if you say so lol

because right now i dont like it xD

@DanZimm that comment was not addressed to you

which one?

physisicts manage to predict the mass of electrons to 10 decimal digits and the position of planets

ass you gain matureness - is this phrase itself a test of readers' maturity?

it is not all goofy

@anon :D

@MarianoSuárez-Alvarez yea im not taking away from that at all, i was totally mistaken to speak ill of physicists

@anon What? Noticing the $ass$ grants maturity?

but my mind is being blown that you can divide by 0

you cannot divide by zero

@GustavoBandeira grants immaturity

if you limit yourself to working with real numbers

then what is happening when 1/0

there is no reason why you should limit yourself to working with real numbers

what numbers is this ok?

@DanZimm I noticed it, I just wouldn't starting joking like "Heh! You said ass! Ah hahaha"

:P

if you introduce $\inf5y$ to the real numbers, you get a consisten system with certain limitations

it is no longer a field

for example, with addition, you do not get a group

but you do get something

@DanZimm Mathematics does not seem to be a religion, it seems to be more something that we build.

if you are careful when you operate you will be safe

@MarianoSuárez-Alvarez totally forgot about this, I completely apologize about this whole debacle

you are used to numbers commuting, for example: $1\cdot 2=2\cdot 1$

but something syou want things not to commute

not just out of fun but because you need them not to

interesting

Heisenberg invented matrix mechanics that way, for example

(this is one of the ways to decribe quantum mechanics)

you are also used to products being an associative operation

but sometimes you want them not to be

and so on and on and on

hrm interesting

there is no reason to limt yourself t the field of real numbers

yea I was being silly

also, there is great intuitional value in infinite and infinitessmals

yea i agree

(and there are variious ways of making them an actual formal theory, like non-standard arithmetic)

hrm

all this does not imply at all that when physicists write 1/0 = infty they are making sense of course :-0

but they might be

further analysis is required to be sure!

but based on the real line with traditional arithmetic, that statement doesnt make sense

and heh yea

you are committing the sin of adscribing to others limitations of yours

why are you assuming they are bounding themselves to operate within the field of real numbers?

im saying if one does only look at the real numbers

(even if they do think they are operating with th field of real numbers :-) )

im not talking about physicists at all anymore

and with traditional arithmetic?

What does "=" mean in this context?

well, in the theory of peano arithmetic there is no -1

let alone $\infty$ :-)

the traditional equality, considering I'm talking about if someone is using the traditional idea of arithmetic

again, not peano

equality is just equality

you are mixing things

There is no problem dividing by zero, just take the ring over $\{0\}$ we can define $ 0 *0=0 $ and $0+0=0$ and hence $0/0=0$

there is no problem dividing by zero if you are ok with not having a field

just as there is no problem with non-commutative division algebras which are fields in which commutativity does not hold

I am born in the city i don't need fields :D

there is no rule that we have to work with fields

but if youre working with the real numbers (meaning a field) then 1/0 = inf doesnt make sense, right?

we work with whatever is convenient to do what we want

@DanZimm of course

no one is saying it does

well jesus thats all i was trying to clarify xD

@DanZimm but it even doesn't makes to calculate with fractions if you are in the natural numbers

well, that is sort of a silly thing to try to clarity

1/0 simply does not make sense

it cannot be equal to anything

included infinity

infinity is nothing

it is not a real number

@MarianoSuárez-Alvarez It is something on non-standard analysis, right?

Like surreal numbers, I guess.

@MarianoSuárez-Alvarez yea thats what i was just clarifying meaning i thought you were saying otherwise at first but you werent ergo i was making a bad assumption and so i need to clarify this was the case

if you put yourself in a context in wich infinity is something, then of course it is equal to itself.

no it is the one point compactifictaion of the real numbers

sometimes im silly :P

@DominicMichaelis who said im talking about the natural numbers lol

@DanZimm the problem with $1/0=infty$ in the context of the field of real nmbers is twofold

theleft hand side does not mean anything, is simply undefined

the right hand side is not a real number

ya i unerstand that

i actualy said that aboove

therefore wondering if the equality holds or not in the context of real numbers is rather weird

@DanZimm no it was an analogy. For you it absoutly makes no sense throwing away the real numbers to take something different. but Some time ago (about cantor) the real numbers was a very silly concept too

the answer is MU

So why try and fool everybody and use the "=" sign?

huh?

Here^

@Danzimm i mean numbers which are no fractions seems kind of weired too doesn't it ?

in a context where both sides of an equality make sense, it of course makes sense to write the equality

@DominicMichaelis right but I was specifically talking about if someone is trying to work with just the real numbers, possibly because theyre ignorant and dont understand these math ideas, which occurs often to many undergrad physics majors

the set $\mathbb R\cup\{\infty\}$ can be endowed with partiaal operations of sum, difference and so on

and there you can operate with care

and there $\infty=1/0$ is a perfectly correct statement, and a true one to boot

that set is certainly not a field, let alone not the field of real numbers

you don't have to go that far. I tried to convince my theoretical physics prof, that the standard scalar product on $\mathbb{C}$ is not antisymmetric but hermitien, I failed ...

yea

thats i guess why i made any statement originally, i often see people around me studying physics to make some pretty bad assumptions, but it wsa improper of me to generalize that to physicists in general

when one iss working with measure theory, and the correspondding integration theory (Lebesgue and friends) one works with $\infty$ as an actual value all the time; there one only adds those things, not multiply them

actually, with $+\infty$ and $-\infty$

@MarianoSuárez-Alvarez I'm confused as to what you're addressing right now

and you have to be careful when you operate because some things are left undefined, line $\infty-\infty$

I am elaborating my point

which one?

tht you should be more lenient to what the physicists do, because often your discomfort is only indicative of your not knowing what they are doing :-)

because the one with regard to infinity I admited I was wrong on and understand that I'm too ignorant to make such assertings

^

if you think I am writing to chastise you for your mistake, you are wrong

i was hoping to explain the point

as you apparently do not care for that, I'll go do something else

im actually interested, just wanted to make sure you werent chastising ;)

@MarianoSuárez-Alvarez ^

Please continue...

all too often online you get people who want to put down not to teach, so I actually really appreciate this

@MarianoSuárez-Alvarez srsly I didn't mean to be a dick, I apologize

@Martin Nothing important, but when I see you in chat I thought I might let you know that a user was trying to contact you here. His ping reached me instead, because of the name conflict.

An occurrence of Agent Smith Calling Agent Smith.

@skullpatrol I feel bad D:

sorry for driving him away

@DanZimm Live & learn pal...

np

yea

best i can do for now

Well do you have a rough idea of what measure theory is about ?

the most i know is theres something called measureability, so no

I hope you know integrals

what do you mean?

thats a vague comment lol

Do you know what an integral is ?

i understand there is a difference between riemannian and lebesgue integrals

yes

at least i have a general understanding

by no means am i an integral expert

I think we don't need to go that far. Integrating in $\mathbb{R}$ works pretty good

But if we want to generalise it to $\mathbb{R}^n$ we gonna face some problems with the properties

yea

At first we could measure the volume of a subset of $\mathbb{R}^n$

ok

I've bookmarned @MarianoSuárez-Alvarez 's wise words.

It would be great if the volume of a subset doesn't change if we rotate it or move it a bit or mirroring it or something like that. Furthermore when we have two distinct sets the volume of their union should be the sum of the volumina

And the unit cube (a cube of length one) should have the volumen 1. But it turns out, that this doesn't work. There is no function with those properties

what do you mean no function with these properties? which properties?

thats why you say we only look at "nice sets" which are measurable. If you don't allow the bad ones you can find such a function which is called measure

OH

totally misunderstood whats going on, i apologize

@DominicMichaelis interesting - this is on a set by set basis, right?

@DaZimm with $f:\mathcal{P(\mathbb{R}^n)}\to [0,\infty]$, $f(A\cup B) = f(A)+ f(B)$ and $f([0,1]^n)=1$

thats the measure of $\mathbb{R}$ ?

it is infinity as $\mathbb{R}$ is really big :)

what is infinity?

@DanZimm A big banana.

@DanZimm how you mean what is infinity ?

@DominicMichaelis lol sorry thats a silly question when it isnt said in my head, what are you referring to as infinity?

I thought the measure was a function

or?

yeah the measure take subset of $\mathbb{R}^n$ and gives them values in $[0,\infty]$, so infinity kind of says that your volumen isn't bounded

right, but what were you referring to when you said "it is infinity"?

The measure of $\mathbb{R}^n$ is infinity because $\mathbb{R}^n$ is so big :D

but I thought the measure was a function O.o

yeah

it is a function on the power set

and $\mathbb{R}^n$ is an element of the power set

oh I follow what you're saying ok, my bad

Well in fact the measure is a function on a subset of the power set

I'm really being an idiot tonight, I apologize @DominicMichaelis and again to @MarianoSuárez-Alvarez :/

ok

Everyone is born as an idion, at least most there are some people where i still have doubts...

@DanZimm You're being it tonight. I've been that my whole life.

heh well I apologize nonetheless

and @GustavoBandeira I doubt this

=D

and I think in this situtation it is pretty clear that $\infty$ is a useful concept

ok yea that makes sense

and right now we don't face problems of $\infty - \infty$ or $\frac{\infty}{\infty}$ because of for the first we don't know if $f(A\cap B)=f(A)-f(B)$ (it isn't in general to avoid $\infty-\infty$ occurences) and for the second there is no interpretation, because what should be a fraction of two sets?

im looking forward to learning measure theory :D

*still feels bad for being an ass to @MarianoSuárez-Alvarez *

I apologize again @MarianoSuárez-Alvarez

well I study physics :D

don; t worry about it

@MarianoSuárez-Alvarez also thank you for going out of your way and explaining that all, I actually do appreciate it

@DominicMichaelis which field particularly? or just in general?

its my second untergraduate year so I didn't specialised in anything right now

ah ok

i should probably keep quiet for a bit so ill see yall later, sorry all again for being ridiculous

oh really don't mind without mistakes you don't learn :)

@MarianoSuárez-Alvarez Is there reason to suspect any sort of "natural" bijection between conjugacy classes and irreps of finite groups to exist? With the symmetric group, we can biject both irreps and conjugacy classes with integer partitions, but that doesn't appear to amenable to working in an abstract way with arbitrary finite groups.

it is usually suspected that there is no sensible bijection

one reason is that conjugacy classes and irreps have "different variance"

conjugacy classes are covariant, and irreps are contravariant, in a sense

depressing but expected. can you expand a bit?

it is the same problem that appears when you wwant a sensible isomorphism betwee a vector sapce and its dual

the symmetric group is quite extraordinary :-)

symilarly, a finite cyclic group $G$ is always isomorphic to $\hom(G,S^1)$

but there is no sensible isomorphism which is natural

I googled conjugacy classes are covariant, and irreps are contravariant and see a nice MO question

(notice tat the conjugacy classes in taht case are the elements of $G$, and the elements of $\hom(G,S^1)$ are the irreps)

@anon yup: it is the standard explanation :-)

probaby someone observed that there is an isomorphism between the space of center of the grou algebra and the ring of representations

magnificent

which is usually called Fourier transofrm

but that is not a bijection of the sets

it is a «quantum bijection»

in that conjugacy classes get maps to "entangled irreps"

has anybody told you you're a tease?

haha

(-:

@MartinSleziak Thank you! I noticed it, but I didn't want to pursue that argument any further. It would in effect be paying too much honor to Mr or Mrs "damn u". Thanks also for opening the meta thread on matrix-equations.

@Martin I don't think that we need Matrix Equations as a tag. This will be a paradise for "I am to lazy to do it on myself" questions

I gonna earn the taxonomist badge today :D

yo

omg, quantum bijection? what is it?

I think it has to do with quantum groups and hopf algebras. Bad bad stuff :D

haha xd

Theorem. (Lusztig) The Grothendieck group of the category of perverse sheaves on the moduli stack of representations of a Dynkin quiver is $U_q(\mathfrak{n}_+)$ for the positive unipotent part of the corresponding Lie algebra.

@MphLee it is just a silly way of saying isomorphim of vector spaces :-)

ah ok

@Mariano what does this theorem tries to say ? :D

supposse you have two sets $X$ and $Y$

and consider the vector spaces $U$ and $V$ which habe $X$ and $Y$ as bases

then a linear map $U\to V$

maps an element of $X$ to a linear combinatio of elements of $Y$, the sort of thing that physicists like to call superposition of elements of $Y$

a "quantum deformation" I gather is an informal way of saying we expand an object to encode more information than before (preferably in a way that is relatable to quantum mechanics, however loosely), such as braid groups encoding "over/under" information not seen in permutation groups. linearization of a set to a vector space would probably count.

Does anyone unterstand the theorem i posted

yeah

Hey someone can explin me these. When i fnind a mathematical theory, is because i found amathematical structure that follow some rules that i've observed in nature..then if the nature has the "axioms" my mathematical structure (for example a set) can be called a model of the nature? it is really confusing me xd ...

@DominicMichaelis no sorry im so noob, i can't

the theorem means that one can recontruct the positive part of the quantum enveloping algebra from the geometry of representations of the corresponding quiver

it is an elaboration of an older theorem of Green which says that thesame envelopng algebra can be obtained as the Hall-Green algebra of the category of reps of the quiver

Lusztig managed to skip the Hall-Green algebra and directly do the construction in terms of the geometry of the representation varieties

(or stalks...)

I have heard the ring of reps of a group called the Green ring. is that the same idea of a hall-green algebra?

the wikipedia page on Hall algebras gives information of a simper situation

yeah

@Mariano that sounds like it is possible for me to unterstand it in some years :D

the idea comes from Hall

it was used by Green and Ringel in the context of quantum groups and quivers

@DominicMichaelis it is a rather difficult result, sadly

lusztig does not come up with other kind of results :-)

i guess lusztig found a proof for the RH but this was not worth mentioning to him ... :D

the theorem is part of his program of understanding reps of finite simple groups of Lie type

one of the great mathematical orgies happing nowadays

some results/theorems are really hard to understand or they are hard only for who doesn't have a good knowledge of the theory?

both

that theorem of lusztig is both technically very demanding

and difficult

at higher levels why mathematicans dont become mad? Did happen? (Im talking outside infinite cardinals xD cantor..)

@DominicMichaelis Perhaps you could post your comment in that thread as an answer - so that people can upvote/downvote to show whether they agree.

hii all

hi

hi

any cocos2d android game developer here?

Are you in the right room?

i'm new to this website..

@jack Try here.

He is a Stack Overflow user

is this room mathematics related to any particular field or general discussion on maths..

hmm yes dude @MphLee

most time we talk about $\mathbb{R}$

I dont get why i can't see mathjax... $\Bbb R$

@MphLee Are you using chatjax?

nope

mm relly idk what it is

@MphLee math.ucla.edu/~robjohn/math/mathjax.html

@Martin did convert the answer

Thanks!

OMG it works, is amazing!

thanks @GustavoBandeira

@MphLee look.

@DanZimm You should look too.

@GustavoBandeira ok, btw why?

@MphLee Why what?

If anyone has a chance... Can you please look at math.stackexchange.com/questions/394292/… for me? I'm really stuck. :(

@PeterTamaroff I was saying "good night"...

@GustavoBandeira interesting question, why you posted it here?Is related to one of my questions?

@skullpatrol hi

@MphLee Nope. It was just for share.

ah

@GustavoBandeira bom.dia

@Charlie Bom dia. =)

@GustavoBandeira :)

What's the meaning of $x_n$ and $x$ in $|x_n-x|<\varepsilon$?

$x_n$ is the term of the sequence, I guess.

@GustavoBandeira xn tends.to x

What is $x$? The number of the sequence? I'm confused because if it is so, I guess it should be $n$.

$x$ is a fixed point, separate from the sequence.

@GustavoBandeira n is the index

@Charlie I understand what it mean - I'm just unable to read that statement.

@user1 Thanks.

@Charlie hi :O

Is it ok to suffer a lot for understanding this? XD

@GustavoBandeira the vterms of this sequence are getting close to x, the distance between them is ssmaller than eps

@skullpatrol ;O

Yep.

So $x$ is the number for which the sequence if converging - if it is converging?

@Charlie I need sleep bye...zzz...

@GustavoBandeira well, if you are used to limits , it gets a bit nicer

@skullpatrol bye..zzzzzz

@GustavoBandeira I would consider the notion of convergence of a sequence fundamental to topology, so the suffering should pay off in the future.

@GustavoBandeira yes

@GustavoBandeira xn is in the interval (x-eps, x+eps)

Last doubt: $x\in \mathbb{N}$?

Or ir could be $\in \mathbb{R}$? (for example)

@GustavoBandeira not.necessarily. Only n, the index. You can have sequences of rationals,e.g

Got it.

Thanks.

@GustavoBandeira anytime :)

Why won't the internet let me sleep?

@GustavoBandeira: Hi.

@BabakS. Hello. Aren't you going to shoot a question, are you?

I'm dumb, and I'll probably be unable to answer.

@skullpatrol haha turn your computer off

@Charlie Zzzzzz..... :O

@GustavoBandeira: No bro. :) I don't wanna irritate you. :D

@skullpatrol ;O

@BabakS. No, you wouldn't irritate me. I'm just stating that at the actual level I'm in, I'm useless.

@GustavoBandeira: How will we can vote the moderator. If the system do that or we will do that?

@BabakS. Here.

@GustavoBandeira: Are you a nominee for that?

I see that

@BabakS. Yes. At the time it seemed that no one wanted to be a mod (I guess mainly due to the meta battles) - then I decided to make myself useful to the comunity. But now there's people more competent than me.

@GustavoBandeira: Yea I see that. But I think it will get more time of you. Won't it?

@BabakS. Were I elected, yes.

I hope you have a good time in that time. It seems to me a very challenging one. :)

@GustavoBandeira:

@GustavoBandeira: I'll choose you definitely. :-) Bye for now

@BabakS. Thanks. Bye for now.

It's weird when I see people choosing me, when my girlfriend says: "I love/want you" - I keep thinking: "Why'd you do that?!"

XD

@GustavoBandeira me too

woot, you got enough candidates for a primary! That's nice. I was worried when there only seemed to be 1 candidate.

}:)

@Charlie I don't know. But I guess my actions in my childhood made people ignore me - now the possibility of people choosing me actually exists.

I'm still not used to the new ways of life.

@GustavoBandeira good

@Charlie The only excerption to that rule is when I'm with the body covered in erotic oil. In such state, it gets obvious I'm the correct choice to make.

tee em eye

@badp Tea with me.

is there any room for cocos2d-android in stackexchange.com site?@skullpatrol

@GustavoBandeira Fine, so long as no erotic oil is involved.

@jack Perhaps, but not on mathematics.se

@jack Try to look for the android.se, stackoverflow.se and gamedev.se.

ok thanks for the reply:)@GustavoBandeira

Sounds like Stackoverflow honestly; Android is for Android users, not devs

@badp I get shinny with erotic oil over my body, dude.

@GustavoBandeira I... I figured that. (Damn you.)

@Charlie If I'm going to your city someday, I'll leave a note on the mathematics department: "Erotic oil was here - be careful, it's wet and slicky".

@badp lol

@GustavoBandeira you crazy

@Charlie :P

@Charlie If it were in portuguese, what word would you choose for "crazy"?

You mean there's more to it than "loco"? POP CULTURE YOU FAILED ME AGAIN

@badp Actually there's a friend's hypothesis in the case...

@GustavoBandeira doido, or louco, or maluco, insano, all above

@GustavoBandeira Ah, the good ole Loco Singularity Paradox Conjecture. If todo is loco then what is loco to begin with?

@Charlie Oh.

@badp :P

@GustavoBandeira sem noção, essas coisas

@jayesh is that you?

@Charlie Is that me what? :P

And definitely no! :P

@JayeshBadwaik ooh!

mh my answer has 5 upvotes and the op accepts hte one with 1 answe, that means with 5 more upvotes i could be a pupulist :D