@user76284 on what page of that paper did you see that notation? I just did a quick scan of it and did not see it.

Hi. I'm aware that mixing is a very general concept, but I'm considering just strong mixing coefficients vs covariance as dependence measures. The former always exists, while the latter doesn't have to. Are there any other advantages of mixing coefficients in this case?

Page 3017?

ok

"Definition 3.3. The Schwarzian derivative of $f$ is the tensor..."

I guess all I have left to figure out is if $A_{ij}^k = \delta_j^k B_i + \delta_i^k B_j$ how can I express $A$ in terms of $B$ in index-free notation.

It's just $I \otimes B$ or something?

Plus the reverse.

How do you write that?

But it's tricky, because you're tensoring different types of tensors.

So we're symmetrizing in the $i,j$ indices. I don't know if I see how to do this invariantly.

Yeah, $A^k_{ij} = \delta^k_{(i}B_{j)}$ in Penrose's notation I think.

Right.

So this lives in $V\otimes V^*\otimes V^*$, and we need a map that switches the last two factors. Call it $\sigma$. Then we're doing something like $I\otimes B + \sigma^*(I\otimes B)$.

@Julius Interesting,

Does zero mixing coefficient imply independence, like distance correlation? If so that would be an advantage

@TedShifrin This looks relevant: en.wikipedia.org/wiki/…

I think that's wayyyyyy off the deep end.

I think what I said is fine. You can check it by evaluating on $(\xi,v,w)\in V^*\times V\times V$. :)

That is still index-free.

You should make sure you know how to do that, btw @user76284.

@student, yes, it implies independence. Other than a couple of advantages specific to my context, I also came up with the fact that mixing coefficients don't need any kind of stationarity of the underlying process.

Hello!

Hello @TedShifrin

How are you?

hmmm, who are you?

maths

LOL, oh

:))

didn't make your name any shorter, I see

ahaha true, I think I have a problem with that, my usernames tend to get longer every time I decide to make a new username

You need to apply a shortening algorithm.

hahaha, I should

@TedShifrin, I have a question about a proof, is it okay to ask you?

Sure.

Proof: If $E_a$ is a collection of open sets then $\bigcup\limits_{a}E_{a}$ is an open set. Proof:Since $\forall$ a $E_a$ is open it follows that x$\in$ $E_a$ $\implies$ $\exists r>0$ : $N_r(x)$ $\subset$ $E_a$. Since x$\in$ $E_a$ it follows that $N_r(x)$ $\subset$ $\bigcup\limits_{a}E_{a}$. $N_r(x)$ is the same as an open ball.

Sure.

Intersection is harder :)

is there something I can do to improve it?

(The last sentence doesn't need to be there. You probably put it there for me.)

oh, yes. I am going to do the intersection right now, and post it here, if you don't mind, and yes, I put it just incase I wasn't clear.

@Ted, this is my proof for the closed one: let x be a limit point of $\bigcap\limits_{a}C_{a}$ hence every neighborhood of x has a point q such that q is not p and q is in the intersection. Assume that x$\notin$ $\bigcap\limits_{a}C_{a}$ that means $\forall$ a x$\notin$ $C_a$ which is a contradiction, because $C_a$ is a closed set.

Oh, I thought we were doing intersections of opens.

This argument with limit points is gonna be tricky.

@TedShifrin for the one I just posted here, or for the intersection of opens?

Yeah, you're right. I was thinking union of closeds.

No, wait, your proof is wrong.

If $x\notin \cap C_a$, that means there's some $a$ so that $x\notin C_a$.

Oh, I see. But if I assumed all the $C_a$s are the same set, then my proof would work, am I correct?

But that never happens.

yeah, I see my mistake.

Thanks so much

Maybe a proof by contradiction isn't the best thing here? Erase that and go back to your first sentence.

Math is life , what a terrible misconception

Hey @TedShifrin ?

?

How come mathematicians tend to be very formal with each other ?

Most don't.

Maybe with just students? Or not even that too?

Nope.

I wonder why I have that false impression then...

Maybe it's you.

It’s definitely me but

Mathematicians tend to be less willing to put up with it than other departments

Which is good

Put up with what? Your weird obnoxiousness?

Ya

You've tested the patience of plenty of the folks here.

And I am pretty damn patient.

Let me ask you this

How come , and it may be a wrong impression , but it seems to me than not many people , atleast in the USA , do research into Index theory?

It seems to me like a very cool subject

It was huge during the 70s and 80s ... maybe not quite so huge now, because things have moved on.

But I'm not an expert on this.

Certainly Atiyah-Singer and such things still show up in research. But Witten theory and Floer homology have taken the forefront.

@Zee living with formal logic makes people very "formal".

@TedShifrin Assuming written theory means seiberg-Witten invariants , isn’t the foundations of that spin geometry ? And that is very much related to index theory? It seems they should be more excited about that...

Sure, but most of the index stuff is older stuff.

Nothing wrong with liking older stuff, but doesn't mean that it's the forefront of research.

@mathsssislife maybe , but one of the least formal people I met was a logician in the philosophy dept

@TedShifrin Oleg Viro , IDK if you know him , told me , “liking old theories is brave , becouse the cream has been skimmed off”

@TedShifrin I think I proved it, but I need to make sure that i'm write about the following: Let A$\subset$B. If x is a limit point of A then x is a limit point of B am I right? It makes sense because A is a subset, but I don't know why I don't feel confident.

Sure, that's right, unless $x$ is an isolated point of $B$. Then it's wrong.

Speaking of the index theorem , Atiyah passed away today...

@Zee it's interesting, because I think that usually when an individual is trained to think a certain way he\she usually applies that way of thinking to other aspects in life

Yes, I made some comments about Atiyah above.

@mathsssislife that’s kinsa true but I think it goes the other way more so , people apply their life to the area they work in , so a visual thinker who went into ring theory would think about rings visually

I meant to say right, not write in my previous post.

That’s an interesting paper , I like fiber bundles...

It's a beautiful paper.

@Zee yeah, I agree. The brain is quite interesting.

@TedShifrin, thats my proof: Let x be a limit point of $\bigcap\limits_{a}C_{a}$. Since $\bigcap\limits_{a}C_{a}$ $\subseteq$ $C_a$ $\forall$ a, it follows that x is a limit point of $C_a$ which implies that x $\in$ $\bigcap\limits_{a}C_{a}$

Because of what I said, I think you need to make the explicit argument for why $x$ is a limit point of $C_a$.

Why couldn't $x$ be an isolated point of $C_a$ and hence not a limit point?

TIL why the Sierpiński triangle is sometimes called the "Sierpiński gasket".

Here's a picture of an actual gasket.

Right, so it's like a version of that, except continuing smaller and smaller holes.

Not too much imagination required.

Yup.

Now I'm tempted to define a "gasket" as a figure which has infinitely many holes in it (each with nonzero area).

@TedShifrin I could consider isolated points, but what I want to show is that the intersection is closed (contains all its limit points), and since $\forall$ a , $C_a$ is closed (by assumption), that implies that the limit point is in the intersection

Well, not a real-world gasket.

But you need to write down the definition of limit point, I think, @maths.

Right, real gaskets are used for sealing gaps.

And then deduce ...

@Tanner, where there are finitely many holes.

In Polish, "Sierpiński gasket" is "uszczelka Sierpińskiego". In case anyone was wondering. :D

And yes, "uszczelka" means "gasket".

@TedShifrin, oh I see. Thanks so much Ted!

You're welcome. I'm leaving for now.

@TedShifrin. Have a great day!

Last night dream, right here in math chat, Balarka, I and a woman user who has a nickname VectorSpacter are exploring a strange question about the trend of the maximum volume of regular polytope for each dimension, where I asked them about whether the nth term of the sequence is computable without having to manually compute all the regular polytopes for each n and then pick the n-polytope with the largest volume. The sequence posted is like this:

> 0, 1, 3, 4, 120 , ...

Now there's just one problem, because the real life counterpart of this sequence makes no sense:

This is because the number of regular polytopes in each dimension is:

> 1,1, infinity, 5, 6, 3, 3, (and so on all threes)

Thus the n-polytopes of largest volume with the largest number of (n-1) facets for each n are:

> 0,2,infinity,12,600, (the rest are n-cross polytopes)

The idea which the nth term of the sequence can only be obtained by having all possibilities for that term, and then select the term that fits the criteria, seemed to be interesting to investigate further. However, it is not very special because the sequence of prime numbers already have this property under the sieve of eratus

What's potentially more interesting, is if in addition to this, you have a sequence whoose nth term not only need to be computed in this manner, but that you also need information from previous terms. That could form some kind of intermediate thing between computable sequence and uncomputable sequence

@TannerSwett another interesting question requiring careful historical analysis. when did the sierpinski triangle 1st become referred to as a "fractal"? ps watched this last nite (thx CA16!) & its cool + has a neat switcheroo, glossed over: peano curve(?) transformed into a snowflake. ted.com/talks/benoit_mandelbrot_fractals_the_art_of_roughness

For the tunnel examples, when you have seen those enough, you will mentally started to see the inner surface to be "squished" when looking from the outside, such that for the observer who is transversing it, it looks flat to them, but as seen from the outside, they look squished

And soon enough, I might be able to see this object as it really is in its native space

Thus globally speaking, they are like hyperbolic manifolds

@MatsGranvik Hi Mats, where are you from if I may ask ?

@KasmirKhaan Finland, but I belong to the Swedish speaking minority.

that is cool ! =p

@KasmirKhaan Where are you from?

@MatsGranvik live in Sweden =P

Vad pluggar du till ?

Jag har pluggat färdigt. Jag studerade till Diplom-ingenjör och blev färdig 2004 från Åbo Akademi.

kemiteknik

Det är ju bra det , du såg mycket yngre ut på fotot

Jag är 31 på fotot. Nu är jag 41.

år

haha okej

forever young...

that is the spirit !

Except, computers in the future, that is now, has zero vacuum tubes and weight less than a can of soda

Things like these does make you wonder, perhaps infinity does exists in some sense:

You are given a bunch of materials in order to approach some limit where almost none of that material is used in the final product, and then the future shows you a completely new way to make the product so that not only you can approach zero, but actually reach zero

A fundamental understanding on what exactly governs when a technological breakthrough occur, may provide us some ideas in mathematics on how we can reach those infinities without first axiomising their existence

Hi chat.

I have a question: is any vector space $V$ over $K$ "something like" $R^n$? More formally, does there always exists an isomorphism between $V$ and $R^\alpha$, for some ring $R$ and ordinal $\alpha$?

@Lucas $K$-vector spaces are "free $K$-modules" (every vector space has a basis) so they look like $K^n$

and the answer is yes

@MatheinBoulomenos I still don't see it for some reason. Can you explain?

Is it true that every geodesic metric space is path connected under the topology induced by the metric?

A geodesic is a path with extra properties, right?

I mean, for example, when you have a two option bet (for example bet about over and under 2.5 goals), if both bets have a greater odd than 2.0, you will have a sure bet and you can bet 1€ to each of them to get money for sure

I would like to know how to compute this, because it is trivial when both has a greater odd than 2.0, but when not I do not know how to know it

Do you have any idea?

Also for three option bets, how could be compute that?

$M$ compact $n$-dim smooth manifold, $\omega \in C^\infty \Omega^{n-1}(M)$, then $\displaystyle \int_M \ \mathrm d\omega = \int_{\partial M} \omega$

I have it

is it true that $(\ell^1, \rVert * \rVert_{\ell^{\infty}})$ is a dense linear subspace of $(c_0,\rVert *\rVert_{\ell^{\infty}})$?

sorry, it was a nonsense question, I solved my doubt ;)

RIP Atiyah

How to prove that $G/K \cong G/N$ where $G=\mathbb{Z}_{4}\times\mathbb{Z}_{2}$, $K=\{0,2\}\times\mathbb{Z}_{2}$, $N=\mathbb{Z}_{4}\times\{0\}$?

well they're both order 2 and there's only one such group! :)

@vesii convince yourself that $G \cong S_3 \times S_3$

@vesii try to take G/K / G/N using third iso thm

@LeakyNun yo leaky , do you know algebraic intergers ?

@KasmirKhaan sure

like how to prove that a number is algebraic ?

hmm

are there algoritms to follow

well you only need to find a monic polynomial which it satisfies

or just straight up following the defintion ?

and "guessing" a poly

well like if x=sqrt(2)+sqrt(3)

then you can go like x-sqrt(2)=sqrt(3), so (x-sqrt(2))^2=3, so x^2-2sqrt(2)x+2=3, so x^2-1=2sqrt(2)x, so (x^2-1)^2=8x^2, so x^4-10x^2+1=0

and you've just proved that sqrt(2)+sqrt(3) is an algebraic integer

actually the algebraic integers form a ring

my question involves cube root

so if you prove that sqrt(2) and sqrt(3) are algebraic then you can conclude that sqrt(2)+sqrt(3) is algebraic

@KasmirKhaan same method

hmm so the process is like that

no deep theory about it ?

or anything that makes one find them easy

and the general method to prove α is algebraic is to find the product of (x-β) where β ranges across all the conjugates of α

of course there's a deep theory behind algebraic integers!

and I also just told you a theorem that makes it easier!

in what course would one take that?

NT?

algebraic number theory

okay thanks !

@LeakyNun math.stackexchange.com/questions/3069049/…

@LeakyNun take a look =p

ok

I offred all what I GOT !

do you still not understand what D means though

its just random

thing

what he did was making D

and also no more questions about algebraic integers?

that process seems to work for sqrt

not cube roots

like take this exa,ple

5+cube root of 2

how does one show that is alg?

x=5+cbrt(2)

x-5=cbrt(2)

(x-5)^3=2

done

@KasmirKhaan ok?

@LeakyNun am not sure i understood the defintion well

algebraic means that we can find a poly

such that once we put that number in the poly

we get a zero right?

yes

s.t the coefficients are integers

hmmmmmmmmm

no that's integral

and my poly is (x-5)^3-2 = x^3-15x^2+75x-127

and I also told you a theorem that should make your life easier

let me try it on other part=p

@LeakyNun how does one choose what to put on the side of x ?

like if we had an expression with complex numbers

if we move "i" to other side

give me an actual example

squaring or cubing it wont help

ok

3+5i

sqrt 3+ 5i

x=sqrt(3)+5i, x-sqrt(3)=5i, (x-sqrt(3))^2=-25, x^2-2sqrt(3)x+3=-25, x^2+28=2sqrt(3)x, (x^2+28)^2=12x^2, x^4+44x^2+784=0

@KasmirKhaan

okay I think I get it now

great

we just have to manipulate the expression

untill we get an integer

and I think you're still misleading me

are we to find an actual polynomial, or just show that it exists?

the optimal part is to get an expression with roots as product

no just proving it is algebraic

so yeah

polynomial

otherwise how would one show that?

because I think you ignored the theorem I gave you

it was an example to follow

you just show that sqrt(3) and 5i are algebraic

seperatly would be nice

alrighty

thanks leaky , sometimes you give me good stuff

other times you give me less good stuff

:D

what times?

it is a joke -.-

instead of saying thanks twice -.-

is there anything you don't understand?

@LeakyNun right now it is clear! ill keep digging and come back later :D

ok

Hi. Is it true that checking integral linear program for infeasibility can't be done in polynomial time?

@MatheinBoulomenos Hi

@TedShifrin Hi Ted!

@TedShifrin I just put a good bounty one the question from yesterday, I know you gave me a good answer but just for curiosity =p

@MatheinBoulomenos sup mathein ? :D

Hi Guys, how to show that a simple and not connectivity graph with 7 vertexes has at most 15 arrows?

I thought of using the Pigeonhole principle but without any success.

@TTaJTa4 Hint: The complete graph on 6 vertices has 15 edges (6*5/2)

@AkivaWeinberger every not connectivity graph is a subgroup of the complete graph?

*connected

And notice that I said 6 vertices and not 7

why 6? so it will not be connected?