The h Bar

Geometric group theory is an area in mathematics devoted to the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act (that is, when the groups in question are realized as geometric symmetries or continuous transformations of some spaces). Another important idea in geometric group theory is to consider finitely generated groups themselves as geometric objects. This is usually done by studying the Cayley graphs of groups, which, in addition to the graph structure, are...

0celo7

Yes, doesn't make it geometry

Danu

You not agreeing doesn't make it less geometric.

(maybe even makes it more so... ;))

0celo7

6:08 PM

you have forgotten something, @Danu

Nov 22 '15 at 21:29, by 0celo7

@Ϻ.Λ.Ʀ. Theorem 1. I am never wrong. Proof. Follows directly from Theorem 1.

Danu

I have not forgotten it, rather I ignored it.

(never saw it)

ACuriousMind

Theorem: The less geometric @0celo7 thinks something is, the more it is. Proof. Obvious for anyone capable of elementary analysis.

0celo7

SAVAGE

jesus christ man

Currently trying to prove the PhD level theorem $a_n\to a\Rightarrow |a_n|\to |a|$

Danu

lol

High school level!

0celo7

what

you people and your fucking PhD level high schools

ACuriousMind

6:10 PM

I think I did that in middle school

Danu

You and your plebs level universities

HDE 226868

My ethics teacher spent 30 minutes the other day ripping on the American education system and praising the German one, especially the vo-tech side.

0celo7

...I don't even know if you're kidding

ACuriousMind

@0celo7 Good

Danu

I'm so excited to start working on my thesis once my exams end!

0celo7

6:11 PM

@ACuriousMind you're mean

Danu

Had the first one on Gauge-Gravity Duality yesterday.

4 more to go

user54412

I remember having to explain to one of my newly-hired European profs that yes, most students at Ivy League schools have never taken a physics course in their lives, and don't know even trigonometry.

Danu

and that they bought their way into the degree? ;D

user54412

Or life-experienced their way :p

Danu

Actually, I'm curious to know what % of people actually get in by "cheap tricks" like heritage, etc.

ACuriousMind

6:13 PM

@Danu Nice to see your enthusiasm is back :)

0celo7

I can't wait to finish this homework

so I can do more homework

user54412

@Danu Less than the general perception. But the university openly admits that intelligence and academic merit are hardly even considerations.

Danu

@ACuriousMind Yupyup!!!

@ChrisWhite Really?

I'd guess that mostly holds for undergraduate

user54412

There was a civil rights investigation recently that concluded that Princeton admits undergrads primarily based on race, and secondarily on geography, and that this is just fine because it's the only way they can achieve the goal of being racist.

2

Danu

@ChrisWhite wut

"the goal of being racist"?

ACuriousMind

6:16 PM

^I second that wut

user54412

www2.ed.gov/about/offices/list/ocr/docs/investigations/…

0celo7

(btw that PhD level theorem was trivial)

(don't think I'm that dumb pls)

user54412

> Here, OCR examined the University’s assessment of whether race-neutral alternatives were sufficient to achieve its diversity goals, of which race was a single though important element. [...] in OCR’s view, there were no race-neutral alternatives that would have worked about as well.

Danu

Sigh...

Btw @ACuriousMind about your earlier ping...

I think it improves the title.

Also the rest of the post.

ACuriousMind

@Danu Both versions are perfectly grammatical. All changes are purely stylistic, and it introduces a spacing error. I don't think we should approve such edits.

Danu

6:23 PM

I didn't catch the spacing error, obviously.

I significantly prefer the wording of the second version.

I also don't think that posts with which nothing is really wrong should not be edited; maybe that's where we start disagreeing.

ACuriousMind

Well, I think that formatting into a better format is fine (such as taking advantage of the text editor's inherent list capabilities), even if you don't correct any errors. Mere rewording, however, seems to me to be too subjective to be a useful edit. Some people prefer that way of wording it, others prefer another. I'll not roll back such an edit, but I'd also not approve it.

0celo7

non-negative or nonnegative

ACuriousMind

anti-negative, obviously

0celo7

I've never heard that before, thanks

Danu

@0celo7 Use the hyphen, IMO.

user54412

6:29 PM

co-operate -> coöperate -> cooperate

0celo7

I hate when people write co-ordinate system

user54412

non-negative -> nonñegative -> nonnegative?

0celo7

@ChrisWhite you drunk

Danu

@0celo7 I also don't like that.

0celo7

@Danu Dude, don't agree with me

People might get the wrong impression

Danu

6:31 PM

It reflects badly on me, I know.

0celo7

I'm just looking out for you

ACuriousMind

@ChrisWhite What's the tilde on the n?

0celo7

@ACuriousMind Some Spanish thing?

ACuriousMind

I know the trema for vowels, but I've never seen that

0celo7

like piñata

Danu

6:33 PM

@ACuriousMind It's not a thing, in English.

user54412

i wanted an umlaut to keep the pattern but that was harder to come by

0celo7

@Danu Well we have it in our Leiwörter from Spanish.

is there an h in that, btw

I can't into German spelling of words I never use

ACuriousMind

It's Lehnwort, not Leihwort.

0celo7

well then

you got the meaning

is leih even a german word

ACuriousMind

@ChrisWhite I see, I thought it was some fancy notation I'd never seen before :/

@0celo7 Yes, leihen means lend/borrow

0celo7

6:36 PM

@ACuriousMind Dude it's used in Spanish

ACuriousMind

@0celo7 Not with this meaning, dude

0celo7

wtf is happening

what meaning

ACuriousMind

Chris wanted to denote that the second n is spoken distinctly from the first, not that it is the Spanish "nj" sound

0celo7

what

how is anyone supposed to figure that out

ACuriousMind

Well..I did :P

0celo7

6:38 PM

whatever

stupid and forgetful, jesus

user54412

Anyway, the point is we are somewhere between non-negative and nonnegative, so we should invent an intermediate word in the evolution

0celo7

what is up with my analysis prof randomly putting periods on the stuff written on the board

he'll write (ii) and put a period, talk a bit

then write something

put another period

I've even seen him write (Sentence.) and put a period outside of the parenthesis

my ODE prof did that too.

yuggib

@ChrisWhite anegative?

0celo7

@yuggib Can the sum of a convergent and divergence sequence converge?

NeuroFuzzy

@ChrisWhite If you co-operate I'll co-ordinate the e-mail of a list of non-negative numbers for you.

yuggib

6:53 PM

@0celo7 if you rearrange the divergent sequence maybe

0celo7

No rearranging

I don't think it can converge, but I can't come up with a proof.

@ACuriousMind Is that for all $(x,y)$ or just $(0,y)$? I'm confused by the order in which the limits are taken, sorry.

GBeau

@FenderLesPaul any news?

FenderLesPaul

@GBeau no sir

GBeau

@FenderLesPaul Let me know....

I'm still in disbelief

I thought I was only going to get into my fallback

I'll probably end up only getting into Stony Brook... :P

FenderLesPaul

hopefully chicago and ucsb send out admissions next week

so I can at least know my fate for sure

yuggib

6:59 PM

@0celo7 I am not sure it is true

but probably is

Danu

@GBeau Applications didn't pan out as hoped?

@yuggib I think it can maybe converge, but I'm not sure.

I thought about some harmonic series shenanigans but it's not so easy

yuggib

@Danu if there's rearranging, it's possible

GBeau

@Danu I was accepted to Stony Brook!

yuggib

if not, I am not sure

0celo7

We haven't done series yet.

Danu

7:00 PM

@yuggib Sure but that's easy.

0celo7

So I doubt you need one of those.

GBeau

@Danu I bombed the PGRE....

0celo7

@Danu Also not allowed.

yuggib

you sum a convergent series and what?

ACuriousMind

@0celo7 Just 0,y

yuggib

7:01 PM

a divergent series or sequence

?

0celo7

sequences $(x_n)$ and $(y_n)$, where $(x_n)$ converges, $(y_n)$ diverges, and $(x_n+y_n)$ converges;

Either prove this cannot be or produce an example.

yuggib

sequences

not series

0celo7

Did I say series?

yuggib

I read series sorry

FenderLesPaul

if I have a 2-form $\beta_{ab}$ and an integral $\int_{\Sigma}d\beta$ then by Stokes' theorem this is $\int_{\partial \Sigma}\beta$ and I have to pullback $\beta$ to the boundary $\partial \Sigma$ correct? So if $\beta_{ab} = v \wedge n$ where $n$ is normal to $\partial \Sigma$ then I can conclude $\int_{\Sigma} d\beta_{ab} = 0$?

0celo7

7:03 PM

@FenderLesPaul It's the pullback by the inclusion.

FenderLesPaul

yeah but isn't $i^{*}n = 0$?

ACuriousMind

@FenderLesPaul Yes. You can conclude what you want.

FenderLesPaul

@ACuriousMind sweet

thanks!

0celo7

why

yuggib

@0celo7 The proof should probably be the triangle inequality with the difference

0celo7

7:08 PM

you get something like $i^*n(v)=g(n,di(v))$

ACuriousMind

In 3D this is just the statement that a vector field everywhere tangent to a (closed) surface has zero integral over that surface

0celo7

for $v$ a vector on $\Sigma$

ACuriousMind

@0celo7 The pullback commutes with the wedge, i.e. $f^\ast(v\wedge w) = f^\ast v\wedge f^\ast w$.

0celo7

@ACuriousMind I know

I don't see how $i^*n=0$

So I'm trying to compute it

ACuriousMind

@0celo7 Because the image of $\mathrm{d}i$ are the vectors tangent to $\Sigma$, and $n$ is normal to it, so $g(n,\mathrm{d}i(v)) = 0$ regardless of $v$.

0celo7

7:10 PM

Right, that's what I just calculated. But why are those the tangent vectors to $\Sigma$?

Danu

Oh... sequence

Then I think it's not possible (something by triangle inequality, like yuggib said)

ACuriousMind

@0celo7 Is it obvious to you, and you just want a "proof", or do you not see that they are?

0celo7

@ACuriousMind A little bit of both.

ACuriousMind

Then keep thinking about it.

0celo7

@yuggib huh

FenderLesPaul

7:13 PM

@ACuriousMind oh wait just to clarify, $n$ is normal to $\partial \Sigma$ and I have $\int_{\partial \Sigma} \beta$ but the conclusion is the same I suppose since it's pullback to $\partial \Sigma$

yuggib

@0celo7 $\lvert a+b\rvert\geq \bigl\lvert \lvert a\rvert -\lvert b\rvert\bigr\rvert$

FenderLesPaul

@ACuriousMind $\Sigma$ was the hypersurface for which I had $\int_{\Sigma}d\beta = \int_{\partial \Sigma}\beta$ with $\beta = v \wedge n$

hopefully that doesn't change the conclusion :p

ACuriousMind

@FenderLesPaul Yes, and from $i^\ast n = 0$ you get $\beta= 0$ so $\int_\Sigma\mathrm{d}\beta = 0$. Seems fine to me.

0celo7

@ACuriousMind Now I'm just confused by what $di(v)$ actually is.

$v$ is a tangent vector to $\partial\Sigma$, right?

@yuggib dude I still don't see what you actually want me to prove

prove that there is no $N$ which satisfies $|x_n+y_n-k|<\epsilon$, etc.?

Danu

You should know the limit of the right hand side of the triangle inequality

FenderLesPaul

7:20 PM

@ACuriousMind thank you kind sir

0celo7

@ACuriousMind Ok, so if $v$ is indeed a tangent vector on $\partial \Sigma$ then $di(v)$ is just the same vector but carried along the inclusion map, so it's a tangent vector $\partial\Sigma$ now regarded as a subset of $\Sigma$, so it's also tangent to $\Sigma$.

ACuriousMind

Wait...I might have messed up $\partial\Sigma$ and $\Sigma$ in one of my statements up there, sorry.

yuggib

@0celo7 like @Danu said

ACuriousMind

Or perhaps not. Not sorting that out now

0celo7

@yuggib I have no clue what he meant by that.

@yuggib should I be looking at $|x_y+y_n|\ge ||x_n|-|y_n||$ or something?

Danu

7:27 PM

I think we gave enough hints.

FenderLesPaul

mmm coffee

0celo7

7:38 PM

@yuggib Suppose $\lim(x_n+y_n)=k$. We must thus show that $\forall \epsilon >0$, $\exists N\in\mathbb{N}$ s.t. $\forall n\ge N$, $|x_n+y_n-k| <\epsilon$. From the triangle inequality, $|x_n+y_n-k|\le |x_n-x|+|y_n-(k-x)|$. Since $x_n$ converges, we may find an $N_1$ such that $|x_n-x|\le \epsilon/2$ $\forall n\ge N_1$. But since $y_n$ does not converge, we cannot find a suitable $N_2$. Thus there exists no $N$ such that the desired inequality $|x_n+y_n-k|<\epsilon$ holds for all $n \ge N$.

Thus the limit of $x_n+y_n$ does not exist.

yuggib

you should use an inequality that gives you an estimate on the other side

but the idea is there

Danu

@0celo7 Not a correct proof

yuggib

use the inequality I gave you with the same idea

0celo7

@yuggib I don't know what $a,b$ are.

Danu

lol

facepalm

ACuriousMind

7:41 PM

@0celo7 If you suppose that $k$ is the limit, why are you after that stating you have to show that it is the limit?

0celo7

Be helpful or say nothing at all, @Danu.

yuggib

$a=y_n - (k-x)$, $b= x_n -x$

Danu

@0celo7 I don't think I'm under any obligation to help you do your homework.

0celo7

@Danu Then don't say anything at all.

Pretty simple.

Danu

@0celo7 That's up to me.

Pretty simple, too.

But I won't tease you any more, don't worry :)

0celo7

7:43 PM

Bye. No reason to talk to you.

Danu

Bye

0celo7

@yuggib I tried that, I don't get anything useful.

Danu

Are you not also in an exam period right now, @ACuriousMind?

0celo7

@ACuriousMind Because I show that $k$ cannot exist?

yuggib

If you subtract something that has a lower bound, can you tell something about the upper bound of the (absolute value of) the difference?

ACuriousMind

7:45 PM

@Danu Yes, but I got only one exam, and its date isn't fixed yet. I mentioned I'm lazy this semester, yes? ;)

Danu

Only one? Lol!

What else are you doing?

0celo7

@yuggib Uh, what

ACuriousMind

@0celo7 The sentence still doesn't make sense. You're saying "Suppose X. We must thus show X."

yuggib

@Danu So have you chosen if you will show up to see some real mathematical physics on wednesday ;-P

ACuriousMind

If you suppose X, there is nothing to show. Then X is true by assumption.

Danu

7:46 PM

@yuggib Would you even recognize me? ;)

@ACuriousMind You have endless patience ;)

0celo7

@ACuriousMind Fine, let $k$ be the limit. For this to be the limit: Insert proof here. Thus $k$ cannot exist.

Danu

What course is the one exam btw @ACuriousMind

0celo7

$k$ is just shorthand for $\lim(x_n+y_n)$.

ACuriousMind

@Danu Ah, well, mostly non-physicsy (or mathy) stuff

@Danu sheaf cohomology

yuggib

@0celo7 $b=\lvert x_n -x\rvert\leq \varepsilon/2$, then $\lvert\lvert a \rvert - b\rvert \geq \lvert\lvert a\rvert -\varepsilon/2\rvert$

Danu

7:47 PM

Whoa you guys have a special course just on that?

Nice.

yuggib

@Danu you would recognize me...

Danu

@yuggib Perhaps :P

0celo7

@yuggib ...what

I still don't get where this is going

or how to prove that

ACuriousMind

@Danu Yes, and it will be continued next semester (although it is then called Topologie singulärer Räume because that's what we're gonna apply it to after we finished building the abstract theory)

yuggib

now substitute $a$ with $y_n - (k-x)$ and use the fact that it is bigger than $\varepsilon$

Danu

7:50 PM

@ACuriousMind That's some 10/10 shit right there.

The mathematics curriculum is not that well-developed here, sadly.

There isn't that much geometry to do :(

0celo7

why is it bigger than $\epsilon$

yuggib

because $y_n$ diverges

0celo7

I'm so confused

what's the motivation for any of this

we haven't done anything this complicated in class

Danu

After next semester I'll have done all the courses that have "geometry" or "topology" in their names, except algebraic geometry.

yuggib

that you want to show that it is absurd to have $x_n+y_n$ converging

ACuriousMind

7:52 PM

@Danu Well, it's not a regular course in the curriculum, it's happenstance that it is given this time

0celo7

yes, so $|x_n+y_n-k|<\epsilon$ cannot be true

but I don't see how you're getting to your crazy inequalities from that

(or how to prove your inequalities)

ACuriousMind

Many of the interesting lectures here are not regular parts of the curriculum, but just something the lecturer wanted to talk about at least once

Danu

@ACuriousMind I see.

yuggib

@0celo7 mine are not crazy inequalities

and you need to think a little bit about that

Danu

The guy who does "cool shit" lectures is taking a sabbatical this semester, I think.

He did stuff like low-dimensional topology earlier

yuggib

7:54 PM

to show that something is not $<\varepsilon$ means proving that it is $\geq \varepsilon$

Danu

There is so much mathematics I'd like to learn :(

@yuggib Stop using that damn $\varepsilon$ shit :P

$\epsilon$ or nothing

0celo7

yes

ACuriousMind

$\varepsilon$ looks cuter, though

yuggib

@Danu $\epsilon$ is awful, it seems $\in$

Danu

Naw

$\epsilon$ is life

$\epsilon$ is love

Haddaway - What Is Love

►

(beat you to it, ACM)

0celo7

7:56 PM

Ok so we want $|x_n+y_n-k|\ge \epsilon$

yuggib

use the inequalities

0celo7

We also have $|(x_n-x)+(y-(k-x))|\ge ||x_n-x|-|y_n-(k-x)||$

I know how to prove that.

So at this point I suppose $|x_n-x|<\epsilon/2$?

yuggib

yes

0celo7

ok, then I get lost

do I then have the above $\ge |\epsilon/2-|y_n-(k-x)||$?

yuggib

$\lvert a-b\rvert\geq \lvert A -b\rvert$ if $0\leq a\leq A$ right?

0celo7

7:59 PM

I guess, I don't know how to prove it.

ACuriousMind

brb, someone reminded me just in time it is Saturday and I can't shop tomorrow.

user54412

@Danu Surface area of a sphere: $4 \varpi r^2$

0celo7

@ACuriousMind One of Germany's only faults tbh.

Danu

@ChrisWhite Unholy symbols!

0celo7

@yuggib I'll try to prove that later, what next

I guess $|y_n-(k-x)|\ge \epsilon/2$ or something?

yuggib

8:05 PM

then you use the fact that there is an $\varepsilon'$ such that $\lvert y_n - (k-x)\rvert \geq \varepsilon'$ since $(y_n)$ diverges

0celo7

hmm

@yuggib I don't see how to continue...

and I don't see how that last statement comes about

I'll just leave it blank I guess

hope he doesn't grade that one

FenderLesPaul

8:20 PM

The Lemon Song live at Filmore

►

so fucking good ugh

yuggib

@0celo7 well, I don't know how to explain it differently

0celo7

@yuggib I don't know what you're trying to explain

yuggib

look if you like this proof better math.stackexchange.com/questions/798710/…

0celo7

I didn't understand it.

ACuriousMind

@0celo7 What do you not understand about it? This might help identify the problem you have

0celo7

8:33 PM

@ACuriousMind I'll tell you once I've done the rest of the problem set

yuggib

I'll try the last time. Take a difference between two positive numbers $a-b$, with $a> b$; do you agree that if $a\geq A$ and $b\leq B$, then $a-b\geq A -B$?

0celo7

@yuggib I agree, but don't know how to prove it.

GBeau

@0celo7 the proof @yuggib links is basically the standard proof

Danu

$a-b\geq A-b \geq A-B$

Just implement estimates one-by-one if everything else is too fast

yuggib

@0celo7 now, choose an $N_1$ such that $\lvert x-x_n\rvert \leq \varepsilon$ for any $n\geq N_1$ (you can by definition since $x_n$ converges)

then, choose an $N_2$ such that $\lvert y_n -(k-x)\rvert\geq 2\varepsilon$ for any $n\geq N_2$

again you can by definition otherwise $(y_n)$ would converge

$k$ is a supposed limit of $x_n+y_n$, and is an arbitrary number

do you agree it is possible up to now?

then, you consider any $n\geq \max\{N_1,N_2\}$, and look at

$\lvert x_n +y_n -k\rvert\geq \lvert \lvert y_n - (k-x)\rvert -\lvert x-x_n\rvert\rvert = \lvert y_n - (k-x)\rvert -\lvert x-x_n\rvert$

since you choose the first on the right hand side to be always bigger than the second

now you use the inequality with $a,b,A,B$ above

0celo7

8:45 PM

@yuggib why is this possible

yuggib

@0celo7 because if else the sequence would converge

by definition of convergence

0celo7

I'm not convinced that that's the converse of convergence.

yuggib

what's the converse of convergence?

ACuriousMind

Ah, negating quantifiers is so fun!

yuggib

Hint: it's this logical statement

0celo7

8:50 PM

I don't see how "there is no $N$ s.t. $\forall n\ge N$ $|y_n-y|<\epsilon$" leads to "choose $N_2$ s.t. $|y_n-(k-x)|\ge 2\epsilon$ $\forall n\ge N_2$"

$k-x$ is just some number

yuggib

$\overline{(\exists k\in\mathbb{R})(\forall\varepsilon >0)(\exists N\in\mathbb{N})(\forall n\geq N)\lvert k-x_n\rvert <\varepsilon}$

0celo7

I'm just too stupid for this...

yuggib

as ACM said, it's just a matter of negating quantifiers :-P

0celo7

@yuggib the negation of "for all" is "there exists"

yuggib

as I always told you, logic is important

yes

0celo7

8:52 PM

I don't see how you get a for all

yuggib

negation of "there exists" is "for all", and you see a nice "there exists" in front of $k$ there

0celo7

I have no idea what you're saying anymore

Danu

This is still going on?

Whoaaaa :D

yuggib

I give up

0celo7

@yuggib consider the series $(-1)^n$

this diverges

can you please show me what $N_2$ is

and what is $k-x$ anyway

Danu

8:55 PM

It doesn't diverge.

It also doesn't converge :P

yuggib

divergent usually means it is unbounded

ACuriousMind

@0celo7 Oh...there is a mismatch of terminology here, I think. "diverges" is not "doesn't converge" but "goes to infinity".

0celo7

@ACuriousMind Definition 2.2.9. A sequence that does not converge is said to diverge.

ACuriousMind

Ah, and now we know where the confusion lies!

yuggib

so the proof above works for unbounded sequences

0celo7

8:58 PM

@yuggib I'm not assuming anything on the boundedness...

Danu

@0celo7 That's a disgusting definition :P

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