@philmcole Yeah, you're gonna want either the entirety of $\Bbb R$, or some ray like $[0,\infty)$

In any case I'm done, and the only non-elementary thing I needed was that it alternates between the positive and negative x-rays finitely many times, which you can prove using compactness somehow I think

I mean your proof is really ad hoc because you're assuming it hits finitely many times on either axis

@BalarkaSen No, it alternates between them finitely many times

yeah that's what i meant

It can hit each one infinitely many times

i don't need that

Yeah but it's true for all paths

i can just floop make it transverse

Yeah but you need differential topology for that or something, no?

Hm.

I guess I don't know a super elementary proof

Uh wait what's the name of the thingy where you have an open cover of a compact thing and there's an epsilon-ball around everywhere that's contained in one of the cover thingies

I guess I need that

Lebesgue covering lemma?

Like, a loop is a function from $[0,1]$ to stuffs

and the plane minus the rays gives us an open cover with two open sets

Wait, so what you're doing is (under your assumption), that you have an alternating sum of +1's and -1's where each sequence of +'s and -'s is finite?

so preimage everything to $[0,1]$, do Lebesgue to find $\epsilon$, break $[0,1]$ into $>1/\epsilon$ pieces, and unpreimage everything

@BalarkaSen Yeah I guess

I am not sure why that is in general finite. 1-1+1-1+1+1-1+1+1+1-1+...

Am I dumb?

and then you have your loop broken into finitely many pieces, each of which avoids one ray

@BalarkaSen There are finitely many pieces

Oh, you're also using that

i seeee

Another way of saying the hypothesis is that I can break the loop into finitely many pieces, each of which avoids one ray

And I can prove that hypothesis with Lebesgue

Wait. What about the Hawaiian earring like I said? You're changing the x-axis to something else?

@BalarkaSen Hm?

So you've moved the earring so that the bad point is not the origin

Just make one piece contained in a small neighborhood of the bad point

Ahh

Alright

It will avoid either the +ve or -ve x-ray

If the bad point isn't on the x-axis it could even avoid both

Right, that's what I meant

You're pushing the H ring upwards a bit

Should I bite into this lemon?

When life gives you lemons you bite into it

Orange-y

Oranginal joke

[insert standard Cave Johnson quote here]

Is that how you could do $\pi_2(S^2)$ as well, I wonder

And by $S^2$ I mean $\Bbb R^3\setminus\{p\}$, but same difference

You'd end up computing $H_2(S^2)$ I think

Oh, fair

Or maybe even $H^2(S^2)$, 'cause forms are cohomology somehow

Just to check: it’s only $\pi_1$ that can be nonabelian?

Yeah

Unless you do relative stuff

Uh, what's the word

Relative homotopy gorups?

I always find that a bit disappointing

Like, $\pi_2(A,B)$, is it just called relative homotopy?

Yeah that thanks

@Semiclassical It means you can't do the Pochhammer contour in $\Bbb R^3\setminus\{p,q\}$

sure

Or any analogue thereof

Equivalently, the hanging picture trick

but up a dimension

@Semiclassical It's disappointing that homotopy groups aren't bad enough?

It’d have to be by analogy since a second homotopy group wouldn’t be about loops

Well, actually, a while ago I asked about maps from general surfaces to $\Bbb R^3\setminus\{p,q\}$, that aren't nullhomotopic but become so under the inclusion of one of the points

and I don't think I got an answer

I guess. Mostly I just like that the fundamental group abelianizes to H_1

'Cause the fact that $\pi_2$ is abelian only means that you can't do that if the surface is a sphere

So you'd need to study H_2(S^2 v S^2)

I mean, all surfaces are just spheres with handles, so maybe you could push away the handles somehow or something so that they don't matter

Handles and capthings

Crosshats

And while there is a relation between higher homotopy and higher homology groups, it’s nothing so clean as abelianization

Well, the maps given by "vanishing p or q" are S^2 v S^2 --> S^2 given by pinching a sphere right

Yeah

So the maps are Z^2 --> Z, projection to one of the coordinates, on homology

The intersection of their kernel is zero

I always forget the name of the lemma connecting pi_n to H_n

Oh, upto homotopy

Not homology

We're explicitly asking for nullhomotopicity

That's harder, yeah. Hm.

I want to say Zassenhous but I’m pretty sure that’s not it

@Semiclassical Hurewicz

There we go

Who's Zassenhous?

Hurwicz vs Hurewicz is also confusing

His name reminds me of a fun story behind the name of Staten Island

Maybe I should compute $[\Sigma_g, S^2 \vee S^2]$ using the Postnikov tower construction that I learnt, uh, yesterday

The Dutch first saw it in the distance, and someone said, in a Dutch accent, "Is tat en island?"

I give up, I don't know the homotopy groups of $S^2 \vee S^2$

("Is that an island?" ...It's hard to do this via text)

RIP

So, uh, Staten Island.

I mostly know Zassenhous insofar as a formula of his shows up on the BCH page: en.m.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula#The_Zassenhaus_‌​formula

Bleh, not sure how to fix that link

en.m.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula#The_Zassenhaus_‌​formu‌​la

Thanks

BCH is handy in quantum physics

You might get some utility out of it in classical mechanics as well, but it’s invaluable for quantum

@BalarkaSen if a fibration has contractible fibers, is that enough for existence of a section?

assume orientable as well.

I can't think of a proof right now

Multifunctions (in the complex analysis sense) don't make sense in $\Bbb R^3$, I guess

Or, at least, branch points don't

Well, you could have a branch loop

btw it's still a open question, what are the homotopy group of $S^2$. As of my knowledge, a recent paper proved that there are infinitely many non-zero homotopy group of $S^2$. That might help for $S^2\vee S^2$.

Like it's a multifunction on $\Bbb R^3\setminus S^1$ or something

@BalarkaSen Hold on, say we have a torus minus a disk, like a genus-1 thingy with boundary

It's compact, so we can draw a sphere around it

or like contain it in a ball

I mean the image of a torus minus a disk in 3-space

Oh wait this isn't gonna work I think

Wait, maybe

Never mind

I have a question. This is the question for which I joined this site but I have been unable to post it because I was afraid I would be ridiculed as being completely stupid. Does anyone mind if I ask it here?

Hold on, if we crunch $S^1\vee S^1\subset T^2$ to a point, we get a sphere, right?

Where the circles are the meridian and longitude

@BalarkaSen

@Adeek when you see this: since you're starting to get into functional analysis, here's a problem I liked quite a lot. Show that the range of a compact operator between Banach spaces is closed iff it has finite rank

So what if we have a map from $T^2\to\Bbb R^3\setminus\{p,q\}$, and through a homotopy we crunch the longitude and meridian to a point

@MohammadZuhairKhan go and ask, though note that people may be caught up with things so responses aren't guaranteed

and we can do that without going through $p$ or $q$ I think?

Something about $\pi_1(\Bbb R^3\setminus\{p,q\})$ being zero, I dunno, I'm assuming it's true

So once we have that, we can factor it as a map $T^2\to S^2\to\Bbb R^3\setminus\{p,q\}$

and since we can't do our thing with spheres, we're done

(Where by "do our thing" I mean "not have it be nullhomotopic, but have it become nullhomotopic once we include $p$ or $q$")

@Semiclassical I give a pretty elementary derivation of BCH here, but I see the conversation is about Zassenhous, not really BCH.

Imagine you have a circle with radius $r$ and centre $(0, r)$. What is the volume of the solid formed when this circle is rotated $2\pi$ about the $x$-axis?

Wait, doesn't latex work in chat?

I am seeing all the $ signs!

@MohammadZuhairKhan There is a link in the room description on how to make it work

@BalarkaSen And that works for other surfaces as well by crushing other stuff

Hey Tobias!

Hi @Daminark

@robjohn neat

Anyway I almost solved it by some substitutions but I was left with a quotient. Is there a quotient rule for integration?

@AkivaWeinberger you're becoming a topological steamroller now jeez

@MohammadZuhairKhan What was the quotient?

TBH when I need to do stuff like $e^{X}Ye^{-X}$ I don’t use full BCH

I instead replace X with tX and hope differentiation with respect eventually gives a nice simplification

@Semiclassical Stein gave a hand wavy proof in class one day, but I had to write up something more concrete. That was a long time ago.

Are these noncommutative?

Yeah @akiva

I vaguely remember doing something that looked like that

@AkivaWeinberger If they commuted, you'd get the standard $e^{X+Y}=e^Xe^Y$

The hope is that repeatedly doing $Z\mapsto [X,Z]$ on $Y$ eventually gives something simple

Like $e^xEe^{-x}f(x)$, where $E$ is the shift operator, IIRC?

It was $\frac {x^4}{r-\sqrt (r^2-x^2)}

Something like that

Which simplifies to $e^xe^{-x-1}f(x+1)=f(x+1)/e$, which is actually probably not what I wanted

Maybe it was $e^xDe^{-x}f(x)$

$e^x(e^{-x}f'(x)-e^{-x}f(x))=f'(x)-f(x)=(D-I)f(x)$

so $e^xDe^{-x}=D-I$

I think that was it, yeah

$e^{yD}f(x)e^{-yD}=f(x+y)$

Or more like $e^{-x}De^x=D+I$

Where D is differentiation with respect to x

I wonder if $e^{-x}D^2e^x=(D+I)^2$

${}=D^2+2D+I$

(I might have the sign wrong)

Probably, or else I wouldn't have been messing around with it at the time

@Semiclassical Isn't $e^{yD}f(x)=f(x+y)$ just Taylor's Theorem?

Oh, wait, that's just $(e^{-x}De^x)^2$, so it's true

@robjohn Yup

Hmm, my version is a bit off then

So the point was $e^{-x}f(D)e^{x}=f(D+1)$

so it's kinda like $(Ef)(D)$ or something but that's probably abusing notation

@AkivaWeinberger Thanks. Actually, that was meant to reply to Semiclassical

typo

We always talk about differentiation with respect to some variable but why can't we ever be disrespectful?

Hmmm

@Semiclassical But, yeah, that was the one thing I did that looks like the that thing

@Daminark "Differentiating with all due respect to $x$…"

Differentiating with a middle finger to x

🖕😎🖕 zoop

I think I’m conflating how functions transform under translation with how operators transform

@MohammadZuhairKhan I imagine some trig substitution would help

@AnubhavMukherjee Yes, I think so. I remember an idea from Mike which goes as follows: Say $E$ is an $F$-bundle over $X$ with contractible $F$. Think of $X$ as a CW-complex. Then by orientability and contractibility $E$ trivilalizes over the 1-skeleton $X^1$, hence admits a section.

By induction, assume $E$ admits a section as an $F$-bundle over $X^n$. Say $s : X^n \to E$ is a section. $X^{n+1}$ can be thought as $X^n$ with $(n+1)$-cells attached to $X^n$. $E$ restricts to the trivial bundle $D^n \times F$ over each of those $n$-cells $D^n_\alpha$. The section $s$ is defined on the boundary $\partial D^n_\alpha$ which you want to extend to all of the disk.

Now the section on the boundary of the disk is nothing but an element of $\pi_n(F)$. So for each of those cells you have an element in $\pi_n(F)$; this gives a $\pi_n(F)$-valued $(n+1)$-cochain of $X^{n+1}$.

Uh, so somehow the obstruction to extending that section is about nonhomologousness of that cochain in $H^{n+1}(X; \pi_n(F))$. I don't remember why anymore.

But your $F$ is contractible, so all those groups are zero

@BalarkaSen even I forgot the exact obstruction argument...but thanks :)

So you can in fact extend $s$ so $X^{n+1}$. Induct; $s$ extends to a section of $X$ if you're a finite CW complex.

I would like to understand the proof of that one step I ignored

If you can find a reference do let me know

@Akiva Oh, interesting idea.

(Also yeah $\pi_1(S^2 \vee S^2) = 0$, so you're okay :P)

@BalarkaSen I forgot to tell you, but I got a nice answer of a question you asked (one year back) , ie. two vector bundle homeomorphic implies vector bundle isomorphic or not.

Well for the genus $g$ surface you have to write down maps from $\bigvee S^2 \to S^2 \vee S^2$ I guess

@Anubhav Indeed?

the answer is false,

Its by milnor ...

Yeah Mike posted an answer

From Milnor's microbundle paper

@BalarkaSen I know it is, I just wasn't 100% sure that that meant I could do what I wanted to do

yes....I read that paper recently...it is very interesting

@BalarkaSen You mean${}\to M_g$, no?

I didn't check mike's answer...I'll probably go and check it

But yeah, you can go from any surface to a sphere by crushing a bunch of loops

I mean, they're all just polygons with edges identified to other edges, yeah?

Crush the boundary of that polygon.

@Akiva Ah, no, I was collapsing the wrong set of loops. I was collapsing pairs (a_g, b_g) plus the loops along which the torii are summed

hello

I see what you mean

OK, that gives you a map $S^2 \to S^2 \vee S^2$

Ah, sure

and finally i can joint the chatroom

Right, and

hm, given that the original map is nullhomotopic once we put in another point (or pop one of those spheres), does that still make it true for this map?

Also @AnubhavMukherjee I didn't really need a full description of the homotopy groups of $S^2 \vee S^2$. I wanted to know the first few and for some reason thought I didn't know how to compute them. But I think I was being a dopey; there should be a way to do it by considering the long exact sequence of $(S^2, S^1)$ and using the $\pi_*(X, A) \cong \pi_*(X/A)$ thing or something, which holds given enough restriction on the connectivity of $X$ and $A$

I don't know homotopy theory :P

Like, maybe the map $f:T^2\to\Bbb R^3\setminus\{p,q\}$ composed with the inclusion map $\iota_p:\Bbb R^3\setminus\{p,q\}\to\Bbb R^3\setminus\{p\}$ is nullhomotopic,

@Akiva if f : X --> Y is null, composing it with Y --> Z is also null, right?

I mean just consider the homotopy $g \circ f_t$

Wait let me finish this

@BalarkaSen i can search for that paper and send you. you might find some interesting ideas there

they used some spectral sequence idea.

And we can write $f$ as $\pi\circ \tilde f$ where $\pi:T^2\to S^2$ and $\tilde f:S^2\to\Bbb R^3\setminus\{p,q\}$

But do we get that $f\circ\iota_p$ is nullhomotopic implies that $\tilde f\circ\iota_p$ is nullhomotopic?

Er, $\iota_p$ should be $\Bbb R^3\setminus\{p,q\}\to\Bbb R^3\setminus\{q\}$, actually; we're putting in $p$

Doesn't matter

I can't parse this. What is $f \circ i_p$?

I wrote it above

The inclusion map that includes one of the points

i_p maps to R^3-{p}. f has domain T^2. ?

The problem is showing that $f\circ\iota_p$ nullhomotopic and $f\circ\iota_q$ nullhomotopic implies $f$ nullhomotopic

@BalarkaSen Yeah

Don't you mean $i_p \circ f$?

…Yes

:P

Dammit

That was actually confusing to me. I didn't mean to nitpick.

Sorry

Np

Okay

So we can write $f$ as $\tilde f\circ\pi$ where $\pi$ maps it to a *sphere

Right?

'Cause we're factoring it through the sphere

Indeed so.

The question is, does the fact that $\iota_p\circ f$ is nullhomotopic mean that $\iota_p\circ\tilde f$ is nullhomotopic?

If so, we have that $\iota_p\circ\tilde f$ and $\iota_q\circ\tilde f$ are nullhomotopic, and since $\tilde f$ goes from a sphere, we know that that means that $\tilde f$ is nullhomotopic

and thus $f$ is nullhomotopic

Let's just write $g = i_p \circ \tilde{f}$

Like maybe there's a nullhomotopy of $\iota_p\circ f$ but it relies on uncrushing the longitude or meridian

should this be proved by induction? :S

Like maybe you can shrink the torus when we get rid of one hole, but you can't do it if you keep the circles crushed

Er

You're asking if $g \circ \pi$ is null implies $g$ is null, right?

Yeah

$g : S^2 \to S^2 \vee S^2$

No, $g:S^2\to S^2\vee\cdot$

Er, yeah

Maybe we can use the fact that the sphere is nice

Of course.

Like, we know $g\circ\pi$ is nullhomologous as well

$\pi_2(S^2 \vee S^2) \cong H_2(S^2 \vee S^2)$ so this might just be true.

(And the isomorphism is canonical, in a sense)

If we show that $g$ is nullhomologous, we know that it's nullhomotopic

$g$ goes to just $S^2$, though, 'cause it's $\iota_p\circ f$ the way you've written it, so we've popped one of the balloons

Yeah I remember

Ah yeah yeah yeah

So we have $T^2\to S^2\to S^2\vee\cdot$

Consider $\pi' \circ g \circ \pi $

Where $\pi'$ pinches one of the spheres

Leaving the balloon we collapsed up there so we remember what that sphere was

Look at a generating class of $H_2(T^2)$

This has to be degree 0 because $g \circ \pi$ is nullhomotopic

The homologues of all three things are $\Bbb Z$, and in the end somehow that element of $H_2(T^2)$ has become zero

Regardless of whichever sphere we pinch.

But the $T^2\to S^2$ map at the start couldn't have killed it

Are we saying the same arguments?

I dunno I was typing too fast to be reading

Just a sec

In the end we get a map $T^2 \to S^2 \to S^2$ which is degree 0, where the first map is degree 1

So the middle map has to be degree 0

Right

And if it's degree zero, it's nullhomotopic as well

Yeah

Hopf degree

So we're done

Now I'm annoyed, though, 'cause it seemed like such an elementary argument when I first said it

So that's the only way you can achieve the nullhomotopy, by pinching circles?

but now it turns out that we had to do degree theory stuff

What do you have against non-elementary methods dude

It's pictorially beautiful

That's what matters

I don't know how to prove that if $S^2\to S^2$ kills everything in the homology group it's nullhomotopic

@Twink induction should work. The $n=1$ case is easy, think about how you'd do the $n=2$ one and generalise it

I don't like things I can't prove

I can send you my .pdf on Hopf degree theorem.

gib email or check discord

hello

happy newyear @robjohn and everyone

anyone could solve this diff eq. : d^2x/(dt)^2-1/t*(dx/dt)+1/x*(dx/dt)^2=0 ?

@MikeMiller I rambled about obstruction theory a bit above. Can you check and explain why it works?

any idea how to begin?

@Student404Mus that’s a nonlinear ODE, which usually means it’s either only solvable by some trick or it’s not solvable in closed form

ther's a hint : if we myltiply by 1/t

does that solve the problem?

And off the top of my head I don’t see any obvious trick. That doesn’t mean there isn’t one, though

Hmm

and the solution to arrive with is x^2=at^2+b ; a b are constansts of integration

Should the last term have 1/x or 1/t in front?

1/x

Hmm

as it written

it seems like an equation of a circle or an ellipse

i tried to solve it in this website: onsolver.com/diff-equation.php ; but the solution is different

@Abra something like that

but in space-time with 2-D

Where I’d start is differentiating the expression x/t*(dx/dt) with respect to t. If you do that and carefully compare the result to your ODE you should notice something

It’s the equation of a hyperbola rather than an ellipse @Abra

@Semiclassical how did you find it ?

By knowing what the equation of a hyperbola looks like?

ah, that requires to be knowledgeable

standard form is x^2/a^2-y^2/b^2=1

(That or with x, y swapped)

The real bit of knowledge I have is that the motion of a relativistic particle with constant proper acceleration is a hyperbola asymptotic to x=t (with c=1)

In the sense of the trajectory traced out in 1+1D spacetine

ok

For details: en.m.wikipedia.org/wiki/Hyperbolic_motion_(relativity) @abra

Am I right to speculate that that’s the origin of this question?

but does that give a hint how to solve this diff. equation?

yes you are right

Not really, no. But the starting point I gave earlier does

That differentiation

Actually, that article does hint at a different approach: under a judicious change of variables (the rapidity section) the equation becomes rather cute. But that comes a bit out of nowhere

@Semiclassical i was just wondering if there is common ways/approaches to solve such equations or is intuition your only lucky charm ?

With nonlinear equations, things aren’t great

yup.

I think Lie theory provides some systematic tools, but tbh I don’t know that stuff

The nice cases are the exceptions when it comes to nonlinear equations.

In this specific case there’s more that can be said, since it’s coming out of special relativity

But I don’t remember it so well off the top of my head

differentiating the expression you gave cannot give the required diff. eq. exactly since the last result is : d^2x/(dt)^2-1/t*(dx/dt)+1/x*(dx/dt)

and the diff . eq. is : d^2x/(dt)^2-1/t*(dx/dt)+1/x*(dx/dt)^2=0

right?