wow, @Calc-Help just joined SE today

orbit-stabilizer, your specialiy4 is gauge field theory?

Is that a joke?

i just guess by your name.

Ah, no. It's taken from the orbit-stabilizer theorem in group theory

I'm only a third year undergrad.

does math department teach group theory in undergraduate school?

Yes? It's a very standard undergrad course.

Which year are you in?

I have taken group theory in my master studies but have never learnt orbit-stabilizer. I learnt this term from some papers after graduation. It often occurs in papers of gauge field theory.

I don't major in math.

Are you in physics?

yes

but group theory is just a selective courses in my graduate school.

Ah, that makes sense. What is your area of research?

gravitation

math.wm.edu/~vinroot/actions415b.pdf

page 5 of this pdf discusses the orbit-stabilizer theorem

Cool! I went to a talk given by the guy who won the nobel prize for discovering gravitational waves yesterday!

What math do you need to know to understand gravity?

differential geomety and group theory are the only courses I took

@CaptainBohemian I think group theory offered by physics dept is geared more towards representation theory, lie groups/algebras, lorentz group, special relativity - at least that's the case here

@orbit-stabilizer very cool

what university do you go to ?

It's a grad course though

UBC

actually I don't learn math specifically. I usually learn math from physics books.

oh nice @orbit-stabilizer fellow canadian

Are you at UofT?

UOFA

I am right next to you I guess

Oh cool. My cousin is doing math there.

wait nvm

what is his name

I never felt the need to study math specifically for physics at least before I graduated from master because physics books always provide required math.

cool

@CaptainBohemian I'm hoping to go through Arnold's classical mechanics book one day

I have never read that book.

this note looks unilike the typical literature in physics

The pdf or the book?

the pdf you just gave me

Ah, well it's aimed at math undergrads learning about actions for the first time... so it's supposed to be introductory

@MaryStar that seems right

Thanks!

@MaryStar wait

@MaryStar yup, looks good. Sorry, I thought I saw something wrong.

Ah ok, no problem!

Hey everyone!

Anyone able to help with this geometric measure theory problem: math.stackexchange.com/questions/2546530/…?

Hey@Daminark

Suppose there exists such an $H$. Consider $S_n/H$. We know $|S_n/H| = d$. Let $S_n $ act on $S_n/H$ by translation. So, we have $\phi: S_n \rightarrow S_d$. Thus, $Im(\phi)|d!$ and $Im(\phi)|n!$. And we also know that $Ker(\phi$) is non-trivial.

@MaryStar Why does 2 map to 4?

2 is mapped to 8 and 8 is mapped to 4, so 2 is mapped to 4. Is this wrong? @orbit-stabilizer

Are you starting with the thing on the right, or the thing on the left?

What was given?

I thought we begin always from right to left, is this wrong? @orbit-stabilizer

No, that's fine. I was asking a different question, nevermind.

Okay, how is 2 mapped to 8?

Oh, you're right - sigh i can't see.

So is everything correct? @orbit-stabilizer

@MaryStar looks like it

Great! Thanks!

@orbit I think I have an idea

So you know how $S_n$ doesn't have any normal subgroups other than $A_n$ for $n\ge 5$, right?

Yeah

Oh.

Yeah

You know this is not trivial since the action on the cosets is transitive

So the kernel isn't $S_n$

Then, $|Ker(\phi)| = \frac{n!}{2} \implies Im(\phi) = 2$.

The kernel isn't trivial either since $|S_n| \nmid |S_d|$ for $d < n$

Meaning the kernel has to be what you said

Sounds good?

Yeah, and that's a contradiction since we have d > 2?

You'll have to say a tiny bit more

The whole thing is this: Let's say you have a map $\phi:S_n\to S_d$

Given by left multiplication on the cosets of $H$ whose index is $d$. The kernel has 3 options: all of $S_n$ (transitive action rules this out), trivial ($S_n$ is larger so that's out), or $A_n$

If it's $A_n$, then the image is order 2, so there's only keeping everything in place or making an order two swap

An order two swap?

I'll say it better one sec

If it's $A_n$, the image is order 2, so a given coset can either stay in place or get sent to one other coset, right?

Right

Meaning the orbit is size 1 or 2

Yes

But that's where we note that because the action is transitive, the size of the orbit of a point is $d > 2$

(Basically the point to note was that the contradiction had to come from being transitive, it wasn't completely automatic)

Hmmm I see...

Wait

Can we not get a contradiction from $\frac{n!}{2}|d!$?

@orbit-stabilizer what does that contradict?

^

Wait, that's not even true. Kernel size does not have to divide the target size.

Hi chat

Hey @Alessandro

Hello

I see there's some group theory going on

Yup

@Daminark I think i get it now, thanks!

@Daminark I'm really bad at working with $S_n, A_n,$ and $D_{2n}$. Do you have any pdfs or notes or practice questions that you found helpful?

How do i prove that there exists an embedding of $S_3$ in $S_6$ as transitive subgroup?