Mathematics - 2017-07-05 (page 3 of 10)

Sha Vuklia

12:13

@Ted ah okay, that's clear. I have one more question, more on notation. Say we have $O(\Vert\vec h\Vert)$. Should we then write: $f(\vec h)=O(\Vert\vec h\Vert)$, or rather $f(\Vert\vec h\Vert)=O(\Vert\vec h\Vert)$?

Astyx

@Sha Depends what the question takes as input

My guess is it takes vectors, so first option

Daminark

I concur with Astyx on this point, this is likely a function mapping $\mathbb{R}^n$ to $\mathbb{R}$ given context, so $f(\|h\|)$ wouldn't work.

Astyx

It would be very vicous to use the notation $\Vert\vec h\Vert$ as a variable if your function is $\Bbb R\to \Bbb R$

Daminark

(removed)

Astyx

Is there any way to make a triple $\vert$ ?

Daminark

12:26

Y tho?

Balarka Sen

what's cookin

Astyx

I've always used $\vert\vert\vert A\vert\vert\vert$ to denote the triple norm $\sup_{\Vert x\Vert = 1} \Vert Ax\Vert$

Daminark

Merp

Tobias Kildetoft

@BalarkaSen Not much. Trying to catch back up with what has been posted on arXiv

Daminark

$\|A\|_{op}$

Balarka Sen

12:27

Gotcha

Dattier

@aminliverpool ùpokjbbvjbggvfghgjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj‌jjjjjjjjjjj

Astyx

@Daminark Did you just provoke me ?

Daminark

@Tobias I'm still on arXiii

3

Tobias Kildetoft

@Daminark lol

Daminark

@Astyx Maybe, your notation is p heathen

user84215

12:31

@Dattier What does that mean?

Daminark

@Tobias what stuff are you looking at in particular?

Tobias Kildetoft

@Daminark I have a feed of the group theory and representation theory categories and I have a look through all the titles.

Then sometimes I will also check the abstract and sometimes even download the paper for later reading

Secret

what interesting infinite groups have you found so far from the reading?

Tobias Kildetoft

@Secret I have not found any such. Not sure why I would find any

Daminark

@Secret wait hold on not all groups are finite?? My experience has been skewed... forgets what $GL(n,\mathbb{R})$ is

Dattier

12:35

it's a litlle boy which have write that

SteamyRoot

Do I hear claims that infinite groups are not interesting?

BLASPHEMY

Astyx

No, they simply don't exist @SteamyRoot

Daminark

I don't know why that's the first example of an infinite group which came to mind instead of like, $\mathbb{Z}$ or something, but lol

SteamyRoot

@Astyx What am I writing a PhD thesis on then? :(

Dattier

@aminliverpool : it's a litlle boy which have write that

Daminark

12:36

@Steamy yeah I was just making a joke that 90% of my time has been spent thinking about finite groups so far

Oh, I mean you're just working on an empty model

Tobias Kildetoft

@SteamyRoot Well, it would not be the first time someone wrote a thesis on a non-existent object

Secret

Tobias: Well you said you have a feed on group theory, that's why I suspect you might came across some infinite groups in the reading

Tobias Kildetoft

@Secret Sure, there are infinite groups in many of those papers. Usually the point is not that the group is "new"

@SteamyRoot What sort of infinite groups are you writing about?

SteamyRoot

@TobiasKildetoft True, but if my groups don't exist, then the entire subject of crystallography is also non-existent

Daminark

in Soug's voice ehhh

SteamyRoot

12:38

(Almost-)crystallographic groups and free nilpotent groups, mostly.

Tobias Kildetoft

neat

user685272

does it have applications in chemistry?

Secret

Oooo and we chemist and solid state physicists use crystallography like crazy

Dattier

@AkivaWeinberger : what's your prefered simple trick ?

Secret

You need those to interpret and solve the neutron diffraction spectras

Akiva Weinberger

12:39

@Dattier I think I found something on why $u_n{}^2-2$ is easier than $u_n{}^2-1$

(or $u_n{}^2+1$)

desmos.com/calculator/izq80r0qia

SteamyRoot

Well, I think most of the properties of crystallographic groups that are interesting to chemists and physicists are known by now.

I doubt you guys care about the classification of such groups in dimensions 7 and higher :P

Akiva Weinberger

In that link, change $a$ to $1$, and slide it to $2$ to see what's up

The iterate of $x^2-2$ looks a lot nicer than the iterate of $x^2-a$ for any other $a$

Daminark

Hmm, is there some particular reason why we're looking at dimensions 7+?

Secret

Well, you will never know, quasiparticles in condensed matter can behave as if they are in higher or lower dimensional space

Astyx

@AkivaWeinberger Reminds me of that Sarkovski thing

Secret

12:41

and dimension 7 is one of those "special" dimensions

Dattier

@Akiva : more a trick is simple and the harder it is to find

Akiva Weinberger

@Astyx That's the "period 3 implies all periods" thing, right?

Well, more general

Astyx

Yup

SteamyRoot

@Daminark Mostly from a mathematics perspective, really.

Sha Vuklia

@Astyx yea it's just in the context of derivatives, so $\vec h$ is the input. I guess if they write "as $\vec h$ goes to zero", that we are safe to use $f(\vec h)$

Astyx

12:42

What they call tent functions

Secret

Condensed matter is weird, collective behaviour can pretty much simulate any wild things you can find in theories

Akiva Weinberger

@Dattier My favorite trick for anything? Not sure; I'd have to think about it

Tobias Kildetoft

@SteamyRoot What is the definition of a crystallographic group again? I only recall having seen the term applied to Coxeter graphs

Daminark

I find it weird when one dimension ends up being just bizarre. Like in manifolds apparently 4 dimensions is just L M A O Z E D O N G while everything else is comparatively well-behaved

Dattier

@Akiva in maths

SteamyRoot

12:43

The $n$-dimensional crystallographic groups are the fundamental groups of the $n$-dimensional flat orbifolds. (and hence, the torsion-free crystallographic groups are the fundamental groups of the $n$-dimensional flat manifolds)

Akiva Weinberger

@Dattier Yeah; for anything in maths

Sha Vuklia

Akiva Weinberger

I'll think about it

Secret

@Daminark For starters: Exotic spheres in $\Bbb{R}^4$ is still open

Astyx

@AkivaWeinberger Are there things other than maths ?

Sha Vuklia

12:43

Anyhow, guys, I’ve never really understood why we can substitute $x=r\cos\theta$ and $y=r\sin\theta$ at the end. Because now we have $f’(r\cos\theta,r\sin\theta)$ in terms of $x$ and $y$, which just seems odd to me.

Akiva Weinberger

@Astyx Music

Sha Vuklia

Like here; how can we have $\dfrac{\partial f}{\partial x}(r\cos\theta,r\sin\theta)$, when there isn’t even an $x$ in the argument? What kind of derivative are you taking then?

Astyx

@AkivaWeinberger Ooh true

Tobias Kildetoft

@SteamyRoot I see. Is this then even related to the term as used for Coxeter groups?

Astyx

@Sha That's typical notation abuse

SteamyRoot

12:44

@TobiasKildetoft A cocompact discrete subgroup of $\mathbb{R}^n \rtimes O(n)$

Astyx

By $x$ you mean the first coordinate

Daminark

Ugh "maths" is sacrilegious terminology

Dattier

who is SteamyRoot ?

Daminark

@AkivaWeinberger Wait that's not harmonic analysis? :P

Sha Vuklia

but there are two variables in the first coordinate, @Astyx?

Astyx

12:45

Yeah, cause the parital derivative is a two variable function

user685272

@SteamyRoot the patterns that the mathematician finds interesting usually show up nature

Secret

I use maths to refer to mathematics and statistics. Some statistics theory are cool (e.g. the various nth moments), it's the data that is ugly and full of crazy numbers

Tobias Kildetoft

@SteamyRoot So is that related to Coxeter groups?

Astyx

For instance if $f(x, y) = x+y+xy + x^2$

Then ${\partial f\over\partial x} (u,v) = 1+v + 2u$

SteamyRoot

@TobiasKildetoft There's definitely a relation to Coxeter groups, but I'm not sure what it is exactly :/

Astyx

12:46

Unless there's something implicit going on which I'm not aware of of course

Tobias Kildetoft

@SteamyRoot heh

Dattier

bye

Astyx

Bye

user685272

cya

Sha Vuklia

oh okay thanks bye

Dattier

12:47

@aminliverpool see you later

Astyx

@ShaVuklia I'm not going right now, if that was to me

Daminark

@Sha that was definitely one of the few times in recent memory that this was actually typed out

Instead of "kthxbai"

Astyx

ktb

Sha Vuklia

oh shit yea I often put half of the chat on ignore, so I don't lose my conversation :P

if there are a lot of people talking

Astyx

Quite a few

Daminark

12:48

@Astyx oh god

SteamyRoot

@TobiasKildetoft I think the affine Coxeter groups are examples of crystallographic groups

Sha Vuklia

@Dami oh hahaha, I've never written that like that

Tobias Kildetoft

@SteamyRoot Not in the definition you gave above. The affine Coxeter groups are countable

Sha Vuklia

I'm civilised

I take the time for people

:P

Secret

I can follow multiple streams of conversations. It's not easy, though

Daminark

12:50

I often say "kthxbai" sorta, not exactly ironically but in the same vein that I will use the spelling "wao" for "wow"

SteamyRoot

@TobiasKildetoft From that definition, it follows that crystallographic groups are countable :O

Sha Vuklia

@Astyx anyhow, so this notation is clear now

thanks to your example

SteamyRoot

A crystallographic group $\Gamma$ fits in an exact sequence
$$1 \to \mathbb{Z}^n \to \Gamma \to F \to 1$$ with $F$ finite

Tobias Kildetoft

@SteamyRoot Ahh, I missed some of the words

Sha Vuklia

actually

user84215

12:52

@Dattier Have a nice day.

SteamyRoot

Anyway, @Tobias, what are you up to nowadays? I think you haven't been here for a while since your postdoc ended?

Sha Vuklia

I'm still confused, also about the first screenshot; how can we replace $r$ and $\theta$ by their (equivalent) $x$ and $y$, when the function is a function of $r$ and $\theta$?

Secret

[This statement is intentionally cryptic] (With Tobias's return to the maths chat, we expect the Algebra Age will rise again. Now because of the dynamics through the beginning of the year 2017 which lead to the draft of the Integral Project, I am kinda curious on how will the Algebra Age interacts with the Integral Age. Hopefully, this time, we can preserve the timeline that leads to The Scroll)

@ShaVuklia just replace all instance of $r$ and $\theta$ (and better with the trigs) with those polar coordinate formulae

Daminark

@Secret Ah, finally, I've been waiting for the return of the algebra age

Secret

I love abstract algebra, but simpleart and waiting and semi let me realise there are abstract algebra in closed form evaluations too. Hopefully this time we can keep them both

user84215

12:56

Is there any rich theory for uncountable groups?

SteamyRoot

@ShaVuklia $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan(y/x)$

Daminark

Okay now wait a second @amin I'm still getting used to infinite groups in general

SteamyRoot

The arctan is a bit of a bother, but you can rather easily calculate $\sin(\arctan)$ and $\cos(\arctan)$

Secret

yup, just use right angled triangles (and some sign fiddling)

SteamyRoot

wait, actually, you probably don't need that since the sine and cosine are always paired with a radius there

Daminark

12:58

@Secret Closed form evaluations... of integrals?

Secret

Integrals, series, and more generally set of counterexamples

Daminark

I see

Secret

but don't worry, pretty much the only person nowadays who will still dump paragraphs on this chat is me, so there will be no issues. Besides, chemistry getting busy means I am not as frequent on this chat as before

Semiclassical

Yeah, I usually default to "draw a triangle" in order to know (for instance) what sin(arctan x) is

Sha Vuklia

@Steamy uhmm okay, so I really just have to think of it as a different representation of the same? so $f(x,y)=x^2+y^2$ is really the same as $f(x,y)=r^2$?

or should I write $f(r,\theta)=r^2$ instead of $f(x,y)$?

Daminark

12:59

phew

:P

SteamyRoot

Well, the point is that when you write $f(x,y) = r^2$

you actually have $r = r(x,y) = \sqrt{x^2 + y^2}$

You write $r$, but you should remember that $r$ is a function of $x$ and $y$ and not a constant

You should definitely not write $f(r,\theta) = r^2$

Sha Vuklia

ohh okay..

$f(x(r,\theta),y(r,\theta))=r^2$

SteamyRoot

I mean, you defined $f$ such that $f(3,4) = 3^2 + 4^2 = 25$

Secret

Physicists like to use the following conversions:
$$\frac{x}{\sqrt{x^2+y^2}}=\cos \theta, \frac{y}{\sqrt{x^2+y^2}}=\sin \theta, r=\sqrt{x^2+y^2}$$

Semiclassical

There's a joke I'm fond of. How do you tell a physicist apart from a mathematician? Write $f(x,y)=x^2+y^2$ and ask for $f(r,\theta)$

SteamyRoot

13:02

But for $g(r,\theta) = r^2$, $g(3,4) = 3^2 = 9$, so $g \neq f$

Yeah, that's another way to see it

Secret

Then $f(r,\theta)=r^2$

Sha Vuklia

@Semi $f(r,\theta)=r^2+\theta^2$ is the correct answer?

SteamyRoot

^

Daminark

@Sha Yup

Sha Vuklia

haha:P

SteamyRoot

13:03

@Secret I guess we can update the joke to include chemists now :P

Secret

O I failed the mathematician test

user84215

What is the most important unsolved problem in pure mathematics?

Dattier

A beautiful result, $$g\in C^2([0,1]), 8(\max(g)-\min(g)) \geq \min(g'')$$

Secret

Well, I am trained in physics back in my undergrad, so

Alessandro Codenotti

random cool stuff $\displaystyle\frac{\sin t}{t}=\prod_{n=1}^{+\infty}\cos(t/2^n)$

Semiclassical

13:03

It's the mathematician's response

Sha Vuklia

I'm glad I'm out of the chaos of physics for the moment, jesus f-ing christ. It's like the best and the worst thing in the world at the same time

Daminark

Oh @Sha we need to make sure you become an algebraist btw

Sha Vuklia

hahahah, well I'll be doing algebra this summer

Daminark

@Alessandro gulp

Sha Vuklia

I failed the test, I just couldn't do it anymore. I'm done :P

Semiclassical

13:04

The physicist response would be $f(r,\theta)=r^2.$

SteamyRoot

The algebra revolution of this chatroom has begun? :D

6

Secret

@Semiclassical yup

Daminark

Yup

Balarka Sen

Topologists are in majority.

Semiclassical

And I have to confess I don't really see that as wrong so much as it requires context to make sense

Balarka Sen

13:06

And will always ever be.

Sha Vuklia

@Semi it's wrong if you don't know it's wrong? otherwise, it's fine:P

Daminark

Oh we'll see about that. Though I guess we can allow the Peter May brand of algebraic topology to persist...

SteamyRoot

Well, @Daminark and me both classify as algebraic topologists for a large part, I guess.

Alessandro Codenotti

@BalarkaSen there's a lot of algebraic topology going on in there though

SteamyRoot

And we apparently both identify as algebrists

Secret

13:07

I am a superposition of a bit of everything, none skillful

but I wish to beef up my topology and abstract algebra

SteamyRoot

@Semiclassical Technically, it's wrong because you use the same symbol $f$.

Balarka Sen

Daminark is a higher category theorist.

Secret

hence the munkres stuff

Balarka Sen

I guess @Steamy is an algebraist.

you and Tobias are the only ones I can list off the top of my head

Alessandro Codenotti

the only thing I know is that I'm not a statistician

Sha Vuklia

13:08

@SteamyRoot he obviously wasn't talking about things being "technically" wrong tho:P

Daminark

@Balarka much as I am currently "working in higher category theory" I'm actually a finite group theorist

Semiclassical

It comes down to interpreting $f(x,y)$ not as "$f$ as a function of x,y is..." but rather "$f$ as presented in Cartesian coordinates is..."

Daminark

@Alessandro same

Balarka Sen

Alessandro is onto foundations, he needs to read the HoTT book

univalent foundations

Semiclassical

Under that convention, $f(r,\theta)$ means "f as presented in polar coordinates "

SteamyRoot

13:11

I disagree. $f$ is the object itself, defined as $f: \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto x^2 + y^2$.

Alessandro Codenotti

We have a grad course here in type theory, I'm definitely going to do it if I don't move for my Master

SteamyRoot

we used variables $x$ and $y$ in that definition, but these variables don't hold any meaning

Daminark

Lol one of my friends is doing an REU project this summer on homotopy type theory

SteamyRoot

If we wrote $f: \mathbb{R}^2 \to \mathbb{R}: (r,\theta) \mapsto r^2 + \theta^2$, this would be exactly the same object

Semiclassical

That's the first interpretation

It's a convention.

Balarka Sen

13:12

i read a few pages from the book a couple years ago

i quite liked what i saw

Daminark

@Steamy I think you could potentially say that the object is that $f$ sends a vector in Euclidean 2-space to its norm, so these are various coordinate representations?

Semiclassical

Function notation does not come down from some authority on high

Alessandro Codenotti

"traditional" type theory is a rather different beast from HoTT I think though

Balarka Sen

yeah

user685272

hi chat

user84215

13:14

for studying set theory, which book is better? Jech's or Kunen's

Semiclassical

Within the context of what a physicist does, the coordinate-system convention is entirely reasonable

Daminark

To spite Alessandro, much as I have no knowledge, I will say Jech

Sha Vuklia

Anyhow guys, I’m still a bit confused. So what do we exactly do when we write $f’(r\cos\theta,r\sin\theta)$ and then subsite by $x$ and $y$? So we transform $x$ en $y$ to $r$ and $\theta$, and then we transform it back, all in the same function. It’s just weird? Is there some abuse of notation here too?

SteamyRoot

@Daminark Well, $f$ is technically not well-defined unless you specify how you represent such vector.

Sha Vuklia

I mean, we've found the derivative of $f$ by using polar coordinates, I just don't see how we are free to go back to $x$ and $y$ just like that

Daminark

13:16

Okay that's true, actually, yeah the physics convention is officially wrong

Sha Vuklia

Like how can $f'(x(r,\theta),y(r,\theta))$ be the derivative of $f$ at $(x,y)$.

Daminark

@Semi :P

SteamyRoot

@Semiclassical You should at least write $f_{\text{polar}}$ or something like that. I know it's easier to write $f(r,\theta)$, but it is undeniably wrong for most functions :P

Alessandro Codenotti

@aminliverpool Jech's covers a lot and is not aimed at beginners at all

user84215

I have read his introduction book

Semiclassical

13:19

Meh. The point is really that when you do physics problens you don't care about arbitrary changes of variables.

SteamyRoot

Well, yeah, I know.

Semiclassical

You generally have either Cartesian coordinates, cylindrical coordinates, or spherical coordinates

Alessandro Codenotti

@aminliverpool Jech, Hrbacek you mean?

SteamyRoot

I'm just very strict because I TA an undergrad course on "mathematical reasoning and proving" where we beat (not literally) these things in students.

user84215

yes

Alessandro Codenotti

13:21

ok, that's a good introduction as far as I know then, I was thinking about another book by Jech

Semiclassical

So to insist in such context that the notation "must be consistent!!" with arbitrary changes of variables seems silly to me

The purpose of notation is to communicate the content in a clear and useful way.

user84215

I mean after this book it is better to continue on Jech's or start kunen's ?

SteamyRoot

@Semiclassical Well, yeah, but if you use $f(x,y)$ and $f(r,\theta)$ interchangably, how do you interpret $f(3,0)$ ?

Alessandro Codenotti

I don't know, you need someone who actually read them to answer

Tobias Kildetoft

@SteamyRoot I am doing a new postdoc now at Aarhus University. But I just started and everyone is on vacation

Sha Vuklia

13:24

@Steamy context

Semiclassical

i don't. But then I typically don't care a whit about a particular value of f.

Daminark

@SteamyRoot Unfortunate example

Semiclassical

What I care about is doing integration/differentiation and understanding functional behavior

SteamyRoot

@TobiasKildetoft That's the largest Danish university, isn't it?

user84215

Unfortunately, math books are expensive.

Alessandro Codenotti

13:25

are you in university? They have good libraries usually

Semiclassical

To insist that notation is wrong because it's not capable of doing all things for all people is silly. It's not intended as such.

user84215

My university library does not pay enough attention to provide books on foundations of mathematics

Semiclassical

Some good discussion of the issue of denoting various coordinate systems is here (MathEducators Beta): matheducators.stackexchange.com/q/5999/1940

SteamyRoot

@Semiclassical Seriously, whatever works for you. But being mathematically correct (which is not always useful, necessary or desirable, for example in physics papers), using these interchangeably is wrong. And that's not a matter of notation, but a matter of definition.

Semiclassical

Notation is a human construction and cannot be 'wrong.' It can be confusing and ill-defined, but there is nothing in the universe that dictates it.

Notation is a tool. Nothing more, nothing less.

Mike Miller

13:34

lol is this an argument that started over Semi's physicist joke?

Semiclassical

yep

Well.

Sha Vuklia

I'm entirely on Semi's side on this:P (for what it's worth, lol)

Semiclassical

The physics joke actually came after some other discussion.

I am channeling my inner pragmatist for this.

Sha Vuklia

anyhow guys, I understand it now. Since we write $f'(r\cos\theta,r\sin\theta)$, we can just see what it does with those arguments "as a whole", and then replace them by $x$ and $y$, since that's how the "Cartesian functions" simply work.

Daminark

Presumably, while it's totally alright to use notation that's convenient, using ill-defined notation is still "abuse", from a purely technical/non-normative standpoint

Semiclassical

13:37

eh. I'd dispute the word 'non-normative' there.

"Using X as such is bad" is plenty normative.

SteamyRoot

@Daminark That's what I mean, yes.

Daminark

Wrong is different from bad, I'd say

Sha Vuklia

I'm going to troll the smartest kid in my class next year and ask what $f(r,\theta)$ is btw:P

Semiclassical

I should note as well that I stole that 'joke' from a prof here who does physics education research :)

Daminark

Using $n$ to mean a real number is, by all accounts, correct, but like can you don't? Ples stahp

Sha Vuklia

13:40

then you're just obnoxious

Semiclassical

@Daminark A somewhat more esoteric instance of that is in special relativity: If you write down $g^{\mu\nu}$ then I'm going to interpret those as spacetime indices $\mu,\nu=0,1,2,3$.

Whereas if you do $g^{ij}$ I'll understand that as purely spatial indices $i,j=1,2,3$.

Secret

List of mathers and their expertise (not necessary maths) I know:
Semiclassical: Semiclassical approximations in condensed matter, complex analysis, real analysis, multivariable calculus, special functions
Balarka Sen: Topology, Manifolds, algebra geometry, transcendental numbers (something else big I forgot)
Alessandro Codenotti: Set theory, cardinals, ordinals, analysis
Steamyroot: Integrals, set theory, analysis (all types), abstract algebra (particularly polynomial rings), ordinals, projective geometry, differential equations, measure theory

2

Alessandro Codenotti

@Daminark if $x:\Bbb R\to\Bbb R$ is the function $x(f)=f^3$, what is $\frac{\text{d}}{\text{d}f}x(f)$?

PVAL-inactive

I definitely win with ?

Semiclassical

I'd strike real analysis from that and trade you differential equations.

Fargle

13:43

@Secret I spot a problem: you put me down as having some expertise.

Semiclassical

b/c I really don't care about foundations of calculus that much.

SteamyRoot

@Secret I know pretty much nothing about set theory and ordinals :P

Daminark

The whole, polar coordinates thing, it's fine to use and is convenient, but the fact that anyone can make sense of it is strictly contingent on a very religious use of r and theta. Otherwise, this is ambiguous

s.harp

@Semiclassical you might say $f(r,\theta) = r^3$ if we are thinking about $f$ as a density

Secret

@Fargle Well you and s.harp are often seen answering finite field related questions and some group theory, I guess that counts?

Semiclassical

13:44

@s.harp The main instance where I remember that being an issue is the Planck distribution

user685272

@Secret jasper?

Fargle

@Secret My twin and I do quite enjoy abstract algebra.

Semiclassical

i.e. it took me a while in undergrad to appreciate that $f(\lambda)\neq f(\nu)$ but rather $f(\lambda)\,d\lambda = f(\nu)\, d\nu$.

(there's also a sign change but w/e)

s.harp

(fuck signs)

Secret

(sorry typo in the fargle reply)

Semiclassical

13:45

(lol)

Fargle

lol you're good, just teasing.

Balarka Sen

(i hate writing chemistry reports)

Secret

steamy and alessandro also tend to reply together which is why I sometimes mix up their expertise

SteamyRoot

Just work mod $2$, $+$ and $-$ are the same thing there.

Fargle

@s.harp I just work in vector spaces over $\Bbb F_2$, it removes any sign issues.

s.harp

13:46

@Secret I dont htink I have ever answered a finite field related question, I know almost nothing at all about abstract algebra

Semiclassical

To paraphrase The Dead Kennedy's: "Sign nazis, sign nazis, sign nazis F*** OFF"

Fargle

snipes Daminark's "SNIPED"

Daminark

Seems like my baby Peter May image is actually taking hold. Gotta keep it up then

Alessandro Codenotti

everything else breaks down in characteristic $2$, but the signs are nice so I guess it's worth it

Tobias Kildetoft

@SteamyRoot No, second largest university.

Fargle

13:47

@AlessandroCodenotti I mean, u rite

Daminark

Rekt @Fargle, didn't even notice it until you pointed it out

Fargle

@Daminark Then you got ultrasniped.

Secret

chat.stackexchange.com/…

but I gues youare right, no finite fields

Balarka Sen

@Fargle Sniped.

Daminark

I dunno if it's ultrasniping, like I wouldn't have said anything so it's sniping a civilian in this context

Balarka Sen

13:48

You will get this comment in the far future when you write "sniped"

2

Semiclassical

The other thing I dislike about the "f(r,theta) notation is WRONG" attitude is that it produces a certain attitude for reading papers in other disciplines

Fargle

@BalarkaSen Calm down Goedel, you're getting too meta for me.

Semiclassical

namely "oh, those silly physicists/enginieers"

Secret

> Danu: Algebraic geometry, topology, (something else I never understood)

SteamyRoot

@Semiclassical Well, the thing is, I don't mind if I see this on a physics paper or exam or thesis.

Semiclassical

13:49

Field theory.

Mike Miller

yeah but I disagree with that attitude and still say oh those silly physicists

Secret

For that bracket statement, ask barlerka and Danu, cause it often involve both of them in the conversaton stream about some kind of high level manifold stuff

Alessandro Codenotti

LOL

Secret

But, I like this one the most:

> Ted Shifrin: Any maths except one math field which I forgot the details

SteamyRoot

But if it's on a mathematics exam or thesis, I'm deducting grades.

Semiclassical

13:51

I'm fine with people pointing out the limitations of the notation and where it becomes ambiguous. But then, a good physicist/engineer should already be aware of such.

SteamyRoot

I also wouldn't have complained about his if this was, like, the h bar. But it's the mathematics chatroom, so ;)

s.harp

"Notation means what I want it to mean"

Semiclassical

That comes to the crux of it, I think. It produces an attitude that physicists/engineers aren't "really" doing math.

s.harp

I think schwinger said that when von neumann pointed out to him in some talk that his definition of unitary was not the same as what the word usually means

Mike Miller

should i get up and go to the gym now or lay in bed for another 20

Balarka Sen

13:53

i would have done the latter

but i suggest the former

even though i don't believe my suggestion

Semiclassical

I'd lay in bed. Which probably means you should get out of bed.

Daminark

Normally I'd take bed on general principle but... I consciously know that gym is better

PVAL-inactive

@Mike Why is any degree 3 alg. curve either a plane cubic or the twisted cubic?

Alessandro Codenotti

every morning to get up I ask myself "do I want to stay in bed with my dreams or get up and chase them?". Just joking, I like the breakfast, cit.

SteamyRoot

@MikeMiller Lay in bed for another 40

Mike Miller

13:54

@PVAL I don't actually know any algebraic geometry.

PVAL-inactive

@MikeMiller I've been lied to.

Mike Miller

Are you going off The List?

Balarka Sen

But, you're listed as an algebraic geometry expert, @MikeM.

Secret

Jason Bourne/Jasper the Loy: Languages, (but no particular maths field spotted)

Daminark

Same @Steamy

Mike Miller

13:56

I don't know what the twisted cubic is.

Balarka Sen

Image of $\Bbb P^1$ in $\Bbb P^3$ given by $(t, t^2, t^3)$ I think.

PVAL-inactive

thats it in affine coords.

lore

code, you should be studying

hi everyone, i'm Lorenzo, an alessandro classmate

Alessandro Codenotti

he lies, I've never seen that person before in my life

Semiclassical

For convenience: en.wikipedia.org/wiki/Twisted_cubic

Mike Miller

13:58

Am I good enough at algebraic geometry to answer your question?

PVAL-inactive

no probably not

It was a troll

Mike Miller

that's what i guessed

Semiclassical

This is neat: "Given six points in $\mathbb{P}^3$ with no four coplanar, there is a unique twisted cubic passing through them."

Makes me wonder what happens if four of them are coplanar.

Mike Miller

then the variety has to contain that plane

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$Mathematics$ Mathematics