Mathematics - 2017-01-10 (page 4 of 6)

2:02 PM

Pet peeve of mine: Paper says "Following standard practice, we define X." But standard relative to who? It's not a definition I'm familiar with, and Google doesn't immediately know it either.

SteamyRoot

Usually standard relative to people doing the same research

Which means you have to follow their references back in time if you want more information

Semiclassical

Yeah, usually.

SteamyRoot

(sometimes through a long series of papers, some of which aren't easily or legally found online)

Semiclassical

For reference, the precise statement is

"Following standard practice, we define a Sturmian, $|S(\omega)\rangle $, as the solution of [equation] where $\tau(\omega)$ is the Sturmian eigenvalue.

Balarka Sen

Terribad notation

SteamyRoot

2:05 PM

Ew, kets.

Balarka Sen

Still bad but I was referring to $|S(v)&

Semiclassical

Eh, it's notation. Just think of it as a column vector and move on.

Secret

Suppose we have $f : \mathbb{R} \to \mathbb{R}$ and $f(x)=e^x$. It is known that $e^x$ is bijective but the image of it does not include the negative numbers. But $f$ as defined here is a bijective function from $\mathbb{R}$ to itself, why is that not a permutation?

Semiclassical

Mostly I just find the terminology weird. Sturmian solution, sure.

Alessandro Codenotti

It's not bijective @secret, what's $f^{-1}(-7)$?

Semiclassical

2:08 PM

probably related to Sturm-Liouville theory, though I don't know how.

SteamyRoot

Damnit. You were supposed to ask why I didn't like the bra-ket notation, so I could make a joke about liking bra's :(

Null

bijective -> surjective in particular

Semiclassical

Ah. I figured it was more the standard mathematician disdain for them.

Secret

that's undefined, but then what do we mean in calculus courses when we often said that $e^x$ is bijective?

Tobias Shuxue Laoshi

I apologize for asking again, but this question probably doesn't deserve a post of its own.

How many sequences $(a1,…,an)\in \Bbb F_q^n$ are there such that $a_i\neq a_j$ for all $1\leq i<j\leq n$ and $a_1+a_2+\dots+a_n\neq 0$? Where $p$ is a large prime.

Semiclassical

2:10 PM

Which is standard and therefore uninteresting.

Secret

and $e^x$ tend to have its domain and codomain defined both as $\mathbb{R}$ in calculus courses?

SteamyRoot

Maybe a bit of that. I do have a bachelor degree in physics too, so I can live with it.

Null

$e^x$ is only injective, or bijective if we observe $\mathbb{R}\to\mathbb{R}^{+}$

Semiclassical

Gotcha.

I don't mind bra-ket notation in most cases.

My main problem with it is that it gets a bit clunky/tedious at times.

Axoren

@BalarkaSen You do differential geometry, right?

Secret

2:12 PM

ok I see. So if we said there is some bijective function from a set $S$ into itself, then the image is guarenteed to be unique for each element, even if $S$ is uncountably infinite like $\mathbb{R}$?

Semiclassical

But it's nice to have notation that makes a visual distinction between vectors and covectors.

Balarka Sen

@Axoren Not particularly. Why?

Alessandro Codenotti

Yes @secret

Null

@Secret that follows from injectivity

Axoren

@BalarkaSen A while back, I feel like I misattributed the wrong field to you. I don't actually remember what you study, but I knew it once.

Balarka Sen

2:14 PM

I study topology.

Semiclassical

Point-set, algebraic, or differential?

Axoren

Ahh, that would explain things.

Balarka Sen

I know a bit of differential geometry but what I know about it can be written at the back of a letterbox.

@Semiclassical Not anything particular :) I like all three of those.

Semiclassical

Fair enough.

Balarka Sen

As well as a lot of other sub-branches of topology.

Semiclassical

2:16 PM

I imagine you're right, but I don't know what the other flavors would be.

Secret

Ok I think I got it now (and by surjectivity, every single element in $S$ will appear in the image)

(sorry guys for those seemly simple question because I often have truoble wrapping my head around uncountable things)

euclid

math.stackexchange.com/questions/1860079/… the first paragraph of second answer

Axoren

I remember being mistaken a month or so back that you were the one to ask about certain geometric things (bivectors, etc). Your strong background in topology and bit of differential geometry is probably what made me mistake you as a differential geometer.

Semiclassical

bivectors \in geometric algebra, doesn't it?

Axoren

Yeah, but they apply in differential geometry, when you start trying to define the exterior product on (over?) manifolds

I'm still not 100% on the terminology involved. I've been struggling for months with understanding the parts of geometric algebras, namely the exterior product. The fact that algebraic geometry and geometric algebra are two different things always amused me.

euclid

2:23 PM

@arctictern hi:)

Semiclassical

Yeah.

arctic tern

hello

euclid

i was waiting for you

Semiclassical

Geometric algebra is basically just Grassmann algebras (or is it Clifford? I forget what the difference is.)

Axoren

Clifford

arctic tern

2:24 PM

grassman algebra is when v^2=0 (so, exterior algebra). geometric algebra tends to have v^2=|v|^2 and clifford algebras have both sign conventions but usually v^2=-|v|^2.

Semiclassical

Ah.

euclid

@arctictern can you explain the first paragraph of your answer? math.stackexchange.com/questions/1860079/…

Socrates

how can I give something back to the community, if the questions that arise are at the moment out of my reach?

arctic tern

@euclid going to have to say which part.

Secret

Ok, I think there should be no holes in me in trying to prove the known result that involution maps are permutations (as part of the study of maps)

euclid

2:26 PM

@arctictern a unit $e$ mod $d$

Secret

uh, those ax=y and x=ay lines are typos. They don't exist

Socrates

@Secret what do you study?

Axoren

inb4 "it's a secret"

arctic tern

@euclid e is an integer mod d which is invertible. equivalently, it is represented by an integer between 1 and d which is relatively prime to d. this is standard terminology in English (not sure what your first language is)

Secret

Currently abstract algebra in some highly disorganised fashion. Later plan to read munkres

Mostly groups at the moment, though interested in general maps in uncountable sets

btw socrates, are you used to be the user null?

Axoren

2:29 PM

@Secret His topology book is nice and well-paced, but I made the mistake of getting an international bargain-bin copy. It was edited in ways I don't even know, but I know that it's at least 20 pages shorter than Ted's copy.

@Secret Also, I'm pretty sure they are.

euclid

@arctictern and $o(\chi ) = d\: \leftrightarrow \:\chi = {\psi ^e}$?

arctic tern

@euclid yes

in any multiplicatively-written cyclic group of order d, if psi is a generator, then an element has order d if and only if it is of the form psi^e where gcd(e,d)=1.

Secret

->besides those listed, the whole chat know that I am still trying to divide by zero (or show that it is impossible for all possible means)

GFauxPas

good morning frands

Axoren

@Secret The same "field" from last night?

euclid

2:35 PM

@arctictern $\{ \chi :{\chi ^d} = 1\}$ has $\phi(d)$ members?

Secret

Axoren: no, that one you discuss with me (the integer like thingy) is not a division by zero

Socrates

@Secret yes, I'm null, changed the nick today

Secret

I see

Socrates

(i keep the avatar to avoid confusion)

arctic tern

@euclid no, $\{\chi:\chi^d=1\}$ has $d$ elements, namely $\{\psi^0,\psi^1,\cdots,\psi^{d-1}\}$ where $\psi$ is a generator

GFauxPas

2:37 PM

Null we see through your disguise

Axoren

@Secret How are you trying to divide by 0?

You could just compactify the positive reals and assign $\frac a 0 = \infty$

euclid

@arctictern we are talking about $\chi :o(\chi ) = d$?

Secret

Axoren: Please refer to this room as the chat right now start to get busy

arctic tern

@euclid we are summing over those $\chi$ with $o(\chi)=d$, which is a propert subset of the $\chi$ with $\chi^d=1$

Axoren

@Secret Is there a reason you remove the @ sign after a few seconds?

Secret

2:40 PM

Truth is, I never add the @ sign in the original message. In fact, I tend to reserve @ for asking questions or some important responses

To me, @ is a valuable commodity

GFauxPas

in some countries, it's even used as currency

Axoren

It's at least two keys more valueable than the $, second only to !

GFauxPas

don't forget ^

Axoren

I don't caret all.

euclid

@arctictern thank you very much

GFauxPas

2:42 PM

rimshot

Sophie

I bought a one of those indian cheap editions of a book and the typesetting is kind of crap

GFauxPas

which book?

Sophie

Hardy's intro to number theory

Axoren

@Sophie I know your pain, in triplicate.

Sophie

Hardy died 70 years ago in december the first this year

...which means his works go into public domain

euclid

2:45 PM

@arctictern please look at this proof in imgur.com/a/d7n7i

arctic tern

@euclid what about it?

Sophie

@Axoren my mother had a gift certificate to spend on a library so I got a translation of Spivak's calculus on manifolds which has a slightly higher quality paper, but the typesetting on the math is terrible

GFauxPas

I have Spivak's Calculus on Manifolds, it was like $20 for the regular version

Adeek

@arctictern Is there a classifying condition for when $a \cap b = ab$ where a and b are ideals ?

Socrates

@Sophie much is simply papercopy, but still maybe it helps?? matematica.cubaeduca.cu/medias/pdf/842.pdf google with filetype:pdf

Sophie

2:47 PM

they converted $x_n$ to $x^n$

GFauxPas

you sure SPivak didnt do that in the original? I know some authors do that, it drives me nuts

Axoren

@Sophie I was just talking earlier about how I got an international version of Munkres' Topology book and I lost at least 20 pages to god knows what editting.

Adeek

@arctictern I know that $ab = a\cap b$ whenever $a + b = (1)$ since we have $(a + b)(a \cap b) \subset ab$.

euclid

@arctictern why summation is on ${e^{\frac{{2\pi iha}}{{{q_i}}}}}$?

Axoren

As for $x_n$ to $x^n$, I've seen many books do that. I think it's a difference of physics background.

Adeek

2:49 PM

But is there a classifying condition for when equality always occur that is an iff ?

arctic tern

@Adeek dunno

SteamyRoot

Hrmf. Someone seems to be randomly old questions of mine today.

Axoren

I've seen $x^i$ to mean the $i$th component way too much and it annoys me as well.

Adeek

I see

Sophie

I can't cope with that. I've been using a PDF instead

Axoren

2:52 PM

I'm always curious as to what people think when they see certain notations without context. What type of structure do they think they're looking at?

What is $\ _c^ax_d^b$?

GFauxPas

aaaaah axoren stop

Sophie

you can try for yourself. Find the proof of FLT and look at the symbols

arctic tern

@euclid If $\psi(g)=e^{2\pi i/q}$ where $\psi$ and $g$ are generators, and $n=g^a$, then $\psi(n)=e^{2\pi i a/q}$ and $\chi(n)=\psi(n)^h=e^{2\pi iah/q}$ with $(h,q)=1$ (i.e. for $h=1,\cdots,q-1$). (Alright I have to leave after this message.)

Semiclassical

I'd read x^i in a differential-geometry context as a contravariant component, I guess.

Axoren

And what do each of $a, b, c, d$ represent in relation to $x$?

Sophie

2:53 PM

personally the only thing I understood was big O notation

SteamyRoot

A clusterfuck.

That's what that is.

Semiclassical

One that's annoying for typesetting are the generalized hypergeometric ${_pF_q}$ functions

which are written something like

GFauxPas

what's the $\LaTeX$ for subscripts before a glyph?

Semiclassical

well, above I did {_p F_q}

Socrates

you mean $_a \delta$?

Axoren

2:54 PM

@GFauxPas There's two ways about it, I personally like "\ _ax" for $\ _ax$

Semiclassical

yeah. the brackets are there just to make sure that Latex doesn't interpret _p as a subscript of something else

GFauxPas

what about \phantom()_a?

Axoren

The "\ " makes sure that the _a doesn't attach to a previous item.

Semiclassical

${_pF_q}$ or $_pF_q$

euclid

@arctictern i wish we could exchange our emails

thank you

Axoren

2:55 PM

"\ " functions like \phantom()

Semiclassical

I think the two work out identically.

Sophie

the translation of math terms is bizarre. Once a teacher started talking about eigenvalues and I thought it was a completely new concept until I realized they're eigenvalues

GFauxPas

good to know

Semiclassical

the \_ tip is a good one

GFauxPas

Sophie, what?

Semiclassical

2:56 PM

I'm guessing they didn't call them eigenvalues?

euclid

@arctictern when you come back again?

GFauxPas

my professor told me he was once in a high school trivia contest and no one understand the question because it was with a factorial in it and the adult reading the question read it like an exclamation point

Sophie

he called them eigenvalues but I didn't know what eigenvalues are called in my native language

GFauxPas

"by convention, ZERO is defined as what?"

Semiclassical

lol

GFauxPas

2:57 PM

something like that

Semiclassical

The saving grace with stuff like x^i as a component rather than an exponent is that you don't expect to see stuff like $(x^i)^3$

the most you'd ever have is $(x^i)^2$, and that's typically taken care of by the summation convention $x^i x_i$.

Axoren

@Sophie Do you have a word for the vectors $x$ along which $Ax = \lambda x$?

GFauxPas

"the power series of which function is the infinite sum of $x$ to the $n$ over $\mathbf{ENN}$!"

Sophie

or when you have an index which has an index $x_{i_3}$

Axoren

Where $A$ is a matrix and $\lambda$ is a scalar?

Semiclassical

2:58 PM

@Sophie Yeah, that gets pretty grotesque.

SteamyRoot

Indexes like $x_{i_3}$ do tend to appear alot in, like, subsequences.

Semiclassical

It's understandable, but it's also a right pain

Sophie

@Axoren eigengenvalue -> autovalor, eigenvector -> autovetor

Semiclassical

autovector instead of eigenvector would be neat.

Axoren

That almost makes more sense in English, given that auto could mean self.

Semiclassical

3:00 PM

Autovalue, though, sounds weird to me. (But then, so did 'eigenvalue' until I got used to it.)

Axoren

An autovector is one that remains on itself when transformed.

An autovalue is associated with an autovector.

I like it.

SteamyRoot

Sounds very spanish/portuguese

Axoren

Might be why I like it.

Semiclassical

If I were to pick a word for eigenvalue, I might do...hmm

Sophie

I had a teacher called Euclid

Axoren

3:01 PM

Eigen means proper in German, right?

Semiclassical

proportion, maybe? Something that signals that an eigenvalue tells you about the relation of an eigenvector to its proportional image.

Axoren

I don't know if there's anything proper about Eigenvectors

Semiclassical

Well, they get mapped to themselves (up to proportionality). That's pretty proper.

Axoren

I don't know. That seems less proper and more noble.

Semiclassical

lol

Axoren

3:03 PM

What about Noblevectors?

Sophie

proper gases

Axoren

They keep to their own, very high society types.

Semiclassical

Noblevectors with associated noblevalues.

Axoren

Noble values

It writes itself.

Semiclassical

As compared with bourgeois vectors

...I can't think of a good joke of what a bourgeois vector would be

Axoren

3:05 PM

Spectral decomposition is just an instance of Noblesse Oblige

@Semiclassical Bourgeois vectors are noblevectors with noblevalues between 0 and 1.

Semiclassical

I guess a proletarian operator would be one that sends vectors to zero. (i.e. an annihilator of noble vectors :P)

Akiva Weinberger

With the vector $\langle256,0,0\rangle$

Axoren

I don't know what to name that vector, but I'll byte.

Sophie

3:23 PM

can I migrate one of my own questions to another stackexchange site or should I just delete it and make a new one?

Alessandro Codenotti

In Italian we do call them "autovettori" and "autovalori"

Axoren

I wish someone with mathematical clout would start calling them noblevectors because I'm absolutely obsessed with that name now.

Balarka Sen

3:38 PM

@AlessandroCodenotti Hi

Socrates

@Semiclassical but low vectors should be send towards something not 0, revolution haha

0 being an exception, obv

because all are equal, but some are more equal ;)

SteamyRoot

"Eigen" indeed means "proper" in German (and Dutch)

In Dutch, you'd translate "mijn eigen vector" as "my own vector"

Semiclassical

own-vector sounds appropriate for $Av=\lambda v$, actually

Sophie

this is my vector!

Semiclassical

so "eigen value" would be "own value"

Krijn

3:49 PM

Actually "eigen value" is also a Dutch word, meaning "self worth" or "self esteem"-ish

Self respect, is probably the best translation

Balarka Sen

"Could you beat David Lynch in an arm wrestle?" "... I think I could take him. Especially if he was meditating at the time." - Cronenberg

lol

Mats Granvik

To be "egen" in Swedish can have also have the meaning "peculiar" or "strange", as a short form for "egendomlig".

Alessandro Codenotti

Hi @Balarka

I saw eraserhead a few days ago in a cinema, didn't like it at all though

Balarka Sen

4:06 PM

I feel a subconscious interest for this genre. I won't say I like it, but I am interested in this peculiar style, and what it tries to communicate.

I haven't watched Eraserhead, but I have seen Lost Highway and Mulholland Drive (yesterday!), which are different in style I suppose

Alessandro Codenotti

I haven't seen Mulholland Drive yet, but Lost Highways (which I've seen in the same cinema) wasn't bad, Eraserhead is much more surrealist and disturbing though, I wouldn't suggest it to anyone

I'd like to see Twin peaks when I'm done with those exams and I have more free time

Balarka Sen

I was referring to the genre of Eraserhead (body horror) in the first message :)

It's also what most of Cronenberg's films are, apparently.

I should emphasize that I don't "believe" in this genre, whatever that means.

Alessandro Codenotti

I don't think I've seen anything by Cronenberg

Balarka Sen

I saw The Fly a couple months ago.

Alessandro Codenotti

I'm looking forward to La La Land, Arrival and Silence which should be in the cinema (in Italy) soon

Balarka Sen

4:19 PM

I want to watch Cocteau's Orpheus trilogy at some point.

But I have been saying this for like half a year so I don't know when

Socrates

can we say this: if $\sum_{k=1}^{n}\frac{1}{(n-k)n}$ diverges for finite n (which is obv), then $\sum_{k=1}^{n}\frac{1}{(n-k)n}$ with $n\to\infty$ diverges too

Sophie

how can a finite sum diverge? And your sum isn't defined

Akiva Weinberger

What do you mean that the first one diverges? That $\lim_{n\to\infty}\sum_{k=1}^n\frac1{(n-k)n}=\infty$?

Then what does the second one mean

Roh

Hi guys

Sophie

@AkivaWeinberger you're dividing by 0

Roh

4:26 PM

Guys, I have a question about Perturbation

ucl.ac.uk/~ucahhwi/LTCC/section2-3-perturb-regular.pdf

In this PDF, the author has solved an ODE using perturbation but I don't know what method he has used

page 2

What's your opinion?

Secret

Hmm, as $n\to \infty$ the sequence $\{\frac{1}{(n-k)n}\}$ tends to zero. Since the denominator is dominated by a power of $n \geq 2$, by the p-series test, the infinite sum should converge I guess...

Mike Miller

4:45 PM

@Balarka There's also the cremaster cycle.

s.harp

5:08 PM

I can also semi-recommend the movie "Zorns Lemma", maybe its not the same kind of modern art you are talking about tho :P

euclid

5:22 PM

@arctictern hi

Socrates

@Secret i guess my question was "how does spinach sound"

s.harp

@socrates wetly

Socrates

the top value can't be included in the terms of an infinite sum, i guess

euclid

@arctictern thank you for being patient with me

Mats Granvik

I found a sign error in the Euler Maclaurin formula for the Riemann zeta function.

arctic tern

5:24 PM

@euclid just sit around here. we can use it like email.

@MatsGranvik there are different conventions for the sign of the first bernoulli number

Mike Miller

@s.harp when i was talking about inducing associativity of H from the quaternion group, i meant the order 8 group

Mats Granvik

@arctictern Oh yes, now when you said it I remember that I have read that on Terry Taos blog.

Secret

.
Well your example only mean the smallest term is of the order of $\frac{1}{n^2}$ and for the largest term, the nth term, one has the indeterminate $0\cdot \infty$. A brief check on their derivatives wrt n both exists, hence we can use L hopital and found that it becomes $-\frac{1}{(n-k)^2}$ thus tends to zero as $n\to \infty$. Therefore your series should converge since any partial sum is between $\frac{1}{n^2}$ and $0$, which by p series test they both converge thus by sandwich theorem, the sum shoudl converge to zero

s.harp

@MikeMiller oh, ok, yes that works without a large mobilisation of effort

Mats Granvik

$$\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}+\sum _{r=1}^{q-1} \frac{B_{2 r} k^{-2 r-s+1} \left(\prod _{i=0}^{2 r-2} (i+s)\right)}{(2 r)!}$$

$$\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}+\frac{k^{-s}}{2}+\sum _{r=1}^{q-1} \frac{B_{2 r} k^{-2 r-s+1} \left(\prod _{i=0}^{2 r-2} (i+s)\right)}{(2 r)!}$$

Just for comparison. This is what I meant.

arctic tern

5:32 PM

you can put it all in one summation and write rising factorials for increased elegance

Adeek

arctic tern

:34682669 start at r=-1 instead of r=1

Adeek

@arctictern I have been trying to figure out why does it suffice to show that there is an x such that phi(x) = (1,0,...0) ?

arctic tern

@Adeek by symmetry you can get different x's to get all the coordinate tuples, in which case you can combine said x's to get any tuple

Secret

Btw guys, a possible group theory question. Suppose I have a set of 3 elements $S=\{1,2,3\}$ and the permutations defined by multiplication as $1=1_{S_3}$, $2=(23)$ and $3=(32)$. Their action on $S$ seemed to form a structure where $(23)$ and $(32)$ forms a group wrt the subset $\{2,3\}\subset S$ which is isomorphic to $\mathbb{Z/2}$ and $1_{S_3}$ act like the identity as usual.
But it is clear that the overall $(S,\cdot)$ is not a group, yet it look like a direct product of two groups. Is this direct product $(S,\cdot)\cong \{1\}\times \mathbb{Z/2}$?

arctic tern

5:36 PM

\cong

@Secret (23) and (32) are the same permutation

Adeek

@arctictern symmetry of what ?

in the equivalence relation on each $a_i$ ?

arctic tern

@Adeek of the argument. one could permute the ideals and do the same argument.

Secret

Yeah its inverse is itself, but I am not sure how to write the overall structure $S$ as a direct product of groups

Adeek

ohh

arctic tern

@Secret S is not a group, it's a set

Adeek

5:39 PM

@arctictern I thought it suffice to show that thing hold just for (1,0,...,0)

arctic tern

@Adeek yes. if you show it holds for (1,0,...,0) with respect to one way of ordering the ideals, then you've show you can get (1,0,...,0) for other way of ordering the ideals, which is equivalent to getting any of the (0,1,...0),...,(0,...,1) for the given ordering of the ideals

Adeek

I see.

Secret

ok let me elaborate. The following semigroup look like two groups being smashed together

Adeek

Yeah I see that makes sense thanks @arctictern

Secret

thus I suspect I can write it as a direct product of two (something), I know one of the bit is isomorphic to $\mathbb{Z}/2$ but I am not sure what the leftover bit resembles

arctic tern

5:42 PM

@Secret well, the subthings {1,2} and {1,3} are both the group Z/2. but {2,3} is not really a group.

it is an interesting way of conjoining groups though

okay, four people talking to me. I have to get food real quick.

Secret

{2,3} will be a group wrt the subset {2,3} of {1,2,3}, or is that a wrong way to think about it?

because under that thought process, 3 will act like an identity for the semigroup formed by the subset {2,3} and 2 will be involutive, hence the structure is isomorphic to $\mathbb{Z}/2$

arctic tern

@Secret since 3x3=2, 3 is not an identity in {2,3}

Astyx

Hi chat

Secret

uh, 3x3=3 according to the table...?

arctic tern

sorry, apparently I'm dyslexic this morning

Vrouvrou

5:49 PM

Have you an idea about this : math.stackexchange.com/questions/2092112/…

arctic tern

your table seems wrong though. shouldn't 3x2 and 2x3 both be 1?

so {2,3} isn't even closed under the operation

Secret

Well, that's the thing, looking at this table just screams it is made of two things, but since the $\mathbb{Z}/2$ looking thing is not a subgroup wrt {1,2,3} (because the 2 and 3 subbox does not give 1) yet it is a group wrt the subset {2,3}, I am at lost on how to write its decomposition

arctic tern

huh?

Secret

ok let me elaborate again (actually, I am not sure if it is a semigroup, might have to check)...

arctic tern

also, the multiplication 1x2=2 is a bit ambiguous, since you defined 2 and 3 by the same permutation (which should force 2=3 but you're declaring by fiat 2,3 are distinct I presume), which means 1x2=2 is a matter of choice

Secret

5:58 PM

(Ok I just checked, the whole structure/3x3 table is a semigroup)
The stuff in the blue box behave like $\mathbb{Z}/2$ but the column and row suggest 2 behave like an identity. Thus I somehow have two 'things' mashed together, but I have no idea how to write a decomposition (if any)

Semiclassical

Something I'm forgetting which I feel silly about.

Secret

(With this elbaoration, we can forget abotu permutations as this is what I tried to say on how the elements 0,1,2 behave when I said those permutations, which might be what confused the issue)

Semiclassical

How long does it take for an arXiv preprint to become visible online? I noticed a paper that was submitted Saturday, but I don't think it would've become immediately available at that point.

iwriteonbananas

@MikeMiller So the $0$-manifold $\{p\}$ has exactly two framings. Can you tell me why $\{p\} \sqcup \{q\}$ where the two points have distinct framings is zero is $\Omega_0^{fr}$ ?

Mahmoud

Hi

arctic tern

6:03 PM

@Secret Given an algebraic structure X with a single operation, we can adjoin an element e with the relations ex=x=xe for all x in X and e^2=2. essentially, adjoining a global identity (which may turn an already-existing identity into just a "local" identity). your structure is this construction on Z/2Z.

Secret

Ah I see. I have been confused in the past by similar structures as they keep popping up in my analysis and I always wonder how to decompose them

Semiclassical

6:16 PM

Hmmmm

I forget. What kinds of transformations preserve incidence of lines?

Obviously affine transformations will do, but I don't think one needs anything that strong.

(seems like projective transformations are incidence-preserving as well.)

Mike Miller

@iwriteonbananas There's a framed path going between them.

Mahmoud

math.stackexchange.com/questions/2092237/… I'm probably going to get down-voted.

SteamyRoot

I'm not sure what you mean with "incidence" but I imagine projective transformations preserve it.

Semiclassical

Incidence geometry is what I had in mind.

Socrates

6:32 PM

can temperature in the air $g$ be viewed continuous? Or is at the atomic level temperature not continuous?

arctic tern

not continuous things can be modeled by continuous things

Socrates

what is even temperature to begin with? how much the molecules "wobble"?

Astyx

Heat is mesoscopic, IIRC

Semiclassical

Temperature is rather like pressure in that it's a macroscopic variable.

Astyx

I have had nice courses on how temperature is defined

Semiclassical

6:34 PM

The pressure exerted on a surface is a convenient macroscopic description of the forces exerted on said surface by the particles in the gas.

Socrates

so, it is not a fortunate model to assume that 100° can be next to 0°?

Astyx

What do you mean ?

Socrates

that in the air, there can be a discontinuity of temperature

Semiclassical

It therefore makes no sense to ask whether or not the pressure can be changed at the microscopic level. It's a useful concept right up until it stops being applicable.

Astyx

This sounds a lot like : "It's useful until it stops being useful." :p

Semiclassical

6:37 PM

To the extent that the concept of pressure is applicable to a given system, it will be continuous.

Yeah, I suppose it does.

Mike Miller

@Balarka under what conditions is compactly supported cohomology the same as the cohomology of the one point compactification?

Socrates

Meh, I don't want to spoil myself AGAIN haha

Mike Miller

this is true for the interior of an oriented compact manifold.

I'm not sure if you need the orientation assumption.

I think the main assumption is probably that the compactified point is still CW/triangulable.

SAWblade

Hey all. :)

Socrates

can/should it be further assumed, that the temperature function is differentiable everywhere? I mean, if we could measure the temperature on a ruler, and plot all those temperatures, should we assume that there are no sharp turns?

Semiclassical

6:48 PM

You mean, like the local temperature distribution in a medium?

Alessandro Codenotti

what does the group of self homeomorphisms of $\Bbb R^n$ look like?

Socrates

@Semiclassical yes

Mike Miller

It's quite big. What do you want to know about it?

iwriteonbananas

@MikeMiller This is true if the added point $\infty$ has a neighbourhood in the one point compactification which is a cone with cone point $\infty$

Semiclassical

It's a bit tricky. If I've got a material with some insulating boundary, then to the extent that we treat that as a perfect insulator then the temperature would not need to be continuous across that boundary.

On the other hand, you can imagine 'zooming in' on this boundary and seeing that the temperature changes rapidly but continuously across it.

Mike Miller

6:51 PM

@iwriteonbananas That's what I was thinking... How do you prove this? Diagram chasing along the various identifications?

Semiclassical

So if the concept of temperature remains applicable once we zoom in, then it'll be continuous.

Alessandro Codenotti

I don't know exactly, I just realized that the set of self homeomorphisms of a topological space is a group so I was wondering what can be said about everybody's most loved space

iwriteonbananas

@MikeMiller I forgot, it was too long ago...it's an exercise in Hatcher

Semiclassical

On the other hand, there certainly are systems where the statistical assumptions necessary for temperature to be applicable simply don't happen.

Balarka Sen

I don't remember it either

I haven't thought a lot about compactly supported cohomology

Semiclassical

6:52 PM

In that case, stuff like energy and particle number will still be applicable. But the temperature won't be.

(This kind of thing is common in non-equillibrium physics. I don't do a lot of that, but my advisor does and I know this kind of thing comes up)

Mike Miller

o well

How y'all doin

iwriteonbananas

Fine I guess

I actually asked about this before: math.stackexchange.com/questions/1661276/…

Balarka Sen

pretty much alright. I am mostly clueless about what I should be doing

Semiclassical

It's worth keeping in mind that the whole reason we care about temperature is that it tells us whether or not energy will transfer as heat from one body to another.

In the temperature distribution case, you make the assumption that each little bit of the medium can be treated as a 'body' in this sense.

But of course, that's a bit silly. If you zoom in far enough, you get to the quantum level and that ceases to be a useful description of the problem.

iwriteonbananas

@MikeMiller Right, I'm just confused by the following: A framing of such a path should be a map $I\to O$ right? But we want this to land in $SO$ at time 1 and in $-SO$ at time 0 right? How's that possible?

Ted Shifrin

6:55 PM

Bananas!!!

Semiclassical

hi @ted

iwriteonbananas

T E D ! T E D ! !!

Ted Shifrin

hi @Balarka, @Alessandro, @Semiclassic

Alessandro Codenotti

Hi @Ted

Semiclassical

@TedShifrin Got a geometry question for you.

Ted Shifrin

6:56 PM

oh, g'night @MikeM

Balarka Sen

Hi @Ted

Astyx

I'm very sad you left me out. I guess I'm undeserving

Anyways hi @Ted

Semiclassical

I was running into some stuff about incidence geometry (in my research, actually)

Ted Shifrin

Indeed. You were also quite silent. Hi @Astyx

So what's the question, @Semiclassic?

Semiclassical

What geometric transformations preserve incidence structure?

Astyx

6:57 PM

Physics tends to have this effect on me

Ted Shifrin

That's way too vague a question.

Semiclassical

I figured.

Socrates

Ok, then I come out of my closet with my excercise. Make a mathematical model to give reason to the following: on the equator there exist at least two antipodal points which have the same temperature on a fixed time. (read: the temperature is "measured" at the same time)

Ted Shifrin

What incidence structure are you talking about?

Semiclassical

If you scroll up a bit to my image (at around 12:16)

Socrates

6:58 PM

I think I just disproved it.

Ted Shifrin

Is this a new name for you, @Socrates?

Socrates

@TedShifrin yop ^^

Ted Shifrin