point-set-topology

[white oxide]

Similarly no open sets of the sorgenfrey line fit in there either, so you got two strictly finer topologies with neither containing another

[tame arrow]

Is the $B^4 = \partial S^3$ a typo? From https://arxiv.org/abs/1710.11562

[gentle osprey]

Light Yagami

[umbral panther]

Why do you think it’s a typo?
The boundary of the 4-ball is the 3-sphere
This paper makes bad choices, sometimes using D for the 4-disk and sometimes B, worse, letting each be a lower dimensional subset of the other. But in this case with the dimension listed, it seems consistent in the quoted paragraph

[gaunt linden]

But shouldn't that be \partial B^4 = S^3?

[umbral panther]

Oh, yeah

[fierce lily]

Can someone help me explain why any basis element containing b1 must contain points of the ai sequence?

[rancid umbra]

a basis element containing b1 = (2,1) is of the form (x,y) with x < b1 < y

what are the only elements less than b1 with this ordering?

[fierce lily]

only ai element, but must there be a point in (x,y) containing b1 because any open set containing b1 must contain a basis element about b1?

[rancid umbra]

could u be more clear?

[fierce lily]

I mean if there is a set Z, and we pick a one point set {1}, then any open set containing 1 do not need to contain other elements from Z. But for this question, we have b1, why a basis element containing b1 of the form (x, y) can tell that the element less than b1 is in (a, b)?

That is, why the only elements ai less than b1 can conclude that ai must be contained in basis element containing b1?

[rancid umbra]

the basis elements in the order topology are of the form (x,y) and [a1,z) where x < y and z > a1 are in {2} x Z_+

consider any basis element (x,y) or [a1, z) containing b1

in the first case, we must have x < b1 < y

but the only elements less than b1 are the ai’s (by definition)

so x must be equal to an for some integer n

now for any m > n, x = an < am < b1

the second case reduces to the first case by considering (a1,z)

[rancid umbra]

what book is this from btw?

[fierce lily]

I search MIT open courseware and type topology, then there occurs the class and the textbook they use is the book written by James munkres.

[fierce lily]

then what if m=n? there will no be the basis element around b1 in this case?

And it only happens when we can determine that it is the order topology?

[rancid umbra]

u have control over m. u don’t want to choose m <= n because it doesn’t allow you to draw any conclusions

and no, the basis element (x, y) contains b1 by assumption

[fierce lily]

I just noticed that n is infinite, so in that case, there must have ai in (x,y) containing b1 right?

[rancid umbra]

n is a finite number

b1 and a1 are also both elements of X; they do not contain each other

[rancid umbra]

(x,y) is an interval here, not a point in X

[fierce lily]

what i mean is that there are infinite ai

Z+ has infinite elements

[rancid umbra]

yes

intuitively, any neighborhood of b1 must contain at least one ai (countably infinitely many actually) since b1 has no immediate predecessor

what i detailed above explains why

[fierce lily]

if we assume that prodcut of U1 and V1 to be B1, product of U2 and V2 to be B2, but it seems that we don't figure out B3 that is the subset of B1 intersects B2? Do we need to find out a B3 here, if we need , how can we find that B3 in this case?

or we can just write (U1 intersect U2) times (V1 intersect V2) to be B3?

[rancid umbra]

yes, because the intersection of finitely many open sets is open

also, this is better known as the box topology

the product topology is defined a bit differently. but in some cases they generate the same topology

[lusty trench]

The product topology is best defined using the subbasis of cylinders.

[earnest ibex]

For the cellular boundary formula in Hatcher, the degrees $d_{\alpha beta}$ in $d(e^n_\alpha)=\sum d_{\alpha \beta} e^{n-1}_{\beta}$ are only defined up to a sign, right? Will it have any impact in computing homology?

[gentle osprey]

ImHackingXD

[unreal stratus]

I believe basically they all are well-defined up to the choice of sign of e_a, e_b in the homology

[radiant holly]

I think it is not possible everytime.

[unreal stratus]

The inverse is also a homeomorphism and sends compact sets to compact sets

[radiant holly]

But every compact space not Hausdorff

[unreal stratus]

I am using the definition that the preimage of a compact set is compact

i guess there are different definitions

[radiant holly]

I mean X={1,2,3} and Y={a,b,c} T|X={X,{},{1},{2},{1,2} and T|Y={Y,{},{a},{b},{a,b}
f(1)=a
f(2)=b
f(3)=c
f:X--->Y homeomorphism but not proper because (Y,T|Y) not Hausdorff @unreal stratus

[earnest ibex]

Yeah exactly, so the choice of sign won´t matter to compute the cellular homology right?

[fierce lily]

I can know that it is open, but it equals B1 intersects B2, is there a requirement that we must find a proper subset B3?

[lusty trench]

The geometric operation that flips the sign of e_a is replacing its attachment map with one that has the reverse orientation. But this also flips the sign of e_a in any equation involving cellular boundaries.

[unreal stratus]

It won't as long as you're consistent

[lusty trench]

So, algebraically, that's no different from leaving e_a intact, and instead replacing e_a with -e_a as a generator of the corresponding cellular chain group.

[unreal stratus]

The way I like to think about the cellular boundary formula is that like

You have a perfectly defined differential from the usual construction

And this formula arises when we identify like H_n(X_n, X_{n-1}) with the free group on the cells, which involves some choices of generators

Having made those choices, you can make everything consistent and it all works out nicely

[rancid umbra]

it doesn’t need to be a proper subset

[fierce lily]

Then if x belongs to intersection of B1 and B2, can we just write it trivially satisfies condition of basis? since it seems that B1 intersect B2 trivially be the subset of B1 intersect B2.

What I mean is the second condition seems unnecessary here because we in fact can just let B3=intersectoin of B1 and B2 if B3 need not be proper subset

[radiant holly]

Oh sorry I understand now

Thanks

[lusty trench]

I think a better way to phrase 2) is to say that, given any finite subset F of B, there exists another subset (possibly infinite) S of B such that \bigcap F = \bigcup S.

Actually, it works for 1) as well, considering the case where F is empty.

Much more elegant than mentioning points when they're unnecessary.

[fierce lily]

Can I conclude that U is open in Y implies that U belongs to topology on Y?

[radiant holly]

I think yes because X={1,2,3} and T|X={X,{},{1},{2},{1,2}} and let (A,T|A) subset (X,T|X) be. And A={1,2} T|A={A,{},{1},{2}} Let U={1} be mean U open in (A,T|A) and (A,T|A) open in (X,T|X) and U open in (X,T|X) and U belongs to topology (A,T|A)

mean why not? lol

[radiant holly]

you try every open in (A,T|A) but every time belongs to topology on (A,T|A) and every open in (A,T|A) open in (X,T|X)

[next crystal]

Its not unnecessary because B1 intersect B2 may not be a basis element. For example take the topology on R^2 generated by open circles. The intersection of two circles B1 and B2 may not be another circle, so B3 cant just the intersection.

Sometimes we can just let B3 be B1 intersect B2 thoigh for example if we use open rectangles instead of circles, then the intersection of two basis element rectangles is another basis element (rectangle)

In your original question in the image its the case that B3 can be B1 intersect B2 but its not always the case for bases of other topologies

[rancid umbra]

yes, B1 intersect B2 is B3.

this is how munkres uses the subset symbol:

[fierce lily]

I just noticed that union of basis element might not be the basis element. so open set is not necessarily the basis element

[gaunt linden]

As a main example to keep in mind, the set of all open disks in R^2 form a basis for the usual topology,

But the union or intersection of two disks is not generally a disk (though it is open alright).

[unreal stratus]

In my view the more helpful definition of basis is just "something generating the topology under unions"

Then munkres' thing is an equivalent reformulation

[glossy wagon]

Hello, is the identity a covering projection?

Any homeomorphism I guess, but then f^-1(x) is just a singleton set so then how can there be a correspondence between Pi_1(A, x) and the members of f^-1(x)?

nvm I guess there doesn't have to and I was drawing wrong conclusions

[unreal stratus]

Yes

[gritty widget]

So Im pretty sure that the unique smallest topology is the topology generated by the subbasis whose collection is the union of all T_a
I just don't exactly know how to proceed with the "smallest" argument

[rancid umbra]

if you have any other topology containing all T_a, then it contains the subbase you suggested

[unreal stratus]

What you said it correct by definition

Like the topology generated by it is by definition the smallest one containing those

[rancid umbra]

alternatively, you could look at the intersection of all topologies containing all of the T_a’s

[gritty widget]

gotcha, thanks potato and c

[rancid umbra]

that’s same idea can be used for the second part

[gritty widget]

Weirdly enough, this result wasn't specified for some reason

[dim perch]

tpolomgy

[jaunty summit]

ignore the crossed square stuff. how do i construct the commutative square in the bottom diagram?

clearly there is some sort of canonical way to define it (or else it would be stated), but i dont know what it is

i know what homotopy fibers are

[dim perch]

so the difficult arrow is pi_1 F(X) -> pi_1 F(f), right (since the rest are all the canoncal map from homotopy fiber to domain)

and for that one, you should try extending the above diagram by adding in homotopy fibers everywhere, using the fact that e.g. the induced map F(g) -> F(a) (together with the canonical maps F(g) -> C and F(a) -> A) makes everything commute

[obtuse meteor]

For Meyer-vietoris computations that aren’t nice (Aka no zeros) but where the boundary maps are sort of computable, how do you do it in a way which isn’t ad-hoc and annoying

Or are we just stuck in the scenario of being sad

[umbral panther]

It can be pretty arbitrary. If you take arbitrary maps from A to B and C and form the homotopy pushout, you get a MV sequence realizing those maps from A to B and C

[radiant holly]

what is pointed space?

[unreal stratus]

A pointed space is a space along with a point in the space

[umbral panther]

The point of things like spectral sequences is to isolate the hard parts of the calculation from the easy parts

[radiant holly]

how? can you give me a example

[unreal stratus]

Idk like any pair (X,x) where x is in X

Giving an example feels silly

[radiant holly]

(X,x) is space isn't it?

[unreal stratus]

X is a space and x is a point in X

[radiant holly]

oh okay mean (X,T|X) topological space and x is a point in X (X,x) pointed space

[unreal stratus]

sure

usually people just write X for the space (X,T|X) in yoru notation

[radiant holly]

but x never a topology mean (X,x) never a topological space it is any space

[unreal stratus]

this is a different use of pairs

[radiant holly]

thanks

[gentle girder]

X is the space. it comes equipped implicitly with a topology. (X,x) is a space further equipped with the data of a point in X

[jaunty summit]

ty

[obtuse meteor]

Sorry I meant to say I know the boundary maps I just don’t know how to do the Z-linear algebra bc I am stupid

[umbral panther]

Yeah, you reduce to linear algebra, which can be implemented on a computer. Basically you just do it. Sometimes there is an issue that there are names that give better answers, which is something the computer cannot see. But usually the answer is to describe more algebraic structure

[rancid umbra]

could i get a hint as to why there is no retract from the mobius band to its boundary?

the boundary is a circle, so i don't think arguing via fundamental groups is going to work

is the intuition that i would have to rip the interior of the mobius band in order to do this?

[opaque scroll]

You can argue using fundamental group or homology, but you need to think about the induced map of the inclusion.

[rancid umbra]

right, so i think i’ve made some progress. the induced homomorphism from the inclusion acts like multiplication by 2
because if a loop goes around the boundary circle once, you can kind of pull the graph down to the center circle and it will have the same homotopy class as the map that goes around the center circle twice

where’s the punch line?

im not quite seeing why this is bad

oh wait, pi(r)(2) = 1, so pi(r)(2) = pi(r)(1+1) = 2pi(r)(1) =1 = pi(r)(1) means that pi(r)(1) is 0

um

now pi(r) is the trivial homomorphism by induction

but this is bad because pi(r) should be surjective

[rancid umbra]

pi(r)(1) has order 2 in Z but no element of Z has order 2

would somebody mind verifying this?

[opaque scroll]

If there was a homotopy retract what would the induced map of the inclusion be

[rancid umbra]

i said it above

[opaque scroll]

I'm not sure I totally follow, but the fact that there is no homomorphism from Z -> Z that maps 2 to 1 is correct.

[rancid umbra]

going half way around the boundary gives you one full loop, which you can pull down to the center circle. completing the traversal around the boundary gives you another loop around the center circle

so the homotopy class of the map going once around the boundary is the same as the homotopy class of the map going twice around the center circle

or rather, if you travel twice around the center circle, you can map one full loop to half of the boundary and another full loop to the other half by pulling the loop around the center apart in a slinky-like fashion

[earnest ibex]

Thank you, and @unreal stratus too

[unreal stratus]

np

[lusty trench]

np

[jaunty summit]

do hopf fibrations have any applications outside of computing the homotopy groups of spheres

[jaunty summit]

i think you can prove this using a klein bottle

if you could retract the mobius strip to its boundary i think it would imply that a klein bottle retracts to a circle

due to how a klein bottle can be decomposed into 2 mobius strips who intersect at their boundaries

[dim perch]

insofar as the homotopy groups of spheres are fundamental to many structures in math, the hopf fibration shows up all over the place

iirc the quaternionic hopf fibration was used critically in milnor's first construction of exotic 7-spheres

[lusty trench]

The Hopf fibration is just nice. It tells you what the quotient space G/H is, when G = SU(2) and H a compact uniparametric subgroup. It's also one of the easiest examples of how the Leray-Hirsch theorem fails when its conditions aren't met. (Not much of an “application”, I know, but...)

[jaunty summit]

interesting. just looked up exotic spheres and im a bit surprised that such constructions even exist in the first place

[jaunty summit]

low key i skipped fiber bundles entirely when i learned alg top but they keep on showing up

i only know them in a differential context (eg. tangent vector bundles)

[lusty trench]

Algebraic topology provides very elegant and powerful tools for studying fiber bundles. I did my master's thesis on a differential geometry problem by 1) reducing it to a problem about vector bundles, 2) ditching differential geometry and calculus, 3) using algebraic topology (more specifically, obstruction theory) to solve the problem.

[jaunty summit]

ive noticed how its one of the most widely applicable fields of math so im not surprised

[lusty trench]

Also, fiber bundles are essential for studying equivariant cohomology. See: Borel construction.

[jaunty summit]

this shit crazy

[alpine nest]

I have no idea what it is or even what it's related to, but just instinctively I'm hating this notation

[jaunty summit]

me actively avoiding anything remotely complicated in category theory because the notation is weird

[ebon galleon]

Trust me, it's much better than some older notation people used for categories.
"Ah yes, I should use script T, script H, script K, and script R for different categories of topological spaces (in my chosen font, they all look the same)"

nLab goes a bit overboard with some of their names, but generally I think the consensus is that it's better to use a bit longer, actually descriptive, names for categories

[ebon galleon]

(not a joke btw; by the time you get to the end of the paper I find it hard to keep track of what's what)

[lusty trench]

Needless to say, don't try to learn from nLab. Learn from a more reasonable source, like Tu's “Introductory Lectures in Equivariant Cohomology”.

[pearl holly]

Lmao

Reminds me of this

[dim perch]

this is a certified peter may monent

only way i can tell it's probably not him: no horrifying diagram models for spectra

[alpine bolt]

Does a topological space in general deformation retract onto its interior? and I guess is ( H_{i}(int(X)) \cong H_{i}(X) )?

[gentle osprey]

AdamCM

[alpine bolt]

where $X$ is a finite CW complex say

[lusty trench]

What do you mean by “its interior”? Any topological space is open in itself...

[gaunt linden]

If we're talking interior relative to some ambient space, then consider this.

[dim perch]

I can think of a variant of this problem: nonempty subspaces need not have nonempty interiors

[tough imp]

Dude what the actual fuck

[gaunt linden]

On the other hand I think a manifold with boundary ought to deformation retract into (but not onto) its interior.

[rancid umbra]

mobius band doesn't

[lusty trench]

If we think of the Möbius band as an I-bundle over S^1, doesn't it deformation retract into the section consisting of each I's midpoint?

[rancid umbra]

like the center circle?

[lusty trench]

Oh, that's a simpler way to phrase it, yes.

[rancid umbra]

yes it does

[gaunt linden]

Sorry, I momentarily forgot what "retract" even is.

[rancid umbra]

i perpetually don't even know

[gaunt linden]

A retraction is required to be the identity on its image.

[pearl holly]

Sup chmonkey

[jaunty summit]

whats a textbook that covers double mapping cylinders?

also, when is a double mapping cylinder a pushout?

[unreal stratus]

lol lol was about to say this

i find his stuff too hard to pass like that

[rancid umbra]

who was the guy that used to come in here spewing shit about lattices and semi-lattices

[jaunty summit]

fuck order theory

[pearl holly]

I think this is supposed to be like a homotopy pushout right? In that case it is an actual pushout iff both of the maps are cofibrations and all spaces are nices right?

[jaunty summit]

yeah its a homotopy pushout

i guess theres an obvious way to verify equality via homeomorphism

[dim perch]

I thought it was enough for all objects to be cofibrant and only of the maps to be a cofibration?

[pearl holly]

Yeah true my bad

[dim perch]

it ees what it ees

[dim perch]

also one riff on this--it's an actual colimit of the original diagram if nice enough, but it's definitely always a colimit of some homotopy equivalent diagram

(one usually first defines homotopy colimits essentially as colimits of cofibrant replacements of the diagram)

[sonic crane]

So, the set of subsets we call the subbasis is not itself a basis for a topology right?

Which is kind of weird

Im using munkres

Oh i see ok. Yeah i guess cuz it just says its a set of subsets whose union is X

So its pretty versatile

The topology generated by the subbasis has as a basis the collection of finite intersections of elements of the subbasis, right?

Ok thanks. The utility is that its just another way of describing a certain topology ? For example, can describe the standard topology on R by using the collection of open rays as a subbasis , instead of describing it using the standard basis of collection of (a,b)

Ok

Do u have any other comments about it?

Considering you just said “sure”

Oki

[pearl holly]

i am eat white dots (am pacman)

[dim perch]

𝔦 𝔠𝔩𝔬𝔰𝔢 𝔪𝔶 𝔢𝔶𝔢𝔰 𝔞𝔫𝔡 𝔰𝔢𝔦𝔷𝔢 𝔦𝔱
𝔦 𝔠𝔩𝔢𝔫𝔠𝔥 𝔪𝔶 𝔣𝔦𝔰𝔱𝔰 𝔞𝔫𝔡 𝔟𝔢𝔞𝔱 𝔦𝔱
𝔦 𝔩𝔦𝔤𝔥𝔱 𝔪𝔶 𝔱𝔬𝔯𝔠𝔥 𝔞𝔫𝔡 𝔟𝔲𝔯𝔫 𝔦𝔱
𝔦 𝔞𝔪 𝔢𝔞𝔱 𝔡𝔬𝔱𝔰 𝔞𝔪 𝔭𝔞𝔠𝔪𝔞𝔫

[gaunt linden]

#chill for weird non-sequiturs and other chitchat, please.

[ebon galleon]

(It relates to the existing conversation about pacman)

[craggy cedar]

Let (X,T) be a path connected space. I wanted it to be true that Every x\in X had to have some nbhd U such that U can be embedded in R^alpha for some ordinal alpha, but the trivial topology on a sufficiently large set is a counterexample. It seems probably true that with enough separation it's true, but how much separation is required (eg, does X being T4 do the trick)?

[merry geode]

How do I compute EX -> BX for X = CP^n

[fading vale]

CP^n is not a lie group for n at least 2 so I'm not sure what this means

[umbral panther]

It’s not a group for n=1, either
But it is a group for n=oo

[merry geode]

Me smh

Sorry, I confused with H_(C*)^(n+1)(CP^n).

I heard E(C^*)^k is some quotient of (C^\infty)^k, but this notion is rather imprecise. How do I make it precise

[umbral panther]

If G acts freely on X and H acts freely on Y, then GxH acts freely on XxY. Thus for any models of EG and EH, their product is a model of E(GxH)
This reduces to k=1

EC* has model C^oo - 0
C^n - 0 / C* = P^n-1
C^n - 0 is 2n-2 connected so C^oo - 0 is contractile

[lusty trench]

A honest to goodness group or a group object in the homotopy category (or whatever it's called)?

[umbral panther]

It’s easy to make it an honest to goodness monoid. There are many values of oo, some of which yield easy groups. I’m not sure about the simplest one

[lusty trench]

Or, right, we could take the multiplicative group C(x) \ {0} and quotient it by the subgroup C \ {0}, right?

But C(x) isn't countable dimensional, because there are uncountably many irreducibles in C[x], all of which we need to invert... So C(x) \ {0} isn't a CW complex...

[merry geode]

May I ask why does EC* have model C^oo - 0

[lusty trench]

EG must satisfy two conditions: it's contractible and G acts freely on it.

Now, C^oo \ {0} is contractible for the reason that @umbral panther said. And C^* acts freely on it by rescaling.

[sly geyser]

wait how is an open set defined for [0, 2pi) anyway?

[rancid umbra]

union of things like [0,ep) for 0 < ep < 2pi and (x,y) for 0 < x < y <= 2pi

you can think of this as either the subspace topology on [0,2pi) inhereted from R or as the order topology on the set [0,2pi) in isolation

they are both the same in this context

[sly geyser]

but [0, 1) isn't open in R?

[rancid umbra]

it is not

[sly geyser]

oh wait....

[rancid umbra]

but [0,1) = [0,2pi) intersect (-infty,1)

[sly geyser]

(I'm new to topology, sorry) yeah I just read the definition of subspace

[rancid umbra]

so it is open in the subspace top

[sly geyser]

I thought it was just a subset S equipped with the all the open sets in X that are subsets of S lol

it seems to be a bit more than that

wait so shouldn't (x, 2pi) where 0<=x<2pi (and the whole set) be open sets too then?

[rancid umbra]

yes, i missed those edge cases, whoops

[sly geyser]

Is there a reason for not defining subspace in this way though?

[rancid umbra]

yes. you want the subspace topology to be the smallest topology such that the inclusion map into the ambient space is continuous

if you work through what this means in terms of open sets, you can "derive" the subspace topology

[rancid umbra]

also, some points just wouldn't be in any open sets if you define it like this. for example, what open set in [0,2pi) would contain 0?

[sly geyser]

oh hmm I just read that on wikipedia and I think I see the link between the two
basically by doing f^-1(U) on the inclusion map you're literally just doing S ∩ U

I see

going back to this

what's the problem with having points both close to zero and 2pi?

wouldn't that just be [0, x) U (y, 2pi)?

well technically they say neighbourhood... (that's beside the point though)

oh wait....

[rancid umbra]

you would have to map things close to (1,0) in the upper half plane to 0 in [0,2pi)

where would you have to map points close to (1,0) in the lower half plane?

[sly geyser]

I see the problem

the problem is that even though [0, x) is open, f[[0, x)] isn't

so it isn't bicontinuous

[merry geode]

Ahh, I see. Thank you eduardo and bw!

[honest shore]

Guys, whats there after learning homotopy, homology and cohomology

[lusty trench]

There are many ways you could go: obstruction theory, cohomology operations, equivariant techniques, spectral sequence stuff, etc.

[iron kite]

K-theory, characteristic classes…

[merry geode]

Is hodge theory relevant

[lusty trench]

Hodge theory is more complex geometry / algebraic geometry, right?

[merry geode]

Hmm

[cedar pebble]

it is but it's a natural place to apply things about cohomology to studying geometry

[merry geode]

Btw how do you endure computing equivariant cohomology without handwaving

Handling EG sounds like

[lusty trench]

If you only need to compute finitely many equivariant cohomology groups, then you can use a finite approximation to EG. Replace the requirement that EG be contractible with merely m-connected for large enough m.

And then your approximation can be a manifold or some other relatively tractable space.

[cedar pebble]

a lot of it comes down to knowing about the cohomology of BG, localization formulas, etc

[lusty trench]

Tu's book is a good place to start. Anderson-Fulton is good too, but a bit more demanding, I think.

At least if your intuition comes from manifolds, rather than from combinatorics, Young tableaux, etc.

[merry geode]

Then, how do you compute churn class there?

[lusty trench]

The Borel construction is functorial. If E -> X is your G-equivariant vector bundle, then you apply the Borel construction to it, getting an ordinary vector bundle F(E) -> F(X), where F(X) = EG x^G X.

And the equivariant Chern classes of E are the ordinary Chern classes of F(E).

[merry geode]

Oh, that's actually simple

[lusty trench]

Conceptually, at least, yeah.

Computations can be annoying.

[merry geode]

I retract that back

c_1(EG x^G E)

And then the space X may become difficult to describe by itself

So you pullback/pushforward from somewhere, and then

Yeag handwaving is too easier

[lusty trench]

It's not like I understand this stuff suuuper well, but in the examples I've seen (and sometimes worked through), X was usually a nice space, like a manifold.

[merry geode]

Yeah, it's simply that I just encountered a case where X = barM_{0,k}(P^4, d)

[lusty trench]

Oh, okay. But then the problem is the fact that you're working with complicated moduli spaces, not equivariant cohomology itself.

[merry geode]

Wouldn't the computation be largely the same?

[jaunty summit]

no way

[dim perch]

i reccomend that, after a study of classical homotopy (e.g. understanding essentially all of haynes miller's 18.905-906 notes), if you're interested in further study, you learn about stable homotopy

to understand a lot of the most modern problems in homotopy theory, one wants to learn how spectra, and in particular the complex bordism spectrum, work

[jaunty summit]

ive been reading through a lot of whiteheads and r browns stuff on crossed modules. would recommend

the paper "van kampen theorems for diagrams of spaces" has been particularly useful

seems like a lot of the ground in this area has already been covered in the 80s but some of the results are like really powerful

[lusty trench]

Is that Brown as in Brown representability, or as in the one who's very enthusiastic about groupoids?

[umbral panther]

How many Browns are there is algebraic topology? 4?

Only 3. Edgar, Ken, Ronnie

[slate bane]

We have the fact that if Y<X, Y is connected and Y<W<Closure(Y), then W is connected.
I was wondering, is this true also for X=R^n and replacing "connected" with "convex"?

The closure of a convex set is convex, but the proof I know can't be equally used for any such W

[opaque scroll]

I guess just think about Y = (0, 1)^2 the open square.

[slate bane]

Oh well, my bad, thank you lol

[dim perch]

anybody here know if people have written about fadell-neuwirth fibrations for equivariant manifolds

[radiant holly]

X={1,2,3} and T|X={X,{},{1},{2},{1,2} ve Y={a,b,c} and T|Y={Y,{},{a},{b},{a,b}} and let f:X--->Y homeomorphism be and h:[1,2]--->X continuous function and
let h(1)=1
h(2)=2 be now
let g:[1,2]---->Y continuous function
g(1)=a
g(2)=b my question (X,T|X) homeomorphic to (Y,T|Y) mean topological isomorphism h is homotopic to g isnt it? ı see functions on same space everytime. ı not see homotopic functions on homeomorphic spaces ı think h is homotopic to g because nothing not change this spaces topological isomorphism

[ebon galleon]

A few problems with this.

1. h and g, as you wrote them, are not fully defined (you need to define them on all real numbers in [1,2], not just 1 and 2);
2. no matter how you try to do that, you can't define these h and g on all of the interval [1,2] in a way so that these maps are continuous. This is because your codomain is discrete but [1,2] is connected, so any function to a discrete space must be constant. Be careful that you are defining maps that make sense and are actually continuous, rather than just writing something down and calling it continuous.
3. homotopy is a relation on maps with the same domain and codomain. Even though X and Y are homeomorphic, you can't ask whether a function h: A --> X and g: A --> Y are homotopic. I agree that f is a homeomorphism though, so what you could as is whether fh and g are homotopic since then they are both functions A --> Y (in your case A would be [1,2] - at least, if your maps made sense)

(Also note that, even though X and Y are homeomorphic, this notion of one map being homotopic to another map after composing with a homeomorphism depends on your choice of homeomorphism. In general, there may be several homeomorphisms between two spaces, and defining a relation like this will typically depend on your choice of homeomorphism)

[radiant holly]

thanks

🫡

[gritty widget]

i am not able to see how the cofinite topology on R would make it a topological group under usual addition

so i just check the continuity of RxR -> R x,y to xy^-1 right

now i did this via checcking the inverse image of closed sets is closed or not

so here any closed set can only be a finite set

and the inverse image in R^2 of a finite set under the map x-y is just a bunch of parallel lines

so i am not able to show that this bunch of parallel lines is closed

[opaque scroll]

A hint might be that a space is Hausdorff iff the diagonal in RxR is closed

[gritty widget]

with the cofinite topology in the product

cofinte topolgy is not hausdorf right

cause like the open sets can only differ at finitely many points

okay so or more context this was the original question

and this is the hint

[opaque scroll]

Bad hint I guess...

[gritty widget]

so does cofinite topology work?

[opaque scroll]

It is not a topological group

[gritty widget]

yeah

i dont even know what the other irrational flow one even is

i was considering a topology on R where bais is of the form [a,b)

then i think it a topological group

uhhhh

we get like a semi closed strip in R^2 as an inverse image of [a,b)

ok i think its open

so its cts

what do u think @opaque scroll?

[opaque scroll]

Seems reasonable, haven't thought it through myself.

[gritty widget]

ok thank you, im dropping this book

[umbral panther]

It is an axiom of a topological group that the inverse is continuous. But it is not continuous in the lower limit topology

[gritty widget]

oh ? but this ?

and i checked with x-y map

[gritty widget]

oh nvm youre right i got the orientation of the strip wrong

yeah i guess it insnt open

[grave solstice]

what would the topology be here?

compact-open? I am not very familiar with topologies on function spaces

or maybe just pointwise? idk

it may be the same, as I said I'm not very familiar with them xD

[rancid umbra]

what book is it?

[lusty trench]

Yes, compact-open. It ends up being this. Write a nonzero real number as sr, with |s| = 1 and r > 0. A continuous homomorphism f : R^x -> C^x has the form f(sr) = s^k e^(z log(r)), for k \in {0,1} and z \in C.

Now, to establish that a sequence of group homomorphisms f_n converges uniformly in compact subsets of R^x, it suffices to establish that f_n converges uniformly in compact neighborhoods of the identity, I think. (I'm still not sure about this step.)

And, if we identify f with the pair (z,k), then this is just saying that (z_n, k_n) converges in the usual way, considered as a sequence in C x {0,1}.

Oh, right. It shouldn't matter. For any compact neighborhood K of R^x that we take (works because R^x is locally compact), f_n converges uniformly in K if and only if (z_n, k_n) converges the usual way.

[grave solstice]

I see, thanks

I was really cought in the topology didn't realize you can just express this continuous maps in an explicit manner

[lusty trench]

Works because we're talking about group homomorphisms only.

[grave solstice]

and continuous, yes

[lusty trench]

Oh, right.

[grave solstice]

There is this similar one

q=p if p odd, q=4 if p=2

[lusty trench]

Ah, I have no clue about p-adic stuff. Sorry.

[grave solstice]

well 1+qZ_p (here the group operation is multiplication ofc) is isomorphic to Z_p, as topological groups. If I'm not mistaken

actually, the left is topologically cyclic

[lusty trench]

Mmm... What would be the dense cyclic subgroup of 1 + qZ_p? Would it be the powers of 1+q?

[grave solstice]

a continuous homomorphism 1+pZ_p-->C_p^x (assuming p odd for simplicity) is completely determined by the image of e^p, for example

[grave solstice]

e^q works

because like (e^q)^n=e^(nq) and Z is dense in Z_p so you will get everything of the type e^(sp) with s in Z_p (I'm using continuity of exp)

[lusty trench]

Oh, nice!

[grave solstice]

and e^(p^n)-->1 as n-->infty so f(e^p)^(p^(n-1)) should go to 1? lol

I wonder if they made a typo, and the image should be the additive group C_p

[grave solstice]

sorry if the right were additive I think this would be continuous homomorphisms Z_p-->C_p, this space is homeomorphic to C_p (you send 1 to whatever constant etc)

[lusty trench]

If it's supposed to be analogous to the previous situation, then it's very likely that they're right and the codomain is C_p^x.

[grave solstice]

yeah, but this left me a bit clueless

Since such a function is completely determined by the value of f(e^p)=y then maybe we can say that the space of such functions is homeomorphic to the space { y in C_p such that for all k in Z (y^k)^(p^n)-->1 as n-->infty }

since that is the only obstruction to continuity, I think

It is worth noting that $\C_p^\times \cong p^\Q\times W\times U_1$ where $W$ is the group of all roots of unity of order prime to $p$ and $U_1={x\in \C_p : |x-1|<1}$

[gentle osprey]

croqueta3385

[grave solstice]

both topologically and as groups

in any case, I think the log still gives an isomorphism C_p^x -->C_p

sry no, it's not an isomorphism

[grave solstice]

I think this property means that if f(e^p)=p^r · w · u then r should be zero and w=1

so the image should just be an element of U1, which is what we wanted

[rancid umbra]

so, i have a question about the dunce cap space. here is the problem statement and relevant info:

i am trying to follow this solution:

what is meant by, "... a map which we can write a + a - a" ?

[hexed steppe]

the boundary of the disk is glued to S1 along 3 arcs

two of the arcs have the same orientation, the third has the opposite

[rancid umbra]

i get that from the arrows, but what do the + and - signs mean (other than orientation) and how do they obey those homotopy rules that the solution uses? i guess, how should i interpret a + a - a as a map?

and how would you decide when to use different symbols for different attachments?

do you use the same letter for things which get identified to the same thing and a different letter for others?

[hexed steppe]

the +/- are the orientation

the addition operation is concatenation

[rancid umbra]

uh

i thought those were for orientation

is the concatenation symbol kind of implicit here?

[hexed steppe]

like a+b means concatenation of a and b

-a means a run in reverse

so if you want its a + a + (-a)

[rancid umbra]

okay, cool

[hexed steppe]

not sure this notation is standard im just inferring

[hexed steppe]

yeah i guess

you either draw a picture of the dunce cap and try to write down the quotient structure

or you try to draw a picture of the space with that quotient structure

but i think most people think in terms of drawings first

and the details of the math second

maybe the thing that is unclear is that you really care about the homotopy class(es) of the attaching map(s)

so using a single symbol is suggestive of the fact that these are the same map

it might be a good exercise to try parametrizing the boundary of the disk/triangle

and writing down f explicitly

[rancid umbra]

i guess im trying to visualize the attatching map as a loop on the torus

i think of the red circle as the domain circle

i can kind of see why the a between 2 and 3 when concatendated with -a is homotopic to just staying on the red line

is that what is meant by a + (-a) homotopic to 0? like, on that arc of the circle, you are path homotopic to the identity (just stauing on the red arc)?

[rancid umbra]

here is a cool video of the dunce cap being realized with a piece of fabric and some zippers
https://youtu.be/34j4CppfRTA?si=ZW-x5EAB0SVqcAB1

[hexed steppe]

not sure i can comment on your picture as its been a while since i thought about these things, but it should be a general fact that a loop concatenated with its reversal is nullhomotopic

[rancid umbra]

right, but here, you interpret it as a path homotopy

not a loop

so the segment [2..3..1] is path homotopic to the identity on that arc

[hexed steppe]

what is path homotopic

does it just mean homotopy equivalence of functions

[rancid umbra]

just a homotopy where the end points of the paths don't have to map to the same thing

[hexed steppe]

ok sure

but in your setting the endpoints will map to the same thing?

[rancid umbra]

not when you start at 2

and end at 1

[hexed steppe]

idk about your picture

but the dunce cap thing

the vertices of the triangle all get identified

[rancid umbra]

yes, that would correspond to the loop passing through its initial point on the circle three times

[hexed steppe]

ok you are right

[rancid umbra]

alr. i just wanted somebody to bounce this idea off of

thank you. this gives me some more confidence and direction

[hexed steppe]

it removes some confidence from me

[rancid umbra]

eh?

[hexed steppe]

i no longer understand the screenshot you posted

at the beginning

[rancid umbra]

there were two

the first one or the second one?

[hexed steppe]

the one you were trying to follow

second

[hexed steppe]

is the orientation for -a backwards?

[rancid umbra]

so, the suspended arrow (not the one on the base of the dunce cap), it has to like, curl out of the screen towards you so that the orientation matches up. then you continuously stretch the dunce around the base of the dunce cap and lay the boundary around

[rancid umbra]

yes. starting at 1, the purple path is going clockwise around the minor circles

[hexed steppe]

mm ok

[rancid umbra]

and at 3 it goes anti-clockwise around until it gets back to 1

[hexed steppe]

oh i see

ok i understand thanks

[rancid umbra]

alr neat

[hexed steppe]

using a torus is a good idea i think

for understanding

[rancid umbra]

yea. its easy to see now that the whole concatenation (attaching map) is homotopic to the identity, since you can deform this map into the red major circle

[hexed steppe]

right

[rancid umbra]

and then you can use question 9 to finish it off

since it will be homotopy equivalent to D^2, which is contractible

thank you for talking this through with me. that was deeply satisfying

[slate bane]

Hi, I can't find an argument for this fundamental group

We have the infinite union of shperes of R^3 with radius 2^(-n) together with the plane z=0

[rancid umbra]

how are the spheres touching?

like, in a line? or a spiral/circle?

well, i don't think that matters actually

[slate bane]

Sorry, spheres centered at (0,0,0) with radii 2^(-n)

[rancid umbra]

ah

it should be trivial

[slate bane]

How would you do it? I can't seem to use Van Kampen

[rancid umbra]

im just kind of visualizing it. you can straighten out any loop going over any of the spheres and then kind of drag it back to the starting sphere by doing that finitely many times

loops going around spheres can be dragged and smushed to the north pole of the largest sphere that the loop bounds

[slate bane]

Oh yes, visually it's clear to me, I was struggling to find a formal argument

[rancid umbra]

ah. i am unfamiliar with the van kampen theorem

[rancid umbra]

i have messed this up a bit. the previous torus not accurately depict the attaching map. here’s the fix

The 1-skeleton of the dunce cap is a circle, so to attach a 2-cell, one needs the first third of the boundary of D^2 to travel around the circle from a base point once counter-clockwise, once again counter-clockwise, and the once more clockwise. but this attaching map is obviously homotopic to the identity, since this is the same as taking the first third of the boundary around once and then mapping the rest to the base point.

the attaching map should actually look similar to this on the torus:

the base point is marked in black and the curve starts at blue angle 0, red angle 0, both blue and red circle oriented counterclockwise

[sly geyser]

why do we use neighbourhoods anyway?

why not just open sets?

[wispy veldt]

sometimes its good to have the set of all neighborhoods be a filter, also it makes certain definitions easier to talk about (ex: local compacntess)

[scenic trellis]

i dont see why the middle column is split exact here (on page 5: https://www.math.purdue.edu/~arapura/preprints/sheaves2.pdf)

where the col exact seqs are defined as

[rancid umbra]

how is concatenation of loops defined when you take a loop in X to mean a map from S^1 into your based space

usually its defined from the unit interval into your space, but you require that the path is the base point at 0 and 1

[scarlet turtle]

a loop is a special kind of path, so if you know how to concatenate paths then you also know how to concatenate loops

viewing the loop as a map from S^1 into X is the same as viewing it as a map from [0, 1] into X, but you require that the values at 0 and 1 agree

[rancid umbra]

then how do you define the concatenation of two maps from S^1 to X

[scarlet turtle]

e.g. taking half of the circle to correspond to the first loop, and then the other half of the circle to correspond to the second loop

[rancid umbra]

so you can define it on the wedge sum of two circles?

[scarlet turtle]

well yes, but you can also define the concatenation to also be a loop from S^1 into your space

on [0, pi] (this is supposed to correspond to the arc of the circle), do your first loop. then on [pi, 2pi], do your second loop

[slate bane]

Equivalently define them as maps from [0,1] to X with same initial and end point, and apply to these two maps the definition of concatenation of paths you already know (it is well defined as the end point of the first is trivially the initial point of the second)

[rancid umbra]

so S^1 —> S^1/(i ~ -i) as a quotient, then then go from that wedge sum

[scarlet turtle]

you dont need to quotient or wedge sum anything

you can define it on S^1 (or [0, 1] for that matter)

[slate bane]

Although to be more precise you would need two maps from the pointed spaces (S1,a0) and (X,x0) I guess

[rancid umbra]

yes

to both

i think its more convenient not to have to go back to [0,1] and making sure that the paths have the same start and end points in some cases

like, for instance, to show that a loop I —> S^1 is homotopic to the identity, you have to show that its homotopic x |-> exp(2pi i x) and then quotient anyway

[scarlet turtle]

possibly, but if you already have a good definition for path concatenation then you may as well take it for loops as well

any reasonable definition of concatenating loops will be equivalent to it anyway

[rancid umbra]

i was just trying to see if i could avoid quotienting my homotopy when viewing it as a map from I x I to S^1 and instead just work with a homotopy S^1 x I —> S^1 instead

[radiant holly]

mean it is not possible?

because if it is possible we cant ask whether function h: A --> X and g: A --> Y are homotopic

[scenic trellis]

nvm

free module splits

[cold estuary]

Is this banach fixed point theorem

[opaque scroll]

Yes, assuming |f| means the sup of the distance between x and f(x) say least

[worthy olive]

What does it mean for an element $q \in S_{n}(X)$?

[gaunt linden]

Context?

[worthy olive]

N chains of a group

I have a textbook that says all elements that can be modeled as the sum of an integer (which is 0 for all but a finite number) times a singular n simplex

[gentle osprey]

CJ_:)

[worthy olive]

(Absurd “q” notation because idk how to import Greek letters)

[white oxide]

That it is a formal linear combination of n-simplexes presumably? Didn't you define S_n(X) or is the definition above used in the same context as S_n(X)?

[worthy olive]

The book I was using defined S_n(X) as that

And idk how to think about linear combinations of singular simplexes

[white oxide]

Okay yeah, I also struggled with that when I first encountered it and I don't think there even is a satisfying answer

My personal answer to that was that it's just a functor that happens to assign groups in a homotopically invariant way which behaves very nicely. It's in some way an arbitrary construction and more homological algebraic than geometric in origin

Homology is often formally dual to cohomology, where things like de rham cohomology is a lot more natural and geometrically intuitive construction

[radiant holly]

X={1,2,3} and T|X={X,{},{1},{2},{1,2} and Y={a,b,c} and T|Y={Y,{},{a},{b},{a,b}} f,g:X--->Y continuous function now let f(1)=a f(2)=b f(3)=c and g(1)=a and g(2)=b g(3)=c be H:X x I --->Y cont. function and H(1,0)=f(1) H(1,1)=g(1) mean f is homotopic to g by definition isn't it?

[night heron]

Well yeah

By definition like you said

[radiant holly]

thank you bro

[night heron]

Who pinged me

[radiant holly]

me but not problem

[night heron]

Oh I see

[unreal stratus]

Tbh ig it more sense conceptually to me if you view them as simplicial abelian groups

Then there are no weird linear combinations of different dimensions ig

[empty grove]

S_n(X) is by definition the abelian group whose elements are formal linear combinations of the singular n-simplices in X

So that's what any element q looks like

[worthy olive]

Yes, but how do you interpret the idea of a linear combination of singular simplices

That was the question

Kerr gave a pretty satisfactory answer

[white oxide]

At one point there was darkness, but then I discovered the universal property of free objects and saw that if I made any group out of simplices, the free abelian one is the most natural. So it is what it is

[unreal stratus]

Oh no I mean like

How you go to the chain complex structure etc lol

And adding different dims

[white oxide]

wdym?

[unreal stratus]

Idk lol what I was saying is irrelevant tbf

Since i realise now that the question was about the individual groups

Lol

[white oxide]

To "sell" simplicial homology, or really most homologies which arent conjured up by derived dark magic: having simplices of any dimensions and the alternating sum differential justified is enough to sell having some notion of complex of groups

But the free abelian group can feel a bit random without the algebra

[jolly umbra]

Let $C_n(X) = {\sigma: \Delta^n \to X \colon \sigma \text{ is continuous}}$, the set of singular $n$-simplices.

[gentle osprey]

mayer-vietorUs

[jolly umbra]

$S_n(X) = { f: C_n(X) \to \mathbb{Z} \colon \text{ for all but finitely many } q \in C_n(X), f(q) = 0 }$, the group of $n$-chains.

[gentle osprey]

mayer-vietorUs

[prime elbow]

Is there any relation between metric space and finite closed topology?

[rancid umbra]

both are hausdorff
tangentially, that is really the only property you needed for your proof in #real-complex-analysis just now

[prime elbow]

In infinite set X, is finite closed topology has Hausdorff property?

[rancid umbra]

oh wait

no

it does not

[prime elbow]

Yes

So only in finite case

[rancid umbra]

right

then i guess i am unsure. that would have been my only guess

[prime elbow]

But I am asking about how these two Topology are comparable?

[opaque scroll]

A topology coming from a metric will always be finer than the cofinite topology, so there's that

Same is true if you replace "coming from a metric" with "Hausdorff"

[prime elbow]

But in that case all closed sets must be finite?

[opaque scroll]

In which case?

[prime elbow]

If metric space is finer than co-finite Topology

[opaque scroll]

So the fact that a metric space is finer, means any finite set is closed, but not necessarily the other way around

[prime elbow]

Are they different meaning of all closed sets is finite and all finite set is closed?

[opaque scroll]

Yes, for example in R all finite sets are closed. But there are closed sets that are not finite (e.g. [0, 1])

[prime elbow]

I think yes

[prime elbow]

Yes

In finite topology, all closed set is finite but not all finite set is closed?

No it has both

But if they have both properties then how metric space is finer than co-finite Topology?

[opaque scroll]

A topology T is finer than a Topology T' if any open set in T' is also an open set in T

[prime elbow]

Yes I was confused with property

[opaque scroll]

In the cofinite topology a set is open iff it is cofinite.

In a metric Topology cofinite sets are open, as well as a bunch of other sets

[prime elbow]

Let U be open in co-finite topology it implies U complement is finite.
Thus U is open in metric topology.

Is it correct?

[opaque scroll]

Yes

[prime elbow]

To show no non-empty finite subset of R is open in standard metric topology.

If any finite subset of R is open then let x in that set.
For x, there exists an open interval (a,b) such that x in (a,b) and it is subset of that finite set.

But (a,b) is not a finite set.

Is it correct?

[kindred cairn]

If they were open they would contain small open balls which are infinite so yea

Also you know finite sets are closed so they cannot be open because the only open and closed sets are the whole space and the empty set

[final grotto]

Is every subset of a discrete topology open?

[gaunt linden]

Yes, that's the defining property of "discrete topology".

[prime elbow]

Okay, thank you

[unreal stratus]

Btw, perhaps slightly pedantic but presumably you mean "every subset of a discrete space"

I mean it's unambiguous but sounds a bit weird to my ears as stated

[hearty condor]

Fun topology fact: you can easily define the closed sets version of a basis for every topological space.

It is immediately useful.

[lusty trench]

A basis of open sets of a topology on X is a set B of subsets of X such that, for every finite subset F of B, there exists a subset S (not necessarily finite) of B such that \bigcap F = \bigcup S.
A basis of closed sets of a topology on X is a set B of subsets of X such that, for every finite subset F of B, there exists a subset S (not necessarily finite) of B such that \bigcup F = \bigcap S.
... Right?

Else I don't see what the analogue would be.

[tender halo]

a basis of open sets is a collection of open sets whose unions generate the open sets, a basis of closed sets is a collection of closed sets whose intersections generate the closed sets

a natural example (that is, somewhat more natural than an open basis) is that zero-sets are a basis for the topology of a regular space

[tawdry valve]

Eduardo was listing the necessary conditions for the set of unions of basis open sets (resp intersections for basis of closed) to form some topology, ie not every collection of subsets generates a topology by unions, but with those assumptions it does (when you define the empty union to be the empty set and empty intersection to be the whole set)

rather than fixing a topology and asking if some collection is a basis

[fading vale]

in Thurstons notes he specifies the coefficients for dehn surgery as being (n1, n2) such that, for the associated diffeomorphism of the torus T^2 -> T^2, when we postcompose with the inclusion into the solid torus S to get T^2 -> S, the kernel is generated by (n1, n2)

This might be a stupid question but does this agree with the usual definition where the coefficients are defined by saying that the meridian m maps to n_1 m + n_2 l?

Because it seems like it just isnt? given the 2nd definition, the kernel of the associated transformation should be i think (n1, b) where b is given by choosing some a, b such that a n1 - b n2 = 1

[red yoke]

The first definition says the preimage of the meridian is n1 m + n2 l right

[fading vale]

well not necessarily, it just imposes that the coefficient of l is n2

right?

at least a priori

[red yoke]

Generated by (n1, n2) means generated by n1 m + n2 l right

[fading vale]

Like if we assume that m is sent to n1 m + n2 l, then since the associated matrix on H_1 should be invertible its clear that l must go to, say a m + b l, where a * n2 - b * n1 = 1 (or -1, it doesnt really matter). Then when we compose with the inclusion into S, the meridians die, and m should go to n2 l, with l going to b l. So if we write this as a map Z^2 -> Z it should send (k1, k2) to k1 n2 + k2 b

[fading vale]

I think thurston means n1 l + n2 m

but lets just swap it to be consistent

and say n1 m + n2 l

If we send m to n1 m + n2 l, then n1 m is going to go to n1 n2 in Z = H_1(S). if we send l to, say, a m + bl, for whatever values of a and b work, then n2 l is going to go to n2 b in Z

er possibly we want n1^2 + n2 b due to a conventions mismatch

But either way it doesnt seem like this should live in the kernel to me

[red yoke]

Why not

[fading vale]

i dont see why it should a priori

[red yoke]

The kernel of T² → S¹ is generated by the meridian

So the kernel of T² → T² → S¹ is generated by the preimage of the meridian
Since T² → T² is a homeomorphism

[fading vale]

i mean like maybe it does given that this agrees but like, given that after you pass to S, the meridinal component of the image of m vanishes regardless, it doesnt matter immediately what you send it to

[red yoke]

I still don't get your point

Another thing is that the former definition should fully describe the Dehn surgery

Since the homeomorphisms preserving the meridian extend to the solid torus

While I don't see how that holds for the latter

[fading vale]

Sorry i realized i am illiterate and was misinterpreting your point, yes i agree which is why im confused

like traditionally, you say that (n1, n2) specifies the image of the meridian

thurston is saying that (n1, n2) specifies the preimage

I guess these are just for sure not equivalent i feel

[red yoke]

Oh hm is it possible that the latter definition glues it like solid torus ← T² = T² → manifold

While the former definition does the opposite

[red yoke]

That would resolve this

[fading vale]

im not sure what you mean by the opposite

[red yoke]

Manifold ← T² = T² → solid torus

[fading vale]

Ah

Maybe?

i feel like theres no way his definition of a (p, q) surgery is different throughout the entire book and swapping the order of everything seems to be the only logical way to resolve that so i think u have to be right

he says to do it the opposite way but like

it has to be that lol

maybe this actually resolves some issues i have with things right after this as well

[red yoke]

Well he apparently also mentions the post-composition with T² → solid torus

[red yoke]

So I just assumed it was this

[fading vale]

well i kind of filled that in for him the actual wording of what he says is "A dehn surgery is when you glue out a neighborhood and glue it back in. You can do this many ways. The generator of the kernel of the inclusion map pi_1(T^2) -> pi_1(solid torus)..."

Thank u arki that was helpful

[red yoke]

Imma go be eepy

[hearty condor]

Yup. Said otherwise given a basis of a topology you can take the complement of every element and the axioms of the basis to generate the closed sets. The conditions to make a family of sets a basis is that:
a. The union of all elements give the space
b. If two elements are in the basis then for every element in their intersection there is a basis element subset of it containing it.

Taking the complements means
a. The intersection of all elements of a basis give the empty set
b. If an element is in the union then something that contains it contains the union.

I mean it is trivial to check that the element wise complement of the topology obtained via the basis gives the closed sets version of the same topology. ( Contains ∅ and X, since the topology contains arbitrary unions of elements of the basis this contains their complement i.e arbitrary intersections of the complements of basis elements, same with finite intersections of basis elements and finite union of the complements of those basis elements.)

This becomes immediately useful when comparing topologies given via closed sets

And its a useful tool to go between the two since every basis for the open sets generates a complementary basis for the closed sets and vice versa

Also

Fun tidbit

All sorts of metastructure remains well behaved as a consequence of the boolean nature of the power set.

Tidbit 2: Try and find a basis that satisfies both sets of axioms.

[hearty condor]

Tidbit 3: fun exercise to prove the metastructure thingy using the joy of category theory

Tidbit 4: As an application look into proving the inherited topology of projective space coincides with the union topology defined on a chain via closed sets.

[fading vale]

New thruston question : so given a link L in S^3 one can take a finite cover associated to it. Thurston says that you can canonically compactify this cover by adding in circles. Intuitively this makes sense but I havent been able to find a rigorous proof of it

Does anyone have a source (or an explanation off the top of their head)

[hollow vault]

With the Munkres book for topology what chapter would you consider the essential chapters to learn for learning topology

[prime elbow]

in standard topology R, Z makes discrete topology with respect to subspace topology, right?

[haughty ledge]

yeah that sounds correct

[rancid umbra]

i would suggest chapters 2 - 6, additionally chapter 1 if you want to strengthen your background in set theory (i found chapter 1 helpful for understanding arguments with ordinals and recursion)

[tender halo]

dont be one of the people this paragraph is written for, topology is a beautiful subject that is worth learning well

[stuck sorrel]

I am only interested in the first half of Munkres, cuz I don’t know how alg topology is applied in statistics

[thin scarab]

i mean here's a whole ass book about it
https://arxiv.org/pdf/2308.10825

it looks like the basic idea is you inflate your data points into n-1 dimensional balls of various sizes until the homology pops out at you

🤓

at some point in the course of growing the balls the homology is "born" and later it "dies"

"6.1.3: Homotopy Shomotopy"

[lusty trench]

As much as I love algebraic topology, what uses does this have from a data scientist's point of view, though?

[merry geode]

I did hear topological data analysis is useful for analyzing types of data.
I guess anything to give useful meaning to a data helps?

[thin scarab]

apparently "homology persistence" can be used to predict financial crises and stuff?

[merry geode]

Wh

[thin scarab]

idk the book mentions some stuff along those lines, just skim chapter 5

[lusty trench]

Sounds big if true. I remain skeptical, though.

[merry geode]

I think H^0 is likely the most useful, as analyzing clusters.

[opaque scroll]

I mean perstistent homology is an invariant of data that isn't really measured in any other way.

And of course having invariants that can discriminant different groups of data is useful in all sorts of settings

[thin scarab]

these are persistence landscapes. i.e. they measure how long each homology class "lives"

[lusty trench]

That's fine from a purely mathematical point of view. From an engineering point of view, however, that's not sufficient justification to use it.

[opaque scroll]

I mean the justification would come down to the specific case.

I guess the question is whether TDA would be more efficient / more accurate than alternative methods.

For example the first thing that comes up when I search is someone using it to distinguish people with diabetes from people without, by the homology of various data collected from them

[lusty trench]

If the data you're analyzing comes from physical measurements (say, in your example, scanning a patient's pancreas), then I believe TDA is much more likely to be useful. After all, you're not pulling a metric for the ambient space out of thin air.

[merry geode]

I did hear TDA labs are effectively converted into AI labs.

[opaque scroll]

Yes I agree. But even if you pull the metric out of thin air, if it gives results it gives results. (I don't know if it does it doesn't or if the jury's still out)

[solemn oar]

What book is this if you don't mind?

[hearty condor]

Pretty neet.

[real notch]

What kind of topological separation properties do we have for ordinals in the order topology?

[prime elbow]

In discrete metric space, all convergent sequences are ultimately constant sequence, right?

[unreal stratus]

Yes, exactly.

[red yoke]

T5 at least?

[prime elbow]

In R^n, if x_k converges to x and y_k converges to y then how can I show that x_k + y_k tends to x + y.

And here they defined convergence by metric d.

[unreal stratus]

which metric?

d(x,y) = sqrt( sum of (x_i-y_i)^2)?

[red yoke]

Probably not T6 though

E.g. first uncountable ordinal + 1

[unreal stratus]

Idek what T6 means

[red yoke]

I'm just going off wikipedia

https://en.m.wikipedia.org/wiki/Normal_space

[prime elbow]

They didn't mention it

[unreal stratus]

Sure, how about we assume the one I had

[real notch]

I hate numbering these things (because I don’t know em )

[unreal stratus]

Hm I think we need to find some nice notation

[prime elbow]

Yes but is it not true for all metrics?

[unreal stratus]

Well it is true for any of the standard ones yes

[prime elbow]

Any hint?

[real notch]

ordinals are funky

[unreal stratus]

Try to estimate the difference and use triangle inequality i think

[red yoke]

Probably more the fact that T6 explicitly involves R

[tender halo]

it doesnt have to

normal + every closed is G_delta is equivalent

[umbral panther]

But that’s defined in terms of countability, which isn’t going to work well with ordinals

T_4 is defined in terms of R, yet does apply to ordinals

[red yoke]

T4 can also be formulated without R or countability though

[real granite]

Perhaps this is one for one of the analysis channels, but I was wondering on the best way to show that this is a metric on the reals.

Positivity and symmetry are obviously true. For triangularity, I just multiplied the required inequality through by $(|x-y|+1)(|y-z|+1)(|x-z|+1)>0$ and made some cancellations to show that this is a metric if and only if $$|x-z| \leq |x-y| + |y-z| + 2|y-z| |x-y|.$$ This condition is obviously true, as $|.|$ is a metric on $\bR$ and $2|y-z| |x-y| \geq 0$. Therefore $d$ is a metric on $\bR$.

[gentle osprey]

Douglas

[unreal stratus]

There's a generalisation I can write up

Consider $\phi: [0,\infty) \to [0,\infty); t \mapsto \frac{t}{1+t}$, so that this metric is $d = \phi \bar{d}$ where $\bar{d}$ is the usual metric. $\phi$ is strictly increasing e.g. by calculus. symmetry and positive-definiteness areobvious. for the triangle inequality we ned to show $\phi(|x-y|) \le \phi(|x-z|) + \phi(|y-z|)$. but $\phi(|x-y|) \le \phi(|x-z| + |y-z|)$ as$\phi$ is increasing, so it's enough to show $\phi(a+b) \le \phi(a) + \phi(b)$. This is another fine computation e.g. by noting $\phi'' < 0$

[gentle osprey]

potato

[unreal stratus]

like $t \mapsto \phi(a)+ \phi(t) - \phi(a+t)$ maps $0 \mapsto 0$ and has derivative $\phi'(t) -\phi'(a+t) < 0$ since $\phi'' < 0$

[gentle osprey]

potato

[unreal stratus]

I guess this is overkill for this specific example, perhaps, but it shows you can systematise things somewhat

like if phi is subadditive and increasing then thisworks

[real granite]

shouldn't the second derivative be positive (so that the function itself is convex)?

actually why does the second derivative even factor into it?

you can just say |x-y| ≤ |x-z| + |z-y| for the standard metric

then since phi is increasing, order is preserved

oh wait i see the issue

yeah without the second derivative you can't "distribute" phi in the way we want to

but yh that looks like convexity to me

[lusty trench]

No, the second derivative being negative is right.

[real granite]

Elaborate...?

Convex means $\phi(tx+(1-t)y)\leq t\phi(x)+(1-t)\phi(y)$ for any $t\in[0,1]$, and this is true if (and only if?) $\phi''(x)>0$, right?

[gentle osprey]

Douglas

[lusty trench]

Yes, that's the definition of convex function. But what phi'' < 0 is the correct condition that makes potato's argument work.

[real granite]

How does $\phi''(x)<0 \implies \phi(a+b)\leq \phi(a)+\phi(b)$?

[gentle osprey]

Douglas

[lusty trench]

Notice that $\phi'$ is a decreasing function, because $\phi'' < 0$. Then $\phi(b) - \phi(0) = \int_0^b \phi'(t) , dt$ is greater than $\phi(a+b) - \phi(a) = \int_a^{a+b} \phi'(t) , dt$. But $\phi(0) = 0$.

[gentle osprey]

Eduardo León

[unreal stratus]

I fixed $a$ and considered $t \mapsto \phi(a)+\phi(t)-\phi(a+t)$. To show this is $\ge 0$, it's enough to show it has derivative $\ge 0$. but the derivative is $\phi'(t) - \phi'(a+t)$. and this is positive as you can write it as $t\phi''(c)$ for some $c$ between $a$ and $a+t$

[gentle osprey]

potato

[unreal stratus]

this is of course equivalent

anyway note $\phi(t) = \frac{t}{1+t} = 1 - \frac{1}{1+t}$ so $\phi'(t) = \frac{1}{(1+t)^2}>0$ and $\phi''(t) = -\frac{2}{(1+t)^3}<0$

[gentle osprey]

potato

[prime elbow]

I think it is not correct

[next crystal]

Yeah should be E\{x}

[opaque scroll]

I think it varies a bit from author to author. I've also seen some people distinguish between limit point and accumulation points

[tawdry valve]

I think the most common convention though is that the above definition (every neighborhood of x meets E) is for an adherent point, and the definition (every neighborhood of x meets E\{x}) is for a limit/accumulation point. The term "adherent point" is often forgotten though, since a point being adherent to E is exactly the same as it being in the closure of E. In most setups people will say "x is in E closure" rather than "x is an adherent point of E"

[alpine nest]

Yeah, which is why I think it makes more sense for "limit point" to mean something diffferent than "a point in the closure"

On the other hand the particular distinction of being a limit point only matters very occasionally, typically when talking about limits of functions

[prime elbow]

But I don't want to continue with this definition

[fierce lily]

Can someone give me some ideas about this question?( it seems that proving A is basis is not a solution.)

[slender glen]

the thing to note is that A cannot contain its boundary

why not?

then this implies A is open. why?

[radiant holly]

let (X,T|X) a topological space be and (A,T|A) subset of (X,T|X) and X={1,2,3} and T|X={X,{},{1},{2},{1,2}} A={1,2} T|A={A,{},{1},{2}} and every x open in U mean U={1,2} and yes U subset of A, and yes A open in X.

[gentle girder]

you seem to really like this example

[rancid umbra]

no need for boundary stuff. a union of open sets is open

[radiant holly]

yes but question say "show"

[radiant holly]

yes ı guess

[rancid umbra]

i’m not giving a full answer, just a hint/suggestion for a more simple approach

[radiant holly]

oh okay

[radiant holly]

but ı think we not need information on about U. Because question say "show that A is open in X and A subset of X"

[rancid umbra]

what r u trying to do?

[radiant holly]

@rancid umbra

[rancid umbra]

yes. r u trying to prove that statement?

[gentle girder]

but you can’t prove it by giving an example

also the example you gave was not enlightening at all

please for your own sake have another example of a topological space in mind

[fierce lily]

does every x open in U refers to {1} and {2} are all open in U?

[rancid umbra]

examples with subsets of euclidean spaces (like R^n) or metric spaces are usually less cumbersome to write out and easier to understand

[rancid umbra]

i wouldn’t pay attention to this atm

[radiant holly]

ı know but ı just give me a idea

[gentle girder]

specifically you’re gonna use (a bit cheekily) that the union of open sets is open

[fierce lily]

Do you mean A equal union of U?

[rancid umbra]

yes

union of all U’s that you get from the hypothesis of the statement

[fierce lily]

if x in at least one of U, then x must be in A. and if x in A, then it must lie in one of U(has given in question)

[rancid umbra]

i think that explanation makes sense.
you are saying that $$A = \bigcup_{x \in A} U_x$$ where $U_x$ is the set such that $x \in U_x \subseteq A$, correct?

[gentle osprey]

c squared

[proven comet]

I think I just proved something, but I am not sure, maybe it doesn't make sense at all. If x is in Rk and y is in Rl the product of any two neighborhouds of both will be a neighborhoud of (x,y) in R(k+l)

[gentle girder]

yes

[proven comet]

Great to know, thanks

[real granite]

$x_n \to x$ if $\forall U\ \text{open} \subseteq \bR\ \text{s.t.}\ x \in U,\ \exists n_0 \in \bN\ \text{s.t.}\ n>n_0 \implies x_n\in U$.

Is that the correct statement of the topological definition of convergence?

[gentle osprey]

Douglas

[unreal stratus]

Yes, for R

[real granite]

Okie. Also, what happens if you omit openness from the definition?

[unreal stratus]

Well then it breaks

And like then it has nothing to do with the topology

Like if this holds for all subsets then take U = {x}

[real granite]

What if it's "open or closed", therefore avoiding issues with sets that are neither open nor closed?

[unreal stratus]

{x} is closed

[real granite]

That is true

[fierce lily]

If there is a topological space X={a, b, c}, Can I say topology={X, empty} is coarser than any other topology on X?

and topology={X, empty, {1}, {1,2}} is finer than topology={empty, X, {1,2}?}, just by determine if one is the subset of the other?

[real granite]

Google says the trivial topology is the coarsest topology

[fierce lily]

Ok, then if we have two topology that contains different element and not subset relation, it will not be comparable right?

ex: {X, empty, {2}} and {X, empty, {1}, {2,3}}

[gentle girder]

it's the same as the subset relation

so like these aren't comparable because neither is a subset of the other

[fierce lily]

OK, I understand, really appreciate that

[real granite]

The discrete topology leads to all subsets being open because we can pick $0<r≤1$ and therefore $B_r (x_0) ={ x_0 }$ which is necessarily in the given subset.

Is that correct?

[gentle osprey]

Douglas

[real granite]

Because we can define a metric space's topology as $U\in\tau$ if $\forall x_0\in U,\ \exists r>0\ s.t.\ B_r(x_0) \subseteq U$

[gentle osprey]

Douglas

[unreal stratus]

Well

what you mean is the discrete metric

In the discrete topology, all subsets are open by definition

But yes agreed

"the discrete metric induces the discrete topology"

[real granite]

I mean, what I said is true if we assume the discrete topology is metrizable right?

(and it is metrizable)

Like what I've said is circular but I haven't actually confused metric and topological spaces I don't think

[unreal stratus]

Well you take discrete topology and then talked about a metric

[jaunty summit]

is the 2 skeleton of a K(G, 1) always homotopy equivalent to some presentation complex of G

like obviously theres a weak equivalence

[languid patrol]

no, you can always glue in 2-spheres which doesn't impose any relation on \pi_1 which you make contractible in the 3 skeleton

so it doesn't have to have anything to do with a presentation

[languid patrol]

if your K(G, 1) has the homotopy type of a CW complex then the two notions agree

[umbral panther]

Those count as trivial relations. You can add trivial relations to the presentation

[jaunty summit]

basically any sort of trivial glueing of 2 spheres can be accounted for i think

[median sand]

This is silly, but I'm blanking for some reason: if X,Y are second countable, why should XxY be second countable? If U_n,V_m are bases of X,Y, then every basic open set UxV is a union of some U_nxV_m, but that doesn't mean the collection {UxV} is countable, isn't its cardinality P(N)?

[opaque scroll]

If S is countable, then S^2 is also countable

[grave solstice]

you can write the basic open sets as unions of U_n x V_m. Any other open set can then be written as unions of basic open sets. You can like unfold the unions then to express it as a union of the U_n x V_m

[median sand]

Yes, but if every UxV is the union of some UnxVm, doesn't that mean there are P(N) many of them? There are N-many UnxVm in total, but each UxV is the union of some, not all, so its cardinality should be P(N), no?

[opaque scroll]

Okay, I just misunderstood what you were confused about. But like croq said, UnxVm is a basis for the topology

If (x, y) is in UxV, then there is x in Un < U and y in Vm in V, so (x, y) in UnxVm < UxV

[median sand]

Yeah, it's better to do it this way, I don't think the successive pulling out of unions works exactly.

[opaque scroll]

It works just fine

Like Union_m UxVm = Ux(union_m Vm)
so
Union_n Union_m UmxVm = (union U)x(union V)

It's just the standard argument for why the x in Um < U is equivalent to being a basis

[median sand]

I think my problem is I tried to show the collection of all UxV is countable, which I don't think it should be. Got all turned around.

[opaque scroll]

Yeah, if you choose Y to be the one point space, then that would just be saying that any second countable topology is countable

[languid patrol]

Second countable doesn’t mean every basis is countable, just at least one is

that’s the important distinction

[real granite]

You as in me or you as in one?

If the former, then that's why I brought up metrizability

[white oxide]

As long as you meant "discrete topology" as in "the topology induced by the discrete metric" it makes sense. Under the "behold! power set" definition of the discrete topology, which is the more usual usage, its a bit awkward

[unreal stratus]

You

[real granite]

former

idk the power set thing

[white oxide]

topology as in, a set of all open sets

[real granite]

idk if its normal to teach it this way but my topology module does metric spaces before topological spaces

[white oxide]

the discrete topology is then just taking all the subsets

[real granite]

oh right yh i see what you mean

[white oxide]

Yeah usually thats done in analysis

[real granite]

my uni is v weird w analysis

we effectively have three (arguably four) modules that do basic analysis

three in first year and one in second year

and i think they should all juast be condensed into one so that we can move onto big boy stuff sooner

[white oxide]

thats a bit odd, yeah

here its real analysis, topology light + multivariable and whatever else that didnt fit into the first semester, measure theory and DEs and lastly complex analysis, which is optional end of the analysis I-IV sequence

[real granite]

we have "calculus I", "sequences and series", "real analysis" as separate modules spread across two yyears

the fourth is calculus II which is vector calc

but the first three could be merged into a single module

[real granite]

yeah i get the impression my university leaves students at least a year out of step with what others are doing

another oddity is that we dont do special relativity (optional) until third year, but apparently most maths students would study that in first year if they choose to

[white oxide]

what does special relativity have to do with here?

[real granite]

not much as it concerns my question

but i mention it bc its another example of my uni doing weird stuff w teaching

[lusty trench]

What's the gentlest introduction out there to spectra and generalized cohomology theories? (I'm familiar with the basics of equivariant cohomology, but from a more pedestrian geometric standpoint. Not from the spectra POV.)

I'm intrigued by statements such as

Integers only have primes p. The sphere spectrum has primes "p", "really p", "really really p", "really really really p", ..., and "p by any means".

From here: https://mathoverflow.net/questions/433863/why-the-sphere-spectrum-is-more-correct-than-mathbbz

[merry geode]

Wow this is neat

[radiant hedge]

Hey, im trying to show by contradiction that (0,1] is connected under the subspace topology.

I know that path connectedness implies connectedness but I'm trying to show that (0,1] is compact in the following way:

[cedar pebble]

I mean generally you want to learn about complex oriented cohomology theories and chromatic stuff I guess

unironically the nlab has very good notes on this from an actual graduate course on stable homotopy theory https://ncatlab.org/nlab/show/Introduction+to+Cobordism+and+Complex+Oriented+Cohomology

[umbral panther]

The gentlest intro is Adams, but that won’t take you that far

[radiant hedge]

Let (U,V) be a seperation of (0,1]. This means that $U \cup V = (0,1]$ and $U \cap V = \emptyset$

[gentle osprey]

Sai

[radiant hedge]

I am trying to show a contradiction. Could I have a hint possibly?

[cedar pebble]

Lurie has some course notes on chromatic homotopy theory that are good. Also Adams is of course good, as is Kochman

[merry geode]

Looking into https://math.stackexchange.com/questions/137764/what-are-e-infty-rings#138677, I guess e-infty ring is a standard infinite category stuff.

[cedar pebble]

Lurie has a gentle lecture that explains this sort of statement in a little more detail https://www.youtube.com/watch?v=FYnHPF5pL4c

[merry geode]

Hmm what would be the chromatic part of this?

[cedar pebble]

of what?

[merry geode]

Like, chromatic homotopy theory.

[umbral panther]

Chromatic is the part that has multiple primes over p

[cedar pebble]

the chromatic story is what comes out of studying complex oriented cohomology theories and their parameterization in terms of formal groups

you get something called the chromatic filtration on spectra

[merry geode]

Ah, cool!

[umbral panther]

You don’t need E_oo rings to get the chromatic story. You can build Morava E-theory without it. I’m not sure you can build Morava K-theory without it. Morava did, but he’s sloppy

[merry geode]

I guess those arise from going infinite. (Tho I need to study what prime is in this context)

Morava E-theory sounds more complex, is it actually less complex

[lusty trench]

Thank you, @cedar pebble and @umbral panther .

[cedar pebble]

you have more "fields" in the setting of ring spectra than you do for ordinary rings

[umbral panther]

E-theory is complicated in some ways, but easier to construct

[merry geode]

Hmmm

[umbral panther]

Regular K-theory (with coefficients localized to (p)) is E(1). Mod p, that’s K(1). But it’s no longer a commutative ring

[cedar pebble]

Morava E-theory and Morava K-theory are tightly related to each other, Morava K-theory is what plays the role of (skew) fields here

[merry geode]

Ahhhh

[merry geode]

When I learned skew fields, it feels like division rings.

[cedar pebble]

yeah that's basically it

[merry geode]

Huh

Terminology difference

[cedar pebble]

every nonzero element being invertible is equivalent to every module being free

this gives you the notion of skew fields

[merry geode]

Yeah

[cedar pebble]

you can make the same definition for ring spectra

[merry geode]

Ah, is it defined from modules being free?

[cedar pebble]

every nonzero homogeneous element of E*({x}) being invertible is equivalent to every graded module over E*({x}) being free

in other words every E-module in spectra is free

this is what it means for a ring spectrum E to be a "field"

[merry geode]

Hmm, what is E^*({x})?

[cedar pebble]

the E-cohomology ring of a point

[merry geode]

Ohhhhhhhhh

[cedar pebble]

classically if you have two skew fields K and K' you say that K and K' have the same characteristic iff K\otimes K' is nonzero

you can do the same thing for fields in the sense of spectra

[merry geode]

Hm, but E^* is a ring, not categorical analogue with mult in categorical settings, right

[cedar pebble]

E and E' have the same characteristic iff E\wedge E' is nonzero (smash product plays the role of tensor product)

[cedar pebble]

yeah I mean you can extract a graded ring by taking E-cohomology but you can work with the spectrum directly too

[merry geode]

$E \wedge E'$

[gentle osprey]

Absta

[cedar pebble]

classically the only characteristics are 0 and p for p prime

you can also ask which characteristics you get in the setting of spectra

and it turns out that you have many more

[merry geode]

I'm sorry for asking much but could you clarify what E-cohomology is?

[cedar pebble]

any spectrum represents a generalized cohomology theory

$E^n(X)=\pi_0\mathrm{Maps}(X,\Omega^\infty\Sigma^nE)$

[gentle osprey]

nGroupoid

[merry geode]

Ahh, I guess I confused spectrum with something else

I guess (generalized) cohomology theory could be easier to handle in some sense.

[cedar pebble]

if you have a cohomological field E (so a ring spectrum which is a (skew) field in the sense I mentioned) then K=E^0({x}) is a (skew) field in the classical sense

and then the possible characteristics are like

if K has char 0 then E has the same characteristic as the Eilenberg MacLane spectrum HQ

so char 0 is not so interesting

if K has char p then the situation is more interesting

[merry geode]

Just to check, we do not have E^(-1) (i.e., trivial) right?

[cedar pebble]

not necessarily

you can have periodic cohomology theories

like complex K-theory KU is 2-periodic

[merry geode]

Hmm, I wonder how E^0 being division ring

Ah wait

E^*(pt) is a skew field right

[cedar pebble]

yeah

in the graded sense

[merry geode]

I was dumb

Anyway

[cedar pebble]

(every homogeneous element is invertible)

if K has char p the funny thing you can do is look at the classifying space X=B(Z/pZ)

[merry geode]

Ah welp, homogeneous. elements.. then different. I guess I need to study in detail to confidently speak something about it

[cedar pebble]

and then ask about E*(X) as an E*({x})-module

if E*(X) has infinite rank over E*({x}) then E has the same characteristic as the Eilenberg Maclane spectrum HF_p

but E*(X) could also have finite rank over E*({x}), it turns out it can only have rank p^n for some n

then you say that E has height n and E has the same characteristic as the Morava K-theory K(n)

so to specify the characteristic of a cohomolgical field E you either just have char 0, or you have to specify a prime p and a height n

this is what explains remarks like "Integers only have primes p. The sphere spectrum has primes "p", "really p", "really really p", "really really really p", ..., and "p by any means".

[merry geode]

Ahhh.

What is "p by any means"?

[cedar pebble]

p^n for any n

or maybe p^\infty which is like HF_p

[merry geode]

Ahhh

Right, HF_p is not like p^n for some n

[cedar pebble]

you can set K(0)=HQ and K(\infty)=HF_p and then the K(n)'s are interpolating between char 0 and char p

[merry geode]

So we have infinitely many primes over p

[cedar pebble]

yeah

[merry geode]

But prime "splitting" behavior itself does not seem that complicated, at least.

No {p, q} kind of funky stuff

[cedar pebble]

it's all nicely labeled by height

[merry geode]

Indeed, considering the module (perhaps length of a module? Or rank, my memory ) ah you said rank

[cedar pebble]

the height here is coming from the heights of formal groups

like the Morava K-theories (and any complex oriented cohomology theory E) are related to formal groups, the height of E is the height of its formal group

[merry geode]

Interesting.

But it seems like sphere spectrum could be a pain to deal with

[cedar pebble]

it can be difficult to compute with but it satisfies some nice universal properties

[merry geode]

I guess it has nicer properties than Z in some sense?

Yeah nice prties

[cedar pebble]

Z is the initial ring, S is the initial ring spectrum

[merry geode]

Idk how to put sphere spectrum in computer

[cedar pebble]

what people often do is study something like S one prime at a time, and one height at a time

[merry geode]

In which settings does these usually arise?

Like typical algebra cannot be done easily with this one, I imagine

[cedar pebble]

I mean you sort of can

it's often harder than the classical setting but there are nice things you can say

[merry geode]

But but you are supposed to compute

Oh!
Well anyway, I mean

Is this alg geo thing?

[cedar pebble]

I mean you have to ask the right sort of questions

this doesn't come up so much in algebraic geometry although it is sometimes related

usually these things are most useful to questions in topology themselves

[merry geode]

Oh so Algtop can go

[cedar pebble]

the Morava K-theories have been used recently by Floer theory people

[merry geode]

Wolfram never works

[cedar pebble]

this is like symplectic topology stuff

specifically they were proving cases of Arnold's conjecture

[merry geode]

Ahh

Fields are so deep

[cedar pebble]

but yeah in principle if you wanted to compute something like the stable homotopy groups of spheres you like

work one prime at a time

and then over each prime you work one height at a time

heights 0 and 1 are pretty classical, everything through height 2 is more or less understood

homotopy theory at height 1 is pretty nice, like if you're working K(1)-locally there is a lot you can say

[merry geode]

What kind of stuff do you prove with these?

[umbral panther]

You compute homotopy groups of spheres. A lot of things can be reduced to them. The Adams spectral sequence, which might be in his book computes the localized at 2 homotopy groups of spheres in terms of the Steenrod algebra. I guess you can glimpse the heights there

You can build lots of counter examples using the Steerod algebra. For example, Atiyah and Hirzebruch produced torsion classes in cohomology that cannot be algebraic cycles. Totaro redid this and strengthened it using more modern technology. chromatic-ish

Adams considered the geometric operation of turning a vector bundle into a bundle spheres. The structure group for homotopy bundles of spheres is close to the stable homotopy groups of spheres. So he considered a map from O to S. Simpler U to S. The homotopy groups of U are torsion free, so they’re easy. But their images are torsion of differing ranks. He made some conjecture about the ranks. This is the first chromatic layer. This produces infinitely many families, in degrees multiples of 2(p-1). That is a sort of wave length, hence the chromatic metaphor

[merry geode]

Hmmm, I see.

[lusty trench]

Turning a vector bundle into a bundle of spheres... Somehow reminds me of the Thom space stuff in Milnor-Stasheff.

[dim perch]

@opaque zodiac you know how like

[opaque zodiac]

no

[dim perch]

"classical" homotopy theory can be viewed as the study of spaces up to some version of homotopy equivalence

[opaque zodiac]

I know what a G-space is and I broadly understand homotopy equivalence but I don't acutally know anything about how the ideas interact

[dim perch]

ok yeah it's like 2 am, ignore the G

that's a spoiler

[opaque zodiac]

now I know we're in the same time zone

[dim perch]

i

you know, if you know @molten dirge well enough, then you might take from context that I'm at the university he graduated from for his phd

anyways!

sometimes you have a group

and it acts

[opaque zodiac]

I only know @molten dirge via the internet

I guess you both got your PhD's from twitter though, right?

[dim perch]

the short story is that there's two "obvious" options for the veresion of spaces in equivariant homotopy theory--homotopy types with a G-action, or topological spaces with G-action modulo equivariant homotopy (up to me slightly lying in the usual way)

[molten dirge]

Incidentally I encountered someone from the Topos Institute yesterday in the wild, turns out they’re real

and by the wild i mean a rationalist meetup

[dim perch]

the problem--on topological spaces with G-action, it turns out maps modulo equivariant homotopy are enough to get the entire homotopy type of the point-set fixed points of your topological G-spaces, which is not an invariant of the underlying homotopy type and its G-action

hence two notions of G-spaces, the "naive" one and the "genuine" one

[opaque zodiac]

right okay

[dim perch]

we've managed to phase out the word "naive" in favor of "Borel" (so-named after "borel-equivariant cohomology''), but "genuine" has unfortunately stuck around