You're older, yeah?

eh?

@AlexYoucis Than who?

He's older than me, say.

But not older than Rob.

If I hadn't skipped going to school for three years, I'd be about a grad student now probably.

It's like I can't speak colloquially haha. You are older than a typical freshman, as you've commented before.

Yes.

Cool beans

thief

@BenW. I lied to you the other day--deciding is the hardest thing in the entire world.

@AlexYoucis So you're not so set on your first decision any more, huh?

@BenW. I don't know :S UCLA is so welcoming!

@AlexYoucis Have you ever physically been to either place? Maybe that will help settle your mind.

@BenW. I have not--I am going to both though, so hopefully that will help. I hope I don't get in anywhere else though haha

What about you? Made up your mind pretty solidly about El Southo, eh?

@AlexYoucis Is the orbit stabilizer theorem the assertion that given an action of $G$ on $S$, then $$|S|=\sum |G:\operatorname{stab} x_i|$$ for a set of representatives of each orbit?

@PeterTamaroff That's what I would call the orbit decomposition thoerem. The orbit-stabilizer theorem says that $[G:\text{stab}(x)]=|Gx|$

Because the author just uses that to get to the conjugacy class equation.

@AlexYoucis Oh, OK.

The author didn't mention that. He did say $|S|=|G:\operatorname{stab}x|$ when $G$ acts trans.

@AlexYoucis I'm not thinking too much about it really. I'll probably wait until all the visits are over. I'm just happy to have some free trips.

And clearly $|Gx|=|S|$

@PeterTamaroff Right

So your claim follows.

@BenW. True, true, true. You going to UCLA?

i.e. every action acts transitively on each orbit.

@AlexYoucis Yeah, I filled out their little email questionnaire today.

Yeah, same. Who did you say you want to meet with?

and more importantly, chicken or veg?

@AlexYoucis Balmer, Tao, Roquier, Totaro, a fifth one I can't remember (oops), and chicken. You?

Chicken good. Totaro, Balmer, Haesemeyer, Hida, Merkurjev

They gave 10 options, I hope 5 isn't too many though.

I picked 5 :S

@BenW. Is Tao a serious meeting?

@AlexYoucis Slight interest in alg. combinatorics, so I put it down. I guess I can edit it later.

@AlexYoucis OK, I notice that under conjugation, the conjugacy class is the stabilizer of $\gamma$.

Moving on...

@BenW. I talked to him once for a while. He's a really, super nice guy. I'm not really interested in any of his work though. Are you into Rep Theory, is that right?

@PeterTamaroff ;)

@PeterTamaroff What's your deal by the way? Undergrad, etc. etc.

@AlexYoucis I start my first year of college in a few weeks.

@PeterTamaroff Very cool! Where at, if you don't mind me asking.

@AlexYoucis I'm in Argentina, Buenos Aires. The university is the UBA

@PeterTamaroff That's where Mariano is, yeah?

As you will note by reading the wiki, I finished the CBC last year (1 year long course to get into the uni=

@AlexYoucis Yes, we met a few times.

@PeterTamaroff You have to take a course to get into the school?!

I want to pay him a visit, it's been a while since we didn't talk and I really enjoy talking with him.

@AlexYoucis It's a possibility, all I'm really sure about is that I want to do something in algebra. (But I'm not sure how sure I actually am.)

@AlexYoucis Well, apologists say it is actually the first year of the career but I just call bullshit on that.

The calculus is OK, but very simple. So is the linear algebra.

Then you have some more general courses depending on the faculty you sign in to. I got Chemistry and Physics.

@BenW. Do you know when we find out logistics about our trip? I.e. which hotel, etc.?

@PeterTamaroff Why does UBA does not accept students directly from school?

@AlexYoucis I think they said they'd email us. They're taking care of the hotel, but I think we might have to get to the hotel from the airport ourselves. Not sure about that though.

@AlexYoucis Technically you are already a student during CBC. Apparently, it is the modern "course to get in". Before, one had an applicant's course to take, but you weren't officially allowed to vote, since you weren't part of the uni... it is politics really.

@BenW. Ah, ok man. I'm out! It's like basically still afternoon over there, it's late over here. Goodnight!

@AlexYoucis 'Night man.

@OrangeHarvester Because they're simply not up to level.

Many people fail the CBC.

@PeterTamaroff You seem like you're pretty well prepared. Your questions and answers seem like they'd be far beyond the average first year student.

@PeterTamaroff Ahh I see. I can understand I guess. So, some sort of levelling ground for people of different demographics. We have a similar concept here for some students, but most of the students who have to take them do not fail.

For example, only 12% of the guys passed the first calc midterm.

And by passed I mean 40% or more.

okay.

I had a calc exam once where 50% was an A.

@BenW. Oh, that doesn't happen here i think.

@BenW. We had a device physics and modelling course where passing was 16% and A was on 55% or something.

@OrangeHarvester I can't understand that.

Can't they just say "Do half the questions"?

@OrangeHarvester Wow, 16% is really low to be passing.

@BenW. Thanks for that. I do work hard, though.

@PeterTamaroff That would be unfair to someone who does score an A I guess (I am not sure how). I myself had 87.

@OrangeHarvester Too smart!

@OrangeHarvester But what's the point on having 16/100.

I find it utterly useless...

That is roughly 1/10

Well, 2/10...

@PeterTamaroff Yeah, as you go higher in math, everyone has to start working hard. When did you first really get interested in math? Like self-studying advanced topics, reading textbooks, etc.

1.6/10

Advanced topics meaning stuff beyond calculus, I guess.

@BenW. not really, I failed in humanities the one time I gave it (and passed later only because of prof). I am very twisted that way.

@PeterTamaroff Well, the courses were graded on the curve. So, it was a $4\sigma$ thing.

@BenW. I can't remember really when I got interested in math.

It was never really hard for me.

I only failed 1 math exam in my life... polynomial long division. =)

@PeterTamaroff Ha!

@PeterTamaroff FUUUUUU

But I think I got into "serious" math in the last 2 years.

@PeterTamaroff Ha, I know the feeling. I failed an exponent laws test in 6th grade.

I mean, the first year was just "getting started"..

I have failed math once because the professor thought I did not write the answers in enough detail even though the answers were correct.

@BenW. Not that serious.

I had to learn about limits and calculus by myself, so I did have to go over that quite a few times, you can imagine @BenW. and I didn't exactly pick the best sources!

@PeterTamaroff Hehe. And then by you found spivak and there was light?

And then along came Spivak!

...and Apostol.

YO! I want a cookie!

@OrangeHarvester It will rot on the way to India!

@PeterTamaroff Whatever.

@BenW. When did you get interested?

@OrangeHarvester "It's not about the cookie, it's about sending a message."

@PeterTamaroff Yes.

Yeah, tell us about you @BenW.

@OrangeHarvester I always liked math in high school, just because it was easy. But it wasn't until I took calculus in my junior year that I found it to be interesting.

@BenW. When did you get interested in math? I myself always found it interesting and did a lot of stuff on my own (number theory, combinatorics all that stuff), but till I was 13-14 I had many different interest to compete with math: football, tennis, programming, video games. At 13-14, I read halliday resnick and got interested in physics, and then by freshman year had shifted to math.

@BenW. I see. Similar thing here I guess.

@BenW. I guess that is how math lures you in. @OrangeHarvester

It was the first time where I encountered math wasn't obvious or intuitive, so I decided to pursue it more seriously after that.

She plays the easy role and gets you with calc!

@BenW. I see.

@PeterTamaroff Yeah! Lures you in, really.

I'm a simpleton; my only real interest was math since the beginning.

And then majoring in math taught me that I don't really know anything about math. But that's the fun thing I guess.

If I can prove that $|S_n :\operatorname{stab}\gamma|=n$ I am done.

I see where this is going. Moving on...

@peoplepower At least you liked it from the start.

@peoplepower You saw light for the first time and had a calculator in your baby hand?

@peoplepower you keep things simple :-) , its a realy headache when you like too many things.

@PeterTamaroff HAHAHAHA.

Gonna head out, goodnight fellows.

@OrangeHarvester I feel like I lacked experiences which are the basis for a lot of intuition, especially at first.

@BenW. good night.

@peoplepower ohh, as in?

@BenW. Nighty night.

@PeterTamaroff Isn't it around 6 am at your place?

@OrangeHarvester I am afraid I cannot remember, but I will instead give a positive example to show the opposite of what I mean: I dabbled with programming, helping me through the logic underlying proofs.

@OrangeHarvester 20 mins to 6 yes.

Maybe I just did not recognize that math is hard and I tried to pass it off for something else.

I see, yes, I think I understand what you mean, I have been in the similar place I think.

@peoplepower peeps

@PeterTamaroff Yes?

Let $S_n$ act on itself by conjugation.

Let $\gamma=(12\dots n)$

Then $|S_n\gamma|=|S_n:\operatorname{stab}\gamma|={S_n}/{\operatorname{stab}\gamma}$

Assuming you mean that we are dividing orders, I agree.

If I can prove that $\operatorname{stab}\gamma=n$ I am done.

So I have to determine how many permutations have the "trivial" parition $n$ of $n$.

These will be the $n$ cycles.

So I have to show there are $n$ $n$-cycles in $S_n$.

Am I right or left?

That would be wrong. Your problem is asking you to prove that the conjugacy class of an $n$-cycle has size $(n-1)!$.

It consists of $n$-cycles, so we hope what you are saying is false.

The conjugacy class would be $S_n\gamma$

Oh, wait.

No, the conjugacy class would be the stabilizer

I had it right the first time

Dunno why I changed it.

It's the orbit since we are acting by conjugation.

@peoplepower OK, so I am right.

The conjugacy class of $\gamma$ is $S_n\gamma=\{\sigma \gamma\sigma^{-1}:\sigma \in S_n\}$

What do conjugates of cycles by permutations look like?

Another cycle.

Of the same order.

Be specific, let $\pi$ be our permutation and our cycle be the given one ($\gamma$).

With the elements permuted as the permutation dictates.

Exactly, so we are just looking for rearrangements of the symbols comprising $\gamma$ which are really just $\gamma$.

E.g. $(23\dots n1)$ is one such rearrangement.

OK, so we're looking at $1\to 2$, $2\to 3$

But that is just $\gamma$...

Cool, $\gamma$ preserves itself under conjugation.

@peoplepower Yes.

All powers of $\gamma$ preserve $\gamma$ under conjugation.

But that is obvious since $\gamma \gamma \gamma^{-1}=\gamma$

Is there anything else that preserves $\gamma$?

Well, say $n=5$ so $(12345)$.

If I give you $(24351)$ you can map $4\to 3$ and fix the others.

So $(34)(24351)(34)=(12345)$

For example.

But it is not the same as (12345).

We want exactly the same permutation function after conjugation.

Well, that's why I wrote $(34)(24351)(34)=(12345)$

Yes, (24351) is conjugate to (12345), but this is not what we are looking for.

We want $\pi$ such that $\pi\gamma\pi^{-1}=\gamma$.

But that is the same as looking for different ways to write $\gamma$ in cycle notation.

Suppose the complex equation $iz^2+(2-i)az-(1+i)a^2=0$ as $a\in \mathbb{C}^{*}$ and $z_1$ and $z_2$ are the solution of this equation and we have also $z_1*z_2 = a^2(i-1)$ how can I prove that $arg(a) \equiv \dfrac{-3 \pi}{8}[ \dfrac{\pi}{2 ]$

You're missing a closing brace.

@peoplepower That is the stabilizer.

@peoplepower where

Aren't we looking at the orbit?

how can I prove that $arg(a)\equiv \dfrac{-3\pi}{8}[\dfrac{\pi}{2}]$

@PeterTamaroff Exactly, we are trying to characterize it so we can find its size. Then the size of the orbit, as you said initially, is just $|S_n|$ divided by the size of the stabilizer.

@pourjour At the very end.

OK, so I said we need the stab to have order $n$.

Which is clear it has.

@peoplepower thanks I fix it

We just make the numbers go merry go round.

ok So any idea would help?

@PeterTamaroff Indeed that guarantees that the size of the stabilizer is at least $n$. :)

@peoplepower At least? Why not exactly?

@PeterTamaroff There could potentially be more ways to write $\gamma$.

Let me think.

ok

@pourjour Sorry, but I'm too busy with this.

You can ask on main, it is not too complex, so you should get an answer fairly quickly.

@peoplepower I don't think there are.

@pourjour I am too ignorant, sorry.

@PeterTamaroff Good, formalize this.

@peoplepower If we change the order of any number in the cycle we obtain another cycle.

@peoplepower haha

I mean, transpose.

@PeterTamaroff It is an "obvious" fact in that it is difficult to prove rigorously. You should probably think of the bijection $\gamma$ to formalize it.

@peoplepower I see. Thinking...

Wait, wait.

The stabilizer of this $\gamma$ is just all the different $n$ cycles. That's what I said earlier. Like $(1234\dots n)$, $(21345\dots n)$.

No.

No?

There are $n$ powers of $\gamma$, so the stabilizer should be comprised of just the powers of $\gamma$.

But I just did tell you that for example $(12)(21345\dots n)(12)=(1234\dots n)$. That is wrong?

The stabilizer consists of things acting upon $\gamma$ but still fixing it.

Oh, again! Silly me.

Sorry.

I return to the orbit, I don't knwo why.

If you do happen to know why the orbit consists of all $n$-cycles, you can count them with some combinatorics.

This is probably the intended approach to the problem.

I see.

So I want permutations such that $(\sigma(1)\sigma(2)\dots \sigma(n))=(12\dots n)$.

This accounts for the powers of $\gamma$, I see now!

For example, we can take $1\to 3$; $2\to 4$, &c

That is $\gamma^2$

Yes.

OK. But $\gamma^n=1$.

K.

Meaning that it is the identity.

So I have the identity plus the $n-1$ powers of $\gamma$.

Good, good.

Now I should prove this covers all possible solutions.

In fact, there is a simple combinatorial arguemnt.

The value on one $\sigma(i)$ determines all the other values.

We have $n$ choices of $i$

Thus we have $n$ possible permutations.

Done.

@peoplepower Am I right or left?

@PeterTamaroff That line of argument is valid, yes.

@peoplepower Melikes.

@peoplepower Do you agree it is clear enough? i.e. no handwaviness?

@PeterTamaroff One would potentially fill in the details here: The value on one $\sigma(i)$ determines all the other values.

How exactly do I determine the rest of the values?

@peoplepower I suspected so.

@peoplepower Suppose $\sigma(i)=j$.

So choose $i,i'$, how must $\sigma(i)$ relate to $\sigma(i')$?

Then $\sigma(i+1)=j+1$ if $j<n$ and $=n-j+1$ if $j\geq n$, me thinks.

And so on.

Obviously if $i=n$ then "$n+1=1$"

Darn, I can't spell obviously correctly.

@PeterTamaroff I will tell you the general correspondence; it looks a bit like equivalence of actions: $\sigma\circ\gamma=\gamma\circ\sigma$.

Notice how that is what you wrote in a shorter form.

@peoplepower Aye.

@PeterTamaroff Anyway, that is how we determine $\sigma$ for all other values; particularly, we use the fact that $\gamma$ is a bijection.

@peoplepower Oh, well, that is what I had in mind yes.

But you owned it. =)

Simple and simpler.

pwned?

Ya

power to the people

@peoplepower Now I will determine all represent

atives in the conjugacy classes of $S_5$

Then find the order of each class

And I have to show with that that the only normal sgrps of $S_5$ are $1, A_5,S_5$

So basicsally show the only nontrivial norml subgrp is $A_5$

$5!=120$.

Oh, boy

Well, you have 25 already taken care of. There are 24 5-cycles and we have one trivial permutation.

I know that $(12345)$ has 23 other representatives, yes. So 24.

@peoplepower So maybe I can choose a $4$ cycle $(1234)$ and work on that one similarily.

Just for the sake of simplicity.

Sure.

In that case I'm looking again at $(\sigma(1)\sigma(2)\sigma(3)\sigma(4))=(1234)$

Yes.

And it is just the same.

There is a caveat, but I will see if you notice it.

Then I can just look at $(1235)$ and all ${5\choose 4}=5$ other classes.

"a modifying or cautionary detail to be considered when evaluating, interpreting, or doing something"

For this case, it is cautionary but not modifying. For other cases, it is modifying.

Well, certainly $\sigma$ must fix $5$.

Indeed.

So the powers of $(1234)$ help again.

It is the same argument (?)

It is the same argument to show that the stabilizer of (1234) is just the powers of (1234).

OK, if we work down to $(12)$, I get $24,30,20,10$

But I'm still missing 36 elements.

Well, you just represented the cycles. There are other permutations.

Sure.

Note the above is $${5\choose k}(5-(6-k))!$$ for each cycle.

But we actually have disjoint classes.

I.e ${5\choose k}$ classes.

OK, so now I can look at...

OK, I ran out of ideas. BRAIN!

...

@PeterTamaroff For instance product of two transpositions.

@peoplepower Yes, but I don't expect a nice patter to show up =P I'll give it a shot though. Did I reach the "caveat" yet?

@PeterTamaroff Your results seem to have taken it into account. I.e. we can do anything we want outside the permutation we are fixing.

OK, for transpositions we need two values only. In general if the permutation is a product of $X$ disjoint cycles, the solution is determined by $X$ initial values.

So we will have... how many representatives? $(n-m)!$? ($m=X$, the $X$ looks ugly)

Then we should take care about possible repetitions.

@PeterTamaroff I put an answer did u see it math.stackexchange.com/questions/315736/…

@pourjour Good =)

now I'm looking for a way to factorise complex numbers

@pourjour Ah?

@peoplepower I am not talking crazy right?

Hm, would it not be $m!$ divided by something, multiplied by $(n-m)!$?

I am pretty bad at counting stuff, btw.

Yes, I noted that is totally off.

I see I exhaust all $k$ cicles $1<k\leq 5$ with the above.

@PeterTamaroff What I got for the class represented by $(12)(34)$ was that the stabilizer has 2*2*2 elements. The first two 2's correspond to permuting within the individual transpositions, the last corresponds to switching them.

I see.

Clearly this is "disjoint" from $(12)(35)$

So these two give two more classes of $15$ elements each?

It is just one class of 15 elements.

I mean together with $(12)(35)$.

This gives another class.

(12)(35) is conjugate to (12)(34), so it is in the same conjugacy class.

Oh, right.

It is not in the stabilizer of the first, though.

In fact every "$2,2$ cycle" should be in the same class.

Disjoint, this is.

The cycle.

yes.

So far I have $24+30+20+10+15$ then.

Now with $2,3$ cycles.

im confused baout this automorphism: x \to x^p in F_p^m

it has order m because p^m = 0?

@user58512 Use dollar signs.

The chat doesn't parse stuff automatically.

why does it have order m

how come it doesn't turn all the elements to 1 and leave them there everntually?

I got it

x^(p^m) = x

Hi @JonasTeuwen how are you?

how many solutions are there to a^2+b^2=1 in \mathbb F_p^m?

What is $n-1$ dimensional affine subspace in $\mathbb R^n$?

Does it mean vectors of the form $(n-1\times 1)$ so basically consists of column vectors, right?

hi

@hhh, why dont you just look up a definition?

@user58512 I think if $m$ is even there must be some solutions.

@awllower, I can prove that every number is the sum of two squares in finite fields by simple counting, but I want to count the number of solutions to this specific equation

But if $p\equiv 3\pmod 4$, then there should be no solutions in $F_p$.

1^2 + 0^2 = 1 in F_3

Oh...

Forgive my stupidness...

my fault using = when I meant \equiv

yeah

So you already have 4 solutions.

So the prroblem is to see whether 1-b² is some square in the finite field.

And I think the number of squares in a finite field is half the number of elements?

seems too difficult :(

I know that's true for the multiplicative part of a finite field, since it's cyclic

Hm

so for the whole field we get (p^m+1)/2 squares

(just have to make sure you count zero once)

Oh

And the number of solutions should be equal to the order of the uniit group in $F_{p^{2m}}$?

By a unit I mean norm 1, over F_p^m

I dont know

Hm, I do not know either...

@TobiasKildetoft, hello

@user58512 hi

@user58512 note that if $p = 2$ then $1 - b^2$ will always be a square

@TobiasKildetoft,if you dont mind I have another question about double cosets: I have proved for transitive group G on X that suborbits (orbits of stab_G(x)) are in correspondence with double cosets stab_G(x) g stab_G(x). I think the orbits partition X and double cosets partition G

is it correct?

sure, the orbits always partition the set acted on

and the double cosets always partition the group

thanks - I want to compute an example in gap that I can see thee bijection: Do you recommend a group I should use?

something like 7 symbols would be good (but that might make the group too big..)

I guess a dihedral group is a good choice, as that is transitive but not doubly transitive on a suitable set

(if it was doubly transitive, this whole decomposition thing would not be so interesting)

yes that seems a good idea I gues ill use D8

thanks!

wait when you fix a point of D8, you get left with C_2?

what do you mean by fixing a point?

stabilizer of one of the corners of the square

ahh, yes

(in my world, $D_8$ has order $16$, not $8$)

of course, if you want to do this partition for a symmetric group it is very easy, and same goes for the alternating groups

what the heck

SL_2(3) is the symmetry group of "chrial tetrahedral symmetry"

but it's 2x2 matrices...

that's weird

@TobiasKildetoft, do you know a decent intro for GAP? I find the manual really difficult to find things in

@user58512 have you had a look at the tutorial on the official site?

Gauß in disguise?

that looks good! thanks

@user58512 it can take a little while to get used to how lists work (when it comes to assigning them to variables and such)

im quite interested in how it works

it must have cool algorithms

hey, any people know their stats here? got a question about an optimal way of checking for how spread out my data is