it's written for people who know some math.

John Baez (Sep 23 2023 at 13:30):

This is mostly stuff I learned from James Dolan. 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.

Chris Grossack (they/them) (Sep 23 2023 at 15:29):

Very cool! And thankfully quite an easy read (at least I thought so). I don't like giving feedback without any suggestions, though, so here's one: If you have an extra footnote worth of space, it might be worth noting that adding the right triangles to your moduli space is a way to compactify it. I'm sure you know this, but there's a cottage industry of people compactifying existing moduli spaces (usually with the technology of GIT quotients, etc) and seeing what the new points on the boundary parameterize. It might be worth adding a sentence of this context to your compactification of the space of acute triangles

Chris Grossack (they/them) (Sep 23 2023 at 15:30):

But I think the article is already great, and I don't have any pointed feedback at all

Chris Grossack (they/them) (Sep 23 2023 at 15:35):

Also I don't know if it was intentional or not, but a nice thing to keep on your radar that I just double checked -- purple/yellow (especially since the purple is dark and the yellow is light) is a good combination for colorblind people. Another common one to have on your radar is blue/orange, which can basically always be distinguished.

John Baez (Sep 23 2023 at 16:18):

Actually throwing in the right triangles isn't quite enough to compactify this space $T$ - there are also the 'infinitely skinny triangles' lurking at $\mathrm{Im}(z) \to +\infty$, $z \to 0$ and $z \to 1$. You see, this $T$ is itself the interior of a triangle in the hyperbolic plane, but it's an 'ideal triangle' with all three vertices at infinity!

John Baez (Sep 23 2023 at 16:22):

In the Poincare disk model of the hyperbolic plane $T$ could be any of the triangles here:

.

John Baez (Sep 23 2023 at 16:26):

But I think the article is already great, and I don't have any pointed feedback at all.

Thanks!

John Baez (Sep 23 2023 at 16:27):

Also I don't know if it was intentional or not, but a nice thing to keep on your radar that I just double checked -- purple/yellow (especially since the purple is dark and the yellow is light) is a good combination for colorblind people.

Nice! That wasn't intentional: I just wanted complementary colors that looked light and dark.

David Egolf (Sep 23 2023 at 16:55):

I haven't tried to read the whole thing yet, but I quite enjoyed the start of the article. The idea of putting isomorphism classes in some kind of topological space to talk about the "nearness" of them is quite cool! That also seems like a very powerful tool to think about collections of things of arbitrarily huge geometric size in terms of a nice bounded region.

This is where I got a bit stuck:

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.

I know that an action of the group $S_3$ is a group homomorphism $f: S_3 \to \mathrm{Aut}_C(x)$ to the automorphism group of some object $x$ in some category $C$. It sounds like $S_3$ is acting on $T$ - presumably then the idea is to place an equivalence relationships on $T$, where we consider two elements in $T$ equivalent if they are in the same orbit of this action.

If $S_3$ is acting on $T$, we are looking for a group homomorphism $f: S_3 \to \mathrm{Aut}_\mathsf{Set}(T)$. To do this, for each element of $S_3$ we want to induce a bijection from $T$ to itself. Each element of $T$ is a point in our purple and yellow region, which specifies an acute triangle (in terms of that point, and the points 0 and 1).

We are given that the action of $S_3$ permutes the three vertices. However, to apply a given permutation I think we need to put an ordering on the vertices of an acute triangle determined by an element of $T$. For example, we could say that the first vertex is the one at 0, the second vertex is the one at 1, and the third vertex is the one at some $a+bi$ in the purple and yellow region (the one that is an element of $T$). Then an element $s \in S_3$ would tell us to take a triangle, and transform it by similarity-preserving transformations until the vertices are appropriately permuted. To take an example, if $s \in S_3$ is $1 \mapsto 2, 2 \mapsto 3, 3 \mapsto 1$, then we would want to apply a similarity-preserving transformation so that the vertex at 0 gets mapped to 1, the vertex at 1 gets mapped to the purple and yellow region at some $a'+b'i$, and the vertex in the purple and yellow region at $a+bi$ gets mapped to $0$. This seems like it could describe part of a permutation of $T$ induced by $s$, namely $a+bi \mapsto a'+b'i$.

It seems like this might be the action of $S_3$ that the article alludes to. But I'm not sure! It seems like it might take some work to show that we can always permute the vertices of the triangle in the way I describe above while preserving similarity.

By the way, the next section of the article reminded me of this video, which provides an enjoyable visualization of Möbius transformations using a "Riemann sphere".

Kevin Arlin (Sep 23 2023 at 18:46):

This is fun! I agree that actually seeing the action of $S_3$ would be a lot easier with a picture or a bit more description. Basically I wish it could have one more page, but I assume that's an external constraint.

Also, don't you want to rotate a triangle about the midpoint of one of its edges to construct a parallelogram, rather than reflect across that edges? I think the reflection generally gives you a kite instead of a parallelogram.

John Baez (Sep 23 2023 at 20:04):

If you have a triangle with its vertices labeled {1,2,3}, the group $S_3$ permutes these labels. More generally if you have anything labelled with elements of some set, permutations of that set act to give new labelings. I think this is standard enough not to require much explanation, at least for readers of the Notices of the American Mathematical Society, where this column will - if accepted - appear.

If there's something really quick that would help explain this idea, I can try it. I don't have room for more pictures, and if I did I would illustrate some of the later, more tricky constructions.

John Baez (Sep 23 2023 at 20:11):

However, to apply a given permutation I think we need to put an ordering on the vertices of an acute triangle determined by an element of T.

Don't think about T at first, think about a triangle in the plane with vertices labeled 1, 2, 3. Permutations of the set $\{1,2,3\}$ act on any such triangle to give a new labeled triangle - the same triangle, just with a different labeling. There's nothing about an ordering involved here, though of course the set 1, 2, 3 is ordered if you want it to be.

T is the set of isomorphism classes of labeled acute triangles in the plane, where 'isomorphism' means that two labeled triangles are related by a similarity (rotation, reflection and/or rescaling) in a way that preserves the labelings, mapping vertex 1 of the first triangle to vertex 1 of the second one, etc.

Since $S_3$ acts on the set of labeled triangles in the plane it acts on the set of isomorphism classes of these.

John Baez (Sep 23 2023 at 20:16):

There's no 'work' involved in anything here... except of course the work involved in learning general concepts. It's all very general: I could replace similarity classes of triangles with labeled vertices in the plane by isomorphism classes of 3d vector spaces with labeled basis vectors and everything would work the same.

John Baez (Sep 23 2023 at 20:18):

The only thing particular to the case at hand is our concrete description of the space T, which we're not using in anything I just said.

John Baez (Sep 23 2023 at 20:19):

Kevin Arlin said:

Also, don't you want to rotate a triangle about the midpoint of one of its edges to construct a parallelogram, rather than reflect across that edges? I think the reflection generally gives you a kite instead of a parallelogram.

Yikes! You're right! Thanks, I'll fix that.

John Baez (Sep 23 2023 at 20:21):

That was an incredibly helpful comment.

Note that if you do this in all 3 possible ways (for the 3 edges of the original triangle), you get 3 new triangles, which taken together with the original one form a triangle similar to the original, but with sides twice as long!

John Baez (Sep 23 2023 at 20:22):

You can then fold up this thing and get a tetrahedron, and that tetrahedron plays an interesting role in the next chapter of this story, which alas I didn't have space to explain.

Niles Johnson (Sep 23 2023 at 20:43):

John Baez said:

think about a triangle in the plane with vertices labeled 1, 2, 3 ...

It's not too hard to realize that transposing the bottom two vertices flips the third vertex across the vertical line Re(z) = 1/2. But is there a short explanation for why transposing a different pair of vertices does ... something else (not sure what; a reflection through a circle of radius 1?) For example, what happens to the triangle whose third vertex is at the point z = (1/2, 1/2 + epsilon)? If that's explainable in a few words, perhaps it would be a useful addion.

I think understanding this requires the reader to realize an implicit detail that I missed the first time: When you say "its first vertex at 0, the second at 1,..." you really mean that the vertex labeled 1 is at 0, the vertex labeled 2 is at 1, etc. So transposing labels means that the triangle has to be repositioned, rescaled, to satisfy this requirement. Maybe rewording to say the explicit label list could clarify? (If you do that, perhaps consider the labels {0,1,2}, so that the vertex labeled 0 is at 0, etc?)

Over on ma(th)stodon I suggested that overlaying the picture you have with a few example triangles might help clarify, and that wouldn't require any extra space. (I don't know if you saw that comment or not; that's the only reason I'm repeating it here; it's of course fine by me if you prefer not to change the picture!) The other comment I made there is that I think the phrase "If we reflect a labeled triangle..." is not explained. My comments above could be considered an expansion of that comment.

John Baez (Sep 23 2023 at 20:51):

But is there a short explanation for why transposing a different pair of vertices does ... something else?

I don't have anything really quick and fun to say about that - perhaps because I haven't thought about it hard enough.

It's clear that we have $S_3$ acting on the moduli space of acute triangles shown in purple and yellow here:

.

It's clear that the point corresponding to the equilateral triangle is fixed by this action. And it's clear from what you said that the transposition 1 $\mapsto$ 2, 2 $\mapsto$ 1 is reflection across the vertical line Re(z) = 1/2. But it takes more work to show that the other transpositions act as (hyperbolic) reflection across the obvious other (hyperbolic) lines separating yellow and purple regions, so I think explaining that would distract people from the main story here.

Niles Johnson (Sep 23 2023 at 20:55):

fair enough; thanks for the clarification though, it will give me something fun to think about :)

John Baez (Sep 23 2023 at 20:55):

Oh, here's something fairly quick, which I still don't want to include in the article!

John Baez (Sep 23 2023 at 20:56):

Think about the portion of the circle |z| = 1 in the colored region of the picture.

John Baez (Sep 23 2023 at 20:57):

Points on this circle correspond to isosceles triangles with vertices labeled 1, 2, 3 where the side 12 has the same length as the side 13.

John Baez (Sep 23 2023 at 20:58):

So these are the points that are fixed by the action of the transposition 2 $\mapsto$ 3, 3 $\mapsto$ 2.

Similarly points on the circle |z-1| = 1 will be fixed by the transposition 1 $\mapsto$ 3, 3 $\mapsto$ 1.

Thanks for raising this issue!

John Baez (Sep 23 2023 at 21:01):

When you say "its first vertex at 0, the second at 1,..." you really mean that the vertex labeled 1 is at 0, the vertex labeled 2 is at 1, etc.

Right, maybe I should have said a bit more forcefully that "first vertex, second vertex, third vertex" are not just off-hand utterances here: I'm using the labeling to say which vertices are the "first", "second" and "third".

John Baez (Sep 23 2023 at 21:03):

I'm not sure I want to pound this point in more forcefully. I'll think about it. I often have the fear, in writing these short columns, that making everything very formal would confuse or bore more people than it will actually help.

These columns are supposed to be entertaining. I like to think that anyone who gets worried about the technical details can stare at what I wrote and figure things out - like you did - while those who don't care about the technical details should not have their noses rubbed in them.

Niles Johnson (Sep 23 2023 at 21:15):

Ah, seeing the circle |z| = 1 as the fixed points of the transposition is great! (Note, if you do decide to put any of that in the article, you've used the labeling {0,1,2} in one of your comments, and then the labeling {1,2,3} in another of them... that's probably my fault, but still... edit: fixed now)

I often have the fear, in writing these short columns, that making everything very formal would confuse or bore more people than it will actually help.

I see what you mean, and I suppose different people will have different preferences for what they consider "a technical bore" v.s. "clear writing". I've been thinking about that kind of thing more generally, so I appreciate your comments about it. I'll be excited to see what you eventually decide :)

John Baez (Sep 23 2023 at 21:19):

I've straightened out my choice of labelings in my comments above, trying to use {1,2,3} everywhere. I see the advantages of using {0,1,2}, though.

John Baez (Sep 23 2023 at 21:23):

Niles Johnson said:

I see what you mean, and I suppose different people will have different preferences for what they consider "a technical bore" v.s. "clear writing". I've been thinking about that kind of thing more generally, so I appreciate your comments about it. I'll be excited to see what you eventually decide :)

I'm pretty sure I'm not going to put in more detail unless a referee forces me. It would be different if this were a paper where I was supposed to be proving theorems. But one reason I like writing a column is that I'm not proving things, just explaining stuff as if I were talking to someone in front of a blackboard. A blog, or forum like this, is even better - since then people can ask questions!

David Michael Roberts (Sep 23 2023 at 22:55):

The 6 yellow and purple regions in T are fundamental
domains

I was slightly confused first time, since I didn't realise the top two regions continued all the way up to infinity. Because I expected the pattern to continue as shown, to cover the whole strip.

David Egolf (Sep 24 2023 at 00:27):

John Baez said:

Permutations of the set $\{1,2,3\}$ act on any such triangle to give a new labeled triangle - the same triangle, just with a different labeling.

Thanks for clarifying what it means to act on labelled triangles! I was trying to understand this action as something complicated and geometric, not just a changing of labels.

I think I am still confused, however. But this is probably just a sign that I would benefit from learning more about the basics of group actions, not a sign that the article is unclear!

John Baez (Sep 24 2023 at 10:11):

David Michael Roberts said:

The 6 yellow and purple regions in T are fundamental domains

I was slightly confused first time, since I didn't realise the top two regions continued all the way up to infinity. Because I expected the pattern to continue as shown, to cover the whole strip.

Okay, I'm too used to this picture:

Fundamental domains of the modular group, as drawn by Dedekind in 1877

John Baez (Sep 24 2023 at 10:12):

I'll say that the regions on top continue up to infinity.

John Baez (Sep 24 2023 at 10:14):

David Egolf said:

John Baez said:

Permutations of the set $\{1,2,3\}$ act on any such triangle to give a new labeled triangle - the same triangle, just with a different labeling.

Thanks for clarifying what it means to act on labelled triangles! I was trying to understand this action as something complicated and geometric, not just a changing of labels.

I think I am still confused, however. But this is probably just a sign that I would benefit from learning more about the basics of group actions, not a sign that the article is unclear!

If you want me to help you straighten things me, go ahead and ask questions. Quite separate from improving my paper, I'm always happy to explain interesting and important math. Doing it makes me think about lots of interesting things... and I hate it when people are confused!

John Baez (Sep 24 2023 at 10:16):

Note that it's possible to think of the process of permuting labels as a complicated geometrical thing: if I have a right triangle where the right angle is at vertex 2, and I switch the labels so now it's at vertex 1, I can also think "okay, now I'm gonna shrink the angle at vertex 2 so it becomes smaller than a right angle, and expand the angle at vertex 1 so it becomes a right angle".

John Baez (Sep 24 2023 at 10:20):

It's like if you and I wanted to trade identities, one way would be for us to trade labels: I could legally change my name to David Egolf and you could change yours to John Baez. Another way would be for me to get plastic surgery to look like you, and learn to dress like you and act like you, and get a gene transplant so I had your genes... and for you to do the same for me. Either way, in the end we'd have a guy named David Egolf with the properties of the old John Baez, and vice versa! But one way is easy while the other is perversely difficult.

But this perversity is actually implicit in the picture here:

.

since in this picture vertex 1 is 'nailed down' at the origin and vertex 2 is 'nailed down' at the point 1 on the real line, so the only way for them to trade properties is for us to morph the triangle by moving the third vertex. We reflect the third vertex across the line Re(z) = 1/2.

John Baez (Sep 24 2023 at 10:22):

I think your mistake may have been to dive in and take the perverse "morphing triangles" approach right away instead of taking full advantage of the much easier "permuting labels" approach.

John Baez (Sep 24 2023 at 10:46):

See, this was fun to think about.

John Baez (Sep 24 2023 at 10:46):

But maybe something completely different was bothering you!

David Egolf (Sep 24 2023 at 15:49):

John Baez said:

I think your mistake may have been to dive in and take the perverse "morphing triangles" approach right away instead of taking full advantage of the much easier "permuting labels" approach.

Yes, I think this was indeed my mistake. For example, you can make some very "skinny" triangles by placing a vertex at 0, a vertex at 1, and third vertex in the sliver of the yellow region that gets very very close to 0. Then, I was trying to imagine how I could obtain a similar triangle to this with third vertex in the top left yellow region. For example if we want to swap the vertex in the yellow region with the vertex at 1, I was imagining a process where:

1. we rotate the triangle about 90 degrees counterclockwise to get the vertex that was in the yellow region down to the negative x-axis
2. we flip this triangle about itself from left to right, so the vertex that was in the yellow region is now on the positive x-axis
3. we then "expand" the entire triangle until the vertex that was in the yellow region is now at 1 (this gives us a very tall triangle, but because the top left yellow region extends infinitely upwards, this is acceptable)

I was led to thinking of this sort of process while I was trying to understand the idea that each of the six yellow or purple regions by itself can let us describe all possible shapes (up to similarity) of acute triangles.

David Egolf (Sep 24 2023 at 16:38):

I think what is still confusing me is this:

• We want to act on labelled triangles by permuting their labels. I can understand this: for each triangle, keep the triangle same, but move labels on the vertices around.
• However, we want to (I think) extend this action to one on $T$. The elements of $T$ are points in the yellow and purple regions, not labelled triangles (although - each such point does represent an isomorphism class of labelled acute triangles).

I'm not sure how to do this, so I'm thinking out loud here. If $L$ is the set of all labelled acute triangles (not necessarily having vertex "1" at 0, vertex "2" at 1, and vertex "3"in a yellow or purple region), then we should have some projection function $\pi: L \to T$ that sends a labelled triangle $t_L$ to its equivalence class of labelled triangles. To take an example, if $t_L$ has vertex 1 and vertex 2 very close to each other, and vertex 3 is far away from these, then the underlying triangle $t$ is a very "skinny" triangle. Then $t$ is similar to other triangles having this shape. Let us assume that $t$ and $t'$ are similar. However, if $t'_L$ denotes a labelled version of $t'$, it is not necessarily the case that $t'_L$ and $t_L$ are isomorphic as labelled triangles: if vertex 1 and vertex 2 of $t'_L$ are not very close to one another (relative to their distance to vertex 3), then $t'_L$ I think is not going to be isomorphic to $t_L$ as a labelled triangle.

Ok, so we have $\pi: L \to T$ that sends a labelled triangle to its isomorphism class of labelled triangles, which is represented by some "standard" triangle determined by its third vertex (in a yellow or purple region). Now let's assume we have some bijection $b:L \to L$ defined in terms of changing labels around. I want to use this to induce a bijection on $T$. To do this, I think I want to introduce a function $\pi': T \to L$ that sends a vertex in the purple and yellow region to the "standard labelling" of the triangle in "standard position" that it naturally induces. That is, $\pi'$ sends $a+bi \in T$ to the labelled triangle $1 \mapsto 0, 2 \mapsto 1, 3 \mapsto a+bi$.

Now we have $\pi: L \to T$ and also $\pi': T \to L$ available to us. Let's assume again we have a bijection $b: L \to L$ which acts by shuffling the labels on each triangle according to some element of $S_3$. Then we can form the endofunction $\pi \circ b \circ \pi': T \to T$. Intuitively, this takes a point $a+bi$ in a yellow or purple region, forms a labelled triangle where $a+bi$ is vertex 3 (and 0 is vertex 1, and 1 is vertex 2), then permutes these labels according to $b$, and then sends this relabelled triangle back to a corresponding point in $T$ by sending it to its isomorphism class of labelled triangles.

It would remain to show that $\pi \circ b \circ \pi': T \to T$ is actually a bijection, and that we can get an action of $S_3$ on $T$ in this way. However, I suspect this relates to your earlier comment, quoted below:

John Baez said:

Since $S_3$ acts on the set of labeled triangles in the plane it acts on the set of isomorphism classes of these.

It sounds like actions on individual things can in general get promoted to actions on isomorphism classes of those things. But this sounds like general theory I can try to understand without invoking specifics involving an action on labelled triangles!

John Baez (Sep 24 2023 at 18:17):

David wrote:

We want to act on labelled triangles by permuting their labels. I can understand this: for each triangle, keep the triangle same, but move labels on the vertices around.
However, we want to (I think) extend this action to one on $T$. The elements of $T$ are points in the yellow and purple regions, not labelled triangles (although - each such point does represent an isomorphism class of labelled acute triangles).

That's right, though most mathematicians would not use the word "extend" here. If I have function or group action on some set $S$ and some larger set $S'$ with $S'$, I might want to "extend" my function or group action from $S$ to $S'$. But here we have a set $S$ and a quotient set $S'$, i.e. a set of equivalence classes of elements of $S'$. So we're more likely to ask if the function or action "descends" from $S$ to $S'$.

(We think of elements of $S'$ as being below those of $S$, like shadows of objects: different objects can have the same-looking shadow.)

Some hardcore category theorists might say "extend" even in this case, since the two cases have this in common: we have a map $f: S \to S'$. Category theorists might ask if a group action "extends along $f$". But normal mathematicians would be put off by this when $f$ is not an injection.

John Baez (Sep 24 2023 at 18:17):

It sounds like actions on individual things can in general get promoted to actions on isomorphism classes of those things. But this sounds like general theory I can try to understand without invoking specifics involving an action on labelled triangles!

John Baez (Sep 24 2023 at 18:18):

Right, it's easier to do things in general. Say you have a group $G$ acting on a set $S$. Say you have an equivalence relation $\sim$ on $S$. And let $S'$ be the set of equivalence classes. Write $[s] \in S'$ for the equivalence class of $s \in S$.

John Baez (Sep 24 2023 at 18:19):

Does the action of $G$ on $S$ descend to one on $S'$? We try to guess a way for $G$ to act on equivalence classes. There's only one plausible guess:

$g[s] := [gs]$

John Baez (Sep 24 2023 at 18:20):

where at right $g s$ means the result of $g \in G$ acting on $s \in S$, but at left is our guess as to how $g$ should act on $[s]$.

John Baez (Sep 24 2023 at 18:21):

Now, whenever you work with equivalence classes you need to check that things are well-defined independent of our choice of representative! Right now that means we need to check whether

$s \sim s' \implies gs \sim gs'$

John Baez (Sep 24 2023 at 18:21):

1) Show that if this is true then the formula $g[s] = [gs]$ gives a well-defined action of $G$ on $S'$.

John Baez (Sep 24 2023 at 18:22):

2) Does this condition hold in the case we care about, where $G = S_3$, $S$ is the set of all labeled triangles in the plane, and $\sim$ is the relation of similarity of labeled triangles?

John Baez (Sep 24 2023 at 18:23):

3) Can you think of an example of a group acting on a set with an equivalence relation where the condition

$s \sim s' \implies gs \sim gs'$

does not hold? What happens then?

David Egolf (Sep 24 2023 at 19:51):

Thanks for explaining things, step by step! Let me see if I can work out the details.

We have a set $S$ with an equivalence relation $\sim$ on it, and a set $S'$ of the resulting equivalence classes. We write $[s]$ to denote the equivalence class of $s \in S$.

We have an action of a group $G$ on $S$ and we wish to investigate whether it descends to an action on $S'$.

We guess that the action can be defined as: $g[s] = [gs]$. If we assume that $s \sim s' \implies gs \sim gs'$, so that $[s] \mapsto [gs]$ defines a function, then does $g[s]=[gs]$ give a well-defined action of $G$ on $S'$?

Let $f: G \to \mathrm{Aut}_{\mathsf{Set}}S'$ be our proposed group action. For $g \in G$, we have $f: g \mapsto ([s] \mapsto [gs])$. First, let's check that the function $f(g): [s] \mapsto [gs]$ is a bijection of $S'$. To do this, we aim to show it has an inverse, which we guess is $f(g^{-1})$. Then $f(g) \circ f(g^{-1}) = [s] \mapsto [g^{-1}s] \mapsto [gg^{-1}s] = [s]$ and $f(g^{-1}) \circ f(g) = [s] \mapsto [gs] \mapsto [g^{-1}gs] = [s]$. So, $f(g): S' \to S'$ has an inverse in $\mathsf{Set}$, so we conclude it is an isomorphism and hence a bijection.

Next, we want to show that $f$ is a group homomorphism. $f(e_G) = [s] \mapsto [e_Gs] = [s]$, so $f(e_G)$ is the identity function on $S'$. Also, $f(gh) = ([s] \mapsto [(gh)s]) = ([s] \mapsto [hs] \mapsto [ghs]) = ([s] \mapsto [gs]) \circ ([s] \mapsto [hs]) = f(g) \circ f(h)$. So, we conclude that $f: G \to \mathrm{Aut}_{\mathsf{Set}}S'$ is a group homomorphism and hence $g[s] = [gs]$ defines an action of $G$ on $S'$.

David Egolf (Sep 24 2023 at 19:56):

Next, we wish to know if $s \sim s' \implies gs \sim gs'$ when $G=S_3$, $S$ is the set of all labelled triangles in the plane, and $\sim$ is the relation on similarity of labelled triangles.

Intuitively, this condition would mean that if we have two similar triangles with "matching labels", then after we permute their labels, they are still similar triangles with "matching labels". Well, certainly if $t$ and $t'$ are similar labelled triangles, then they are still similar after we change the labelling of their vertices. It remains to show that the "labels still line up". I'm not sure how to show this formally, but intuitively if we transform one triangle to lie on top of the other (by similarity-preserving transformations) such that the original labels line up, then doing this will make the new labels line up as well, because we permuted the labels in the same way on each triangle.

David Egolf (Sep 24 2023 at 20:15):

Finally, we wish to consider a case where a group acts on a set so that $s \sim s' \implies gs \implies gs'$ does not hold.

As an example, assume our set $S$ is the set $\mathbb{Z} \times \mathbb{Z}$ of pairs of integers, having elements of the form $(n,m)$ with $n,m \in \mathbb{Z}$. We place an equivalence relationship on $S$ by saying $(n,m) \sim (n',m') \iff n=n'$. So, two tuples are equivalent exactly when their first coordinates are equal. We now let the group $S_2$ act on $S$ by permutation of coordinates. Then if $g \in S_2$ is the permutation that swaps two things, we can have $(n,m) \sim (n',m')$ but $g(n,m) = (m,n)$ and $g(n',m') = (m', n')$ are only equivalent if $m=m'$. So in general, in this case, $s \sim s'$ does not guarantee that $gs \sim gs'$.

If we try to extend our action of $S_2$ on $S$ to equivalence classes in this case using the formula $g[s] = [gs]$, we will run into a problem. Namely, that $g[s]=[gs]$ and $g[s']=[gs']$ might be different even if $[s] = [s']$. For example, trying $g[(1,2)] = [g(1,2)] = [(2,1)]$ and $g[(1,3)] = [g(1,3)] = [(3,1)]$ shows that $g[s] = [gs]$ does not define a function: $[(1,2)] = [(1,3)]$ but $g[(1,2)] \neq g[(1,3)]$.

David Egolf (Sep 24 2023 at 20:23):

So, in summary, a group action on a set does not always descend to a group action on equivalence classes of that set under an arbitrary equivalence relationship. However, if $s \sim s' \implies g(s) \sim g(s')$ holds, then I think that the action does descend to an action on the equivalence classes. And I think this condition holds for our case of interest with labelled triangles, so our action on acute labelled triangles can descend to an action on the isomorphism classes $T$ of these acute labelled triangles.

dusko (Sep 25 2023 at 09:42):

John Baez said:

I wrote a little article explaining the concept of 'moduli space' through an example:

• The moduli space of acute triangles.

This article is due October 1st so I'd really appreciate it if y'all could take a look and see if it's understandable. It's just 2 pages long, and it's written for people who know some math.

my modest contribution to the conversation is to say that this is fantastic, john! you learned from james dolan and i learned from you. i also learned from you about octonions, so long ago that in the meantime i forgot.

you really go in the corners. i don't know how the future archeologists (if they exist) will measure this, but you do make a difference :pray:

John Baez (Sep 25 2023 at 18:38):

Great, @David Egolf! You crushed those problems!

John Baez (Sep 25 2023 at 18:43):

David Egolf said:

Next, we wish to know if $s \sim s' \implies gs \sim gs'$ when $G=S_3$, $S$ is the set of all labelled triangles in the plane, and $\sim$ is the relation on similarity of labelled triangles.

Intuitively, this condition would mean that if we have two similar triangles with "matching labels", then after we permute their labels, they are still similar triangles with "matching labels". Well, certainly if $t$ and $t'$ are similar labelled triangles, then they are still similar after we change the labelling of their vertices. It remains to show that the "labels still line up". I'm not sure how to show this formally, but intuitively if we transform one triangle to lie on top of the other (by similarity-preserving transformations) such that the original labels line up, then doing this will make the new labels line up as well, because we permuted the labels in the same way on each triangle.

To prove it formally I think we'd have to define the set of triangles in the plane a bit more formally, and the set of labelled triangles, and the relation of similarity of labeled triangles, and the action of $S_3$ on labeled triangles. Then we could just "grind it out" and check that $t \sim t' \implies gt \sim gt'$. This process should be completely mechanical, uninspired... and uninspiring.

It's important to know how to formally define concepts like this and work with them in an ultra-precise way. It comes in handy when you want to be very sure about something, especially when a more intuitive approach fails. But in this particular case I don't think it's worth bothering. Indeed, it's also good to practice just "seeing" things. (Here I include spotting when a plausible-sounding fact is not really true.)