point-set-topology

[empty grove]

(The map that you attached the 2-cell along restricted to the faces)

Pick any of the 3 faces of the 2-cell. The natural map from that to S^2 is not the same as the natural map from one of the 1-cells to S^2 (because there are no 1-cells anyway)

[silver umbra]

i see, that totally makes sense

[coarse night]

non injective inclusion

[empty grove]

[tiny ridge]

What is homotopy limit of quasicategories

Just for the pullback diagram X -> Z <- Y say

Where can I find an explicit description

Do I do a Kan fibrant replacement on Z <- Y and then take the actual pullback of SSets? That sounds wrong

Am I working in the wrong model structure? Do I need to know what the Joyal model structure is?

[empty grove]

Probably. Joyal is the right model structure for quasicats

The weak equivalences are categorical equivalences of ssets (a generalization of equivalence of quasicats), the generating acyclic cofibrations are the inner horn inclusions, and the cofibrations are the monomorphisms, so the fibrant objects are exactly the quasicats.

[tiny ridge]

Ah good that makes sense

[gentle osprey]

darqube

[broken nacelle]

is this proof of the above proposition correct?

ping me if you need me to elaborate

oh forgot to mention N subset U

[coarse night]

isn't that by definition of LCH space

what's your definition

[broken nacelle]

every point got a compact neighborhoud

and it's hausdorff obviously lel

[coarse night]

oh, thaat one

there are a lot of definitions of LC, which turn out to be equivalent for H spaces

[broken nacelle]

do you think my proof is correct tho?

I haven't done much math in a while so I'm looking for a sanity check

[coarse night]

yes

I just really wanted to use Hopkins

[broken nacelle]

what's hopkins

[coarse night]

ok wait

I mistook it as a different question, mb mb

ok looks correct

[broken nacelle]

thankyu lol

[coarse night]

lol

[coarse night]

#groups-rings-fields message

[unreal stratus]

If $X,Y$ are $G$-spaces, is $X \times_G Y$ just $X \times Y$ equipped with the diagonal action or do we quotient out by anything?

[gentle osprey]

potato

[empty grove]

Diagonal action quotiented by the action

It's a non equivariant space

(X × Y)/G

[marsh forge]

ive always disliked this notation

[unreal stratus]

Oh okay sure lol

[empty grove]

Matches the M ⊗_R N for modules

[unreal stratus]

Oh this is funny cause I was thinking for a while about smth and this analogy happened to come in handy

Well it was like, $X^i \times Y^j$ is equipped with an obvious $\Sigma_i \times \Sigma_j$-action and they construct $(X^i \times Y^j) \times_{\Sigma_i \times \Sigma_j} \Sigma_k$

[gentle osprey]

potato

[unreal stratus]

which must be directly analogous to extension of scalars, like promoting it to a sigma_k-space in the obvious way

[gritty widget]

einstein summation

[empty grove]

Oh are you looking at symmetric spectra?

[unreal stratus]

lol

Nah this is actually just some configuration spaces stuff

So like lots of sigma_k-equivariant maps lol

In particular, this is a lemma which formalises like

[empty grove]

Yep

Has the same universal property

[unreal stratus]

A choice of k points in X disjoint union Y is equivalent to a choice of i points in X and j in Y where i + j = k lol and this is how we're keeping track of the Σ_k-actions

Thanks Moldi

Is there anywhere good to look for equivariant stuff lol probably like May's stuff for homotopy theoretic tings

[empty grove]

Maybe May's equivariant homotopy and cohomology theory lol

Idk what you mean by homotopy theoretic rings though

Like ring spectra?

[marsh forge]

one interseting thing is that model categories are still the bread and butter of most equivariant work

so the OG references are all still good

(i assume you mean homotopy theoretic things)

[empty grove]

They exchanged the normal and super reaction buttons and now my muscle memory is to always super react

[marsh forge]

lol

[unreal stratus]

I said tings lol

Yeah things

[unreal stratus]

Nice

[tiny ridge]

I have no understanding of Bredon equivariant cohomology

Feels like unnatural futzing around just to make the apparatus equivariant homotopy sensitive

[marsh forge]

I actually kind of agree, I've never really understand the "genuine" equivariant category

someone once told me a compelling justification for wanting to invert representation spheres but i forget

[cosmic socket]

Are compact objects in Top or CW, compact spaces?

[tiny ridge]

I think they're just finite discrete sets lol

Direct limit is terrible in Top

[umbral panther]

Direct limits are fine in Top. Categorically compact objects are the compact spaces. Maybe CW complexes don’t have direct limits

[marsh forge]

Are you sure https://mathoverflow.net/questions/288648/what-are-compact-objects-in-the-category-of-topological-spaces

[tiny ridge]

Direct limits are not at all fine in Top. Nothing recognisable represents colim Hom(X, Y_i). This is why I had to work with quasitopological spaces for my research lol (which is a slightly civilized subcategory of Presh(Top))

[marsh forge]

oh ive never heard the term

what are they

[tiny ridge]

Presheaves of sets on Top such that gluing holds for open covers and finite closed covers

Notion is originally by Spanier Whitehead

[umbral panther]

Oh

[tiny ridge]

You may enjoy this post https://mathoverflow.net/questions/124616/why-are-quasitopological-spaces-needed-in-sheaf-theoretic-approaches-to-the-h-pr

@bw. It is very easy to come up with examples of direct limits of perfectly reasonable spaces which are indiscrete for instance

Say, S¹ -> S¹ doubling map

[marsh forge]

so slightly stronger than asking for a sheaf?

[tiny ridge]

Right

[marsh forge]

interesting

[umbral panther]

The first reason for genuine G spectra is that they come up. The obvious definition of K- theory in terms of equivariant bundles satisfies Bott periodicity for complex representation spheres

The second is Poincaré duality. Better, Atiyah duality. You want a Thom isomorphism for the tangent bundle for a manifold with isotropy. The fibers of this bundle are representations, so it’s like saying you want representation spheres to be invertible like ordinary spheres

[marsh forge]

Ah yeah the equivariant thom isomorphism was the one I had in mind

[plain raven]

let's fuckin gooo

i just really like spanier's book on alg top so i'm compulsively responding to this

[tiny ridge]

it's a good one

everyone used that in the olden times

[novel acorn]

Ibsen do you just know all of topology history

[tiny ridge]

That sounds like a backhanded insult

[coral pivot]

You do topology, that is a backhanded insult you did to yourself 🙂

[tiny ridge]

Truly the times are telling when a noncommutative geometer says your field is a self inflicted insult

[dire moon]

sorry to interrupt with a potentially dumb question but are there cases where the topological pushout of 2 closed subsets of a space by their intersection is not their union ?

[unreal stratus]

This is interesting cause iirc someone claimed this was true in a talk I went to lol

[marsh forge]

lol

its a common misconception

and very understandable given the name compact

[unreal stratus]

No, like say X is union of A and B, then the map A disjoint union B -> X is a closed map (as A, B are closed) and is surjective and identifies exactly those points in the intersection

So it induces the desired homeo

[dire moon]

Ah of course ! thanks

[unreal stratus]

np

[quick bough]

how do i construct such a path homotopy?

can't seem to get it

[thorny agate]

Yea you were right

thanks!

[empty grove]

Compact spaces are compact only relative to closed T_1 inclusions

[marsh forge]

you just mean like

a filtered colimit in which every map is a T_1 inclusion?

[empty grove]

Ye a closed T_1 inclusion

[marsh forge]

oh missed a word yeah

okay before i send this let it be known that i have a migraine and you need to be nice to me

is there a relationship between having the homotopy type of a finite CW complex and having the homotopy type of a compact space

im so sure theres like a super obvious counter example

[empty grove]

Homotopy type of a finite CW complex should imply homotopy type of compact space since finite CW complexes are compact

[marsh forge]

right

what about the other way

im mostly curious about whether compact spaces are compact in the infinity category of spaces

[empty grove]

The Hawaiian earring works since its fundamental group is uncountable

[marsh forge]

i honestly don't know which of those two the hawaiin earring is

not a finite cw i imagine

[empty grove]

Ye

[marsh forge]

hmmmmmmmm

Top just sucks then

i give up

[empty grove]

[coarse night]

[quick bough]

can anybody help?

[abstract saffron]

First time?

[empty grove]

Take the straight line homotopy between the 2 paths on the square itself, and then horizontally compose with H tilde

[umbral panther]

A simpler counter example is the one point compactification of the integers, {1/n} cup {0}

[marsh forge]

Not even close lol

[quick bough]

is there a counterexample for a space that is not hausdorff, but every compact set is closed?

[gritty widget]

god i hate topology

[marsh forge]

Lol

Any finite set should do. Every subset is closed and every subset is compact.

Oh but that’s hausdorff

god i hate topology

[gritty widget]

lol

[tiny ridge]

What about the line with two origins

[quick bough]

elaborate

[tiny ridge]

Do you know the line with two origins

[quick bough]

no hahha

[tiny ridge]

R x {0, 1} modulo equivalence rel (x, 0) ~ (x, 1) iff x nonzero

[umbral panther]

In the line with two origins, the origin is compact but not closed

[tiny ridge]

Why not

[umbral panther]

Oops
I meant a compact neighborhood of the origin. [-1,1]x0

[tiny ridge]

Ah, good

Funny

[gritty widget]

lol i said this example earlier but deleted it

[quick bough]

hmm

[tiny ridge]

Try R discrete topology with two origins

Same construction above but replace R by R_d

[umbral panther]

That’s discrete

[quick bough]

^

[gritty widget]

https://topology.pi-base.org/ just maybe...

[tiny ridge]

Interesting

Q with two origins

Lol

I think that works

[umbral panther]

Probably implies Hausdorff. You’d have to check the details carefully, but if x and y are not separated by opens, a net that converges to x probably converges to y. The image of that net plus x is probably compact but doesn’t contain y

[umbral panther]

{1/n} cup {0}

[tiny ridge]

Dang

I'm out of ideas

[quick bough]

hmm

but it’s definitely not true, right

[thorny agate]

Can we do this with finite sets?

[quick bough]

like compact => closed implies hausdorff

[tiny ridge]

bw says it is

[quick bough]

no

has to contain the cofinite topology

[tiny ridge]

That's an interesting point

Why does the cofinite topology fail

Ah I see why

[thorny agate]

Fuck

[tiny ridge]

What about the cocountable topology on R

[umbral panther]

This is a terrible argument. The net could just be constant. I guess you could solve that by taking a net indexed by open sets containing x and require each point to be in that open but not equal to x

More importantly, why is the image of a net compact? Is this really true?

[unreal stratus]

helo

[unreal stratus]

hm

[tiny ridge]

Spam enough example ideas until it's false or until you're dead

[empty grove]

If you find a counterexample you can alter the statement a little to make it not a counterexample

(like put a not in it)

[quick bough]

i mean such a space is at least T_1

[tiny ridge]

Nash equilibrium point set topology exercise

[quick bough]

i think

[quick bough]

wouldn’t that already apply that it’s not hausdorff if there’s a net that converges to x and y

since in a hausdorff space limits of nets are unique

[umbral panther]

I’m trying to prove the contrapositive
I’m assuming it’s Hausdorff and trying to find a compact that is not closed

[unreal stratus]

that's not the contrapositive right

[umbral panther]

No

In my first sentence I stated one thing. In the rest, I tried to prove the contrapositive of that thing. But I didn’t state that thing very clearly to show that it was equivalent to what was asked

[unreal stratus]

I'm confused what you're trying to prove

We're trying to prove that not Hausdorff doesn't imply not (compact => closed)

[unreal stratus]

I don't see how this relates to it

Indeed in a Hausdorff space every compact subset is closed

[umbral panther]

I left out the word not
I’m assuming not Hausdorff and finding a compact not closed

[unreal stratus]

Ah

Makes sense so you're actually proving that the statement given was true nice

hm

[tiny ridge]

Take an infinite set X in R with cocountable topology. Consider a countable infinite subset Y, and consider an open cover consisting of R \ Y, and open neighborhoods of each point of Y

[unreal stratus]

Oh I googled it and ofund the answer lol

And it is what Ibsen just said

lmao

[tiny ridge]

Oh cool

[unreal stratus]

Well not quite actually but uh

Okay nvm sure yes

I thought you meant X was the space for some reason lol but yes

[tiny ridge]

Lewis talked about cofinite topology above which made sense to me

[unreal stratus]

Uncountable set with cocountable

Works

[tiny ridge]

But cofinite fails for stupid reasons

[unreal stratus]

The compact subsets are precisely the finite ones

and hence all closed

Kinda interesting hm

Seems co-X big for a space of cardinality > X is interesting

Idk how often these spaces are used for things besides counterexamples though

[tiny ridge]

Almost never

[unreal stratus]

Idk what like analytic / point set topologists do lol

[tiny ridge]

Bing's work from the 50s is important

He did awful point set topology

Of horrifying spaces

He tried to disprove the Poincare conjecture by point set topology

But instead the machine was used to prove topological Poincare conjecture in 4D later

[umbral panther]

What is this an example of?

[tiny ridge]

Compact implies closed but not Dusseldorf

Hausdorff

[thorny agate]

Dusseldorf?

[umbral panther]

Oh, Y is a subset of X, reaching a contradiction, showing that X is finite

[tiny ridge]

Yes

[sonic zinc]

Hello everyone,
Is it is possible to use homotopy theory to study if a geometric object (like vector field, differetial form, or Poission bracket) is well-define globally on manifold?

[tiny ridge]

The keyword is "obstruction theory"

Dunno what is your exact situation though

[sonic zinc]

Okay so I am dealing with something simialr to a poission bracket
it is called getzler-soloviev bracket which is define on supermanifold and my supervisor told to study if this bracket can globazlied.
I tried the navie approch which is to check if it is ivariant under local coordinate change, but it is not
So my supervisor told look for homotopy methods to see if bracket is globalazable

[tiny ridge]

Is it a locally defined object on the supermanifold

Ie you write it in coordinates

[sonic zinc]

yes

$((f,g))=\sum_{k,\ell=0}^{\infty}\left(\partial^{\ell}\left(\partial_{k,\phi}f\right)\partial^{k}\left(\partial_{\ell,\phi^{\ddagger}}g\right)+(-1)^{\mathrm{p}(f)}\partial^{\ell}\left(\partial_{k,\phi^{\ddagger}}f\right)\partial^{k}\left(\partial_{\ell,\phi}g\right)\right)$

[gentle osprey]

beep boop

[sonic zinc]

where
$\partial_{k,\Phi_{i}}=\frac{\partial}{\partial\left(\partial^{k}\Phi_{i}\right)}$

and

$\partial^{\ell}=\frac{\partial^{\ell}}{\partial t^{^{\ell}}}$

[gentle osprey]

beep boop

[sonic zinc]

\phi are the smooth section of vector bundle

[sonic zinc]

So I should learn about obstruction theory?

[tiny ridge]

This looks like if it isn't coordinate independent it should be a section of a different, naturally defined bundle

Ie the way it changes under coordinate transforms should be set as the new transition functions

In which this would be a section of

Idk how you'd apply homotopy methods to this problem beyond, like, vector bundle theory

One should determine what the natural vector bundle here is

[sonic zinc]

Yes, with like navie clacluation
like I took simplest case where I have one variable t
and two variable on the supermanifld \phi and \phi^dagger
but it is not invariant
even at the simplest term I got something like this

$\frac{\partial}{\partial\tau}\left(\frac{\partial f}{\partial(\frac{\partial\phi}{\partial\tau})}\right)\frac{\partial}{\partial\tau}\left(\frac{\partial g}{\partial(\frac{\partial\phi^{\ddagger}}{\partial\tau})}\right) =\frac{\partial}{\partial t}\left(\frac{\partial f}{\partial(\frac{\partial\phi}{\partial t})}\right)\frac{\partial}{\partial t}\left(\frac{\partial g}{\partial(\frac{\partial\phi^{\ddagger}}{\partial t})}\right)+\left(\frac{\partial t}{\partial\tau}\frac{\partial^{2}\tau}{\partial t^{2}}\right)^{2}\frac{\partial f}{\partial(\frac{\partial\phi}{\partial t})}\frac{\partial g}{\partial(\frac{\partial\phi^{\ddagger}}{\partial t})}
+\frac{\partial t}{\partial\tau}\frac{\partial^{2}\tau}{\partial t^{2}}\frac{\partial}{\partial t}\left(\frac{\partial f}{\partial(\frac{\partial\phi}{\partial t})}\frac{\partial g}{\partial(\frac{\partial\phi^{\ddagger}}{\partial t})}\right).$

[gentle osprey]

beep boop
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tiny ridge]

I don't know if I can help you with the actual calculation

Differential geometry isn't my strongest suit, especially when it comes to coordinate computations

[tepid sparrow]

I'm currently studying Max Karoubi's book "Introduction to K-Theory" and I'm feeling a bit lost with the formalism he uses to define K-theory groups and why he gives a very general construction on banach categories. I'm struggling to understand why he chooses this approach and what advantages it offers compared to other books like Allen Hatcher's vector bundle and k theory or the chapters on k theory in the book "fibre bundle" by husemoller ?

[umbral panther]

Maybe there’s a point, but if you don’t know what it is, go with Hatcher

If you want to study K theory of operator algebras, you’ll need functional analysis, but you’ll know you need it

[sonic zinc]

Thank you
I just wanted to know if homotopy methods can be used to study this type of problem (while waiting for the co-author of my supervisor to provide me with some refernces)
I think u gave me a decent idea

[grizzled ibex]

forcing, topological games etc...

[tidal lynx]

they need to find a hobby fr

[civic verge]

Guys I am proving that X=int(X) I did the part (=>) , I wanted to read opinions or my writing is wrong.

it should be noted that I'm translating the text

I did something simple

[bitter smelt]

What is X?

Some open subset of R?

[silver umbra]

so part (a) i have now, but part (b) i rly have no idea how to do

https://math.stackexchange.com/questions/1295134/question-about-identifying-pairs-of-edges-of-disjoint-2-simplices

here's a relevant stack post

[empty grove]

Not sure if this works, but here is my attempt. We know what the vertices, edges and faces in this Delta complex structure are supposed to be, and we only need to orient the edges and faces. In particular, we know what the 1-skeleton is. Viewing the 1-skeleton as an (as of now) undirected graph, we wish to direct all its edges such that the faces can be glued on along the triangles in this graph correctly. I think a sufficient condition for this is for the graph to be directed acyclic, meaning that there shouldn't be a directed path from a vertex to itself. This is sufficient because if you pick any triangle in this graph, it must then be directed as the ordered set {0,1,2}, and then there is an obvious way to orient and glue any 2 simplices on this triangle to it. So the problem reduces to, can we always turn an undirected graph into a directed acyclic one? My guess is that this is not possible in general but here we will have some extra conditions on the graph that make it possible.

[empty grove]

Btw for the purposes of alg top, as far as I know, Delta complexes aren't that useful (I have never had to use them). Instead we have theory that works almost equally well for CW complexes and simplicial sets, so it's not like restricting to the special case of a Delta complex makes things that much nicer.

So I'm not sure how much return you're going to get from putting this much time into Delta complexes

Hmm maybe they are useful in low dim topology or for combinatorial problems, I wouldn't know about that

[silver umbra]

https://math.stackexchange.com/questions/3215307/prove-that-every-undirected-graph-has-some-orientation-that-is-a-directed-acycli

according to this stack post every undirected graph has an orientation that makes it a directed acyclic one

i also see how that conclusion would work to make it a delta complex

but like

[empty grove]

Oh convenient

[silver umbra]

we havent used an extra conditions on the structure imposed by

having to identify pairs of faces

between simplices

(and this condition is what guarantees that we get a closed surface)

[empty grove]

True

Maybe Hatcher had a different proof in mind? Or maybe this graph argument is simpler for the kind of graphs we get using this condition?

[silver umbra]

ok im pretty sure this proof doesnt work

i think its assuming that you dont have edges going from a vertex to itself straight up

but in this construction, ur definitely going to have that

if im not wrong, we we consider a delta complex structure on the torus like this

the below is what the graph actually looks like?

but the argument that every graph has an acyclic orientation

is just to give an arbitrary ordering on each of the vertices

and for each edge, orient it so that it goes from the smaller vertex in this ordering to the bigger one

[empty grove]

Ah yes

Won't work

[silver umbra]

rly not sure what the strategy would be then... i mean i definitely think its going to be a combinatorial argument but

i dont really have a sense of what the 1-skeleton treated as a graph with distinct vertices

should always look like given the imposed conditions

[empty grove]

Yeah I am not sure either

[silver umbra]

maybe it produces some kind of condition for the degree of each vertex?

(the undirected degree)

[empty grove]

Nothing straightforward at least since arbitrarily many 2-simplices could share a vertex

Has to be at least 2 I guess

But can be any number >=2

[silver umbra]

damn yeah

thats true

[empty grove]

Might not be the right direction to begin with

Or maybe we can define directed 3-acyclic as a directed graph which has no directed cycles of length 3 (but length 1 and 2 are ok) and prove that any finite undirected graph can be given this structure

Not sure how this would go lol

Actually I think you will get a condition on the loops allowed on each vertex

No wait

Can't you orient loops arbitrarily after you have oriented all the other edges according to some order on the vertices

The only sus triangles will be the ones for which all 3 edges are loops on the same vertex

But then if a 2-simplex is being attached to such a triangle, can we say that the attaching map on one of the faces can be flipped without changing the topological space

Yes we can, the attaching map remains continuous and if we attach the 2-simplices one by one, we can see that this change in the attaching map gives us something homeomorphic

Does that make any sense

[silver umbra]

im kind of getting it

but is the graph ur considering the one with distinct vertices

or nondistinct ones

[empty grove]

What do you mean by distinct vertices?

Like all 3 vertices of each 2-simplex are distinct?

[silver umbra]

like, in what i drew above

u have two different graphs

one where the vertices in ur graph actually are distinct points in ur topological space

the other where like

they're not

well tehy are but like

as a graph structure

they're not

[empty grove]

But we only have 1 topological space

I am not picking a cover where they are distinct if that is what you are asking

[silver umbra]

okay

what do you mean by a loop then

just a cycle in the graph?

[empty grove]

Are you comfortable with pushouts?

[silver umbra]

ummm

i havent heard of them

[empty grove]

A single edge whose endpoints coincide

[silver umbra]

oaky

[empty grove]

It's a kind of gluing of spaces, which is what we are doing here. I bring it up because I think I have an argument using those

Let me try and draw a diagram for this

[silver umbra]

ok

[gentle osprey]

Moldilocks1337 ✓

[silver umbra]

ok

[empty grove]

In this case, we are given a space X constructed as

[gentle osprey]

Moldilocks1337 ✓

[empty grove]

Where X_1 is the 1-skeleton, and we are taking the disjoint union of X_1 and a bunch of copies of Delta^2 and gluing them along attaching maps from each boundary to X_1

This is what is called a pushout square

Now a standard categorical fact is that if you are attaching a bunch of disjoint copies, you can attach them one at a time, like in the following sequence

[gentle osprey]

Moldilocks1337 ✓

[empty grove]

(Assuming only 4 2-simplices are being attached here)

Does this make sense?

[silver umbra]

sure, this makes sense

[empty grove]

Ok nice

So now we have oriented each edge of X_1 as follows:
Pick an ordering on the vertices. Every edge with distinct end points is directed from the smaller to the larger vertex. An edge with the same end points is directed arbitrarily.

[silver umbra]

sure

[empty grove]

Now suppose we have a 2-simplex being attached at some stage

It's attaching map, from its boundary to X_k for some k, actually factors through X_1 because it is supposed to be attached to the 1-skel

Look at what edges its being attached to

If there are no loops among those edges, then we can of course orient this simplex appropriately

Because then it is being attached to an acyclic triangle

[silver umbra]

right

[empty grove]

Otherwise, it has either 1 or 3 loops

Because if it has one loop from vertex x to x, and another edge is from x to y for some y, the last edge is also x to y

The case of 1 loop is also good, because the non-loop edges are both directed the same way - either towards the vertex with the loop or away from it, and in either case they form an acyclic triangle with this loop

Is this ok?

[silver umbra]

makes sense, the other vertex is gonna be bigger or smaller

[empty grove]

Ye

[silver umbra]

so yeah, both of them r gonna go toward it or away

[empty grove]

The case of 3 loops is more complicated because we will now have to tweak the pushout

I should have mentioned earlier, we are going by induction on the number of 2-simplices

[empty grove]

So suppose in this diagram that up to X_4 we have tweaked whatever necessary so that X_4 is a Delta complex structure without disturbing the directions of the edges

And we want to show the same for X_5

And we have tackled the cases of 0 or 1 loops

And now want to tackle it for 3 loops

So suppose that the attaching map in this case is attaching along a cyclic triangle

[silver umbra]

ok

[empty grove]

I claim that we can replace this attaching map with another attaching map without changing X_5 up to homeomorphism

The change in the attaching map is: you pick one of the three edges and traverse it in the opposite direction

Should I define this more formally or is this description understandable?

[silver umbra]

so ur swapping the orientation of one of the loops

to make it acyclic

[empty grove]

This doesn't break continuity because the end points of the edge are the same

Yep, but I am not changing the directed graph

Only the attaching map

[silver umbra]

thats a little hard to grasp

isnt the attaching map determined by the orientation of the edges in the 1-skeleton

[empty grove]

Oh no, I am assuming that we have this above diagram but it is not quite a Delta complex structure, the only obstruction being the possibly wrong orientations of the attaching maps

And we know we have this not-quite-a-Delta-complex structure on the given space because it is obtained by identifying pairs of edges

[silver umbra]

ok

[empty grove]

So the attaching map is determined by 3 maps from [0,1] to the 1-skeleton whose endpoints match in the appropriate way to make a triangle

And I am reversing one of these 3 maps, which doesn't break continuity because its end points are the same

[silver umbra]

are we sure it doesnt break some other rule of delta complexes tohugh

[empty grove]

Ye that can be checked. I don't like thinking about Hatcher's definition of Delta complexes because it gives a lot more to think about, but if you phrase it in terms of attaching maps as I did the other day, this is fine.

Idk how convincing that is lmao

But it's another argument with pushouts and their categorical properties: You have a collection of 2-simplices all glued along a bunch of edge inclusions (this forms something like a generalized pushout diagram) and you can then decompose it into a sequence of pushouts of this kind

I doubt Hatcher intended for it to be this complicated

[silver umbra]

i mean i honestly have no idea

there are some problems in hatcher that are just insane

[empty grove]

Lol

What I mean by a generalized pushout here is

[gentle osprey]

Moldilocks1337 ✓

[empty grove]

This is the pushout you will take when you want to glue 2 2-simplices along one edge

[silver umbra]

sure

[empty grove]

But if you have 3 2-simplices, you would have something like this

[gentle osprey]

Moldilocks1337 ✓

[empty grove]

Here the "pushout" is defined as taking a disjoint union of all 3 simplices, and then you glue them along the 2 maps. So in this case, simplex 1 is being glued to simplex 2 along an edge and simplex 2 to simplex 3 along possibly another edge depending on what the maps from I are

But each simplex has 3 edges and for example another edge of simplex 1 could be glued to another edge of simplex 2 or 3 and so one

[empty grove]

But the point is that we can decompose this as first gluing all the copies of I to each other to form a 1-skeleton, and then glue the 2-simplices to this 1-skeleton as in this diagram

I have made too much of a mess to stop now

[silver umbra]

im getting a little lost

[empty grove]

Ok then I should stop

[silver umbra]

also like

the fact that the graph is a triangulation of a closed surface

we still havent used

[empty grove]

I don't think that will matter

[silver umbra]

hmm

yeah i mean im definitely not seeing why it would

but also like this is insane lol

[empty grove]

Ye fair

[shadow charm]

I wonder if maybe a dual strategy is easier to work with?

Like assign to each triangle the number 1 if it is cyclically oriented and 0 otherwise: this corresponds to a labeling of the dual graph on your surface

And now you want to get all labels down to 0 by flipping bits on adjacent edges successively

[marsh forge]

honestly this problem is extra fucked bc i have never been convinced that delta complex structures are actually well defined

like every time ive tried to sit down and algorithmically explain how to label the verticies

[shadow charm]

Based

[marsh forge]

ive run into so many issues

it always works in practice but ive never actually figured out how to explain it lmfao

[shadow charm]

I know exactly what you mean

[marsh forge]

esp the orientation stuff

like there always seems to be a choice of labeling that works out but people have asked me how to do it in general and i always get tripped up

[fading vale]

its not just me!!

[marsh forge]

i would honestly challenge anyone to try to write a concrete expository on delta complexes without contradicting themselves lmfao

(and not just bc i think it wont work i would genuinely be happy to understand it if it does make sense)

new uchicago reu project

[fading vale]

No thank u

[quiet thorn]

i remember trying looking for an reu paper on this and couldn't find one

[marsh forge]

thats because its impossible

[unreal stratus]

Lol

[empty grove]

I think of them as semisimplicial sets so it's not too bad

They are equivalent

Semisimplicial sets being the presheaves on the semisimplex category, which is the category of finite totally ordered sets but only with injective order preserving maps

And to give a space a Delta complex structure is the same as giving it the structure of the geometric realisation of a semisimplicial set, so you can come up with the semisimplicial set that you expect to get and then check that the realization does give you what you want

Though I am not sure why they are useful at all. Is it just to define simplicial homology? But then cellular homology is better in every way that I can think of, so why?

[marsh forge]

delta structures are maybe slightly easier for beginners

but yeah clearly CW is the obvious choice for most people in practice

[empty grove]

Is that really it? I have seen way more confusion with them than with CW complexes lol

Because of all the orientation stuff

I was guessing that they'd be useful in some low dim topology or combinatorial topology or topological combinatorics

[marsh forge]

I think the boundary map for CW complexes confusees some people? idrk

[empty grove]

I can believe that, but it just seems like a weird reason to define a whole in between thing and do a bunch of more confusing stuff with it, especially since CW complexes are in Hatcher ch0 and assumed for the rest of the book anyway

[marsh forge]

yeah idgi honestly

its also useless above like dimension 2 lmfao

[quick bough]

does the forgetful functor from the category of pointed spaces to Top preserve all colimits?

[marsh forge]

i think no?

This would suggest that the forgetful functor has a right adjoint

and I don't think theres any way to make that work

[quick bough]

yeah, i’m trying to prove that it doesn’t have a right adjoint

but i was trying to use that it doesn’t preserve all colimits

but it kinda seems to me it does

actually, does the converse hold: if a functor preserves all colimits, then it has a right adjoint?

[marsh forge]

if the categories satify the hypothesis of the adjoint functor theorem

[quick bough]

is this the case here?

[marsh forge]

honestly Top is so awful i always for get what is true 1-categorically

at a guess yes

[quick bough]

hmm

how does one show that it doesnt have a right adjoint then

can’t think of a counterexample for a colimit

[marsh forge]

you might be able to show it can't work with hom sets directly...

but that seems hard

[quick bough]

oh wow, that seems tedious

[marsh forge]

oh i see it actually

there is a (very simple) colimit that doesn't get preserved

actually a bunch haha

what types of colimits can you think of?

[quick bough]

is the wedge sum a colimit here

[marsh forge]

yes

[quick bough]

okay, yeah

[marsh forge]

wedge sum is the coproduct of based spaces

[quick bough]

that was my first thought

[marsh forge]

the simplest example is the initial object

[quick bough]

but it gets preserved or am i just not thinking right haha

[marsh forge]

initial objects are colimits of empty diagrams

[marsh forge]

whats the coproduct in spaces

[quick bough]

disjoint union

[marsh forge]

so if i apply forget to a wedge do i get the disjoint union of the two spaces?

[quick bough]

well, you still get the quotient

where you identify the base points

[marsh forge]

I think you are confused on what it means to preserve a colimit

it doesn't mean you get the same space

[quick bough]

yeah, i know that

[marsh forge]

it means the functor sends coproducts to coproducts

[quick bough]

but wouldn’t that still be a pushout

like

[marsh forge]

oh thats the misunderstanding

preserves colimits means preserves like

the shape of the colimit

so it can't just send a coproduct to a pushout

it has to send a coproduct to a coproduct

[quick bough]

oh okay, i was overthinking it a little bit

got it

thank you tho

[marsh forge]

np

[plain raven]

it does not preserve binary coproducts.

ok nvm this seems to have been resolved.

my bad

[plain raven]

People should learn simplicial homology. simplicial complexes are important

both in homotopy theory and outside of homotopy theory

many combinatorial models for spaces can be thought of as simplicial complexes

For example, Kan complexes/infinity groupoids

many interesting properties of homology can be thought of as arising from purely combinatorial properties of simplicial complexes and not necessarily from anything to do with spaces per se.

Singular homology is a special kind of simplicial homology; it is the simplicial homology of the simplicial set of singular simplces of a space.

CW homology is useful for thinking about CW complexes. Simplicial homology is helpful in thinking about basically every area of math which has intercourse with algebraic topology.

The category of semisimplicial sets is of fundamental categorical interest. It is the free cocomplete monoidal category with a distinguished pointed object 1 -> M

[golden jackal]

What 💀

[empty grove]

Could you explain how to think of them like that?

[empty grove]

Homology for a simplicial set and homology for a simplicial complex seem like completely different things to me. I can't see how learning about the latter helps with the former at all

[empty grove]

So far I haven't been convinced that this property of the (semi)simplex category has anything to do with how they act like spaces. It feels to me more like a coincidence with minimal use than some extremely useful way to view topology

[sick elbow]

Is there some Chinese in this group,i want help

With a math video that is in Mandarin possibly
It's the only video on that topic that i can find.....

https://youtu.be/AP8Wvn9X6v8
This is the video

The time duration where i need clarification is from 15:00,from 25:45 and from 27:20...

Here he discusses
1)Some Rhombus sort of structure,i want to know with what arguments did he arrive at that conclusion

2)He talks about Cauchy Crofton Formula,what context is he talking about it.....(like why is using that there),i have rough idea of what Crofton Formula is...

3)third in 27:20 one,He talks about evaluating integral(it comes up in his slides) integral(1/(1+x^3)) how does it come up there

Help with these,even if you can just translate and type what he said,i would be very thankful

I am posting here because this seems three of the most relevant channels for this topic

[thorny agate]

Sort of confused by identification maps as well as the relations

What exactly is the relations ~ being described in each case?

and then "the identifications" makes no sense

doesn't identification depend on the definition of f and g?

this is part of my confusion

"the quotient topology" doesn't make sense to me

since it depends on the relation right?

[novel acorn]

for D^2 the relation is that any two antipodal points on the boundary are identified

while for S^2 it's that antipodal points are identified

[thorny agate]

ELI-dumb the difference?

[novel acorn]

The boundary of D^2 is S^1

so in one you identify the antipodal points of a circle

and in the other you identify the antipodal points of a sphere

[thorny agate]

oh I can't read

I thought they were both circles

Ok ok that clarifies my first question

still confused on the whole "the" situation

[coarse night]

where are you confused?

[thorny agate]

I guess I'll start with the definition of identification map

how can we say Y has "the" quotient topology

what is the relation?

is it somehow induced by the map?

[coarse night]

antipodal on the bounday

that is, x ~ -x for all x on the boundary of D^n

[thorny agate]

I meant in the general definition

[thorny agate]

here

[coarse night]

oh yeah

it's not 'the' quotient topology but the quotient topology inducecd by the identification map 'f'

[thorny agate]

ah

ok that makes more sense

I think I'm all good now

[coarse night]

cool

[thorny agate]

ok this is just a dumb question

how do you regard a unit disk as the upper hemisphere

[marsh forge]

?

Just like

the upper hemisphere is a disk

imagine cutting a sphere in half

[thorny agate]

yea so the upper hemisphere is a half sphere, not a disk

do they just mean take that plane where we made the cut and call that D^2?

[marsh forge]

the disk is the half sphere

[thorny agate]

what

[marsh forge]

up to homeomorphism

flatten the half sphere

[thorny agate]

oh fuck me

💀

thanks

[mint rose]

I think that's thai actually

[sick elbow]

ok ok i didn't know

[civic verge]

Yeah

Is a subset of X, I have to prove that X =int(X)

[thorny agate]

How is $g^{-1}(k(U))$ the saturation of $f^{-1}(U)$?

[gentle osprey]

spamakin

[marsh forge]

by consutrction right?

[thorny agate]

wdym by construction

[marsh forge]

g sends an element to its equivalence class

taking preimage is taking all elements of equivalence class

Like do you agree that g^{-1}(U) for any U is saturated?

[thorny agate]

I don't see that immediately

[marsh forge]

What does it mean to be saturated?

(by saturated i mean equal to ur own saturation)

[thorny agate]

oh nvm

ok yea I see it yea

[marsh forge]

cool

does the whole thing make sense then

[thorny agate]

yea use commutativity and the assumption

I don't see where we use the non-trivial intersection though

[marsh forge]

Well intuitively if A is missing an equivalence class then there will be some point in X/~ not in the image of A/~ -> X/~

[thorny agate]

ah

[marsh forge]

Oh yeah thats where it is

in the very last sentence

"clearly ... onto"

is where you're using this assumption

[thorny agate]

just so I'm understanding

here U doesn't have to be open or anything right?

[marsh forge]

no no

any subset

[thorny agate]

cool

that's what I thought

[bitter smelt]

idk what you were doing but here's how it goes:

Assume X is an open subset of R. You want to show X = int(X).
By definition, int(X) \subset X. So you must show X \subset int(X). That is, that all points in X are interior points.

At this point we need to know how you define open set. If it is a set for which all points are interior points, there is nothing to show. If you are just using the standard basis of R by open intervals, you proceed as follows (much like what you did).

Suppose WLOG X = (a,b). Let x \in (a,b). Then a < x < b, so that there exists \epsilon > 0 such that |x - a| > \epsilon, |x-b| > \epsilon. Then x \in (x-\epsilon, x+\epsilon) \subset (a,b). Thus x is an interior point. Since x was a generic point in X, all points in X are interior points, so X \subset int(X).

And that's pretty much it. Hope this helps. I would have walked you through it more, but I couldn't really follow your work and I could tell the language barrier would be an issue.

I think that's the direction you were attempting to prove. Lmk if you actually wanted X = int(X) implies X open

[uncut fulcrum]

Does anyone have a reference about how to come up with the 4 fibrations of spheres?

[thorny agate]

mapping cylinders and cones are weird 💀

that is all

[golden gust]

given a subspace $i : A \into X$, what's the terminology for a map $r : X \to A$ such that $r \circ i$ is only homotopic to $id_A$ ?

[gentle osprey]

(m+p)akka

[marsh forge]

homotopy retract

[thorny agate]

Is there a "isomorphism theorem" type thing for quotient spaces

Bredon started throwing around the term "factors through" without formally defining it

And it feels familiar to how people talk about factoring through quotients in algebra

But a formal definition would be neat

[marsh forge]

he's hinting at the universal property of the quotient space

[thorny agate]

Ah

I'll look at that then

[marsh forge]

essentially the idea is that if I have a continuos function X->Y and an equivalent relation ~ on X such that f respects ~

then the function should factor through X/~

i.e. there is a map X/~ -> Y making the triangle commute

[trail charm]

called “passing to the quotient” i think

or “descending to quotient”

[marsh forge]

theses are all mostly english rather than like

offiical definitions or something

[plain raven]

At first approximation I mean this in the most direct and very superficial sense, like almost all combinatorial models for spaces are basically just different ways of defining simplicial complex, some of which have technical advantages.
However I can try and give you theorems which justify my POV.
I'll try to put this in formal terms just so we can get some interesting mathematics.
An abstract simplicial complex is a set V of vertices together with a set K of finite subsets of V called simplices. K is closed downward. (K, V) represents a space, a simplicial complex, where if sigma \subset V is a simplex with n+1 points then in the realization we have the n-simplex which is formed by the convex hull of those elements. A morphism (K, V) -> (K', V') is a function V -> V' sending simplices into simplices.

On the other hand consider the category of symmetric simplicial sets, i.e., presheaves on the category of finite cardinals and not-necessarily order preserving maps. There is a nerve-realization adjunction between symmetric simplicial sets and abstract simplicial complexes given by sending an ASC (K,V) to the SSS of all maps from the standard n-simplex (P([n],[n]) to (K, V).
Theorem: The right adjoint is fully faithful and exhibits A.S.C.'s as a reflective subcategory of S.S.S.'s.
Proof: Hopefully a straightforward exercise of showing that the composition of left and right adjoint is the identity on A. S. C.'s.

We could do a similar theorem for simplicial sets maybe but it is uglier and a little less natural. There is a nerve realization adjunction between SSets and ASC's as before but you wouldn't get the original ASC back. If you want to fix this, consider the category whose objects are A.S.C.'s where V is partially ordered and the restriction of the ordering to any simplex in K is a total linear order; the morphisms are order preserving maps. Then this embeds as a fully faithful subcategory of SSets.

For semisimplicial sets, same as simplicial sets but maps have to be injective on simplices.

My main point here is the obvious one, that if you look at what the geometric realization of a simplicial set is doing, it's building a simplicial complex where the formal simplices in X_n serve as labels for copies of the n-simplex. Thus, simplicial sets are combinatorial schemas for simplicial complexes.

If you want a deeper claim, observe that a simplicial complex is, by definition, built by gluing simplices together. The word "gluing" reflects that colimits are involved somehow. A simplex here can be understood as gotten by repeatedly taking the join of the singleton space with itself, 1 * 1 *1 * 1 = \Delta^3. Thus simplicial complexes are freely generated by starting with the singleton space and taking the closure under monoidal products (joins) and colimits (gluing along permissible maps). The free cocomplete monoidal category generated a distinguished object together with certain special maps between objects precisely formalizes this intuition: in the case of semi-simplicial sets you are allowed to glue along face inclusions, for simplicial sets you can glue along faces and degeneracies, for symmetric simplicial sets add permutations.

In this way I argue that simplicial sets categorically capture what simplicial complexes are: the smallest class of spaces containing the singleton and closed under joins and colimits along certain permissible classes of gluing maps.

[plain raven]

If you have a simplicial complex, and you want to compute its simplicial homology, its associated chain complex is gotten exactly by taking the associated simplicial set (or semi-simplicial set or symmetric simplicial set depending on your choice of convention), taking free Abelian groups and looking at the Moore normalization, which is the the unique cocontinuous monoidal functor from simplicial Abelian groups to chain complexes sending the singleton space to the chain complex 0->Z ->Z->0. This is in my view the most elegant way to motivate the definition of the chain complex of a simplicial complex; it is not at all an ad hoc construction, it is completely naturally categorically.

I completely disagree with your perspective here that they are completely different things, they are actually almost exactly the same thing, not just in a derived category sense or some shit but up to isomorphism of chain complexes, but you have to understand how simplicial complexes and simplicial sets are the same thing before you can understand how their associated chain complexes are the same.

[empty grove]

Haven't read it yet, but aren't simplicial complexes different from Delta complexes

[empty grove]

"Same" feels like too strong a word for what you're describing, I only see how it's a generalisation

[empty grove]

Ok I see how that monoidal cat universal property is not entirely unrelated

I've seen ASCs being used in combinatorics and such but I still don't see how the perspective you've given is one people should study for homotopy theory. For one thing it seems more of hand wavy intuition in places, and in other places seems like saying not much more than that simplicial sets are what you get when you generalize in a certain way.

[empty grove]

Sure, but in that sense singular homology of a space is also "the same" as the homology of a simplicial set. Why bother defining simplicial complexes or "the nerve of a simplicial complex" or whatever it is called in that case? The same link holds for general top spaces

By "they seem unrelated" what I meant was that they seem just as related as singular homology and homology of a simplicial set

[empty grove]

And coming back to this, a simplex in a simplicial complex is uniquely determined by its vertices while this isn't the case in a Delta complex

[median sand]

is the following correct: if S\subset R is not discrete, then there is a sequence x_n\in S with all elements distinct and x_n-> a\in S.
Since S is not discrete, there is an a\in S such that every r-ball contains an element of S distinct from it. Using this, construct a sequence of distinct elements converging to a: for n=1, take the element x_1\in S that's in the 1-ball around a, then inductively take r=min(|x_1-a|,...,|x_n-a|,1/n) and pick x_n+1\in S from the r-ball around a (this ensures that x_n+1 is distinct from all the previous elements).

[stable kite]

I've been wondering if the following is true:
Let p : Y to X be a local homeo and U a domain in Y such that p limited to U is injective. Then already p limited to U is a homeomorphism.

[umbral panther]

Yes, an injective local homeomorphism is a homeomorphism onto its image

[limpid fern]

was about to say it has to be bijective lol but you sniped me

[tiny ridge]

If you have a map f : A -> B and restrict to a subset A' of A, we call that restriction. What about f : A -> B' where B' is a smaller subset of B containing f(A)? Should you call that corestriction?

[limpid fern]

try proving it @stable kite

[tiny ridge]

I have wondered this sometime.

[limpid fern]

is there no actual term for that

[tiny ridge]

People say "it factors through"

But that's not a verb

Verb lol I mean noun

[limpid fern]

"restriction of the codomain"

[umbral panther]

Corestriction is for B -> C
You can restrict or corestrict anything
For C -> B or A -> C it’s a property of f that it extends

[tiny ridge]

I agree with this

[umbral panther]

I think that substantive difference should explain the grammatical difference

[tiny ridge]

There are very few scenarios I can remember where treating a map f : A -> B as f : A -> C where C is a proper subset of B properly containing f(A) has been actually useful for me

Exactly once I think

Conjecture: most of the time "mapping onto image" is enough

[stable kite]

Oh, I guess for each point in U find a nbhd in U with p restricted to that nbhd a homeo -> gives us an open cover of both domain and image space, we get a family of homeos in the other direction that are all compatible and thus define a cont. map from the image to the domain?

[limpid fern]

The one I had in my head is a little simpler

[stable kite]

I do have the visual intuition that this only works for connected domain

[stable kite]

Which one did you have in your head? :)

[limpid fern]

the restriction of p to U is injective so p: U-> p(U) is bijective

just gotta check that p is open and continuous

[stable kite]

Ohh openness makes huge sense yeah

[limpid fern]

p is obviously continuous

openness follows quite quickly too

this is just using the definition of homeomorphism

[stable kite]

Yeah it does, thank you kindly!

[tidal lynx]

is the order topology on a totally ordered set (X, <) just the topology generated by sets of the form {x | x < y} and {x | x > y} for each y in X ?

I ask this because Wikipedia says this

if X has a greatest element p then the topology generated by the second definition contains no neighborhoods of p

which confuses me

for example the extended reals are certainly not generated by just sets of the form (a, b)

the the infinities would be lonely (no neighborhoods for them to live in)

[limpid fern]

maybe try looking at X= [0,1]

I don't see how there are no neighborhoods of 1

[tidal lynx]

Ok so for every a, b in [0, 1] we take the set (a, b) to be a part of our subbase

even when b = 1 intervals of the form (a, 1) don’t contain 1

[umbral panther]

The subbasis is sets of the form (a,oo) and (oo,b). The basis is finite intersections of sub basic sets and includes (a,b). But, as Wikipedia says, it also has to include the infinite intervals, for exactly the reason you say, in case there is a maximal or minimal element

[tidal lynx]

oh yeah I didn’t read after (not included in my screenshot)

this was all one big whoopsie

[unreal stratus]

As someone dipping their feet into spectra beyond the Spanier Whitehead category for the first time - how much should I worry about the different models for categories of spectra when learning? Many places seem to focus on sequential spectra and I imagine knowing about them should be enough until I bump into smth that specifically requires some other model?

Ofc the stable infinity categories are equivalent so for that stuff it shouldn't matter lol

[marsh forge]

I think you really don't need to worry much about this at all

you should learn the sequential model

and you should accept the existence of the smash product

[silver umbra]

in the last equation here

is hatcher suppressing a sum over the index t

[white oxide]

Writing the v-hats or w-hats would be a bit awkward without splitting the sum

[silver umbra]

right but

shouldnt there be a second outer sum over t

[white oxide]

Oh that

You have to interpret j bigger than t as ordered pairs

You could re-express that as two separate sums, sure

[silver umbra]

ahh okay that makes sense

[white oxide]

It helps being lazy with sum notation and free-styling it for recognizing this sorta stuff

[unreal stratus]

Lol sure thanks

[marsh forge]

once you learn the infty formalism youll be able to prove it exists easier anyway

[unreal stratus]

Yeah I'm reading more HTT over next couple months

[unreal stratus]

Didn't know this was the case actually, nice

Cheers

[marsh forge]

yeah its the universal product (w properties) such that the infinite suspension functor is symmetric monoidal

everything else is determined

[unreal stratus]

Sure nice

[marsh forge]

but yeah the two facts that matter

are that inf sus is symm monoidal

and that the tensor product preserves homotopy colimits in each variable

i.e. for fixed X the functor $-\otimes X$ preserves colimits

[gentle osprey]

themaxj

[marsh forge]

(this is also true of other tensor products)

[unreal stratus]

Sure nice

[feral copper]

Hey! Given a complex line bundle $L\to X^4$ and a surface $\Sigma\subset X$, the fact that $c_1(L)[\Sigma]=\frac{i}{2\pi}\int_\Sigma F_A$ for some connection $A$ is Chern--Weil theory, right?

[gentle osprey]

matplotlib

[feral copper]

What happens if $\Sigma$ is non-orientable? Can I relate $\Sigma\cdot\Sigma$ to the integral of the curvature in some way?

[gentle osprey]

matplotlib

[umbral panther]

If it’s non orientable, then the intersection number is only defined mod 2, so you can’t relate it to a real number like the curvature

[feral copper]

I take $\Sigma\cdot\Sigma$ to be the Euler class of the normal bundle, this is an integral lift of the mod-2 intersection number

[gentle osprey]

matplotlib

[feral copper]

(and in fact, for oriented surfaces, it is equal to the value of the intersection form on the surface)

[coarse night]

Can anyone explain how to interpret the lemma as being “concentrated on the diagonal”

[feral copper]

Somebody's working on the Wu formula it seems 😄

[coarse night]

Soon enough

[feral copper]

That's because the value of u'' cupped with (a,1) is the same as (1,a)

(P.S.: for a LaTeX-friendly version of Milnor--Stasheff: https://aareyanmanzoor.github.io/assets/books/characteristic-classes.pdf)

[coarse night]

Still lost

[feral copper]

It doesn't tell you the distribution around the diagonal, it could well be an inverted Gaussian, but you know that if you take some value over (a,1), you take the same at (1,a)
There's symmetry around the diagonal

[coarse night]

I kind of get that, so it’s “symmetric along diagonal” thing

[umbral panther]

Yes, there is a Chern-Weil version of the Euler class. It is the Pfaffian of the curvature. Presumably this ends up somewhere weird if the bundle is not oriented, but I’ve never studied it. In this case, for a 2d bundle, the curvature is just a skew 2x2 matrix, so the Pfaffian is just an entry, and you can concentrate on the orientation issues

[feral copper]

Do you have some reference in mind dealing with this stuff? 🙂

Well nvm, I found this: https://math.stackexchange.com/a/2768279/259363

I think I can try to work things out from there!

Except I'm doing Seiberg--Witten theory and I'll have more struggle XD

[umbral panther]

Here are two books I love for other reasons:
Bott & Tu
Chern, Complex manifolds without potential
The second covers Chern-Weil. The first covers Euler classes. I don’t know if either combines them

[feral copper]

So In my case, the Euler class of $\Sigma$ is $e(\Sigma)=\frac{-1}{2\pi}\mathrm{Pf}(F_A)$ for $F_A$ the curvature of a connection on the normal bundle $\nu\Sigma$?

[gentle osprey]

matplotlib

[umbral panther]

I guess

[shadow charm]

i imagine this is a difficult to answer question, but a relatively common one: what's the geometric interpretation of Poincare duality?

I can see pretty clearly what happens for the correspondence between H^0 and H_n but that's not particularly interesting, and I'm struggling to understand what's going on in other dimensions

[feral copper]

You have a nice geometric interpretation/consequence in a 4-manifold, by looking at the intersection form. This also works for curves on a surface, although it gives a skew-symmetric pairing

You also have b_k=b_{n-k} for all k

Fun fact: take my favourite field, Z/2, then you don't even need orientability, it always holds (as long as you're closed)

[umbral panther]

If homology is geometric, so is PD
Assume oriented manifold
1 Field coefficients: for every cycle not homologous to zero, there exists a cycle of complementary codim which is transverse and has intersection number 1
2 Z coefficients not torsion: you figure it out
3 torsion: for every torsion cycle there is a cycle of one less than complementary dimension disjoint with linking number 1?

[shadow charm]

Do you think you could elaborate a bit on the intersection interpretation for field coefficients

[umbral panther]

The result is surprising, but is the statement not clear? Do you want to know about transversality? How this is connected to other statements of PD?

[shadow charm]

im not sure what you mean by "is transverse": transverse with what?

and what PD has to do with intersections in the first place 🤔

[umbral panther]

We start with one cycle. There exists another cycle transverse to it with intersection number 1

[shadow charm]

intersection number in this case meaning what?

just intersects it in a single point (nvm this wouldnt make much sense)?

[umbral panther]

You can achieve that. I’m not sure how much harder that is. I mean the signed count of all intersections. A cycle is made up of simplicies. By general position, all the cycles are in the interior of the full dimensional simplicies. So it’s like an intersection of vector spaces and there are orientation issues

[shadow charm]

right okay

they intersect in finitely many points and you sum +- 1 depending on orientation

yeah that is pretty surprising

[umbral panther]

I guess there are three versions of PD over fields: cohomology, homology, and both. One is that the cup product on cohomology is a perfect pairing into the top dimension, which is F (if connected, closed, oriented). For homology, we use the geometric intersection pairing. Since homology and cohomology are dual, we can compare cup and intersection and they are the same. The third version is the cap product gives an iso from cohomology to homology. This works integrally

[shadow charm]

so that's what people mean when they say cup is poincare dual to intersection (struggling to actually see this though)

wdym by perfect pairing?

[umbral panther]

V tensor W —> F is called a pairing
It induces a map V -> Hom(W,F). If this is an iso, the pairing is perfect. (Assume finite dim, then this is symmetric in V, W. If infinite, be careful about perfection)

[shadow charm]

right okay

so basically generalised non degeneracy

[craggy pendant]

Oi

[shadow charm]

okay kind of off topic but you can build the cohomology ring using the cup product. With all this duality business cant you do the same on homology but with the intersection? I guess you'd get a coalgebra or smth though

[umbral panther]

The homology of any space is a coalgebra. But why would you study that when you can dualize? Coalgebras were first invented to study the (co)homology of Lie groups, so you can’t avoid it by duality

[shadow charm]

yeah fair enough just wondering if it was used that way

i dont like coalgebras

[unreal stratus]

Oh didn't realise coalgs came up first then

[naive hare]

is there some way to generalize semidirect product from group theory to homotopy theory

φ : A→(B≅B)
A ⋉φ B

[feral copper]

Idk, I usually think of vector bundles as semi-direct products, because of the exact sequence 1⟶F⟶E⟶B⟶1 with the canonical zero-section B⟶E

Or the converse actually: I try to think of semi-direct products as bundles

Not sure that helps though x')

[naive hare]

also wat i wrote isnt right, as A is a space with paths(groupo elements) and B=B is a path space

[feral copper]

Actually, I think some semi-direct products of Lie groups are genuine bundles

[naive hare]

Aut(x):=sum_{a}|a=x|
φ : A→(Aut(B))
A ⋉φ B

this would be more acurate

strange seeing groups and automorphism groups in this way tho

groups as spaces where group operations are paths

[umbral panther]

Yes.
If G acts on X by sufficiently coherent equivalences, then you can form an X bundle over BG. Call the total space E. You get a fiber bundle X->E->BG, which you can think of as an extension of groups loop X -> loop E -> G. If the action of G on X preserved the base point, then there is a section BG -> E corresponding to a splitting of the group extension
Loop E = loop X semi G
(Loop = Omega = based loops)

[naive hare]

i dont understand

[umbral panther]

The homotopy category of based, connected spaces is equivalent to the homotopy category of topological groups. The inverse functors are Omega, based loops, and B, the classifying space.
Once you know that, you can import any concept from group theory into homotopy theory

[marsh forge]

Do you have a reference for this claim?

its true that (infinity categorically) grouplike E_1 spaces are equivalent to connected spaces

but I don't know if this implies what you've said

In particular I assume you need to impose homotopy categories on both sides, at the very least?

[thorny agate]

this k is due to the universal property of the quotient right?

[marsh forge]

it follows from the universal property but also its not hard to show there is only one map making that diagram commute

[thorny agate]

wait how else would you show it?

[marsh forge]

by hand

[thorny agate]

wouldn't any other way just be making the universal property just explicit?

[marsh forge]

yeah, but you don't need to know its universal

[thorny agate]

fair enough

ok

[marsh forge]

like, as sets there is only one map making it commute

and you can show this is continuous

Actually wait

[thorny agate]

it's just k([x]) = [i(x)] right?

[marsh forge]

Sorry, I am a bit out of it today. I don't think this follows from the universal property because its not a priori obvious the two equivalence relations are compatible

[thorny agate]

wait what

[marsh forge]

Oh wait

yeah no its fine

sorry sorry

[thorny agate]

yea

[marsh forge]

I was thinking the wrong map

[thorny agate]

nah you're fine lol

[marsh forge]

its the universal property applies to g\circ i

[thorny agate]

ye

[marsh forge]

i was thinking just i

okay

anyway yeah

[thorny agate]

so this yea

neat neat

[marsh forge]

yes

[thorny agate]

universal properties are always the best part of learning any part of math

oh that same proof says that S^2 / ~ is "easily seen to be Hausdorff"

on an actual sphere, that makes sense

but with the relation described how do I actually show this?

[marsh forge]

I think theres a common trick

where you show that like

the action identifies neighborhoods

so like

[thorny agate]

action?

[marsh forge]

oh sorry identification

So like, I have two classes in S^2/~ i want to show are separated by open sets

and I do this by picking lifts

and picking compatible open neighborhoods in S^2

and then projecting back down

and arguing they are still disjoint

[thorny agate]

ah makes sense

[marsh forge]

I think the fact that S^2 is a metric space makes this particularly nice

[thorny agate]

it's a metric space?

oh it's a subspace of R^2

errr

[marsh forge]

yeah all i mean to say is like

[thorny agate]

R^3?

[marsh forge]

you can consider like

a small disk on the surface of S^2

[thorny agate]

yea pretty much

that's what I figured

[marsh forge]

and the distance between the two points

[thorny agate]

yea the distance part was throwing me off cause I didn't realize it was a metric

so I was trying to figure something out

[marsh forge]

I think you don't actually need to invoke it

[umbral panther]

So use that statement instead. Morally, grouplike monoids are groups and you could substitute them for any application, such as semi direct products

It should be fairly easy to prove that the inclusion of topological groups into group like monoids is fully faithful, leaving the question of what is the image. But it includes K(pi,n), so it has everything

Or you could construct a version of loops valued in groups. That’s the Kan loop group

[marsh forge]

Well, a grouplike E_1 space isn't a topological space

its a homotopy type with structure

So there's not an "inclusion" of topological groups

Basically I don't know how to make any statement of this kind at the level of topological spaces, only up to coherent homotopy

[umbral panther]

A group isn’t a property, either

[marsh forge]

huh?

My point is just that I think you lose all "up to homeomorphism" type data

I only know the statement at the level of infinity categories (or model categories)

I'm mostly asking bc I am curious if there is a strictification of this result its not something ive seen

[umbral panther]

E1 is a structure. A monoid is also a structure, in fact an E1 structure, so there is an inclusion of monoids in E1 spaces

Are you aware of strictifying E1 spaces to associative monoids? You just need an associative structure on Omega, namely the Moore loops

[marsh forge]

Yeah, you can take Moore loops, but I don't think this gives the equivalence of categories you wanted?

I was a bit confused about the claim, honestly.

[marsh forge]

This isn't my objection it was about whether you are dealing with topological spaces or not

Like, one one side I am suppose to have the homotopy category of topological groups and on the other side I am supposed to have 1-connective topological spaces?

(also E_1 spaces are not the same thing as loop spaces)

[umbral panther]

The homotopy category of pointed connected spaces

[marsh forge]

Okay, so the claim is that $\Omega: \mathrm{hTop}_{\geq 1}\to \mathrm{hTopGrp}$ is an equivalence?

[gentle osprey]

themaxj

[marsh forge]

I guess its not even clear to me that functor lands in hTopGrp at all

[umbral panther]

There are two steps. I separated them. One step is strict associativity. The other is inverses. That step is the Kan loop group. But you complained about the other step as well, and about ten other things, none of which I could tell what it was

[bitter smelt]

[marsh forge]

Sorry I am still just trying to understand the claim, and along the way you've said some stuff I couldn't quite parse

But maybe this isn't worth spelling out if you're not wanting to clarify

I was just curious bc your original claim seemed much stronger than the result I knew

[umbral panther]

You know that there’s a map in the other direction, B. You know enough to prove that it’s an equivalence without even using the Kan model

[marsh forge]

My issue is understanding what categories we are meant to land in because this is kind of technical

So I start with a topological group BG and I can assign to it a connected space BG, I guess we are saying the model for B isn't all that important

I can also go backwards by taking loops

But I am not sure these two functors are inverse without at least being more careful about the domains and codomains

Like I don't think that B is essentially surjective onto connected spaces?

I'm also not sure what model for loops takes a connected space to a strict topological group

this latter point might just be ignorance

[umbral panther]

First show that it is fully faithful. Then show that it hits K(pi,n). Use some closure property to show essentially surjective

[marsh forge]

Do you have a reference? I am saying I don't see how this works.

[umbral panther]

I’ve told you the name of the model more than once

No, I don’t believe in references

[marsh forge]

lmfao what

[umbral panther]

I believe in exercises

[marsh forge]

thats unhinged

anyway, you've said both kan loops and moore loops, I am not sure what Kan loops are. Is that the correct thing to google?

I'm not even saying you're wrong I just am trying to figure out the claim you've made lol

the only references I can find to kan loops are about simplicial sets?

[umbral panther]

Then prove that simplicial groups are homotopy equivalent to topological groups

[marsh forge]

Wait, so we aren't actually landing in topological groups?

You could've mentioned lmfao

[umbral panther]

Sure we are

[marsh forge]

Can you explain your functor from connected topological spaces to topological groups then?

[marsh forge]

I am actually not sure if this is true either

[quiet mirage]

hey there, anyone free ? i have a question

[thorny agate]

Just ask your question

[quiet mirage]

Is there a general way to represent knots on double torus? We know it for knots on toruses of genus 1. similarly is it possible for double torus knots?

[feral copper]

Wdym? Embeddings of circles on a genus two surface? Or maybe the genus 2 handlebody?

[novel acorn]

Yes!

it's called the wirtinger presentation
Tho it's more general than what you mention

[quiet mirage]

yeah its for the fundamental group right?

[novel acorn]

yes it's related to that

[quiet mirage]

I'm aware of it for knots on genus 1 torus. What is it for knots on double torus or in general genus-n body?

like i just need a direction to think, so that I can find an answer. any help appreciated

[novel acorn]

Maybe look at how many times knots cross around each of the four circles in T^2#T^2?
So it would be parametrized by like 4 integers

[quiet mirage]

hmm, I'll try that. thanks for the idea

[acoustic dome]

Hello I need some help 😭
Why is this funny formula called cocycle? More specifically, what is the "chain complex" associated with this formula, thus living up to its name of cohomology group.

[umbral panther]

Appendix C should answer the specific question

[acoustic dome]

Unfortunately Appendix C only provides routine and unenlightening def of cohomology for both group and lie group, and no mentioning of what's Ad* doing right here at all 🥲

[umbral panther]

That’s just the name of the representation

Ad means the representation of G on g. Ad* means the representation of G on g*

It is also the name of the action of G on G by conjugation. There is only one action of G on g, but in some descriptions of g it looks like conjugation. If you think of g as left invariant vector fields, it doesn’t look like conjugation, but if you think of it as the tangent space at the identity, the easiest way to see an action is conjugation

[uncut fulcrum]

What are the maps associated to each of the hopf fibrations? How does one prove that R^n is a division algebra over R --> S^n-1 is an H-space?

[unreal stratus]

Don't you need a normed division algebra?

But then you can just restrict the thing to S^(n-1)

I'm not sure if you can prove that implication directly

[uncut fulcrum]

yea normed. does there exist a nice reference to read about the proof?

[marsh forge]

Proof of which statement exactly?

[opaque scroll]

The hopf fibration S^2n+1 -> CP^n just associates S^2n+1 with the unit sphere in C^n+1, then applies the standard projection.

[uncut fulcrum]

R^n is a normed division algebra over R --> S^n-1 is an H-space

[marsh forge]

Ah I think this is actually slick let me see

I think you just use the assumptions to note that restricting the product to the unit sphere gives the desired structure

[obtuse meteor]

To further answer @uncut fulcrum’s q: to get say the map S^7 -> S^4 with fiber S^3, you can see it as

S^(4n+3) -> HP^n where HP is quaternionic projective space. n = 1 gives the desired map

Similarly for octonians

(These four maps for R,C,H, O with n = 1 are generally what I see people referring to as the Hopf fibrations)

[uncut fulcrum]

Oh yes this helps a lot, thank you so much!

Im not the most familiar with topology so this actually really helped me get an algebraic view

When I was looking at the S^3 -> S^2 situation i viewed as like 1 -> Stab_G X -> G -> G \cdot X -> 1, and was mostly looking to see if all of these things are like that

Like through orbits and such

I'm actually still getting a bit stuck: I've computed that S^0, S^1, S^3 are the only spheres that are Lie groups (is this right?), so i dunno what to do with S^7

[obtuse meteor]

S^7 is a Lie group sooo

[uncut fulcrum]

wait wut. what'd i do wrong. I got that H^1(S^7, R), H^3(S^7, R) are both zero.

hmm ig then it works

[umbral panther]

S7 is not a group. The octonians are not associative, so they don’t form a group. Writing down the Hopf map and OP2 is tricky

[uncut fulcrum]

Does there exist a way for me to view it more algebraicly?

Or should I just use the regular way of projection maps

[obtuse meteor]

Oop this was my fault!

The fact that non-associative spaces exist…

Cursed

Interesting fact about spheres in this same vein: spheres are the only H-cogroups which are also manifolds

(Can be weakened co-H space actually)

[hazy lotus]

can I show that f:[0,2pi)->S^1 defined by f(x) = (cos(x), sin(x)) is continuous just from the categorical properties of the product topology and subspace topology?

Since cos(x) and sin(x) are both continuous from [0,2pi)->R, we get that h=(cos(x), sin(x)) is continuous from [0,2pi)->R^2. Additionally, the inclusion map from the subspace S^1 to R^2 is continuous.

I feel like f being continuous should follow immediately from h being continuous to the product topology and the range of h being contained in S^1

[umbral panther]

Yes
It’s not enough to know that the map from S1 to R2 is continuous. That would be true even if you have S1 the discrete topology. What you need to use is that the topology on S1 is the subspace topology

[hazy lotus]

it being the subspace topology is equivalent to the inclusion map being continuous

I know that if f:A->B is continuous and f(A) \subseteq C and i:C->B is continuous, then h:A->C defined as h(a)=f(a) is continuous by the topological argument that h^{-1}(U_c) = f^{-1}(U_b \cap C) = f^{-1}(U_b), but I want to be able to phrase this in a more general categorical language

i.e. just proving universal properties about f, i, h using topology and letting the rest follow categorically

I'm not really experienced with category theory so I'm not sure how it would look

[hazy lotus]

also if g is f^{-1], how would I show that g^{-1}([0,pi)) is not open. We can compute g^{-1}([0,pi)) = f([0,pi)) = {(cos(x),sin(x)): x \in [0,pi)} and we would try to show (1,0) is not an interior point. Except it gets a bit messy when considering open balls and having to actually choose a point in the open ball and outside of the arc. I would like to try to avoid this problem.

[tawdry widget]

I know that knot theory is about embedding S^n into S^(n+2). Are there results about other situations? Like are there results about classification of g-torus Σ_g embedded into S^4? studying how torus are knotted. Also are there classification of 1-knots S^1 on g-torus Σ_g?

[umbral panther]

Yes, there’s a great theorem about embedding of CPn in CPn+1

[tawdry widget]

What is the theorem called and what about situations I mentioned? I mean g-torus in S^4, and S^1 in g-torus?

[umbral panther]

Knot theory usually means codimension 2
Codimension 1 is a lot easier. S1 in a surface is indecomposable elements of the fundamental group

[umbral panther]

I don’t know a name for the theorem, but it is that there is only one embedding of CPn in CPn+1 with the standard map on homology. In particular, if you take a knotted S2n in S2n+2 and take connected sum with CPn it unknots. It cancels anything

[tawdry widget]

Sounds very interesting, thank you, do you know some reference about those you said?

[umbral panther]

I think it’s an h-cobordism exercise

[tawdry widget]

Thank you, I will check it out

[tawdry widget]

What about torus in S^4 though… it’s of codimension 2. I just searched for some knot theory book, they discuss S^2 only. Any result about how torus is knotted in S^4? I mean there certainly exists knotted torus in S^4, I just wonder if there is some kind of classification of them.

[naive hare]

https://media.discordapp.net/attachments/503654482484396053/1123284158735720528/215424215199776768.png

https://media.discordapp.net/attachments/503654482484396053/1123284501464875038/215424215199776768.png

https://media.discordapp.net/attachments/503654482484396053/1123284807120584775/215424215199776768.png

How to define this semidirect product

?

[novel acorn]

idk what the other images have to do with anything lol

but take two groups G and H

and a map phi: G -> Aut(H)

[naive hare]

Here I'm using spaces as groups where paths are group operations.

[bright acorn]

What do you mean by "using spaces as groups"?

[novel acorn]

I don't think that's well defined if you don't take the homotopy classes of the paths

[naive hare]

More like using a space as a model for a group via paths

[novel acorn]

(Here I assume you're talking about the fundamental groupoid of spaces)

[bright acorn]

Yeah, just considering the loop space over a pointed topological space doesn't give you a group.

[unreal stratus]

Infinity groupoid

[bright acorn]

because if you take a loop c : S^1 -> X

concatenating with its inverse loop doesn't give you the constant function at the basepoint

something similar happens with associativity

concatenation is only associative up to homotopy

[unreal stratus]

H-group rather than group ig lol

[bright acorn]

yup

Which is still pretty cool imo

[unreal stratus]

Having stuff work on the nose would be too boring

[naive hare]

Honotopy groups are a neat generalization of groups

[unreal stratus]

They are examples of groups not a generalisation tho

[novel acorn]

OK continuing on
Take groups G and H and a homomorphism H -> Aut(G)
Then the semidirect product of G and H via phi is the group whose underlying set if G x H and whose group operation is (g,h)(g',h') = (g phi(h)(g'), hh')

[bright acorn]

they are not really a generalization of a group

[unreal stratus]

I mean ig all groups are homotopy groups lol

[bright acorn]

but a particular case

[novel acorn]

homotopy groups aren't a generalization of groups?

they are groups

[bright acorn]

yeah, since every group G is the fundamental group of the eilenberg maclane space K(G,1)

[unreal stratus]

For G commutative if n > 1 ofc

[bright acorn]

ah yeah

[naive hare]

Well spaces are generalizations I mean

[unreal stratus]

In what way do you mean

[naive hare]

Paths within the space form a group, but it's generalized by having non identical identities between these group operations.

More structure

[bright acorn]

[novel acorn]

homotopy classes of paths form a group*

it's called the fundamental groupoid of a space

[naive hare]

Ok

Construct the Tits group as a space