Mathematics - 2024-03-23

CowperKettle

00:03

Misha Gromov - 1/4 Beauty of Life seen through Keyhole of Mathematics

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psie

01:24

I'm reading the following paragraph in my notes on Lebesgue measure:

> We first construct an outer measure on all subsets of $\mathbb R^n$ by approximating them from the outside by countable unions of rectangles...This approach is somewhat asymmetrical in that we approximate sets (and their complements) from the outside by elementary sets, but we do not approximate them directly from the inside.

What does it mean to approximate a set "from the outside" (or even inside)?

Thorgott

smaller and smaller sets containing the given set approximate it from the outside

larger and larger sets contained in the given set approximate it from the inside

psie

ah ok, makes sense

Jakobian

@psie literally take infimum of all sizes

Of sums of their volumes

psie

I see

copper.hat

05:26

@psie You can always try asking directly in the comments. @Jakobian Thanks you for answering.

leslie townes

i dunno, asking about anything only makes things worse

copper.hat

06:01

i always ask myself that

user 85795

06:31

💯% ✅

Peter

12:19

Users able to close-/delete-vote : Please help us in CURED

Lucky Chouhan

@copper.hat Hello monsieur, how are you doing?

Lucky Chouhan

12:33

what is the difference between integral on [a,b] and (a,b)? because we evaluate integrals in both interval in the same way :')

one potato two potato

13:47

@Thorgott Let $M$ be a mapping torus with monodromy $f:X\to X$. Then $\pi_1(M) = \pi_1(X)\rtimes_{f}\Bbb Z$. So if I abelianize, I get $H_1(M) = H_1(X)^{f}\times\Bbb Z$ where $H_1(X)^{f}$ means an element of $H_1(X)$ which is invariant under $f_*$ action.

If $X$ is an $n$-manifold, then using PD, I can say $H_1(X)^f\simeq H^{n-1}(X)^f$?

I wonder if the $f$-invariance is preserved under PD map

Alessandro Codenotti

I am, once again, extremely confused by ordinal arithmetic. Is it true that if $\beta\leq\alpha$, then $\alpha+\beta\geq\beta+\alpha$?

Seems to be true by induction on $\beta$

Thorgott

15:30

@onepotatotwopotato you have to take coinvariants, not invariants

@onepotatotwopotato Poincaré duality takes the $f$-(co)invariants of $H^k(X)$ isomorphically onto the $\pm f$-(co)invariants of $H_{n-k}(X)$, where the sign depends on whether $f$ preserves or reverses orientation

one potato two potato

what is coinvariants?

oh the MO post I found a few days ago says coinvariant hmm

Thorgott

its the quotient by the subgroup of elements of the form $x-f_{\ast}x$

one potato two potato

15:50

Oh so the resulting $H_1(X)^f$ is invariant under $f_*$-action

Thorgott

notation-wise, $H_1(X)^f$ is invariants and $H_1(X)_f$ is coinvariants

one potato two potato

Ah, so $H_1(X)_f\times \Bbb Z$ is more correct.

copper.hat

@LuckyChouhan good thanks

Alessandro Codenotti

16:08

@AlessandroCodenotti ah this is actually false

Jakobian

17:02

@AlessandroCodenotti I was just going to say to look at the Cantor's normal form

i.e. find non-zero naturals $n_1, ..., n_k$ and $m_1, ..., m_l$ with $\beta = \omega^{\beta_1}n_1 + ... + \omega^{\beta_k}n_k$ where $\beta_1 > ... > \beta_k$ and $\alpha = \omega^{\alpha_1}m_1 + ... + \omega^{\alpha_l} m_l$ with $\alpha_1 > ... > \alpha_l$

If $\beta_1 < \alpha_1$, its clear that $\beta+\alpha = \alpha \leq \alpha+\beta$

So the only possibility is to $\beta_1 = \alpha_1$ and so $\beta+\alpha = \omega^{\alpha_1}n_1+\alpha = \omega^{\alpha_1}(n_1+m_1)+\omega^{\alpha_2}m_2 + ...$

and similarly, then we will have $\alpha+\beta = \omega^{\alpha_1}(n_1+m_1)+\omega^{\beta_2}n_2+...$

but if $n_1 = 1$ and $m_1 = 2$ lets say, then coefficients of $\beta$ namely $n_2, ..., n_k$ can be pretty much anything

For example, let $\beta = \omega+1$ and $\alpha = \omega \cdot 2$

psie

18:29

Consider the definition of the Lebesgue outer measure on $\mathbb R$, namely $$\mu^\ast(A)=\inf\left\{\sum_{i=1}^\infty\mu(R_i):A\subset \bigcup_{i=1}^\infty R_i, R_i\in\mathcal{R}(\mathbb R)\right\},$$ where $\mathcal{R}$ is set of all $1$-dimensional rectangles, e.g. closed intervals, and $\mu$ is simply the length of those rectangles, i.e. the right endpoint minus the left one.

Consider now $E=\mathbb{Q}\cap [0,1]$ and let $\{q_i:i\in\mathbb N\}$ be an enumeration of the points in $E$. Given $\epsilon>0$, let $R_i$ be an interval of length $\epsilon/2^i$ which contains $q_i$. Then $E\subset\bigcup_{i=1}^\infty R_i$. Then we have $$0\leq \mu^\ast (E)\leq\sum_{i=1}^\infty \mu(R_i)=\epsilon,$$ and since $\epsilon>0$ is arbitrary, $\mu^\ast (E)=0$.

Now, in my text, they remark that if we had covered $E$ with a finite number of intervals, then this union would have to contain $[0,1]$. Why is this?

Or put differently, does the countable infinite cover not include $[0,1]$?

Thorgott

it does not, otherwise the length of its intervals would sum to at least $\mu([0,1])=1$

@psie use that the rationals are dense in the reals

psie

18:45

hmm ok, I will have to think about it some more. In this context, I struggle with how density implies that the finite cover has to contain $[0,1]$...

Derivative

Why do we usually study homology with coefficients in rings such as $\mathbb Z$ or $\mathbb Z_m$? Does anything interesting happen if we choose a funny ring e.g. $\mathbb Z[\sqrt 2]$?

Balarka Sen

@psie This sounds wrong. Consider $[0, \sqrt{2})$ and $(\sqrt{2}, 1]$. These cover all the rationals in $[0, 1]$, but their union is not $[0, 1]$.

Namely, it misses $\sqrt{2}$.

Xander Henderson

@BalarkaSen Those are not closed.

The definition psie gave is that $\mathcal{R}$ is the set of closed intervals.

Balarka Sen

That part was confusing to me. They said "1-dimensional rectangles, e.g., closed intervals"

"e.g." to me is distinct from "i.e."

Xander Henderson

Yeah, they are giving a definition which is really written for $\mathbb{R}^n$, but applying it to $\mathbb{R}$.

Three is a lot of extraneous nonsense.

But lots of people also don't know the difference between "i.e." and "e.g." It is a common mistake.

But the only way in which the comment makes sense is if the intervals are meant to be closed.

Thorgott

18:51

yeah, they ought to be cloesd

the point then is that a finite union of closed intervals is again closed (an infinite one typically is not)

Xander Henderson

Indeed.

Thorgott

@Derivative homology with coefficients in $\mathbb{Z}$ determines homology with any other coefficients by the universal coefficient theorem

Xander Henderson

@Derivative :( $\mathbb{Z}_m$ is bad notation. ):

Is $m$ even prime, bruh?

Thorgott

conversely, studying a problem with coefficients in all $\mathbb{Z}/p$ and $\mathbb{Q}$ is typically enough to get results with coefficients in $\mathbb{Z}$

Balarka Sen

$\Bbb Z_{(p)}$

@Thorgott Not to contradict you or anything but FYI: there are serious usage of homology (in particular, chain groups) with coefficients in $\Bbb R$ which are not "formally recoverable from $\Bbb Z, \Bbb Z/p$ or $\Bbb Q$."

The idea is that $\Bbb R$ is a nice topological field.

Thorgott

18:55

the Gromov stuff?

Balarka Sen

Yep

Thorgott

this is true, but it requires data on the chain level, as you indicate

Balarka Sen

indeed true

in a similar but distinct spirit, the rational cochain complex has a stronger structure on them than the integral one (that of a cdga).

Thorgott

I'm also not saying other coefficients are irrelevant, to be clear, e.g. you can get all sort of weird coefficients appear in the Serre spectral sequence (granted, in most applications, these will be just f.g. abelian groups again), but the reason it's typically just $\mathbb{Z}$, $\mathbb{Z}/p$ or $\mathbb{Q}$ is that that's where the homological information is concentrated

@BalarkaSen commutative?!

Balarka Sen

yes, on the nose commutative

granted, the space has to be a simplicial complex

one builds a notion of a PL differential form, with coefficients in Q

then wedge product is on the nose (graded) commutative

numdam.org/article/PMIHES_1977__47__269_0.pdf

Ch 7

Thorgott

19:02

man I'll die if I ever have to learn rational homotopy theory

Balarka Sen

@Thorgott Read Ch 7 up until Example (i) "Polynomial forms"

It's 2 pages, beautiful stuff

The proof of the key lemma is not due to Sullivan but someone called Bruce Renshaw, as Sullivan mentions in the footnote

page 296

Derivative

@Thorgott hi Thorgott, so the thing is I can't understand that from the statement of the universal coefficient theorem

it overall feels very obscure to me

I'm trying to read Hatcher by myslef

Thorgott

@BalarkaSen "spaces", "cells", etc.

this is some wild non-formalism

@Derivative what precisely are you not understanding?

Balarka Sen

@Thorgott IMO, he is thinking categorically without saying it.

Derivative

ok so for example, wikipedia says there's a short exact sequence involving the Tor functor and the sequence splits

how does that imply that we can get the homology with arbitrary coefficients from the $\mathbb Z$ homology?

and second: why on an intuitive level does $\mathbb Z$ have such a property, given that rings can have very weird algebraic properties?

Balarka Sen

19:17

The point is for any ring $R$, there is a homomorphism $\Bbb Z \to R$.

Thorgott

the outer two terms in that short exact sequence involve only homology with coefficients in $\mathbb{Z}$, the middle term is homology with arbitrary coefficients

so knowing homology with coefficients in $\mathbb{Z}$ => knowing the outer two terms => knowing the middle term => knowing homology with arbitrary coefficients

Derivative

nice

Thorgott

@Derivative cause the singular chain complex is defined over $\mathbb{Z}$ already

the singular chain complex with other coefficients is just obtained by tensoring

and it stands to reason that the homology before tensoring determines the homology after reasoning

Derivative

not sure that's intuitive

Thorgott

not sure if I would say it is intuitive, but it's true

there's a conceptual explanation for this sort of phenomenon, but it's a bit more sophisticated

Derivative

19:22

one other thing that kind of sticks out is that there are so many homology and cohomology theories

Thorgott

my point with regards to how your original question was phrased is moreso that this is not first and foremost because $\mathbb{Z}$ has some special property, but because our construction of the singular chain complex is over $\mathbb{Z}$ to begin with (the formula for the boundary of a simplex only involves coefficeints $\pm1$)

Xander Henderson

Okay, so, suppose that $\varepsilon < 0$...

Thorgott

(granted, $\mathbb{Z}$ being a PID or at least hereditary helps at some point if you compare this version of the UCT to more general ones)

Balarka Sen

I contend that the reason is there is always a homomorphism $\Bbb Z \to R$. This allows you to compare $H_*(X; \Bbb Z)$ and $H_*(X; R)$. Namely, take the map from the first to the second by changing coefficients. This map extends naturally to one from $H_*(X; \Bbb Z) \otimes R$ to $H_*(X; R)$.

The "miracle" is the cokernel is nice to describe.

That is a homological algebra fact, essentially because tensor behaves sufficiently naturally when applied to exact sequences.

Thorgott

I think the reasons go hand in hand

$\mathbb{Z}$ being the initial ring is kind of why the singular chain complex is defined over it in the first place

Balarka Sen

19:29

IMO the reason singular chain complex is defined over integers is because we think about numbers before thinking about arbitrary rings when we want to weight stuff (eg, simplices)

Thorgott

we really just wanna think about orientations, yeah

$\mathbb{Z}$ is the simplest/universal ring in which we can make sense of that

Balarka Sen

Erm, not sure I agree. 2 times a simplex is an important notion, not related to orientations.

It's going around the same loop multiple times

Thorgott

fair

user10478

I'm surprised this doesn't appear to be on Wiki. Is there a way to exploit linearity to shortcut solving linear programs? If I'm thinking about the geometry correctly, scaling by constants is unproblematic ($(max_x f(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = 5$ entails $(max_x 2f(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = 10$ whether or not $f$ is linear, or even if the constraints are made nonlinear), but do you also get...

...$(max_x g(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = 4$ and $(max_x h(\vec x)$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = 3$ jointly entail $(max_x (g(\vec x) + h(\vec x))$, $A\vec x \leq \vec b$, $\vec x \geq \vec 0) = 7$ when $g$, $h$, and the constraints are linear?

Thorgott

ultimately, it makes both geometric and algebraic sense to take singular chains with coefficients $\mathbb{Z}$ first and foremost

Jakobian

19:36

@user10478 what about $g = -h$?

user10478

@Jakobian Doesn't any non-trivial instantiation of that violate non-negativity constraints?

Jakobian

which constraints? $x\geq 0$?

user10478

Yeah

Jakobian

why would it violate it?

Thorgott

@Balarka btw I spent like the last 2 weeks trying to learn how to calculate $\pi_{n+2}(S^n)$ without using spectral sequences, Steenrod squares or differential methods and I have utterly failed at grokking it

user10478

19:41

Hmm, maybe not actually.

Jakobian

Right. In your notation, $$(\max_x (g(x)+h(x)), Ax\leq b, x\geq 0)\leq (\max_x g(x), Ax\leq b, x\geq 0) + (\max_x h(x), Ax\leq b, x\geq 0)$$ but the equality doesn't have to hold

Xander Henderson

@Thorgott You are dealing with dark magicks, known only to Grothendieck (and, of course, Satan himself).

It is no surprise that you failed.

Balarka Sen

@Thorgott The 2nd stable stem you mean?

Thorgott

this actually happens to be disjoint from Grothendieck

@BalarkaSen all the groups individually, though it's not really easier or harder than the stable result

Xander Henderson

Why can I never spell that dude's name... (though I don't know where the "d" came from).

Balarka Sen

19:45

Because you somehow know it stabilizes fast

Thorgott

it stabilizes one step earlier than predicted by Freudenthal (cause there is a map of Hopf invariant $1$ in dimension $3$)

Jakobian

Its easy, you just have to remember that the second part of the name doesn't spell an obscenity

Xander Henderson

@Jakobian Only the whole name all spelled out...

Thorgott

so the question is really just whether $\pi_4(S^2)\rightarrow\pi_5(S^3)$ is the zero morphism or not

Xander Henderson

Got it!

(Grothendieck is a four letter word in my house.)

Thorgott

19:47

and this question turns out to be really annoying depending on how you go about it

user10478

@Jakobian So, if $g = -h$, then every $\vec x$ in the input space maximizes $g + h$ with objective function value $0$.

Thorgott

also, historical fun fact you may or may not know already, but Pontryagin originally announced the $2$-stem was $0$ and was corrected later

Balarka Sen

lol

Thorgott

took $12$ years to correct, too

Pontryagin was doing it the differential way and apparently just missed one framing of the torus or something

Jakobian

@user10478 sure... if you take $A = I$ and $\mathbb{R}^n = \mathbb{R}$ lets say, $b = 1$, and $g(x) = x$, then...

Balarka Sen

19:49

@Thorgott Ah yeah that one's pretty annoying

I never understood framed cobordisms of surfaces.

Thorgott

it was Whitehead (G.W. not J.H.C.) who proved the correct result in 1950

Balarka Sen

You need to use the Arf invariant somewhere

So fuck that

Jakobian

$(\max_x g(x), Ax\leq b, x\geq 0) = 1$ but $(\max_x h(x), Ax\leq b, x\geq 0) = 0$

Xander Henderson

@BalarkaSen Watch your fucking language!

Thorgott

and he just does it by writing down explicit formulae for everything

it's ridiculous

Balarka Sen

19:50

@XanderHenderson Oh sorry, I mean A** invariant

Thorgott

like, this is the definition making the crucial part of the proof work

Xander Henderson

@BalarkaSen Much better. Thank you.

Balarka Sen

lol @Thorgott

Xander Henderson

@Thorgott What is it that the kids say about mathematics like that?

Ah... yes...

That is "cursed".

Thorgott

and $Q$ is some map that is defined 30 pages earlier

Jakobian

19:53

@XanderHenderson I think the kids say its cool these days

Thorgott

all these 50s homotopy theory papers are like that

Toda's are even worse in terms of readability

Balarka Sen

JHC Whitehead is clearly the chaddest of the two

Thorgott

they fix explicit representatives for all canonical identifications and give them letters in the first section that they will proceed to use like 30 pages later and then prove every claim by writing down explicit formulae

it's absolutely mind-boggling

Jakobian

today the kids learn category theory and become homotopy theorists or algebraic geometers

Thorgott

@BalarkaSen oh his papers aren't more readable

however, he was a big pig enthusiast, so he's very much a chad

Balarka Sen

19:56

JHC was a topologist through and through

user10478

@Jakobian Okay, I think I see the problem. So is there any notion of adding two LPs (either in objective function, constraints, or both) if you play with the details (change $max g$ to $min -g$, switch back and forth between primal and dual, and so forth), or can LPs only be scaled by constants and not added?

Thorgott

said his mathematical inspirations came from scratching his pigs necks

Balarka Sen

lmfao

i didnt know this

Thorgott

63

Q: J. H. C Whitehead (and his pig)

I am actually completing Master's Thesis on Lawson Homology. In order to do this, I am writing an appendix on Higher Homotopy Groups. Now, as you know, one of the most important Homotopy theorists ever is J. H. C. Whitehead and much of the standard material on homotopy groups etc. is due to him. ...

ho.history-overview

Jakobian

@user10478 Its not my field of expertise but my guess is "no there isn't"

user10478

19:58

Okie, thanks

Balarka Sen

@Thorgott Do you know how JHC proved every closed oriented n-manifold is a branched cover over S^n branched over some subcomplex?

Jakobian

misspelling "but" as "because" is my new record

Balarka Sen

took a triangulation of M, took the standard triangulation of S^n by two n-simplices, proceeded to mark the two as 1 and 2, then marked the triangles in M alternately as 1 and 2, sent all the 1's to 1 and all the 2's to 2, "folding M to S^n" in the process

Thorgott

G. W. Whitehead was great too, though, I've been learning a lot from his textbook

the craziest guy was Toda, though. managed to compute everything throughout the $19$-stem all the way back in 1962 and it's all borderline impossible to read

Balarka Sen

Or maybe that was Alexander, actually

@Thorgott yeah i have seen his papers once or twice, closed tab immediately

Thorgott

20:01

and the unstable computations have only improved marginally since his days

Jakobian

@Thorgott Is there a name for a ring such that if $P, Q$ are prime, then $P+Q$ is prime or whole ring?

Thorgott

not sure if his computations have ever been replicated in full at all, even

@BalarkaSen awful font, too

@BalarkaSen "alternatingly" requires a lot of care, I think

also, I believe this was Alexander, yeah

at least for $3$-manifolds

I think I briefly looked at that paper when I was preparing my talk on knot theory

Balarka Sen

@Thorgott yeah i never figured out the details lol

Thorgott

ah yeah, I think it was in "Note on Riemann spaces"

Balarka Sen

also for 3-manifold it shows every closed oriented 3-manifold branches over S^3 with branching set a 1-complex and Alexander just says "we can wiggle it a little to make it a link"

which is not at all clear lol

Thorgott

20:04

lol, topology back then was crazy

Balarka Sen

still is man

Thorgott

do you know how Freudenthal proved his suspension theorem originally?

Balarka Sen

nope

Thorgott

cause it's not any of the methods you would find in a reasonable textbook

I tried skimming his paper once, I think what he does is use simplicial approximation, then use some Hopf invariant and simplicial chain considerations to find a point whose preimage is just a single point, cut the target sphere into a small ball around the point and its complement, take preimages of that and "makes it look like a suspension"

Balarka Sen

Lmao

Based

Thorgott

20:07

I don't understand how they did anything pre-40s

not enough formalism, too much ingenuity

Whitney's 1936 manifold paper was also so ahead of its time, it's ridiculous

Balarka Sen

Whitney was a genius

Thorgott

1950's is the only era where I understand some papers, though the homotopy theory papers (as indicated) are way too formulaic for me to grok

anything 60s and after is too complicated

this line of thinking is depressing, I should stop it

Balarka Sen

Read the 2 pages in Sullivan's paper man

It's great

Thorgott

I did, what he's getting at seems to be the comparison theorem for (co)homology theories

Balarka Sen

Did you see Example (i)? It's genius IMO

That piecewise polynomial forms on simplicial complexes computes singular cohomology.

The key lemma is a polynomial form on $\partial \Delta^n$ extends to a polynomial form on $\Delta^n$ -- this is what guarantees acyclicity of the simplex wrt "rational polynomial forms cohomology".

Thorgott

20:23

it's doesn't strike me as surprising, but it very much strikes me as impressive

Balarka Sen

It's so simple!

Just like that you have a cdga $A^\bullet(X; \Bbb Q)$ (consisting of rational piecewise polynomial forms on a simplicial complex $X$) which is quasi-isomorphic to $C^\bullet(X; \Bbb Q)$.

The main theorem of rational homotopy theory says $A^\bullet : \mathrm{Spaces} \to \Bbb{Q}\mathrm{CDGAs}$ is an equivalence of homotopy categories, restricted to the subcategory of $\mathrm{Spaces}$ consisting of nilpotent (e.g., simply connected) spaces whose integral homology groups are finite-dimensional rational vector spaces.

Thorgott

yeah that's some crazy shit

also nilpotent is a scary hypothesis, never really needed that before

Balarka Sen

20:40

It makes so much sense though, because all the "higher operations" that come from failure of cup product commuting at the cochain level (eg, Massey product, Steenrod squares, ...) just go away

$A^\bullet(X; \Bbb Q)$ has nailed down the homotopy type of $X$ to some degree. Turns out it has only nailed down the rational part of it, which is fair enough.

Jakobian

There was a terrorist attack in Moscow, 60+ people died and like 150 were injured

Balarka Sen

All of Sullivan's works are just that: he writes big scary tomes but there's some 2 pages somewhere in the paper which explains everything

Thorgott

tbh I don't even know what a nilpotent operation is

Balarka Sen

I don't know either

Thorgott

ha

Balarka Sen

20:44

Where did you see that?

Thorgott

see what?

Balarka Sen

"nilpotent operation"

leslie townes

wisecracking voice there's nothing rational about that homotopy theory

Thorgott

I mean, that's what a nilpotent space is, isn't it?

$\pi_1(X)$ is a nilpotent group and operates nilpotently on all $\pi_n(X)$

Xander Henderson

@Thorgott pretty sure it is a space in need of Viagra.

Thorgott

20:55

nil-, not im-

@Jakobian I would expect this to fail even in $1$-dimensional non-domains

Balarka Sen

@Thorgott Oh, you mean a "nilpotent action"

It's a technical condition which attempts to bypass the hold that the fundamental group has over the space, to reduce to a more abelian, homological scenario

It is perfectly fine to assume $\pi_1(X) = 0$ for starters. Simply connected spaces with a map between them which induces isomorphism on homology is a weak homotopy equivalence.

The analogous result holds for nilpotent spaces, hence the usage

Thorgott

yeah, apparently it was developed by Dror for that reason link.springer.com/chapter/10.1007/BFb0060891

didn't know this, pretty cool

I was forced to learn simple Whitehead recently, didn't know we can do even better

Balarka Sen

The geometry of nilpotency becomes clear if you think in terms of the co-CW decomposition, i.e., the Postnikov tower. Nilpotent iff the Postnikov tower can be refined to a tower where the $K(\pi, n)$-fibrations at each step have abelian fibers: $K(A, n)$, $A$ abelian.

Thorgott

21:11

that can't be quite it, right?

the fibers in the Postnikov tower are $K(\pi_n(X),n)$s and the higher ones are abelian anyway

the first step in the Postnikov tower is itself a $K(\pi_1(X),1)$, so I get this can be decomposed into fibrations with $K(A,1)$s as fibers, $A$ abelin, if $\pi_1(X)$ is nilpotent

but where does the action come into play?

Balarka Sen

@Thorgott I mean to say that the monodromy of the principal fibration is trivial

That is to say, $\pi_1$ acts trivially on the homology of $K(A, n)$ as well

Thorgott

they won't be principal in general

@BalarkaSen this makes sense, I guess

my understanding is that "localizing a space" works in the nilpotent case because you can take a Postnikov tower like that and just localize the groups $A$ describing the Eilenberg-MacLane fibers to obtain a new Postnikov tower

this is in May's More Concise, but that book scares me like hell

Balarka Sen

That's correct, but it is nowhere close to the intuitive picture of localization Sullivan had :) Famously, he looked at Felix-Halperine and said "This is good, but it isn't rational homotopy theory"

Thorgott

what's the intuitive picture lol

Balarka Sen

Don't work with co-CW (Postnikov), but with CW itself.

You know how to define the rational sphere?

Thorgott

21:21

nope

Balarka Sen

Take $S^n$, consider the inverse system $\cdots \to S^n \to S^n \to S^n$, where the $d$th arrow is the degree $d$ map.

Take mapping telescope construction.

That is $S^n_{\Bbb Q}$.

Similarly, one can define $D^n_{\Bbb Q}$. That's contractible, but nevertheless it is a space, and there is a natural cofibration $S^n_{\Bbb Q} \to D^n_{\Bbb Q}$.

Take a simply connected space $X$ and replace all cells by rational cells inductively

That's $X_{\Bbb Q}$

The central pillars of the theory are stupidly simple. A baby could come up with it. It takes an adult to prove some theorems ultimately, but this simplicity is the central point.

See Theorem 2.2. here, from where I learnt whatever little I know

Thorgott

22:04

clever, but not so clear that this has the desired properties

wait

you don't mean a mapping telescope, do you

something's going in the wrong direction

I guess we should take a system $S^n\rightarrow S^n\rightarrow\dotsc$ and mapping teslecope, so that we get a colimit on homotopy groups

Balarka Sen

@Thorgott That's right, I had the arrows wrong. $\Bbb Q$ is the direct limit of $\Bbb Z$'s, not the inverse limit (that would give completion of spaces as opposed to localization)

You do not get colimit on homotopy groups. $S^n_{\Bbb Q}$ for odd $n$ is just $K(\Bbb Q, n)$ (!! but this is related to Serre's finiteness)

You get colimit on homology groups.

Thorgott

@BalarkaSen I do not see how this is a contradiction

homotopy group of a mapping telescope is certainly the colimit of homotopy groups

any map or homotopy of maps from a compact space factors through a finite step in the telescope

Balarka Sen

S'pose that's correct

@Thorgott It's clear that $H_*(X_{\Bbb Q})) = H_*(X) \otimes \Bbb Q$

The nontrivial point is "homology localization implies homotopy localization". the proof here requires going to back to the co-CW point of view again

Thorgott

ah, that's non-obvious, I think

the subtlety is that it's not quite clear what the degree $d$ map induces on higher homotopy groups

though this phenomenon is pretty well-studied, there should be a direct proof the colimit vanishes

Balarka Sen

22:21

You can find it at the end of Chapter 2 in Sullivan's notes above

Thorgott

ah, my suggestion works: given $\alpha\in\pi_k(S^n)$ and writing $\iota$ for the identity of $S^n$, we have $(d\iota)\alpha=d\alpha+{k\choose2}[\iota,\iota]h_0(\alpha)-{k+1\choose3}[[\iota,\iota],\iota]h_1(\alpha)$, where $h_0,h_1$ are Hilton-Hopf invariants

but $(n\iota)((n-1)\iota)\dotsc(3\iota)(2\iota)\alpha=(n!\iota)\alpha$ since $\iota$ is a suspension

so for $n$ getting arbitrarily large, this element becomes arbitrarily divisible in $\pi_k(S^n)$, hence vanishes

Balarka Sen

What are you proving?

Thorgott

that the colimit of the homotopy groups vanishes

in the case $X=S^n$

Balarka Sen

@Thorgott Why is this identity true and what does it mean?

Thorgott

you don't wanna know why it's true, it's awful

it's a qualitative version of the Hilton-Milnor theorem that generally quantifies the failure of a map between spheres to be a co-H-map plus two pages worth of auxiliary calculations

Balarka Sen

22:33

seems like a good reason to think in terms of rational spaces, then

Thorgott

my point is that it guarantees that if we take an element of $\pi_k(S^n)$ and apply the degree $d$ map enough times, the element becomes a multiple of an arbitrarily large integer (wrt divisibility)

so it vanishes since that group is torsion

and the colimit is $0$

except I need a different argument for $\pi_{2n-1}$ if $n$ even lol

Balarka Sen

if you're going to use Serre's finiteness anyway, it is not difficult to show the degree $d$ map on $S^n$ naturally translates to the degree $d$ map on $K(\Bbb Z, n)$ under $S^n \to K(\Bbb Z, n)$.

it essentially follows from the proof of Serre's finiteness

Thorgott

that's just tautology

cause it's an iso on $\pi_n$

Balarka Sen

I'm saying $S^n_{\Bbb Q} = K(\Bbb Q, n)$ follows from proof of Serre's finiteness (for odd $n$, and some added junk for even $n$)

After which it is obvious that the degree $d$ map is torsion

Thorgott

ah, you're saying the localizatrion is $S^n\rightarrow K(\mathbb{Z},n)\rightarrow K(\mathbb{Q},n)$ ($n$ odd)

though I guess this doesn't explain why the colimit construction works

Balarka Sen

22:38

sort of, i have to open hatcher to see what fiber sequence i had in mind

Oh I forgot Hatcher talks about localization

Thorgott

I don't think he ever explains how it works

though he references it in the SS text when construct the EHP SS

I ought to learn that at some point

Balarka Sen

@Thorgott You're right, my reasoning is circular. Somewhere I need to apply rational homology iso implies rational homotopy iso to get that d acts nilpotently from Serre finiteness

But that's what we're trying to avoid

Oh, hold on.

1. Proof of Serre finiteness implies $S^n \to K(\Bbb Z, n)$ is an isomorphism on rational homotopy.
2. $(S^n \stackrel{d}{\to} S^n) \to (K(\Bbb Z, n) \stackrel{d}{\to} K(\Bbb Z, n))$ commutes, as you said, by tautology
3. Therefore, we get a map $S^n_{\Bbb Q} \to K(\Bbb Q, n)$, and the codomain is $K(\Bbb Q, n)$ because, as you said, since hocolim commutes with $\pi_*$
4. By (1), $S^n_{\Bbb Q} \to K(\Bbb Q, n)$ is an isomorphism in rational homotopy, because hocolimit and rational homotopy groups commute

$n$ is odd above.

Therefore, $\pi_i(S^n_{\Bbb Q}) = 0$ for $i > n$. This means the degree $d = 1, 2, 3, \cdots$ maps eventually killed everything in $\cdots S^n \stackrel{d}{\to} S^n \cdots$

You don't need that silly formula

psie

22:55

I'm totally baffled by this sentence, maybe another pair of eyes can see what I can't see. It's a bit out of context, but in these notes on measure theory (proof of proposition 2.6, p. 13), there is the following sentence:

> Each rectangle $R_i$ [...] is an almost disjoint union of rectangles $S_{j_1j_2\ldots j_n}$, and their union contains all such products exactly once...

By all means, what does the author mean by "products"? I'm very puzzled.

Balarka Sen

Don't blindly apply Serre finiteness but use the proof

The proof pretty much takes the hofiber $F$ of $S^n \to K(\Bbb Z, n)$, and runs the spectral sequence on $K(\Bbb Z,n-1) \to F \to S^n$

Rationally things pair up and kill each other in the $E_\infty$ page, proving rational acyclicity of $F$.

I'd remembered thinking through this point before, but forgot -- it's been a while.

Thorgott

ah OK the missing piece was that $K(\mathbb{Q}.n)=\mathrm{hocolim}K(\mathbb{Z},n)$

so we can work functorially

Balarka Sen

the best kind of math is tautological math

easiest to keep in mind than random formulas

Thorgott

all math strives to be tautological

lol I didn't have that formula in mind, but I do have a copy of G. W. Whiteheads Elements of Homotopy Theory lying next to me

not that I was seriously thinking this was the right method, I just wanted to say it could probably be done

Balarka Sen

yeah fair

technical feats dont impress me much these days

math should be simple

Thorgott

23:03

I don't understand either simple or complicated

Balarka Sen

simple can be extremely hard to "see", but once you see it, it's obvious

imo

do you remember any differential geometry? eg what the sectional curvature means

and the scalar curvature

Thorgott

I've worked with them before, but I have little to no intuition for them

my mind somehow doesn't work well with structures that are too rigid

Balarka Sen

23:18

Sectional curvature of $(M, g)$ is a function $K : Gr_2(TM) \to \Bbb R$ from the Grassmannian of 2-planes on M which takes a 2-plane $P \in Gr_2(T_p M)$ at a point $p$, and spits out $K_p = Gauss.Curv.(\mathrm{exp}_p(P))$, Gaussian curvature of its exponentiation onto $M$, which is a small patch of a surface with induced metric from $g$

$scal : M \to \Bbb R$ is a scalar function given by $scal_p = \int_{Gr_2(T_p M)} K_p(P) dP_p$ where $dP_p$ is the natural (Haar) measure on the Grassmannian of 2-planes

This is called the scalar curvature -- average of sectional curvature

one usually normalizes scalar curvature so that $S^2$ has constant scalar curvature $2$.

It should be 1, because $S^2$ has sectional curvature 1 on any plane in the Grassmannian (which consists of one point only). But we still normalize it to be 2

@Thorgott The question then is what comes to your mind when you think "spaces of positive curvature everywhere"? Let's say sectional curvature, at all points, at all planes, are positive.

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