point-set-topology

[rapid lagoon]

However, the part I am stuck on is showing that this smallest topology containing all of T_i is actually uniquee

I start of by asserting that $\exists A, \bigcup T_i \subseteq A$ and $\exists A, A \subseteq \bigcap \tau_\beta$

[gentle osprey]

dulg

[rapid lagoon]

How do I go from here to show that $A = \bigcap \tau_\beta$?

[gentle osprey]

dulg

[onyx raft]

Can you post which problem in the book it is?

[rapid lagoon]

Sure

4b

[onyx raft]

Unique smallest topology containing all the collections T_\alpha? It'll be the finite intersections

An open set is an 1-ary intersection so it contains each T_\alpha

And if T contains each T_\alpha it is closed under finite intersections and so contains the finite intersections

Do you see why this is true? @rapid lagoon

If it's for a homework you'll have some holes to fill in the argument obviously

e.g verifying that this is indeed a topology

[radiant cedar]

They should be. How do you conclude that X is contractible for n = 1 when we are removing the diagonal?

[coarse night]

it's homotopy equivalent to S¹, I was mistaken. See later msgs

[radiant cedar]

I see, sorry didn't check the later messages. Will the homotopy I described work here?

[quick bough]

so i want to prove the universal coefficient theorem, but i’ll need to cover all the basics of algtop for this (i want to write like a small thesis on this), do you guys have any good literature on this?

[nimble portal]

@urban zinc Do you mind checking my written proof for the second-countability of the line with two origins? Just writing down everything I wrote last time here

[novel acorn]

Word

[nimble portal]

quiet.

too lazy to install a tex editor and learn how to format things..

word easy...

[novel acorn]

Overleaf moment

[nimble portal]

uggy but easy...

[novel acorn]

also here [p] means the image of p?

[nimble portal]

Huh? WHere

s [p]

[novel acorn]

[(q,y)]

I meant p as an arbitrary point

[nimble portal]

p is just an arbitrary point in M

The points in M are equivalence classes

[gritty widget]

learn it

[nimble portal]

(M is the line with two origins)

[gritty widget]

i'm goint to tell you every single time you post something written in word

[nimble portal]

LMAO

It's just the formatting thing that gives me pain Going onto newlines and making text a certain size and bla bla bla

Speaking in TeX is the easy part

[quick bough]

tex >

[novel acorn]

Tbh I'm not too sure what ur doing here

[nimble portal]

Here's M if that helps

[novel acorn]

you first show that there exists an set indexed by rationals that contain p

that's okay

[nimble portal]

This is the "standard second-countable basis for R" I was referring to

[novel acorn]

Okay yeah
Your second paragraph leaves a lot of details out (like showing that what you claim is actually a basis)
Also you only say that x is in B_i
Where does the y come from?

[nimble portal]

The y comes from {-1, 1}

[novel acorn]

Oh wait

I get it now yeah

[nimble portal]

How did I not show it was a basis?
I first showed that given a point p in U subset M we can find a set V in U containing p

[novel acorn]

yeah

[urban zinc]

The wording of the second paragraph is kinda confusing

[novel acorn]

and
That is not the only condition for a basis?

[urban zinc]

Wait what

[nimble portal]

So if we just consider arbitrary points in M

Well the x coordinate of the equivalence class representing the point is gonna be in an open interval (with rational endpoints) in R

[novel acorn]

she said she showed that for any x in open U there exists V in the basis such that x in V subset U
But that isn't the only condition for a set to be a basis

[urban zinc]

I think a set of open sets B is a basis for the topology of M iff for every p in M and every open set U containing p, there's a set in B which contains p and is contained in U?

[nimble portal]

Oh technically I didn't do the reverse inclusion

[nimble portal]

So if we just index all those intervals

[novel acorn]

isn't there a condition that if x is in the intersection of two elements of the basis there is a third element that contains x and is entirely inside the intersection

[urban zinc]

That's implied by this

[novel acorn]

ah wait

that might be equivalent

yeah

[nimble portal]

Really?

Dang

[novel acorn]

I was doing a Munkres where he names them as two different conditions

[urban zinc]

Yeah an equivalent one is that the sets of B cover M and the intersection thing you mentioned

[nimble portal]

Maybe I'm a bit biased but I thought it was pretty clear

Why si it confusing?

[novel acorn]

man depression is actually impacting my ability to do math wtf

[urban zinc]

Okay tbh it's fine if you write V_i,j = {[(x,j)] : x in B_i} for j = 1 or -1

Currently you have two different sets named V_i

[nimble portal]

Oh

[urban zinc]

Otherwise I think it's good

[nimble portal]

I thought I had one named V and others named V1, V2, V3, ...

[urban zinc]

i just needed to take a while to read it

[urban zinc]

You have two different sets named V1, one corresponding to y=1 and one for y=-1

[nimble portal]

OH

GOOD CATCH.

Fuck me I made the same mistake I did yesterday too 😭

[urban zinc]

I was mostly confused about that haha otherwise it's good

[nimble portal]

keep forgorring I need to include the indices for each y

[urban zinc]

(Just make sure you can spell out the details easily if prompted, like why is that set countable, why is that set open, why can we choose a and b to be rational, etc)

[nimble portal]

Yep I totally can

[urban zinc]

Great!

[nimble portal]

Is this bad notation?

I think it looks pretty clean but maybe not the most immediately readable :p

[urban zinc]

It seems good!

[nimble portal]

poggers

[nimble portal]

I think this does the trick?

There's the case that O_1 and O_2 are singletons but then they wouldn't be open in quotient top so it doesn't matter anyways

Maybe instead of "must also be a member" I should specify that it's a common member

[next crystal]

what does [(0,1)] mean

the brackets

[gritty widget]

equivalence class

[abstract saffron]

Dw, I have seen far worse

[urban zinc]

Yeah this is fine, if you want to be more explicit I would specify that the preimage of O1 has to contain the intersection of a ball with radius eps1 with X, and similarly with O2 and eps2 so choosing x with 0<|x|<min(eps1,eps2) does the trick

Also use LaTeX!!! Overleaf is free and easy to use (of course it's not the only LaTeX setup option)

[nimble portal]

😭

After this pset

whch will be a while I’m gonna get to the stereographic proj one in a bit

[urban zinc]

I'm starting a diffgeo reading group with some friends soon 🥺 should be fun

[urban zinc]

That one's not too bad dw

[coarse night]

cool

[nimble portal]

INVITE? :3

[urban zinc]

It's not online sorry :(

I can tell you what sections we do though

I think we're gonna go somewhat slowly and idk if we're using Spivak or Lee

[gritty widget]

spivak DG or spivak CoM?

[abstract saffron]

As a rule of thumb, Spivak

Although, for diff geo, I'd recommend Sternberg

It's a bit horrible with all the technicalities, but you'll survive

[urban zinc]

DG

For context the other ppl in the group are physics grad students

[coarse night]

Gl

[unreal stratus]

GL(n,K)

[coarse night]

I heard a physics guy asking what a soft manifold is

[abstract saffron]

Ahh, so maybe something GR?

[urban zinc]

I think they just want to review diffgeo, not specifically physics rn

[thorny agate]

how is this red-underlined part not always an equality even in the infinite union case?

I can't think of a counterexample

[urban zinc]

Is this an arbitrary topological space

[thorny agate]

yea

[coarse night]

it's not usually equality, it's equally when the collection is locally finite

for a counter example, look at A_n=(1/n, 1]

[thorny agate]

what does locally finite mean?

[coarse night]

good exercise to show the equality when the collection is locally finite

locally finite means each point has a nbd that intersects only finite many members of the collection

[thorny agate]

Got it

ok I'll try that counter example + that locally finite condition

[coarse night]

got the counter example?

[thorny agate]

uh haven't had a chance to write it out lol

[coarse night]

okay

[thorny agate]

the difference is that 0 is not in the union of the closures of the A_n but 0 is in the closure of the unions of the A_n?

[coarse night]

yes

[thorny agate]

I have a ``proof'' of the reverse inclusion, which of course is false but I can't find the mistake.

Suppose $x$ is in $\overline{\bigcup A_i}$. Then for all open sets $U$ such that $x \in U$ we have that $U \cap \overline{\bigcup A_i} \neq \emptyset$. Then we have that for some $A_i$, $U \cap A_i \neq \emptyset$. Thus $x \in \bigcup \overline{A_i}$. I can't see the error.

[gentle osprey]

Spamakin🎷

[coarse night]

U ∩ (∪A_i)= non empty

and you get the intersection with the union is nonempty meaning one of the U_i intersect noj trivially

[thorny agate]

yea

[coarse night]

doesn't mean that the same set will intersect any open nbd of x

[thorny agate]

right but at least one set will, might be a different A_i for each U

but at least one

oh

nvm nvm I see

[coarse night]

cool

[restive kernel]

does anyone have recs for a good algebraic topology textbook?

[abstract saffron]

I guess #book-recommendations

[thorny agate]

what happens if I change "metric space" to just "topological space"

I can still show that second countable => countable dense set

but my proof for the reverse direction relies on me forming epsilon balls

is there a way to get around this or is there some counter example? I can't think of one

[urban zinc]

there is a counterexample

[thorny agate]

ok

[urban zinc]

can't remember what it is rn but there is one...

[nimble portal]

Hmm

Is it something like this for the north pole side?

Compute the line through x = (x^1, ..., x^(n+1)) and N

Set that equal to any point on the linear subspace where x^(n+1) = 0 to find the intersection

The intersection will be (x^1, ..., x^n, 0) = (u, 0)

Then compute sigma(x), noting that since x^(n+1) = 0, you get sigma(x) = (x^1, ..., x^n)

So sigma(x) = u

[tender halo]

plenty of counterexamples, 2^c (product of c copies of a discrete space with 2 points) is not second countable, neither are niemytzki plane or sorgenfrey line, but they all have countable density

sorgenfrey line is what you want to look at probably, its the easiest example of the bunch

R/N is another easy example

there are countable spaces with no countable bases too

there is a countable dense subset of 2^c which has no countable base at any of its points

there is another construction i can think of but its too long to be put in sentence or two

[urban zinc]

yeah what's the equation for this line?

[nimble portal]

(0, ..., 1) + t(x^1, ..., x^(n+1) - 1)

And if it passes through the subspace then x^(n+1) = 0 and you get t = 1 to satisfy the eqn

[urban zinc]

Random thought, interesting that your question is different lol must be a different edition

[urban zinc]

not quite t=1

You want the last coordinate of (0, ..., 1) + t(x^1, ..., x^(n+1) - 1) to be 0

so what value of t should you choose

[nimble portal]

1 + t(x^(n+1)-1) = 0 => t = 1/(1-x^(n+1))

I was lazy with my math hehe oop

[urban zinc]

Yeah that's good

And that shows that sigma(x) is what you want it to be

And then do basically the same thing for \tilde\sigma

[nimble portal]

Indeed

Do me a favor?

[urban zinc]

What lol

[nimble portal]

Send the whole problem from Lee’s book? I’m away from my desktop now

[urban zinc]

I have a different edition lol but I'm sure it's just as good an exercise

btw did you also do this one? it's cool and quick

[nimble portal]

dw those were the next questions anyways

[coarse night]

R with right limit topology, aka Sorgefrey line

[thorny agate]

yea i gave up and googled an example and found that

[coarse night]

I never solved Lee beyond chaper 1

[radiant cedar]

I'm going through the proof of the Jordan-Brouwer separation theorem and I have some troubles with the part where they show that Σ is the set of boundary points of the bounded component U_1 and the unbounded component U_2. Specifically how is it possible that the curve gamma intersects Σ when we are essentially removing a slice of the circle (I'm thinking S^1 here) and creating a path that goes through the removed slice? The only way that R^n \ A becomes path connected is if there is a "hole" in the space homeomorphic to S^1 through which we can go through with gamma.

[abstract saffron]

I never read Lee beyond chapter 1

[nimble portal]

zamn y’all some haters

[lean knot]

Is it possible that in a metrizable topological group G, there's an open set U and an element x ≠ 1, s.t. U = U ⋅ x ?

[ocean narwhal]

U = G

[lean knot]

thanks

[coarse night]

Lol

If your U is non the entire thing then it’s impossible however @lean knot

[lean knot]

how come?

[ocean narwhal]

the empty set

[coarse night]

Hint: WLOG let U be a nbd of 1 then can you find an element of U.x which is not in U?

Try to generalise

[lean knot]

i get it now, thanks

[lean knot]

Let X be a metric space, G be the set of all elements in X × X of the form (x, x). Then is G necessarily a G_δ set in X × X?

Yes, it is

Sorry to ask some questions that are silly to you, but I do not major in math. It's not trivial to me

[abstract saffron]

Strangely this feels more measure-theoretic than topology

I am tempted to say yes, but I don't see a rigorous proof. My intuition can be wrong

[abstract saffron]

It's not trivial at all

Not your fault

[abstract saffron]

Yes, my intuition is correct

[nimble portal]

Any hints for a starting point to derive the formula for the inverse of the stereographic projection map?

[abstract saffron]

Draw a figure, and see geometrically what is happening

The rest is analogous for higher dimension

[umbral panther]

It could be an open subgroup, like Z/2 in Z/4, or SO(2) in O(2) or Z_p in Q_p

[coarse night]

Ok fair, W LOG

Hmm thought for all x for some reason

[quick bough]

is there an easy counterexample for a net such that the filter generated by it yields a different net?

or more like what's the intuition why it doesn't necessarily yield the same net

[white oxide]

@coarse night well?

[coarse night]

Think of iteratively building spaces from ground up

You start with points then add lines between them then fill some lines and so on

You keep attaching higher “cells” to the previous “skeleton” of your space

For example S1 is a point and you add a line from that point to itself

[white oxide]

I was just trying to joke that your category of Top spaces has CW-complexes as its objects : (

[coarse night]

Jokes on you then

I mean that’s somewhat true

[white oxide]

Actually for a serious question, can you make a category out of CW-complexes or how much do you have to extend it?

[coarse night]

If you are willing to throw away “some spaces”

[white oxide]

Smallest category with CW-complexes as objects nice enough to allow for the homotopical constructions of choice*

[coarse night]

Category of CW complex is a real thing where your maps are “cellular” meaning n-skeleton goes inside n-skeleton

[hidden crag]

i mean

you don't really need cellularity

[coarse night]

Also every space is weak homotopy eqv to a CW

[hidden crag]

you can use cellular approx and consider it a subcat of top

by just using cont. maps

[coarse night]

I wasn’t considering upto homotopies but sure

Same thing anyway

[coarse night]

This allows you to just work on CW spaces in homotopy theory

[white oxide]

Every topological space meeting what criteria?

[coarse night]

Just existing

[white oxide]

That's pretty interesting considering how batshit stuff can go

[unreal stratus]

OwO

[hidden crag]

me

[coarse night]

I said “weak homotopy equivalence”

[white oxide]

Ah...

Looking up weak homotopy equivalence that sounds more like more of a fact about homotopy groups and their structure than the underlying spaces

[coarse night]

Yeah homotopy groups are almost enough to classify CW spaces

Ok the almost is too vague nvm

[radiant cedar]

I'm trying to find an example of pointed spaces (X, x_0) and (Y, y_0) and a map f : (X, x_0) -> (Y, y_0) such that the induced map on the fundamental groups f_* : pi_1(X, x_0) -> pi_1(Y, y_0) is surjective, but not injective. Will taking X to be the torus S^1 x S^1 and Y to be S^1 with f being the projection work here?

[coarse night]

Yes

[radiant cedar]

I'm quite lost on figuring out how to verify that the kernel of this map is not zero. I can see that it's surjective since for any loop in Y we can take the corresponding loop on either one of the factros of S^1 x S^1 and use the projection. How should this injectivity be shown?

[coarse night]

On pi1 level you have ZxZ to Z

[white oxide]

The loop has to on the side you are projecting onto

[radiant cedar]

Or can I just say that for any loop in S^1 there is at least two loops in S^1 x S^1 which map to it

[coarse night]

Which can never be injective

[white oxide]

*potentially easier to realize why Z->Z^2 can't be surjective which implies Z^2->Z can't be injective

[coarse night]

How does that implication hold?

[radiant cedar]

I guess the same spaces works if we want to find the spaces for which the induced map is injective and not surjective. Taking X to be S^1 and Y to be the torus S^1 x S^1 and letting f be the inclusion map?

[white oxide]

Existence of left/right inverses and identity being bijective?

[coarse night]

That works for sets

But then there is a surjection of Z to Z2 as sets

Unless I’m missing something trivial

[radiant cedar]

If this is true then what I proposed doesn't work

With the inclusion I mean

[white oxide]

Right, I was assuming the part where surjectivity has been shown here for the homomorphism which would imply that it is an isomorphism, but there being no epimorphism contradicts that.

[coarse night]

You can try much simpler example like take a 2pt set and a one pt set

If you are struggling

[radiant cedar]

The fundamental group of these sets are trivial both?

[coarse night]

Also like kerr mentioned, say you are projecting onto the first factor then any nontrivial loop in the second factor goes to 0 failing injectivity

[coarse night]

Oh on pi1 level. That works for pi0

[white oxide]

Suppose f_* : Z^2 -> Z is injective and surjective, then f* is an isomorphism and hence there exists an isomorphism Z->Z^2 of groups. Show that no surjective homomorphism Z -> Z^2 exists

^ that wouldve been what I meant

[coarse night]

Injective does that mean surjective

There are special cases when this is true, like say a surjection from Z^n to Z^n implies injection hence iso

[white oxide]

f* being an isomorphism means that there exists an inverse isomorphism, but that would be contradicting the fact that no epimorphism Z to Z x Z can exist.

[coarse night]

I don’t actually get what you are doing

What are you assuming and trying to show?

[white oxide]

Given: f star is surjective. Assume: f star is injective. Implies: there exists a surjective homomorphism Z -> Z^2.

The last thing is false, hence the assumption that f star is injective must be false as well

[coarse night]

oh I get your poiny ˡ now

[trail charm]

i dont fully understand what #4 is intuitively

what is {omega}

as in, what is the “set of the set of the naturals”

[novel acorn]

it's the set that contains the set of all natural numbers

[trail charm]

(for context, it is an example of a topological space)

yeah but whats the importance of it

this notation shows up here too

[novel acorn]

I think it might be there to represent like an element at infinity?

[trail charm]

so i guess 4 is supposed to mean “X = all natural numbers and the set of natural numbers itself?”

oh yeah that makes sense

[novel acorn]

I assume that's needed for some technical reason to make the topology he chose work

(probably something to do with those complements of finite sets)

[trail charm]

(wonder why he doesnt just use N for naturals and omega_1 for first uncountable ordinal)

[novel acorn]

old notation probably

[trail charm]

published 1993

[umbral panther]

Open subgroups are boring. But there’s another example. Take a nontrivial normal open subgroup. Take U a coset and x a point in the subgroup. Then Ux=U. Eg, U is the reflections in O(2) and x is a rotation

[urban zinc]

omega is standard notation for the natural numbers (as an ordinal)

[trail charm]

gotcha

[urban zinc]

I guess it's just to emphasize that you should be thinking about it as an ordinal?

Do you know the set theoretic construction of the natural numbers?

So like 0 = {}, 1 = 0 \cup {0}, 2 = 1 \cup {1}, etc.

[trail charm]

yea

ok yea that makes sense

[urban zinc]

Okay yeah, omega \cup {omega} is the next ordinal after omega

(we call it the successor ordinal of omega)

there are essentially three types of ordinals: zero, successor ordinals, and limit ordinals

0 is zero (obviously), 1 is a successor ordinal, and omega is a limit ordinal

note that the successor of omega is still countable, the first uncountable ordinal is denoted by either omega1 or Omega

(and it has cardinality aleph-one)

[trail charm]

i see i see

interesting ty

[fickle hamlet]

If Y is a subset of X. How does A and B disjoint nonempty sets whose union is Y and the limpoints of A and limpoints of B are all different imply closA intersect with B is empty?

Wouldn't something like Y=(0,1]union (2,3) subset of R, and A =(0,1), B = (2,3)union{1}, be AunionB =Y and A intersect B be empty, but 1 is a limpoint of A which is in B

[novel acorn]

Assume it's non empty

Then there would exist a point x such that any nbhd of x intersects both A and B

But that would imply that x is a limit point of both A and B

Which is a contradiction

[novel acorn]

This example violates your second assumption

Namely that the limit points of A and B are different

[fickle hamlet]

but limpoints of A is [0,1] and limpoints of B is [2,3]

[novel acorn]

Okay wait lol hm

Okay sorry yeah

[novel acorn]

Do you mean cl(A) intersect int(B) by any chance?

[fickle hamlet]

no, it just says closA int B

[novel acorn]

I believe what you wrote doesn't hold in full generality

[fickle hamlet]

this is the lemma

im lookin at this part

the bottom picture is just for context

[novel acorn]

Can you send the full proof

[novel acorn]

Tho the case might be I'm just being a massive dunce

[fickle hamlet]

i mean this is just for the first part

and thats pretty much all

also this is right before it

o wait did i misunderstand the lemma maybe

i think it means A doesnt conatain limit points of B

and B doesnt contain limit points of A

then my example fails

ok then I think it works because, A' be the set of limpoints of A, closA intersect B =(AintersectB) union(A'intersectB)

and A int B is empty so it must mean A'intersect B is nonempty if we assume closA intersect B is nonempty

ya that makes more sense

[thorny agate]

5f, how do I show the subspace is connected

[urban zinc]

First show that {0}×[-1,1] and {(x,sin(1/x)) | x>0} are connected so that if there were a separation, the former would have to be in one open set, and the latter would have to be in the other open set

And then show that those two sets can't be disjoint by openness

[coarse night]

There’s a much easier way, show A connected then so is A closure, show that your set is the closure of {(x,sin1/x)}

[thorny agate]

Ok yea this was much easier since I already showed the first part

How do you recognize that this is the closure though

[urban zinc]

Show that 1) it's closed and 2) {0} × [-1,1] are all limit points of {(x, sin(1/x)) | x>0}

[coarse night]

hint: exercise

tbf you don't actually have to show it's equal to the closure

just show all the points on Y axis is a limit point of your set, that's enough

[thorny agate]

Well yea but I mean like looking at that set

I've never proved before that set is the closure of {(x, sin(1/x)) | x > 0}

So I would have never thought to do that direction of proof

[coarse night]

btw A connected then for any set with A ⊂ B ⊂cl(A) is connected

[coarse night]

how did you prove it?

[thorny agate]

Prove which part?

[coarse night]

cl(A) connected

there's a proof I like

[thorny agate]

I used discrete valued maps since they seem overpowered

ok so I'm just using the fact that any continuous discrete valued map is constant on a set iff that set is connected.

Let $X$ be a connected set and let $f\colon \overline{X} \to \{0, 1\}$ be a discrete valued map. 
$X$ is connected so $f(X)$ is constant so WLOG say $f(X) = \{0\}$. 
$\{0\}$ is closed and $f$ is continuous. 
Thus $f^{-1}(0)$ is closed and contains $X$ and $\overline{X} \subseteq f^{-1}(0) \subseteq \overline{X}$.
Since $f$ is a discrete valued map and constant on $\overline{X}$, $\overline{X}$ is connected.

[gentle osprey]

Spamakin🎷

[thorny agate]

@coarse night

[coarse night]

ah cool that's the one I like as well

[thorny agate]

what's the proof you had in mind

oh nvm I misread what you said

ye it's slick

discrete valued maps seem like an overpowered tool but very nice

[whole ridge]

Can we speak of "homeomorphisms" for something other than topological spaces?
Like can we say that the symmetries of a geometric figure are homeomorphisms?

[thorny agate]

homeomorphisms need some sort of continuity

And the symmetries of a geometric figure, while they form a group, don't really have any continuity

What would be the open sets?

Unless I'm missing something

[coarse night]

you might want to look at continuous group actions

[urban zinc]

It's just that you need to use the fact that the sin(1/x) bit gets arbitrarily close to the y-axis

i.e. the y axis are limit points

if you could separate them by some finite distance then it would no longer be connected

[coarse night]

Anyone have some motivation for these relations?

[whole ridge]

I think I should add some context:
I work on Tic-Tac-Toe grids and one could say that some grids are "equivalent" or "similar" except for a few deformations. For example these two grids.

In fact, the deformations that can be applied to a grid without really "changing" it (without the position of the pawns between them changing). Exactly match the symmetries of a square.

[coarse night]

These are just group actions, you don’t need any topology for this. ask in #groups-rings-fields

[urban zinc]

It seems like you want to look at groups

^ what they said

[whole ridge]

Can we call these deformations homeomorphisms even if there is no notion of "continuity"?

[urban zinc]

No they're just group elements

There's no use of introducing topology here

[coarse night]

Doesn’t sound a very useful thing to ti define

[whole ridge]

Ok thanks, but actually I just wanted to know what word to use 😅...

[thorny agate]

Symmetries of a shape are a classic example of a group

Which would go into #groups-rings-fields

[umbral panther]

There are two answers to what is the steenrod algebra that are nicer than the original. One is Milnor’s description of the dual. The other is the formal group of automorphisms of the additive formal group

But they’re not a lot nicer

[coarse night]

Can you link or refer some?

Milnor char classes?

[grizzled ibex]

Any big list like the biglist from toronto, but for algebraic top?

pls send 😭

[umbral panther]

Is the steenrod algebra in that book? I don’t think so, but maybe I forget

The key miracle is that the steenrod algebra acts faithfully on the cohomology of RP^oo. This was known to Adem, if not Steenrod. Everything follows from that. In particular, it implies the Adem relations. Of course, this is circular. You have to prove them generally before you can conclude that it acts faithfully, but you could guess that to motivate the relations

The steenrod algebra is acts by hopf algebra endomorphisms of the cohomology of RP^oo. The first miracle is that it injects. The second, maybe less surprising result is that it surjects. This is maybe more conceptual and you can create cleaner calculations. But then you’re still left with the question of where did the squares come from and what their crazy relations are

[tiny ridge]

I find this a bit backwards. Steenrod algebra is by definition End(HZ_2) to me. Adem relations are a computation of the algebra.

But yes, I agree that the point is all mod p cohomology operations are some combination of the Steenrod squares aka they generate

I think there might be a geometrically minded way of going about doing Steenrod squares. Write Sq of a class in a manifold in terms of the Stiefel Whitney class of the normal bundle to an embedded representative, where you can take one of the various many definitions of Stiefel Whitney classes. Then, to deduce Adem relations, use Wu’s formula, which you can deduce from splitting principle.

Of course, this only works with classes which are PD to homology classes admitting embedded representatives

Even immersed.

[jaunty glade]

im going to start my undergrad thesis this coming semester and ik i want to do it on something in topology or algebraic topology but im unsure of what to do it on. my advisor and i talked about doing it on normality of topological spaces (particularly on urysohn's lemma and the tietze's extension theorem) but i dont think i want to do it on normality since he said its been done many times before. any tips?

[novel acorn]

how much algebra do you know

[novel acorn]

for a less algebraic thing
Maybe you can talk abt topological groups?
Or LCH spaces

[abstract saffron]

Math aside, if you wanna do something original with an undergrad thesis, you probably have chosen the wrong discipline 😄

[jaunty glade]

ik a little bit of algebra. i took an intro course on algebra (95% group theory and 5% rings) and a second course on algebra (which was 90% group actions and 10% ring theory, which still wasnt a lot)

[novel acorn]

you can talk abt the fundamental group of a topological space

[jaunty glade]

ive been reading through a text i like to help me get ideas (thats where i got the idea to do it on urysohn's lemma/tietze's extension theorem) and the main reason why im unsure of whether i want to do it on that is cos the theorem is very early in the text (in the topological prelims chapter of it)

[tiny ridge]

Here’s a cool result that takes some work to write a proof of. Let X, Y be compact metric spaces. If X x R is homeomorphic to Y x R, then X x S^1 is homeomorphic to Y x S^1. This has significance in the solution of the 4D topological Poincare conjecture. You can consider presenting a proof of this.

Just a thought though

[novel acorn]

This feels like a covering spaces thing

[tiny ridge]

The converse sort of would be

The point is, given a homeomorphism X x R -> Y x R there’s no reason its periodic in the time direction. Ie, say f : X x R -> Y x R is a homeo, g(x, t) : X x R -> Y is projection to the first coordinate. If f(x, t+1) = (g(x, t), t+1), you’re done. Just pass to the quotient by the natural Z actions on both sides.

time = R

[novel acorn]

topological groups sound like a fun thesis
Tho depending on how you approach them they can also be hellish

[jaunty glade]

they could be hellish? 💀

[novel acorn]

look up haar measure if you wanna see

realistically there are some very nice results abt topological groups that aren't that bad you can prove

regarding like how they relate to compactness and separation axioms

nothing groundbreaking but semi interesting in my opinion

[jaunty glade]

topological groups sound pretty interesting

[novel acorn]

There's a part in Munkres abt them

[jaunty glade]

the text i have been reading introduces them in the second chapter

but it uses a little bit of category theory

[novel acorn]

and in Folland's real analysis book in the last chapter where he develops analysis on them he starts by proving topological results abt them and that's very readable

[novel acorn]

ah that's not that bad haha
Good practice to learn more category theory

[jaunty glade]

yeah i never got any exposure to category theory except for the soft intro to it in one of the appendices of the text

[novel acorn]

As somebody I know once said "category theory is something you learn privately"

It's good to have exposure but no need to force urself

I think Folland's text gives some nice results around them so you can look in there if you want

[jaunty glade]

that sounds like a good read

[novel acorn]

Ah wait it actually only has one

hmmmm

The other book I know that does them is called Fourier Analysis on Number fields
But the proofs are very terse

tbh now I'm worried I don't have a good reference for you

[jaunty glade]

i dont mind terse proofs

the text i referred to before is called 'manifolds, sheaves, and cohomology' by t. wedhorn

[novel acorn]

oh lol I actually ordered that like yesterday lmaooo

[jaunty glade]

xddd

[novel acorn]

Okay then look at section 1.1 in Fourier Analysis on Number fields by Ramakrishnan

[abstract saffron]

this sounds utterly familiar...

[novel acorn]

wow yeah who would say what a coincidence

[abstract saffron]

I think I saw this exact title twice today

[novel acorn]

My hand slipped during the springer sale and accidentally ordered 4 books

[jaunty glade]

your hand slipped and ordered 4 books

[abstract saffron]

yeah, slipped

totally buying it

[novel acorn]

look bro it's 15$ per book that's like if a dude came up to you and said free crack
You aren't gonna say no

[abstract saffron]

me, never bought a single book my whole life : no

[jaunty glade]

i would say no

cos crack is wack

[novel acorn]

I mean neither did I but like why not
I have now slowly accumulate more books until my bookshelf is a very peculiar shade of yellow

oh wow I just now realized springer books are piss yellow

what a world

[jaunty glade]

i get free pdfs of the springer books on the site xd

[abstract saffron]

Thing is, I will move quite often, and it's not wise to carry Library of Congress with me

[abstract saffron]

Ssssshhhh

[jaunty glade]

wot

[novel acorn]

True
Last time I moved I had to carry two books in my hand for an 11h flight cuz my bags were too heavy lmao

[abstract saffron]

there's the site, and there's the site

[jaunty glade]

i meant the springer site

xd

[novel acorn]

springerlink my beloved

[jaunty glade]

yes

i love it

[novel acorn]

Tbh while I was at my last uni I just religiously borrowed books from the library

I had like 15 books out at any given time

[abstract saffron]

I am doing exactly that

[jaunty glade]

i havent been to my school's library in a long time

[abstract saffron]

having 6 books surrounding me at the moment

[jaunty glade]

but sometimes a prof would let me read through the extensive library in his office

[novel acorn]

Well I'm probably gonna be localized in a single continent for the time being so no need to worry abt taking books that far

[abstract saffron]

sigh my prof doesn't have a lot of books

[jaunty glade]

one of my profs entire office is nothing but full bookshelves xd

and he lets me read through it sometimes

[abstract saffron]

Idk why but except the classics, profs I know like to shit on books

[jaunty glade]

shit on books-

[abstract saffron]

Like "oh, this is wrong btw, everyone thinks so but it's actually wrong"

[languid steeple]

anybody have an idea?

[novel acorn]

what is the closure of S

[jaunty glade]

imagine using that notation for subset

[novel acorn]

I think drawing the set would help a lot

[abstract saffron]

try to draw it out. it's drawable, and the answers are immediate once you draw it out

[jaunty glade]

yes

[languid steeple]

okok tysm

my guess is {(1/n, 0): n ∈ N} U {(0, 1/m): m ∈ N}, am i correct?

o

nvm

[novel acorn]

you're missing a point

[languid steeple]

missing a point?

like a point on a grid?

i think i got it

thanks

[civic verge]

b) es vacio

[fresh thunder]

Here is my answer: (a) is S, (b) is empty set

I can explain in details what you wany

[bitter smelt]

if you combine all the answers given here for part (a), along with the hint, you have it

the power of friendship fr

[tulip bluff]

Can someone here explain what is meant by some homology groups having torsion elements?

[solemn oar]

Torsion elements are elements of finite order.

[tulip bluff]

What does this mean in the context of homology, does H_1(S^1) have torision elements as even though it's isomorphic to Z that has no finite order elements, but going looping around it multiple times is essentially the same as looping around once?

[cosmic socket]

As you said Z has no elements of finite order so it is torsion free. I don’t know what you mean by looping around multiple times is the same as looping around once.

[solemn oar]

Z has no torsion.
Torsion elements are detected by torsion groups, Tor^1(G,Z/p^k)

[tiny ridge]

Looping around multiple times is not the same as looping around once, the winding numbers of these loops are different, even though they have the same image.

That’s basically why H_1(S^1) is Z and not Z/nZ

[tulip bluff]

This makes sense. Thank you (:

[tiny ridge]

On the other hand, consider H_1(RP^2) = Z/2Z. Can you see why the generator of the first homology is 2-torsion, geometrically?

[tulip bluff]

I haven't computed the homology for the projective plane, but I guess it would be due to identifying the anitpodal points on the 2-sphere?

[tiny ridge]

Correct. Do the computation, and then geometrically explain this phenomenon.

It’s a neat exercise

[tulip bluff]

I'll do that (:

My algebra is a bit rusty, but what is the definition for 2-torsion? Is it just having elements of order 2?

[tiny ridge]

Yep.

[tulip bluff]

I might be very wrong here, but say I have a group G that has elements of order two, is there necessarily a subgroup of G that is isomorphic to Z/2Z?

[tiny ridge]

Yes, that’s correct. Can you give a proof?

(What if G has an element of order n, where n is some arbitrary integer?)

[tulip bluff]

It probably generalizes to Z/nZ. There is no requirements for G to be abelian?

[tiny ridge]

None.

[tulip bluff]

Sure. I'll try to workout these and the homology for the real projective 2 space, thank u (:

[unreal stratus]

Here's a like funne thing I was thinking about uh like

Say you take a wedge of two spheres, or you take two circles and attach them using a line between them

The two are homotopy equivalent, but the main way I know to say this is to appeal to the theorem that we're quotienting out a contractible subcomplex of a CW complex

Is there an obvious homotopy equivalence though

[tiny ridge]

Both are deformation retracts of R^2 with two points removed. Is that an acceptable answer?

Deformation retracts are annoying to write down but can be pictorially described

[unreal stratus]

Okay fair enough lol. But coincidentally I was only doing this construction since it's easier to write an explicit deformation retraction of R^2 \ two points onto the latter

(I think)

Like continuity is easier to show

It also feels like this is the sort of thing that should be really easy lol

I'll think about it

[tiny ridge]

Ok. Say X is wedge of two circles, Y is a pair of spectacles. f : Y -> X is the quotient map. Ill describe a candidate homotopy inverse g : X -> Y

[unreal stratus]

Sure

thank

[tiny ridge]

Consider the midpoint of the nosepiece of the spectacle, and break it into two lollipops O-. Call this Z. Y is Z v Z, where the wedge point is the bottom of the lollipop stick. O—O = O- U -O

[unreal stratus]

Understood

[tiny ridge]

There’s a map S^1 -> Z which can be defined as follows. Let A be an arc in S^1. Fold the arc by identifying points to the right and to the left of the midpoint of A.

Wedging this map with itself gives a map g : X = S^1 v S^1 -> Z v Z = Y

[unreal stratus]

I'm having trouble sort of visualising that folding map hm

Oh I misunderstood nvm

V cool thanks

[tiny ridge]

To be 100% clear, if the arc A = [-eps, eps], the folding map is quotienting by x ~ -x for all x in A

[unreal stratus]

I imagine there's another way to do this so like

[abstract saffron]

Guys, I have a question

I'm giving a series of lectures on homology. Can I skip the treatment of reduced homology?

[unreal stratus]

You basically think of sending the path traversing the spectacles to one that travels a shorter distance along the nosepiece and then goes more quickly around the edge

Reduced homology is v important and should be pretty quick to describe right

[abstract saffron]

I have seen no applications, and it appears to me more like a side note

[abstract saffron]

It should be pretty quick indeed, but also annoying

[tiny ridge]

Just say H_0(connected space) is Z in unreduced, reducing means setting it 0

One sentence

Then move on

If you’re doing relative homology it doesn’t matter really

Because H_n(X, pt) is exactly reduced homology

[abstract saffron]

Exactly my point

[tiny ridge]

Yeah this makes sense

[tiny ridge]

H_n(X, my point)?

[abstract saffron]

good one

[solemn oar]

Surely you'll talk extensively about relative homology, so then defining reduced as a special case is easy.

[coarse night]

empty space

good luck convincing H_(-1)( Φ)= Z

[wicked yew]

is there a simple counter example I can use to show this is not true for every norm? I've been trying for a while but couldn't think of anything.

[tiny ridge]

I was shocked to realize a while ago that the empty map \emptyset -> X is a fibration

[coarse night]

lmao what

[coarse night]

so what's the lift of a map from any non empty set?

,tex \being{tikzcd}
& \varnothing \ar[d] \
{p}\times I \ar[ur, "?"] & X
\end{tikzcd}

[tiny ridge]

There’s no lift but thats the point right?

[coarse night]

oof NVM

How's that a fibration then?

[tiny ridge]

Fibration means if you have an initial lift, and a homotopy of the base, then the homotopy is liftable

But there’s no initial lift at all

[coarse night]

oh right start with a intimal

[tiny ridge]

So vacuously, the condition holds

[coarse night]

lmao

[tiny ridge]

Yeah…

[coarse night]

apparently these are called cursed fibration

gimme a slick argument as to why orientable real line bundle on a "nice" space is trivial

By nice I mean possibly paracompact

[tiny ridge]

I was going to say, choose a fiberwise Riemannian metric and then take the unit S^0-bundle. This is disconnected by orientability, hence a section of this double cover gives a section of the original bundle

[coarse night]

https://math.stackexchange.com/questions/4565706/an-orientable-line-bundle-is-always-trivial
I don't like this

[coarse night]

that's a cool one

but why the S⁰ is disconnected by orientability?

I know it's true but can't make a convincing argument

[tiny ridge]

Depends on your definition of orientability. If that double cover is not disconnected, its nontrivial, which means there’s a nontrivial monodromy over some nontrivial homotopy class of a loop on the base. This monodromy map R -> R switches 1 and -1, so its orientation reversing

[coarse night]

ah I see

cool argument

[tiny ridge]

What this proves is a three way bijection: real line bundles on X <-> double covers of X <-> Hom(pi_1(X), Z/2) = H^1(X; Z/2)

First is unit S^0-bundle wrt some metric, second is monodromy of the double cover

[coarse night]

last one comes from the classifying space?

[tiny ridge]

I didn’t use the classifying space anywhere

It also comes from that, sure

[coarse night]

otherwise I don't see the bijection

[tiny ridge]

By last one you mean (1) <-> (3)? That’s just composing (1) <-> (2) and (2) <-> (3)

These two are bijections, so is their composition

[coarse night]

ohh, yeah I meant 1 <->3

cool

[tiny ridge]

If you pick a line bundle L from (1) and push it all the way to (3), what you get is w_1(L), the first Stiefel Whitney class

So w_1 is just checking if a line bundle is trivial on a loop on the base or not.

For a general vector bundle E, you can go to the orientation bundle or the top exterior power \Lambda^k E (k = rk E) which is a line bundle, and then take its w_1

Thats what w_1(E) is

[coarse night]

yeah so It just checks the orientability

[tiny ridge]

Yup

And if its not orientable, along which loops is it not orientable

There’s a similar interpretation for w_2, but an order of magnitude more complicated

It checks if a bundle is “spinnable” or not

[unreal stratus]

I might try to find what I had lol

[wicked yew]

Ty

[left merlin]

Do i just try different points and see if they fit the criteria for limit points

[unreal stratus]

Well it should hopefully be clear what the answers are if you draw a lil picture or smth

So then you just need to prove specific points are in it / not in it yeah

[left merlin]

as in, draw something cartesian, like a [0, 1] by [0, 1] sqaure

[tiny ridge]

Is the “ordered square” [0, 1]^2 with order topology lol

Or just Euclidean?

[left merlin]

yup

but im slightly confused because an answer im looking at says that the only limit point is {(0, 1)} which im having difficulty parsing

[tiny ridge]

Ok. Drawing a picture helps, but be mindful of what your open sets are. They’re horizontal/vertical intervals (but something a little more subtle at the edges)

[left merlin]

hm i see

wouldnt every neighborhood of the point (0, 1) not intersect any point of A

[tiny ridge]

This is why I said you should be careful with the edges

For example, (0, 1) < (1/2, 0). And everything in between forms an open set.

Draw these sets

Theres nothing interesting happening on the other side, stuff smaller than (0, 1).

[left merlin]

Open sets in the ordered square are made up of open intervals (a, b) x (c, d)?

would appreciate some more guidance

am self studying so probably am missing some fundamental things

[tiny ridge]

No, that’s just the product topology

[left merlin]

would drawing this set look like a horizontal interval and a vertical interval overlapping at (1/2, 1)? or am i completely missing it

Ah i see

[tiny ridge]

Open sets are generated by intervals in the lexicographic ordering

In the lexicographic ordering, (0, 1) < (1/2, 0), yes?

[left merlin]

right, this is the dictionary ordered set

Yup

[tiny ridge]

So what’s the interval which has those as endpoints

[left merlin]

[(0, 1), (1/2, 0)]

That's helpful already

[tiny ridge]

But draw it.

Or describe what the points in that set are explicitly

[left merlin]

In 1d?

Ah

(a, b) where a is larger than 0 but smaller or equal to 1/2, and b is 1 ?

[tiny ridge]

Are you sure?

Hmm, yeah ok that sounds right

I’m a bit confused about the b = 1 part

It’s union of all the vertical segments {a} x [0, 1] where 0 < a < 1/2, together with the endpoints, no?

[left merlin]

Ah yup

What does {a} x {0, 1} mean precisely? I can sort of guess what you mean but am confused

How does that resemble a verticle segment

And side question - in the product topology open sets would resemble overlapping rectangles geometrically right

[tiny ridge]

{a} x [0, 1] means all the points whose first coordinate is a and the second coordinate is anything between 0, 1 (0, 1 included)

[left merlin]

Got it

[tiny ridge]

So thats a vertical segment right?

[left merlin]

Yup

[tiny ridge]

Well anything that can be written as an arbitrary, possibly uncountable, union of (possibly overlapping, yes) rectangles

Even the open disk is an open set

It’s the same as the Euclidean topology, which you can prove

[left merlin]

Hm i see

[tiny ridge]

Now, write down all the open intervals in the lexicographic ordering containing (0, 1) and show that they necessarily contain points of the form (1/n, 0) possibly for large n

Unlike the Euclidean topology, the open intervals sort of hit the top of the edge of the square and reappear out of the bottom edge of the square and keep doing this ad infinitum

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSN6YxpkrD4T7fgDlxU-kS9tnFtdHc2LfF3qA&usqp=CAU

Like this is an open interval in the ordered square. The vertical segments are uncountable in number, indexed by real numbers in between the first coordinates of the endpoints

[coarse night]

so these come up naturally or just counter examples

[left merlin]

Ok i think i got it

[tiny ridge]

Perfect!

[left merlin]

Every neighborhood of (0, 1) can be written as [ (0, a), (b, c)]. a has to be smaller than or equal 1 to contain (0,1), and b has to be greater than 0. c can be arbitrary.

each of those neighborhoods intersects with A at some point other than (0,1), and therefore is a limit point.

[tiny ridge]

Now that you mention it, I’m getting some funny ideas. I’ll say in a bit

[coarse night]

don't tell me they come up in foliation

[left merlin]

What I was confused about is if (0, 1) was a limit point, why (0, 1/n) wouldnt be one too. but if (0, 1/2) is x, then there exists a neighborhood of that point that doesnt intersect A at all and therefore isnt a limit point

[tiny ridge]

Exactly

[left merlin]

Thank you so so so much

You easily just made my day

[tiny ridge]

No problem!

[tiny ridge]

OK so

Take the strip 0 <= y <= 1 in Euclidean topology, and consider the foliation by irrational slope line segments y = sqrt(2) x + c, c in R

Grr, this doesn’t work!

I was going to try to produce the order topology on that domain by some foliations construction but I don’t think this is possible anymore

[tiny ridge]

I tried for a bit to find a natural occurrence, but couldn’t. So, not sure.

They’re kind of pretty though

[coarse night]

foliations lol

[left merlin]

Actually I think im missing one other thing

Every neighborhood isnt encapsulated by [ (0, a), (b, c)], because what about sets like [ (0, 1/2), (0, 1)] which is an open set of I^2 that has (0,1), but doesnt intersect A at all

[tiny ridge]

That’s not an open interval

Closed intervals contain endpoints, open intervals dont

The valid open interval would be ((0,1/2), (0,1))

[left merlin]

ah but that doesnt have 0,1

i see

[tiny ridge]

précis

[left merlin]

right - because closed intervals are never open sets at least in this order

ty again

[abstract saffron]

Does anyone here do knot theory?

[tiny ridge]

Knot really

What’s the question?

[abstract saffron]

I have an oral exam where I have to talk about a theorem of my choice, and was wondering if there's any particularly deep and profound result.

[tiny ridge]

This is pre-Thurston knot theory or post-Thurston

ie Rolfsen type thing or more geometry infused

[rough cedar]

the last time I answered this question my advanced access was revoked

[tiny ridge]

If the former, I was going to say Papakyriakapoulos’ theorem

Corollary: Knot complements are K(G, 1)

But thats a very old theorem

[abstract saffron]

I know there are Floer, Khovanov and Khovanov-Rozansky ring

[tiny ridge]

Oh you want to do some modern knot theory

[nimble portal]

WHAT DID YOU SAY

[abstract saffron]

But to me it's only for classifying knots. Not particularly profound, idk.

[nimble portal]

Dude I have like 1/18th of your brain cells and even I haven’t gotten my Advanced removed (somehow)

Wtf did you say for that to happen 😭

[abstract saffron]

I should specify. They're all insanely profound, no doubt

[rough cedar]

"not to brag, but I know how to tie my shoe laces with a knot. so yes, I would consider myself a knot theorist nvm this is adv"

[abstract saffron]

But I need a precise statement. Something that accumulates to one theorem

[hidden crag]

mad unfunny

[rough cedar]

your mum is mad unfunny

[coarse night]

Remove adv access again

[rough cedar]

I will drive over timo's dog if he does that

[tiny ridge]

Well, Thurston’s hyperbolization of knots which arent torus knots or have satellite components is super important and part of geometrization conjecture

but hardly something you can present in one lecture…

[abstract saffron]

Lol, I have 15 minutes 😄

And no, this time it's not a lecture. It's an oral exam.

I'll be murdered tested on the blackboard.

I can assume any background I want, just that I wanna impress the examiners a bit.

[tiny ridge]

I’m confused. You’re presenting a specific topic, and they’re going to grade you on that?

[abstract saffron]

They're gonna see if they admit me or not

Grad school application stuff, you know

[hidden crag]

[tiny ridge]

Oh ok

Yeah so you should probably present things which are way more modern than that

3-manifold topology is over

[abstract saffron]

I heard some guy presented Lefschetz's fixed-point some years ago, so apparently I can expect a very high level

[tiny ridge]

Oh ok thats not too bad

[abstract saffron]

It's OK 😄 not like I'm defending my PhD

[coarse night]

That’s not high level imo

[abstract saffron]

Exactly, it's lame

But I can expect everyone else to think the same

[tiny ridge]

How many years do you have to prepare

[abstract saffron]

a month

[tiny ridge]

Hmm

[abstract saffron]

Dw, I learn fast

And I have background in alg topo and basic knot theory.

Euhm, "basic"

There's also this recent result by Bayer-Fluckiger for an integer to be a signature of a 4q-1 knot in S^(4q+1), but I'm afraid I'll get myself annihilated with high dimensional topo

[tiny ridge]

How about Milnor fibrations? Certain knots are fibered, as in their complement can be written as a surface bundle over a circle with fibers being Seifert surfaces. They arise sometimes as singularities of complex polynomials eg z^2 = w^3 in S^3 \sub C^2 is a trefoil knot, and the map S^3 -> S^1, (z^2 - w^3)/|z^2 - w^3| is the fibering

Theres some extensive results on which knots are fibered.

[tiny ridge]

So specific

[abstract saffron]

Ohhh, I heard about this one today, like building a knot from singularities of curves or something

[tiny ridge]

Yeah

[abstract saffron]

Hmm, this might work actually

[tiny ridge]

Milnor has one or two nice papers

[abstract saffron]

In general he has an archive 😄 this man writes really well

[tiny ridge]

Agree!

Here’s a high dimensional knot theory result. Every knot S^2 -> S^5 is the unknot

Theres a critical dimension for S^n -> S^m

You need the h-cobordism theorem for this

[abstract saffron]

Hmm, this' gonna be a tough one to choose

[tiny ridge]

Do you know the slice-ribbon conjecture

[abstract saffron]

I certainly don't

[tiny ridge]

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

[abstract saffron]

Keep in mind that I just started learning knot theory 2 weeks ago, don't expect much

[tiny ridge]

Its all good, slice ribbon is a very important still open problem

Since you were annoyed about having so many invariants and not knowing what to do with them, I think Alexander polynomial detects if a knot is slice

[abstract saffron]

There's also one about if Jones' polynomial detects an unknot

But they go really, really deep, and I can't find good intro to the problems

[tiny ridge]

Jones is a very modern and hard invariant, IMHO. I don’t really understand what it means.

I forgot that its open if Jones detects unknots

Alexander polynomial is a super nice invariant. Lots of geometry involved

[abstract saffron]

There are many interpretations of it

But for jones', if I wanna go beyond skein relation, I have to dig very, very deep

[tiny ridge]

yeah

either gauge theory or quantum algebras

messy

[abstract saffron]

Yes

Also, isn't it ironic that ppl were correct the whole time? Knots don't represent elements, but they do represent(-ish) the fundamental particles.

[tiny ridge]

Uhm, do they?

[abstract saffron]

don't string theorists love this stuff?

Iirc, the idea is small strings loop, knot, vibrate somehow to make the particles

[tiny ridge]

I think they don’t care particularly about knottedness of strings. I think strings are just models for a fundamental particle inasmuch that they have vibrational modes which means that the energy spectrum is discrete (roughly), and that there’s some conformally invariant measure associated to their trajectories, which are 1+1 = 2 manifolds so Riemann surfaces

And then they can do their QFT crap with this conformally invariant theory

I’m not a physicist but this is my impression

[abstract saffron]

I wonder how feather would react to this

This is sure as hell a lot of buzz words

[tiny ridge]

I can explain what I mean by the words but itll only make me sound more stupid

[abstract saffron]

As if we be not already

[tiny ridge]

Take an actual physical string, hold the two ends. The waves made here satisfy the wave equation, and theres fundamental harmonics right

These are the discrete energy levels

Thats essentially why one would model a particle on a string, because energy is quantized in this hypothetical model

Plus, when a string traverses the spacetime, it lays out a worldsheet (a particle traversing in spacetime lays out a worldline in special relativity)

In SR or GR in general the worldlines are geodesics, satisfying the geodesic equation. So they should satisfy something in string theory also, yeah?

Essentially, they should be minimal surfaces.

This is the basic idea

Except, there’s no definite trajectory for anything after quantum mechanics. So every worldsheet is possible its just that some are weighted more based on if they’re close to satisfying the minimal surface equations or not

This is the “path integral measure” e^iS/h d\sigma on all possible worldsheets, where S is something that tells you what this weight is (its called the action)

[abstract saffron]

Holy 😄

[tiny ridge]

If you have to compute probability of certain physical events, you chug its pdf in the Feynman path integral, integrate with respect to e^iS/h d\sigma, and get your probabilities out

[abstract saffron]

If someone not well-respected tells me this, I'll say sober up and go touch some grass.

Hard to believe one can formalise all of this

[tiny ridge]

One can’t

[abstract saffron]

Intuitively it makes sense, but I can't imagine the math involved

[tiny ridge]

Much of QFT and by extension, string theory, is complete batshit

[abstract saffron]

Well, they proved something along the line of the dimension of the strings must be of a particular number for the theory to be consistent

So I guess there's some logic in there

[tiny ridge]

Dimension of the spacetime the strings are in yeah. That falls out of having to have a certain path integral converge iirc lol

You need an extra 11 complex dimensions, along with the 4 existing

[abstract saffron]

Lmao, that's it?

[tiny ridge]

I think so

[abstract saffron]

I thought it was something more profound

[tiny ridge]

Theres a factor of (d-11) as coefficient in front of some badly diverging path integral, if I remember correctly

So they just set it to be 0 lol

[abstract saffron]

Bruh, and ppl actually hope to get something useful out of this

[tiny ridge]

The cool thing of course is that if you do assume all of this, you get that the spacetime is R^4 x CY where CY are extremely special 11 dimensional complex manifolds

Theres some neat math that comes as fallout but idk about the way they get there lol

But again, I’m no string theorist

[abstract saffron]

Actually, where did it come from?

I know the Kauffman bracket interpretation, but it seems somehow arbitrary

[tiny ridge]

It comes from Witten successfully computing a specific path integral for Chern-Simons theory

[abstract saffron]

Wait, what?

😄

Nah, I'm not asking about string theory

[tiny ridge]

Yeah, thats the origin of Jones polynomial

[abstract saffron]

Wait, really??

How the hell...

[tiny ridge]

Yep, you cam look at the original paper

[nimble portal]

How do I go about deriving this formula for the inverse in part (b)??? Just a hint for where to start off...I don't really understand what it's doing geometrically

I can prove bijectivity but I just don't understand where that formula comes from

[abstract saffron]

More drawing, feather, more drawing

A very good picture can be found from Milnor

[nimble portal]

I mean like

The way I'm trying to visualize it is

[abstract saffron]

John Milnor, Analytic Proofs of the "Hairy Ball Theorem" and the Brouwer Fixed Point Theorem. The American Mathematical Monthly, Vol. 85, No. 7 (Aug. - Sep., 1978), pp. 521-524 (4 pages)

[nimble portal]

I know what the map from S^n{N} -> R^n does geometrically

It just takes a point on the domain across the line segment intersecting the point and the subspace where x^(n+1) in R^(n+1) = 0

So I'm trying to invert that in my head

[abstract saffron]

Yeah, well, that's it

Make a drawing, it helps a lot

[nimble portal]

I'm making mental drawings? 😭

[tiny ridge]

So youll join a line on the hyperplane x^n = 0 to the north pole, and intersect that with the sphere

[nimble portal]

Maybe I should try it on S^1 instead of S^2 lol

S^2 is my goto when I think sphere

[tiny ridge]

Good idea, S^1 is better

[nimble portal]

So I just need to invert the secant line between a point on the unit circle and the x axis? O.o

[abstract saffron]

Here's my rough drawing of it

[nimble portal]

YOUR rough drawing?!??~?~?

first of all

YUO DREW THAT?!

second of all

YOU DREW THAT JUST NOW?!?

[abstract saffron]

forgive my hand writing, it was from an old lecture of mine on Poincare-Miranda theorem

[urban zinc]

Find the equation for a line through (u,0) and N

And find where that line intersects the sphere

And then you get the formula for sigma^-1

[abstract saffron]

When you lock yourself with geometry long enough, you become proficient at drawing

[nimble portal]

Oh duh…

[coarse night]

I can see your proficiency

[tiny ridge]

Ok I looked the OG paper up, heres what the Jones polynomial actually is youll love it

[nimble portal]

Idk why I was trying to consider the line between the preimage on the sphere and (u, 0)

The whole point is the image is relative to the north pole

[abstract saffron]

Apparently I didn't lock myself long enough

Here's another though

[abstract saffron]

I'm patiently waiting

[tiny ridge]

Take the trivial bundle S^3 x SU(2) -> S^3. For any connection A on this principal bundle, consider the (Chern Simons functional) quantity CS(A) = \int_{S^3} tr(A \wedge dA + A \wedge A \wedge A). For any knot K in S^3, define the Wilson loop operator (known as holonomy to mortals) W(K; A) = Tr e^{i\int_K A}.

The Jones polynomial, upto appropriate normalization, is Z(K; t) = \int e^{i t CS(A)} W(K; A) dA

You integrate over all connections lmao

This is the path integral

[abstract saffron]

Wheel of Fortune, final round: find the following word
ho_o_o_y

[tiny ridge]

Its the expected holonomy of a randomly sampled SU(2)-connection along the knot LOL

[patent quarry]

is the right perspective that this is a partition function?

ah ok yes.

[tiny ridge]

where no one knows what expectation means

[tiny ridge]

exactly

[patent quarry]

what's the interpretation of the parameter t?

something like inverse temperature?

[abstract saffron]

Huh, this is not so bad somehow

Yeah

straight from statistical thermodynamics theory

[tiny ridge]

I think this sounds right

Its the same as the Boltzmann distribution except this ones nonsense

[patent quarry]

ok no i'm not sure it's quite right.

[abstract saffron]

you have the Boltzmann distribution to be summing over e^(-1/(RT) * something simple)

Where R is the gas constant and T is the absolute temperature

[patent quarry]

or maybe it is like a quantization of a boltzmann distribution?

[tiny ridge]

Yeah, I think physically this t is integer

[patent quarry]

does negative integer t have physical meaning too?

[tiny ridge]

This has to do with CS(A) being well defined upto integer multiples of 2pi for gauge transformations

[abstract saffron]

I'm surprised that somehow this can be boiled down to Kauffman's bracket for dummies

It surely took a genius

[tiny ridge]

I think the main theorem in Wittens paper is that the skein relation is true

He proves that without computing the path integral

which of course no one can do because shit dont make sense lol

[abstract saffron]

No shit

[abstract quail]

I haven't read the whole conversation, but the historical origin of Jones polynomial I believe is via von Neumann algebras. He was creating a tower of von Neumann algebras, and each algebra was generated by some generators, which obeyed certain relations. Similarity among these relations and the relations among the generators of the Artin braid group was pointed out by a student to Jones during a seminar. Jones then thought about it, corresponded with Birman, and then arrived at his polynomial by defining a trace function on his algebras.

[abstract saffron]

Also, why should it be a knot invariant? The most standard proof is via Reidemeister's moves, which is in no way enlightening

[tiny ridge]

Oh wow

[patent quarry]

doesn't "expected holonomy" answer this?

[abstract saffron]

I certainly don't know what a holonomy is

[patent quarry]

or do you want an actual proof?

ah.

[tiny ridge]

if you have a vertical piece of wire and an electron does a loopy about it then its “phase changes”

[patent quarry]

well look at this integral defining W(K;A).

[tiny ridge]

look up the Abramanov Bohm effect

idk if the spelling is right

this is what holonomy means

[abstract saffron]

And here I am, thinking i'm done with physics

[tiny ridge]

mathematically its just integrating a closed not exact form about a loop

and getting a nonzero answer

[patent quarry]

idk quantum physics or knot theory, so here's a naive question: is it fruitful in math and/or physics to study "free energy" log(Z(K; t))/t in this setting?

[tiny ridge]

youd have to ask someone whod know, maybe physics stackexchange

i have no idea but sounds like a good q

Im just here to trash talk dA

[abstract saffron]

Interesting, I don't even know there's such a thing as free energy log(Boltzmann)/t

Only know the formulae because it shows up in Molecular Biology

[patent quarry]

oh idk what physicists call it.

[tiny ridge]

log Z is free energy yeah

some people call it pressure which is ridonkulous

physicists are mad trash

[patent quarry]

idk i think path integrals are good.

[abstract saffron]

What's even an interpretation of that in general setting then?

[patent quarry]

no idea.

[abstract saffron]

I have no background in statistical physics, so...

[tiny ridge]

[abstract saffron]

Well, I guess I know what I'll spend my summer on

[patent quarry]

i just know that it shows up in math for somewhat uninteresting reasons: if the partition function looks like exp(constant/temperature) then you need to take log and divide by 1/temperature to get something asymptotically nontrivial.

[abstract saffron]

as t tends to infinity?

Well, no biologists have thought about that for sure

[patent quarry]

my feeling is that since mathematicians have had some success defining "infinite-dimensional lebesgue measure" in various contexts, there should be some good intuition to be had by formally computing with path integrals.

[coarse night]

thought it's a topology channel

[tiny ridge]

Im wrong, derivative of log Z wrt \beta is energy

[abstract saffron]

It got spilled over from knot theory

[patent quarry]

by “defining” i mean “working around the inherent meaninglessness of.”

[tiny ridge]

It seems its up on the air what exactly is allowed when computing a path integral, and these physicists just ignore everything and throw their bag of tricks at it

[abstract saffron]

I have lost count the number of times I got mad over their non-rigour

[tiny ridge]

Which is why one gets crazy stuff like 1+2+3+… in their crap

[patent quarry]

yes but i’m saying it seems ok to me to sometimes take their physical intuition on faith.

[abstract saffron]

Jacobi notation flying everywhere

[patent quarry]

i think renormalization shouldn’t be so scary to mathematicians!

[abstract saffron]

Yeah, they showed that it is actually -1/12 and consistent with Casimir effect

from 2006 iirc

[tiny ridge]

i have seen renormalizations you wouldnt believe

NON INTEGER dimensions

come on

theres a limit to craziness

[abstract saffron]

Sobolev?

[patent quarry]

lol i've seen that stuff too.

[tiny ridge]

No like, literally, integral d^s x

s not integer

wtf right?

[coarse night]

s is complex?

[patent quarry]

well aren't they interpreting "dimension" as "scaling exponent?"

[tiny ridge]

yeah think so

[abstract saffron]

maybe they overheard Mandelbort

and thought it might be cool to bring some fractals

[coarse night]

lol you are good friends with physicist and talking trash behind their back

[patent quarry]

i think that should be fine for a mathematical audience, within reason. i'm of course not advocating for abandoning rigor in math, just saying that i think suspension of disbelief can be useful for getting some feeling for what "should" be true.

[tiny ridge]

it goes both ways

i say all of this openly of course

[coarse night]

imagine physicist trash talking maths

[tiny ridge]

sometimes its well deserved

[abstract saffron]

I don't mind the intuition approach to math or physics, but it has to make sense and brings some reasonable, testable result

[patent quarry]

a lot of what we’re discussing is experimentally verified lol.

[abstract saffron]

Euler solved Basel's problem by doing what he was not supposed to do, but this is a new level

Yeah, but they don't justify why it's true

[tiny ridge]

im ok with a bit of trolling tbh

i just think i should get the pass to trash talk also

[patent quarry]

what?

[tiny ridge]

as long as thats the deal i dont care if physicists are writing d^s f where f varies over all smooth functions and s is a quaternion

[coarse night]

how do they actually define it though?

[hidden crag]

interdisciplinary trash talk is goated

[coarse night]

but maths have the free pass to trash talk anything

[abstract saffron]

This is why I learn enough of everything to shit talk about anything

[coarse night]

anything nok rigor

[patent quarry]

"define" seems like the wrong mindset, no?

[coarse night]

[tiny ridge]

i agree tho

[patent quarry]

afaict the entire point is to develop models with predictive power. if you write down a formula that matches real-world observations, within some tiny error, you win the nobel prize in physics.

[tiny ridge]

mathematicians would do well writing things less precisely and talking about it like its experiments

arnold style

[abstract saffron]

Also, I just remember we also have fractional calculus