Mathematics - 2018-06-09 (page 2 of 2)

I seem to remember making up a proof of something related to what you're asking for submanifolds of $\Bbb R^N\subset S^N$ by choosing a generic point far away and taking the vectors from the submanifols to that point. By dimension count, this will generically be nontangential, so gives a section of the normal bundle. What if you now choose more generic such points?

BTW, @Balarka, have you any thoughts on this? It got posted when Mike was in here yesterday.

Neither of us could see how the fixed point would be relephant. The fact that Lee Mosher (whom I've known forever, even if he doesn't remember) backs me up is heartening :P

Balarka Sen

Yeah... that's strange

Ted Shifrin

Maybe that hypothesis somehow allows one to avoid my sledgehammer.

Balarka Sen

I think your sledgehammer is easier to avoid. If $f : M \to M$ is a degree 2 covering map it has a nontrivial deck transformation that has no fixed points as it switches every pair of fibers.

Ted Shifrin

Ah, for my "chord" construction to work, I'll need $N>2n$, so I'll never get enough sections of the normal bundle.

Balarka Sen

6:20 PM

Does that work or am I bullshitting?

Ted Shifrin

Hmm ...

Balarka Sen

It seems to work

Ted Shifrin

That seems to rely on the number $2$. I don't see how that would generalize.

Balarka Sen

Yeah I don't think this necessarily generalizes

Ted Shifrin

Hmm, if you have a $3$-fold cover, the group of deck transformations has to be cyclic, doesn't it? So maybe it does generalize.

Balarka Sen

6:27 PM

I still don't see why $f$ having a fixed point was so important though!

Ted Shifrin

But your deck transformation remark makes it work in the $C^0$ category, no smoothness needed at all. So that's better, if it's right. :)

I'm surprised Lee didn't point that out. I'm trying to find an error ...

I suppose you have to argue (in the smooth category) that deck transformations act smoothly.

Balarka Sen

Well, you'd need $f$ to be a smooth covering map in that case

Oh, it's a local diffeo

Wait, wouldn't this mean total space of any double cover admits a fixed point free homeomorphism?

Ted Shifrin

Connected?

Balarka Sen

If it's not connected then it's just two copies of the same thing

Just flip those

Ted Shifrin

Well, my argument definitely needs it to be a self-cover.

Lefschetz FPT will perhaps contradict you if you don't have $\chi = 0$.

Balarka Sen

6:39 PM

Yeah your argument shows that you can actually get the fixed point free diffeo as an isotopy of the identity diffeo

LFPT contradicts only if it's also homotopic to the identity map

The Lefschetz index of the identity map is $\chi$

But the nontrivial deck transformation is rarely homotopic to identity

Ted Shifrin

So are you saying it's true in generality?

Balarka Sen

Eg consider the surface of genus 3, three torii squashed togather. Consider the axis going through the hole of the middle torus "perpendicularly". Rotate by 180 degree. This is a fixed point free homeomorphism of $\Sigma_3$

Daminark

Hello!

Balarka Sen

But $\chi(\Sigma_3) \neq 0$.

Ted Shifrin

hi Demonark

Yeah, that's the standard example for any odd genus, Balarka.

Balarka Sen

6:44 PM

@TedShifrin I am saying my argument is a triviality, because if there is a double cover $p : M \to N$, then $\Bbb Z_2$ acts freely on $M$ by homeomorphisms. Consider the action of $1$; that's a self-homeomorphism of $M$ which has no fixed point.

It doesn't require the base space to be $M$.

Alessandro Codenotti

Hi @Dami

Ted Shifrin

Ah, so does that generalize to any degree?

Balarka Sen

Hm, I doubt.

Ted Shifrin

What about prime?

Balarka Sen

Point being, for $n > 2$ there are nontrivial permutations of $\{1, 2, \cdots, n\}$ for which has a fixed point :)

It's unclear to me when I can pick a deck transformation which is fiberwise a derangement.

Ted Shifrin

6:49 PM

Of course. When the degree is prime, the deck transformations have to act by cyclic permutation, don't they?

My brain feels like it's missing stuff it used to know.

Balarka Sen

First you have to assume that the cover is normal. Prime index subgroups need not be normal subgroups anymore if your primes are bigger than 2.

If it is, then the deck transformation group is order p, yeah

It's cyclic

Ted Shifrin

So my argument wins :P

Balarka Sen

It is definitely an interesting argument

Ted Shifrin

hides from Kasmir

So I wonder if there's a $C^0$ counterexample when it holds more generally for $C^1$.

Danu

HI TED!

Ted Shifrin

6:59 PM

Hi Danu!

Were your ears burning? I mentioned you a few days ago.

Danu

how are you doing?

Oh, huh, I don't think I got a ping

2 days ago, by Ted Shifrin

When I visited Danu in München, he pointed out to me that most of the graduate lectures for his courses (maybe even all) were in English.

this?

You didn't @ me

Ted Shifrin

I didn't want to bother you :P

Danu

Oh well

I'm writing the abstract for my first talk outside my own university!

mercio

hi danu

Danu

Wanna read it?

hi @mercio

Ted Shifrin

7:01 PM

How far are you traveling?

I'm about to break for lunch, @Danu, but you can email it to me.

Danu

Not far, it's Stuttgart (5 hours by train?)

Ted Shifrin

That's quite a schlep. But I guess they're paying for you to travel and visit?

Danu

Yea :-)

Ted Shifrin

Exciting!

Danu

I wanna go there anyways, my grandparents live there

Yeah

And teaching Analysis II is also pretty fun atm

Ted Shifrin

7:03 PM

Teaching or TAing?

Danu

I finally properly learned the inverse and implicit function theorems hahaha

TAing

Ted Shifrin

I was about to complain you needed to learn some stuff :P

Danu

My own work is also going well

Ted Shifrin

You can always tell your students to watch my videos :P

That's great! Glad you're doing so well.

Danu

I've got this really good collaboration going on with this other student of my supervisor

Lots of fun

Ted Shifrin

7:04 PM

Oh, that's optimal.

Danu

And we're getting some results now too

nothing too huge, but decent stuff

Ted Shifrin

Guess you're not regretting your decision from 2 years ago too much.

Danu

Ghehe

Pretty happy doing math

Abcd

$f(x)= \log_2(x^2+5x)- 2\log_2(ax+1)$ where, a>0
$\lim_{x\to \infty}(f(x)+2)= 0$.
Find a.

Attempt:

Danu

Last week though there was this talk by a physicist (Gukov, pretty famous guy) doing mathy physics stuf fand it sounded really epic too

Though everybody seemed to agree it was mostly very conjectural/flashy without really backing it up

Abcd

7:06 PM

$\lim_{x\to \infty}(\log_2\dfrac{4x^2+20}{a^2x^2 +1+2ax})=0$

Danu

He promised to give one unified method of getting a WHOLE bunch of 3-manifold and 4-manifold invariants

Ted Shifrin

Well, sometimes stuff like that leads to interesting work, @Danu.

Danu

including Seiberg-Witten invariants

Ted Shifrin

Typo, @Abcd?

Danu

sounds too good to be true

Ted Shifrin

7:06 PM

I wouldn't expand out the squaring, @Abcd. Bad habit.

Danu

It was about this and this

Abcd

a=2

Ted Shifrin

Where did $4x^2+20$ come from, @Abcd?

Danu

Somehow associate to each 4-manifold a vertex operator algebra

Abcd

@TedShifrin log manipulation

Danu

7:08 PM

and that VOA gives all kinds of invariants if you compute certain numbers from it

Ted Shifrin

Why not just look at $\dfrac{x^2+5x}{(ax+1)^2}$ and ask what the limit is as $x\to\infty$? You make things too hard, @Abcd.

Abcd

I have got the answer, Was making silly mistake earlier

@TedShifrin its 1/a^2

Ted Shifrin

So you need $\log_2(1/a^2) = -2$.

Abcd

yeah right that would be straight forward way

Ted Shifrin

Try to be intelligent with your algebra ... don't expand things you don't need to expand.

Abcd

7:09 PM

I had converted 2 to $\log_24$ then used formula of log a+ log b

@TedShifrin Okay.

Ted Shifrin

I spent so much time in the university math classes trying to break bad habits from high school!!

Balarka Sen

@Daminark DEATH GRIPS IS ONLINE

Ted Shifrin

@Danu: It's way beyond me. I'm about to go eat lunch, but email me your abstract. :)

Daminark

Good Lord

Ted Shifrin

Bye, all, for now.

Balarka Sen

7:11 PM

See you, @Ted!

Daminark

See you

Danu

@TedShifrin Sent it!

I mean, that stuff is also way beyond everybody else haha

and like I said, nothing in there is actually proven

more like strong evidence from physics

but even if it's true it might be completely intractable etc

Abcd

How to calculate dual limits?

Like: $\lim_{x \to 0- , m \to \infty}\{3(\cos^{2m}x)-1\}$

I have reached 3e^{mx^2 }

Balarka Sen

7:53 PM

@Daminark u here

Vulthuryol

10

Q: Question about statement of Rank Theorem in Rudin

Theorem Suppose $m,n,r$ are nonnegative integers, $m\ge r, n\ge r$, $F$ is a $C^1$ mapping of an open set $E\subset \mathbb{R}^n$ into $\mathbb{R}^m$, and $F'(x)$ has rank $r$ for every $x\in E$. Fix $a\in E$, put $A = F'(a)$, let $Y_1$ be the range of $A$, and let $P$ be a projection in $\mat...

Can someone elaborate how to apply inverse function theorem to conclude that A(V) is open?

Ayesca

8:21 PM

Hello people! I am studying the topic: vector spaces and I have a confusation about finding bases of vector subspaces, which is how do I work to find a base of a vectorial sub-space according to the following cases ?:

a) find a base of a C-vector space (that is, vector space on the body of complex numbers) and the elements that make up the vector space are complex numbers

b) find a base of a vector R-space and the elements that make up the vector space are complex numbers.

Leaky Nun

it depends, and it depends

Ayesca

of what?

Mancala

8:43 PM

How many orientations can you put in any smooth manifold?

Mike Miller

If it's connected, 0 or 2

Therefore if it's disconnected, 0 or 2^(# of components)

Daminark

@BalarkaSen now I am

Mancala

9:11 PM

@MikeMiller What would be an example of a connected with no orientation?

MatheinBoulomenos

@Mancala $\Bbb R\Bbb P^2$ or the Klein bottle

Mancala

@MatheinBoulomenos Does the Möbius track also, yes or no?

MatheinBoulomenos

yes, that also works: the Möbius strip is not orientable

(but the Möbius strip has a boundary, and typically when you say "smooth manifold", you mean without boundary, that's why I didn't mention it)

Mancala

Thanks!

MatheinBoulomenos

Hi @MikeMiller

mick

9:46 PM

Reopen or fuse ?

-1

Q: Tommy’s integral $\int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx = \frac{5 }{36}\pi ^2 $

Consider Tommy’s integrals: $$a) \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4}\, dx = \frac{5 }{36}\pi ^2 $$ $$ b) \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^5}\, dx = \frac{\zeta(2)}{2} + \frac{-1}{4} \big(\log^{\,2}(3) +2\text{Li}_2(1/3) - 5\zeta(2) ) $$ $$ c) \i...

calculus integration special-functions closed-form

MatheinBoulomenos

@EnjoysMath every equivalence of categories has an inverse up to natural isomorphism, that's the $F$ you want

mick

Any ideas ?

Whoops posted link again

geocalc33

@anon

anon

→ 1 message moved to Trash

yeah?

geocalc33

10:01 PM

just getting back to the question about the loop

So my question was: can you assign a group structure to points on a loop

and there would be infinite points

nbro

Hi!

Does anyone here have a solid understanding of Dirac's bra-ket notation and tensor products using this notation?

Mike Miller

Hi @MatheinBoulomenos

mick

10:19 PM

Hi all

Alessandro Codenotti

Hi @Mike

MatheinBoulomenos

@MikeMiller I think I need to learn more homotopy theory. I don't even know why a fibration gives you a spectral sequence on cohomology (I only know spectral sequences for abstract stuff like composition of derived functors)

Hi @Alessandro

Alessandro Codenotti

Hi @Mathei

MatheinBoulomenos

@MikeMiller this book is on sale right now: springer.com/de/book/9783319234878 from the table of contents and reviews, it looks good

@AlessandroCodenotti this is really basic logic, but I want to make sure that I made no mistake. If $K$ is a real closed field of cardinality $\mathfrak{c}$, then $K$ is complete, right?

Alessandro Codenotti

I have a question which is probably easy but it's about things I don't know much about. I have two smooth manifolds $X_1$ and $X_2$ and their universal covers $Y_1$ and $Y_2$ which are also smooth manifolds. If I have an embedding $Y_1\to Y_2$ can I use it to get an embedding $X_1\to X_2$? I see it works locally but can I glue them together well?

MatheinBoulomenos

10:26 PM

because the theory of real closed fields is categorical, so there's an order-preserving isomorphism from $K$ to $\Bbb R$

and although completeness is not a first-order statement, it only depends on the order, so we get the completeness of $K$ from the completness of $\Bbb R$

@AlessandroCodenotti this feels like there could be some subtlety. We're using the properties of a first-order theory to show some second-order statement

but I don't see anything wrong

Alessandro Codenotti

@MatheinBoulomenos Are you sure it's categorical?

MatheinBoulomenos

@AlessandroCodenotti ah, it isn't

Alessandro Codenotti

Actually I don't think it is, via compactness we can get a non archimedean model of RCF and with Löwenheim-Skolem we can make one of cardinality $\mathfrak c$ if I'm not mistaken

MatheinBoulomenos

yeah

I was confusing completeness and being categorical and having quantifier elimination etc.

geocalc33

Hey

MatheinBoulomenos

10:33 PM

my logic knowledge is almost non-existent

Alessandro Codenotti

RCF has QE and is both model complete (which follows from QE) and complete, but not categorical

MatheinBoulomenos

can you remind me of the difference between model complete and complete?

geocalc33

If you have an infinite set of points on a closed loop can you assign a group structure to it?

ant

Say $L|K$ is a field extension, and $\alpha\in L$ is algebraic over $K$, I can find a monic poly in $K[x]$ that has $\alpha$ as a root by taking $$\prod_{\sigma\in \text{Aut}_K(L)} (x-\sigma \alpha).$$
Is there a reason it is preferred to say "Where $\sigma$ ranges over all embeddings of $L\hookrightarrow \overline{K}$"? Or is there a one to one correspondence between automorphisms of $L$ over $K$ and such embeddings?

MatheinBoulomenos

@geocalc33 you can assign to every set a group structure (this follows from Löwenheim-Skolem), but the question is if you can find a group structure that says something interesting about the context from which the set comes from

geocalc33

10:35 PM

Oh

Thanks

Alessandro Codenotti

@ant Think about $L=\Bbb Q(\sqrt[3]{2})$ and $K=\Bbb Q$

MatheinBoulomenos

@ant the reason why it is stated that way is because $L/K$ may not be normal

Alessandro Codenotti

@MatheinBoulomenos "every set admits a group structure" is actually equivalent to AC over ZF

MatheinBoulomenos

if $L/K$ is normal, then there is a correspndence between embeddings and automorphisms (actually that's if and only if), so if you take a non-normal extension, this will fail (as e.g. the example of Alessandro shows)

@AlessandroCodenotti wow, I didn't expect Löwenheim-Skolem to be this strong

I thought you may could get by with just the ultrafilter lemma

geocalc33

@MatheinBoulomenos do the primes admit a group structure

MatheinBoulomenos

10:37 PM

@geocalc33 any set does

but that doesn't really tell us anything useful

well, I shuold say, every non-empty set

geocalc33

so maybe someone could find a group structure with the primes?

that's useful?

Alessandro Codenotti

@MatheinBoulomenos Hmm I'm not sure exactly how much choice is needed for compactness, Löwenheim-Skolem and similar model theory results, I just assume AC for everything if I'm doing model theory :P

MatheinBoulomenos

@AlessandroCodenotti unless I'm wrong (which is plausible, since I'm talking about logic), the ultrafilter lemma implies compactness and compactness implies upward Löwenheim-Skolem

and the ultrafilter lemma is weaker than AC over ZF

so somehow downward Löwenheim-Skolem is what makes the statement so strong from a reverse-mathematical PoV

@AlessandroCodenotti apparently, downward Löwenheim-Skolem only needs dependent choice

Alessandro Codenotti

Hmm, there's something weird going on then

MatheinBoulomenos

oh no, this is for the classical statement of downward Löwenheim-Skolem which only implies the existence of a countable model

geocalc33

10:42 PM

would it be useful to map an infinite set of numbers such as the integers onto a finite segment of a loop

MatheinBoulomenos

so actually what you said was correct, full downward Löwenheim-Skolem implies AC

Alessandro Codenotti

But I'm not even sure what's the right formulation of L-S without choice or with weak versions of choice, what does "for all $\kappa$" mean? Do we take well-orderable cardinals or are the wonky ones included?

MatheinBoulomenos

good point

I don't really worry that much for this kind of stuff, I'll just apply AC to a proper class or even use the universe axiom which is equivalent to some large cardinal and call it a day

geocalc33

would it be useful to map the primes (at least a finite number of them) onto a finite section of a loop

and then study a group structure

Alessandro Codenotti

The only large cardinals I know something about are the measurable ones which look rather innocent when approached with ultrafilters/measures but are apparently very strong

MatheinBoulomenos

10:46 PM

fun fact: if you trace the references (and the references of the references etc.), most modern papers in arithmetic geometry refer back to some SGA volume of Grothendieck, where he proves Grothendieck duality for sheaves on a site and because he did everything in extreme generality and some of this stuff is awkward in ZFC, this doesn't work in ZFC

Alessandro Codenotti

woops

MatheinBoulomenos

there hasn't been a case where you couldn't make it work in ZFC though

but technically, that's something you have to do on top of the other stuff

Alessandro Codenotti

Can you get the general version to work in some reasonable extension of ZFC?

Mancala

@MikeMiller If it is connected can not have only one orientation?

MatheinBoulomenos

yes, this works in Grothendieck-Tarski set theory

which is implied by ZFC+ most large cardinal stuff some people use

Mike Miller

10:50 PM

@MatheinBoulomenos That book is famous but I have never read it.

MatheinBoulomenos

for example, it works if you have ZFC+ there is a measurable cardinal

Mike Miller

A Serre fibration, whose base is a CW complex, gives you a spectral sequence because you can filter the total space by $\pi^{-1}(X^{(p)})$ and identify the subquotients

(think about the induced filtration of the assoc'd chain complex)

MatheinBoulomenos

okay sounds reasonable

Alessandro Codenotti

@MatheinBoulomenos That's not too bad

Mike Miller

Fomenko-Fuchs is probably quite good, I don't know precisely what to reccomend you

Horrific spelling ugh

MatheinBoulomenos

10:53 PM

When we did homotopy theory in my alg top course, we were going really fast, so I feel that I'm lacking in the basics

Mike Miller

I learned at least 50% but probably more of my math through... I dunno, chasing references? Looking up things that interested me? Reading random chapters of things?

Definitely my homotopy theory (I read Hatcher's book, sure, but that's a fraction of the field)

MatheinBoulomenos

@MikeMiller this looks like a good book to read random chapters

Mike Miller

I just mean that I am not well-prepared to recommend books on the subject

MatheinBoulomenos

okay, sure. Just wanted to ask if you have an opinion. If it's famous, it can't be that bad

Mike Miller

Have you been trying to read something and had trouble?

MatheinBoulomenos

10:55 PM

I didn't like Bredon

Mike Miller

Those are funny in conjunction

MatheinBoulomenos

what is funny in conjunction?

Alessandro Codenotti

Do you still get a ping if I answer with an empty message?

MatheinBoulomenos

no, it just looks like ":n" where n is some number

Alessandro Codenotti

I see

MatheinBoulomenos

10:57 PM

I didn't really have trouble with reading something, but I did have trouble with the way my prof covered homotopy theory

we spend a lot of time on cohomology and duality before that so that it just felt very rushed

geocalc33

Take an infinite set such as the integers, rationals or primes and say you were able to map the elements of the set into a finite space.

For example maybe you could map each element of the integers or primes onto a line or loop of finite length. In order to do this my guess is that you would need the elements of the set to really converge rapidly such that if you took the limit of the set tending towards infinity you get a finite result. Would this be a useful thing to do? Obviously you can't literally map each element of an infinite set but you can always map the next one, and the fact that it converges might be helpful.

MatheinBoulomenos

@geocalc33 would the map $\Bbb Z \to \Bbb Z/n \Bbb Z$ be an example of what you have in mind?

these maps are certainly very useful and they map the intergers in to some finite set

geocalc33

is there a name for that

so i could read more about it?

MatheinBoulomenos

the keyword is "modular arithmetic"

geocalc33

could you explain the above notation

sorry

Alessandro Codenotti

11:09 PM

Answering my own question that doesn't work with $X_1=\Bbb R\Bbb P^2$ and $X_2=\Bbb R^3$

MatheinBoulomenos

@geocalc33 maybe start with the wikipedia article

Modular arithmetic

In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus (plural moduli). The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones Arithmeticae, published in 1801. A familiar use of modular arithmetic is in the 12-hour clock, in which the day is divided into two 12-hour periods. If the time is 7:00 now, then 8 hours later it will be 3:00. Usual addition would suggest that the later time should be 7 + 8 = 15, but this is not the answer because clock time "wraps...

this is covered in every book/notes/course on elementary number theory

geocalc33

@MatheinBoulomenos for example $ \Bbb Z \to \zeta(Z) $

$ \Bbb Z \to \zeta(Z)\ $ where Z $ ne1 4

Kasmir Khaan

11:24 PM

@MatheinBoulomenos mathein :D Hi :D

@MatheinBoulomenos sorry was not on chat when you texted me few hours ago

geocalc33

11:35 PM

map the integers to the integer values of a zeta function

ant

11:56 PM

@geocalc33 You want to restrict the zeta function, so that it's codomain is within the integers? Or you want to take $\zeta|_{\Bbb Z-\{1\}}:\Bbb Z-\{1\}\to \Bbb C$?

geocalc33

map the integers to the zeta function for integer values

sorry i'm not great at math

ant

Meaning literally just applying the zeta function to integers?

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$Mathematics$ Mathematics