Mathematics - 2015-05-22 (page 4 of 5)

Soham Chowdhury

14:01

Oh, I think I can guess what you mean. Not really positive as much as non-seriously.
Anyway, I have to get back to my pdf (because there are few things better than "aha!" moments that come when you learn new things)!

BBL

Chris's sis

@SohamChowdhury It was a good thing, I mean you're positive!

Soham Chowdhury

I know.

Chris's sis

;)

Soham Chowdhury

Go do advanced analysis instead of making random strangers on the internet smile, lol. ;)

God, where is @BalarkaSen when you need him?

Chris's sis

hehehe, the truth is that I learned a lot of stuff, but these days I'm mainly focused on a certain type of problems. Maybe I should take some pictures with the books on my table right now.

Soham Chowdhury

14:04

Hey @iwriteonbananas

iwriteonbananas

whazzzzzappp

Soham Chowdhury

(super basic) group theory, finally. I got finished with the first chapter of my book.

iwriteonbananas

cooleo

now do exercises

Soham Chowdhury

I've done a few (~60%)

Balarka Sen

@SohamChowdhury nah.

iwriteonbananas

14:05

good

Soham Chowdhury

@BalarkaSen :(

Balarka Sen

hey @iwriteonbananas

iwriteonbananas

soham, you have summoned balarka

sup balarka

Balarka Sen

@SohamChowdhury i don't have a motivation to do universality.

i "know" about those, and that's about it.

Soham Chowdhury

@BalarkaSen I see.

Are you familiar with any functional programming language?

Balarka Sen

14:07

@iwriteonbananas no fun algebraic topology question on the mains.

no.

Chris's sis

@Hippalectryon I'm not evil but sometimes I get angry too fast (when not needed)! :D

iwriteonbananas

@BalarkaSen im afraid i dont have one right now either

Chris's sis

Back to my research.

Balarka Sen

that's a shame.

iwriteonbananas

in the lecture today we showed that $H_n(S^n) \cong R$ for any homology theory w/ values in $R$-modules though

pretty cool

induction argument

using mayer vietoris

Balarka Sen

14:08

right

Soham Chowdhury

Now you're speaking Balarka's language

iwriteonbananas

homology theory satisfying dimension axiom i meant

Balarka Sen

which reminds me, you told me you'd do $H_\bullet(CP^n)$, @iwriteonbananas

Soham Chowdhury

yo @TedShifrin

Ted Shifrin

hi bananas, @Balarka, @Soham

Balarka Sen

14:09

homology theory satisfies the dimension axioms : extraordinary homology theories don't.

hello @Ted

iwriteonbananas

@BalarkaSen you told me i should do it, i dont recall saying i would :D

i saw hatcher wrote something about it in the cellular homology seciton though

Balarka Sen

don't read it

Ted Shifrin

@SohamChowdhury We rarely need him :P

iwriteonbananas

lol

Balarka Sen

you just need 2.1.22 to do it

iwriteonbananas

14:11

and example 0..6

Balarka Sen

heh

Soham Chowdhury

@TedShifrin Haha

@iwriteonbananas grad student?

iwriteonbananas

@SohamChowdhury no

are you a undergrad student?

Soham Chowdhury

HS

Chris's sis

@SohamChowdhury What is your best to compute the elementary series (without pen and paper) $$\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n}$$?

Ted Shifrin

14:13

glares at bananas for the omission of greeting

2

Soham Chowdhury

Don't try, really
I suck at these things, bad

iwriteonbananas

sorry ted

<3

@Chris'ssis log(2)

Chris's sis

@iwriteonbananas Yes.

@SohamChowdhury OK

Soham Chowdhury

I knew the answer. No clue how to do it.

Except by somehow transforming it into an integral.

Balarka Sen

taylor

Chris's sis

14:14

@SohamChowdhury too advanced

@BalarkaSen too advanced

Balarka Sen

i don't care

Soham Chowdhury

Taylor how?

Balarka Sen

expand log(1 + x)

Soham Chowdhury

I don't even remember what $log(1+x)$ looks like

Ted Shifrin

But you need Abel's Theorem to know you can plug in $x=1$, @Balarka.

Balarka Sen

14:16

grunts

Soham Chowdhury

@Chris'ssis Ask WolframAlpha.

Easy peasy

Chris's sis

@SohamChowdhury Did you ever hear of Catalan's identity? Sometimes called Botez-Catalan identity?

Soham Chowdhury

I think so.

iwriteonbananas

@Chris'ssis too advanced.

Chris's sis

@SohamChowdhury Use that.

Soham Chowdhury

14:18

Can't remember what it is. Something to do with the generating function of $C_n$?

Balarka Sen

lol @iwriteonbananas

Chris's sis

@iwriteonbananas The identity is taught in middle school btw.

iwriteonbananas

@BalarkaSen so $H_n(\Bbb{C}P^{2n})$ is free on $2n$ generators?

Balarka Sen

no.

go use exercise 2.1.22

Soham Chowdhury

@Chris'ssis What a cool middle school.

iwriteonbananas

14:19

ok hold on

@Chris'ssis no

Chris's sis

@iwriteonbananas Yes, it is!

;)

iwriteonbananas

i guess maybe in romania

Chris's sis

@iwriteonbananas Yeah (I learned about it).

iwriteonbananas

cool

Soham Chowdhury

Okay, the Fibonacci numbers thing?

Chris's sis

14:22

I don't remember if I learned about it in the middle school.

iwriteonbananas

@BalarkaSen okay. $H_i(\Bbb{C}P^n) = 0$ for $i> n$. $H_n(\Bbb{C}P^n)$ is $\Bbb{Z}$ if $n$ is even, else it's $0$.

Balarka Sen

and how did you prove it?

iwriteonbananas

exercise 22

Balarka Sen

ok, but how did you apply it?

iwriteonbananas

using the cell structure on $\Bbb{C}P^n$

Balarka Sen

14:25

be explicit, please.

iwriteonbananas

$\Bbb{C}P^n$ has one k-cell for each even $k\leq n$

and no k-cell for odd k

Balarka Sen

so what? i am not asking you to do it with cellular homology, note that. you need to use the long exact sequence somewhere.

iwriteonbananas

im only using the cell structure of $\Bbb{C}P^n$ and exercise 22

Balarka Sen

i'm not following your logic. how does having a single k-cell for even k and no cells for odd n imply that k-th homology is Z and 0, respectively?

in particular, i am not sure how you used exercise 22

iwriteonbananas

$H_n(\Bbb{C}P^n)$ is free w/ basis in bijective correspondence to the n-cells if there are no cells of dimension n-1 or n+1.

note that i havent computed the k-th homology for $k<n$ yet

Balarka Sen

14:30

ohh, i didn't see that.

sorry, sorry.

iwriteonbananas

np

so, long exact sequence for the rest?

Balarka Sen

mhm

most of exercise 22 is useless : what's important is the observation $X^n/X^{n-1} \cong \vee S^n$

iwriteonbananas

yeah

i just used that

Balarka Sen

great job, what does that give you?

iwriteonbananas

hold on

what's the kernel of the boundary map $H_n(X/X^{n-1}) \to H_{n-1}(X^{n-1})$?

$H_{n-1}(X) \cong \Bbb{Z}/im \partial$

hmmm

Balarka Sen

14:42

well, the snake map is the whole hard part. i don't recall, i guess the snake map is zero.

iwriteonbananas

no wait

Balarka Sen

oh, no, it's not

you don't need to know about the snake map

iwriteonbananas

i think it's 0

Vrouvrou

i.sstatic.net/KyF2o.png @TedShifrin i found this

Balarka Sen

give me a second

Vrouvrou

14:42

@TedShifrin what about $||u||_{L^p}+||\nabla u||_{L^p}$

iwriteonbananas

is $H_{n-1}(X) \cong \Bbb{Z}$?

Vrouvrou

@TedShifrin is there a relation between the rwo norms please ?

iwriteonbananas

in fact, is $H_i(X) \cong \Bbb{Z}$ for $i\leq n$?

Balarka Sen

right, i don't think you need to know about the snake map. note that $H_k(\Bbb CP^n/\Bbb CP^{n-1}) = 0$ for $k \neq 2n$

Ted Shifrin

Yes, there should be. The norms should be equivalent, @Vrouvrou. This is one of those standard inequality exercises using convexity.

Balarka Sen

14:46

no, that's definitely not right, @iwriteonbananas

N3buchadnezzar

@BalarkaSen hard

iwriteonbananas

btw. ive been treating the case that $n$ is even this whole time, i forgot to say

Balarka Sen

i'm confused with your notations, then.

what is hard, @N3buchadnezzar?

N3buchadnezzar

complex

iwriteonbananas

oh i made a mistake

Vrouvrou

14:48

oh ok and where i can found the proof ? @TedShifrin

Balarka Sen

i guess, @N3buchadnezzar. i've never seriously studied analysis.

iwriteonbananas

is $H_{n-1}(X) = 0$?

Balarka Sen

what the hell is $X$?

iwriteonbananas

lol sry

$X = \Bbb{C}P^n$ for some even $n$

Balarka Sen

you don't need to partition $n$ into even and odd cases

iwriteonbananas

14:49

true

Balarka Sen

anyway, yes, $H_{n-1}(\Bbb CP^n)$ is $0$.

iwriteonbananas

ok

Balarka Sen

super-false

iwriteonbananas

god damn it, i was working with the mistake i made earlier

$H_{n-2}(\Bbb{C}P^n) \cong \Bbb{Z}$

Balarka Sen

indeed.

if $n$ is even, that is

iwriteonbananas

14:52

yeah

gotta run, bbl

Balarka Sen

bye

Soham Chowdhury

@BalarkaSen are you guys doing Hatcher?

Hey @MikeMiller

Balarka Sen

bananas is doing hatcher, yes.

morning, mike.

Soham Chowdhury

And you are?

Balarka Sen

not right now

Soham Chowdhury

14:56

Mm.

Mike Miller

morning

Soham Chowdhury

Free groups are cool.

Balarka Sen

they sure are

Soham Chowdhury

Finally!

We agree on coolness at last.

:P

Balarka Sen

well, they are. free groups are complicated groups.

as an example, can you find out which groups appear as subgroups of free groups?

Ted Shifrin

14:59

Goodnight @Mike

Tobias Kildetoft

@BalarkaSen Well, for the rank one case, it is easy (not that the answer is complicated in the other cases, just harder to prove).

Balarka Sen

sure.

N3buchadnezzar

@TedShifrin heya what is a bidisk?

Ted Shifrin

The product of the unit disk with itself in $\Bbb C^2$.

N3buchadnezzar

aight, thanks

Ted Shifrin

15:04

sure thing

wonderful result from elementary several complex variables, @N3B: not holomorphically equivalent to the unit ball in $\Bbb C^2$ !!

N3buchadnezzar

Working on some Runge domains, and complex convexity

Looking particularly at a result from Wermer

Ted Shifrin

ah, cool

N3buchadnezzar

"There is a bounded domain in C^2 which is analytic equivalent (holomorphic?) to the bidisk, but which is not a Runge domain"

The result above pops up as in regard to the following lemma "If f is a function that is continuous on the closed unit disc in C and holomorphic on the interior of the disc, then the graph of f in C^2 is polynomically convex"
Then the author says that this does not neccecarily hold for biholomorphic images of polydiscs.

Not quite sure why they bring up biholomorphic functions...

Mats Granvik

Saved by an epsilon!

Don Larynx

Hey guys

If you're learning linear algebra, check out my app play.google.com/store/apps/details?id=com.joc.matrixfree its meant to serve as a double-checker

Mats Granvik

15:16

Epsilon is no longer needed, works fine by removing the equal sign from the inequality.

Mike Miller

morning to you too @TedShifrin

Soham Chowdhury

15:29

@BalarkaSen I'll try. Right now I'm just skimming the groups chapter. (There's another, after rings/modules)

N3buchadnezzar

@MatsGranvik You know complex analysis right?

Abdou Abdou

it helps gettin ur right path in night too

Mats Granvik

@N3buchadnezzar Not really.

N3buchadnezzar

=/

Soham Chowdhury

@BalarkaSen: I always found this very illustrative of how I work. Don't know about other people.

As in: I prefer the second one.

Mats Granvik

15:30

@N3buchadnezzar I have problem with the definition that a complex number plus infinity is infinity. It does not make sense to me.

N3buchadnezzar

@MatsGranvik The rieman sphere <3

Abdou Abdou

thats DWT compression rate @SohamChowdhury

Ramanewbie

Anybody knows about programming language python ?
Or knows where I can get live help ?

Soham Chowdhury

@Ramanewbie Find one of the SO chatrooms

Ramanewbie

@SohamChowdhury What's SO ?

Abdou Abdou

15:38

there s lot of people ready to help here codereview.stackexchange.com

or here codegolf.stackexchange.com

Soham Chowdhury

Wow. Automatic image-linking.

Ramanewbie

@SohamChowdhury What's the difference between SO and Stack Exchange ?

Soham Chowdhury

Stack Exchange is a bunch of question-answer sites

SO is one of them

Abdou Abdou

SO is the eldest

Soham Chowdhury

Oldest*

Sorry. It's sort of reflexive.

Ramanewbie

15:40

@SohamChowdhury I searched already, no room about Python on SE

@AbdouAbdou Thanks for the links

Abdou Abdou

nvm

Ramanewbie

@AbdouAbdou ?? Why nevermind ?

Balarka Sen

@SohamChowdhury Don't. It's a nontrivially hard problem.

Ramanewbie

How to access the chat for codereview.stackexchange.com ?

Abdou Abdou

i said no mention it

Soham Chowdhury

15:43

@BalarkaSen mhm, ok

Balarka Sen

@Soham Fact : Every subgroup of a free group is free.

Abdou Abdou

N3buchadnezzar

Any idea how $K = \{ (z,\bar{z}) \mid \, |\text{Re}\,z<|\leq 1\,,|\text{Im}\,z|<\leq 1 \}$ Looks?

Soham Chowdhury

^ Imaginary

Abdou Abdou

SirPython's Mansion

A place for SirPython to test out his servant, SirAlfred

Balarka Sen

15:45

This is called Nielson-Schreier theorem. It takes pages of algebra to prove it using algebraic machinery, but there is a nice, short proof using covering spaces.

Soham Chowdhury

@BalarkaSen I see.

Abdou Abdou

there s lot of people wellversed about python in codegolf

go to their chat , and keep ur questions as significative and brief as possible

there s nt integer division tag in mah.se why ?

Soham Chowdhury

Test $f^{-1}$

Abdou Abdou

there s lot of helpful leads here ..... yay

math.stackexchange.com/questions/tagged/floor-function

iwriteonbananas

16:11

@BalarkaSen do you have a cool problem for me?

Karim Mansour

morning guys

I have a question can you have xy - yx = 1 where x,y $\in$ B(H) ?

is it possible to have this group presentation?

@BalarkaSen?

Mike Miller

I can't parse the question. What is H? What is B(H)? What group presentation are you looking for?

Karim Mansour

B(H) is bounded linear operators of Hilbert space

I mean does such a group exists ?

Mike Miller

Are you asking if you can find bounded operators $X,Y$ such that $XY-YX=I$?

Karim Mansour

yeah

Mike Miller

16:17

It's certainly not possible on a finite dimensional vector space. I would sort of be surprised if it was possible.

Karim Mansour

yeah I thought so aswell just wanted to double check with someone too

Mike Miller

Nevermind, that's not true. Every group can be written as a group of operators on some Hilbert space.

This is essentially the Gelfand-Naimark theorem, along with knowing the construction of a group C*-algebra.

Ted Shifrin

The Heisenberg Uncertainty principle is basically a commutator equation like that, @Mike @Karim

Chris's sis

Ah, not really ...

;)

Karim Mansour

oh I see

yeah I heard about gelfand naimark theorem

anon

16:20

@SohamChowdhury that is in fact the reason I began calling it the starboard

Ted Shifrin

I have to decide about boundedness, however.

Karim Mansour

I see thats interesting @TedShifrin

Mike Miller

There's probably a more human way to actually write down a group of operators, though.

Ted Shifrin

$X = d/dx$, $Y=\text{multiplication operator}$.

Mike Miller

@Ted: Bounded on what Hilbert space, precisely?

anon

16:22

@MikeMiller I think it's also possible if the characteristic divides the dimension.

Ted Shifrin

Yes, @anon.

Karim Mansour

1 moment I will have to go get milk fast and come back

brb

Ted Shifrin

That's the issue, @Mike. If we do $L^2$, then differentiation is unbounded.

Mike Miller

@anon yeah, but i was working over $\Bbb R$ or $\Bbb C$, since we said Hilbert space... :)

@Ted: The construction outlined above spits out groups of bounded operators, if you're worried about there existing such a thing.

Ted Shifrin

But if I do a Sobolev space, presumably they'll be bounded.

Mike Miller

16:24

@Ted: I think you have to decrease the index to get a bounded operator.

Ted Shifrin

Hmm, yeah.

Mike Miller

If you do, then the norm of the operator is 1.

SMF

Hi there, sorry if I'm interrupting any conversations...I just wanting to know if anybody had insight for a question I asked that has received barely any attention (no answers or comments). Here's the question: http://math.stackexchange.com/questions/1291315/how-to-visualize-mathbbc2
Thanks

Soham Chowdhury

@SMF: Someone is surely going to point you to something a certain bloke called von Neumann said.

@anon :)

@SMF It's a nice question, though.

I'd like some sort of answer myself, similar to the question on visualizing determinants.

SMF

Hmmm...but I can only get used to / remember concepts that I at least partially understand!

Anyway, I have to go, but hopefully somebody spots and can answer my question without me having to but a bounty on it to get more attention...

Balarka Sen

16:38

@iwriteonbananas i'll have to think.

Mike Miller

@TedShifrin This is a nice question. I'm going to write an answer when I get to my office.

Karim Mansour

back

Soham Chowdhury

with or without milk?

that is the question

Karim Mansour

operator algebra is actually very interesting field

Mike Miller

oh yeah, absolutely

Abdou Abdou

17:08

i was wondering if there s some commutative associative group (abelian) (G,°) where ° isnt addition,multiplication , or any more that 2 degree equation like x°y=x²+y²

i think there s not

but any proof available ?

Karim Mansour

you can always construct a table that is a group and is commutative

and the table determines how elements are constructed

that is trivial one

Abdou Abdou

table of possible elements inside G ?

Karim Mansour

yeah

Abdou Abdou

it should require lot of work

i prefer bruteforcing it

it works sometimes

Karim Mansour

I mean cayley table for example suppose G = {a,b,c,d} then define ab = d etc

and you can construct it in such a way

such your group is commutative

that is one simple way of doing this

Abdou Abdou

17:13

ok thx

any relative links in math.se ?

Karim Mansour

Cayley table

A Cayley table, after the 19th century British mathematician Arthur Cayley, describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplication table. Many properties of a group — such as whether or not it is abelian, which elements are inverses of which elements, and the size and contents of the group's center — can be discovered from its Cayley table. A simple example of a Cayley table is the one for the group {1, −1} under ordinary multiplication: == History == Cayley tables were fir...

Abdou Abdou

no i meant answered questions about it

thanks for wiki article

Karim Mansour

I think you could also define some group operation explicilty too such your group operation is commutative

well you could ask on mse I don't know if there is question probably.

Abdou Abdou

thanks

Paul Plummer

What do you mean by isn't addition? @AbdouAbdou

Abdou Abdou

17:23

i meant not naive x+y operation

Karim Mansour

like he doesn't want group operation to be integer addition

can you allow modulo addition ?

Abdou Abdou

a%b != b%a

im thinking of xoring

Karim Mansour

because ($Z_n$,+) also works

Abdou Abdou

wait ... is xor associative ?

Karim Mansour

I guess this is related to cryptography somehow ?

xor what it does it flips the binary from 1 to 0 right ?

Abdou Abdou

17:25

yes im workin on cryptographic project

yes

and vice versa

Karim Mansour

let us denote xor by U

Paul Plummer

xor is basically addition over $\Bbb Z/ 2 \Bbb Z$, or vector spaces over that field

Karim Mansour

yeah

and you can check that by its truth table you will see its cyclic of order 2 so it must be isomorphic to Z/2Z

how do you know that Z @PaulPlummer ?

Paul Plummer

All finitely generated abelian groups (this is the fundamental theorem of finitely generated abelian groups) Are direct sums of of modular groups and the integers, so depending on what you mean by addition, all finitely generated ableian groups are addition over modular groups and $\Bbb Z$. @AbdouAbdou

@KarimMansour Huh?

Karim Mansour

the integer Z

how do you write it in latex

Paul Plummer

17:28

\Bbb Z, or \mathbb Z

Abdou Abdou

in fact its not addition in proper sens ; its a+b-ab

Chris's sis

"Even geniuses have to push themselves very hard. Maybe they are geniuses because they do so." - Donald Knuth

anon

@AbdouAbdou xor is a+b, regular or is a+b-ab

(mod 2)

Abdou Abdou

thanks @PaulPlummer and @anon

and @KarimMansour

anon

@PaulPlummer modular groups sounds a lot more advanced than integers mod n

Paul Plummer

17:32

When you start looking at things that are not finitely generated things get a lot more complicated, sadly I don't know much about infinite abelian group theory

Chris's sis

Donald Knuth - "I was scared and wanted to prove myself"

►

@r9m ^^^

anon

indeed the modular group is ${\rm PSL}_2(\Bbb Z)$

Paul Plummer

@anon Lol, yah when I reread it I thought I should have said quotients of $\mathbb Z$

anon

just say cyclic

Paul Plummer

Well I wanted to stress the connection to "arithmetic", since he was asking about whether or not there are abelian group that are not addition (which I am still not sure what @AbdouAbdou means by that completely) @anon

Abdou Abdou

17:36

well i thought each commutative group is abelian

Paul Plummer

That is true

since they are the same thing

anon

Groups of units in domains are written multiplicatively. For instance the positive reals with multiplication form a group, and we don't use the addition symbol to denote the group operation there.

Paul Plummer

But I don't really know what you mean by the operation not being addition

Abdou Abdou

ok lemme diggin out an example

Paul Plummer

Huh "The additive group of a ring is an abelian group, but not all abelian groups are additive groups of rings (with nontrivial multiplication)..."-wikipedia. That is interesting

Too bad there are no sources to go along to that section

anon

17:40

by using something called "transport of structure," you can define the operation $a\circ b:=a+b+ab$ on the set of numbers strictly greater than $-1$ - that seems related to what you were talking about a second ago @AbdouAbdou

Mike Miller

@PaulPlummer: Try to make $\Bbb Q/\Bbb Z$ into a ring.

Paul Plummer

I think, in a way, that means there are abelian groups, with no sane way to interpret them as "addition" groups, since one would expect for addition groups to form a non trivial ring.

@MikeMiller Ah, yes

Willl definitely have to work out that example in detail

Mike Miller

@TedShifrin: I gave a partial answer to that question I linked. I couldn't give a full answer :(

Soham Chowdhury

I am tired of every book starting with set theory

Ted Shifrin

@MikeM: My immediate reaction was to ponder the Thom-Pontryagin construction and think about framed submanifolds of $S^n$.

Paul Plummer

17:50

Just skip the section, if you know the important stuff in there @SohamChowdhury

Ted Shifrin

@SohamChowdhury Interesting. I've written four books and in none of them did I do that :P

Mike Miller

Mine too, @Ted. But is there a good reason to believe that my $M$ is parallelizable?

Ted Shifrin

Why would you need that (and what's $M$?).

We're just looking at framed codimension-2 submanifolds.

The framing is of the normal bundle.

Mike Miller

Silly. Yes.

Ted Shifrin

But, as you point out, the submersion gives you the extra structure of a fiber bundle.

Mike Miller

17:53

Yes, trivial normal bundle comes from being a fiber bundle.

Ted Shifrin

Well, the Pontryagin construction gives you one regardless.

Mike Miller

@TedShifrin: I have to admit, I don't really understand what PT does in the unstable range. Let me think about that.

Ted Shifrin

That was something that I got led through by Rob Kirby on my geometry/topology oral. I actually didn't know it. And then I taught it in my undergraduate diff top class one time I taught the class. But I haven't thought about it in 35 years now.

Mike Miller

I learned about it recently but forgot...

Ted Shifrin

Well, if you get somewhere, let me know.

Mike Miller

17:59

I'd really like to do this problem without hunting through a classification of 4-manifolds. :P

SirPython's Mansion

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