point-set-topology

[red yoke]

Yea the Euler characteristic is the alternating sum of the number of n-cells

F - E + V in this case

[radiant pollen]

Ahh okay I see

[obtuse meteor]

Does anyone know a categorification of the classifying space approach to bundle theory

I.e given a bundle map E/B -> E’/B is there a corresponding thing happening in [B, classifying space]

I suspect there is but I can’t work it out

On second thought this is probably just pullback stuff

[fierce lily]

Can someone explain why we can not prove uniform topology is coarser than product topology in the same way as we prove box topology is finer than uniform topology

And second thing is, for box neighborhood U, can I change 1/2 epsilon into epsilon?( does it matter?)

[swift fjord]

Because they define the box nbhd as an infinite product

You don't have such elements in the product topology

[left eagle]

is there an easier way to prove the equalities in problem 2 than A subset B and B subset A?? seems really tedious to do

[fathom steeple]

I have that X' and X'' are both covering spaces of X, say with projections p1 and p2. How can I show that if, for some x' and x'' with p1(x')=p2(x''), there is a covering space map p: X' \to X'' with p(x')=x'', then it is unique?

I think it suffices to show that p1 = p2 \circ p but i dont know how to show this

oh also i have that X is locally simply connected

[rancid umbra]

its not as tedious as it appears. its almost tautological

[left eagle]

yeah i realize that the two definitions of each side are practically identical

but i cant rlly say "oh these two definitions are equivalent therefore the sets are equivalent".... can i?

[rancid umbra]

if it appears clear to you and is acceptable by whoever is reading it (if anybody), then yes, that method works

[thin tide]

what book is this from?

[left eagle]

Not a book my problem set 😅

[thin tide]

oh lmfao

formatting looks nice

[uneven scroll]

Hi I have a question regarding this problem:

let f: [a,b] -> M be the path from x to y. If there doesnt exist an t s.t. f(t) is in the boundary of A then there must be a discontinuity there. Im not quite sure how to prove that.

[proven hollow]

should this be infinite hausdorff spaces? edit: sorry, I think I misinterpreted what a limit point is, I figured it out

[rancid umbra]

you can decompose M into the interior of A, the boundary of A, and the exterior of A (the interior of M\A).

if either x or y is in the boundary of A, then you are done. so WLOG, assume x is in int(A) and y is in ext(A) = M \ cl(A) = int(M\A)

(since, if x is in A, its either in int(A) or bd(A), similarly for y in ext(A))

you can show that if there is no path between x and y intersecting the boundary of A, then the path component of x and y is disconnected

using the decomposition of M will help

[rancid umbra]

it is necessarily true that in a hausdorff space if a subset has a limit point, then the metric space is infinite. due to the hausdorff property, points are isolated from finite sets

[uneven scroll]

well thats kinda the problem. I did that and basically got

i dont really know what do do with that

i know that E and I have to be disjoint and since [a, b] is connected, the thing i just wrote down is connected aswell

well i know that it cant be connected because the bpundary pf A is missing but i wasnt sure how to argue that

[silver ridge]

I'm quite struggling to show the first inclusion here. I tried to go by the definition of incusion (x in LHS implies x in RHS) and by the characterisation of closure as being the smallest closed set containing whatever and I've gotten stuck both ways. Can someone give a starting point please? (pls ping)

[rancid umbra]

you are close. Let me give you another push:

let C be the path-component of x and y. You know that int(A) is an open nbhd of x disjoint from y, and ext(A) is an open nbhd of y disjoint from x.

how can you disconnect C using int(A) and ext(A)?

[silver ridge]

I figured it out using another characterisation of closure

[uneven scroll]

What's nbhd

[rancid umbra]

neighborhood

[uneven scroll]

Oh okay

Im not quite sure what you mean by disconnect

[rancid umbra]

do you have a working definition of what it means for a set to be connected?

[uneven scroll]

There is a continuous path s.t. the path is completely in the set

[rancid umbra]

let me think for a bit, since it seems like you are unfamiliar with the standard definition of connectedness

[uneven scroll]

Oh

Sorry

[rancid umbra]

all g

just need to figure out how to explain it to you

[uneven scroll]

My problem is that intuitively it makes perfect sense, it's just that I dont know how to formulate it

[rancid umbra]

suppose that f : I --> M is a path from x to y, with x in int(A), y in ext(A).
assume for contradiction that the image of f never intersects the boundary of A

consider the following numbers:

a = sup{t in [0,1] : f(t) in int(A)}
b = inf{t in [0,1] : f(t) in ext(A)}

first, verify for yourself that the above supremum and infimum even exist.

next, how do a and b compare? is a > b? is a < b? are they equal? maybe you can only say that a <= b or a >= b?

finally, consider the intervals [0,a) and (b,1]: how are points in [0,1] which trace out the path f distributed among the sets

{t in [0,1] : f(t) in int(A)},

{t in [0,1] : f(t) in bd(A)},

and

{t in [0,1] : f(t) in ext(A)}.

put this all together and reach a contradiction

@uneven scroll

[uneven scroll]

Sorry, I'm on my phone atm, I'm almost home. I'm thinking about it while walking. I'll respond in a few minutes if that's okay

[rancid umbra]

all good, i was just pinging to make sure you saw

[uneven scroll]

1. I would say a and b exist since I is bounded in both directions. 2. Intuitively a = b. Im not sure how ti justify that tho.

3. Why (b,0]?

did you mean (b,1]

[rancid umbra]

meant (b,1]. (b,0] is empty

[uneven scroll]

is that what you mean by " distributed among the sets"

[rancid umbra]

i mean, with the assumptions we made and with regards to the decomposition of M as int(A) U bd(A) U ext(A), which sets can the path f lie in?

[uneven scroll]

intA and extA

[uneven scroll]

thats what i meant by this, maybe i used wrong notation

[rancid umbra]

the point is that {t in [0,1] : f(t) in int(A)} and {t in [0,1] : f(t) in ext(A)} union to [0,1]

[uneven scroll]

we know that those sets are not closed

but [0,1], is

[rancid umbra]

?

any space is closed in itself

draw a picture

of what is going on here

[uneven scroll]

wait maybe i read it incorrectly

[uneven scroll]

not sure what you mean. the interval [0, a) isnt closed

for example

[rancid umbra]

why are you focusing on not closed sets unioning to the entire space?

what does that gain you?

[uneven scroll]

{t in [0,1] : f(t) in int(A)} = [0, a), no?

[rancid umbra]

yes!

and what about the other one?

[uneven scroll]

yeah im very confused

thats what i meant by that

[rancid umbra]

forget about them being open or closed or what not

[rancid umbra]

what is the corresponding statement for (b,1]?

[uneven scroll]

well i wanna explain my thought process first:

that was my thought process before

[rancid umbra]

gamma([0,1]) is a subset of M, how are you concluding that it is a subset of [0,1]?

[uneven scroll]

just forgot gamma

[rancid umbra]

gamma([0,1]) != M cap gamma([a,b]). you are missing everything on the path going from 0 to a and b to 1

i think your intuition is there tho

[uneven scroll]

no?

since gamma is in M

per definition M>A

maybe im misunderstanding something

[rancid umbra]

this is the picture you want to have in mind and is the intuitive situation

[uneven scroll]

that feels wrong

[rancid umbra]

the situation we are imagining here is impossible if f is to be continuous

[rancid umbra]

why?

sorry

[uneven scroll]

i would imagine that a = b

[rancid umbra]

a and b should be in the same spot

but other than that

[uneven scroll]

also gamma [0,1] < M

[rancid umbra]

yes, and? its a path in M

[uneven scroll]

yes

[rancid umbra]

whats your point

[uneven scroll]

this

i dont see how thats untrue then

[rancid umbra]

no, you wrote gamma([0,1]) = M cap gamma([a,b])

[rancid umbra]

this is true

[uneven scroll]

oh sorry my notation went fromt the interval [a,b] for the path to your notation [0, 1]

i just mixed up the notations

instead of a b put 0 and 1

[rancid umbra]

anyways, this is the right idea.

what i was trying to show you was that
[0,a) = {t in [0,1] : f(t) in int(A)}
(b,1] = {t in [0,1] : f(t) in ext(A)}.

combined with the fact that [0,a) U (b,1] = [0,1] and that a <= b, you reach a contradiction

because you are missing the entire interval [a,b]

[uneven scroll]

wenn i already kinda knew that, my problem is that i dont know how to show a <= b

just me saying it must be true is not really enough

[rancid umbra]

show that b is an upper bound for {t in [0,1] : f(t) in int(A)}

what would go wrong if there were some t in [0,1] with f(t) in int(A) and t > b?

[uneven scroll]

t would be bigger than a

[rancid umbra]

how do you know?

[uneven scroll]

lemme think for a sec

[rancid umbra]

yes. this is bad since int(A) cap ext(A) = empty

the last thing to show is that [0,a) is actually equal to {t in [0,1] : f(t) in int(A)}

[uneven scroll]

^

is this correct?

1. i mean

[rancid umbra]

both sets are non-empty and bounded above. the set corresponding to a is bounded above by 1 (and a tightly by b), the set corresponding to b is bounded below by 0 (and tightly by a)

so yes

[uneven scroll]

so by the same argument we can show that a>=b and thus a=b

[rancid umbra]

i don't actually think you have that much control over a and b

all you can say is a <= b

[uneven scroll]

mhm

wouldnt that same argument work for a>=b

[rancid umbra]

no, because we are showing that b is an upper bound for the set corresponding to a. unless im missing something that you want to explain in more detail

showing that a is a lower bound for the set corresponding to b is just a <= b again

[uneven scroll]

yeah juzt realized that

okay so basically i wanna show that the sets you mentioned look like [0,a) and (b,1]

[rancid umbra]

yes

one inclusion is easy

[uneven scroll]

what do you mean by one inclusion

[rancid umbra]

{t in [0,1] : f(t) in int(A)} is a subset of [0,a)

[uneven scroll]

oh

okay 1 sec

[rancid umbra]

im a bit worried about one thing that im hoping won't be a problem

[uneven scroll]

lemme write soemthing down

likewise for (b,1]

[rancid umbra]

yes

now we need the other way

[uneven scroll]

wait

t<a

ofc

[rancid umbra]

i don't think thats quite right

[uneven scroll]

not <=

[rancid umbra]

i think i made an oversight

[uneven scroll]

https://tenor.com/view/cat-explosion-stairs-gif-26161925

[rancid umbra]

what we're running into is this issue

where the path can skip back and forth

and thats why {t in I : f(t) in int(A)} is easily seen to be a subset of [0,a), but not necessarily the other way around

tbf this is a lot easier if you know the definition of connectedness

this was my attempt at an ad hoc solution avoiding it

[uneven scroll]

are you by any chance german

[rancid umbra]

i am not

are you, by chance?

[uneven scroll]

do you have a chatgpt acc

[rancid umbra]

yea, why?

[uneven scroll]

thats our def of connectedness

[rancid umbra]

yea, this is what's usually called path connectedness

ein metrischer is something like a metric space?

[uneven scroll]

yeah metric space

we also know that if A is connected then f(A) is also connected for f continuous

[rancid umbra]

right. so. in a metric space M, M is said to be connected in there are two non-empty, disjoint, open subsets of M which union to M.

this is useful here because we don't have to worry about this situation

[inland laurel]

A Is it a topological space?

[uneven scroll]

A is a subset of M

and M is a metric space

idk if thats a topological space

i dont think we defined it that way

[rancid umbra]

i was trying to avoid that

and just keep it self-contained

[inland laurel]

i think if M a topological space and A subset of M and yea if M path connected and a map go to another a topological space if f not homemorphism not f(A) path connected

[uneven scroll]

maybe we can avoid it

[uneven scroll]

if we just stick with a

[rancid umbra]

we can use this 🙂

define the function F : int(A) U ext(A) --> {0, 1} as F(x) = 0 if x in int(A) and F(x) = 1 if x in ext(A).

look at the path F o f : [0,1] --> {0, 1}

if the image of this function is to be connected, what can it be? ({0,1} has the discrete metric)

[uneven scroll]

{0,1} as in the boundary of [0, 1]?

[rancid umbra]

sure

that works

[uneven scroll]

whats sure

:c

[rancid umbra]

i mean, its fine to think of it like this

[rancid umbra]

if you can show that F is continuous, then you should be making good progress

[uneven scroll]

How is F continuous if you jump from 0 to 1

[rancid umbra]

try working it out using the definitions

int(A) and ext(A) are both open subsets of int(A) U ext(A)

[uneven scroll]

lets take intA as an example

wait

like this

[rancid umbra]

what are you trying to show

[uneven scroll]

continuity for functions

for f:D -> N

[rancid umbra]

great. so fix your epsilon > 0. if your epsilon larger than 1, then any delta works since U_epsilon(x) = {0,1} for any x in {0,1}

so you only need to consider epsilon <= 1

[uneven scroll]

that might be a stupid question but why "U_epsilon(x) = {0,1} for any x in {0,1}"

[rancid umbra]

when epsilon > 1, that is true because for x in {0,1}, |x - 0| = |x| <= 1 < epsilon and |x - 1| <= 1 < epsilon

[uneven scroll]

oh

[rancid umbra]

not a huge deal

[uneven scroll]

well my x is not allowed to leave intA

[rancid umbra]

i specified that the x here was in {0,1}

[uneven scroll]

Im so sorry my brain is not made for topology, linear algebra is way easier :c

what im thinking is

if the green line is [0,1]

and the black line if F(x)

then i cant choose delta s.t. a is in Udelta

[rancid umbra]

that picture can't happen

the function F o f will end up being constant

[uneven scroll]

oh

i was thinking of F

not F o f

[rancid umbra]

then why add the green line?

[uneven scroll]

i was just visualizing [0,1]

[rancid umbra]

i dont follow your picture

[uneven scroll]

well im realizing that i misunderstood what F was

[rancid umbra]

F is just like an indicator function. it tells you if you are standing on int(A) (F has output 0) or ext(A) (F has output 1)

[uneven scroll]

yeah

[rancid umbra]

so for epsilon bigger than 0, we can look at U_epsilon(0) in {0,1} first

we need to find delta > 0 such that F(U_delta(x)) is contained in U_epsilon(0)

for a fixed x in int(A)

are you following so far?

[uneven scroll]

i think so yes

[rancid umbra]

how are we going to use the fact that int(A) is open to get our hands on a delta?

[uneven scroll]

f(U_delta(x)) meaning F o f (U_delta(x))

[rancid umbra]

just capital F

i made a typo

trying to show that F is continuous

[uneven scroll]

well in an open set you can always find a U

around some point t

[rancid umbra]

mhm

[uneven scroll]

im so sorry i feel very stupid atm

okay

[rancid umbra]

the forall t needs to come first

[rancid umbra]

yea

neat

[uneven scroll]

same goes for extA

[rancid umbra]

right. so can you see how F is continuous?

[uneven scroll]

tbh no. I dont understand why it should be continuous if you can jump from 0 to 1

somewhere inbetween

[rancid umbra]

int(A) and ext(A) are, separated. so its okay to jump between function values wherever the domain has gaps

its the same reason that g : (-infty, 0) U (0, infty) --> R defined as g(x) = 1 if x > 0 and g(x) = -1 if x < 0 is continuous

if that seems unfamiliar to you, i would suggest taking a detour to try and prove that g is continuous on (-infty,0) U (0,infty)

[uneven scroll]

I proved it in a different way.But thank you so much for painstakingly running me through your ideas. Im very sorry for being stoopid ><

[uneven scroll]

just as a rough scetch: I used the a you defined earlier and used the continuity of gamma to show that a cant lie in intA and extA

because those sets are open and gamma continuous gives us a U_delta region

[sonic crane]

Anybody here knows about topological data analysis?

[gritty widget]

https://media.discordapp.net/attachments/811392135885750322/1254709076101697548/Screenshot_20240621-120642.jpg?ex=667a7a63&is=667928e3&hm=68f0c958a0896190d2a77903a9ec629d27d019c8732c4616fcd29b5041ed0318&

I need some help on Ex 1.28 at the bottom

The S^1 case i think can be shown by considering a small rotation

I don't have any progress on the higher odd dimensions

The excercise wants me to use the 'Facts without proof part'

[inland laurel]

how ı find connected surfaces of a topological space ?

[opaque scroll]

I know some people that work with TDA if that counts

[sonic crane]

It looks cool!

[red yoke]

Consider R^2n = (R²)^n

[gritty widget]

Oh okay

So that works like i pair up coordinates?

[red yoke]

Ye

[gritty widget]

Wow okay very cool 👍

[red yoke]

(did you already solve it)

[gritty widget]

Yeah I got a sol in another server to find a transformation such that inner product is 0

[sonic crane]

Another server!?

What is it

[gritty widget]

Oh the homework help one?

[fierce lily]

why finite product can not assume a form of elements like product (x_ -1/2 epsilon, x_+1/2 epsilon)? I mean how does it matter to finite and infinite for the existence of this element.

[swift fjord]

Because there's infinitely many sets in that product that aren't the entire space

I'm not sure what you mean

What's the definition of the product topology

And how does it differ from the box topology

[fierce lily]

finite product of topological space?

[fierce lily]

and box is infinite product

[quartz horizon]

So concretely, given topological spaces X and Y, what is the product topology on X x Y?

[fierce lily]

it should have basis element U product V, U is open in X and V open in Y

[quartz horizon]

Mhm

[red yoke]

And what are the basis elements of the box / product topology of ∏i Xi

For infinite index set

[fierce lily]

I think I might say if basis element of a product topology can be product of U_ and we let index be finite? and in this case, why we can not prove that uniform topology is coarser than product topology by saying finite product of ( x_-1/2 epsilon, x_+1/2 epsilon) is basis of product topology

[red yoke]

For a finite product, all three topologies are the same

But for an infinite product they are different

[fierce lily]

ok I see, really appreciate that

[warm hedge]

is there a difference between the statements
"a function f : G x X -> X is continuous "
and
"a group action f from the topological group G to topological space X is continuous" ?

[unreal stratus]

Usually the latter is defined to mean the former

[warm hedge]

if an action is continuous that doesnt mean that every open set has an open image, right ?

[inland laurel]

if a continuous group action, let G be a topological group and X top space (g,x)--->g . x for G x X---->X if this is working yea is a continuonus and you right action is continuous that doesnt mean that every open set has an open image by definition meaning this is not important for continuous group action

[warm hedge]

i am trying to understand why an equivalent relation which is induced by a continuous action of a group G on a topological space X. gives its topological properties to its equivalent classes (orbits) . for eample if the e.r is G delta then its orbits are G delta on X.

i guess it has to do with the continuity, but i am not sure how

[quick bough]

does anyone have a pdf of the book of donald yau on colored operads?

[unreal gull]

i found that u can piece together two paths via a common point to find a path between any two points in $\cup A_\alpha$, but I have struggling to show that the resulting path is continuous

[gentle osprey]

Henry

[unreal gull]

in particular, the union of topogical spaces is not neccesarily a topology itself, so how can u even define continuous in this case?

[unreal stratus]

Wdym by necessarily a topology itself

The union of subspaces is still a subspace

[unreal gull]

but the A_\alpha are not subspaces of some larger space A

oH

WAIT

💀

ty

[inland laurel]

i think this is not working (ı talking on about image)

[unreal stratus]

This is the correct approach. It is continuous by what is often called the gluing lemma

But all you need fro this is that given a function [0,1]->X, if it is cntinuous on [0,1/2] and [1/2,1] then it is continuous

[inland laurel]

sorry that working

[unreal gull]

in the second image (the second part of the proof), isn't he assuming that $C$ is connected already when trying to prove that fact? In particular, he uses the fact that for a point $x_0$ of $C$ and each point x of $C$, $x_0 \sim x$

[gentle osprey]

Henry

[rancid umbra]

C is the connected component of x0. by definition, C is the collection of points x such that x ~ x0, i.e., x0 and x are contained in some connected subset. this connected subset is a subset of C

[unreal gull]

right

but then why do we need to show that C is connected

dont we alr know that?

[inland laurel]

What is finite-sheeted cover?

[sour seal]

It's a covering space in which each fiber has a finite cardinality

[unreal gull]

why is the topologists sine curve connected but not locally connected?

like how do u see those 2 facts

[inland laurel]

Then a finite covering space ( from compact space go to compact space) a finite sheeted cover?

[sour seal]

Yes because infinite discrete spaces cannot be compact.

[inland laurel]

Thank you Apoc

[sour seal]

Hatcher has many interesting exercises on this topic.

[inland laurel]

I think because this covering space a locally homeomorphism and every point in X is finite in another space. No idea on about it is enough for proof

And I saw on here

[limpid fern]

connected because its closure of the image of a connected space under a cont function

as for locally connected im less sure

i believe it's something to do with the origin

[sour seal]

For me, sometimes a space is connected only because you have few open sets, and then it may fail to be locally connected. Broom space is another nice example

[rancid umbra]

whenever you zoom in near zero small enough, you see a bunch of disconnected line segments, which is not something one would hope for if it were to be locally connected

for instance, if B is a nbhd basis of the origin, then the open box (-ep,ep) x (-1/2,1/2) intersected with the topologist's since curve contains a connected open subset B of the topologist's sine curve in B.

But B cannot be connected, since it is a disjoint union of line segments

[rancid umbra]

you do not know that connected components are connected subsets just from the definition of connected component in isolation since there are no assumptions imposing that a connected component actually be connected

it is a consequence of the definition and the theorem that you proved that connected components are connected

[unreal gull]

this is the def given in munkres:
\textbf{Definition - (Connected) Components}: Given $X$, define an equivalence relation on $X$ by setting $x \sim y$ if there is a connected subspace of $X$ containing both $x$ and $y$. The equivalence classes are called the \textit{components} (or \textit{connected components}) of $X$.

[gentle osprey]

Henry

[unreal gull]

so doesn't C being a connected component mean that it is connected?

oh i see it means that there is a connected subspace of C that is connected?

[rancid umbra]

yes

the latter

[unreal gull]

hmmm, but then in this next proof, at the very end, it uses the fact that C is a component => C is connected

[red yoke]

It does, but not by defn

[unreal gull]

ohhhh i see

sorry i didnt see that message

[plucky veldt]

in case of this definition, it means that there's a connected subspace of X containing any two points in C

now in the those theorems you showed, he proves that by consequence of the definition, each of those subspaces is contained in C and C is a connected union of those subspaces

so then he uses the proven fact that C is connected in the following theorems

[proven comet]

Is $B_1(0) \subset \mathbb{R}^2 \setminus {(x,0) | x \in \mathbb{R} \setminus{0}}$ simply connected?

[gentle osprey]

Gangster Spongebob

[red yoke]

Yes

[scarlet turtle]

you can deformation retract that space to the origin

[wise sun]

I know that some closed curves are not homotopic to a constant curve, because inside the closed curve there is missing an element. Are there any other reasons for why a closed curve is not homotopic to a constant curve?

[red yoke]

It could take too long to homotope to the constant curve

Like Hawaiian earring but each circle has its own cone

[red yoke]

(don't take this literally)

[obtuse meteor]

Maybe I am being stupid, but in case I'm not: For n >= 4 The mapping class group of S_{0,n} is generated by half-twists about simple closed curves which surround 2 of the marked points on one side and n-2 of the marked points on the other side.

Right?

[wise sun]

I see, thanks. That means, I need to use more abstract arguments in my assignment..
Also don't worry, I can distinguish intuition from rigour

[obtuse meteor]

If I'm doing my computations right (for another thing) this fact is

1. Deeply fortunate
2. Deeply Unfortunate
3. Trivializes part of a problem I was thinking about, but in a weird way
4. Seems to furnish a contradiction in math

so...joy???

me when I am confused:

[obtuse meteor]

Notably I mean if I have any fixed n choose 2 of these (one for each pair of points), then that's enough

[chilly mica]

this reminds me, the wikiped pag for the hawaiian earring mentions $( \prod_{i = 0}^{\infty} \mathbb{Z} ) / (\bigoplus_{i = 0}^{\infty} \mathbb{Z} )$ as a summand of its first homology group, and that group is like, my favorite group at the moment, its fun to think about

[gentle osprey]

terdragontra

[gentle girder]

truly an emotional rollercoaster

[sonic crane]

Can someone explain how the algebra encodes topological information ?

Context: i know some algebra up to a bit of modules, and i learned some intro topology just the basics (up to like defn of hausdorff space and homeomorphism stuff in munkres)

[cedar pebble]

do you mean like algebraic invariants of topological spaces?

like fundamental groups/homology/cohomology

[sonic crane]

I guess so. Is that what it is?

I guess im just not totally sure what the algebra part in algebraic topology is trying to do

[cedar pebble]

right so here's the example you usually start with

[sonic crane]

Oh ok well its capturing some aspect of the topology using algebraic structures and going back to topology to interpret it there

Right

[obtuse meteor]

It continues. I thought I resolved this this afternoon. But now I don’t believe my resolution of the problem

Deeply unfortunate

[cedar pebble]

for any (reasonable) space X with a basepoint x you can consider the fundamental group \pi_1(X,x)

this is the group of homotopy classes of loops in X based at x

two loops are homotopy equivalent roughly speaking if one can be continuously deformed into the other

\pi_1 is a homotopy invariant of spaces: if two spaces are homotopy equivalent then their fundamental groups are isomorphic as groups

[sonic crane]

What are loops in X based at x

A sequence?

[cedar pebble]

a loop in a space X is a continuous map from the circle S^1 to X

if you like it's a continuous map f:[0,1]->X from the unit interval to X such that f(0)=f(1)

[sonic crane]

Ohhh

Bruh

[cedar pebble]

the basepoint of the loop is such that f(0)=f(1)=x

[sonic crane]

This is crazy 😂

[cedar pebble]

one way you can use this is like

if you have two spaces X and Y that have genuinely different fundamental groups then they cannot possibly be homotopy equivalent, in particular they cannot possibly be homeomorphic to each other

more generally you can consider the group of homotopy classes of maps S^n->X and this defines the n-th homotopy group \pi_n(X)

in general for n>1 these groups are insanely hard to compute

there are other much easier to compute invariants like homology and cohomology, and in a similar way these assign Abelian groups/modules to spaces in a way that is homotopy invariant

they are useful because they let you "linearize" problems in topology

like in many cases doing (linear) algebra is much easier than doing topology directly

[sonic crane]

That is cool

[unreal stratus]

well any lol

[lusty trench]

Linear algebra over Z hits a sweet spot. With Z coefficients, you can detect the orientability of a compact manifold (unlike Z/2Z coefficients), but you can also assign a nontrivial top class to a compact non-orientable manifold (e.g., RP^n for even n). But Z-modules don't have excessively long free resolutions, and you can use Künneth's theorem without lots of annoying higher-order Tor terms or spectral sequences...

[sonic crane]

Wow i dont have enough math maturity yet to understand this but come back to me in maybe a year! Lol

Tbh i havent learned anything about manifolds for some reason

I would like to so im not ignorant

[umbral panther]

Work over all fields at once. Easier than working over Z. Less information, but don’t worry any that until you have an example where it isn’t sufficient

[cedar pebble]

just because you can doesn't mean you should

[pearl trail]

I’m confused. How is [a, b) open due to it being a basis element that generates R_l. [a. b) surely doesn’t look open.

[lusty trench]

[a,b) isn't open in the Euclidean topology of R, but it's open in the new topology they're defining.

In fact, one important point is that you can endow the same set with different topologies.

[pearl trail]

I need to show R_l and R_k is not comparable. If I pick [0, 1) \in R_l, this is not in R_k, right (need to justify because R_k elements are of the form (a, b) \ 1/n \in Z+)? Then I can consider (-1/10, 1/10) - K.

Should I add that we cannot use the basis B'' to generate [0, 1)?

[lusty trench]

To show that R_l and R_K's topologies are incomparable, you need to exhibit two things:

• An open of R_l that isn't open in R_K.
• An open of R_K that isn't open in R_l.

[pearl trail]

thinking how to form the argument rigoriously

[lusty trench]

By definition, [3,4) is open in R_l. However, it's not open in R_K. To see why, notice that, the topologies of R and R_K are indistinguishable on (2,5). So [3,4) is open in R_K if and only if it's open in R, which isn't the case. That solves the first part of the problem.

[pearl trail]

How are these indistinguishable on (2, 5)?

the topologies of R and R_K are indistinguishable on (2,5)

Because 2, 5 are not in 1/n, where n \in Z+?

[lusty trench]

No, that's not enough.

It's because the whole interval (2,5) doesn't contain any point of K.

So any open of R_K inside (2,5) must be a union of whole intervals (a,b), i.e., it must be open in R as well.

[pearl trail]

hmmm

[lusty trench]

By definition, $U = (-1,1) - K$ is open in $\mathbb R_K$. Now suppose $U$ were open in $\mathbb R_\ell$. Since $0 \in U$, there must be a basic open $[a,b)$ such that $0 \in [a,b) \subset U$. In particular, $a \le 0$ and $b > 0$. But, for large enough $n \in \mathbb Z^+$, we have $a \le 1/n < b$, meaning that $1/n \in [a,b) \subset U$, contradicting the fact that $U \cap K = \varnothing$. Hence $U$ isn't open in $\mathbb R_\ell$.

[gentle osprey]

Eduardo León

[lusty trench]

And that solves the second part of the problem.

Actually, without even mentioning U, you can simply notice that any neighborhood of 0 in R_l contains points of K. (In fact, infinitely many points of K.) But, by definition, there are neighborhoods of 0 in R_K that don't contain any point of K.

[pearl trail]

I follow this more. For part 2

[pearl trail]

Can’t we use a similar argument for part 1?

[lusty trench]

I used the same kind of argument for both parts.

[pearl trail]

I don’t get this last sentence

I understand now!

[lusty trench]

Ah, good.

[pearl trail]

Brilliant arguments

[pearl trail]

So we could use many different examples right, any overlapping with 0 example for part 2 (-3, 3) for example, or any inclusive set for part 1, such as [2, 3)

[lapis monolith]

This makes a lot of sense, thanks @lusty trench!

[lusty trench]

No problem. You're welcome!

[lusty trench]

Yes for part 2. If by “inclusive set”, you mean “half-open interval not containing 0”, then also yes for part 1.

[left eagle]

How could one prove that [0, 1) is not an open set in the K-topology over R (or is it an open set)?

[pearl trail]

The argument is above

[left eagle]

nvm im dumb

[zenith bay]

i readd #❓how-to-get-help and im ready please help

[lusty trench]

It's quite simply, actually. The Riemann sphere $\mathbb CP^1 \cong S^2$ is the quotient of the unit sphere $S^3 \subset \mathbb C^2$ by the action of the unit circle $S^1 \subset \mathbb C$ by simultaneous rotations: $t \cdot (z_1, z_2) = (tz_1, tz_2)$. The Hopf fibration is the projection $S^3 \to S^2$.

[gentle osprey]

Eduardo León

[lusty trench]

And how do you generate the images? Consider $\mathbb R^3$ as an open subset of $\mathbb S^3$ using the stereographic projection, and then restrict the $S^1$-orbits in $S^3$ to $\mathbb R^3$.

[gentle osprey]

Eduardo León

[zenith bay]

S1-orbits?

i havent read any algtop in like 6-7 years i struggle big time with this

[lusty trench]

If a group $G$ acts on a space $X$, then the $G$-orbit of a point $p \in X$ is the set of points of the form $g \cdot p$ for $g \in G$. In our case, $G = S^1$ and $X = S^3$.

[gentle osprey]

Eduardo León

[zenith bay]

sure

but I struggle to translate this into an image that seems to consist of a bunch of rotated tori

[lusty trench]

It's not a bunch of rotated tori. It's a fibration of S^3 by circles.

[zenith bay]

circles, yes

[red yoke]

Hmm what happens if you take cross sections of the sphere

Do you get a bunch of tori in S³
Plus two circles at the end
I think this gives you the solid torus + solid torus = S³ construction

[zenith bay]

i am aware that "the hopf fibration" is not just this single image and that this is a projection of some subset

but i wish to compute these circles

rather, find some parameterisation of each of these circles

[left eagle]

Would this be a valid proof for R_k not being finer than R_l... or would i need to elaborate more??

To me it seems a bit wishy washy and not very concrete but maybe im wrong

[red yoke]

Ig this gives the parametrization

[zenith bay]

for example, this image appears to simply be a set of circles that have been angled and rotated around some common central axis (like z=0 or whatever) - is this correct?

[red yoke]

Yea

[lusty trench]

That's telling you that the central axis in R^3 corresponds to the circle in S^3 that passes through the North pole - the point not covered by the stereographic parametrization.

[zenith bay]

interesting

and each of the circles in R^3 corresponds to a point on S^3?

these are fibres, right?

[lusty trench]

Each circle in S^3 corresponds to a point in S^2. One of these circles is the vertical axis in R^3, and it corresponds to the north pole in S^2. The remaining circles are still circles in R^3 and they correspond to the other points in S^2.

[lusty trench]

Oh, right, missed that. I'd have to check, but it's plausible.

[red yoke]

In particular the R³ visualization is given by (x, y, z, w) → (x/w, y/w, z/w), so rotating about the z/w-axis corresponds to complex maps (s, t) → (zs, t) in C² (I think)

[zenith bay]

oooooooooooooooooooooh

right okay

[zenith bay]

so i suppose this image is some set of equidistant equatorial points on S^2

and you're saying the "axis" they are rotating around is actually just some degenerate circle of infinite radius

yeah i can grind out the algebraic part from this

nice, thanks

[red yoke]

More like a circle perhaps

[zenith bay]

yeah like a circle of points around the equator

just a guess

seems probably right

[red yoke]

Not necessarily equatorial

[zenith bay]

the "angle" of each circle is 45 degrees and the chirality probably comes from the "handedness" of the projection

[zenith bay]

sure

[red yoke]

Are you making a new mathcord banner logo

[lusty trench]

What's a “mathcord banner”?

[zenith bay]

bigger plans than these my friend, bigger plans than these...

[lusty trench]

Vector bundles are ubiquitous in geometry and topology, and characteristic classes are a powerful tool to measure how far a vector bundle is from being trivial. Do we have a similar theory of “algebra bundles”, where fibers are (not necessarily unital or associative) algebras over a field? For example, given a “Lie algebra bundle” (is this a thing?), I'd like to know if there are any obstructions to reconstructing a principal bundle (or maybe some other more general kind of fibration) from it.

[left eagle]

Im stuck with (b) it seems impossibly long and tedious... is there a simple way to do it that im missing??

the smallest topolog on X containing T_a is the topology generated by sub-basis S = Union of T_a. from there i have to prove any topology containing T_a contains the T_S

but that just seems so lonnnggg

[alpine nest]

Indeed, that's not the quickest route.

Hint: use (a)

[left eagle]

intersection T_a is a topology on X (a) and its the largest topology contained in all of T_a but i dont see how that helps with the smallest topology containing T_a?

[alpine nest]

Well, smallest means "it's a subset of every other topology containing all of T_a"

[left eagle]

ye

[alpine nest]

So every set that's in that smallest topology will also be in all topologies containing T_a

Which is where the notion of intersection comes in

[left eagle]

So u can construct the smallest topology from the topologies containing T_a? or smth else?

But from part a the intersection of topologies isnt necessarily a topology

[alpine nest]

Yes, the starting point is the collection of all topolgoies that contain all of T_a

[alpine nest]

It is, and you've already used that property. It's the union deosn't have to be

[left eagle]

OHHHH oops i got it backwards

[left eagle]

So then the smallest would be the intersection of all those topologies

[alpine nest]

Depends what you mean by "those"

[left eagle]

all topologies that contain all of T_a

[alpine nest]

Yep

[left eagle]

but then how would i prove that its "unique"?

[alpine nest]

Probably worth noting that this family is nonempty, because the discrete topology contains every other one.

[alpine nest]

If T_1 and T_2 are both the "smallest" topology, then each of them is a subset of the other, so they are identical

Incidentally, this is an argument worth putting a pin in, because something similar shows up in other contexts (such as generating vector subspaces or generating sigma-algebras in measure theory/probability)

[left eagle]

oh because definition of smallest is that every topology satisfying blah is a subset of it

[alpine nest]

Yep

[alpine nest]

And the approach every time is "take the intersection of all the subspaces/sigma-algebras/etc that contain your original collection"

[left eagle]

ohh yeah i recall my professor saying something about intersections being nice and unions not so much

How would i construct the largest topology contained in all of T_a then? with intersections too?

[lusty trench]

(a) literally gives you the construction.

[alpine nest]

I thought that was the thing you had already done

[alpine nest]

This is correct

So one is the intersection of all your starting topologies, and the other is the intersection of all the topologies that contain all of your starting topologies

[left eagle]

wait im confused

the intersection of all topologies that contain starting topologies is the smallest topology containing all of the starting topologies

but isnt the intersection of all starting topologies smaller than each topology? How is that the largest topology contained in all those topologies

[lusty trench]

Largest topology contained in every T_a = Intersection of the T_a's.
Smallest topology containing every T_a = Intersection of the S_b's, where {S_b} is the family of topologies containing every T_a.

[alpine nest]

Well yeah, if it's contained in each of them then it has to be smaller than each of them. But out of all those small topologies, one is the largest.

[left eagle]

oh i think i confused contained in all of T_a for union of T_a instead of the topology is contained in each T_a

(^ is that what the problem is trying to say/ask)

[alpine nest]

Yes, you're looking for the largest topology T such that T is contained in T_a for every a

[left eagle]

kk tysm u saved like an hour of wasted work lol

[alpine nest]

And you'd correctly identified that one as the intersection of all the T_a's

[left eagle]

*by accident i think but hey ill take it at this point

[pearl trail]

Isn't this basically definition of intersection?

Largest topology contained in every T_a = Intersection of the T_a's.

[left eagle]

yeah but unless u prove it it aint true -probably

[alpine nest]

You do need the fact that the intersection of topologies is also a topology, but the rest is indeed mostly just definition of intersection.

[silver ridge]

Does bredon mean "A space D equipped with the discrete topoloy" by "A discrete space D"?

[red yoke]

Yes

[silver ridge]

thanks

[pearl trail]

Is this correct?

[red yoke]

Yea, but {a, b, c} is redundant

[pearl trail]

why?

[red yoke]

That's X

[pearl trail]

oh, thanks!

[proven hollow]

is is true that $$lim(A) \cap A^c = \partial A \cap A^c$$?

[gentle osprey]

asfda

[opaque zodiac]

What it A^c? Complement?

[proven hollow]

yes

[alpine nest]

What about the lim?

[hearty condor]

Okay so I have encountered an interesting idea.

Take this. Or the equivalent interior operator.

These always give you a topology on X.

Now throw K3 into the bin. Instead ask of your operator to be eventually idempotent. I.e some n in N s.t c^n+1=c^n

This

Is a discrete dynamical system

The spaces equipped with this operator are eventually topological spaces

Or some sort of almost topological

[proven hollow]

limit point of A (x is a limit point iff every open neighborhood U_x contains a point in A that is not x)

[red yoke]

It is if eventually means c^∞

[red yoke]

Actually even without this c^∞ works I think

[hearty condor]

( \partial A \cap A^c= (\bar{A}-A°)\cap A^c= \bar{A}\cap A^c - A°=lim(A)\cap A^c - A°=lim(A)\cap A^c )

Where A° denotes opening

[hearty condor]

No

I have not checked those

I don't even know how to do some other stuff

But cool idea

[gentle osprey]

God Of Bushes

[hearty condor]

It precisely means after some number of applications of closure or interior operator the space behaves like a topological space

[red yoke]

By c^∞(X) I mean the union of all c^n(X)

If you choose the same n for all subsets then it might not be possible for c^n to yield a topology for any n

But if they can vary then it's doable (equal to c^∞)

[silver ridge]

Why do we care about quasi components?

[opaque zodiac]

they're a bit better behaved in some senses

[silver ridge]

They're not necessarily connected so how is that possible

[unreal stratus]

||Do we care about quasicomponents||

[opaque zodiac]

ehhh well you know

I'm thinking from the point of view of like set-theoretic topology and stuff, but honestly if you're in a context where components and quasi-components are the same thing, you've made some kind of mistake in your life in the first place

[unreal stratus]

They only seem important if you care about more advanced pointset for its own sake - they've never come up for me lol (as a "user" of pointset topology)

[silver ridge]

Right

[opaque zodiac]

There's a connected metric space with the property that none of its non-trivial separable subspaces are connected

And that's a metric space

not some arbitrary weird topological space

[silver ridge]

Mhm I see

[unreal stratus]

metric spaces can be weird tbf lol

[silver ridge]

The general impression is that we don't care about them so I won't either

[unreal stratus]

c.f. topologist's sine curve

[opaque zodiac]

yeah but I work with metric spaces all the time, so I have to be okay with them

[unreal stratus]

yeah it is all relative

[opaque zodiac]

I get really annoyed when dipshits on reddit are like 'uhh, Cantor space isn't really a space cuz not CW-complex'

[unreal stratus]

Lol

[silver ridge]

First mistake is using reddit

[tender halo]

cantor space is the only real space

[unreal stratus]

I mean I am partially of that view and partially against it lol

"Not being a cw complex" is kinda silly reason not to like a space

[tender halo]

did you know, every compact space is an image of a closed subset of the cantor space

[lusty trench]

I was going to say that. Isn't Cantor space, like, one of the most fundamental metric spaces?

[unreal stratus]

But i guess it is something that is pathological/weird from pov of alg top (for some things)

At least it has a renaissance from condensed maths lol

[lusty trench]

For example, you can construct the compact interval [0,1] as a quotient of the Cantor space, and that's related to the binary (or any other base) expansion of a real number.

[opaque zodiac]

need to be metrizable

every compact Hausdorff space is a quotient of 2^kappa for some kappa

[tender halo]

well i just said cantor space instead of cantor cube of weight kappa

cuz i didnt want to explain what cantor cube is

and you ruined it

[opaque zodiac]

I'm sorry

[tender halo]

[opaque zodiac]

I'm just still mad about the fact that Lubos Motl answered a question about Hilbert spaces on the physics stack exchange with 'all Hilbert spaces are isomorphic' with no size qualification and despite comments pointing out that the answer is wrong, it's still up

[lusty trench]

I miss his blog. It was a fun read.

[unreal stratus]

Entire field of study in shambles

[opaque zodiac]

Perhaps but he was a sexist asshole

[lusty trench]

Maybe physicists have no reason to care about non-separable Hilbert spaces?

[opaque zodiac]

So they shouldn't although they sort of in principle show up in quantum field theories

but those have like superselection rules that (iirc) should restrict the 'real' part of it to something separable

[tender halo]

shouldve went further

all vector spaces are isomorphic

[opaque zodiac]

I mean Cantor space never stopped being relevant in logic

[lusty trench]

How relevant is logic for “normal” mathematics? By the way, point-set topology looks like a part of logic to me.

[unreal stratus]

I mean it is certainly important to different areas lol

I mean that like it is now relevant to more people perhaps

[silver ridge]

This is morally true

[amber zephyr]

pointset in the modern day is descriptive set theory which people would say is logic

[amber zephyr]

it shows up every now and then but if you want to avoid it you can

[unreal stratus]

But not really too seriously since cantor space already turned up via p adics lol

[tender halo]

extremely powerful in entirely useless ways

[amber zephyr]

like model theory heavily intersects my own work

[silver ridge]

Are you a model theorist perchance

[amber zephyr]

no

[opaque zodiac]

Well okay so the two parts of (classical mathematical) logic that end up touching grass most often are probably model theory and descriptive set theory. Results from those fields are still sometimes used to solve things in other fields. And both of them use Cantor space a lot

[lusty trench]

I mean, it's not like Z-linear algebra, which is completely unavoidable if you do geometry or topology, in the honest geometric sense of algebraic or differential topology, rather than point-set topology.

[opaque zodiac]

oh nice, what's the connection?

[amber zephyr]

I work in operator algebras and quantum information intersection

[opaque zodiac]

aaah

I do continuous logic and I used to do physics

so in principle I should be equipped for that kind of thing but I'm not

[amber zephyr]

and some very hard chains of equivalences (MIP*=RE if u know these words) have been trivialized by some cont logic

[opaque zodiac]

yeah I know about this

[tender halo]

they build models of platonic solids out of papier mache, the art of building them is called model theory

[opaque zodiac]

I've talked to Isaac Goldbring about it quite a bit

[amber zephyr]

oh nice u know isaac?

[opaque zodiac]

yeah

[amber zephyr]

wait r u ultra lol

[opaque zodiac]

No

I know Ultra though

[lusty trench]

Oh, okay.

[amber zephyr]

there is a good chance i know u also

[opaque zodiac]

And I also know Bradd Hart

hmmm

are you Isaac's student?

[amber zephyr]

no im in waterloo

[opaque zodiac]

hmm

[amber zephyr]

but i feel like i have met all of isaacs students?

[opaque zodiac]

I wasn't Isaac's student

[amber zephyr]

ah ok

but yeah me and ultra will be cooking using some model theory to find non-hyperlinear groups soon^TM

[opaque zodiac]

nice

you know you can actually do ™️ in discord with ™️

[amber zephyr]

lol

[opaque zodiac]

what the fuck it also changed it in the quote block

:tm:

[lusty trench]

$\texttrademark$

[gentle osprey]

Eduardo León

[lusty trench]

What?

[opaque zodiac]

probably a font thing

[tender halo]

noo the basterds trademarked punctured neighborhood

[unreal stratus]

Lol

[lusty trench]

Would it be fair to say logic is the part of mathematics that physicists would be indifferent to if mathematicians suddenly decided to change the rules of how it works? (By contrast, suppose we changed the foundations of analysis or differential geometry. Physicists would be up in arms!)

[tender halo]

it is unclear to me what that entails

like if we chose a different foundation from ZFC? i dont think most people would care

[hearty condor]

Oh smort.

[gritty widget]

I’m not so sure they’d be up in arms. I’ve talked to physicists/chemists/etc who can’t say what a hilbert space or group representation is

[opaque zodiac]

Look I used to be a physicist and they don't give a shit about the way calculus is currently formalized

[lusty trench]

I think it's fair to say that the real numbers are intended to behave the way certain physical objects behave. Like, if your definition of the real numbers didn't uphold the intermediate value theorem, then you'd reject it, because the physical counterpart does uphold the intermediate value theorem.

[opaque zodiac]

sure

although I guess like

honestly

I kind of can't see physicists jiving with intuitionistic logic

[quartz horizon]

You used to be a physicist?

[opaque zodiac]

yes

[quartz horizon]

huh..

[opaque zodiac]

I was in a physics PhD program for a bit more than 4 years before switching to math

[quartz horizon]

wow

I’m just about to enter a physics phd program…

i mean

i did a math undergrad

but I’m not sure I’d want to or could do math

[opaque zodiac]

what kind of physics do you want to do?

do you know?

[quartz horizon]

condensed matter physics

[gritty widget]

I mean, you can assign the name “real numbers” to whatever. At some point, you’ll want to talk about limits in an ordered archimedean field, and then you’ll get the reals whether you like it or not

[quartz horizon]

cause it’s just so bizzare and weird and cool

[opaque zodiac]

some people get kind of mad about the whole 'reals as a set of points' vs. 'reals as an abstract locale' thing, but that would also be almost entirely irrelevant to physics

[opaque zodiac]

ah nice, yeah there's a lot of neat stuff there

[quartz horizon]

mhm

[opaque zodiac]

I did string theory, incidentally

[quartz horizon]

apparently some category theory gets used

[quartz horizon]

i see

i think that’s mostly something I’m trying to avoid as long as i can

[opaque zodiac]

oh I wouldn't recommend engaging with string theory

I think that a lot of the criticisms people use against it aren't really founded, but it is also somewhat dying as a field

there's a lot more jobs in condensed matter

[quartz horizon]

yeah it’s like

hmm how to say

i feel like I’ve gotten unreasonably lucky with my interests…

this happened in my undergrad too

i really liked all the applied math courses

but i also happened to score very well on the exams for them

[opaque zodiac]

are you more interested in experiemental physics or theoretical physics?

[quartz horizon]

theoretical, i’d say

but I’m not opposed to experiment

[opaque zodiac]

okay

[quartz horizon]

I’m just not good with my hands

[opaque zodiac]

Are you interested in actually trying to go into academia?

[quartz horizon]

i don’t know…

[opaque zodiac]

I don't really know what the academic job market is like in condensed matter physics, but I think it's probably better than high energy stuff

you don't need nearly as much money to do condensed matter experiments

[quartz horizon]

mhm

and a lot less money than string theory experiments…

[opaque zodiac]

yes

ugh the urge to well actually is overwhelming

[quartz horizon]

it’s fine you can

im just some random masters student after all

well, former masters student

graduated today

[opaque zodiac]

You didn't really say this, but one of the things I find annoying is that people peg the experimental difficulty with string theory on string theory itself when it's really a fundamental problem with quantum gravity research in general. All of those problems apply just as much to loop quantum gravity, for instance

but string theory gets all of the shit for it

[quartz horizon]

mhm, i see

i remember my standard model lecturer mentioning this

that the incompatibility between GR and the standard model is kinda overblown

you can couple gravity to the standard model just fine, and it works

until you get to very extreme situations

but these also happen to be situations where we don’t exactly have a lot of experimental data anyway

[opaque zodiac]

And some of the criticisms are just actually wrong. People will sometimes say string theory is just like 'curve fitting' where it's so flexible you can make it do anything, but this is the exact opposite of its problem. It's far too overconstrained, and they had a lot of difficulty finding vacua that even broadly resemble reality (they found a lot of anti-de Sitter space, but reality looks more like de Sitter space)

[quartz horizon]

i guess doing a phd in it really
makes you know a lot about it..

[opaque zodiac]

yeah

[quartz horizon]

i hope i won’t suffer from a knowledge problem for my phd

[opaque zodiac]

wdym by a knowledge problem?

[quartz horizon]

like…

i’ve only really gotten into the maths discord in the last couple months

and i’ve met so many people who’ve done a ton of self-study

it feels like their gaps are bigger than my entire knowledgebase

it’s just not the kind of thing i’ve done at all

[opaque zodiac]

well almost all of the physics stuff I'm saying now is stuff I picked up in my PhD

[quartz horizon]

mhm, right

I do know it’s something I have to get used to

my mode of learning has been

idk much, but what i do know i know very well

as opposed to having a lot of breadth

but I’m not sure i can really keep this up in a phd

[opaque zodiac]

it is something to be concerned about, a phd can be pretty intensive

[quartz horizon]

yep

though i did ask a phd student here about their experience and

they said it was definitely easier than the masters here

so that’s kinda reassuring

[opaque zodiac]

The difficulty of graduate classes is all over the place. Some intro classes can be really tough but then in a higher level class, especially one where the prof doesn't want to grade, there can often be very little homework

[tame arrow]

In principle, string theory should be able to predict new particles or other experimentally accessible physics. Its failure to do so may mean string theory is wrong or it may mean that string theorists made a serious error in their approach. (Maybe Calabi-Yau compactificaiton is wrong etc.)

[opaque zodiac]

Why should it be able to do that in principle?

[quartz horizon]

im not doing my phd in the us so i wonder how much this changes things

[opaque zodiac]

Like why does the underlying physics of reality need to occur at energy scales that can happen in particle accelerators that fit on the Earth? I don't see any reason a priori why that needs to be true

[tame arrow]

Any compactification will predict particles. It should be expected to reproduce the standard model and something new we can test.

[opaque zodiac]

I don't think string theory is right but I also don't think your argument is valid

[tame arrow]

It's very heavy higher category theory mixed with tools used in low-dimensional topology.

[opaque zodiac]

The Planck scale is just really big

Using a standard particle accelerator you'd need one on the scale of the solar system to probe it. This isn't a unique failure of string theory, it's a fundamental barrier to experimentally testing quantum gravitational stuff

So no

Its failure to do so may mean string theory is wrong or it may mean that string theorists made a serious error in their approach. 

this is just an incorrect statement

[quartz horizon]

uhhhh

that sounds really complicated…

gosh

[opaque zodiac]

Like does this mean that 'the theory of gravitons' is just necessarily incorrect since we can't reasonably build a particle accelerator that can produce them?

[tame arrow]

Sorry, I replied to the wrong user.

[tame arrow]

Originally, the hope was that there were only a small number of compactifications. The one corresponding to our universe would reproduce the standard model exactly etc. This failed becaue it turned out there were a huge number of potential compactifications (10^500 was thrown around) and people turned to speculating about statistics of the swampland (the weak gravity conjecture (which antedates that is one example), but there are several others. https://en.wikipedia.org/wiki/Swampland_(physics)

[tame arrow]

I feel like the higher category theory isn't that complicated and the topology isn't that hard either. What confused me is that it wasn't obvious how to plug it into other physics. Namely, if I wanted to add fractional spins to a supersymmetric string Lagrangian, how would I do it and what would my options be.

[opaque zodiac]

Okay but I still don't understand your point here. I wasn't saying that string theory didn't end up failing to give a clear realisitic model (I pointed that out explicitly)

[quartz horizon]

i mean you say the words “higher category theory” and it’s already waaaaay over my head

[opaque zodiac]

I'm saying that statements like In principle, string theory should be able to predict new particles or other experimentally accessible physics. aren't really substantiated by anything other than like naive experimentalism. We're pretty confident on theoretical grounds that gravitons should exist and we already know that they're not going to be realistically visible to things like particle accelerators, so what 'principle' dictates this?

[umbral panther]

Higher category theory means chain complexes

[opaque zodiac]

It does?

[tame arrow]

I think it should be some sort of "don't give up on obtaining experimental predictions; if some ideas haven't panned out consider new approaches within string theory" principle. It's quite possible that the next big discoveries are far beyond the reach of our technology, but if it looks that way we should probably act like we've done something wrong.

[opaque zodiac]

Why does that mean we've made a mistake?

I'm not saying people shouldn't look for experimentally viable theories, but like I said we more or less know that we don't know how to directly observe gravitons. Is that a mistake?

[tame arrow]

No, I don't think it is. But string theory predicts a lot of particles. We should be able to observe some.

[opaque zodiac]

Why?

[tame arrow]

Can all models of higher categories be expressed as chain complexes even the strict ones?

[umbral panther]

No, this was hyperbole. The main point of (oo,1)-category is that it is the right structure on the category of chain complexes. But there are triangulated categories that are not chain complexes, such as spectra. Categories enriched in spectra are the (oo,1) version of additive categories, but there is also the nonlinear case

The first place this arises in string theory is homological mirror symmetry, which is chain complexes

[tame arrow]

You can just think of a category and imagine the Hom-sets are categories themselves (and there are two composition rules) and build upward from there with the higher Hom-sets being their own categories.

[tame arrow]

For (\infnty,1)-categories the idea is more like spaces with the higher levels being homotopy.

[quartz horizon]

Yeah the issue is like

I am not an algebraist

[quartz horizon]

You say it like it’s so easy..

[umbral panther]

It’s not a story a physicist would tell you

[quartz horizon]

I am a physicist, so..

[umbral panther]

Chain complexes are the main topic of this channel. Topology, not just algebra. It’s grad math, but low level and widely studied. Higher categories are cutting edge research, but it’s mainly guardrails for using the chain complexes correctly

[quartz horizon]

huh…

that’s really weird to me

why would chain complexes matter so much

[umbral panther]

(Chain complexes are pure algebra, but they are often applied to topology)

[quartz horizon]

also like

i have spent a while in this channel and not needed chain complexes

hmm what even is a chain complex

lemme look it up

right…

i mean i can parse the def

but not sure i really understand it

welcome to higher-level math i suppose..

[umbral panther]

The first example of a chain complex is differential forms and the exterior derivative

[naive trench]

My fav ones

[quartz horizon]

Ah yeah, i guess I’d need to learn diffgeo for that

I’ve only ever really worked in coordinates

And not like abstract differential forms

[tame arrow]

Googling, there doesn't seem to be an algorithm to determine whether a knot is slice or not. Is there some barrier to developing an algorithm? Why can't you just use PL knots and try out all possible PL disks? Is it impossible to lower bound the size of the simplices needed?

[fading vale]

How would you get a bound on the number of necessary simplices?

[umbral panther]

Compared the theorem that unknotting is decidable (in fact in NP)

[tame arrow]

I would try to argue that the bounding disk can't be much more complicated than the knot. The slice-ribbon conjecture is a big deal. Someone must have thought along those lines before.

[obtuse meteor]

Benson once said he didn’t understand why ppl did knot theory or HF since the classification problem was solved

Sometimes he would be a bit dumb

[fading vale]

if you measure complexity in terms of number of simplices clearly you can always take a barycentric subdivision of the disk and get something messy so i think you want some argument to establish bounds on the minimal number of such simplices, or some other notion of complexity, right?

But I dont know how you would do this in R^4

in S^3 Haken's algorithm uses the fact that normal surfaces in triangulated 3 manifolds come from points in an integer lattice

so i guess you would want a similar theory/set of results for an arbitrary triangulation of S^4

maybe these exist but i dont know of them if they do

[obtuse meteor]

A good reason to care about chain complexes imo:

Rank nullity is one of the best theorems about vector spaces. At a (very) abstract level. It tells you if you have maps f : V -> W and g : W -> Z with the kernel of g the image of f (imagine a projection from W to V^perp) then W is V + Z

This fails for more abstract things (like abelian groups) but the structure is still useful. The kernel = image means exact. The fact W =V + Z is to say this splits

Chain complexes encode sequences that have similar properties. And attached to chain complexes are algebraic invariants which detect things like

1. how far off is it from exactness (this is homology/cohomology)
2. how badly does splitting fail (this is…also homology/cohomology LOL, but of a different type, e.g group cohomology)

Another way to view chain complexes is just as a big bundle of algebraic data that’s convenient to give a name to

Why it shows up so much in topology is sort of a minor miracle

[unreal stratus]

A (connective) chain complex is an abelian space

[fading vale]

i feel like it makes sense because basically the only method we have of understanding topological spaces is stratifying them into data of various levels and then understanding how the levels combinatorially relate to one another

Well i should not say only

But one of

[obtuse meteor]

Basically the only

Everything has to boil down to combinatorics eventually

Humans only deal with finite data after all :)

[unreal stratus]

But on a more serious note like you can model spaces with simplicial sets and replacing set by an algebraic category gives you (up to equivalence) (connective) chain complexes

[obtuse meteor]

The fact that out of these you can build actual chain complexes satisfying d^2 = 0 is the “minor miracle.” It’s not clear from the defn of chain complex that these would arise from combinatorial data in topology

[fading vale]

yeah

[unreal stratus]

Well that is less natural than the simplicial pov

imo

[fading vale]

i suppose one way to think about it is that d^2 = 0 is basically an algebraic expression of some kind of compatibility condition

e.g. of cocycles or whatever

[obtuse meteor]

And often finding a good chain complex to consider is how significant progress gets made in developing new homology/cohomology theories or doing computations

[unreal stratus]

Indeed basically all the "ad hoc" signs and d^2 = 0 just seem to come from more natural simplicial objects lol

[obtuse meteor]

Seething at the category theorists being right

[fading vale]

for example in floer homology it and its proof demonstrates the compatibility of a hammer with my skull

[hearty condor]

about it.

[lusty trench]

How is the simplicial point of view "more natural"?

[coral pawn]

By the simplicial pov, do you just mean singular homology/cohomology defined on simplicial sets?

The one where you take the free ab group on n simplices and define maps using alternating sums

[lusty trench]

Well, sure, singular homology factors through a functor from spaces to simplicial sets.

[coral pawn]

I'm just a bit confused what exactly potato meant by simplicial pov

[coral pawn]

Because if it's this, I didn't see why the alternating signs follow from the simplicial pov

[lusty trench]

I can't tell for sure, but it's probably something along the lines of "we care about simplicial gadgets for their own sake, rather than as tools to solve other problems we actually care about".

[coral pawn]

Somehow I always thought of the alternating signs as being the condition we need to get d^2 = 0, which is needed to compute homology/cohomology

[lusty trench]

Indeed.

[coral pawn]

Sure but at some point you'll have to start alternating signs and I can't think of a justification other than d^2 = 0

[red yoke]

Surely alternating follows from topological intuition for boundary

[coral pawn]

Yeah

Topological/combinatorial

[red yoke]

Or like Stokes' theorem

[coral pawn]

But I wouldn't say one is more natural than the other

[lusty trench]

Maybe if you're someone whose idea of homotopy theory is the nLab's skewed version of it, maybe in that case you would think simplicial gadgets are more natural because they're used in actual constructions of infty-categories or some such?

[coral pawn]

Anyway, I had a question about the stabilization of infinity categories. Is the following characterization correct: given a pointed infinity category C, the stabilization of C is a pointed category Sp(C) with a functor C ---> Sp(C) such that the functor preserves suspensions, Sp(C) is stable (i.e. suspension is an equivalence), and this functor is (homotopy) initial with respect to this property

The other definition is stabilization is the one on nlab

Which presents it as a homotopy limit

One direction is straightforward I think. Given something that satisfies the definition on n-lab, we can infer our characterization by noting that filtered homotopy limits commute with finite homotopy colimits (like the suspension)

The other direction seems trickier. We basically want to show that the functor C ---> Sp(C) not only preserves suspensions, but also limits

I think this characterization might be incorrect because I can't think of a way to show that but it seems like it should be the obvious characterization in terms of universal properties

[silver ridge]

Can someone sanity check me, we can just define a map f: [0,1] -> X by f([0, 1/2)) = {p}, f(1/2) = {q}, f((1/2, 1]) = {p,q} and this is continuous by open set definition right?

The only issue is that f([0,1]) should be connected too

oh

maybe {{p} ,{q}, {p,q}} is connected

[alpine nest]

I don't understand your definition of f

Especially f((1/2,1]); what's f(3/4) for instance?

[silver ridge]

{p,q}

[alpine nest]

But {p,q} isn't an element of the space

You want a function, not a multifunction

[silver ridge]

oh I think I see

[alpine nest]

Your space is X = {p,q}

[silver ridge]

right

[alpine nest]

And it has a topology in which the open sets are {emptyset, {p}, {p, q}}

[silver ridge]

And I want to show p and q are pathconnected

[alpine nest]

And that [0,1] is homeomorphic to its image in your space

[unreal stratus]

Dold-Kan

[silver ridge]

I think f([0,1)) = p, f(1) = q works then

[alpine nest]

Well, it's continuous

[silver ridge]

the inverse of open sets is open as [0,1] is clopen

[alpine nest]

Actually hang on, they can't mean arcwise connected in the usual sense, it must be used a synonym of path-connectedness

Because you can't have an embedding of [0,1] into a two-point space

[silver ridge]

I only know one definiton I'm afraid

[alpine nest]

https://en.wikipedia.org/wiki/Connected_space#Arc_connectedness

[alpine nest]

Yeah, that's path-connectedness

Arcwise connectedness often has another meaning

It's a stronger condition

[silver ridge]

is it on the wiki page?

oh thats what you linked

[alpine nest]

Yes, I even linked directly to that section

[silver ridge]

Yes I didn't realise

[alpine nest]

But your space definitely isn't arcwise connected because an embedding is a bijection, and you can't have a bijection between [0,1] and a two-point set

[silver ridge]

right

[alpine nest]

It is path connected (so in the sense of your material it is arcwise connected, but it's worth remembering that the terminology might vary)

[silver ridge]

Gotcha, thanks for clearing this up 👍

[alpine nest]

Basically welcome to topology, where each notion has several dozen slightly varying definitions that aren't at all consistent in literature

[silver ridge]

lol. even when I do know what the terms mean they're still wacky

quasi-components not necessarily being connected was not obvious from the definitions givenb

thank you random user on MSE for the counterexample 🙏

[alpine nest]

There's a great book "Counterexamples in Topology"

Worth getting

[silver ridge]

Already pirated it 🫡

[umbral panther]

Consider the case of spaces. If you take compact spaces and apply your definition, you get the Spanier-Whitehead category. If you take all spaces you get finite desuspensions of suspension spectra, not all spectra

If you define your target category as compactly generated categories, then you get ind(SW), which is the right answer. But that’s kind of magic and it would be better to have a more explicit construction, and that’s why limits

[coral pawn]

Is there a way to characterize the stabilization in the standard universal "it is the initial/final object satisfying blah?"

[umbral panther]

It is the inverse limit of the loops functor. Didn’t you mention this characterization, but didn’t like it?

[coral pawn]

I was thinking if there is a characterization of stabilization as the initial/final category where looping/suspending is an equivalence + other conditions

As you demonstrate in your example, just requiring looping/suspending to be an equivalence isn't enough

I guess you could just unpack the limit definition into that form

But that feels like cheating

[dim perch]

both the "spanier whitehead category" and "omega-spectrum objects" constructions occur as (co)limits of ∞-cats

[unreal stratus]

Every time i think of suspension i get 4.3.2.15 fear

iykyk

[dim perch]

in a sec I'll find the HA citation

[umbral panther]

It is the direct limit of the suspension functor, in the category of compactly generated categories. That’s a great definition. The only problem is computing in this category

[coral pawn]

Fair enough

[dim perch]

ok i actually don't get it, is this sag or something?

[unreal stratus]

HTT. The point is that this is a very general proposition in HTT (on Kan extensions, roughly) that gets applied a lot

[umbral panther]

In fact, the definition of spectra as ind(SW) is Adams definition, before omega spectra. But it was unwieldy then

[dim perch]

ah i was looking for HA.1.4.2.24

[unreal stratus]

I think when I found it in HTT I was like eh not worth learning it, seems rather specific, but then it is used a lot

[dim perch]

oh right

this is the one for the people who are extremely keen on citing like

[unreal stratus]

Well to talk about subcategories of diagrams admitting colimits, roughly

[dim perch]

every little HTT thing

[unreal stratus]

Iirc it is applies to say that essentially the functor from pushout squares to arrows (taking one of the "original" arrows) is a Kan fibration, and a trivial Kan fibration if cofibers always exist

[dim perch]

you know, homotopy theorists come in two types

[unreal stratus]

so that you can have a sigma boi

[dim perch]

the 4.3.2.15 types and the 7.3.2.1 types

[unreal stratus]

what is 7.3.2.1, HA?

(so algebra in the stable homotopy category?)

[dim perch]

relative adjunctions

[unreal stratus]

or

ah

there are also the non-Lurie model category enjoyers i guess

But we don't talk about them

[dim perch]

ah right, the "my citation for this fact is this entire 100 page paper from peter may" crowd

also tbqh I'm still holding on for like

the post-HTT/HA higher categorical people to reign supreme

slowly but surely we will replace HA with things that are at least on the arxiv

[umbral panther]

Yet a third definition is Goodwillie’s definition as homogeneous linear functors. Basically Brown representation

That’s the equivalence between small categories and compactly generated categories: Yoneda embedding. Colimits in the domain of definition become limits in the categories of functions

[ebon galleon]

ummm hello? hiiii? haiiii?

[unreal stratus]

Lol

[dim perch]

https://tenor.com/view/excited-kitten-kitty-hi-hii-hey-gif-21229532

[unreal stratus]

What if we were homotopy theorists but we just shared cat memes

[alpine nest]

You'd betcome category theorists

[ebon galleon]

do homotopy and category theorists do anything else?

[unreal stratus]

Real

[dim perch]

C2 equivariant homotopy is quite trendy nowadays

[inland laurel]

hi ı find different definitions of Cell decomposition, ı need a topological space and a open cover that is true meaning X a topological space and E a open cover if (X,E) a cellular decomposition provide axioms this is a true definition right?

[lusty trench]

A cell decomposition of a space X isn't given by an open cover of X. It's given by a partition of X's set of points, such that

(a) For each part Y, there exist a number n \in N, called the dimension of Y, and a continuous map f : D^n -> X, called Y's attachment map. Here, D^n is the closed unit ball in R^n.
(b) The restriction of f to the open unit ball is a homeomorphism between said open ball and Y.
(c) The image of the unit sphere is contained in the union of parts of dimension < n.

[inland laurel]

thank you

[livid escarp]

Is there a conceptual reason for why cohomology allows for richer multiplicative structure than homology? As far as things go I would expect them to be equivalent

[lusty trench]

The diagonal map d : X -> X x X and the projections p1, p2 : X x X -> X induce the product on cohomology.

[livid escarp]

Yeah, the technical reasons are clear. Maybe there is like a phylosophical (?) reasoning to it, like the Hom functor being more expressive or maybe I'm just trying to intepret too much.

[silver ridge]

Can someone check my argument please?

Let $f : X \to R$, $g : X \to R$ be continuous functions such that for $p,q \in X$, we have $f^{-1}(0) = p$, $g^{-1}(0) = q$. Let U, V be open sets in $\mathbb{R}$ containing 0. $U \cap V$ is then too an open set containing 0. Consider $f^{-1}(U \cap V) \cap g^{-1}(U \cap V)$. These are open sets containing $p,q$ respectively. If they are never disjoint (if X is not hausdorff) (not disjoint for arbitrary U,V open containing 0), then $\bigcap_\alpha f^{-1}(U_{\alpha} \cap V_{\alpha}) \cap g^{-1}(U_\alpha \cap V_\alpha) \neq \emptyset$ but this infinite intersection is simply $f^{-1}(0) \cap g^{-1}(0) \neq \emptyset$ which implies that $p=q$, a contradiction. Thus X is hausdorff.

[gentle osprey]

swifteeee

[quartz horizon]

I don’t understand how you get from the pairwise intersection not being disjoint to the full intersection not being disjoint

also wow i was trying to come up with a counterexample before I read the exercise

neat that this is a true result

[hoary breach]

use the fact that R is Hausdorff

[alpine nest]

And note that if q is not equal to p, then f(q) is not 0

[alpine nest]

Yeah, I didn't know that one

[pulsar junco]

how to describe surface such as at the pic? without using Bézier curves on glued rectangles

[quartz horizon]

Ok this is maybe too much of a hint..

[red yoke]

Twisted rectangles

[silver ridge]

This has thrown me off, they can be not disjoint in different ways so there's no guarantee that the inifnite intersection should be nonempty

[quartz horizon]

Mhm

[silver ridge]

if q is not equal to p, then f(q) is not 0. There exist disjoint open sets in R, U, V containing f(p), f(q) respectively as R is hausdorff

Lemma: $f^{-1} (V) \cap f^{-1} (U) \subset f^{-1} (U \cap V)$.
Pf: if x $\in$ LHS, then $f(x) \in V$ and $f(x) \in U \implies f(x) \in U \cap V \implies x \in f^{-1}(U \cap V)$.

Then, as U,V disjoint in R, $U \cap V = \emptyset$, so $f^{-1}(U \cap V) = \emptyset$, and by the lemma above, $f^{-1}(V) and f^{-1}(U)$ are disjoint open subsets containing p,q respectively. Thus, X is hausdorff.

[gentle osprey]

swifteeee

[quartz horizon]

Yeah this should work

Though one thing that’s nice to know is preimage preserves basically every set operation

[alpine nest]

That's OK, although your inclusion is actually an equality: $f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(V)$

[gentle osprey]

Outsider

[quartz horizon]

Mhm

[alpine nest]

Yeah that

[quartz horizon]

Basically cause preimage has both a left and right adjoint

[alpine nest]

And you don't need continuity or any other properties of f for this

It holds for every map.

[silver ridge]

I'm always sceptical about preimages so I only stated as much as I could prove

[alpine nest]

Preimages are much better-behaved than images

[quartz horizon]

They preserve union, intersection, complements

[silver ridge]

I think you need continuity to make sure pullback preserves openness

[alpine nest]

That's why continuity and measurability are defined in terms of preimages

[quartz horizon]

I mean it’s basically cause subsets are naturally contravariant

[silver ridge]

tbh I'm sceptical about any function operation on sets

[alpine nest]

Yeah, I mean you don't need continuity to have the fact that preimages are compatible with set operations

[silver ridge]

oh right

[quartz horizon]

You can identify subsets of X with functions from X to {0, 1}

And this is contravariant in X

Preimage is really just function composition

To me that’s morally why it’s so nice with subsets