Hello@AlexClark

Well is the null set an open set or a closed set?

clopen

it's both, @Rem.

@BalarkaSen there?

yep

can I look at a free group as the group of all combinations of elements of a group, and their inverses?

whoa, same crowd

is it true that $H_n(X\times Y) \approx H_n(X) \times H_n(Y)$ ?

haha

@iwriteonbananas no :)

oh :(

example?

@BalarkaSen and mine?

$(X, Y) = (S^1, S^1)$.

@SohamChowdhury meh

yes or no?

@SohamChowdhury what d'you mean?

free group is group of all combination of elements of a set rather than a group

yes.

informally, the free group of some set is the set of all "words" of powers of elements of the set

i know that

@iwriteonbananas $H_2(S^1 \times S^1) = \Bbb Z$, while $H_2(S^1) \times H_2(S^1)$ is trivial

but you're thinking about good stuff.

@BalarkaSen good point :P

question : can I compute $H_n(X \times Y)$ knowing homology of $X$ and $Y$ in all dimensions?

think about it, it's a nice problem.

everyone's here now

hey, @AlexC

@AlexClark yes

how'd it go?

cooleo

:P

@BalarkaSen mayer vietoris?

principal*

blah. $(fg) \subset (f)$

no need to do fancy stuff.

oh, sure, you need the fact that it's a PID

@BalarkaSen did you say that to me?

@iwriteonbananas choosing an open cover of $X \times Y$ can be messy.

no, i said that to Alex.

ok

it's a hard problem. there's even a name, but I am not gonna tell you that.

you might try and think about $X, Y$ which can be given a CW-structure

it's very irritating to have to refer back to arbitrarily numbered "Lemma 6.4"s

anyone got any tips for dealing with those?

ok, will do

@iwriteonbananas that's one problem I have seen where using some general nonsense naturally simplifies much work. it's essentially about symmetric monoidal structure on $\mathsf{Ch}_\bullet$ and computing homology of $\otimes$-product of chain complexes.

yikes, okay

try having a stab at it for a bit, and let me know when you don't want to think about it anymore. i'll give you the keyword.

okay

@AlexClark i made a schedule for myself today, btw, after you went away to your whiteboard

mostly Aluffi and schoolwork, an hour or two of Munkres

i'll know what i'm supposed to be doing at any given time of day

that's the reason why i'm doing it.

aren't those SE icons ? pasteboard.co/KD8XIwN.png I found those on source forge

meaning?

@iwriteonbananas why don't you try something for simpler cases? for example, take $X \times S^n$. you know the stuff you have to know to compute homology of this thing.

it does work for me, they're the reason i got by somehow during my class 10 exams

@AlexClark good luck!

@alex so they're not only used here ? Ok then...

@alex Ok fine.

I've finished the subgroups section. (I told you it would be smooth sailing for me after free abelian groups :P)

@alex No I didn't say you were

oh, @iwriteonbananas, you know when the long exact sequence of homology becomes a (split) short exact sequence at every level, right?

@alex Does that look like I'm angry when I say 'OK fine ' ?

might've

they're weird over in 'straya

@soham Who are you answering to

@alex You're from Australia ? What city ?

@AlexClark what did I say about straya? :P

@marcelolpjunior very similar

hey @anon

@SohamChowdhury I will read, thank you.

@BalarkaSen probably, but im not sure what u wanna hear

morning guys

@SohamChowdhury heya

Morning@KarimMansour

@marcelolpjunior Consider where the poles of $\frac1{\sin(\pi z)}$ are and what the residues are at each pole.

@SohamChowdhury Does that help to show why that is true?

@iwriteonbananas tell me what you have in mind.

@BalarkaSen o/

hi

hi

@BalarkaSen i forgot, im trying to compute the homology of the torus using MV for arbitrary an homology theory

it's very confusing

i don't care about arbitrary homology theory if you don't have the dimension axiom.

sorry, dimension axiom is fulfilled

forgot to mention

then it's ok

i find it hard to see what the maps actually do when we're in a general homology theory (satisfying dimensino axiom)

homology of torus?

yeah

cover the torus by two cylinders

right, sure.

then the intersection is a disjoint union of two circles

$H_\bullet(\Bbb T; R)$ is easy.

and the two sets that cover the torus are h-top equivalent to circles

@BalarkaSen what's R?

eh, some ring.

ok

well

so what's bothering you?

no, i dont think so

consider this chunk of MV

no, i dont think so

$0\to H_2(T)\to H_1(A\cap B) \to H_1(A)\oplus H_1(B)$

where A,B are the cylinders

i'm sorry, i am a bit busy. i'll get back to you after a few minutes.

i.e. the sequence is $0 \to H_2(T) \to R\oplus R \to R\oplus R$

np

how do we figure out what the last map is?

it's the map $(i_*, j_*)$ where $i: A\cap B \to A$ and $j:A\cap B \to B$ are the inclusions, but what is it explicity?

$H_2(T)$ is the kernel of that map

@DanielFischer Do you know what a Y-triangle is ?

No, sorry. Never heard of that.

ok...no problem..

hey everyone

@iwriteonbananas: A homology rheory that satisfies the dimension axiom must be singular homology with respect to some group (at least on finite CW complexes).

You prove this by looking at the proof that cellular homology is isomorphic to singular and going "Oh hey that works for my theory too"

ugh, i havent read that proof yet

is there a way of finding out what the map $(i_*, j_*)$ does if i dont want to use that fact?

Yes, I wasn't answering your question, just telling you something related.

And also yes.

I have faith that you can find it.

i pretty much did, but i was thinking about the generating loops of the two circles and the induced map on singular homology

i.e. the generators of each circle are mapped onto (singular) homologous loops in the cylinder

it's not clear to me why that's the case for any ordinary homology theory

@iwriteonbananas so essentially you want to know that the inclusion maps $i, j : S^1 \hookrightarrow S^1 \times [0, 1]$ taking $S^1$ to the top and bottom copies of the cylinder respectively induce the same maps $i_* = j_* : H_1(S^1) \to H_1(S^1 \times [0, 1])$. this is not hard : compose both $i_*$ and $j_*$ with the isomorphism map $H_1(S^1 \times [0, 1]) \to H_1(S^1)$ induced by the deformation retraction of the cylinder onto the center circle. the resulting composed maps are the same.

now right-cancel.

I need to determine whether or not $\mathbb{Q}[x]/(x^3-1)$ is a field. I know that this is a field iff $(x^3-1)$ is a maximal ideal iff $x^3-1$ is irreducible.

I know that I can write $x^3-1 = (x-1)(x^2+x+1)$ but I can't tell whether or not either of these are units.

Also, how can I check that $x^3-1$ itself isn't a unit?

@user112495 consider $x^3 - 1$ in $\Bbb Q[x]$. quotient map $\Bbb Q[x] \to \Bbb Q[x]/(x^3 - 1)$ takes this to a zero divisor. a field doesn't have a zero-divisor.

or, indirectly, $(x^3 - 1) \subset (x - 1)$, which implies it is not maximal. thus your quotient ring isn't a field.

sorry, it should have been $x - 1$.

mis-edit.

@BalarkaSen Oh yeah, of course! Thanks.

no problem.

hello, @bolbteppa.

Hey @BalarkaSen absolutely fascinated by path integrals if you've seen them?

I have heard about those, but never really studied. I'm listening if you want to talk about it :)

Well too busy to do so now tbh but definitely soon, enjoy! :D

sure.

@MikeMiller Isn't it true that homotopy equivalent spaces always retract to homotopy equivalent spaces?

Pretty sure it is : just making a sanity check.

I mean, if $X \simeq Y$, and $X$ retracts to $A$, then mustn't there always be a retract $Y \to A$?

I guess composing with the homotopy equivalence does the trick.

nevermind.

hello, @Ted.

hi, @Balarka ... It's way past your bedtime, but I imagine it's still beastly hot :( Our public radio station had a long story on the heat wave there.

it's very hot. quite a few people died out of heat.

Very sad. I hope you drink plenty of water.

Want to borrow the rain I have outside?

I have to. Otherwise I'll dehydrate. That's how the people are dying outside.

I want that rain for California, thanks.

California is so screwed.

we're turning into California, I guess. lots of quakes.

there has been one of 5.4 the day before yesterday.

Well, California is particularly screwed because I'm moving back there.

haha

Mommy Nature is totally pissed off at the world.

Anyone know if this (pumj.org/docs/Issue1/Article_2.pdf) template is simple to do (i.e., as amsart) or if it is custom?

Who knows? It's a totally unrecognizable font, I will say. I have bought a few extra TeX fonts, but don't recognize that one.

Morning, @TedShifrin. Where can I read about jet bundles and generalized PDEs?

In particular, well, just jet bundles. I don't know anything about them.

Goodnight, @MikeM. Hirsch does jets.

OK, thanks. I'll take a look at that tonight.

It sounds scary, but it isn't.

It doesn't sound too scary, I just don't know what it is. I want to understand Gromov's h-principle PDR stuff at some point; it's near the top of my pie-in-the-sky reading list.

And of course it's all stated in terms of sections of jet bundles.

I even had jet bundles in one of my papers ... so that's proof that they're not that scary.

LOL, ok. I believe you.

Gotta get back to work. See ya. Thanks for the tipoff on Hirsch... I need to just buy that book already.

@AndrewThompson looks like Iwona font, or a closely related font. Looks like all they did was change the font and did something to make the pdf file stupidly large.

Skimmed through bits of groups acting on the circle last night, I was very surprised to see that $PL_+([0,1]) $ does not contain a free group of rank 2! (At a tablet right now so probably wont respond)

Please if i have an expression that is true for all $v\in X$ can i say that is true for $v=v_n$ where $(v_n)\subset X$ ?

@TedShifrin ?

have you an idea please

my text says $\frac{d^2\psi}{d\xi ^2}=\xi^2 \psi$ has the approximate solution $\psi (\xi )\approx Ae^{-\xi ^2/2}+Be^{\xi ^2 /2}$, particularly valid for large $\xi$. Checking manually suggests this is a good approximation, but what might have motivated finding such an approximation?

That is, how might one arrive at that approximation?