Mathematics

4:27 AM

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

4:30 AM

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

4:51 AM

seems its called "Hall theorem" as well.

Drew Brady

5:21 AM

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

5:36 AM

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

5:52 AM

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

6:08 AM

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

6:16 AM

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

6:32 AM

@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

6:35 AM

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

6:56 AM

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

7:22 AM

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

7:45 AM

PROGRESS!

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$Mathematics$ Mathematics