point-set-topology

[chrome ridge]

?

[umbral panther]

What do you mean by linking number?

[chrome ridge]

https://en.m.wikipedia.org/wiki/Linking_number

[umbral panther]

I only know how to use cup product to define a different linking number, that of torsion classes in a closed manifold

[abstract saffron]

maybe check Gauge Fields, Knots, and Gravity by Baez and Muniain

there's a section about physical interpretation of some knot invariants. I don't recall if there's anything useful there, but it might be worth a look

[stone timber]

Why is there no concept of “conservative tensor field” coming from “vector potential”?

[lime sable]

https://en.wikipedia.org/wiki/Solenoidal_vector_field is this what you want

[distant lichen]

I don't believe this is what they're asking, I think they're wondering if there's some sort of generalization of a conservative vector field to larger order tensors

[brave escarp]

Is the map $i_*$ allways injective ?

[gentle osprey]

Gibzen

[novel acorn]

no

[feral copper]

Of course no; take the inclusion of the hollow torus into the filled torus. Then, on the level of the H_1, the meridians of the torus are killed in the solid torus! That means: there is kernel

Or even simpler: the inclusion of the circle inside the disc

For higher degree homology: the inclusion of the n-sphere inside the (n+1)-ball gives kernel on degree n

[brave escarp]

thank you !

[umbral panther]

i_* is injective if and only if delta is zero. If it were always injective, delta would not have a special name

[heady skiff]

how does this make sense? what if $x = t_0a_0 + t_1a_1 + \dots + t_{j - 1}a_{j - 1} + t_{j + 1}a_{j + 1} + \dots + t_na_n$ so that $x \in \text{Bd } \sigma$? and then we can use the disjointness of $\text{Bd } \sigma$ and $\text{Int } \sigma$ to show that $x$ cannot be in any face contained in $\text{Int } \sigma$?

[gentle osprey]

okeyokay

[heady skiff]

anyone?

[knotty vine]

I dont understand, Int sigma contains no faces at all

@heady skiff Assuming all t_i > 0 (except for t_j of course), then x is in the interior of the face s spanned by all a_i except a_j

[heady skiff]

how though, if the boundary of sigma consists of all points x of sigma such that at least one of the barycentric coordinates is zero

because here tj is zero

so by their assertion shouldn't it be in the Boundary of sigma

or is it in the intersection of the boundary of sigma and the face s spanned by all a_i except a_j

[knotty vine]

ya

By definition, the boundary is the union of proper faces

[heady skiff]

hm i guess i got lost then

alr that helps, thanks

[heady skiff]

did i miss something in point-set? if Y is a subspace of X, how does showing that C closed in Y iff C closed in X show that Y is a closed subspace of X?

and by closed subspace, i'm assuming that if we consider Y as a subset of X, it's closed

[ebon galleon]

Consider: Y is closed in Y

[heady skiff]

oh bruh

nice

do we even need the condition that each sigma is a subspace of K? i thought if you have any continuous mapping f: X --> Y and say A is a subset of X, then the restriction of f to A is automatically continuous

• we're working in R^n right

[knotty vine]

subspace / subset all the same thing

[heady skiff]

lol ok i'm being picky

[knotty vine]

actually, subspace is correct, cause just a subset doesnt necessarily imply the subspace topology

[heady skiff]

yeah i'm getting mixed up between the topologies lol

[chrome ridge]

How to show that they have different cup product structures? Not sure how to compute the cup product here.

[feral copper]

I guess this can help: Linking number and cup product

[stone timber]

Exactly what I am wondering. There is no formal mention of it in the literature

[chrome ridge]

This assumes alexander duality; not introduced in the book yet

[cedar pebble]

in this case you can get away with noting that one of these deform retracts to S^1xS^1 while the other deform retracts to a wedge of S^1's and an S^2; they have the same cohomology groups, but the cup products are obviously different

but in general you would use Alexander duality to detect linking numbers in terms of the cup product like this

this is also how things like Massey triple products and things like this come to be

[chrome ridge]

How would one do the same deformation retraction trick with Boroomean rings compared to the 3-comp. unlink? Not sure how to visualizes this.

[cedar pebble]

I mean if you wanted to do more complicated examples like this you would probably just use Alexander duality instead of doing explicit deform retracts like this

[heady skiff]

i don't really understand how this implies that t_v is continuous - didn't they just reiterate how t_v is defined 💀

moreover, i'm trying to see it in the open-set continuous definition but can't

[heady skiff]

anyone?

[knotty vine]

Whats Lemma 2.3?

Dont use the open set definition of continuous, it's just a piecewise linear function

[heady skiff]

just the glueing lemma

i.e. if it's continuous on each of the simplices then it's continuous on the entire complex

[knotty vine]

Clearly the constant zero function is continuous, right?

And the barycentric coordinates are linear in x

Linear functions are continuous

[heady skiff]

oh

is that how piecewise continuity works

💀

[knotty vine]

idk what piecewise continuity is

But a piecewise defined function is continuous if it is continuous on the pieces and continuous at the boundaries of those pieces

That gluing lemma is a generalization of that

[silver umbra]

does anyone here have some familiarity with Kleinian groups

[normal herald]

What’s the motivation for point set topology? What is the basic intuition? I reads this;
Topology is the art of reasoning about imprecise measurements, in a sense I'll try to make precise.

In a perfect world you could imagine rulers that measure lengths exactly. If you wanted to prove that an object had a length of l you could grab your ruler marked l, hold it up next to the object, and demonstrate that they are the same length.

In an imperfect world however you have rulers with tolerance. Associated to any ruler is a set U with the property that if your length l lies in U, the ruler can tell you it does. Call such a ruler RU.

Given two rulers RU and RV you can easily prove a length lies in U∪V. You just hold both rulers up to the length and the length is in U∪V if one or the other ruler shows a positive match. You can think of RU∪V as being a kind of virtual ruler.

Similarly you can easily prove that a point lies in U∩V using two rulers.

If you have an infinite family of rulers, RUi, then you can also prove that a length lies in ⋃iUi. The length must lie in one of the Ui and you simply exhibit the ruler RUi matching for the appropriate i.

But you can't always do the same for ⋂iUi. To do so might require an infinitely long proof showing that all of the RUi match your length.

A topology is a (generalised) set of rulers that fits this description.

[gaunt linden]

That doesn't really sound like a mainstream intuition, and I'm not sure how useful it is.
My intuitive idea of imprecision in a ruler definitely isn't that it's something that will give a crisp yes/no answer to every "does this object fall into is such-and-such range of lengths" question -- but that's how it needs to behave if we want to use it as an analogy for "open set".

[fringe cypress]

I think the closest to a concensus definition I've found is only simple discontinuities + only finitely many discontinuities. Tao in his first analysis book defines it like that, Paul Lamar in his notes does the same, and Ross in Elementary Analysis defines it as taking a finite partition P = {t_i} and knowing that the function is uniformly continuous in the open intervals between the points of P, though I found a MSE answer saying its that the function only has simple discontinuities and the set of discontinuities has only isolated points

[fringe cypress]

A friend of mine wrote a neat (free) substack article on how neighbourhood spaces might be a better intro to general topology that's also equivalent to the usual presentation https://mathematicallyforward.substack.com/p/what-do-topological-spaces-actually?sd=pf

Maybe those are more intuitive

Though maybe your question is why we use the standard presentation at all/how that came about

[ebon galleon]

Intuition for the ideas maybe but topologies in terms of open sets is much nicer to work with than the neighborhood definition

[fringe cypress]

Ye, that's the conclusion of the article, basically. The open sets definition is much easier for proofs

[limber wren]

You'll get different answers, my perspective is that point set topology provides a very general framework (general in that it's purely set theoretic), for talking about continuous change. Normally you think of functions as changing continuously as meaning you can make the change in outputs arbitrarily "small" by controlling how small your change in inputs are. This requires a notion of when points are considered close together, so some people may describe topology as a generalization of describing when points are close to one other, or "in the same neighborhood", or "close enough together to meet some threshold". The problem is that in general topology doesn't always describe "closeness" the way we think of it intuitively. But you can study metric topologies, and in those situations the topological idea of "closeness" better matches our intuition

[ebon galleon]

Plus, the part which really makes that definition into a topology and not a weak notion like a pre or pseudotopology is v). Which is very unintuitive imo

Like 1-4 are clear. But 5 is awkward

[fringe cypress]

To add to this, I remember reading there was some controversy on whether we should assume topological spaces are Hausdorff because the non-hausdorff spaces generalise too much from the intuitive image of closeness. Idk if there's really controversy but that's an example where stuff is too general

[tribal palm]

in topology this notion of “two points are close if they are in the same neighbourhoods” does not quite match up to your everyday intuition of closeness… for example it is not symmetric, in the sierpinski top {ø, {a}, {a,b}}, b is by the above notion close to a (and in fact inseparable from a), while a is at the same time also isolated from b— so b is “close” to a, but a is not “close” to b

[gaunt linden]

It seems to be the moral equivalent of "actually forget about all that, we need sufficiently many of the neighborhoods to be open, so openness needs to be part of the definition".

[ebon galleon]

Basically yeah, it asserts every neighborhood contains an open neighborhood

[fringe cypress]

I have an urge to disagree that it's awkward — certainly less clear than the others — but I probably have that urge because my main exposure to topology was metric spaces and I can imagine shrinking a blob/ball I'd view 5 as like generalising a property of euclidean spaces that you'd notice if you work with them enough. Maybe it's much more awkward if I didn't have that initial exposure

[ebon galleon]

Or in english, "Every neighborhood of x contains a smaller neighborhood which is a neighborhood of each of its points"

[ebon galleon]

I have an urge to disagree with your disagreement... That axiom is precisely why this definition is so awkward to work with, and it's why you don't commonly see it nowadays in textbooks unless you are working with more general notions of what a "space" is.

[limber wren]

it's not even so much saying points are close to one another, it's more like saying an open neighborhood around a point is a "closeness threshold". Like X should be a closeness threshold (possibly the weakest), the union of closeness thresholds should be a closeness threshold (possible a weaker one, since it's larger), because continuity requires saying "I want f(x) to be within this threshold of closeness of f(y), and I can do so by making values of x withing "some" closeness threshold of y"

but with such a general definition, you get terrible topologies lol!

where you don't have many useful neighborhoods to have interesting continuous functions

[tribal palm]

i recall seeing that brown’s textbook has nice intuitive explanations for all of these… but i don’t remember his explanation for 5

[knotty vine]

Nice survey!

[ebon galleon]

Like 1-4 asserts that you have a neighborhood filter assigned to each point of X, which contains x in its intersection (i.e. is contained in the principal ultrafilter at x). Okay. That is workable, and turns out to be equivalent to a notion called a "pretopology" or a "Cech closure operation". The latter is a pretty nice definition, and has a fairly intuitive interpretation of what it means for a closure to be topological (which I would say is honestly a pretty good alternative to the usual. Exactly: It just means the closure is idempotent, or in other words it takes sets to closed sets).

[tribal palm]

you’ve aint seen nothing in characterizations of top spaces until you’ve seen the one terms of the closure operation

[limber wren]

nice looking book!

[ebon galleon]

And that is a weakening of Kuratowski's closure axioms (when you add idempotency, you get that)

[fringe cypress]

How much experience do you have working with neighbourhood spaces off the top? I haven't and I'm mainly taking my friend's word that they're more annoying to work with + I tend to use the open set properties more often

[tribal palm]

hm?

[ebon galleon]

See my comment about cech closure spaces

[tribal palm]

ohh pretty neat

[knotty vine]

Closure operations are nice cause theyre like monads and i like those

[ebon galleon]

I have some experience working with them, but I generally avoid it. I've worked with generalizations of topologies which take alternative forms as their basis (closure operations, neighborhood filters, limits of filters), and the neighborhood filter definition is annoying to work with (limits of filters is also pretty annoying for some things), albeit unavoidable for some things

[fringe cypress]

Wow, when are they unavoidable?

[knotty vine]

Im guessing in the world of pretopologies

[ebon galleon]

Well, although I like cech closure spaces, they are a bit of a nightmare when you want to describe something like a product of closures lol (more generally, limits in Cl)

[ebon galleon]

Because for that, you need to go into the world of pretopologies to describe it, since there's no simple direct formula for products in terms of the closure

[tribal palm]

i have glanced at this exercise in munkres a couple of times, before promptly deciding to do other things, it goes something like “show you can define 14 different unary operations solely by composing the closure and complement operations”

[ebon galleon]

I've ranted about the various ways you can define what a topology is a few times now lol

[ebon galleon]

(also due to Kuratowski)

[tribal palm]

ofc

[tribal palm]

it’s fascinating though

[ebon galleon]

(altho pretopologies actually aren't too bad since you drop axiom 5. Still way better than trying to describe them in terms of limits of filters lmao)

[violet summit]

Any ideas on how to construct the set W in the hint?

[tribal palm]

god i spend way too much time idly flipping through books i don’t understand half of… i’ve seen filters mentioned in Bourbaki but i forgor what they are precisely

oh well it’s 4 am here, i should get out of bed

[fringe cypress]

You obtain it from the fact p is continuous

[violet summit]

I feel like I need to use Y compact as well, since i'm not sure how to show the preimage is contained in U

[fringe cypress]

An upward-closed join- (or lower-)semilattice of P(X) containing X but not containing ∅

[tribal palm]

hold my beer while i run to the uni library for a book on order theory to chech what is a lattice again

it’s just some 40 min away

[fringe cypress]

The wiki for filters generalises it to posets and the filter on a set is a particular instance on P(X) ordered by inclusion

[tribal palm]

actually order theory sounds kinda fun, i think i will see what they have

[fringe cypress]

Yaay

Another convert

[ebon galleon]

Consider: Since U is open, X - U is closed. And since U contains p^{-1}(y), ||y is not in p(X-U), which is closed since p is closed. So Y - p(X-U) is a neighborhood of y, with its preimage contained in U||

[violet summit]

Oh so I needed the closed fact too

Thanks

[ebon galleon]

Well yes, you'll need to use all the assumptions you are given generally

[violet summit]

Yeah, just didn't expect it to show up there

[heady skiff]

dude what the fuck does this even mean

my book on the real projective plane: "this is formed by taking the Mobius strip and a disc and sewing their boundaries together"

????

how is that in any way rigorous

does anybody know of any good sources that have constructions that actually have some math in them

in particular of the real projective plane

[tribal palm]

i haven’t gotten to them yet, but isn’t that just a quotient space?

[heady skiff]

wdym

[ebon galleon]

4 am

[tribal palm]

oh wait nvm the möbius strip and a disc

ye lol

[heady skiff]

hold up

get out of bed?

you david goggins or smt

[tribal palm]

no idea who that is

[tribal palm]

(i didn’t see the “and a disc” part…)

[tribal palm]

i have fallen into a biphasic sleep cycle again; i sleep in the early night and then again around noon… vastly impractical

[knotty vine]

Have you tried making this rigorous yourself? You know what a mobius strip is and that its boundary is a circle

You also know that the boundary of a disk is a circle

Now glue the two together along that circle

[fringe cypress]

Section 22 of Munkres's Topology book covers quotient topology, which is the formalism for glueing stuff together. For projective plane idk

[tribal palm]

or the disjoint union perhaps

[knotty vine]

With some parts identified

Which is called a pushout

[tribal palm]

wait does that make sense

isn’t the whole point of disjoint unions that there is no overlap

then wouldn’t the approriate formalism rather some quotient of the product of the strip and the disc

[knotty vine]

You take the disjoint union and then quotient

[fringe cypress]

You can always basically take the disjoint union by doing 1×X ∪ 1'×Y where 1 and 1' are different singletons. Usually, 1×X is basically the same as X

[tribal palm]

does the product also work?

[knotty vine]

No

I mean maybe with a bunch of extra work, but generally no

[ebon galleon]

trying to mix limits and colimits

[tribal palm]

right, thanks, i’m just tripping, probably from lack of sleep and nutrition

living that sweet life

i should see a doctor

[ebon galleon]

I think

you should eat and then do eep

[knotty vine]

NO! First see a doctor! But before that, worry about what the doctor will say!

Then worry if your doctor is actually qualified!

Are they actually a doctor?

Wait, did you actually have an appointment?

[tribal palm]

[heady skiff]

what exactly does c) say precisely? under what quotient map are we giving the union the quotient topology?

something like injecting each simplex into the complex?

bro what does it mean to "take them separately" 😂

[knotty vine]

For each simplex s in K, take the union of |s|

[heady skiff]

so we're considering $\bigcup_
{\alpha \in A} |s_\alpha|$?

[ebon galleon]

Let consider the sum of spaces $\coprod_{k \in K} |k|$ of the simplices $k$. The obvious map $\coprod_{k \in K} |k| \to |K|$ is a quotient map.

[gentle osprey]

Ryx Theory E-Boy

okeyokay

[heady skiff]

oh ok

[trail charm]

i'm stuck on figuring out how they got H^1 and H^2

i'm also not really sure where they got the 0 in the complex

anyways, for H^1, im(Z -> Z^3) is isomorphic to Z since that map is injective, and so i guess it reduces to solving that ker(Z^3 -> Z^2) is isomorphic to Z^3, then H^1 = Z^3 / Z = Z^2

[lime sable]

it's not injective, it's 0, which is why H^0 = Z

[trail charm]

oh oops yeah true

i just got thrown off by the thing looking like a short exact sequence

ok so then need to show ker(Z^3 -> Z^2) isomorphic to Z^2

take $\begin{pmatrix} x \ y \ x \end{pmatrix} \in \bZ^3$. then we have
[ \begin{pmatrix} 1 & 1 & -1\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x + y - z \ x + y - z \end{pmatrix} = 0 ]

[gentle osprey]

ana(functor)mono(morphism)

[trail charm]

which implies that x + y - z = 0

but im not sure why this implies that the kernel is Z^2

[warm quiver]

what is the rank of the matrix

[trail charm]

1

[knotty vine]

then youre done!

[warm quiver]

the domain has rank 3 so what does rank-nullity tell you

[trail charm]

yeah so rank nullity gives that it has nullity 2

but then i get an issue with H^2

because then that means that ker(Z^3 -> 0) is isomorphic to Z^2

oh wait

shit notational error

lol

i should've had ker(Z^2 -> 0)

oops

[warm quiver]

yeah, so another rank argument gives u H^2

[trail charm]

yea

i just had 3 and 2 in the wrong spot

ty

[warm quiver]

oof yeah that’ll make a difference

[fringe cypress]

If $K\subseteq \bR^n$ is compact and $E\subseteq K$ has only isolated points, does it follow that $\operatorname{cl}E$ is countable?

[gentle osprey]

BlaKaligula

[fringe cypress]

No, also I hate this https://en.m.wikipedia.org/wiki/Isolated_point#Two_counter-intuitive_examples

Very strange

With the Cantor set example, you can modify it to work with the middle fourths Cantor set and get a closure with positive measure from a set of isolated points

[high hill]

https://youtu.be/lwWeRMmXIoU?si=-uLfExCR72IJsLQu
how does one convince themself tubing doesnt affect unknotting

oh so the tube is allowed to intersect itself

hmm...

[knotty vine]

wdym? I dont think it has to. Certainly it cant intersect the knot

Actually no, it cant intersect itself otherwise you wont preserve knot B

[trail charm]

where does $\bZ_2$ come from? the rank of $\begin{pmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \end{pmatrix}$ is 2 and $\ker(\bZ^2 \to 0) = \bZ^2$, so wouldn't $H^2(\bR P^2, \bZ) = \bZ^2 / \bZ^2 \cong 0$?

[gentle osprey]

ana(functor)mono(morphism)

[high hill]

Then im stuck with my original block

[steel glen]

are you ok with the fact that deformation retracts are homotopy equivalences?

[high hill]

alright...

[steel glen]

are you saying you are good now, or do you want me to keep going

[high hill]

keep going

i dont follow at this point

[brittle rapids]

Z^2 is a submodule of Z^2

proper

[trail charm]

how is Z^2 a proper submodule of itself

[steel glen]

we have a deformation retract from the tubed knot to the original knot.
perhaps more importantly, you have a deformation retract from X - K to X- T(K), where X is the ambient space, K is the knot, and T(K) the tubed knot

[high hill]

but u also have a retract to the null knot...?

[steel glen]

being an unknot requires X - T(K) to be homotopy equivalent to X - T(K_0)

[steel glen]

a retract is not a homotopy equivalence

[knotty vine]

I havent read the original question, but 2Z + Z is a submodule of Z + Z for example

[steel glen]

unless you were referring to a deformation retract shuri

[high hill]

yes..

[steel glen]

then no, not necessarily

why should we have a deformation retract from an arbitrary T(K) to K_0

[high hill]

if u can swallow A, u can swallow B also

[steel glen]

i don’t follow

[brittle rapids]

the Z^3 -> Z^2 map has image spanned by (-1, 1) and (1, 1)

[high hill]

Theres too much jargon for me to continue down this explanation

[steel glen]

i have used no jargon

[high hill]

topology

[steel glen]

are you coming into this from just the conway video

if so then yeah sorry i did use too much jargon

[high hill]

Yes. Id have to spend a while revising to understand

[steel glen]

i have to go, but if you want me to flesh this out later when i get a chance let me know

[trail charm]

true but i dont see how we have a similar thing here tbh

[trail charm]

right

[brittle rapids]

which is proper

think about the lattice generated

[trail charm]

i have no idea what a lattice is

[brittle rapids]

the grid in Z^2

[high hill]

The way I see it. If "swallow A" is a def retract, then so is "swallow B". And both.

[brittle rapids]

ok there's another way to see it ig

the coordinates add up to 0 mod 2 for both generators

so it'll stay that way for all linear combinations

[trail charm]

oh gotcha

[brittle rapids]

the other coset will be the subset of Z^2 whose coordinates add up to 1 mod 2

so you have Z_2

[trail charm]

yeah true

ok ty

also just curious what would the lattice look like

[knotty vine]

Did you read the comment by "unique_two" on the youtube video?

Quick explanation of the last few sentences because the quality isn't great. By taking the tube through the deformation we get a 3d shape at t=1, from which we obtain a 1d shape by taking the boundary and intersecting with the vertical plane. Part of that 1d shape will be a curve going from left to right, which is what he draws in green, note that this curve isn't knotted. Take this curve back through the deformation to t=0, then it still sits on the boundary of the tube, hence it is knotted the same as the knot B. This is a contradiction the knotted curve cannot be deformed into an unknotted curve.

Theres also a fantastically funny comment by "jeffwalters1749"

which I wont quote here...

[high hill]

Im happy with that. Im not happy with the fact tubing does not potentially get in the way of any unknotting process.

[knotty vine]

Yeah, so I think the tube just follows whatever process we assumed we had

[high hill]

But A and B (presumably) have to come together and so im convinced the tube will intersect with itself to do stuff

[knotty vine]

I think I was wrong before

The tube just follows

So it could intersect itself

[high hill]

So the process is a homotopy

formally

[knotty vine]

Not really

[high hill]

i mean any unknotting process is, isnt it?

homotopy from the identity map to the continuous map between your knot and unknot

[knotty vine]

Ive never really studied knots, but wikipedia says its a continuous [0,1]-family of homeomorphisms R3 -> R3 so that the first one at 0 is the identity and the last one at 1 carries the one knot to the other

Just a plain homotopy is not going to work because any knot is homotopic to a circle

[high hill]

right homotopy allows intersection

nvm about that

If instead you look at the inverse space, X - K as maximo suggested then maybe it works?

[knotty vine]

Im not actually sure if there may exist inequivalent knots that have homotopy equivalent complements

Again I dont know anything about knot theory specifically

Apparently the Gordon–Luecke theorem shows that homeomorphism of the complements is the same as equivalence of the knots

But this is not an easy fact and its in fact not true for links

[high hill]

ok think ive convinced myself

Like. Each point on the string corresponds to a cross section of the tube.

This point must always stay inside when you unknot

other than that, the tube is allowed to intersect anything

And then I think I can follow his result

[high hill]

Ah wait this sucks for A

hmm

I guess you need to convince yourself any unknotting process that can occur in R3 can also occur inside the tube

[knotty vine]

You dont need the cross section idea I think

[knotty vine]

Also dont think so

[high hill]

[knotty vine]

At least, that comment makes me think so

Clearly (since we have a homeomorphism), that tube in the original picture gets carried to a tube in the final picture where the knot is unknot

The unknot fits entirely inside a 2d plane

This plane intersects the tube in the final picture

Since its a tube, there must be some path from left to right (there may be other crap like Conway draws, but we can ignore it)

This path doesnt intersect itself (since we have a homeomorphism) and it also doesnt intersect the unknot

Now carry this path back in time to the original picture

There it sits in the "boundary" of the tube

Hence it follows the knot B

But that shows that B is actually also the unknot!

So the only way for A+B to be unknot is if A and B are both unknot

[high hill]

Ok ive convinced myself of one thing. Take a double loop. Like 2S1. Wrap this up using a torus. Now you cant "undo" this into S1

So the tube must be allowed to intersect itself for the purposes of this proof

[high hill]

In fact, Im not sure what his argument is supposed to do to such a construction. Hmm.

hmm maybe not

huh so the tube need not intersect itself... for this

[knotty vine]

Yeah, the tube doesnt intersect itself ever, since we have a homeomorphism R3 -> R3 at each point of the process and those are always injective

[heady skiff]

why exactly are these boundaries not included in Lk v2?

is the print just weak?

[knotty vine]

No, read the text defining Lk again

[heady skiff]

oh is it because we consider the pink segments to be the interiors of the simplices containing v_2

[knotty vine]

Yep

[heady skiff]

okey thanks

is g continuous, since it just involves multiplication and addition?

[knotty vine]

Yeah, it a bit like linear interpolation

[heady skiff]

hm i'm not sure what that is

but at this point whenever i see arithmetic in a function that's claimed to be continuous that's my go to mental justification lol

[knotty vine]

The composition of continuous functions is continuous

[heady skiff]

indeed

[knotty vine]

Although, we do need some gluing in this case

[heady skiff]

wdym

oh

oh yeah right after they use glueing lemma

to finish proof

[heady skiff]

don't E^J and R^k coincide for some natural number k lol

[ebon galleon]

When J is finite sure

E^J ~= R^|J|. But for infinite J they are different.

E^J would be the direct sum of R, |J| times, whereas R^J would be the product of R, |J| times

[heady skiff]

i see okay

thanks

i also don't really understand what they mean by "each such subspace is a copy of R^N for some N". do they just mean that each subspace is isomorphic to R^N for some N?

that follows pretty easily

[obtuse meteor]

there's a fun proof of this called the Eilenberg-Mazur Swindle

but it is decidedly less geometric than conway's idea (though IDK how exactly to make conway's work...)

[umbral panther]

Barry Mazur is a week older than Conway

The Mazur swindle applied to knots involves passing through wild knots. I’m suspicious that I really know the foundations of wild knots

[knotty vine]

Oh right I remember this

[heady skiff]

so here i'm trying to prove the => direction of this iff - so obviously I set up $\sum_{i = 1}^n t_i(a_i - a_0) = 0$ and am trying to prove that $t_i = 0$ for $i = 1, \dots, n$, but to use the geometric independence of the $a_i$, am I allowed to say that $\sum t_i = 0$? or am I only allowed to use the condition that $\sum t_i = 0$ and $\sum t_ia_i = 0 \implies t_i = 0$ for all $i$

[gentle osprey]

okeyokay

[heady skiff]

so then I would have to prove that $\sum t_i = 0$ otherwise

[gentle osprey]

okeyokay

[heady skiff]

in order to use that condition

which seems fucking annoying

ohw ait nvm

i'm trying to get some intuition as to geometric independence in terms of {a_i - a_0 | i \in I} being linearly independent

i know it has to do with them not being a scalar multiple of each other

but i'm trying to visualize how a_i - a_0 being lin independent implies the a_i are geometrically independent and i can't really see it lol

[knotty vine]

To see that some set of points in geom. indep. you translate them so that one of the points becomes the origin and then check lin. indep of the remaining points

In this case you translate so that a_0 becomes the origin

[heady skiff]

oh so you're talking about the case that a_0 = 0?

i guess that may be easier to visualize then

[knotty vine]

No, im saying, translate everything by -a_0

[heady skiff]

oh

[knotty vine]

Or: move the origin to a_0

[heady skiff]

ok lemme try to see this in R^2

[cloud flax]

Hello, what is the Alexandrov compactification for ]0,+inf[?

I think it's S^1, although not entirely sure

[ebon galleon]

Yeah that's correct

[cloud flax]

And for C^n?

[red yoke]

S^2n

[grave sun]

Any tips on this? In the previous part I showed that {\N \cup \infty} with the order topology is compact, but I have no idea what neighborhoods of infinity would even look like here

[safe torrent]

Weird question but is there any "prime number theorem" for prime knots?

[fierce lily]

Prove that the intersection of a compact set and a closed set
is compact. I only know that compact set is closed and bounded, and intersection of two sets must be closed, but don't know how to prove the intersection must be bounded? Can anyone give me some suggestions for this question?

[gritty widget]

any subset of a bounded set is bounded

[fierce lily]

We can state that because the intersection of one compact set and one close set is the subset of that compact set, so we can prove that the intersection is bounded, is that corret?

[gritty widget]

yes

[fierce lily]

Got it, really appreciate that

[cedar pebble]

yes there is a prime factorization theorem for knots under connect sum

[safe torrent]

Where can I read about this?

[cedar pebble]

pretty much every basic knot theory textbook covers this

if you google "prime knot decomposition theorem" there are loads of expository notes on this

[safe torrent]

How about the distribution of prime knots?

[cedar pebble]

I mean the prime knots don't decompose further but yes that phrase will probably pull up similar notes

[safe torrent]

So there is a prime factorization theorem but probably no "prime number theorem"

Constructing such a thing sounds very nontrivial

[heady skiff]

can i get a hint to show that the real line with the finite complement topology is not a Hausdorff space? so far, I'm letting any r1, r2 in R, letting U be any neighborhood of r1, and trying to show that it intersects r2. i'm going for contradiction - so that if r2 is not in U, then in particular r2 is in R - U, so that r2 is in a finite set. other than that i'm stuck lol

oh

it's just that we can do the same thing for r2

let U be any neighborhood of r2 and suppose that r1 doesn't intersect it, then r1 is in R - U which is finite

but this contradicts the fact that if we take U = O then R is in O and R - O

[unreal stratus]

This isn't what Hausdorff means

What you are doing is showing that the topology is indiscrete, since you are saying any open set containing r1 contains r2 (which was arbitrary) i.e. contains the whole space

[heady skiff]

wait wdym

i'm trying to show that it's not hausdorff

or do you mean the approach that i'm taking

show's that it's indiscrete

instead of not hausdorff

[unreal stratus]

Well you seem to have the wrong notion in mind I mean rather than Hausdorffness

[heady skiff]

maybe i got the negation wrong

a space is not hausdorff iff for every two distinct points x1 and x2

every neighborhood of x1 contains x2 right

[unreal stratus]

No

[heady skiff]

hm

[unreal stratus]

That condition is equivalent to being indiscrete

[heady skiff]

ohhh

wait hausdorff involves two neighborhoods

[unreal stratus]

Yes

There's also T0 and T1

Between indiscrete and hausdorff

[heady skiff]

oh okay

so the negation of being hausdroff would be that

for any two distinct points x1 and x2, for every neighborhood of U1 and every neighborhood of U2 their intersection is nonempty

hm is there something i'm missing as to why the cofinite topology is not hausdorff

[lime sable]

it would be "there exists a pair of points such that all pairs of their neighborhoods have nonempty intersection"

[heady skiff]

this seems rather nontrivial and the author treated it as trivial

oh right forgot the "for each pair"

[lime sable]

no it's just that it has to fail to some pair of points, not every pair of points

[heady skiff]

right

[gaunt linden]

For any total order A with a largest element m, the neighborhoods of m (in the order topology) are exactly the subsets of A that contain { a in A | a > b } for some b < m.

[lime sable]

try some examples out

what do you notice

[heady skiff]

well all the neighborhoods are infinite

i don't think that tells me much tho lol

[lime sable]

what about the intersections mentioned in the hausdorff condition

[heady skiff]

uhh

[gaunt linden]

(But it's a strange exercise if it (a) doesn't expect you to be familiar enough with ordinals to just write "omega_1 + 1" for its \hat\omega_1 but (b) does expect you to be familiar enough with ordinals to have an idea what to do with omega_1 in the first place.

[heady skiff]

hm i mean i guess if it's hausdorff, then that would imply that for every pairs of points x1 and x2 every infinite neighborhood of x1 and every infinite neighborhood of x2 would have empty intersection

which seems pretty false

well infinite neighborhood is a loose characterization of the open sets

[gaunt linden]

(In general you can prove by transfinite induction that every successor ordinal is compact under the order topology).

[heady skiff]

i'm not sure if that's a proper characterization

[lime sable]

not every neighborhood, some neighborhood

[heady skiff]

oh right 🤦‍♂️

keep forgetting

[lime sable]

sure it's false, but we've basically just restated the condition

but maybe something clicked for you

get an example of two neighborhoods (in this case, each of distinct points)

and ask yourself what you know about their intersection

because you want to show that it's nonempty for all pairs of neighborhoods

so you need to first figure out what you know about it before you show more properties about it

[lime sable]

or even just with an example, take their intersection and look at its concrete properties

and then you can do this again with another example if you need to

for example, i'll take the points 0 and 1 with the cofinite neighborhoods R - {1/2, 3, 5} and R - {10}

and then think about their intersection

[gaunt linden]

For the cofinite topology it might be useful to restate Hausdorffness in terms of closed sets instead: The space X is Hausdorff is for any distinct points x and y, you can cover all of X by two closed sets such that one of them avoids x and the other avoids y.
Now what do closed sets in the cofinite topology look like? Especially, what do closed sets that manage to avoid some point look like?

[merry geode]

So I have this conumdrum:
Consider X to be the plane R^2, and A be finite points on the plane.
Then, one has LES
... -> H^i(A) -> H^i(X) -> H^i(X, A) -> ...
where H^i (A) = 0 for i >= 1.
That means H^i (X) = H^i (X, A) for i > 1.
Yet IIRC this gave a contradictory result in my exam. So is my reasoning abt this correct?

RIP, I guess I got it wrong because I wrote H^0 (A) = 0

Why am I like this 🥲

[fickle elm]

And if it is the cohomology, the LES should be H^i(X,A;G)->H^i(X;G)->H^i(A;G)->H^{i+1}(X,A;G) because you need to apply Hom(-,G) and it reverse the arrows in homology.

[merry geode]

But that's cohomology

[hidden crag]

H^k usually denotes cohomology

was the task to compute the homology of R^2 - finite amount of points?

[candid wyvern]

A simple example of a pair (X, A) with A closed for which the homotopy exten-sion property fails is the pair (I, A) where A = {0, 1,1/2,1/3,1/4, ···}. It is not hard to show that there is no continuous retraction I×I→I×{0} ∪ A×I . The breakdown of homotopy extension here can be attributed to the bad structure of (X, A) near 0. With nicer local structure the homotopy extension property does hold, as the next example shows.

I don't quite understand why this pair doesn't have a homotopy extension

[merry geode]

Yes.

Sorry about me confusing subscript with superscript

[hidden crag]

That’s homotopy equivalent to the |A| fold Wedge sum of circles

[merry geode]

..My bad, did not know it was that easy

[wispy veldt]

just to be sure is hatcher trying to say this ?

[gentle osprey]

James Banach

[wispy veldt]

or am i missunderstanding something

[drowsy moth]

hello anyone

is here

[red yoke]

Yes

[hidden crag]

This is elaborated on in the part of chapter 4 that introduces fibrations

[feral copper]

https://dontasktoask.com/

[fierce lily]

Is there any suggestions about how to find the boundary points and interior points of a cantor set?

[median sand]

This is more of an analysis question. Afaik the interior should be empty, since the Cantor set has measure zero, so because the set is closed it should follow that all the points are boundary points.

[feral copper]

It's about as much topology as it is analysis; this question is suited for this channel

And yeah, C has empty interior

[ebon galleon]

Hatcher just-draw-the-fucking-diagram challenge impossible

[fierce lily]

Therefore, because the set has measure 0 and it is closed, so all points are boundary points, is that correct? And we can know that its boundary points are also empty?

[median sand]

Well yes, that's exactly what I said. Are you asking why the set of boundary points is non-empty? What I said shows that the Cantor set equals its boundary, so to show the latter is non-empty you have to show the Cantor set is non-empty. If it's the classic Cantor set ([0,1] with the middle thirds recursively removed), then 0 and 1 belong to it, so it's non-empty. I don't remember the definition of general Cantor sets to speak for them.

[fierce lily]

You mean because its length is zero while it is closed, so the set is equivalent to its boundary? Is that correct?

[feral copper]

$$\partial E=\overline{E}\smallsetminus\mathring{E}$$

[gentle osprey]

Matplotlib

[feral copper]

This is why

[fierce lily]

Ok, I got it

[feral copper]

(here, the interior E° is empty)

[fierce lily]

the set consist of all interior points and boundary points

So if its interior points are empty, can I make conclusions that accumulation points are also empty?

[gaunt linden]

What does it mean to say that an accumulation point is empty?

[fierce lily]

I mean if i get the conclusion that interior points of cantor set is empty, can I also say that its accumulation points are empty?

[gaunt linden]

For that matter, what does it mean to say that an interior point is empty?

I don't get what it means for a point to be empty.

[feral copper]

I assume they mean that the set of accumulation points is empty?

And the set of interior points is empty

[gaunt linden]

But that is clearly not true.

A compact infinite set must have accumulation points.

[feral copper]

I'm just translating what they mean, not giving an opinion

Here, the Cantor set is also equal to the set of its accumulation points; it has no isolated points

(or, as you'd say, the set of isolated points is empty)

[fierce lily]

I mean in a cantor set, I have know that the set of interior points is empty, the set of boundary points is equivalent to a cantor set, its accumulation points must contain all its limit points, so it should be a cantor set, I think what i said is wrong before

I just assumed that accumulation points must be in the set, but it was wrong

[feral copper]

A set is always partitioned by its subset of limit points and its subset of isolated points

(by definition)

And for a closed set X, the subset of accumulation points of X is equal to the set of limits of sequences of elements of X

[gaunt linden]

You're right that because there is no interior points of the set, every point in the set is a boundary point.
In general there can also be boundary points outside the set, (namely limit/accumulation points don't need to be in the set), but that doesn't happen for the Cantor set because it is closed.

[feral copper]

(I'm sorry if I'm getting your point wrong; it's hard to decypher what you mean...)

[feral copper]

As Troposphere said, because of this

[gaunt linden]

(At least for metric spaces, which happily includes all R^n and their subsets).

[warm crown]

may i ask how do i attempt this problem? asking for a friend btw

[feral copper]

Draw the line between x and f(x), and this is indeed a line if f(x)=/=x. It will cut the circle somewhere: that's the image of the map from D^2 to S^1

I'm not sure I follow. What is the "limit point of a point"?

[fierce lily]

I mean accumulation points means every neighborhood of points contain at least one point in the given set

[fierce lily]

because we know that the set of boundary points of C is set C while the set of interior is empty, choose a random element x from C, then every neighborhood of element x must contain at least one element from the set C if we call x to be the accumulation point. But its interior is empty, so I am confused how to determine its accumulation points

[ebon galleon]

what

[fierce lily]

I mean a cantor set

[ebon galleon]

Yeah that wasn't the part I'm confused on.

You can explicitly check: if you take a point in the cantor set, for every ε > 0 there's some other point in C with distance less than ε

Interior and accumulation/limit points aren't necessarily related

[fierce lily]

it seems that every elements in a cantor set must be accumulation point

And it also seems like cantor set should consist of many closed intervals, but I don't know if it is a correct statement

[alpine nest]

It's not, but the statement about accumulation points is true

[fierce lily]

You mean the sentence every elements in a cantor set must be accumulation points?

[alpine nest]

Yes, every element of the Cantor set is its accumulation point.

But the Cantor set contains no interval, closed or otherwise

[fierce lily]

I rememberd that the cantor set are divided by three parts and remove the middle part, and keep repeating, why it can not be regarded as containing many intervals?

[alpine nest]

Because you keep repeating it infinitely many times, and the cantor set is what you get in the limit.

At every step along the way you've got the union of finitely many intervals, but in the limit you lose that.

[fierce lily]

You mean because of length 0?

[alpine nest]

You could think of it like that. Alternatively, any interval subset of [0,1] will eventually not be entirely covered by the intervals you get when constructing the cantor set.

So it can't be a subset of the final intersection.

[fierce lily]

So you mean every subset of [0,1] will not be entirely contained when keep constructing cantor set infinitely many times?

[heady skiff]

is this also how we define open sets in |K|?

[ebon galleon]

Yes

Consider De Morgan's laws

[heady skiff]

oh right

hm how would I show that the open sets coincide here?

so i know that if $A$ is open in $|K'|$ then $A \cap \sigma'_i$ is open in $\sigma'_i$ for all $i$, and I want to show that $A \cap \sigma_i$ is open in $\sigma_i$ for all $i$. so since every simplex of $K$ is a union of a finite number of simplices of $K'$, we know that every $\sigma_i$ contains $A \cap \sigma'_i$ for all $i$

[gentle osprey]

okeyokay

[heady skiff]

the issue is just looking at $A \cap \sigma$ - is it true that $A \cap \sigma = \bigcup_{i \in I} A \cap \sigma'_i$?

[gentle osprey]

okeyokay

[heady skiff]

anyone?

[abstract saffron]

Munkreas?

I remember I jotted down the proof somewhere, but it's been a while

[fading fern]

what does this notation mean

saw it in an algebraic top book

[hidden crag]

could mean several things

send a screenshot of where you found it

[fading fern]

will try to draw it

pc got no camera

[knotty vine]

Do you just mean $(\ZZ,n)$ ?

[gentle osprey]

Semer

[fading fern]

yes that

[knotty vine]

Doesnt mean anything specific to me

[fading fern]

do i still send drawing

[knotty vine]

Do you have the textbook?

[fading fern]

yes

[knotty vine]

Which one is it?

[fading fern]

"Lectures on Algebraic Topology"

by Albrecht Dold

[languid patrol]

this depends on your definition of geometric realization |K|, but however you define it it should be a tautology that if f: K \to K' is a morphism of finite simplicial complexes and f: |K| \cong |K'| as sets then f is a homeomorphism

[fading fern]

it appears regularly on chapter 2

[knotty vine]

By Bott?

Or Dold

[fading fern]

dold

that sequence appears in chapter 2

[knotty vine]

Oohhh Its a chain sequence

[fading fern]

yeah

[knotty vine]

There should be a C everywhere tho

[fading fern]

apparently its something about cones i think

yeah its cones

C(Z,n) refers to the cone of (Z,n)

[languid patrol]

the issue with infinitely many simplices might arise if you write some n-simplex as a union of infinitely many (n-1)-simplices in a perverse way

[knotty vine]

Huh Im actually not sure what it means

[hidden crag]

I'm not finding this in chapter 2 of the pdf

[fading fern]

weird

i found the sequence in page 23

[heady skiff]

hmm okay thanks

[hidden crag]

i clicked the wrong "lectures on algebraic topology" lmao

[knotty vine]

Is it maybe the chain sequence with only Z at index n ?

[fading fern]

maybe

[knotty vine]

idk if that makes sense

[fading fern]

i thought it was the poset of integers thought as a category

and kernels and images defined in a categorical sense

[knotty vine]

Found it

[fading fern]

oh im blind

thanks

[knotty vine]

p 17

[fading fern]

found it

[knotty vine]

Somewhat confusing notation tho

[fading fern]

now i can finally find some concrete examples on the book

[knotty vine]

Seems like a nice book looking at the contents, but maybe not as an introduction to algebraic topology. Seems to be mostly focussed on homological algebra

Idk what youre using it for

[heady skiff]

nobody in my class was able to understand why this inequality holds and the prof said "anyone who gets this is a fucking genius"

can somebody explain to me why it holds?

[novel acorn]

hatcher proves this via induction iirc

ah no he proves something similar

[heady skiff]

oh someone found a stack exchange post

[novel acorn]

sorry but what's K^1 here?

and what's the ordering of the simplices

[heady skiff]

the first subdivision of K

[novel acorn]

like B is a subcomplex of A?

[heady skiff]

uh barycentric subdivision

[novel acorn]

ah he does prove this

[heady skiff]

ye

or face of

i think

[knotty vine]

Yeah, B is a subsimplex of A, meaning theres an edge between their barycentres in the subdivision

[novel acorn]

the idea I think comes down to the fact that for an n-simplex [v_0, ..., v_n] we have that b (barycenter) = 1/(n+1) v_i + n/(n+1) (barycenter of the face without the vertex v_i)

[knotty vine]

Heres an idea, the "largest" n-simplex with diameter 1 is the regular n-simplex and the longest edges in its subdivision are the ones from a vertex to the barycenter of the whole simplex

This length is $\sqrt{\frac{n}{2(n+1)}}$ which is $\le \frac{n}{n+1}$ when $n \ge 1$

[gentle osprey]

Semer

[knotty vine]

So we could in fact do somewhat better than the book

Not the most rigorous proof tho

[pseudo ocean]

Is there a comprehensive list of Cobordism classes like homology groups?

the literature around cobordism theory seems to be scattered

[novel acorn]

Are you looking for the classification of manifolds up to cobordism?

[pseudo ocean]

yep pretty much

[novel acorn]

https://en.m.wikipedia.org/wiki/Classification_of_manifolds

[novel acorn]

cobordism classification is most useful in dim > 4 since there the classification up to diffeomorphism is impossible

and then the cobordism classification uses surgery theory techniques

[pseudo ocean]

I wanted to get into the nitty gritty of computing cobordism groups

[novel acorn]

There was a good book on surgery theory I read that went into this lemme see if I can find it

I think it's called The Geometric Hopf Invariant and Surgery Theory by Ranicki

[hidden crag]

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

[pseudo ocean]

i see

I had this little primer on surgery theory by ranicki too

apart from milnor, this guy seems to be a pillar in the theory

[hidden crag]

https://people.math.harvard.edu/~kupers/notes/orientedbordism.pdf

this has a geometric computations of the low dimensional oriented cobordism groups

in case you don't want to dive into stable theory

[umbral panther]

Not really. Cobordism calculations are almost purely homotopic and are input into surgery, not vice versa

Very early in the history, Roklin tried to use cobordism to compute pi^s_3 and got the wrong answer. Someone ran the argument backwards and constructed the Rokhlin invariant

[abstract saffron]

what inequality in particular?

There are two versions, one in Alg Topo, and one in Convex theory. Iirc, both are quite similar, but the proofs differ

[knotty vine]

The inequality that shows that the maximum edge length in subdivision gets smaller

And the proof seems not too crazy to me #point-set-topology message

[heady skiff]

yeah this book is funny

sometimes he mentions things in his proof and doesn't give justifications

even though they're entirely nontrivial

what does it mean to be a linear function?

surely it doesn't mean in the usual sense, like y = mx + b

because we don't even know how s is defined

on the vertices

[feral copper]

It's the restriction of an affine map to a simplex as a subspace of euclidean space

And you ask that if you have two maps on two adjacent simplices, then they 'glue nicely' on the meeting face

So it's like 'affine by parts'

[heady skiff]

hm what's an affine map?

my book didn't define it

[feral copper]

AX+B

Like, a matrix * a vector + a vector

[heady skiff]

oh ok i see

thanks

[heady skiff]

if $s: |K| \to |L|$ is a simplical approximation of $f: |K| \to |L|$ but for some vertex $v$ of $|L|$, $f^{-1}({v}) = \varnothing$, can we say anything about $s^{-1}({v})$?

[gentle osprey]

okeyokay

[knotty vine]

Following Armstrong: v is its own "carrier"

[heady skiff]

@narrow cairn you're self-studying from Lee right

[narrow cairn]

i was, ive moved on to rotman now

[heady skiff]

oh damn nice

[narrow cairn]

i only got through the first four chapters and part of 5

[heady skiff]

oh ok that makes sense

i was gonna ask what's ur thoughts on it for point-set lol

cuz honestly i was going to go back and review point set over the break

but i'm kind of tired of munkres and want something new lmao

and then learn alg top from lee

or maybe hatcher or smt

[heady skiff]

is there any easy way to see that the interior of a simplex is open?

i know that it consists of all points where each barycentric coordinate is positive

this is often the case in topology (at least for me) where it's intuitively/visually obvious

but i don't know how to approach proving it formally

[hidden crag]

is that not just defined as the topological interior of a simplex

[heady skiff]

wdym

it's defined as the complement of the boundary yea

of the simplex

[ebon galleon]

perchance the boundary is closed

[heady skiff]

but i guess that requires showing that the boundary is closed

[hidden crag]

well the boundary is closed

[heady skiff]

which is like another problem lol

oh

wait so the boundary of a simplex is just the union of all the proper faces right

ok now i need to see how this is closed lmao

[hidden crag]

the boundary of a set is the intersection of two closed sets

[heady skiff]

wait wdym

even if that was the case they define boundary here in a different way tho

[ebon galleon]

great minds think alike

[nova fjord]

Is there a simple proof that the spaces $\mathbb{R}^{\infty}, \mathbb{S}^{\infty}, D^{\infty}$ are not metrizable?

[gentle osprey]

Eternal Way

[umbral panther]

There are lots of different spaces that go by those names, but most of them are metric

[nova fjord]

The topology on R^inf is given by saying that a set is open in R^inf if it's intersection with each R^n is open. Similar deal for the infinite sphere and ball in this problem

[red yoke]

Pick a metric d. If d gives the desired topology, then for each n we can choose a_n = (0, …, 0, 2b_n, 0, …) (in the nth component) such that d(a_n, 0) < 1/n and b_n > 0. But [(-b1, b1) × (-b2, b2) × …] ∩ R^∞ is open.

[nova fjord]

Apologies, but I'm not seeing the contradiction here. Is that this sequence has infinitely many non-zero terms?

[red yoke]

This sequence converges in the metric but not in the desired topology

[median sand]

So a quasiabsolute value is defined as something that has the usual properties of an absolute value, except instead of the triangle inequality it satisfies the inequality below for some real C>0. I want to show that a QA induces a topology with its open balls U_r(x) as a basis, but I was having trouble for whatever reason showing explicitly that the intersection of any two balls contains a third, so I thought to do this in a roundabout way:

It is known that every QAV is a power of an AV. If | | is the power of the AV | |', then the balls U and U' of the QAV and the AV are nested (for any x and r there is an s such that U_s(x)<=U'_r(x) and vice versa). Since the AV balls U' are a basis for a topology, it follows that the U balls also are a basis for a topology, the same one in fact.

Does this work?

[nova fjord]

Ah ok, thanks!

[novel acorn]

this is the idea yeah

[median sand]

The way it's plainly stated here made me think I must be missing something simple and it's possible to do this directly from the definition.

[feral copper]

Uhm, if |x|'=|x|² for instance, then what about balls of radius less than 1 or greater than 1? Something happens here! (regarding who is nested in who)

The conclusion should remain the same, but the details require a teeny tiny bit more work

[tribal palm]

this book is proving absolutely fantastic

[hidden crag]

Looks fun

[tiny ridge]

john roe is a fantastic author

[distant lichen]

What does it cover?

[tiny ridge]

I bet it talks about the Toeplitz index theorem

I know John Roe because of his text on Atiyah-Singer index theory, it's a great textbook

[distant lichen]

This is quite cool

Surprising breadth

This gives me the impression of a kinda fun topics course?

[median sand]

The radii are not necessarily the same. The point is that for every r there is an s such that U_s<=U'_r and vice versa.

[feral copper]

Ah yeah true

[last marlin]

If I'm given a surface with this identification, how do I find the fundamental group for it with basepoint circled v? (Do not use the fact that this is RP^2)

[feral copper]

You can compute an easy presentation for it

Seifert Van Kampen helps

(you don't really care who the basepoint it, because it's path-connected)

[last marlin]

<ab | (ab)^2=1>

Which is Z_2, which should be correct

[feral copper]

ab is the generator yeah, it's not two generators a and b

[last marlin]

But how do I say definitely that ab is a generator?

[last marlin]

Yes

[feral copper]

a and b are not loops, they don't start and end at the same point

v=/=w

Do you see why ab is the loop spanning the thing?

You SvK by saying it's pi1 of the punctured thing (puncturing in the middle of the square) glued together with pi1(disc)=1

[last marlin]

The thing is, why am I only looking at the boundaries? What if I take a loop starting with a and then take it in the interior instead of going along b?

[feral copper]

Enjoy my terrible computer mouse drawings

Wait, is that correct on the left?

(brain fart moment)
On the left, it should be a Mobius strip

[obtuse meteor]

these are homotopic

[obtuse meteor]

For example in this application of SvK to be entirely technical you would need to take the basepoint in some small nbhd of that cut out disk (and not on the edges), and the generator would be "move to the bottom left vertex from that pt, then go up through a,b (free to move in interior as well, just in the general shape of this) and then once you get to top right return to the base pt"

but this distinction is kind of meaningless in terms of the calculation

but it might make thigns more clear as to "why you're only looking at the boundaries"

[median sand]

But that identification is not the Mobius strip though. Mobius strip identifies one of the edge-pairs with the same orientation, this doesn't. This should be the projective line or smth related iirc It's the projective plane.

[novel acorn]

Wait a second this doesn't look like Boys surface 🤨

[silver umbra]

is X x X - diagonal necessarily disconnected?

where X is any topological space

[cedar pebble]

it's connected if X is empty or 1 point

[unreal stratus]

lol

well i would argue in both of those cases it is still disconnected

:)

[cedar pebble]

the empty set is connected

[unreal stratus]

No

Anyway uh pretty sure it needn't be disconnected

e.g. if you take X any set of cardinality > 1 and then give it the indiscrete topology

the product still has the indiscrete topology

subtract the diagonal and uh pretty sure that still has indiscrete topology

so it's connected

[gaunt linden]

If X is a circle then X×X is a torus, and removing the diagonal leaves it connected.

[unreal stratus]

nice

Actually we can think about this geometrically - this is just the space of configurations of two points in our original space.

[ebon galleon]

The empty set is connected.

[unreal stratus]

So e.g. if X = R, we get something at least not path connected e.g. since no matter how we move the particles continuously, their order will be preserved

But if X = S^1 then it's easy to see we can move our pair of points wherever we want

(just rotate them around together until one of them is where you want it, then move the other appropriately)

:)

[knotty vine]

Yeah its the configuration space of two points!

[gaunt linden]

So it's "removing the diagonal disconnects" that's the exceptional case.

[unreal stratus]

I'm not too sure what the general criterion is, at least for X a manifold

(i know a little more about unordered configurations)

[gaunt linden]

If it's of dimension 2 or higher, it's easy to see with your argument that X²\diagonal is still connected.

[unreal stratus]

Yes

[gaunt linden]

And there are not many connected 1-manifolds to begin with.

[unreal stratus]

Yeah I mean more generally like you can view each point as living in X \ a point

if we think of moving one particle then the next

Of course, the point you remove does matter in general

So yeah i guess that's why for dim X > 1 you just use the fact X \ a point is still path connected if X is?

[gaunt linden]

Yeah.

[knotty vine]

Damn thats a massive nlab page https://ncatlab.org/nlab/show/configuration+space+of+points

[unreal stratus]

yeah i just did a summer project on them hehe

[gaunt linden]

I wonder what happens if the original X was only connected but not path connected?

[unreal stratus]

True

[potent sky]

I have a 3-manifold M with boundary $\partial M$ and $H_1(M; \mathbb{Z}_2) = 0$. How can I show that each component of $\partial M$ is a 2-sphere?

[gentle osprey]

*-algebra

[potent sky]

i think im supposed to look at the long exact sequence for the pair $(M, \partial M)$ with the coefficients i say there

[gentle osprey]

*-algebra

[potent sky]

oh yeah my M is compact w/ boundary

[hot locust]

How do I show that the set of dyadic rationals are dense in
[0, 1]?

I was thinking of finding a continuous mapping, should that work?

But somehow I can't think of an example of a map that works.

[solemn oar]

You can try adapting the proof that the rationals are dense in R.

[rain sentinel]

Not sure what you mean by mapping but the usual proof goes: Let a,b in [0,1] with a<b then there’s an n in N such that 2^(-n) < b-a. Show then there exists an integer multiple of 2^(-n) in (a,b).

[hot locust]

Ah sorry. By mapping I meant a continuous function, that was a mistake.

[unreal stratus]

Then it's sort of unclear what you mean

[rose vale]

My uni doesn't offer a point set topology course, but offers a research project where you basically teach yourself topology

The person who runs the project is a topologists so I think it just ends up being a less structured topology course

Is it worth doing a "self study" project, rather than research project?

It still gives credits but I dont know if its worth taking over another course

[tough jay]

Hi, I'm currently trying to compute the non reduced homology groups of the Klein Bottle using Mayer-Vietoris sequence, it was fine for the case where n>2 but for n=1 I am stuck because I end up with the following exact sequence : H1(K) → Z → Z+Z → Z → 0 while on a detailed example I found on a lecture note gives H1(K) → 0. I don't know where I messed up and I hope I am not confusing between reduced and non reduced homology...

[unreal stratus]

yes they seem to be using the reduced Mayer-Vietoris sequence

At least in my experience, usually the reduced is more useful for computations

since then all your Z's there vanish

:)

@tough jay

[tough jay]

Ahhh I thought so too... now it makes sense, thanks !

[silver umbra]

i'm trying to concretely wrap my head around the difference between isotopy and homotopy, and i figured that it'd be easiest to start with embedded curves

so two embedded curves can be homotopic, but not isotopic, if you cant homotope one curve to the other via embedded curves only

[feral copper]

Isotopy is a homotopy where at all times it's an embedding

[silver umbra]

right

[feral copper]

Homotopy, you allow crossings and shit to happen

[silver umbra]

right, so what would be an example of embedded curves

where the only way to homotope one to the other

[feral copper]

Then there's ambient isotopy too

[silver umbra]

is having some kind of crossing

[feral copper]

Hopf link is homotopic to unlink, but it shouldn't be isotopic to it (it actually is, it isn't ambient isotopic to it)

So the homotopy allows for this:

In the middle, it's not an embedding (it's an immersion here, but it needs not even be an immersion at all times, it can be weirder)

[silver umbra]

ahh i see

wait, so then aren't all links homotopic to the unlink

[feral copper]

They are, if you allow components to cross themselves

But when people say link homotopy, they want distinct components to remain disjoint at all times

But yeah, all knots are homotopic

[silver umbra]

i see

[fervent root]

I need some guidance on this question from Lee. To show it is a basis I know it suffices to show that the intersection of two basis elements is contained in another basis element, but I dont know which two to pick.

[hot locust]

I was looking for a continuous function from a dense set to dyadic rationals in [0,1] so that I could show that it's dense. But I couldn't think of a map so I was asking for examples.

[unreal stratus]

Ah okay that makes more sense yes

[hot locust]

Do you think the map
(a, b) ----> a/2^b

Is a continuous map?

Ig I ended up trying to make a stronger statement that dyadic rationals are dense everywhere in R.

[unreal stratus]

yes sure but if b is a rational then 1/2^b needn't be rational

so i'm not sure what you mean

[hot locust]

Nah as in, a and b are positive integers

[knotty vine]

The map has to be surjective for the lemma you want to apply to apply

As in, surjective to [0,1]

[unreal stratus]

well every map out of the integers is continous

so yes

i think it's easiest just to do this by hand or i'd say it's obvious if you are fine assuming decimal or binary expansions exist

[hot locust]

But I am specifically looking at dyadic rationals, all of them, in [0, 1]

That won't work?

[knotty vine]

What is the lemma?

Is it that the image of a dense subset of a surjective continuous map is dense?

[hot locust]

Ahh even my prof said that. I'd try that if this doesn't work out.

[hot locust]

I think so yea

Wait I'll check

[knotty vine]

But the map ZxZ -> [0,1] you defined isnt surjective

[hot locust]

Right

I'll try the other way then.

[knotty vine]

It would be interesting to use that lemma but I dont have an idea for it

The other way is certainly easier

[hot locust]

But this is just what popped up in my head when I saw the problem so I kept trying, ig. I might come up with something later maybe. Thanks anyway.

[astral idol]

I want to show that for a topological space (X, T) if a set A intersects with every dense set of X then the interior of A is not empty

Its supposed to be an equivalency and I already showed the inverse implication. But I'm kind of stuck on this one. I'm only supposed to use the elementary definitions of the above that just make use of open sets only (no sequences or anything)

I'm not yet used to reasoning on general topological spaces. With a distance function I can sort of "visualize" these properties but once you get rid of distances or norms I can't really visualize it that much. So any additional pointers/interesting ressources would also be appreciated !

[unreal stratus]

Well one hint is to express "interior of A not empty" in terms of closed sets

||actually once you do that it becomes easy lol rip||

[unreal stratus]

the point is often like density is defined in terms of closed sets, interior is defined in terms of open sets, so try to put them on the same footing by expressing everything in terms of closed sets or everything in terms of open sets

[astral idol]

ah.. damn, yeah, I think I can see it

well, I haven't had math in like 2 years and the course document thing only had the definition of interior points and adherence points

so I didn't think about expressing it that way

|| ||

[unreal stratus]

Hey yeah dw

Just pointing out it's a useful idea for topology problems

[haughty yew]

I was trying to compute the first homology group of RP2. I got Z x Z2, but shouldn't it be Z2? Where did I go wrong?

[red yoke]

The vertices are not all v

[haughty yew]

Oh

Thanks!

[red yoke]

(Attempt at rigour)

Generally what level of rigour is expected for doing alg top?

Or is this overly pedantic

[feral copper]

Generally, if you can draw a fine picture with a somewhat-fine explanation which doesn't seem hand-wavy, it's good enough (although that has to be correct and convincing)

(yes, I'm caricaturing the reality of the situation)

[white oxide]

Is this the right place for questions about the topology of the space of ultrafilters over N?

[naive trench]

Im trying to prove that an arbitrary intersection of compact sets is compact in a hausdorff space.

Since we are in a Hausdorff space every compact is closed so the arbitrary intersection of closed sets is closed but this intersection is in each compact set and bc we are in a Hausdorff space then the intersection is compact. Is this right?

[white oxide]

Yeah, thats right

[naive trench]

Thanks ^^

I should putted some notation but I think its clear

[feral copper]

Yes! Also, "Don't ask to ask, just ask" (although I myself know nothing about those )

[feral copper]

An intersection (whatever it is) of closed sets is closed, and any closed subset of a compact set is compact

So, yes 😉 (what I said uses Hausdorfficity, if that's a word)

[safe torrent]

What are the conditions for something like this:
Let X be a topological space and let A,B be subsets of X s.t H_n(X) = H_n(A) ⊕ H_n(B)? I found some counter examples for this but I feel like this proposition is true iff you have special conditions. I'd like to know these conditions.

[feral copper]

For all n? Or fixed n?

[safe torrent]

For some n

This proposition won't work for n=0

[feral copper]

Doesn't Mayer--Vietoris basically tell you the condition?

[safe torrent]

I think it works for n>0

[feral copper]

Like, it spits out an iso iff an arrow is injecitive/surjective somewhere

[safe torrent]

I haven't learned that yet

So that's the way to go about it?

[feral copper]

Computes homology of a reunion in terms of homology of each piece (a homology SvK basically)

Instead of giving out an amalgamated product, it gives a LES

[feral copper]

Unless X is the disjoint union of A and B

[safe torrent]

Think of genus 2 surface

[feral copper]

Also, one sufficient condition is that the intersection is contractible

But the general condition has to be weaker; again, check MV, it should give a condition in terms of the homology of the intersection of A and B

[safe torrent]

Average topological nonsense

Alright I'll take a look at that thanks

[feral copper]

That's not a disjoint union though

[safe torrent]

True...

I need another reference for alg top

Hatcher is

[feral copper]

(I couldn't agree more, but that's the kind of opinion which is far from being shared by others sadly)

(on the contrary, people swear by Hatcher like it's their Holy Bible)

[hidden crag]

It’s a common opinion here

That Hatcher is bad

[red yoke]

How does tom Dieck compare

I'm looking for a quick review of approximately Hatcher Ch.1-3
But tom Dieck looks lengthy

[tiny ridge]

Tammo Tom Bombadillo

Hatcher is for to-be geometric topologists.

It's a fantastic book for those who care about spaces

[feral copper]

I consider myself a geometric topologist and I still hate it

[tiny ridge]

With some exceptions, mayhaps

[paper wedge]

whats geometric topology

[tiny ridge]

Study of topology of manifolds

[paper wedge]

cool

probably some fucked up stuff happens over there

[red yoke]

So everything is nice and metrizable

[feral copper]

If you know what a handlebody decomposition is, what Kirby calculus is and like geometric operations, you're a geometric topologist

[feral copper]

Any manifold is metrizable

[tiny ridge]

There are some 3-manifold topologists who don't know Kirby calculus

They understand surgery diagrams

[feral copper]

Isn't that basically the same thing?

(also, I was kind of reducing the field, but ofc it's far more than what I said!)

[tiny ridge]

Non-integer surgeries allowed 😅

[feral copper]

You can always go back to an integer surgery

Just more messy sometimes

[tiny ridge]

Yes but loses important geometry

[paper wedge]

surgery lmfao

[tiny ridge]

Good luck figuring out if this 20 component link diagram is hyperbolic Seifert fibered

[feral copper]

Technically it doesn't since it's the same manifold, but I see where you're going
Also, there are people doing Heegaard splittings/diagrams and trisections for instance, that's geometric topology too

[gritty widget]

this is why i have the latex bot blocked

[tiny ridge]

Lol

[elder loom]

THERE

[feral copper]

@elder loom there's #latex-testing if you need to update it 50 times

[tiny ridge]

Delete the OG message

[gentle osprey]

sergeEmbedding

First, I apologize for any weird formatting, this is on my phone. 

I have a question about convex polytopes.

We have for finite sets $S$ that their convex hull $C$ is a polytope.

We also have a tower:

$$C_n=\text{ convex hull of } S$$
$$C_{n-1}= \text{ convex hull of Int}  (S- (\partial C_n)).$$

In the Euclidean plane, this gives rise to a *non-intersecting* polygonal chain, as any $\partial C_n$ has no crossing line segments, and any $\partial C_i,\partial C_{i+1}$ has a connecting segment between the two closest vertices.


How does this generalize to multiple dimensions? Is there an $n$-dimensional equivalent to the line segment connecting $\partial C_i$ and $\partial C_{i-1}$?

[red yoke]

They're called surgeons

[paper wedge]

yeah so fucking funny

i wonder tf is surgery

in math

so cool that some guy may tell u "oh yeah man i just had surgery"

[feral copper]

It's just removing a solid torus and gluing it back with a diffeo on the boundary torus

[paper wedge]

and then it references some math shit for u

[feral copper]

Fact: you can build any 3-manifold from the 3-sphere by a finite number of such procedures

[naive trench]

Hausdorffness I would say and thanks. This type of exercises look trivial but I must be careful

[paper wedge]

procedueres being "removing" tori?

[feral copper]

Removing and re-gluing

[paper wedge]

whats the boundary torus

[feral copper]

A solid torus $\SS^1\times\DD^2$ bounds a hollow torus $\SS^1\times\SS^1$

[gentle osprey]

Matplotlib

[red yoke]

I'm assuming compact 3-manifolds

[tiny ridge]

@feral copper These hyperbolic 3-manifold topologists don't care about surgery diagrams at all (but they know what it is)

[feral copper]

Closed even

[paper wedge]

ahaa

i only dealt with the hollow torus

lmlfao

[feral copper]

I've never seen it before but I'm 150% positive it works with boundary too

[elder loom]

there are doctorates specializing in surgery from the University of Chicago that are essentially guaranteed to kill you every time if they need to perform surgery on somebody

[paper wedge]

funny u know about geometric topology but not basic percentages.. haha

[tiny ridge]

Yeah you don't need compactness do perform surgeries really

Just need a knot in your 3-manifold

[paper wedge]

how do umath this

like how do u write this in math

[tiny ridge]

Scoop it out, put a horrible twist, fill it back in

[red yoke]

Is it like a mutation of knots

But manifolds

[feral copper]

For the operation yeah, but can you obtain any 3-manifold from either S^3, B^3 or R^3 in finitely many surgeries? (I guess yeah, but Idk)

[tiny ridge]

Kind of yeah

[tiny ridge]

Nah

[feral copper]

Ah!

[tiny ridge]

There are geometrically non-finite 3-manifolds

Like Whitehead manifold

[feral copper]

I knew there was a catch

When I asked closed, I meant the existence theorem

[tiny ridge]

Of course

[feral copper]

Thanks, now I'll spend a few hours (probably) reading on the Whitehead manifold(s?)

[tiny ridge]

I mean it's an extreme example

But all manifolds obtained from R^3 by surgeries would have cylindrical ends

[feral copper]

On a totally unrelated topic, but since you're there: how are you doing? And also, I submitted a v2 of my paper including the proof of the NO Thom conjecture!

[tiny ridge]

Great!!

[feral copper]

Sure!

[tiny ridge]

I'm thinking a lot about high-dimensional contact topology

[feral copper]

Idk if the proof interested you, but I should update the arXiv once it gets accepted (hopefully)

[tiny ridge]

I can take a look for sure

[feral copper]

What are you interested in? Cuz usually people care more about the 3d case

[tiny ridge]

High dimensional contact topology has unexpected flexibility which is not available in dimension 3. Currently I'm thinking about fillability and symplectic cobordisms

Fun fact: For 2n+1 >= 5, overtwisted S^(2n+1) has a concave filling.

[feral copper]

Really?

That's quite surprising

[tiny ridge]

Yeah super bizarre

It can't have a convex filling, overtwistedness obstructs that

[feral copper]

We do have counterexamples to this in 3d, right?

[tiny ridge]

I think it's true for S^3 also but don't take my word for that.

In 2n+1>=5 the proof goes as follows. One shows that there's a Liouville structure on S^(2n+1) x [0, 1] where the bottom (x 0) has overtwisted structure and top (x 1) has tight structure, bottom is concave top is convex

Then one caps it off by M \ Darboux ball on top where M is your favorite symplectic manifold

This construction doesn't work in dim 3 but I think something else does

There's some work by Etnyre

[feral copper]

I honestly lack the sufficient background to make sense of this, so I'll just nod gently and take your word for it

[feral copper]

This I'm not surprised

[tiny ridge]

Ok I have a question for you

[feral copper]

Go ahead, I just hope I can answer it

[tiny ridge]

In a Weinstein trisection, do you demand the contact structures on the boundary of the Weinstein pieces match whenever they intersect or nah

If not, can you ensure they do? Can you ensure they are at least homotopic (as plane fields)?

[feral copper]

You don't ask for contact-type things, unless I'm missing something

Wait, there is contact stuff happening indeed, but you'll make more sense of this than me:

[tiny ridge]

Oh you only trisect symplectic 4-manifolds

[feral copper]

Of course!

[tiny ridge]

With the same symplectic form on each Weinstein piece as the big guy

We bisect all 4-manifolds