Mathematics - 2017-08-29 (page 6 of 6)

Kasmir Khaan

19:00

@LeakyNun oh :/ okay, ill keep working on some exercices for an hour or so, ill send you how I worked things so you can correct them other time i see you here =P

@LeakyNun Thanks alot for your help ! was really usefull :D

@LeakyNun Thanks!:D

Leaky Nun

I'll delete that message.

Kasmir Khaan

saved

usukidoll

Would the limit points be 0 and the square root of 2? @gian

Leaky Nun

[Building the third Fibonacci group]
$ij=k$
$jk=i$
$ki=j$
$i^2=j^2=k^2=ijk$
How am I supposed to prove that $i^4=1$...

gian

@usukidoll, those are the bounds of the interval.

However, think back to the definition of a limit point.

Leaky Nun

19:03

@Secret you might have some idea

gian

It is a point whose open neighborhoods contain at least one other distinct point.

So applying this to your example.

Let's say we could iterate through every point in $[0, \sqrt{2})$.

Then for every point $p$, we consider all open intervals containing $p$.

And, as it turns out, every open interval containing $p$ must also contain another distinct point, say $q$.

Because this is true for all points $p$ in $[0, \sqrt{2})$, EVERY point in the interval $[0, \sqrt{2})$ is a limit point.

And so is $\sqrt{2}$.

Leaky Nun

@AlessandroCodenotti ti chiamo

gian

Keep in mind that a point does not need to be in the interval for it to be a limit point.

Does that make sense @usukidoll?

Leaky Nun

@AkivaWeinberger te llamo

usukidoll

So the limit points are 0 and sqrt 2 because the open interval must also contain another point

gian

19:09

Read above. I said EVERY point in the interval is a limit point.

Semiclassical

$(i^2)i=i(i)i=i(jk)i=(ij)(ki)=kj$

so $i^2$ sends $i=jk$ to $kj$.

Leaky Nun

@Semiclassical what sorcery

usukidoll

Like 0, 1/2, 3/4 ... All the way to the square root of 2 which is 1.41

Semiclassical

using @Secret's equations

gian

Yes, all of them. @usukidoll

But do you understand why?

Semiclassical

19:11

and $(i^2)j=i(ij)=ik$, so $i^2$ sends $j=ki$ to $ik$

usukidoll

They're open neighborhoods that contain at least another point

gian

Right, all of these points' open neighborhoods contain another distinct point in $[0,\sqrt{2})$.

Leaky Nun

$i^5=i^2kj=iikj=jjkj=jij=jk=i$ Oh My God

Semiclassical

what I'd like to then say is that, since $k=ij$, $(i^2)k=ji$

usukidoll

So how do I find the closures?

Like intersections

Leaky Nun

19:14

@Semiclassical and then?

Semiclassical

and that amounts to $j=ik$.

gian

Well there are two ways @usukidoll.

Semiclassical

@LeakyNun eh, can't really be an "and then?" if I can't actually justify the implication

something's weird.

usukidoll

Alright

gian

The first one that you are referring to is the intersection of all closed sets containing $[0,\sqrt{2})$.

Leaky Nun

19:15

$iiiii=ijkijki=kkkj=kijkj=jjkj=jij=jk=i$

$\implies i^4=1$

gian

But this can be a bit difficult to digest.

The second method is a bit more intuitive.

Semiclassical

eh, that assumes $i^{-1}$ is well-defined.

usukidoll

What's the second method?

Leaky Nun

@Semiclassical it is

I'm dealing with group presentations

Semiclassical

Ah.

gian

19:16

The second method is just to take the union of the interval $[0,\sqrt{2})$ with all of its limit points.

Semiclassical

then yeah, that's fine

Leaky Nun

basically $F(i,j,k)/\langle ijk^{-1},jki^{-1},kij^{-1}\rangle$

Semiclassical

ahh.

usukidoll

Like

$[0,\sqrt{2})$ U all the limit points

gian

So what did we say were the limit points of $[0,\sqrt{2})$?

Leaky Nun

19:17

@Semiclassical it's fun. Its generalization is called the Fibonacci groups.

Tobias Kildetoft

@LeakyNun Well, the normal subgroup generated by those elements

gian

Yes, exactly.

usukidoll

One of them was 0

Semiclassical

so proofs like $ijk=i(i)=(k)k$

Leaky Nun

@TobiasKildetoft hmm?

usukidoll

19:17

Other one is sqrt{2}

gian

@usukidoll, we said that all of the points in $[0,\sqrt{2})$ were limit points.

Tobias Kildetoft

@LeakyNun I don't think those elements generate a normal subgroup on their own

usukidoll

Oh right ahhh

All of the points 1/2, 3/4 1 those guys

gian

So all of the points in $[0,\sqrt{2})$ are limits points, and so is $\sqrt{2}$.

Leaky Nun

@TobiasKildetoft you're right, I need to take the normal closure.

Akiva Weinberger

19:19

@LeakyNun Mau?

Leaky Nun

@AkivaWeinberger el problema esta resoluto ahora :P

gian

Therefore, we're essentially just tacking on the $\sqrt{2}$.

Akiva Weinberger

Nice

gian

So the closure of $[0,\sqrt{2})$ is $[0,\sqrt{2}]$.

Kasmir Khaan

@LeakyNun It was easy ! xy=e iff yx=e
yx=e =zx , by right cancellation law we get y=z

:D

Semiclassical

19:19

ugh, dangit. i think i broke my mathematica

Leaky Nun

@KasmirKhaan that's overkill, but it's a nice idea to build proofs based on previously-proved theorems

Semiclassical

or at the very least gave it a headache

Kasmir Khaan

overkill as in i did more than needed?

usukidoll

Ohhh okkk

Kasmir Khaan

@LeakyNun Yes I thought I use what I just proved =p

Leaky Nun

19:21

@KasmirKhaan no, overkill means you used a nuke to kill a fly

Kasmir Khaan

haha :D

gian

@usukidoll, does that seem intuitive enough?

Kasmir Khaan

Ehm I ll find another way to prove it =p

Leaky Nun

So it's $F_{\{i,j,k\}}/\langle ijk^{-1},jki^{-1},kij^{-1}\rangle^G$, right? @TobiasKildetoft

Tobias Kildetoft

@LeakyNun Right, where $G$ is that free group

Leaky Nun

19:22

@TobiasKildetoft right

Dattier

Hi, $$f\in C^3(\mathbb R) \text{ with }f' \times f'''<0 \\
\text{Is it true that :} \forall a,b\in \mathbb R^2, |f(a)-f(b)| \leq |f'(\frac{a+b}{2})|\times |a-b| $$

Tobias Kildetoft

But much easier to write it as a presentation like you did earlier

gian

@Dattier, did you already ask this on the main site?

Leaky Nun

$\langle i,j,k | ijk^{-1}, jki^{-1}, kij^{-1} \rangle$ @TobiasKildetoft

Tobias Kildetoft

yep

usukidoll

19:23

So the closure is just that interval because all of the limit points are inside the interval like 1/2, 3/4, 1 up to 1.41

Leaky Nun

TIL Tits group (named after Jacques Tits)

usukidoll

:0 wut language lol

Dattier

@gian I can't, and it's an enigma (I know a solution)

Semiclassical

"The Tits group is thin and finite."

(I am easily amused)

gian

@usukidoll, the closure essentially "fills" in the boundary. At least that's intuitive in the context of real numbers.

Krijn

19:24

There's also the Tits theorem

And of course the Cox-Zucker Machine, but that was done on purpose

The Tits theorem is quite interesting

usukidoll

Alrighty.

Leaky Nun

(Trivial) challenge: identify $\langle a_1, a_2, a_3, a_4 | a_i^2, [a_i,a_j], (a_i a_{i+1})^3\rangle$ where $|i-j| \ne 1$

gian

@usukidoll, are you an undergraduate student?

Semiclassical

$[a_i,a_j]$ = commutator of $a_i,a_j$?

Tobias Kildetoft

@LeakyNun that is a neat one

usukidoll

19:27

Was... Got the ba in math...postbach Studies

Leaky Nun

@Semiclassical yes

Tobias Kildetoft

@LeakyNun Someone recently asked on main about the version with the commutator replaced by that last relation also for the rest of the indices

Semiclassical

hmm

Tobias Kildetoft

(I won't name the last relation to not spoil the exercise)

usukidoll

Anyway, If I were to find the intersection do I take the intersection of all the limit points to the interval $[0, sqrt{2})$?

Leaky Nun

19:28

@TobiasKildetoft :O

Kasmir Khaan

@LeakyNun I think you mean something like , xy=e means that y=x' , zx=e means that z=x' , since the inverse is unique z=y

Leaky Nun

@usukidoll \sqrt2

usukidoll

Oops

Leaky Nun

@KasmirKhaan no, I don't mean that. That's a theorem you need to prove.

usukidoll

Sorry on a phone

Kasmir Khaan

19:29

@LeakyNun hmm but I meant more like that approach =p

Leaky Nun

@usukidoll no, the closure of [0,r2) is not [0,r2).

gian

@usukidoll, the alternative definition is to take the intersection of all closed sets containing the interval $[0,\sqrt{2})$.

Kasmir Khaan

@LeakyNun I could not find easiar way

Leaky Nun

@KasmirKhaan and I mean not that approach.

Kasmir Khaan

@LeakyNun damn it >< Ill keep trying =p

Leaky Nun

19:30

@KasmirKhaan you have xy=e and zx=e.

gian

A good exercise would be to prove that the union of a set with its limit points is the same as the intersection of all closed sets containing it.

Leaky Nun

Fiddle with these two equations

Kasmir Khaan

okay !

usukidoll

How do I find the closed sets?

Leaky Nun

@usukidoll how do you define closed set?

gian

19:32

@usukidoll, think about a closed set in terms of limit points.

Semiclassical

@LeakyNun oh. is that supposed to include $a_1 a_4=a_4a_1$, or it taken as cyclic?

Leaky Nun

@Semiclassical it isn't.

Semiclassical

well, $|4-1|=3\neq 1$, so the definition you wrote doesn't quite match that

Leaky Nun

(I'll skip the $a$'s) $11,22,33,44,131^{-1}3^{-1},242^{-1}4^{-1},121212,232323,343434$.

usukidoll

Closed set - a subset A of a topological space X is said to be closed if X-A is open

Semiclassical

19:34

right.

maybe $[a_i,a_j]$ when $|i-j|=2$ is simplest.

Leaky Nun

@Semiclassical you're right.

Tobias Kildetoft

@LeakyNun What? I thought you did mean to include $[a_1,a_4]$

Semiclassical

hah, controversy

Leaky Nun

oops!

usukidoll

But then that would mean that the limit points in that interval would be closed? @gian

Tobias Kildetoft

19:35

Otherwise the answer is probably not what you think

Leaky Nun

$11,22,33,44,131^{-1}3^{-1},242^{-1}4^{-1},141^{-1}4^{-1},121212,232323,343434$

Semiclassical

Heh, so now there's two questions

gian

@usukidoll, a subset of a topological space is closed if and only if it contains all of its limit points.

Leaky Nun

@Semiclassical ignore the one without $[1,4]$

Semiclassical

i'm guessing 414141 isn't in this definition either

gian

19:36

Consider your example :$[0,\sqrt{2})$.

Leaky Nun

@Semiclassical it isn't

414141=444111=41

Semiclassical

right.

gian

This interval is not closed because it does not contain all of its limit points, namely, it does not contain $\sqrt{2}$.

Leaky Nun

@TobiasKildetoft are you acquainted with Fibonacci groups?

Tobias Kildetoft

@LeakyNun Well, if you replace $[a_1,a_4]$ by $(a_1a_4)^3$ you also get a "nice" group

Leaky Nun

19:37

@TobiasKildetoft hmm, that's a problem for me!

Tobias Kildetoft

No, not really. I have heard the name, but nothing more than that

Leaky Nun

@TobiasKildetoft $F(2,4) = \langle a,b,c,d | abc', bcd', cda', dab' \rangle$

Tobias Kildetoft

Ahh, makes sense

Leaky Nun

$F(2,5) = \langle a,b,c,d,e | abc', bcd', cde', dea', eab' \rangle$

Tobias Kildetoft

(much easier to see the connection to Fibonacci when written as equalities rather than as elements)

Leaky Nun

19:38

A good exercise is to find $F(2,m)$ for $m=1,2,3,4,5,7$, and prove that it is infinite for other positive integer $m$

@TobiasKildetoft it's shorter :P

Semiclassical

I imagine one could generalize this in the following way.

Tobias Kildetoft

@LeakyNun Note that when I wrote "nice" group above, I did not mean a finite one

Leaky Nun

@TobiasKildetoft :(

Semiclassical

Let $G$ be some graph on $n$ vertices

Leaky Nun

Semi is getting somewhere.

Semiclassical

19:40

1) Take $a_i^2=e$ for each element $i=1,2,\cdots n$

gian

Is anyone here acquainted with persistent homology?

usukidoll

Oh the interval is open because it doesn't contain the square root of 2

Semiclassical

2) If $i,j$ are adjacent, then $(a_ia_j)^3=e$

gian

Well the interval is half-open. @usukidoll

Semiclassical

3) If $i,j$ are not adjacent, then $a_i a_j=a_j a_i$.

gian

19:41

But it's definitely not closed.

Tobias Kildetoft

@Semiclassical That is a good idea. Now put numbers on the edges

Semiclassical

hmm!

I don't know where this is going but it's neat.

Tobias Kildetoft

@Semiclassical And you get what are known as the Coxeter graph of a Coxeter group

Semiclassical

oh damn

and now I get flashbacks of the ADE classification

usukidoll

Oh ok so how do we find the closures of a half open interval?

Leaky Nun

19:42

@TobiasKildetoft if this is the final answer then we have different answers... or I'm wrong

Tobias Kildetoft

Coxeter groups are those with a presentation where all generators are of order $2$ and all relations are by giving the order of a product of two elements

Semiclassical

Though for that I guess I have in mind Dynkin diagrams rather than Coxeter graphs

Tobias Kildetoft

@LeakyNun final answer for what?

gian

@usukidoll, the closure for any interval is the same.

Leaky Nun

I know next to nothing about graph theory

@TobiasKildetoft the group in question?

Tobias Kildetoft

19:42

@Semiclassical Yeah, the Dynkin diagrams are related strongly to certain Coxeter graphs

@LeakyNun Which one of them?

gian

It does not matter whether its open or half-open. The closure is computed in the same way.

Leaky Nun

7 mins ago, by Leaky Nun

$11,22,33,44,131^{-1}3^{-1},242^{-1}4^{-1},141^{-1}4^{-1},121212,232323,343434$

Semiclassical

Right. But the Coxeter graphs are more generic?

Tobias Kildetoft

@Semiclassical Basically, since the Weyl group of a Dynkin diagram does not depend on which of the roots is bigger, types B and C become one

Semiclassical

huh.

usukidoll

19:43

How to find? Is it taking the intersection of limit points and the entire interval?

Semiclassical

I actually know some of that, or did...about 8 years ago

gian

Union @usukidoll.

Semiclassical

It was for a physics REU where I (somehow) got slotted into a string theory research problem

Tobias Kildetoft

and then the order of a product is $2$ more than the number of edges between them (or $2$ more than the number we put on the edge)

usukidoll

Limit points U interval

Semiclassical

19:44

which was honestly pretty silly in retrospect

Krijn

There was this uproar about that paper Hironaka wrote on resolution of singularities in char $p$

Tobias Kildetoft

@Semiclassical There is a full classification of which Coxeter graphs lead to finite Coxeter groups, and among these we have the Dynkin diagrams

Semiclassical

right

Krijn

About three months ago or so

Has there been any news?

Balarka Sen

@Krijn Is that paper correct?

Semiclassical

19:45

x-x-x-x-x is the An series

Krijn

@BalarkaSen That's what I'm trying to find out

Balarka Sen

oh, so you don't know

Tobias Kildetoft

@LeakyNun Well, I haven't given a full description of that group here, so there is not enough data for us to disagree yet :)

Balarka Sen

I was actually wondering about that

Semiclassical

x
|
x-x
|
x
|
x

gian

19:45

Yes @usukidoll.

Semiclassical

is another kind

Leaky Nun

@TobiasKildetoft agreed.

gian

I must go now @usukidoll, good luck.

Tobias Kildetoft

@Semiclassical Those are type D

Semiclassical

kk

I honestly don't remember it well

Leaky Nun

19:46

@MikeMiller hi

Tobias Kildetoft

B and C are the ones with a double edge at one end

Leaky Nun

we are talking about group presentations

usukidoll

Thank you for everything @gian :)

Leaky Nun

when it suddenly became graph theory

Tobias Kildetoft

@LeakyNun Well, you were the one to bring in a Coxeter group (even if you did so unknowingly)

Krijn

19:47

@BalarkaSen You'd think someone somewhere must have written something about it, right?

Semiclassical

Well, the connection to graph theory is pretty natural if you note that 1234 has 12 adjacent, 23 adjacent, and 34 adjacent. that's just the path graph on 1234

Krijn

Can't find anything dated later than 3 months ago or so

Tobias Kildetoft

@Semiclassical So the upshot of all of this is that the original group mentioned is the Coxeter group in type $A_4$, which is in fact a Weyl group

Balarka Sen

@Krijn Weird, huh

Semiclassical

if you add an edge from 1 to 4, you get the other one

hmmm

Tobias Kildetoft

19:48

@Semiclassical Right, which is called the affine Weyl group of type $\widetilde{A}_4$

Krijn

@BalarkaSen Very strange

Leaky Nun

@TobiasKildetoft how do you prove that a group presentation is infinite

Tobias Kildetoft

@LeakyNun Depends a lot on the presentation

Leaky Nun

@TobiasKildetoft Take $F(2,6)$ as an example.

Semiclassical

To any Coxeter group, we can associate a Coxeter graph (which is in general a weighted graph).

Tobias Kildetoft

19:50

I only know methods that work for Coxeter groups, and this is because these have some very nice extra properties

Semiclassical

Does the other direcition hold as well?

Tobias Kildetoft

@Semiclassical Yes, both directions hold. Most of the groups are just infinite

Semiclassical

Right.

Tobias Kildetoft

@Semiclassical Also, most of the finite ones really are just the Weyl groups. There are only $2$ extra exceptions, plus an infinite family of very well-known ones.

Semiclassical

@TobiasKildetoft and to A4 we can associate the Lie group SU(5)

though I forget exactly how/why

Tobias Kildetoft

19:54

@Semiclassical I prefer $SL_5$

Semiclassical

kk

Tobias Kildetoft

We can also associate the quantum group $U_q(\mathfrak{sl}_5)$

Semiclassical

I mostly bring that up because for quite a while there were hopes in particle physics that the various forces could all be unified under an SU(5) symmetry

Tobias Kildetoft

Basically, the association is by recreating the Lie algebra which has that Dynkin diagram associated to its root system. The Lie group is then the one for the Lie algebra

Semiclassical

alas, the failure to observe proton decay killed that idea

Tobias Kildetoft

19:56

(there is an explicit way to get generators and relations for the Lie algebra - or the quantum group - from the Dynkin diagram)

Semiclassical

yeah.

in the context of the research project, my vague recollection is

Tobias Kildetoft

or rather, for the quantum group it really goes the other way and one defines the quantum group using such relations in the first place

Semiclassical

we had a way to systematically search through a parameter range of string theory models

and each output would be a whole bunch of roots

by looking at which roots were connected (in the Dynkin diagram sense) to others, we could associate a product of Lie groups to that

e.g. SU(2) x SU(3) x SU(10)

so Dynkin diagrams were essential to implementing that classification as an algorithm

(my main recollection from it, tbh, is that the old code was written in Fortran and he wanted us to remake it in C)

(which led to me figuring out how to replace a 14-deep nested loop with one line of code. was pretty proud of that, lol)

Leaky Nun

I'll go now. @Semiclassical ping me if you want to continue and if you come up with anything.

Will Hunting

20:23

I wanted to post an answer about the x^2+x+1=0 has no real roots question then realised that so many answers have been given that mine is just a small variation of a few of them, so deleted it, lol.

Topologicalife

Hi.

Regarding to this question math.stackexchange.com/questions/128082/…: how did he get the equation $(a \cos \phi + b \sin \phi)(u_x \cos \phi + u_y \sin \phi) = -u$?

Topologicalife

20:40

Got it.

Will Hunting

LOL

I have heard of clopen for open and closed sets and circline for circle or line. Are there any more such terms?

Topologicalife

Discontinuous: a continuous and not continuous function.

(?)

Will Hunting

What? LOL

Topologicalife

:D

Will Hunting

Now you just want to confuse me!

Will Hunting

21:09

I just had a muffin, LOL.

Mitch

21:46

@WillHunting I've heard 'line' for either a row or a column (not mattering which). That's not exactly what you're thinking of though.

Will Hunting

@Mitch WTF LOL

Mitch

'clopen' is for closed and open at the same time, which is allowed given their technical definitions

Akiva Weinberger

22:00

This is a great puzzle video. It presents three puzzles with really clever solutions. (It starts with a bit of historical background)

Simple Math Problems To Fool The Best

►

Mitch

@AkivaWeinberger nice

Balarka Sen

22:52

@Akiva Problem 2 is super easy with a little bit of calculus; it's going to be half the height (distance between the two parallels).

Akiva Weinberger

Yup

Balarka Sen

But goddamn those Russians they gave these stuff

Akiva Weinberger

And once you have that it's easy to rederive it without the calc

Will Hunting

I am going to sleep now. You all continue to misbehave.

Akiva Weinberger

throws chair

Balarka Sen

22:55

I haven't figured out 1 yet but I'll think about it a little. Let's look at 3.

Oh hell what new kind of cancer is problem 3

Hm

Daminark

*brave new kind

Leaky Nun

create a nowhere continuous function

Balarka Sen

$f(x) = \text{Wait, but why?} \, \forall x \in \Bbb R$

Leaky Nun

also, let $G = F_{\{a,b\}}$. Express $a^2b^2a^{-2}b^{-2}$ in the form $ghg^{-1}$ where $g \in G$ and $h \in \langle [a,b]\rangle$.

@BalarkaSen do we need the rational indicator?

Balarka Sen

That is an example, sure.

Leaky Nun

23:06

btw I've been thinking about the problem for as long as I left this room

I couldn't sleep.

and I still haven't slept.

It's 7am.

Leaky Nun

23:18

@BalarkaSen any idea?

Balarka Sen

On what?

Leaky Nun

my problem about group

Balarka Sen

I am not thinking about it.

Leaky Nun

ok

@BalarkaSen solve pi_1(X) = Z5?

what the hell would the space look like

Balarka Sen

google lens spaces

Leaky Nun

23:21

ok thanks

Balarka Sen

bit more easier but messy looking are dunce caps

I drew a straightline on my notebook and it became an arc of a circle

#StellarGeometrySkills

skullpatrol

@Mitch?

Welcome, pal.

Leaky Nun

23:36

@BalarkaSen any visualization skills?

Balarka Sen

23:47

@AkivaWeinberger I figured out 1 by the way. I forgot that fact about trapeziums.

Can't do 3. I tried a bit of trigonometry to no luck :P

Akiva Weinberger

@BalarkaSen (Re: #1) It's easier when you realize it's all invariant under affine transformations

so you can assume the triangle is equilateral

Balarka Sen

Oh that's a cool way to think about it

Akiva Weinberger

The solution to #3 is honestly brilliant

I didn't get it, but it doesn't involve any trig

Balarka Sen

I was thinking about fiddling with the law of cosines.

or sines I guess lol

Ok, yes, solution to 3 really is a brave new kind of cancer

Ted Shifrin

Good grief, @Balarka — you've derailed your sleep schedule yet again.

Balarka Sen

23:56

Truly.

Transcript for

$Mathematics$ Mathematics