Mathematics - 2018-08-24 (page 1 of 3)

Rudi_Birnbaum

04:27

Hi all, I need to understand the following group theoretical statement (which is made in the context of chemical applications but is of course independent of those):"According to Group Theory the existence of an $n-1$ fold degeneracy* is related to the existence of a set of n identical sites which form a doubly transitive orbit of a symmetry group."

*)degeneracy here means dimensionality of the irreps over $\Bbb R$ (which in turn can of course be anytime translated into language of irreps over the algebraically closed $\Bbb C$ in observing that pairs of complex conjugate irreps are to be subsumed under 2-dimensional irreps.) Any help appreciated!

Drew Brady

Hey anyone able to help me with this seemingly obvious measure theory problem?

Trying to show that a line in R^2 has Lebesgue measure (in R^2) zero.

Let's write the line as L = {x + tv, t in R}

WLOG we can assume x = 0, since Lebesgue measure is translation invariant

So we just need to show that for unit vectors v, span(v) has measure zero.

Mike Miller

Isometry invariant, in fact, so may as well let L be the x axis

Drew Brady

Yeah, it's easy if isometry invariant

because then you just fit rectangles of size eps/2^n around the intervals (n, n+1)

Mike Miller

Yup

Drew Brady

which sum to eps, and your done.

Mike Miller

04:30

Certainly volume of cubes is an isometry invariant! So that argument then applies to any line by rotating the sets you choose

Drew Brady

Yeah, I assume that's what Cohn was looking for.

Same argument for a circle in R^2, I guess.

Reduce to S^2 (scaling translation)

Then take B(0, 1 + \eps) \ B(0, 1 - \eps).

oops, I guess, S^1

my bad.

Rudi_Birnbaum

Hi @TobiasKildetoft maybe you could have a look at this chat.stackexchange.com/transcript/message/46383712#46383712 ?

Rudi_Birnbaum

04:51

seems its called "Hall theorem" as well.

Drew Brady

05:21

Hey, anyone able to help me with this measure theory problem?

Suppose m is Lebesgue measure on the real line. Find a Borel set B such that 0 < m(B \cap I) < m(I) for all bounded open subintervals I \subset \R.

I bet it's some sort of equivalence relation/aoc construction, but not quite sure.

hint would be nice.

Aditya Kumar

Does balarka hang around here anymore?

Drew Brady

It's kinda annoying because it can't be R \minus a set of measure of zero

Alessandro Codenotti

@DrewBrady There's no AC needed here

Think about how to make such a set in $[0,1]$ rater than $\Bbb R$

Secret

05:36

fat cantors

but anything more exotic than fat cantors?

Drew Brady

hm you can't remove any interval

Alessandro Codenotti

True, but you can fix the removed intervals later

Drew Brady

maybe you can do some kind of refinement thing.

you do like (0, 1/3) \cup (2/3, 1) and then add (3/9, 4/9) and (5/9, 6/9)

but at this point your still screwed, because you can't have a complete interval either.

it's gotta have measure less than 1, else you're screwed by the interval (0, 1) also.

Secret

05:52

is B a subset of R or a subset of I?

Drew Brady

R

Secret

and I is just one fixed interval or we need it to hold for all possible I s?

o wait I misread

It's $B \cap I$, not $B \setminus I$

Drew Brady

all I

Secret

How about a fat cantor that tiles the space, like for every gap, place a fat cantor, then you will always be left with some gaps, but they are never large enough to form intervals thus every bounded open interval will guarantee to intersect with it

Alessandro Codenotti

06:08

You can do this with any nowhere dense of positive measure, the fat Cantor set being the most intuitive example: you start with it, then for every interval disjoint from the set you fit a smaller copy of the set inside that interval, iterate this process

Drew Brady

Hm does the following work?

Let R = \cup_n (p_n, q_n) where p_n, q_n are rational

Now let B = \cup_n ComplementOfFatCantor(p_n, q_n)

Basically, you write R as a disjoint open rational cover

then you stick a complement of fat cantor inside each one

Alessandro Codenotti

The complement of a fat cantor set contains plenty of intervals

Drew Brady

oh right...

Alessandro Codenotti

@AlessandroCodenotti Important: if you start with a fat Cantor set of measure $1/2$ and do this iterative construction to construct a set $A$ it might look like that for all intervals $I$ we have $m(A\cap I)=\frac12 m(I)$, but this can't hold

Drew Brady

I see, so I can just enumerate rational intervals. Add cantor to the first one

why not?

oh never mind. if m(A \cap (0, 1)) = 1/2 m((0, 1)) then obviously it cannot also satisfy m(A \cap (0, 2)) = 1/2 m(0, 2)

oops nevermind

Alessandro Codenotti

06:16

Why not? $A=(0,1/2)\cup(1,3/2)$ does

The problem is in all intervals, it contradicts the Lebesgue density theorem

Drew Brady

so How about this.

we enumerate all rational intervals, (p_1, q_1), (p_2, q_2), .. etc

we start by putting a fat cantor in p_1, q_2.

call this set B_1

*p_1, q_1

then we check if m(B_1 \cap (p_2, q_2)) = 0

If it is, then we take B_2 = B_1 \cap a fat cantor set on (p_2, q_2)

If not, B_2 = B_1.

Proceed.

Take B = \cup B_n.

Zee

What are you guys trying to do

Drew Brady

we're trying to construct a Borel B such that 0 < m(B \cap I ) < m(I) whenever I is a bounded open interval

Alessandro Codenotti

06:32

@DrewBrady Yup, this works

Zee

The right side is obv trivial

Alessandro Codenotti

It's a strict inequality

Zee

Oh I see

This sounds fun , I hope am not too drunk for this

Are we on the real line ?

Secret

yup

Zee

Good to see you again

Secret

06:35

I wish we have uncountable data visualisers, because it will be quite cool to see how spongy these sets are

(Yeah, had been busy in PhD)

Drew Brady

I see. So with this construction m(B^c \cap I) > 0 since I contains an interval (p_n, q_n) for q_n \neq p_n, and B \cap (p_n, q_n) <= (q_n - p_n)/2.

thus m(I) = m(B \cap I) + m(I \ B) > m(B \cap I) >= m(B \cap (p_n, q_n)) > 0, with the last inequality by construction.

*and m(B \cap (p_n, q_n)) <= (q_n - p_n)/2.

Zee

Is there a hypothesis that I and B need to contain some fixed x?

Drew Brady

No

mick

06:56

2

Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

Let $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins , and prime twins grow like $O (n * ln(n)^2 ) $ as i...

Any ideas ?

Zee

This should work , take the intervals [0,1] and so on . Then do the cantor construction except you remove 1/4 at each step

wait

Never mind

Drew Brady

@AlessandroCodenotti I think something is wrong

Alessandro Codenotti

07:22

Why? @DrewBrady

Drew Brady

Well, here's the construction right: Let Q_n denote an enumeration of the rational endpointed-open intervals. Let B_1 = fat cantor on Q_1. Let B_n = B_n-1 if m(B_n-1 \cap Q_n) > 0, else B_n = B_n-1 \cup fat cantor on Q_1.

Let B = \cup_n B_n,

right? @AlessandroCodenotti

Then if I is an open interval bounded then for some n, Q_n \subset I, right?

So then m(I \ B) >= m(Q_n \ B) = m(Q_n) - m(B \cap Q_n) >= m(Q_n)/2 > 0, right?

Then m(I) = m(B \cap I) + m(I \ B) > m(B \cap I) >= m(Q_n \cap I) > 0, right?

Agree?

Alessandro Codenotti

@DrewBrady how do you get the last inequality here?

Drew Brady

which inequality

Alessandro Codenotti

>=m(Q_n)/2

Drew Brady

^@AlessandroCodenotti

Oh, yah, that's not necessarily true...

sorry I was thinking that there was some sort of ordering on the rational open intervals so that m(B \cap Q_n) = m(Fat cantor set on Q_n) or m(set strictly smaller) but that's not necessarily true.

like there could be a lot of intervals before which contribute a little which add up to more than m(Q_n)/ 2.

Secret

07:45

PROGRESS!

Leaky Nun

07:59

> One way of overcoming the difficulty of undecidable questions is to extend the
theory ZFC, to obtain a richer theory which provides more information about
sets. (An alternative solution is simply to accept as a fact of life that some questions
have no answer.)

I like this way of life

Zee

08:21

@AlessandroCodenotti @DrewBrady this is a little hand wavy but you construct a cantor set in each interval of size 1/4 then take the the complement in each interval and then again construct a cantor set of size 1/16 and then so on , your gonna have a set of measure 1/3 in each interval which would also have positive measure when intersected with an interval but also smaller

abenthy

08:51

Is every action of a connected Lie group $G$ on an oriented manifold $M$ necessarily orientation preserving? The argument I have in mind is that $G$ maps to a (path-)connected subgroup of the group $Diff(M)$, every element of which is connected to the identity by a path and hence must preserve the orientation. Is this correct?

Secret

09:44

> but any extension of ZFC will necessary involve a lot more impredicability, as well higher and higher levels of infinity that is even less likely to have a real life counterpart

Vrouvrou

@Perturbative here you have changed the condition $\exists i_0,\forall j\in I \Omega_{i_0}\cap \Omega_j\neq \emptyset $ by $\bigcap_{I\in I}\Omega_i \neq \emptyset$

Rudi_Birnbaum

10:34

@Secret I always thought maths is about what can be thought and not about what you actually find in "real life" (whatever that might be).

Leaky Nun

@Rudi_Birnbaum scheena dog

Rudi_Birnbaum

@LeakyNun meassi dia aa

(whoa perfect dialect!!)

Leaky Nun

wat betekent dit?

Secret

@Rudi_Birnbaum that is true, but higher order infinities are so localised in the domain of set theory that they don't have practical uses even within pure mathematics

Rudi_Birnbaum

@LeakyNun meassi = "merci", dia = "dir", aa="auch"

Leaky Nun

10:38

:o

wat sprikst du, franzoesisch?

Rudi_Birnbaum

baierisch

Its a common lean word in Bavaria, we have quite a few of them

Leaky Nun

you mean loan word

Rudi_Birnbaum

paraplü, plafond, duchet

yes, sry

Leaky Nun

what do those mean?

Rudi_Birnbaum

umbrella, ceiling, bed blanket

Leaky Nun

10:41

in which language does plafond mean ceiling?

Rudi_Birnbaum

I thought French?

Leaky Nun

indeed

ik heb een neuer franzoesisch woorde gelerd

Rudi_Birnbaum

I maent to say Baierisch (especially the Bayern brand) has many French loan words.

Leaky Nun

I've heard that a speaker of an Alemannic dialect met a speaker of Baierisch

and communicated in English instead

Rudi_Birnbaum

@LeakyNun Well for the one Alemannic and Bavarian are more closely related than both to the Hochdeutsch (we call actually "Schriftsprache"). So keeping in Dialect would be one option. That requires a bit of practise. But often strong dialect speakers are not that good in the HOchdeutsch, then English might be better.

I know for sure my Grandma could not understand my wifes grandma (who is from close the Danish border). And I have to admit I did the same sometimes in northern Germany (use English...)

Leaky Nun

10:49

I personally think that they should all speak plaatduutsch instead

because plaatduutsch is beautiful

Rudi_Birnbaum

@LeakyNun yes it is

but my all-time fav. Germanic language is Frisian

Leaky Nun

how non-germanic do you need to get to say "ich"

"ik" is so much more germanic

@Rudi_Birnbaum Gothic :P

Rudi_Birnbaum

but its a natural shift somehow for Germanic

Leaky Nun

sure

the un-germanic "high germanic consonant shift"

that softens virtually every consonant

Rudi_Birnbaum

There are quite a few brands of frisian and I like the eastern brach which is almost extinct

yes

Leaky Nun

10:52

you might as well eald aenglisc specan

Rudi_Birnbaum

@LeakyNun :-) yes thats also cool!

@LeakyNun Whats your mother tongue? English I suppose?

Leaky Nun

Cantonese

Rudi_Birnbaum

Oh!

Can you speak any other Chinese languages?

Leaky Nun

Mandarin

and a very little bisschen of Hokkien

Rudi_Birnbaum

How closely are those three related?

Leaky Nun

10:56

they are mutually unintelligible

they branched off like 2000 years ago (or more)

Rudi_Birnbaum

Like German and Italian?

(though these brached off a bit earlier I think)

Leaky Nun

right

Rudi_Birnbaum

How many languages do you speak?

Leaky Nun

but Cantonese and Hokkien borrowed a lot of words from what would be Mandarin 1000 years ago

(so did, you know, Japanese and Korean and Vietnamese)

Rudi_Birnbaum

OK!

Leaky Nun

11:00

@Rudi_Birnbaum no idea

Rudi_Birnbaum

What about the intersection between Sanskrit and Mandarin? I heard both (though not related) borrowed quite a bit from each other is that true?

@LeakyNun :-)

Leaky Nun

well just the terms related to Buddhism, I would suppose

Rudi_Birnbaum

Ah Ok.

Leaky Nun

I don't know how much Sanskrit borrowed from Mandarin

Rudi_Birnbaum

Linguistics is fascinating. I am intrigued by the idea that all languages could be related. Though linguists commonly accepted zero-hypotheses seems to be nothing is related with nothing unless there is valid proof.

which is conservative and makes sense scientifically but can be boring as well.

Leaky Nun

11:03

I'm more intrigued by the idea that every living organism is related lol

Rudi_Birnbaum

Me too! Just recently thought about it!

I think that we are related to trees is more or less a fact.

Leaky Nun

sure

Rudi_Birnbaum

But I don't know what the current Lehrmeinung in gerenal is.

Fargle

Linguistics is pretty amazing. I'm still awestruck by the mere existence of language. Imagine an existence without communication this precise (as imprecise as it is). Much stock is given to fire, agriculture, and the wheel, but I'd argue that the development of language is pretty directly responsible for the success of mankind.

Leaky Nun

@Rudi_Birnbaum I think virtually every biologist believes in common descent

Rudi_Birnbaum

11:05

I mean there are like Archeae and stuff, and I don't think we have a complete tree.

Leaky Nun

Rudi_Birnbaum

In gross and general sure. But there might be alternative ideas for the very origins.

Say when life was not really life already.

What if the "soup" was divided then

Just my thoughts.

Leaky Nun

> The standard hypothesis states that the ancestor of the eukaryotes diverged early from the Archaea,[78][79] and that eukaryotes arose through fusion of an archaean and eubacterium, which became the nucleus and cytoplasm; this explains various genetic similarities but runs into difficulties explaining cell structure.[76] An alternative hypothesis, the eocyte hypothesis, posits that Eukaryota emerged relatively late from the Archaea.[80]

Rudi_Birnbaum

I see!

@Fargle Is of course possible. Though we have really just one example of such a species.

Fargle

There's also the fact that bacteria, archaea, and eukaryotes not only have the same chemical to code their genetics (DNA), but the code itself is almost entirely identical as well (e.g. thymine-adenine-thymine almost always means "put tyrosine in this protein").

Rudi_Birnbaum

11:11

And then it depends how you define "success"

Fargle

Yeah, absolutely. I don't know that a language in the abstract would necessarily always have the effect it has had upon us.

Rudi_Birnbaum

@Fargle I have been talking to a colleague of mine who does DNA research on a chemical bases. And we came up with the hypothesis that it might be the case that DNA indeed is optimal in a sense, that any small chemical variation might spoil its properties such that it would be useless for encoding biological structures.

That would mean its at least some "local minima" but would be also in agreement with the spectacular idea, that its without alternative ... Though highly unlikley but the more fascinating

Fargle

That's possible. In which case the fact of the similarity of the "programming languages" across species wouldn't any longer be evidence of universal common descent

But would be heuristic evidence of the fact that evolution, despite not (necessarily) being guided by an intelligence, is most certainly not random.

Rudi_Birnbaum

Yes and it would make the tree of life more opaque ...

in particular concerning the beginnings (which where RNA not DNA btw).

Fargle

Right.

Simply Beautiful Art

11:20

@LeakyNun can try to explain inaccessible cardinals better if you want.

Fargle

There may have been multiple forms of "ur-life" which didn't have the same origin, but came about in the same circumstances with the same available elements, and so had the same structure. These could have developed into domains we know today, and may have intermingled both before and after becoming "actual" life (wherever we draw that line). And it becomes much harder to know for certain at that point.

Leaky Nun

@SimplyBeautifulArt sure

Rudi_Birnbaum

@Fargle Yes, exactly.

Fargle

That's thought-provoking. I hadn't considered that.

Simply Beautiful Art

@LeakyNun any particular place?

Leaky Nun

11:23

@SimplyBeautifulArt what do you mean

Simply Beautiful Art

Any particular place you would like me to explain in?

Leaky Nun

here if you want

we need to indoctrinate everyone with inaccessible cardinals

3

Simply Beautiful Art

lol

Leaky Nun

(I'm sure Asaf will approve :P)

Simply Beautiful Art

Well they are regular limit cardinals

Their ordinal counterpart being regular limits of omega ordinals.

Leaky Nun

11:27

and by that you mean they are cardinals c such that cf(c) = c and c=aleph_lambda for some limit ordinal lambda

Simply Beautiful Art

Mhm

Leaky Nun

do you have an example? :P

Simply Beautiful Art

You can't really write one out in terms of smaller ordinals (hence the term "inaccessible")

Leaky Nun

fair enough

Simply Beautiful Art

Essentially $\alpha$ is regular if for any $\beta<\alpha$, there is no sequence of ordinals $\alpha[\eta]$ such that $\alpha[\eta]<\alpha$ for all $\eta$ and $\sup_{\eta<\beta}\alpha[\eta]=\alpha$.

Leaky Nun

11:30

is every successor cardinal regular?

Simply Beautiful Art

Yes

But to be inaccessible, it has to be a limit cardinal

For this to occur, it has to be an aleph fixed-point ($\mathfrak c=\aleph_{\mathfrak c}$)

Leaky Nun

why?

well we know many aleph fixed points there are... exactly as many as the ordinals

Simply Beautiful Art

These are the only points where there are $\mathfrak c$ cardinals below them.

Leaky Nun

is c also the c-th fixed point?

Simply Beautiful Art

Yes, it must also satisfy that

Leaky Nun

11:34

what the hell

how meta does this go

Simply Beautiful Art

Since those are the only points with $\mathfrak c$ aleph fixed-points below them

Leaky Nun

ok go on then

Simply Beautiful Art

It must also be the $\mathfrak c$-th fixed-point of the $\alpha$-th aleph fixed-point.

Leaky Nun

no i mean go on with the inaccessible cardinal

Simply Beautiful Art

Well that's just about most of the interesting bits

It must be this really big fixed-point of fixed-points of fixed-points of fixed-points...

Leaky Nun

11:37

can you go even more meta?

Simply Beautiful Art

You mean like how ZFC cannot prove/disprove their existence?

Mr. Xcoder

\o! This conversation reminds me of SBA teaching ordinals :P

Simply Beautiful Art

:P hello

Leaky Nun

no, like you said fixed point of fixed point of fixed point

Simply Beautiful Art

Oh

Leaky Nun

11:39

I’m wondering if you can do the c-th description

Simply Beautiful Art

Well you know the Veblen function?

Alessandro Codenotti

Veblen's functions is the keyword

Simply Beautiful Art

:P ninja'd

Leaky Nun

no

Alessandro Codenotti

Damn what a ninja

Simply Beautiful Art

11:40

ouch

So essentially we could make a function

$\varphi(0,\alpha)=\aleph_\alpha$

$\varphi(1,\alpha)$ is the $\alpha$-th fixed-point of $\varphi(0,\alpha)$

$\varphi(2,\alpha)$ is the $\alpha$-th fixed-point of $\varphi(1,\alpha)$

etc.

$\varphi(\mu,\alpha)$ is the $\alpha$-th cardinal which is a fixed-point of $\varphi(\eta,\alpha)$ for all $\eta<\mu$.

Leaky Nun

I see I asked the right question

Simply Beautiful Art

Inaccessible's are fixed-points of all of these

Also $I=\varphi(I,0)$

($I$ denoting an inaccessible cardinal)

Leaky Nun

what?

Simply Beautiful Art

hm?

Leaky Nun

ok

how about a double veblen function

Simply Beautiful Art

11:43

You can generalize this to a 3 variable Veblen function

Leaky Nun

you win

Simply Beautiful Art

xP

$\varphi(0,\alpha_1,\alpha_2)=\varphi(\alpha_1,\alpha_2)$

Alessandro Codenotti

You might find this old question I asked interesting math.stackexchange.com/questions/2405052/…

Simply Beautiful Art

$\varphi(1,\alpha_1,\alpha_2)$ is basically the same thing, except instead of starting with $\phi(0,\alpha)=\aleph_\alpha$, you start with $\phi(0,\alpha)=\varphi(\alpha,0)$

(i.e. build the same hierarchy off of the old one)

and you continue constructing in this manner

The inaccessibles are fixed-points for like all of these

and any recursively built extension of fixed-points of these

Leaky Nun

how about the c-th veblen function

Simply Beautiful Art

11:47

(gonna go eat)

Not entirely sure what that's supposed to mean

but if you can build it recursively

then yes

Leaky Nun

why is V[k] a model of ZFC?

Alessandro Codenotti

The idea is that $V_\alpha$ already models most of the axioms (for $\alpha>\omega$ a limit), regular limit gives replacement and strongly inaccesible powerset

Simply Beautiful Art

Note that I've described weakly inaccessibles. Strongly inaccessible cardinals are regular cardinals that are limits of beth cardinals instead of aleph cardinals.

Leaky Nun

I see

does L[k] model anything then

Simply Beautiful Art

ZFC implies V[k] is a model of ZFC

ZF implies L[k] is a model of ZFC

Leaky Nun

12:00

I see

Simply Beautiful Art

(according to wikipedia)

Leaky Nun

why dont we all just assume V=L

do you have statements independent of ZFC+(V=L)

Simply Beautiful Art

Probably

Leaky Nun

?

Simply Beautiful Art

12:15

V = L is the axiom of constructibility and it implies things like GCH is true

Leaky Nun

so i’m asking if you have independent statements

Simply Beautiful Art

well idk

But it appears to be inconsistent with certain large cardinal axioms, so I dunno if I really like it ;P

Leaky Nun

> In this chapter we define the constructible universe and develop its elementary
theory

proceeds to spend 52 pages talking about it

Simply Beautiful Art

oof

ttyl8r

Leaky Nun

extensionality, union, infinity, power set, foundation, comprehension, collection

which of them are satisfied by a transitive set?

> One form of the reflection principle in ZFC says that for any finite set of axioms of ZFC we can find a countable transitive model satisfying these axioms.

I don't see how this can be true

vasili111

12:47

In sentence "SAT-based Sudoku solver" What abbreviation "SAT" stand for?

user1732

Standardized admissions test

vasili111

@user1732

user21820

@vasili111 Super Advanced Troll.

3

More seriously, it refers to the NP-complete problem called boolean SATisfiability.

Boolean satisfiability problem

In computer science, the Boolean satisfiability problem (sometimes called propositional satisfiability problem and abbreviated as SATISFIABILITY or SAT) is the problem of determining if there exists an interpretation that satisfies a given Boolean formula. In other words, it asks whether the variables of a given Boolean formula can be consistently replaced by the values TRUE or FALSE in such a way that the formula evaluates to TRUE. If this is the case, the formula is called satisfiable. On the other hand, if no such assignment exists, the function expressed by the formula is FALSE for all possible...

vasili111

@user21820 Thank you :)

user21820

@vasili111: But I'll say that directly writing a sudoku solver is so easy that it is quite pointless to solve it via a SAT-solver.

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$Mathematics$ Mathematics