Mathematics - 2015-11-24 (page 6 of 6)

user174558

23:00

@morphic Rotman wrote lots of algebra books, and it seems he combines bits and pieces into his grad algebra text, omitting some parts and some proofs here and there.

Owatch

:>

Think I'll shower now.

Ted Shifrin

Night, @Owatch.

Owatch

The bathroom I share is a former public one.

And is also kept at outside temps.

Ted Shifrin

Yikes!

Owatch

That being, about 5 degrees C.

user174558

23:01

@morphic But what I like about his abstract algebra book is that he proves the cubic and quartic formulas in full. Not that many care about them, but this is extremely rare nowadays.

Owatch

Waiting for the hot water seems to be endless.

Oh well.

I have more stories about that place for later.

See you all! Thanks a lot for helping me Ted!

Ted Shifrin

You're welcome, @Owatch.

Chris's sis the artist

@JasperLoy have you watched the one below (motivational video)?

DESIRE - Motivational Video

►

morphic

@TedShifrin Dennis Sullivan is teaching an advanced algebraic topology course here next semester

Ted Shifrin

He's fantastic, but it'll surely be very advanced and indecipherable :)

morphic

23:05

Yea I read the syllabus and it's super hard

Ted Shifrin

What are the main topics?

L33ter

hi @TedShifrin

morphic

Something about generalized homology/cohomology theories and axioms

Chris's sis the artist

@JasperLoy watch all the series by Mateusz M, it's very nice.

Ted Shifrin

Yeah, that's like a third-year grad course, @morphic.

hi Karim

Chris's sis the artist

23:08

OK, time to finish a paper.

BBL

L33ter

so I have been doing some algebraic topology for my project

I wanted to ask you a question @TedShifrin

can you classify the topological spaces that have the fundmental group being abelian ?

i am sorry

topological spaces *

morphic

We are learning about the homology functor in class these days

Ted Shifrin

@MikeM may have a good answer for you. He's the topologist :)

L33ter

alright

Ted Shifrin

True for any topological group or H-space, Karim, but I don't know if there is a classification.

Dan Rust

23:11

Hi all

Ted Shifrin

Hi @DanRust. Actually, Karim, Dan may know the answer to your question.

L33ter

yeah I was thinking for example for a square if we identify opposite sides of the square

Ted Shifrin

You said topological spaces, Karim. That's super general. If you restrict to compact surfaces (2-manifolds), then it's well known.

L33ter

then it will give a two circles that are glued together

Dan Rust

It's a good question but my intuition is that it's unlikely.... there's just too many abelian groups

Ted Shifrin

23:13

No, no, the square is filled in, Karim.

L33ter

yeah for filled in one

yeah I was thinking if we don't fill it in

then it will not be abelian

Ted Shifrin

So the 1-skeleton is two circles. But they're not filled in the way you think.

L33ter

I se

I see

Dan Rust

I actually came to ask a question of my own which isn't too disimilar

Does anyone know of a torsion-free group (finitely presentable) whose abelianisation has torsion?

equivalently, a CW complex whose fundamental group is torsion-free, but first homology has torsion

Ted Shifrin

That sounds impossible, but if you think about presentations, it seems plausible that you could have $a,b$ that don't commute and have some complicated relations, but when you make them commute you get something like $a^2=1$.

Dan Rust

23:17

yes, my attempt found a good candidate, but I can't show it's torsion-free

L33ter

what does it mean torsion free group ?

Ted Shifrin

Yeah, that seems challenging. Page @anon. :)

Dan Rust

$\langle a,b,c \mid a^2 = bcb'c' \rangle$

torsion-free means there are no elements of finite order

if anyone can show that group is torsion-free I'd be happy :P

L33ter

what is abelianisation?

is it the commutator ?

Ted Shifrin

I think such things are hard, @DanR.

Dan Rust

23:19

the abelianisation is the quotient of G by its commutator subgroup

Yeah I know @Ted .... :(

Ted Shifrin

Perhaps @MikeM or @PVAL have some topological ideas ...

L33ter

I see nice question

Ted Shifrin

Or @Balarka ... but he's in conference and not around ...

Dan Rust

If an example exists, it'd debunk a paper which recently went on arxiv... but I have my doubts the paper is correct

L33ter

I gotta go study I will see you guys :D

Ted Shifrin

23:21

My immediate instinct was to say no, @DanR. A real algebraist should know this immediately.

Bubye, Karim.

Dan Rust

see you

morphic

what is $b'$ and $c'$

Dan Rust

inverses of $b$ and $c$

I found a nice example of a group which is isomorphic to a familiar group in a 'non-obvious way' in my search

Ted Shifrin

This reminds me of a ring theory question I stumbled over long ago: Can two elements in a commutative ring generate the same ideal without being associates?

Dan Rust

the subgroup of $\langle a,b \mid a^2=b^2=1\rangle$ of all words of even length

Ted Shifrin

23:27

What is that, $\Bbb Z$? :P

Dan Rust

yep

haha

Ted Shifrin

This is why I'm not an algebraist :P

Dan Rust

thought it'd make a nice question on an intro to group theory exam

Ted Shifrin

Exams aren't the place for treachery, but it's a good homework exercise !!

Dan Rust

haha true

Ainz Ooal Goal

23:28

@TedShifrin Btw, is $\mathbb{N}^\mathbb{N}$ in bijection with $\mathbb{N}$ or $\mathbb{R}$ ?

Dan Rust

reals

Ainz Ooal Goal

(or something else ? idk)

Ted Shifrin

I was going to say YES, but @DanR gave it away.

Dan Rust

sorry :(

Ainz Ooal Goal

Is there an easy way to show that ?

Ted Shifrin

23:29

Hint, @Ainz: What about $\{0,1\}^{\Bbb N}$? Do you recognize that?

Ainz Ooal Goal

Ooh binaries

Ted Shifrin

Yuppers.

Dan Rust

hint: continued fractions

Ainz Ooal Goal

that's cool :D

Dan Rust

@ted's hint is probably more useful

Ted Shifrin

23:31

@OldJohn!!!

Old John

@TedShifrin Hi Ted!

Ted Shifrin

Lovely to see you!

Ainz Ooal Goal

Well then, i'm off

Old John

@TedShifrin Just dropping by (briefly) - spend more time these days in Music SE

Ted Shifrin

How is your mountain climbing?

Old John

23:32

@TedShifrin Coming on nicely - 81 of my intended 214 climbed :)

Ted Shifrin

So impressive!

I've now retired, but I won't be climbing mountains.

Dan Rust

what else is retirement for if not golf and mountain climbing?

Ted Shifrin

I am an abysmal failure, then, @DanR.

2

Old John

@TedShifrin Congratulations on the retirement - you willnot regret it!

Agawa001

@robjohn thanks, i jumped directly to numerical part lol, but i expected something like an irrational number expressed with roots and pi

Old John

23:36

Dammit - have to go. Bye for now Ted - should be back soon(ish)

Ted Shifrin

bubye, @OldJohn

Old John

@TedShifrin waves!

Agawa001

@AlecTeal as a remark of latest incident, it was just a measly notice about ur website no need to go offensive against me 2/ im not the one who flagged/starred anything of maybe one whole page perimeter of that deleted message 3/ plz dont say another unnecessary thing after this

morphic

Now he's going to say something

Clarinetist

Lol thanks @TedShifrin . @Nickolas I don't know of an extremely general theorem. It depends on a lot of things, such as how you're measuring MSE, what your functional form is, etc.

and how many variables you have

In the univariate case, it's simple

In the $p$-variable case, you need matrices

and some matrix calculus

PVAL

23:51

@DanRust Is there torsion in $\pi_1$ of the Klein bottle? That's one of the more obvious candidates. Here $H_1$ is $\Bbb Z \oplus \Bbb Z_2$. I kind of think Gordon told me there are left-orderable groups with torsion in $H_1$ but I forget the examples.

Dan Rust

yeah, the Klein bottle has $\pi_1$ and $H_1$ isomorphic

PVAL

No it does not.

It's <a,b | aba=b^-1>.

Kasper

hi guys, I've been experimenting with different ways to present a proof using prezi

I'm wondering if you guys think this makes a proof understandable (without someone narrating the presentation)

prezi.com/fafg-rxjqhsz

PVAL

@DanRust On page 2 of these notes Rolfsen states that the Klein bottle group has $\pi_1$ LO which in particular implies Torsion-free.

Clarinetist

@Kasper I'm not a fan IMO. I think the narration is necessary

PVAL

23:56

I'm certain someone across the hall knows the proof of this better than anyone else.

Clarinetist

@Kasper I would prefer that you just use Beamer instead. The moving around that Prezi is so well-known for is quite distracting IMO

Kasper

@Clarinetist I was thinking about this, today I was doing in this style with a prezi for a presentation (so with narrating), and they found the prezi very helpful for keeping in touch with the big picture of the proof

but I guess without narrating it is a different story

Mike Miller

No surface except $\Bbb{RP}^2$ has torsion in its fundamental group.

Dan Rust

@PVAL sorry you're right. Klein bottle has a weird fundamental group $\langle a,b \mid ab = b'a\rangle$

Kasper

@Clarinetist do you think also in this prezi, I'm moving around too much? I tried to limit it a bit

Dan Rust

23:59

hah, don't know how I missed that

Clarinetist

@Kasper Honestly, yes. I don't think it's necessary. I like how Beamer does pausing

Mike Miller

@DanR: Equivalently, because they're all isometric quotients of the Euclidean or hyperbolic planes, any isometry of these of finite order has fixed points.

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$Mathematics$ Mathematics