point-set-topology

[feral copper]

I mean for general trisections, no, but for this geometric stuff, yeah

[tiny ridge]

We Weinstein bisect 4-manifolds!

[feral copper]

If you've got a non-symplectic 4-fold, it won't have a Weinstein trisection

[feral copper]

But then it's not induced by any structure on the 4-fold

[tiny ridge]

Nope. But the two pieces have a Weinstein structure and the contact form on the boundary can be made to agree along the intersection

Essentially it's equivalent to an R-invariant contact structure on (4-manifold) x R

But yeah in that case the contact structures obviously agree along the boundary

[feral copper]

Ah!

[tiny ridge]

Very fishy that Weinstein bisections and Weinstein trisections serve sort of orthogonal purposes

So here's a question. Let M be a 4-manifold. Suppose you have a symplectic structure w on M and an R-invariant contact structure a on M x R. What's an interesting way to couple the two?

Can we do like C w + da where C is so massive that this is still symplectic

Or maybe C should be a dynamic parameter

In IR, C -> 0 and you recover a Weinstein bisection

In UV, C -> infty and you get a Weinstein trisection of the symplectic manifold

Can we make these structures talk to each other

[feral copper]

I've never seen this stuff before! I know someone I can ask, but I myself am clueless (don't overestimate the teeny tiny bit of reading and conference attending I had related to contact/symplec geometry)

[tiny ridge]

It was just a kind of funny thought

I don't know what one would do with this kind of coupling

But since the above gives an interpolation between w and da, there may be a natural way to do handlemoves on a Weinstein trisection to get a bisection

One should try on CP^2

[heady skiff]

i don't really understand why we consider S to carry P onto R^n x 0, if we're sending the basis to the standard basis of R^n. where's the embedding? why do we even consider the embedding?

[feral copper]

But CP(2) is kind of a bad toy example, because everything always works so well; at least, take some number of blow-ups of CP(2), and if you're brave enough, take a K3 surface (recall that a minimal genus trisection has genus 22)

An even less trivial example than blow-ups of CP(2) could be an algebraic surface in CP(3), like CP(1)xCP(1)

[tiny ridge]

Good call

Isn't there an algorithm to turn a toric manifold into a trisected one

Take the Delzant polytope and cut it appropriately

[feral copper]

Not an algorithm per say, but yes of course take the Delzant polygon and trisect it; if you're lucky, it gives you a trisection by pre-imaging everuything. However (and the case of CP(1)xCP(1) is actually such an example), the pieces of the trisection of the polygon need not be connected!

X2 is the union of X2a and X2b

[tiny ridge]

Fair, I don't mind disconnected pieces. How do you decide how to cut?

[feral copper]

That's the fun part: there is no algorithm per say (none that I'm aware of), and you've gotta try a few cuttings until one works

[tiny ridge]

Huh

[feral copper]

Because here the obvious choice of cutting the square into three wedges doesn't work for CP(1)xCP(1)

The case of CP(1)xCP(1) has a 4-fold symmetry however, so it's more suited for a quadrisection

You might wanna read this: https://arxiv.org/pdf/2206.04161.pdf

Just to get a flavour of why this is a difficult question to answer in all generality

[tiny ridge]

Suppose you find a trisection using the toric picture. It needn't be Weinstein, yes?

For that you need additional handle moves unrelated to the toric action to make all the 2-handles have the correct framings, I'm guessing

[feral copper]

Yeah that's the thing: it needs not be Weinstein indeed. Lambert-Cole had this issue originally, but it happens that you can approximate each sector of the toric trisection by Weinstein domains (iirc, I hope I'm not missing something crucial)

I think he discusses it in §3.3 of https://arxiv.org/pdf/1807.10131.pdf (page 15)

[tiny ridge]

It makes sense. The only obstruction for a 2-dimensional handlebody to be Weinstein is that the 2-handles have framing tb-1 where tb is Thurston Bennequin number of some Legendrian realization of the attaching curve

This is a theorem of Eliashberg. The moment you have this it's Weinstein

[feral copper]

I see

[tiny ridge]

Meh, it's easier than Eliashberg's theorem, who proved this is also equivalent to Stein-ness

[feral copper]

Also, I suspect you mean 4-dim 2-hb?

[tiny ridge]

Yup

Sorry

[feral copper]

No biggie, just making sure I'm following!

[heady skiff]

i'm going to attempt to prove that a 3-simplex is a tetrahedron; for this, would it suffice to show that any point in the 3-simplex can be represented as a point of the line segment joining one vertex to some point in the face opposite it (a 2-simplex), and conversely, any point on a line segment joining a point on the face to a vertex is a member of the simplex?

[tiny ridge]

@feral copper So, in some sense, if all your 2-handles have highly negative framing you'd be done: one can keep adding zigzags to the attaching Legendrian knot to balance f = tb - 1 from f <<< tb - 1

Zigzag decreases tb by 1

[feral copper]

Yeah

Of course

[tiny ridge]

Now I suppose one has to fuck around with a triple of these Legendrian attaching curves because trisection...

[feral copper]

Yeah you'd have to deal with all three at once

Tweaking one might screw up even more the other two, like a Rubik's cube basically

But wait, the sectors of the trisection are 1-hb, not 2-hb

[tiny ridge]

Oh, funny. All 1-handlebodies are almost canonically Weinstein, maybe the point is to get union of two sectors right at a time.

I don't understand what's happening anymore very well

There's sort of 0 effort needed to get a 1-handlebody to be Weinstein.

[feral copper]

Yeah, but for a Weinstein trisection, what you want is that the Weinstein structure is induced by the ambient symplectic structure; equivalently, everything should glue nicely into a symp. structure on the 4-fold

[tiny ridge]

What's confusing me is that in a 1-handlebody, there cannot be too many different Weinstein structures. Start with D^4, you attach a Weinstein 1-handle. The symplectic structure is totally determined by how you attach it. But all possible ways of attaching a 1-handle are isotopic, because a 1-handle attaches along a pair of contactly embedded Darboux balls in S^3

And all such pairs are contact isotopic

It seems to me that there's no effort in making a single one out of three of those Weinstein sectors with the induced \omega as the symplectic form. Where exactly is the fuckup in trying to make all three at once?

After making one we must focus on the contact boundary #^k(S^1 x S^2), divide it into two #^k(S^1 x D^2)'s intersecting along a fixed #^k(S^1 x S^1)

And then fix all this data

[feral copper]

Really? There's only one symp struct on a 1-hb of genus 1? That seems so wrong

I mean, regardless of this, I'm sure there something very much non-trivial, because showing that any symp 4-fold has a W trisection relies on the fact that such a 4-fold is a branched cover of CP(2), and you put the locus in bridge position wrt to the W trisection of CP(2), and then you lift, and you've gotta carefully check it's all good, which took Lambert-Cole quite some effort

[tiny ridge]

I know this proof but why do we do it this way

What goes wrong in the pedestrian approach

[feral copper]

I suppose PLC could answer that I'm just a kid playing with legos, this is beyond my pay-grade

[tiny ridge]

In a subcritical (strictly less than half dimensional handles) Weinstein handlebody, the symplectic structure is only dependent on the algebraic topology

[feral copper]

You could always send him an email, he's a very cool and chill dude, he'd be glad to answer I'm sure

[feral copper]

Interesting!

[tiny ridge]

Essentially because of h-principle for subcritical isotropics

Any two subcritical isotropic embeddings are isotopic iff formally isotopic

Formal isotopy: just a smooth homotopy of the maps and a homotopy of the plane field and stuff

But in the 4D case it's just pedestrian. There's a unique Weinstein D^4, a unique Weinstein 1-handle (upto isotopy)

[feral copper]

That's quite a strong statement

[tiny ridge]

You attach along a pair of balls

[feral copper]

And that holds in 4d too?

[tiny ridge]

Yes but kind of trivially

Coz boundary of 4d is 3d and subcritical for 4d means 1d whose boundary is 0d

So bunch of points in a contact manifold

Of course, the contactomorphism group acts k-transitively for any k

Just move around the points

Do you agree that once we construct a single Weinstein sector, the problem boils down to constructing a bisection for the leftover symplectic 4-fold with boundary

A bisection that bisects the boundary in a very specific way

[tiny ridge]

Like this

The leftover symplectic 4-manifold (the "2/3rds") should really be thought as a sutured manifold

The boundary breaks into two pieces, one is convex the other is concave

They meet along the surface of genus g (k in my notation above)

[feral copper]

In any case, I don't understand why the fact that there's essentially one way to give any sector of a given trisection a Weinstein structure imply that we're done. Because this 'proof' seems like you can give any 4-fold a symp struct by just taking a trisection of it. Something has to happen when you glue the sectors together (stating the obvious here, but still)

[tiny ridge]

I'm not saying we're done!

I'm saying we can construct a candidate 1 out of 3 of the sectors, symplectically embedded inside the symplectic 4-manifold

By hand

And then remove it and reduce it to a relative bisection problem for the rest of the sutured 2/3rds

[feral copper]

Ah

I didn't quite catch that

Okay, so you bisect relatively this remaining bit, and you want to glue it back to the X1 you removed

[tiny ridge]

Once you bisect relatively, we can definitely glue back. The problem must be finding this relative bisection, no?

Maybe this isn't always possible or something unless your by-hand 1/3rds was nice enough (stabilized enough?)

[feral copper]

You said it's always possible, didn't you?

[knotty vine]

What is even the difference between a 3-simplex and a tetrahedron? A 3-simplex is the convex hull of 4 points in general position so both of those statements certainly hold

[tiny ridge]

Any closed smooth 4-fold can be Weinstein bisected. I haven't thought about relative bisection

[feral copper]

Ah I see, yeah existence of this depends on the contact struct you have on the bdry of the removed bit, and that's in general not true you can extend it

(?)

[tiny ridge]

Yeah

There's some boundary conditions one has to sweat over

@feral copper There has to be a proof of Weinstein trisectability this way

It's nice to apply Donaldson's theorem but it completely obfuscates the problem

[feral copper]

PLC's proof relies on Donaldson at some point?

[tiny ridge]

That symplectic 4-folds branch over CP^2 is essentially Donaldson

[feral copper]

Auroux's work uses that? Really?

[tiny ridge]

Here's the idea. If you have a symplectic 4-manifold, you can produce approximately holomorphic sections to large powers of the line bundle corresponding to the symplectic form

Construct 3 linearly independent onces (need a huge power of the line bundle)

Call them s0, s1, s2

[feral copper]

Ah, when you mean Donaldson, you mean his work, not the diag theorem

Cuz I didn't see the link
(reading you now)

[tiny ridge]

Define M -> CP^2, x -> [s0(x) : s1(x) : s2(x)]

Oh yeah I meant the Donaldson divisor theorem

"Approximately holomorphic geometry and symplectic submanifolds" is I think the paper name

[feral copper]

Yeah yeah of course then, and yeah Auroux's thing is indeed a convenient black-box that hides some of the tricky details

[tiny ridge]

I read Donaldson's divisor theorem quite carefully, it's a fantastic theorem but one can soften it up at places

It's a shitload of analysis which is not geometric in any way

But there are alternate methods to prove certain results (like his symplectic Lefschetz hyperplane theorem) which are geometric

[feral copper]

Then you got yourself something to write down

[tiny ridge]

I don't care enough to, I want to understand

[feral copper]

Pro gamer's move

[tiny ridge]

Topology cannot be done by a hammer analysis blackbox!

[feral copper]

Modern 4d topology can, unfortunately...
I mean, things like Thom and Seiberg--Witten invariants' corollaries are inevitably going to have to use the smooth structure somewhere. But it's really disgusting that it does it in such a convoluted analytic way...

[tiny ridge]

Well at least one should know why it's necessary

Or if at all

[feral copper]

Same goes with Heegaard--Floer homologies: if you've gotta use Atiyah--Singer, then it's clearly that you're doing something not geometric enough to my taste

For Thom, for instance, we know it is necessary to use the smoothness of embeddings, because the locally flat Thom conjecture is entirely different

Also, you know you have to use smoothness when you have an exotic pair for instance

[tiny ridge]

Yes but doubt that finding Weinstein trisection is as hard a problem.

[feral copper]

Yup definitely

[tiny ridge]

Here's a philosophical comment, though. What good is having a bunch of hard analysis smoothness detection invariants if you run out of manifolds to use them on?

In some sense this is why Akbulut corks are famous. An exotic pair of smooth 4-manifolds always differ by a very specific Stein piece, carved out and glued back in differently.

You know what to look for

[feral copper]

Because you're never gonna run out of 4-manifolds, there are at least an many as there are finitely presented groups

(because handle decompositions)

[tiny ridge]

Knowing they exist isn't enough to construct them. I can draw a random Kirby diagram and can claim I know the 4-manifold. But I in fact do not

[feral copper]

You mean that all exotic pairs are twisted pairs? No, this is wrong in 4d

Ah, a Stein piece, not necessarily a ball

[tiny ridge]

It's not an algorithmic problem to look for an exotic S^4.

[feral copper]

Then maybe, idk

[tiny ridge]

Indeed

A contractible Stein 4-manifold

[feral copper]

One can always dream

[heady skiff]

oh well i guess my book defined it differently

[tiny ridge]

What I meant is it has become easier to detect if two things are homeomorphic but not diffeomorphic, but "which two?" is still a topological problem and not an analytic one

You have to have a gut feel to be able to guess exotic pairs

Checking ensues afterwards

[feral copper]

I don't see how people get any intuition at all on these

For the life of me, what is an exotic S^7 I wish I knew

I wish I could visualize the standard S^4

[tiny ridge]

Before Akbulut, there wasn't many good strategies!

There's an operation called a Gluck twist

This existed

I don't know many more

[feral copper]

Yup, and this produces... the standard S^4 x')

The most recent 'twisting' operation was (iirc) the Price twist, which is also standard

Funnily enough, all exotic spheres in dimensions >=5 are twisted spheres...

[tiny ridge]

On S^4, but Gluck twist can produce exotic pairs elsewhere, no?

[feral copper]

I doubt this is true for S^4, if there is an exotic S^4 that is

[tiny ridge]

Akbulut's thing is called a cork twist, btw

[feral copper]

Ah, yes of course

Yeah Akbulut's cork, right!

That indeed produces exotica

[tiny ridge]

In fact all exotica are produced by cork twists

It's a very easy to prove, topological theorem. It's in his book

[tiny ridge]

I don't think that can be true for S^4, yes? Diff(S^3) deformation retracts to SO(4) by Hatcher's theorem

There's no nontrivial sphere twists in dim 4

[feral copper]

No no no, people are actively looking for twisted exotic S^4's

[tiny ridge]

Maybe I misunderstand by what you mean by a sphere twist

It's not just removing a ball and gluing it back in

[feral copper]

Remove a ball and re-glue it

[tiny ridge]

?

No then that cannot possibly be true. The boundary is an element of Diff(S^3)

Therefore homotopic to a linear diffeomorphism

[feral copper]

Aaaah no sorry, I checked

Gluck twist spheres are constructed by cutting out a tubular neighborhood of a 2-sphere S in S4 and gluing it back in using a diffeomorphism of its boundary S2×S1. The result is always homeomorphic to S4.

[tiny ridge]

Yeah I thought Gluck twists are still not known to be standard on S^4

[feral copper]

I didn't know this result, but that should've rung a bell for me, sorry!

[tiny ridge]

But yeah my point is: how can you stop believing that there has to be topology if you're doing topology?

That's like selling your soul to the devil lol

[feral copper]

I'm not sure I get this right

[tiny ridge]

It's a half joke but like, at the end of the day we want to understand manifolds. Moduli space of instantons won't tell me what the manifold looks like! That doesn't mean instantons are useless but we should always remember they're really secondary, and the analysis works because there's some deep topological reason it should work

Understanding topology of manifolds at the end of everything is still a pedestrian cut and paste problem

At least, that is my conviction

[heady skiff]

to show that (a, b) and (0, 1) are homeomorphic, would the function x |-> x/b suffice?

[feral copper]

Ah! Yes okay, I see what you mean now! I entirely agree that what matters the most is the geometry and topology of the manifold

[tiny ridge]

In all cases we can't get eg instanton Floer homology to do what we want it to do is because some surgery formula is lacking, yes?

[feral copper]

Wdym?

[heady skiff]

like would it be a homeomorphism between the two

[feral copper]

I don't know anything about instanton Floer, and I know a little about SW

[unreal stratus]

No

[feral copper]

Between the two what?

Ah intervals

[unreal stratus]

I mean you are assuming a/b = 0 right lol

[feral copper]

No, shift, then shrink

[unreal stratus]

Also what if b =0

[tiny ridge]

I don't know much either, but even in SW, the crucial point is we have all these nice surgery formulas and whatever we can compute we compute by using these

[unreal stratus]

Just use a picture ig

[feral copper]

$$x\in{}]a,b[{}\mapsto\frac{x-a}{b-a}$$

[gentle osprey]

Matplotlib

[unreal stratus]

European spotted

[tiny ridge]

At the end of the day it's just a homology long exact sequence we need

[heady skiff]

wait how

[unreal stratus]

Why else would it be bijective

[feral copper]

Yeah, you only care about invariants that come from something geometric (like the pseudo-holomorphic SW invariants)

[tiny ridge]

I was talking to an instanton guy a while back and he was talking my ears off about how he couldn't get his surgery exact sequence to do what he wants it to do

[feral copper]

These you know don't vanish, because there's geometry at play

[tiny ridge]

That's life, you know

[unreal stratus]

Matplotlib you are very helpful

Thanks for being a good library

Nice bio btw lol

[feral copper]

Ah, enshittification

[heady skiff]

i'm tired

true

[unreal stratus]

Oh sorry I meant

The cohomology

[feral copper]

Yeah well, it's not like you can force the manifold to be different; that's just how it *is, otherwise it's no longer the same

[heady skiff]

oh

would x - a/b - a work

[unreal stratus]

Ye

[heady skiff]

i only saw "shift and shrink"

guess that makes sense ya

[unreal stratus]

Shift so a = 0, then do what u did

Essentially

[feral copper]

This basically

[unreal stratus]

Ok this is great

[feral copper]

Funnier exercise: find a homeo between (a,b) and (c,d)

Shift, shrink, and shift again

[unreal stratus]

Find a homeo between (a,b) and [a,b]

Fr though a fun exercise is to fun a bijection between (a,b) and [a,b)

[tiny ridge]

Incidentally, pseudoholomorphic curves is to me the most conceptual of all the modern analytic technology. The details can get very hairy but the ideas are super robust and clear and intuitive

[feral copper]

Am I missing something? One is compact, the other isn't?

[unreal stratus]

Yeah was a joke

Dw lol

[tiny ridge]

Of course, it's because Misha Gromov discovered it and he cannot even write a single inequality correctly, much less a PDE

[feral copper]

Yeah it's basically a flexible topological notion that encapsulates algebraic things

So of course it's difficult, because it's rich. But also it's good, because it's rich

And at the end of the day, you understand algebraic things the best, because that's what everything you think of is: algebraic

(saying algebraic as in defined by algebraic equations, not as in abstract algebra)

[tiny ridge]

I personally think of a minimal surface when I think of a pseudoholomorphic curve, than an actual algebraic curve

[feral copper]

I'm based, I acknowledge that

[heady skiff]

bet

[feral copper]

Btw, aren't you <&268886804013383681> Ibsen?

Dang, I don't remember how to link roles

Honourable

[tiny ridge]

Anyway, I'm going to try to think about what goes wrong in the pedestrian approach to the Weinstein trisection business. It's not clear at all to me how to relatively bisect a sutured symplectic manifold

[feral copper]

The yellow one

[tiny ridge]

Nope!

[feral copper]

I think there lies your obstruction: extending the contact structure on the boundary to a Weinstein in the whole thingy. Which makes me think: because there's not so many Weinstein structures on a 1-hb, and the relative thing is actually a gluing of two 1-hb's, it may be possible to find a basic obstruction in terms of just that, and that could be enough?

Idk if it's true or still open that a symp 4-fold has positive Euler char, but since the genera of everything are expressed in terms of that Euler char, that could at least give you a hint at what fails when this is violated?

[tiny ridge]

Note that the boundary #^g(S^1 x S^2) should be thought as union of two contact manifolds-with-boundary #^g(S^1 x D^2), with the Liouville vector field pointing out of one and into the other

[heady skiff]

so a homeomorphism between $(a, b)$ and $(c, d)$ (assuming wlog $a < c$) would just be given by $x \mapsto \frac{(c - a) + x}{d - c}$

[tiny ridge]

This is what I meant by the suturing

[gentle osprey]

okeyokay

[tiny ridge]

The theorem says it's possible to do this if g is sufficiently large so the obstruction, if exists, must vanish after sufficient stabilizations

The theorem being "Weinstein trisection exists"

[feral copper]

Nein, but close enough: $$x\mapsto\frac{d-c}{b-a}(x-a+c)+c$$

[gentle osprey]

Matplotlib

[tiny ridge]

Using aposteriori knowledge

[feral copper]

No need to assume a<c

[feral copper]

Sure, we're interested in another proof, so why not also use the result to guide us!

[tiny ridge]

We want to essentially find an embedded 3-dimensional handlebody bounding the suture (the surface #^g(S^1 x S^1)) which serves as the bisection

Can of course start by picking a smoothly embedded bounding handlebody.

[feral copper]

But then, even if everything is Weinstein, it must still glue to a trisection, and that needs not be 100% obvious

[tiny ridge]

If we find cuts which cut everything into Weinstein pieces, then of course they glue back in along the cutting regions.

We're sort of trying to construct the cutting region by hand

Yes OK, there's a topological obstruction. After deleting the embedded Weinstein 1-handlebody we constructed by hand, the remaining "2/3rds" need not have the topology of two copies of #^g(S^1 x B^3)'s glued along #^g(S^1 x D^2)

[feral copper]

Wait, I just looked at it and it's wrong! @heady skiff It should be: $$x\mapsto\frac{d-c}{b-a}(x-a)+c$$

[gentle osprey]

Matplotlib

[feral copper]

Yeah exactly, the complement of a 1-hb is not nec a 1-hb

No

That's not the reason I'm reading you wrong

Because they are, you start with a trisection...

,ti

[gentle osprey]

The current time for matplotlib is 11:20 PM (CET) on Wed, 22/11/2023.

[tiny ridge]

Ah, yeah, that's a good idea. Start with a topological trisection

I missed that.

[feral copper]

However, you can always assume that the trisection is balanced and the three sectors have the same number of 1-handles

[tiny ridge]

Then arrange one sector to be Weinstein (or rather, symplectically embedded, with the canonical Weinstein structure on a 1-handlebody)

[feral copper]

So you get rid of even more issues with balancing it

[tiny ridge]

Ok, let's do that

[feral copper]

The complement needs not be symp embedded anymore though

[tiny ridge]

Then all the three boundaries are topologically same

[tiny ridge]

Right, the complement certainly has a symplectic structure, just restrict \omega. It doesn't agree with the canonical Weinstein for 1-handlebody and stuff

[feral copper]

And the thing is: if you fix the trisection, you're killing all freedom you had in extending to a bisection of the 2/3rds

So you can only start with the (possibly not Weinstein) bissection that comes from the ambient trisection, and you're only allowed to perform inner stabilizations of it to make it Weinstein

That's not gonna work out I'm afraid

[tiny ridge]

Why exactly? Is there some monodromy issue with the contact structures on #^g(S^1 x D^2)?

The three prongs which separate the three sectors

Loopy around

[gentle osprey]

Matplotlib

[tiny ridge]

Right

[gentle osprey]

Matplotlib

[tiny ridge]

Yes, Z has a symplectic structure

\omega restricted to Z

[gentle osprey]

Matplotlib

Matplotlib

[tiny ridge]

Yes, and delZ has a fixed suturing, ie the standard Heegaard decomp

Because delZ = delX_1

[gentle osprey]

Matplotlib

[tiny ridge]

H_12 and H_13 both admit a contact structure

One convex, one concave

[feral copper]

Indeed

[gentle osprey]

Matplotlib

[tiny ridge]

Yep

[gentle osprey]

Matplotlib

[tiny ridge]

So the question is can we symplectically bisect the sutured synplectic cobordism (Z, w) between H_12 and H_23

So that the two pieces are Weinstein wrt w

Weinstein 1-handlebodies, in fact

[feral copper]

Uhm, relative compression bodies (topologically it's the same), that is a punctured surface cross I

No, sorry

I'm confusing stuff here

(again)

[tiny ridge]

I get you though 🙂

It's H_12 x I

Where you collapse the boundary delH_12 x I to delH_12

[feral copper]

I mean it's not just 1-hb's, it's sticked to the boundary of Z in an annoying way

Yeah this

[tiny ridge]

Yup, sutured cob, most annoying

But it has a totally wild symplectic structure w. No control on this

We nevertheless want to find a middle level set in this cobordism

[feral copper]

So the question is: can you inner-stabilize the bisection of Z you already have to obtain what we want? (because if we build a rel bisection from scratch, there's no real hope, is it)

Cuz if we want it built from the ground up, we can manage one rel handlebody, but its complement need not be one too

[tiny ridge]

I'm more open to the idea of building it from scratch, but maybe one should first try to stabilize the inner guy, and isotope it around

That's a more concrete idea

Do some kind of h-principle on the already existing topological bisection

Wiggly wiggly wiggly

And stabilize stabilize stabilize

[feral copper]

We could want to perform boundary stabilizations though; then the entire trisection of X is stabilized, but we must ensure that X1 and Z remain symplectic in the process

I'd actually be surprised if boundary stabilizations were not needed in general

Cuz that means that basically, any trisection of X can be made Weinstein by only stabilizing in one sector

[tiny ridge]

🙂

[feral copper]

I'm sure there must be stabilizations everywhere, it has to be global

[tiny ridge]

I want to understand why that's not true first, so I'd try exactly that.

BTW

[feral copper]

Sure

[tiny ridge]

Isnt it open that two Weinstein bisections are related by stabilizations

Or is it solved now

[feral copper]

Maybe?

[tiny ridge]

Basically if we understand this concretely we'd prove it

Because topologically it's true

[feral copper]

Yes indeed

However, that hides under the rug why you need to stabilize, and how and how many times

And you wanted a pedestrian construction

[tiny ridge]

And we're trying to prove "Weinstein trisections are obtained from topological ones by stabilizing and wiggly h-principle on the facets separating the sectors"

[tiny ridge]

Correct, the need for stabilization is something I want to understand. I'm certain this is some h-principle man

We'll prove it. Let's think about it

[feral copper]

You're actively doing it, I'm only throwing random ideas but I'm no geometer, don't get me wrong

[tiny ridge]

I'm going to head to bed now but I'll think more in the morning

[feral copper]

Sure, have a nice sleep!

I need to finish preparing exercise sheets now, great xD

[tiny ridge]

Nah, I don't know anything about trisections, you're helping a bunch

[feral copper]

Was very nice talking again!

[tiny ridge]

I'm sure h-principle for Weinsteinizing trisections is true

Someone just has to prove it

[feral copper]

I think it's an interesting research topic that could indeed work

[tiny ridge]

Here's the point: The Donaldson-Auroux divisor theorem is an h-principle

It's a very analytic one but it is an h-principle

One takes an almost symplectic divisor then wiggles it a lot until it becomes symplectic

Morally

Good night!

[feral copper]

Good night

[unreal stratus]

Gn

[white oxide]

Step 1: equip discrete topology

[umbral panther]

Not enough if a=b

[ebon galleon]

edge cases don't exist

[gaunt linden]

Then (a,b) and [a,b] are the same set. 🤓

[tiny ridge]

@feral copper Is the "binding surface" X1 \cap X2 \cap X3 a symplectic submanifold of the ambient symplectic 4-fold?

[feral copper]

I think not necessarily I don't see anything non-trivial which you can say about this surface in general even (like: what is its self intersection, is it lagrangian/symplectic, etc)

But maybe I'm missing the obvious

[feral copper]

In the case of the toric trisection of CP(2) and CP(1)×CP(1), can't you say something precisely because it's the preimage of an interior point in the moment polygon? What do you get?

[tiny ridge]

@feral copper Yes, good point. Actually the torus is Lagrangian in the standard toric trisection of CP2

The binding surface cannot be symplectic in fact

It's nullhomologous

[tiny ridge]

It's going to be very rarely Lagrangian as well, because Lagrangian surfaces have highly specific normal neighborhoods

Kind of mysterious

[feral copper]

Of course, it bounds something...

So that also tells me that the self-intersection vanishes

You even have a global section of the normal bundle, by using the fact that it bounds a 3d 1-hb and taking the inwards pointing normals

[tiny ridge]

@feral copper so the only way it can be Lagrangian is if Euler char is 0 ie it's a genus 1 trisection

Hyper rare

[feral copper]

Might interest you: https://arxiv.org/abs/2212.13576

[feral copper]

Those are entirely classified: it's either one of the genus one trisections of S^4, or the standard one of CP(2) or -CP(2)

[tiny ridge]

Oh interesting

[tiny ridge]

Makes sense

This looks like an ugly mess

[feral copper]

Genus two trisections are classified as well, but it's far from easy. Genus three is open, and most likely more difficult than genus one Heegaard splittings

[tiny ridge]

Like what on earth do those criteria mean

[feral copper]

Probably something to a geometer?

[tiny ridge]

Yeah this is not at all meaningful topologically

I have a question

You could potentially define a generalized Weinstein trisection where you have a sutured Weinstein 4-manifold W with boundary dW = Y+ U Y- having a fixed Heegaard splitting, and define a trisection as union of three copies of W cyclically pasted, yes?

The Weinstein sectors now have a bit more topology

[feral copper]

The thing is: for a relative trisection of a 4-fold with boundary, you don't get a HS on the boundary but an open book

[tiny ridge]

I'm talking about trisection of a closed 4-fold, but with three sectors being more topologically involved

But what is a relative trisection?

[feral copper]

you have a sutured Weinstein 4-manifold W with boundary dW = Y+ U Y-
Am I missing something?

[feral copper]

A trisection of a 4-fold with boundary

[tiny ridge]

Yes but what happens at the boundary? Or rather, what's an example picture?

[tiny ridge]

Sorry, I mean let M be a closed symplectic 4-fold. Define a W-trisection of M to be M = W U W U W

W = subcritical Weinstein handlebody returns usual notion

[feral copper]

The 3d 1-hb's are relative compression bodies (surf with boundary x I u 2-handles), and the 4d things are glued in a way that is compatible

It's a mess I've never dived into, there's a ton of details and conventions...

[tiny ridge]

Ah OK

[feral copper]

Ah okay I see

[tiny ridge]

Yes then it makes sense that there is an open book at the boundary

[feral copper]

That surf with boundary is the trisection surface too

[tiny ridge]

And the relative stuff works symplectically also?

Because that sounds interesting, the boundary contact structure is supported by some open book by Giroux correspondence. Maybe this one

[feral copper]

I don't think anyone was brave enough to tackle the symplectic relative case, especially because you'd need a relative version of Auroux

[tiny ridge]

Lol

Fair enough!

[feral copper]

Ah, people have looked at this and found no relation between Giroux's contact structure associated with the obd and the symp one

[tiny ridge]

Infuriating

[feral copper]

But also, people didn't look far enough

If you wanna read about relative trisections, gl, because there's pretty much only one reference (Nick Castro) and it's a mixture of hand-wavyness and over-detailing

[tiny ridge]

The following is true along those lines: every Weinstein manifold admits a PALF.

Positive Achiral Lefschetz Fibration

But maybe manipulating Lefschetz fibrations to get a symplectic trisection isn't easy

Otherwise that's how they'd do it in the closed case

Rather than use Auroux

[gritty widget]

is there a theorem or a counterexample that if we have paths that start at x1 and end at x2 (x1 not equal to x2) then those paths are homotopic?

[red yoke]

They are homotopic as maps from [0,1] but may not be path-homotopic

[gritty widget]

is there a counterexample?

[red yoke]

Take two points on a circle

[gritty widget]

oh

understandable

I have another question

what if these paths "cross the same road"?

so if we look at the image they're the same set

there should be a path homotopy between them right?

[feral copper]

This is obviously not true in general, contrary to what Arki said!

Take a map from the point (1,0) to the point (-1,0) on a circle, one going overpositive y-coords and the other negative y-coords

Those are not homotopic

[gritty widget]

yes

what about the second question?

we are interested in existence of a path homotopy between paths that have the same image

[feral copper]

Still not true, take the same example with the wedge of two circles and you ask them to meet at the crossing point

Ah, again, still not true

Take a loop on the circle (a special kind of path), one +1 loop and one -1 loop

They both have the same image: the whole circle, and are not homotopic

[gritty widget]

one last question

if the starting point is x_1 for both paths, the ending point is x_2, but x_1 is not equal to x_2

and the image is the same for both paths

they are path homotopic right?

[feral copper]

Take a +1 loop on the circle, and add either of these examples before or after, your choice. This defines surjective maps that are not homotopic

So again, no!

Basically: do a full turn, then do half-turn either from above or from below

These are not loops, but not path-homotopic

[gritty widget]

that's a good example

thanks

[feral copper]

The circle is the best first example for homotopy xD

[safe torrent]

fundamental group of a circle

[feral copper]

The fact that it has pi1=Z is really incredible

You don't even know how much yet

(what's more interesting is that it has a contractible universal cover, but I digress)

[safe torrent]

Stuff gets interesting when you go to covering spaces

[feral copper]

This is also why the n-torus is amazing

[safe torrent]

Based

[gritty widget]

now what if we exclude self intersections?

[feral copper]

Mmmh not really

[feral copper]

Don't even try, I will always find a counter-example. Maybe not in the circle, but somewhere-else for sure

[gritty widget]

again same starting point and same endpoint not equal to each other and same image

[feral copper]

I think the torus is enough even

[feral copper]

And injective in the interior?

[safe torrent]

Higher homotopy groups for weird spaces get weirder and weirder

[gritty widget]

injective everywhere

[feral copper]

Ah yeah different endpoints sorry

[gritty widget]

yes

[feral copper]

Then I think that should be restrictive enough to say that one is a re-parametrization of the other, but I honestly doubt that it actually holds

[gritty widget]

I believe there should be a proof somewhere, or a counterexample

[feral copper]

Yeah one is gonna have to be a (continuous) re-param of the other

[gritty widget]

how do we perform the re-param

[feral copper]

I believe not, because it's not very interesting, it kills all the fun in homotopy

[feral copper]

Probably look at the lift in the universal cover

Not sure it's even needed

[gritty widget]

I'll need to learn what universal cover is first

(I don't know algtop)

[safe torrent]

Stupid question but can you extend n homotopy groups to n being anything not just integers

[feral copper]

What's a half-dimensional sphere?

[safe torrent]

[feral copper]

(that's pretty much the same question)

For integer-dimensional things, there's only one choice: the unit sphere. For half-dimensions, you have all those fractal shenanigans

Definitely not unique

[safe torrent]

Yeah

It's ugly

[gritty widget]

what about complex numbers

[feral copper]

Wdym?

[gritty widget]

well it's probably nonsense

dimension can't be a complex number

[ebon galleon]

Complex dimensional spheres

[feral copper]

Yeah because dimension measures how you scale volume wrt to length scaling

However, one random thought. If you can somehow define \pi_n in terms of properties of a differential operator of order n, then using fractionnal calculus you might be able to interpolate between homotopy groups?

But I doubt it will carry anything meaningful geometrically

[gritty widget]

any idea how the reparam can be done or sources I can read?

[feral copper]

No idea for a formula, but try the obvious and if that doesn't work then it's probably not true in general

Also, no idea for a source, because I don't think anyone really cares, since, as I said, it kills all the homotopy fun

[tiny ridge]

https://mathoverflow.net/questions/212375/is-there-a-generalization-of-homotopy-groups-to-fractional-dimensions

[feral copper]

Obviously π3/2(X) is half abelian.
More like it should be... sesqui-abelian

[tiny ridge]

Barely understand the answer tbh

By barely I mean not at all

[knotty vine]

If the space is Hausdorff, then an injective continous map is a homeomorphism on its image, so we have two homeomorphisms between [0,1] and the image (which are even "orientation preserving"?) Can we do something with that?

[feral copper]

Yeah same, I know nothing about K-theory

[tiny ridge]

I should probably learn something at some point

[gritty widget]

I'm not sure

I'm trying to do a reparam

[knotty vine]

Theres no way to do an explicit reparam if you know nothing else about the paths

Unless you know the actual functions of course

[gritty widget]

there seems to be a natural one that would work for any pair of functions

[knotty vine]

Yeah, so we have two paths $f,g : [0,1] \to X$ with are both equal to $x_0, x_1$ on the endpoints. We also know that $f$ and $g$ are injective and that $I \coloneqq \im(f) = \im(g)$. Assume that $X$ is Hausdorff, then $f$ and $g$ are both homeomorphisms\linebreak $[0,1] \iso I$. Hence we get a homeomorphism $g^{-1}\circ f : [0,1] \to [0,1]$ which is the identity on the endpoints. I think this is your reparametrization.

[gritty widget]

oh

this should work

[gentle osprey]

Semer

[gritty widget]

I'm wondering how we get the path-homotopy from this

wait

[umbral panther]

What it should have said is that there is a part of the homotopy groups of spheres, called the image of J, which are cyclic groups of order the denominators of Bernoulli numbers. These numbers are p-adically continuous, so you can interpolate them. The answer skips this motivation and goes to trying to promote these numbers into groups

[safe torrent]

[tiny ridge]

Ah, this I understand

Thanks!

[gritty widget]

is this correct ?

[knotty vine]

Yep looks good

[gritty widget]

well can i ask a question

[knotty vine]

thats what this server is for

[gritty widget]

here the professor considered epsilon f(x0)/2

what if epsilon is 2f(x0)

[knotty vine]

Say f(x0) is 1

then 2f(x0) is the interval (-1,3) right? But then the neighbourhood U we get might also map to points < 0

I guess it depends on the precise definition of continuity you use

[gritty widget]

do you want to tell you the definition we use?

[knotty vine]

I guess its: a function f : X -> R is continous at a point x_0 if for every epsilon, there exists a neighbourhood U of X such that f(U) is in the open range (f(x0)-epsilon, f(x0)+epsilon)

[gritty widget]

yes

epsilon positive

[knotty vine]

ye

[gritty widget]

it is like for every epsilon positive , there exist alpha such as , d(x,x0)<alpha implies d(f(x),f(x0))<epsilon

which i think is equivalent to what you said

[knotty vine]

Oh wait, X is a metric space?

The question was about topological spaces

[gritty widget]

the professor said it is the same

[knotty vine]

Yikers

[gritty widget]

like we can consider both metric and topological

[knotty vine]

Sure

The solution given here works for topological spaces. Topological spaces are more general than metric spaces

There are topological spaces which are not (induced by) a metric space

[gritty widget]

if in a metric space it doesn't work?

[alpine nest]

It does work regardless.

[alpine nest]

Then you'd get a neighborhood which doesn't necessarily get mapped entirely into the positive numers, but that's not an error in the proof.

The statement is that there will be some neighborhood that gets mapped entirely into the positive numbers.

Not that every neighborhood will

To get the correct neighborhood you need to pick a good epsilon (in this case, f(x0)/2)

[gritty widget]

isn't epsilon for the neighbourhood of f(x) , and alpha for the neighbourhood of x?

[alpine nest]

Yes

Continuity (in the metric version) says that for every epsilon.... there exists an alpha...

So to prove continuity you need an argument that will work for every epsilon

But to use continuity, this means you can pick any epsilon you want or need.

In this case you know that f is continuous at x_0, so you're free to pick the epsilon according to your needs.

And continuity will guarantee the existence of a neighborhood (or an alpha) with specific properties related to that epsilon

So what happens for some other epsilon is neither here nor there, you just need one epsilon that will give you what you need.

[gritty widget]

okay thank you

i'm still convinced that it should satisfy all epsilon though

[alpine nest]

Well, take y = x (as a function from R to R)

It's continuous at x = 1, but for epsilon = 2 you'll get a neighborhood (-1,3) on which f is not always greater than 0

[alpine nest]

But that fact contradicts neither the statement nor the proof that you've posted here

The statement only requires you to find one neighborhood of x_0 that will have the desired property

[gritty widget]

okay got it

[merry geode]

How do you compute for X = [0, 1] and A = {1/n : n int}, homology of H(X, A) and H(X/A)?

X/A seems like infinite wedge sum but I am not sure.

(Well I guess it is not considering open nbhd of 0)

[obtuse meteor]

For the first long exact sequence of a pair should tell you this

[merry geode]

Can you do that?

[unreal stratus]

yes

[merry geode]

Why do you have C(A) -> C(X) as an injection?

[unreal stratus]

long exact sequence of a pair always exists though

[merry geode]

Don't you need
"0 -> C(A) -> C(X) -> C(X, A) -> 0"
to be a short exact sequence to obtain the long exact seq of homologies?

I guess failure of injection H(A) -> H(X) is measured by H(X, A) afterwards.

[unreal stratus]

sorry i said smth silly

but i mean, A is a subpace of X

you always get a long exact sequence for a subspace inclusion

just think about what C(A) -> C(X) "is"

[merry geode]

Hmmm

Idk why this felt wrong, but I see how this is possible.

Now, how do I compute H(X / A)?

[obtuse meteor]

Are you trying to compute it exactly or just show it’s not the same as H(X,A)

You can compute H1 pretty explicitly I think which is probably really what you want

[merry geode]

Yeah, I guess I can stop at H^1(X / A)

But how?

[obtuse meteor]

You can show H_1(X / A) is Not countably generated

Which is enough bc H_1(X,A) is

[merry geode]

Hmm

H_1(X, A) ~= H_1(X) from exactness, right

[obtuse meteor]

Feels off

Oh one moment lemme think

You should do just A = {0,1} as a test example

[obtuse meteor]

Bc there this is a good pair and H_1(X, A) = H_1(X/A) = Z, so your computation of this must be incorrect

The idea should be you inherit something from H_0(A) bc this is not connected

[unreal stratus]

yes you are assuming H0(A) = 0 or smth

which ain't the case

[merry geode]

I mean when A is a set of disconnected points

..wait

[obtuse meteor]

:)

[merry geode]

Where does H_0(A) come from?

[obtuse meteor]

It’s generated by the path components

And it’s in the LES of a pair

[merry geode]

Isn't H_1(A) -> H_1(X) -> H_1(X, A) -> H_1(A) --

[unreal stratus]

no

[merry geode]

Sorry, I suck at counting numbers.

[unreal stratus]

it goes down to 0

at the end

[merry geode]

I was like 1 - 1 = 1

[unreal stratus]

lol dw

[obtuse meteor]

You should use the reduced version to simplify this computation

[merry geode]

True

[obtuse meteor]

This will tell you H_1(X,A) = Z^{# of connected components of A - 1}

For your choice of A this just says you’re countably generated free abelian

[merry geode]

This uses H_1(X) = 0 (reduced or not), right

[obtuse meteor]

Yes. And H_0(X) reduced is 0

[merry geode]

Ah, makes sense! Thank you, I realized where I was wrangled into iffy computation.

[fading fern]

https://www.maths.ed.ac.uk/~v1ranick/papers/dold4.pdf

page 40, proposition 4.6

i dont get what it means with it saying eta is an homotopy equivalence

eta goes from a complex to another

not topological spaces

[unreal stratus]

so page 34

There is a notion of homotopy equivalence of complexes

often called "chain homotopy equivalence"

A chain homotopy between maps $f,g: C_* \to D_*$ of chain complexes is a collection of maps ${s_n: C_n \to D_{n+1}}$ such that like $ds + sd = f-g$ (where i'm kinda abusing notation slightly but this is in the degrees such that it makes sense lol

[gentle osprey]

Potato E-Girl

[unreal stratus]

Then you define chain homotopy equivalences in the same way as you would define homotopy equivalence of spaces, if that makes sense

like f:C* -> D* is a chain homotopy equivalence if there's a map g going the other way such that fg and gf are homotopic to idntity

[fading fern]

wait

d means the boundary operator right

in ds+sd

[unreal stratus]

ye

[fading fern]

alright

[unreal stratus]

ofc i've abused notation there like i guess rigorously i should say like

s_n : C_n -> D_(n+1)

followed by the differential D_{n+1} -> D_n

and similarly the differential D_n -> D_(n-1) followed by s_{n-1}

and f_n and g_n

[fading fern]

wait think i got your definition

but doesnt eta apply to 2 different complexes

instead of chain maps

[unreal stratus]

wdym

[fading fern]

SP and (Z,0)

[unreal stratus]

isn't eta a map between complexes

Yeah

[fading fern]

yes

it is

[unreal stratus]

So this

[fading fern]

oh didnt see that

sorry im used to skimming text a lot

thanks

i think i got past the wrong category problem

gonna keep rereading though

alright i got it formally now

not intuitively but something is something

thank you

[merry geode]

What is the motivation for the homotopy s of "ds + sd = f - g"

It seems quite out of place to me.

[unreal stratus]

well, you can view the proposition 4.6 as good motivation for that!

Like they correspond to homotopies in spaces via that

You can also create an "Interval" in the category of chain complexes and work out that this notion of homotopy is basically the same as in spaces when you translate between the different types of cylinders

[merry geode]

Ohh

Thank you!!

[heady skiff]

sanity check: if X is a finite topological space, then the discrete topology and the cofinite topology coincide right

[gritty widget]

you are sane

[heady skiff]

good to know 👍

[tiny ridge]

Boundary of (simplex x I) = top - bottom - (boundary of simplex) x I

ds = f - g - sd

s is the operator which crosses with I

[gentle osprey]

sergeEmbedding

[elder loom]

I know I already asked this

but I figured it had been a few days and ended up buried with no response

[tribal palm]

what is an example of a space that is regular but not normal?

[merry geode]

Ugh {1, 2, 3} is not a collection of sets

[steel pond]

oops

silly me

but then the intersection is still the empty set right?

if A={1,2,3}, and curlyA = power set

or is curly A not always the power set

ohh is this general notation for any set of sets, not necessarily the power set of a set

[merry geode]

Yes

Any "set" of sets.

[red yoke]

Long ray × closed long ray is LCH (thus completely regular) but not normal

Munkres p.204 has a nice proof

[tribal palm]

thanks! though i’m not familiar with this long ray space, and i can’t find where in the book it’s introduced

[knotty vine]

I dont really understand what this connecting segment is, arent the boundaries disconnected? Also, why are you taking the interior of a finite set?

[elder loom]

Oh, sorry. The interior shouldn’t be there. It should be

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

[gentle osprey]

sergeEmbedding

[knotty vine]

I still dont understand the connecting line segment part. This is what your question is about right? The rest clearly works in all dimensions

[tiny ridge]

Question, and I'm too lazy to Google it so hoping someone can just tell me why: every 3-manifold is a branched cover over S^3 branched along a link (Heegaard decompose, Z/2 fold the two handlebodies, glue). Allegedly this can be improved to branching set being a knot ie connected

What's the easiest way to see this

Oriented closed blah blah is of course assumed

Bonus question for @feral copper: Surely a trisection tells you every 4-manifold is a branched triple cover of S^4 branched along a surface

[elder loom]

right, that’s what I don’t know how to prove in general.

With any two convex hulls $C_i,C_{i+1}$ in the plane, the space $X$ bounded by $\partial C_i, \partial C_{i+1}$ has at least one line segment $xy$ from a vertex $x \in C_i$ to a vertex $y \in C_{i+1}$. The closest two vertices seem to give such a pair, but I don’t know how to prove that they do so.

[gentle osprey]

sergeEmbedding

[knotty vine]

Oh I see, youre talking about the annulus bounded by those two boundaries

I dont think the closest two vertices will work

Is this a counterexample?

excuse the ms paint

You can prove that there must be a point in Cn to the left of x:

[knotty vine]

Ah yes that works

But seems a bit harder to prove i feel

[elder loom]

This would work for any convex subset of the region red to blue

containing one vertex from both

by definition

My original idea was to show at least one simplex does not intersect the interior of blue

[elder loom]

with at least one vertex from both

[feral copper]

Ha! I didn't know the proof using Heegaard splittings, so I don't know. Give the week-end and I'll think about it

Also, for the knot thing in the 3d case, I don't think it's easy to prove you only need a knot...

Either you can prove there's a universal knot such that any 3fold is a branched cover of it (iirc the fig8 works)

Or you try from a link and tweak it to a knot, but it seems like it's not natural enough :\

https://math.stackexchange.com/a/4492029/259363
This is interesting, seems like you can get a knot from a HS
Actually, we know there are 4folds that cannot be branched over a connected surface, so there's something special to 3d (probably it uses standardness of splittings of the sphere, or Schoenflies or something like that)

Wait, if you can show that Schoenflies implies Hilden's knot result, then you actually disprove Schoenflies (or another 4d result, that's just an example)

I'll read that carefully tomorrow evening! Thanks for the food for thought, and I'll free the channel again

[knotty vine]

Sure, the whole annulus can be triangulated

[tiny ridge]

Aha. I'll have to read Hilden's proof a little carefully

[merry geode]

Is linear chains and barycentric subdivision stuff important?

[hidden crag]

Yes

[merry geode]

Hmm, where will it be used after the excision proof?

[pearl holly]

ye I guess it's also used in simplicial approximation

[merry geode]

Hmm

[heady skiff]

this is not a topology right

(union of an arbitrary number of open sets need not be open)

[pearl holly]

ye

[heady skiff]

if ${\mathcal{T}\alpha}$ is a collection of topologies on a space $X$, $\cup \mathcal{T}\alpha$ need not be a topology right

[gentle osprey]

okeyokay

[heady skiff]

just consider the cofinite topology and the discrete topology, and take an intersection of their open sets

or wait hmmm

because then it wouldn't be open under the cofinite topology

but it would be open in the discrete topology...

well i guess it needs to only be an open set under one topology for it to be an open set in the union of the topologies

so never mind that argument doesn't work

in that case taking the discrete topology as a counterexample will never work lmfao

[gritty widget]

this is not topology

the topology in munkres's book doesn't start until chapter 2

[ebon galleon]

Does it even start then?

[gritty widget]

so true catboy!

[coral pivot]

lol

yeah its better to ask in discrete math

[steel pond]

oof alright

[merry geode]

Bbbbbut point set..

[steel pond]

im ngl i dont know what point set means

[radiant cedar]

anyone here know a good introduction to fibrations and cofibrations? hatcher doesn't have a very good take on the subject

[hidden crag]

Mays concise course

[hot locust]

Given a set X, a topology on it is a collection of neighbourhoods of all points in X and not union of it right? 'cause if it's a union then even though in two topologies the neighborhoods of a point x might be different their union can be the same. Am I right about this?

[unreal stratus]

I'm not really sure what you mean

If you take the union of a topology on X then you just get the entire space

[outer harness]

If you’re still wondering, I think you could take X = {1,2,3}, T_1 = {ø, {1}, X}, T_2 = {ø, {2}, X}, and the union of these is {ø, {1}, {2}, X} which is not closed under unions

[heady skiff]

Lol yea I literally did basically the same example

[gritty widget]

let $B=\stackrel{\circ}{A}$ , we have $B\subset A\implies \overline{B}\subset \overline{A}\implies \overline{B}\backslash B \subset\overline{A}\backslash B $. $B$ is open by definition so it is equal to its interior, then $\overline{B}\backslash \stackrel{\circ}{B}: \in \overline{A}\backslash \stackrel{\circ}{A}\implies \partial B \subset \partial A \implies \partial\stackrel{\circ}{A}\subset \partial A$ is it correct ?

is it correct ?

[rancid umbra]

it looks like you meant to say B is open, not closed. this looks fine

[hot locust]

No I meant the union of the family of neighbourhoods for all points in X.

[gentle osprey]

john.

[unreal stratus]

Isn't that the same thing

X is always a neighbourhood of any point

[hot locust]

Wait I was thinking along the lines of this-

This example

If I take the union, both topologies here are the same

But the neighborhoods for different points are different

[ebon galleon]

"Neighborhood" is a property which is determined at a single point

The point they are trying to make I guess is that you really need to know which sets are neighborhoods aat a given point, for each point in X. Rather than knowing which sets are neighborhoods of some point

[rancid umbra]

need a sanity check: the graph of the absolute value function is homeomorphic to R. Is the subspace topology on the graph of the absolute value function the same as the topology induced by the induced homeomorphism?

[gaunt linden]

The graph of any continuous function is homeomorphic to its domain.

[rancid umbra]

right

but is it the same as the subspace top

inhereted from R^2

[gaunt linden]

Which other topology on the graph could we be considering?

[rancid umbra]

you get a map from R to the graph which induces the same topology as R on the graph

there is also the topology coming from R^2, when we think of the graph as a subset of R^2

those should be the same i think

[gaunt linden]

The map from R (i.e. the domain) to the graph (with the topology inherited from R^2) is the map I'm claiming is a homeomorphism!

[rancid umbra]

okay neat. so, the inclusion map of the graph should be a topological embedding into R^2

[gaunt linden]

Yes.

[rancid umbra]

that map is also an immersion, correct?

[gaunt linden]

Hmm, is "immersion" even used about non-smooth manifolds?

[rancid umbra]

there is topological immersion (in Lee's) but im asking about smooth immersion

its... kinda tripping me up since I can't seem to prove using derivations that the graph of the absolute value is not an embedded submanifold of R^2

[gaunt linden]

Right, so it doesn't even make sense to ask whether it's an immersion, because the range doesn't have a differentiable structure already.

[rancid umbra]

i thought it was an immersed submanifold of R^2 when you think of it as being homeo to R

[gaunt linden]

Alternatively, if we consider R^2 to be the codomain, the map isn't smooth at 0 (or in general at points where the function is not differentiable).

[gaunt linden]

Yes, you can transfer the differentiable structure from the domain -- but then it is trivial that the function you transfer the structure along preserves it!

[gritty widget]

you can make it into a smooth manifold, but the inclusion map into R^2 won't be an immersion because of the singularity at the origin

the typical way to prove these kinds of statements is to look at tangent vectors

[rancid umbra]

okay, thank you for clearing that up. i think curves would be the best way to show this then

[merry geode]

Why was a diagonal morphism continuous?

[red yoke]

Generally the diagonal map is the map A → A × A induced by identity maps A → A via universal property of products

[merry geode]

Ah then I should rather ask about its existence.

[red yoke]

The product of topological spaces is the product object in Top

And by construction the diagonal map sends x to (x, x)

See e.g. Munkres 19.6

[merry geode]

Does any product object have a diagonal map?

Oh wait. That's obvious. Duh.. sorry.

[heady skiff]

this is for my real analysis hw, but what a suitable topological proof for G/F being open (in the case that G n F is nonempty) would be to just say that X - F is open in R so just view G as a subspace of X - since G/F = X - F n G then G/F is open by the def of the subspace top

[white oxide]

How does this topology separate p from arbitrary closed sets? That would seemingly mean that any closed set in Y is bounded
https://math.stackexchange.com/a/975878

[heady skiff]

oh wait this proof simplifies immediately to the standard top on R lmao

[white oxide]

If X is the original space G\F = G \cap X\F and similarly for the other set

[heady skiff]

yee i figured that out lmaoo

oh yea i didn't even need to involve balls and shit lmfao

[white oxide]

unless you are interested in the completeness of the category Top, the answer is that you show that the diagonal map has pre-image the intersection of pre-images of the images of pi_1 and pi_2

[heady skiff]

potentially silly question, but what's the easiest way to see that the complement of an open set is closed? i understand that the definitions are iffs (in particular the definition of a closed set in a topological space) but the negation of closed is not open

[white oxide]

containing all limit points is one alternative def, mind you not in every topological space can limit points be determined using just sequences

but in analysis they are good enough for the job

[heady skiff]

i see, that makes sense

thanks

[white oxide]

anyone 🥺

Oh hold on. If a closed set doesnt contain p then its complement must contain a neighbourhood of p already, hence said closed set is bounded

[maiden oracle]

https://mathoverflow.net/questions/110595/minimal-number-of-cells-of-a-cw-complex-up-to-homotopy I have a question about the line "There is an epimorphism π1(X1)→π1(X) from the 1-skeleton which is a connected graph and whose fundamental group is a free group on a subset of the 1-cells of X [Hatcher, 1A.2]." I don't really know why this is an epimorphism. Intuitively it is clear what that homomorphism does, but i'm honestly not exactly sure about the precise definition and i don't understand why it's surjective? I would be glad if someone could help me :)

couldn't there be loops in pi1(X) that live outside the 1-skeleton of the complex?

i would think this map might be injective but not surjective

[tiny ridge]

It follows from Seifert van Kampen theorem that pi_1(X) is a quotient of pi_1(X_1)

See Hatcher chapter 1

There can be loops which live outside the 1-skeleton, yes, but they're all homotopic to loops inside the 1-skeleton

Roughly speaking: one can take an arbitrary loop in X, homotope it a little so that it misses the center of each 2-cell, then delete the centers of the 2-cells to get something homotopy equivalent to the 1-skeleton, compose with the homotopy equivalence to homotope the loop into the 1-skeleton

The crucial fact here is that, given a 2-complex, deleting a point from the interior of each 2-cell gives something homotopy equivalent to the 1-skeleton

The higher cells do not matter (also a Seifert van Kampen exercise)

[maiden oracle]

ok that makes sense, thank you :)

[craggy kraken]

Have knots inside >3 dimensional foliated manifolds been studied much?

If so, is there a good reference?

[eager vigil]

Is this argument valid? I get that s is linear, but K and L are not vector spaces from what I can see, so can we use the dimension theorem from linear algebra to conclude that s is not onto?

Or does this perhaps not use the theorems from linear algebra...?

Could one use that the image of a simplex by s must be of equal or less dimension that the simplex to conclude that the image of the simplicial complex must be of equal or less dimension and thus not onto?

[hidden crag]

Yes

[eager vigil]

Oki, thanks

[calm ravine]

Hi ! I am struggling in my computation of the Frolisher Spectral sequence for a complex manifold.
More precisely the question is Which differentials can be non 0 at the second page of the Frolisher sequence.
First I have try to compute the term Ep,q2 and I find :

$E^{p,q}_2 = \frac{{(x,y)\in A^{p,q}+A^{p+1,q-1}/\overline{\partial}x=0,\overline{\partial}y=-\partial x }}{d({A^{p-1,q}+A^{p,q-1}+A^{p+1,q+2}/\overline{\partial}x=0})}$

[gentle osprey]

Critotomatic

[calm ravine]

So the second differential must be something from $A^{p,q}+A^{p+1,q-1}$ to $A^{p+2,q-1}+A^{p-3,q-2}$
By a degree argument it seems clear that the first component of the differential must be 0 and the second one can only have the $\partial$ part.

But with this I cannot find another relation who imply that $d_2=0$ for some p,q

[gentle osprey]

Critotomatic

[floral bear]

is the additive group of (continuous) k-vector fields the double dual of the chain group?

since its dual, $\Omega^k$, is also the dual of $C_k$?

if so, what's that isomorphism look like? does it even canonically exist?

[gentle osprey]

cosín™

[umbral panther]

The dual of differentials are called currents. They are like chains, but a lot bigger

These spaces are not usually reflexive, but maybe if you use the right topology

[heady skiff]

does anybody know if this is a definition, or something that I could (attempt) to prove

[safe torrent]

Is this correct? The topological space D that I have doesn't have 2 simplex so the ker of del 2 is 0

What remains is ker del 1

There should be /0 at the end but I forgot about it

[unreal stratus]

Looks good to me

But not really enough context

[heady skiff]

is this a typo? should it be $A_{\sigma(0)}$ instead?

[gentle osprey]

okeyokay

[unreal stratus]

Probably should have a hat on the A_sigma(i-1) tbf

[heady skiff]

wdym, so it should be the hyperplane spanned by $\hat{A}{\sigma(0)}, \hat{A}{\sigma(1)}, \dots, \hat{A}_{\sigma(i - 1)}$?

[gentle osprey]

okeyokay

[heady skiff]

i don't really know how to see this, can i have a hint please?

i'm assuming for contradiction that we can write it in the span of those points

ohh, it's just that it would contradict the uniqueness of expression of $\hat{A}_{\sigma(i)}$ as a vertex of the k-simplex spanned by $\hat{A_0}, \dots, \hat{A_k}$

cool

[gentle osprey]

okeyokay

[heady skiff]

would the unique smallest topology just be the union of the collection T_a along with all unions of an arbitrary number of elements of T_a and all possible finite intersections of elements of T_a?

[ebon galleon]

Close: arbitrary unions of finite intersections of elements of U T_a

[alpine nest]

I'd go cheeky and just take the intersection of all the topologies containing all the collections

[ebon galleon]

This too

[heady skiff]

oh yeah lmfao

[ebon galleon]

But for a more explicit description, this is the topology generated by the union U T_a, regarded as a subbasis.

[heady skiff]

i feel like the largest topology is going to be the trivial topology

[heady skiff]

DAMMIT

i thought of that but i went for my original answer

[ebon galleon]

Yeah

My point is just that the order you do those in matters

[hexed steppe]

discrete you mean?

[alpine nest]

I like that one more than mine, more descriptive

[heady skiff]

oh oops

that's better

LOL

[hexed steppe]

but why

would that be true

[ebon galleon]

I like yours more than mine. Easier.

[hexed steppe]

cf the name “descriptive set theory”

[alpine nest]

As opposed to prescriptive set theory

[heady skiff]

wait, how would this not be equivalent to my original answer?

[hexed steppe]

actually i imagine this is basically the etymology

you need to take unions

of finite intersections

you did not allow that

[heady skiff]

oh yea, i guess in my answer i implicitly assumed that we're just taking the arbitrary unions of open sets

[ebon galleon]

You said that you take the union U T_a, let's call that T, and then took arbitrary unions of subsets of T along with finite intersections in T

That's not a topology

You need to close it under finite intersections first, then arbitrary unions.

[heady skiff]

ah, it's not a topology because we don't consider the union of two finite intersections to be open huh

if we're defining it in my original way

[hexed steppe]

yes

[ebon galleon]

This is the general process for generating a topology from a collection of subsets. So just check that this is a topology. The only special thing about this case is which collection we are generating from which is T

[heady skiff]

ah ok, makes sense

thanks

[heady skiff]

wait nvm, i don't think this is right - because, say working in a fixed topology T_ai, it doesn't have to be discrete, so T_ai doesn't have to contain every single subset of X

the trivial topology wouldn't work either, since it doesn't account for proper subsets...

ok wait this second part is confusing me. i don't really see how we can take a topology larger than T_a which is contained in T_a, the only one would have to be T_a itself

..maybe that's the point

nvm

i'm hella dumb

it's just the intersection isn't it

of the topologies

[ebon galleon]

yes

[heady skiff]

cool 👍

[hollow geyser]

This proof seems right, but the wording feels a little off. Would any of y'all change anything about it?

[naive trench]

Algebraic topology book with basic stuff? Maybe next semester I have a seminar about it and I didnt remember anything of alg topology from my 2nd year of the degree

[tiny ridge]

@hollow geyser Some proofs are meant to be done via induction

I'd rewrite it completely with induction because then I don't need to adjust every neighborhood simultaneously

[hollow geyser]

Induction was my first guess, but does ignoring it make my above proof incorrect? Big interescrtion loosely is already induction, is it not?

Wouldn't induction just be extra work compared to big intersect?

[tiny ridge]

Your proof is not incorrect. It lacks tedious detail like why is U_i and U_j disjoint

This detail is just not present in the inductive proof.

[hollow geyser]

You're right

[tiny ridge]

You only need to do exactly 1 big intersection

Not all of them at once

[hollow geyser]

I wrote the reasoning earlier, then deleted it because I did not like the wording.

That was the bit that was sussing me

[heady skiff]

is A a basis for each topology on X which contains A?

i'm a little bit confused on the wording

[alpine nest]

Certainly not, since the discrete topology contains A

And A is definitely unlikely to be a basis for it

[heady skiff]

oh right, A would just have to be all one point sets then huh

[alpine nest]

But A is a basis for some topology (the one it generates)

[heady skiff]

ye, just trying to figure out the converse inclusion rn

[umbral panther]

Every collection of subsets is a subbasis for the topology it generates. Not every subbasis is a basis

[alpine nest]

True, but the statement starts with "if A is a basis for a topology" 😉

But you're right to point out that fact that not any family of sets is a basis

[umbral panther]

A set of subsets is a basis if every finite intersection is covered by its members

[heady skiff]

cool, thanks for the tips

[heady skiff]

what's a linear extension?

[knotty vine]

linear interpolation

[heady skiff]

oh so it's just the set of all straight lines joining the vi

[heady skiff]

for two edge paths ${v v_1 \dots v_{k - 1} v}$ and ${v w_1 \dots w_{l - 1} v}$, would you define $\alpha: I \to |K|$ to be the linear interpolation given by $\alpha(0) = \alpha(1) = v$ and $\alpha(\frac{i}{l + k}) = v_i$ for $1 \leq i \leq k - 1$ and $v$ for $i = k$, and $\alpha(\frac{i}{l + k}) = w_i$ for $k + 1 \leq i \leq l + k - 1$?

[gentle osprey]

okeyokay

[heady skiff]

i feel like this is a case of it being intuitively clear, but at the same time tedious to verify so authors tend to say it's clear and not show it

[knotty vine]

wdym? Whats the relation between the screenshot and what you wrote?

[heady skiff]

oh yeah sorry that wasn't clear, i was just trying to figure out the image of {vv1...vk-1vw1...wl-1v} under alpha

and rigorize how it was homotopic to the product of the two respective loop classes

but it just seems like a total pain

[knotty vine]

Yeah that seems right, but it could always be off-by-one with these things idk

[knotty vine]

Isnt it exactly equal to the concatenation of two representatives?

[heady skiff]

i guess so?

[twin siren]

for this question, what kind of indicators tell me how rigorous i have to be in my working

i didn't do real analysis (doing metric spaces rn), and i have a difficult time understanding how rigorous to be, what clues to look for about what they want, etc

becauseo bviously exams are time limited so i cant waste time

[charred sundial]

it depends of the professor, but in general it would be ideal if you are convinced that you would be able to write a formal proof if you had the time

[twin siren]

TY,

if i were to decide to formally prove this

eg when proving b A = {a, b, c}, would i just assert that for e>0 (a-e, a+e) intersects both A and R \ A

and repeat for b and c

or is that sometimes not rigorous enough?

[wicked pollen]

(generally I would write that as dA, if I had to say it without latex)

[charred sundial]

that's totally fine

[twin siren]

ok, ty

[charred sundial]

as a grader, I would be convinced that you could write the whole formal proof

this is the only important step, the rest is bookkeeping

[twin siren]

Ok, I see, so that's the standard to try hit 🙂

[shy tartan]

hi friends, this is going to be a strange question but i'm looking to figure out what the prof is asking me to do for his homework question LOL. i've been sick with covid and was absent from class for the last 2 weeks and i emailed him to ask for the homework in which the only question he assigned was to "identify the relationship between taking the interior and closure of a set", but is the relationship not just the contrast between definitions? unions vs intersections and open vs closed subsets? this is following the lecture of compactness and having the finite intersection property being equivalent so i have a feeling hes trying to point me somewhere in that direction but I'm lost and I would like to give this to him tomorrow when i return to class

[tribal palm]

as i’m sure you know, opennes and closedness are not logical opposites (you may have that a set is either, neither or both), but they are in fact dual concepts, which you make think of informally as that every statement about an open set U can be translated to a statement about the closed set X \ U, and vice versa

[shy tartan]

oh 😭 sometimes the brain tries way too hard haha. i think i can see where to go, thank you

[tribal palm]

i should sleep but i’m sure some here’ll be of help if you need another hint

[rapid lagoon]

If we have two closed sets A and B

will A\B always remain closed?

[ebon galleon]

No

Consider [0,2] \ [0,1] = (1, 2]

Now, if F is closed and U is open then F \ U will be closed again (b/c it's F n (X\U)) and vice versa

[rapid lagoon]

Of cours

[unreal stratus]

Indeed if X is a metric space and C a closed subset, then the boundary D of C is closed and C \ D is the interior which is open (and so usually not closed)

e.g. [0,1] \ {0,1} = (0,1)

[silver umbra]

i want to show that a particular subset of R^2 is totally disconnected

would it suffice to show that it doesn't contain any open disks?

[red yoke]

Counterexample: line

[silver umbra]

what would be a general strategy of showing that a subset of R^2 is totally disconnected then?

[red yoke]

By showing the only connected subspaces are points

[silver umbra]

is there some more concrete way of showing that

that's literally just the base definition

every example ive seen so far are just subsets of R, where the only connected subspaces are intervals

so the proofs all revolve around not showing that the set contains an interval

[red yoke]

By showing any connected subspace with more than 1 point has points in the complement?

Do you have a specific problem in mind

[silver umbra]

it's pretty advanced, i hardly understand the constructions being used

https://indico.ictp.it/event/a14294/session/34/contribution/142/material/0/0.pdf

example 1.19 here defines the schottky group

and they mention in passing that one could show its limit set is a cantor set

it would suffice to show it's totally disconnected

but idk how to do that

[silver umbra]

im just looking for a general strategy though

like the standard argument for subspaces of R is: choose two points in ur subspace, then show that theres a point between them that isnt in your subspace