Mathematics - 2020-08-05

12:08 AM

Congratulations to former chatter @Xander on being elected moderator. It was a tight fight, I see.

1:26 AM

math.stackexchange.com/questions/3778134/… – I wonder if I should post an answer which simply points out that all non-discrete finite topological spaces are also non-Hausdorff.

Technically that doesn't answer the question at all; in order to have a complete answer to the question, you'd also need to have a "naturally occurring" example of a finite topological space which is not discrete.

Balarka Sen

@TerranSwett There's actually plenty depending upon what is "natural". Take any finite simplicial complex, collapse the open faces to points.

That's a finite topological space which models the simplicial complex, in the sense that they have isomorphic homotopy groups.

Balarka Sen

2:37 AM

@Lelouch It might be interesting to note what happens to quasi-isometries "in the limit". Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $f : (X, d_X) \to (Y, d_Y)$ is a $(K, \varepsilon)$-quasi-isometry, so $K^{-1} d_X(a, b) - \varepsilon \leq d_Y(f(a), f(b)) \leq K d_X(a, b) + \varepsilon$.

Scale the spaces $X$ and $Y$ by some large constant $\lambda >> 1$, so $d'_X = \lambda d_X$ and $d'_Y = \lambda d_Y$ are the new metrics. Then $f$ becomes a $(K, \varepsilon/\lambda)$-quasi-isometry with respect to this scaled metrics

If $\lambda \to \infty$, then supposing $(X, d_X)$ and $(Y, d_Y)$ converge in appropriate Gromov-Hausdorff sense to some limit $(X_\infty, d_X^\infty)$, $(Y, d_Y^\infty)$, $f$ induces a $K$-biLipschitz map between them

Eg scaled limit of the Cayley graph of $\Bbb Z^n$ is $\Bbb R^n$, so any quasi-isometry of $\Bbb Z^n$ gives a bilipschitz homeomorphism of $\Bbb R^n$ in the limit.

Heh I guess this gives a short proof that $\Bbb R^n$ is not quasi-isometric to $\Bbb R^m$ if $m \neq n$; both the spaces are stable under scaling

This is not a good perspective for hyperbolic space, because if you scale a $\delta$-hyperbolic space by $\lambda$ you get like a $\delta/\lambda$-hyperbolic space, so as $\lambda \to \infty$ you get a $0$-hyperbolic space... what is known as an $\Bbb R$-tree

Horrible objects in general

But anyway in the limit a quasigeodesic will be a bounded Lipschitz distortion of a tree path in this $\Bbb R$-tree, or something like this

And that is why Morse lemma is true I think

Yuvraj

3:35 AM

@robjohn yes sir

Alessandro Codenotti

6:34 AM

@BalarkaSen or you can compute their asymptotic dimension

Balarka Sen

7:43 AM

True @Alessandro

Infinity

8:15 AM

What is an example of a sequence which has a limit but does not converge?

Leaky Nun

8:39 AM

@Infinity could you define the two terms?

Infinity

For example you have a_n = (1/n) which has the limit of 0 as n approaches infinity. Ist ist enough to claim that a_n is convergent?

In order to prove that a sequence is convergent one should prove that it is 1. bounded and 2. monotonic. Does 'having a limit' contains both of these conditions in itself?

Leaky Nun

8:55 AM

those are not definitions

@Infinity do you know the precise definitions for "has a limit" and "converges"?

robjohn

9:06 AM

@Yuvraj was that a question on an exam?

Infinity

@Leaky Nun Convergence means: A function approaches a final and constant term as the input approaches infinity and limit is that constant value the the function approaches

Now I see that having a limit automatically means that the sequence converges

But then comes up another question: Why in math textbooks it is usually said, in order to prove convergence we should prove that it is 1. bounded and 2. monotonic? Why not simply say: If a_n has a limit then it converge?

Yuvraj

9:34 AM

@robjohn yes

user434058

10:17 AM

@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.

Yuvraj

10:30 AM

@FakeMod he former condition implies that it's a constant function, meaning $f(x) = 3$. The integral evaluates to $6$ which is the last option

how to prove that ?? I saw that will be constant due to even no s are coming to constant but I didn't answered because I can't prove it for odd

16 mins ago, by FakeMod

@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.

@robjohn hi sir

Leaky Nun

@Infinity using the theorem "bounded + monotonic => converges" can avoid actually finding the limit

@Yuvraj $\lim_n f(x/2^n) = \lim_n f(x) = f(x)$, $\lim_n f(x/2^n) = f(\lim_n x/2^n) = f(0)$

Fargle

@Yuvraj Note that this condition also implies that $f(x) = f(x/k)$ for all $x$ as well. (Why?) Therefore you can assume without loss of generality that $k < 1$. From there, you can consider $f(k^nx)$ for higher and higher $n$, and use the continuity of $f$, and I leave the rest to you.

Leaky Nun

so $f(x) = f(0)$

it's easier to see if you try to picture it

Yuvraj

10:45 AM

where limit n tends to infinity?

Fargle

Indeed.

Yuvraj

got it

@Fargle what about this?

$f(x)=\sin(\log_a(x))$,

can we prove same fo this

Fargle

No, it doesn't satisfy that property

And it's obviously not constant: $f(a^{\pi/2}) = 1 \neq 0 = f(a^\pi)$

Thorgott

12:17 PM

Is there a good way of seeing that $\mathbb{R}[X,Y]/(X^2+Y^2+1,aX+bY+c)\cong\mathbb{C}$ (where $a,b$ aren't both zero) without doing the ugly computations

Lelouch

1:08 PM

@BalarkaSen Nice, but at the risk of being wrong, doesn't this work to show that $\mathbb{R}^m$ and $\mathbb{R}^n$ are not quasi-isometric ? It suffices to show that $\mathbb{Z}^m$ and $\mathbb{Z}^n$ are not quasi-isometric. So basically, assume $m < n$. consider the ball of radious-$r$, in $\mathbb{Z}^n$. By the ifrst property of quasi-isometry, the image of $\mathbb{Z}^m$ inside this ball contains $\Theta(r^m)$ lattice points.

On the other hand, by the second properrty of quasi-isometry, this contains $\Theta(r^n)$ lattice points. Which is possible iff $m = n$

it's probably equivalent to what you said, I can't say becaues I don't understand how are you defining convergence of metric spaces

Leaky Nun

@Thorgott WLOG $a \ne 0$, then $X = b' Y + c'$, so $X^2 + Y^2 + 1$ becomes $(b'Y + c')^2 + Y^2 + 1$

which obviously has no real solution, so it is an imaginary quadratic

so the quotient is C

@Lelouch the risk of being wrong is always a risk worth taking

Balarka Sen

@Lelouch Yes, growth rate of a group is a qi-invariant

That is another way to do it, what I said is different

Lelouch

how is the metric space convergence being defined in this case

Balarka Sen

pointed Gromov-Hausdorff convergence

Lelouch

thanks, i'll look it up

Balarka Sen

1:16 PM

You say two compact metric spaces $X$ and $Y$ are $\varepsilon$-close if there is a map $f : X \to Y$ which is a $\varepsilon$-isometry, i.e., $d_Y(f(x), f(y))$ is bounded above/below by $(1 \pm \varepsilon) d_X(x, y) \pm \varepsilon$ and is $\varepsilon$-surjective, meaning every point of $Y$ is in $\varepsilon$ distance from $f(X)$

A sequence of compact metric spaces converge Gromov-Hausdorff to some metric space X if for any 1/n the tail of the sequence is 1/n close to X

Lelouch

Basically $f$ is $(1 + \epsilon, \epsilon)$ quasi-isometry ?

Balarka Sen

yes

For complete metric spaces $(X, p)$ and $(Y, q)$ with a choice of a basepoint you compact balls around $p$ and $q$ in the respect metric spaces

Thats pointed-GH

So for example $(1/n \,\Bbb Z, n |\cdot|)$ converges pointed Gromov-Hausdorff to $(\Bbb R, |\cdot|)$, centered at the origin

Easily checked from above definition

Thorgott

yeah, that's a nice way of putting it without making the computation explicit

guess that's as straightforward as it gets

Balarka Sen

$(1/n \,\Bbb Z, 1/n \, |\cdot|)$ I mean

Also above I should have written $d_X = \lambda d_X'$ or something

Billy Rubina

1:41 PM

If you downvote something, it shows:

But if you upvote something:

It doesn't show: "Please consider adding a comment if you think this post can be worsened".

8

Alas, the hypocrisy.

Thorgott

truly, 'tis a society we reside within

Balarka Sen

Lmao

Mike Miller

You should have run for mod elections

We need big new ideas around here

Balarka Sen

1:57 PM

Nechayev intensifies

William Oliver

2:44 PM

Hello, is it generally true that a principal ideal of a ring (left, right, or two sided) are the same as cosets of the multiplicative group of the ring in itself?

Fargle

I might not be thinking of the same thing you're thinking of, but the answer looks like "no": $\Bbb Z^\times = \{1,-1\}$, but principal ideals in $\Bbb Z$ are $n\Bbb Z$, which aren't cosets of $\{1,-1\}$.

Billy Rubina

@MikeMiller It would be unfair with the competitors: How would they compete with such ideas from the future? (2:25)

Thorgott

coset in which sense?

the units aren't an additive subgroup

orbits under the natural multiplication action, I guess

William Oliver

3:03 PM

I suppose we would have to assume the ring has multiplicative inverses

Thorgott

So it is a skew field?

Sorry, I'm not really following

William Oliver

3:21 PM

I don't think I know enough about this stuff to ask meaningful questions haha, thanks for your help. Got to learn more before I ask this again

hash man

4:58 PM

https://math.stackexchange.com/questions/3780293/calculate-int-0-infty-frac-log-x-dxxaxb-using-contour-integra
I have a question regarding taking the limit with residue

in this example at z=0, the residue is 0, so no problem with taking the limit when epsilon->0

on the selected answer @ (1) the author just has written that the integral vanishes as epsilon tends to 0.

my question, how to handle a case when the circle is not 0. I mean the limit won't be 0.

Ted Shifrin

5:22 PM

@hashman: No, $0$ is a branch point of the function, so residue doesn't even make sense. I don't understand your question. The whole reason the residue theorem works is that if you have a meromorphic function $f(z)$ with a pole at $a$, then the integral of $\int f(z)\,dz$ around a little circle centered at $a$ is precisely $2\pi i$ times the residue of $f$ at $a$.

hash man

5:44 PM

so what the residue theorem won't work at 0?

@TedShifrin? I can't understand why? for each circle there is an environment with a pole

around 0

Ted Shifrin

No. The log function is not a well-defined function in any neighborhood of 0 (deleting 0). You need to understand branch points.

That's why they use the keyhole contour.

Alessandro Codenotti

6:16 PM

Hi Ted

Balarka Sen

Hi @Alessandro

and @MikeMiller

Hey @Balarka

Howdy, nerds.

Hey @Fargle!

Do you want to think about some topological dynamics with me? I don't know if I'm missing something or the book is missing an hypothesis (I assume it's the first)

Fargle

6:19 PM

How goes it?

Balarka Sen

I don't know anything but tell me

Alessandro Codenotti

So the setting is a continuous $G$ action on $X$, $G$ is a topological group and $X$ an Hausdorff space. The action is called (topologically) transitive if any open set has dense orbit, and point transitive if there is a point with dense orbit

It's obvious that point transitive implies transitive

Balarka Sen

Ya ok

Alessandro Codenotti

And the book says that if $X$ is second countable then the converse also holds

I see that with some extra assumptions ($X$ compact, or more generally Baire)

But the book so far has been very careful in stating explicitly when $X$ is assumed compact and it's not an underlying assumption

The point is that if $Y\subseteq X$ is the set of transitive points (points with dense orbit), then $Y=\Bigcap UG$ where $U$ ranges over a basis for the topology of $X$, so if $X$ is second countable $Y$ is actually a comeager $G_\delta$ set, but I don't see why must it be nonempty without extra assumptions

Balarka Sen

Hm yeah this is not clear to me I'd have to think

Alessandro Codenotti

6:27 PM

But I don't have a counterexample either so it's probably true because of reasons I'm missing

Ted Shifrin

Hi, demonic @Alessandro, a @Balarka, @Fargle

Fargle

Heya @Ted, how's your morning going?

Tobias Kildetoft

Helloes

Ted Shifrin

Just fine, thanks. Just back from a walk.

Heya @Tobias

Thorgott

isn't it non-empty, because you assumed there is a transitive point to begin with?

Alessandro Codenotti

6:36 PM

No I'm assuming the action is transitive and I want to show it is point transitive

Hi Tobias

Ted Shifrin

clears throat

Alessandro Codenotti

I greeted you earlier!

Ted Shifrin

Oh, you greeted me when I wasn't here :D

Edward Evans

clears throat

Ted Shifrin

Just my last message was there.

Better get that checked, @Edward.

Quit the cigs.

Edward Evans

6:40 PM

I don't smoke cigarettes!

haha

also yo everyone

Ted Shifrin

clears throat back at Edward

Edward Evans

hope you're wearing a mask

user434058

Yo!

Edward Evans

Hi @Fakemd

mod

@Fakemod

user434058

Hi @Edw

Ted Shifrin

6:42 PM

Yo, @FakeMod

user434058

That will ping you. (Only three letter are needed, so you can ping me by calling me @Fak)

Edward Evans

@Fak off

weeey

Why must macadamia nuts be so expensive

user434058

@TedShifrin Hi @TedShifrin, how's your day goin'?

user434058

@EdwardEvans Which accent is that? :P

Ted Shifrin

Oh, double ping. Doing great, thanks, and you? What are we here to discuss today?

Edward Evans

6:44 PM

@Fakemod I guess a cockney or essex accent

user434058

@TedShifrin Well,... I just came here to "Yo" back to Edward.

Ted Shifrin

O, yo.

Thorgott

ok, I see why Baire would suffice

but there must be something missing otherwise, no?

Alessandro Codenotti

7:16 PM

That's what I'm wondering

I might ask on MSE

Thorgott

7:31 PM

for $U$ open with dense orbit, $UG$ intersects any other $U^{\prime}G$, of course, but it doesn't seem clear why they all intersect simultaneously

geocalc33

8:21 PM

I'm having trouble understanding which shape you get when identifying points on one torus to another torus

Thorgott

Please be less specific

geocalc33

gluiing the outer surface of a torus to the outer surface of another torus bijectively

Balarka Sen

8:50 PM

@Alessandro Yeah I think this is just false. Take $f : S^1 \to S^1$, $f(z) = z^2$, and restrict the system to the dyadic rational points on $S^1$; then it's still a chaotic system but every orbit is periodic I believe

As in, it's topologically transitive

Alessandro Codenotti

But that's not invertible

There is no implication in either direction for semigroups actions

Balarka Sen

Oh you want invertible

I dunno

Alessandro Codenotti

Yeah semigroups are too ugly

For semigroups you don't even have that point transitive implies topologically transitive, look at $\{0\}\cup\{1\n\mid n\in\Bbb N\}$ and $f(1/n)=1/(n+1)$

Balarka Sen

Sure yeah

Alessandro Codenotti

9:06 PM

I'll think about it some more and just ask on MSE if I can't figure it out

I found it in another book too but with super strong assumptions (like compact Polish $X$)

Balarka Sen

Ok

Alessandro Codenotti

Uh, there isn't even a topological-dynamics tag on MSE

There is one on MO, with a grand total of 26 questions lol

I'm having troubles finding any example of an action by a group which is topologically transitive but not point transitive

Balarka Sen

There's a wild homeomorphism $f : S^1 \to S^1$ which leaves a minimal Cantor set invariant, that might be relevant

I don't know if it is

Everything outside the Cantor set has dense orbit I think

Alessandro Codenotti

Uh where can I read about that?

Balarka Sen

Don't; it is probably not going to be helpful

Alessandro Codenotti

9:18 PM

Maybe but it sounds interesting regardless

Balarka Sen

It's called Denjoy's example or Denjoy homeomorphism

Thorgott

the Denjoy is short for "Don't enjoy"

Edward Evans

It's don't enjoy but with a scottish accent

Alessandro Codenotti

lol

Balarka Sen

hilarious

Alessandro Codenotti

9:21 PM

Which part of the UK are you from? @Edward

Edward Evans

@Alessandro The South West from a city called Plymouth, which is on the border with Cornwall (more people seem to know what Cornwall is)

Alessandro Codenotti

@BalarkaSen thanks, I'll add it to the list of "weird topological nonsense that I want to learn about"

@EdwardEvans what if I don't know where either is? Let me look that up on google maps

Edward Evans

hahaha

Alessandro Codenotti

I was in the UK only once, and I stayed in Aberdeen the whole time

Edward Evans

Cornwall is historically its own country, but it's now just a county in England

Also fair, Aberdeen is like "nice Scotland"

along with Edinburgh

basically everywhere other than Glasgow is nice Scotland

controversial

I've signed up for 3 lectures and 2 seminars for the Wintersemester

but one of the lectures is just modular forms again, and I should pass it this time lol

Alessandro Codenotti

9:24 PM

Ah I see, it's on the southwest tip basically

Edward Evans

Right

Balarka Sen

@Alessandro Can I say some crazy nonsense

Alessandro Codenotti

Oh this is unrelated, but have you seen Peaky Blinders? @Edward

@BalarkaSen sure! Well maybe, what kind of nonsense?

Balarka Sen

What about $S^1 \times [0, 1]/(z, 0) \sim(z^2, 1)$, and then acting by $\Bbb R$ on this space, where $s \cdot (z, t) = (z, s + t)$?

Edward Evans

I haven't, but it's fairly popular lol

Alessandro Codenotti

9:28 PM

I wanted to know if people in Birmingham really talk like that lol

Balarka Sen

Look at only the periodic orbits

Edward Evans

Loool the Birmingham accent is like.. the most hated accent in the UK

Balarka Sen

isn't that action topologically transitive as well

Alessandro Codenotti

Hmm ok let's see

Edward Evans

@Alessandro yeah they actually speak like that

just watched a video lol

Alessandro Codenotti

9:30 PM

That's amazing thanks

Edward Evans

Also, it's referred to as the "Brummie" accent

Alessandro Codenotti

@BalarkaSen Ok so this is like a twisted torus and the action moves point along the big circles?

Balarka Sen

yeah this is some nonsense like this

Alessandro Codenotti

Ah ok and then you restrict to the periodic orbits because the full action is point transitive

Balarka Sen

right

I have no idea if this is actually topologically transitive though

Alessandro Codenotti

9:36 PM

I'm not sure either

Balarka Sen

I'd hazard a guess that it may be? flowing a little interval on the meridian makes it larger and larger as it returns to the meridian

because $z^2$ stretches arcs

Alessandro Codenotti

The problem is that it sounds plausible, but for compact Hausdorff spaces topologically transitive does imply point transitive

(because they are Baire)

Balarka Sen

but the union of the periodic orbits is not compact

Alessandro Codenotti

Oh right, sorry

Makes sense

I had already forgotten that we restricted the action

Balarka Sen

yeah; you can alternatively start with the 2^blah roots of unities in $S^1$ than $S^1$ itself

z^2 is topologically transitive on that

if it's true it should be easy to prove that if $T : X \to X$ is top transitive then the $\Bbb R$-action on the mapping torus of $T$ is top transitive

if it's false it would be false lol

Alessandro Codenotti

9:39 PM

Right lol

Seems like a very good candidate to me, but I'll need to think about it

Balarka Sen

same

Alessandro Codenotti

I'm afraid I need to sleep first, my brain doesn't work anymore

Balarka Sen

I am going to bed as well lol

cya

sorry for terrible candidate counterexample

Alessandro Codenotti

It's an ugly question, I'm expecting an ugly answer

Good night

AbstractAlgebraLearner

10:16 PM

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Got a good one for you

nbro

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