Mathematics - 2012-07-20 (page 2 of 5)

Kannappan Sampath

04:00

@PeterTamaroff Nope!

Peter Tamaroff

@KannappanSampath What does it unify?

@KannappanSampath Whoooops!

anon

number theory and representation theory and galois theory ...

all that good stuff

Kannappan Sampath

@PeterTamaroff Geometry, Number Theory, Algebra, Analysis -- but have I not said everything?

Peter Tamaroff

@KannappanSampath WOW

robjohn

@anon I used that $10^{3^n}-1=\left(10^{3^{n-1}}-1\right)\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)$

Steven Li

04:01

Let A_1, A_2, ... , A_n and B be well-formed formula of propositional calculus. If B is a deduction from A_1, A_2, ... , A_n, then A_n is a deduction from A_1, A_2, ... , A_n-1 implies B.

Peter Tamaroff

@anon Could you try and explain what it is about? Or is it too complicated?

anon

@robjohn Basically the same thing, because the factor on the right is $\Phi_{3^{n-1}}(10)$.

robjohn

@anon and that $\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\equiv0\pmod{3}$

anon

@PeterTamaroff It's too complicated.

Steven Li

How do I prove that?

Peter Tamaroff

04:03

@anon Do you understand it?

anon

no

Kannappan Sampath

But, yeah, there are nice expository articles on my to-read list.

anon

Isn't that just a rule of how deductions work? I'm not sure.

Steven Li

It's a so-called metatheorem

It's different from a normal theorem

So I'm not sure how to prove a metatheorem

Kannappan Sampath

@anon Noted. Will go through. Thanks for telling me that.

robjohn

04:06

:5439350 putty.exe - it turns your computer into a lump of clay?

anon

hey, it's not my ftp. I can't vouch for strange .exe's...

J. M.

@anon People should remember to have protection on hand when they go in strange places, anyway...

Kannappan Sampath

I hear putty is something like torrent business.

Eugene

putty is ssh for windows

anon

interesting

robjohn

04:10

@Eugene like $\text{pu}\textbf{tty}$

Peter Tamaroff

@HenryT.Horton I'm looking at that now

J. M.

PuTTY.

Eugene

yup

Kannappan Sampath

@anon Is it just me or there are lot of German titles (as compared to English titles)?

Peter Tamaroff

I assume you're thinking about $[0,2\pi)$ as a subset of $\mathbb R$ with the Euclidean metric.

Eugene

04:11

are you all talking about something else though?

anon

I haven't browsed that much.

Eugene

one of my students call erdos the jesus of math recently

irked me a little

J. M.

@Eugene how so?

Peter Tamaroff

@HenryT.Horton Let $(x,y) \in S^1$, say $x=(\sin \theta,\cos\theta)$ and $y=(\sin \alpha, \cos\alpha)$. Then $d_S(x,y)=2\left|\sin \left(\frac{\theta+\alpha}{2}\right)\right|$

J. M.

Because he was itinerant?

Eugene

04:13

@J.M. he said it's because he spread math to the world.

as in gave a bunch of public lectures and stuff

J. M.

A bit of a stretch, though. I see churches a lot, but not that many math groups...

Peter Tamaroff

Now the inverse of $f$ should be $g:S^1\to [0,2\pi)$ such that $g((\sin \theta,\cos \theta))=\arccos \theta$?

@HenryT.Horton Why is it not continuous in in the Euclidean metric?

anon

inverse trig function of an angle?

Peter Tamaroff

@anon What are you asking?

Henry T. Horton

The inverse maps $(\cos t, \sin t)$ to $t$...

Look at the point $(1,0) \in S^1$

anon

04:17

wooooosh

Henry T. Horton

What does the inverse map a neighborhood of $(1,0) \in S^1$ to?

Peter Tamaroff

@HenryT.Horton Right. It tears it apart in some sense,

anon

it's actually been linked in chat a number of times already

Peter Tamaroff

@HenryT.Horton Yes, stupid of my part.

Kannappan Sampath

@anon Oh, OK. My bad. :(

Henry T. Horton

04:19

So that's not continuous, you map some things real close to $(1,0)$ to points nearly $2\pi$ apart

Peter Tamaroff

@HenryT.Horton Interesting example. Thanks!

anon

gluing two ends of a string together is continuous, cutting a string is not

Henry T. Horton

String theory

CLarue

If the codomain of a continuous bijection is $\mathbb{R}$, then it is monotone, that order property need not hold or make sense for other codomains.

Peter Tamaroff

@HenryT.Horton I'm given in the text that a circle of radius $1$ is topologically equivalent to a circle of radius $2$. That is proven by a map $x\mapsto x/2$, right? (Like just shrinking one into the other, and inversely)

However, it claims they are not metrically equivalent.

That is, the distances aren't preserved through the tranformation.,

Henry T. Horton

04:23

$x \mapsto x/2$ turns it into an ellipse

Shrinks the $x$-dimension but not the $y$

anon

he might be talking about $x$ as a vector in R^2

Peter Tamaroff

@HenryT.Horton I meant $x$ as an ordered pair. Like ${\bf x}=(x,y)$

Henry T. Horton

Yeah right!

Peter Tamaroff

@anon Yes, thanks.

user19161

@anon Strange that you use dollars around x but not around R2.

2

Peter Tamaroff

04:25

@JasperLoy He's too lazy to code \Bbb R^2

anon

@JasperLoy I began typing with discipline, but lost it very quickly

Henry T. Horton

Then you would want $\mathbf{x} \mapsto \mathbf{x}/\sqrt{2}$

If I did my arithmetic correctly, which is rare

Peter Tamaroff

@HenryT.Horton Right. I'm sloppy today.

anon

Wait, what?

Kannappan Sampath

@HenryT.Horton I guess, this is not that rare situation! Sadly, you were wrong. Peter was right.

Peter Tamaroff

04:26

@HenryT.Horton Just that the book says radius $4$ I think, sorry. I miswrote that.

Anyways.

anon

To map the circle of radius $r$ to the circle of radius $1$ centered around the origin, we need ${\bf x}\mapsto{\bf x}/r$.

sqrt(2) is when a square gets involved

Henry T. Horton

aaaaaahhhhhhhhhhhhhh

anon

or if you're talking about scaling area

Henry T. Horton

$x^2 + y^2 = r^2$, not $x^2 + y^2 = r$

Peter Tamaroff

@HenryT.Horton BUUUUUUUUUUUUURN!

Kannappan Sampath

04:28

@anon Right. The equation of a circle $\|\bf{x}\|=r$

Peter Tamaroff

XD

Henry T. Horton

I'm crying

Peter Tamaroff

@HenryT.Horton Oh, noes.

anon

I'm crying too, but for a different reason.

Kannappan Sampath

What happened @anon?

I wish anon had called himself by his name. :(

anon

04:29

I'm saying it's so funny I'm cryi.. you get the idea.

Henry T. Horton

Listen guys I just do theoretical stuff, none of these concrete calculations!!!

Peter Tamaroff

Today I saw those facebook pollings with something like $7-3\times 0+1=?$. A guy answered wrongly and another was spelling the correct answer step by step. I thought to myself: "Blah, why do you even bother about that." Maybe I should.

@HenryT.Horton ¿"Know that feel, bro"?

user19161

@HenryT.Horton Theoretical stuff actually involves many concrete calculations.

J. M.

@PeterTamaroff He's following the footsteps of the great Grothendieck. ;)

Kannappan Sampath

@J.M. no-example-kind.

user19161

04:32

@PeterTamaroff Even if you change your username I can identify you using the inverted question mark!

J. M.

@PeterTamaroff Just link them to the math.SE thread... >:) Gerry's answer, I suppose.

Peter Tamaroff

@JasperLoy ¡I'm marked!

I also have the "eñe" =D

user19161

@PeterTamaroff That rhymes with I'm fuqed!

Kannappan Sampath

@anon Have some interesting questions?

(say in representation theory?)

user19161

@KannappanSampath You lack exercises?

anon

04:36

I'm studying group theory offhand. Do you mean like, research-esque questions, or textbook-esque questions?

I have an interesting one about trees and commutator subgroups.

Kannappan Sampath

@JasperLoy Not really. But looking for research-esque questions as @anon puts it.

@anon Please state it for me.

user19161

@KannappanSampath You seriously thinking of research as a freshman?

Kannappan Sampath

@JasperLoy Well, I always liked trying my hand at technically hard things.

(I have been successful at times; sometimes I don't go anywhere.)

user19161

@KannappanSampath I just feel that you should study more topics first before plunging yourself into it. But others may disagree.

anon

k. In the same way that trees correspond to ways to fully parenthesize strings, we can use them to designate iterated commutators, like [[G,[G,G]],G]. Call it [G;T] for a given tree T. Then for what trees T and S is [G;T] a subgroup of [G;S] for all groups G? This vastly generalizes the lower central series embedding into the derived series.

For appropriately loose interpretation of "embedding."

Kannappan Sampath

04:41

@anon Hmm, I get the embedding part. And, it is very interesting.

I should review many things before attempting this.

anon

On the same tack: we can use ordinal numbers to extend group series out beyond infinity via transfinite induction. I wonder if there's a good way to describe transfinite extension of these tree things.

Kannappan Sampath

Did you formulate the question yourself?

Peter Tamaroff

@HenryT.Horton Hey

anon

I also had one inspired by percolation theory, about viewing inverse limits with random group theory by viewing the indexing set as something to percolate in that I mentioned to Eugene.

@KannappanSampath Yes.

Kannappan Sampath

@anon I know no percolation theory. :(

anon

04:45

@KannappanSampath my knowledge of percolation theory is about equivalent to the AMS article I read on it

just the basic idea of percolation, wherein certain group extensions are activated with some probability, and then studying the limiting probability that the inverse limit has so-and-so properties.

Henry T. Horton

@PeterTamaroff Hello, Peter, how are you doing, today,

Peter Tamaroff

@HenryT.Horton I was going to write something about that $1$-sphere thing.

Henry T. Horton

Here in the states, we just call it a "circle"

Peter Tamaroff

Give me a sec

@HenryT.Horton I know that $f$ is continuous iff for every neighborhood $M$ of $f(a)$ there exists a neighborhood $N$ of $a$ such that $f(N) \subset M$.

I want to use that to prove the inverse is not continuous.

Kannappan Sampath

Also, I wanted to ask if you know some place where (most)all properties about character table or those properties of group that can be read off from there. @anon

anon

04:49

I've only studied rep thry a little, but what's up?

I mean, obviously, conjugacy classes.

Kannappan Sampath

@anon Well, there are more trickier ones I know of (and is standard). I am sure there are many more hear-say things. :(

@MarianoSuárez-Alvarez May be you have some input here?

Henry T. Horton

@PeterTamaroff Take a neighborhood of $0 \in [0, 2\pi)$ and a neighborhood of $(1,0) \in S^1$

anon

you can derive character tables for tensor / symmetric / alternating powers of a rep from the rep's original char table

via theory of symmetric polynomials essentially

Kannappan Sampath

@anon I see. I am not aware of this. I'll read this up.

Henry T. Horton

????

A neighborhood of $0 \in [0, 2\pi)$ is of the form $[0, \varepsilon)$

Peter Tamaroff

04:55

@HenryT.Horton Sorry, I was thinking about $0$ in $\Bbb R$. My bad.

anon

@KannappanSampath here's a starting point

Peter Tamaroff

@HenryT.Horton And a nbhd of $(1,0)$ in the $1$-sphere is an arc, right? But going both ways.

anon

well, contains an arc

Henry T. Horton

Yeah mang

Kannappan Sampath

@anon reading Thank you.

Peter Tamaroff

04:58

@anon But now I'm considering the metric space $(S^1,d_s)$ where $d_s$ is the restriction of the Euclidean metric to $S^1\times S^1$. Thus the nbhd is an arc.

@HenryT.Horton Who are you answering to?

anon

a nbhd can be more than just an open ball

a nbhd of x is just any set containing an open set containing x

Peter Tamaroff

@anon I'm in metric spaces now. Open sets are defined via nbhds which are defined via open balls.

anon

lame

Henry T. Horton

You hear that Peter? anon thinks your balls are lame

Peter Tamaroff

@anon Come at me, bro.

Kannappan Sampath

05:02

LOL

Peter Tamaroff

@HenryT.Horton But srsly now. The image of $[0,\epsilon)$ under $f$ is the arc starting and containing $(0,1)$ and running clockwise.

Henry T. Horton

Oh, you want to prove continuity of the forward map before proving the inverse is not continuous?

CLarue

What makes induction more difficult to understand than, say, contradiction?

Peter Tamaroff

Dread, you're right!

I was looking it the other way around.

Henry T. Horton

Owned

anon

05:08

Can one search through one's favorites list in a specific tag category? I have 14 pages of them..

aha! @KannappanSampath here is an appropriate generalization

Henry T. Horton

@PeterTamaroff Well Peter? are you ready to stop losing and start winning?

Peter Tamaroff

@HenryT.Horton What do you mean? I feel like I'm winning ¬¬

(Whatever "winning" means)

anon

urbandictionary.com/define.php?term=winning

Kannappan Sampath

@anon Nice. I am still engrossed in symmetric poly.

Henry T. Horton

@PeterTamaroff Did you finish the problem then?

Peter Tamaroff

05:16

@HenryT.Horton Yes, I'm just writing it out.

Henry T. Horton

Oh ok

I will go home in a moment then

Peter Tamaroff

@HenryT.Horton You're in your office?

Henry T. Horton

Yes

Peter Tamaroff

05:32

@HenryT.Horton Can I say $g$ "tears" $S^1$ at $(1,0)$?

Or that $S^1$ is closed but $[0,2\pi)$ is not!

Henry T. Horton

No.

BenjaLim

@PeterTamaroff what are you doing now?

Peter Tamaroff

@BenjaLim Drawing the cutest pictures of $S^1$.

BenjaLim

serious.

Henry T. Horton

He's polishing my trophies

Get back to work, slave

Peter Tamaroff

05:34

@BenjaLim Supah-serious.

J. M.

@HenryT.Horton Damn, I misread the "halls" as "balls"...

BenjaLim

no what are you trying to do with $S^1$

Peter Tamaroff

@BenjaLim I swear I didn't touch her, I swear!

BenjaLim

I'm in no mood for jokes

Peter Tamaroff

@BenjaLim OK.

@BenjaLim Why?

BenjaLim

05:35

Don't worry.

What are you trying to prove now?

Henry T. Horton

He's making a circular argument.

Peter Tamaroff

@BenjaLim I'm writing a little discussion about why the map $f:[0,2\pi)\to S^1/f(x)=(\cos x,\sin x)$ is continuous, one-one and onto.

But it's inverse is not.

BenjaLim

ah.....

:D :D :D :D :D

Peter Tamaroff

@BenjaLim Yay!

Former_Math_Addict

In a bar:
dumped guy: "My girlfriend left me because I told her I loved her!"
other guy: "What a witch!"
dumped guy: "No, it's all my fault, she's a mathematician, and I should have said 'I love you and only you'".

BenjaLim

05:36

@PeterTamaroff Read number 2: math.stackexchange.com/questions/169876/…

Peter Tamaroff

@Former_Math_Addict HAHA

BenjaLim

@PeterTamaroff read number 2

Henry T. Horton

What's a girlfriend?

Peter Tamaroff

@HenryT.Horton LOL

@BenjaLim Yes!

BenjaLim

Sorry i meant number 1

Peter Tamaroff

05:37

@BenjaLim Yes, sure. I understood the first time

@BenjaLim In my case $(1,0)$ is the problem

BenjaLim

yes

@PeterTamaroff Now prove that $S^1$ is not homeomorphic to $[0,1]$

Peter Tamaroff

It's like $g:S^1\to [0,2\pi)$ cuts $S^1$ at $(1,0)$

@BenjaLim Don't you mean $[0,1)$? BTW, I only know about topological equivalence in emtric spaces.

BenjaLim

that's definitely not homeomorphic to $S^1$

one is compact the other is not

but why is $[0,1]$ not homeomorphic to $S^1$?

Peter Tamaroff

@BenjaLim I don't know about homeomorphisms yet. Remember, only metric spaces now.

I'm only one section away from Topological Spaces though.

BenjaLim

@PeterTamaroff doesn't matter

Peter Tamaroff

05:40

@BenjaLim ¿¿??

@HenryT.Horton How can I say correctly that $S^1$ is transformed into $[0,2\pi)$ by cutting it at $(1,0)$ so this transformation isn't continuous?

Just colloquially.

Points close to each other are set apart.

BenjaLim

@PeterTamaroff you're still thinking about continuity in the wrong way.

Peter Tamaroff

@BenjaLim Why?

BenjaLim

You need to think about it in terms of open sets

your function $f$ gives a bijective map from $[0,2\pi)$ to $S^1$ yes?

To see the inverse is not continuous

Peter Tamaroff

@BenjaLim Yep

BenjaLim

look for example at $[0,\pi/2)$

this is open in the subspace metric on $[0,2\pi)$ yes?

Peter Tamaroff

05:44

@BenjaLim Yes.

BenjaLim

Because it is $(-\pi/2,\pi/2) \cap [0,2\pi)$

Peter Tamaroff

It is the underlying set of the metric so it is open.

@BenjaLim ?¿?¿

BenjaLim

Now the image of this under $f$ is the quarter strip of $S^1$ from $0$ to 90 degrees, excluding $(0,1)$

yes?

Peter Tamaroff

@BenjaLim Oh, $\pi/2$...

BenjaLim

@PeterTamaroff I made the correction

Peter Tamaroff

05:46

@BenjaLim Yes

BenjaLim

it should have been the top point excluded

Now suppose that this strip is open

Look at the point $(1,0)$

Peter Tamaroff

@BenjaLim It isn't, right?

BenjaLim

yes

but that's what you want to show :D

because showing that this is not open

Peter Tamaroff

@BenjaLim Right.

BenjaLim

proves that $f^{-1}$ is not continuous :D :D

Peter Tamaroff

05:47

@BenjaLim Precisely.

@BenjaLim I guess you're no longer under the weather!

BenjaLim

well yes

Now peter

Peter Tamaroff

@BenjaLim Sorry, I meant to use "under the weather" as "blue" but it meant "ill".

You're ill?

@BenjaLim Yes?

BenjaLim

explain to me why you can never find a ball $B$ about $(1,0)$ such that this $B \cap S^1 \subseteq f\big([0,\pi/2) \big)$

no

just had an argument with someone

anyway

@PeterTamaroff draw a picture!

Peter Tamaroff

@BenjaLim I already did! =D Wanna see it?

BenjaLim

show the pic

that's a good enough proof

And here's justification for why we can just draw balls and not any open set: The balls form a basis for the Euclidean topology on $\Bbb{R}^2$

Peter Tamaroff

05:51

@BenjaLim Yeah. =D

BenjaLim

@PeterTamaroff GIVE ME THE PICTURE GODAMMIT

Peter Tamaroff

@BenjaLim I have to scan it, Speedracer.

J. M.

I don't think we can win here...

@PeterTamaroff It can mean either, depending on context.

Peter Tamaroff

@BenjaLim OK, scanning in progress.

Done.

Uploading...

BenjaLim

ok

what I basically wanted to tell you

if you want such a ball $B$

it first of all has to be entirely above the $x$ - axis

is

otherwise it can't be such that $B \cap S^1$ is contained in the image

but if it is entirely above the $x - axis$

it misses $(1,0)$

that's where constructing such a ball is impossible

Peter Tamaroff

06:01

@BenjaLim You're still talking about $x\mapsto (\cos 2\pi x,\sin 2 \pi x)$ right? Becasue my pic is about $x\mapsto (\cos x,\sin x)$

BenjaLim

yes

well same thing really :D

Peter Tamaroff

@BenjaLim So my pics are ok, no?

BenjaLim

I'm a bit confused with $f$ and $g$

Peter Tamaroff

@BenjaLim $f$ is $f:[0,2\pi)\to S^1$ and $g$ is its inverse.

Oh!

Where it says

$f(N')$

$g(N')$

it should be

BenjaLim

what is $N'$

@PeterTamaroff do you get my explanation above?

Peter Tamaroff

06:04

@BenjaLim The red arc that has $N'$ next to it.

@BenjaLim It is a nbhd of $(1,0)$ in the metric space $S^1$

BenjaLim

I think you have the pics the other way round

because if $N'$ is a neighbourhood about the point on the circle

talking about $f(N')$ doesn't make sense

Peter Tamaroff

@BenjaLim I just said it should read $g(N')$

=D

BenjaLim

ok

Peter Tamaroff

@BenjaLim I think I do.

BenjaLim

good

Peter Tamaroff

06:09

@BenjaLim So how is univ going for you?

BenjaLim

starting on monday

Peter Tamaroff

@BenjaLim What are courses are you taking?

BenjaLim

algebraic topology

special topics in algebra

and

several variable analysis

Peter Tamaroff

@BenjaLim "Special topics"?

BenjaLim

decided by the lecturer

Peter Tamaroff

06:14

@BenjaLim Oh. I need to study algebra.

BenjaLim

hahahahahaah

Peter Tamaroff

@BenjaLim Seriously, dude.

@BenjaLim Remember I started with linear algebra a while ago?

BenjaLim

yes?

Peter Tamaroff

@BenjaLim Well, I didn't really move on.

BenjaLim

ok

why

Peter Tamaroff

06:22

@BenjaLim Algebra was boring. But you and t.b. told me it is important.

BenjaLim

ahahahahahahahahahahahahahahaha

complete opposite of me

Peter Tamaroff

Maybe later on it will get interesting.

@BenjaLim Why?

BenjaLim

i like algebra

Peter Tamaroff

@BenjaLim I'm not saying I don't. Just that it hasn't caught me up yet.

@BenjaLim Topology exercise:

BenjaLim

yes>

user19161

06:27

@PeterTamaroff I remember you got A survey of modern algebra.

Peter Tamaroff

Prove that if $f:X\to Y$ and $g:Y\to X$ are inverse functions with $X$, $Y$ metric spaces, then the following statements are equivalent:

$(1)$ $O\subset X$ is open $\iff$ $f(O)\subset Y$ is open.

$(2)$ $F\subset X$ is closed $\iff$ $f(F)\subset Y$ is closed.

I'll prove that and call it a day.

user19161

@PeterTamaroff Did you leave out continuity there?

Peter Tamaroff

@JasperLoy Not $f,g$ being continuous is equivalent to those, too.

I actually have four equivalencies

Oh, sure.

user19161

@PeterTamaroff Oh I misread the question. I see now.

Peter Tamaroff

I have these other two, however:

BenjaLim

06:31

I think this is a triviality from the fact that complements distribute over taking preimages

Peter Tamaroff

$(3)$ $f,g$ are continuous
$(4)$ For each $a\in X$ and $N\subset X$, $N$ is a nbhd of $a$ $\iff$ $f(N)$ is a nbhd of $f(a)$

I have as a theorem that $(4)$, $(3)$ and $(1)$ are equivalent.

And I have to prove $(1)$ and $(2)$ are.

I think I can use $(3)$ as a middle-man.

It follows by the characterization of continuity in terms of open and closed sets.

$f:X\to Y$ is continuous $\iff$ the preimage $f^{-1}(F)$ ($f^{-1}(O)$) of a closed (open) set $F$($O$) in $Y$ is closed (open).

This long theorem about those four equivalences says that $X$ and $Y$ are topologically equivalent iff there exist inverse functions that establish a one-one correspondence between the open sets / closed sets / complete systems of nbhs of the two spaces.

BenjaLim

aka homeomorphism

@PeterTamaroff (1) and (2) are equivalent

Peter Tamaroff

@BenjaLim Yes.

I have to prove it.

BenjaLim

simply by the fact that for any $U \subset Y$ you have $f^{-1}(U^{c}) = (f^{-1}(U))^c$ :D

Peter Tamaroff

@BenjaLim That's all there is to, right?

BenjaLim

06:42

but write it out

because now

the preimage is just $g$

so you actually have $g(U^{c}) = g(U)^c$

Peter Tamaroff

I think I got it: $O$ is open iff $O^c$ is closed $\subset X$, by $(3)$ (which I have as a theorem is eqiv to $(1)$) iff $f^{-1}(O^c)=f^{-1}(O)^c$ is closed iff $f^{-1}(O)$ is open which is true by $(3)$. So $F=O^c$ does the job.

Conversely $O=F^c$ should work just the same.

@BenjaLim

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$Mathematics$ Mathematics