Mathematics - 2023-08-03

Ted Shifrin

00:00

Rest up and stay away from people.

Semiclassical

00:29

so this is sorta neat

came across a very good paper recently. i'm still trying to absorb it, but one line did stick out

"The (Qnx) boundaries are directly expressed by the surface area of the elliptope. This area is known to be $5\pi$, by a direct calculation found on math.stackexchange.com." arxiv.org/pdf/2111.06270.pdf

well....

25

Q: What is the surface area of the 3-dimensional elliptope?

The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-ellipt...

i've kept my MSE profile non-descript b/c i've prefer to keep some distance, but in this case i suppose that had a small cost. (i mean, it's not as if i did anything besides pose the question. David H is the one who figured out how to actually compute it)

one potato two potato

00:48

1

Q: Dehn presentation has a linear isoperimetric inequality

The following lemma is from the book Discrete groups by Ohshika. Lemma 1.15. If a group $G$ has a Dehn presentation, then $G$ has a linear isoperimetric inequality. Proof. Let $G = \langle S\mid R\rangle$ be a Dehn presentation. We can assume $R$ is closed under cyclic conjugation (which means ...

geometric-group-theory

Semiclassical

i guess i did conjecture the (correct) result, but eh

one potato two potato

I tagged GGT but actually anybody can understand the post

Semiclassical

@onepotatotwopotato to clarify, is the proof the one from Ohshika as well?

one potato two potato

yes

Semiclassical

okay. it might be worth clarifying that, e.g., "The following lemma and its proof are from..."

one potato two potato

00:52

I edited but most people would think the proof is also from the book. I'm questioning the given proof

I only read the first few pages but I think the book is good though.

冥王 Hades

I can’t even bore myself to sleep by reading history textbooks like this

Ted Shifrin

01:12

@Semiclassical Good grief!

Thomas Finley

02:20

In any vector space, ax = bx implies that a = b.

Is this statement true or false?

This was a question from an excercise in freedberg, insel,spence's book on linear algebra

There is some ambiguity in this statement.

They haven't mentioned which is scalar or a vector are it might well be true that both are vectors . But my intuition says, a,b is a scalar and x is a vector

Oh, wtf

I posted a wrong question.

In any vector space, ax = ay implies that x = y.

This is the one.

Ok, in here I feel this statement is true.

But the book says false.

My logic is, I feel a is a scalar and, x,y are vectors.

So, a^{-1} exists

Multiplying both sides of ax=ay by a^{-1} gives, x=y.

No info whether a,x and y are vectors or scalar is given.

What d'ya guys think?

Am I the winner? Or the book?

----------------------- This ends my post.

Semiclassical

@TedShifrin i still shake my head a bit at how involved that integration is

i guess it wouldn't have been impossible for me to do, given that i do have some knowledge of hyperelliptic integral shenanigans

but i was glad to have someone else do it instead :P

Ted Shifrin

03:16

@ThomasFinley The book. Read your argument critically.

leslie townes

thomas: implcitly x, y are elements of an abstract vector space and a is a scalar. the implication would be true with the additional hypothesis that the scalar a is nonzero (so that a^(-1) exists and your proof would work). without that restriction it is false, because 0x = 0y will always hold even when x = y does not.

i guess it's again true in the case of the {0} vector space, but then only for a very uninteresting reason.

Thomas Finley

03:36

@leslietownes Oh! That was the trickery...

Ted Shifrin

03:53

@Thomas Good beginning exercise. Using the axioms (definition) of a vector space, prove that $0\vec x = \vec 0$ for every vector $\vec x$.

copper.hat

04:10

Hi @TedShifrin

Ted Shifrin

Howdy, @copper.

copper.hat

Need a break from my mind numbingly mundane work.

Thomas Finley

@TedShifrin I feel this to be simple: $0\vec x=(0+0)\vec x=0\vec x+0\vec x.$ Now, as, $0\vec x\in V$ so, $-0\vec x$ exists and adding it on both sides of the preceeding equation we get, the desired result i e $0\vec x=\vec 0.$--- Isnt it?

leslie townes

fuses, timers, slop of black paint on neighbor's drive, more fuses, more timers

Ted Shifrin

@Thomas Good job.

This trips lots of people up.

It shows how important the distributive laws are.

A similar exercise is to show that $(-1)\vec x = -\vec x$.

Many people try to do that without the one you already just proved.

@copper.hat We can numb minds with the best of ‘em.

copper.hat

04:19

I read category theory when I need complete numbing.

Ted Shifrin

I can’t bring myself to read that… ever.

Thomas Finley

@TedShifrin Yes.

@TedShifrin I think the more generalized version of this statement is, $-(a\vec x)=(-a)\vec x$ (?)

Ted Shifrin

Sure. It boils down to the first plus the fact $-a=(-1)a$ for real numbers. All the same thing.

Thomas Finley

The proof in here looks less complicated for: We have $a\vec x+(-(a\vecx))=\vec 0$ Also, $a\vec x +(-a)\vec x=(a+(-a))\vec x=0\vec x=\vec 0.$ Now, as the inverse of an element in a Vector Space is unique, we must have, $-(a\vec x)=(-a)\vec x$ . I think this is what you meant by the proof @TedShifrin?

Ted Shifrin

Or the associative law $(-1)ax = (-a)x$.

one potato two potato

07:08

research.wmz.ninja/projects/phd

2

Phd simulator!

rogerroger

09:03

what do people think about complicated proofs which prove relatively niche mathematical theorems

Jakobian

10:00

what kind of people

I don't understand the question

like, what's the context here

is it like you're trying to show your proof to the world and wonder what people will think about it?

that's my best guess

one potato two potato

surprise surprise!

Dannyu NDos

Do you mean like, the computer-assisted proof of four-color theorem?

nathdwek

10:22

Hi guys I have a quick notation question

I am writing an article that involves complex numbers, with a lot of "separate" treatment of real and imaginary part

in a field where this is not super super super usual, in the sense that I have not found a lot of existing notational conventions in the literature

Do you have a recommendation for a "standard" way of writing real and imag part of a number/vector that is less "heavy" than Re() and Im()?

In my written note I have a tendency to use $x^R$ and $x^I$ to indicate real and imag part of $x$

But wondering how common that is, and also makes it a bit cumbersome with actual powers (I deal with quite a lot of norms so would have to write ${x^R}^2} and hope it's immediately clear to a reader

missing sorry. should be ${x^R}^2$

right subscript is already quite busy. I was thinking about right superscript. Have you guys seen that used in the past?

Jakobian

is there a reason why you can't write $x = a+bi$ or something?

nathdwek

well then it basically makes the subscript of a and b very busy

and puts the "actual" information in the subscript?

because to distinguish between the a and b of x and the a and b of y I would write something like $a_{x}$

Jakobian

I think it's a bad idea leaving something in notation for the reader to figure out

nathdwek

I appreciate the time and attention you took to answer. I don't want to give the impression of attacking your idea (and it's quite standard indeed) but as I said the right subscript of $x$ is quite busy as well, so putting it all together I would have something like $a_{x_{f}}$

mmmmh

@Jakobian yeah but between one sentence to say "right superscript indicates real and imag part" and one sentence to say "we consider x=a+bi", the text is not very different and the mental effort of the reader not so much either

Sorry if there are weird duplicate messages or anything, dept connection is very up and down with HVAC works.

Jakobian

Instead of subscripts you could also do something like R[x] and I[x]

nathdwek

10:37

mmmh, yeah

Well to ask it, more directly and without introducing personal bias/existing solution:

Here's an actual example

${\left \lVert \bm{\rho} \right \rVert}_1 = \sum_j \lvert \rho_j \rvert = \sum_j \sqrt{{\Real\!\left(\rho_j\right)}^2 +{\Imag\!\left(\rho_j\right)}^2}$

I 'd like the rightmost side to be written more compactly and elegantly

I understand maybe there is not a lot/nothing to be done

but if anyone has an idea, or can refer to an area/paper/literature where a different convention than Re/Im is used

It would help me greatly :)!

Jakobian

12:06

I've discovered those vege food lunchboxes, and they just taste so good

nathdwek

Bon appétit

porridgemathematics

12:44

does anyone know why if you glue together sobolev functions in a piecewise linear way, the derivatives on the common parts of the pieces coincide almost everywhere? More precisely, I'm asking about this question I posted math.stackexchange.com/questions/4746913/… , but it really comes down to what I say roughly

I feel like there should be some general theorem I could just cite here..

user4539917

14:36

@robjohn is the app at all related to what you're working on, sir?

robjohn

15:23

@user4539917 It relates very tangentially. That app seems to be for organizing statements and data, but not determining whether the statements and their data are true. Some of the software I wrote helps a student learn how to prove statements logically.

user4539917

That sounds interesting.

geocalc33

15:41

$\mathrm{SO}^*(2)$

Is this already notation for something?

I know $\mathrm{SO}(2)$ is the special orthogonal group

one potato two potato