Mathematics - 2015-05-23 (page 2 of 4)

8:22 AM

@anon are you around?

yeah

I have a group homomorphism $f : G \to H$. This naturally extends to the profinite completion $\widetilde{f} : \widehat{G} \to \widehat{H}$. Can you give me an $f$ which is not an isomorphism, yet $\widetilde{f}$ is an isomorphism?

> f is not, yet f^~ is not

I guess there must be some kinda trivial counterexample with small, simple groups, but I can't find one.

@BalarkaSen it's true that $S^2/A$ where $A$ is a finite set of points is a wedge sum $S^2\vee S^1... \vee S^1$ right?

8:27 AM

yeah

ok

how would you prove that?

@BalarkaSen $G\to\hat{G}$ for (noncomplete profinite residually finite) $G$?

e.g. $\Bbb Z\to\Bbb Z_p$. also aren't finite groups already profinite complete groups?

@iwriteonbananas prove it for $S^2/\{pt, -pt\}$ first and then induct

that's $S^2/S^0$ and (CW complex, subcomplex) is always a good pair

@anon eh. well, how about groups $G$, $H$ which aren't profinite?

8:30 AM

wait no the profinite completion of $\Bbb Z$ is not $\Bbb Z_p$. bleh

yeah, it's $\prod \mathbf Z_n$

$\cong\prod\Bbb Z_p$

what's $(S^1\times S^1)/A$ ?

@iwriteonbananas Do you want to prove that $S^2/A \cong S^2 \vee S^1 \vee \cdots \vee S^1$ using an explicit homotopy equivalence or something?

do you mean \A ? or is A a group acting?

8:33 AM

@BalarkaSen no, that's not possible, i dont think they are homeomorphic

@iwriteonbananas $(S^1 \times S^1) \vee S^1 \vee \cdots \vee S^1$

@anon $A$ is a discrete set of points on $S^1 \times S^1$. the $/$ is quotient of topological spaces

@iwriteonbananas giving an explicit homotopy equiv would be maddening. it's easy to "see" it, though.

@BalarkaSen no, i tried doing that before but couldnt come up w/ one

exactly

so i don't know what's your question. can't you see it?

so X/A means X/~ where ~ identifies points in A? didn't know that.

@BalarkaSen ok, thanks that's what i figured

i can, it's all good. just wanted to ask u if it generalizes to any finite number of points

8:37 AM

@anon yeah, people use that terminology because it's too much work to write down X/~ and ~ : x ~ x', x \in A each time you want to say something about quotients.

@anon yeah

@iwriteonbananas are you doing that exercise in chapter 2.1, btw?

i vaguely recall that hatcher had a problem about computing homology of $(S^1 \times S^1)/A$. if that's what you're doing, don't cheat by finding an explicit homotopy type of the space!

the point of the exercise is to make the reader comfortable with the long exact sequence

yeah thats the one im doing

i thought the point of the exercise was to realize that (X,A) is a good pair and then identify the homotopy type of X/A

(i am lecturing you for no reason at all : the way i did it in class was by finding an homotopy type :P)

haha

ok fine i'll do LES too

8:49 AM

right.

ok

do you remember part b) of that exercise?

yeah, something about torus with the longitudal circle pinched, right?

@BalarkaSen In class?

well, yeah

orientable surface of genus 2

oh, so double torus mod circle?

8:56 AM

yeah

it's again easy to do it using an explicit homotopy type : it's just homotopy equivalent to torus wedge circle !

:P

but i recommend you not to do it

no, wait

you dont know what circle im talking about

it's the longitudal circle, right?

oh, nevermind, i just opened my copy of hatcher

if you pinch A, it's torus wedge torus.

if you pinch B, it's torus wedge circle

i dont see that

first pinch B to a point. it's like a horned sphere with two horns wedged along the pointy edge

now smoothen the test of the rest of both of the horns continuously.

8:59 AM

yeah

ok

is there a more rigorous explaination?

in particular, horned sphere with two horns identified along the pointy edge is equivalent to sphere with two horns and a 1-cell attached with boundary points pasted to the two pointy edges

@iwriteonbananas well, it's like i said : double torus with B collapsed is homotopy equivalent to torus connected sum cylinder with the two opposite boundary circles attached to a point

and this is homotopy equivalent to torus connected sum cylinder with the two boundary circles attached to the boundary of a 1-cell

this is in turn homotopy equivalent to torus wedge sphere wedge circle, which is torus wedge circle.

this is about as rigorous as it gets. if you want something more rigorous than this, use the long exact sequence :P

please write down an explicit homotopy equivalence at each step

just kidding :P

phew

i can live w/ that explanation

btw. how can we express the loop A in terms of generators of the two tori?

you mean generators of the punctured torus?

9:06 AM

two holed torus has 4 generators, right?

what's A in terms of these generators?

yeah.

oh, you want to express A in terms of generators of the double torus?

call the generators of (fundamental group of..) the left torus a,b and of the right torus c,d

yeah

you don't mean fundamental group. you mean the first homology group.

Sheesh. You study at IISc, Balarka?

you are forgetting about the basepoint, @iwriteonbananas

@Soham no, why would I?

9:10 AM

ok, let the base point be the wedge point of the 1-skeleton

I dug up a few chats (I was too curious about "In class"). IISc Bangalore?

Wow.

You were invited there?

nah, i was invited, but didn't go there

@BalarkaSen So what's "In class"?

@iwriteonbananas i have no idea. not bothering to think about it.

ok np

9:11 AM

I know for a fact that KC Nag doesn't cover algebraic topology in 10.

:)

:)

@BalarkaSen so you won't say?

i.. don't think i should.

You take classes at some uni then, I guess. Wow.

Wow.

9:17 AM

Is this some sort of mirror bot?

Wow! That's unheard of!

In India, more or less.

not regularly. i just took the alg top class.

Where?

In Kolkata?

@iwriteonbananas no, you already know i do.

9:17 AM

Mirror bot?

in kolkata, but i'd'nt like to say where.

ok, gotta go.

@BalarkaSen i know, i was joking

I thought you were repeating the last words of my messages.

Last words of your messages?

@BalarkaSen I understand. The only reason I wanted to know was, you know, because I'd like to do so too.

@Huy Yeah, like the smiley and the "wow". Don't mind me, I'm sort of distracted and sad right now. :)

9:19 AM

Why? :(

Because I've always wanted to do what Balarka's doing, but I have no idea how he's managed to. In India, it's sort of hard.
Bye guys, I'll be back in an hour or two.

I guess it comes down to being Balarka. :)

user147690

@SohamChowdhury It would be dangerous to compare yourself to Balarka. Some people start earlier than others, some people start and are already amazing, see Terrance Tao: "at the age of eight, Tao began to teach high school calculus at Garfield High School after attending calculus courses when he was only seven years old." - Wiki

user147690

@SohamChowdhury You are already way way ahead in terms of mathematical ability by age

user147690

I started caring about math when I was almost 19

I'm not comparing myself to him per se, it's more about taking classes at an university, which I doubt I would ever get an opportunity to. I've always wanted to (and I know it's very common in the US, for example), but it's AFAIK not possible over here. He's managed to do so somehow, and I have no idea how to, so that's the thing that makes me sad.
I stopped comparing myself to people who're better than me at something a long time back, but this is different.

Nevertheless, it's super kind of you, @AlexClark. Thanks. :)

Hey, can a group of order n contain elements of every order from 1 to n?

I know $S_3$ does.

9:35 AM

I second what Alex says, @Soham. I started mathematics from a young age (although I really knew nothing about a year ago, and still got lots of holes in my understanding -- I am studying some linear algebra right now since I never did it, for example), but being me has it's downside. I am hopelessly bad at my school studies, for one. Also, I am super-slow at grasping new concepts.

@BalarkaSen "hopelessly bad at my school studies" too

Can you answer my group question, please?

As for the uni, I'd'nt elaborate on how I got there, but I assure you I didn't ended up going there because I wanted to go there. Got lucky, I guess. You'll get plenty of time to study mathematics after you graduate, though :)

@Soham Good question. $S_n$ works, I guess.

Not sure.

@SohamChowdhury by lagrange's theorem, there would have to be no numbers less than n which are coprime to n

how many numbers like that do you know?

yikes, yes.

oh, group of order $n$. $|S_n| = n!$.

Ah. :)

@anon Factorials!

9:40 AM

nope

Then?

Primorials?

nope

I have no clue.

primorial doesn't work.

Gimme a hint.

9:41 AM

by the way S3 has order 6 and does not contain any element of order 5

So what's the answer?

the only such number is 2

there are no such groups except Z/2

n - 1 doesn't divide n for all n > 2

Okay, that makes sense, but S3?

Oh, order 6.

it has no elements of order 5, as anon already said

9:44 AM

So do symmetric groups of order n have elements of all orders from 1 to n?

symmetric group on n letters do

$S_n$?

yeah

ok

for each 1<=k<=n, the symmetric group S_n contains the obvious cycle on {1,...,k}

which has order k

9:47 AM

an interesting question you could ponder on would be : what's the largest k such that a given finite group G has elements of all orders between 1 to k. it essentially boils down to some number theory.

N3buchadnezzar

What is $\mathbb{T}$ in regard to complex analysis?

circle group most likely. possibly the torus.

context-dependent, @N3buchadnezzar

@anon Right.

@BalarkaSen I'll try that.

it might or might not be a hard question. i haven't thought about it.

i guess there is more to it than some number theory. i take back what i said.

N3buchadnezzar

9:49 AM

@BalarkaSen Look again :p

using the usual linear order on N is a bad way of organizing orders of group elements.

@BalarkaSen you will be so proud of me!

user147690

@AlecTeal Why should Balarka be proud of you?

That's between me and @BalarkaSen

user147690

@AlecTeal Sure

user147690

10:04 AM

@AlecTeal Maybe you want to tell us so we can be proud of you also?

Oh, I've been doing number theory stuff and I don't dislike it >>

user147690

@AlecTeal I am proud of you for giving it a shot!

Also it wasn't from a book

user147690

@AlecTeal What number theory stuff?

I needed a result for a project Alex, rather than just "trying it for a few values and hope" I proved it!

user147690

10:11 AM

@AlecTeal Have you written it up?(online for us to ponder)

Yes, just.

I'm happy with my proof though (which is also rare!) so ... :P

@AlexClark Do you find that latexing your answers to exercises helps?

user147690

@SohamChowdhury Yes I do - but only because I am very quick at typing them up

I know you do. I've always wanted to, but feared it would be a waste of time.

@SohamChowdhury it can be

user147690

10:12 AM

@SohamChowdhury It is unless you are doing proofs. Usually the time between writing down the proof far exceeds the time to type it. So it doesn't slow you down at all

@AlexClark Hmm

user147690

@SohamChowdhury Mainly I hate writing on paper since the page ends and I have to start on the new one. When I go to the new page I can't help but write the relevant information at the start... Then I end up losing pages or making a mess

user147690

That being said, I would still rather just use a whiteboard, but I am not often at home(where one sits directly at my PC)

@SohamChowdhury the proofs on maths.kisogo.com/index.php?title=Ring#Important_theorems that one (expand them) were BORING to write because they're so trivial, a lot of the todo tags are just "I cannot be bothered"

Haha

user147690

10:17 AM

@AlecTeal I know that feeling actually this one wayyyy more specifically

@AlexClark how much of the book have you done?

That looks like it was fun to type!

user147690

@SohamChowdhury Much more than what is up. But not much from those early chapters(since I had learned much of the stuff from other textbooks)

user147690

@AlecTeal Oh it was

You're doing rings, I guess?

Polynomial rings?

user147690

10:20 AM

@SohamChowdhury Yep and yep

@AlexClark I will win this "who has it the worst" battle (if winning is the right word :P) check out the top two proofs for maths.kisogo.com/index.php?title=Normal_subgroup I've only done half of one

I do them in so much detail to make them link-worthy for "newcomers"

"Simple proof"

It is a simple proof

Yes, but it looks funny on the page. :P

user147690

@SohamChowdhury That one is actually. But it is really nice the first time I looked at it

10:22 AM

I gotta go now. You guys keep comparing how bad you're off. :)

user147690

@SohamChowdhury Hahahaha

I think I won

Unless Alex is working on a wiki filled with definitions and theorems, in which case we ought to combine!

user147690

I dunno. I had to do the phi function for 30 values, which is just mindless plug and chug :P

user147690

@AlecTeal Haha but you are right. You probably have done this sort of thing 100 times

@AlexClark if you'd told me I would have written a program for you!

user147690

10:24 AM

@AlecTeal I'll check how long it took me in the revision history

maths.kisogo.com/index.php?title=Norm is a good page, see nice definitions,

/me checks

I don't know how

user147690

Oh it took me 13 min, what am I whinging about...

user147690

Well 13 min from phi(4) to finished

That's not to bad!

user147690

@AlecTeal Well I better get back to work now, talk later![the norm page is pretty nice btw]

10:27 AM

@AlexClark it was math.stackexchange.com/questions/1295186/… BTW

Also @AlexClark when you start posting solutions to textbooks and whatnot, tell me. I've started putting those on the wiki.

11:28 AM

@AlexClark, you there?

user147690

@SohamChowdhury Now I am

There's this problem I was struggling with:

Show $\Pi_{g\in G}g = f$ if $G$ is a finite group with a single element of order 2, called $f$.

I later discovered from the errata that the group is abelian. -_-

I still haven't solved it. (I was writing up the financial aid req for the summer program.)

What's the entire problem?

What I posted there, with the added info that $G$ is abelian.

Don't solve it for me, please.

11:43 AM

I'm not even sure it is true.

Hahaha

Must be.

What's the context?

Oh okay!

if $f$ is of order 2 what is the inverse of $f$

@SohamChowdhury

I think 2 as well? That's what's giving me pause.

No hints please, though.

....

No what is the inverse of $f$

Not the order of the inverse

Oh, shit.

$f$.

Haha, okay.

All other elements "cancel out".

11:46 AM

So if we have $efabcd...$

Yeah

You can take the elements of order 3,4,5,6,.... and move them around so they're in order - they do not CANCEL OUT as such, they just form $e$

That's what. So they line up as $aa^{-1}bb^{-1}\cdots f\cdots = f$.

It's a sort of cancelling out to me.

Oh okay, I was going with generators!

But yes, that's the gist, all the others are distinct from their inverses (except e)

It's not really a hint, but yes...

12:15 PM

@SohamChowdhury: Are you studying Group theory?

user147690

12:40 PM

@Balarka Are you here?

I am

user147690

@AlecTeal When is $(f)$ is prime iff $f$ is irreducable true?

user147690

The quotient ring k[x]/I is a field iff f is irreducible. k[x]/I is a field iff I is a maximal ideal. So (f) is maximal iff (f) is prime, in all unital rings maximal ideals are prime, but in what domain is the first statement true in?

user147690

Sorry I changed notation arbitrarily it was originally a facebook message I was about to send. Let me make it less terrible

user147690

In what domain(if any) is the statement $(f)$ is prime iff $f$ is irreducible true?

I know that the quotient ring $k[x]/(f)$ is a field iff $f$ is irreducible$ - I also know that $k[x]/(f)$ is a field iff $(f)$ is a maximal ideal.

So then from those two facts we obtain the statement $(f)$ is maximal iff $(f)$ is prime. I know that in all unital rings that maximal ideals are prime, but the converse isn't true in PIDs I believe

user147690

12:52 PM

-----------------------------spacebreak-----------------------

In what domain(if any) is the statement $(f)$ is prime iff $f$ is irreducible true?

I know that the quotient ring $k[x]/(f)$ is a field iff $f$ is irreducible - I also know that $k[x]/(f)$ is a field iff $(f)$ is a maximal ideal.

So then from those two facts we obtain the statement $(f)$ is maximal iff $(f)$ is prime. I know that in all unital rings that maximal ideals are prime, but the converse isn't true in PIDs I believe

user147690

Sorry didn't render latex until after editing was locked

user147690

@Paul could you comment on everything after the space break, everything above is redundant

Yah, I will look at it in a minute or two @AlexClark

user147690

@PaulPlummer Thanks, also where have you been? Something happen today?

Sleeping :D

Getting sort of "on track", and I only got a couple hours of sleep the night before so I went to bed early

user147690

12:56 PM

@PaulPlummer Oh haha, nice. Look at your chat graph, your chat day is starting where it normally goes towards ending

user147690

@AlecTeal Are you writing a wiki response haha?

@AlexClark ~~I wasn't~~ No, I am not.

user147690

@BalarkaSen Ahhh could you comment on the stuff under spacebreak above? Everything above spacebreak is redundant

yikes, what the hell.

oh, the text below spacebreak looks much cleaner.

user147690

@BalarkaSen Indeed lmao

1:03 PM

It is true for $F[x]$, where $F$ is a field, for example.

user147690

True

user147690

But it is true in UFD's?

Oh, you mean prime ideals.

I misread.

It's true whenever $R[x]$ is an integral domain.

user147690

Is it true in general for integral domains?

user147690

Someone on facebook is trying to clarify but I don't know either

1:05 PM

Yes, prime ideals of integral domains are irreducible.

user147690

Iff?

user147690

I can see one one way but not the other

huh, good point.

No

What @Paul says.

$R = \Bbb Z[\sqrt{-5}]$ is a counterexample.

That ring is pretty much a counterexample to anything :P

user147690

1:09 PM

What does $\Bbb Z[\sqrt{-5}]$ mean? It just means all polynomials with the same coefficeint $\sqrt{-5}$?

It's the ring of elements of the form $a + b \sqrt{-5}$ where $a, b \in \Bbb Z$

It is like $\Bbb Z [i]$ except not

user147690

Ok haha that was my other guess

user147690

why is the notation $\Bbb Z[x]$ different from expectation?

what do you mean by different/

user147690

1:10 PM

Well when it refers to polynomials I mean?

user147690

Why does $\Bbb R[x]$ mean all polynomials with real coefficients for example

$\Bbb Z[\sqrt{-5}]$ is the ring of polynomials in $\sqrt{-5}$ with integer coefficients.

user147690

But it is polynomials of order 2, I think I am missing something here

$(\sqrt{-5})^n + a_0 (\sqrt{-5})^{n-1} + \cdots + a_n$ reduces to an element of the form $p + q\sqrt{-5}$, note that.

would appreciate if you stop swearing everytime you mess up something :P

user147690

Would you actually?

1:13 PM

I sure would.

user147690

If so my bad lol

user147690

Is that abnormal where you live?

nah.

it's more than normal, on the other hand.

user147690

@BalarkaSen Actually that was the only instance of the f word from this account from a non-referencing use. I only have two s words to you even :P

thanks, but no thanks for that info.

:P

user147690

1:17 PM

Would you accept a formal apology?

no need to apologize :P it was just a passing comment/suggestion

user147690

I feel bad now though for some reason haha

user147690

So it was true in a UFD, but fails in an integral domain, and I don't yet know what an integrally closed domain is

integral closure of a domain is the ring of elements in your domain which are roots of monic polynomials with coefficients in the fraction field of your domain

and an integrally closed domain is something whose integral closure is itself

user147690

Okay to the second line. For the first line I don't fully understand, fraction field of my domain refers to a quotient ring correct, if so, what quotient ring?

1:26 PM

fraction field of an integral domain is just like constructing $\Bbb Q$ out of $\Bbb Z$

you take your domain $R$

and then construct $R \times R/\sim$ where $\sim$ identifies $(x_1, y_1)$ with $(x_2, y_2)$ if $x_1y_2 = y_1 x_2$ (verify that this is an equivalence rel)

you can verify that this is a field.

1:43 PM

@Huy I am, hope to finish the groups chapter this week. Next up is rings/modules. God, I'm excited.

Why are you excited?

Because I'm learning new things, that's why. As good a reason as any.

:)

Was really busy with non-math stuff till now. Back to work.