Hmm

I think it's coming back a little bit

Say $f : \Bbb C \to \Bbb C$ is holomorphic and $U$ is open in the domain; pick $z \in U$ and look at $w \in f(U)$. I want a small disk around $w$ that's contained in $f(U)$.

True that

So, uh, for any $v$ near $w$ let $F(u) = f(u) - v$. Notice that $F(u) = (f(u) - w) + (w- v)$; I think the idea is if the $|w - v|$ bit is small enough then $f(u) - w$ would have the same number of zeroes as $F(u)$, but then $f(u) - w$ would have at least one since so does $F$ (dude, $F(z) = 0$)

That'd say for all $v \in f(U)$ with $|w - v| < \epsilon$, $v$ is in the image of $f$.

I think that bit is accomplished by Rouch\'e

Working with the linear homotopy $\{(f(u) - w) + t(w - v)\}_{t \in [0, 1]}$

@Fargle: Do not abuse proofs by contradiction!

@TedShifrin :(

But hi :)

Hi @Ted

Hi @Balarka

You can prove a stronger statement than open mapping. You can prove that if $f-w_0$ has a zero of order $k$ at $z_0$, then near $z_0$ it takes on every value $k$ times.

hello handsome ppl

@TedShifrin Hello :D

Of course, I personally think this is easier just from the power series expansion, but I do love Rouché.

hi @Kasmir

@TedShifrin I got something to ask ._. feel free to say no

@TedShifrin Right, I think the proof I remember the best is to understand that upto change of coordinates $f$ locally looks like $z \mapsto z^k$. That's a branched covering, hence open map. Your stronger statement is a corollary of this, of course.

Yes, but Rouché will prove the statement quite easily.

@Kasmir: Since when do you sit on ceremony?

doesnt roché has that theorem about finding roots?

More or less, yes.

@TedShifrin sit on ceremony ? ><

dont know what that mean

The English phrase is "stand on ceremony." :) I modified it for you.

Rouché is black magic of a very high level

still dont get it :D

Google it

and why do you need to modify it for me

@Alessandro It's topologically very apparent

Nah, Alessandro, there's a nice generalization of it and the argument principle in the differential topology setting.

cuz am black ?

we proved this fact using Rouché in my complex analysis class, it was pretty neat

LOL, @Kasmir. I had absolutely no idea.

@BalarkaSen what do you mean?

I modified it for humor.

This obviously worked well. If I can't teach you math, I try to teach you humor. :P

haha

am not black but just like saying that

Its funny because many black ppl say that if they feel rejected by a girl =p

You're in Sweden, you told me. So what is your ethnic origin?

or by anything , am not rasist ofc ,just find that perticual thing funny in movies

french and tunisian

that is why I can speak french =p

and kasmir khan is not my real name

Ah, very cool. And you landed in Sweden. Très intéressant.

Well, I knew that much.

oui oui =p

Anyhow, so what is your question of the moment?

And you should google "stand on ceremony." It's a nice phrase in English.

well its not a question really , I got 5 exercices that iv done

i want you to look at them and say if they are ok or not

I mean it's not that magical after you see the argument principle

ill figure out the "not" alone

oh i got it now :D

Ted did not expect me to be so polite?

grrrrrr

No, I'm not going to grade your homework for you. That truly is beyond what I'm willing to do on MSE.

@TedShifrin So may I email it to you or you busy ?

If you have a focused question, of course, I'll try to help.

Okay =p

@Alessandro The argument principle says the winding number of a curve is the number of zeroes of the function. Rouch\'e just says that winding number is a homotopy-invariant. (The homotopy of the corresponding functions needs to be carefully done correspondingly)

this homework is graded

It was bad enough grading topology assignments when I voluntarily taught a topology course to 2 of my former students after I retired.

I get from it 0.5 points

I understand that :D

Ted once gave me an exercise which proves a version of the argument principle in the smooth category.

I wanted to try my chances >< am happy either way

@Balarka: I think that is not a correct representation of Rouché.

Yes, and you never did it!

Which university are you reading at, @Kasmir?

@Ted False. I did and sent it to you :P

It's needed for Hopf Degree Theorem, among others.

*studying

Oh, well, in that case I'm old and forgetful.

I quite liked the exercise.

@AndreasAlmgren Stockholm university

I see @Andreas is also a denizen of Sweden.

@AndreasAlmgren är du svensk ? =p

Ja

haha najs vad pluggar du?

Ted snipped me again with that comment

snipe ≠ snip :P

Inget ännu, läser sista året på naturvetenskap. Funderar starkt på ren matematik. Hur är undervisningen på Stockholms universitet, om du känner för att berätta? @Ted, yes

Lots of Swedish people invading the room. Maybe not the best time to post Swedish curses.

Probably not, Balarka. Stick to Bengali.

Someone is having fun downvoting me ... sort of at random. In one case, it was merited because the OP completely changed the post after reading my answer ... and then my answer was no longer appropriate.

Why care?

Reasonable question.

But I can understand your feelings on one side.

I have been the victim of a vendetta in the past, but I doubt that's it this time.

xD

@AndreasAlmgren helt generellt är det bra, fast man får inte mycket under föresningarna , man ska plugga mycket själv =p

Ok, tack!

I recently encountered someone whom I know well from answers in geometry. I have a feeling he and I disagree on style of appropriate answers (like he writes very complete ones and I often give hints). This system isn't perfect.

@AndreasAlmgren Lund och SU är bäst inom matte ( tror jag )

Np :)

So, @Andreas, what level of math person are you? :)

Ultimate noob

Ah ... is there such a thing? :P

atm at least

Kasmir used to be one of those. :)

haha =p

I haven't finished high school yet

Ah, cool. I'm currently teaching some high school kids.

@Balarka: This is quite neat.

You like it?

Andreas I recomand that you take "förberedande kurs i matematik"

Why?

Well, I spent 40 years or so teaching university, so it's different. But so far I think it'll be fun. It's a special school mostly for math geeks.

it is a good course before learning higher math

because they wont do anything basic from the start

Kasmir doesn't know what language(s) to speak in now.

15 lectures then you take an exam

well

I figured its not nice for you

not understanding what we saying

Well, I speak French to the French folks and it's not nice to others. :P

it took me a while to get that , but I got there

Well Ted because you are very nice guy also , that too

Well, that sounds like something @TedShifrin. @KasmirKhaan I will look it up

So, if you're here, @Andreas, presumably you're significantly interested in mathematics?

@TedShifrin Ah, the deRham cohomology hint helps.

That is a nice fact

We have a number of high school students in this chat ... Balarka, Akiva, MeowMix, Heather (is she high school yet? I think so) ...

yeah she's in high school now

i think

Yes I certainly am! It's one of the few things in school that is not boring.

My high school kids pretty much all claim they're bored in school. Makes sense, given that they're doing more advanced stuff outside of school. At least one of them may be bored even in this more advanced course; but I'm prepared to throw him harder stuff.

@Andreas: Here's a (not unusual) question I like to give. Maybe you know it; maybe you don't. If I choose $n+1$ integers from $1$ to $2n$, must it be that one of them divides another?

@TedShifrin you gonna be here for a while ? =p am planning on doing some exercices and it would be good if i could ask you about what i dont get =p

I'll be sorta around for a few more hours, then disappearing.

okay :D dirichet is not something that they teach in sweden =p

at least not in highschool

Dirichlet what?

In my case I find school boring in large part because I don't like most of the things I am forced to read. I think that I should intuitively know the answer to your question to be able to answer it - the way you are asking - but I do not. I can tell you one thing and that is that I am certainly not one of the geeks you are teaching. My math interest is quite recent.

@TedShifrin pigeonhole principle i think its called too

LOL ... I like the question because there are different ways to do it, and because, like a lot of basic number questions, you can experiment and try lots of examples to develop intuition.

Oh, @Kasmir, that's only one of the ways to do it.

thats the first that come to mind for me =p

It wasn't for me. :)

if i see that kind of Q's

haha =p

hey guys, can somebody help me out please

@Andreas: I don't need to give you questions. I'll shut up :)

That's a bit vague, @Sylent.

We all have to begin somewhere.

i need to get from $2(cos(x)^2 - sin(y)^2) - (cos(y)^2 - sin(x)^2)$ to ... $0.5*(cos(2x) + cos(2y))$

I never liked the number theory type stuff that so many young people love. I developed a respect for elementary number theory for teaching purposes at the undergraduate level.

If somebody can point me in a direction that'd be greatly appreciated

im a bit stuck

Have you started with the double angle formula, Sylent?

Just started

What does it say?

erm, ive deviated from the question quite a lot

purely because im exploring something

BTW, I don't believe it's correct.

wolframalpha says its true

Let's try $x=0$ and $y=0$.

What's $2\cos^2x + \sin^2 x$ when $x=0$?

the original question, prove the identity $cos(x-y)cos(x+y) = cos(y)^2 - sin(x)^2$

Aha ....

but i substituted using the identity $sin^2 + cos^2 = 1$

You messed up somewhere.

and got to basically an identity that goes 2=2=2

so I thought, lets use the second two 2's to get the first 2

if that makes sense

its a mini adventure tbh

albeit not actually answering the question

So this question is correct. :)

You need to use the formula for $\cos(x-y)$ and for $\cos(x+y)$. Did you do that?

i can answer the question if i really wanted to

i just wanna complete my mini adventure

Well, did you try $x=y=0$ as I asked you to in your "mini-adventure" formula?

hold on

Hey @Ted!

sub x = y = 0, got 1

cant be righjt

right*

Hi Demonark.

what are the guidelines to posting links?

@TedShifrin I wonder if there's a messy hands on way to do that problem by triangulating $M$ and homotoping $f : N \to M$ to inclusion of a subcomplex of dimension less than $\ell$. If $df$ has rank less than $\ell$, locally it's like the inclusion $\Bbb R^{< \ell} \subset \Bbb R^n$ after all.

Hmm, how do you get from local to global, Balarka, but, yeah, probably.

I mean you can produce triangulations on $M$ and $N$ and homotopy $f$ to be a simplicial map by simplicial approximation theorem.

@TedShifrin ^

It then remains to show that actually maps to a subcomplex of dimension less than $\ell$, which is not entirely obvious but I am definite that has to be true.

@Sylent: So your original thing simplifies easily to $\cos^2 x - \sin^2 y$, and I highly doubt that is $\frac12(\cos 2x + \cos 2y)$, but let me think.

Check the links, both equations simplify to that, I just dont see how

It's still amazing how it takes a one-line argument using forms. increase in RESPECC

Well, let's see. $\cos 2x = 2\cos^2 x - 1$ and $\cos 2y = 1-2\sin^2y$. Does that give it?

Yup, it does.

You were right. I was wrong.

OOOHHHH

I completely forgot that identity

D'oh

I asked you about double angle formulas :P

@Balarka: It's about damn time!

I was completely over-looking those ones XD, thanks man

Sure.

@Balarka finally you've joined the dark side

Nifty.

tfw you have to watch a 30 second ad before watching a 20 second meme

nice

@Daminark I have sided the dark joint, rather

@SimplyBeautifulArt Have you thought of what to teach tomorrow?

@Balarka: It's actually not so astonishing that a differential topological condition plays nicely with forms.

Hi @Jasper.

@Jasper Idk, I kinda have to see where people are at

@TedShifrin Hello, I should take some cooking lessons lol.

Ted you should write "Differential forms: A homotopic approach"

I'm done writing, but thanks.

do Carmo has a book Differential Forms, not well known.

:P

He's not a natural differential forms user, so it's weird.

Yeah I've heard, I intend to look at that at some point

Henri Cartan has a very nice book on forms.

Differential forms using model categories

That's more like Griffiths/Friedlander/Morgan (rational homotopy theory).

Ah yes I have heard of that

@TedShifrin That book is actually volume 2 of a course. I can't find volume 1 anywhere.

I still have a handwritten version of that book.

Maybe he wrote the forms book to learn about them?

Hm, I wonder what nlab has to say about differential forms

I never saw another volume, Jasper. I owned Cartan's book when I still had books. What was in volume 1?

Demonark, he had the same adviser I did, so I'm sure he knew about them. Just never naturally used them in his books or research, that I'm aware of.

@TedShifrin It's like differential calculus. I think what was translated into English got out of print.

@TedShifrin What is your opinion on this

puts Balarka on permanent ignore

When we ignore someone, we still can see the person's icon in small.

And when we navigate the rooms, we still see the icon in big.

There should be a way to make it totally unseen.

@Ted I didn't write that stuff!

You asked my opinion?

Also Urs Schreiber has discussed extensively below that statement in a small wikibox

@BalarkaSen Is he a famous mathematician?

He's the founder of nlab

I have never visited nlab and don't intend to.

@TedShifrin a is equv to be mod H , where H is a subgroup of G

@TedShifrin can you clearify what this is saying ?

h in H , a= bh

Ted: Do you recommend Cartan over DoCarmo for forms?

@Daminark By the way, I was just listing the books. I didn't recommend anything for forms, lol.

I don't remember details, Demonark. In general, I recommend a book that has true content and not just formal stuff ... so differential geometry with forms would be good.

@Kasmir, $a\equiv b\pmod H$?

@TedShifrin yes

I thought that Demonark was DogAteMy until I realised that DogAteMy is Akiva, lol.

@TedShifrin am doing cosets and quotient groups =p now I understand the notation for modulos, Z/nZ make sense now :D

Z/nZ = n

Typically, @Kasmir, that should mean $a^{-1}b\in H$, so, yeah, $b=ah$ for some $h\in H$. Notice you can solve for $a=bh^{-1}$.

the idea of coset

we take a subgroup H

the elements of the coset are those of the from bh right?

I dont really get the process

A coset is just $aH = \{ah: h\in H\}$ for some $a\in G$.

do we fix one element of G

I think doCarmo does diffgeo of curves/surfaces in that book using forms. I'll check out Cartan and let you know

@TedShifrin so we do fix one element a in G right ?

Also @Balarka lol at nlab stuff

Read my sentence, Kasmir.

@TedShifrin we multiply a by elements of H , either on the left or the right and we get the coset

morning ted

Tes Dhifrin @Faust

Well, you work with only one side, @Kasmir, most of the time.

glares at Demonark

@Faust sup faust7

there the same if H is normal

Lol morning everyone

Yup, normality is equivalent to every right coset's being a left coset and vice versa.

okay but what is not clear to me

do we have to fix one element of G

or we do this for all elements?

like fix a , aH is the coset

one coset is generated by a

I dont see the use of this ><

we picked a subset of G

H is also a subgroup

we took element of G and we did the product

You will write $G$ as a union of these cosets when $H$ is a subgroup. Equivalence relation/partition.

aH

and the intersection is either the whole set or phi

Am very eager to see what is the use of such structure

-.-

Well, keep learning.

let me keep thinking and reading and ill come back to ya :D

Heisenberg group

is it commutative?

@Faust like the meth cook ?

Nope, not when $n>2$.

why cant i find it

When $n=2$, it's pretty boring.

@Faust it is not =p matrices are not in general

i know but when i tried to show that it wasnt it didnt go well

Do XY , and YX

the entry 1.3

wont be the same

i did but in the example i picked it worked out

Put in general letters, Faust.

one last stupid question

i did

And make sure it's not $2\times 2$. In that case the group is isomorphic to $(\Bbb R,+)$.

i made a stupid assumtion

$S_4$

it has 24 elements

the biggest is order 4 right?

yes

i cant seem to think of how to list them

well

easy way is to it by type

(123) , (134) , (124) ,(234) , (132) ,(143) ,(142) ,(243)

(12)(34) , (13)(24) ,(14)(23)

thee only one four cycle?

there should be 4 right?

no those are transpostions

1 maped to 2 and 3 mapped 4

@Kasmir: He knows. He's asking about 4-cycles.

no im asking about the four cycles lol

How many are there, Faust?

oh ><

4? + identity?

I do it systematicly

identity has order 1

written (1)

ah ok think i can list them all now

thansk @KasmirKhaan

@Faust :)

Count carefully, Faust.

if you get number that is not divisible by 24

then you did something wrong

possble types are 1 , 2,3,4,6,8,12

everything is order 4 3 2 or 1

in $S_4$

we talking different things =p

i means the number of elements of each type

ah

3 cycle you find 8 of them

@TedShifrin whats a chair of a department do/mean

In charge of the department.

hello

Manages everything, makes some tough decisions.

Hi, Nate.

wierd ok

Most organizations need bosses, right?

Or else it's total anarchy.

yeah i guess

Sorry, coming in halfway through a conversation. But would a department chair be listed as something like "Professor, Chair"

Something like that.

I've been trying to figure out who runs the math department around here.

aH is the left coset or the right?

I was listed as Professor, Associate Head of Mathematics Department for years.

I call it a left coset, Kasmir. Read your damn book!

wow.

i didnt actually find out until today who the chair of our math department was

@TedShifrin okay sorry =p