point-set-topology

[grim knot]

pff

[tough hamlet]

the discrete topology has every subset open

[grim knot]

yes sure, thank it makes sense

[tough hamlet]

np

[grim knot]

if two sets are homeomorphic does it mean that they share the same topology(?)

[ebon galleon]

Fora homeomorphism f:X-->Y, if you identify elements of X with their image in Y, then yes

Strictly speaking though, they are likely different sets

So the topologies are probably not equal

[grim knot]

Okay, because for the exercises above, we are looking at continuous bijection between the above spaces

[ebon galleon]

Continuous bijections are not homeomorphisms in general

[grim knot]

what does fail here?

[ebon galleon]

A homeomorphism is a bijection f:X-->Y such that f and f^{-1} are continuous

Unlike with algebraic structures, where f being a bijective homomorphism implies (in many cases at least) f^{-1} being a homomorphism

You can have bijections where only one (or of course neither) direction is continuous

[broken nacelle]

Bump

[grim knot]

for example, we have a continuous bijection from $(\mathbb{R}, \mathfrak{O}_ {\text{disk}}) \rightarrow& (\mathbb{R}, \mathfrak{O}_ {eukl})$

(can somebody tell me how to do a subscript without getting a curl?)

[ebon galleon]

could do _{\text{disk}}

I think

Yes, this is an example of one such bijection

You have one too many $

[grim knot]

anyway, to the above message, if we have a continuous bijection in the other direction we would have a contradiction since one space is compact(right) and the other is hausdorff

[ebon galleon]

Before the second (\mathbb

[gentle osprey]

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

[ebon galleon]

Not quite, spaces can be both compact and hausdorff

(think of [0,1]: it's hausdorff, and it's compact as a closed and bounded subset of R)

Any finite discrete space also works

[grim knot]

yes but in this case since$ (\mathbb{R}, \mathfrak{O} {eukl})$ is Hausdorff, and $(\mathbb{R}, \mathfrak{O} {disk})$ is not, there should be a contradiction(?)

[gentle osprey]

damn_guuurl

[ebon galleon]

Discrete space is Hausdorff

wiat I was wrong hehe

nvm

[grim knot]

I took the wrong space....................

gonna write it in the old way

so, we have a continuous bijective function from R,O_eukl to R, O_cofin

which could be the identity

[ebon galleon]

Yes

[grim knot]

now if we had one in the other direction, we would have an isomorphism, which can not exist, since one space is Hausdorff and the other is compact

[ebon galleon]

(in fact, and map is continuous, so any bijection works for this)

[grim knot]

R, O cofin is not Hausdorff, is it right?

[ebon galleon]

don't combine the two. You could either say that it cannot exist since cofin is compact and eukl is not (R is not bounded with standard topology), or you could say cofin is not hausdorff by eukl is hausdorff

[ebon galleon]

correct

[grim knot]

Now I have it, thank you

[cursive tendon]

Let $S \subset \mathbb{R}$, $L(S)$ be the limit points of $S$. Let $p \in L(S) \setminus S$, is it true that there always exists a neighborhood $N$ of $p$ such that $N \cap (L(S) \setminus S) = {x}$?

I see this is not correct in $\mathbb{R}^2$

[gentle osprey]

Mattuwu

Mattuwu

[cursive tendon]

oh, not really true. I think S = (0, 1) \ {1/n} and p = 0 will work as a counterexample

[dire moon]

Hello ! For a simplicial set $S$ let $|S|$ be its realization. I think I saw a claim that $\pi_0(|S|)$ is in bijection with $S_0$ modulo the relation generated by
$x_0 \sim y_0$ iff there exists $x_1 \in S_1$ such that face maps map $x_1$ to $x_0$ and $y_0$ respectively.
I'm not sure I understand that claim. If points $x_0$ and $y_0$ in $S_0$ are path connected in the realization, can one show they belong to the same class for this relation ?

[gentle osprey]

plougue

[thorny agate]

I don't get how $U$ is such that $\overline{U} \subseteq V$

[gentle osprey]

Spamakin🎷

[ebon galleon]

What do you mean? Think about open sets in R. For example, $\overline{(0,1)} \subseteq (-1,2)$

[gentle osprey]

Wizard King Ryx

[ebon galleon]

Open sets can contain closed sets. In fact, for locally compact hausdorff spaces, we may always find a pair of such neighborhood around any point

[thorny agate]

that's not my question

my question is how is the U constructed in the proof such that it's closure is contained in V

ofc I know open sets can be contained in closed sets. Also local compactness has nothing to do with this

[thorny agate]

I am referring to this image I sent here

[ebon galleon]

Yes I know, I was just stating. My bad for misunderstanding which part you were looking at

Since g is 0 on U, then since g is continuous, $g(\overline{U}) \subseteq \overline{g(U)} = {0}$ is disjoint from ${1}$. If $\overline{U}$ was not a subset of $V$, it's image would have to intersect ${1}$

[gentle osprey]

Wizard King Ryx

[thorny agate]

oh it uses the continuity of g

ty

completely forgot about that

[eager herald]

so to use van kampen on $S^1 \vee S^1$, we choose to break up the set into two pieces: $S^1$ with a small arc in the other circle (that can deformation retract to a point) and likewise for the other part. Clearly van kampen says that since their intersection is path connected (and we only have 2 sets so the triple intersection thing is automatic) and since the intersection is simply connected, we get that $\pi_1(S^1 \vee S^1) = \mathbb{Z} \ast \mathbb{Z}$. my question is what goes wrong if we just make the two parts one circle and the other?

[gentle osprey]

sean

[eager herald]

their intersection (the singleton on the basepoint) is path connected and obviously have trivial fudamental grp

[ebon galleon]

The two sets need to be open. If you just take each of the circles S^1 separately, they are not open in S^1 \vee S^1

[eager herald]

ohh that makes sense

[umbral panther]

In practice you can thicken up closed regions to open regions in reasonable spaces, so it applies to closed covers. But if you glue two copies of the Hawaiian earrings at the basepoint, it doesn’t apply

[grim knot]

I know that this is in german, but I think it should be still understandable

Can somebody tell me what it means to glue these two spaces through the given function phi?

Does somebody also have easier examples?

[trim chasm]

can't both be true?

[hidden crag]

We’ve talked about that before, were you able to work it out back then?

[grim knot]

I have too many things in my head, and I do not understand it anymore

I just do not see in my head what this looks like

is the idea that you glue A and f(A) together?

[hidden crag]

Yes

Now think about what z^15 does and work through the definitions

Take your time

[grim knot]

I forgot complex analysis

does it mean that if $z = re^{i{\theta}}$, that then we would have a more streched out number? $z = r^{15}e^{i15{\theta}}$

[gentle osprey]

damn_guuurl

[hidden crag]

And what does that mean

[grim knot]

I mean it looks like we identify the whole C to f(C) besides from the disk $D^{2} = {z| |z|<1}$

[gentle osprey]

damn_guuurl

[hidden crag]

Yeah X is just that subset

You identify x in X with f(x) after taking the disjoint union

Now think about what this will result in

[grim knot]

If I do not see it wrong, these spaces are glued for |z|>1, which means that we identify them in that case

but we have like a ball for |z|<1

or wait a sec

I'm gonna make a drawing

It's hard to draw, but i think I see it now

[hidden crag]

I mean you’re still gluing along f

But you might mean the right thing

[grim knot]

Jeez its so ugly

I know it's wrong

[feral copper]

Heyo! So I have two submanifolds X and Y of some ambiant manifold, and I assume that Z=X∩Y is also a submanifold. Moreover, for all point p∈Z, I have that codim(TpX∩TpY)=dim(Z). I have been told that this type of intersection is called 'neat' (the case dim(Z)=0 is a transverse intersection). Is that standard terminology?

[umbral panther]

You mean dim = dim, not codim = dim, right?

Neat is kind of standard, but rarely useful. Transverse is very standard, but this definition is wrong. You can have high dimensional transverse intersection. The definition of transverse is that the two tangent spaces together span the ambient tangent space. Sometimes people say that the sum is the whole tangent space. Equivalently the codim of the intersection is the sum of the codim of the two manifolds

[feral copper]

Yes my bad, dim=dim!

Yes you're right I could have transverse circles in a manifold with bigger dimension than 2 🙂

In my case, I have neat intersections everywhere, and it's not obvious how to perturb things to become transverse... (neat intersections of surfaces along curves in a 4-manifold)

Thank you!

[umbral panther]

I think neat should be locally equivalent to transverse in a submanifold

[feral copper]

Wdym?

As long as the sum of dimensions is less than that of the ambiant space, I can always perturb to be transverse

But if I have, say, two spheres meeting neatly along a circle, that doesn't tell me anything regarding the transverse intersections of the perturbation, right?

(again, I'm only thinking of knotted surfaces, not anything other-dimensional 🙂 )

[umbral panther]

If X and Y intersect neatly in M and M embeds in N, then X and Y intersect neatly in N. This is supposed to be obvious. If X and Y are transverse in M they are no longer transverse in N because there’s more room to wiggle

I’m suggesting the converse, that if a curve and a surface intersect neatly in points in a 4 manifold N that near one of the points of intersection the curve and surface should be contained in a 3 manifold M in which they are transverse

[feral copper]

'room to wiggle' idk, I'd say that an empty intersection is transverse
Aaaah I see, okay! Yeah that makes sense, although I wouldn't trust it to be so easy 🙂

There are always bizarre 'jokes' in topology, and I've learnt not to be decisive about anything and try to find counter-examples to everything!
Anyways, thanks a lot for your output 🙂 I couldn't find anything anywhere talking about 'neat intersections'

[umbral panther]

It’s not something people prove theorems about. You just observe that a particular example is neat. Whereas transversality is useful because it creates neat intersections. If you have a neat intersection it’s because you already understand your example pretty well

[fallow ginkgo]

Dumb question but

How do I "quickly" check that certain maps that come from homotopy are infact continous.
Eg : $h,h' : X \to Y$ and $k,k' : Y \to Z$ be pairs of homotopic maps. Then $k \circ h$ and $k' \circ h$ are homotopic.

Proof : Let $H,K$ be the homotopies. Then consider the homotopy $F:[0,1] × X \to Z$ defined by $F_t = K_t \circ H_t$.

[gentle osprey]

Noob666

[fallow ginkgo]

I feel I'm missing something very simple

2nd question, how do I put this situation into a "functoriality" statement. Like composition of maps is corresponding to something about homotopies that "looks like composition".

[unreal stratus]

Tbh I would usually do it just by not thinking of homotopies in that way lol

Well like just gluing maps of the form A x I -> B directly

[fallow ginkgo]

what way would you recommend?

[unreal stratus]

Here you can just give a nice formula for the homotopy and it is easily seen to be contjnuous

Well like you want a map X x I -> Z

[ebon galleon]

Well F is a composite of continuous maps, so F is continuous. If you check when t=0 and t=1, it should give you kh and k'h'
But what potato is suggesting might be clearer

[unreal stratus]

You can just do like (x,t) -> K(H(x,t),t)

[unreal stratus]

That's not enough

Well

F isn't a complosition as given

So you should just write that out imo lol

[ebon galleon]

The way you've written it is a composite

My bad

[fallow ginkgo]

Ohhh yes that makes sense

[ebon galleon]

That's what I had in mind

[fallow ginkgo]

Also, I keep seeing people mention "the space of all paths on X". Any source where I can read about this perspective?

[unreal stratus]

There are a few things that can mean, but likely the loop space is meant

[ebon galleon]

This statement should essentially be that composition of homotopy classes of maps is well-defined.
We can take a category hTop, whose objects are spaces, and morphisms are homotopy classes of maps. This gives us our composition rule (since we know that the assignment [f] \circ [g] := [f \circ g] is well-defined)
This in itself is not so much about functoriality afaik

[unreal stratus]

This is showing the obvious functor Top -> hTop is actually a functor ig

[ebon galleon]

ig, but that's obvious provided hTop is a category lol

[formal tide]

rereading munkres I just noticed this definition does not work for a one element set :p

is there a nicer way to define the order topology that works? other than adding a 4th clause

[languid patrol]

@formal tide I mean you’re only including one extra space… I don’t see a natural way to extend it

[thorny agate]

how do you have an order on a 1 element set?

[formal tide]

hmmmm I think there should be some universal property but I'm not finding it

[formal tide]

a <= a; that's it.

[gaunt laurel]

Hi, would I need to know field theory or galois theory for chapters 1 to 3 in Hatcher?

[novel acorn]

not really

[gaunt laurel]

Ok, thank you

[unreal stratus]

No

[formal tide]

It seems that "the coarsest topology having all closed intervals as closed sets" works but that's still not a universal property

nvm doesn't work for non bounded tosets

[nocturne matrix]

I'm not sure this is the right channel for this question, but here it goes

Is it true that with the statement "a set $S$ has $\infty^n$ elements" we mean that there exists a bijection $\phi:S \to \mathbb{R}^n$?

[gentle osprey]

Plagionotus

[tough hamlet]

no

[feral copper]

Does that even mean something? Also, is that #point-set-topology?

[tough hamlet]

also wrong channel

[nocturne matrix]

R is in bijection with R^n so it's dumb

what channel should it be in?

[feral copper]

It depends, did you read this sentence in a book/lecture note? Because if not, it's general math discussion or a general question
If you did, what course/level? Also, if you did, that's a bad teacher

Maybe that's linear algebra, and this is a clumsy way to say your space has dimension n?

[nocturne matrix]

well I was taking Geometry 2 so we did some affine, euclidean and projective geometry. The notation was used to quantify how many planes are in a sheaf of planes and the teacher said it can be roughly interpreted as "the set is in bijection with R^n" but it was just some qualitative reasoning and she said we would give a proper meaning to it in topology. I know the notation is sometimes used to say a vector space has dimension n but I have never actually used it in that context

[feral copper]

The set is in bijection with R^n
You agreed yourself that n is irrelevant here, so most likely the teacher meant something more than a mere 1:1 correspondance. I have honestly never seen this notation, and I'd say it's a terrible notation just because I cannot imagine any reasonable meaning behind it. Not that I'm knowledgeable or anything, just that it should be a bit obvious what it should mean

[nocturne matrix]

We said that a fan of planes has $\infty^1$ planes while the set of all planes passing through a given point in space (idk the english name) has $\infty^2$ planes.

[gentle osprey]

Plagionotus

[feral copper]

Oh a pencil of planes? But I've never seen this \infty^n notation before

Maybe try #algebraic-geometry? But don't tell them I sent you

[nocturne matrix]

pencin, sheaf and fan of planes should all be the set of planes containing the same given line

idk the name of the other

[languid patrol]

@formal tide you could just add the condition that X \in \tau, as long as the whole space is in there you’re fine.

[formal tide]

yup, I ended doing that for my notes, ty

[balmy field]

Does anyone know a crazy, accessible fact (intro topology and good complex analysis) about the Riemann sphere? I’ve seen the conformal mappings. Wondering if there was anything crazy though

[tawdry widget]

Hi, I need help with the definition of regular neighborhood, and a question about linking number.
F is an orientable, compact, connected genus g surface with n boundary components, with 2g+n-1 generators of its first homology group as shown in the first picture:

My question is about the second picture. What is the definition of the regular neighborhood V of F in this case?
I am not sure I got it correctly, which leads to another question, why is the linking number lk(e_i, f_i)=1? Since I thought e_i, f_i are parallel, why 1 not 0?

[broken nacelle]

you need only some ring theory for cohomology

[abstract saffron]

lol, another knot theorist fellow 😄

[tawdry widget]

I don’t study knot theory I am just interested in it…. It’s gtm 175…

Should be contained in basic algebraic topology so I asked here…

[abstract saffron]

I like this book, but tbh, it has the least transparent proof of this prop

I read it a couple times, never cracked it either 😄

[tawdry widget]

You know other references? Of this proposition?

[abstract saffron]

man, it's been some time since I studied knot theory, not sure if I remember anymore

[tawdry widget]

I see

I will probably figure out what V is, I am still in need with explanation why lk(e_i, f_i)=1 though

[abstract saffron]

for minor-related things, try a short intro to alexander polynomials by massuyeau
https://massuyea.perso.math.cnrs.fr/notes/Alex.pdf

which is a dirty version of Intro to Combinatorial Torsions by Turaev

Lickorish is very good, but when it comes to Alexander polynomials, I really recommend going through other books and combining multiple sources. Lickorish didn't do it justice.

[tawdry widget]

Thank you I will check it out. It seems like it’s more general, Alexander polynomial for CW complexes. I didn’t know that, thank you so much.

[abstract saffron]

Another one with good intro to Alexander polynomials is Knot Theory & Its Applications by Murasugi.

Albeit you kinda need to know Seifert surfaces first, but if you do, the link to Alexander polynomials is transparent with this book

[abstract saffron]

yes, Alexander polynomials is the characteristic polynomial of Khovanov homology, iirc

[tawdry widget]

You know textbook about it?

[abstract saffron]

Or Floer homology. Never remember which one is Alexander's and which one is Jones'

[tawdry widget]

Jones polynomial has generalization too?

[abstract saffron]

Turaev is de facto the bible of advanced knot theory.

It introduced a lot of advanced tool. But you really need to be good at Alg Topo for this one

[abstract saffron]

Oh yes

Wait till you see you can combine both homologies to get what's called HOMFLY homology

the characteristic polynomial is like, the combined of both Jones and Alexander

[tawdry widget]

I see

[tawdry widget]

What book are you referring to?

[abstract saffron]

this one

it's lecture note from a course in Knot Theory at ETH back in 1990s

[tawdry widget]

I see, thank you very much.

[tawdry widget]

I know the definition of Seifert surface. I don’t know why the standard way of constructing one for a given knot gives you the Seifert surface with minimal genus though.

[abstract saffron]

... because it doesn't?

the construction of seifert surface depends on the knot diagram. Different diagrams give different surfaces, with different genuses.

It's only for some certain classes of knot, that it's proven that certain types of diagrams give rise to surfaces with minimal genus.

[tawdry widget]

What certain types?

[abstract saffron]

iirc, alternating knots are one of the classes

In general, computing minimal genus is NP-complete. That should show how hard it is.

[abstract saffron]

yes, I was correct. Proven by Crowell. See page 60 of Lickorish.

[tawdry widget]

I see. Thank you.
Do all of those we discussed, have generalization in more general case? Like higher dimensional? K (homeomorphic to S^n) in S^(n+2), and this higher knot K has a Seifert surface F of dimension n+1 whose boundary is K. Something like that? And Alexander matrix, Alexander module, Alexander ideal, linking number used to calculate Alexander matrix… these kind of things?

[abstract saffron]

So Alexander polynomial is the Euler characteristic polynomial of Knot Floer homology.

We study the homology groups from here, and less the polynomial. Euler characteristic is just one tool, and not the most powerful too. We have a far bigger arsenal at disposal from Homology Theory.

recently, there has been a paper about constructing Seifert surfaces from differential geometry's point of view, so we can expect an even bigger arsenal from Analysis.

One example about higher dimensional is perhaps slice knots, which are the boundary of D^3 (in comparison with trivial knot, which is the boundary of D^2).

[tawdry widget]

Yeah. The thing is I saw some articles talking about those things in the case of dimensions 2, just like dimension 1, I thought maybe they are some general thing for any dimension.

Like this one, Alexander modules for dimension 2, those things, linking numbers , just look similar to the dimension 1 case. Also some articles talking about coloring, reidmeister moves, of knotted surface. It really makes one think they are general things…

[abstract saffron]

knot theory is a deep, deep rabbit hole 😄

[tawdry widget]

I see.

If I could know that there is no such generalization yet, it would also be a relief .

[abstract saffron]

I mean, we do know how to build these things for knots, it's almost straightforward to do the same for other things. Reidemeister moves -> colourability -> Alexander matrices and polynomials -> homology -> ??? -> profit

I expect there are generalisations, but again, knot theory is very young. We only have the most powerful tools, e.g. Khovanov-Rozansky triple-graded rings and Rasmussen's s-invariant, for almost 20 years.

Rasmussen is still alive, for example

[tawdry widget]

I see. I need to search harder. Speaking of coloring, coloring of a classical knot involves a quandle. (mappings from arcs of the knot diagram to a quandle). There are higher knots, but are there things like quandle , for higher dimensional cases?

[abstract saffron]

I didn't even know there was a thing called a quandle 😄

Strange, I never saw it mentioned anywhere

I guess there are, otherwise it'd be difficult to define more advanced tools. But again, I'm not well-versed in higher dimensional generalisations like this one

[tawdry widget]

I see. Last thing, do you know reference of proving any knot is either torus, satellite or hyperbolic?

[abstract saffron]

Oh man, lol. That fact is really, really deep

[tawdry widget]

I see. Never mind then. I am only interested in algebraic things for now… if it involves lots of analysis I will pass…

[abstract saffron]

the original proof of that is by Thurston

William P. Thurston. Three-dimensional manifolds, Kleinian groups and hyperbolic geometry.
Bull. Amer. Math. Soc. (N.S.), 6(3):357–381, 1982.

this was a huge breakthrough in knot theory back in 1980s

[tawdry widget]

Yes yes, they all say it’s proven by this mathematician Thurston. I just wasn’t able to find his article. Thanks a lot

[abstract saffron]

man, he basically won his Fields Medal for this 😄 so good luck

[tawdry widget]

🥲

No wonder they say “ Thurston famously proved that…”

I will save it for a day in the future far from now…. But thank you.

[abstract saffron]

Just to show its power, in 2020, they used this fact to classify all prime knots up to 19 crossings

https://drops.dagstuhl.de/opus/volltexte/2020/12183/

That's also how I learned about the fact, and who proved it

[tawdry widget]

My god…

[abstract saffron]

I told you Knot theory is very young 😄

[tawdry widget]

I see. I am just interested in it, only looking for the most general results, that fit all dimensions for now. I am not ready to read any advanced result about specific things like “classification of knotted sphere up to 6 crossing”

[abstract saffron]

ehh, nothing is gonna fit for all dimensions

There's a reason we have knot theory, as opposed to higher dimensional topology

[tawdry widget]

I thought you said Alexander this Alexander that… those are general…

[abstract saffron]

low dimensional stuff is just weird

[abstract saffron]

to dimension 4, maybe. But I don't think there are those things for higher dim. We have a lot more tools when it comes to higher dims.

[tawdry widget]

I see. When you say dimension 4, do you mean 2-dimensional K in S^4? Or do you mean 4-dimensional K in S^6?

[abstract saffron]

very good question 😄 I'm not well-versed in this, I only know classical knot theory

gotta let someone else answer this

[tawdry widget]

No you have already told me many things I have never heard of. I have a lot of searching to do after this.

Thank you again, especially for those references. Before this conversation I only know Lickorish 😂

[abstract saffron]

A good resource to accompany Lickorish is a series of lectures on knot theory on YouTube. The most famous one, really

[tawdry widget]

With 10 episodes right

[abstract saffron]

"Math at Andrews", yes

[tawdry widget]

I don’t like him though, he rarely gave any proof…

[abstract saffron]

it's very informal, but also gives lots of insights and where to go, where to search

I only understand knot theory with intuition built from that series. Ain't no way I could grind through abstract gibberish by Turaev

another one is the series of lectures on knot theory from Princeton's summer school, given by Rasmussen himself. Also a lot of insights and intuition, but you kinda need to be somewhat well-versed in Alg Topo and Ring theory.

Finally, there are some lectures on hyperbolic knots too, mostly famous after Lisa Piccirillo proved that Conway knot is not smoothly slice.

Now these ones require some intuition and well-versed with geometry

https://www.youtube.com/watch?v=Rk2LiHzUpps

https://www.youtube.com/playlist?list=PLldN_DpkXL3ZuSpCA7e2yNqn8lZ4YsJsc

[tawdry widget]

Thank you so much. The preliminary knowledge you mentioned.
Algebra: I have finished most of basic algebra by Jacobson. I probably have the algebra background needed.
Algebraic topology: I read gtm 119 rotman. But I feel lost when people talk about duality. Alexander duality, Poincaré (? Idk) duality… all those dualities…
Geometry: I don’t know what kind of geometry I should read. Like I have no idea about “surgery”, in idk, differential topology?

[abstract saffron]

Idk what should be required in Geometry. I studied Differential Geometry and am familiar with surgery, and it comes up from time to time, but only as an intuition and not really involved in proofs.

I guess if you have strong intuition with 3- and 4-dimensional objects, then that should be enough. For example, you should know how to build Seifert surfaces from knot diagrams with surgery.

For Algo topo, duality is not really that fancy 😄 I'd recommend Nakahara for intro (but I guess you passed that point already) and Munkreas. Although almost everyone in this channel would recommend Hatcher.

Idk how much algebra there is in Jacobson tho.

[tawdry widget]

I probably wouldn’t worry about algebra, or homological algebra. I guess I need to learn more about topology.

[abstract saffron]

also shout out for spelling Poincaré in French

[tawdry widget]

Idk, it was just auto spelling of the keyboard.

[abstract saffron]

another good test is to visualise geometrically and to draw out what's going on in Chapter 2 of Lickorish

in particular Theorem 2.10. That one took me good a good hour, def putting intution to the test.

[tawdry widget]

I see. I will definitely continue with Lickorish. I am attracted by the way he introduces Alexander polynomials, in a IDK, geometric way…

That X^infty thing

[abstract saffron]

do couple it with this book

[tawdry widget]

Yeah, already downloaded it

[abstract saffron]

There are at least 4 ways to define Alexander polynomials

If you can't find at least 3 ways, then you haven't understood it yet.

[tawdry widget]

How many have you told me.

[abstract saffron]

😄

the question is how many ways do you know

[tawdry widget]

Only one… which I am still reading…

[abstract saffron]

so, in that intro series to knot theory on YouTube by Andrews, he defined Alexander polynomials as a generalisation of k-colourability.

[tawdry widget]

Yeah some matrix with elements like -1, 1-t things

[abstract saffron]

iirc, a knot is k-colourable if and only if k divides Alexander polynomial, evaluate at 1

that's the first way

[abstract saffron]

this one defines Alexander polynomials from Seifer surfaces, as the det of that matrix of linking numbers with some minus of transpose to make it invariant

that's the second way

I'll let you find one more by yourself

[abstract saffron]

i.e det of Seifert matrix

I see Lickorish defines the same, although somehow I found it less transparent

[abstract saffron]

the fourth way I know is in this series, although I didn't understand a single thing

there. minute 17, first lecture

[tawdry widget]

I see. That’s awesome. It definitely helps to know equivalent ways to define it.

[abstract saffron]

proving that they are equivalent is worth a Master thesis 😄

knot theory is a very deep rabbit hole

[tawdry widget]

God… then I only hope in the future I will be able to understand the equivalence between two of those four…

Thank you so much , will keep reading. Big help today.

[abstract saffron]

U r welcome

Glad that my studies of knot theory helped. No one in this server except me studied knot theory anyway.

😭

[tawdry widget]

😂 I hope they discuss more about it, since I don’t view it as some very specific topic… it sounds like a general, main, broad subject to me

[scenic trellis]

does anyone have recommendations for textbook (or lecture notes or talks) on cerf theory

[scenic trellis]

https://math.stackexchange.com/questions/4104171/recommendations-for-cerf-theory

[tiny ridge]

What do you want to know

If you want to learn the pseudoisotopy theorem I would recommend Sanders Kupers' book "Diffeomorphism Groups" available online

[scenic trellis]

[scenic trellis]

I want to the case if I have a family of functions between two morse functions on manifold M that is parameterized by smoothly moving a point on another manifold N

mmm i think it is basically same as cerf's one-parameter family case (right?)

[nimble portal]

A punctured closed ball in R^2 is open right?

[ebon galleon]

Like a closed unit ball (norm <=1, let's say), but a point is removed? No

[steel glen]

does it still contain some boundary points?

[nimble portal]

Ah wait yeah it’s neither open nor closed right?

Yes it does

[ebon galleon]

Right

[nimble portal]

It’s punctured

ie centerpiint removed

It’s not closed because it doesn’t contain all of its bd (as the point we puncture at becomes a bd pt, but it won’t contain that)

And it’s not open because not every point has an open ball around it contained entirely in the set (take the points where x^2 + y^2 = 4, for example)

[ebon galleon]

Right

[nimble portal]

Any point on this set necessarily has elements outside of the set for all epsilon > 0

ugh

This is funny

This is a pretty good example

Really natural construction to show how sets can be neither open nor closed (and I’m sure there’s a good example for clopen too)

[steel glen]

in general, if you contain some, but not all, your boundary points, you are neither open nor closed
clearly the set isn't open, and the complement will also not be open for similar reasons

[ebon galleon]

empty set be clopen

[queen prism]

easiest example of a not-closed-nor-open set would be [0, 1) in R

[nimble portal]

Yeah

[novel ember]

i am finally

understanding svk

[steel glen]

is svk van kampen?

[novel ember]

yes

[nimble portal]

And the entire space in the discrete topology :p

[ebon galleon]

Always the entire space

any subset in the discrete topology

[steel glen]

i think we know they meant every subset

[steel glen]

very good theorem

S tier

[nimble portal]

Topology contains the entire space

So it’s definitionally open

Its complement is then the empty set which is also open

Hence it’s closed as well

Yes?

[ebon galleon]

Right

[nimble portal]

Likewise for empty set

Coolio

tan(Q) both

[hoary breach]

you're welcome

[coarse mist]

I have to show that the collection

B = { (a, b) | a < b, a and b rational}

is a basis that generates the standard topology on R

But isnt the topology generated by this basis courser than the basis used to generate the standard topology?

Because the standard topology includes basis items like (-sqrt2, sqrt2), and this one doesnt

[gritty widget]

"topology generated by B" is not just B

[ebon galleon]

They produce the same topology still, even though one basis is a subset of the other

[gritty widget]

it includes a little more than that

[coarse mist]

Even if they generate the same topology, isnt there still some open sets in the standard topology that arent in the mentioned topology?

[gritty widget]

"even if they generate the same topology, aren't they different?"

[ebon galleon]

No. The statement that they generate the same topology means that the topologies generated have exactly the same sets

[coarse mist]

Right, but how does the first collection of bases “attain” sets that have irrationals as their endpoints

if what im asking doesnt make sense feel free to just tell me and ill move on and come back to it

somethings j not clicking so it’ll probably help anyways

[ebon galleon]

So the proper way to show that would be to show that, for bases B and B', for any $x \in U \in B$, there is a $U' \in B'$ with $x in U' \subseteq U$, and vice versa (iirc at least). (Or show that each basis element of one is open in the others topology)
In this case, you can think about it as approximating the intervals with irrational endpoints, with ones that have rational and points. For instance, for the interval (0, sqrt(2)) we can find a monotonically increasing sequence ${x_n}$ of rationals that converges to sqrt(2). Then the union $\bigcup_{n=1}^\infty (0, x_n)$ shows that (0, sqrt(2)) is open with respect to the basis with rational endpoints

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

Sorry for any typos, typed on phone :/

[queen prism]

your basis doesn’t include intervals with irrational endpoints but they can still be contained in the resulting topology
so in some sense including intervals with irrational endpoints is redundant

[vast stump]

Hii pls fill this form out it doesn't take much time its a survey pls its for english class

https://forms.gle/KHLjj5QKQ1EbYkK57

[limber wren]

every interval (a,b) is the countable union of of intervals with rational endpoints, so in terms of the topology the two bases generate, rational/irrational endpoints won't make a difference.

[feral copper]

@unborn lotus might wanna remove this spam. Also, Happy Birthday

[feral copper]

Where can I read more about this Euler class that is an obstruction to extend sections?
https://mathoverflow.net/a/175846/137265

[languid patrol]

@feral copper there is a good general discussion of obstruction theory in characteristic classes by Milnor

It has this fact and much more

[feral copper]

Oh
Oh yes, §12 xD I guess I missed this chap when I studied it back then... Thanks!

Wow they even do the Gysin sequence, amazing! Thanks 😄

[languid patrol]

Yep it’s a classic, good read for you to do

Enjoy!

[coarse mist]

My assumption is that each basis element is itself an open set, is that wrong?

[gritty widget]

not every open set is a basis element

but every basis element is an open set

the open sets in the topology generated by a basis are the unions of the basis elements

(-π, π) isn't an element of the basis consisting of open intervals with rational endpoints, because it doesn't have rational endpoints, but it is still in the topology generated by it because you can write it as a union of intervals with rational endpoints

it's the union of (-3, 3), (-3.1, 3.1), (-3.14, 3.14), (-3.141, 3.141), ...

[coarse mist]

Ohh I see

Thats super helpful

Thank you

dont know how I didnt see that

thanks to everyone who helped

[next crystal]

Another way to view it is that every point in (-pi,pi) has a basis element containing that point and contained in (-pi,pi)

[limber wren]

Yeah, that's part of the definition of a topology being generated by a basis, the open sets in the generated topology can be written as unions of basic sets, so in particular each basic set is also open in the generated topology.

alternatively, the topology generated by a collection of sets B is the smallest topology containing each element of B

[cerulean oriole]

Why (if it's even true) is the commutator [a,b] = aba^{-1}b^{-1} continuous in each of a and b separately in a semitopological group (group where inversion is continuous and multiplication is continuous in each variable separately)?

[lime sable]

if you fix a, then as a function of b it's just a composition of continuous functions, right?

oh wait i see the issue

[cerulean oriole]

Yeah, b “appears twice in the expression” so that doesn't immediately work.

[lime sable]

does the inner automorphism group get a topology or nah

[coarse night]

you can give it the compact open topology

I haven't read the context tho

[grim knot]

guys, I'm trying to find the foundamental group of the torus with seifert van kampen, and in the solution they say that $\partial (Q)$ is a strong deformation retract of $Q = [-1, 1]^2$, please tell me this is wrong

[gentle osprey]

damn_guuurl

[ebon galleon]

\partial [-1,1]^2, as in a square without interioir? Yeah that's wrong lol

[grim knot]

I have it in german, but I think you see the symbols

I'm not misinterpreting right? I mean if we take away our point p then it's true for sure

[tough hamlet]

that's probably what they meant yeah

[grim knot]

okay thank you, it really confused my intuition on deforetracts for a minute

thanks guys

[tough hamlet]

np

[tepid hazel]

@steel glen Because it states that tau is the collection of all subsets of X so X must be within tau because X is a proper subset of itself

[steel glen]

no

where did you get that from

i have to go. feel free to post the full question or just ask back in #math-discussion
sorry about that

[tepid hazel]

You're good thank you!

[ebon galleon]

Okay, so now we need to check the union requirement. So if ${U_i}{i \in I}$ is a collection of subsets that satisfy this half-open interval requirement, then for $U = \bigcup{i\in I} U_i$, and any $x \in U$, we can find a half-open inerval such that $x \in [a,b) \subseteq U$.

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

@tepid hazel

What does it mean for x \in U? If we know that an element is contained in a union of sets, what can we say?

[tepid hazel]

That it's an element of at least one of the sets?

[ebon galleon]

Right! And we know that then we can find a half open interval with x \in [a,b) \subseteq U_i, where U_i is one of the sets containing x

How does this then get us a half-open interval contained in U?

[tepid hazel]

That's the part I'm not sure about

Because it's the union of all U_i?

[ebon galleon]

RIght

And since [a,b) \subseteq U_i

We also know then [a,b) \subseteq U

All good so far?

[tepid hazel]

Yep we're good

Sorry I got a phone call

[ebon galleon]

Aight

[limber wren]

I've only seen anything like this in the context of c*-algebras/non-commutative geometry, and it was a subgroup of automorphisms that "was analogous to inner automorphisms", but outside of that I've never encountered it

[ebon galleon]

So the last thing to check then is that, if ${U_i}{i=1}^N$ is a finite collection of subsets for which this property holds, then it also holds for $U = \bigcap{i=1}^N U_i$. So if $x \in U$, then we need to find a half-open interval $x \in [a,b) \subseteq U$. To start off with, since $x \in U_i$ for each $i$, we can find open intervals $[a_i, b_i)$ with $x \in [a_i, b_i) \subseteq U_i$.

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

Where do you think we can go next?

[tepid hazel]

Im guessing that since U_i exist within U there is a [a_i, b_i) that exist within U as well?

[ebon galleon]

Well we don't know that there is a U_i with U_i \subseteq E (in fact, this happens iff U_i = U since intersections)

There mayb be bits of each [a_i, b_i) that essentially "stick out" of U. How can we fix that? (by considering something smaller than each of the half-open intervals, of course. What's a natural guess for the set contained in all of hte intervals [a_i, b_i)?)

[tepid hazel]

R?

[ebon galleon]

R is larger than (contains) all of the U_i 's. Think in terms of operations on sets (i.e. unions, intersections, complements..)

[tepid hazel]

Yeah I'm pretty lost with this one

[ebon galleon]

If you intersect each of the $[a_i, b_i)$ 's, then you'll get a set $S$ that's contained in each half-open interval. And in particular then, it's contained in each $U_i$, hence in $U$. By construction then, $x \in S \subseteq U$. The only detail to check is that $S$ is a half-open interval $[a,b)$

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

Take $a = max_{i \leq N} (a_i)$ and $b = min_{i \leq N} (b_i)$. Check that $[a,b)$ is equal to the intersection $S$

[gentle osprey]

Ultrawizard King Ryx

[tepid hazel]

This was my original thought but I discarded it because isn't that just U_i exactly?

OH! wait I think I got it

[ebon galleon]

Nope. Essentially what we're using here is that if $A_i \subseteq B_i$ for each $i \in I$, then
$$
\bigcup_{i \in I} A_i \subseteq \bigcup_{i \in I} B_i \qquad\text{and}\qquad \bigcap_{i\in I} A_i \subseteq \bigcap_{i \in I} B_i
$$

[gentle osprey]

Ultrawizard King Ryx

[tepid hazel]

So you're intersecting every value x such that a <= x < b as opposed to intersecting every interval itself

[ebon galleon]

No, we are intersecting the intervals themselves. Explicitly, this is the set of everything that's in each of the intervals

[tepid hazel]

Okay what do you mean "in" each of the intervals

I think thats whats tripping me up here

[umbral panther]

The inner automorphism group is the quotient by the center. Surely the center is closed? Are you talking about topological groups or was this part of the conversation about semi topological groups?

[ebon galleon]

$$ \bigcap_{i=1}^N [a_i,b_i) = {x \in \mathbb{R} \mid x \in [a_i,b_i) \forall i}$$

[gentle osprey]

Ultrawizard King Ryx

[tepid hazel]

Okay so it's each x within the interval

[ebon galleon]

each of them, but yeah

[tepid hazel]

OH OKAY

Cool

I got it

Is that it?

I feel like theres a part at the end

[ebon galleon]

No that's it

[tepid hazel]

Oh okay so are there any examples of things that arent Topologies?

[ebon galleon]

uhh i mean generally it's more interesting when things do form a topology. An easy example however is if you just ignore one of the axioms (i.e., a collection that does not include the empty set, or the entire space, or is not closed under intersections)

[tepid hazel]

Oh yeah I suppose so lol

Thank you for the help!!

[misty parrot]

Tricky but cute question from MO: https://mathoverflow.net/questions/452140/does-there-exist-a-continuous-open-map-from-the-closed-annulus-to-the-closed-dis

[grim knot]

Ihave a question again to this exam question. Is it wrong if I define $A = (\mathbb{R}^2 \times {0}) \setminus (1,1,0)$ and $B= ({0} \times \mathbb{R}^2 ) \setminus (0,1,1)$

[gentle osprey]

damn_guuurl

[grim knot]

I though that both sets are open and path connected and A n B is the y axis right?, which is also path connected

[steel glen]

yeah it looks fine. did they say that was wrong?

[grim knot]

They have another solution, I think my sets are not open

[steel glen]

hm yeah they aren’t open near the y axis

or on the y axis

so maybe you can just do your A union {0} x R x (-1,1)

and a similar idea for your B

[grim knot]

There is a problem because they are not open in the subspace topology?

[steel glen]

yeah any open ball around (0,0,0) for example

in your subspace, it would have to intersect both the xy plane and the yz plane

do you see why

[grim knot]

If I understand it correctly A and B have to be open in Y right? So we need a set s.t A= U n Y

[steel glen]

what's Y

[grim knot]

opps miswrote

X

OH WAIT

[steel glen]

this is what an open ball translates to in X

you can see if you center any open ball around the y axis, it will have part of it on the yz plane and part of it on the xy plane

[grim knot]

A is open in X if there exists U in R3 s.t. A = U n X

[steel glen]

yes that's the definition of the subspace topology

[grim knot]

I could not see it because I continuosly thought that X was the whole space

jeez

not R3

[steel glen]

what do you mean

X isnt R^3

[grim knot]

yeah exactly, but I was completely ignornig R3

and the meaning of A open in X with the subspace topology

and was justing thinking that having a plane without a point is an open set

so I get this now, but do you know how I could get an isomorphism from pi_1(X, x0) to G?

[steel glen]

what is G

[grim knot]

Z * Z

the fundamental group, they ask after an explicit isomorphism

[steel glen]

are you able to show that A and B are both deformation retracts of S^1?

[grim knot]

I mean, both are homotopic to {0} x B_eps(1,1) (without){p}

[steel glen]

not sure what you mean

[grim knot]

I know that both deformate to S1

I have that

[steel glen]

so if r is the deformation retract, are you aware that r_* is an isomorphism of fundamental groups?

[grim knot]

yes I agree

Oh okay, wait I didn't do it formally, but the best way is to explicitely find a deformation retract?

and then get the induced isomorphism?

[steel glen]

if they want an explicit isomorphism then i think so

did you compute pi_1(X,x_0)?

[grim knot]

yes is Z * Z

[steel glen]

ok im confused then

what is G

[steel glen]

so they are both equal

[steel glen]

when you say pi_1(X,x_0) is Z*Z, you mean it is isomorphic to it right?

[grim knot]

Yes, it's not equal sorry

and they want the isomorphism

[steel glen]

then yeah you can construct an explicit isomorphism by finding an isomorphism
pi_1(X,x_0) -> pi_1(S^1 v S^1)

[grim knot]

sorry, I din't want to confuse you

[steel glen]

no problem

though i am still a little confused as to what kind of answer they are looking for

[grim knot]

I can translate it:
b) find the isomorphic type of $\pi_1 (X, x_0)$ which means, find an algebraically defined group G, which is isomorphic to $\pi_1 (X, x_0)$

[gentle osprey]

damn_guuurl

[steel glen]

and c asks for an explicit one?

[grim knot]

yes

[steel glen]

ok. i think if you can say pi_1(S^1 v S^1) = <a,b>, then you can use the induced homomorphisms from the deformation retracts to get an explicit isomorphism

im saying to use deformation retracts so you can argue that there's 2 generators

[grim knot]

okay so I have to find the deformationretract

[steel glen]

shouldn't be too bad. what were your final A and B

[grim knot]

homotopic to? S1

both

I see that S1 v S1 is a deforetract of X though

[steel glen]

yeah

writing an explicit deformation retract maybe is not as easy as i thought haha

but i can't see any other way to approach the problem

[grim knot]

this is what the gave(?)

and I'm like, yes sir, got it

[echo berry]

im a bit stuck on showing that the map $g\colon X \to G$ is a homeomorphism given that we have topological spaces X and Y, $X \times Y$ given the product topology, and the subspace $G = {(x,y) \in X\times Y : y = f(x)}$ of $X \times Y$ where $f \colon X \to Y$ is continuous.

[gentle osprey]

Eso

[echo berry]

ive just defined g to send x to (x,f(x)) , but im confused how the open sets in G look

[tough hamlet]

do you know why g is continuous

[echo berry]

no

id of course attempt to figure that out but im not sure about the open sets of G

[tough hamlet]

G just has the subspace topology

so continuity as a map into G is the same as continuity as a map into the product

[echo berry]

sorry what do you mean

[tough hamlet]

the product being X×Y here

[echo berry]

hmmm

do you mean i can use the universal property of the product

[tough hamlet]

yes

[echo berry]

ahh okok

thx

[tough hamlet]

np

[grim knot]

Why is it not possible to flatten S2 to D2?

[tough hamlet]

flatten as in what

[grim knot]

to take the sphere in R3 and make it to the disk D2 in R2

[tough hamlet]

what do you mean by make

find a homeomorphism?

[grim knot]

yeah might be an homeomorphism or an homotopy

I don't know how to explain it, but also continuosly moving it to D2

[tough hamlet]

because the disc is contractible

and the sphere is not

the sphere has nontrivial second homology

[grim knot]

oh, I'm not that far yet in topology

[tough hamlet]

actually you know about fundamental groups right?

[grim knot]

yes exactly

[tough hamlet]

you can do it with those now I think about it

if you remove the middle point from the disc

you obtain something homotopy equivalent to the circle

[grim knot]

yes makes sense

[tough hamlet]

so with fundamental group Z

if you remove any point from the sphere

[grim knot]

then you are contractible

[tough hamlet]

yeah

[grim knot]

so does it have to do with that?

[tough hamlet]

this is one way to show it

you can remove a point on a disk and obtain something with fundamental group Z

but removing any point from the sphere you get something with a trivial fundamental group

so they aren't homeomorphic

[grim knot]

okay that makes sense

[tough hamlet]

(with homology you can do better and show they aren't homotopy equivalent)

[grim knot]

can you tell me how it works, I'm really curious now

or maybe if you know a webpage with that

[tough hamlet]

what homology?

[grim knot]

how to show it with homolgy

[tough hamlet]

well H_2(D^2)=0 because it's contractible

and you can compute H_2(S^2)=Z

and homology much like the fundamental group is homotopy invariant

[grim knot]

Oh well this is really nice

the problem with fundamentalgroups is that it is the same for both right

[tough hamlet]

yeah

they're both simply connected

[grim knot]

this is really beautiful, thank you for explaining it

[tough hamlet]

np

[grim knot]

yeah, this is why I had a struggle for a sec

[unreal stratus]

Here's a funny proof that they can't be homeomorphic: every map D^2 -> D^2 has a fixed point (standard fact), whilst the antipodal map is a map S^2 -> S^2 without fixed points

The nice thing about this is that it only requires knowledge that π1(S^1) is non-trivial

[grim knot]

oh with brower's thm, this is nice!

[nimble portal]

What are interesting examples of manifolds in other areas of math? Algebraic varieties with manifold structure are the only example I know lol

There’s Lie groups

Any structures or spaces which arise in analysis that are naturally manifolds?

[tough hamlet]

uh

[nimble portal]

Plas get out of here stop spoiling yourself nerd

[tough hamlet]

well ig the (n-)torus is a big one

Fourier series

[nimble portal]

FOURIER SERIES?

Torus is like

not what I’m looking for

that sounds so dickish lmao sorry

[tough hamlet]

$(S^1)^n$

[gentle osprey]

Daybroken🥭

[tough hamlet]

uh what else ig there's

[nimble portal]

I mean like it’s just very universal space

Whereas

Varieties and Fourier series

Those are much more unique and interesting examples

How are Fourier series manifolds?

[tough hamlet]

no the torus is

fourier series study the functions on it

[nimble portal]

Oh

Is this because heat equation? LOL

Oh wait

After edit nvm

[tough hamlet]

a 1-periodic smooth function on R is just a smooth function on the circle

[nimble portal]

[tough hamlet]

similarly a Z^n invariant smooth function on R^n is just a smooth function on the torus

can replace smooth with a lot of adjectives

[nimble portal]

So 1-periodic functions on R^n are smooth functions on T^n?

[tough hamlet]

1-periodic is a bit weird in R^n

[nimble portal]

yeah wait what does that even mean

[tough hamlet]

but yeah invariant under Z^n translation

[nimble portal]

LMAO

I’m such a silly goofy dummy!!1!!1

[tough hamlet]

there are examples in PDE as well

like characteristic/noncharacteristic hypersurfaces of an equation

[nimble portal]

AHAH now that’s more like it

I’ll check that out

[oblique heath]

Banach Manifolds

Any infinite dimensional manifold is not inherently geometric in the regular way

[rain ether]

Fréchet manifolds

[tough hamlet]

yeah banach/frechet if you allow that

in particular invertible elements of a banach algebra are a banach lie-group

[oblique heath]

Some ppt I had saved a while back on infinite dim manifolds

https://www.mat.univie.ac.at/~michor/Madrid-2016.pdf

[rain ether]

nice, michor

[tough hamlet]

I had a project that involved solving a PDE by looking at the derivative of Banach-Lie group action

so yeah nonlinear functional analysis in general but again idk if you'd count it

[steel glen]

is there an ambient isotopy that takes T^2 to itself, but that swaps the generators for its fundamental group?

[coarse night]

depends on where you embed. In S³ there is, thik why

[tiny ridge]

There is no such ambient isotopy of the standard torus in R^3 but if you puncture the torus then suddenly there is

[steel glen]

does this amount to turning the torus inside out

[tiny ridge]

It does.

[steel glen]

im trying to justify to myself why the extra point in S^3 lets us do this

[tiny ridge]

Think very hard about the boundary of D^2 x D^2

[coarse night]

if you know Hopf fibration, that might help

[steel glen]

i dont know hopf fibration haha

this just came into my head because a lecture i was watching mentioned torus knots T_{p,q}, and i was wondering if this was the same as T_{q,p}

[tiny ridge]

Yup, good question

[steel glen]

oh wow

i think i can kind of visualize it

sheesh ok yeah that kind of makes sense lol

thank you both

[steel glen]

i just noticed the similiarity of reversing T^2 - pt and what i thought up in S^3

this makes a lot of sense now

[tiny ridge]

It's the same thing. Send a point on the torus to infinity

[steel glen]

yeah

then just bring it back

[coarse night]

ye but the reversing is no longer "one to one" feels like

[tiny ridge]

You can think of it as tugging at a ring

The ring runs along the central circle of the handle

If you pull it it becomes this

The ring is metallic, and it's stuck to a elastic material which just sort of pops to the other side of you pull hard enough

Imagine if there were door handles like this

tugs at the door handle, ends up everting the door

[coarse night]

you get inside the room by turning it inside out

[tiny ridge]

It's more like getting inside the room by redefining the inside as outside and vice versa

In general the following question is interesting. Let S be a surface (not closed, otherwise it's boring) in R^3. Which self-diffeos of S are realised by an ambient isotopy?

There are knotted annuli in R^3 whose boundary components cannot be swapped by an ambient diffeomorphism

[coarse night]

how do you see there isn't any?

[tiny ridge]

I didn't flesh out the details but suppose F is a diffeomorphism of R^3 which does this. F preserves the standard torus T \sub R^3, restrict dF to TR^3|_T.

Mod out by the tangent bundle, dF passes to a fiber preserving diffeomorphism dF : T^2 x I -> T^2 x I of the normal bundle

Restricted to T^2 x {0} it reverses orientation because x, y goes to y, x. This forces the I factor to reverse

Aka F sends inside of the torus to the outside. This is impossible because the exterior and the interior are not diffeomorphic

You can probably now ask me why S^1 x S^1 -> S^1 x S^1, (x, y) -> (-y, x) isn't realised. That preserves orientation so the above argument doesn't seem to apply.

Ok, I mean if a diffeo F : R^3 -> R^3 extends a diffeo of T^2 then F must preserve say the interior. This means F restricts to a diffeo of S^1 x D^2. All such diffeos are isotopic to S^1 x D^2 -> S^1 x D^2, (t, z) -> (t, e^{i k t} z), k integer

None of them have the desired effect.

That is to say, on the boundary it is just a Dehn twist about a meridian. The swap map is not isotopic to a Dehn twist about a meridian

Induce different maps in homology

Even easier, a diffeo of S^1 x D^2 cannot restrict to a swap map on the boundary because the meridian is nullhomotopic in the solid torus whereas it is not so after the swap when it becomes the longitude.

But the above argument actually says there's no nontrivial diffeo of a closed orientable embedded surface in R^3 realised by an ambient diffeomorphism.

[urban zinc]

anyone know if the # of topologies on a set of size n is known?

either up to homeomorphism or not

[gritty widget]

wikipedia says there are 8977053873043 distinct topologies on a set with 10 points

[unreal stratus]

Seems only upper/lower bounds are known given the references I've found

and some congruences lol

[tiny ridge]

There's a OEIS entry

I don't think good formulas are understood

[obtuse meteor]

Congruences???

That feels weird af

New generating function just dropped

[river granite]

yes it's an open problem even for not-so-large n

[grim knot]

guys, if we already did covers and all that stuff in topology, what is it left for us to do in algebraic topology? I'm scared ngl

I also wanted to ask, what do you think are the most important spaces of which I should now how to calculate the fundamental group?
I thought about: S1, Sn, Torus, Möbiusstrip, Klein Bottle, RPn space

[hidden crag]

As in what else there is in the subject algebraic topology?

[grim knot]

jup

[hidden crag]

A lot

Sehr sehr viel even

Coverings and pi_1 scratch the surface

If you’re interested in further topics you could pick up a textbook on it

[grim knot]

oh okay, this scares me now haha

[grim knot]

https://link.springer.com/book/10.1007/978-1-4757-6848-0
I think this is the book qe are gonna use

[hidden crag]

Oh are you taking a second course in it?

[grim knot]

what I asked until now was the mandatory course topology.
I may not be the smartest, but I loved the topics, so I took algebraic topology, which takes place next semester.

(I still don't understand coverings, and my exam takes place on Wednesday and I can't really study them, cause I have probability and statistics tomorrow

[hidden crag]

AT is very cool, right choice

Good luck with your exams!

I have a numerical analysis exam tomorrow

[grim knot]

oh that's cool, I also have one about it next monday on how to solve ODE's with num ana

[hidden crag]

I kinda disagree with the „cool“ part

[grim knot]

HAHAHHAHAHAHAHA yeah well, numeric is not my thing either, but can be appreciated anyway

[tawdry widget]

Gtm 175 lickorish

Second paragraph, what does “algebraic sum of the transverse intersection points” mean? I’ve never seen such expression….

[umbral panther]

A more standard phrase is “algebraic count of intersections,” but it’s not a count, because some intersections are positive and some are negative. It’s the sum of these numbers, all of which are +1 or -1 (without transversality they could be more general)

[tawdry widget]

Oh, is it like counting how many time f pass through F? With one direction of F chosen since F is orientable, f is also oriented, +1 if f pass through F by this direction, -1 if it’s the opposite direction?

[umbral panther]

Yes

[tawdry widget]

Thank you very much, got it☺️

[sullen nimbus]

This is from Lang algebra from the monoids section i have taken an introductory course in topology but i couldn't quite understand it. I would appreciate any help or explanation

[novel acorn]

what exactly is confusing you (tho this is more algebraic topology oriented as this is somewhat related to the classification of closed surfaces)

[sullen nimbus]

The glue part

glue S-D_o to S-D'_o by identifying C with C'

And what does it mean by sth to be independent of homeomorphisms

[ebon galleon]

As it, it's well defined on the homeomorphism classes. Since it's an operation that, a priori, depends on which object in the homeomorphism class we choose

[gaunt linden]

It says independent up to homeomorphism, not "of".

[ebon galleon]

And they're saying that it doesn't actually matter what the choice it

[rapid lagoon]

Hi, a simple question born out of curiosity: how does topos theory generalize the fundemental group?

I heard someone snark that the fundemental group in point-set topology is merely part of the real clique, and that topos theory does it "better"

[thorny agate]

I feel like I'm being dumb with this question

If $G$ is a topological group with identity $e$, $U$ any neighborhood of $e$, and $n$ any positive integer, then there exists a symmetric neighborhood $V$ of $e$ such that $V^{n} \subseteq U$.

[gentle osprey]

Spamakin🎷

[thorny agate]

if n is even then I have an answer. Basically use a map setting (g, h) -> gh^-1

get a preimage of U, call it V1 x V2. Let V = V_1 cap V_2

[tough hamlet]

if n is odd

do it for n+1

V^{n+1} contains V^n

[thorny agate]

oh duh

ok ty

[tough hamlet]

np

[wispy berry]

A global cross section \sigma: G/H \to G of the projection map \pi:G \to G/H is always G-equivariant? By definition?

[tough hamlet]

no

[broken nacelle]

how accurate is this? lol

[wispy berry]

is there a specific category of groups for which this holds?

[tough hamlet]

ok after some thought

any G-equivariant map (from a nonempty set) into G is obviously surjective

so this happens if and only if H is trivial

i.e. there are no interesting examples

[unreal stratus]

i mean it's a nice textbook

not rly much point using much category theory at an earlier level imo right until it becomes more complicated

[broken nacelle]

oh I wasn't looking for that

I was just wandering if it could give math textbook recommendations

[thorny agate]

what does "continuous relations" mean here

[tough hamlet]

both sides are continuous

[thorny agate]

?

both sides of what

[tough hamlet]

the left side is continuous

the equation

[thorny agate]

A A^t is continuous? what does that even mean?

[tough hamlet]

yeah it's a continuous function

[thorny agate]

from what to what

f(A) = A A^t?

[tough hamlet]

yes

[thorny agate]

ah

and then f(A) = I is the other function

[tough hamlet]

yes

[thorny agate]

ty ty

[tough hamlet]

np

[balmy field]

Need some help with conformal mappings. Want to say circle/orientation preserving implies holomorpic, but can’t find an easy proof online

[ebon galleon]

The statement at the end "since $U_i \cap V_i = \emptyset$" is false

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

This does not hold in general

I would suggest proving this statement in two steps:
First, prove that for a closed subset F, and a point x not in F, you can find open neighborhoods that separate them
Second, use this fact to prove that you can separate two closed subsets with neighborhoods, as in the question

[ebon galleon]

In the second paragraph, you want $V_x = \bigcup_{i=1}^n V_{y_i}$, so that it covers it (this union is still disjoint from $U'_x$ though! And intersecting the $U_x$ 's was right).
Other than that, looks good!

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

For compactness proofs along this vein, you'll often want to union one of the neighborhoods, and intersect the other

Also, it can be helpful to think of compactness, not as something within F, but within X. What I mean by this is that you don't really need to worry about writing $F \subseteq \bigcup F \cap U_y$. Rather, just consider $F \subseteq U_y$. The reason this works is because for every subset $V$ of $F$ that's open in $F$, there is always an open set $U$ in $X$ that $V = F \cap U$

[gentle osprey]

Ultrawizard King Ryx

[ebon galleon]

Sorry, that should be the union of U_y's, not just a single U_y lol

[steel glen]

i was given the following information in a lecture im watching

k is a knot, X = X(k) the knot exterior, G = pi_1(k)
since H_1(X) = Z, there is a surjective map phi : G -> Z such that ker(phi) = [G,G]
[G,G] is a normal subgroup of G, so the covering space \tilde X corresponding to [G,G] is a normal covering space
the lecturer then mentions that the deck transformation group is Z
how do we know the deck transformation group is Z?

[languid patrol]

Because G/[G, G] = Z by the existence of phi

[steel glen]

so if a normal covering space corresponds to a normal subgroup H of G, the transformation group is G/H?

[languid patrol]

Yep, that’s part of covering space theory

[steel glen]

alright thank you!

[languid patrol]

In general if H isn’t normal the deck transformations are the normalizer of H modulo H

[abstract saffron]

yay, another knot theorist 😄

[red yoke]

Why are the two isomorphisms isomorphisms? If I'm interpreting this correctly, the first isomorphism restricts to H^n(∪X) ≈ ∏H^n(X) on the factors, but the H^n(∪X)'s generate H*(∪X) while the ∏H^n(X)'s do not generate ∏H*(X). The same question for the second isomorphism.

Say, an element φ0×φ1×… ∈ ∏H*(X) where 0 ≠ φi ∈ H_i(Xi).

[languid patrol]

They are obviously isomorphisms of groups right? So the only thing to check is that the intersections of cycles from different spaces in the wedge sum is trivial in reduced homology and that there are no nontrivial intersections in the disjoint Union

[red yoke]

I don't see how it is even surjective

[languid patrol]

Which map?

[red yoke]

The one on the first line

[languid patrol]

Are you satisfied if the index set is finite?

[red yoke]

Yes

I'm also satisfied about Hn instead of H* for infinite index sets

[languid patrol]

Meaning Homology vs cohomology or?

[red yoke]

Meaning H^n(∪X) ≈ ∏H^n(X)

[languid patrol]

That is a very confusing statement

Do you mean H^n on both sides or H_n on both sides?

[red yoke]

Oh wait they're both cohomologies

But the question still stands

[languid patrol]

Yes okay then I’m very confused. You understand why this is true in each degree but not why it is a ring homomorphism yes?

[red yoke]

I understand it for Hn but not for H*

[languid patrol]

H^* is just the direct sum of all the H^n as a group

[red yoke]

Like why is this in the image

[languid patrol]

That is not in H^*

H^* is \oplus_n H^n

[red yoke]

I mean in the image of the isomorphism in the first line

Say a wedge sum of n-spheres for n ranging over N, and an element whose n-th coordinate is a generator of H^n(n-sphere).

[languid patrol]

I guess this just depends on how Hatcher defines H^* when the cohomology of the space is infinite

I.e exists in infinitely many degrees

[red yoke]

How is it usually defined?

[languid patrol]

I don’t remember any particular convention.

[red yoke]

But does this make sense

[languid patrol]

Yes if H^* is the direct sum of the H^n then this isomorphism cannot possibly hold for the reason you mentioned

[red yoke]

Then replacing ∏H* with the direct sum of ∏Hn should fix the problem ig?

[languid patrol]

Yes that seems to be the right solution

All a bit ugly though.

It all looks a bit nicer on the level of homology but then you don’t have this remark about the ring structure

[grim knot]

Can somebody tell me how to find coverings and covering spaces? Cause I get it, but no idea on how you come up with them. Do you just dream about them?

[novel acorn]

there are two ways
There's the algebraic way and then there's the topological way

[grim knot]

Could you state both?

[novel acorn]

now idk if you've learned this but covering spaces correspond to subgroups of the fundamental group
So you could generate covering spaces if you know their fundamental group

[grim knot]

How would it be done then?

[novel acorn]

so let's say you have a CW complex
Now the topological way to go about it is to notice that an n-sheeted covering space has n times all the k-dim parts of a CW complex
So you can draw like all the 0-cells of the covering space and connect them in a way such that they locally look like the original space

[grim knot]

Like if I have S1, does it mean that all possible covering spaces are of order nZ?

[novel acorn]

Here's the exact theorem

[grim knot]

so do I understand it correctly, that $p_*(\pi_1(\tilde{X}, \tilde{x}_0))$ is always a subgroup of $\pi_1(X,x_0)$

[gentle osprey]

damn_guuurl

[feral copper]

Yes, this p_* gives the 1:1 correspondance

Because an equivalence of coverings descends to a conjugacy of the associated subgroups

[grim knot]

does it also mean that for any subgroup we can find a covering?

[feral copper]

Yes

And if it has finite index, then the index is the number of sheets

Fun exercise: try to draw all coverings of the torus corresponding to all possible index two subgroups

(step0: how many 2-fold coverings are there, using this correspondance?)

Also, there is an important word in the theorem: path-connected coverings

[grim knot]

isn't just one up to groupisomorphism?

[feral copper]

It's conjugacy classes of subgroups, not mere isomorphisms

Recall that here the whole group is abelian 😉

This exercise is good to practise this theorem!

[grim knot]

Why are these exercises so hard

[feral copper]

Do you want me to explain it a little? 🙂

[grim knot]

It's just that I don't get it really fast, I don't see it as fast as I should

Like can we cover The torus with S1x S1?

[feral copper]

I'll ask a simpler question first. Do you know a double cover of the circle S(1)?

[grim knot]

I'd say S1->S1 t-> t^2

[feral copper]

Yes!

Now, if X->W is a p-sheeted covering, and Z->Y is a q-sheeted covering, can you construct a covering of WxY in an obvious way? What is the number of sheets?

[grim knot]

It should be the cross product of both or not(?)

[feral copper]

Yes it is. If I let f:X->W and g:Z->Y denote the two coverings, then you may define h:XxZ->WxY by h(a,b)=(f(a),g(b)).

I'll let you check the details that it gives a covering space.

[gentle osprey]

Matplotlib

[feral copper]

(I'll let you try to fill the details in.)
Now, back to the link with the 'classification theorem'. How many connected coverings are there of the torus? I claim there are 3. Because there are exactly three subgroups of index 2 of Z², and they are not conjugate (because Z² is abelian).
Check that these two coverings are corresponding to the subgroups Zx{0} and {0}xZ (I may have swapped the order, you do the computations!).
There is a remaining subgroup: {(x,y)∈Z² | x+y≡0 (mod 2)}. This one is a bit more tricky to draw.
To find it, use the 'square with opposite sides identified' representation of the torus. Here is a hint:

Try to represent all three coverings by this 'square' method; you should have two units of area on the left, and one on the right.

[unreal stratus]

Given a (not-necessarily-smooth) manifold M, what is the orientation line bundle on M?

I've seen a construction for smooth manifolds in e.g. Bott-Tu but it's defined completely in smooth terms

(Or perhaps the source I'm reading isn't actually working in the topological category., though that seems implied ..)

[umbral panther]

It’s probably better to talk about the orientation double cover, rather than the orientation line. But line bundles are the same as double covers

If you can define the degree of a local homeomorphism between connected open subsets if R^n, you can define the orientation double cover. If you have a bunch of charts for your manifold, ie, embedding if R^n into it, you can build a new manifold with similar charts, but only identify them if the transition function between the charts is orientation preserving (with respect to the orientation on R^n)

[unreal stratus]

Okay I know about the orientation double cover - how does that relate to the line bundle?

In this case it's cause I'm reading a paper where they talk about just taking the orientation line on a manifold and doing stff with that

But perhaps I can also (for concreteness) assume they are talking about smooth manifolds, since the main applications are towards them lol

[umbral panther]

Double covers are principal bundles with structure group Z/2. Line bundles are equivalent to principal bundles with structure group GL_1(R), which is homotopy equivalent to Z/2. Choose a metric to get a bundle with structure group O(1), which is Z/2

[unreal stratus]

Oh sure, thanks

[frozen pebble]

hello
is anyone free to help me with some problems in general topology?

prove that a normal space X can be functionally separated.
problems like this

[bitter smelt]

https://dontasktoask.com

[outer harness]

I’m a bit confused about the definition of a boundary. Munkres defines it as $\partial A = \overline{A} \cap \overline{X\setminus A}$. But doesn’t this mean the circle $S^1 \subseteq \mathbb{R}^2$ has a boundary? I thought it didn’t.

[red yoke]

The circle is its own boundary

At least for the topological boundary ∂A = cl(A) \ Int(A)

[outer harness]

When does something have no boundary then?

Oh wait yeah I put the definition wrong

[gentle osprey]

LeftCosetOfANormalSubgroup

[red yoke]

Any clopen set has no boundary for example

Actually that seems to be iff

[outer harness]

I see. Why does it say on here https://www.quantamagazine.org/how-mathematicians-use-homology-to-make-sense-of-topology-20210511/ for example though that the triangle has no boundary

[red yoke]

That's another definition of boundary

[outer harness]

Huh. What’s that definition?

[red yoke]

First you define the boundary of an n-simplex to be a (alternating) sum of its "faces"

Eg the boundary of a triangle is a sum of its edges

[outer harness]

wdym by sum

[coarse night]

formal sum

[red yoke]

Then you define the boundary of a map from an n-simplex to a space X to be the sum of the maps from its faces to X

[red yoke]

I.e. it's just symbolic

I.e. Free Abelian group with basis being the simplices

[coarse night]

you can make it precise by looking at free ab group generated by the simplices

[outer harness]

alr that’s too advanced I hope I’ll get there soon

why are there multiple definitions of boundary tho lol

[red yoke]

The same reason we have multiple definitions of closed

[outer harness]

wait what

like in a topological space?

[red yoke]

Which can mean a subspace being topologically closed, or a curve having its starting point being the same as its endpoint, or a manifold having no boundary

It's just cuz we have too few words

[coarse night]

nothing adv or fancy, think of simplices as paper cuts then the su becomes some stack of papers,

[outer harness]

ok so how would one distinguish in words between these two definitions of boundary

One is the boundary of a subset of a topological space and the other is ?

[red yoke]

It's usually clear from context

The other is the boundary of a (singular) simplex

Or a homomorphism

[outer harness]

I see. thanks!

[grim knot]

In this case I'd say again that it is a 2-sheeted cover, I might be wrong though

[grim knot]

this is really a nice exercises, thank you so much for going into the details

I think it is really hard to somehow identify these covering maps at the first glance

[grim knot]

are finite set closed with respect to the euclidean topology? I'd say yrs, but I'm not sure

[feral copper]

My pleasure! It's an exercise I haven't seen anywhere, and yet it's an important one

Try doing the same with 3-fold covers now!

Or do 2-sheeted covers of the 3-torus T³

[languid patrol]

I think it might be easier to visualize covers of graphs

like to classify all degree <= 5 covers of the figure 8 is a nice exercise and kind of shows you the gist of how you might come up with a covering space

[feral copper]

Ah correct, I have completely forgotten that not everything is a manifold 😛 #based

[languid patrol]

Just looking at a shape and finding it's covers will be impossible in practice if you have a big CW complex

But for graphs it's easy and it will give you the intuition for what the geometry of covering spaces looks like

Another source of covers is certain bundles over manifolds M

but that's another story

[feral copper]

I'm going to make a talk to survey Topology (in a broad sense) next Fall, mostly to PDE and Probability students. I'm looking for 'puzzles' and 'tricks' that have a topological interpretation behind; does anybody think of anything?
An example I have in mind is the following puzzle (see image attached). It was shown impossible to solved; the idea is that the rope is a non-trivial loop in the fundamental group of the complement of the metal thingy.

This puzzle is fun because it helps me introduce the concept of a fundamental group, and I'm going to talk about 'counting holes' and all that jazz

[languid patrol]

Picture hanging puzzles?

[hidden crag]

How about the

One we talked about a while ago

Yeah that one topos

[feral copper]

Ah yes that can do indeed!

[languid patrol]

I.e. how to hang a picture on a wall so that removing one nail drops the picture

[hidden crag]

Yeah we talked about the generalization as well right

[feral copper]

We discussed the k choose n nails version a while ago 😛

[hidden crag]

Iirc

[languid patrol]

That too! But it might be a lot for a little talk

[feral copper]

Yes that's a good idea (I'd stick the 1 in 2 version)

[languid patrol]

2 in 3 was also nice iirc

[feral copper]

I'd have liked something that involves tricolorability of links too; like showing something is not the unlink by counting the number of tri-colorings

(Because I'd like to talk about knots a tiny bit)

Anyway, I think that puzzle and hanging pictures is going to be enough if I want to introduce the objects! Thanks, I didn't think of it

[languid patrol]

Another good one is 3 utilities problem on paper vs on torus

[abstract saffron]

trefoil is the classical example I think

[feral copper]

Yeah I thought about graph genus, but besides saying 'hey you can draw it on a torus' it doesn't tell much about Topology as a whole 🙂

[abstract saffron]

maybe the pearl division problem?

[languid patrol]

I think the proof using the euler characteristic of the triangulation of the plane by the graph is pretty thematic in topology

[abstract saffron]

that one is a cool application of Borsuk-Ulam theorem

but anyone watching 3b1b would know that one already

[feral copper]

Yes I like this

[abstract saffron]

Another one coming to mind is the game Hex. It turns out that its determinacy is equivalent to Brouwer fixed-point theorem

[feral copper]

But I'm based, the \chi is my favourite invariant

Thanks for your insight folks, I think I have more than enough now actually! 😄

[abstract saffron]

Ohhhh, how about the rent division problem?

[feral copper]

What's this one?

[abstract saffron]

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

it's solved with Sperner's lemma, which is also equivalent to Brouwer fixed-point theorem

I guess it's topology-ish in a discrete sense

[feral copper]

Funny how so many things are using fixed-point theorems!

(which makes perfect sense)

[grim knot]

correct me if I'm wrong, is it enough for A to be a deformation retract, s.t. the above function is an isomorphism?

(induced function on the fundamentalgroup)

[feral copper]

If X is homotopy equivalent to A, then this map is an isomorphism. A deformation retraction is a special kind of homotopy equivalence

[unreal stratus]

Well more specifically if rho is a (pointed) homotopy equivalence

[grim knot]

thank you guys!!

[haughty yew]

what are the prerequisites to read hatcher’s algebraic topology?

[unreal stratus]

Patience

[rough cedar]

I think I'm going insane

{ 0 } is clopen in Z_p?

I know that it's closed

I'm not sure about open

[ebon galleon]

Is its complement closed?

Equivalently, is there a sequence of nonzero elements that converge to zero?

[novel acorn]

lol the profinite topology is always fucked

[rough cedar]

do spheres of radius 0 exist

[ebon galleon]

No

Well, we don't really care about them at least

[novel acorn]

if it's open then Z_p is discrete

[ebon galleon]

Open sphere of radius 0 is empty, closed sphere of radius 0 is single point (in a metric space)

[novel acorn]

they're both a single point?

[ebon galleon]

No

[novel acorn]

oh open sphere

okay that's true

[ebon galleon]

Nothing has distance < 0 🙃

Unless you are working with some degenerate version of metric spaces

[novel acorn]

yeah I was thinking you were talking abt S^-1 and B^0

[ebon galleon]

S^{-1} is empty

[novel acorn]

fuck okay today is not my day it seems