@MikeM: Aren't you high in the air or something?

I'm the air. The flight might be better if I was high, too.

Oh, shelling out money for WiFi, eh? :D

Hopefully there's no one misbehaving as on the Delta flight yesterday (which cost Delta a bundle since they refunded all ticket purchases but for the troublemaker).

Hi @MartinS

Hi!

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

A bounty room, huh? I've never played much with bounties, but maybe I should be less antisocial :)

Well, so far not much content was posted there. But maybe if more people know about the room it will get going.

I'll check it out :)

In short, the intention is that people could post there questions for which they offered bounty but did not receive sufficient answer.

Potentially, if enough people do this, the room might contain list of non-trivial (interesting ?) questions.

I guess we would all have to go check in there and see if something interests us or is in our sphere of competence. Do the tags work as usual in there, I assume?

Well, you can search the room for occurrences of the tag you're interested in. So far, there were two questions about topology. Nothing with tags containing the word geometry.

@TedShifrin

hi

As I said, only three questions have been posted there so far. If more people know about that room, maybe eventually it will have more content. (But maybe I am too optimistic.)

Hi Karim.

Hey @TedShifrin :)

Hi, @Danu.

I was wondering for theorems like this one. Let X be normeld space $Y \subset X$ then for all $f \in dual(X)$ $\| f|_Y \| = dis(f,perp(Y)).$

@MartinS: I will have to figure out an easy way of getting to the bounty room to look at it frequently.

I mean the proof I understand but how do people figure this stuff out ?

do they see geometry in infinite dimensional or what haha

I'm indulging myself aesthetically with TikZ :)

Results will come soon!

Handlebody stabilization

Karim: Normally, as with finite dimensions greater than $2$ or $3$, one derives some intuition (sometimes wrong) from pictures one can see and things in finite dimension. Is $Y$ a closed subspace of $X$ in your statement?

Yeah

@Danu: I've never progressed past Mathematica and Adobe Illustrator and \includegraphics :P

yeah I mean finite dimensional picture makes sense

Karim: So a linear functional is given by dot product, for example, and then it is just basic geometry in finite dimensions. Now you have to figure out why it generalizes.

I see

@TedShifrin Preview:

No, screw it

I'm only going to show the end product

(In finite dimensions or in Hilbert space.) So you're moving to less structure. Perhaps a bit surprising then ...

I won't look, ever, @Danu.

@TedShifrin There are various ways to access chat. One of them is to display your favorite rooms. You can add a room to your favorites by clicking on the star in the room info/room description.

Have you given your Euler class talk yet?

No, next friday. But I prepared it now

Ah, thanks for the "adding to favorites" tip, @MartinS. :)

I'm actually going to prove that e is PD to generic zero section.

I think that using favorite rooms is a relatively easy way if you want to know about several rooms whether something new was posted in them.

I'm afraid I'll lose all of the audience in that proof though. Cause it's a bit technical.

Excellent, @Danu. I'm with MikeM that that is what people should come away remembering.

Yeah, me too. That's why I wanted to prove it.

Evening @Ted.

Well, Thom class is technical. But it's all about intersection theory.

Also, I didn't know how to prove it a priori.

We already have the Thom class. But even from there, it is annoying to get to e

@Ted hey

I see @TedShifrin it would be awesome to see the intuition for infinite dimension I mean in some particular cases I can see the intuition directly but mostly not.

Before Bott/Tu did it in their book, the folklore result that cup product is Poincaré dual to transverse intersection appeared in no book. My algebraic topology prof spent a week trying to prove it (because I kept bugging him) and his proof collapsed the last day. @Danu

Good times.... :P

Karim: Often things fail in infinite dimensions, especially without Banach/Hilbert assumptions, so it's interesting that it generalizes, I suppose.

How do Bott & Tu prove it?

@Danu: Thom class :D

The proof I am doing is a bit different from what appears in MS for instance

It of course uses Thom class

But in a different way

You should look at Bott/Tu, btw. I should have mentioned that earlier.

I have

Not much.

But I might do it more at some later point. Not now.

They have much better proofs than MS.

Of course, they base things on deRham, not Steenrod squares (ICK).

Is there a good book that talks about this kind of stuff like when things generalize to infinite dimension ? I am interested to read something like that ?

I honestly don't know, Karim, since I don't think about that sort of thing much.

@TedShifrin But de Rham theory doesn't work with $\Bbb Z$ coefficients.

Mostly I don't care, @Danu :D

Meh

I do feel like it's preferable to keep everything topological rather than differential, from a first principles POV.

Griffiths, Morgan wrote a really interesting book on rational homotopy theory (based on lectures in the 70s in Italy) using deRham.

Hmm

I have it in the original handwritten notes form :) Dunno how the final Springer book looks.

Believe it or not, for some perverse reason, I scanned all those pages to .pdf when I was burning everything in my office.

Typewriter :\

@Ted Hi

hahaha :) Can't avoid sentimentality

No, handwriting, @Danu :D

Salut, @Astyx.

You actually burned your stuff?

Not literally. But it went to the recycle dump in the sky.

You can be sentimental when I'm gone, @Danu. I don't believe in it. :D

Hmkay

@TedShifrin When my grandfather died earlier this year, it turned out he had hoarded a lot of stuff. That was really not nice.

My father was severely disappointed, on a personal level

You wouldn't believe how much got thrown out at both my house and my office, and I am not a hoarder. ... And I gave away many thousands of dollars' worth of books.

Good on you, is what I'm trying to say.

You can still be sentimental when I die :P

Speaking of sentimental, BEHOLD

fuck yes.

Hello chat

Gotta bike home now but I'll be back later @TedShifrin!

I've actually got one small questoin about the Euler class left

Damn good, @Danu. I could never draw that ...

hi @Fargle!!

@TedShifrin How goes it?

@TedShifrin 2 hours of work, even with already having the handles pre-defined (and considerable experience).

My ex-colleague Will Kazez draws amazing topology diagrams using Illustrator. Absolutely amazing.

Nice. I'll check it out

OK, @Fargle .. you learning anything this week?

@TedShifrin Still going through Rudin, also going through D&F for some interesting exercises.

Well, I'm waiting for your next Rudin exercise set :)

hi @Fargle

It's taking longer than I expected, haha.

And hello, @Adeek.

I want to go through D&F and allufi chpter 0 problems once this semester ends.

@Fargle: Focus. (Ha ha ...)

@TedShifrin would you like to see my elliptic curve project?

You can email it to me, Karim, but I don't have time to look at it for a while. Despite retirement, I'm still buried in grad school (and other) recommendations.

@TedShifrin at your email that is posted on mse ?

Yup.

sent it

if I have continous function such that $F(Span(A)) = 0$ does it mean that $F(clos(Span(A)) = 0$ ?

Hello everyone, what are you talking about?

@TedShifrin I am pretty sure this is correct that must be a different characterization of continuity in terms of closure.

You should know that from basic analysis/topology. Go think.

I mean it is intuitively obvious. I keep forgetting some characterization of contuinity.

yeah ok

it is from the fact that $f(clos(A)) \subset clos(f(A))$

@Danu oh hey, genus three. nice

brb school

@TedShifrin Ultimately, the best approach is just to calculate the cohomology of $BG$ and work with that... and if you like, you could do that with differential forms.

@MikeM: Not.

Hi @Semiclassic.

hi @ted

Your approach is literally contained in this, since an invariant polynomial in the curvature gives a differential form on $BG$, I rather believe.

Maybe even harmonic?

@MikeMiller Maybe you know the answer to the question I asked Balarka yesterday (or maybe it was the day before). Take the unit cube in $\mathbb{R}^3$ and identify each pair of opposite sides with a $90$ degree twist. This gives $8$ "different" spaces depending on the choices of direction of the twists. How many of these are there when we identify them up to homeomorphism?

Also, what is the smallest $n$ such that the space embeds into $\mathbb{R}^n$ (and does this depend on the choice of twists).

Isn't $BG$ infinite-dimensional, @MikeM?

@TedShifrin Yeah. So?

@Tobias: This sounds a lot like a problem in Hatcher. But not the counting different ones part.

@TobiasKildetoft Do you always have the same fundamental group?

@MikeM: The construction I know for $BG$ isn't even giving a smooth manifold. I pass.

@TedShifrin Yeah, it was inspired by recalling an exercise about that space from Hatcher (though I could not recall what the exercise was. Balarka reminded me that it was to calculate the fundamental group when all twists were counterclockwise).

@MikeMiller No idea

So, as Mike suggests, different fundamental groups would tell you different homeomorphism type :)

My guess is you'll probably get different fundamental groups but I don't have predictions on how many.

Trick Balarka into calculating them.

@MikeMiller Good idea

LOL ... He's got real school exams.

@TedShifrin If you have a free smooth action of $G$ on a Hilbert space, you're done. Certainly this is immediate for $O(n)$ and $U(n)$. Does every compact Lie group embed into one of them?

I could construct one in general if I worked harder but I won't.

Yeah, sure. I guess the only construction I remember for $BG$ is the infinite join.

Agreed that one's not so pretty.

To be honest, I have no idea how to write down differential forms on Hilbert manifolds.

Much as my life is built on differential forms, they live on finite-dimensional creatures (sometimes with singularities).

Nah, differential forms are fine. Still sections of $\Lambda^k T*M$. There's no top form anymore, but we're strong folks and will survive even this.

Anything that depends on local coordinates computation is out the window. I'll have to think about my moving frames set-up and see what I think.

Heya, DogAteMy.

Hey what's up

Lunchtime for this bonzo.

@TedShifrin Local coordinates should be fine on a Hilbert manifold, pick an ONB and expand in that. Perfectly good definition of $d$ this way. Alternatively work locally in each finite-dimensional slice. But I will admit that my preferred definition of $d$ is the invariant one (even if I never compute that way).

I've never actually worked out the story of harmonic forms on $BG$ but maybe I ought to.

I have never thought about this, @MikeM. I don't know how much of the usual structure equations stuff (which is how I ordinarily do Chern-Weil, etc.) will make sense in this universality. If it does, I'm sure Bott already did it.

Part of my current project is to do away with differential forms so I can use more general coefficients, anyway. So I'm maybe a ilittle biased.

If you do away with differential forms, you never speak to me again! :D

In equivariant Floer homology, ya bozo.

I didn't think there were any differential forms in that story.

That's the only current kind of equivariant instanton homology - via differential forms (see Austin-Braam, the paper I once had that's now been torn to shreds via travel in my backpack at all times).

homology via cohomology?

interesting ...

Sometime you'll have to give me a lecture. Maybe on an actual blackboard.

I can tell you the story when I actually have the freaking paper done.

Outta here for now. Bye all.

later, @ted

Bye

Only contains numbers from $\mathbb{Z}$?

$(\sqrt{2})^2=2...$

You'll find that $\sqrt{2} \in A$

Nevermind, but then I'm unsure

You might start with something simpler, then: $A_k=\{x\in \Bbb R:x^k\in\Bbb Z\}$

SemiC's hint is the way to go

^

I'm clearly not made for this stuff, as I see literally no semantic difference

With $A$, I have to consider every $k$ at once.

It's countable is it not ?

$k$ is fixed

I know there are $\sqrt[3]{2}^3=\sqrt{2}^2$, does this yield to anything?

that means that $\sqrt{2}\in A_2$ and $\sqrt[3]{2}\in A_3$.

Yes

So each $A_k$ is certainly countable.

But we also find $-\sqrt{2}\in A_2$

You could also break up $A_k$ even further

But I can see why that doesn't matter

Yeah. $A_k^+$ and $A_k^-$, for instance.

into sets $A_{k,n} = \{x \in \mathbb{R} \mid x^k - n = 0\}$.

god, I'm so lost with this... I really wonder why we don't have a tutorial-course

where $n \in \mathbb{Z}$ is fixed.

It'd be simpler had they said $x^k\in\mathbb{N}$.

Wait, nm

wrong statement

it doesn't matter

Yeah.

Since there is a bijection from $\mathbb{Z}$ to $\mathbb{N}$

How many $x$ can be in the set $A_{k,n}$, at most?

That's a good point @SteamyRoot

As for every $A_k$ it holds that $\vert A_k\vert=\vert\mathbb{Z}\vert$

Since we can just get the $k$th root

Well, there could be two $k$th roots. But that's not a problem.

Yes, I was too lazy, sorry :P

It'd be a bit more interesting if it was $x\in \Bbb C$ instead of $\Bbb R$

Yeah - if you want to, you can prove the following

Then each $A_{k,n}$ would always contain $k$ elements.

@Semiclassical But not that much different

That shouldn't change much, should it?

That's probably right, but I wonder.

As you'd have at most $k$-solutions

Which is still countable

this is just a special case of the countability of the algebraic numbers

Well, the reason I wonder is that if you now take a union---say, $A_{k,1}$ through $A_{k,n}$---then you'd end up with $\frac{k(k+1)}{2}$ solutions

damnit I was about to type that as an exercise :P

Wait, I'm still being silly

I meant $A_{1,n}$ through $A_{k,n}$.

@Semiclassical Don't worry, probability is very high that I am much more silly than you

So it'd grow like $k^2$ not like $k$.

Maybe that doesn't make a difference, though.

yeah, that shouldn't make a difference. $|\Bbb Z \times \Bbb Z|=|\Bbb Z|$.

yep

So it doesn't make a difference, no.

Aaanyways.

You should end up with $A$ being a countable union of countably infinite sets.

In fact, as far as I understood, $\vert\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}\vert=\vert\mathbb{Z‌​}\vert$ holds

For finitely many factors, yes.

And you can replace $\mathbb{Z}$ with $\mathbb{Q}$ there as well

Yeah.

Yup

What I don't remember is if $|\Bbb Z^\Bbb N|=|\Bbb Z|$.

@Semiclassical no, that's $|\Bbb R|$

okay.

but that might not be what you mean

That's the set of functions Z to Z, it has the cardinality of R

Hmm.

the set of all countable sequences ($\Bbb Z^{\Bbb N}$ has the cardinality of the reals

That fits with decimal representations being uncountable, sure.

the set of all countable sequences with all but finitely many zero is countable

Heh

I'd have thought $\Bbb Z^\Bbb N$ is a countable union of countably infinite sets.

@Alessandro What I didn't understand, how do I know how many functions exist from $\mathbb{Z}\to\mathbb{Z}$?

Is there some choice required in saying that the set of finite parts of an infinite set has the same cardinality of the set?

For any $z \in \mathbb{Z}$, how many images can it be given?

There are ways to show that there are as many functions $Z\to Z$ as there are real numbers

$\{1, 2\}\to\{a, b, c\}$ should have $\vert\{a, b, c\}\vert^{\vert\{1, 2\}\vert}=3^2=9$ possible functions, shouldn't it?

Hm, wait, nevermind my previous question, I'm pretty sure you need choice

axiom of choice?

Yep, for my question about finite parts

Hello

I'm still a bit perplexed. Looking elsewhere, it seems like a union of countably many countable sets is countable. But then how is Z^N uncountable?

The difficulty is that anything symbolic needs a symbolic math engine behind it (in the javascript).

You correctly sorted 16 numbers. ouch

So maybe stick with numbers and any binary feature?

@Mitch basically any type of "special" numbers

Smith numbers

Taxicab numbers

Catalan numbers

Mersenne primes

but presumably not with small probability

Triangular numbers

@Semiclassical How is that a countable union of countable sets?

so Mersenne is sorta out

...derp

$\cup_k \Bbb Z$ is different than $\prod_k \Bbb Z$.

maybe restrict the possibilities so that they're 50-50 yes/no

Could do it as something non-numerical as well. As a simple example: Show a polygon and ask "is this convex"

@Semiclassical $\vert\bigcup^\infty_{k=1} A_k\vert=\vert\mathbb{Z}\vert$ with any $\vert A_k\vert=\vert\mathbb{Z}\vert$, but $\bigcap^\infty_{k=1}A_k=\emptyset$?

That seems to be true, yes. (Not that I remember how it is proven)

@Semiclassical Right (and for the latter one, if instead of a specific set we just took a collection of non-empty countable sets, we would need choice to even know that the product is non-empty)

Basically, I should prove it

But I donÄt know how, so fuck it

Pretty much, yes.

You can definitely find that question on MSE, though.

Uhhh

@Semiclassical good direction...easy to do javascript graphics... but the concepts seem ... basic.

wouldn't the intersection of all $A_k$ contain at least $\mathbb{N}$?

Point.

or $\mathbb{Z}$ even

yeah. $\Bbb Z^k \subseteq \Bbb Z$, after all (usual multiplicative product here)

good afternoon

@Semi you mean like $\{z^k \mid z\in \Bbb Z\}$ ?

@WillHunting yes, i have. i have not, however, studied any rigorous analysis, besides a little bit of complex analysis (cauchy's integral formula, laurent series, residue theorem)

Yeah.

image of $\Bbb Z$ under $z\mapsto z^k$.

Complex analysis > real analysis

im interested in complex algebraic geometry :)

Though that may be because real analysis is a foundations course and complex analysis typically isn't.

hello, @AkivaWeinberger

I always found that a bit weird, if I'm honest.

Calling it complex analysis rather than complex calculus.

I'd argue with hilbert's hotel

calculus courses are usually a bit less rigorous

Hi

i mean, it depends if its an undergraduate or graduate course, but either way, i think its just much more rigorous (even if not very rigorous) than the calculus sequence

Yeah, but in complex analysis you do a lot of stuff that would fit more in the calculus side than the analysis side

e.g. contour integration, residue theorem

I guess you could argue that any calculus course after real analysis should be an analysis course

Guys

Last time you said $\{y\in\mathbb{R}:\exists a,b,c\in\mathbb{Q}\text{ such that }ay^2+by+c=0\}$ is countable, because you can represent the coefficients as tuples

However, what if $a=b=c=0$?

Then it holds for any $x\in\mathbb{R}$

and thus it's uncountable

That's true

They meant that at least one is nonzero

The task doesn't say anything about that, so I assume that all may equal zero

:)

I'm pretty sure that's an oversight :P

hey guys, need some help with maths revision

well, further maths

how would you plot arg(z -a)=arg(z-b), like what does it look like?

because i completely forgot

What's a ?

bonjour toute le monde

hey kaj :)

would you care to play a game of chess?

Sure @meow-mix. Was that your friend request I got recently?

yes; finap13

lazy questions are annouying

i don't even think he tried on that one.

She, based on the name, but yes.

Indeed @Semiclassical. It's funny. People have gotten so used to / tired of that, that if one writes a question with even a little effort, it typically gets showered with upvotes.

See my questions, e.g. haha

sorry didn't realize

No title, question text is in an image, no work shown

not even the tags are right

LOL

lol once i saw a problem about solving the midpoint and one of the tags was "algebraic-geometry"

i actually dont blame them though, they probably thought it was literal geometry w/ elementary algebra

Yeah, I can understand those (though ffs look at the meanings of the tags before you use them)

> representation-theory

^

What're your preferred time controls @meow-mix ?

yeah, ok, missus.

@KajHansen i usually play 30 | 0

Cool cool. That's fine with me

Nice, @MikeM. Have a +1 from me.

Time controls for what?

chess

Ah.

I sent a challenge @meow-mix

@Kaj received :)

Any ideas how I could show that $A_1\times A_2\times \cdots\times A_n\text{ countable}\Leftrightarrow A_i\text{ countable, with }i\in\{1, 2, \cdots, n\}$?

Specifically, I have trouble showing $\Rightarrow$

What happens if $A_k$ is uncountable for any $k$ @NaCl ?

@KajHansen That doesn't matter. I need to show that both holds iff the cross-product is countable, then $A_i$ is countable

Playing chess atm @NaCl; I'll give it a closer look in a bit

Thank you!

@TedShifrin Thanks for brining up Kazez: His paper on introductory contact topology seems nice!

Lots of pictures!!

@NaCl hint: use cantors diagonal

I can tell they're not native TikZ. They're very nice though.

to prove that for $A,B$ countable, $A \times B$ is countable

@meow-mix eh heh, take a look at the tag she added after the question got closed

Should be doable by induction

ahahaha

@Semiclassical humorous

yeah, had to chuckle at that

@meow-mix thanks

well, sorry karla :P

we're not going to do your homework

not if you don't put some effort into it first

now this is a nice answer: math.stackexchange.com/a/2035525/137524