Mathematics - 2017-10-22 (page 2 of 5)

« first day (2636 days earlier) ← previous day next day → last day (2387 days later) »

2:01 AM

@MatheiBoulomenos hope you did not forget about my question ! :D

@MatheiBoulomenos conjugation is the automorphism $\varphi: \zeta_p \mapsto \zeta_p^{p-1}$

MatheiBoulomenos

@KasmirKhaan you can break it down to 2 simpler statements, which are together equivalent to that one you mentioned

what i dont understand here is that

if $p \equiv 1 \pmod 4$, then $p=4k+1$, then $\zeta_p^{p-1} = \zeta_p^{4k} = (\zeta_p^{2k})^2$ is a quadratic residue

lets say that order of G is p^m *r

the order of K here is p^m ?

maximal subgroup or does not have to be ?

MatheiBoulomenos

2:04 AM

one is every subgroup whose order is a power of $p$ is contained in a $p$-Sylow subgroup, the other is that all $p$-Sylow subgroups are equivalent

so if i understand right

we are guranteed that we have a p^m subgrouo

MatheiBoulomenos

yes

that's Sylow 1

what I said is wrong

if we have a subgroup of order p^l , where l < m

we need $p-1=4k$ to be quadratic residue itself

2:05 AM

then all those subgroups must be conjuagte of one another

so order of H = p^l , if there is another subgroup K, K = gHg' for some g in G

K here also have same order as H , since we have a bijection

hmm

$a$ is a quadratic residue mod $p$ iff $a=c^{2m}$ for some $m$ where $c$ is a primitive root of $p$, and $c^{p-1} \equiv 1$

if we have an example with numbers, order of G = 5^3 * 7

H is a subgroup of order 5^2

then all subgroups of order 5^2 are conjugate

MatheiBoulomenos

Sylow says nothing about groups of order 5^2 being conjugate in this setting

hmm :(

What does it say in this example?

MatheiBoulomenos

But is says that all subgroups of order 5^3 are conjugate

2:08 AM

so we still working on maximal subgroups?

$a$ is qr mod $p$ iff $a^{(p-1)/2} \equiv 1$

MatheiBoulomenos

and it says that if you have one subgroup of order 5^2 $K$ and one of order 5^3 $H$, then there is some $g$ such that $K \subset gHg^1$

if $p=4k+1$, then $(-1)^{(p-1)/2} = (-1)^{2k} = 1$. If $p=4k+3$, then $(-1)^{(p-1)/2} = (-1)^{2k+1} = -1$.

@MatheiBoulomenos have I proved it?

Therefore $\varphi : \zeta_p \mapsto \zeta_p^{p-1}$ is in the normal subgroup of order $2$ iff $p \equiv 1 \pmod 4$

Therefore the degree $2$ field is fixed by conjugation iff $p \equiv 1 \pmod 4$

Therefore it is real iff $p \equiv 1 \pmod 4$

MatheiBoulomenos

yes, this works

thanks

MatheiBoulomenos

2:13 AM

I like your solution, it's very number-theoretic, there's a more field-theoretic approach as well

@MatheiBoulomenos could you give me a hint?

what do you mean by K is a subset of gHg^-1 ?

@KasmirKhaan it means every element of $K$ is also an element of $gHg^{-1}$

hmm i get that

But i thought i understood it in one way

but turns out it was wrong ><

MatheiBoulomenos

@LeakyNun what's the degree of $\Bbb Q(\zeta_p) \cap \Bbb R$ over $\Bbb Q$?

@LeakyNun but you shouldn't worry too much about finding exactly the same solution I was thinking of. You already gave me two valid solutions (quadratic residues and discriminants). I can just tell you what I had mind

2:19 AM

@MatheiBoulomenos alright, tell me then

MatheiBoulomenos

I was thinking like this: complex conjugation generates a subgroup of order 2, thus $\operatorname{Gal}(\Bbb Q (\zeta_p) \cap \Bbb R: \Bbb Q)$ is cyclic of order $\frac{p-1}{2}$ as a quotient of a cyclic group, thus $\operatorname{Gal}(\Bbb Q (\zeta_p) \cap \Bbb R: \Bbb Q)$ contains a subgroup of index 2 iff $2 | \frac{p-1}{2}$ which is of course equivalent to $p \equiv 1 \pmod{4}$

Richard Dawkins roasts people bad lmao

bad as in

the roast is bad

or he does it hard

the latter of course

2:38 AM

@MatheiBoulomenos wow, that's much faster than qr

and neater, sicherlich

@MatheiBoulomenos still here? :)

MatheiBoulomenos

@Kasmir yes

:D

so from what i understood now

We have H is a maximal subgroup

ie a p sylow subgroup

we conjuagte H by g, ie gHg^-1 we get another p -sylow subgroup H'

then H' intersect K ( K subgroup of G , with order divisible by p ) this intersection is also a p sylow subgroup

MatheiBoulomenos

I think you copied the statement of Sylow 2 wrong

oh

MatheiBoulomenos

2:47 AM

it's overly complicated

can you give me a better statement?

MatheiBoulomenos

Version 1: - For any subgroup K whose order is a power of $p$ and every $p$-Sylow subgroup $H$, there is some $g$ such that $K \subset gHg^{-1}$

Version 2: - Any subgroup $K$ whose order is a power of $p$ is contained in some $p$-Sylow subgroup

-And all $p$-Sylow subgroups are conjugate

how does one use that?

i mean what is the use of it?

Akiva Weinberger

http://collegeadmissions.uchicago.edu/apply/essay/past-essay-questions

MatheiBoulomenos

well, proving Sylow 3 from Sylow 2 is not difficult, for one thing

Akiva Weinberger

2:52 AM

> So where is Waldo, really?
–Inspired by Robin Ye, AB'16

Find x.
–Inspired by Benjamin Nuzzo, an admitted student from Eton College, UK

>
Have you ever walked through the aisles of a warehouse store like Costco or Sam's Club and wondered who would buy a jar of mustard a foot and a half tall? ...
–Inspired by Katherine Gold

MatheiBoulomenos

Here's a fun exercise: Show that there is an action of $S_5$ on a set of 6 elements with only one orbit @KasmirKhaan

@MatheiBoulomenos okay working on it :)

@MatheiBoulomenos isn't it too easy lol

Akiva Weinberger

Lemme think about that one as well

Spoiler no spoiling

Hmm I guess I have an answer but it's probably not the intended one

There's a very visual construction.

2:57 AM

so we can do qr with group theory as such: $-1$ generates a subgroup of $G$ with order $2$, so $H = (\Bbb Z/p\Bbb Z)^*/\langle-1\rangle$ is cyclic of order $\dfrac{p-1}2$. Now, $-1$ is a qr iff $-1$ is in $Q$ the subgroup with index $2$. Let $\varphi : G \to \langle -1 \rangle$ be the natural homomorphism. If $-1 \in Q$, then $\varphi[Q] = \langle -1 \rangle$. We find $\varphi|_Q^{-1}[\{1\}]$ is a subgroup of $Q$ of index $2$, which is a subgroup of $G$ of index $4$. @MatheiBoulomenos

Akiva Weinberger

Wait what the hell

SteamyRoot changed his name (and I'm a month late)

hmm is the group really S_5 ?

not S_6 ?

@KasmirKhaan yes

@AkivaWeinberger It was me :3

MatheiBoulomenos

For $S_6$, the exercise would be trivial

2:59 AM

okay idont see yet how the 6th element would be hit by any element of S_5

but let me think

yeah i thought so too =p

Akiva Weinberger

@BalarkaSen ...What?

just taking (123456) would hit all

yet another argument goes like this: if $-1$ is QR, then we can build the (not really exact) sequence $G \overset2\longrightarrow QR \longrightarrow \langle -1 \rangle \overset2\longrightarrow \{1\}$, QED.

@MatheiBoulomenos have you seen my two arguments above?

@MatheiBoulomenos Just to clarify, this can be done using the dodecahedron and some cubic trickery, right?

i sorta want to look up the Cayley diagram for S5

3:00 AM

(I am purposefully not being precise)

@Semiclassical Meh, Cayley graphs of finite groups are boring objects

@BalarkaSen stares

Akiva Weinberger

@BalarkaSen Ah OK yeah that's the answer I came up with

maybe, but they don't look boring: commons.wikimedia.org/wiki/File:CayleyGraphSymmetricGroup5.png#/…

@Semiclassical say what

MatheiBoulomenos

@Balarka I guess you can. I'm not to good at symmetry groups of geometric objects. I take A Non-Geometric Approach$^{TM}$ to group theory

3:02 AM

@AkivaWeinberger Cool, we're bros. BTW this is what I was referring to: chat.stackexchange.com/transcript/message/40118475#40118475

Akiva Weinberger

Algebra: The Actively Non-Geometric Approach

oh. this does all remind me of a problem I saw this week

@LeakyNun It is true. Cayley graphs are only interesting upto quasi-isometry, and finite groups are quasi-isometric to trivial groups.

@MatheiBoulomenos can you give me something that can help ?

Proof: Take big big finite graph, pinch pinch to small small and floop

3:03 AM

posed in non-permutation language, alas, but it's basically a problem about S4:

MatheiBoulomenos

@BalarkaSen I like Caley graphs of free groups as they give an geometric description of the universal cover of a bouquet of circles

Akiva Weinberger

Also they look pretty

when we have trigonometric, algebraic, logarithmic and exponential functions, what's the convention to order them when they're multiplying each other? for example I've always seen $e^x \cos x$ and not $cos x e^x$

You must build a very specific tower out of four differently colored pieces that can be stacked in any order. But when you start building, you don’t know what the correct order is.

Upon assembling the pieces in some order, you can consult an architectural oracle (he goes by Frank) who will inform you if zero, one, two or all four pieces of the tower are in the correct position. Your tower doesn’t count as finished until the oracle confirms your solution is correct. How many times should you have to consult the oracle, in the worst case, to assemble the tower correctly?

Akiva Weinberger

(The usual plane embedding anyway) (Re: CG of FG)

3:04 AM

I have no issue with Cayley graph of $F_2$.

It's a hyperbolic group; the graph is negatively curved tee-hee

@BalarkaSen that was so rigorous my nonexistent cat did a backflip

Vacuously

Akiva Weinberger

@Twink I guess $\cos xe^x$ looks like $\cos(xe^x)$; $\cos(x)e^x$ looks potentially ambiguous; and $(\cos x)e^x$ looks annoying

@Daminark But it is the proof!

MatheiBoulomenos

@Kasmir Hint: I asked you this right after we were talking about Sylow 2, that may or may not be a coincidence

You just flooooooop and you're done

3:05 AM

what about $\ln x \cos x e^x$?

@Twink $e^x \ln x \cos x$

then what's the convention?

@MatheiBoulomenos haha smart! :) okay ill keep working, but please give me solution if you gonna sleep , i wont look at it untill i try hard :)

algebraic, exponential, logarithmic and trigonometric?

@Twink use brackets

Akiva Weinberger

3:06 AM

"Don't write confusing stuff" @Twink

when I don't wanna use brackets

@AkivaWeinberger :P

Akiva Weinberger

"If it looks ugly and you can make it look not ugly do that"

I mean real talk though that was very unrigorous, you don't know if "big finite" numbers even exist

What if there aren't any numbers bigger than 7?

there must be a rule or something

3:06 AM

@Daminark does TREE(3) exist?

@Daminark we must be in $\Bbb F_7$

Akiva Weinberger

Besides for, like, "be consistent", I don't think there's any rule

or convention

Akiva Weinberger

I don't think anyone cares enough

to make a convention

MatheiBoulomenos

@BalarkaSen ah, the rigour of topological arguments! I also like drawing a picture that shows that it's obvious for a sphere or something like that and then say this proves it for any (insert reasonable properties) spaces

13 mins ago, by Leaky Nun

@MatheiBoulomenos isn't it too easy lol

I gladly take it back.

3:08 AM

there is one, that's why we all write $xe^x \cos x$

@LeakyNun TREE(3) is just aleph_7

i don't think there's strict conventions

it is too easy; just take a dodecahedron and 6 cubes and juggle juggle juggle juggle you're done

@LeakyNun

Akiva Weinberger

@LeakyNun Did your idea not work?

e.g. sin(x) cos(x) vs. cos(x) sin(x)

3:09 AM

@AkivaWeinberger I just found an element of order $6$ and wrongly thought I'm done

Akiva Weinberger

Yeah I got what Balarka's got

i'd tend to write the former, but i can't think of a reason why the latter is impermissible

@AkivaWeinberger Let $f : \Bbb R \to C(\Bbb R, \Bbb R)$ be a map, consider the function $f(x)$

@BalarkaSen "function $f(x)$"

<= to be continued thug life music starts

Akiva Weinberger

3:10 AM

@LeakyNun But it is a function, oddly enough

since the codomain is a function space

@AkivaWeinberger why?

oh, lol!

the trickery

C(R, R) is a space of functions

lmao

you got me

I learnt this from Daniel Fischer

Akiva Weinberger

Wait is there more to this @BalarkaSen

3:11 AM

does any real-life example exist though

@LeakyNun Sure

@LeakyNun How can real life exist if the real numbers don't?

@Daminark God

Just take a family of functions with a varying parameter

@BalarkaSen like $a \mapsto x \mapsto \sin(ax)$ lol

3:12 AM

Ya

Akiva Weinberger

Take any function $\Bbb R^2\to\Bbb R$ really

The function $f(a)$

A function $\Bbb R \to C(\Bbb R, \Bbb R)$ is also the same as a function $\Bbb R \times \Bbb R \to \Bbb R$

@AkivaWeinberger combine our ideas and you get currying

Akiva Weinberger

and instead of the variables being $x$ and $y$ call them $x$ and $a$

3:12 AM

Which Akiva beat me to

Akiva Weinberger

@LeakyNun is hungry

lol

In fact, you can look at the space $C(\Bbb R, C(\Bbb R, \Bbb R))$

MatheiBoulomenos

luckily, $\Bbb R$ is compactly generated

<= to be continued

No, I'm kidding, I meant to finish the sentence

Akiva Weinberger

3:13 AM

Wait so that's $\Bbb R^{\Bbb R^{\Bbb R}}$ @BalarkaSen

@MatheiBoulomenos hast du meine arguments gesehen?

Akiva Weinberger

Oh wait no

With a reasonable topology, it's the space $C(\Bbb R^2,\Bbb R)$

Homeomorphic to the space, that is

Akiva Weinberger

That's $(\Bbb R^{\Bbb R})^{\Bbb R}$, I meant

which is $\Bbb R^{\Bbb R\times\Bbb R}$

Yep

Akiva Weinberger

3:14 AM

so stuff makes sense and we get to pat ourselves on the back for coming up with a good notation

And cochains are elements of $\Bbb Z^{X^\Delta}$

(Chains are elements of $\Bbb Z^{X^\Delta}$ with finite support)

Wait hold on I might have it backwards

MatheiBoulomenos

@LeakyNun yes, you're arguments work

Akiva Weinberger

Oh no I don't I'm good

Just write $\hom(C(X), \Bbb Z)$ for heaven's sake.

Akiva Weinberger

$\rm N^o$

$\hom(C(\Delta,X),\Bbb Z)$, anyway

if $C$ means the continuous maps from blah to blah

Well, no $C_*(X)$ is a $\Bbb Z$-module.

Not a space of continuous maps

Akiva Weinberger

3:18 AM

Ohh I see

It's the $\Bbb Z$-module generated by singular $*$-simplices in $X$

Akiva Weinberger

Right right right yeah

Yep

Akiva Weinberger

(btw that last comment is pronounced "riririye" approximately)

could you tell me the solution lol

3:22 AM

juggle juggle juggle

The point is there is a way to inscribe 6 distinct cubes in a dodecahedron.

MatheiBoulomenos

too much geometry for me

And isometries of the dodeca-dawg permutes them

Transitively!

Hm, the only thing might be that A_5 is the (oriented) symmetry group of the dodecahedron

And the full symmetry group is A_5 x Z/2, not S_5

can I just map $(12345)$ to $(12345)$ and $(12)(345)$ to $(123456)$?

They should generate the whole $S_5$ (because the latter cubed is $(12)$

and there should be no collision because their orders are co-prime

I should repeat this beautiful equation once a day: $$Q_8 \ast Q_8 \cong D_8 \ast D_8$$

Is * the free product

@BalarkaSen no, the central product

3:28 AM

"oh"

"interesting"

"vary naice"

Akiva Weinberger

@LeakyNun I don't know if I believe that all nontrivial relations between the first two are true for the last two

erm, I mean

MatheiBoulomenos

@LeakyNun if that argument was valid, you could easily show that $S_3 \cong \Bbb Z /(6)$

Akiva Weinberger

nontrivial relations between x = (12345) and y = (12)(345), and between x = (12345) and y = (123456)

But yeah the set of isometries of the dodecahedron is $S_5$ so just identify pairs of opposite sides and you're done

why is it $S_5$?

@Akiva No I don't think the isometries of the dodecahedron is S_5

It's A_5 x Z/2

Akiva Weinberger

3:31 AM

Wouldn't be a direct product

That's what I am worried about

Akiva Weinberger

I think it is S_5 lemme check

I mean the orientation preserving isometries are A_5

MatheiBoulomenos

According to wikipedia and some other random site I googled, it's $A_5 \times \Bbb Z/(2)$

1 --> Z/2 --> Iso --> Iso^+ --> 1

splits as a direct sum

Akiva Weinberger

3:32 AM

Oh crap you're right

Hmm

Damn

I was so sure it was S_5

Oh duh 'cause with the inscribed-cubes construction, central inversion does nothing

so that wouldn't work

Yeah

Good call

ugh mathematica

Akiva Weinberger

google.com/search?q=cubes+in+a+dodecahedron&tbm=isch @LeakyNun

one of the irritating little holes in its graph functionality is that, while it knows a catalog of them, the catalog it says it knows doesn't always match the catalog it can actually draw from

Akiva Weinberger

The orientation-preserving stuffs on the dodecahedron are A_5 because they permute five cubes you can shove in there

3:34 AM

So we only get a transitive action of A_5 on the six cubists or what?

in particular, it lists the transposition graph on 4 elements among its catalog of known graphs...but it can't actually load that graph if i ask it to

mathematicaaaaaa

Akiva Weinberger

Transitive graph?

Oh, wait, I am confused. You have 5 cubes which inscribe; their edges just intersect each face on a pentagram. Duh.

Akiva Weinberger

Yeah the you use the pentagram to demonically summon A_5

well if you have a transitive action of $A_5$ to $S_6$

since $A_5$ is a normal subgroup of $S_5$

you easily get a transitive action of $S_5$ to $S_6$

3:37 AM

The 6 comes from the opposite faces of the dodeca-dawg

MatheiBoulomenos

@LeakyNun how?

@MatheiBoulomenos good question

lol

Akiva Weinberger

Random thingy

nvm, discard it

3:41 AM

There's a sequence of these exceptional isomorphisms that I am forgetting

PGL2(5) = S_5 = some shit ...

Akiva Weinberger

Right so the point is I no longer know how to shove S_5 onto [6]

er put'em in a blender or someth

imma doze off and not thinkabaoutit

Akiva Weinberger

a little bit of brooklyn advice

hey i was looking at a hot network MO question earlier today and i cant find it anymore

where did you go you little

Akiva Weinberger

fuhgeddaboudit

MatheiBoulomenos

3:44 AM

the exercise is really not that hard

I already gave a hint

Akiva Weinberger

Wait did you I missed it

i refuse to think about it in algebraic terms @Mathei

i refuse

Akiva Weinberger

If it involves silov I don't know it

sylow

you cannot make me think about it

esp i'm very sleepy now, almost on a Tommy Wiseau high

i did not hit her, i did not, i did nawt

MatheiBoulomenos

it involves Sylow

Akiva Weinberger

3:45 AM

You what

oh hey I found the MO question

Akiva Weinberger

Did you just watch The Room then

I need to see that

@AkivaWeinberger What rock do you live under??

Akiva Weinberger

I'm partway through The Disaster Artist (book)

How can you be in the internet and not watch The Room

Akiva Weinberger

3:46 AM

@BalarkaSen Just 'cause I know about it doesn't mean I've seen it

I know the important bits

He didn't hit her, what a funny story Mark, hi doggy

MatheiBoulomenos

41 mins ago, by MatheiBoulomenos

@Kasmir Hint: I asked you this right after we were talking about Sylow 2, that may or may not be a coincidence

Akiva Weinberger

Ah fair

Yeah I don't really know Sylow

@AkivaWeinberger Ok good you're the perfect meme boy then

Akiva Weinberger

I mean I saw "We Are Number One but it's explained by Bill Wurtz" and I kinda feel like that's peak meme already

WANO is a classic

So is Billy

Akiva Weinberger

3:49 AM

I'm sorry, "Billy"? :D

the MLG Bill Wurtz, who else?

Oh, I was watching a video that day about the meme songs through the ages.

Really good video

Here it is, for educational purposes

Akiva Weinberger

@BalarkaSen I know you meant bill wurtz but I found it funny you called him Billy

I am on a very loose mode right now, please excuse my unprofessional phrasings and language

Akiva Weinberger

"Hall of the Mountain King" is very memey

It only gets actually fun after 1860

Akiva Weinberger

3:54 AM

I mean I'm sure but also

like

that's the most memey a classical piece can get

(Technically not "classical" I'm sure but whatever)

MatheiBoulomenos

@Balarka if you truly want it geometric, just let $PGL(2,5)$ act on the set of $1$-dimensional subspaces of $\Bbb F_5^2$ ...

Oh I mean Hall of the Meme King is after 1860

Akiva Weinberger

Oh OK I wasn't paying attention

@Mathei Oh. OH.

@MatheiBoulomenos hmm i want to see if i understood the question right

MatheiBoulomenos

3:55 AM

now you just need to prove that $S_5 \cong PGL(2,5)$

geometrically :P

Akiva Weinberger

PGL = ?

P^1 F_5 have 6 points

Akiva Weinberger

Projective gemema line

we have S_5 action on {1,2,3,4,5,6} and we want to show that this action is transitive right?

Akiva Weinberger

Projective gasomething linear

Projective General Linear?

MatheiBoulomenos

3:56 AM

@Kasmir no, you want to show there exists a transitive action of $S_5$ on a 6 element set

@AkivaWeinberger You should watch that video

@MatheiBoulomenos okay :)

Akiva Weinberger

I'm watching the video

Oh.

How far in are you?

Some late 19's shit?

Akiva Weinberger

And I Will Always Love You / Serbia Stronk

3:58 AM

ahahah serbia stronk is great (thanks to @Sha for introducing me to that)

My dreams are getting super weird ever since I study those foundation of mathematics stuff. Nearly 90% of the content of those dreams I am unable to trace it back to any events that occurred within past 3 days of the dreams

Akiva Weinberger

the fuck is tunak tunak

I mean, a safari in a desert that is in the middle of a country park?

Akiva Weinberger

In the middle of Cleveland

Last time I been to something remotely similar is at least 6 years ago

3:59 AM

tunak tunak is like the best meme that has ever come from the Indian subcontinent

watch the original video

it's damn good

« first day (2636 days earlier) ← previous day next day → last day (2387 days later) »

Transcript for

Oct '1722

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile