Mathematics - 2015-08-31 (page 3 of 3)

8:00 PM

Cellular approximate $\Bbb RP^2 \to \Bbb CP^\infty$ to get a map $\Bbb RP^2 \to \Bbb CP^1$. By a similar argument, this is well-defined on homotopy classes. The proves surjectivity.

Daniel Fischer

@Moses Yes, if you have $(E_1 + E_2)X = X$, then you must have $u = I_X$.

Mike Miller

Ok, you're done. Let $X$ be an $n$-dimensional CW complex, $Y$ a CW complex, no other conditions. Does this tell you anything general about $[X,Y]$?

Balarka Sen

oh, of course you are right.

I think we need $Y^{(n+1)}$. 'Cause we need to cellular approximate a map from $X \times I$, which is $(n+1)$-dimensional.

So my claim would be $[X, Y] = [X, Y^{(n+1)}]$.

Mike Miller

Correct. The map $[X,Y_{n+1}] = [X,Y]$. And what can you say about the map $[X,Y_n] \to [X,Y]$?

Moses

@DanielFischer One more thing. Where did you get $E_{k}X = E_{k}Y$ in your explanation?

Owatch

8:06 PM

Hm, how might you factor $3x^{2}-2$?

Daniel Fischer

@Moses From $Y = u(X)$ and $E_k = E_k \circ u$.

Mike Miller

Oops. The map ... is a bijection is what I meant to say.

Balarka Sen

huh?

Daniel Fischer

@Owatch Over $\mathbb{R}$: $(\sqrt{3}\cdot x - \sqrt{2}) (\sqrt{3}\cdot x + \sqrt{2})$. Over $\mathbb{Q}$ it's irreducible.

Balarka Sen

oh, you wrote down $=$ when you wanted to specify that map is a bijection. That's fine, I get it.

Mike Miller

8:09 PM

Anyway, quick answer for my question there?

Owatch

@DanielFischer Thanks. What is Q?

Balarka Sen

OK, it's surjective.

Mike Miller

Good. Now assume all n-manifolds are n-dimensional CW complexes and prove the Hopf degree theorem.

Daniel Fischer

@Owatch The field of rational numbers.

Owatch

oh

Balarka Sen

8:11 PM

@MikeMiller <--- dunno what degree of a map between manifolds is yet.

know it only for spheres.

Owatch

@DanielFischer I'm trying to solve $\int{\frac{1}{3x^{2}-2}}dx$ using partial fractions.

But I guess it won't work.

As it's a technique for solving rational expressions

And there are no rational roots?

Balarka Sen

(heh. I am learning some good homotopy theory for free)

Daniel Fischer

@Owatch Why wouldn't it? It's a rational function that you want to integrate.

Owatch

Alright alright...

Mike Miller

@Balarka: $H^n(M) = \Bbb Z$ for an oriented manifold $M$. Degree is what number you're multiplying by.

Balarka Sen

8:16 PM

Ah, yeah, I get it (plus, I know the proof of that now).

Mike Miller

Ok, so prove the Hopf degree theorem.

Balarka Sen

Also, what was the statement of the Hopf degree theorem again?

Any two map between finite dimensional manifolds are homotopic if they have the same degree?

Mike Miller

Maps from an oriented n-manifold to $S^n$ are homotopic iff they have the same degree.

Absolutely not between manifolds.

You can disprove that after you're done with Hopf.

Balarka Sen

hm, ok.

Would it be ok if I try that tomorrow morning?

Mike Miller

Sure.

Balarka Sen

8:20 PM

OK. Thanks so much for all this, @MikeMiller.

Moses

@DanielFischer What is the significance of these two decomposition results?

Daniel Fischer

8:36 PM

@Moses Basically, that you can split $a$ into parts corresponding to the components of the spectrum, and those are independent. So when proving things, you can (often) assume that the spectrum is connected.

Balarka Sen

8:47 PM

@MikeMiller Out of curiosity, is there any direct way to compute $[\Bbb RP^2, S^2]$ (and thus $[\Bbb RP^2, \Bbb CP^\infty]$), without resorting to the bijection with second integral cohomology group?

Hmm. Actually, why can't we use the Hopf theorem here?

Eck, $\Bbb RP^n$ is not oriented. Oh well.

Wait, that's actually the same thing.

Hopf's theorem asserts that for oriented $n$-manifolds $M$, homotopy classes of maps $M \to S^n$ are in bijection with $H^n(M; \Bbb Z)$.

Mike Miller

Well, it's more specific than that. What you said is just that they're countable. (Of course, this is sort pedantic, and I know what you mean.)

@BalarkaSen: That's not what we proved.

Balarka Sen

oops, sorry

@MikeMiller Yeah, the map is specified in Hopf's theorem.

It sends a homotopy class of maps to the degree of a representative.

Mike Miller

Right. Anyway, can you fix your proof?

Balarka Sen

I'm trying this right now.

It'd have been great if some analogue of that theorem worked for all manifolds, in say, $\Bbb Z_2$ coefficients.

Mike Miller

It's just absurdly completely false for manifolds in general. Not even close to true. Out the window silly false.

You can make a version of it work for $M \to S^n$, where $M$ is a non-orientable closed $n$-manifold.

Balarka Sen

9:00 PM

Hmm. I don't know how to incorporate the piece of information that the bijection map is degree in whatever proof I come up with. It's best if I think about this tomorrow, now that I have a good (?) idea.

@MikeMiller ~~oh?~~ which is?

Mike Miller

Meh, you'll get it.

You shouldn't need so much work to fix your argument.

Balarka Sen

still, I don't want to think about it right now. it's midnight here.

Mike Miller

ok

Kevin Driscoll

@robjohn I've started reading the Arxiv every day now to keep up with papers appearing in my field and the ChatJax script is really improving the quality of my life. People put LaTeX in their abstracts but its not rendered on the Arxiv pages, so today I just tried starting ChatJax and it renders all the Tex, making things much more readable

Chris's sis the artist

@robjohn I'll also add to my book this particular problem: calculate in (a reasonable) closed-form $$\int _{\large \gamma}^1\int _{\large \gamma}^1\frac{1}{(x+y) (1+x y)}\ dx \ dy$$

Moses

9:13 PM

@DanielFischer Lastly, just to confirm one thing, when we say that $\mathcal{A} \subseteq \mathcal{L}(X)$ how do you define the mapping $E_{1}$ where $E_{1} \in \mathcal{A} \subset \mathcal{L}(X)$?

Daniel Fischer

@Moses $\mathcal{A}\subset \mathcal{L}(X)$ means every element of $\mathcal{A}$ is an element of $\mathcal{L}(X)$, so it is a (linear) mapping from $X$ to itself.

Moses

@DanielFischer So you aren't using the result that every $\mathcal{A}$ has an isometric embedding into some $\mathcal{L}(X)$? Which we define as let $X = \mathcal{A}$ and $L_{a}b = ab$ for every $b \in \mathcal{A}$?

Daniel Fischer

@Moses We started from the specific situation that $\mathcal{A} \subset \mathcal{L}(X)$. No need for abstract theorems that we can embed.

Kevin Driscoll

@Chris'ssistheartist Do you check whether a CAS like Mathematica or Maple can do these problems before working on them, or is it irrelevant to you?

Chris's sis the artist

@KevinDriscoll To be honest I care so less what Mathematica, Maple can do. :-) I'm on them (these integrals) even id they (CAS) fail again and again.

robjohn

9:25 PM

@Chris'ssistheartist Does that have a nice closed form? I reduced it to a single integral and Mathematica spat back some pretty ugly formula.

Chris's sis the artist

@robjohn Of course. :D

Kevin Driscoll

@robjohn Yea mine gave it in terms of derivatives of the gamma function, I think

robjohn

@KevinDriscoll Yeah... there were Digammas in several places.

Moses

@DanielFischer I'm just trying to figure out (and I might be missing something simple) how composition and multiplication seem to be used interchangeably like when you showed that $(1-E_{1})(E_{1}X) = (E_{1}-E^{2}_{1})(X)$?

robjohn

@Chris'ssistheartist Oh, you've actually put the same limit for both integrals... and I assume that $\gamma$ is the Euler-Mascheroni constant?

Chris's sis the artist

9:29 PM

@robjohn Yes! :-)

Kevin Driscoll

@robjohn I tried to make Mathematica do the double integral directly (I didnt' question whether or how that works in this case) and it gave me back a sum of 2 PolyLog functions

Daniel Fischer

@Moses Because composition of maps is the multiplication in the algebra $\mathcal{L}(X)$.

Kevin Driscoll

With a numberical value around 0.072

Daniel Fischer

But of course, you can also take an abstract Banach algebra and then identify each element with the corresponding left translation, @Moses.

Moses

@DanielFischer Bare with me, I only started this recently. But I thought that question might be dodgy :)

@DanielFischer Yes that's what I have seen. The translation stuff.

Daniel Fischer

9:32 PM

@Moses I might bear with you, but I certainly won't bare with you ;)

Moses

@DanielFischer sigh We use 'bare' here cough

Daniel Fischer

@Moses So … you're not from Greenland, I suppose?

Balarka Sen

@DanielFischer What does it mean to "bear with someone"? Have a glass of bear with that someone, tolerate that someone, or handing that someone with a teddy bear, or does it mean to let your pet bear bearpile that someone?

English is so confusing.

Moses

@DanielFischer If they reject social conventions such as proper use of the English language...then quite possibly.

Daniel Fischer

@Moses It's just too fricking cold to take off the clothes there.

Moses

9:41 PM

@BalarkaSen What's your first language?

Daniel Fischer

@Moses Like everybody: unarticulated cries when hungry.

Balarka Sen

lol

Chris's sis the artist

@KevinDriscoll @robjohn $$\int _{\large \gamma}^1\int _{\large \gamma}^1\frac{1}{(x+y) (1+x y)}\ dx \ dy= \chi_2\left(\left(\frac{1-\gamma}{1+\gamma}\right)^2\right)$$

Moses

:)

Balarka Sen

@Moses Something pretty unheard of.

Mike Miller

9:43 PM

@BalarkaSen: It's beer, not bear. A glass of bear would probably be gross.

Balarka Sen

Oh, fair enough, haha.

Moses

@BalarkaSen What's it called?

Balarka Sen

Bengali.

Moses

@DanielFischer Look to your right. I got no credit for the bear comments. What's up with that?

Balarka Sen

Arrow-pings don't stick to starred comments.

Daniel Fischer

9:47 PM

@Moses Starfleet is elsewhere currently.

Moses

@BalarkaSen Keel. What are your feelings regarding the Bengal tiger?

Balarka Sen

@Moses That I'd never want to face one of those...?

Moses

Correct

@BalarkaSen You watch cricket?

Balarka Sen

@ "feeling" : I certainly wouldn't cry my eyes out because we have only 1500 of those remaining or something.

@Moses Sure. We have lots of them in our garden.

Moses

@BalarkaSen You watch cricket?

My chat is rejecting and accepting comments, hence the repeat.

Mike Miller

9:51 PM

@DanielFischer Ah, that's my only language.

Balarka Sen

I'm perfectly willing to watch crickets, if only I'm given a pair of earplugs.

lol, @Mike.

Moses

Good day/night to all

Daniel Fischer

Have a diaphanous nightie.

dREaM

10:10 PM

@Rememberme what's up.

@AlexClark See this link: math.stackexchange.com/questions/1415418/…

Mike Miller

@dREaM: He was explicitly considering $G$ and $A_7$ as a subgroup of $S_7$.

Did he ask about the general case?

Balarka Sen

@MikeMiller Arbitrary thought of a sleepy person : We have proved that cohomology is dual to homotopy. This should somehow mean that (a dualized version of) homological algebra can be used to study homotopy types too, right? What I am saying is that, there must exists something at a general nonsensical level which puts homotopy theory, cohomology theory and homology theory at the same footing.

Mike Miller

No. Go to bed.

See here. Then go to bed.

Balarka Sen

I'm going, I'm going. But whatever you say, I'm gonna think about this. I don't know yet, but I believe there's something nontrivial at play.

Mike Miller

Read the thing I linked.

Balarka Sen

10:25 PM

Heh. I don't understand a thing.

I'm going to bed.

Mary Star

10:40 PM

We have the following Lemma:
"If the characteristic of $F$ is other than $2$, then the solutions of $$X^2-(t^2-1)Y^2=1$$ with $X$ and $Y$ in $F[t]$ are given by $$(X, Y)=(\pm x_n, y_n)$$ where $$x_n+\sqrt{t^2-1}y_n=(t+\sqrt{t^2-1})^n$$ fr $n$ in $\mathbb{Z}$. "

Can we write each element $x$ of $F[t, \sqrt{t^2-1}]$ in the form $x=a+b\sqrt{t^2-1}$ where $a,b \in F[t]$ due to the lemma? Or is this always true?

anon

10:50 PM

the lemma is unnecessary to prove every elt of $F[t,\sqrt{t^2-1}]$ is of the form $a(t)+b(t)\sqrt{t^2-1}$

show that numbers of that form are closed under multiplication and addition, and that t and sqrt(t^1-1) are both of that form

Chris's sis the artist

@KevinDriscoll @robjohn I'm also pretty interested in that integral but in 3 dimensions. Not sure yet what it yields.

anon

or you could argue any monomial in t and sqrt(t^2-1) reduces to a power of t times a power of (t^2-1), possibly then times sqrt(t^2-1). these monomials are of the given form, which are closed under addition.

robjohn

@Chris'ssistheartist What is $\chi_2$?

Chris's sis the artist

@robjohn Legendre's chi function - mathworld.wolfram.com/LegendresChi-Function.html

robjohn

@Chris'ssistheartist Well, if you drag enough special functions into many problems you get a nice form. :-)

Chris's sis the artist

10:53 PM

@robjohn lolll, that's true. :-)))

Mary Star

We have that $t=t+0\sqrt{t^2-1}$ and $\sqrt{t^2-1}=0+\sqrt{t^2-1}$.
We also have that
$+ \ \ : \ \ x_1+x_2=a_1+b_1\sqrt{t^2-1}+a_2+b_2\sqrt{t^2-1}=(a_1+a_2)+(b_1+b_2)\sqrt{t^2-1}$

$\cdot \ \ : \ \ x_1\cdot x_2=(a_1+b_1\sqrt{t^2-1})(a_2+b_2\sqrt{t^2-1})=a_1 \cdot a_2+a_1 \cdot b_2 \sqrt{t^2-1}+a_2 \cdot b_1\sqrt{t^2-1}+b_1 \cdot b_2 (t^2-1)=(a_1 \cdot a_2+b_1 \cdot b_2 (t^2-1))+(a_1 \cdot b_2+a_2 \cdot b_1)\sqrt{t^2-1}$

Is this correct?

So, we conclude that each element of $F[t, \sqrt{t^2-1}]$ can be written in the form $a(t)+b(t)\sqrt{t^2-1}$, right? @anon

anon

yes

I Want To Remain Anonymous

@PedroTamaroff How do you take notes? It seems you take a lot...

eaglgenes101

um

someone that knows something about numerical integration here?

GBeau

I keep having to double-take when I read a set is "finite" after introduction of internal set theory

Pedro Tamaroff

11:40 PM

@IWantToRemainAnonymous Yes. I usually pick a book and flesh out what seems relevant.

I Want To Remain Anonymous

@PedroTamaroff Uh, everything that's relevant? Like all the def, proofs, motivation, intuition,...?

I Want To Remain Anonymous

11:58 PM

Wow

Pedro Tamaroff

@IWantToRemainAnonymous Usually definitions, proofs, and perhaps some discussion. But this last thing rarely, unless it is too intricate that I cannot remember.

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