Mathematics - 2020-07-04

12:41 AM

What's the name of the field that studies how global properties of manifolds are determined by properties of solutions of differential equations on those manifolds, or something along those lines?

Mike Miller

Do you have an example in mind because I don't know what you're thinking of

Thorgott

12:53 AM

Hodge theory?

Ted Shifrin

1:25 AM

Sounds like parts of global analysis.

Hodge theory is too specific, methinks.

Knight

1:36 AM

@TedShifrin Hello Sir!

If a periodic function have a period $T \in \mathbb Z$, then it's Fourier series is
$$
f(t) = a_0 \sum_{n=1}^{\infty} \left(
a_n \cos ( \omega n t) + b_n \sin(\omega nt)
\right)
$$

Where $\omega = \frac{2\pi}{T}$. Considering our function continuous, we have
$$
a_0 = \frac{1}{T} \int_{0}^{T} f(t) dt
$$

Balarka Sen

Grumble I can't sleep. Also I have to read probability

Mike Miller

probably the name they're looking for is "geometric analysis"

that's not really what i would call the thrust of that field, but i don't think there's any field by that description

Knight

For $a_n$ we have
$$
a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(\omega nt) dt \\
a_n = \frac{2}{T} \cos(\omega n t) \int_{0}^{T} f(t) dt + \frac{2}{T}\int_{0}^{T} \sin(\omega n t) \int_{0}^{T} f(t) dt ~~dt \\

a_n = 2a_0 \cos(\omega n~t)\bigg|_{0}^{T} + 2a_0 \cos(\omega n t)\bigg|_{0}^{T}$$

What's the mistake?

Knight

2:02 AM

I got my mistake

Drathora

Hey guys ^^

user69608

3:42 AM

@Drathora hi

Mary Star

5:29 AM

Hello!! Let $1\leq m,n\in \mathbb{N}$ and let $\mathbb{K}$ be a field.
For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$

Let $\lambda$ be the eigenvalues of $\mu_a$ then we have that $\mu_a(c)=\lambda c$ so we get that $ac=\lambda c$, or not? But what does this mean? That $\lambda$ is also an eigenvalue of $a$ with eigenmatrix $c$ ?

Ted Shifrin

@MaryStar: Where are you getting the equation $\mu_a(c)=\lambda c$? This only holds if $c$ is a matrix whose columns are all multiples of the eigenvector of $a$ with eigenvalue $\lambda$.

Knight

If Fourier series for $f(t)$ is $$f(t)= a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos (\omega n t) + b_n \sin(\omega n t) \right]$$ then what will be the Fourier series for $f(t+a)$ ?

Mary Star

@TedShifrin Do we not use this definition : en.wikipedia.org/wiki/… ?

Ted Shifrin

What is the period of $f$ and what is $\omega$, @Knight?

I know what an eigenvector is, @MaryStar. Where are you getting your equation?

Knight

5:44 AM

@TedShifrin $f$ is a continuous function with period $T$ and $\omega = \frac{2\pi}{T}$

Mary Star

We have in this case $T=\mu_a$ and $v=c$, or not? @TedShifrin

Ted Shifrin

Oh, I misread, @MaryStar. I apologize. I thought $\lambda$ was an eigenvalue of $a$.

Leaky Nun

@TedShifrin clearly $f : \Bbb R/(\frac{2\pi}\omega)\Bbb Z \to \Bbb C$ /s

Mary Star

@TedShifrin Ah ok! So is it correct that from $\mu_a(c)=\lambda c$ we get that $ac=\lambda c$ ? Does this mean that $\lambda$ is also an eigenvalue of $a$ with eigenmatrix $c$ ?

Ted Shifrin

So, yes, @MaryStar, if $\lambda$ is an eigenvalue of $\mu_a$, then there is some nonzero $c\in\Bbb K^{m\times n}$ with $\mu_a(c)=\lambda c$. You didn't state quantifiers.

Then, as I stated, the columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$. I don't know what an eigenmatrix is.

Mary Star

5:50 AM

Ah ok! What I want to show is that $\text{Sp}(\mu_a)=n\text{Sp}(a)$. With the above we have shown that an eigenvalue of $\mu_a$ is also an eigenvalue of $a$, right? What is left to show to get the desired result?

Ted Shifrin

What does that equation mean?

Mary Star

I also a little confused with that, I haven't seen that notification before in the notes. Is it maybe meant the number of $\text{Sp}(\mu_a)$ is equal to $n$ multiplied by $\text{Sp}(a)$ ? @TedShifrin

Ted Shifrin

Sp is the spectrum, i.e., the set of eigenvalues?

Mary Star

@TedShifrin Yes

Knight

Ted when you get free please give your views on how we can go on for simplifying the Fourier Series for $f(t+a)$

Ted Shifrin

5:58 AM

Then the equation makes no sense to me.

Mary Star

@TedShifrin Ok, so it cannot be meant the cardinality, can it?

Ted Shifrin

You never answered my question ages ago, @Knight. You need to translate your variables and use addition formulas for sin and cos. But I'm leaving now.

I have no idea, @MaryStar.

Good night, all.

Mary Star

Ok.. Thank you!! Good night!! @TedShifrin

Knight

@TedShifrin I answered it sir

15 mins ago, by Knight

@TedShifrin $f$ is a continuous function with period $T$ and $\omega = \frac{2\pi}{T}$

Mary Star

Just one thing.... Now I looked again... "sp" is not the spectrum..it is the trace @TedShifrin

Leaky Nun

6:16 AM

Spur (Mathematik)

Die Spur (Spurfunktion, Spurabbildung) ist ein Konzept in den mathematischen Teilgebieten der Linearen Algebra sowie der Funktionalanalysis und wird auch in der Theorie der Körper und Körpererweiterungen verwendet. == Die Spur in der linearen Algebra == === Definition === In der linearen Algebra bezeichnet man als die Spur einer quadratischen n × n {\displaystyle n\times n} -Matrix A {\displaystyle A} über einem Körper K {\displaystyle K} die Su...

aha that's why

Mary Star

Yes.. But how is the trace of a linear transformation defined? Again the sum of eigenvalues? @LeakyNun

Leaky Nun

sure

it's also the sum of the diagonal

Mary Star

For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$ We have that an eigenvalue of $\mu_a$ is also an eigenvalue of $a$. So what do we need to get the result? Do we have to check the multiplicity? @LeakyNun

Mary Star

6:48 AM

Is the multiplicity of an eigenvalue equal to the number of eigenvectors that correspond to that? @LeakyNun

Knight

Can we simplify the Fourier series of $f(t+a)$ to get something like the Fourier Series of $f(t)$. $f$ is a continuous function with period $T$ and $a \in \mathbb R$.

Mary Star

7:03 AM

Because if the multiplicity of an eigenvalue is equal to the number of eigenvectors that correspond to that, then since the columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$, we would have that the multiplicity is equal to the number of columns, i.e. n, and so we would get the desired result, or not? @LeakyNun

Calvin Khor

8:18 AM

@Knight yes its a simple change of variables, you end up needing to multiply by a complex exponential so its not only the fourier series of $f(t)$ but that's life for you

Knight

@CalvinKhor That’s really what I found !

We should write Fourier Series in form of exponential not in trig form

But what’s the exponential form of Fourier Seires lol

Calvin Khor

8:34 AM

instead of $c+\sum_{n=1}^\infty a_n \cos(n x) + \sum_{n=1}^\infty b_n \sin(n x)$ (I'm ignoring the period) you write $\sum_{n\in\mathbb Z} c_n e^{inx}$. You can convert exponentials to trig using $$e^{\pm inx} = \cos(nx) + i (-1)^n \sin(nx)$$ and trig to exponentials using $$\cos(nx) = \frac{e^{inx}+e^{-inx}}2, \ \sin(nx) = \frac{e^{inx}-e^{-inx}}{2i}$$

@Knight

lmao oops instead of $(-1)^n$ i should have wrote $\pm$, that was some weird autopilot typing

Knight

@CalvinKhor So, we have $$ f(t)= \sum_{n=-\infty}^{\infty} c_n e^{i ~n ~t} \\ f(t+a)= \sum_{n=-\infty}^{\infty} c_n e^{int + in a} = \sum_{n=-\infty}^{\infty} c_n e^{int} e^{ina}$$

Am I right?

Calvin Khor

yes

Knight

But I still cannot see how to write $f(t+a)$ in form of $f(t)$, that sigma operates on the whole product.

Calvin Khor

yes, you can't, but you can write one in terms of the other like above, that's probably as far as you can go in general

sorry misunderstood you

Knight

No, you understood very correctly. Thanks for your help Cal

Calvin Khor

8:43 AM

if you prefer the computation directly for the coefficients
$$ \mathcal F[f(t+a)](n) = \int_0^{2\pi} f(t+a)e^{-int} dt = \int_a^{2\pi+a} f(\tau) e^{-in(\tau-a)} d\tau = e^{ina} \mathcal F f(n)$$

i forgot to divide by $2\pi$

Knight

What would $a_0$ correspond to in this exponential expression.

Is it $c_0$ ?

Calvin Khor

that's just $n=0$

yeah

Knight

Okay

Calvin Khor

np!

Knight

9:15 AM

@CalvinKhor Are coefficients $c_n$ same for both $f(t+a)$ and $f(t)$?

36 mins ago, by Knight

@CalvinKhor So, we have $$ f(t)= \sum_{n=-\infty}^{\infty} c_n e^{i ~n ~t} \\ f(t+a)= \sum_{n=-\infty}^{\infty} c_n e^{int + in a} = \sum_{n=-\infty}^{\infty} c_n e^{int} e^{ina}$$

Calvin Khor

the symbol $c_n$ there is the same, but the fourier coefficients of the tranlated function are not those of the original function

rather its $\mathcal F[f(t+a)](n) = e^{ina} \mathcal F[f(t)](n)$

I write it this way instead of $c_n$ so that its clear which function's fourier coefficient you're talking about

but i gtg hope thats clear ttyl

Knight

Bye

Thanks you so much

Manolis Lyviakis

9:52 AM

@BalarkaSen is that formal enough? and how would i procceed for n>2

feynhat

@MikeMiller I agree. The join model of EZ/2 gives S^\infty. But I want to construct EZ/2 with the simplicial model.

Abdullah O. Alfaqir

Hello. Could anyone please help me with understanding the Proof of change of variables theorem from Spivak's calculus on manifolds?

feynhat

@BalarkaSen How though? EZ/2 has 2 0-simplices, 1 1-simplex, and all higher simplices are degenerate. For higher simplices, say, [0, 1, 1], we identify glue the face [0, 1] with the 1-simplex [0, 1], right?

If so, then EZ/2 has no higher simplex. How can the quotient BZ/2 have a non-degenerate simplex in each dimension?

i m confooz

@ManolisLyviakis What are you doing?

Manolis Lyviakis

10:07 AM

im trying to prove munkres 53.1 theorem

of his topology book

leme posted

but its esay to waave hands in homotopy theory i want to be formal i think i proved it for n=2

for n greater than 2 my $g$ function will be the same its restriction to the different n pieces of $I$ $g=p_i( of something ) $ where $p_i$ is the linear positive map of $I$ to $[a_{i-1} , a_i ] $

im running the positive lineara maps faster thats all im doing

what you think @feynhat if you have time check my proof if it makes sense and it is formal enough.

Thorgott

Why is this not clear? $f$ is the same thing as the concatenation $f_1...f_n$ up to a piecewise linear reparametrization that fixes the endpoints, so homotopic

Manolis Lyviakis

its clear

im asking if what i wrote above is an ok way to formulate what you just said

orientablesurface

11:04 AM

hello

If $g\circ r^1$ is smooth, where $r^1$ is a standard coordinate on $\mathbb{R}^2$, does it follow that $g:\mathbb{R}\rightarrow \mathbb{R}$

is smooth?

Thorgott

$r^1$ restricts to a diffeo $\mathbb{R}\times\{0\}\cong\mathbb{R}$

orientablesurface

aha , thanks

so yes

$\mathbb{R} \times \{0\}$ is a smooth manifold?

Thorgott

11:21 AM

it's a product of smooth manifolds

orientablesurface

oh right, singletons of $\mathbb{R}^n$ are smooth

Knight

If $f$ is a positive continuous function, and if we have $$\int_{0}^{1} f(x+a) -f(x) dx \lt 0$$ $a \in \mathbb R$ then can I conclude that $$f(x+a) \lt f(x)$$ for all $x$ in domain?

Thorgott

does a function having negative integral over some interval imply that it's everywhere negative?

draw a picture in case the answer isn't clear

Knight

11:35 AM

@Thorgott Okay! I got it, all we can conclude is that area below the $x-$ axis is more than the area above the $x-$ axis under the graph of $f(x+a) -f(x)$ in the interval $[0,1]$. Right? But how to write this information analytically?

Thorgott

$\int_0^1f(x+a)-f(x)dx<0$

that's precisely what the integral means

Knight

@Thorgott Can’t we get anything more than that?

I mean from that can we conclude anything more? $f$ is always positive is given

Thorgott

What do you want to conclude?

Knight

$f$ is a continuous periodic positive function with period 1. I want to prove that the average value of $f(x+a)$ is more than that of $f(x)$

More than or equal to, I meant

Thorgott

they will in fact be equal, so try proving that

Knight

11:50 AM

So, I attempted to arrive at a contradiction, $$ \int_{0}^{1} f(a+x) dx \lt \int_{0}^{1} f(x) dx $$

@Thorgott They can be equal

Thorgott

I'm saying they always are

the periodicity assumption is necessary for this, of course

Knight

Really? Can you show me the path for proof? I’m pretty much excited to see and it’s about to dawn and we would have enough sunlight

Thorgott

The claim is that $\int_0^1f(x+a)dx=\int_0^1f(x)dx$. Try starting with a substitution and then use periodicity.

Knight

Let $x+a=u$, we have $$\int_{0}^{1} f(u) du = \int_{0}^{1} f(x) dx $$ What’s next?

Thorgott

That would already be the conclusion

Mike Miller

11:59 AM

@feynhat You still haven't explained why you thought the result from some other method is RP^n.

Knight

Yes I understand $u$ and $x$ Are dummy variable but if antiderivative will exist then I think $u$ will appear as $x+a$. I believe in you Thor but can you please help me a little more in convincing myself about the conclusion?

Mike Miller

I still don't understand the claim "EZ/2 has no higher simplex".

Thorgott

Well, you need to perform an actual calculation to convince yourself of this conclusion

Knight

:)

Mike Miller

Actually, I'm going to bow out, because it's not very interesting to me. Sorry.

Shouldn't have pulled the bait and switch.

CowperKettle

12:23 PM

feynhat

@MikeMiller Sorry. I thought all RP^n (n > 1) were K(Z/2, 1). My bad.

Knight

@CowperKettle I admire your jokes but I never get them :-)

Would you explain this one?

feynhat

Any n-simplex of $EG$ can be written in the form $[g_0, g_0g_1, \dots, g_0g_1\dots g_n]$, where $g_i \in G$. But our $G$ has only two elements. So any 2-simplex will degenerate into a 1-simplex right? Like [0, 1, 0] or [0, 1, 1]

@Knight Its the graph of y=tan(x).

By 'no higher simplex', I meant all higher simplices are degenerate.

Knight

12:39 PM

@feynhat yes but why only the graph of tan not any other function?

feynhat

uhh... tan line?

Knight

Oh okay!

user193319

Let $\mu$ be a Haar measure on a locally compact group $G$, let $U$ be any open subset with finite measure, and let $H$ be an $\sigma$-compact open subgroup. Why does $U$ intersect only countably many cosets $xH$?

Mike Miller

@feynhat Well, $S^n$ is not contractible.

I don't really follow why [1,1,1] is a degenerate simplex.

I tend to feel like if you want something concrete than build a concrete model. The simplicial model is abstraction meant to work for formal reasons. So you should work with it formally.

feynhat

Wouldn't the geometric realization of [1, 1, 1] be a 0-simplex?

feynhat

1:05 PM

@MikeMiller I just want to see it for one simple case (that's why I chose Z/2).

Mike Miller

1:21 PM

??? The 2-simplices are by definition triplets $[g_0, g_1, g_2]$. So their geometric realization (before quotienting etc) are 2-simplices. But

If you want to argue that the image of that in the geometric realization (after quotienting) is a single point you'd have to make an argument.

Charlie

2:16 PM

What the name given to functions that are $\in C^\infty(\Bbb R)$, it's an "X function" where X is the name of someone, I can't quite remember

Leaky Nun

smooth

Charlie

There's some other name I swear

Something like "Jacob functions", with someone's name

Thorgott

never heard of that

Charlie

damn

this is going to drive me crazy

I must be wrong about them being smooth

fwiw I was thinking of Schwartz functions

Thorgott

2:32 PM

ah

shelton Benjamin

Can we use the differential equation adx=vdv for calculating freefall velocity in case of large height?

Can we use the differential equation adx=vdv for calculating freefall velocity in case of large height?
Like putting a=-Gm/(R+h)^2 and then solving by taking limits of integral as h to h-delta h to calculate velocity after particle has travelled delta h height

Balarka Sen

@feynhat I don't understand why this means every higher simplex is degenerate

I mean, I know for a fact they aren't. But what is the argument?

I mean, [0, 1, 0] should be a 2-simplex which is non-degenerate.

The geometric realization of the 1-skeleton of EZ/2 is an interval (0 ----> 1), and [0, 1, 0] is a 2-simplex which you glue to this by pasting the edges [0, 1], [1, 0] and [0, 0] to the edge 0 ---> 1, to the edge 0 ---> 1 but oriented oppositely, and to the degenerate edge 0 ---> 0 (i.e. collapse to 0) respectively

The triangle is very much not degenerate. You might be forgetting what a simplicial complex is

You have simplices of the form [0, 1, 0, ba-dee-daa, 0, 1, lalalala, 0, 0, 1], sure, there are lots of vertex repetitions. That means when you actually attaching the simplex in the process of making EZ/2 it's faces are glued in a very complicated fashion to the lower skeleta

The interior of the faces aren't getting squished

Or rather, note that this EG is a delta-complex not a simplicial complex so it's faces doesn't determine what the simplex is

I have to run and read probability to prepare for a meeting tomorrow; ping me if you have q I'll respond later

user435118

4:54 PM

I have a math/programming question (mainly maths), can I ask here?

Leaky Nun

yes

user435118

I am given a number n, I want to find the next number after n where the sum of the digits in even positions in its number is equal to the sum of the digits in odd positions.

user435118

But looping through all the digits one-by-one is too long, apparently there is a quicker way to do it

user435118

n can go up to n to the power of 100

user435118

My current code loops through every number after n, splits into 2 lists (1 even index numbers and one odd), checks if they’re equal. If yes it returns the number if not it keeps going

user69608

5:22 PM

find the number of 7 digit numbers the sum of whose digits is even.

Knight

Does a continous peridoic function have a convergent Fourier Series?

feynhat

> Does 'a' continous peridoic function have a convergent Fourier Series?

Yes. Many do.

But many also don't.

Balarka Sen

This is a famous question of Fourier. The answer is no in general.

Leaky Nun

@Knight mathcounterexamples.net/…

Thorgott

depends on your notion of convergence

Balarka Sen

5:29 PM

By a famous theorem it converges almost everywhere anyway

feynhat

@BalarkaSen Thanks a lot for this.

This cleared up a lot of things

I was actually confusing the nerve construction with the delta complex construction

Balarka Sen

@feynhat Yeah actually I think I simply misspoke there. I think 0 --> 1, 0 --> 0, 1 --> 0, 1 --> 1 are all non-degenerate edges in the 1-skeleton.

Everything is nondegenerate

The homotopy gets killed by how higher simplices are attached at each stage: It's a massive contractible complex.

Leaky Nun

@BalarkaSen I'm degenerate

Balarka Sen

Every Lean coder is

Thorgott

if your function is integrable and has left- and right-hand limits at a point, then it's Fourier series converges to the average of these limits in Cesaro mean

Balarka Sen

5:32 PM

Nobody cares, Thorgott

Thorgott

this is actually important

Balarka Sen

No

In fact, you are unimportant.

Arjun

@BalarkaSen , btw, that's mean

feynhat

I should study simplicial complexes. I actually don't know much beyond the definitions.

...and stop thinking about concrete example for simplicial construction of EG.

Balarka Sen

I think it was a fine thing to think about

feynhat

5:37 PM

Thanks for your time though @BalarkaSen @MikeMiller

Balarka Sen

@Arjun Everyone is unimportant though

feynhat

@Arjun Cesaro mean, yes.

3

Balarka Sen

Lmao

Thats pretty good

I have to read probability Jesus Christ

Thorgott

The reason why this is important is cause it tells you that if you want the Fourier series to converge pointwise, its limit has to be the average of its one-sided limits (granted, I have no clue what happens with functions where these don't exist, but that probably is actually unimportant). This explains part of the behavior of Fourier series at jump discontinuities. This result can also be adapted to give a rather quick proof of uniqueness of Fourier series (for sufficiently nice functions).
If you impose additional conditions on the one-sided derivatives of the function, you can improve c…

Balarka Sen

Do you seriously think anyone is going to read that paragraph Thor

Arjun

5:39 PM

@BalarkaSen ,depends on perception, if you feel so, good.don't demotivate others

Thorgott

no, but that also didn't stop me from proving the projection formula in here

so why would it stop me now

Balarka Sen

I read your proof of the projection formula

you tricked me to

Mike Miller

@Arjun If that sentence demotivates you, then you've ascribed too much importance to Balarka

2

Balarka Sen

LMAO

Mike Miller

Taking his statement seriously means that he, too, is unimportant, his words leaves in the wind

Thorgott

5:41 PM

you deserved reading it :)

Balarka Sen

Too far, Thorgott, too far

Calling me a dirty algebraist

Thorgott

that's what you call me all the time

Balarka Sen

Because thats what you are

I am an honest man

Knight

@LeakyNun Our function is periodic

Thorgott

2/3.5 things I'm doing this semester are analysis/topology

Knight

5:43 PM

@Thorgott Point wise convergence

@BalarkaSen Really? But you said above that in general it is not true.

Balarka Sen

"almost everywhere pointwise convergence" is different from "pointwise convergence"

Thorgott

yeah, won't happen in general

Balarka Sen

what is up with these probabilists by the way

Thorgott

also Balarka forgot to mention that the famous theorem has an additional hypothesis

Balarka Sen

a random metric space??? the hell

yeah I did mean L^2

Thorgott

5:46 PM

L^p for p>1 suffices

Balarka Sen

NOBODY CARES

Knight

Doesn’t periodicity ensures convergence?

Thorgott

lol, Hunt sure did care

Balarka Sen

you think a man named Hunt is important?

totally ridiculous, just look at his name

Thorgott

I'm not saying he's important, I'm saying he cared

Knight

5:48 PM

Who is senior Thor or Bala?

Balarka Sen

yeah because he was a nobody

nobody cares

checkmate atheists

Thorgott

curiously, for functions on the real number line, you need L^p for some 1<p<=2

I guess p>2 isn't trivial anymore cause you don't have the L^p space inclusion anymore, but I wonder why 2 is the exact barrier

Balarka Sen

@LeakyNun Drawnish Giri drew 3 perfectly winning games

Mike Miller

6:33 PM

Why does it matter who is senior

If they're talking about something purely mathematical then you can read the justifications and figure out which is correct or incorrect

If it's non- or pseudo-mathematical then you can read their informal arguments and decide which informal argument you find more convincing and take that one more seriously

user435118

Anyone got any ideas?

Balarka Sen

given earlier conversation I think it's safe to say we are both 8th graders calling each other names

user435118

2 hours ago, by Daniil

I am given a number n, I want to find the next number after n where the sum of the digits in even positions in its number is equal to the sum of the digits in odd positions.

Thorgott

yeah, 8th grader sounds about right

Balarka Sen

6:51 PM

Hm. Suppose $(\Omega, \mathcal{A}, \mu)$ is a probability measure space, $T : \Omega \to \Omega$ is an ergodic transformation and $f \in L^1(\Omega, \mathcal{A}, \mu)$.

Think of the iterates $X_n = f \circ T^n$ as random variables. What's the probabilistic interpretation of $T$ being ergodic in terms of the stochastic process $\{X_n\}_{n \in \Bbb Z_+}$?

It must be something like, $X_{n+k}$ becomes asymptotically independent to $X_k$ as $n \to \infty$

That's like mixing, which is stronger, I guess

$\Bbb P(X_n \in A, X_0 \in B) = \mu(T^{-n}(f^{-1} (A)) \cap f^{-1}(B)) \to \mu(f^{-1}(A)) \mu(f^{-1}(B)) = \mu(T^{-n}(f^{-1}(A)) \mu(f^{-1}(B)) = \Bbb P(X_n \in A) \Bbb P(X_0 \in B)$

Ergodicity implies strong mixing so yeah

Any function on some chunk of the stationary process and some other function on a shift of the chunk taken earlier become asymptotically independent as we shift more

Weak mixing would translate to asymptotic uncorrelation

Sophie

7:06 PM

I have an infinite group generated by $\{t_1,t_2,t_3,t_4,t_5\}$ and also $t_1^{10}=t_2^{10}=t_3^{10}=t_4^{10}=t_5^2=e$ and also things like $t_2t_1t_2t_2t_1t_2=e$. Does anyone know if this group might be isomorphic so something interesting?

Balarka Sen

What a horrendous group

Sophie

it's the orientations obtained by placing regular pentagons side by side in the plane

which means I can make pretty things with it like this i.imgur.com/jJoku7L.png

Balarka Sen

Ah yeah i was wondering if it's something like that

Sophie

the pretty picture more than makes up for the ugly algebra imo

Balarka Sen

these x^p = y^q = z^r = 1 etc stuff appears when you take oriented isometry groups of geometric figures on the plane

i was thinking of what you get if you do it with a triangle, eg the von Dyck group

these are not crystallographic groups on the plane though

user193319

7:13 PM

Conjecture: Let $G$ be a topological group. If $C \subseteq G$ is path connected subset, then $\langle C \rangle$ is a path connected subgroup. I believe I have a counterexample: Let $G = \Bbb{Z}$ with the discrete topology and let $C = \{1\}$. The set $C$ is trivially path connected, because it is a singleton, but $\langle C \rangle = \Bbb{Z}$ is not path connected.

Does this sound like a genuine counterexample?

Thorgott

ye

user193319

yeet

Balarka Sen

@Thorgott you know youre in big trouble when the name talagrand pops up in what youre reading

user193319

yeet bois

Thorgott

I just spent an entire minute thinking about "minimal maximal ideals" until I realized what I was doing

@Balarka I only stumbled upon his concentration inequality once, I think

Vrouvrou

7:43 PM

please someone have a reference for this :math.stackexchange.com/questions/3745349/…

user193319

What's the definition of a compact neighborhood? Is it an open set which is compact?

Thorgott

a neighborhood which is compact

neighborhood being any set containing an open set about the point

being compact and open is a somewhat unlikely property

user193319

@Thorgott Ah, okay. I see. How do I prove that in any locally compact group, there is a compact unit-neighborhood?

Nevermind...

Thorgott

@Vrouvrou Clearly, $|E|\le|G|$. On the other hand, $|G|\le\sum_{J\subseteq E,J\text{ finite}}|\mathbb{Z}^J|=|\{J\subseteq E\colon J\text{ finite}\}|\cdot\aleph_0=|E|$. By Cantor-Schröder-Bernstein, $|E|=|G|$.

Drathora

8:12 PM

Hey guys

Thorgott, did you get any time to look into weakening the constraints on the coarea formula?

I've been looking around a lot but struggling to get away from the need for the function to be Lipschitz

Thorgott

not yet, I'm afraid

been kinda busy

there was a comment in Federer about only requiring bounded limsup of the difference quotient a.e. or something

Drathora

8:32 PM

Yeah, a lot going on here also >_<

I found a lot of things about rectifiable sets but that condition still required a Lipschitz map

Vrouvrou

@Thorgott I don't know algebra and I want to help someone that's why I'm sourcing a reference

why $|E|\le|G|$?

Thorgott

you can easily come up with an injection from $E$ into $G$

Vrouvrou

8:47 PM

I don't know. how ?

Thorgott

the most natural choice of map will work

Vrouvrou

9:04 PM

$G=\{f:E\to Z, |supp(f)|<\infty\}$ can i define the identity $id:E\to G$ ?

but how it can be defined ?

@Thorgott

Thorgott

no, an identity map always goes from a set to itself, but $E$ and $G$ are different sets

Vrouvrou

ohh yes I don't see it

I don't know what I can define

Thorgott

9:22 PM

I give you an element of $E$. Can you give me a map $E\rightarrow\mathbb{Z}$ that has something to do with this element?

Vrouvrou

the absolute value ?

Thorgott

what's the absolute value?

Vrouvrou

for $a\in E$ then |a|\in Z

Thorgott

that doesn't tell me what the absolute value is

Vrouvrou

9:39 PM

$|.|: E\to Z, a\mapsto |a|$

Thorgott

those are symbols, not definitions

Vrouvrou

a norm if the element of f. are vectors in general

Thorgott

I do not understand this sentence

Vrouvrou

9:54 PM

the map that define from E to Z is the norm map

Thorgott

what is the norm map?

Vrouvrou

sorry I can't choose the norm map because ||a||\in R^+

I don't how to define the map from E to Z

Thorgott

I give you an element $x\in E$. Can you give me a map $E\rightarrow\mathbb{Z}$ that does one thing with $x$ and a different thing with all the elements of $E$ that are not $x$? (The reason for this requirement is that we want to have different maps for different $x\in E$ in order to get an injection from $E$ to $G$)

Vrouvrou

no I don't know

Thorgott

10:14 PM

then I don't know how to help you, sorry

Vrouvrou

I need a book

or I can define $\chi_a(x)= \begin{cases}1 \, if\, x=a\\0\, if\, x\neq a\end{cases}$

here $a$ is fixed on E

Thorgott

10:30 PM

yup, that works!

Balarka Sen

10:56 PM

Hm, I forgot how to couple random variables. Let $X, Y$ be random variables with $\Bbb P(X \leq t) \leq \Bbb P(Y \leq t)$ for all $t \in \Bbb R$, I want to modify them without changing the distribution such that $X < Y$ (pointwise on the probability space). What do I have to do again?

Oh nevermind I just simulate

$U$ be uniform on $[0, 1]$, $X = F^{\leftarrow}_X(U)$, $Y = F^{\leftarrow}_Y(U)$ where $F_X, F_Y$ are the distribution functions

That does the trick

This trick never gets old

Balarka Sen

11:35 PM

The inequality is flipped above but ok

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