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1:59 AM
Can you elaborate on what additional context my question needs?
-1
Q: Is any space with this property homeomorphic to the three-torus?

Obie 2.0If 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...

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…
4:15 AM
Thanks Martin
 
 
4 hours later…
8:07 AM
@MartinSleziak I am going to delete my questions this site is so rude
 
 
3 hours later…
10:55 AM
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 $2$
 
11:10 AM
And it answers my question about the spectral radius and norm from yesterday
 
 
2 hours later…
1:16 PM
@RyanUnger Whiteboy strikes back as Blackboy
When will this end
 
1:34 PM
uhm, never?
 
How would you define a graph that moves? (nodes move through an ambient space)
Or would this simply become a vector field
 
@BalarkaSen did you read the paper on beauty?
 
with nodes acting as like particles or smeothing
 
1:57 PM
@BalarkaSen I sent you a big question....
 
I'm so confused
can I @ someone?
 
@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.
 
2:15 PM
@BalarkaSen I said regular neighborhood of the 1-skeleton.
 
Your first line was missing 1-skeleton, Ryan
OK, let me see.
 
(removed)
\o Mike
 
Any two triangulations of R^3 should be isotopic. What is the example where they have non-homeomorphic 1-skeleta?
 
(removed)
 
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.
 
2:25 PM
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.
 
You should be able to slide handles inside a ball of radius $n$ and then let $n \to \infty$
 
The claim is that the regular neighborhoods of any two triangulations are isotopic
 
Hi chat
 
(Ambiently)
Hi
 
2:34 PM
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.
 
Handle calculus?
 
Sliding handles
These are handlebodies. You can pick a handle up and move it around
 
Ok sure
I don’t see how to do this isotopy
 
Luckily you have time
 
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
 
2:46 PM
@Semiclassical how's it goin?
 
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
 
yeah, the suspense must be tough
 
It’s not without anxiety no
 
Good luck!
 
Thanks
 
2:50 PM
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...
 
3:22 PM
hey guys, would substituting $(dx)^2$ for $c(dx)^2$ be acceptable?
since they both vanish really fast.
 
What
 
in an equation
-1
Q: $v^2(dt)^2=cv^2(dt)^2$?

LuywThis 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...

 
Choose $c=1$
 
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 $0$?
 
3:48 PM
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 $\mathcal{A} = \sigma(A) = P(A) = \{\varnothing, 0, 1, \{0, 1\} \}$
 
4:00 PM
That's one possible sigma algebra
 
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)
 
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?
 
The sigma-algebra generated by $\{A\}$ is the trivial sigma algebra, $\{\emptyset, \{0, 1\}\}$.
It's the smallest sigma-algebra that contains $A$.
 
ah, ok
yeah I got it now, thanks
 
4:16 PM
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 $10/3\notin\Bbb{Z}$ :((()
 
5:06 PM
I don't think induction is a good way to approach this (precisely because of the failure of your inductive step)
 
@Thorgott well, I thought I have made a mistake, but now it does not seem... Thank you!
 
 
2 hours later…
7:32 PM
off topic: Got any movie recommendations?
 
Haha
These are actually very good movies
 
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
 
Absolutely that'd be great idea
 
oh, that list is nice
at least theones i regoxnize i usually liked
 
7:55 PM
Yeah that list is amazing
 
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
 
Are you complaining about von Trier as a director or as a person?
 
As a director. I don't know person
 
I mean he is famous for some controversial statements in interviews
 
I see
 
8:02 PM
Like getting banned from Cannes
 
Oh, wow. Didn't know that
 
(even though he was allowed back recently)
 
Klaus Kinski is known for like beating his children and yet still was an exceptional actor
 
I think his pretentiousness comes off apparent from the movies he makes
LOL
I love Klaus
 
I hear Melancholia is great
 
8:04 PM
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
 
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
 
lol
i think i pass
 
My bad Kinski sexually abused his children, I don't know if he beat them
 
rip
 
Not exactly what I'd call a great improvement
 
8:07 PM
Herzog threatened to kill him a few times but never did it
Nobody would have disagreed
 
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.
 
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?
 
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"
 
Seems probable.
The lollipop
 
Ya good description
So I'd expect it comes from the Hurewicz $\pi_2 X \to H_2 X$?
 
8:16 PM
Certainly the restriction to pi_2 does
 
Ya obviously
By definition of Hurewicz
 
But what about the rest?
 
If you have a semidirect product how do you say what the abelianization is
 
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.
 
8:18 PM
Probably it specializes to H_1(X) + (fixed points in pi_2)
 
So you're looking at the part of $\pi_2 X$ on which $\pi_1 X$ acts abelian-ly?
 
Sorry not fixed points. Quotient by the action
 
If $X$ is a simple space this is the full thing
 
Set gx = x for all g,x
I expect Hurewicz always factors through this quotient
 
Wow interesting
 
8:22 PM
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?
 
Good point, maybe. Those are torii which bound solid torii, intuitively
All the "meridians" are nullhomotopic
 
Yeah
 
So image of $[S^2, X] \to H_2(X)$ and $[T^2, X] \to H_2(X)$ are the same???
 
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
 
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
 
8:35 PM
I don't understand at all
It's the Hurewicz image.
The proof was correct
 
Also $[T^2, X]$ would be larger; it's based at a fixed loop.
 
An element of pi_1(LX) is a map from T^2/(S^1 x *)
 
$[T^2, X]$ has like a $\pi_2 X$ and a $[S^1 \vee S^1, X] = \pi_1(X) \times \pi_1(X)$ in it
Yeah
 
For the case X = T^2 we can't overcome the pinch
 
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
 
8:44 PM
Anyway it's now clear these all come from Hurwicz image
 
$[T^2, X]$ 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…
11:00 PM
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 $\pi$?
 
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...
 
I skimmed that already but didn't see how to find points where $\theta$ is also rational (or $\theta pi$ is rational, in radians).
 
whats cos of pi/4?
ah nevermind, square root of 2 is not rational
 
yup
 

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