Mathematics - 2013-02-27 (page 1 of 5)

skullpatrol

00:03

A skull plushie.... looking for his comic book ;-)

Peter Tamaroff

@people peeps

peoplepower

@PeterTamaroff Yellow

Jonas Teuwen

@JacobBlack Good.

Charlie

You know one thing I like , @skull? Gummy bear

skullpatrol

@Charlie Me too...Look what they did to Mickey

Peter Tamaroff

00:17

@peoplepower Simple question.

If two permutations are congujate they determine the same partition of $n$

This stems from $\alpha(i_1\dots i_r)\alpha^{-1}=(\alpha(i_1)\dots \alpha(i_r))$

peoplepower

Um... what do you mean partition? As in enumeration of sizes of cycles?

Ah, that has to be.

Peter Tamaroff

Just $\mu\in S_n$

On the other hand, the author says that if two permutations det. the same parition they are conjugate, and he says we can see that from that last formula too.

Charlie

@skullpatrol oh dear!

Peter Tamaroff

That is, both the if and only if can be proven from that @people

For example

peoplepower

@PeterTamaroff Yes. One is more immediate than the other, but it is a true statement.

Peter Tamaroff

00:20

Let $(135)(24)(76)$ and $(753)(12)(46)$

Then if we use the permutation $(71)(26)$ we can get from one to another

skullpatrol

@Charlie here

peoplepower

@PeterTamaroff Don't you need (35) as well?

@PeterTamaroff Just list like cycles below one another, remove parentheses and you have defined the desired one-to-one correspondence.

Charlie

@skullpatrol how cute!

skullpatrol

@Charlie bye

Peter Tamaroff

@peoplepower Oh, true.

@peoplepower Well, it is not that simple. One must order them, somehow.

Charlie

00:26

@skullpatrol bye :(

peoplepower

@PeterTamaroff There are multiple solutions, one for each allowable ordering.

skullpatrol

@Charlie I'll talk to you later.

Peter Tamaroff

@peoplepower Ok, but you mean one must order them by magnitude for example? Because an arbitrary order can lead to an impossibility.

Charlie

@skullpatrol okay

peoplepower

@PeterTamaroff Yes, allowable means that n-cycles must line up for each n.

Peter Tamaroff

00:29

OK but in my example, take

$(135)(23)(76)$
$(753)(12)(46)$

Then having lined up $1,7$ together, this forces us to line up $(76)$ with $(12)$

peoplepower

Why?

Peter Tamaroff

@peoplepower Because else it isn't a bijection O.o

The $(23)$ should be a $(24)$, my bad on that one.

peoplepower

Oh.

It still works (1742).

Peter Tamaroff

Ah?

peoplepower

I do not see where it would fail to be a bijection. Give a specific point of failure.

Peter Tamaroff

00:36

I interpret that you map the upper number to what goes down, right?

peoplepower

Yes, $1\mapsto 7$; $2\mapsto 1$; etc.

Peter Tamaroff

Ie $1\to 7$, $3\to 5$, $5\to 3$, $2\to 1$, ...

MSEoris

Hey does anyone know what a trivial open set is? is it just the null set?

Peter Tamaroff

@MSEoris Probably $X$ and $\varnothing$

Can be considered the "trivial" open sets.

$X$ stands for the underlying set

MSEoris

thanks @PeterTamaroff !

Peter Tamaroff

00:38

Ok, @people, but if we map $7\to 4$ we don't get the conjugation to work, I think.

Oh, duh.

I was misundertanding the procedure.

@peoplepower I see.

Thanks.

My mind's light bill increased by 22W per hour.

peoplepower

@PeterTamaroff No problem; I had not thought of it this way before either.

Peter Tamaroff

@peoplepower But now, how does it follow from that conjugation formula?

peoplepower

@PeterTamaroff We defined $\alpha$, now apply it to get what you want.

Peter Tamaroff

@peoplepower Yes, of course. I was thinking about a more "direct" result from the formula.

Nevermind.

user19161

01:07

@robjohn Congrats on getting 50k! You are now halfway to 100k!

Peter Tamaroff

I made a Rock of Ages playlist

Look at that, I got accepted @Jacob

Peter Tamaroff

01:40

@peoplepower You there? I got to the class equation =) Finally.

Gustavo Bandeira

@PeterTamaroff 0 upvotes.

Peter Tamaroff

@GustavoBandeira I know. All because of you!

Gustavo Bandeira

@PeterTamaroff No. 1 upvote because of me.

Peter Tamaroff

@GustavoBandeira Good. Now your Karma returns to balance.

Gustavo Bandeira

@PeterTamaroff lol

Charlie

01:46

My name is Charlie...naah it's

Not good

Peter Tamaroff

@Charlie Whu?

Oh, the "Bond, James Bond" thing?

Charlie

@PeterTamaroff yeeah...

robjohn

@JacobBlack Thanks for the half a compliment ;-)

Charlie

I want a plushie

Peter Tamaroff

@Charlie A dog is even better.

If you're feeling lucky, aim for a partner.

$\text{bf}\geq \text{dog}\geq \text{plushie}$?

2

Charlie

01:50

@PeterTamaroff I already have a dog, I.want a plushie

@PeterTamaroff I love plushies

Peter Tamaroff

@peoplepower The notation is getting me dizzy.

$\operatorname{stab}(axa^{-1})=a\operatorname{stab}xa^{-1}$

The $axa^{-1}$ is actually $a$ acting on $x$, but how does $a^{-1}$ hop in?

The action is from the right AFAIK

peoplepower

@PeterTamaroff The first line looks like a typo.

Eric Gregor

Is there a rough way of saying what an Eisenstein series is?

Peter Tamaroff

@peoplepower Which one?

Eric Gregor

easy to state way that may be sloppy

Peter Tamaroff

02:03

That is what the guy is saying.

peoplepower

We have stab(ax)=a^{-1}(stab(x))a.

Peter Tamaroff

@peoplepower Oh, that's why I couldn't friggin prove it!

The formula looked nice!

Thank you.

The guy writes that formula and just says "The proof is clear".

=)

peoplepower

@PeterTamaroff Oops, I also erred. It should actually be $\rm{stab}(ax)=a(\rm{stab}(x))a^{-1}$.

Peter Tamaroff

@peoplepower Yes, yes, I got you.

I made a Rock of Ages playlist

user51819

Is anyone familiar with Moebius inversion?

Peter Tamaroff

02:09

@user51819 Barely, why?

peoplepower

@user51819 Yes.

user51819

Trying to figure out the Moebius transformation for $\sum_{k=2}^{N}k^2\varphi(k) f(\lfloor \frac{N}{k} \rfloor)$ where $f(x)$ is an arithmetic function

example $f(x) = 2x^2 + 3x + 11(2^x+1)$ but it can be anything really, just showing you the kind of stuff inside it

user51819

02:25

did I kill the room...

Dan Brumleve

02:36

@user51819 i would start with the euler product? not sure how to handle \frac though... i doubt there is any simplification for all arithmetic functions, maybe f needs to be multiplicative?

Peter Tamaroff

@user51819 This is the MSE police. You have the right to remain in silence. Everything you say can and will be used against you in court.

Dan Brumleve

err i mean \floor not \frac

is that a federal offense?

Peter Tamaroff

You have the right to an attorney. If you can't afford one, the state will prov.... Oh, @DanBrumleve blew it.

Dan Brumleve

good timing. anything you say...

en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula moebius inversion is defined in terms of a sum over divisors but there are #Generalizations that may be relevant

Peter Tamaroff

Whenever I see someone use $\sqrt{a+b}=\sqrt a +\sqrt b$ my reaction is

user51819

02:40

@DanBrumleve Yeah I looked at that already unfortunately

hitting a brick wall with this thing

Dan Brumleve

well you don't get a unique g from a particular f because the floor function is not injective

user51819

Scroll down a bit in Generalizations

It can be used with floor function

also en.wikibooks.org/wiki/Famous_Theorems_of_Mathematics/…

Dan Brumleve

i am looking at that part but i'm not sure what you are asking anymore

user51819

Just trying to solve $\sum_{k=2}^{N}k^2\varphi(k) f(\lfloor \frac{N}{k} \rfloor)$ using Moebius inversion

Dan Brumleve

moebius inversion is undefined for general arithmetic functions

multiplicative ones yes i think so

Peter Tamaroff

02:49

@anon Is there any notation for "$G$ acts on $S$"?

anon

yes

Peter Tamaroff

Good! Which is...?

anon

let me find the latex for it

Dan Brumleve

i mean, not only is it impossible to define moebius inverse on a complex-valued arithmetic function, it can't even be done for an integer-valued one. only multiplicative ones en.wikipedia.org/wiki/Multiplicative_function#Convolution

robjohn

@PeterTamaroff square root is not linear?

anon

02:52

@PeterTamaroff things like $G\curvearrowright X$ or $G \circlearrowright X$ are the ones I prefer

user51819

well I mean I am trying to evaluate that expression so is there a way I can separate them and calculate them separately?

Dan Brumleve

why is everything its own square root today?

anon

today we celebrate the trivial ring, where everything is its own square root

Dan Brumleve

@user51819 the sum formula is a way to calculate it, right? what do you mean by separate?

Peter Tamaroff

@robjohn Wha?

@anon Thanks.

anon

02:55

indeed, not only is the square root linear, it is constant

user51819

Like to calculate parts of the expression a different ways

if Moebius inversion won't work for all of it

robjohn

@PeterTamaroff Hello

Dan Brumleve

do you mean calculate exactly? if f is complex-valued how do you know the sum will evaluate to an integer?

user51819

not sure what that means

f is just a polynomial-type expression

everything involved is an integer

Peter Tamaroff

@robjohn Haha, hello.

@robjohn Did you happen to see my question about uniformly distributed sequences?

Dan Brumleve

02:59

i guess i am just re-asking your question. i hadn't noticed you were convolving with phi maybe there is a specific answer.

robjohn

@PeterTamaroff no, is it on main?

Peter Tamaroff

@robjohn Nope. Let me find it for you.

Dan Brumleve

anyone know much about en.wikipedia.org/wiki/… ?

Peter Tamaroff

Let $(a_n)_{n\in \Bbb N}$ be a sequence. Define $N(a,b;k)$ to be the number of integers $j \leq k$ such that $a_j\in [a,b]$ for $a<b\in [0,1]$. We say that $(a_n)$ is uniformly distributed on $[0,1]$ if, for any choice of $a<b$ we have that $$\lim\frac{N(a,b;k)}k=b-a$$

Prove $$\frac 1 2,\frac 1 3,\frac 2 3,\frac 1 4,\frac 2 4,\frac 3 4,\frac 1 5,\dots$$ is UD over $[0,1]$, @robjohn

robjohn

@PeterTamaroff So you populate all the $\frac kn$ for each n and move on?

Peter Tamaroff

03:04

@robjohn Hehehe, yes.

anon

double-take

robjohn

@PeterTamaroff It seems as if we want to prove it for $k = \frac{n^2-n}{2}$ and fill in intermediate cases

@PeterTamaroff I don't see anything difficult, just getting the notation and all the cases covered. Busy work.

@PeterTamaroff Do you have a slick solution that bypasses the busy work?

Peter Tamaroff

@robjohn I noted that $$a_{T_n}=1/n$$ but I didn't give it much thought.

Where $T_n=\frac{n(n+1)}2$

I also noted one can determine exactly how often a fraction repeatss.

anon

isn't there some clever way to test UD with complex exponentials?

that could bring in analysis related to roots of unity

Peter Tamaroff

@anon Something Fourierish?

anon

03:11

right

Peter Tamaroff

@robjohn if you set the numbers like this

$$\eqalign{
& {1 \over 2} \cr
& {1 \over 3},{2 \over 3} \cr
& {1 \over 4},{2 \over 4},{3 \over 4} \cr
& {1 \over 5},{2 \over 5},{3 \over 5},{4 \over 5} \cr} $$

robjohn

@PeterTamaroff That's why I gave $n^2-n$; that many entries covers the denominators from $1$ to $n$

Peter Tamaroff

then standing at $k/n$ means the number will repeat $n$ rows down and $k$ columns left

@robjohn Yes.

robjohn

I prefer$$
\frac12\\
\frac13\ \frac23\\
\frac14\ \frac24\ \frac34\\
\frac15\ \frac25\ \frac35\ \frac45
$$

the spacing is not quite right

Peter Tamaroff

@robjohn My coding knowledge doesn't =P

Oh, it is just \\

Thought it was some align environment.

Did anyone know Bear Grylls was a British special ops?

And he climbed Everest.

He's a total badass.

robjohn

03:22

@PeterTamaroff I'll try not to piss him off.

Peter Tamaroff

@robjohn He does it himself, no worries!

BAD DUM TSS

anon

answer + comments

user51819

:|

Peter Tamaroff

@anon HAHHAHAHAHA

Orange Harvester

@PeterTamaroff Icefall

Peter Tamaroff

03:25

LAWL

@OrangeHarvester 7 mins?

Orange Harvester

@PeterTamaroff Well, this is as good as 20 min bear grylls documentary.

Peter Tamaroff

@OrangeHarvester Oh, but the Bear's accent makes it pay off.

@OrangeHarvester When does the guy fall?

Orange Harvester

@PeterTamaroff boo....

Peter Tamaroff

Or the ice fall?

Or the ice and the guy fall together?

Orange Harvester

@PeterTamaroff no person falls, there is nothing dramatic in that, except for the ice moving away leaving the rifts behind.

Peter Tamaroff

03:29

@OrangeHarvester What are rifts?

Orange Harvester

@PeterTamaroff are you even watching the video?

Peter Tamaroff

@OrangeHarvester Kinda. I am also trying to finish watching the hunger games.

Orange Harvester

I see.

The ice gets deposited in between the gorges in the mountains.

Peter Tamaroff

Dem big ass killer dog be lose

Killin errbdy around.

Hide yo children, hide yo wife.

Orange Harvester

and when you move near it, it gets unstable and moves out very fast (in like 4-5 minutes)

and leave you with the gorges just like that.

Peter Tamaroff

03:30

@OrangeHarvester Damn.

Orange Harvester

One can be unlucky to fall in one, if he is on of the ice sheets when it starts moving.

@PeterTamaroff yeah, that is why mountaineers wear ropes connecting all of them, and carry ladders around.

Peter Tamaroff

@OrangeHarvester "If I fall you all fall, bitches!"

Orange Harvester

@PeterTamaroff Hahahah, yeah.

Peter Tamaroff

OK, all this "holographic come to life" thingy in the Hunger Games made it a little lame.

anon

@PeterTamaroff that actually wasn't in the book

there are invisible hovercrafts, artificial lightning etc. so whatevs

Peter Tamaroff

03:35

@anon There is a limit...

anon

I didn't think it was too overdone. This is far in the future, and presumably most of north american wealth is put into the hunger games, which is a lot of money for technology.

It's not like everyone and their banana has a hovercraft.

3

Peter Tamaroff

Yes, I did notice.

I don't like the aesthetics proposed in the movie though.

anon

I think the biggest issue with the movie, as with any movie adaptation of a book, is that the thought process of the characters is more opaque and unexplained.

Peter Tamaroff

The arrow should have gone through his friggin hand... damn.

Peet should have died there.

Except... she aimed for the bone. Which is nearly impossible.

Even if she hit the bone, the arrow would probably bump to a side.

An arrow to the eye should have been better.

anon

which part are you talking about?

Peter Tamaroff

03:45

@anon They are atop the ship and "Cato" is chocking Peet.

And she shoots him in the hand.

But nevermind @anon I have an algebra question.

Suppose $G$ acts on $S$.

anon

ah. how long was cato left to the dogs before catnip mercy kills him? in the book I think it was hours, but iirc the movie makes it quick.

Peter Tamaroff

Oh, the movie was just seconds.

user19161

Guys, I need to do something very important tomorrow, wish me luck...

anon

:luck:

Peter Tamaroff

Then if we partition $S$ into $G$ orbits, each orbit $A$ is stabilized by the action in the sense that $gA=A\in \pi_G(S)$

@JacobBlack I wish you success, not luck.

Orange Harvester

03:48

@JacobBlack Wish you success.

Peter Tamaroff

@anon Yet the author says that we call an action prmitive if the only stabilized partitions are $\{S\}$ and the singletons of $S$. But apparently this partition is always stabilized too.

user19161

By the way, if I am thinking of switching from gmail, should I try hotmail or yahoo mail?

Orange Harvester

gmail is better

much much better

yahoo interface is as slow as a snail

user19161

I just want something new.

Orange Harvester

and hotmail interface is similarly stupid too.

anon

03:50

@PeterTamaroff Right. This clearly rules out any action with more than one orbit as imprimitive.

Orange Harvester

@JacobBlack try zoho.com

user19161

Also, I hate how the google accounts are all connected.

Alexander Gruber

@JacobBlack good luck

Peter Tamaroff

@anon So primitive implies transitive.

Orange Harvester

@JacobBlack try mail.zoho.com. Its interface is slow too, but the accounts are not connected. Use Mail client to get around that.

user19161

03:51

@OrangeHarvester OK, but between my two listed options?

Orange Harvester

@JacobBlack hotmail and yahoo are similarly connected.

anon

@PeterTamaroff Yes, but not vice versa, because actions may "permute" the cells of the partition, in the sense that for some $g\in G$ and $A,B$ in the partition we may have relations like $gA=B$, for instance.

Peter Tamaroff

@JacobBlack What is this thing you're gonna do?

user19161

@PeterTamaroff It's a secret.

Peter Tamaroff

@JacobBlack What is it related to?

anon

03:52

My bet is on romance.

Orange Harvester

@PeterTamaroff Knowing as much as I do about Jasper, there has to be a girl somewhere.

user19161

@PeterTamaroff It is related to my health and my military training.

Peter Tamaroff

@OrangeHarvester Two bets for romance already... aaaaand they're gone

Orange Harvester

@anon Boo.

Peter Tamaroff

@JacobBlack I see.

Orange Harvester

03:53

@PeterTamaroff yeah.

anon

@PeterTamaroff But his health and military training are for romance, you see...

Orange Harvester

@JacobBlack Ohh, okay, you mentioned something about that. Hope it goes well.

user19161

@anon I have no romance in my life now. I only hope to recover and move on with my life.

anon

As you wish, captain.

skullpatrol

Make it happen.

user19161

03:54

However, if I were a young girl, I may fall in love with anon.

anon

I am the master of my fate:
I am the captain of my soul

user19161

Actually speaking of romance, I have recently fallen in love with yet another user on SE, but I will keep her identity a secret.

Orange Harvester

how long ago was that?

And I think I already know. :P

user19161

It's not anyone who comes to chat.

Peter Tamaroff

I wonder how singleton is translated to spanish.

It is a pretty cool name.

Orange Harvester

03:55

@JacobBlack I think I know.

anon

foreveraloneset

user19161

@OrangeHarvester It's not anyone who has ever come to chat...

Peter Tamaroff

@anon HAHAHA

Orange Harvester

@JacobBlack Yeah, even then I think I know.

user19161

@OrangeHarvester WTF? Are you a god?

anon

03:57

m-m-m-ma poker face

Orange Harvester

@PeterTamaroff ningún amigo?

skullpatrol

lol

user19161

Haha OK @ora you may send your guess to my email now...

Orange Harvester

@JacobBlack sent.

user19161

Guys, @ora guessed correctly...

Orange Harvester

04:00

@JacobBlack :-)

Peter Tamaroff

@OrangeHarvester Tell me, tell me. Do you know my mail?

Orange Harvester

@PeterTamaroff I do not know your mail, but Jacob told me not to, so.... (I know spoiler spoiler. ) :P

Peter Tamaroff

Damn you Jasper.

user19161

Hehe, I don't even know anything about this girl, haha!

anon

The word love is well-diluted then.

Peter Tamaroff

04:03

@anon Good one, sire.

user19161

Yes, very diluted.

Orange Harvester

@anon As always with Jasper.

user19161

Haha.

Peter Tamaroff

@OrangeHarvester Shame on you.

Orange Harvester

@PeterTamaroff wat?

Peter Tamaroff

04:04

That.

anon

thank thor for diagrams

Peter Tamaroff

@anon Man. If for every $x$ in $S$, $\operatorname{stab} x$ is a maximal subgroup, then $\operatorname{stab}x$ is the same for any $x\in S$ right?

I mean, since we can't have proper inclusion on any direction which would violate maximality.

anon

I don't think so

Consider G acting on the disjoint union of coset spaces $G/M\sqcup G/N$, where $M,N$ are distinct maximal subgroups of G

Peter Tamaroff

"...there exist no subgroup $H$ such that $\operatorname{stab}(x)\subsetneq H\subsetneq G$."

But every stabilizer is a subgroup-.

Oh, stupid me.

Nevermind.

I was being idealistic again.

anon

the stabilizers of two different points are not generally comparable in terms of the partial order on the lattice of subgroups

Peter Tamaroff

04:10

Yes, yes.

anon

hmm. groupprops has no proof information on its profinite sylow theorems entry.

I am disappoint.

Peter Tamaroff

Create them yourself.

Be the change you want to see in this world man.

It is like "Changing the World 101" stuff.

@anon Is it normal that some theorems feel totally unnatural at first? Like this primitivity criterion

anon

Yes. The definition of a primitive action tripped me up the first time I read it - it made no sense.

Peter Tamaroff

Hehehe, OK.

anon

I believe someone should write a nice counterexamples in algebra, with a nice part I on group theory.

It surprises me that for infinite groups it's possible for a subgroup to properly contain a conjugate.

Hmm, that gives me an idea: a conjugal closure of a subgroup H<G could be one that is maximal within the partially ordered set of subgroups S that are conjugate and comparable H.

Peter Tamaroff

04:23

You mean that if $G$ is a subgroup, it properly contains $^a G$ for some $a$?

anon

yes

Peter Tamaroff

The math jargon used here is interesting youtube.com/watch?v=TZJLtujW6FY

Around 1:30 if you're lazy.

anon

did some background research on that 9gag post did you?

Peter Tamaroff

@anon Where?

anon

oh, it was more than a week ago, I couldn't give you a link. it was just a youtube video of paperman (I think; it wasn't really titled)

Peter Tamaroff

04:27

Oh, I saw Paperman a while ago, but I wanted to see it again and this came up.

It is a really cute story.

Like the first minutes of UP, which is actually more sad than it is cute.

Dominic Michaelis

huhu

Peter Tamaroff

@anon Man

anon

Mmhmm.

@DominicMichaelis huehuehue?

Dominic Michaelis

i only know hue from the funny coloring function in mathematica is there something different ?

Peter Tamaroff

I just realized, and correct me if I'm wrong. By if $G$ acts on $S$ and a partition $\pi(S)$ is stabilized by this action, then in some sense the induced action of $G$ on the power set "acts transitively" on the partition.

julien

04:38

@Stephan Smith Hello.

Peter Tamaroff

@julien You have some pretty nice answers.

Orange Harvester

freaking hotels be expensive.

Dominic Michaelis

@julien hi :)

julien

@Peter Thanks!

Dominic Michaelis

@julien yeah i really like your answers

julien

04:40

@Dominic Hello! Thanks!

Peter Tamaroff

@DominicMichaelis This

anon

@PeterTamaroff That is not at all clear to me. I would not expect it to be true. It just means that the partition is stabilized (ie a union of orbits) of the action induced on the power set.

Peter Tamaroff

@anon I mean, whenever $A$ is in the parition and $g$ is in $G$, $gA=A'$ for some element of the partition.

That is what the definition says, doesn't it?

anon

What's to stop for example a partition with cells A,B,C, that may permute A and B but always fixes C? That would not be transitive, for there would not exist a g for which gA=C, nor for which gB=C.

@PeterTamaroff No, the action would be transitive on the partition if for any two cells X,Y in the partition, there exists a g such that gX=Y

julien

Do you guys know a good reference (with full proof) of the Jordan normal (canonical) form of a real/complex matrix?

Peter Tamaroff

04:42

@anon Yes, I see my mistake.

Thanks

Dominic Michaelis

@julien that {{0,1},{0,0}} stuff ?

I should have it in my script of the last semester

julien

Y@DominicMichaelis Yes {{a,1}{0,a}} etc

Dominic Michaelis

but it is a german script, can you speak german ?

Peter Tamaroff

@julien I think I might have something in Spanish.

Sanchez

@julien, what do you need it for?

Peter Tamaroff

04:44

@anon (I meant the defintion of primitivity, not of transivity here, just for the record)

julien

@DominicMichaelis I took 4 years of german in highschool and I can barely say hi

Peter Tamaroff

@Sanchez You be the Interpol of notes?

julien

@PeterTamaroff And my spanish is even worse...

Sanchez

@Peter, no. Just curious.

Dominic Michaelis

We did proof it constructive, but that algorithm really sucks :D

julien

04:46

@Sanchez Because of this thread on the connectedness of matrices with positive determinant. I came up with a solution using Jordan normal form, and the Op wants a good reference.

anon

@PeterTamaroff it should be possible for the induced action (of an imprimitive action) on a partition to be either itself primitive or imprimitive. however a primitive action does not admit a partition to have an induced action on in the first place.

Sanchez

@julien, I see.

Peter Tamaroff

@anon Blimey. "We observe first that $G$ acts imprimitively on a set $S$ (by hyp. $G$ acts trans on $S$, $|S|>1$) if and only if there exists a proper subset $A$ of $S$ with card $\geq 2$ such that for any $g\in G$ either $gA=A$ or $gA\cap A=\varnothing$."

I'll try to prove that as a side lemma.

That's how the proof for a primitivity criterion starts.

Dominic Michaelis

@julien I guess the algorithm of the jordan normal form is the only thing i can't from linear algebra which we made in teh 2 semesters ...

julien

@DominicMichaelis I confess I forgot it too... What did you like best in these linear algebra classes?

anon

04:51

@PeterTamaroff hint for the hard direction: partition G into cosets of A

Peter Tamaroff

@anon OK. I'll work on it.

It is kinda nasty how he just lays it out there.

We observe...

Dominic Michaelis

As i heared it it was the eigenvalue and eigenspace. In my first semester i only heared linear algebra for physics, but now i love that more algebra related things like semigroups and groups, rings fields ...

julien

@DominicMichaelis I like diagonalization too. Here is an exercise/result I found striking when I first encountered it: a matrix A is normal if and only if its adjoint A* is a polynomial p(A).

Dominic Michaelis

Normal is that $A^\ast A=A A^\ast$ right ?

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$Mathematics$ Mathematics