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

I'll delete that message.

saved

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

[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 $i^4=1$...

@usukidoll, those are the bounds of the interval.

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

7:03 PM

@Secret you might have some idea

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 $\sqrt{2}$.

@AlessandroCodenotti ti chiamo

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?

@AkivaWeinberger te llamo

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

7:09 PM

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

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

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

@Semiclassical what sorcery

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

using @Secret's equations

Yes, all of them. @usukidoll

But do you understand why?

7:11 PM

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

They're open neighborhoods that contain at least another point

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

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

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

So how do I find the closures?

Like intersections

7:14 PM

@Semiclassical and then?

and that amounts to $j=ik$.

Well there are two ways @usukidoll.

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

something's weird.

Alright

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

7:15 PM

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

$\implies i^4=1$

But this can be a bit difficult to digest.

The second method is a bit more intuitive.

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

What's the second method?

@Semiclassical it is

I'm dealing with group presentations

Ah.

7:16 PM

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

then yeah, that's fine

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

ahh.

Like

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

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

7:17 PM

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

@LeakyNun Well, the normal subgroup generated by those elements

Yes, exactly.

One of them was 0

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

@TobiasKildetoft hmm?

7:17 PM

Other one is sqrt{2}

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

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

Oh right ahhh

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

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

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

7:19 PM

@LeakyNun Mau?

@AkivaWeinberger el problema esta resoluto ahora :P

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

Nice

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

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

:D

7:19 PM

ugh, dangit. i think i broke my mathematica

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

or at the very least gave it a headache

overkill as in i did more than needed?

Ohhh okkk

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

7:21 PM

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

haha :D

@usukidoll, does that seem intuitive enough?

Ehm I ll find another way to prove it =p

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

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

7:22 PM

@TobiasKildetoft right

Dattier

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

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

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

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

yep

7:23 PM

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

TIL Tits group (named after Jacques Tits)

:0 wut language lol

Dattier

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

"The Tits group is thin and finite."

(I am easily amused)

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

7:24 PM

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

Alrighty.

(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 $|i-j| \ne 1$

@usukidoll, are you an undergraduate student?

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

@LeakyNun that is a neat one

7:27 PM

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

@Semiclassical yes

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

hmm

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

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

7:28 PM

@TobiasKildetoft :O

@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

@usukidoll \sqrt2

Oops

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

Sorry on a phone

7:29 PM

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

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

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

@LeakyNun I could not find easiar way

@KasmirKhaan and I mean not that approach.

@LeakyNun damn it >< Ill keep trying =p

7:30 PM

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

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.

Fiddle with these two equations

okay !

How do I find the closed sets?

@usukidoll how do you define closed set?

7:32 PM

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

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

@Semiclassical it isn't.

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

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

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

7:34 PM

right.

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

@Semiclassical you're right.

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

hah, controversy

oops!

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

7:35 PM

Otherwise the answer is probably not what you think

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

Heh, so now there's two questions

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

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

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

7:36 PM

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

@Semiclassical it isn't

414141=444111=41

right.

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

@TobiasKildetoft are you acquainted with Fibonacci groups?

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

7:37 PM

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

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

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

Ahh, makes sense

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

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

7:38 PM

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

I imagine one could generalize this in the following way.

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

@TobiasKildetoft :(

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

Semi is getting somewhere.

7:40 PM

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

Is anyone here acquainted with persistent homology?

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

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

Well the interval is half-open. @usukidoll

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

7:41 PM

But it's definitely not closed.

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

hmm!

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

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

oh damn

and now I get flashbacks of the ADE classification

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

7:42 PM

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

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

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

@LeakyNun final answer for what?

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

I know next to nothing about graph theory

@TobiasKildetoft the group in question?

7:42 PM

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

@LeakyNun Which one of them?

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

7 mins ago, by Leaky Nun

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

Right. But the Coxeter graphs are more generic?

@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

huh.

7:43 PM

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

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

Union @usukidoll.

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

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)

Limit points U interval

7:44 PM

which was honestly pretty silly in retrospect

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

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

right

About three months ago or so

Has there been any news?

@Krijn Is that paper correct?

7:45 PM

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

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

oh, so you don't know

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

I was actually wondering about that

x
|
x-x
|
x
|
x

7:45 PM

Yes @usukidoll.

is another kind

@TobiasKildetoft agreed.

I must go now @usukidoll, good luck.

@Semiclassical Those are type D

kk

I honestly don't remember it well

7:46 PM

@MikeMiller hi

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

we are talking about group presentations

Thank you for everything @gian :)

when it suddenly became graph theory

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

7:47 PM

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

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

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

@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

@Krijn Weird, huh

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

hmmm

7:48 PM

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

@BalarkaSen Very strange

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

@LeakyNun Depends a lot on the presentation

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

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

7:50 PM

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

Does the other direcition hold as well?

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

Right.

@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.

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

though I forget exactly how/why

7:54 PM

@Semiclassical I prefer $SL_5$

kk

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

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

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

alas, the failure to observe proton decay killed that idea

7:56 PM

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

yeah.

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

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

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)

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

8:23 PM

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.

Hi.

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

8:40 PM

Got it.

LOL

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

Discontinuous: a continuous and not continuous function.

(?)

What? LOL

:D

Now you just want to confuse me!

9:09 PM

I just had a muffin, LOL.

Mitch

9:46 PM

@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.

@Mitch WTF LOL

Mitch

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

Simple Math Problems To Fool The Best

10:00 PM

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

►

Mitch

@AkivaWeinberger nice

10:52 PM

@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).

Yup

But goddamn those Russians they gave these stuff

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

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

throws chair

10:55 PM

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

create a nowhere continuous function

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

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?

That is an example, sure.

11:06 PM

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.

11:18 PM

@BalarkaSen any idea?

On what?

my problem about group

I am not thinking about it.

ok

@BalarkaSen solve pi_1(X) = Z5?

what the hell would the space look like

google lens spaces

11:21 PM

ok thanks

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.

11:36 PM

@BalarkaSen any visualization skills?

11:47 PM

@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

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

so you can assume the triangle is equilateral

Oh that's a cool way to think about it

The solution to #3 is honestly brilliant

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

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.