point-set-topology

[trail charm]

how much does it cost to get these books printed lol

[rugged quiver]

Need help with mayer vietoris of a sphere. Why it follows from this that H1(S^n) = 0 ?

[hidden crag]

can you post more context?

i mean ok i guess i the picture is enough here

you can see this by taking a closer look at the maps in your sequence since you actually know what they are instead of just that they exist

[rugged quiver]

This if full. I think the argument is without taking a look on boundary map, boundary map is very complicated.

Nvm its the difference of pullbacks of inclusion maps, something like that, not boundary map

[hidden crag]

your map from R to R \oplus R is injective

[lime sable]

if you understand how we get that sequence, then you can just count dimensions here since everything is a vector space over R

[hidden crag]

this plus exactness and dimensions gives you your result

[rugged quiver]

u mean the alternating sum of dimensions = 0? is that the argument of conclusion?

[hidden crag]

you can see that R \oplus R -> R has to be surjective (why?) and then you're done

[rugged quiver]

okay let me think abit, thank you

[rugged quiver]

i followed f is a zero map, since everything = R is in kernel.
image of zero map is zero and kernel of last map is H1(Sn). is that correct?

[tidal cedar]

I paid like $110 w/ shipping for hardcover

[hidden crag]

yes

why is f the zero map though?

[rugged quiver]

g is injective to 1 dimensional vectorspace, so its surjective. image g = R = ker f. so everything is in kernel for f, so its zero map?

[rugged quiver]

does it sound right?

[hidden crag]

sounds correct

[rugged quiver]

thank you alot<3

[hidden crag]

yw

[heady skiff]

i mean i'm not really sure if this works, but i can't really come up with anything else

can we just apply brouwer fixed point theorem lmfao

i feel like we can't because then the problem would be too easy and we're not even using the fact that it's a retraction

[swift fjord]

this is proving a special case of brouwer's fpt...

[heady skiff]

well

[swift fjord]

you can't apply a theorem that would trivialise the problem

[heady skiff]

i'm using contradiction

f(x) - x \neq 0

[swift fjord]

ok well, try using pi_1

[heady skiff]

so this is kinda hinting that we can divide by f(x) - x

yeah i tried that

well

[swift fjord]

and?

[heady skiff]

not fully

well so i know that we can embed pi_1(A, a) into pi_1(X, a) right

so i'm trying to come up with a contradiction

the tricky thing is though that it's hard to kinda associate our given continuous map f: A --> A with this

since it's not a loop

[swift fjord]

first of all, an embedding doesn't necessarily become an embedding of fundamental groups

but that doesn't matter

[heady skiff]

oh

[swift fjord]

you know that A is a retract tho

[heady skiff]

ye

[swift fjord]

so you have an embedding i:A\to X and a map r:X\to A such that ri=id_A

[heady skiff]

what i meant is that the homomorphism of fundamental groups induced by the inclusion is injective

yeah

and so like

[swift fjord]

yes

[heady skiff]

i wrote rif = f

and tried toying around w that ig

[swift fjord]

ok here's my final hint

B^2 is contractible

this should allow you to finish the proof

[heady skiff]

nullhomotopic to a point?

[swift fjord]

yes

nullhomotopic means homotopic to a point btw

[heady skiff]

oh

right

[swift fjord]

the important part is that pi_1(B^2)=0

[heady skiff]

oh huh i didn't know it was contractible

[swift fjord]

well

it's convex

try and show how you can contract every loop to a point

[heady skiff]

oh right

straight line homotopy woohoo

[hidden crag]

Did you remove my reaction shin

[swift fjord]

no?

it's still there

[hidden crag]

Yeah I put it back

But it was gone

[ebon galleon]

mod fight

[trail charm]

hmmm inspiring

[heady skiff]

i'm getting my ass kicked by this problem holy

here's my scratch work

so i know in particular that we must have $j_*: \pi_1(A, a) \to \pi_1(B^2, a)$ injective

[gentle osprey]

okeyokay

[heady skiff]

but since $\pi_1(B^2, a)$ is trivial, we must have $\pi_1(A, a)$ trivial

[gentle osprey]

okeyokay

[heady skiff]

otherwise it wouldn't be injective

now... to involve f...

oh

nvm

Kill me

this problem is HAUSDORFF

(hard asf)

[distant lichen]

What

[hazy orbit]

what is wrong with this?

[gritty widget]

what do you think

[hazy orbit]

I dont have a clue

[gritty widget]

better start picking apart the "proof" then

you should share your own thoughts on problems when you post them

[hazy orbit]

I think it should be all Aa not for some Aa

U intersect all Aa

Not U intersect some Aa

1st line is good but the 2nd line seems faulty

It should be U must intersect all Aa, so that x must belong to the closure of all Aa

@gritty widget

[coral pawn]

What did J. Frank Adams mean by this?

Is there an element of pi_n(X_n, A_n) canonically associated to S^1 ^ E^m?

[next crystal]

That parts fine, the issue is that different nbhds of x might intersect different A_alpha’s so we cant conclude that x is in the closure of some A_alpha

[hidden crag]

Remind me what E^m is here again?

https://people.math.rochester.edu/faculty/doug/otherpapers/Adams-SHGH-latex.pdf#page151

Also here’s a cleaner pdf of adams

[unreal stratus]

Is every ENR locally compact?

[languid patrol]

Yes

It’s a closed subspace of a Euclidean space

[unreal stratus]

Hm I didn't think they needed to be closed

Oh it seems definitions disagree lol

[languid patrol]

Well either way

[unreal stratus]

Hm how does the more general thing work

[languid patrol]

So what is the more general thing? What do you require of the map?

[unreal stratus]

X is an ENR if there exists an embedding i into R^n for some n and an open set U of R^n s.t. i(X) -> U admits a retraction

So i needn't be closed

So U is clearly locally compact

[languid patrol]

What is an embedding topologically though? Just an injective map? Or more?

[unreal stratus]

Homeomorphism onto its image

[languid patrol]

Okay

Well anyway the image is going to be a set X

Contained in U

With this map r: U \to X

Take an x in U and a y in X such that x is not in X

[unreal stratus]

Yes

[languid patrol]

Then there exists a pair of neighborhoods separating x, y and the pre image of the neighborhood of y intersected with X and the neighborhood of x must be open and not intersect X

So then X is closed

So okay

In general if you have a Hausdorff space then the locus where two maps agree is always closed

This is basically what we’re doing

So since X is a retract of a Hausdorff it can be cut out as the loci where the retract and the identity map agree

So it’s closed

In U

Which is all you need

Now U is locally Euclidean so any closed is locally compact etc

[languid patrol]

Does this make sense to you?

[unreal stratus]

Yes that is very pretty thank you

Take the empty set with the unique metric

But for a more serious answer, I'll think lol

[tiny obsidian]

f(x)=x+1 is a contraction in this case, and hence would have a unique fixed point by some theorem...

[gaunt linden]

You'd need your subset to be closed under adding 1, for the requirement to even make sense.

What does "complete" mean there? That the subset must be complete under the metric we construct?

[feral copper]

Hei hei! So in my paper, I describe one of the three double coverings of the 2-torus, and the referee said they didn't understand the picture very well. More precisely, they said, regarding Figure 11:

This figure is hard to fully understand.
Is it really? How could I explain it better? The covering in question is the only one which does not induce the trivial covering on one of the circles in the product S^1xS^1. It is obtained by projecting the square on the left (rotated 45°) to the square on the right.

The Figure 12 also gives a different view of what happens...

[languid patrol]

They are just a really lazy referee. But perhaps it would help if you write the index 2 subgroup of the fundamental group which you are using

Perhaps you already do this above idk

if figure 11 is your only explicit description of the map p tilde then I agree with the referee

[feral copper]

I don't think they're a lazy referee necessarily; they thoroughly reviewed the entire thing by writting an average of 3 comments per page! Maybe they didn't give it enough thought, because there are occurences of something I write down and they say is missing, or don't understand when it's really obvious.
Yeah I did write the index two subgroup indeed!

There are other pictures describing what the covering is not (mainly I'm proving the covering is this one), and there should be enough examples and evidence for the reader to understand, idk :\

I just don't know if I should leave it as is, or if I should do something; maybe they'd get upset or take it personnally if I didn't change a thing there?

[languid patrol]

ah perhaps they’re annoyed that you re-oriented the original square

So it’s harder to understand the map

[feral copper]

Ah, yeah maybe

[languid patrol]

Maybe they would prefer them to be both diamond orientation

That’s very reasonable

[feral copper]

Okay, I'll do just that then. Thanks for your output!

[unreal stratus]

nice

[broken nacelle]

I thought this defined a topology but it actually defined a topological space derp

@ebon galleon this doesn't prove you wrong but I still want you to see this

[ebon galleon]

Yes, I am familiar with Kuratowski's closure axioms

[alpine nest]

Yeah, you can define a topological space in a lot of ways

[ebon galleon]

They do end up giving you a topology

Closed sets are precisely the fixed points of the closure operation

[languid patrol]

you can also define it as a sheaf on totally disconnected sets

[ebon galleon]

And I know the point is that it's a closure operator on a (complete, Heyting) lattice of subsets so the entire set is the top element (the 1), but I hate calling the set itself 1 >.<

[broken nacelle]

That's the only thing you hate?

[ebon galleon]

What else is there to hate?

It's a perfectly fine definition

[ebon galleon]

There are, ahem, worse definitions you can use.

[coral pawn]

I think it's just a spectra

[unreal stratus]

Let $A$ be a simplicial $F$-vector space where $F$ has characteristic $0$, and $\mathrm{Sym}$ denote the free simplicial/graded commutative ring functor. Then $\pi_* \mathrm{Sym} A \cong \mathrm{Sym} \pi_* A$ by, for example, using the Quillen equivalence between cdgas and simplicial commutative rings over $F$. Is there any easier / more direct way?

[gentle osprey]

n-truncated potato

[unreal stratus]

I guess this basically amounts to a Kuenneth-type theorem as well as stuff about group actions lol

so i imagine it can't be too easy

[umbral panther]

If there were an easy proof, it would not care about the difference between chain complexes and simplicial vector spaces. But it is wildly false in positive characteristic chain complexes. (Eg, an acyclic chain complex has nontrivial homology of sym)

For positive characteristic simplicial vector spaces, I think sym is a homotopy invariant functor, but more complex than sym

[unreal stratus]

Yeah I am more familiar with the positive characteristic situation ironically since there is a good amount of literature on it e.g. Bousfield's unpublished notes and stuff

[unreal stratus]

Yeah fair enough

Yeah I guess there can't be any particularly clean proof as then it would also be more likely characteristic-agnostic

Basically I am writing up smth and said I'd go via cdgas, but my advisor suggested trying to supply a direct proof. that seems sort of impossible without a fair bit of work

It may be possible just to do it via power operations analogously to the positive characteristic case though - that'd be cute

[coral pawn]

taking the smash product of a space with the singleton is just the singleton right?

Is the unit object with respect to smashing the two element set?

[umbral panther]

aka S^0

[coral pawn]

Got it

Also, what is the group structure on the $0^{th}$ level of $[\sum^\infty X, \sum^n E]$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

Here X is a pointed space and E a spectrum

For higher levels we just consider the map S^m --> S^m v S^m

[umbral panther]

For any spectra X, Y, [X,Y] is an abelian group. You don’t have to use that one started as a space. And shifting by n is an equivalence of categories, so that doesn’t change what structures you get

But if you know how to do it for higher levels, just shift this to higher levels to get the structure

[unreal stratus]

(note though that some questions in this vein will depend on your model of the category of spectra btw)

[coral pawn]

What would the group law look like on [X,Y] on the n^th component?

[coral pawn]

Is the idea that since takng suspension is invertible, we might as well consider [suspension(X), Y] instead of [X,Y]?

[coral pawn]

Is there a canonical map X ---> suspension(X) where X is a spectrum? I'm trying to understand how exactly we're making suspension invertible. If such a canonical map existed, i assume localizing would do the trick

[distant lichen]

Suspension of spectra is a shift of the indices

[unreal stratus]

well, [susp^2(X), susp^2(Y)] is most convenient as then it's clear you get an abelian group

[coral pawn]

Why take the suspension of Y

[unreal stratus]

well you need to suspend both

suspension is an equivalence of categories

but things aren't automatically isomorphic to their suspensions

[umbral panther]

Have you studied the Spanier-Whitehead category? It’s a subcategory of spectra, but a lot easier to build

[coral pawn]

I have not

[coral pawn]

Is it an equivalence even for the naive category of spectra (i.e. no homotopy classes of morphisms)?

[unreal stratus]

which category do you mean

[umbral panther]

If you don’t take homotopy classes, then there are many models of spectra with different answers

[unreal stratus]

oh yeah this is the adams one pain

[coral pawn]

Objects are spectra and morphisms are level wise making the appropriate diagram commute

[unreal stratus]

"objects are spectra" lol

what you mean by spectra is my question

but i think this is the Adams one right

[coral pawn]

Sequences of spaces indexed by naturals with maps S^1 ^ E_m --> E_m+1

Based spaces

[unreal stratus]

oh i've just worked out what you meant by this lol

[umbral panther]

SW category is pairs (X,n), where X is a based finite complex and n is an integer. Thought of as the n-th suspension of X, the point being that n might be negative.

Maps from (X,n) to (Y,m) are the colimit over i of based maps from S^n+i X to S^m+i Y. You can take homotopy classes either before or after the colimit

[hidden crag]

This gets refined later on

Oh I didn’t see your msg bw mb

[umbral panther]

One model for spectra is called Omega spectra. En -> Omega En+1 is a homeomorphism. Then shifting the indexing is an equivalence before taking homotopy

[coral pawn]

Isn't the general rule to take homotopy classes as late as possible?

[umbral panther]

There’s trade off. The earlier you take homotopy classes, the easier. The later, the more correct

Proving that you can take them early is convenient

SW is the category of small spectra. It doesn’t matter what you do, you get the right answer. There are all sorts of technical issues that come up when you try to expand to all spectra

[umbral panther]

Questions like this make it sound like you’re confused. You should study the SW category to focus on the important core. Once you understand what spectra are supposed to be, you can add technical details

[unreal stratus]

Also let me point out to FMB that the fact we hve this colimit means we have suspension being an equivalence of cats

[umbral panther]

Oh, yeah, that was the main point. The equivalence is just changing the number n

[unreal stratus]

ye

[trail charm]

who decided that cohomology should have something called a "cohomology theory"

that is so dumb

couldn't have called it anything else

[feral copper]

How would you have called it?

[trail charm]

didnt think that far

but surely there's something better than "cohomology theory"

[feral copper]

Again; what term would you coin?

[tiny ridge]

Why is it a bad terminology

[trail charm]

it just seems silly

[tiny ridge]

Why does it seem silly

[trail charm]

if someone asked me what cohomology theory was i would assume they were asking about the theory of cohomology

[feral copper]

The thing is: there are more than one theory

[tiny ridge]

That's the point, "which theory?"

[trail charm]

oh my god this is horrible

i stand corrected

[feral copper]

Oh btw Ibsen, I have proved my non-orientable Thom conjecture wrong!

[tiny ridge]

Oh nice!

[feral copper]

What's even weirder: the smooth case is the same as the locally flat case

That is: maximal genera of orientable or non-orientable surfaces behaves differently accordingly to the smooth structure

[tiny ridge]

that's very strange

You should probably write that down

[feral copper]

Indeed; the reason is: local surfaces (thanks Whitney--Massey)

[trail charm]

this article is driving me insane

dear lord

[feral copper]

I'm going to add it to my paper indeed, I have had the referee report for about a month now

[trail charm]

ok nevermind i guess this is the same as all the named rings

[feral copper]

Welcome to the cohomology zoo!

Please be respectful; don't feed them covariant functors

[tiny ridge]

Here's a funny one: not all things called "X cohomology" is a cohomology theory

[trail charm]

???

that's ridiculous

nuh uh i am never doing algebraic topology again after this semester

this has just ruined any good thoughts i had toward alg top

[feral copper]

Said by every algebraic topologist ever

[trail charm]

dont try to pull me into your cult

[feral copper]

Oh no don't get me wrong, I don't eat this bread, I'm a geometric topologist!

[tiny ridge]

Yeah we are not complaining about your distaste for algebraic topology

[trail charm]

good

also can this be generalized to something more than CW complexes

[feral copper]

That's Eilenberg--Steenrod (?)

[tiny ridge]

The crucial thing here is two cohomology theories match if they match on points. This cannot necessarily be ensured if you don't restrict to CW complexes

The proof uses cellular cohomology

[trail charm]

ah ok

good to know i will write that down

is that iff or just if?

[feral copper]

Well, if they match, they match on points...

[tiny ridge]

iff. If they already match of course they match on points

Sniped

[trail charm]

oh true yeah

forgot about that

ok sick

[hidden crag]

I really don’t get what’s so insane about this

Cochain complexes can be built from a lot of stuff, the ones in topology you see at first are just awfully specific aren’t they

[feral copper]

#group-cohomology

[hidden crag]

and even that's from a less general pov

[fickle elm]

Why would algebraic K-theory be considered as a cohomology theory?

[hidden crag]

I think it's just a list of related topics

which modern treatment of algebraic k theory is

[umbral panther]

Etale cohomology is a generalization in one direction. Topological K-theory is a generalization in another. (Higher) algebraic K-theory of schemes is a generalization in both directions

[abstract saffron]

It's not even complete

[warm quiver]

Quantum cohomology isn't even functorial lmao

[pseudo ocean]

does anyone have a resource that gives a detailed description of handles? The literature is scattered and the wikipedia page only either redirects me to Surgery Theory (yeah ik the application of handles) or handlebody decomposition, which only gives a broad overview on handles. From how I understand it, a handle is basically a "cylinder" meant to be attached to a manifold. Is there a more broad definition? Do handles have an underlying structure like Cells?

[distant lichen]

There's not one definition of a cohomology theory anyways

If you take full ES axioms, almost nothing is

[umbral panther]

A k-handle for an n-manifold is a nice copy of D^k x D^n-k

An attachment of a k-handle is an embedding of S^k-1 x D^n-k in the boundary of an n manifold. Then you can glue on a handle to get a new manifold with boundary

Morse theory shows that every smooth n-manifold can be built up from the empty set by handle attachments. For PL manifolds it’s even more trivial. For topological manifolds of dimension >4 it’s true but difficult

By taking a handle decomposition and modifying it, you can prove theorems like h-cobordism

[unreal stratus]

dimension axiom is cringe

ascend

[distant lichen]

Crystalline cohomology isn't even homotopy invariant in the sense you'd want for algebraic geometry

A^1 homotopy invariant, that is.

[distant lichen]

Yes

Though lots of interesting things aren't even extraordinary cohomology theories

Differential cohomology is not homotopy invariant at all

I like the approach using \infty-toposes but not everything fits into that narrative either

[unreal stratus]

yes

[distant lichen]

Actually, is there a nice framework where one can fit Hochschild & cyclic homology and algebraic K-theory

Probably something with colimits of functors constant over some object

[unreal stratus]

okay you have cool interested catscradle

what do you work on? i see you have G+ hehe

[heady skiff]

where can i learn about simplical complexes/complexes or whatever

not tryna learn about them in our assigned textbook

and idk if munkres has material about them

[hidden crag]

why

[heady skiff]

i think he introduces them when he talks about triangulations

but anyways i just hate how he explains things, it's not rigorous at all

you can take a look

[ebon galleon]

did you just... upload an entire textbook in response

[heady skiff]

yes.

[hidden crag]

Is this freely available?

[gritty widget]

yeah it is he just posted it freely available to everyone here

[heady skiff]

any chads wisheth to recommendeth some alt'rnate books which has't valorous sections on simplical complexes and such?

[distant lichen]

this is a good question

Broadly geometry and topology

[unreal stratus]

||just do simplicial sets instead||

[unreal stratus]

I guessed xd

[heady skiff]

what is yond?

[distant lichen]

I like algebraic geometry and CFT stuff at the moment, but I've not really settled down

[unreal stratus]

Yeah sure thing

[distant lichen]

Also wanted to learn some more on the analytic side of mathematical physics

[unreal stratus]

by that do you mean class field theory or conformal field theory since it could be either lol

now you say mathematical physics i assume the latter

[gritty widget]

based!

[distant lichen]

Sorry, conformal field theory yes

[unreal stratus]

Cool :)

[distant lichen]

Especially using elliptic cohomology

[unreal stratus]

it's funny cause i know a major guy in mathematical CFT stuffs so there is probably a very short path from me to you academically

i guess academia is just a small world

[distant lichen]

Ah, I wouldn't be surprised

[unreal stratus]

André Henriques lol

[hidden crag]

@heady skiff since you didn’t answer I deleted and I’m gonna ask you to not just send books here

[heady skiff]

wdym

[hidden crag]

Come on potato

[unreal stratus]

Okay I thought it could be legally on there

but now realise no

[hidden crag]

???

[heady skiff]

like didn't answer

my fault bro

[hidden crag]

If this is freely available because I cba checking

But yeah we gotta enforce discord tos concerning piracy here

[heady skiff]

gotcha

[distant lichen]

I'm not good with names but it rings a bell

[knotty vine]

Anyways, I see that Spanier has a chapter on complexes but I havent read it

[heady skiff]

ohh cool i'll check it out

thanks

[knotty vine]

What a weird book.
Chapter 1: Categories, functors and groupoids (finally an AT book that starts right!), Chapter 3: Simplicial complexes ???, Chapter 9: Serre spectral sequence !!!

[heady skiff]

why exactly does $x \in a(U)$ imply $x \in U$?

[gentle osprey]

okeyokay

[gritty widget]

it doesn't. where are you reading that?

[heady skiff]

well

so I'm trying to show $U \cup a(U) \subseteq p^{-1}(p(U))$

[gentle osprey]

okeyokay

[heady skiff]

and if i'm not going insane, $p^{-1}(p(U)) = {x \in S^2 \mid p(x) \in p(U)} = {x \in S^2 \mid x \in U}$

[gentle osprey]

okeyokay

[heady skiff]

if $x \in U$, then the inclusion $U \cup a(U) \subseteq p^{-1}(p(U))$ is obvious, so I'm assuming $x \in a(U)$

[gentle osprey]

okeyokay

[heady skiff]

(well obvious if my set theory is functional)

[gritty widget]

got a proof of that second equality?

[heady skiff]

well everything in p(U) is of the form p(x) for x in U right

so it should just be all x in U

[gritty widget]

i'm not following you

got a precise proof?

i.e. show one inclusion and then the other

[heady skiff]

uh i could type one up later but for rn i don't really wanna get stuck on this lol

is there a counterexample?

[gritty widget]

okay let me say it more plainly

you are going insane, that is wrong

[knotty vine]

The second equality here is where youre going insane

[heady skiff]

LMFAO

thanks

oh

it's just because other elements on in U can be mapped to the image of U under p

?

[gritty widget]

"other elements on in U"?

you mean "not in U"?

[heady skiff]

ye

[knotty vine]

p is surjective, its fibre has 2 points

[gritty widget]

then yes, that's right. the other elements not on U which get mapped to p(U) anyways are...

(write it out. assume p(x) is in p(U). this means p(x) = p(y) for some y in U. that means...)

[heady skiff]

i think i'm going insane

"frustration is ignorance exiting the body"

provocative

[heady skiff]

how exactly do we know that p is continuous and open?

oh

quotient maps are continuous

DUH

ok wait hold up

they contend that p(a(U)) = p(U)

how exactly is this true, if flipping a point in U takes us out of U

yeah i don't understand

let $x \in p(a(U))$, then $x = p(x')$ for some $x' \in a(U)$. but they argue later that $a(U)$ and $U$ are disjoint, so how can $x'$ be in $U$

[gentle osprey]

okeyokay

[unreal stratus]

I'm not sure what your issue is

[unreal stratus]

well a(x) and x become the same under p

[heady skiff]

oh

very true

[unreal stratus]

by the way, the same proof works for 2 replaced with n for any n

(and 3 replaced by n+1 etc lol)

[tiny ridge]

This is strange. You're saying there's no notion of functoriality under symplectic immersions (M, w) -> (M', w')?

What about Lagrangian correspondences?

[warm quiver]

I’m not sure about the symplectic category, but in the algebraic category it isn’t functorial I’m pretty sure

[tiny ridge]

Oh ok fair, I don't know how to define quantum cohomology in the algebraic world

[warm quiver]

Yeah, it’s a lot more intersection theory from what I gather

I think that’s probably where functoriality breaks down

Although there is a pretty amazing theorem about quantum cohomology of a flag variety being isomorphic to the T-equivariant homology of the corresponding affine grassmanian where the ring structure is a pontryagin product structure or something along those lines. For complete flag manifolds Fl(n), the corresponding affine grassmanian has the homotopy type of the loop space of SU(n) which I think is pretty cool and probably another way to interpret Bott periodicity.

[tame arrow]

Does anyone know of a good survey article on surface braid groups (2-dimensional braid groups), that is, surfaces braided in R^4?

[umbral panther]

How do they form groups?

[tiny ridge]

So QH*(G/P) is H*_T(G/B)?

Sounds like mirror symmetry dogwhistle

[warm quiver]

I’m not entirely sure if G/B is the affine grassmanian, but see the OG reference: https://arxiv.org/pdf/0705.1386.pdf#page23

This (apparently) has a relation to mirror symmetry: https://arxiv.org/pdf/2210.17382.pdf

Actually this seems like a good set of slides: https://freemath.xyz/chowslides.pdf

and there’s some symplectic stuff in there for you

[tiny ridge]

Thanks, I'll take a look. Looks quite complicated

[warm quiver]

If G/B=K/T for K a maximal compact subgroup of G, then QH^(G/B)=H^T_(ΩK). I think this is one of the corollaries.

I forgot how to asterisk

Also, this might only be true after localizing at some classes

[tame arrow]

I would assume by the horizontal product of their motion pictures as described by Kamada (2002) Braid and Knot Theory in Dimension 4 .

[pseudo ocean]

define "nice copy"

[chrome ridge]

What does it mean fort this pairing to be "unomidular", "integral" bilinear form? Looked up online defintions of these words but not sure if they make sense in this context

[umbral panther]

There’s no real definition of a handle in isolation, only a handle attachment, which I defined, and a handle decomposition, which is a sequence of handle attachments

[tribal palm]

what is some nice condition on a space X such that ø and X are the only subsets that are both open and closed

[queen prism]

i think that's called connectedness

[broken nacelle]

yep

good ol connectedness

[edgy tree]

Connexité stuffs (Idk the English wording)

[tribal palm]

whaat, i’d have expected it to be some separation axiom

no wait the separation axioms are purely local properties aren’t they

[ebon galleon]

Usually locally, but they can have some more global definitions (e.g. (X,T) T_1 is precisely that id: (X, T) --> (X, cofinite) is continuous)

[dry jolt]

Integral just means that it's an integer-valued bilinear form on a free Z-module, unimodular means that it defines an isomorphism from said module to its dual. Equivalently, if you fix a basis for the module, a matrix representation for a unimodular form has determine +1 or -1

[chrome ridge]

How does unimodular differ from non-singular then ?

[dry jolt]

You can be non-degenerate but not unimodular

[chrome ridge]

Wait so unimodular is only defined when the bilinear form is defined on one module rather than product of two modules ?

[umbral panther]

No

If you take a unimodular pairing and multiply it by 2, it is still non degenerate but it is no longer unimodular

[dry jolt]

I don't understand this, a bilinear form is defined on an R-module M; it's a map M \otimes M -> R

[chrome ridge]

This is the def I have for being a non-singular bilinear pairing; I am just trying to understand how this is different from what you said about being unimodular

[umbral panther]

I’m not sure how standard it is that a bilinear “form” is a pairing of M with itself. If you have a pairing of two different modules (like H^i and H^n-i), you can turn it into a self pairing by taking the sum of the two

[dry jolt]

Ah, yeah that's fair, I'm being overly restrictive.

[dry jolt]

Given this definition, then yes they are equivalent. I guess I haven't seen non-singular before so I was thinking non-degenerate

[chrome ridge]

Wait now I am a little confused, I was understanding that non-singular and non-degenerate that are the same thing

[dry jolt]

non-degenerate says that if B(x, y) = 0 for all y, then x = 0. Equivalently, the map A -> Hom(B, R) is injective

[umbral panther]

In the case of Z, this is becomes perfect after tensoring with Q

Non singular is not a standard term and if I heard it I would assume it meant non degenerate, but that’s not how your book defines it

[chrome ridge]

on a related note, how to do this arugment for the H^*(M, Z) with the torsion factored out?

I am assuming there is a trick with applying the universal coefficient theorem but note sure how it goes

[dry jolt]

Universal coefficients splits, and the torsion part of H^i is Ext_1(H_{i-1}, Z), so once you factor out the torsion, the map h is an isomorphism

[chrome ridge]

But that would give $$ H^{n-k}(M; Z) / Tor \to^h Hom(H_{n-k}(M), Z) \to^{D^*} Hom(H^k(M), Z)$$ which is not quite how the argument goes. We need $$H^k(M,)/Tor$$ in the last Hom. Am I misunderstanding anything ?

[gentle osprey]

ru0xffian

[dry jolt]

The last term is H^k / Ext, again by the splitting of universal coefficients

[chrome ridge]

But doesn't D^* map The dual of H^k to the dual of H_(n-k) isomorphically?

[haughty cedar]

is it because R is not in it?

[gritty widget]

sure, that's one reason. another is that you can't write sets like (0, 1) union (1, 2), which should be open in your proposed topology, in the form (a, b)

you may be confusing topologies and their bases

[haughty cedar]

yeah, fair enough

[heady skiff]

here, does this work because if we assume that if $h$ has no fixed point then $H(x, t) = tx + (1 - t)h(x)$ is a homotopy between the inclusion $j: S^1 \to \mathbb{R}^2$ which is a contradiction since $j$ is not nullhomotopic, and if we assume that $h$ maps no point to its antipode then $F(x, t) = \frac{tx + (1 - t)h(x)}{\norm{tx + (1 - t)h(x)}}$ is another suitable homotopy between $h$ and the identity map of $S^1$, which is another contradiction (since it's not nullhomotopic)

[gentle osprey]

okeyokay

[coarse night]

Yes

[heady skiff]

cool thanks

[coarse night]

There’s something modification required for the first part @heady skiff

The map is not correct

[heady skiff]

how

[coarse night]

Where are you using the fact that there’s no fixed point?

[heady skiff]

oh i guess that's true

[coarse night]

|| the map should look similar to F||

[unreal stratus]

also uh something incomplete here

"is a homotopy between the inclusion " and what?

and also yeah you want to modify it, since in fact any map into R^2 is null homotopic

[heady skiff]

and h

which is nullhomotopic

[unreal stratus]

ye

well h maps S^1 -> S^1

which suggests how you should modify j to make H make sense

[heady skiff]

ah okay

thx

[unreal stratus]

np

[heady skiff]

could i get a hint for a please? i'm assuming for contradiction that it is nullhomotopic

i'm struggling a little bit because the things that i know about retracts have don't have a converse for if something is not a retract

so i don't really know where to start, plus all the theorems in this chapter were based off of S^1

oh it's just a generalization of (1) => (2) isn't it

[edgy tree]

some context:

(C(E, d) /~, d^) is a metric space, such that:
C(E,d) is the set of sequences (elements are in E) which sequences are Cauchy for d distance.
~ is an equivalence relation defined by: (x_n)n ~ (y_n)n <=> lim d(x_n, y_n) = 0, for n -> +inf

I have to prove that we can define d^ distance on C(E,d) /~
At first, I thought I had to show d^ is a distance but it isn't what the questions asks me. Some hints would be welcome, stuck since yesterday

[pallid delta]

1. prove d^ is well defined, i.e. it doesn't depend on the choice of representants of the equivalence classes
2. prove it verifies the metric properties and is therefore a metric
   The hardest one should be the triangle inequality:
   prove it using the triangle inequality on (E, d) by then taking the limit

"I thought I had to show d^ is a distance but it isn't what the questions asks me"
It is
"définir une distance" c'est définir quelque chose que l'on peut légitimement appeler distance, pas juste une fonction quelconque

[edgy tree]

la preuve pour montrer que c'est une distance, je l'ai déjà en tête.
Par contre le premier point me tracasse, avec " le choix ".

[pallid delta]

Soit x = (xn) fixée, y = (yn) ~ y' = (y'n) deux suites équivalentes
d(xn, y'n) <= d(xn, yn) + d(yn, y'n) -> d^(x, y)
Peux tu finir la preuve ?

ah oui, et si c'est pas déjà fait, montrer que la limite existe aussi, sinon la fonction ne peut même pas être mal définie

[edgy tree]

pour ça, j'ai:
xn de cauchy, yn de cauchy donc le couple (xn,yn) de cauchy.
Comme d est uniformément continue, d(xn,yn) de cauchy
et comme R est complet, d(xn,yn) converge

[pallid delta]

(xn, yn) est de Cauchy dans quel espace ?

en vrai norme 1 ça passe, ok

[edgy tree]

(R, d distance usuelle)

[pallid delta]

oh laa
d(xn, yn) oui, (xn, yn) non

[edgy tree]

quoique j'ai un doute maintenant en relisant les hypothèses

[edgy tree]

si j'ai bien compris pour ça:
je passe à la limite:
lim d(xn,y'n) <= lim d(xn,yn) + lim d(yn,y'n)
or yn ~ y'n <=> lim d(yn,y'n) = 0
d^(x,y') <= d^(x,y)

je décompose aussi d(xn, yn) pour obtenir l'égalité d^(x,y) = d^(x,y') ?

[pallid delta]

par symétrie, on a aussi d^(x, y) <= d^(x, y')

[edgy tree]

les deux premiers points sont good pour moi.
Je dois revoir pour l'existence de la limite.

[pallid delta]

donc ça ne dépend pas de la classe d'un argument. Je te laisse montrer que ça passe pour 2 arguments

[pallid delta]

do you mind the baguette ?

nice eyes btw

[hidden crag]

i'll say it's okay to do this one convo in french

[edgy tree]

ty, it would be worse to understand in english 🙇‍♂️ 💀

[pallid delta]

does the statement about it being fine in the help channels not really extend to the advanced channels ?

[ebon galleon]

Timo n'aime pas le français >.<

[pallid delta]

c'est pas bien

[gritty widget]

french.... ....

[heady skiff]

so if the projective plane is defined as the quotient space obtained from S^2 obtained by identifying each point x of S^2 with its antipodal point -x, would it be appropriate to kind of view it as glueing opposite ends of the sphere together in some way?

[ebon galleon]

Yes

Quotient = glueing, essentially

[heady skiff]

also, if $x \in a(U)$, is $x \in p^{-1}(p(U))$ since $x \in a(U) \implies -x \in U$, and since $p(x) = p(-x) \in U$, we see that $p(x) \in p(U)$ which gives $x \in p^{-1}(p(U))$?

[gentle osprey]

okeyokay

[quiet thorn]

What does topological glue taste like

[ebon galleon]

.< Glue is not for eating, much less topological glue

Probably plasticy

[hidden crag]

Nice try fed

[quiet thorn]

I always imagined it tasted like frosting..

[heady skiff]

anyone?

[gritty widget]

there's a small typo but yes

[heady skiff]

what's the typo?

[gritty widget]

i won't tell you

:3c

[quiet thorn]

[heady skiff]

why is it that $p(a(U)) \subseteq p(U)$? I got $x \in p(a(U)) \implies x = p(a(y))$ for some $y \in U$, so that $x = p(-y) = p(y)$, or $x \in p(U)$

[gentle osprey]

okeyokay

[heady skiff]

does that work?

[heady skiff]

i have

🍪

if you answer

[hidden crag]

a maps x to -x and p identifies x and -x

[heady skiff]

🍪

thanks

[heady skiff]

does anybody know where i can find explicit equations for this homeomorphism? i could (possibly) do it but i'm too lazy.....

also, how exactly is it injective?

because wouldn't this point get mapped to f(a, b) as well?

[knotty vine]

no

[knotty vine]

From wikipedia

where R = 1, r = 1/3

rewrite theta and phi from angles to cartesian coordinates and you're done

idk if the orientation is the same as in the text

[spice remnant]

I just needed confirmation on something basic: we can't view R^2 as a universal cover for the double torus, right? And that's why we move to hyperbolic space to find a simply connected cover?

[umbral panther]

Topologically R^2 and H^2 are the same

[spice remnant]

That is true. I should have specified. I want to know if the following is correct: if we want to go the route of simply connected coverings and deck transformations, since we can't tile R^2 with octagons, it's not easy to find the group of deck transformations

Or should it be put another way

[pseudo ocean]

ah

[normal herald]

Hi, why can a disk have any number of verticies?

Along its boundry

[merry geode]

Why not?

[languid patrol]

Put them at angle 2j\pi/n and thank me later

[normal herald]

Because I thought a vertex symbolised a point which holds together glued edges?

[merry geode]

Hwat

[normal herald]

Tf is a vertex in a topological sense and tf does it have to do with Euler characteristic

[knotty vine]

There's multiple definitions of "vertex" but since you're talking about euler characteristic, I assume you mean vertices in an abstract polyhedron/polytope?

[normal herald]

Yes, especially in regards to a disk

[hidden crag]

If you’re confused about the definitions and basic results maybe revisit the source you’re reading from first

[normal herald]

Like I’m not taking a topology class I’m taking a enumerative combinatorics class but this concept came up and they recommended reading and I was curious

[knotty vine]

In that case a vertex is just an abstract element of some set V, together with a set of edges E, and a set of faces F. Each of these sets are related by incidence. These sets and relations are abstract, theres no space yet.
A geometric realization of an abstract polyhedron (V,E,F) is a topological space consisting of a point for each element in V, a line segment for each element in E, and a polygon for each element in F, all connected along the boundaries according to the incidence relation (which I left out), and not connected in any other way (no weird intersections etc).
An abstract polyhedron is a disk if its geometric realization is homeomorphic (or even homotopy equivalent) to a disk.

If V,E,F are all finite, we can compute its Euler characteristic. It turns out that all abstract polyhedra which are disks have Euler characteristic 1

[normal herald]

Aight so when I cut a square outta surface I remove a face and 4 verticy, but when I cut a disk I’m not sure about how many verticy I remove, is it one

Like how does a connected sum work with a tube. You have 2 vertices on the boundary of the tube that you needa glue to the cutout disks. What happens to the Euler characteristic. The case with the rectangular tube with 4 faces is straightforward

[knotty vine]

Topologically speaking, a space is a disk whenever it is "homeomorphic" to a disk. Meaning there exists a continuous bijection between the two, whose inverse is also continuous. A square/triangle/octagon/whatever are all homeomorphic to a disk!

You can show that when the geometric realization of two abstract polyhedra are homeomorphic, then they have the same Euler characteristic

[normal herald]

HAHAHAHHA

[knotty vine]

Glad you like it!

[magic geyser]

What’s a subset A of R and a continuous f:A —> R that doesn’t extend to an open U containing A?

maybe doesnt exist. idk

[unreal stratus]

That's an interesting question actually, since it's way easier if U is closed instead

[magic geyser]

is there a non paracompact subset of R?

[magic geyser]

yeah you can just use a partition of unity. which made me think of paracompactness, but i know nothing about non paracompact spaces or where they live

[ebon galleon]

No; every metric space is paracompact and every subspace of a metric space is again a metric space

[magic geyser]

ah ok, then that should answer my question i guess

ty

wait i was assuming i could extend locally... am i losing my mind? continuity should let you extend locally right? but my brain cant bring up the proof

also actually paracompactness of the subset might not be enough because i would need the subcover to be locally finite at points outside of the subset as well... i would need every open cover of A to have a subcover thats locally finite at every point of R...i think...

[magic geyser]

right the closed case is Tietze extension thm

[west brook]

Abstract polyhedra cannot be disks

[knotty vine]

I loosely defined an abstract polyhedron to be a disk if its geometric realization is homeomorphic (or homotopic) to a disk

[west brook]

And that’s impossible

[knotty vine]

A filled-in triangle doesnt work?

[west brook]

That isn’t an abstract polyhedron

Abstract polyhedra need exactly two faces at each edge

[knotty vine]

Ok I didnt properly define an abstract polyhedron

My vague definition will differ from your precise one

I think the original question was also really about circles and not disks but im not sure

A circle is a polyhedron right?

[west brook]

No

[knotty vine]

Ok a polytope then

[west brook]

Still no

[knotty vine]

A non-filled in triangle has two edges at each vertex?

[abstract saffron]

I'm surprised to see some baguette in the chat

[west brook]

And?

[knotty vine]

Could you define a 1-dimensional polytope for me?

[west brook]

A line segment

[knotty vine]

I dont know a thing about polygons. Can you explain?

I only know simplicial sets/complexes

[west brook]

Explain what?

[abstract saffron]

Essentially it's polytope, but you don't specify the points and the edges

well, faces

I vaguely remember that under some very mild conditions, an abstract polytope admits a geometric realisation. In most cases, you can think of the geometric one, but keep in mind the axioms when you prove stuff

Same goes with polyhedron

[knotty vine]

Ok I was sorely mistaken using the words "abstract polytope"! I guess I shouldve said simplicial complex.
Then again the Euler characteristic applies to simplicial complexes which seem to be more general? Are they?
What is an example of an abstract polytope without geometric realization? I guess they can have transfinite dimension

[west brook]

Hemi-octahedron

[abstract saffron]

It's fucked up, but there are some examples I think

[knotty vine]

I guess were talking about a different kind of geometric realization

[knotty vine]

I seems to me this is a projective plane, right?

[west brook]

Yes

[knotty vine]

So why isnt the projective plane a geometric realization of the hemi-octahedron?

[west brook]

Because the projective plane isn’t a polyhedron

[abstract saffron]

The thing is, it means the object must be embeddable in Euclidean space, and I'm not sure how you can embed projective plane

[knotty vine]

Ohhh I get it, I was just thinking of topological spaces, but you want the realization of an abstract polyhedron to be a "concrete" polyhedron

[west brook]

Yes

[knotty vine]

I suppose polyhedra are mostly studied for their combinatorics (and symmetry groups?) Not their topology

[abstract saffron]

it has its own theory. Most common textbook on this is Ziegler's Lectures on Polytopes.

It has links with linear programming (obviously), but their topology is not that interesting afaik

There's research regarding the geometric aspects of it, but not abstract topo that we know

[knotty vine]

Cool!

[tiny ridge]

The keyword is "geodesic polygon"

There's a tiling of H^2 by regular geodesic 4n-gons. Geodesic in the hyperbolic metric!!

Then, as you said, the group of deck transformations can be explicitly nailed down as a subgroup of PSL(2, R)

[fervent root]

is this proof from munkrees proving the uniqueness of the characteristic of quotient spaces?

[spice remnant]

Right. I was just wondering if the same isn't possible at all in R². Maybe we can tile R² with non-regular octagons? Or somehow use something other than tiling to treat it as a covering space?

Sorry my question is so vague!

[umbral panther]

You can tile H^2 by regular octagons. Then apply a homeomorphism if H^2 with R^2 to get a tiling of R^2 by irregular octagons

[tame arrow]

Can that be generalized to higher dimensions H^3 and R^3 etc.?

[umbral panther]

Yes, H^3 is homeomorphic to R^3. I think you can tile it be regular dodecahedra

[west brook]

Yes, in two different ways

[gaunt linden]

Congruent tiles in H^n don't map to congruent tiles in R^n, though, so the resulting tiling is not all that interesting, is it?

[heady skiff]

here, if i'm using the fact that $h(x)$ has no fixed points, would i have a homotopy $H$ between $h$ and the anitpodal map $a$ given by $\frac{(1 - t)h(x) - tx}{\norm{(1 - t)h(x) - tx}}$? i'm assuming that would give my contradiction given that $a$ is not nullhomotopic (if that's true), which i would have to prove of course

[gentle osprey]

okeyokay

[heady skiff]

how the fuck am i supposed to see that it's not nullhomotopic

i could show that a* is a nontrivial homomorphism of fundamental groups, which boils down to showing that for some loop class [f], [a o f] is not homotopic to the constant loop [ex0], but showing things are not homotopic is not very fun

[hexed steppe]

homomorphism f : Z -> Z is determined by f(1).

so figure out which homotopy class corresponds to the integer 1, then compute its image under a*.

“showing that things are not homotopic” is the entire purpose of algebraic topology.

[heady skiff]

oh so it's just going to be sent to the loop which winds in the clockwise direction

which is just -1 in Z so therefore it's nontrivial

could i get a hint on this pls 😭 i've been stuck on it for a while i know, but i literally can't make the connection between $f$ and the fundamental groups $\pi_1(A, a_0)$ and $\pi_1(B^2, a_0)$. so I know that $\pi_1(B^2, a_0)$ is trivial, and since $A$ is a retract of $B^2$ we must have $\pi_1(A, a_0)$ trivial. the only connection I can make between $f$ (which is not a loop) and these groups is looking at the induced homomorphism $f*$ given by $h \mapsto f \circ h$, and i'm assuming for contradiction that $f(x) \neq x$ for all $x \in A$, so this gives me $(f \circ h)(s) \neq h(s)$ for all $s \in [0, 1]$??

[gentle osprey]

okeyokay

[heady skiff]

i think i should just use brouwer's fixed point theorem, someone on here suggested looking at the fundamental groups but this might be easier/i've been stuck on their approach for a bit

lol yea that was way easier holy

[unreal stratus]

wll the case A = B^2 is Brouwer's fixed point theorem so ye

[heady skiff]

so I'm given the fact that for each $n$, there's no retraction $r: B^{n + 1} \to S^n$. can I get a hint to show that the inclusion map $j: S^n \to \mathbb{R}^{n+ 1} - 0$ is not nullhomotopic? i'm cheating a little bit and using the result from the next chapter, namely that their fundamental groups are isomorphic

[gentle osprey]

okeyokay

[heady skiff]

this chapter established a lemma stating that if $h: S^1 \to X$ is nullhomotopic then $h_*$ is the trivial homomorphism of fundamental groups, but I'm unsure if this is generalized to $S^n$

[gentle osprey]

okeyokay

[heady skiff]

maybe that's the point of this exercise

[hexed steppe]

yes. look up “degree” in hatcher/wikipedia etc..

[hexed steppe]

not for arbitrary X

it is only true because R^n+1 - pt is homotopic to Sn

find a suitable retract etc.

[tiny ridge]

Sure, you get a completely terrible highly irregular tiling of R^2. How do you use that to construct the covering space?

[spice remnant]

Yeah, that was essentially my question.

So you have a discrete subgroup G in PSL(2,R) such that H/G is the double torus (H is the upper half plane). Conjugating the action with the homeomorphism f: H to R² gives an action on R², quotienting out by which should give you the double torus again--R² is a nice universal covering space for the double torus through this action.

Does this work or did I mess up somewhere?

[tiny ridge]

Sure, @spice remnant, but what is the purpose of the homeomorphism to R^2?

It's not providing you with any extra insight

The group G that is acting does not have anything to do with the geometry of R^2

[spice remnant]

Yes, it isn't. The insight comes only from H (and in fact things look better in the disc model)

[tiny ridge]

Yep

Here's one construction of the universal cover which may be more insightful. Take the Euclidean regular 4g gon with edges marked by the letters A1, B1, A1^-1, B1^-1, etc whose quotient is the surface of genus g after identifying the boundary according to the word [A1, B1] ... [Ag, Bg]. Now take infinitely many copies of this Euclidean regular 4g gon and paste them as follows: take a polygon and attach another polygon by a side preserving the letter and the orientation. Do this for all sides. You should obtain a 2-complex made up of 4g+1 polygons now, with a central polygon and the rest pasted to it's edges

Take a noncentral polygon in this 2-complex and repeat the process.

After doing this infinitely many times you're left with a 2-complex made up of Euclidean regular 4g-gons where each pair of polygons share an edge, ie there are no "open edges" left

This has a piecewise Euclidean metric: each face is a Euclidean regular 4g-gon, put the Euclidean metric there.

Each edge is isometric to an interval, etc

The group of deck transformations act naturally on this object (how?), the quotient being the surface of genus g

What you have constructed is a piecewise Euclidean model of the hyperbolic plane. This can be immediately seen by computing the total angle at a vertex: there are 4g Euclidean regular-4g gons pasted at a vertex, each of which has Euclidean internal angle ((4g - 2) * π)/(4g)

Total angle: (4g - 2)π

Angle defect (subtract from 2π): (4 - 4g)π

Since g >= 2, the angle defect is negative. This is reminiscent of negative curvature.

Indeed, the angle defect here is 2π*(2 - 2g) and 2 - 2g is the Euler characteristic of the surface of genus g

This should remind you of the Gauss-Bonnet theorem

What's happening here is that if you quotient by the deck group, you'll end up with a surface of genus g made up of a single Euclidean 4g-gon, which has a "sharp" corner where all the vertices meet, and the angle defect there is 2π(2 - 2g). This gives a metric on the surface which is flat everywhere except at a single point

That single point is where all the curvature is concentrated in

Ie the curvature of this "metric" is a Dirac delta at that vertex, multiplied by 2π(2-2g)

Such metrics are often called "orbifold metrics". They are singular Riemannian metrics, with rather controlled point-singularities

[heady skiff]

do these vi need to be distinct/linearly independent?

also i don't see how this implies that the vectors are linearly independent 💀

[unreal stratus]

you can just remove it and get the same hyperplane ig

[heady skiff]

are you talking about being in general position => v_i - v_0 are linearly independent

[unreal stratus]

uhhh

wait i may have been dumb

ignore

[heady skiff]

lol no worries

[unreal stratus]

anyway

[gaunt linden]

(It's sloppily written -- it really does need to be "... if any proper subset of them spans a strictly smaller hvperplane"; otherwise the condition is impossible to satisfy).

Suppose they are not linearly independent so there is a nonzero linear relation between the (vi-v0). Then one of the coeficients is nonzero -- say the one for (v5-v0) -- and you can divide through by that coefficient and rearrange to find v5-v0 = (some linear combination of (vi-v0) for i != 5).
But then v5 = (linear combination from before) + 1·v0 = (some linear combination of v0, v1, ..., v4, v6, ... vk, where the coefficients sum to 1).
This means that any linear-combination-with-coefficient-sum-1 of all the vi's, you can use this representation of v5 to rewrite it to a linear combination with unchanged coefficient sum that doesn't mention v5. So in this case the vectors without v5 don't span a strictly smaller hvperplane.

[heady skiff]

That doesn’t seem so easy to see

[gaunt linden]

The dificulty is mostly in getting everything straight while explaining it.

The basic idea is the general principle "if there is a nontrivial linear relation, it generally means you can do without one of the vectors, by using the linear relation to replace it with the rest", and from there it is just a matter of working out how that works in the concrete case we're looking at.

[tacit blaze]

imo it's easier to proceed by contraposition. If the vectors are linearly dependent, then you can say $v_j-v_0=\sum_{i\ne j}\mu_i (v_i-v_0)$ for some $j$ and $(\mu_i)$. Now we can prove that the hyperplane spanned by all is the same as the hyperplane spanned by all except $v_j$.\
Let $x$ in that hyperplane. $$x=\sum_{i=0}^n\lambda_iv_i,\quad \sum\lambda_i=1.$$ Then, since $v_j=v_0+\sum\mu_iv_i$, you can rewrite $$x=\sum_{i\ne j}\lambda_iv_i + \lambda_j \left(v_0+\sum_{i\ne j}\mu_i(v_i-v_0)\right).$$ This is a linear combination of all vectors but $v_j$ and you can show that the sum of coefficients is still $1$.

[gentle osprey]

upheaval

[gaunt linden]

That looks like essentially the same argument I gave.

[tacit blaze]

oh I did heavily missread what you said, read too quickly

thought it was something else

[heady skiff]

do you think it's possible to learn about simplical complexes without cell complexes and cw complexes? my book introduces it before introducing manifolds or whatever, and although the pictures are nice i just feel like it's a bit overwhelming

like it just says

"ok a 0 simplical complex is a point, a 1 simplical complex is a line, etc..." and things without justification

as in it's mostly text for the proofs ykwim

like "talking it out"

[gaunt linden]

Simplical complexes are the conceptually simplest notion and should probably be understood first.
I can see some possible arguments that cell complexes (which are formally a generalization of simplical ones) might feel more natural, but that is at least fairly debatable.
CW complexes are a further generalization, with certain (arguable) technical advantages, but should probably not be the first one to learn.

[heady skiff]

I see

Okay thanks, I’ll continue reading then

[unreal stratus]

this is also what i did

[heady skiff]

uh is there any rigorous proof that a "closed line segment" is a 1-simplex? i.e., if we have a in the "closed line segment", we can find a_1 and a_2 such that a_1 + a_2 = 1 and a_1v_0 + a_2v_1 = a?

or is the side of topology where i'm supposed to believe things without rigorous proof

[ebon galleon]

Yeah, that would be the definition of the closed line segment

[heady skiff]

ok intuitively i understand what it means for "whenever two simplexes of the collection intersect they do so at a common face", but what does this mean rigorously? say A and B are simplexes in this collection, such that A n B is nonempty. does this just mean that A n B = C where C is a simplex whose vertices are a subset of A and B?

[formal tide]

If c is not isolated, does there exist a sequence xn -> c with all xi =/= c?

[gaunt linden]

Yes.

[gaunt linden]

You'll need to give more context for that question.

[ebon galleon]

I believe you need first countability for that

Or maybe being a sequential space is enough, not sure

[formal tide]

yeah first countability is enough, wonder if it holds in general or in sequential spaces

[hexed steppe]

this doesn’t exist, except in the sense that no mathematician communicates/thinks exclusively in terms of formal proofs.

[heady skiff]

idk man, this book is starting to make me think otherwise

[hexed steppe]

read a better book then

[tiny obsidian]

this is more the side of topology where rigorous proofs that are mostly painful are dropped because they provide very little

like sure they could go general and give you an explicit homeomorphism of any closed line segment to the one in R^2 from (1,0) to (0,1), but it's not particularly relevant

[hexed steppe]

right

though it is important to think about how you would supply the details yourself if needed.

[tiny obsidian]

yes

[gaunt linden]

It can be disorienting for students who are used to (e.g. from analysis) "rigor" meaning that everything is explicitly reduced to symbolic argument and algebraic definitions of all the freshly constructed functions, though.

[tiny obsidian]

I'm more clarifying for okey that whenever intuitive or non-proven statements are given at this point, it's usually because no one actually wants to prove them for time+effort reasons, but that doesn't mean that it can't be actually checked properly that everything works out

[gentle girder]

I really have to profess my love for Allen Hatcher with respect to the "rigor level" in his books

I think he got it just right

[tiny obsidian]

speaking of hatcher

[coarse kestrel]

What does homotopic through maps of pair (X,A) -> (Y,B) here mean

[tiny obsidian]

it means that the homotopy works well with A, i.e. that each $H(\cdot, t):X\to Y$ has $H(A,t)\subseteq B$

[gentle girder]

a (homotopy of) map(s) of pairs just means that the homotopy phi_t(A) subseteq B for all t

[gentle osprey]

Edward II

[gentle girder]

basepoint-preserving homotopy for example is a homotopy through maps of pairs (X, x_0) -> (Y, y_0)

[gentle girder]

(from hatcher)

[heady skiff]

I see, yeah that’s a good point. I guess I’m just a little unclear on when I need to rigorize and prove things, and when I’m allowed to just see/intuit things in my head

[hexed steppe]

you're allowed to do whatever you want

you ideally would do both. "just" seeing/intuiting is somewhat underselling it. often such things are far harder than supplying a rigorous proof.

see https://terrytao.wordpress.com/career-advice/theres-more-to-mathematics-than-rigour-and-proofs/

[tiny obsidian]

omg it was terry

I've been wondering where I'd seen the idea of those three stages

[tribal palm]

are there metric spaces where the space of loops Ωp X is path connected but X is not??

[steel glen]

take something like (X = D^2 \cup {(2,0)}). then (\Omega_{(0,0)}X) is path connected ((D^2) is simply connected) but (X) is not

[gentle osprey]

maximo

[steel glen]

that should say D^2 U {(2,0)}

[tribal palm]

D is the disc?

[ebon galleon]

Yes

D^n = n dimensional (closed) disk/ball

[tribal palm]

right also Ωq X with q = (2,0) is trivially path connected

[languid patrol]

it's just a difference between based and unbased loop spaces, i think here \Omega_p is presumably loops based at p

for unbased loop spaces your intuition is correct

but those are.... not based 😎

[alpine nest]

The theory of cringe loops still has a lot of open problems.

[tribal palm]

ok ye lecturer is currently going through unbased loops and their many variants and are clearing leading towards the notion we will stick with

[tribal palm]

$$\overline A = \bigcup_{x\in X} \bigcap_{\substack{U\in\mathscr N(x) \ U \cap A \neq \emptyset}} U $$

does this check out

[gentle osprey]

Jens

[coarse night]

looks right but why would you write this

just write intersection of all closed subsets containing A

[tribal palm]

practice

or actually in this problem i did use this logical form of the closure (that of the set of adherent points), i just wrote it out sensibly in words, turned out nice

[coarse night]

X= {a, b} and the topology is{ ϕ, {a}, {a,b}}. What's the closure of {a}?

@tribal palm

[tribal palm]

X

[coarse night]

do you get X by using your definition?

actually you do, I thought first union is over A and not X. nvm

[tribal palm]

yes, phew, took me a min to write out

[coarse night]

everything good

[tribal palm]

well tbh i’m still not confident it will hold in general

but it looks close enough for the moment so i will just pretend everything is good and go on

[heady skiff]

what does it mean for three points to be non-collinear? that none of them is a scalar multiple of the other? and what about non-coplanar?

obviously it means they don't lie in the same plane

but as i n, how can we express that mathematically

[knotty vine]

By saying theyre geometrically independent?

[tacit blaze]

3 points a0, a1, a2 are not colinear if they aren't aligned, and equivalently it means a1-a0 and a2-a0 are independent

likewise 4 points aren't coplanar if no plane contains them all, and that's equivalent to a1-a0, a2-a0, a3-a0 being independent

[knotty vine]

I.e., geometric independence provides the definition of non-co(linear/planar)

[heady skiff]

as in they're equivalent

or they're defining it as such

[knotty vine]

Defining ig, just check that you intuition for the word "non-coplanar" coincides with the definition and your good

[heady skiff]

ok thanks

[knotty vine]

Or you could take non-coplanar to mean #point-set-topology message and then prove that this is just geometric independence for 4 vectors

which is trivial

[heady skiff]

why does S not carry P onto R^n instead of R^n x 0? also, by unit basis vectors they just mean the standard basis vectors right?

[knotty vine]

R^n x 0 is just R^n embedded into R^N in the first n coordinates

Also, yes

[heady skiff]

oh ok so there's an implicit injection

oh wait so geometrically independent => linearly independent

since linear independence is one of the conditions

oops

or is that not true

[knotty vine]

4 points in R3 can be geometrically independent, but never linearly

But lin indep => geom indep

[heady skiff]

so to show that the $t_i$ in $\sum_{i = 0}^n t_ia_i$ are unique we couldn't just do an argument like $\sum_{i = 0}^n t_ia_i = \sum_{i = 0}^m s_ia_i$ and subtract from both sides and assert that the $t_i - s_i = 0$?

[gentle osprey]

okeyokay

[knotty vine]

Is this refering to something? Idk the context

[heady skiff]

oh yea my fault

wdym

like for example if n = m and t1a1 + t2a2 = s1a1 + s2a2 then (t1 - s1)a1 + (t2 - s2)a2 = 0?

or am i trippin

[knotty vine]

Oh wait youre right, im dumb

Yeah and then you gotto use independence for the last step

[heady skiff]

ye that makes sense

cool thx

nah youre chillin lol

[knotty vine]

Im gonna delete that embarrassing msg

[heady skiff]

lol dw it wasn't embarassing

ok now i'm confused as to how an affine transformation preserves geometrically independent sets - an affine transformation is not necessarily injective right

[knotty vine]

Still not sure how to do the last step though, since a_i are only geometrically independent

[knotty vine]

It is, the definition ask for non-singular transformations

i.e. injective ones

[heady skiff]

because if we have $t_1 + t_2 + \dots + t_n = 0$ and $t_0T(a_0) + \dots + t_nT(a_n) = 0$, then I tried to use linearity, but then $T(t_0a_0 + \dots t_na_n) = 0$ and that doesn't tell us anything/we can't use the geometric indepedence of the $a_i$ since $T$ is not necessarily injective

[gentle osprey]

okeyokay

[heady skiff]

oh does non singular mean injective

i thought it just meant non square matrices

in any case my linear algebra is shit so i'll believe you

[knotty vine]

The sum of the t_i's is 1 no?

[heady skiff]

oh i thought to show that smt is geometrically independent you set up these equations and then show that the tis are 0

[knotty vine]

Oh I thought you were talking about an n-plane, nvm

[heady skiff]

oh yea my b

[chrome ridge]

any references for defining the linking number in terms of the cup product ?

[heady skiff]

i don't really understand how these two equations imply that the expression in brackets represents a point p of the line segment joining a1 and a2

because i don't see how you can factor out (t1 + t2)/lambda

oh i see

it's just adding the coefficients which have to add up to 1

wait what

wait what's the difference between an n-simplex and a n-plane

they're the exact same definition

oh

we just require that the ti are nonnegative in the definition of n-simplex

ok how the fuck do you show that barycentric coordinates are unique

like how do you even begin to use the condition $\sum_{i = 1}^n t_i = 1$

[gentle osprey]

okeyokay

[novel acorn]

wait sorry what are you trying to show

[heady skiff]

that for any point x in a n-simplex, its barycentric coordinates are unique

[novel acorn]

that t_i = s_i for all 0<=i<n

[heady skiff]

ye

[novel acorn]

god okay what's the definition of barycentric coordinates that you're using

so you choose n+1 affinely independent points in R^{n+1}

and you wanna show that only one set of t_i's determines each point in the simplex that this set of points determines

[heady skiff]

i'm assumign affinely independent means geometrically independent

[heady skiff]

i thought the geometrically independent set {a_0, \dots, a_n} is defined to span R^n

let me post my book's definition rq

lowkey might just take it for granted

[fringe cypress]

Suppose $$\sum_{i=0}^nv_it_i = \sum_{i=0}^nv_is_i,$$ so $$\sum_{i=0}^nv_i(t_i-s_i)=\sum_{i=1}^nv_i(t_i-s_i) + v_0(t_0-s_0) = 0.$$ Because the $s_i$ and the $t_i$ sum to 1, $$s_0 = 1 - \sum_{i=1}^ns_i,\qquad t_0 = 1-\sum_{i=1}^nt_i.$$ By assumption, ${v_i-v_0}_{i=1}^n$ is linearly independent. Go from there.

[gentle osprey]

BlaKaligula

[heady skiff]

oh shit thanks ur a g

wait this is not the usual definition of the unit n-ball right? isn't it usually the set of (x_1, \dots, x_n) in R^n that satisfy x_1^2 + x_2^2 + \dots + x_n^2 \leq 1?

and he defines the sphere in a similar matter

[unreal stratus]

that is the standard ball

note that like

if a >= 0 then like

a <= 1 iff a^2 <= 1

if that is the confusion

[heady skiff]

oh right you can square both sides

duh

[fringe cypress]

Thank god 1² = 1

[heady skiff]

LOL

hold up

so according to their definition, S^0 = {-1, 1}

nvm

[heady skiff]

i don't really understand how x necessarily needs to be in the interior of a face, for what if it's on the boundary?

[heady skiff]

anyone?

🦗

also how does this work/how is this logically sound? didn't they establish earlier that $\mathcal{R} = {w + tp \mid t \in [0, a)}$?

[gentle osprey]

okeyokay

[heady skiff]

so if $b > a$ then by definition $y \notin \mathcal{R}$

[gentle osprey]

okeyokay

[merry geode]

The reasoning is explained in the next sentence I think

Hmm it does look a bit strange in how it's saying "x \in Int s".

[heady skiff]

yeah

oh well, will also take that for granted

thanks

[merry geode]

This does look obvious, though.

Ah wait

[heady skiff]

wdym 😭

[merry geode]

Maybe it's supposed to say b < a

Oh nvm

It clearly cannot intersect with $Bd U$ at $b < a$, considering $\mathcal{R} \cap U$.

[gentle osprey]

Absta

[merry geode]

@heady skiff do you see?

[heady skiff]

yeah i see that, but that's assuming that b > a is a typo right

[merry geode]

No, that's not a typo.

[heady skiff]

oh are you going for contradiction or smt

oh

like that's the only other possibility

[merry geode]

"x lies between w and y on the ray" agrees with "y = w + bp for some b < a"

y cannot be closer than x to w, if you consider what boundary is.

[heady skiff]

oh right they're saying that the intersection only consists of w + tp where t \in [0, a) for the intersection with U and not the closure of U - U

ah i misread

ok thanks that helps

[merry geode]

Good that you got it!

[heady skiff]

ok wait but how is this not a typo 😭

unless i can't do basic arithmetic

[merry geode]

Now that's..

Hell is this book?

Mfw t sudddenly becomes 1-t and vice versa

[heady skiff]

LOL

munkres, elements of algebraic topology

[merry geode]

Ah, maybe this is why my class did not use munkres for algebraic topology.

[heady skiff]

well this is not the main textbook, it's armstrong which is HORRIBLE

worse textbook i've ever encountered

but then again i'm two pages into this book so i can't rlly judge

[merry geode]

Neither used armstrong as well

Just used Hatcher

[heady skiff]

does hatcher assume familiarity with point set?

/is it purely an alg top book

[merry geode]

Yep, it does assume familiarity IIRC

[heady skiff]

ah ok, i'll probably review a lot of point set then just start alg top all over again lol

[merry geode]

Hmm, are you yet to do some point set topology?

I think knowledge of the basic concepts are generally enough.

Like, up to separation axioms.

[heady skiff]

i don't even know what the separation axioms are

we did cover point set

[ebon galleon]

you've seen Hausdorff, surely

[heady skiff]

but i felt like we sped through it and i didn't learn it fast enough

oh yeah

lol i saw "you"

and got scared

[ebon galleon]

Hausdorff, T_1 (frechet), T_3 (regular), T_4 (normal) ... so on are the separation axioms

[merry geode]

Yep

T_1 had a name? TIL

[heady skiff]

ok now i'm even more confused

how does a) imply that the restriction of f to the Boundary of U is a bijection?

okay i understand that rays originating at the origin span all of R^n

but like.

i don't know what U is...

[merry geode]

The equation f(x) is the key here.

[heady skiff]

right, i guess injectivity follows from just being injective on all of R^n - 0 lol

surjectivity is a bit strange though

[merry geode]

I mean

The intersection of (a)'s logic and that paragraph is in the def of function f(x).

Can you identify the fiber f^(-1) (y) ?

[heady skiff]

true but i kinda got lost in the proof of (a) 😭

uh what's a fiber again

just preimage ig

ok well

i guess intuitively it makes sense

[merry geode]

$f^{-1} ({ y })$

[gentle osprey]

Absta

[heady skiff]

like something like this:

[merry geode]

Yep

[heady skiff]

and i guess U is bounded and convex and open

so we can't have anything like two distinct points that will break f

[merry geode]

Now only need that intuition to carry over onto precise logic

[heady skiff]

yea, i'll rigorize on my second reading i think

i'm spending like an hour per page 😭

[merry geode]

Maybe Munkres was also a bad choice, after all

[heady skiff]

munkres is a lot better than armstrong, much more rigorous imo

lee was another choice but a little too technical, want to come back to it at some point tho

[merry geode]

Hmm

[heady skiff]

i'm excited to grind hatcher over winter break tho/go back and review things

i hate it when i feel like i haven't learned some well enough then we move on to the next topic, but that's my fault

[merry geode]

Yea, I think sometimes you have to go through it

[heady skiff]

ye

[ebon galleon]

hmm what will i grind over winter break

[merry geode]

A thesis?

[ebon galleon]

me does not have to do thesis

[merry geode]

How abt beginning it now?

[unreal stratus]

ye

i am do thesis over winter brak

😭

thanks fro reminding me

[merry geode]

You seem to have studied a lot of your field.

[heady skiff]

what's your thesis gonna be on

[unreal stratus]

Good question, I barely know

[ebon galleon]

I mostly know stuff for my other courses this term, so I mostly work on homotopy theory and category theory rn lol

[heady skiff]

here k4 is a simplical complex right

[unreal stratus]

But the field is uh chromatic homotopy theory

[heady skiff]

real

[ebon galleon]

[unreal stratus]

so like between homotopy theory and algebraic topology

but also uses algebraic gometry

[heady skiff]

damn

sounds fancy

[unreal stratus]

and rep theory

[heady skiff]

that's cool

[merry geode]

Ah so you learned lots of category upfront

[unreal stratus]

so i thought woul dbe fun

[ebon galleon]

potato you should learn topos theory with me

i think that will be my winter reading

[unreal stratus]

higher topos theory or like 1-categorical stuff

i don't feel i have much motivation beyond general sheaves nonsnse

[merry geode]

Homotopy is now given colors? Woo

[unreal stratus]

hm

[ebon galleon]

1-categorical stuff, as grounds for higher topoi

I don't want to jump straight into inf-topoi, I think it'll be helpful to have intuition and motivation for it from regular topoi

[merry geode]

I am like, what even is a topoi

[umbral panther]

Wavelength. Red shift