hi chat

@Ted I saw your message from a couple days ago and was hoping to find you online yesterday, but you logged in shortly after I went to sleep... the timezones aren't working in my favour here

This thing is quite quiet.

Sorry about that.

It's OK. I'm to blame just as much.

I'm glad you guys took the heat for it, I didn't want to be at fault

need some help any1 here?

I want to prove that if |z|=1 the sequence ${z^n}$ does not converge except z=1. Proof: if argument of z is rational multiple of π then it is periodic if it is irrational multiple then $z^n=e^{iqnπ}$ where n=natural q=irrational , and simply say that this goes to infinity as n goes to infinity?So it diverges.

That last case does not go to infinity as $n \to \infty$

Because it stays on the unit circle

Is it obvious---without homology---that $\mathbb R^m\setminus S^{n-1}$ is connected for $n<m$?

I would say yes?

the OP I asked said "yes because $\Bbb R^m$ is a vector space"

I don't understand that explanation. Do you?

Hmmm... that sounds a bit weird, but it may match my intuitive reasoning

What is your reasoning?

My reasoning is that $S^{n-1}$ can be seen as the set of unit vectors of $\mathbb{R}^n$

So?

I guess you can explicitly produce a path from the origin to infinity

$\mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{n-m}$

Yeah okay

@SteamyRoot you want $m-n$ since $n<m$.

Oh, right, I switched them

Yeah, my idea is to identify $S^{n-1}$ with $S^{n-1} \times {0} \subset \mathbb{R}^n \times {0}$, and then it becomes easy to construct a path between any two points

^that was my thought too, in the above message

Well... I'm not convinced this approach is limited to vector spaces... nor that it only works for vector spaces

So I don't find the explanation you were given satisfactory

More generally, can subtracting a codimension 2 compact submanifold disconnect the embedding space?

I want to prove that if the ${z^n}$ sequence converges then z=1 or |z|<1 im stuck on the z=1 case.

can you help me?

No clue

If $z\ne 1$, what does the series look like?

@ManolisLyviakis The intuition is that if $z$ lies at some rational angle, it's a root of unity, so after every so many rotations, you end up back at z

So it can't converge, because no matter how far out you go, you're ending up somewhere different on the circle than $z$

If it's at an irrational angle, you will densely fill out the unit circle

what is z does not lie in a rational angle?

if*

ohhh thanks

cant i use cauchy convergance to prove it?

I would use Poincare recurrence...I think. But I study geometry, not analysis.

@ManolisLyviakis Let's see

Do you want to write down a formal $\epsilon-N$ proof?

ye but i cant trying to use Cauchys or standard congergance definition

Write down those definitions please

in chat

ohh sure

Just so we're on the same page @ManolisLyviakis

$z_n$ sequence converges to $a$ iff for every e>0 there exists $n_o$ such that for every $n\geq n_0$ $|z_n-a|<e$ first definition

cauchys definition $z_n$ converges iff for every e>0 there exists $n_0$ such that for $n,m \geq n_0$ $|z_n-z_m|<e$ for every e.

Right, and you've proved that $\Bbb C$ is complete?

yeap.

use \epsilon btw

so the first definition

i know that my $z_n$ converges then i get $|z| \leq$

$|z| \leq 1$

Yup

for $|z|=1|$ now i have to prove that z must be z=1 and not any other complex on the unit circle

yes

and im stuck :P

suppose argz=Θ Θ not zero

So, for the formal proof, I guess assume $z=\mathrm e^{\mathrm i\theta}$, $\theta\ne 0\mod 2\pi$.

ye

Then $z^n=\mathrm e^{\mathrm in\theta}$.

yeap

now if i use any definition the modulus are always 1's i dont get anywhere

I think the trick is to look at $|z^n-z^m|$

Show that it fails the Cauchy definition

If you fix $n$, you should be able to increase $m$ so that $z^n$ and $z^m$ are on "opposite" sides of the unit circle

Or maybe use the definition of $|\cdot|$?

Try to get this in terms of real numbers?

ok ok

just a sec

@ManolisLyviakis you need some argument that if the arguments of your complex numbers differ by more than $\delta$, the modulus $|z^n-z^m|$ will be greater than $\epsilon$

cant get it yet

You could just look at the sequence $n\theta$ with $\theta \in [0,2\pi)$ where everything is taken modulo $2\pi$, and define a metric as in math.stackexchange.com/questions/11073/unit-circle-metric

@Danu Depends on what S^{n-1} means.

Any embedded sphere? Then no. That's Jordan curve theorem.

Unit sphere inside some n-dimensional subspace? Then yes.

Hello @BalarkaSen

hi

@BalarkaSen I have a nice differential topology problem, want to hear it?

Not really.

Shame

Then take a converging sequence, which is always a Cauchy sequence. Then from some $N$ onwards, $\|(n+1)\theta - n\theta\| < \epsilon$ for all $n > N$.

I solved it, I'm not asking for help...

Hence $\|\theta\| = 0$ and thus $\theta = 0$. So $z = e^{i\theta} = 1$.

i used cauchys convergance

for n and m=2n

@0celo7 No, never. Take any two points, a path between them. Make it transverse to the submanifold. That's disjoint from the submanifold.

and got $|z^n-1| \leq 1$ so $z^n$ but go to 1 not that z=1

@BalarkaSen Ah, of course

$ \eq e$ **

$|z^n-1| \leq e $

now that i proved that it convgerges to 1 when |z|=1 im stuck again.

need z=1 not $z^n --> 1$

Stop using $z^n$ and $1$ then :P

Use $z^{n+1}$ and $z^{n}$

ok then i get |z-1|<e

thats a circle at the point (1,0) with radious e .how do i prove it is a single point?

Hence $z = 1$.

Because it must hold for all $\epsilon > 0$.

i can just say that? no proof for it?

$(\forall \epsilon > 0: \|x-y\| < \epsilon) \implies x = y$

ohhhh have to check that somewhere didnt know

x,y can be complex?

Yup

$\mathbb{C}$ is a Banach space, in particular a normed vector space. And for a normed vector space we have that $\|x\| = 0 \iff x = 0$

and to prove that property

|x-x|=0

if |x-y|<e for every e then x=y

In general for any real number, $x<\epsilon$ $\forall\epsilon>0$$\implies x=0$.

@BalarkaSen I told you the condition on the dimensions :P

morning chat

@Danu Oops. So $S^n$ is an arbitrary embedded sphere in a manifold of dimension $m$, of codimension less than $1$? (aka $m > n + 1$)

Then, modulo assuming smoothness, my transversality proof says complement is always path connected

Yeah

It was already settled anyways

kind've annoyed by the back-and-forth in these comments: math.stackexchange.com/a/1869978/137524

@Danu Huh? SteamyRoot seemed only to prove it for a sphere inside a lower dimensional subspace.

For arbitrary embeddings this is not that easy?

I mean, I can understand if my answer wasn't sufficiently helpful. But that entire set of comments isn't really relevant to what I did.

@Semiclassical I proved something wrong that was on a poster of ours at that conference

whuh-oh

i hope you didn't point it out at the conference :)

Not the poster I showed you

@Semiclassical No, but they're presenting that poster this week again

hrm.

how major of an issue is it?

@BalarkaSen I was thinking about this, too.

But everybody seemed so convinced...

There's a blip on the spectrometer reading that our failed grad student took months ago

I am not :)

I redid all of his measurements last week

I found the same blip, but it's not what he thought it was

oh?

It might be interesting, but it's not as interesting as we had hoped for

You need transversality if everything is smooth and well. For non-smooth things, I don't even know how to do it.

ahh, that's too bad.

Basically, he thought it was Ga oxide, but I ran some pure Ga oxide and it's not a Ga oxide peak

So now I have to create a mixed phase Ce oxide

@BalarkaSen I think the OP only asked about a standard embedded sphere in the subspace though, so it's alrightl.

Cue the 1000C furnace music >:3

Scrap that, I do know how to do it topologically.

Alexander duality.

@Danu OK, then it's quite clear.

Is Alexander duality a big step from the stuff that I know?

Nah. It's on section 3.3 Hatcher.

@Semiclassical Also, we apparently had some super pure mixed phase Ce oxide laying around, I proved it's not mixed, which is strange.

Also, it's funny to notice that the cup structure on cohomology actually gives one lower bounds on the number of charts needed to cover a manifold.

It's a totally different color than what it should be.

weird

Are you all professors?

It should be peach, but it's carbon-colored.

nope

@ManolisLyviakis Prof Emeritus, PhD, Dr., and knighted.

At your service.

haha

Ah, Alexander duality is the generalization of that theorem for embeddings of disks and spheres. Nice

@Danu Hm.

@0celo7 Who just so happens to also be working as an undergraduate research assistant :P

@ManolisLyviakis Nobody here is

i wish that ill be some day a Dr

@BalarkaSen I realized that after doing the topology exam ($T^3$ cannot be covered by three charts)

I'd count Ted, were he around at the moment, though in his case he's retired

so ur not Dr's are u undergraduats?

I'm a grad student, albeit in physics.

im on my last year at a math university Crete-Greece

I'm doing a phd in Group Theory

I'm currently figuring out how to pulverize extremely hard metals

have 13 classes yet to pass

Without breaking anything

LN2!

I can freeze them and make them brittle

when im at the math exchange i feel such a noob on math

@Danu Curious, I don't know how to show this. I'll think about it.

@ManolisLyviakis Ah, nice. Πού είσαι?

crete (hrakleeio)

you?

Είμαι στό Μόναχο

(correct me if I make spelling/grammar mistakes please)

Ohh you aint greek

Nope

so far so good with your spelling haha nicely done

what are you doing at Munich

Μαθηματικά

phd? or are u a Dr

Speaking of, there was a nice problem which asked to show the torus (T^2) can be covered by 2 charts. It took some time to see it.

@ManolisLyviakis Είμαι φοιτητής

ohhh πως και εμαθες ελληνικα ?

Κάνω το Master Theoretical & Mathematical Physics στο πανεπιστήμιο LMU

aah you'r a post grad

@ManolisLyviakis ΄Εχω πολλούς Έλληνες φιλλούς

φίλους* :)

Εσύ? Τι σπουδάζεις?

@ManolisLyviakis Ah of course, the emphasis was wrong

Also what is the right verb for "what do you study"? I got this from google translate but I never heard any of my friends say "σπουδάζω"

Math im on my 4 year going to 5ith the standrd undergrad years here are 4.

@ManolisLyviakis Ναί, ξέρω

What kind of things do you like to study?

the verb is " σπουδαζω"

So it is the correct word? Okay

σπουδαζω, σπουδαζεις, σπουδαζει,... I, you, , it....

Σπουδάζεις πολλαπλότητες?

I prefer algebra .

@ManolisLyviakis :(

Εγώ δεν ξέρω τίποτα για άλγεβρα

@BalarkaSen I misremembered, this is probably not true. The torus can be covered by 3 charts instead.

But that too is not so easy to see.

@BalarkaSen This is false

By the cup structure on cohomology

Lol

T^2, I mean.

i tried to learn some basic for manifolds but they are hard

That can be covered by 3 charts.

Yeah

Oh, cannot be by 2, you mean?

Yes

manifolds are wierd to understand as a concept

OK, I'll try to see how to prove it. Give me some time.

@BalarkaSen The $n$-torus cannot be covered by $n$ charts, which follows from [do you want a hint?]

No, don't want a hint.

Thanks though.

@ManolisLyviakis Τι σας αρέσει για άλγεβρα;

At least I have a lower dimension model (aka T^2) to ponder on now.

The $n=3$ case was on my exam :P

(but phrased much more concretely)

i got that manifolds are n-forms the exterior product of basis of the space with the linear functions something like that

I heard. I mean, I'll try to get something for n = 2 (which would be easier, probably) and try generalizing for n = 3.

That's my strategy, spelled out loud.

@ManolisLyviakis Wow that sounds a bit like word salad to me ;P

yea i know

Manifolds are spaces that locally are homeomorphic to Euclidean space but not necessarily globally

i dont remember very well

ohh thats a better aproach

yeah that was the whole concept that locally you can use euclidean geometry

Manifolds are just an example of schemes.

@BalarkaSen Sure, that works. You'll probably end up proving something more general

duck

@ManolisLyviakis So there is a notion of calculus (note that derivatives only need local information) etc

Very beautiful subject

i like solving diophantine equations using ring or field extensions. but i mot there yet.

Ah, really? I don't know anything about that

Sticking with the Greek roots

:P

hahaha kinda

you can solve a diophantine equation on a smaller ring instead of the "rich" C .where you dont have to investigate all the cases you may have in C

also you can see diophantine equations as curves and try find integer roots .but that is something algebraic geometry does. Im not close to that either.

:"(

the fermats theorem was solve with a connection of elliptic curves and diophantine eq.

Θέλεις να μείνετε στην Ελλάδα; Η θέλεις να πάω στη Γερμανία η σε άλλη χώρα;

nah i dont want to stay in greece but im a poor guy so.. dont know if i hvae alot of options abroad.

@ManolisLyviakis I also want to learn some basic algebraic geometry, but mostly for the geometry

@ManolisLyviakis Ah... Are you afraid that the living expenses will be too high?

yeap.

They are kind of high here in Munich ;| But I think there are a lot of other places where it could be better...

it is already hard for me to study here.

And, of course, a part time job here pays pretty well

ill have to work at the same time

Though it doesn't sound pleasant to have to work while studying

germany is considered a good option i have alot of friends at berlin

yeap

Yeah, lots of Greeks here

There is a big community

btw i was trying to understand n-differential forms in order to understand what exactly those dx,dy's are when you integrate

Right

what i got is that they are an exterior product of the basis of the vector space of linear maps which is the maps kroneckers delta or something

any intuition as to understand what exactly those dx's are except that they are a really small distance going to zero?

@ManolisLyviakis This sort of physical understanding has not been very helpful in my understanding of the real mathematics, but maybe I can say some things

yeah i dont want the physical approach

@ManolisLyviakis There is this thing called the wedge product on forms $\wedge$, and you can construct $n$-forms by taking wedge products of lower degree forms

yeap

So to get a definition of a differential form it's quite useful to have the notion of a tangent space

Do you know hwat the tangent space at a point of a manifold is?

ohh

yeah thats where my intuition starts to crash

it is a space

where i can move the starting point

@ManolisLyviakis Intuitively, $dx$ is like something which eats an infinitesimal cube inside $\Bbb R^n$ and spits a number.

offf

perfect

ohh*

guys i got to go

But I find the intuitive notion not very helpful to understand the "real formalism"

Alright

danu add me on facebook

΄Γειά σού

if ya want to chat about math

Indeed, differential forms are "dual" to vector fields. They eat vector fields (well, tensor fields) and spit out numbers.

γεια σου

@Danu Well, that's what the real formalism means :P

manolisjam my fb name

@BalarkaSen I don't know...

"means" in what sense?

They are elements of the $n$-fold exterior product of $T^*M$. Aka, are alternating multilinear maps $TM \otimes \cdots \otimes TM \to \Bbb R$. That is, it eats $n$ vector fields, and spit out smoothly varying scalars.

The "alternating" thing can be interpreted as it encodes signed area of infinitesimal cubes ($n$ vectors) while eating them.

I know these things, but I'm just saying I don't find that to be actually useful in practice

OK. For me, that intuition is essential to make sense out of things.

It's probably because of my physics background

In physics, the above is all one learns, with no rigorous backup whatsoever.

Nah, I don't really know much about differential forms. Just what I know from reading Ted's calculus book.

When confronted with real mathematics, one finds out how powerless one is with just such vague ideas

At least, that is my experience (and I've seen it in other TMP students too)

Fair enough. Doing manipulations are essential too, along with geometric intuition.

OK, finished dinner.

I guess this partially explains the attitude of some physics students who switch to math: They try to be very "rigorous" (end up being pedantic) as a sort of reaction to the years of vague explanations

Makes sense.

good evening everybody

@Alessandro hello. good morning ;-)

@Danu I know so many students who have switched from physics to maths due to lack of rigour and none going the other way around :(

@Danu Mhm, I see.

If charts $U, V$ covers the 2-torus, then I have the relative $H^1(X, U) \times H^1(X, V) \to H^2(X, U \cup V)$, the relative cup product.

$U$ and $V$ are contractible, so H^1(X, U (or V)) are the same as H^1(X) by long exact sequence of a pair.

$U \cup V = X$, so it says the cup product map on H^1 is zero. That's garbage.

Clearly this generalizes.

@Semiclassical Do you know if the Jacobi theta function $\vartheta_{3}(0,q)$ has an asymptotic expansion as $q \to 1^{-}$ that would help explain why the integral $\int_{0}^{1}\vartheta_{3}(0,q) \, \mathrm{d}q $ converges?

can't say I do

@Danu Eh, I guess one needs to rigorously check that under all the identifications made, the relative cup product map is the same as the actual cup product on $H^1$. But that's not hard: the map $H^1(X) \times H^1(X) \to H^2(X, X)$ is just given by inclusion of $C^1(X) \times C^1(X)$ into $C^1(X, U) \times C^1(X, V)$, composed with the usual cup product map, to $C^1(X, U + V)$, with the inclusion to $C^1(X, U \cup V)$ (which is how the construction of rel. cup product goes).

The composition of the first two maps is clearly the same as the usual cup product. Postcomposing that with the third map doesn't matter, as it's an isomorphism on homology.

So, yeah, there you go.

hi again @Alessandro: I guess we missed each other again ... G'day @Balarka

@TedShifrin Hullo.

How's your day?

@Ted I'm still here! How are you?

I noticed yesterday there's an user named "Teddy" and now I'm afraid i just pinged the wrong person

Hi, Alessandro.

good evening @balarka

@Alessandro: Yeah, I guess you need to do TedS to get me.

@TedShifrin makes sense

so, I was saying, how are you? I haven't been here for a while

@BalarkaSen Yeah, nice right

and it works for $n$ charts on the $n$ torus because you have a cup of $n$ generators of first cohomology that you know is not trivial

Hi @Ted!!

.

So a slightly stronger statement probably is that it cannot be covered by $n$ open, acyclic subspaces

Why am I invisible on the right pannel?

You're not.

What have you been up to @TedShifrin?

I don't see myself there.

I can't even edit my messages.

@Danu Right.

(I'll return when things are fixed, hard to write without having the option to edit)

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