Mathematics - 2015-05-30 (page 3 of 4)

@BalarkaSen i've now proved that $H_i(X\times S^n, X) \approx H_{i-1}(X\times S^{n-1}, X)$, so inductively $H_i(X\times S^n, X) \approx H_{i-n}(X)$. hence $H_i(X\times S^n) \approx H_i(X) \oplus H_{i-n}(X)$. only thing i havent proved yet is that there is actually a retraction $r:X\times S^n \to X$ as you claimed yday

Soham Chowdhury

@Rememberme romantic music

Remember me

lol

how do you write in italics @Soham

on chat

Because I dont want answers like "By a pen"

iwriteonbananas

@Rememberme i spent 8 years laying a fiber connection from the nearest town that has internet to my hut

Remember me

Do you live in amazon??

Soham Chowdhury

8:35 AM

@iwriteonbananas I smell a graph theory problem

But, really, where do you live?

Remember me

Nice sense of humor @Soham

Hi@DanielFischer

iwriteonbananas

@SohamChowdhury it's a secret

Daniel Fischer

Hi @Rememberme

Remember me

I think @iwriteonbananas you live in Amazon

iwriteonbananas

@Rememberme good guess

Remember me

8:39 AM

Like how i found that Jasper lives in Malaysia @iwrite

Thats a long story

are you from france?

Who??@Iwrite

you

No no.....

Bad guess

:p

But mine is not a secret

iwriteonbananas

so...?

Remember me

8:44 AM

I am from the place where spices are very famous....@iwrite

iwriteonbananas

hmm, france?

Remember me

I just said a no

Okay one more clue

iwriteonbananas

ok hurry up i gotta go

Remember me

I am from the place which would have been the second most safest place if the world had ended in 2012 @iwrite

iwriteonbananas

perhaps france?

Remember me

8:47 AM

Nope dammit ...

Where does Balarka live?

Where does Soham live?

Thats the place I live

@Iwrite

iwriteonbananas

i think you're from france

i gotta go now

Remember me

I think I am from India

@Iwrite

Balarka Sen

@iwriteonbananas good work!

well, the retract is just obtained from pinching the copy of $S^n$ in $X \times S^n$ to a point.

now, now. rubs hands. $H_i(X \times S^n) \cong H_i(X) \times H_{i-n}(X)$.

what can you tell about the index of the two copies of homology in the direct sum?

CBenni

is bumping a question allowed, and if yes, how?

Soham Chowdhury

9:06 AM

@Rememberme God, that's unbearable.

@CBenni don't know, I believe edits/comments trigger a jump to the top.

CBenni

Its always painful when a question leaves the front page, rarely will you ever get even a comment on it in that case ...

Thing is, I dont really know what to change about the question @SohamChowdhury ^.^

user147690

How is the study going @Soham?

Soham Chowdhury

I've actually been worrying about some other things.

user147690

@SohamChowdhury Oh non-math?

Soham Chowdhury

But everything is easier on the second pass, so the chapter seems easier now.

@AlexClark Not strictly non-math. I basically have one shot at the national math olympiad this year (for some weird reason, you're not allowed to sit the national one in 12th grade unless you clear it in 11), so I'm considering how to spend my time.

user147690

9:19 AM

@SohamChowdhury Ahh I see. Mathematics olympiads never occurred at my terrible highschool.

Balarka Sen

<-- mehs at IMO.

Soham Chowdhury

It's just that once you have a taste of "real" math, olympiad math is not that exciting. But (unless you are Balarka) colleges love olympiad medals, and I want to get into a really good one.

user147690

@BalarkaSen Agreed, it all seems very mechanical.

JC574

yo

Balarka Sen

hello, @JC574

user147690

9:21 AM

Hi JC

JC574

I have another question

Soham Chowdhury

@AlexClark So it's kind of a problem for me, trying to "choose", as it were.

user147690

@SohamChowdhury What is the dilemma? Regretting not doing the olympiad vs what? Regretting wasting heaps of time preparing for it?

JC574

Let $f:X \to Y$ be a nonconstant holomorphic map of Riemann surfaces, degree $2$. I want to find a non-trivial holomorphic homeomorphism $g:X \to X$ such that $g^2$ is the identity and $f \circ g = f$.

I get the general idea of $g$, but showing it's a holomorphic homeomorphism...

Soham Chowdhury

No, not regretting.

JC574

9:23 AM

when I say $g^2$ I mean $g \circ g$

Soham Chowdhury

I can prepare from now on, but it'll take up all my time and I can't study all this wonderful math properly.

I have a shot at it this year.

I'm confused about how to allocate my time.

Balarka Sen

do what you'd like to do

Soham Chowdhury

if I had a clear answer, I wouldn't be asking.

JC574

Since $f$ is degree $2$ we can define a map taking any ramified point to itself, and any non-ramified point to the other point in the preimage

$f^{-1}(f(p)) = \{p,q\}$ and $p = q$ exactly when $p$ is a ramification point

user147690

I can't give a relevant opinion unfortunately. Australia has endless opportunities to get into great universities with no risk and hence olympiads matter not at all.

Balarka Sen

9:26 AM

@SohamChowdhury are you going to have fun doing olympiad math? if not, don't.

user147690

We have the 25th best math uni in the world and you can get into it with above average grades from highschool

Balarka Sen

envies @AlexClark

JC574

so $p \mapsto q$ satisfies the composition properties we want for $\sigma$

OH

I get why it's holomorphic

I think..

sorry guys

cya

user147690

@BalarkaSen My uni has the 4th best psychology degree in the world, and you can get into it with literally average grades (OP 13 out of 25)

Soham Chowdhury

@AlexClark well, not from outside Straya, I'd guess

user147690

9:29 AM

@SohamChowdhury Well you can, but you would need to pay heaps of money up front.

user147690

Half of my university are Asian students

Soham Chowdhury

envies @AlexClark

I'm a poor Asian, which is one of the best things ever to be, haha.

And living costs are high, I've heard.

user147690

Living costs are high and my family were in the bottom 5% financially I would imagine

Soham Chowdhury

Hm.

I've been very worried about college recently, actually. :(

Balarka Sen

don't worry about it. the unis in kolkata are slowly progressing on math, @Soham

ISI's got a pretty good number theorist recently, for one.

she solved the very old open problem of Zariski, I heard

this. I am not surprised that there is a counterexample : it's false in the standard topological setting with R^n.

JC574

9:38 AM

that's cool

Balarka Sen

sure is

JC574

thanks for linking that!

Balarka Sen

(the counterexample with R^n involves Siefert surfaces, iirc)

@JC574 no prob

Remember me

@BalarkaSen you don't like olympiads??

Balarka Sen

nah

Remember me

9:42 AM

Why??

Balarka Sen

bunch o' ad-hoc problems

2

Soham Chowdhury

@BalarkaSen agreed

Remember me

I didn't get you

Balarka Sen

google the meaning of ad-hoc

...

Remember me

ad-hoc?

user147690

9:43 AM

@Rememberme They are mechanical techniques to memorise essentially

Remember me

Ahh... Yes that is something I hate....

Well @AlexClark hi!! I am at last 15 today :p

user147690

@BalarkaSen How do I show that every maximal ideal of the ring of continuous functions from the interval $[0,1]$ to $\Bbb R$ is of the form $M_c = \{f\in R|f(c)=0\}$ for some $c\in[0,1]$?

user147690

@Rememberme Oh Happy birthday! You are old now.

user147690

Hey @Ashwin

Balarka Sen

@AlexClark ew

user147690

9:47 AM

@BalarkaSen I have shown that $M_c$ is a maximal ideal through the first isomorphism theorem of rings

user147690

But I don't know how to extend that to every ideal

Balarka Sen

dunno, don't care

user147690

Are you alright?

Anonymous

@AlexClark My friend!!!!

Balarka Sen

ring of continuous functions is a sad ring

2

user147690

9:47 AM

@Ashwin How are you doing man?

user147690

@BalarkaSen Unfortunately it is on my assignment, even though it does indeed seem sad

Anonymous

@AlexClark My data package might end at any moment LOL

user147690

@Ashwin Date package?

user147690

Oh shittt, that sucks

user147690

@Ashwin When will you get to recap?

user147690

9:49 AM

@BalarkaSen Is this less terrible? Showing $M_c$ is not finitely generated

Balarka Sen

less terrible, sure, but still pretty sad

Anonymous

@AlexClark Anytime and I got enough money for the recharge xD

user147690

@Ashwin Next semester hopefully I'll be TA'ing, then I'll have money :). I'll be the wealthiest I have ever been xD

Anonymous

@AlexClark I need a party,ok?:p

user147690

@Ashwin :). When is your birthday?

Anonymous

9:51 AM

@AlexClark 7 September

JC574

Is there some way of using limits of polynomials to do it? that'd be cool

user147690

@Ashwin Maybe I can send you something :P. If I am tutoring I'll have money at that point xD

Anonymous

@AlexClark nwm,I don't celebrate my birthdays

user147690

Nor do I

user147690

But other people celebrate them for me hahaha

Anonymous

9:52 AM

@AlexClark Every day is as insignificant as other days

user147690

Indeed^

user147690

Except the day all my assessment is done hahaha

JC574

@AlexClark maybe this will work:

set $e_c : C[0,1] \to \mathbb{R}$ to be evaluation at $c$

user147690

@JC574 It is meant to be an algebra problem, but it seems it has to be solved with compactness

JC574

now $\ker e_c$ is what we want to be maximal

user147690

9:55 AM

Wait is this for not finitely generated or for showing all ideals are of that form @JC574?

JC574

no sorry

user147690

@JC574 Yep that was how I proved $M_c$ was maximal

JC574

cool

umm now..

user147690

$R/\ker e_c \cong \Bbb R$

Balarka Sen

ideals of $\mathcal{C}([0, 1])$ are pretty mucked up

JC574

9:57 AM

how does the other way work when we do it in commutative algebra

user147690

@BalarkaSen Well the maximal ones are all of that form ᕙ༼ຈل͜ຈ༽ᕗ

JC574

we have $C([0,1]) = \ker e_c \oplus \mathbb{R} $ don't we?

Anonymous

@AlexClark I dunno how much I am studying these days,but my freestyle soccer skills are exponentially increasing.

user147690

@Ashwin Well that is something to be happy about - I don't know how much I am studying now either, since I stopped recording finally

user147690

@JC574 Hmmm

Balarka Sen

9:59 AM

no, I mean, we have so much ideals that that quotient-is-field condition doesn't help

JC574

ah fair enough

user147690

@JC574 I think by the third isomorphism theorem

Balarka Sen

@JC574 you claim that $0 \to \text{ker} \to \mathcal{C}([0, 1]) \to \Bbb R \to 0$ splits?

JC574

oh i dunno

Balarka Sen

because of the natural vector space structure?

JC574

10:00 AM

f(x) = (f(x) - f(c)) + f(c), and the intersection has to be trivial right?

Balarka Sen

you might be right, I am just not thinking hard enough.

it probably does indeed split

JC574

I think it does

wait do we want any ideal not finitely generated?

isn't there an easier way to do this by some ascending chain or something

user147690

I have two problems

Anonymous

@AlexClark See you soon!

Balarka Sen

@iwriteonbananas did you see what I pinged you?

user147690

10:06 AM

One was showing that all maximal ideals of the ring of continuous functions from $[0,1]$ to $\Bbb R$ are of the form $M_c$ for some $c\in[0,1]$

user147690

@Ashwin See you later!

user147690

The second was showing that $M_c$ is not finitely generated

user147690

$M_c=\{f\in R|f(c)=0\}$

user147690

I've shown $M_c$ is maximal

user147690

Now I just need to show that all maximals are of the form $M_c$

user147690

10:07 AM

I am reading this atm:

user147690

3

Q: Maximal ideals in the ring of real functions on $[0,1]$

Assume $S$ to be all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$? Thanks in advance.

JC574

oh that's very nice

yeah I was trying to think how to get a common root, should have tried to use compactness like you said

Balarka Sen

you misunderstand me. we have proved that $H_i(X \times S^n) \cong H_i(X) \times H_{i-n}(X)$.

user147690

@JC574 I don't actually understand the answer yet actually haha. I'll read it again in a min, just need to stretch my legs

Balarka Sen

so what's special about the index of the homologies in the direct product copy?

one is $i$, the other is $i - n$, right?

iwriteonbananas

10:11 AM

well

if $i<n$, it's just $H_i(X)$

Balarka Sen

yes, under the convention that $H_{-\text{blah}} = 0$

iwriteonbananas

the other special case is $H_n(X) \oplus \Bbb{Z}\approx H_n(X\times S^n)$

@BalarkaSen is that a convention?

Balarka Sen

now you're just nitpicking irrelevant details :P

iwriteonbananas

:P

Balarka Sen

anyway, have a look yourself : en.wikipedia.org/wiki/K%C3%BCnneth_theorem

this is called Kunneth formula. I think hatcher proves it for CW-complexes geometrically somewhere on the extra chapters in cohomology.

iwriteonbananas

10:15 AM

oh, cool

Balarka Sen

but there is a general nonsense proof using $\otimes$-structure on $\mathsf{Ch}_\bullet$

iwriteonbananas

meh, ok. i dont really understand much on that wiki page

the exercise is cool though

i need to study degree stuff now

Balarka Sen

@iwriteonbananas most of the wiki pages are incomprehensible. have a look at Hatcher if you're interested

degree theory is good stuff

user147690

@Balarka In a UFD, such as $\Bbb Z[x]$, all prime elements are irreducible and vice versa, so I have proved that $<f>$ where $f$ is irreducible is a prime ideal, so then I already have $<p>$ where $p$ is a prime element is a prime ideal right

iwriteonbananas

@BalarkaSen will do

Balarka Sen

10:19 AM

sure, @AlexClark

user147690

@BalarkaSen Oh sweet.

user147690

7 hours ago, by Alex Clark

@anon $ab\in(f)\implies ab=rf$. $f$ is irreducible. Either $a$ contains $f$ or $b$ contains $f$. So then $a=fd$ or $b=fd$ for some $d\in\Bbb Z[x]$ hence $fd\in(f)\implies a\in(f)$ or $b\in(f)$

user147690

Anon said that was fine, do you agree^?

Balarka Sen

yeah

user147690

Awesome

Soham Chowdhury

10:27 AM

@BalarkaSen there are times when one must swear, young Padawan

Balarka Sen

@iwriteonbananas I think Hopf's result is pretty fascinating.

I have never read the proof, though. Guess it involves some differential stuff.

Silent

@robjohn, how to show that there is an inductive set $S$ for which $p\notin S$ and $n<p<n+1$?

iwriteonbananas

11:02 AM

@BalarkaSen yeah, just read about it. quite cool

robjohn

11:19 AM

@Silent I have never dealt with inductive sets, sorry.

JC574

11:44 AM

When talking about Riemann surfaces we'd require isometries to be holomorphic right?

nvm

iwriteonbananas

12:15 PM

@BalarkaSen proof of brouwers fixed point theorem w/ degree theory is neat

Silent

@robjohn, ok

@robjohn, will you please let me know how to show that if $m,n$ are integers and if $m<n$, then $m+1\le n$?

Balarka Sen

12:40 PM

@iwriteonbananas which proof are you referring to, exactly?

robjohn

@Silent I'd start by showing that $n-m\gt0$ and that $n-m\in\mathbb{Z}$. Then noting that the smallest integer greater than $0$ is $1$. Thus, $n-m\ge1$ which is $m+1\le n$

:-)

Balarka Sen

@iwriteonbananas the one I know is this: extend the map $f :D^n \to D^n$ to a map $S^n \to S^n$ which takes upper hemisphere to upper hemisphere by $f$ and lower hemisphere to upper hemisphere via $-f$. this is nullhomotopic, so has degree $0$. this means $f$ has a fixed point. image of $f$ on the lower hemisphere is nullset, so the fixed point must be attained somewhere in the upper hemisphere. restrict to that disk.

JC574

you can also do it by induction (i think)

@Silent

Silent

12:55 PM

@robjohn thank you so much!! :)

fubal

1:32 PM

a book of mine states $\mathbb{E}_x[N(y)] = \sum_{k=1}^\infty P(N(y) \ge k)$ for $N(y) = \sum_{n=0}^\infty 1_{X_n = y}$

can someone explain to me why that holds?

ok I see that you can write $P(N(y) \ge k) = \sum_{j=k}^\infty P(N(y) = k)$ and then reorder it to get the normal representation of $\mathbb{E}[N(y)]$

is there some way to generalize this? because the book author (Durret) doesn't say anything when doing this, so it seems to be something used often?

evinda

1:45 PM

Hi :)

Could you take a look at my question?

1

Q: Find solutions of the differential equation $3x^2y''+5xy'+3xy=0$.

Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$. That's what I have tried: Since $x>0$ the differential equation can be written as follows. $$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$ $$p(x)=\fra...

differential-equations

Chris's sis

2:22 PM

@robjohn I just finished the alternating version of the series problem I showed you these days and you solved. Its closed form is very nice too, simple.

It's also interesting to see that some of the approaching ways lead to hard-to-calculate series and integrals. The experience has the last word, indeed.

evinda

2:55 PM

@robjohn Hi... Are you familiar with solutions of differential equations in the form of a Frobenius series?

I have this question:

0

Q: Solution of differential equation - We find only one

I want to find all the solutions of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n, x>0 (m \in \mathbb{R})$ of the differential equation $x^2 y''+ xy'+x^2y=0$. I have tried the following: Since $x>0$ the differential equation can be written as: $$y''+ \frac{1}{x}y'+y=0$$ $$p(x)=\frac{1}{x}, \...

differential-equations recurrence-relations power-series

Soham Chowdhury

3:18 PM

@BalarkaSen I saw a very, very similar proof of this in a biography of Dirac.

Chris's sis

I also finished the quadratic version of that series. So, I have a whole family at the moment.

(My creativity is epic these days - maybe it's time to publish some new paper)

Soham Chowdhury

@Chris'ssis that is a great, if rare, feeling

Chris's sis

@SohamChowdhury If you attend integrals, series and limits and work on them very (extremely) hard, do research, and you're ready to dive into the craziest stuff, that feeling just comes to you after a while. :-)

Soham Chowdhury

not just with these.

Chris's sis

@SohamChowdhury Definitely, you're right! :D

Soham Chowdhury

3:29 PM

the feeling of being really creative is rare and super-enjoyable.

and it lasts for a few hours at a time, in my experience.

Chris's sis

@SohamChowdhury Yes, it is! I mean it makes you feel so fulfilled!

@SohamChowdhury True! If I manage to find something interesting I simply wake up the next day so happy, full of joy, ready to retake the research. I even have dreams about that.

Andrew Thompson

Are all discrete subgroups of (R, +) classified (up to isomorphism)?

Nevermind, I'm an idiot.

Chris's sis

@robjohn again, just to point out this thing, I'm shocked and overwhelmed by the beauty of the alternating series.

The date with such a beauty might require some recovery time aferwards. :-)

TOO NICE TO BE REAL!

Balarka Sen

3:58 PM

@SohamChowdhury yeah, well, it was in Perelman's "what is mathematics?", one of the first book of mathematics I ever read.

but I don't care about these puzzles anymore

robjohn

4:42 PM

@evinda have you tried plugging in the series to the differential equation and setting the coefficients to $0$?

@Chris'ssis Great! I'd like to see it sometime.

Chris's sis

@robjohn Ah, I think I didn't send you the solution to the version you solved (calculated).

robjohn

@Chris'ssis I have something from a few days ago.

Chris's sis

@robjohn Yeah, but that one is the solution to the Knuth's problem.

@robjohn Just a few seconds.

@robjohn Sent in db.

robjohn

@Chris'ssis Got one that is dated yesterday. I will look at it later. We are cleaning and organizing our house (every Saturday) -_-

Chris's sis

@robjohn Yeap. OK. Look at it whenever you want to (the solution is just straightforward). :-)

robjohn

4:49 PM

@Chris'ssis Thanks for sending it. I won't be back on line much before dinner time.

Chris's sis

@robjohn My pleasure. OK. I'll also be away for some hours. :-)

iwriteonbananas

4:59 PM

@BalarkaSen exactly, up to minor modifications, that's the proof i was referring to

evinda

@robjohn I try to now. we have the sum $\sum_{n=2}^{\infty} \frac{(-1)^n x^{2n}}{4^3 \prod_{j=3}^n (2j)^2}$. If we differentiate it, will we write the sum from 3 or isn't there a problem if we write it from 2?

Gato

@robjohn Hi, do you how can a find a function (necessarily not continuous ) $f$ defined on $\Bbb{R}$ such that $f\circ f(x)=-x$

evinda

5:17 PM

@robjohn I found that $x^2y''=-\frac{a_0}{2} x^2+ a_0 \sum_{n=2}^{\infty} \frac{(-1)^n 2n (2n-1) x^{2n}}{4^3 \prod_{j=3}^n (2j)^2}$

$xy'=-\frac{a_0}{2} x^2+ a_0 \sum_{n=2}^{\infty} \frac{(-1)^n (2n) x^{2n}}{4^3 \prod_{j=3}^n (2j)^2}$

$x^2y=x^2 a_0-\frac{a_0}{4} x^4+ a_0 \sum_{n=2}^{\infty} (-1)^n \frac{x^{2n+2}}{4^3 \prod_{j=3}^n (2j)^2}$

@robjohn Have I done something wrong? Because when I calculate the sum of these three terms, I do not get 0...

Mike Miller

@PaulPlummer: I just read your posts. They're very fun. I liked your most recent one more, but maybe that's just my taste in games.

As a note it took me a while to realize why your goal was to show that "the group is $C'(1/6)$: I read that as the subgroup of $F_2$ (which of course doesn't really parse) rather than the group with presentation $\langle x, y \mid a_1, a_2, \dots \rangle$.

Does player 2 win for every finite group? How about for the game where you replace normal generators with generators?

Paul Plummer

Ah Thanks, I will take a look at it, and try to make it more clear. Player 2 does win for every finite group, including replacing with generating sets. Basically you will always choose an element to make your group larger (as otherwise every element in the set being presented would be an element of the group you can generate, contradicting the set generating the group) @MikeMiller

Mike Miller

Ah, right. This is morally the same as your $\Bbb Z$ example. Nice!

(The complaint I had about your goal of showing that player I wins $F_2$ could probably just be solved by saying "We will prove otherwise by showing that $\langle x, y \mid a_1, a_2, \dots \rangle$ is $C'(1/6)$")

Paul Plummer

More generally, this take into account finite groups, and infinite groups like $\mathbb Z$, if every strictly acending chain of infinite index groups is finite then you eventually get to a finite index subgroup, and you can always start widdle down the index @MikeMiller

Mike Miller

I gotcha, that's great

Paul Plummer

5:28 PM

So if player two can win, play two will actually be "done" in finitely many innings, at least for finitely generated groups

Paul Plummer

5:49 PM

I think the interesting thing about the previous post is that it shows whether or not there is a winning strategy for that game is independent of the usual axioms. But I could see how the game is not everyones cup of tea. And I did update it. @MikeMiller

Mike Miller

I found it fascinating! I was just saying I liked the new one even more.

evinda

Can we differentiate $L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})$ as follows?

$$\frac{d}{dx} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+\frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})$$

$$\frac{d^2}{dx^2} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+\frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})$$

Or have I done something wrong?

Paul Plummer

Well that is good to hear. This one has been on my mind for a while

Balarka Sen

6:25 PM

@MikeMiller assuming you're not busy, have you see the message on bordism homology I pinged you?

@iwriteonbananas yeah, it's a cool one.

Mike Miller

No, I haven't.

Balarka Sen

this

I haven't checked if the dimension axiom holds, admittedly.

Pretty sure excision does (which, combined with long exact sequence, would give the Mayer-Vietoris axiom).

Mike Miller

Change that to say "Compact manifolds", and you mean because boundaries are closed, not because boundaries are compact. Your first description doesn't give you anything I recognize. Your second one needs to be modified in some way, I think, because I don't see why $[X]+[Y]$ is going to be $[X \sqcup Y]$.

The dimension axiom shouldn't hold. If you're trying to mimic cobordism groups, why would it?

Balarka Sen

Well, I have no rigorous proof that it is indeed cobordism theory. I said "haven't checked if the dim axiom holds" not because I want it to hold, but use it as a litmus test that my theory indeed has higher probability of being cobordism theory.

@MikeMiller yeah, whoops

typo

Mike Miller

If you set this up so that $[X]+[Y] = [X \sqcup Y]$ you've almost just written down the definition of bordism groups.

Balarka Sen

6:33 PM

? last time I talked with you about this, I think you told me that you don't know if bordism groups can be realized as homology of any chain complex

Mike Miller

Of anything interesting. Let $\Omega_n(X)$ be the $n$th unoriented bordism group of $X$. Then $\Omega_n(X)$ is the $n$th homology of the chain complex $\Omega_*(X)$ with trivial differential.

Balarka Sen

ok, I get you :P

evinda

Could you take a look at the edit part of my question?

1

Q: Differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$

The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given. Show that the equation has $0$ as its singular regular point . Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$ Show that if $a=n$, where $n \in ...

differential-equations

Balarka Sen

@MikeMiller well, if it's indeed the bordism groups, then proof of bordism theory being a (generalized) homology theory gets not-so-hard.

i bet homology of (nice?) chain complexes always satisfy the mayer-vietoris axiom

Mike Miller

How so?

Uh, what does that even mean? What does it mean that a random chain complex "satisfies the Mayer-Vietoris axiom"?

Balarka Sen

6:42 PM

*homology of

Mike Miller

That doesn't help.

Balarka Sen

well, ok, let's start with something simpler.

homotopy invariance, say.

(duh : of course satisfying the mayer vietoris axiom for hmlgy of a random chain complex doesn't make sense. you need a functor Top --> Grp first)

so, ok, to prove that it's a homotopy invariant, we need to show that an htpy equiv $f : X \to Y$ induces a htpy equiv at the chain complex level.

let $f, g : X \to Y$ be homotopic. if we can prove that $f_\#, g_\#$ at the chain level are homotopic, then we are done.

we can just mimic the usual prism operator, I guess.

right, $\sigma : M^n \to X$ be a singular manifold, and $F : X \times [0, 1] \to Y$ be a homotopy between $f, g$. then $F \circ (\sigma \times \text{id}) : M^n \times [0, 1] \to Y$ is indeed a map from a manifold-with-boundary to $Y$

linearly extend to get a map $C_n(X) \to C_{n+1}(Y)$. i haven't checked, but surely it's a chain homotopy

guess i'll just write this up somewhere instead of filling the chat with nonsense

Mike Miller

6:58 PM

Suppose your groups are, or are appropriately modified to be, the unoriented bordism groups. Then any proof you provide here will be trivially translatable to a proof without mentioning your chain complex because the definition you provide is essentially the same thing. e.g. as you noticed above for homotopy invariance.

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