Mathematics - 2019-07-20

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Adamant

01:59

Can you elaborate on what additional context my question needs?

-1

Q: Is any space with this property homeomorphic to the three-torus?

If you have a three-dimensional, locally Euclidean space such that any path along a coordinate direction is closed (you always eventually come back to your starting point), is this necessarily homeomorphic to the three-torus, i.e. the product of three circles? I think it must be but I'm not cert...

general-topology

I know it's a bit straightforward for anyone who understands the topic well, but I didn't, so that's why I asked....

The comments answered the question, so clearly it was comprehensible to them.

1 hour later…

ABStkokes

03:07

very importanat math.meta.stackexchange.com/questions/30495/…

Martin Sleziak

03:18

A bit more readable form of the above link: Should'nt people belonging to this community check properly before claiming it is from LIVE contest? It seems to be a follow-up of: Reason for putting question on hold and when I should move my question to MathOverflow and how.

ABStkokes

04:15

Thanks Martin

4 hours later…

Secret

07:57

academic.oup.com/philmat/article/23/1/87/1432455#

ABStkokes

08:07

@MartinSleziak I am going to delete my questions this site is so rude

3 hours later…

jublikon

10:55

Hi; can anyone help me out with a task?

Let $n \in \mathbb{N}$ and $A = (a_{ij}) \in \mathbb{R}^{n \times n}$ where $a_{ij} := i +j \quad \forall 1 \leq i, j \leq n$.
Find the rank of matrix A.

i already noticed that these matrices are symmetrical

but the rows are not linear dependent by just looking at them

is there some technique that can help me out here?

Thank you in advance :)

I also noticed by trying some matrices, that, except for dimension $1$, that rank is $2$

Alessandro Codenotti

11:10

this is one ugly operator @Ryan

And it answers my question about the spectral radius and norm from yesterday

2 hours later…

Balarka Sen

13:16

@RyanUnger Whiteboy strikes back as Blackboy

When will this end

skillpatrol

13:34

uhm, never?

Ultradark

How would you define a graph that moves? (nodes move through an ambient space)

Or would this simply become a vector field

skillpatrol

@BalarkaSen did you read the paper on beauty?

Ultradark

with nodes acting as like particles or smeothing

Ryan Unger

13:57

@BalarkaSen I sent you a big question....

Ultradark

I'm so confused

can I @ someone?

Balarka Sen

@RyanUnger Where

@skillpatrol No but seems interesting

@RyanUnger Oh this? What do you mean by "regular neighborhood of a triangulation"? I understand regular neighborhood of a simplicial complex.

Ryan Unger

14:15

@BalarkaSen I said regular neighborhood of the 1-skeleton.

Balarka Sen

Your first line was missing 1-skeleton, Ryan

OK, let me see.

skillpatrol

(removed)

\o Mike

Balarka Sen

Any two triangulations of R^3 should be isotopic. What is the example where they have non-homeomorphic 1-skeleta?

skillpatrol

~~(removed)~~

Balarka Sen

Not isotopic (this is in fact not true), but concordant. Proof: Hauptvermutung holds in <= 3. Pass to a common refinement by barycentric subdivision. These are concordant moves.

Mike Miller

14:25

It's not too bad to cook up non-homeomorphic 1-skeleta. For one cubulate R^3 in the obvious way and then decompose each cube into simplices in the obvious way. This will have constant valence. Now just barycentricallt subdivide some simplex a million times

Now the second graph does not have constant valence

It's likely that the statement of course was not that there's an ambient isotopy taking one triangulation to the other

Regular neighborhoods should be homeomorphic still.

Balarka Sen

You should be able to slide handles inside a ball of radius $n$ and then let $n \to \infty$

Ryan Unger

The claim is that the regular neighborhoods of any two triangulations are isotopic

Semiclassical

Hi chat

Ryan Unger

(Ambiently)

Hi

skillpatrol

yo

Balarka Sen

14:34

Nonhomeomorphic graphs can easily have isotopic regular neighborhoods, @Ryan. Take Z^3 in R^3 and Z^3 with a barycentrically with a vertex at (1/2, 1/2, 1/2) joined to all the 8 of it's nearby vertices.

By easy handle-calculus they have isotopic regular neighborhoods.

Ryan Unger

Handle calculus?

Balarka Sen

Sliding handles

These are handlebodies. You can pick a handle up and move it around

Ryan Unger

Ok sure

I don’t see how to do this isotopy

Mike Miller

Luckily you have time

Ryan Unger

Are you just supposed to slide that thing off to infinity

You can move the handles so it’s all bunched up at one place

And then just sent it along a line

skillpatrol

14:46

@Semiclassical how's it goin?

Semiclassical

Pretty well. Did a phone interview for an adjunct Physics instructor job at a nearby liberal arts uni yesterday

Dunno how I did at that

skillpatrol

yeah, the suspense must be tough

Semiclassical

It’s not without anxiety no

skillpatrol

Good luck!

Semiclassical

Thanks

skillpatrol

14:50

in other news: the raiders have increased their budget for the new las vegas stadium to $1.9 billion; for 65, 000 seats that comes to about $29,000/seat. Only the rich and famous can afford that...

Luyw

15:22

hey guys, would substituting $(dx)^2$ for $c(dx)^2$ be acceptable?

since they both vanish really fast.

Ryan Unger

What

Luyw

in an equation

-1

Q: $v^2(dt)^2=cv^2(dt)^2$?

This kind of related to another question I posted here but it's more general. $|\vec{v_1}|=|\vec{v_2}|$ and I am trying to find $a$ after the decrement in it (after the particles have moved). $\vec{v_1}$ is parallel to the left line, and $\vec{v_2}$ is parallel to the bottom one, the angle $\varp...

calculus geometry

Thorgott

Choose $c=1$

Luyw

it's not about finding $c$..

I mean, since both vanish to zero, can I consider them to be equivalent for any $c \neq 1$ and $0$?

Andrii Kozytskyi

15:48

hey I'm having a some trouble with some basic intuition about measure theory, can you help me out please?

suppose we have a set $A = {0, 1}$. what would be its $\sigma$-Algebra? is it just its power set, $\mathcal{A} = \sigma(A) = P(A) = {\varnothing, 0, 1, {0, 1}}$?

jax broken, I meant $A = \{0, 1\}$ and $\mathcal{A} = \sigma(A) = P(A) = \{\varnothing, 0, 1, \{0, 1\} \}$

Alessandro Codenotti

16:00

That's one possible sigma algebra

Balarka Sen

A set doesn't come with a canonical sigma-algebra.

You can consider the trivial sigma algebra consisting of the empty set and the full set, eg

But incidentally these are the only two sigma algebras possible on the two-point set: the trivial sigma algebra, and the power set algebra (or the discrete sigma algebra)

Andrii Kozytskyi

hmm yeah that would make sense

but the above power set would be the only possible $\sigma$-Algebra for $\mathcal{A} = \sigma(\{A\})$, wouldn't it?

Balarka Sen

The sigma-algebra generated by $\{A\}$ is the trivial sigma algebra, $\{\emptyset, \{0, 1\}\}$.

It's the smallest sigma-algebra that contains $A$.

Andrii Kozytskyi

ah, ok

yeah I got it now, thanks

manooooh

16:16

Hello all!!

I would like to prove that for all $n\in\Bbb{N}$, $$\frac{(2n)! } {(n!) ^2}$$ is an integer using induction, but I am not able to make a proof about a variable is in a set like $\Bbb{Z}$

Base step: for $n=1$ we have $2!/(1!)^2=2$, which is an integer, so base step holds

For inductive step: $$\frac{(2n)!}{(n!)^2}\in\Bbb{Z}\implies\frac{(2(n+1))!}{((n+1)!)^2}\in\Bbb{Z}$$

Done so far: $$\frac{(2(n+1))!}{((n+1)!)^2}=\frac{2(n+1)(2(n+1)-1)!}{(n+1)^2(n!)^2}=\frac{2(n+1)(2n+1)(2n)!}{(n!)^2(n+1)}=\underbrace{\frac{(2n)!}{(n!)^2}}_{\in\Bbb{Z}}\underbrace{\frac{(n+1)+(3n+1)}{n+1}}_{(*)}$$

I am not able to prove that $(*)$ is an integer (which is indeed false, since for $n=2$ we have $10/3\notin\Bbb{Z}$ :((()

Thorgott

17:06

I don't think induction is a good way to approach this (precisely because of the failure of your inductive step)

manooooh

@Thorgott well, I thought I have made a mistake, but now it does not seem... Thank you!

2 hours later…

s.harp

19:32

off topic: Got any movie recommendations?

Mike Miller

math.ucla.edu/~cm/movies.html

Balarka Sen

Haha

These are actually very good movies

Mike Miller

Yes, I wasn't joking

He has some rankings wrong I think but those are pretty exceptional films

Some of his students feel we should make our own such pages but we have been lazy

Balarka Sen

Absolutely that'd be great idea

s.harp

oh, that list is nice

at least theones i regoxnize i usually liked

Alessandro Codenotti

19:55

Yeah that list is amazing

Balarka Sen

A little too much Lars von Trier on the list, is my one complaint, I guess

Not an actual complaint just that I hate that guy lmao

Alessandro Codenotti

Are you complaining about von Trier as a director or as a person?

Balarka Sen

As a director. I don't know person

Alessandro Codenotti

I mean he is famous for some controversial statements in interviews

Balarka Sen

I see

Alessandro Codenotti

20:02

Like getting banned from Cannes

Balarka Sen

Oh, wow. Didn't know that

Alessandro Codenotti

(even though he was allowed back recently)

Mike Miller

Klaus Kinski is known for like beating his children and yet still was an exceptional actor

Balarka Sen

I think his pretentiousness comes off apparent from the movies he makes

LOL

I love Klaus

Mike Miller

I hear Melancholia is great

Balarka Sen

20:04

Kinski was just plain mad. He was a paranoid schizophrenic. After Aguirre he like went off on a tour where he would blabber nonsense to the crowd pretending he is Jesus

Alessandro Codenotti

Essentially von Trier said in a press conference in Cannes that he can sympathise with Hitler thinking about him in the bunker at the end, which of course wasn't received very well

You can find the press conference on youtube if you want to see it by yourself

Balarka Sen

lol

i think i pass

Mike Miller

My bad Kinski sexually abused his children, I don't know if he beat them

Balarka Sen

rip

Alessandro Codenotti

Not exactly what I'd call a great improvement

Mike Miller

20:07

Herzog threatened to kill him a few times but never did it

Nobody would have disagreed

Balarka Sen

lmao

Melancholia is good, but that's one of the few things on his list of movies that are

The usual strife, Nymphomaniac, Antichrist, Breaking The Waves etc, are not good IMO.

Mike Miller

Well I've never had those recommended to me so I don't feel a need to go to bat for them

@BalarkaSen Can you identify the image of the canonical map $\pi_1(LX) \to H_2(X;\Bbb Z)$ sending a loop in the free loop space to the corresponding torus homology class?

Balarka Sen

Sounds scary. There's a split extension $1 \to \pi_2 X \to \pi_1 LX \to \pi_1 X \to 1$ by taking the section of constant loops.

So it's a semidirect product. The action has to be the usual $\pi_1$-action on higher homotopy, right?

Given by composing with the map $S^n \to S^n \vee S^1$ "squishing the sphere into a circle"

Mike Miller

Seems probable.

The lollipop

Balarka Sen

Ya good description

So I'd expect it comes from the Hurewicz $\pi_2 X \to H_2 X$?

Mike Miller

20:16

Certainly the restriction to pi_2 does

Balarka Sen

Ya obviously

By definition of Hurewicz

Mike Miller

But what about the rest?

Balarka Sen

Hm

Mike Miller

If you have a semidirect product how do you say what the abelianization is

Balarka Sen

If $G$ is $N$ semi $H$, you just add the commutators of elements of $N$, of $H$ and of $N$ and $H$. Follows from the presentation.

Mike Miller

20:18

Probably it specializes to H_1(X) + (fixed points in pi_2)

Balarka Sen

So you're looking at the part of $\pi_2 X$ on which $\pi_1 X$ acts abelian-ly?

Mike Miller

Sorry not fixed points. Quotient by the action

Balarka Sen

If $X$ is a simple space this is the full thing

Mike Miller

Set gx = x for all g,x

I expect Hurewicz always factors through this quotient

Balarka Sen

Wow interesting

Mike Miller

20:22

But what does the H_1 factor do??

Oh, it's zero along the constant sections isn't it

So this is really just the Hurwicz image?

Balarka Sen

Good point, maybe. Those are torii which bound solid torii, intuitively

All the "meridians" are nullhomotopic

Mike Miller

Yeah

Balarka Sen

So image of $[S^2, X] \to H_2(X)$ and $[T^2, X] \to H_2(X)$ are the same???

Mike Miller

In the geometric chain model indeed they are zero because they have small image

@BalarkaSen I guess so

This seems weird

Lol how about T^2 = X

Balarka Sen

lmao

Take $X$ to be a group so that $\pi_1 LX$ is literally abelian (because $LX$ has loop multiplication). Then $\pi_1 LX \cong \pi_2 X \oplus \pi_1 X$. This is the case when $X = T^2$.

So it's certainly not zero on the second factor

This is weird and too hard for me

Mike Miller

20:35

I don't understand at all

It's the Hurewicz image.

The proof was correct

Balarka Sen

Also $[T^2, X]$ would be larger; it's based at a fixed loop.

Mike Miller

An element of pi_1(LX) is a map from T^2/(S^1 x *)

Balarka Sen

$[T^2, X]$ has like a $\pi_2 X$ and a $[S^1 \vee S^1, X] = \pi_1(X) \times \pi_1(X)$ in it

Yeah

Mike Miller

For the case X = T^2 we can't overcome the pinch

Balarka Sen

Right, I'm too sleepy. It's clearly $[S^2 \vee S^1, X]$

And that makes a lot of sense. You can take $S^2 \to S^2 \vee S^1$ and then compose to get an element of $H_2$ represented by a sphere.

And how many times you lollipop the sphere around the circle doesn't matter because it gets killed in homology since it's like a map Torus -> S^1 -> X and that bounds a map Solid Torus -> X

Maybe I'm belaboring/blabbering the point

Mike Miller

20:44

Anyway it's now clear these all come from Hurwicz image

Balarka Sen

$[T^2, X]$ has something smaller than $\pi_1(X)^2$ in it. It's got all the commuting pairs of elements of $\pi_1(X)^2$

I can't quite figure out what else it exactly has

Intuitively feels like it's in bijection with $\{(a, b) \in \pi_1(X)^2 : [a, b] = 1\} \times \pi_2(X)$

OK, the disk gives a nullhomotopy of $[a, b]$. So we're counting nullhomotopies of $[a, b]$ upto homotopy.

Nullhomotopies of loops in $X$ upto homotopy surely corresponds to $\pi_2(X)$

2 hours later…

user10478

23:00

Are there any values of $\theta$ where $\theta$, $cos(\theta)$, and $sin(\theta)$ are all rational and nonzero?

...or, $\theta$ is a rational multiple of $\pi$?

s.harp

Group of rational points on the unit circle

In mathematics, the rational points on the unit circle are those points (x, y) such that both x and y are rational numbers ("fractions") and satisfy x2 + y2 = 1. The set of such points turns out to be closely related to primitive Pythagorean triples. Consider a primitive right triangle, that is, with integer side lengths a, b, c, with c the hypotenuse, such that the sides have no common factor larger than 1. Then on the unit circle there exists the rational point (a/c, b/c), which, in the complex plane, is just a/c + ib/c, where i is the imaginary unit. Conversely, if (x, y) is a rational point...

user10478

I skimmed that already but didn't see how to find points where $\theta$ is also rational (or $\theta pi$ is rational, in radians).

s.harp

whats cos of pi/4?

ah nevermind, square root of 2 is not rational

user10478

yup

s.harp

23:15

maa.org/sites/default/files/pdf/upload_library/22/Evans/…

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