Mathematics - 2018-11-13 (page 2 of 6)

(Folland also is about the locally compact abelian case, for the non-abelian case there is a book by Taylor called "Noncommutative Harmonic Analysis", that book is great but hard (and also more about representation theory than fourier transforms) )

Mike Miller

i am surprised you can say anything about the nonabelian case

does he also assume compact?

s.harp

@Mike I was being a bit too ambitious with that book

He doesnt do the same kind of analysis, but he does look at non-compact groups

Mike Miller

is this the same as PDE taylor

s.harp

for example the representation theory of SL_2(R) and SL_2(C) are in it

Leaky Nun

what is it with $\widehat{K} = \Bbb A/K$...

s.harp

1:22 AM

I believe so, but I have never looked into PDE Taylor, this one is called Michale E Taylor

Mike Miller

i think they are the same

i like his pde series a LOT

Leaky Nun

for a more serious note, I really don't like how the Fourier series guys here use sin cos instead of exp

I really don't care about trigonometric functions

Ted Shifrin

It depends whether you're working with complex Lie algebras or real forms, I imagine.

Mike Miller

ok

@s.harp added that book to my list

looks very cool

s.harp

Looking through it there is no non-abelian fourier transform in it tho

Ted Shifrin

1:28 AM

Go to sleep, @Oskar.

CaptainAmerica16

@TedShifrin I know it has something to do with the fact that $a<b$ is $b-a$ and $c<d$ is $d-c$. They're greater than 0.

s.harp

@Ted funnily enough, my name is Oskar, and I'm now going to heed your advice and go to sleep

Ted Shifrin

LOL @s.harp

@CaptainAmerica: If $c<d$, what have you proved in an earlier part (I believe) about $-c$ and $-d$?

If you haven't, prove it now.

Mike Miller

@s.harp yeah the book looks v crazy

taylor is a solid author

CaptainAmerica16

I think you're referring to the fact that they're $\in P$.

Ted Shifrin

1:31 AM

@CaptainAmerica, stop saying nonsense like that.

CaptainAmerica16

OMG, how else should I word it?

Do you get what I mean?

Ted Shifrin

No.

CaptainAmerica16

;-; give me a second.

Mike Miller

@TedShifrin what are these exercises on?

Ted Shifrin

Spivak's early exercises on manipulating inequalities.

Chapter 1, early exercises.

Mike Miller

1:34 AM

got it

CaptainAmerica16

I mean that $-c$ and $-d$ are positive - unless I've misunderstood something.

Ted Shifrin

Sorry, phone.

Sank

2

Q: Prove the existence of uniformly distributed sequence of reals in $[0,1]$

A sequence of reals $\{x_n\}$ in $[0,1]$ is uniformly distributed iff for every $a<b$ in [0,1], $$\lim_{n\to\infty}[\sum_{j=1}^n 1_{(a,b]}(x_j)]/n=b-a$$Prove that such a sequence exists. (Hint: Show that it suffices to show the above result holds for rational values $a$ and $b$. Here is my a...

Would like some feedback!

Leaky Nun

@TedShifrin I'm very confused right now

Daminark

I was told this was the linguistics chat

Leaky Nun

1:39 AM

Serre's group rep book claims that he can get Fourier series by looking at representations of $S^1$

Daminark

It says math but I shall trust my insider info

Mike Miller

@Daminark It's a glitch, you got it right

Anyone know the etymology of lord?

Need it for a project and this, the linguistics chat, seemed like the right place to ask

Leaky Nun

ok it's 01:45 AM and I should go to bed instead of thinking about ~~maths~~ linguistics

so see you guys

Ted Shifrin

@CaptainAmerica: You've misunderstood. I tell you $-1<5$. And what are you telling me about $-(-1)$ and $-5$?

Hi, linguistic Demonark.

CaptainAmerica16

Scrap everything I said earlier. Scrap it all

Daminark

1:47 AM

@Leaky so the brief version that I know about is that you can kinda have the circle act on itself by addition and this leads to a representation $S^1 \to L^2(S^1)$. Decomposing this rep into irreps is the problem of writing a function as a Fourier series

The details evade me, and good night!

Ted Shifrin

It's all very simple. We already said it, @CaptainAmerica. $a<b$ means $b-a\in P$. That's ALL.

How's it going Ted?

that's the detail

LOL @details evade.

I guess, just that I haven't actually sat down and proven that $e^{inx}$ are precisely irreps of $S^1$

Ted Shifrin

1:48 AM

Character $n$?

Mike Miller

@Daminark $S^1$ acts on $\Bbb C$ by multipliying by $z^n$

exercise: an abelian compact Lie group acting on $\Bbb C^n$ splits into a sum of 1D actions

Ted Shifrin

That reminds me of something Schur.

Mike Miller

harder exercise: the same is true if they act on a hilbert space

@TedShifrin It should remind you of a linear algebra fact... :)

maybe involving two matrices

Ted Shifrin

Well, it's all linear algebra.

Daminark

Ah compactness didn't come to mind

Mike Miller

1:52 AM

@Daminark it's about the existence of a finite set of "topological generators"

Ted Shifrin

Two matrices? yes.

I'm well acquainted.

Mike Miller

I think you probably see the argument now.

Ted Shifrin

Provided I see why one decomposes.

Mike Miller

Ah, ok, I see your objection.

Compact Lie group ~ may as well assume unitary action.

Ted Shifrin

Yeah, OK, extra trick. :P

Mike Miller

1:56 AM

I didn't think of that point until you made it.

Ted Shifrin

I've taught some of this stuff, but that doesn't mean I remember it.

Mike Miller

The existence of finitely many topological generators is probably not obvious, but I know proofs.

It's certainly clear for $S^1$. :)

("Topological generator" = the subgroup they span is dense)

Daminark

So if the action is assumed to be unitary isn't this immediate? Irreps of abelian groups are 1-D by Schur's lemma and then unitary reps should just decompose for free into irreps

Before you said that I was starting to think, at least based on what my rep theory prof has been alluding to, that you'd define $[v,w] = \int_G \langle \varphi(g)v, \varphi(g)w \rangle \omega$ where $\omega$ is some volume form, and that's kinda the compactness part

Érico Melo Silva

wow i am m a d

Daminark

What happened?

Érico Melo Silva

2:08 AM

more duke and stony shenanigans

i keep trying to submit technical support tickets but apparently the mechanism to submit them is also broken

Daminark

r i p

Mike Miller

you could just give up

Érico Melo Silva

none of my letter writers received the recommendation requests from them

Mike Miller

@Daminark how does schur's lemma work? is it fine for arbitrary abelian groups, not just finite?

i never remember that proof

ever

Daminark

I think it should work in general, since it just relies on the $G$-map having an eigenvector since we're in $\mathbb{C}$, and then the eigenspace is a subrep

Mike Miller

2:12 AM

ah I see

so here's the argument I had in mind, which works for Hilbert spaces, which might be a bonus

1) $G$ is a compact Lie group, which means it has a Haar measure with volume 1 / invariant volume form, which means you can average. So one can average out the inner product and the result will be an inner product for which $G$ acts unitarily

2) Unitary means diagonalizable.

CaptainAmerica16

@TedShifrin OK, everything is positive: $b-a$ and $d-c$. From this we can say $(d+c)- (b+a)$ is within the postive numbers. SO: $b-a + d-c > 0$ therefore, $a+c < b+d$.

That seems so convoluted.

Mike Miller

3) diagonalizable and commuting means that you're simultaneously diagonalizable.

4) In a compact Lie group you can find finitely many elements which generate a dense subgroup. Simultaneously diagonalize that action. You have now diagonalized the action of all of $G$. This is what you wanted.

Daminark

Ah, so is the point that you only have to choose a basis for finitely many people at one shot?

Daminark

2:35 AM

Okay I'll talk a bit of AG out loud. So if $X\subset \mathbb{P}^n$ is some variety such that $I(X) = (f)$, its Hilbert polynomial should be $h_{(f)}(l) = \binom{n+l-1}{n} - \binom{n+l-\deg(f)-1}{n}$

Or wait no I'm dumb we're in projective space

$h_{(f)}(l) = \binom{n+l}{n} - \binom{n+l-\deg(f)}{n}$

Yahya

is the inverse of an open continuous map also continuous?

Mike Miller

@Daminark Yup that's it

Daminark

And I just realized the easy way to see it is hypersurface intersection formula, if $F$ has degree $d$ then $h_{I+(F)}(l) = h_I(l) - h_I(l-d)$

Mike Miller

Your argument also seems good but I would need to understand how to decompose a Hilbert space rep into irreps

MatheinBoulomenos

2:54 AM

@MikeMiller there's a Zorn's lemma argument that the existence of stable complements is equivalent to being a direct sum of irreps

Daminark

Yeah the idea I had in mind in finite dimensions once you managed to make the rep unitary was inductive, you split it up, split those up underneath, etc so in Hilbert spaces that wouldn't quite work out nicely

Hey Mathein!

MatheinBoulomenos

Hey Daminark

Mike Miller

Ah I see

Nice!

Shaun

I just got my eleventh gold badge on MSE, the Socratic badge! Hooray! :)

Yahya

nvm the inverse might not exist

Shaun

2:56 AM

My last message was all I came here to say. Carry on!

MatheinBoulomenos

I wrote down the argument in the general case (for modules) in my blog, if anyone is curious: prop 3.14 here: wlou.blog/2018/06/25/semisimplicity-and-representations-part-1 the only Zorn's lemma that one needs for the direction we're interested in reduced to the fact that every proper left ideal is contained in a maximal left ideal

oh no wait, we need Zorn's lemma twice

Mathematics

Does anyone know any software where I could plot a velocity-time graph?

CaptainAmerica16

3:18 AM

@Shaun I starred your message to help you celebrate.

Shaun

3:38 AM

Thank you, @CaptainAmerica16 :)

CaptainAmerica16

Sure :D

Secret

link.springer.com/chapter/10.1007/978-3-642-31977-8_1

wut?

Pig

hi @MatheinBoulomenos - nice blog!

MatheinBoulomenos

hi @Pig thanks!

Secret

3:58 AM

sciencedaily.com/releases/2018/10/181012082713.htm

Wait a minute, then what is the point of the denominator in the natural frequency format?

it never get factored into the calculation

ninja hatori

4:45 AM

Anyone know local criteria for flatness ; I want some discussion on it?

Secret

5:01 AM

what else do you need besides the jacobian?

Liad

5:20 AM

@lush yes i get what you mean. thanks!

Sharath Zotis

Why does a norm of 0 imply a spectral radius $|\rho(A)|$ <1

Secret

ncbi.nlm.nih.gov/pmc/articles/PMC200816

Min other wirds, knew your reference class

Buddhini Angelika

6:00 AM

A graph is 2-connected if removal of any one vertex will not make it disconnected, right?

Ted Shifrin

6:14 AM

@CaptainAmerica16 Note that by part (ii), $-c<-d$. By part (i), $a+(-c)<b+(-d)$, so $a-c<b-d$.

Liad

@MatheinBoulomenos hi , remember the question with $Y=\{(t,t^2,\dots,t^n) : t\in K\}$ ?

Mike Miller

6:30 AM

@SharathZotis norm in what sense...? An operator with operator norm 0 is just the 0 matrix

Zee

7:33 AM

@OskarTegby your crazy for not wanting to be distracted by them

Zee

7:48 AM

I really ought to stop drunk emailing my professors

Alessandro Codenotti

8:24 AM

@Liad what's the question?

lush

8:40 AM

yoyo

Secret

8:53 AM

arxiv.org/abs/1004.2440

So...

Elementary evaluation of integrals in a nutshell = expand the integrand into a power series, and hopefully find a rearrangement that is in terms of finitely many special functions

Word problem for groups

In mathematics, especially in the area of abstract algebra known as combinatorial group theory, the word problem for a finitely generated group G is the algorithmic problem of deciding whether two words in the generators represent the same element. More precisely, if A is a finite set of generators for G then the word problem is the membership problem for the formal language of all words in A and a formal set of inverses that map to the identity under the natural map from the free monoid with involution on A to the group G. If B is another finite generating set for G, then the word problem over...

So er... if you like, finding close form is the word problem in the domain of all infinite series

Leaky Nun

9:10 AM

2 days ago, by Liad

im trying to show that $Y= \{ (t,t^2,\dots,t^n) : t\in K \}$ is a closed subset. so an element of $Y$ is of the form $\sum_{i=1}^n \alpha_i t^i$ where $\alpha_i \in K$. how can i find a polynomial s.t $Y$ is the of zeros of this polynomial? (i thought maybe showing $I(Y)$ is radical, but it didn't work as well.. ) @AlessandroCodenotti

lush

@LeakyNun btw, I'm meeting "the god" almost every day :D just to make you jealous lol

Leaky Nun

@lush :(

lush

sorry, couldn't resist

Leaky Nun

lol

Alessandro Codenotti

9:52 AM

@LeakyNun I meant what's the new question he has now on the same problem

towc

here's another weird resource answer:

Yuriy S

Hey everyone! If you don't mind me asking, why is there so much group theory discussion in this chat? I'm definitely not objecting, just curious?

towc

> Three hardcover books and 5 paperbacks are placed on a shelf. How many
ways can the books be arranged if all the hardcover books must be together
and all the paperbacks must be together?

they say "2! × 3! × 5!", but my answer was simply "2"

imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf q11

I said 2 because if all the As must be with As, and Bs must be with Bs, then you either have all Bs on one side, and all As on the other, or the other way around

oh wait, maybe in that answer, it assumes that an A is different from another A

Secret

@YuriyS Because recently the algebra squad have gathered and they are all discussing intensely about their algebraic studies

towc

actually, I'm still lost on 14, 15, and 17: imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf

maybe they missed some factorials in there somewhere?

I completely fail to see the logic in 17

I thought it should just be 8! (not 8!², or 8*8)

s.harp

10:07 AM

did you know that operator algebraists are actually analysts?

Secret

they work with continuous space, of course they are analysts

s.harp

topologists and geometers also work with continuous spaces

Secret

::argument fail::

Suddenly realised I don't know what analysis is

s.harp

analysis is everything involving $\Bbb C$ or $\Bbb R$ algebras that nobody else claims

so sometimes even infinite-dimensional sub-manifolds is analysis

non-commutative geometry? thats analysis

Secret

hmm, both $\Bbb{R}$ and $\Bbb{C}$ are complete (forgot word) infinite spaces. Is it safe to say that analysis is the study of complete (forgot word) infinite spaces in general?

(I think I want to say dense, but it does not sounds quite right. There is a property unique to reals and complex numbers, but I forgot what it is)

Leaky Nun

10:16 AM

@s.harp man, the book you gave me is dense

s.harp

@LeakyNun which book was that? Folland?

Leaky Nun

A Course in Abstract Harmonic Analysis

how long do you think do I need to chew 4 chapters?

s.harp

well, what's your functional analysis knowledge?

let me look at the chapters again

Leaky Nun

I know what L^2 is... I hope :P

alirejali.ir/afiles/up/other/book5/…

s.harp

ive got the book infront of me, chapter 2 you can just skim through, trying to understand the consequence of the theorems, ie the whole 8 pages of Haar measure can be boiled down to basically one sentence: Locally compact groups have a translation invariant measure, for compact groups the entire space has finite volume

you don't need to comprehend those proofs, though it can be fun, i remember the part where Tychonoff enters into existence of Haar measure that was quite nice

after that use chapter 2 as a reference

to be honest, you can do the same with chapter 1

get the basic idea of the gelfand transform and move on

chapter 3 is just basic definitions

Secret

10:20 AM

ok, I did not knew that:

1

Q: complete ordered set with least upper bound property

Is there any order that can make complex numbers complete ordered set with least upper bound property? I came up with $x+yi > m+ni$ if $x+y>m+n$ and if sum is equal then the complex number with greater real part is greater. will this work? if not, are there any examples?

complex-numbers order-theory complete-spaces

$\Bbb{C}$ violates lub

s.harp

then you can read chapter 4 using the other 3 as reference material

so depending on your tempo you can do it all somewhere between 2 days and a month

Leaky Nun

@s.harp heh...

so just skip all the proofs lol

Danu

@MikeMiller I see

s.harp

@LeakyNun something an old instructor used to say

Danu

Thanks!

Yuriy S

10:26 AM

How does one find a solution for linear recurrence looking like this? I mean, proving it was not that hard, but how could anyone guess it in the first place? math.stackexchange.com/q/2994975/269624

Secret

ok I am not so sure anymore, I thought least upper bound property unites both reals and complexes, but I just found out today that complex number have no such property. I might need to recheck to see what is the essence of real numbers

Leaky Nun

@s.harp ???

s.harp

@Secret how were you envisioning a sensible order on $\Bbb C$?

@LeakyNun it means "don't be afraid to skip complete understanding certain things and take them as given"

Secret

uh... I never thought about the least upper bound property applied to "more than one dimensions" I am so used to view $\Bbb{C}$ as a two dimensional object and thus think of the lexicographic ordering alot. This is why I do not knew lub does not hold because lub demands that every subset has a least upper bound, which you cannot if the set cannot be totally ordered

s.harp

@Secret lexigraphic ordering is not an acceptable ordering for $\Bbb C$, as you have that the square of positives can be negative

Secret

10:35 AM

Also this link explains how one of my intuition of $\Bbb{C}$ as a two dimensional vector space is sometimes misleading:

30

A: Total ordering on complex numbers

If we had an order on the complex numbers, then either $i \prec 0$ or $0 \prec i$. If $0 \prec i$, then $$0i \prec ii \implies 0 \prec -1$$ Then since $0 \prec -1$, we see that $0 \prec (-1)^2 = 1$. Using (iii) we get $$0 \prec -1 \implies 1 = 0 + 1 \prec -1 + 1 = 0 \implies 1 \prec 0 \prec 1...

If you think of $\Bbb{C}$ as like $\Bbb{R}^2$ but with some multiplication structure between two vectors, then it might be tempting to think that one can fit a hilbert curve to linear order all complex numbers. But the issue of this is then you will change the multiplication structure as $i^2 = -1$

thus that is impossible

Leaky Nun

@s.harp I'm very afraid, for the record

Tobias Kildetoft

Is AC necessary to show that all ideals in a local ring are contained in the maximal ideal?

Leaky Nun

@TobiasKildetoft I don't think so

you mean all proper ideals, but I don't think so

given a local ring, the maximal ideal is the ideal of non-units

hmm...

that statement might need AC?

Secret

9

Q: A confusion about Axiom of Choice and existence of maximal ideals.

The proof of the theorem of statement that every ring has a maximal ideal uses zorn's lemma or the axiom of choice.Now, the defintion of ring as well as the definition of maximal ideal don't depend on the axiom of choice.Is that true? So, given a ring I should be able to check whether it has maxi...

ring-theory axiom-of-choice axioms

If no AC, you cannot even find a maximal ideal

(I wish I can elaborate more but my knowledge on ideals are terrible)

Leaky Nun

@TobiasKildetoft I mean, I'm tempted to (re)define local rings as rings where the sum of two non-units is a non-units, just to avoid all this

@TobiasKildetoft you might want to ask it on main lol

s.harp

10:49 AM

@LeakyNun What do you mean?

Leaky Nun

@s.harp it's a joke response to your "don't be afraid ..."

Secret

And I do not exist under Berlin's Rule

Tobias Kildetoft

@LeakyNun I was mainly curious, because I has implicitly assumed it was not necessary when I posed an exercise this week

Precisely in showing that local iff non-units form an ideal

Leaky Nun

11:04 AM

@TobiasKildetoft P.486 (P.517 of PDF) constructive commutative algebra

Tobias Kildetoft

@LeakyNun Well, that was a highly non-enjoyable read which told me very little, other than that it does seem like AC is probably needed.

Leaky Nun

lol

2constructive4you

Tobias Kildetoft

exactly

s.harp

It seems to me that "Do you need AC for X?" will always result in a highly non-enjoyable finisher if you get into details

Tobias Kildetoft

@s.harp Well, unless the answer is just "yes, that statement is equivalent to AC"

or of course "no"

s.harp

11:15 AM

@Tobias but thats not a proof, ie details

honestly I don't think I've ever even seen the proof that Zorn's lemma $\iff$ AC

Leaky Nun

given a surjection, I think you can look at the poset of local sections, and then use Zorn to build a global section, so this proves =>

<= is standard :P

Tobias Kildetoft

@LeakyNun What do you mean "given a surjection"?

sections of surjections does not imply AC

Secret

16

Q: Zorn's Lemma And Axiom of Choice

How can I prove Zorn's lemma is equivalent to Axiom of choice?

elementary-set-theory axiom-of-choice

Leaky Nun

@MikeMiller and he never came back

s.harp

<= relies on the fact that there are more ordinals than any set can contain, which is nasty

Leaky Nun

11:19 AM

@TobiasKildetoft does it not?

I thought AC is equivalent to saying that every surjection has a section

Tobias Kildetoft

@LeakyNun I am fairly certain Asaf said at some point that it is (slightly) weaker than choice

Leaky Nun

ok...

@s.harp there's a long proof that doesn't require that

Tobias Kildetoft

@LeakyNun Ahh, I think I recalled the wrong statement

Leaky Nun

so, it's a choice either way

Tobias Kildetoft

surjections in each direction implying bijection is weaker, but sections of all surjections does imply AC

Leaky Nun

11:21 AM

ok cool

Secret

Attempt at a lame joke: Does zorn's lemma holds for the class of constructive proofs that Zorn's lemma is equivalent to AC

(I guess the first thing is: wtf is a well ordering of proofs)

s.harp

you only need a partial ordering, in this case you can take symbol length of the proof

however it is not so that every chain of proofs has an upper bound

Leaky Nun

so it holds vacuously...

2

@TobiasKildetoft Is a $\Bbb C^\ast$ algebra an algebra over the ring $\Bbb C^\ast$?

Secret

lol that's unexpected

Tobias Kildetoft

@LeakyNun No

Secret

11:24 AM

I would be able to contribute more to this discussion had I have a more firm grasp of prime ideals. Anything with the word "prime" just confuses me like hell

Tobias Kildetoft

also, that makes no sense since $C^*$ is not a ring

Leaky Nun

@TobiasKildetoft that's the joke

Tobias Kildetoft

(note that it is not $\mathbb{C}^*$-algebra, it is just $C^*$-algebra)

I see

I seem to have left my sense of humor at home today :)

Leaky Nun

so $C$ for continuous?

Tobias Kildetoft

no, C because it comes after B

Secret

11:27 AM

B is boolean algebra?

Tobias Kildetoft

Banach

Leaky Nun

@TobiasKildetoft ok seriously

Secret

right

Leaky Nun

were you serious 0.0

s.harp

@LeakyNun there were C* and B* algebras

then it turned out they were the same

Tobias Kildetoft

11:29 AM

I am not actually sure if there is any truth to it, but I heard someone claim it at some point. That originally some people studied Banach $*$-algebras, but they were missing an additional condition to get what they were needed, which led to $C^*$

s.harp

now there are only C* algebras

Leaky Nun

rip markdown

s.harp

note that a Banach-* algebra is not the same as B* algebra

Will Hunting

Rudin's Functional Analysis uses B* algebras.

Leaky Nun

wat o.o

Tobias Kildetoft

11:29 AM

@s.harp Ahh, I see

Leaky Nun

okok information overload

@s.harp plot twist

Will Hunting

But most other books use C* algebras.

Leaky Nun

🅱* algebras

Secret

The first time I heard about C* algebra is the operator theory in quantum mechanics

Will Hunting

Therefore, we have proven that B=C. As a corollary, a bat is really the same as a cat.

5

Secret

11:31 AM

sooooo tempted to star that, but we need one more animal

checks if there exists F algebras

Will Hunting

There are V* and W* algebras as well.

Secret

F-algebra

In mathematics, specifically in category theory, F-algebras generalize algebraic structure. Rewriting the algebraic laws in terms of morphisms eliminates all references to quantified elements from the axioms, and these algebraic laws may then be glued together in terms of a single functor F, the signature. F-algebras can also be used to represent data structures used in programming, such as lists and trees. The main related concepts are initial F-algebras which may serve to encapsulate the induction principle, and the dual construction F-coalgebras. == Definition == If C is a category, and F: C...

lol wut, goes straight into abstract nonsense

Will Hunting

kthxbai

s.harp

@Secret part of the story B* = C* is the understanding that every C* algebra is a sub-algebra of B(H) for some hilbertspace H

@LeakyNun "The term C*-algebra was introduced by I. E. Segal in 1947 to describe norm-closed subalgebras of B(H), namely, the space of bounded operators on some Hilbert space H. 'C' stood for 'closed'."

Leaky Nun

ok...

now what is C in C^infty

Tobias Kildetoft

11:36 AM

@LeakyNun "continuously differentiable"

Leaky Nun

fair...

Secret

@s.harp I see

Unknown MathMan

@LeakyNun But I want to know, how the definition of homomorphism came, given the requirement that by isomorphism we need to preserve all group theoretic properties? (chat.stackexchange.com/transcript/message/47567908#47567908) ?

Asaf Karagila

@TobiasKildetoft Actually, it's open if it's weaker. But we (read: I) expect it to be weaker. There's also "If there is a surjection, then there is an injection in the other direction [but not necessarily a section]" which is stronger than what you mention, but also it is open about whether or not it is weaker than choice. Again, I expect the answer to be that it is weaker.

s.harp

how did you figure out people were talking about AC?

Unknown MathMan

11:58 AM

@LeakyNun In other words, how you should proceed to solve this problem - "List the properties of a map need to be satisfied by two groups to be identical under group theoretic properties".

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$Mathematics$ Mathematics