point-set-topology

[quartz edge]

there is an easier way to do this

with a labeling

aba^-1c

[urban panther]

sorry, that eludes me a little - is that from the perspective of some fundamental polygon

[quartz edge]

yea

for a, b, c you will need a 1-cell each

the attachment of the 2-cell is made clear from drawing out that polygon

and labeling the edges and observing how the interior attaches

[urban panther]

this is the construction youve explained, right

[quartz edge]

yes

[urban panther]

ok, cool. damn

[quartz edge]

you can also look at it this way:

this clearly shows all 3 1-cells

and it shows the 2-cell as well

just doing the gluing together of identified edges as usual, you can see what happens to the 2-cell

[urban panther]

is there some algorithmic way to construct a fundamental polygon from a cellular chain complex?

[quartz edge]

hm?

uhh

[urban panther]

i see! thanks

[quartz edge]

not really no

the other way around, yes

but cw complexes are much richer in structure

[urban panther]

oh, no, i just mean that its not intuitively clear to me how to construct the fundamental polygon of a space

[quartz edge]

because the attachment maps are a lot less opinionated

[urban panther]

got it. thanks!

[quartz edge]

np

[urban panther]

im just wondering then

by the same construction

if i dont attach one loop

[quartz edge]

i just knew the above because cylinders are easy to construct by edge labeling of a square

[urban panther]

is that resulting space a cone

[quartz edge]

mmm

yes

it is

but you get a triangle

for the fundamental polygon

[urban panther]

ah, yes , ok.

[quartz edge]

aa^-1b

[urban panther]

read my mind on what i was gonna ask

[quartz edge]

so only 2 1-cells obv

theres a nice thing that happens actually

[urban panther]

yea ?

[quartz edge]

if you have an edge labeling that identifies all vertices of the fundamental polygon with one another

then the edge labeling tells you the fundamental group of the resultant space immediately

you dont even have to compute anything

[urban panther]

oh, i think ive heard of this one -- the first homotopy group can be read off 'as relations' right?

[quartz edge]

yep

[urban panther]

very cool!

[quartz edge]

so take for example the klein bottle

pretty hard to compute the fundamental group

but we know the edge labeling is

... let me think

it's the other one from the torus

so let me look for just a sec

yeah ok

aba^-1b

[urban panther]

haha, i see where youre going -- the vertices should identify, so we can use that result right?

[quartz edge]

now the fundamental group

is the group with two generators ab

and aba^-1b = 0

and no other relations

[urban panther]

very cool. thank you for this reminder

i'll keep it in mind

[quartz edge]

ye

fun fact

you can abelianize that group

and get the first homology group

[urban panther]

ahh, homology is a little bit away for me at the current moment (not sure what it means properly yet)

[quartz edge]

it's actually much simpler than homotopy groups

like if someone explained it properly you would immediately understand it

[urban panther]

i see. im vaguely aware that it also measures loops (intuitively)

[quartz edge]

it measures the failure of cycles to have properly filled out interiors

aka it measures holes

[urban panther]

right, which is the exactness of chains

[quartz edge]

or incompleteness in attached stuff

yeah

[urban panther]

haha, im having a puerile moment

[quartz edge]

:^)

[urban panther]

is there a good way to think about homology groups then? the kinda learn-curve that intimidates me is that theres some sort of variance here, where theres relative homology, and also* simplicial and cellular homology (?)

[quartz edge]

i dont really know any abstract homology theory

but

nor do i know cellular properly

but ive just read up on simplicial and singular homology

[golden gust]

I heard from a postdoc that in hindsight he thinks the best way to be introduced to homology is via morse homology

[quartz edge]

dope

i just saw simplicial homology first

[urban panther]

morse from morse functions re cw complexes ?

[quartz edge]

in a physics textbook

yeah

[urban panther]

ok, will read up on that

thank you!

appreciate the help @quartz edge

gonna go ahead and fiddle with more cw complexes

[quartz edge]

🐲

[unreal stratus]

Apparently this is not path connected (according to a mark scheme) but I've discussed it with a couple of people and we seem to all agree it should be.

Surely given any two points I can use the path $\gamma(t) = \begin{cases} (0,0) & t = 0 \ (t,\sqrt{t} \sin(1/t))& t > 0 \end{cases}$ to connect them?

[golden gust]

yeah I asked my tutor about that qn, he asked the lecturer and it really is path connected

[gentle osprey]

potato

[golden gust]

so you're good

[unreal stratus]

Oh okay

Phew lol

This question was bizarre anyway imo because it seemed they basically just wanted you to say yes/no for whether all the things are connected or not right?

[golden gust]

yeah

[unreal stratus]

not me trying to prove them all

[gritty widget]

Does the serre spectral sequence imply that

if we have serre fibrations F to Y to B and F to X to B

and in both cases the action of pi_1(B) on the cohomology of the fiber is trivial

that the cohomology of Y and X is isomoprhic?

[empty grove]

Ye this should follow from Zeeman's comparison theorem

Hmm you might need to have a map Y to X that fits in with the identities on either side

I am not sure actually lol I am not very used to this stuff rn

[gritty widget]

this is very silly

but i thought this should be true just because

everything before the arrow only has B's and F's

[empty grove]

Lol but we need the differentials to behave nicely too

Wait lemme check Zeeman's theorem once

I don't think this applies lol

You don't have these morphisms f_r

[empty grove]

Like the 2 spectral sequences will be isomorphic via the identity map as bigraded modules, but not as spectral sequences I think

[gritty widget]

ah okay

that was my next thing

[empty grove]

Because for that, the identity maps need to commute with the differentials, and the differentials would then have to be the same

[gritty widget]

can we not just take the the identity

okay i should go look at the differentials

[empty grove]

Yeah I think differentials are gonna depend on the total space

Where are you studying from btw?

[gritty widget]

hatcher but i don't like it so any recommendations would be nice

[empty grove]

Ye I studied from Hatcher for the most part too, it is bad

You can try McCleary

[gritty widget]

poor hatcher

[empty grove]

User's guide to spectral sequences

[gritty widget]

gets such bad wrap

[empty grove]

lol deserved ngl

[gritty widget]

i had heard this had ltos of mistakes?

[empty grove]

idk I studied only a few things from it

Look at this shit

This is why

Hatcher bad

[marsh forge]

Even if the spectral sequences were isomorphic

[gritty widget]

okay no this is true

[marsh forge]

you shouldn't get an isomorphism of cohomology rings

due to hidden extension issues

[gritty widget]

oh

nevermind

[marsh forge]

E_infinity being an isomorphism is not the same as the convergent thingy being the same

its close

[gritty widget]

oh but there are no hidden extensions if we have

coefficents in a field

so if you have cofficents in a field

and both converge

and on some page there is an isomoprhims every where on that page

[marsh forge]

I don't think this comment about fields is true...

[gritty widget]

then they converge to the same thing

[marsh forge]

Hidden extension is a kind of informal term, but any time you have a filtration and look at associated gradeds

you risk having multiplicative extensions

basically the issue occurs when like

"ab" enters the filtration before a and b

so ab = 0 in the assgr

so the SS doesn't know about the multiplication in its entirety

[gritty widget]

okay so we can have two spectral sequences E and E'

and isomorphisms on the infinity page

but the things they converge to aren't isomorphic?

[empty grove]

Ye, we just know that they have filtrations such that the associated graded modules are isomorphic

Over a field, they will then be isomorphic as vector spaces

But Max is talking about when everything is algebra

Then the algebra structures may not be the same even over a field

[gritty widget]

ah okay so

[marsh forge]

There are some limits as to "how bad" the hidden extensions can be

i.e. if you can rule out any possible filtration jumps

you can conclude they dont exist

[gritty widget]

H^p(X) is isomorphic to H^p(Y)

but we don't get an isomorphism on the graded ring structure automatically

[marsh forge]

yeah

over a field anyway

[gritty widget]

what do you recommend for learning about spectral sequences

[marsh forge]

honestly I just like

stared into the void of papers

until i understood them

idk if thats a good strategy

If you learn a little bit about the Adams SS

Dan Isaksen has some good talks about like

how to work with it practically

and this extends to other SSs

[empty grove]

I can't find them

[gritty widget]

they are on the echt webpage

[empty grove]

oh

[gritty widget]

in the mini courses

[empty grove]

got it ty

[marsh forge]

someone at a conference called it

"eChat"

and ive never been able to read it any other way

[empty grove]

I spent the day reading ch2 of hovey model cats

It's so cool

But also who comes up with this shit didn't expect to see ordinal arithmetic show up like this

[marsh forge]

oh i mean a lot of what I assume you're referring to is like

okay this construction clearly makes sense modulo set theory BS

and so now i have to fix the set theory bs

[empty grove]

Hmm idk if it's just modulo set theory bs

It seems kinda like an inherent thing

[marsh forge]

I'd have to read the relevant stuff

[empty grove]

Did you see hovey? He says that other places just do the finite case

Like he defines this factoring property for transfinite sequences of morphisms

of some cardinality

[marsh forge]

right okay

[empty grove]

And an object is kappa small if for all cardinals lambda of cofinality greater than kappa, all lambda sequences have that factoring property

or that object does wrt those

I need to sit with the proof for another day to see how someone came up with this as a sufficient condition

It is just so wacky

All the other conditions are the obvious ones

[gritty widget]

i do think the right way to think about this is

its set theory bs

you should think about why the smallness in the small object arguement is important

but then the kappa-small stuff is just exactly condition you want to look at

when you want to deal with things that are not small but

small enough

and keeping track of the kappa lets you keep track of the small enough

[empty grove]

Ye like I understood the proof

I just wanna reconstruct it backwards sort of lol

[pearl holly]

tbh I skipped all of that shit lmao

[empty grove]

That is not ✓

[pearl holly]

it get's so technical I just couldn't do it

I should have my ✓removed

[empty grove]

Ye it really helped to already be familiar with ordinal stuff

But reading the proof also helped me with some other stuff in model cats

Like all the nice properties that retractions and pushouts and direct limits have wrt liftings

[pearl holly]

also today I heard that there are exactly 9 model structures on set

[glossy pine]

here are some common definitions of the boundary of a topological space. could you also have some sort of definition of something that goes like this?

choose a point s in the space and shoot vectors (or rays) out of it in all directions. then, for each vector, find a point b such that a small step epsilon>0 toward s will be in the space and a small step epsilon>0 away from s will be out of the space. then b is an element of the boundary. if you do this for every possible vector, you will end up with the entire boundary

[gritty widget]

can you draw a picture for what you mean

[glossy pine]

sure gimme a sec

here, but you do this for every possible vector

actually im now realizing it must be only for a convex shape

without any holes in it

ok I guess that limits this a lot lol

[unreal stratus]

Well you'd have to define what you mean by 'directions'

I guess you're basically working inside a normed vector space which is very much a special case of top spaces

[marsh forge]

this will not work

At least not for a fixed choice of "s"

I think it will work if you allow s to vary though, and you are like, a connected shape

[glossy pine]

makes sense

and probably using the word ray is better since it is not necessarily a vector space

[glossy pine]

oh I didn't see this, yeahhhhh realizing that now

[glossy pine]

im just giving a vague description but yeah

[marsh forge]

wait ofc you dont need connected

im big dumb

[marsh forge]

ray still doesnt make sense

[glossy pine]

oh

alright

[bitter smelt]

How does the fundamental group of $S_g$, where $g \geq 2$, act on $\mathbb{H}^2$?

[gentle osprey]

Migillope

[bitter smelt]

In particular, I am trying to understand what this might be referencing "If $S$ has empty boundary and has a hyperbolic metric, then its universal cover $\tilde{S}$ is a simply connected Riemannian 2-manifold of constant curvature $-1$. It follows that $\tilde{S}$ is isometric to $\mathbb{H}^2$, and so $S$ is isometric to the quotient of $\mathbb{H}^2$ by a free, properly discontinuous isometric action of $\pi_1(S)$"

[gentle osprey]

Migillope

[bitter smelt]

deck transformations?

[hollow harbor]

Yes.

That should be fine

[empty grove]

😵‍💫

[empty grove]

Every point not in A \ B

So just every point in B?

Why is it phrased like that, are we supposed to also consider points in X? But then closed in Y makes no sense/coincides with closedness of point of f(X)

[hidden crag]

[gritty widget]

I wouldn't worry about it. I guess it's phrased like that to point out it doesn't have to be closed when taken from Y\f(X)

[empty grove]

ok lol

[bitter smelt]

How can I prove this:
A conjugacy class of a primitive element of $\pi_1(S_g)$ has a simple representative implies each pair of representatives in the class are not linked at infinity

[gentle osprey]

Migillope

[bitter smelt]

(From pg. 230 primer on mapping class groups)

says to use the (unproven) statement in proof of prop 1.6:

So if we call $\alpha$ a simple representative of $[a]$, the conjugacy class of a primitive $a\in \pi_1(S_g)$, then $\alpha$ has the properties given in that proposition

[gentle osprey]

Migillope

[bitter smelt]

but I'm not sure how to relate lifts of $\alpha$ to any two elements of the conjugacy class. Are boundary points of lifts of $\alpha$ invariant under conjugation of $\alpha$ in the fundamental group? I don't think its that simple

[gentle osprey]

Migillope

[bitter smelt]

something along the lines of let $\beta$, $\delta \in [a]$. Then they are conjugate, in particular, to $\alpha$. BWOC assume their lifts to $\mathbb{H}^2$ intersect. Then... somehow this implies two lifts of $\alpha$ intersect. Maybe?

[gentle osprey]

Migillope

[feral copper]

Hi! So if you take a covering p:Y→X which is nice, then the Euler characteristic behaves nicely: \chi(Y)=n\chi(X), with n the number of folds. Now, what happens for branched coverings? Can I relate \chi(Y) with \chi(X), the number of folds and the Euler characteristic of the branching locus?
Actually, there's a lot more I'd like to know regarding branched covers: fundamental groups, deck transformations, euler numbers, necessary and sufficient conditions for existence of an n-folds branched covering with prescribed branch locus... Anyone having a reference in mind?
Thanks!

(Riemann-Hurwitz is only for ramified covers between Riemann surfaces and with branching locus a finite set of isolated points)

[rancid umbra]

i’m having some trouble with exercise 5c and i’m having trouble understanding some of the solutions online. they state that if p is the quotient map, then for an open set U of G, we have p[U] is the union of all uH for each u in U. this seems incorrect to me, as p[U] = {uH : u in U} is not the same as the union of all such uH. they are even different objects, as the union is a subset of G, while p[U] is a subset of G/H

[cursive flume]

https://proofwiki.org/wiki/T3_Property_is_Hereditary

This proof seems false to me. in particular this statement does not hold.

from F being closed in the subspace, it follows that F=H cap W for some W closed in T

i tried fixing the proof. does this seem OK?

[rancid umbra]

y is not in W, so there are disjoint open nbhds U and V of y and W respectively, U intersect Y and V intersect Y are disjoint open (in Y) nbhds of y and F respectively, qed. this proof seems overly complicated @cursive flume

[cursive flume]

wait

i'm unsure what's the difference between what you wrote and what I have

[rancid umbra]

maybe overly complicated was a bad description. its just verbose imo

but thats just me, i like short proofs

[cursive flume]

yes,i write much,as i like to spell out all details explicitly,learning now

later on,when i will omit all details explicitly

[cursive flume]

but do you agree this is false though?

[rancid umbra]

what that a subspace of a T3 space is T3?

or that the proof linked is false?

[cursive flume]

the proof linked is false

a subspace of a T_3 space is T_3

however,the wiki proof is false

[rancid umbra]

the only part that seems false is when they state that F is closed in T

[cursive flume]

yep

[rancid umbra]

we arent given that H is closed

[cursive flume]

precisely

that was my issue

[rancid umbra]

nice catch

[cursive flume]

Does anyone know an example where path connectedness hereditarity fails?

[marsh forge]

just take two disks in a plane

[cursive flume]

Does (0,1) union (2,3) subset R work?

[marsh forge]

disjoint disk

sure

that also works

[rancid umbra]

bumping since im still unsure how the proof of this is supposed to go

[tired quail]

ok so

basically the way im doing this is by proving that this topology obeys each step of the definition of a topology

aka for the first one i proved that both the empty set and X are in the topology, because yk since its a collection of topologies then their intersection has of course X and the empty set

but how would i prove that any union of open sets of the intersection of the collection of topologies is an open set

sorry if this is a dumb question its 4 am so i cant rlly think straight

[marsh forge]

You need to prove that if a bunch of open sets are in the intersection

then their union is in the intersection

you can do this in a pretty straighforward manner unraveling definitions

idk how much of a hint you want

[rancid umbra]

just a start: take any collection of sets in the intersection. since each set is in T_a for each a, then ||so is the union||

[tired quail]

ohhh yeah i understand now, i proved it now but honestly writing it all out explicitly without using notation here at 4 am isnt really the greatest idea i think id probably pass out

but thank you so much!! you both helped

also for this part, to prove that there is a unique smallest topology that contains all of the family id just have to state that there is a topology that is the union of 3 sets: the union of all topologies, all possible unions of its elements, all possible finite intersections of its elements. and the union of those 3 would be the unique smallest topology right?

or do i have to prove it in a more concrete way

[rancid umbra]

think about defining a sub-basis to do what you want for the smallest topology containing all of them.
alternatively, you could consider all topologies containing each T_a

think about part (a) for the largest topology contained in all of them

[tired quail]

hm ill think about it thank you though

[tired quail]

oh that was easy

the union of the whole family of topologies is the subbasis

[rancid umbra]

ye

[rancid umbra]

the alternative was to intersect all topologies containing each T_a

[coarse night]

just take indiscrete

[coarse night]

still need help?

[rancid umbra]

yes bro

[coarse night]

U ↦p[U] is open

[rancid umbra]

as a function?

[rancid umbra]

wait im confused on what this means exactly lol

[coarse night]

you said p[U] ={ uH | u ∈ U} does not sit right with you

[rancid umbra]

p[U] = {uH : u in U} is the right thing

i dont agree that p[U] is the union of all uH such that u in U

but thats what some online solutions have said

and i cant really follow them

[coarse night]

are you ok with saying p(u)=uH

[rancid umbra]

yes

[coarse night]

if yes then it's same as saying imf = ∪ (x ∈ domain) {f(x)}

[rancid umbra]

thats not the same as union f(x) for all x in the domain tho

[coarse night]

you are taking union in the quotient G/H not as a subset of G

[rancid umbra]

https://dbfin.com/topology/munkres/chapter-2/supplementary-exercises-topological-groups/problem-5-solution/

does the answer for part (c) make sense to you?

[coarse night]

you mean it should be ∪{uH} and not ∪uH? I can definitely see that

well aren't quotient maps supposed to be open maps by definition?

[rancid umbra]

not in munkres

p : X --> Y is a quotient map if it is surjective and U subset Y is open if and only if p^{-1}[U] is open in X

[rancid umbra]

yes

i think what they mean is that p^{-1}[p[U]] is the union of all uH such that u in U

that i agree with

[coarse night]

I believe what they are doing is showing the inverse image of UH in G is open amd from that they are concluding the image must have been open

that is ∪uH is open, thought the way it's written is wronf

p^-1 p (U) = ∪ uH

[rancid umbra]

yes

cool, thanks for talking that out with me man

[coarse night]

and that just means p is open map?

[rancid umbra]

i dont think so. just because V is open in X doesn't mean that p[V] is open in Y.

example in munkes : take A = {(x,y) in R^2 : x >= 0 or y = 0} and consider p = pi_1 from A into R

p is a quotient map that is neither open nor closed

[coarse night]

there's an "iff"

[rancid umbra]

its more like strongly continuous, not open

if that makes sense

[coarse night]

[coarse night]

page?

[rancid umbra]

section 22 exercise 3 page 145

like, if i am the preimage of some subset of Y and i am open, then im open when taken to Y by p.
but there can be open sets in X which arent complete preimages of sets in Y

[coarse night]

hmm

[gaunt salmon]

If I take the unit disk in C and I create a quotient space by gluing points z whose magnitude is 1 (they lay one the edge of the disk) if their argument is k*2pi/n apart. Fixed n in N{0},. And k varies in Z. What is the fundamental group of this space? I think it is Z/nZ. How would one prove this?

[sharp frost]

I think this is homotopic to the n-holed sphere? You can inductively use Van Kampen’s theorem

[pastel linden]

does anyone have a reference for general point-set stuff for fiber bundles?

I've been working through the constructions myself but this is very time consuming

[opaque cloud]

guys

can a manifold have a genus of infinity?

[sleek thicket]

Yes

[opaque cloud]

a genus of uncountable infinity?

[sleek thicket]

Look up "infinite type surface"

No

The fundamental group of any manifold is countable

Maybe this convinces you that the genus can't be uncountably infinite, if you figure out how to define things so that that makes sense

[opaque cloud]

the fundamental group is related to homotopy right?

like deforming curves on the space to measure how many holes it has?

[sleek thicket]

yeah

[opaque cloud]

what exactly are the elements of the fundamental group? the curves? like the f: [0, 1] -> T (T being the space)

[sleek thicket]

Almost

So we want to require f(0) = f(1) = p for a fixed "basepoint" p

So it's loops based at p

[opaque cloud]

right

[sleek thicket]

But then the tricky bit is that we mod out the set of all these loops by an equivalence relation to get the group structure

[opaque cloud]

equivalance relation between...?

[sleek thicket]

between the loops

[hidden crag]

Two loops are set to be equal if there exists a homotopy between them

Graphically speaking if they can be continuously deformed in one another

[gentle osprey]

Arrow

[gritty widget]

Does my solution work?

[gentle osprey]

Arrow

[sleek thicket]

this works, although for clarity it's better to say "open subset of X contained in U" than "open subset of A"

The latter sounds like you mean open in the subspace topology on A

[gritty widget]

Oh yeah that makes sense.

[gentle osprey]

Arr0w_04

Arr0w_04

[gritty widget]

using lots of set theory relations

try writing out what should be true here and seeing if it is

you can discover the relations yourself

Hm OK, I'll give that a go.

either way you should be able to follow the definitions and do this

as you can for 90% of general topology problems

(this is a hint)

[gaunt salmon]

Can you explain more? I disagree with the n-holed sphere thing. Wouldn't that give Z as fundamental group?

Or the free group on n generators

[gritty widget]

n-1 generators

[gentle osprey]

Arr0w_04

[gritty widget]

Does that work in regards to set up? This is kinda tricky to visualize for me....

[gaunt salmon]

But the unit disk construction. Can you explain to me how it is homeomorphic to holed sphere?

[gritty widget]

no one said homeomorphic (except for you just now)

anyway, I'm not interested

Like I am reading Munkres right now, and there aren't that many definitions to use to make this work

[dry jolt]

you want to show that given a collection of open sets in the intersection of all the topologies, their union is also in the intersection of all the topologies

[gritty widget]

Right.

[dry jolt]

what does it mean for their union to be in the intersection of all the topologies

[gritty widget]

It means that given an indexed family of sets that consists of open sets in the family of topologies, the union of that indexed family is contained in the intersections of those topologies?

[dry jolt]

right, ig i should've emphasized the word intersection. What I'm getting at is that you should show that for any of the topologies, the union of all the open sets is contained in that topology

[gritty widget]

you will not get a homeomorphism, but rather a homotopy equivalence ignore this i didn't read

[marsh forge]

This seems right to me. In the case n=2 you recover RP2 and you can prove this by viewing your space as S^1 with a disk glued around n*generator

Basically if you draw the picture you can prove this from van kampen

[gritty widget]

Walter: yeah that makes sense, now I am stuck on how I could go about doing that. The only thing I can think of doing is using a basis to show this, since that is kind of intertwined with the notion of an open set. But that doesn't even make sense to do that since basis shouldn't be involved.

[dry jolt]

there's no basis involved

[gritty widget]

yeah

[marsh forge]

@gaunt salmon

here's the case n=3

[marsh forge]

you shouldn't get either

[gritty widget]

i didn't even read the original question lmao

i assumed they were asking things about n-holed spheres

i had my own question in mind

[dry jolt]

@gritty widget ok think about it like this. We have a collection of open sets U_a contained in the intersection of all our topologies. In particular, this means that all of the U_a are contained in each of our topologies. We want to show that the union of the U_a is contained in the intersection of all our topologies. To do this, we fix a topology T in our collection and show that the union of our U_a is contained in T. Can you fill it in from here?

[marsh forge]

I was being dumb.

[gritty widget]

Maybe, I'm thinking of how I can do this. Does a proof by induction work here?

Like if prove the union of U_1 and U_2 is in T, then induct

[dry jolt]

a proof by induction is not necessary, nor would it work here since we want arbitrary unions to be in the intersection of the topologies, not just countable unions

we know that all of the U_a are contained in T, right?

[gritty widget]

then take transfinite induction

oh right the unions are arbitrary 🤦‍♂️

[gritty widget]

right

[dry jolt]

then what can you say about the union of the U_a and T?

remembering that T is a topology

[gritty widget]

well each U_a can be expressed as a union of basis elements in T nvm that idea sucks

[dry jolt]

I think you're over complicating it

no need for bases, remember the most basic properties/definition of a topology

[gritty widget]

well it has to be in T since T is a topology

[dry jolt]

exactly, since T is a topology a union of open sets in T is contained in T

now, we've shown that the union of our U_a is contained in T, but we chose T arbitrarily from our collection of topologies

[gritty widget]

[gritty widget]

so it applies to the whole family of topologies, so unions are proved?

[dry jolt]

just to spell it out completely, yes we've shown that the union of the U_a is contained in every topology in our collection, hence by definition it is contained in the intersection of our topologies

[gritty widget]

OK that makes so much more sense now. Thanks Walter, I think you're right, I was really overcomplicating this.

[dry jolt]

definitely try working through it on your own for intersections

[gritty widget]

yep I will

[dry jolt]

glad i could help though, point-set stuff can be deceivingly straight-forward at times

[gaunt salmon]

Ok thanks a lot. The image you drew is simlar to what I drew. It makes sense. The paths that dont cross the edge are null-homotopic. The ones that cross the edge and pop out 2pi/n clockwise rotated are one homotopy class, and the ones that pop out 2pi/n counter-clockwise are another, and they annihilate eachother kind of.

I haven't read van kampen, but it was the next thing I was gonna study, so that makes sense.

And then I can apply it to this and try to understand the computation you did

[marsh forge]

Yeah there are two proofs I can think of, and Van Kampen is the simpler of the two

[gentle osprey]

Arr0w_04

[gritty widget]

I solved this one on my own but I wanted to post my solution to verify

[gentle osprey]

Arr0w_04

[gritty widget]

Does that work out?

sorry for asking so many questions, I'm brand new to topology and I wanted to make sure I was getting this stuff right initially.

[golden gust]

I think this also works:

let $\tau_\mathcal{A}$ be the topology generated by $\mathcal{A}$, and let $T$ be the collection of all topologies that contain $\mathcal{A}$. then $\tau_\mathcal{A} \subseteq \tau$ for all $\tau \in T$, so $\tau_\mathcal{A} \subseteq \bigcap_{\tau \in T} \tau$. but also $\tau_A \in T$, so we have equality

[gentle osprey]

Average J∘du=du∘j enjoyer

[golden gust]

wait oops. reading this again, it looks like you show that the topology generated by A is a subset of the intersection, but not the reverse? in fact the reverse is true because the topology generated by A appears in the intersection, but strictly speaking you'd need to say that

I should've read your soln properly before posting sorry

[golden gust]

but I guess the point of this is that you don't need to look at elements and individual basis elements, you can just work with the topologies themselves

[gritty widget]

Oh yeah that makes sense, I actually did prove the reverse

Thanks for the help, that was really helpful.

[grave maple]

Question about convergences. According to the pic, the supremum of an empty family of convergences is the intersection of an empty famility, which is just X. This makes sense. However, the infimum of an empty family is the union of an empty familiy, which is empty. And this makes no sense because every x in X, the limits of the principal ultrafilter on {x} always contains x and so cannot be empty. Am I missing something?

[rancid umbra]

Let ${X_{\alpha}}{\alpha\in J}$ be an indexed family of connected spaces; let $X$ be the product space $X=\prod{\alpha\in J}X_{\alpha}$. Let $\mathbf{a}=(a_{\alpha})$ be a fixed point of $X$. Given any finite subset $K={\alpha_1,\dots,\alpha_n}$ of $J$, let $X_K={\mathbf{x}\in X:\alpha\in J\setminus K\implies x_{\alpha}=a_{\alpha}}$. Then $X_K$ is connected.

There is an obvious bijection $f:X_K\to \prod_{i=1}^nX_{\alpha_i}$. If this map were a homeomorphism , I would be done, but I am failing to see how to show continuity of $f$ and $f^{-1}$. Anybody have any hints or nudges in the right direction? This feels kind of simple, but I can't seem to work through it right now.

[gentle osprey]

c squared

[rancid umbra]

oh wait, $f=\pi_{\alpha_1}\times\pi_{\alpha_2}\dots\times\pi_{\alpha_n}$ so $f$ is continuous

[gentle osprey]

c squared

[rancid umbra]

each of the projections is open and the finite product of open maps is open

okay

nvm

[gritty widget]

It is a homeomorphism

[rancid umbra]

yeth

[gritty widget]

That projections are open, doesn't mean it will be open after restricting to a smaller subset

But direct calculation works

[rancid umbra]

not sure what you mean here

restricting the domain of the projections?

[gritty widget]

Yes

It generally wouldn't work. It works here

[rancid umbra]

direct calculation of?

[gritty widget]

You can check that pi_i restricted to that subset is an open map by calculating it on a subbasic open set

[rancid umbra]

hmm. tbh i’m having a bit of trouble doing that. if you don’t mind, i can show what i have so far

[gritty widget]

A subbasic open set here is of the form $\pi_j^{-1}(U) \cap X_K$ where $U\subseteq X_j$ is open. Show that $$\pi_i(\pi_j^{-1}(U)\cap X_K) = \begin{cases} X_i\text{ or }\emptyset & i\neq j \ U & i = j \end{cases}$$

[gentle osprey]

Blitz

[high haven]

Are C*-algebras and operator algebras ever of use in algebraic topology or TDA?

[grave maple]

Bump.

[proud herald]

It's by definition of empty intersection / empty union.
Some reading here:
https://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union

[grave maple]

Right. So in the union case, lim F = emptyset. But this is a problem because when F is the prinicipal ultrafilter on {x}, it's supposed to contain x.

In other words, constant sequences always have a limit regardless of the convergence.

[dim meadow]

@grave maple where are you reading this?

I've encountered these ideas in passing and thought about a lattice of convergence but I've never actually seen it in a book

By convergences I'm assuming you mean filters or something

[grave maple]

Yes. This is from some PDF I downloaded. I guess it's from a book.

[dim meadow]

Can you post their definition of a notion of convergence?

This sort of thing isn't that standard maybe, idk

[grave maple]

Here.

So the (CENTERED) law is the problem.

[dim meadow]

Okay this makes sense

Good to see a survey on this actually

Fun rabbit hole to go down at some point in the future

Okay so do you get what this is trying to kind of axiomatize?

I'll give some exposition

I guess

If you want

@grave maple

[dim meadow]

Because this isn't quite my interpretation I guess

So here's an example of a theorem that's really nice

[grave maple]

That's what it means though.

If you take the constant sequence x, x, x, .... and take the filter made up of all its tails, that's your {x}^up arrow filter.

This filter converges to x by CENTERED.

[dim meadow]

Suppose that you have a compact metric space. Then every sequence has a convergent subsequence. By fixing an ultrafilter that extends the cofinite filter you can pick a limit for every sequence

That's a motivating theorem

For this sort of thing

[grave maple]

I understand this theory. You don't need to explain it to me.

[dim meadow]

Sorry

I don't know your background

And I got excited to talk about this

[grave maple]

I'm just wondering about the definition of the inf.

[dim meadow]

Oh okay

[grave maple]

No worries. I do research in this area actually.

[dim meadow]

Oh neat

[grave maple]

I'm formalizing my research in a theorem prover and this definition of inf just doesn't work.

[dim meadow]

I guess my interpretation of a principal ultrafilter based at x is it selects the x indexed element

But I guess I'm a little confused because I'm thinking about sequences lol

Sorry

[grave maple]

The ultrafilter containing {x} represents the constant sequence x,x,x,x....

We obviously want this to converge to x.

[dim meadow]

I mean in terms of the function it plays in terms of convergence

Oh I see

I guess I'm not sure what the issue is

With respect to inf and sup

[grave maple]

It seems to me that the inf defintion has to consider the case when the family is empty.

[dim meadow]

Can you explain again

[grave maple]

Sure. Let C be a family of convergence structures.

We want to define inf C.

To do so, you have to define what lim_{inf C} F is for every filter F.

If that filter happens to be the ultrafilter containing {x}, then x in lim_{inf C} F must hold.

However, if you define lim_{inf C} F as Union_{c in C} lim_c F, you run into problems when C is empty.

[dim meadow]

Is there an order structure on the set of convergence structures?

[grave maple]

Yes. If c and c' are convergence structures, then c <= c' if convergence in c implies convergence in c'. In the PDF however, the write it as c >= c'.

In terms of lim, this means lim_c F is a subset of lim_c' F.

[dim meadow]

Okay I see, so that's why inf is defined the way it is I guess

Do theorem provers not know how to handle empty unions?

[grave maple]

Indeed, but it's wrong because they are ignoring the case where the family C is empty.

[dim meadow]

There are conventions for empty unions

[grave maple]

The union of an empty family is the empty set.

[dim meadow]

Yeah

So what's the problem with that here I guess?

[grave maple]

The problem is that lim_{inf C} F cannot be empty when F is the ultrafilter containing {x}.

It has to contain x as a limit.

[dim meadow]

Why?

[grave maple]

That's the CENTERED law.

[dim meadow]

How does that fit in here at all?

It doesn't seem like it should to me

[grave maple]

inf C has to be a convergence structure.

That means it has to obey the CENTERED law.

[dim meadow]

Okay so you are saying you haven't identified what the inf over the empty set should be here

As a convergence structure

[grave maple]

I know how to fix it.

I just wanted to get some feedback from someone here about the mistake in the PDF.

[dim meadow]

I see

[gaunt linden]

Sounds like you'll need to interpret the union in the class of centered convergences, so an "empty union" is the one that maps a filter to its intersection.

[dim meadow]

Yeah that makes sense

That you would need to work in both places

[grave maple]

In the C = empty set case, inf C should be the discrete convergence structure, i.e. the one where only the constant sequences converge.

[gaunt linden]

(How do sequences relate to filters in this setting?)

[grave maple]

Every sequence gives rise to a filter whose base consists of the tails of the sequence.

[dim meadow]

That's not a filter then

I was actually kind of concerned about this

[grave maple]

What's not a filter?

[dim meadow]

Oh no I'm being silly

A empty relation between X and FX is all you need

[grave maple]

Huh?

[dim meadow]

The centered law is only talking about principle ultrafilters

Or principle filters

Oh I see

[grave maple]

Yes.

[dim meadow]

Never mind

I am being silly

Okay so you just want the pairing (x, {x}^\uparrow)

Then

Is what you're saying

Seems reasonable

Sorry for the confusion lol

[grave maple]

Yes. As I wrote before, you want constant sequences to converge.

[dim meadow]

Took me a while to parse the definitions

Yes

Okay so that's it then

I guess

[grave maple]

So do you see the mistake?

[dim meadow]

What do you study in this area btw?

I mean I haven't seen where they define the inf

Oh you posted it initially

Yeah that's a mistake

[grave maple]

Yes.

[dim meadow]

Whenever I see inf and sup over empty sets I guess I always think initial and terminal

And it makes sense that that's the initial thing here

[grave maple]

Yeah. The convergence structures on a set form a complete lattice.

[dim meadow]

Yeah makes sense

It's definitely the thing generated by the empty union

At least

Lol

[grave maple]

At the moment, I'm writing a paper generalizing results about topological groups acting partially on a topological space to the setting of convergence spaces.

[dim meadow]

Oh that's really neat

I'm very interested in looking into this I guess, I feel like I want to know about the relationship between filters and convergence beyond the basic stuff I've thought about

[grave maple]

I can recommend you some books if you want.

[dim meadow]

Yeah that would be great

[grave maple]

That PDF that I'm posting excerpts from is called "An introduction to convergence spaces".

The books that I usually consult about this stuff are:

• Convergence Structures and Applications to Functional Analysis (Beattie, Butzmann)
• Function Classes of Cauchy Continuous Maps (Colebunders)

This last one requires some knowledge of category theory.

[dim meadow]

Okay that sounds neat

Thank you

[grave maple]

They don't mention the inf/sup stuff though.

They just delegate it to papers.

[dim meadow]

Yeah that's fine lol

I mostly read topos theory stuff, Computability theory stuff, and logic I guess

Can you tell me about some of the results in this area?

[grave maple]

Should I tell you why we care about convergence spaces to begin with?

[dim meadow]

Sure

I guess I have motivation from the example of filters extending cofinite filters

But I'd like to hear your take

[grave maple]

Well, if you know enough analysis, you know that they've come up with all kinds of notions of convergence.

[dim meadow]

Sure

[grave maple]

Unfortunately, topology is not enough to handle all these different notions of convergence.

A good example is convergence almost everywhere with respect to some measure.

There is no topology that will produce that kind of convergence.

[dim meadow]

Sure that makes sense

Although I guess you can go higher in the borel hierarchy

Relative to that space

[grave maple]

Another problem with topologies is that, if you want to talk about the continuity of higher-order functions, you always have to work with the compact-open topology.

In categorical terms, the category Top is not Cartesian closed.

So, given these drawbacks, it makes sense to search for a nicer category of spaces to do analysis in.

[marsh forge]

are convergence spaces cart. closed?

[grave maple]

Yes.

[marsh forge]

crazy

[grave maple]

Of course, the algebraic geometers want more than this - they also want it to be abelian. The category Conv of convergence spaces is not abelian though.

[dim meadow]

So you are enhancing the category Top?

By enriching it with filters or something

Or is that a bad take

Does the category of convergence spaces form a topos?

[grave maple]

I don't know what you mean by "enriching", but Top is basically embedded in Conv.

I don't think it is a topos. I know it is a topological category.

[dim meadow]

What embedding do you use?

[grave maple]

Convergence is Top is basically determined by the neighborhood filters.

[dim meadow]

Sure

Right

Makes sense

Nice

[grave maple]

I am going to bed. We can chat some more later if you want.

[dim meadow]

Yeah I'd like to, probably should read more of the basics before I do though

[plain raven]

Folks, I have a diagram of cohomology groups

And I have to prove that this here diagram commutes

and let me tell you something

I am absolutely having the damnedest time

[bitter smelt]

Does Euler characteristic less than or equal to 0 of a surface imply the universal cover is contractible

[wise vigil]

euler just explained complexes to me.
I'm copy pasting my study notes here
I just want to ensure that I understand this correctly.

PS: I'm very bad at math, not advanced at all, but the topic is advanced (I think) that's why I post here. If this is not appropriate, please inform me.

is this correct?

[marsh forge]

There is more nuance than your notes suggest about the attachment

for example, a simplicial complex has strict rules about how you attach new simplicies

a CW complex has very weak rules

Also like

I think every single example here is more general than a manifold

manifolds are pretty simple spaces

[wise vigil]

thanks

[wise vigil]

I see, what I should have written instead?

[marsh forge]

I would say

"Topological spaces in general can be unwiedly to work with an extraordinarily complicated. For this reason, we often restrict our attention to spaces that can be fashioned out of simpler ones according to certain rules as to how they may be attached to eachother"

[wise vigil]

ok, I get that part. I mean, what other "objects" do topologist usually play with besides manifold?

[marsh forge]

oh

i mean most spaces are not manifolds

its like asking what numbers do number theorists care about that aren't divisible by 5

[gritty widget]

Topologists in general? All of them

[marsh forge]

I would say almost all topologists probably confine themselves to studying CW complexes, but even this has a lot of counterexamples

[wise vigil]

btw, that's a good rephrasing, thanks

[gritty widget]

People that study topological groups, I feel like would care the most about locally compact and pseudocompact spaces

[marsh forge]

I feel like most topological groups are CW but maybe thats just a bad list of example in my head

Like all Lie Groups, all discrete groups

[gritty widget]

People that study continuum theory would care about compact metric spaces, continua and Hausdorff continua the most, probably

[marsh forge]

Yeah that was the most notable counterexample I had in mind

Any time I try to make such a statement people seem to protest

but it was my best attempt to answer the question of "what spaces do most topologists play with" and I think most topologists are either studying manifolds, homotopy theory, or something closely related to one of those

[surreal lantern]

speaking of topological groups

https://cdn.discordapp.com/attachments/780487998940250152/986668609696825354/unknown.png

[wise vigil]

I think topologist study topologicla groups (such as manifold), that makes more sense right?

[surreal lantern]

does anyone know how the A^-t/2 is defined

[wise vigil]

btw, my motivation for learning cw-complex https://michael-bronstein.medium.com/a-new-computational-fabric-for-graph-neural-networks-280ea7e3ed1a if that helps

[surreal lantern]

where A is a matrix and t in [0,1] (this is a proof concerning homotopy equivalences of topologcial groups that's why i posted it here)

[gritty widget]

Manifolds are not topological groups in general

For example, S^4 is not a topological group

To study something you need to narrow your point of view. I think that's natural. People don't study just one kind of space

[surreal lantern]

am i maybe better off posting this in #diff-geo-diff-top idk

[gritty widget]

Can't you diagonalize a positive definite matrix?

[wise vigil]

yes, it is just that manifold is a topological group that I am kinda familiar with, so that's the example im putting in my notes haha

[surreal lantern]

you can

[gritty widget]

and then such matrix has positive eigenvalues, no

[surreal lantern]

yes

is this something related to the matrix exponential

[gritty widget]

you define the power by calculating the power on each of the elements on the diagonal after the diagonalization then

If A is a non-negative-definite matrix, you write A = BDB^-1 where D is diagonal with >= entries

D^t is defined in a natural way, by taking power of each element on the diagonal

then A^t = BD^tB^-1

[surreal lantern]

ohh okay thank you, i haven't seen that before

[gritty widget]

👍

this is more linear algebra tbh

[surreal lantern]

ye you're right

i just didn't know what it was gonna be ig so i went with the channel that fits the book where its from

[dry jolt]

the mapping cone of a map of CW complexes is contractible iff the map is a homotopy equivalence, right?

actually, I'll just add my full question here: when working with chain complexes, it's not hard to show that the mapping cone of a homotopy equivalence is exact. Idk much about algebraic topology but i expected this to work for the singular chain complex of the cone of a map between spaces, but I'm confused because the singular homology of a point is Z in degree 0

so in particular, if you have a homotopy equivalence of spaces, the mapping cone is contractible, hence its homology is isomorphic to that of a point, right? but then its singular chain complex isn't exact? I also don't really know anything about singular homology so feel free to lmk if I'm completely wrong about something here

[marsh forge]

okay a few questions here

the first statement isn't an iff, you could have the cofiber (mapping cone) be trivial but the fiber nontrivial

I don't have a great example off the top of my head

In the category of spectra your statement is true however

a good analogy is that a cokernel can be trivial without the kernel being trivial

[dry jolt]

ah, that's fair

[marsh forge]

As for the exactness statement no we don't expect the singular complex of a contractible space to be exact

The right way to think about this is the following

we expect mapping cones (cofibers) to be "homotopy equivalent" to 0 in the category where they are taken

So we expect the mapping cone of a chain complex to have 0 homology

and we expect the mapping cone in spaces to be contractible

[dry jolt]

ohhh, zero object in the category of spaces

[marsh forge]

but this doesn't mean that after appyling a random functor you still get the 0 object

pointed spaces technically but yeah

[dry jolt]

that helps so much, tysm

[marsh forge]

np

[dry jolt]

ok i'll have to think about this more then because now i find the passage from spaces to chain complexes a bit confusing

that'll teach me to learn alg top before all this homological business 🙃

[gritty widget]

is this true?

let f: X to R^n be a map that is never zero

view S^n as the one point compactification of R

is the map X to S^n smash X induced by f

homotopic to the map constant map

or sorry

the map that just sends x to (0,x)

[marsh forge]

What map is this? The map that is trivial on the X component?

[bitter smelt]

Reposting since it's been buried (and is simple to state)

Is the universal cover of a surface of Euler characteristic less than or equal to 0 necessarily contractible?

[marsh forge]

oh man i think this is like

a qual problem ive seen

[honest narwhal]

Hmm

So we are assuming connected here

If we're assuming closed orientable then this is just classification of surfaces + uniformization

If we don't want to cite the latter

Then we can prob also use Poincare duality. The universal cover will then be a non-compact orientable surface

H_2 of that surface is gonna be H^0_c

Which should be 0

Also I guess working mod 2 we shouldn't need to worry about orientability (or maybe we do tbh)

The scary case is with boundary

[bitter smelt]

Say, without boundary

or punctures

[bitter smelt]

Could you elaborate on this?

[honest narwhal]

A simply connected Riemann surface is biholomorphic to either the sphere, the unit disk, or the plane

So in particular those 3 are the universal covers of any closed orientable surface

But this involves a fair bit of complex analysis and requires knowing that closed orientable surfaces can be made into Riemann surfaces

So I think arguing by Poincare duality is better

[marsh forge]

is the claim that the universal cover of a surface is the sphere or disk up to homotopy hard

you have so much control over the algebra

[bitter smelt]

I was hoping to avoid homology, many of the first and second years haven't seen it yet, but I'll investigate these two ideas some more, thanks!

[marsh forge]

oh

i think this would be quite hard to prove without homology

or higher homotopy groups

[honest narwhal]

Yeah I was thinking Hurewicz basically

[marsh forge]

since the universal cover is simply connected you can use whitehead-for-homology

but I guess we need integral coeffs in that case

annoying

still, you get a lot of control from P.D.

[honest narwhal]

I guess you could give surfaces as polygons and talk about tilings?

I'm not sure how that looks exactly though

[bitter smelt]

Is this a generally known fact? It was stated as though it was free

[marsh forge]

i did not know it

[sharp frost]

Hi, is there something analogous to intermediate value theorem which I can use to deduce that it's impossible to stretch a loop that contains a hole into a loop which does not contain a hole? This is in 2 dimensions btw

[tawdry valve]

have you seen the fundamental group before? and by stretch, do you mean a homotopy of the inclusion of the loop?

[sharp frost]

yeah I just want to show that the left loop isn't homotopic to the right loop

and yes I know about fundamental group

[tawdry valve]

nice nice

so the pro strat

is to use the fact the a homotopy equivalence induces an isomorphism of fundamental groups

so the space we're looking at is R^2 - {0}

and a useful thing to do is to see that R^2 - {0} is homotopy equivalent to the circle S^1 by radial projection

[sharp frost]

ah yes, I see

thanks mr Joseph

[tawdry valve]

no problem 🙂

there's a lil fiddling you have to do at the end depending on what exactly you want. The fact that radial projection induces an isomorphism tells us that these two loops are not loop homotopic (fixing basepoint). You probably want that they are not homotopic, where we don't care if the base point moves or not

but you could make an argument that if you (supposing to the contrary) had a homotopy moving the basepoint, you could construct one that fixes the basepoint

[sharp frost]

yeah so I can pretty much take the retraction $(x,y)\to \frac1{\sqrt{x^2+y^2}}(x,y)$ and see how the two loops look once they get retracted under this map

[tawdry valve]

yep

the key is knowing that the induced map on fundamental groups is injective here, so knowing they are not homotopic after retracting tells you that they are not homotopic before retracting

and it's injective because you have the homotopy inverse which is just inclusion of the circle into R^2

[gritty widget]

yes

x goes to (f(x),x)

[sharp frost]

is it possible to prove that the Euler characteristic of an odd-dimensional manifold is 0 using only homology (and not cohomology)?

[sharp frost]

how do you get them to cancel out though

the proof in Hatcher (cor 3.37) mixes a bit of cohomology in there

[sharp frost]

yes, I'm aware of the definition of the definition of Euler characteristic in terms of homology groups. I'll be more precise in my question: How do you prove, using homology only, that a closed odd-dimensional orientifold has zero euler characteristic?

if it can be done then it should be easy, given that it's a small question on one of the previous years' exams for my algtop class (5% of a 2 hour exam)

[sharp frost]

nvm I got it

[foggy heath]

sketch of the proof:

1. for 'good enough' spaces try to find its triangularization
2. using singular homology, show that spaces realized by a triangularization can always have Euler characteristic zero, if they're compact, closed and of genus 0. (-)
3. (not easy) showing that each space satisfying (-) is homeomorhpic to some of the spaces given in 1. Then we get our result as homology is invariant under homeomorphism.

not sure if this works and maybe it's the way Hatcher used

[swift fjord]

I assume Hatcher's proof uses poincare duality and the isomorphism of homology and cohomology over a field?

I think that's the most straightforward way

[foggy heath]

poincare duality is mainly used for corresponding homology to cohomology. But nevertheless, one have to use at least one of them to prove the Euler characteristic thing. So maybe that's not the key point here

[swift fjord]

The idea in the proof is that poincare duality for an orientable manifold will give you H^i(X,Q)=H_n-i(X,Q), but a result about homology over fields gives
H_n-i(X,Q)=H^n-i(X,Q)
So it gives you that H_i=H_n-i for all i, and from this you can deduce the result by the odd dimensionality

But I think bobbicals wants a direct proof without going through this

[wet pewter]

I have a very possibly stupid question. But I have trouble understanding why a sup always have to exist and I think deep it comes down to one question. Do you always know whether an arbitrary element exist in a well defined set?

[swift fjord]

Not really a topology question, but existence of an element is exactly an asseetion that a set is nonempty

I'm not sure what you mean by 'well-defined set'

[wet pewter]

Sorry I am asking not as clear but I meant like say you know it’s non empty and you know there are some elements but then here comes a random element and do you knows if that element is in the set or not

And also I don’t know what it means of “well defined” that’s also one of the question I want to ask

[wet pewter]

Also feel free to recommend a place for me to ask this question

[plain raven]

Nobody, it's silly to say that an element without a predetermined containing set is a pathology. I don't know how that helps the student, either

[wet pewter]

So this is the context of where I have my doubt. I am mostly not understanding why a sup of Fn exist. Thanks to how it is defined is so complicated lol

Also to put it more simple terms, in step 1 of this wiki page, I am not understanding why step1 is a legitimate. As isn’t the act of “checking” an countable thing but the sets might be uncountable. So how whether is (A+B)/2 an upper bound can always be checked

[stable kite]

You don't need the result of this check at all - all you need is that the given numbers is either an upper bound or not, for the argument to work

[wet pewter]

And that is precisely what I don’t understand, why do you always know if an element is an upper bound. (Yes I do understand that there does exist an upper bound)

When I look at this in the context of the proof of vitali covering lemma, Fn is a set that is not defined by its length. Of course I know there is an upper bound of l(E) but how do I know an arbitrary element I will be a upper bound or not, the definition of Fn doesn’t mention at all

[stable kite]

Again, when you have an upper bounded subset of the reals then you always have a supremum in the reals as well, this is the property used in the proof you photographed.

That property follows directly from every construction of the reals, an example proof you found on Wikipedia, and that proof works - even though I do understand that it can be hard to wrap one's head around the proof logic

I think your questions can be addressed much better in #proofs-and-logic to be honest :)

[wet pewter]

The funny is I know exactly where my head can’t wrap around

[wet pewter]

Ok time to bother those guys now thx for the help 🙂

[stable kite]

Best of success in learning! :D

[wet pewter]

Thank you!

[bitter smelt]

What is the fundamental domain of $\pi_1(S)$ for $S$ a hyperbolic space on $\mathbb{H}^2$ acting by deck transformations

[gentle osprey]

Migillope

[bitter smelt]

Somehow, I feel it ought to "strips" of the space (taking the disk model), but I don't see how this follows from the algebra

For example, with each color being a distinct domain

But I just don't see it. The fundamental domain is defined as follows (according to wiki)

Take one point from each orbit of the action. This constitutes a fundamental domain.

But I don't see why this even has to be connected in this case. Is it just that we "can" take a fundamental domain to look like this, and thus do whenever we mean "take some fundamental domain"?

If this is what we mean, how do we know it is possible? Is it necessarily true that we can always pick representatives like this? A stronger statement would be 'is it possible to pick, for any given element in the disk, some deck transformation that sends the point to any given strip'

This might be the case: hyperbolic elements of the fundamental group (in fact, all elements of S_g, g>=2 which is the surfaces I care about) have deck transformations which act by translation on some axis. It probably is the case that we have an axis moving sufficiently fast in any given direction, but I do not know this to be true for certain

[bitter smelt]

I figured it out

[foggy heath]

what's your thought?

I would simply think like this:

1. The deck transfromation is a discrete over H^2. So it looks like a lattice.
2. Then we can think about H^2 modulo by this lattice. Since it's discrete, the quotient is then 2-dimensional and compact.
3. so the quotient would be 'similar' to a small region.

not sure if I'm right

I think the key is the orbit over H^2 of each deck transform induced by an element of the fundamental group of S, is discrete.

[foggy heath]

so I think if your way of thinking, it seemed that you treated the orbit as a continuous thing. that's not the case imo

[bitter smelt]

I'm not sure what you mean by the deck transformation being discrete over H^2. It'll act by translation on an axis connecting two fixed points on the boundary of the disk, so that seems continuous to me?

The orbit is certainly discrete though, I agree

[foggy heath]

maybe I was wrong on saying the transformation looks like a 'lattice' because I was thinking something like how you construct a torus from plane. But anyway, it's a fact that the deck transformation is a discrete group.

[foggy heath]

I was thinking that the deck transformation is induced by some elements of the fundamental group of S. So 'all possible deck transformations' forms a group

[bitter smelt]

Of course, it is in fact a group isomorphic to the fundamental group, since H^2 is the universal cover

(of a genus g>=2 surface)

so each element of the fundamental group has a corresponding deck transformation

I think that we can find a tiling of H^2 by some collection of fundamental domains, which we can probably make look like strips if we desire

but it isn't necessarily true that all choices of fundamental domains might be disjoint (ignoring boundaries) I think

[foggy heath]

let me think about the defination of fundamental domain...

[bitter smelt]

There's a lot of freedom with fundamental domain, and that the wiki page says "Typically, a fundamental domain is required to be a connected subset with some restrictions on its boundary, for example, smooth or polyhedral" make me think we probably can

(this is, not really a sound argumentation of course)

[foggy heath]

so, for example, for construction of a torus, one take the quotient of a plane by a lattice. The fundamental domain in this case will be the 'unit' parallelogram right?

I mean there's will be some choices but all look like equivalent

[bitter smelt]

A fundamental domain

In the quotient space?

Are we considering the group action of Z^2 on R^2 by translations?

[foggy heath]

then, in the g>2 case, I think the idea will be similar that a fundamental domain of H^2 acted by that group can be obtained and it's a compact region

[foggy heath]

yes

[bitter smelt]

Then yeah, we can choose a fundamental domain to be the unit square

This is probably a canonical choice even

but we could also choose something horribly disconnected, it seems, just from the definition of fundamental domain

[foggy heath]

not necessarily square as it depends on the lattice, but anyway it's the unit shape given by the two generators

[bitter smelt]

if its Z^2 acting on R^2 by translations it would be a square, no?

[foggy heath]

on R^2 you take two independent vectors than you get a Z^2

not necessarily orthogonal

[bitter smelt]

oh okay fair, yeah depends on the lattice

mm, still should be a square, just rotated maybe?

doesn't really matter

anyway, what's the analogy?

For reference, the fundamental group acts on H^2 properly discontinuously, so we already have that the quotient is compact

[foggy heath]

so based on this idea I was assuming theirs 'only one' fundamental domain up to isomorphism, as different choices of them can be related by a group action

[bitter smelt]

err, wait let me check that

[foggy heath]

that's my guess, hope it's right

[bitter smelt]

ok maybe not

[foggy heath]

I think yes

as geometrically the quotient is nothing but a small region surrounded by the dots

so it's like you cut them off from the original plane

[bitter smelt]

what is the relation?

Between, for example, if we take the typical integer lattice, a fundamental domain of unit square, and another fundamental domain of top half of unit square and bottom of some other square far away

we can act on the bottom half of the other square by whatever the translation is between the two squares to get the unit square back, but this is an action on only half of the second fund. domain

there's certainly a bijective correspondance between any two fundamental domains, as they each have one element from each orbit

[foggy heath]

yeah

[bitter smelt]

we can just map orbit rep of x to orbit rep of x on each domain

for each x

[foggy heath]

and moreover I think usually we choice the fundamental domain so that it's a connected region on the plane

[bitter smelt]

We do, but the fact that we can/must means the collection of fundamental domains dont tile

we can make any intersection we'd liek

[foggy heath]

I mean in picking the representatives from each orbits you don't have to very discontinuously taking them

[bitter smelt]

even if we require connected

[bitter smelt]

huh?

[foggy heath]

https://en.wikipedia.org/wiki/Fundamental_domain

for exmaple check this picture?

[bitter smelt]

the polygon one?

[foggy heath]

yeah

[bitter smelt]

what about it

[foggy heath]

I think tile is not very confusing here. Yes, by tile I mean there's some point in two different choice of fund. domains. But on the other hand, in each fund. domain the point represent The same orbit. A key fact is orbits cannot intersect with each other.

[foggy heath]

oh sorry, I was thinking that 'in each fund. domain the same point represents different orbit'

[foggy heath]

that's right, but would this reduction be helpful for computing the Betti numbers from the remaining dimensions?

for example to compute a 2-manifold (i.e. surface) one considers the Betti numbers of dim 0, 1, 2.

After the reduction it's now sufficient to consider dimension 0,1 or dimension 1,2.

But still, you have to choose one of them to compute

[sharp frost]

does anyone know what is meant by the action of a group on a CW complex

[foggy heath]

it depends on how the group action is given. Without a specific action, one can say nothing but the definition of group action itself.

[sharp frost]

let me see whether there's any context

[foggy heath]

free action means 'no fixed point'

[sharp frost]

oh this must be the group action

this isnt a current exam btw just in case you were wondering

[gritty widget]

no one would have suspected this in the first place

[sharp frost]

lol

[foggy heath]

in fact that's no about the group action. It's the properties of free group

but let me think

[sharp frost]

the only group action I've heard about on a space is the group of deck transformations

but it never specifies anything about deck transformations here

[foggy heath]

and the kernal of the homomorphism \pi is usually called the syzygy. And if the syzygy is 0, then the group is called free

[sharp frost]

polish words smh

[foggy heath]

I heard that's also an English word. Rarely used though...

I'm not a native speaker so I don't know

[tawdry valve]

for groups, people often just call elements of the kernel "relations" on those generators

[foggy heath]

yeah and that's a fancy word describing relations

I think the fact is: once you get the 'relations', then the relations is then again a stucture. One may then consider the 'relation of the relations' and it's usually not trivial.

So there's a chain of relations. In this case, the word syzygy maybe more suitable.

https://www.ams.org/notices/200604/what-is.pdf

check this for syzygy

[tawdry valve]

I think for groups in particular higher syzygies aren't really a thing because every subgroup of a free group is free

so the kernel will be free

but yeah for modules over non PIDs i can imagine it's very interesting

[foggy heath]

okay, I think the idea to this problem will be lile: Let this CW-complex be X.

1. for a group action G on X, the orbit is either 1-dimensioal or 0-dimensional. In cases the orbit of x in X in 0-dimensional, it's a fixed point of G.

[foggy heath]

I'm not sure about the next steps. Is this really a theorem? or just condition?

for example consider a Z^1 action which is induced by the generator a sends a point to its next point....(for example we arrange the 0-dimensional subcomplex in order)

then it's transitive?

but how about X has some singularities?

then you consider there's an action on each 'directions'

[foggy heath]

yeah mainly it's interesting on modules

here we regard an abelian group a module over Z

[foggy heath]

so at least I know : this is true for X being a loop.

[sharp frost]

oh okay so they're not assuming any particular group action

[marsh forge]

this is true

[foggy heath]

while the group is Z

[marsh forge]

Or wait

Sorry I didn't see 1 dimensional

still true I guess. are you asking for a proof?

[sharp frost]

I guess so, but mainly I was just wondering what is meant by the group action on a CW complex

[marsh forge]

Oh like

a group action on a set

[sharp frost]

and whether it's just an arbitrary group action on the space

yeah

[marsh forge]

except the action is continuous and probably cellular

[sharp frost]

cellular?

[marsh forge]

Just to be sure you know what a CW complex is?

[sharp frost]

yep

[marsh forge]

cellular would just mean it takes cells to cells so like

it acts sort of on the cells rather than on points

i mean it does act on points

but the structure is on the cells

[sharp frost]

oh right

that simplifies things

[foggy heath]

I was assuming the action is determined by it's action over the 0-dimensional subcomplex

because everything in 1-dim face (edge) can be then induced

but the action over 0-dim things are like a permutation?

[sharp frost]

I definitely should have said this before but there's a hint about using the "classification of covering spaces"

not that I know how covering spaces are classified either

[foggy heath]

I always imagine this simply 'curved simplex'

[marsh forge]

I would say that is probably too naive haha

CW complexes are sort of like

a priori wildly more complicated

[foggy heath]

yeah, many ways. when you would like to involve the idea of continuous mappings in the gluing

but that's technical

[marsh forge]

well

also just because the gluing rules are far less strict

and the resulting spaces can look a lot more complicated

[foggy heath]

this reminding me of something not related, yet very intereasting. Do you know this is related to Milnor's famous solution to the question of 'can one hear the shape of a drum'?

[swift fjord]

That's the cool part, you don't have to compute them! You have an even number of terms and every pair of terms has opposite sign because of odd dimensionality, so everything cancels out

[foggy heath]

Here what we talked about is the construction of a torus via the quotient of R^2 by a lattice. And this definition can be generalized to higher dimension cases where a general torus can be defined.
The question arises: From two different choices of lattices, what are the relationship between the induced tori? Of course as topological spaces they're homeomorphic, but that's not enough for our interest. Milnor asked: are there rotations of n-dimensional space so that the two torus can be associated to each other?

And the general answer is no. And the counter example is used for showing that: there're two 'drums' which are not the same, but having the same sounds! The latter has a math description.

[foggy heath]

and it turned of for a closed oriented surface the most important homology is of 1-dimensional

[foggy heath]

the solution is very short as it only deals with the math problem after traslated from physics. So the original problem stated by Kac maybe more interesting:

[foggy heath]

yeah, I realized that you're right. I was only thinking about a general approach to Euler characteristics. In the odd case, yeah, you're right, the computation would be very easy.

So it turned out for even dimensional spaces like surfaces, things would be complicated and need further discussion.

[foggy heath]

maybe the original question-and-answer would be hard to be a first read, so I would recommend ones who are interested to read some survey articles on this problem. It's a great example on how math connects different areas.

https://en.wikipedia.org/wiki/Hearing_the_shape_of_a_drum

[foggy heath]

I remember the excatly same argument is used to show that a Kahler manifold always has odd-dimensional Betti numbers 0. Because of the duality from H^{p,q} to H^{q,p}.

[gritty widget]