Mathematics - 2021-03-05 (page 1 of 2)

OneCold Ruben

00:06

Anyone familiar with the discrete method of defining a distance on a metric space?

Mike Miller

00:23

@Thorgott I don't think it's annoying. Give an atlas consisting of charts either contained in the interior of the two sides, or consisting of charts obtained by gluing a boundary chart on one side to a boundary chart on the other via f.

OneCold Ruben

Why do math texts ask you to prove things instead of telling you how to prove things?

Semiclassical

@user2103480 dunno, i don't deal with fluid mech that much, especially not involving the stress tensor

Thorgott

@MikeMiller I think writing the glued charts is annoying. First of all, you need to restrict the domains so that they actually glue together to an open set in $M\cup_fN$ and then you also want the charts to have image a box or something in $\mathbb{H}^n$ (because you need to propagate the twist by $f$ through the entire chart) and both conditions need to be fulfilled simultaneously. At least I don't see a better way.

@OneColdRuben cause math is something you ultimately only learn by doing yourself, not by watching others do it

of course, any sensible text will provide you with the tools necessary to prove the things it asks you to prove

user2103480

@Semiclassical ok ok

OneCold Ruben

I find a complete lack of explanation, in math explanations at a higher level. Just read the book, Just read the book carefully, Just read the book 6 times and more carefully.

user2103480

00:31

@leslietownes I just noticed the guy with the bleached hair wears a tracksuit bearing the logo of a german football club lol

Mike Miller

Yeah that's what I had in mind Thorgott.

Notice that the resulting atlas is canonical, though ugly

Thorgott

actually, after second thought, I'm not sure if that works at all

why is gluing two boundary charts going to be smooth at the boundary

Mike Miller

Definition chase, to say that a map from a half-space is smooth means it extends to a smooth map on a slightly larger open space

That's definitely the correct recipe

It's the atlas consisting of charts contained in both sides and those whose intersection with $\partial M$ is a chart and intersection with $M, N$ are half-space charts

Thorgott

but these extensions may not agree with the other chart

what I'm saying is that a smooth function on upper and lower half-space respectively don't always glue together to a smooth function on all of Euclidean space

this atlas definitely works in the topological category, but I'm not convinced it does in the smooth one

Mike Miller

Why doesn't your collar idea run into the same issue

Global choice so no compatibility problems?

(I agree that this is an issue)

OK, yeah, I think it's because of this global choice thing.

Your collar idea is fine

One is then reduced to showing that collar neighborhoods are unique up to isotopy, though I think all you actually need is that their germs are unique up to isotopy, which I believe is very straightforward and analagous to what you did for discs

Thorgott

00:40

Taking union of two collars give me an open subset homeomorphic to $\partial M\times(-1,1)$. This, together with the embedded copies of the interiors of $M,N$ gives an open cover of $M\cup_fN$ by three open subsets that carry compatible smooth structures by transport, hence they give a smooth structure on $M\cup_fN$

Mike Miller

Yeah

Definitely not the language I would use but it's surely the same and the atlas junk is just unpleasant to talk about

Thorgott

it's a global choice of straightening out the boundary, so to say

Mike Miller

yeah

Thorgott

This explains why Lee does the explicit atlas construction in his topological manifolds book, but uses collars in the smooth manifolds book

I thought the atlas would still work and he was just lazy the second time around, but there actually is a substantial difference

Mike Miller

My collar isotopy argument: Fix a "standard collar". Take any other collar, shrink it into this standard collar, isotope it so that it is linear with a derivative argument. Then observe that the space your derivatives take values in is contractible

(I'd like to say "the derivatives are all diagonal block matrices which are all-1s except the last entry, which is positive", so that the derivatives live in the space R_{>0}, but that's not quite right; the matrix need not be diagonal; the argument is still fixable)

Then use that to isotope to your standard collar

Thorgott

00:45

what's a "linear collar" in this context?

it's not like it lies in one chart

Mike Miller

Dunno, whatever the proof forces it to be

What is written above might actually isotope the collar to be fiber-preserving

your standard collar is M x [0,1) so the fiber factor is a line and then linearity makes sense

robjohn

@TedShifrin either would be an interesting statistic

@OneColdRuben $d(x,x)=0$, $d(x,y)=1$ when $x\ne y$

Thorgott

yeah ok, I buy that doing this fiberwise works

sweeping uniformity issues under the rug, but it ought to work

Mike Miller

Yeah I get more paranoid if something is noncompact.

Ted Shifrin

01:00

@robjohn i think I skew the mean age in chat up about 4-5 years on a typical day. ;)

robjohn

Combined, we do more than that

leslie townes

01:30

then i show up and lower the mean with my extreme youth

Karim Mansour

02:19

Hi

leslie townes

howdy

Karim Mansour

I haven't studied deformation theory yet but does anyone know if you start with smooth projective variety. In particular it is a manifold. What if you vary the almost complex structure associated with S. Do you get a family of smooth projective varieties ?

I think you might actually leave the projective category if you do that

anyway I know next to nothing about this but yeah worth to know if anyone is willing to discuss.

leslie townes

i don't even know what a projective variety is. i mean, i could guess, but i missed that by a mile in my own study.

Karim Mansour

@leslietownes oh np. Consider first is $\mathbb{P}^n = (\mathbb{C}^{n + 1} - \{0\}) / \sim$.

After that consider homogeneous polynomials whose zero lie in $\mathbb{P}^n$ and those define projective varieties.

Hi, @TedShifrin

Mike Miller

02:54

There's no reason to even believe that the deformed ACS is even a complex structure. But even if you're deforming among integrable complex structures the resulting complex manifolds need not be projective.

Look into the moduli of complex structures on complex tori and on K3 surfaces.

Karim Mansour

I should say subject to integrable condition. But yeah you escape projective category.

Do you suggest a book about this ? I want a deformation say that is not algebraic that capture a lot of properties in families and doesn't escape projective.

Thanks for the suggestion about K3 surface I have Huybrecht has a book on K3 surfaces also found his book about complex geometry does what I am saying so I will read both @MikeMiller

OneCold Ruben

03:49

@robjohn This was regarding an interval and and the function F(x) = x^2 trying to find the preimage of the interval [-1,0).

robjohn

04:25

@OneColdRuben okay, I guess I don't understand where the discrete comes in.

OneCold Ruben

04:35

That was a different question.

robjohn

I was responding to

5 hours ago, by OneCold Ruben

Anyone familiar with the discrete method of defining a distance on a metric space?

Mike Miller

06:05

No, I never read a book on this material @KarimMansour, since it's not really my specialty. I'm just aware of results.

I probably learned the story about complex tori from stackexchange or just from conversations. The K3 story I learned from Joyce's book on manifolds with special holonomy. I love that book but I strongly suggest against it for you, since it points in a very very different direction then you are looking in (a lot of it is about very hard analysis of PDE but you are more interested in the complex and algebraic-geometric side).

Huybrechts good might be good, I do onot know.

Ted Shifrin

I already pointed out Morrow-Kodaira and, indeed, Kodaira's published papers, as good sources. The key thing to get started is to understand that the infinitesimal deformations come from $H^1(X,\Theta_X)$, which is the algebraic geometer's notation for the sheaf of sections of the tangent bundle.

Karim seems not to have looked for any of it, though.

Mike Miller

Amusingly you see variants of the same idea in gauge theory or really anywhere you'd like to deform something.

Ted Shifrin

06:26

Sure. Kodaira rules :)

Matthew Christopher Bartsh

06:58

I posted an answer in math.stackexchange.com/questions/65760/… a few hours ago. My answer discusses binary and hexadecimal notation and pronunciation and proposes a new unified set of names for all powers of two. Do you know of a forum where I could post it and thus have some discussion of it?

The name of any power of two can be deduced from a small set of rules, so memorizing the new names is unnecessary.

fido9dido

09:59

I want to know the meaning of the symbol ^, r in random number uniformly distributed

Leaky Nun

@fido9dido "and"

fido9dido

thx

porridgemathematics

11:00

Is it true that $\{(x,y,z) : x^2 + y^2 + z^2 = 1 \} \cup \{(0,0,t) : -1 \leq t \leq 1 \}$ is homotopy equivalent to a wedge sum of $|\mathbb{R}|$ many circles?

via collapsing the straight line connecting the antipodal points

love_sodam

My proof that $M_f$ is Hausdorff is roughly 2 pages

Hope I'm not the only one..

Jakobian

11:35

@LeakyNun um. It could mean minimum

Leaky Nun

@Jakobian yeah but look at the entire picture

Jakobian

this notation is preferable in probability

Leaky Nun

the things to the left and to the right are statements

@porridgemathematics no, because in the wedge sum the individual loops won't be connected

Jakobian

maybe, probably

porridgemathematics

@LeakyNun ah yeah, thats true

would squashing down that line give a homotopy equivalence to some space?

wait thats a stupid question, of course it will, some nice space I mean?

i think it would be the wedge sum of two cone like things

user2103480

11:44

@porridgemathematics It's a torus where you identified the meridial circle, dunno if that helps

Mike Miller

This space is homotopy equivalent to $S^1 \vee S^2$. You can find an argument in terms of collapsing contractible subcomplexes in Ch0 of Hatcher

user2103480

@porridgemathematics if this were a wedge sum you could disconnect it by removing a point

ah okay sorry

homotopy equivalence is of course not homeomorphism

Leaky Nun

12:06

@porridgemathematics or you can just pull the string out

to see that it is S1 V S2

user2103480

@LeakyNun or collapse a meridian

love_sodam

12:22

I don't know why most of the students in my school hate silverman's complex analysis

anakhro

13:42

@love_sodam I find that complex analysis needs an overhaul in general, let alone textbooks on the subject.

StudySmarterNotHarder

0

Q: $N \subset \bigoplus_{n \geq 2} \Bbb{Z}$ is a graded ring with homogeneous elems of degree $i$ that are linear relations of $i$-fold prime products.

Let $M = \bigoplus\limits_{n \geq 2}^{\infty} \Bbb{Z}$ be the graded ring. However, we will not be using componentwise multiplication. Instead we define a new multiplication. Define $P_1 \subset M$ to be the set of all (prime-indexed) "linear relations" in the primes: $a = (a_p) \in P_1 \iff \s...

StudySmarterNotHarder

13:58

E.g. $2 + 3 - 5 = 0$ so $(1,1,-1, 0, \dots) = a$ is the relation. The set of all such relations where we replace prime by products of primes, forms a graded ring with a special multiplication defined.

StudySmarterNotHarder

14:30

@JackOhara

BigSocks

15:08

0

Q: Is there a notion of "product" that fits this description?

So I want to take an algebraic structure $A$ (given as a set and a function, maybe some relations eventually) and I want to make sure it is (countably) infinite. If it's already infinite, we do nothing. If it's finite, I want to take $\omega$ many copies of it and stitch them together in the most...

logic model-theory universal-algebra

this is probably just the free product, but I am kinda hazy

anakhro

@BigSocks how do you quantify "in the most boring way"?

BigSocks

Sort of “adding no information” or “adding the least info possible”

So you don’t get something too different afterwords. I want to be able to say that the original structure can be recovered basically

anakhro

Well using the finite words structure that we discussed that last time with your semigroup problem, you can get a countably infinite structure, and then an obvious projection onto words of length 1.

BigSocks

Yeah, the finite words structure is kind of what I have in mind for what it looks like

anakhro

This is just the direct sum usually.

$\bigoplus$

BigSocks

15:21

Yeah, and direct sum is usually coproduct right? And algebraic categories always have that I think

I guess I just wanted to see if that reasoning held for any sort of algebraic category

Also in a way that was “presentation independent”- just talking about sets and functions defined on them

anakhro

I think algebraic categories don't necessarily have coproducts.

BigSocks

ncatlab.org/nlab/show/algebraic+category

I thought “having all binary coequalizers” got you coproducts in particular

But I am rusty

anakhro

Not sure.

BigSocks

Need some categorypilled folk

Thanks for putting thought into it btw

anakhro

What ways does the coproduct fall short of your idea, @BigSocks?

BigSocks

15:32

I think it actually matches my idea! But I am not so sure how to express it generally mostly because of the confusion we share with respect to the category theoretic language

Like, "$\bigoplus_{i < \omega} A$ is always of the same type as $A$" is also my worry

anakhro

What do you mean by "type" here?

Just belongs to the same category?

BigSocks

Like if $A$ is a monoid, $\bigoplus_{i < \omega} A$ is a monoid. If $A$ is a vector space, $\bigoplus_{i < \omega} A$ is a vector space. If $A$ is a algebraic guy, $\bigoplus_{i < \omega} A$ is the same kind of algebraic guy

@anakhro If this were what I meant, the answer would be obviously yes if "having all binary coequalizers" implies we get the coproducts we want

and also I hope it is equivalent to what I mean

Matthew Christopher Bartsh

What is a good forum for proposing and discussing a new system of names for for all powers of two that are whole numbers?

BigSocks

And it seems totally obvious that this should be true. I just wonder if there is some cheeky counterexample that ruins if for this particular kind of algebraic category of guys, like if they have different sorts or have special constants things break (?)

@MatthewChristopherBartsh what do you mean by "power"

Matthew Christopher Bartsh

I mean 2^n

BigSocks

15:43

where $n$ is what kind of number

Matthew Christopher Bartsh

Any whole number

BigSocks

so all the negative ones don't fit your description and all the nonnegative ones do, right?

Matthew Christopher Bartsh

Any whole number, including negative whole numbers.

BigSocks

aha, but when $n < 0$, $2^n$ is not a whole number

anakhro

maybe he means "all powers of two where the powers are whole numbers"?

Rather than the power of two being the whole number?

Matthew Christopher Bartsh

15:48

Correct, anakhro

BigSocks

yeah maybe we should clear up "$2^n \leftrightarrow$ a power of 2" and "$n \leftrightarrow$ the exponent" or something

Matthew Christopher Bartsh

I don't follow.

What does $ mean here?

BigSocks

ah

anakhro

$ is just "math mode" for latex.

BigSocks

just ignore it basically - makes math look nice

Matthew Christopher Bartsh

15:51

Do I need to use it here?

BigSocks

nah

anakhro

I'd probably call such numbers "dyadic integers" or something like that. :P

BigSocks

@anakhro yea

Matthew Christopher Bartsh

Why?

anakhro

Well the "dyadic rationals" are numbers of the form a/2^n

(a,n integers)

So I'd be comparing it to this, but if I just wanted the powers of two, I'd call them "dyadic integers".

Matthew Christopher Bartsh

15:55

That's interesting.

anakhro

Not a perfect name, but most math people would be able to guess what it is.

BigSocks

also for fun you could think of it as all the numbers you can write in binary by just writing a 1

and then filling the rest with 0

Matthew Christopher Bartsh

2^3, -2^3, 2^-3, -2^-3 would be called what?

BigSocks

@anakhro in light of what @MatthewChristopherBartsh just said these would be like the positive "dyadic integers"

anakhro

@BigSocks probably

But doesn't help the question he just asked entirely since he's introducing all the dyadic rationals where a=\pm 1

@MatthewChristopherBartsh at that stage I would not give them a name, because it's more useful just to write it with set notation.

BigSocks

16:02

@anakhro yea

Matthew Christopher Bartsh

How do you write it with set notation?

BigSocks

$\{ \frac{a}{2^b} \vert $ $ a \in \{ -1,1 \} \wedge b \in \Bbb Z \}$

anakhro

{±2^n | n ∈ Z}

Or what BigSocks wrote

BigSocks

yeah they're the same set, but anakhro's can be easily copypasted

Matthew Christopher Bartsh

What about the negative powers of two like - 2^3

*?

BigSocks

16:08

what about em

Matthew Christopher Bartsh

My bad. I missed the plus minus

Are some dyadic rationals negative?

anakhro

Yes.

As a set, they are exactly {a/2^n | a,n ∈ Z}

Matthew Christopher Bartsh

Wikipedia, Dyadic rationals says,

In mathematics, a dyadic rational is a number that can be expressed as a fraction whose denominator is a power of two. For example, 1/2, 3/2, and 3/8 are dyadic rationals, but 1/3 is not. These numbers are important in computer science because they are the only ones with finite binary representations; they also have applications in weights and measures and in musical time signatures.

anakhro

Yes. No conflict.

Matthew Christopher Bartsh

No mention of negative numbers anywhere in that, nor in the rest of the article. Is Wikipedia at fault?

What is not conflict?

Jakobian

16:11

also they seem to have importance in some constructions in probability and analysis because they're dense in R, and have this "graded" structure

anakhro

No, wikipedia mentions it more generally when it discusses p-adic in the next paragraph you missed copy and pasting,

"integer" --> element of Z = {...,-2,-1,0,1,2,...}

You can just read the first sentence of what you copy and pasted, too, and notice it does not conflict with my set: "a dyadic rational is a number that can be expressed as a fraction whose denominator is a power of two."

They only can give you finitely many examples, so don't let that discourage your imagination. :P

Matthew Christopher Bartsh

But they should have included a negative number, surely?

And what about imaginary numbers?

anakhro

@MatthewChristopherBartsh they did include negative numbers.

And no, imaginary numbers are not included.

Matthew Christopher Bartsh

What would you call the set if imaginary numbers were included?

How would you specify it with set notation?

anakhro

How are you including imaginary numbers?

Matthew Christopher Bartsh

16:23

@anakhro I meant in the examples there was not one negative number.

anakhro

@MatthewChristopherBartsh in the examples there also wasn't one with the denominator 16, but such dyadic rationals still count.

Matthew Christopher Bartsh

i times 2^3 is 8i which is an imaginary number, right?

anakhro

Take a look at our examples of set notation, and see if you can take a stab at this one yourself (in particular, BigSock's might help here).

BigSocks

@MatthewChristopherBartsh remember, when $a$ is negative, the whole thing is negative

Matthew Christopher Bartsh

I am not good at maths.

anakhro

16:29

Well it's a good thing that being good at mathematics is not a pre-requisite to doing math. :)

BigSocks

it's ok, don't be too hard on yourself

Matthew Christopher Bartsh

If Z is the integers, what is the set of real integers plus the imaginary integers?

That came out wrong I think

BigSocks

kind of not the standard way to say it, but check out the Gaussian integers

Matthew Christopher Bartsh

If Z is the integers, what is the imaginary integers?

What is the symbol?

anakhro

Yeah, the Gaussian integers is notated as Z[i] and is the set {a + bi | a,b in Z}

leslie townes

16:33

Some people write Z[i] for the Gaussian integers. this is a special case of a general ring theoretic construction

but it can also just be the name of a set

Matthew Christopher Bartsh

Is an imaginary number a complex number?

leslie townes

side note, i love when math concepts generalize something and the terminology follows suit. e.g. not all gaussian integers are integers. and algebraic integers also don't have to be integers. you forget how weird that seems to some people, although it is common for other reasons in other areas of English usage (e.g. an attorney general is not a general and a vice president is not a particular kind of president)

anakhro

People tend to stray from the terminology "imaginary", reserving it for "imaginary part" which is the part "b" of a complex number "a+bi".

Mike Miller

Note that your question is answered by the first sentence of the wikipedia article on imaginary numbers

"An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i"

Matthew Christopher Bartsh

Is there a way I can look at our conversation and reread it carefully later? Or do I need to take notes as we go along?

anakhro

16:40

@MatthewChristopherBartsh you can always refer back to it in the chat later (using the search part, or scrolling back that far).

Matthew Christopher Bartsh

Okay. I was copypasting your formulas into Notepad like crazy.

Does the conversation stay here forever?

anakhro

It stays here long enough that you don't have to worry about it. :)

leslie townes

i have mixed feelings about wikipedia as a math resource. it is certainly a lot better than it used to be. some topics are still kind of a randomized dump of partial answers to questions that nobody asked. and the encyclopedia premise is somewhat at odds with mathematical practice (e.g. different people defining the same things slightly differently, changing the status of things like definitions and theorems). it is better than nothing, particularly with a lot of libraries closed.

Matthew Christopher Bartsh

The math articles in Wikipedia are nearly always incomprehensible to me. It's like they are in foreign language.

How do I get your attention next time I come here?

Mike Miller

One should get a book, they are better than wikipedia. But first step due diligence is googling. I just had to comment something to the same effect to an MO poster, where the point is much more relevant, as that site is in principle for people doing research.

Matthew Christopher Bartsh

16:47

What is MO?

anakhro

@MatthewChristopherBartsh there is almost always someone here to help, so just drop by and ask.

MO = "Math Overflow".

leslie townes

math overflow is a funny site. i understand why it has the name, but it always makes me think of something undesirable that would happen after a plumbing malfunction. uh oh, the math is backing up and leaking all over my tile floor.

Matthew Christopher Bartsh

But what if I have a question about something someone said?

@leslie lol

anakhro

@MatthewChristopherBartsh you just ask. You can also hover your mouse over the message you wish to reply to, then click the arrow on the right that looks like "|->"

Matthew Christopher Bartsh

Math to me is like a minefield.

Mike Miller

16:50

I don't much care for the name. But I also don't much care about the name.

leslie townes

i tend to participate anonymously on websites. i can't even remember what pseudonym i used to post there. which is a shame because i think i answered at least a few good functional analysis questions.

Matthew Christopher Bartsh

@leslietownes Like this?

That seems convenient.

Mike Miller

I think the only thing attaching it to your name is good for is to keep yourself honest. If I participated pseudonymously I would probably never look back at old posts someone asked about.

Matthew Christopher Bartsh

@leslietownes Do mean original thinking that you could publish?

anakhro

@leslietownes your name sounds like an writer's pseudonym already. ;)

leslie townes

16:53

i think the norms have changed around this somewhat. participation is more often seen as a positive thing. when that site was getting off the ground i thought it was weird to see people with actual teaching responsibilities at the beginnings of their careers obviously spending more time on MO with their friends than preparing for classes.

anakhro

@MatthewChristopherBartsh it was nice meeting you anyway. Enjoy your day!

BYE EVERYONE

leslie townes

enjoy yours too

Matthew Christopher Bartsh

Bye everyone

Jakobian

17:31

books are often more specific, directed, and you often need to be acquinted with the topic, unless you want to spend time reading a book

so there are pluses and minuses

leslie townes

a bad book can also definitely be worse than a mishmash of internet references

user2103480

@leslietownes haha I remember before I studied math and thought "man these wikipedia articles are indecipherable"

now I honor wiki for what it is. It is quite good for math I think, although it gets less complete for grad classes

And my blood pressure rises when I have to read statistics or computer science articles that aren't formulated in mathematical language

Edward Evans

Adele ring

In mathematics, the adele ring of a global field (also adelic ring, ring of adeles or ring of adèles) is a central object of class field theory, a branch of algebraic number theory. It is the restricted product of all the completions of the global field, and is an example of a self-dual topological ring. The ring of adeles allows one to elegantly describe the Artin reciprocity law, which is a vast generalization of quadratic reciprocity, and other reciprocity laws over finite fields. In addition, it is a classical theorem from Weil that G {\displaystyle G...

the most complete wiki page I've ever seen

user2103480

amazing

leslie townes

the entry for eigenvalues and eigenvectors is also bizarrely long

my favorite math wikipedia genre is the very high quality graphic, often animated, that is more intricate and confusing than the underlying idea

user2103480

17:37

Fourier transform

In mathematics, a Fourier transform (FT) is a mathematical transform that decomposes functions depending on space or time into functions depending on spatial or temporal frequency, such as the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of space or time. The Fourier transform of a function of time is a complex-valued function of frequency, whose magnitude (absolute value...

this one's my top contender

Jakobian

yeah, I often get frustrated when I'm reading a relatively good book, but then it starts being incomprehensible/non-formal/sweeping under the carpet/completely wrong

leslie townes

maybe i'm just not a graphical guy, but the stuff that passes for illustration on wikipedia mostly confuses the hell out of me

Jakobian

but sometimes it's on me, which just makes it more frustrating

I still don't fully know how to deal with those kind of situations though

user2103480

@EdwardEvans I wanted this week's friday song for you to be another obscure russian thing but now I found this

Franzl Lang - Einen Jodler hör i gern (Hardstyle Buamz Remix)

►

Edward Evans

oh gott

der Jodlerkönig at it again

user2103480

17:44

it is written

Edward Evans

lmfao

I was anticipating the beat

knew exactly when it was gonna drop

Alessandro Codenotti

@EdwardEvans en.m.wikipedia.org/wiki/Completeness this one is also very complete

Edward Evans

Bei den Antworten zum zweiten Kommentar "vallah das wird auf tuningtreffen gepumpt" hahaha

user2103480

@EdwardEvans lmfao

the comments are top quality there

Edward Evans

@Alessandro I was like "mate this isn't that complete"

and then it hit me

Alessandro Codenotti

17:48

I'm very sorry

I'm just bored at the airport

user2103480

....

Edward Evans

do math next to a Karen and see if you get arrested

user2103480

@EdwardEvans "get the descriptive set theorist!"

Alessandro Codenotti

Eh I wouldn't even be the first Italian to do that

Edward Evans

lmao

user2103480

17:50

@AlessandroCodenotti get arrested?

Edward Evans

that was a story right

Alessandro Codenotti

@user2103480 make another step and I'm going to... Uh.... Describe some sets

Edward Evans

some guy was doing PDE on a plane and got arrested

geocalc33

yeah^

Alessandro Codenotti

@user2103480 get arrested for doing math on a plane

Maybe he was an economist actually, I'm not sure

user2103480

17:50

@AlessandroCodenotti "I will start properly discontinuous actions"

Alessandro Codenotti

snopes.com/fact-check/italian-economist-removed-terrorism apparently it's been fact checked by snopes as well

Edward Evans

hahaha "doing math that looked like terrorism"

Balarka Sen

All economists should be arrested

Nothing they say work

Edward Evans

do you reckon he flew economy class

..

kms

Astyx

it's not like people listen to economists anyway

Alessandro Codenotti

17:55

I just googled him, I was expecting him to look much more arabic after this tale, but I guess that passenger was just incredibly paranoid somehow

Balarka Sen

@EdwardEvans good joke

Edward Evans

ty xo

Mike Miller

@leslietownes I think it's probably more accurate to say that they spend their free time doing it. i certainly don't do that in lieu of, or more time than, I spend on teaching or research. It's an addition. A timepass.

Timepass is a great indian english word.

Balarka Sen

tp

leslie townes

there are apocryphal stories of books not being allowed into the soviet union for touching on sensitive topics of political advocacy, such as the highly political movement to Free Abelian Groups

Balarka Sen

17:56

i like it

Alessandro Codenotti

In any case I already landed, and nobody called security even though I was reading a paper on the plane, I guess that topological dynamics doesn't look as threathening

Mike Miller

Pastime suggests it's a pleasant activity, whereas timepass suggests you're literally just doing it to kill the time

(To me)

Balarka Sen

yeah lol

Alessandro Codenotti

Or maybe terrorists don't LaTeX their stuff (they really should)

leslie townes

@MikeMiller I think you're right, in general, although for at least one person i knew, he was very much MOing all day and not even preparing for class or working on his thesis. it does make sense that people who love math are going to spend their free time on it. i do love "timepass."

Balarka Sen

17:57

Today I learnt

Mike Miller

Yeah that's probably true

Alessandro Codenotti

@EdwardEvans to be completely honest I also feel afraid when I see a PDE though

Balarka Sen

The following

Edward Evans

the curly d looks like terrorism, if he was doing ODEs he'd have been fine

Balarka Sen

If A is a fiinite subset of R^n, then seen from any arbitrary point x in A, most of the set A is concentrated around at most a single point y in A

More details after dinner

Alessandro Codenotti

17:59

This smells of Gromov

Balarka Sen

There is a connection

user2103480

@BalarkaSen brownian motion was derived by a financial mathematician 5 years before einstein. it's not a reasonable description of stock markets but it is certainly a useful insight coming from economics

Balarka Sen

This is due to Schramm and Benjamini

Transcript for

$Mathematics$ Mathematics