point-set-topology

[gritty widget]

Oh, well, maybe I used the wrong word. I basically mean obsession but not in a bad way

I thought I'd be more fancy with my wording

[prisma arrow]

perverse hobbyist

[jagged sage]

Is cl(B_((1,0))) union B_1((-1,0)) path connected

I think not but no real rationale

[gritty widget]

why not just p(t) = (2t-1, 0)

What kinda manifold does the polygonal word $\langle a,b,c\vert abacb^{-1}c^{-1}\rangle$ represent

[gentle osprey]

cocoomba

[plain raven]

yeah i misread the question. nvm my hint is nonsense

[lunar yoke]

doing the gluings for b and c you get a torus with a hole, and on the boundary of that hole you are left with the gluing for a

the orientations will not match up, i.e. you get something nonorientable

[gritty widget]

how did you get this

[lunar yoke]

drawing pictures

[gritty widget]

My pictures must be drunk

[lunar yoke]

maybe im wrong idk

[gritty widget]

idk what im doing so its all you

i trust

[gritty widget]

How are you gluing b and c together? The hole tells me that this is the sharp of 2 surfaces

[lunar yoke]

ok this is the best i got

couldnt find the pen for my drawpad so i had to draw this with my finger lmao

you dont glue different letters

[gritty widget]

That is a beautiful picture

[lunar yoke]

you only glue the same letter

and the orientation on the initial polygon is given by wether you put a or a^{-1}

[gritty widget]

I got the torus part and then I was wondering where is that a

I see now it’s literally the hole

[lunar yoke]

so yeah there is definitely some closed form expression for this manifold but i dont know enough about that to tell you which one it is

maybe its the klein bottle or smth

[gritty widget]

So would “torus with a hole” not be good enough of an answer

[lunar yoke]

https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces

well there is this classification of closed surfaces

all of them are either the sphere, a connected sum of tori, or a connected sum of projective planes

since ours is not orientable, it will be a connected sum of projective planes

idk how many though

[gritty widget]

How do we know it’s not a projective plane and a sphere

[lunar yoke]

thats not on the list wdym

[gritty widget]

Ah

I see

There is supposed to be a formula for these connected sum things

I gotta remember it

I tried classifying based solely off Euler characteristic but couldn’t find that formula

[lunar yoke]

ah ok so the name of this hole thing is a crosscap

conway has a beautiful proof of this classification result

with lots of pictures of zippers

https://www.maths.ed.ac.uk/~v1ranick/papers/francisweeks.pdf

[gritty widget]

Yeah that’s a lot pages… can’t I just use this

[lunar yoke]

to be fair for such a classification result that is really a low number of pages

cmon its 6 pages with lots of images

but yeah ofc you dont need to read this to solve your exercise

was more of a "neat to know"

[gritty widget]

I am basically cherry picking parts I find interesting from the article

[lunar yoke]

ok its a sphere with 3 crosscaps

so i guess its the connected sum of 3 projective planes, which is in "normal form" in the sense that this appears on the clasification list above

neat

apparently this is called Dyck's surface

(these are the same surface)

although if you want this to be a manifold i guess you should smoothen the edges in the second picture

[gritty widget]

...

ok but how do I get here? I dont think we ever did crosscaps

Like it's fun and all but... you know... assessments

[lunar yoke]

well thats for you to figure out lol

what do i know what you're allowed to use

[gritty widget]

basically diagrams and words and the theorem about connected sums as well as their euler characterstic

thats what im trying to make work

[lunar yoke]

well you know the answer now

it has one side, is of genus 2, and has euler char -1

[gritty widget]

oh thank god I got the right Euler characteristic

[lunar yoke]

homeomorphic to P^2 # P^2 # P^2 = T^2 # P^2

i mean yeah the euler char and the fact that it is not orientable already uniquely determines the surface by the classification theorem

so technically that should suffice

[gritty widget]

And how did you determine if it was orientable or nonorientable again?

It had to do with inverses but I don’t know what it has to do with in detail

[lunar yoke]

well we saw in our example that since b and c occur both normally and inverted in the word, we can glue them just fine, but a occured normally twice, and we got something nonorientable

so without knowing any theorems i'll go out on a limb here and conjecture that whenever you have a letter occuring only normally (without its inverse) in the word defining your polygon, it will not be orientable

[gritty widget]

This jogged my memory

I’ve definitely heard that before (in class)

Thanks

I'm trying to find the closure of the subset S={(x, x^2): x ∈ [0, 1] \ Q} in the unit lexicographical order topology and I think it is S U {{0}x[0,1)} U {{1}x(0,1]} is this right?

[obtuse meteor]

Anything which contains a Möbius strip is nonorientable

In fact this is an if and only if

And is equivalent to what you have just said if the letters occur at most twice

[rancid umbra]

every non-orientable surface contains a mobius strip?

[rancid umbra]

is your square ordered so that (x,y) < (u,v) means x < u or (x = u and y < v)? or is it the other way around?

[gritty widget]

yes precisely that

[hidden crag]

Yes

[gritty widget]

oh wait actually I think 0x0 wouldn't be a limit point as it has a nbh disjoint from S

[rancid umbra]

then cl(S) should be equal to S U ((0,1) x {1}) U ((0,1) x {0})

So cl(S) is S along with the top and bottom edges of the unit square

[gritty widget]

I don't see it can you please explain why?

[rancid umbra]

If (x,y) is a limit point of S that is not in S, then y must be 0 or 1. try to show this.

[peak crystal]

How to construct an example in which topological space is T4 space but there is subspace which is not T4. We know that closed subspaces of normal space(T1 space) is normal (T1 space) so we have to find not closed subspace.

[gritty widget]

Take something T3.5 but not T4 and Stone-Čech compactify it

Alternatively take Tychonoff plank and a deleted Tychonoff plank

That is [0, omega_0] x [0, omega_1]

We delete it by removing point (omega_0, omega_1)

Spaces that are T3.5 but not T4 are usually hard to find iirc so don't expect anything prettier

[coarse night]

@peak crystal

[gritty widget]

This isn't Hausdorff

[gritty widget]

topology was a mistake

[coarse night]

i see

[gritty widget]

people sometimes write normal/regular but what they mean is that it doesn't have to satisfy, T1 axiom iirc

[coarse night]

op mentioned T4 explicitly so

[gritty widget]

sometimes it's reversed

so your normal might mean their T4

[coarse night]

🛞 is it for the wheel theory meme btw?

[peak crystal]

for normal is it true but for T4 we need T1 also

for normal/regular T1 is not necessary, This definition I used.

[gritty widget]

By the way. I use the convention that normal = T4 and regular = T3. So spaces which are normal but not T1 don't exist for me

[pseudo coral]

question, is anyone familiar with nets from topology?

I need to prove the product of two compact spaces need be compact using the notion of nets which I am not too familiar with.

[gentle osprey]

MyMathYourMath

[pseudo coral]

how do I use projection maps which are open and continuous to conclude this net has a convergent subnet in the product

[gentle osprey]

MyMathYourMath

[pseudo coral]

indexed over some directed indexing set

Im new to this notion of "nets"

[gritty widget]

$z_\lambda = (x_\lambda, y_\lambda)$ and we can take a subnet $f:S\to \Lambda$, $x_{f(s)}$ of $x_\lambda$ which is convergent. Now you can take a subnet of $y_{f(s)}$ using $g:M\to S$, $y_{f(g(m))}$ which is convergent. Then $(z_{f(g(m))})_{m\in M}$ is convergent.

[gentle osprey]

Blitz

[pseudo coral]

so $x_\lambda,y_\lambda$ are in $X$ and $Y$ respectively?

[gentle osprey]

MyMathYourMath

[gritty widget]

yes

you should check details so to make sure that what I said here makes sense

[pseudo coral]

okie doke!

[void gazelle]

Hi, guys, what are the topologies on C^n and C^{n+1}\{0}?

[gritty widget]

C is complex numbers and you meant C^(n+1)\{0} ?

[void gazelle]

yes

[gritty widget]

topology is the one induced from the Euclidean metric

C^n is just R^(2n)

[pseudo coral]

its the standard topology of all open balls right

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

right?

[gritty widget]

I mean... yes, the topology from polydisks and the topology from the Euclidean metric is the same

[pseudo coral]

a cartesian product of different radius disks

[void gazelle]

Thank u!

[winged viper]

does anyone know how to understand this functor from may's book? O(G) is the orbit category (objects are the G-sets G / H and morphisms are G-equivariant maps), and E(G / H) is just the covering space E / H, where E is the universal cover. I'm confused where E(-) sends the morphisms though (and the proof doesn't seem to explain it). I assume E(alpha) is just the map satisfying the last sentence in the theorem, but is that even well-defined? There isn't a canonical isomorphism between the fiber F_b and G / H right? And May says that functoriality follows from the previous lemma--I imagine one would have to show that E(alpha) is exactly the map described in the proof of the lemma? I don't see how that's obvious...

here's the previous lemma

[gritty widget]

Just blackbox it

there's nothing insightful about the proof of Tychonoff theorem

I think it'd just be more efficient to read the proof if you want to

[hidden crag]

Why are you trying to avoid the general case?

[lunar yoke]

A finite product of discrete spaces is discrete, and compactness is equivalent to finiteness for discrete spaces. Is this what you want?

[gritty widget]

No, he probably means the Cantor set

[unreal stratus]

Just use ultrafilters smh

[cursive vigil]

Can someone give me a simple easy to understand definition and distinction between homology and cohomology

[lunar yoke]

for singular (co)homology, you will find definitions in any textbook on algebraic topology or even on the respective wikipedia pages

there are lots of different (co)homology theories (singular, cech, sheaf, group, ...), and very general definitions in terms of infinity categories probably don't help here

one of the main distinctions is that cohomology usually has more structure and exists for a larger variety of things

in some cases like singular (co)homology over fields they give you the same results though

i guess all of this is explained on the wikipedia page https://en.wikipedia.org/wiki/Singular_homology

[cursive vigil]

Ah thx

[winged viper]

thanks! ill take a look

[long grail]

why is a space with a countable dense subset called “seperable?”

[gritty widget]

https://mathoverflow.net/questions/51494/why-the-name-separable-space

[long grail]

hmmm

[rapid olive]

how would one prove p_1 on R^2 sends closed sets to closed sets

yes

what's a counterexample

i can't think of any

[gritty widget]

yes

[rapid olive]

it's very slow

this makes sense thx

[gritty widget]

for the projection X x Y -> X to be a closed map you want Y to be compact

the proof uses the tube lemma

in fact, if this is true for all spaces X, it follows that Y is compact

[unreal stratus]

It's cute

[gritty widget]

If x y is a limit point of S that is not

[odd flame]

does anyone have any intuition for the pasting lemma

[gritty widget]

when one continuous function and another love eachother very much...

[odd flame]

from what i understand it seems say that if two continuous maps "act" equivalently on the intersection of two closed sets in X, the baby of those two maps is a cont function from X -> Y

well ok that's just a restatement of the thing

[gritty widget]

continuity is local

this is one of those things that i struggle to produce intuition for that isn't just the statement, because it's already so simple

i feel like it's already intuitive as stated

[lunar yoke]

i thought they meant pushout / pullback pasting first and got so confused with this lmao

[odd flame]

perhaps im just missing exercises with it

thank u anyways tterra

[gritty widget]

maybe to build intuition you could look at what happens if you omit "closed"

or why it only asks for two (i.e. finitely many) closed sets

with something so simple like this, i think the best way to build intuition is to try and break it

[odd flame]

im not sure i see what breaks if they're not closed

but ty for the general advice i'll keep it in mind

[unreal stratus]

you can have insane examples like covering [0,1] by singletons

with open sets you get more 'wiggle room' ig

[gritty widget]

It's a lemma which tells you a sufficient condition for a function to be continuous, which is useful because you often want to overlap two or more functions and not think too hard about why is the result still a continuous function

Note that pasting lemma holds for locally finite closed coverings as well

Note that covering [0, 1] by singletons is point finite but not locally finite

So basically, it holds for infinite families of closed sets too as long as they don't overlap too much

[odd flame]

munkres notes that it holds if A and B are open in X tooo

[gritty widget]

It holds for any open cover

[unreal stratus]

ye with open covers it works sexily

[gritty widget]

I feel like it's more used for closed sets

[unreal stratus]

rly? i feel like stuff with bundles/manifolds/schemes &c. uses some form of pasting all the time

just ig the closed set case is when it's more subtle

[gritty widget]

something something gluing property of sheaves

[unreal stratus]

fr

[gritty widget]

Yeah because that's whats usually applied in explicit examples

Open version is probably "too obvious"

[odd flame]

what is the space R^X here

that does look like a capital X right

and not \times

i'll rephrase to asking about the uniform metric/topology and what it "looks like"

is it a metric defined in terms of a metric which is defined in terms of a metric

[unreal stratus]

the set of functions X -> R

[odd flame]

im going off of these two defn's from munchers

[bitter smelt]

Has anyone read (portions of) Ratcliffe's Foundations of Hyperbolic Manifolds?

[woven sinew]

The function $\overline{\rho}$ is only a metric in the usual sense, if the functions are bounded - but the corresponding topology is still fine, so maybe not important.

To get a sense of what the topology looks like, note that the $\overline{\rho}(f,g)$ is the maximal (supremum, really) possible distance $|f(x) - g(x)|$ between $f$ and $g$ evaluated at a point $x$, so on a graph of $f$ and $g$, you would look for the biggest vertical distance between them. What functions are contained in the open ball $B_f(r)$ for some function $f$? The function $g$ is in the ball, if the maximal (sup) possible distance between $f(x)$ and $g(x)$ is strictly less than $r$.

[gentle osprey]

TimeTravellerOne

[supple heart]

can someone help me understand this definition?
so the empty set belongs to O, the set X belongs to X, whats the union and intersection parts supposed to mean?

the union is pretty obvious if X belongs to O

[gritty widget]

What they did is first replaced metric on $\mathbb{R}$, so that it's bounded. This makes $\overline{\rho}$ a metric in the standard sense.

[gentle osprey]

Blitz

[wet crow]

why ?

O is a family of subsets

it's not a subset of X

[supple heart]

i just dont understand how all of that creates a topological space

[wet crow]

I mean you have to know what a open set in the metric or R^n case is to understand

the idea is that a subset in X is either in O or not in O

not "a subset of" but "in"

[supple heart]

yeah it said a set in a metric space is open if every point has a neighbourhood

in that set

[gritty widget]

It creates a topological space because we defined it so. Don't get me wrong, your average topological space can be pretty pathological. But it's what's a good model of a space for most purposes

[wet crow]

that's a quite bad definition if that's actually supposed to be one

but anyways, what O is is a collection of subsets of X, which are the opens

and unions have to be still opens

[supple heart]

so its a collection of open sets?

[wet crow]

that's in the paragraph yes

I mean, it's basically a definition of what a open set is in X

we say E c X is open if E € O

[supple heart]

oh

i think i understand now, i watched a youtube video about it

[echo berry]

how would one apply the inductive step on an orientable surface of genus g to compute its homology via Mayer Vietoris

[obtuse meteor]

You can represent a genus g surface as a genus g-1 surface and a genus 1 surface with disks removed from each which are glued together along a cylinder

this will basically allow you to apply the inductive hypothesis, but for this you need to know what the genus g surface with a disk removed deformation retracts to

(maybe this is not really induction, but it is the spirit of things)

[woven sinew]

Ah, I missed that the bar over the rho was actually an operator on metrics.

Or something like that. I see that the rho with the bar has been defined to be a particular metric.

[gritty widget]

I'm wondering if Alexander subbase theorem holds for Lindelof property, being countably compact, and other generalizations of this spirit

I feel like it probably should but I don't feel like going through the details of the proof rn

[echo berry]

ok im kinda done with this. im having now issue to create a map alpha from H(A \cap B) in the MV sequence 0 -> H_2(X) -> H(A \cap B) -> H_1(A) \oplus H_1(B) to get its kernel, which is isomorphic to H_2(X). Here A is the surface with a hole pinched to it, which is homotopic to the wedge sum of 2g S^1 spaces. B is a disc from the surface, and so its intersection is homotopic to S^1. X is the whole surface, of course.

[obtuse meteor]

The map alpha has a particularly simple formula, especially since B has zero homology (being contractible)

it may be best to draw things IMO

namely can you deform the loop which bounds B to something when you envision everything a polygon gluing

[sturdy notch]

For h_* an ordinary homology theory, is the homology of the empty set always 0 by convention or is this somehow supposed to follow from the axioms?
I typically don't care about the empty set but in my exercise, this seems to matter quite a bit

Ig it would follow from additivity, but I'm specifically looking to deduce it without supposing additivity of h_*

[swift fjord]

Shouldn't this follow from LES too? Take the pair (*,\varnothing) for a one-point space *, and the LES gives you that the homology of the empty set is 0

[unreal stratus]

perhaps it relies on specific properties of the sections i have? idk

[gentle osprey]

potato

[sturdy notch]

ah yes, seems so, thanks a lot, that makes my life a whole lot easier.

[swift fjord]

This only works for h_i for i>0 tho, i'm not sure how you'd show it for h_0 without additivity, tho there might be a way still

[sturdy notch]

yes I noticed this just after sending the message. I'll assume this for now and go on with the exercise/life as is, there might be a better way to solve this specific problem (I don't want to share the exercise as it is graded)

[lunar yoke]

I think taking the LES of the pair (empty, empty) gives you H_n(empty) = H_n(empty,empty) = 0 for all n

[swift fjord]

Why is it the case that H_n(X,X)=0? Necessarily

[lunar yoke]

no you get the 0 from the LES

maybe i wrote that in a bad way

[swift fjord]

Not sure I see why

[lunar yoke]

by definition you have H_n(empty,empty) = H_n(empty) right?

[swift fjord]

Yea

[lunar yoke]

so you get an exact sequence of the form X = X = X -> 0 where X = H_0(empty)

[swift fjord]

Right

And then inductively it's 0 for all n

Since the identity is mapped to the identity by H_n

[lunar yoke]

oh but this works too btw

by the argument you just said

[swift fjord]

True

When we say ordinary cohomology theory are we also assuming H_n=0 for n<0?

Or generally assuming this for (co)homology theories

[lunar yoke]

i think the terminology is not too agreed upon

but there are cohomology theories with non-zero negative values

[swift fjord]

In this case we assumed they're 0

For your argument to work

[lunar yoke]

no not really

[swift fjord]

You need the LES to end at a 0 tho at some point otherwise you won't have this part

[lunar yoke]

If X = X = X is exact, then X = 0

[coarse night]

is it true that $(\mbb{R} \setminus 0)^{\infty} \simeq S^1 \vee S^1$?

[swift fjord]

Oh, right

True

[swift fjord]

So yea that works

Nice

[coarse night]

also is $(\mbb{R}\setminus\mbb{Z})^{\infty} \simeq$ Hawaiian earring?

[coarse night]

᪲^ ∞ means one point compaction.

[ornate berry]

I don't think so

[coarse night]

I do tho

kind of matches with the observation that any nbd of 0 in the ring contains infinitely many circles, similar to saying there's infinitely many nbd (n, n+1) outside any compact subset of R\Z

[ornate berry]

Oh I read this as R/Z

Kinda makes sense now, but not sure

[coarse night]

yeah I didn't prove it's homeo tho

just 'makes sense'

It can't be infinite wedge of S^1 as that's not compact

[gritty widget]

It's enough to check that if you remove the point bounding all the rings, then they are homeomorphic to R\Z

Now it's enough to observe that each of those left-over rings are open in the Hawaiian ring and homeomorphic to R

[wide egret]

i struggle

does anybody have a projection of the 5_2 knot with bridge number 2

i have been fucking around with string for like 30 mins total now and i can't seem to find it

[wicked yew]

This is what I have

with this question what do I assume is the metric-distance function in the reals? Or is it arbitary? I'm trying to work the question with an arbitary metric but then I get d(d(x1,x2),d(x3,x4)) and I don't know where to go from there

[unreal stratus]

Standard metric on the reals yes

[wicked yew]

as in the euclidean emtric?

metric?

[unreal stratus]

Ye, the one induced by the standard (in fact any) norm

btw imo this question is easiest if you just use the characterisation of continuity in terms of sequences so you don't have to mess around too much with balls and what not

though it doesn't really matter

[wicked yew]

but since this is from X x X

do I have to do a sequence (xn)x(yn)?

[unreal stratus]

well ((x_n, y_n)) or smth but ye

Also as a sidenote, though besides the poitn of this question: you can equivalently define the product metric in the naive way i.e. distance from (a,b) to (c,d) is just d(a,c) + d(b,d) and then it's fairly straightfroward

but ofc harder with this because they defined the product metric using euclidean norm on R^2

[wicked yew]

this would just be another valid metric in X x X right?

[unreal stratus]

ye but the two are equivalent

which means d cts with one iff with the other

[wicked yew]

oh ok

so how do I like the sequences d((x_n,y_n)) to the product metric

link*

[unreal stratus]

bound $|d(x_n,y_n) - d(x,y)|$

[gentle osprey]

potato

[wicked yew]

i used the fact that x_n tends to x and y_n tends to y then added and subtracted the same term

something like that right?

[icy schooner]

is there a way to find all index 3 subgroups of Z*Z?

[swift fjord]

Is that a free product

[unreal stratus]

yup

[icy schooner]

yeah

I got the 3-sheeted covers of S^1 ^ S^1 by looking at 3 vertex valence-4 graphs

was thinking maybe I can also get them by finding index 3 subgroups of Z*Z

[gaunt linden]

Those graphs sound like a promising ways to the index-3 subgroups too. :-p

[icy schooner]

true they give the relations of the generators

[coral pivot]

@bright acorn

let me tell it to you hear as ng and yamin are speaking in that channel

so the idea is that K theory talks about vector bundles right

so $K(X)$ is like you consider the set of isomorphism classes of vector bundles over $X$, give the operations $\oplus,\otimes$ and do a "grothendieck" completion to make it an additive group under $\oplus$ right

[gentle osprey]

JohnD.S.

[coral pivot]

(lmk if ur familiar with this)

[bright acorn]

Yup

[coral pivot]

so let me rephrase this in another way

[bright acorn]

This definition would be the K0 functor tho, right? Iirc for higher K groups you need a different construction.

[coral pivot]

well the otimes gives you the K ring, like the whole cohomology ring

but for intuition i think K^0 is fine

like in operator theory the main thing we care about is K^0 really

(well maybe not main thing but like, theres major classification programs based on it)

[coral pivot]

so back to this

one thing is every vector bundle over nice enough X is a subbundle of a trivial bundle

and this is basically like, i have the trivial bundle whose fibres are R^N

and i am choosing a subspace for each fibre

choosing a subspace is the same thing as specifying a projection matrix right

[bright acorn]

Oh, I think this is the reason why for K theory we usually work with loc compact hausdorff spaces?

[coral pivot]

yeah it might be

[bright acorn]

At least most K theory books I see do K theory only for these.

[coral pivot]

yeah i think its because we have nicer vector bundle over these

(you can extend it to the whole of top tho btw)

[coral pivot]

so now a vector bundle is the same as the data of attaching a projection matrix (continuously) to each point of the space right

or that is, you take choose something from M_n(C(X)). you need to assign some equivalence relations and what have you but at the end it boils down to doing this

so what you can do then to define K theory is look at the infinite matrix group M(C(X)) and make it equivalence classes of projections from this

does this make sense?

[bright acorn]

Wait, I have seen this before

There's this result

[coral pivot]

oh is this efton park?

[bright acorn]

Where they prove that for nice enough spaces K0(X) is the same as a group of idempotent operators on X

Idem(C(X))

[bright acorn]

Yeah

That's the main book I used for basic K theory stuff

[coral pivot]

nice efton park is an operator theorist so he wrote it very suggestively for operator k theory

(i think this is true dont quote me lol)

yeah so we are somehow dealing with projections (or idempotents as park puts it) of M(C(X)) right

so now you do like, the most obvious thing

you know how there is a duality between compact haussdorf spaces and unital commutative C* algebra

where X corresponds to C(X)

the obvious thing is replace C(X) with a general unital C* algebra

[bright acorn]

Gelfand Naimark?

[coral pivot]

mhm

[coral pivot]

and the exact same construction gives you K theory for unital C* algebras

then the 1 dimension unitization of a non-unital gives you the K theory for non-unital C* (this is dual to one point compactification of LCH spaces so it makes sense)

[bright acorn]

Ah

So K0 for unital commutative C* algebras is the same as K0(X) where C(X) ≈ A (A C*-algebra)?

Or maybe the more explicit construct where K0(A) is a certain group of projection matrices M_n(A).

[coral pivot]

yes to both

i think this was the first time i became truly convinced of "C* algebras is non-commutative topology"

[bright acorn]

Pog, what sort of "purely C*-algebraic" results can you prove using these ideas?

[coral pivot]

because thats literally what you are doing here

[bright acorn]

That's really cool

[coral pivot]

the main thing i know is that it is useful for classification

the elliot program is a good example

i could not give you anything more concrete as i havent had to use it really

[bright acorn]

that's fair, have you looked into Blackadar yet?

I have it downloaded and it looks really interesting

[coral pivot]

i havent but I will read a dedicated operator K theory book like that at some point in grad school i imagine

my knowledge is like a chapter of a book and talks i have been to as of now oof

[bright acorn]

What about algebraic k-theory? The thing with that is that I have heard defining higher k groups in this setting was really non-trivial.

[coral pivot]

i know a bit about it

i think only just K0 really

like have you ever heard of serre-swan?

it basically says f.g. projective modules are the generalizations of vector bundles

its kinda nice

[bright acorn]

Yeah, I haven't seen a proof of it tho. Karoubi goes into it in depth but I haven't taken a look into the chapter yet.

[coral pivot]

yeah i havent seen the proof either

i know in C* some people care about the algebraic K theory of C* algebras

like there is some knowledge to be gained in comparing algebraic K groups to topological K groups of C* (there is some map called the comparision map)

i read a paper on it, seemed interested but also way above my league

lol

[icy schooner]

Construct nonnormal covering spaces of the Klein bottle by a Klein bottle and by a torus.

Can anyone give some hints

we want to find a nonnormal subgroup of <a,b|aba^(-1)=b^(-1)>

that I guess is somehow pasted together from a klein bottle and a torus

[odd flame]

can i get a nudge on this pls

[river granite]

write the definitions, since the uniform metric should be a sup of things you should be able to note that e.g. |f_n(x)-f(x)| \leq sup (...) for all x in X

[odd flame]

im a lil behind in my understanding is the uniform metric a metric on a metric on a metric

i think i asked about this earlier

[gritty widget]

[odd flame]

going off of these two definitions

perhaps im abusing the term "metric on ___" but the uniform metric is defined in terms of a different metric

[odd flame]

ok i understand the standard bounded metric i was overcomplicating that

but the uniform metric defn up there - is R^J supposed to just be a finite product of R's?

[deep ibex]

J should be able to be an arbitrary index set. But the supremum can be unbounded in some situations, so it may have to take values in the extended reals unless there are some other conditions on the set R^J.

Recall that elements of $\mathbb{R}^{J}$ are simply functions from $J$ into $\mathbb{R}$.

[gentle osprey]

384920

[odd flame]

oh? perhaps i read over that part

another stupid question, what in gods name does the ~ mean here

i have to show that they're homemorphic but idek what they are

[odd flame]

what did you mean to go in the (...) here? im guessing the diameter of the set maybe?

[hidden crag]

Equivalence relations

[odd flame]

i figured

[hidden crag]

You’re identifying points in your space

[odd flame]

by quotienting by a relation?

[hidden crag]

Yes

Those are all the torus

[odd flame]

so in the case of b), is it literally just saying (0,1) \times (0,1) where 0 = 1

[hidden crag]

I denotes the closed Intervall

But yes

[odd flame]

[hidden crag]

You take the Intervall glue together your endpoints and get S^1

[echo berry]

how would one construct a map, and particularly unique, from A \oplus B / Im(f,-g) to some other Z that both make the diagram commute, i.e showing that the first object mentioned makes a pushout in Ab

[odd flame]

given two products X \times X and Y \times Y, is showing that X is homeomorphic to Y suffice to show that the products are homeomorphic too

[gritty widget]

Yes

A x B is homeo to X x Y whenever X is homeo A and Y is homeo B

And it holds for arbitrary products too

Arbitrary coproducts

[odd flame]

[gritty widget]

I think this is just a fact from category theory

Yeah. This is obvious from all the diagrams

[odd flame]

can i get a nudge on this? intuitively i see what timo said about them all being the torus, that'ss obvious from S^1 \times S^1 but im having a hard time coming up with the maps

im being asked to show that those 4 are homeomorphic

[gritty widget]

I'd just spam this theorem: continuous bijection from a compact to Hausdorff space is a homeomorphism

[odd flame]

we technically havent defined compactness

[gritty widget]

Okay no, there's easier theorem to use here

[odd flame]

this looks useful

[gritty widget]

Saves me on typing

So a continuous surjection from a compact space into a Hausdorff space is a quotient map

Because image of compact is compact, meaning such map is closed, hence a quotient

Now you have the parametrization [0, 1] to S^1 using p(t) = exp(2pi i t)

[odd flame]

im a little held up on the thm tho

what is X* supposed to be that seems like a weird way of defining it

[gritty widget]

Using it you easily see that b, c are honeomorphic and a, c are homeomorphic (the map g(s, t) = (p(t), p(s)))

X* is the quotient by relation ker g

Where ker g = {(x, y) : g(x) = g(y)}

[odd flame]

also t,s backwards there no?

oh no they're not nvm

[gritty widget]

If you know group theory then it's similar but there they can simplify it to a normal subgroup {x : g(x) = e} but here we don't have any "zero" or "identity"

[gritty widget]

Yes. That's a typo

If you write d) without a parametrization then that's a map (a, b) x (x, y) to ((2+a)x, (2+a)y, b) from S^1 x S^1 to that set

It's clearly continuous and all you need to prove it's bijective

I mean bijective but not onto R^3 but onto your set

So injective

[odd flame]

the big ugly to write subset of R^3 yeah

[gritty widget]

The b coordinate gets fixed so that's good

[odd flame]

im understanding most of this but this only a question due to my lack of intuition for the e^ 2 pi i think (i learned this like a week ago somehow) - in p(t), what is t? it's a map from [0,1] to S^1 so is it only one coordinate?

[gritty widget]

((2+a)x)^2 + ((2+a)y)^2 = (2+a)^2 so since 2+a is always positive we also get equality on the a coordinate

Dividing by 2+a gives you equality on x, y coordinates

So it's injective

[gritty widget]

t is a real number

From [0, 1] here

It's a standard way in which we can parametrize the circle, and I guess, not completely obvious

In the book "Complex made simple" they derive this parametrization from scratch in the appendix

If you care to look at it

[river granite]

I meant the first defn. here by the (...) earlier, just too lazy to write and I'm on phone rn

[odd flame]

i got that one

[odd flame]

gotcha, was making a silly mistake, will def take a look at that tho

[gritty widget]

It's also the only way I know how to define cosine and sine functions without adhering to the concept of angle - which is not defined before giving, say, parametrization of the circle using exp

I think

Geometry giving us false sense of comfort

[odd flame]

also still talking about that map 2 -> 3 - does anything need to be be said about the equivalence relation in the defn of the map?

[gritty widget]

Just that p(t) = p(s) iff s = t or s, t are in {0, 1}

That is, we identify 0 with 1, or as written there, 0 ~ 1

[odd flame]

might i ask how you came up with this map

[gritty widget]

I identified a with one cosine and b with sine

Etc.

[odd flame]

[odd flame]

why is surjection obvious?

as always, i appreciate your patience

also different question but this should be a mobius strip right

wait no this is in R^2 lmao

oops

is it "compressing" points to the first and third quadrants of R2?

[gritty widget]

because the image of this map is what it is, after the parametrization has been done

with sines and cosines

[gritty widget]

if you draw it, what sort of equivalence classes do you get?

I get 4 different equivalence classes

looks like just some weird 4 point space, not discrete

[odd flame]

it's one point in each quadrant

the top right of the circle and bottom left get squished

[gritty widget]

no, it's not

[odd flame]

oop

ohhh it's xx'

one sec one sec

[gritty widget]

the relation gives identified points with first coordinates satisfying some condition

[odd flame]

is it the entire right half and entire left half, except for the x intercepts

[gritty widget]

Coset of points with x < 0, with x > 0 and two points (0, 1) and (0, -1)

[odd flame]

that's what i meant but you said it right lol

[hidden crag]

how so? you still have a lot of classes where x is greater than 0 and x' is smaller than zero

or am i being stupid

you're identifying points where either both x coordinates are positive or both are negative

[gritty widget]

If x>0 and x'<0 then xx' < 0 so they aren't related to each other

yes, exactly

hence the cosets of points with x>0 and points with x<0

[hidden crag]

but there are more than 4 points that stay distinct under that relation

i think i'm misunderstanding you

[gritty widget]

like what?

I think you're thinking of relation (x, y) ~ (x', y') iff xx' < 0 or (x, y) = (x', y')

[hidden crag]

i was being stupid

[gritty widget]

maybe not, that's not an equivalence relation actually

[hidden crag]

don't mind it

[gritty widget]

I don't, I know you're a smart person

dw

[hidden crag]

[odd flame]

does S^1 have a "standard" topology on it

[odd flame]

thinking about this question but the open sets of the quotient depend on S^1 no?

[gritty widget]

subspace topology

[odd flame]

[gritty widget]

oh, I heard of the pseudocircle before. Nice

[odd flame]

but that's still this though isn't it?

i've never heard of it

[gritty widget]

I'm not sure what I was doing at the time but probably something homotopy related

[gritty widget]

yes, if you have quotient of a topological space, then its topology depends on the topology of the original space

fortunately here the space after quotienting only has 4 points so seeing which sets are open should be trivial

those are those subsets which preimage by the quotient map is open

the preimages here are just unions of arcs

[silk tapir]

When someone says genus g surface, do they always mean connected sum of g tori? Or does this also include things like the Klein bottle, which happens to have genus 2 but (I think) can't be realized as the connected sum of 2 tori?

[rugged swan]

It's the connected sum of g tori (topologically)

[silk tapir]

So, not stuff like the Klein bottle?

[rugged swan]

Yeah

Usually you consider genus g curves over C so it's automatically an oriented R surface

[echo berry]

.

what am i missing

A, B are the objects that get mapped from f and g respectively from a common domain X

all in ‘Ab’

Its essentially just to show what the pushout in the category of abelian groups is and to show that it works

[gritty widget]

Uh... please put brackets around A \oplus B next time

[echo berry]

so that for any other object, if it exists, it will have a unique mapping from the oplus thing

[echo berry]

sorry

i thought it was obvious

[gritty widget]

well, now I know what you mean but it wasn't obvious for me before

you just want to prove that it's the push out in Ab

[echo berry]

yep

[gritty widget]

okay so suppose we have some maps x:A to Z and y:B to Z which make the diagram commute
Define u:(A(+)B)/im(f, -g) to Z by u((a, b)+im(f, -g)) = x(a)+y(b)

You want to prove that this map is well-defined

(a+f(m), b-g(m)) gets mapped to x(a+f(m))+y(b-g(m)) = x(a)+y(b)+xf(m)-yg(m)

and from commutativity of the diagram the last term is = 0

xf(m)-yg(m) I mean

you get that part?

but yeah... it's just checking that everything works fine

[echo berry]

yep makes sense

ty ❤️

its way more strisght forward than assume most of the time

[wicked yew]

how can the metrics be equivalent? I'm kinda confused since I thought the norm is derived from (l_infinity)

oh is it saying that the limit of the supremum will be equivalent to the sum of the squares, square rooted?

[gritty widget]

it's definitely not equivalent, otherwise l^infinity would contain a subspace isomorphic to l^2

and l^infinity isn't reflexive

[wicked yew]

so how would i start disproving this?

[wicked yew]

what do you mean by it isnt reflexive

[gritty widget]

That the map into the double dual isn't a isomorphism

[wicked yew]

is this the answer to my disproving question?

yeah i don't have the machinery for that proof

[gritty widget]

well, it's not a completely elementary argument

yeah, I thought you'd like something more on the basic level

[echo berry]

cant u just find an example like sequence that is constnt no?

[wicked yew]

oh yeah a counter example would probably be fine

thanks

[echo berry]

or 1/sqrt(x)

well ye

[wicked yew]

how far into a metric spaces course is this

[gritty widget]

it's functional analysis

not metric spaces

[wicked yew]

ah

[odd flame]

yall say munkrees or munkers

important topology question

[swift fjord]

neither

[gritty widget]

for an elementary argument you'd like to find a sequence convergent in one norm but not the other

something like the standard unit vectors should work

wait no

[echo berry]

ye i already answered i think

say (1/sqrt(n))_{n}

[gritty widget]

yeah, 1/sqrt(n) on the first n coordinates

l^2 norm is 1 but converges to 0 in l^infinity

[echo berry]

you mean in l^2 divergent?

its norm would be divergent

[gritty widget]

I mean, yes, it's divergent in l^2 but it's enough to notice that it doesn't converge to 0 specifically

[echo berry]

oh

[wicked yew]

wait wouldnt the sup be 1

or?

[echo berry]

uh also

[gritty widget]

no, 1/sqrt(n)

[wicked yew]

oh i thought x_n was 1/sqrt(n)

[gritty widget]

to see that it's divergent in l^2 you can notice that it converges pointwise to 0
and convergence in l^2 implies convergence pointwise

it doesn't converge to 0 so it doesn't converge at all

[wicked yew]

what did you define x_n to be?

[gritty widget]

Eso defined x_n to have 1/sqrt(n) on the first n coordinates, and 0 otherwise

[wicked yew]

but isn't xn in the reals

[gritty widget]

no, it's a sequence in l^2

meaning it's a sequence of sequences

[wicked yew]

wait so the bold x defined in the question is not just one sequence?

damn i'm confused

[gritty widget]

it is

[wicked yew]

oh what you did was the distance function between two sequences?

[gritty widget]

I'm not using the notation in your question

x_n I'm taking to be some sequence of points in l^2

see

[wicked yew]

hmm ok

[gritty widget]

you can change it to y_n if that bothers you

or x(n)

[wicked yew]

does the supremum in my question mean the supremum of all the terms within the sequence?

[gritty widget]

yes

[wicked yew]

ah i think i understand the sequence you created

[jagged sage]

If the boundary of a sub space is connected doesn’t the sub space itself have to be connected

[gritty widget]

union the open balls of radius 1 centered at (1, 0) and (-1, 0) in R^2

the subspace is disconnected, but its boundary is connected

[jagged sage]

That makes sense thanks. the converse is false though, right? If A is connected then boundary doesn’t have to be connected right

[unreal stratus]

Ye correct

[jagged sage]

Thanks. If the interior of a subspace is connected, the subspace itself is not necessarily connected right? Don’t have an example in mind but feel like it’s true

[unreal stratus]

Yeah you can take any connected open set disjoint union some disconnected set with empty interior

(Then the interior is that open set and is connected, whilst the whole set won't be connected)

For example, take (0,1) and add in {2}

[wicked yew]

is this example trying to say you can "stretch" the circle to form a square?

and i'm assuming the lines given are the lines you would stretch each quarter of the circle along

[gritty widget]

this actually holds for any compact convex set with non-empty interior

you can always stretch them out onto the closed disk

[unreal stratus]

But ye the way they are doing it is to split the square and disk into 4 regions in a nice way (so that the "arc" of each of the quarter circles is mapped onto one of the sides of the square)

Then your 4 maps glue together to form a continuous map from the closed disk to the square

[gritty widget]

I'm doing a problem which asks whether the sequence convergences in the product, uniform and box topologies for a_n_m ={n/m where n<=m , 0 otherwise} . For my proof I said that as m goes to infinity, the sequence converges to 0, then I argued that given that the sequence is pointwise convergent it converges in the product topology. For the uniform topology, I took an open set U containing 0, then there is a basis element B(0,ε) = (-ε, +ε) so choosing N > 1/ε, then a_n_m ∈ B(0,ε) for n,m ≥ N so the sequence converges in uniform topology. For the box topology I said it does as well converge because it's eventually 0 but I'm not sure how to formally show that using open bases. Also is my work so far right?

[bitter smelt]

point-set Q

if $\overline{f}$ is the extension of $f : B^n \to B^n$ to $\overline{B^n}$, where $f$ is continuous, and $\overline{f}$ is an open map on $\partial B^n = S^{n-1}$, does this somehow imply $\overline{f}$ is continuous

[gentle osprey]

{ }

[bitter smelt]

im X Ying but the full context is too much

so if this isnt true ill just figure it out in a dream or something

[lament needle]

This is from my knot theory textbook they keep using the term "unkotted arc" but this is the only part that tries to define it and this defenition is not comprensible to me, does anyone know what it means by
"ball arc pair" I assume it means the ball and the arc intersecting the ball but then i dont know what it means for that to be "piecewise linear homemorphic to the product of an interval with a disc-point pair" and i dont know what a disc point pair would be

[gritty widget]

Wdym extension? As a map from closed ball to itself?

Take f(x) = x on the ball and f(x) = -x/2 on the sphere

[bitter smelt]

Sorry. AN extension

I.e.: just a map when testricted B^n, is identically f

But no, your counterexample is fine

Thank you

I'll dream the solution to my problem instead

[shadow rampart]

in a compact space is every open set a union of finitely many basis elements?

[coarse night]

not necessarily

[gaunt linden]

Surely not. [-1,1]² is compact, but if we take the basis consisting of open rectangles intersected with [-1,1]², then finitely many of them cannot create e.g. the open unit disk.

[rapid olive]

struggling on this q

not sure how to use the connectedness here

[lunar yoke]

well do you maybe know some theorem relating connectedness and continuous maps?

[rapid olive]

if f:X→Y is a continuous surjection and X is connected then Y is connected

[lunar yoke]

correct

[rapid olive]

oh

wait

[lunar yoke]

a classification of connected sets in R would help too

[rapid olive]

we take img f

right?

[lunar yoke]

yes

[rapid olive]

ok thx

[bitter smelt]

i figured out my problem. not in my dream but as soon as i woke up

[opaque cloud]

now i know 0 measure theory and only super basic topology but can you do integration on topological spaces? like the most basic top space (not even hausdorff)??

[tidal cedar]

not really

[wet crow]

you can't do integration without a measure

and a topology doesn't induce a measure

but you can do integration on plain sets without a topology though

[opaque cloud]

oh I see so you need literally no structure on your set? all you need is a measure?

by a measure you mean a map from the sigma algebra to [0, inf) ?

[little hemlock]

ye, correct

a non topological example:

[opaque cloud]

damn actually sick!

measure theory's so cool

[little hemlock]

When ur measure space is the natural numbers with counting measure, integration is just infinite summation

[opaque cloud]

oh damn

too bad it's insanely difficult to do stuff like that in infinite dimensional spaces like making the feyman path integral rigorous

[cedar pebble]

one situation where you have a very nice theory of integration is for any locally compact topological group

in this situation you have a unique up to constant measure which is translation invariant

[cerulean oriole]

Given a topological space X and a uniform space Y, can locally uniform convergence on the space of all functions from X to Y be described by a uniform space structure?

[shadow charm]

on borel algebra or something bigger?

[cedar pebble]

on Borel subsets yeah

[jagged sage]

If A is open in X and B subset of X for X topological space, is it true that A \cap cl(B) = cl(A \cap B)

I think no that there is some counterexample but everything I try works

[uncut surge]

uuuh usually the rationals in R give good counterexamples for this kinda stuff

A = rationals, B = irrationals, then left hand side is the rationals, right hand side is empty

[jagged sage]

Aren’t the rationals not open in R

With the usual topology

[uncut surge]

oh fuck damn yeah

i can't read lmao

[jagged sage]

Ur good any other ideas

[unreal stratus]

Take A = B

and then this is A vs cl(A)

[jagged sage]

Oh yeah that is obvious. Thanks I don’t know how I didn’t realize

[unreal stratus]

Np, just yeah trying fairly degenerate cases like A = B, one of them empty/whole space or rationals/irrationals (as mentioned above) is usually good for these sorts of things

[bitter smelt]

What does it mean for a simplex to be of maximal volume

In particular, the context is the statement: An n simplex in B^n (Poincaré disk model) is of maximal volume if and only if it is regular and ideal

(Obviously with this statement you could say a simplex of maximal volume is one which is regular and ideal. This is also obviously not the definition I’m looking for)

Is it just a simplex of dimension n for which there exists no other simplices of the same dimension with strictly greater volume?

This is what the name implies but I cannot find a source for this

[odd flame]

im having trouble seeing how the definition of the product topology leads to the "and" in the last part of this thm

any pointers

[gritty widget]

Munkres proves this in the discussion immediately prior.

[ocean narwhal]

as ngroupoid pointed out, when the group is topological and locally compact you have a theory of integration (Haars theorem). unfortunately this leaves out some very important topological groups (namely infinite dimensional topological vector spaces (a theorem says a topological vector space is locally compact iff it is finite dimensional)). as a result getting measures on Banach/Hilbert/etc spaces is very hard. (one example of such a measure is the Wiener measure on the space of continuous functions. you can construct it using donskers invariance principle)

[ocean narwhal]

you can always take the borel subsets and add the null sets

[void gazelle]

Hi, guys, is it true that in a subspace topology Y\subset X, A is closed in Y iff A= Y\cap A', where A' is closed in X?

[coarse night]

that's by the definition of the subspace topology

[void gazelle]

i thought by the definition is for open

[gritty widget]

take complements, you get the same thing

[void gazelle]

thanks!

[jagged sage]

If I have X, Y topological spaces and A in X and B in Y are sub spaces, with f:X to Y homeomorphism with f(A)=B, how can I prove g: A to B and h:X\A to Y\B homeomorphisms as well

It seems like proving g homeomorphism isn’t bad directly, bijectivity seems simple and continuity and inverse seems trivial. H not sure

[coarse night]

if you are saying it's tivial...

[jagged sage]

What do you mean

[little hemlock]

what do you mean what do you mean

It seems like proving g homeomorphism isn’t bad directly, bijectivity seems simple and continuity and inverse seems trivial.
it sounds like you know how to do it already lol

[jagged sage]

For g, I think know how. For h, though, not sure. I also feel like there might be more elegant solution than directly proving both but idk

[little hemlock]

ah okay i see what you mean

you don't have to prove both

basically X \ A can take the role of A and X \ B can take the role of B since f(X \ A) = Y \ B

just a matter of swapping out the subspaces in question

i.e. just prove that g is a homeomorphism

[jagged sage]

Yeah u are right idk how I didn’t realize complements did that. Thanks

[little hemlock]

things are nice since f is a bijection. npnp

[limber ravine]

If ${O_i}$ is an open cover for $A$. Then $A \subset \bigcup_i O_i$

[gentle osprey]

mns

[limber ravine]

And so, we can intersect both sides of the inclusion yielding $A \subset \bigcup_i(A \cap O_i)$

[gentle osprey]

mns

[limber ravine]

Correct?

[plain raven]

yes

[normal holly]

What are the key/important theorems of (pure) lattice theory?

[coral pivot]

ok hello @frigid patrol @gritty widget @cold temple , I am gonna type out why C* algebras is "non commutative topology". I think I was talking to you 3

so historically

C* algebras arised from the study of operators on hilbert space, which was studied due to quantum physics

So maybe it makes sense why C* would describe kind of "non-commutative" objects

like think about a classical physics system, you can think about functions on it

say the position and momentum functions

and these commute

and really this whole function space should commute for the same reason.

now lets think about a quantum system

the position and momentum functions no longer commute, so the function space of a quantum system is no longer a commutative algebra right

so thats maybe the first intuition, is that you have this non-commutative(quantum) system, and that is reflected in its function space

Ok so thats physical intuition, lets get into the math

---

So we start of with this result of Gelfand-Naimark

for a commutative C* algebra, you can form what is called its spectrum

this is the set of all maximal ideals which is also the set of all characters, and this can be given a nice topology to be a LCH space.

and basically gelfand Naimark says that a commutative C* algebra A is isometrically isomorphic to C_0(Spec(A)

and what this means really is every commutative C* algebra is a function space of some LCH space, and vice versa

and this is almost an equivalence of categories, so we can say commutative C* algebras is basically topology

and then it motivates saying that C* algebra in general is non-commutative topology

The reason this is justified is how we have managed to extend a lot of topology/geometry to this setting

The way to think of this imo is that you are identifying a space with its function algebra, and the non-commutative function algebra corresponds to some abstract space (there is an analytic spec like in AG but i wont go into that)

my blog will tell you how compactifications have an analogue in unitizations, and how you can use this for a nice characterization of stone-cech compactification. I think this is an accessible example

---

What sold me was K theory, essentially K theory is about vector bundles, turns out all vector bundles are contained in a trivial bundle

so you can think of a vector bundle as assigning projections at each point (to project to the subspace)

i.e some projection matrix in M(C(X))

turns out operator K theory just replaces C(X) (which are all the commutative unital C*s) with a general C*

and that this is extremely useful

like you are literally just changing a C(X) for a C* algebra

it quite literally is just making matrix entries non-commutative

---

I also know I was asked for an explicit example of a non-com top space

so let me talk about non-commutative torii

so basically look at the regular torus T^2 right

lets think about the functions on it, C(T^2)

my claim is that this is the C* algebra generated by 2 unitaries u,v that commute.

the proof? use Gelfand-naimark, and note that a character is determined by where u and v are sent. Since these are unitaries these land in the unit circle. So its a S^1 x S^1 choice right

so the character space is S^1 x S^1.

[gentle osprey]

computations

[coral pivot]

This is no longer a commutative algebra however notice its basically just the torus with this small twist rotation factor

so it makes sense why this is a non-commutative torus

you can generalize this to higher dimensions, i encourage you to think of that

you can do stuff like the irrational flows etc on this non-com torii too which is cool and maybe shows that this is still behaving like a top space

I close out by saying that non-commutative topology is just pure C*

you can decide to add extra structure on it and do what we call non-commutative geometry

alain connes basically built this field up and all your favourite things in commutative geo probably has an analogue in non-com geo

I havent learn too much of this yet, mainly because connes book is a monster

one example i have seen is a spectral triple which is a triple of a * subalgebra of B(H), a hilbert space and a distinguished operator satisfying some property

and the commutative case of this corresponds to basically a spin^c manifold with its dirac operator.

so thats a non-com spin^c manifold right there

There is also non-commutative dynamics

you get to study dynamics of C* algebras and group actions of C* algebras by nice groups

and for the latter you can form a new C* algebra called the crossed product which encodes info about this action

this can be used to study the action very well, and in particular it can be used to study the commutive case

i.e topological dynamics

the last thing is non-commutative measure theory

a special class of C* algebras called von nuemann algebras (or W* algebras)

in the commutative case are L^infty(X) for some measure space X

so studying these is non-com measure theory

i am not really too big into measure theory so i cant say much beyond that but i know its something people have studied.

thats it thanks if u made it to the end of this giant wall of text

#point-set-topology message

link to the top lol

[gritty widget]

Hello

What are some (non-trivial/non-boring) examples of:

(1) spaces that are homeomorphic to interval [0,1) ?

(2) spaces that are homeomorphic to { cos(2pi*x) + i sin(2pi*x) | x in [0,1) } ?

is "topologist's sine circle" homeomorphic to (1) or (2) ??

https://i.stack.imgur.com/a9KHY.jpg

also there is "Warsaw circle" to consider too... https://ncatlab.org/nlab/show/Warsaw+circle

https://ncatlab.org/nlab/files/warsaw.gif

[gritty widget]

1 and 2 are the same, and both Warsaw circle and topologist ... are different (they are compact for example)

[gritty widget]

do you mean (1) (2) are homeomorphic? I think (1) is non-compact but (2) is compact...

My bad. 2) is a circle
Still, all 4 spaces are not homeomorphic

Reason is local connectedness

That the last 2 are not homeomorphic would require me some work, but the idea is that the string needs to map to itself, while approaching from one side you get a limit for topologist ... but for Warsaw circle, you don't get a limit

[gritty widget]

😀

[golden gust]

all vector bundles are contained in a trivial bundle
is this always true? I can only find references which deal with compact base spaces

[unreal stratus]

doesn't seem to hold in general but for the interests of K-theory ig you usually work with stuff likek compact hausdorff spaces or the homotopy type of finite cw complexes or maybe paracompact spaces anyway

and the statement posted above holds for the first two types of spaces at least

[fervent bridge]

suppose X = {0, 1} and Y is a topology on R generated by the opens (-oo,0) and [0,oo). X and Y aren't homeomorphic, but they seem pretty much the same from the topological point of view because the continuous functions Y -> A are in 1-1 correspondence with the continuous functions X -> A. is there a name for this relationship between spaces?

[gritty widget]

X is the Kolmogorov quotient of Y

Which is the T_0-fication of a space

[lunar yoke]

if you really had a bijection Hom(X,A) = Hom(Y,A) natural in A, then X would already be homeomorphic to Y by abstract nonsense

[gritty widget]

I think they had that A is good enough somewhere in their mind, but didn't write it

T0 is enough I think

[fervent bridge]

interesting thanks

yeah i figured it wasn't gonna be natural in A

[rapid olive]

struggling on a proof that the intersection of two compact subsets of a haussdorff space is compact

i know that a compact subset of a haussdorff space is closed, but I'm not sure how to use that

[gaunt linden]

Suppose you have an open cover of the intersection.

[rapid olive]

yes

[gaunt linden]

Extend it with the complement of one of the compact subsets, yielding an open cover of the other one.

[rapid olive]

hm

uh

yes i get it

then that has a finite subcover, but the only potentially infinite part of the cover is on the intersection

so if you remove that complement again you get a finite subcover of the intersection

[gaunt linden]

Or in other words, that complement is obviously useless for helping cover the intersection.

[rapid olive]

yeah

[coral pivot]

yeah that might be the case, the vector bundle construction is on compact spaces tho so its fine.

[gritty widget]

the intersection is a closed subset of one of the sets so compact

[rapid olive]

i thought that but i couldn't find the theorem anywhere in hatcher

so I wasn't confident it was a thing

[gritty widget]

1. compact subsets of Hausdorff space are closed
2. closed subsets of compact space are compact

[unreal stratus]

Also just to give you a nice counterexample for the nonhausdorff case: take the real line and adjoint two points A and B, whose only open neighbourhoods are the entire space X and X minus the other point

||Then any open cover of X minus {A} has a subcover of cardinality 1 and similarly for X minus B, so each id compactz but the intersectionis just R||

I guess this space is also T1

[ocean narwhal]

you can find it in Rudin (though maybe its an exercise)

[ocean narwhal]

smh my head all topological spaces are hausdorff

[little hemlock]

I don't see why because X x I deformation retracts separately onto X x {0} and X x {1}, we can conclude i is surjective?

i know that i is surjective if X x I deformation retracts onto X x bd(I), but that's not what we have

well in fact i would be an isomorphism in that case

[little hemlock]

nvm i see. they are just identifying Hn(X x bd(I)) with its natural direct sum decomposition

[cerulean oriole]

Anyone?

[little hemlock]

let $f(z) = z^n$. How does one rigorously see that $f_* : H_1(S^1) \to H_1(S^1)$ has degree $n$?

[gentle osprey]

kxrider

[cerulean oriole]

It maps the curve γ = (t->e^2πit) to γ^n right?

And fundamental group of S^1 is cyclic group generated by γ, so a homomorphism mapping γ to γ^n is the nth power map

[little hemlock]

sure, if all is right in the world, we'd expect gamma to generate H_1(S^1). this comes down to showing that n(gamma) - gamma^n is a 2-boundary, which is where im a little stuck

[cerulean oriole]

Wait, why did I think H_1(S^1) was the fundamental group?

Define a map F from [0,1]^2 to S^1 such that
F(s,t) = e^{2pi(n-t)s}
Then F(s,0) = γ^n(s), F(s,1) = γ^{n-1}(s), F(0,t) = 1, F(1,t) = γ^{-1}(t) where γ^i(s) is notation for e^{2pi i t}.
Does this show γ^n = γ + γ^{n-1}?

(From which γ^n = n γ by induction, at least for n > 0.)

[unreal stratus]

Hurewicz here i guess aha

[jagged sage]

If I have x1,…,x_n distinct points in Hausdorff space X, to show there exists pairwise disjoint xi in Ui for all n=1,…,i can I just take arbitrary point x and by definition there exists disjoint sets containing x and xi for all i 1 thru n

Seems obvious intuitively but not sure about rigorously explaining. Seems like it should follow almost directly from def

[gritty widget]

what do you mean

just take the average over all vertices?

[gritty widget]

Induction on n

[unreal stratus]

Take disjoint neighbourhoods pairwise and intersect/take unions appropriately, say

[wicked yew]

I know that A is composed of elements of form (1/9)^j * (1/10)^i

does this help with this question in anway

i'm struggling to find the intersections since I end up excluding some elements of A

[empty grove]

Look at the first n digits in the decimal expansion

[wicked yew]

in a general decimal expansion?

[cerulean oriole]

I haven't actually read this stuff properly
I just know some definitions

If adding some numbers and dividing by a known constant is computation-intensive, what isn't?

[gentle osprey]

potato

[unreal stratus]

I can't seem to find a nice proof for whatever reason; ig the key idea is it's like the orbit space of the action of U(n) on the (orthogonal) Stiefel manifold of n-planes in C^infty, with that mfd being contractible ig?

oh okay so ig it boils down to applying this? This appeals to a given construction (Milnor's) tho

[jagged sage]

If I have connected subspace A of X and B supspace of X, how can I showed B connected if A subset B subset A closure?

Tried saying there exist U,V non empty open subsets that form seperation of B. Then A subset of U or V since otherwise their intersection will form seperation. But now stuck

[gritty widget]

If A is a subset of U, try taking closures and see what you can get from there.

[jagged sage]

I think I can get a contradiction if I have V intersect U closure is empty. Not sure if this is true though

[gritty widget]

It's true. U is closed in B, so it is its closure intersected with B.

So its closure cannot intersect V.

In symbols, $U = \overline{U} \cap B$, so $$\emptyset = U \cap V = \overline{U} \cap B \cap V = \overline{U} \cap V,$$ since $V \subset B$.

[gentle osprey]

rakko

[jagged sage]

Makes sense thanks

[wicked yew]

still kinda stuck

[unreal stratus]

Yeah so a cool thing is to follow the hint moldi gave one considering the first n digits of the decimal expansion and like

Using that to define the nth set in ur intersection

[wicked yew]

but within the nth set would it be a finite union of singletones?

[unreal stratus]

But yes I remember finding this question annoying due to nonuniqueness of decimal expansions

[unreal stratus]

No

[wicked yew]

0.1 = 0.10000000...... right?

[unreal stratus]

Well the nth set should be infinite

Since A is infinite

[wicked yew]

would it just be an intersection of the sets [first n digits are 0 or 1]

for n>= 1

[unreal stratus]

That is one way of doing it

I think there's actually an even easier way of doing it actually

Rather than worrying about the first n digits, you can just worry about the nth digit

[wicked yew]

thats what i was thinking before

but isn't that an infinite union of singletons for every set

and so we cant be sure if it is still closed

[unreal stratus]

what does this mean i mean

[wicked yew]

like say we focus on digit 1

[unreal stratus]

I mean every infinite set is an infinite union of singletons

[wicked yew]

hmm im kinda confused

[unreal stratus]

but yes like if we allow for all the numbers with, say, first digit 0 or 1

[wicked yew]

cuz i thought an infinite union of closed sets is not closed

[unreal stratus]

not necessarily

[wicked yew]

well it may not be closed

[unreal stratus]

It is just yes it may not be closed

But yes you can show that the set of all reals in [0,1] with first nth decimal places 0 or 1 is closed

e.g. by considering it as the preimage of a closed set under a continuous function, say

also as a like health warning uh

don't get too bothered about rounding; I think the best interpretation is that whenever you have trailing 9s you just round up (that's what I did)

[wicked yew]

ok so I don't need to consider 9's?

[unreal stratus]

uh well you do but like

I was just gonna say don't get too caught up about whether e.g. 0.19999.... is in A lol

ig like write a convention (I mean the most obvious to me is that if a finite expansion exists we pick the finite one)

but yeah I remember this question giving me a bit of a headache due to possible ambiguity xd

[wicked yew]

did you mean 0.0999999....?

[unreal stratus]

No

Oh

sorry yeah sure

and 0.999.... of course

[wicked yew]

ah yeah that makes sense

[wicked yew]

is there a trivial function for this?

[unreal stratus]

yes i think nope

like you want a function that checks what the nth digit is

[wicked yew]

would it be continuous? Since the function would just map to integers

[unreal stratus]

ye lol i messed up but it does work if you do first n digits actually

lol i did that originally in the sheet and thought it might work to just consider nth digit, sorry mb

[wicked yew]

np it helps me think

would an arbitary f : Z -> R work

where Z is restricted to [0,1,....9]

[unreal stratus]

not sure what you mean by "arbitrary" here, and I assume you have the domain and codomain the wrong way round

[wicked yew]

i was thinking of mapping the nth digit to something in the reals

[unreal stratus]

I don't get why that is useful

the poitn is that you want a function from [0,1] -> Z or [0,1] -> R, so that you can conclude a subset of [0,1] is closed

[wicked yew]

oh yes my bad

[gritty widget]

Mapping [0, 1] into Z is pretty trivial

But mapping Z into R is pretty trivial too

First case the maps are all constant

Second case the maps are all continuous

Don't you want something like... {0, 1, ..., 9}^omega to [0, 1] maybe?

Then f({0, 1}^omega) is compact

[wicked yew]

would omega be like the first n digits?

in your notation

[gritty widget]

omega = natural numbers

= first countable ordinal

= first countable cardinal

So it's just a countable product of discrete spaces

Homeomorphic with the Cantor set

[wicked yew]

lost me ngl 🤣

[wicked yew]

could I find a continuous map with this

[gritty widget]

Mapping interval to Z?

Sure, map everything to 0

[wicked yew]

would that show that [first n digits in decimal expansion either 0 or 1 ] is closed?

f is continous and is mapped to a closed set so must the preimage?

[gritty widget]

No

If you insist on doing it this way then show it's a finite union of closed intervals

The proof using countable product of {0, ..., 9} is nicer imo

[wicked yew]

i have not worked with these definitions yet

ill try the other way

[unreal stratus]

I am wondering if there is some meme way of doing this problem tbh, might have a think

[gritty widget]

Meme way?

[unreal stratus]

Oh dw i had an idea for an alternative approach but doesn't work

[odd flame]

problem is trying to show that the product of path connected spaces is path connected - isnt the projection map here backwards?

reading a SO post on it, isnt this answr projecting from X the product space into X_i the component of the product...?

[gritty widget]

Projections go from the product to its components. There is nothing wrong with this answer.

[odd flame]

my confusion is that in order to show that the product X is path connected, dont we need to build up from each X_i being path connected

[gritty widget]

That's what's being done, though.

Paths in each component are being used to build a path in the product.

They start from two points x and y in X. Project them down to x_i and y_i in X_i and use its path-connectedness to get a path \gamma_i in X_i from x_i to y_i. Then the definition of the product gives you a path \gamma in X which projects down to \gamma_i under each \pi_i.

That's a path joining x and y in X, which is what you were looking for.

The "building up" you require is the "the definition of the product gives you..." part.

[odd flame]

so you're also using that the inverse projection map is continuous...?

i mostly see it tho

[gritty widget]

What is the "inverse projection map"? The projections are not invertible.

Which step?

[odd flame]

seeing the path for any pair of points in X_i is easy enough, literally just follows from the definition

so i guess "definition of the product"

projection map is X -> X_i and the path is X_i <- [0,1]

feel free to sully me i could just be silly

[gritty widget]

It's basically the universal property of the product.

More concretely: define $\gamma\colon[0, 1] \to \prod_{i\in I}X_i$ by $$\gamma(t) := (\gamma_t(t))_{i\in I}.$$ This is continuous since each $\gamma_i\colon[0,1]\to X_i$ is, and $\pi_i\circ\gamma=\gamma_i$.

[gentle osprey]

rakko

[gritty widget]

This is a theorem in Munkres, if you want a reference.

Theorem 19.6.

[odd flame]

damn you knew that really fast

will look tho

[gritty widget]

I have a copy of it.

[odd flame]

so i guess a somewhat adjacent question, is there a space that highlights the difference between box/product topology well?

[gritty widget]

Any infinite product, since that's where those topologies will differ.

On the same page as the theorem I cited is an example using the countably infinite product of R with itself.

The diagonal function f(t) = (t, t, ...) from R to the product is continuous if the codomain has the product topology, but it's not continuous if given the box topology.

The exercises have more.

[prisma arrow]

Explicitly if I am given two functions and I dont know if they are in same homotopy class how do I know if there exists a homotopy between them?