Mathematics - 2018-12-13 (page 2 of 2)

Secret

17:00

yup

Semiclassical

Are two 2-adic numbers considered to be close if their last few digits are the same? obv. with base-2 numbers it's the opposite

Jasper Loy

@Secret THat is the most interesting news I read this year actually.

Here comes @MikeMiller!

Semiclassical

@JasperLoy a common variation on that question: "Where do you see yourself in 5 years?"

hate that question

Secret

> Where do you see yourself in 5 years?

This question make no sense as we can no longer predict deterministically where we are in even 2 days later

Rithaniel

I think I'm seeing messages out of order. Either that or I've lost the flow of the conversation.

Secret

17:02

the society have gone so complicated that prediction is no longer possible

Jasper Loy

@Rithaniel Is it because you are ignoring some users?

Rithaniel

No, I don't ignore people, generally.

Semiclassical

i think this conversation has just been a bit disjointed

Rithaniel

Well, to date, I've not ignored anyone on this chat.

Jasper Loy

Then maybe the flow of the conversation is just free association or random. =)

Secret

17:04

another possibility is that sometimes lag can cause messages to switch order at the cilent's end

Jasper Loy

I mean, there is no need to stick to one topic strctly at one time right?

Semiclassical

@Secret I suppose the point of the question is more for you to indicate your life goals and comment on how much progress you hope to have made

Jasper Loy

@Secret Yes, some messages might not display on some computers till hours later...

Rithaniel

Yeah, I think Semiclassical was saying that he hated the "5 years" question, but he said it before the question was posted on my screen.

Semiclassical

no, that's the order in which it happened

Jasper Loy

17:05

The 5 year question is just an interview question, standard fare...

Semiclassical

I said I hated that question, then Secret repeated the question and indicated why it was nonsense

Jasper Loy

Some of these interviewers will ask you stupid questions, because they are stupid.

However, I think the 5 year question is an OK question.

Rithaniel

Ah, gotcha

Secret

If there is a question that ask the interviewee to ask the interviewer a question, I will ask them the following:

Semiclassical

I'm not sure it's a bad question, it's just one I really don't like because I don't have that mindset

Secret

17:07

> What do you think of the typical interview questions and their textbook responses. Can you think of an interview question that is resistant to textbook solutions?

My experience told me people tend to go O_O whenever they heard meta statements

and then they enter some kind of trance state

Jasper Loy

I think the 5 year question is to see if your personal goals align with those of the company that is hiring you.

Semiclassical

probably

Jasper Loy

Like you want to be CEO in 5 years, OK, we will make you CEO in 5 years.

Semiclassical

(or "no, gtfo")

Secret

One of my wild desires is to ask someone a question and then said person enters nirvana as they tried to answer it

that will be a cool sight to behold

Jasper Loy

17:11

@Secret That sounds incredible. I mean, I didn't expect you to say that.

Semiclassical

hi @ted

oh, two fun things relating to actual previous research of mine

one is that there's another paper out which cites ours, which is fun

Jasper Loy

@Semiclassical Ours? You cowrote?

Secret

@JasperLoy that I say is a desire means no I don't have that ability (yet?)

Semiclassical

yeah: me, my advisor, and a collaborator (in that order on the paper)

the other is that, on a hunch, I went to the probability seminar last week

Jasper Loy

I see. Me, mine, and I, lol.

Semiclassical

17:13

and the talk turned out to be directly related to stuff I presented in my thesis defense talk

Secret

But to understand the ultimate limit of human knowledge it is to comprehend these:

Ineffability

Ineffability is concerned with ideas that cannot or should not be expressed in spoken words (or language in general), often being in the form of a taboo or incomprehensible term. This property is commonly associated with philosophy, aspects of existence, and similar concepts that are inherently "too great", complex or abstract to be communicated adequately. A typical example is the name of God in Judaism, written as YHWH but substituted with "the Lord" or "HaShem" (the name) when reading. In addition, illogical statements, principles, reasons and arguments may be considered intrinsically ineffable...

Semiclassical

in fact, one of the papers the guy wrote (and which he mentioned on one of his slides) cites our paper :)

Jasper Loy

@Secret Are you a student of Buddhism?

Secret

nope, but I do have some buddhist friends

Semiclassical

Now if only I'd realized that ahead of time...

Jasper Loy

17:14

@Semiclassical Oh so did you tell him that?

Semiclassical

yeah, after the talk

Secret

and just recently I was reading the wikipedia article on buddhism cosmology

Semiclassical

had I realized beforehand that he'd be doing so, I'd have dragged my advisor along

Secret

because I want to get more information about the concept of nothing

Jasper Loy

@Secret If you are, I recommend you get a copy of the 4 main Nikayas published by Wisdom.

Semiclassical

17:15

(and had i realized it a few days in advance, I'm sure we could've set up a meeting. missed opportunities)

Jasper Loy

@Semiclassical Sometimes, when you miss one chance, a better one comes. =)

Semiclassical

a different one comes, anyways. not sure 'better' is always a good metric

Jasper Loy

Anyway @Semiclassical I stopped taking meds for a few months now. I am not sure if what I am feeling now are withdrawal symptoms or not, but I didn't feel any different for a month or two.

Rithaniel

I would argue that "better" doesn't even constitute a metric. A pseudo-metric, perhaps.

Semiclassical

@Rithaniel a poset, maybe

Jasper Loy

17:17

Yeah because often one doesn't know if something is good or bad.

Sharath Zotis

Is $<x^2+5>$ a prime ideal in $\mathbb{Z}_7$

Semiclassical

@JasperLoy i just picked up a refill on one of mine this morning

Rithaniel

Because two things might have a "betterness distance" of 0, but not be the same element.

Semiclassical

namely, the one which I take for sleep stuff

that's one I notice if I forget to take

Sharath Zotis

It is reducible as 3^2 +5 = 14 which is 0 in $\mathbb{Z}_7$

But I am having trouble finding what it reduces to

Jasper Loy

17:20

@Semiclassical I am going to do more reading on meds and therapy before deciding on anything again. There is a lot of information from different places saying different things. =)

Semiclassical

sounds right

Sharath Zotis

Anybody have an idea?

Ted Shifrin

@Sharath: If you know $3$ and $-3$ are roots of $x^2+5\in\Bbb Z_7[x]$, then you know how to factor.

Sharath Zotis

ah yes so (x+3)(x+4) and since both are in $\mathbb { Z } _ { 7 } [ x ]$ then it is prime ideal

Ted Shifrin

Whoa, slow down.

In $\Bbb Z$, is $\langle 6\rangle$ a prime ideal?

Review the definition of a prime ideal.

Sharath Zotis

17:24

yes as 2*3 = 6 and 2 or 3 is in $\mathbb{Z}$

Ted Shifrin

No, write down the definition of a prime ideal.

Sharath Zotis

The definition is: Let $R$ be a ring and $I \subseteq R$ be an ideal. We say $I$ is prime if $ab \in I$ then either $a \in I $ or $b \in I$

Ted Shifrin

OK, now read that and understand it. Then read what you typed to me above.

Sharath Zotis

Oh yes neither 2 nor 3 is in $<6>$

What about 12 ? $12 \in <6>$ and 6*2 = 12 and $6 \in <6>$

Semiclassical

Needs to be true for all.

Sharath Zotis

17:28

I seee

Ted Shifrin

Wait, wait. Where did 12 come from?

Oh, I see ... Right. Whenever $ab\in I$ you must know that $a\in I$ or $b\in I$. That's a "for all" statement.

Semiclassical

Just because there exists some $ab\in I$ such that $a\in I$ or $b\in I$ doesn't mean that it's true for every $ab$

Sharath Zotis

ok makes sense

so to correct the thing I said about my problem. neither (x+3) nor (x+4) are in $<x^2+5>$

so it is not a prime ideal

Ted Shifrin

Right.

Semiclassical

Hmm. Seems like that should be pretty generic, i.e. if $p(x)$ is reducible in some polynomial ring then $\langle p(x)\rangle$ isn't a prime ideal

Sharath Zotis

17:31

Now with maximal ideals, $<x^2+5> \subset <x+3> \subset \mathbb{Z}_7$

Mike Miller

@Semiclassical reducible element in a ring

Sharath Zotis

this implies that $<x^2 + 5>$ is not maximal, corect?

Ted Shifrin

@Semiclassic: I would rather say "isn't a prime ideal in that polynomial ring" since you brought that phrase into it.

Mike Miller

And that <p> isn't prime is as you say definitional

Ted Shifrin

Correct, @Sharath.

Semiclassical

17:32

@TedShifrin yeah, fair

@MikeMiller ah, so that's the terminology? not surprising that I'd flub that tbh

Ted Shifrin

@Sharath: Can you see that any maximal ideal must be prime?

Mike Miller

@Semiclassical no, I am just saying polynomials are not actually relevant

Semiclassical

ahhh

yeah, fair

hmm, how big of a sparse matrix do I need to make

3-by-3-by-n-by-n-by-m...welp

Sharath Zotis

hmm @TedShifrin. why is that @TedShifrin

Semiclassical

(m versions of a 3-by-3 block matrix with n-by-n submatrices)

Ted Shifrin

17:35

Try proving the contrapositive, @Sharath.

Sharath Zotis

does it stem from the fact that a maximal ideal is not a subset of any other ideal

other than the ring itself

Ted Shifrin

Yes, of course.

Sharath Zotis

hmm I see we can use the theorem that $R/M$ is a field given $M$ is maximal ideal in ring $R$

Semiclassical

nope, sagemath doesn't like what I'm doing

Ted Shifrin

Don't use that theorem. Just do it directly, like you saw with the examples we discussed.

Sharath Zotis

17:38

Ok

Ted Shifrin

If $I$ is not prime, show it cannot be maximal.

Alessandro Codenotti

18:06

Hi @Ted

OneRaynyDay

Hi guys, following the proof on why $l^p$ is first category in $l^q$ if $p < q$: math.stackexchange.com/questions/1097869/…

I'm a bit confused on why we need to show that the individual $A_n$'s used in the construction needs to be closed?

So what if it's open? $l^p = \cup_{n \in \mathbb{N}} K_n$ where $K_n$ is the open ball centered around zero with radius $n$

To show $K_n$ is nowhere dense in $l^q$, we just need to show that there exists some sequence $x \in l^q$ such that $||x||_q < n$ and $||x||_p = \infty$ right?

Then simply take some $x \in (l^q \cap (l^p)^c)$, then $||x||_q = C$, then divide the sequence elementwise by some large enough constant and you get it's within $n$ radius of the origin

Yeah... not really seeing where the closedness is important here

Isa

18:40

@Palimondo Seems like you are using 🤰 to denote two different things. You said, 🤰 is a pregnant woman and then you write this '👤 ≡ 👤: {🤝,📜}' which is supposed to be equivalent to 'and equal in dignity and rights.' ?

OneRaynyDay

Ah, for anyone also wondering - it's because for a set $A$ to be nowhere dense, $(\overline{A})^o = \emptyset$

In this case if $K_n$s were open we would need to first take their closures, so finding closed balls were just 1 step-expedited

Palimondo

19:03

@Isa I don’t know what you mean. The 🤰 is there just once as part of the triple for “born” on first line. Then there’s kind of second statement on the second line…

What is tripping you up?

Ted Shifrin

@AlessandroCodenotti Hi, demonic Alessandro ;)

Isa

aha in the second line you wrote again the square 🤰, why? Or is it that I am seeing another square but it's not? Sometimes emojis 'change' depending on the system you work

Alessandro Codenotti

@PaulPlummer Interestingly the wikipedia proof uses HB instead of a nonprincipal ultrafilter, so amenability of $\Bbb Z$ should be a very weak statement. Do you know about the strength of "every Abelian group is amenable"?

Also a sanity check: The Cayley graph of a finite group with $n$ elements is quasi isometric to the complete graph on $n$ vertices right? Because I can take the whole group as generating set and get a complete graph as Cayley graph wrt this generating set

Fargle

19:24

Hello everyone.

Semiclassical

just asked Sage to compute the intersection of the 62-dimensional simplex in 63-dimensional space with a 56-dimensional polytope

let's see how much it hates me

apparently, it hates me too much to actually finish that computation :)

Paul Plummer

19:46

@AlessandroCodenotti I have no idea, seems like an interesting question. Maybe integers being amenable (where Folner is not the definition)could be independent of ZF?! I don't really know much about amenable groups.

Also, yes they are quasiisometric

Alessandro Codenotti

I wouldn't be surprised if it were independent

Paul Plummer

Finite groups are quasiisometric to points, and this holds for any bounded space

Alessandro Codenotti

I might ask about it on main

Paul Plummer

Yah, seems like a cool question, I would be interested in an answer

Alessandro Codenotti

Amenability is the most interesting thing we've seen so far in my opinion!

Paul Plummer

19:49

It is definetly cool, but it is in some sense opposite of hyperbolic behavior

Alessandro Codenotti

I've seen those "universe of groups" pictures, but I don't know why they're opposite

Paul Plummer

Well one thing is that(non elementary) hyperbolic groups contain free subgroups which is an obstruction

@AlessandroCodenotti As to the quasiisometry question, it is interesting that some infinite groups(not finitely generated) have no cayley graph where they are unbounded

Alessandro Codenotti

Uhm that's weird

But in that case the properties of the Cayley graph also depend heavily on the generating set I think?

Paul Plummer

Cayley graph always depends on the generating set, but groups like that you end up where the quasisometry class does not depend on the generating set

in most of GGT you restrict to finitely generated groups because the quasiisometry class does not depend on finite generating sets (for example you could choose the whole group as a generating set to get a bounded diameter graph)

Alessandro Codenotti

Yeah, sorry, I wasn't very precise, with infinitely generating sets you might even get cayley graphs which are not quasiisometric, right?

Paul Plummer

20:05

Yah

But there are groups where you always get quasiisometric graphs, like the ones where every cayley graph is bounded diameter, even though it is infinitely generated

Alessandro Codenotti

Ah, cool

Paul Plummer

20:31

The comments to this question point out that it is independent of ZF @AlessandroCodenotti

Alessandro Codenotti

I see, interesting!

Now I wonder about its strength though, it sounds like it should be in between ZF and ZF+HB

It surely is stronger than ZF and weaker but possibly not strictly than HB

But since it only involves countable objects I don't see how could it be equivalent to HB

Paul Plummer

20:52

Another interesting question is ZF+shift invariant finitely additive probability measure on integers can prove the equivalence of the Folner sequence and shift invariant finitely additive probability measure for discrete countable groups

basically is it enough to for integers to get the common equivalent definitions?

Alessandro Codenotti

Good question indeed, I have no idea

Balarka Sen

@AlessandroCodenotti Whoosh collapse it to a point

and the foosh explode it to any finite graph you want

qed

Alessandro Codenotti

Paul sniped you

Balarka Sen

did he specifically say woosh and foosh

otherwise it's not a true snipe

Alessandro Codenotti

He didn't use the whoosh which is clearly the correct technical terminology though

Oh come on you can't snipe me saying you weren't really sniped

Balarka Sen

21:06

rip

Alessandro Codenotti

That's not how it works

Semiclassical

welp. my pattern for vertex counts so far is 4,40,576

sooo my next one probably has thousands of vertices...which probably explains why Sage is taking so long :P

Balarka Sen

@PaulPlummer Is any sufficiently reasonable metric space with two ends quasi-isometric to $\Bbb R$?

My mental picture is, I end-compactify it, then take an arc going between the two components in the end

Semiclassical

(I think I may be making things a touch more difficult than I need to, tho.)

Balarka Sen

I should be able to collapse the space to the arc but how do you do this, formally?

Leyla Alkan

21:14

Hi all! Is this proof correct? How does the underlined sentence follow by the previous one?

Semiclassical

Actually, quick sanity check

Balarka Sen

@Semiclassical You are insane

Semiclassical

damn

Suppose I've got two convex sets and I intersect them. Is this preserved by linear transformations?

i.e. if $C=A\cap B$ for convex $A,B$ and $F$ is some linear transformation, should $F(C)=F(A)\cap F(B)$?

Paul Plummer

@BalarkaSen Depends on what you mean by sufficiently nice. For example n^3 for integers n is not quasiisometric to the integers (with induced metric).

Maybe if there are enough quasiisomeries, or it is geodesic

Balarka Sen

Oh, because the distances grow far

Cannot linearly scale it down

Paul Plummer

21:17

Yah

Balarka Sen

rip. Good example

Paul Plummer

Actually geodesic is not enough

Semiclassical

according to a random internet paper: "For any linear transformation T and convex closed sets A and B, T(A ∩ B) = T A ∩ T B if and only if A ∪ B is path-connected with respect to ker(T)."

now to figure what that actually means...

MatheinBoulomenos

@Semiclassical take $A=\langle (1,1) \rangle$ $B=\langle (1,0) \rangle$ and let $F$ be the projection onto the first coordinate

Semiclassical

Hmm. So $A\cap B=\{(0,0)\}$

then $F(A)\cap F(B)=\langle (1)\rangle \cap \langle (1)\rangle=\langle (1)\rangle$ whereas $F(A\cap B)=\{(0)\}$

checks out (as a counterexample, I mean)

in the context of the statement I just gave, that makes sense because $A\cup B$ is certainly not convex

So my approach is probably not going to be suitable

Paul Plummer

21:28

Something like being uniformly quasigeodesically "connected", and every triangle is in a uniform neighborhood of one of the uniform quasigeodesics should get it. whoosh foosh.(but that is basically saying it is quasiisometric to a line so that isn't really interesting) @BalarkaSen

But I think that is isolating the main features

MatheinBoulomenos

quasiisometry seems like a very coarse notion

Paul Plummer

Oh and have two ends

Balarka Sen

@PaulPlummer Aha, I see

Paul Plummer

@MatheinBoulomenos Haha

Liad

22:47

Hi, $X$ is a normed space. i want to show that $x_1,\dots,x_n$ are independent iff for all $\alpha_1,\dots,\alpha_n \in \Bbb F$ there is $\phi \in X^*$ s.t $\phi(x_i)=\alpha_i$. im stuck on this question, any ideas?

$\Bbb F\in \{\Bbb R , \Bbb C\}$

Richard

can someone help me with morphological gray-scale gradient ?

Ted Shifrin

23:11

@Liad: Surely one direction is easy. Where are you stuck?

Ultradark

Are most 3-manifolds hyperbolic

Ted Shifrin

Certainly not.

I have no idea what "most" means, anyhow.

Balarka Sen

23:26

@TedShifrin A random mapping torus is hyperbolic, if I recall correctly

Ted Shifrin

I know very little about 3-manifolds, actually, since I live in the complex world by choice :P

Or lived.

Hi, a @Balarka.

Balarka Sen

In the sense that there is a dense subset of the mapping class group of a surface such that the corresponding mapping torus has a hyperbolic metric

Ted Shifrin

With the Thurston classification of 3-manifolds, I realize hyperbolic structures play a huge role. But a random negatively curved manifold isn't hyperbolic, at any rate.

Presumably you mean a surface of genus $g\ge 2$?

Balarka Sen

Yeah I believe so

Ultradark

Is an orbit another name for an integral curve

Ted Shifrin

23:28

No.

You need to learn some mathematics systematically and not just come in here asking random questions.

Ultradark

In dynamical systems the integral curves for a differential equation that governs a system are referred to as trajectories or orbits

Balarka Sen

An integral curve is an example of an orbit

Ted Shifrin

The word "orbit" is far more general.

Ultradark

Just out of curiosity, what is geodesic flwo

Ted Shifrin

Ultra: I'm not going to play this game.

Go read books.

Balarka Sen

23:39

@TedShifrin I'm planning to give a student talk on group cohomology the coming semester, and learn something about it in the process

Ted Shifrin

Cool, @Balarka. You probably already know more than I do.

Is this a grad student seminar?

Balarka Sen

Nah, just an undergrad math club

Ted Shifrin

That's pretty fancy for undergrads.

Unless all the undergrads are like you, which I know they aren't.

Balarka Sen

Haha yeah I don't plan to go into the derived functors formalism

Ted Shifrin

But do they even have any idea what cohomology is?

Balarka Sen

23:42

No. I plan to explain what $H^2(H; N)$ is and why group extensions of $H$ by $N$ are classified by it

Pretty concrete stuff

I'll draw some pictures for the semidirect product

Ted Shifrin

Ah, all formal algebra. You see?! You have reverted! :D

Balarka Sen

I said I'll draw pictures, Ted!

Ted Shifrin

I typed what I typed before you typed that.

:P

Balarka Sen

Lol

Ted Shifrin

So semidirect products are like nontrivial fiber bundles.

Balarka Sen

23:43

Ya exactly

Ted Shifrin

I'm not sure I know how to "draw" a simple example.

Balarka Sen

The general picture I have in mind is $N$ semidirect $H$ looks like an $N$-bundle over $H$ (picture a lots of sticks, all the cosets of $N$, over $H$), and I'll explain why the multiplication $(n_1, h_1)(n_2, h_2) = (n_1 \varphi_{h_1}(n_2), h_1, h_2)$ in the group the way it is by basically explaining that the action $\varphi : H \to \text{Aut}(N)$ that's conjugation inside the semidirect product is giving monodromy on the fibers of my bundle picture

Ted Shifrin

So is there a concrete topological picture we have for some examples? I guess the Euclidean group is the easiest non-discrete example.

Balarka Sen

Right, that's $O(2)$ semidirect $\Bbb R^2$

Ted Shifrin

I always tried to draw (at least schematic) pictures for $G\to G/H$ as a bundle ... even in my algebra book.

Balarka Sen

23:48

Well, the isometries of the plane, I mean

Ted Shifrin

Right, sure.

Balarka Sen

@TedShifrin Yeah I just don't think normal subgroups and quotients are explained well enough in first year algebra

Ted Shifrin

That's why I started with rings and quotients, where you don't have to stress over normality at first.

Balarka Sen

That's a good cause

Fargle

I know I definitely had trouble with normality and quotients in my first pass through the subject

Ted Shifrin

23:50

well, hello @Fargle

Paul Plummer

@BalarkaSen In the random walk models this has been proven, although interesting in "counting" models it is still open

Ted Shifrin

Anyhow, @Balarka, you'll have to send me your LaTeXed notes when you give the talk :P

Fargle

Hey @Ted

Balarka Sen

@TedShifrin Uh oh

Ted Shifrin

LOL

Or not.

Balarka Sen

23:51

I can't draw pictures in LaTeX, I'm not Danu

Ted Shifrin

I can't, either. I draw them otherwise and input them as .eps into my LaTeX document.

Well, scanned handwritten notes will suffice, @Balarka. Or not ...

Fargle

I just use Paint. Professionality be darned.

Balarka Sen

@PaulPlummer What are those models

Ted Shifrin

I'm just curious to see what you do.

Balarka Sen

@TedShifrin Cool, I can do that

I just didn't want to LaTeX that's all :P

Ted Shifrin

23:53

Understood.

checks to see if @CaptainAmerica is afire

CaptainAmerica16

No, just in pain. I just got back from the dentist.

He was stabbing my gums.

Ted Shifrin

At the dentist at 7 PM?

CaptainAmerica16

No 4:30

Paul Plummer

@BalarkaSen I am not super familiar with random walks, but basically you take random walks in the Cayley graph, as the walks get longer the probability you land on a pseudo Anosov goes to 1. counting model would be taking balls in the cayley graph and counting the proportion of pseudo Anosov's. It is open that this goes to 1 as the radius goes to infinity(it is known that at least a positive proportion is pA though)

Ted Shifrin

Oh, "just."

CaptainAmerica16

23:55

I "just" got in the door.

Ted Shifrin

still confuzled

CaptainAmerica16

Words mean nothing >:C

Fargle

@CaptainAmerica16 What? I don't know what you mean.

CaptainAmerica16

I don't mean anything.

Fargle

Huh?

CaptainAmerica16

23:57

Exactly >:|

Balarka Sen

@PaulPlummer Oh cool shit

CaptainAmerica16

So, I found out my brother can count to 10. He's only 2 years old.

Ted Shifrin

Soon you'll be trying to teach him limits.

CaptainAmerica16

Hehe. I already told my mom that I plan on teaching him addition and subtraction before he starts kindergarten.

Ted Shifrin

Don't ruin his life.

CaptainAmerica16

23:59

You think he won

*won't like it

?

Transcript for

$Mathematics$ Mathematics