point-set-topology

[opaque totem]

and when you need to use ... use \ldots

oo also \left( \right) around bigcup

that'll make the whole sentence look beautiful

[frigid river]

I'm sorry, but I am still not sure why I am trying to show this?

[empty grove]

This will show that under the homeomorphism, a basis goes to a basis

So if the basis is countable in the domain, then its image is a countable basis of the codomain

[gentle osprey]

festina lente, cucurbita

[empty grove]

That will not matter

Try to prove that if a set {B_i} generates a topology by unions, it is a basis

That is the reason we define a basis the way we do. It's just a way of saying that the family generates a topology by unions

[pearl holly]

Okay so I am stuck on this exercise:

(Exercise 3)

[hollow harbor]

So in the plane it's an infinite popsicle, and you're squishing it down to the x axis I think?

[pearl holly]

and my initial thought was to show that q is a quotient map using just the definition. But that didn't work. So then I tried to show that q maps open saturated sets to open sets. But I can't seem to express the saturated sets in a good way. So then I tried to construct quotient maps f and h such that q = f o h (the composite of f and h). There is an earlier exercise about retractions and how they are quotient maps so I tried construction a retraction. So first we can construct a function h: A -> R X {0} such that h(x, y) = (x, 0). But now I am completely stuck on showing that h is continuous!!! If I show that it is continuous then it is a retraction and hence a quotient map. Then I just have to figure out a different quotient map f such that f o h = q and then I'm done. But I'm already stuck at the first part lmao

[pearl holly]

yeah I think so

[hollow harbor]

Ok well

Open sets in R x {0} are made of open intervals

What is the preimage of an open interval

Under h

[pearl holly]

well just the same open interval?

because h(a) = a for all a in R X {0}

so the inverse is the inclusion map

[hollow harbor]

Yes, but there are other points which map to a too

h is a projection

[pearl holly]

I honestly don't know lmao

but wait, what topology does R X {0} have? The product topology?

[hollow harbor]

It's just the topology from R

Product topology with a point is the same topology

(this is a good thing to check I guess)

Anyway, if a = (x, 0) and if x < 0, then we're on the handle of the popsicle. So there's only one point which gets squished down to a, a itself like you said.

If x ≥ 0, then now we're in the popsicle and A includes all the points (x, y) for any y in R. All of these points map down to a = (x ,0).

So to show continuity, we have three checks:

1. Preimage of an open interval completely contained in x < 0
2. Preimage of an open interval completely contained in x > 0
3. Preimage of an open interval containing 0
   The third case is where you have to be careful and remember that we are talking about the subspace topology on A.

(since the open intervals in R are a basis for the topology, openness of preimages of other open sets follows from checking openness of preimages of open intervals. To see this, remember that preimages of unions are unions of preimages)

[pearl holly]

yeah okay this will get me started. I will think about this!

[frigid river]

[pearl holly]

hmmm but don't I know this: the inverse is the inclusion mapping and it is continuous. Therefore, h is an open map and since h(a) = a for all a in R X {0} then it must be continuous?

wait wait... Isn't f: R -> A defined as f(x) = (x, 0) a right continuous inverse to q? Since q(f(x)) = q(x, 0) = x? So then by an earlier exercise q is a quotient map?

[hollow harbor]

I don't know what you man by inverse

Ok, yeah, that might be a fine proof that q is a quotient map

[pearl holly]

okay well then the exercise was not that hard, I just looked at the wrong places all the time lmao. But thank you so much for the help!

[frigid river]

How would I prove this?

Let $X$ and $Y$ be topological spaces, and $f:X \to Y$ be continuous.
$\Longrightarrow f(\bar{A}) \subset \overline{f(A)}$, for any subset $A$ of $X$.

[gentle osprey]

festina lente, cucurbita

[hollow harbor]

Suppose x is in f(A bar). Then what does it mean to claim that x is in f(A) bar? It means that we want to show any open neighborhood of x intersects f(A).
Well we know there is a y such that f(y) = x and any open neighborhood of y intersects A. So take an open neighborhood of x. What's the first thing you can do with it?

[verbal wraith]

You don't need the concept of neighbourhoods for basic things like this. I personally wouldn't bring up neighbourhoods without introducing the neighbourhood filter.

[empty grove]

based

[gritty widget]

[honest terrace]

I'd love to see a topology book written by nikita tbh

[cerulean oriole]

Neighbourhood filters 😍

[cerulean oriole]

++

[empty grove]

👀

[marsh forge]

wdym the "concept of neighborhoods"

neighborhoods are the second most basic concept in topology after topologies themselves

[empty grove]

second most basic concept is of continuous functions as functions on the topological spaces that induce lattice morphisms satisfying the first isomorphism theorem

[marsh forge]

honestly "neighborhoods bad" might actually be a worse take than the lattice thing lmfao

[empty grove]

[obtuse meteor]

+1

[fading vale]

hot pedagogy takes

[gritty widget]

reminds me of the time i was explaining how derivatives are all about linear approximations to a highschooler in VC, and somebody interrupted to think out loud about whether or not you could model differential calculus using infinity categories

[marsh forge]

analysts HATE him

[gritty widget]

i got a little angry

[marsh forge]

yeah i mean, its kind of an interesting question for after the explanation is over or another time

but interrupting for that is just like

:iamverysmart:

speaking of

we need an emote of that variety

[hollow harbor]

is that not

[marsh forge]

hmm i guess it works

[hollow harbor]

the so true meme is very iamverysmart

[obtuse meteor]

[honest terrace]

ok but what's the answer

[gritty widget]

i let them ramble and then told them that they got the target audience wrong

[fathom cave]

Im confused with this defintion. does it mean X is zero dimensional if it has a basis that has more than 1 clopen subset? not every member of this basis has to be clopen set right

[empty grove]

all the basis elements should be clopen

[fathom cave]

is it? it doesnot say every subset is clopen tho?

[cloud owl]

clopen 👎
opposed 👍

[empty grove]

thats what basis consisting of clopen set means presumably

[marsh forge]

yeah this is common phrasing in math

it means only clopen sets

[gritty widget]

usually "consisting of x" means "all elements are x," even though you can interpret it as there being only one x

[marsh forge]

if we don't take this approach then every disconnected space is zero dimensional haha

(assuming we respect the plurality)

[fathom cave]

ok ty

[marsh forge]

oh no wait

it would be every space duh

empty set

always forget the empty set

[empty grove]

do you mean the least element of the lattice

[fathom cave]

i will need time to process what max said lol
im still learning

[marsh forge]

oh uh

its not deep

every space has at least two clopen sets, the whole space and the empty set

we can always throw extra open sets into a basis and then get a new basis after making sure intersections unions etc are all added too

so if being zero dimensional is just about having some basis with some number of clopen sets

then any space would be zero dimensional

Ultra has declared it to be so

[pearl holly]

Okay so this is an unusual question, but there are supplementary exercises on topological groups. Do topological groups have anything to do with algebraic topology? Is it worth doing the exercises?

[marsh forge]

Uh

[gritty widget]

it's worth doing them regardless

[marsh forge]

Okay the first question

yes they are super important

the second question

maybe

depends

[empty grove]

You won't see them used in munkres I think, if your aim is only alg top

[marsh forge]

like at some point you should work with topological groups if you want to do alg top

[empty grove]

From munkres

[marsh forge]

does that have to be right now? or soon?

no

Like honestly you won't work with topological groups seriously until after you cover basically everything in hatcher

idek if concise mentions them either

[gritty widget]

i am lie group pilled so i am legally obliged to tell you to do them

[pearl holly]

oh crap, well then I have a long way to go lmao

[marsh forge]

I mean thats very different imo

[obtuse meteor]

Discrete topology on a group is the only one 😌

This post inspired by covering space theory

[marsh forge]

lie groups are a very specifically behaved subset of top groups

[pearl holly]

okay this sounds interesting. I think that I am going to do some of them at least

[marsh forge]

but most of the topological groups you will ever care about are the matric groups, tori, and discrete groups

honestly my suggestion

[obtuse meteor]

Is Z/4Z a topological group with the weird circle topology

[marsh forge]

don't bother with the really point-sety stuff

if the exercise statement itself interests you

do it

otherwise you can probably safely ignore it

[pearl holly]

okay great! Thank you!

[marsh forge]

no idea

and i refuse to think abt it

[obtuse meteor]

But imagine

[cerulean oriole]

…

[obtuse meteor]

“My finite topological group has the same homotopy type as S^1”

[cerulean oriole]

How many topologies does it even have?

[marsh forge]

afaik "how many topologies are on a set with n elements" is an open question

[cerulean oriole]

Does it have any other than discrete and indiscrete?

[obtuse meteor]

Idk

[cerulean oriole]

n=4 tho

[marsh forge]

topologies as a group or topologies as a set

[obtuse meteor]

Topologies as a group

Is what we’re concerned about

[marsh forge]

i mean

you can write down the topology and open sets easily enough

just check if they are preservered

[obtuse meteor]

But checking things

Sucks

[marsh forge]

lmao faye

[obtuse meteor]

And I don’t have paper

[cerulean oriole]

The closure of {1} is a subgroup right?

In a top group

[marsh forge]

if i have learned any skill that made be better at math in the past year or so

[cerulean oriole]

So only three possibilities I think

[pearl holly]

but you do have a brain

[marsh forge]

its that actually buckling down and checking things is worth it

[obtuse meteor]

Oh I agree

But

I don’t have paper lol

Or a pen

[marsh forge]

fair

anyway

[cerulean oriole]

Use bren

[pearl holly]

USE

[obtuse meteor]

I’m having breakfast

[pearl holly]

BREN

[marsh forge]

the only worthwhile topology on Z/4

is discrete

[cerulean oriole]

YES

[cerulean oriole]

Or any finite group?

[obtuse meteor]

My sister is getting a breakfast shot

[marsh forge]

Yes

well

any cyclic group

im worried ill say something too strong if i say any finite group

[cerulean oriole]

IDT this was about worthwhile topologies

[marsh forge]

but i can't think of a counterexample

well it would be cool

[cerulean oriole]

Do you want it to be Hausdorff?

[marsh forge]

if Z/4 with the homotopy type of a circle was a topologyical group

it would be interesting to me

not interesting enough to check

but interesting

[cerulean oriole]

Or T_0 for that matter

Then it's discrete

[marsh forge]

hausdorff is a super unnatural condition if you want to work with finite spaces

i guess i mean "against the point of studying them in the first place"

bc you want like

to model homotopy types using as few points as possible or at least finitely many

so of course you're sort of going to want to "trick" your topology

its a good example i think if you are like interested in the study of homotopy types themselves

its not super interesting if you're just using homotopical stuff to study spaces that we already care about

[obtuse meteor]

That’s totally an open question and that’s such a scary one

I give you a homotopy type

What’s the min # points you need

To make it

[cerulean oriole]

OK so the topology of a finite group is uniquely determined by the closure of 0 which is a normal subgroup.

Any normal subgroup should do.

[marsh forge]

i mean

is that surprising

that its open

homotopy does not behave well at the point-set level at all

R^n and a single point have like nothing in common lol

[obtuse meteor]

Oh it’s not surprising

It’s just

Scary

Like

I’m imagining some combinatorial homotopy theory ever becoming relevant some day

And this being a relevant question to ask

And crying

[marsh forge]

ultra ur not allowed to make me feel uneducated in this channel

its my safe space

no

[fathom cave]

this is my proof for Q is zero dimensional
any comments?
so I want to show Q is zero dimensional under usual topology.
we know that $B={ (a,b) : a , b \in \mathbb{I}}$ is a basis of $\mathbb{R}$
. Basis element $A$, of $\mathbb{Q}$ is $Q \cap (a,b)$ where (a,b) is an element of B
then, $ \mathbb{Q}-A =[(-\infty, a] \cup [b, \infty)] \cap \mathbb{Q} = [(-\infty, a) \cup (b, \infty)] \cap \mathbb{Q} $
which shows A is clopen. This is true for any basis element of $\mathbb{Q}$ hence Q is zero dimensioinal

[gritty widget]

put backslashes before your set brackets to make them not disappear

[gentle osprey]

Zero0

[fathom cave]

I is the set of irrational numbers

[marsh forge]

@potent shadow to give a more detailed explanation of why your question is getting

Mathematical tools are by their nature both incredibly precise and very restricted to their contexts

It doesn't make a lot of sense to ask such a vague question because it is mathematically speaking meaningless

[empty grove]

Works

[marsh forge]

Like there is no answer either way

[gritty widget]

fadey?

[cursive flume]

is the only difference between inner product and non-degenerate symmetric bilinear form the positive definiteness?

I mean there is a claim I heard that only riemannian metrics induce inner products on tangent spaces

other metrics do not induce inner products

metric=a (0,2) tensor field which is symmetric and makes the musical isomorphism invertible

or,in other words,a (0,2) tensor field,which is symmetric and non-degenerate

then one can define the signature of the metric and the claim is only the riemannian signature metrics induce inner products on tangent spaces

so what do lorentzian metrics induce then?

non-degenerate symmetric blinear forms and nothings else?

or how to name the object they induce

the only thing is that we weaken the condition on inner product

that it can be negative

and my question is if that has a specific name

pseudo inner product

right

but is the statement true then, that every non-degenerate symmetric positive definite bilinear form is an inner product?

idk I think so,I heard this term before so it must exist lol

wait sec i'm a bit confused

if it need not be positive definite that's true

[gritty widget]

i've seen those called scalar products, with the terminology "inner product" used to mean positive definite. lee uses this, i don't know how standard it is

[cursive flume]

but since having -es, won't it be also degenerate?

like the 'inner product induced by the pseudo riemannian metric' will be degenerate,or?

[gritty widget]

non-degeneracy is part of the definition for a pseudo riemannian metric

[cursive flume]

oh right,sorry my bad

it is the definition for any metric independent of signature

I just stated it above lol

[cursive flume]

[frigid river]

How wrong is this?

I only care about "$\Longrightarrow$" for now.

[gentle osprey]

festina lente, cucurbita

[marsh forge]

I think your middle arrow is doing far too much work haha

you might as well just assert the result is true

[empty grove]

It just seems a bit incomplete because the second step is a big jump

[frigid river]

[ivory dragon]

proof: exercise for the TA

[marsh forge]

theres a weird tension when self learning between not needing to fully explain yourself for a grade

but also running the risk of lying to yourself about what you understand

if spelling out the details scares you I would suggest you need to do it

[gritty widget]

soapy water

[pearl holly]

CATenoid

[gritty widget]

there is an entire section in lee's riemannian manifolds on them

chapter 8, riemannian submanifolds

anyways uh channel's free i'm just posting something neat

(i saw someone typing i didn't want them to think it was occupied)

so learn it lol

xd

IRM is more entertaining imo

[obtuse meteor]

All math books are super boring imo

It’s kinda criminal how badly they’re written

[marsh forge]

idk of an alternative tho

ive never seen an entertaining math textbook but that might be for a reason

[obtuse meteor]

Yeah rip

I just don’t read books

[marsh forge]

tbh I can't say that I've learned all that much from front-to-back style book reading

knowing the basic structure and content of the books related to what you are learning is super useful but i feel like its rare to actually read them that way without lecture or guidance and have it be productive

[obtuse meteor]

Ye

I am educated by Wikipedia, discord, and the nlab

(As well as lectures + psets)

(And the occasional paper)

[marsh forge]

yeah

i think eventually

you can't get away with ignoring reference texts

as a lot of material basically only exists in them

esp. proofs

[frigid patrol]

Based

Except for nlab

Unbased

[cursive flume]

nlab vibes

[obtuse meteor]

Based definition

[gritty widget]

a kth order partial differential equation on a manifold M is a closed embedded submanifold of the kth jet bundle of the trivial bundle M x R -> M

oh come on that one's at least vaguely geometrical

[obtuse meteor]

It’s actually an object in the eilenberg-Moore category of the jet bundle comonad

[opaque totem]

you said let x\in\bar{A}, but what is A or \bar{A}?

[obtuse meteor]

This is so illegal

I mean the context here is specifically making fun of the nlab so

[cursive flume]

where did you get the urs emote frm?

:urs:

but where could you get it from if you had nitro? or it doesn't exist yet and you'd make it?

[quaint agate]

Hey, anyone up for giving their take on an idea I had for a knot theory project?

[gritty widget]

i'm knot sure many people here are tied up in knot theory

[quaint agate]

Maybe not, hm

well I guess it wouldn’t hurt to post anyway

computing the sum of the Alexander polynomials is what I mean

[chrome dew]

you make a pretty strong claim that the red path is unique, you should probably test this out on more than 1 example, try doing this on multiple knots, namely the same knot in different crossing patterns and in non minimal crossing diagrams to see if the claim holds up to just this basic line of inquiry

[cedar pebble]

Yea this is unreasonably strong, as is the pattern with most of your knot theory ideas

[gentle osprey]

ProphetX

ProphetX

[cursive flume]

I think the LHS is the diffgeo perspective, and the RHS is the vector calc perspective,right?

[hollow harbor]

yes, this is correct

by linearity of evaluation maps

[gentle osprey]

ProphetX

ProphetX

[gritty widget]

yes the dot product is C^infty linear

u can do this

it is legal

[silver umbra]

what does the topology of the orthogonal group O(n) look like?

what do the open sets look like?

if you have a linear transformation g in O(n)

is the open ball of radius r centered at g

the set of all linear transformations h in O(n)

such that the operator norm of g - h

is less than r?

[little hemlock]

is the equality in i) by any chance a mistake?

since im pretty sure showing X_fg \subset X_f \cap X_g requires showing that for a prime ideal P not containing fg, we have P does not contain f nor g.

[true robin]

Hint you only need p to be an ideal in this case

[little hemlock]

ah right, thanks

[true robin]

Np

[cursive flume]

I just realized something: a Lie algebra of a lie group is either the tangent space at the identity or isomorphic to it. since this is a vector space,then the lie algebra is in particular a manifold,right?

so each tangent space of a manifold is a manifold?

[sleek thicket]

yes

This matters sometimes

[cursive flume]

this is shocking

[sleek thicket]

Do you know the exponential map?

[cursive flume]

yes

[sleek thicket]

Well that's actually smooth

And its derivative at the origin is the identity, so by the inverse function theorem it's locally a diffeomorphism from the origin to the identity

[fathom sandal]

Not really sure if this goes in here but uh what does this region even look like? I'm having trouble trying to imagine it

[empty grove]

#calculus

[river granite]

Picture a solid sphere of radius 2, take the quarter-sphere that consists of the points such that 0≤z,x, then slice that quarter sphere twice: once at the plane z=x (and take the lower half since these points verify z≤x), then again at y=z, taking the points that verify y≤z. Maybe play around on https://www.geogebra.org/3d to get a better picture.

If you have further questions about this ask in #multivariable-calculus, I think it's a better fit than this channel

[fathom sandal]

Oh ok

Yea it showed up in my geometry course is why 😅

[river granite]

I thought it seemed like a typical multiple integration problem, but I guess the volume can also be found through elementary arguments (it'll end up being a fraction of the original sphere's volume)

[wanton marsh]

I wouldn't want to see anyone trying to get it by a triple integral

[fathom sandal]

Wait why

@river granite how do you know that it'll end up being a fraction of the original sphere's volume?

it does say to go and treat it as a union of tetrahedra

[craggy slate]

Hi there, I've got a reference request: looking for a full, fleshed-out construction of the standard charts on a general embedded submanifold, and a proof that they're smoothly compatible. Anyone know of a good source? I've looked through Lee's Smooth Manifolds and Boothby, but I might have missed it. Thanks.

[hollow harbor]

You should really think of the topology on just about any subspace of matrices / finite dim linear operators to be "matrices are close if they're close in every entry", i.e. the topology inherited from R^(n^2). Because any two norms on a finite dimensional vector space induce the same topology, this means the topology on O(n) is induced by operator norm balls, or by balls for the euclidean norm (square all entries, sum, square root) or by balls with taking the max of the absolute values of the entries, etc. Any usual way of describing the topology on R^(n^2) restricts to any space of finite dim operators. This is useful because visualizing that two operators are close in operator norm is a little wacky (at least to me).

Of course if you're dealing with O(V) for some fin dim abstract inner product space V, then you would need to pick a basis in order to make sense of this "close at every entry" business. And if you're dealing with infinite dimensional inner product spaces it's not as easy to understand the topology on orthogonal groups. But fortunately for R^n it's pretty visualizable.

[vocal anchor]

I'm trying to visualize the 3-dimensional handles, and wanted to see if I got the k = 0, 2 cases right, since I find them harder to visualize than the other two:

• k = 0: put a 3-ball next to the manifold (e.g. disjoint union); its core is the origin; its cocore is the sphere bounding the ball; attaching sphere/region are empty; belt sphere is the same as the cocore (a peculiarity only happening here, because the 0-ball is just one point)

For k=2 there'll be a picture:

Imagine the simplest 3 manifold, e.g. the 3-ball, and punch a cylindrical hole through it. The result will be a three-manifold into whose boundary the cylinder can be embedded. Glue the handle into this cylinder

blue = belt sphere; purple = cocore; dark green (in the middle) = core; red = attaching sphere; light green = attaching region

[vocal anchor]

This picture is from encylopedia of maths, and apparently the right handle is supposed to be a 2-handle for n=3. However, by definition the handle should be a (solid) cylinder, which is not homeomorphic to that annulus-cap-thingie, is it?

[hollow harbor]

you know these things?

when you flatten one out against the ground

you get a pancake (which is a short cylinder with a large radius)

i should hope so!

[vocal anchor]

thank. why am I so terrible at this? 😦 dang

[hollow harbor]

You are not terrible at it

This is not supposed to be easy to understand

[vocal anchor]

shh, I'm trying to be dramatic 😄 Not terrible, but in all honesty, I should be better at this by now, given my level. Then again, I'm an algebraist and not a topologist

[hollow harbor]

As many household objects as I can google image search to explain my topology ideas, I will never be able to understand what is going on in the five lemma. So enjoy your skillset and enjoy getting to think with other skillsets once in a while too, even if you don't feel comfortable with them.

[vocal anchor]

math is just, like, manipulating symbols on a page, dude

rep theory, i guess. or maybe let's say "quantum algebra" as a fancy term for hopf algebras and theirs rep theory

[empty grove]

As a fictionalist I unironically agree

[vocal anchor]

Anyway, is this picture of a two-handle also correct? I guess the one from encyclopedia of maths is more natural when one wants to attach handles to a sphere, right?

too much topology for me 🙈

thanks, at least a bit of confidence regained. and I don't know jack about locally compact quantum groups and C* algebras - I was a physicist once, and so am lacking in a lot of basic math. er... not trying to imply lcqg and c* algs are basic math lmao! just that I don't have broad knowledge, only very specialized one

[hollow harbor]

[pseudo crane]

[void ingot]

[sleek thicket]

[sacred gust]

[tight agate]

[pearl holly]

[rugged swan]

Let X be a notherian sheaf, does every ideal sheaves on O_X need to be coherent ideal sheaves ?

[quaint agate]

[random slate]

https://twitter.com/solidangles/status/1402062560820613123

Bill Shillito (@solidangles)

Hey computational AG people! What are some cool elementary applications of Gröbner bases, stuff that would be accessible to high schoolers?

I'm looking for a few good "okay so what?" topics to have my GHP students explore (maybe in groups?) after we do Buchberger's algorithm.

Twitter · Jun 8, 2021, 12:38 AM

[marsh forge]

whoah

you have high schoolers learning about grobner bases?

[random slate]

I'm going to be teaching them this summer

[chrome dew]

are these gonna be online or recorded? I wanna go back to highschool 😎

[random slate]

Nah XD

[chrome dew]

oh well lol

[random slate]

It's gonna pretty computational instead of proof-y, playing around with stuff, using SageMath, etc ...

So what I'm looking for is just neat problems you can solve with GB's.

Unfortunately none of the ones I have are particularly visual

I'd been thinking of doing something with linkages maybe but I don't know much about them.

[tawdry valve]

Dummit and Foote has some exercises about n-colorings of graphs via grobner bases

maybe that's approachable? I know nothing about grobner bases, I just remember seeing that those problems existed

[silver umbra]

how many holes does a pair of pants have

i've looked through various answers on MSE and wikipedia and there doesn't seem to be an agreed upon answer

https://en.wiktionary.org/wiki/pair_of_pants

wiktionary says that it has genus 2

wikipedia says that it has genus 0

[sweet wing]

the 2d surface pair of pants has genus 0

it's basically a S^2 with 3 punched holes

[cedar pebble]

^

[sweet wing]

if you thicken it

[cedar pebble]

yea it's genus 0 as a surface

[sweet wing]

bleh

you prob get something nonzero if you thicken

[fading vale]

In what sense is the defn of a Galois cover p: Y -> X (meaning Y is connected and Aut(Y|X)\Y is isomorphic X) similar to the defn of a Galois extension?

[sweet wing]

the fixed field of L over the action of Aut(L/K) is K iff L/K is galois

if L/K is not galois the fixed field is some extension of K

similarly if Y->X is not a galois covering then the orbits is some cover of X

[fading vale]

I see

so its more saying that theres no non trivial factorization through intermediate fields/covers

this makes sense

[little dew]

the way I think about it is that if you imagine pants that have a super large waist and really thin legs, if you were to look at it from above and then deform it by squishing it nearly flat against the ground, the big "hole" on the top won't be a hole anymore and you'd end up with a shape that looks like this -- meaning that with thickness it'd be genus 2

though I'm kind of a newbie, so don't totally trust what I say

[lean marten]

I think it'll depend on what you mean by genus right?

Cause the strict definition is the largest number of nonintersecting closed curves which dont separate the space

In which case it's definitely genus 0

But if you use homology or fundamental group it should be Z^2

[wanton timber]

The two different answers I see online seem to stem from either taking a sphere and subtracting 3 disks (genus 0) or taking the boundary of a solid ball minus a y-shaped tube (genus 2)

[lean marten]

So rank 2

[wanton timber]

I would say that if you're not looking at compact surfaces (without boundary), you should probably tell people what you want your definition of genus to be 😋

[lean marten]

^

[wanton timber]

If you know some algebraic geometry (scheme theory), then L being Galois over K should be the same as spec(L) having a map to spec(K) and Aut(spec(L)|spec(K))\spec(L) will be isomorphic to spec(K), so there is a "space" version of the statement that is exactly analogous to that of traditional topological spaces

[tight agate]

If you have any intuition for differentials, it might be worth showing that the sheaf of relative differentials is 0 when the extension is galois

or just separable

(im assuming the book eventually talks about etale covers)

[tight agate]

actually, ignore that

the jacobian calculation should give you some more intuition

[gritty widget]

Why are sections of a bundle different from functions on the base manifold?

[pastel pier]

The bundle can be non-trivial

So global sections may not exist

[gritty widget]

Aren't sections defined as maps from the base space to the total space, satisfying a compatibility condition with the projection map?

[pastel pier]

Yes, but depending on the kind of fiber bundle that you have, global sections may or may not exist. For vector bundles you always have global sections, for principal bundles global sections exist if and only if the bundle is trivial

[snow shadow]

what is an easy argument why a line and a cross is not homeomorphic?

[empty grove]

removing the intersection point from the cross leaves 4 connected componenents, but removing any point from the line leaves only 2 (or 1 if there are endpoints)

[snow shadow]

yeah i was just thinking about that, because i got a comment on my assignment about this...

[flint cove]

what comment?

[gritty widget]

Gotcha, vector bundles being only associated bundles rather than principal bundles.

[marsh forge]

a topological space is a space that transforms like a tensor

[pearl holly]

So this exercise is kinda sus. Given two proper subsets A, B of two connected sets X, Y, respectively, I need to show that (X x Y) - (A x B) is connected. Isn't this trivial since (X x Y) is connected and (X x Y) - (A x B) is a subset of (X x Y), which is connected?

[floral gust]

{0} union {1} is a subset of a connected set [0,1] but {0} union {1} is not a connected set

So the premise "Subsets of connected sets are connected" is false

[pearl holly]

Oh no sorry. I confused myself with topological spaces. (X x Y) doesn't necessarily need to be finer than (X x Y) - (A x B), right?

"finer" in the topological sense

[empty grove]

what do you mean finer?

you cant compare topologies on different underlying sets

[pearl holly]

yeah that's true, I just confused myself lmao

[marsh forge]

toki draw some pictures

always draw pictures

[pearl holly]

okay sure, I will draw a banana

[floral gust]

draw bollz

[marsh forge]

i guess what this question is asking is like

okay so if X and Y are connected of course XxY is. But what if AxB is so much nastier than A or B that it somehow ruins it

[pearl holly]

hmm yeah I understand what you mean. I will try to draw a picture of this scenario and see what it will give me. Thank you!

[stray bane]

Can anyone help me find an example of a non-compact metric space which isn't seperable?

[empty grove]

discrete metric on an uncountable set

[stray bane]

Thank you!

[pearl holly]

Moldi studying

[empty grove]

[stray bane]

Just to be sure, this would induce a basis in which all of the basis elements are singletons or the whole set X right?

[empty grove]

any basis of it would need to contain singletons yes

[stray bane]

Ah yes, naturally there could be more than one

[empty grove]

yep

[pearl holly]

Have I understood it correctly?

bruh

let me take a screenshot

[empty grove]

the horizontal line you have drawn is not homeomorphic to X

[pearl holly]

Why not?

[empty grove]

what is the homeomorphism?

[pearl holly]

well idk lmao. Is it because the "X axis" can be infinetly long?

[empty grove]

No

horizontal lines in the product are isomorphic to the first factor usually because they are just X x {y_0} for some fixed point y_0

and the homeomorphism to X is "ignore the second entry in each pair"

but here you dont have a complete horizontal line

it gets cut off by A x B

so it wont necessarily be homeomorphic to X

[pearl holly]

yeah okay I get it, I will try a different approach then. Thank you!

[empty grove]

[marsh forge]

oh wait

i might have given toki a dumb task

i think drawing this might be too hard lol

@pearl holly

you need a X and Y to be 2D which means XxY is gonna be hard to draw lmfao

on no you dont

[empty grove]

That solution is very close to an actual solution tho

[marsh forge]

you can use two circles

i dont think theres a valid picture proof here moldi

[empty grove]

I see what you mean

[marsh forge]

the best bet here is to try out the torus example

[empty grove]

close to a valid proof in the sense that you ||pick the lines carefully||

[marsh forge]

and then prove it in a more real way

[pearl holly]

hmm okay I will think about it

[empty grove]

btw instead of trying to express the whole space as a union of intersected connected spaces, you can try to show that given any 2 points, there is some connected subspace in which they both appear

which I find is easier to handle, and also it seems more intuitive if you replace connectedness with path connectedness

[pearl holly]

hmm okay I will keep this in mind!

[empty grove]

actually I dont think you do

because X-A and X-B are not given to be connected (at least thats how I was interpreting your message)

like the statement applies when X=Y=R and A and B are any proper subsets

[pearl holly]

Moldi, you saw my picture that I posted, right? What if I just remove the lines that lie between the origin and the box? You said that the horizontal line was not homeomorphic to X, but what if I just remove those lines and have verticle lines there instead? Then the union of all of these will be connected this time and will also equal (X x Y) - (A x B)

[empty grove]

I don't get what you mean by remove all the lines "between the origin and the box"

Like all the horizontal lines that get cut off by A x B?

[pearl holly]

yes exacly

my discord crashed so I lost all the things that I wanted to write

[empty grove]

oof

Right so let's call lines which don't intersect A x B, good lines

[pearl holly]

So I just look at the union of all {(x, y)} such that y is not in B. Then I look at the union of all {(x, y)} such that x is not in A. The union of these sets will be connected and also equal to (X x Y) - (A x B)

It's kind of handwavy but doesn't it work?

[empty grove]

Hmm those 2 unions are not connected

Because you're taking everything in X x (B complement)

So if X = R and B is a point it's not connected

But you are very close

[pearl holly]

I honestly don't know how to proceed now lmao. It really feels like the union of those 2 are connected because they share at least one point but if it's wrong then I have no idea lmao

[empty grove]

Yeah the union would have been connected if they were connected lol

So instead of unioning these 2 big sets

Just take the union of a good horizontal and a good vertical line

This will always be connected

And then you vary the horizontal line

And take union of all of these "pluses"

And all these pluses have the same vertical line so are connected

And now in this larger thing you vary the vertical lines over all good vertical lines and take a union

[pearl holly]

yeah that is actually what I wanted to write down lmao

[empty grove]

Lol

[pearl holly]

and then it's proved right?

[empty grove]

You have to show that every point lies on some good line

Btw what I was suggesting you do was to take any 2 points, a and b, and find some good lines that they lie on. If you get good lines with different orientations ie 1 vertical and 1 horizontal, then the union of these 2 is connected and contains both points, and therefore in the whole space, these 2 points must be in the same component. Otherwise if you get that both a and b are in some good vertical lines, then you can find at least 1 good horizontal line, and that will intersect both of these good vertical lines so the union of these 3 is connected. So in the end you have proved that any 2 points of the space lie in the same connected component, so there is only 1 connected component

And this feels more intuitive to me because it's a lot like drawing rectangular paths from a to b, avoiding A x B

[empty grove]

In any case, you have to show this (I assumed this in the beginning of my proof)

[pearl holly]

okay well that isn't too hard to prove. It's intuitive and can be easily written down formally I think. But then what?

[empty grove]

Then my proof or in your proof you will use it to show that the union you have constructed is in fact the whole space

Since you only union the good lines

It's equivalent to showing that every point lies on one

[pearl holly]

okay and then you just use this long post and the proof is completed, right?

[empty grove]

Yeah

Like, either of the 2 proofs works

[pearl holly]

okay great! I won't write this down formally, it's just too long and boring. But this is the idea that I tried to write down earlier in my second attempt but it didn't work lmao. But thank you so much!

[empty grove]

[pearl holly]

btw, are you seriously studying? If so then I'm just disturbing you by asking dumb questions lmao

[empty grove]

Lol no I decided to just have the role permanently because I was wasting too much time in general

[pearl holly]

oh okay I understand you. I kind of feel the same now lol

It's easy to get distracted

[empty grove]

Yes

[pearl holly]

So I'm stuck on this exercise: Let p: X -> Y be a quotient map. Show that if each set p^-1({y}) is connected, and if Y is connected, then X is connected. I honestly don't even know how to start. I can't come up with a direct proof since connectivity hasn't any "nice properties" that I know of, except that a space if connected iff the only clopen sets are the empty set and the whole space. But I can't find a way of using it here. When I see p^-1({y}) I think of saturated sets and that a quotient map maps saturated open sets to open sets, but I can't find a good way of using this fact. I also thought that the contraposition of this statement would make things easier since then I could assume that X equals a union of two disjoint open sets, but then what? A contradiction doesn't seem to give me anything either. I looked at theorems and exercises that I've done (on quotient maps and in this current section) but I can't find anything to use here. Any hints?

[gritty widget]

if A, B is a disconnection of X then each fiber of p lies in exactly A or exactly B, by connectedness (lemma 23.2 in munkres). maybe that could be useful

[pearl holly]

hmm what is a fiber?

[gritty widget]

fancy word for preimage of a point

p^(-1)({y})

that's the first thing that comes to mind for the problem

[pearl holly]

yeah I thought about that as well but how do I know that p^-1({y}) is a subspace?

[gritty widget]

define subspace

[pearl holly]

well I define it to be a subset A of a space that has the subspace topology, i.e every open set is of the form A cap U where U is open in X

and to prove that lemma 23.2 you need a subspace I believe

[gritty widget]

what would "p^(-1)({y}) is connected" mean if not with respect to its subspace topology

[pearl holly]

yeah that's what I was trying to figure out lmao. But I guess that I can simply assume that it has the subspace topology

[gritty widget]

you can and should

[pearl holly]

okay great! Thank you!

[gritty widget]

ah this is a nice exercise

[pearl holly]

wait, doesn't this become quite trivial then? Because if I assume that X = C u D then p^-1({y}) is completely contained in one of those, say C. But then the union of these must be a subset of C but the union of these fibers is X. But then D is empty?

this might be really stupid tho lmao

I didn't even use the fact that Y is connected lol

[gritty widget]

how do you know the preimage of every point ends up in the same set C

[pearl holly]

yeah that's true

[gritty widget]

you might end up coming to that conclusion if you try to do a proof by contradiction, so don't write it off completely

just saying it's a bit of a jump

[pearl holly]

okay so there's a chance that every single preimage is in the same set?

[gritty widget]

depending on how you want to prove this maybe, i'm not sure

i proved the contrapositive

[pearl holly]

oh okay!

[cedar pebble]

@marsh forge do you know much about explicit computations in Morava K-theory, even just K(1)?

[pearl holly]

Okay so I think that I got something real this time. So just like before, each preimage must lie in either C or D (where C and D are the separations of X). We also know that the union of all these preimages must be X. Some of these preimages will be completely contained in C and some will be contained in D. So C will be a union of preimages and so will D. These are therefore complete preimages of subsets in Y and hence saturated. Now p(C) and p(D) are open since p is a quotient map. Furthermore, these are disjoint (p(C cap D) = p(C) cap p(D) by a theorem in my book) and the union of these will also be Y, which is a contradiction since Y is connected. Does this sound right?

[marsh forge]

No sorry I've been meaning to actually learn this stuff

like tbh even just the K(1)-local homotopy groups of S which should just be like the image of J are a computation that I didn't quite understand when I've seen it

[cedar pebble]

right I suppose I'm more interested in like

understanding what say K(1)-cohomology looks like for really basic spaces

stuff like punctured curves or configuration spaces or whatever

I know some people have computed 2-periodic K(n) cohomology for some configuration spaces?

[marsh forge]

well if you ever find anything worth reading send it my way

I'd love to look at it / compare notes / whatever

(unless its about curves all my homies hate curves)

[cedar pebble]

curves are nice 😦

[cold vine]

Why is contractible weaker than saying the space deformation retracts to a point? These seem like the same thing to me?

[flint cove]

Who claims that is weaker? seems like the same thing to me, too.

[marsh forge]

I guess that you can prove that something is contractible without exhibiting it as a deformation retract

so technically the deformation retract thing might be a priori stronger

[flint cove]

But being contractible by definition asserts that there is a homotopy equivalence to a point

and composing the X→∗-morphism with the ∗→X-morphism of the homotopy equivalence should give one the retract, right?

[marsh forge]

A deformation retract requires that the homotopy fix that point for all t

Like I think the two notions should end up being equivalent for reasonable spaces

I won't bet my career on them being the same in all cases but I'd be surprised

[flint cove]

Ah, so just claim that you're working over CW complexes and hope the non-hausdorff-gang doesn't notice

(at least that's my heuristic so far)

[marsh forge]

For decent connected spaces you can assume you have only one 0-cell in which case this is easy by CW approximation etc etc

[cold vine]

Hatcher: "Contractible: This amounts to requiring the identity map of the space be nullhomotopic, that is, homotopic to a constant map. In general, this is slightly weaker than saying a space deformation retracts to a point"

[marsh forge]

yeah i believe this

does hatcher give a counterexample

it will almost certainly have to be pathological

[cold vine]

Maybe it's this one?

This zig-zag in the exercises apparently is contractible but does not deformation retract onto any point

[marsh forge]

can you send the defns

[cold vine]

Yeah second

[hollow harbor]

I think a weak deformation retraction allows the subspace to move

Might be wrong

[hollow harbor]

Yeah, it just need to be a subset.

At all times

Ohhhh I remember how this problem goes now

[cold vine]

yeah

[cold vine]

theres the defs

[marsh forge]

this is gross

but i get it

Side note: I remember this folklore result that nice spaces should be replaceable by spaces with a single 0 cell but I cannot remember what the technical requirements are

or find a source for it

does anyone know it

[hollow harbor]

It's not gross!! It's such a nice picture

You drag it to the right half a period, letting the fronds follow their trails and then fold in against the stem.

And clearly there's no deformation retract since fronds close to the stem will always shear away from the stem if you hold it still

[marsh forge]

I actually get it less now

oh no i get it again

I don't see your nullhomotopy though

no i see it now

[frigid patrol]

i have no idea what ure tryingto say here

[marsh forge]

okay so the issue is like

if you try to move the little guys without moving the big guy

you will have little guys arbitrarily close to the big guy and you'll have contintuity issues

but if you "flow" the big guy with the frongs

fronds

you're fine

I still think this is gross

because its like

just a bad space

[frigid patrol]

i still dont geti t

[marsh forge]

how much time did you spend trying to picture it

apparently less than a minute

try harder lol

[obtuse meteor]

Ahhh I see

This space bad

Don’t quite see the weaker one

[marsh forge]

ok so if you see the contuinity issue

the solution is kinda the naive one

if the stem has to move with the fronds

then move it with the fronds

[hollow harbor]

it's so pretty!

[frigid patrol]

what do you mean fliw

[hollow harbor]

imagine it's an infinite tube full of water, and you turn it vertical

and the water falls down

through the little tubes into the main tube, and then keeps falling

[frigid patrol]

i see, so thats the deformation retraction in the weak sense?

[marsh forge]

yes

[flint cove]

oh wow, thanks for posting this. That's a really interesting counterexample.

[coral pawn]

If X = U U_i is an open cover of X, then C is closed in X iff C intersect U_i is closed in U_i for all i.

This is true right?

[small obsidian]

Obvious if there's finitely many such intersections haha. Let me think about the infinite case

[hollow harbor]

i'm also not immediately convinced in the infinite case

[coral pawn]

Suppose $C \cap U_i$ is closed in $U_i$. Then $U_i - C$ is open in $U_i$. This means that $U_i - C = O_i \cap U_i$ where $O_i$ is open in X. Taking union on both sides over i gives us that X - C = union of open sets. Thus C is closed

[gentle osprey]

Have a Banana, Bitch

[hollow harbor]

ok, each point in X\C is in some U_i, and has an open neighborhood in U_i which does not intersect C cap U_i. This open neighborhood is also open in X, and does not intersect C.

that's another proof!

[coral pawn]

Thanks

It's been a while since point set

What's the obvious proof for finite covers?

[hollow harbor]

union of finitely many closed sets is closed

union the U_i cap C's

[coral pawn]

But U_i cap C's are closed in U_i

Not X

[hollow harbor]

true

good point, so you need to do something like this either way

[coral pawn]

Maybe Kaynex has some ideas?

The case I'm concerned with is finite

and the proof I'm reading does not bother mentioning it

So there might be some trivial way to see this

Or it might just be math books being math books

[small obsidian]

Oh yeah no I think I fell for the same trap

[coral pawn]

Thanks for the help

[marsh forge]

I realize I am supposed to be the "expert" in the sense that I am an REU mentor

but does anyone have any suggestions for topology-flavored topics that someone who might well only know calculus could approach over a summer

I've been thinking and it seems like there aren't a lot of cool ones, but I think ppl do sign up for that

[frigid patrol]

Look at A Mathematical Gift by Ueno Shiga and Morita

[marsh forge]

Oh! This is cool.

[hollow harbor]

perhaps something with complex analysis and riemann surfaces? idk if that would be too much.

[marsh forge]

Maybe! Do you have ideas? Not really my area

[hollow harbor]

if they know multivariable calc, maybe there's some way to do basic morse theory / morse homology

uhhh

the stuff i did in an REU was about dynamics of flows on flat riemann surfaces with singularities (called translation surfaces, or dilation surfaces if you add some holonomy) which definitely didn't require much background at all but might be too far from what you're hoping to do since it was quite dynamics flavored

there are lots of topological questions one can ask about these surfaces though

in particular, i think people are really interested in the topology of the moduli spaces of these surfaces, but idk if that's too crazy.

[marsh forge]

I like dynamics! it's good to have a variety bc I havent met my students yet

and I want them to feel like they have options

There's some cool interplay between homology and dynamics too

but idk if they will know enough algebra

for me to teach them abt this

(I think it probably can be done rationally?)

[hollow harbor]

sure, so you may want to look at this
https://arxiv.org/abs/math/0609392

there's a lot of nice stuff going on here about answering questions about, say, billiards, or illumination problems, where the toolkits combine topological dynamics, complex analysis, and a couple interesting pieces of algebra (mostly with groups of linear transformations, which i think is pretty grasp-able)

it's also cool because maryam mirzakhani's work is really closely tied to this subject (the more advanced wing of it about dynamics of teichmuller flows on moduli spaces of flat surfaces).

[marsh forge]

Thanks!

[hollow harbor]

here's another paper on the subject about a particular problem which lends itself to nice, easy-to-understand examples and really friendly introductions to the interplay between topology/geometry/algebra:
https://arxiv.org/abs/1407.2975

but they end up using heavy duty machinery in that paper

(the eskin-mirzakhani stuff)

[marsh forge]

maybe u should be an reu mentor

[hollow harbor]

here's a nice set of lecture notes as well for another topic in the field of flat surfaces, "square tiled surfaces" or "origamis" which are really easy to define but admit a lot of very nice, understandable algebraic and topological questions.
these correspond to these lectures, which are good but imo a little choppy: https://www.youtube.com/watch?v=PB8k-1MwshU

[marsh forge]

I actually never met eskin

only heard amie talk about him

[hollow harbor]

Wtf you knew Amie???

Why didn't those two accept me

[robust zodiac]

Suppose `$R(\vec{v})$ is a 3d rotation that tranforms the $\hat{z}$ axis vector into the $\vec{v}$ vector, and $A$ is an arbitrary 3d rotation.
Consider then:
$$
R^{-1}(A \vec{p}) A R(\vec{p})
$$

[gentle osprey]

ConfusedReptile

[robust zodiac]

now, I note that if I consider how this acts on $\hat{z}$, I get that it gets transformed into $\hat{p}$, then to $A p$, then to $\hat{z}$ again - so it's unchanged, meaning this is a rotation around the z axis.

[gentle osprey]

ConfusedReptile

[robust zodiac]

Now... how would I got about finding what angle is this a rotation by, in terms of p and A?

This is a small part of a larger problem related to Wigner little groups, and yet it's the one I'm stuck on...

[honest narwhal]

@marsh forge you thinking apprentice/160s?

[marsh forge]

I'm assuming like

IBL

yeah

i dont want my kids to just produce the same "fundamental group and SvK" paper

that everyone always writes

or basic covering theory paper

[honest narwhal]

Yeah better to do something that's not usually done in a class lmfao

[marsh forge]

but i also wanna avoid finite spaces tbh

so idk

[honest narwhal]

Yeah I sympathize with that

I buy dynamics as a possible topic, I think to do something interesting with Riemann surfaces you might want more background?

Maybe geometric group theory type business?

[obtuse meteor]

max doesn't believe in subjects which are not algtop 😌

[honest narwhal]

I think Max is fine with dynamics lol

It's more like

[elder yew]

Are you retaking alg top at UCSD max?

[obtuse meteor]

(joke)

[elder yew]

Or you just gonna go in and try to yeet on the quals

[honest narwhal]

We want something specific

Or idk I assume in this regard Max thinks likes me

It shouldn't just be a couple definitions and baby propositions here and there

Ideally you want a target theorem that shows within the paper that what you're doing is cool

[marsh forge]

if i can

which should be possible

my real ideal strat

would be to like

TA for alg top lol but idk

Yeah topological dynamics is cool

[elder yew]

I applied to work under Justin Roberts at SD

[marsh forge]

So as I understand it and maybe @sleek thicket can correct me

The uchi reu kids get to pref topics

and one of the topics they can pref is strictly alg top

[sleek thicket]

i believe so

[marsh forge]

yeah so if i get one of those kids idk what ill do

but if i get just topology

[honest narwhal]

Persistent homology time

😛

[marsh forge]

i think top dynamics is good

oh

honestly

tda could be worse

[sleek thicket]

i decided i regret my feedback on the u chicago form

for preferences

[marsh forge]

lmfao

what did you put

[sleek thicket]

I didn't put logic at the time and now I want to do model theory

ultra warned me

[honest narwhal]

What happened to you

[marsh forge]

oh that might be rough sham

[sleek thicket]

yeah its an oof

[marsh forge]

what did you pref tho

[sleek thicket]

no diea

[marsh forge]

no no this is for my kiddos

[sleek thicket]

it was posted on here and on twitter a while ago

[marsh forge]

who might not know enough to do serious algtop

[sleek thicket]

you did thouhg

[marsh forge]

yeah so

[sleek thicket]

😔

[marsh forge]

im a young person

[sleek thicket]

also max i have still heard nothing

[marsh forge]

so i will get paired with an apprentice

i kno sham

me too

[sleek thicket]

oh sick

[marsh forge]

lol

[sleek thicket]

i no longer have to worry about max being assigned to mentor me

[marsh forge]

ive told u like 3 times

i wont be mentoring u

[sleek thicket]

X

i have no memory of this

[marsh forge]

no im gonna get assigned to like

a uchicago 2nd year who switched majors

at most

everyone else should be in the full program i think

and they'll get a real grad student

[sleek thicket]

i feel like emailing and asking what's up

but maybe that would be rude

[marsh forge]

I wouldn't

I talk to him weekly and he says it will be done soon weekly

[sleek thicket]

rip

[marsh forge]

the real strat if i have a uchicago kid

is to walk them through a really impressive reu paper in topology

so that peter adopts them

[sleek thicket]

awww

[obtuse meteor]

adopts :P

that is precious :P

I need a mathematician to adopt me

[gritty widget]

guillemin and pollack might be readable

[hollow harbor]

[vast current]

aren't you adopted already?

[obtuse meteor]

not really?

[vast current]

ask your advisor that took you to that REU :c

[verbal wraith]

Here's another type of combinatorial complex that I came up with if anyone is interested.

[obtuse meteor]

these are pretty well known iirc

similar to a simplicial or cw complex

just modeled on the topological n-cubes

even have a wiki page

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

[verbal wraith]

Are these really the same thing?

So these kind of complexes that I came up with are a bit more nuanced in that they don't model things that look like manifolds.

[cedar pebble]

rip 1 hour too late to appear in Sham's twitter screenshot

[verbal wraith]

They're more general in the sense that the boundary of the boundary is not in general empty.

And you can also have reflection surfaces: so you can connect an interval to itself by connecting one of its end-points to itself.

I didn't actually define a complex or write up the homology theory for them this is just a basic idea.

[marsh forge]

If d^2 is not zero why do you expect to get a homology theory

[verbal wraith]

You can still define a quotient that looks a lot like homology

[marsh forge]

I am not sure this makes them more general? What are you actually comparing

[marsh forge]

I don't see how but I guess I have to take your word for it

[verbal wraith]

Here we have a travelling operator and the homology is just chains modulo travelling.

The travelling operator is kind of nice in itself so it's not immediately obvious that there's still a need to construct homology groups.

[coral pawn]

Open sets of irreducible spaces are irreducible right?

[honest terrace]

yh

two nonempty disjoint open subsets O1, O2 of an open subspace U of X would also be two nonempty disjoint open subsets of X

[wanton timber]

The statement is correct. This argument is not. You can be reducible but not disconnected.

I'm lazy, so I'll just link you to https://proofwiki.org/wiki/Open_Set_of_Irreducible_Space_is_Irreducible

[honest terrace]

What ?

I'm not proving that the subspace is connected

[wanton timber]

Two nonempty disjoint opens is a disconnection

[honest terrace]

I think there's a misunderstanding

I'm using the fact that a space is irreducible iff every two nonempty open sets have nonempty intersection

[obtuse meteor]

so @marsh forge bc I'm a bimbo I'm struggling with this problem https://estrogen.fun/i/4wsn.png

my idea is to steal an idea from local homology

the first step

• show that if it has one zero (which we can just relocate to be at the center point by coordinate transformation shenanigans) then it has degree = multiplicity of that 0

now

for finitely many zeros what I wanna do

is surround each 0 by a small ball

[marsh forge]

what chapter is this from

[obtuse meteor]

chapter 1

fundamental gorup

[marsh forge]

yeah

then thats the wrong machinery

hahaha

[obtuse meteor]

well yeah I won't use local homology

but stealing the idea

so then like surround each 0 by a small ball

and show that the degree of the total thing is the sum of the degree around each boi

by like

the outer loop can be realized as a composition of paths around each zero

by just kinda squishing it down until it's "tight" around these small balls I've chopped out

and then I can map each part of those guys

@empty grove what's wrong?

[empty grove]

I just read it without context

[marsh forge]

this is very much not my kind of problem and I am not sure how this approach would work. Im at the gym but ill chew on it and get back to you

[obtuse meteor]

it's very much not my kind of problem either lol

I do algtop to take big theorems and smash them into things not to prove actual concrete reasonable results about complex polynomials

[obtuse meteor]

OH GOD

[marsh forge]

i wonder if peter is assuming you know complex analysis here lol

[honest narwhal]

Just saw this and no, what's a quantum symmetric space?

[marsh forge]

dami do u have an idea for this

[honest terrace]

I'm not really familiar with all the concepts involved, but can't you just apply the same idea than the homotopical proof of the fundamental thm of algebra ?

[honest narwhal]

Oh the problem from Concise chapter 1?

[marsh forge]

yes

[obtuse meteor]

soph / someone said something about this being possible with pure algtop

I mean like

[marsh forge]

im sure it is

[obtuse meteor]

the solution with complex analysis should be like

"contour integral along homotopic bois is the same"

and then like

compute the contour integral

waves hands

[honest narwhal]

Let me think about this

[empty grove]

You could probably also integrate f'(z)/f(z) on the circle

[marsh forge]

yes i think thats not allowed

that would be my idea + argument principle

[empty grove]

That counts the number of roots inside with multiplicity

[marsh forge]

et and cetera

[honest narwhal]

Write $p(z) = (z-a_1)\ldots (z-a_k) (z-b_1) \ldots (z-b_l)$ where $|a_i| < 1$ and $|b_j| > 1$.

[gentle osprey]

Sloth King Daminark

[honest narwhal]

Let's say monic because who cares

[obtuse meteor]

I don't want a solution really

more like

is my idea reasonable

[true robin]

Isn’t it just homotopic to z^n?

[honest narwhal]

Oh okay I just started thinking out loud

[marsh forge]

it is the type of idea where I would need to see it done faye

everything is homotopic to everything in C saketh

[true robin]

Well ok, i mean in an easy way.

[honest narwhal]

Oh so you're saying the degree is the sum of local degrees inside the ball?

[obtuse meteor]

yes

[honest narwhal]

Basically

[obtuse meteor]

I want to prove that and then like

just prove it for one zero

which well proving it for one zero

[honest narwhal]

I could buy something like that working but you'd need to also show that adding zeroes outside the ball doesn't change anything

[obtuse meteor]

is basically just the computation of pi_1(S^1)

[honest narwhal]

So you'd want a part saying something like

q(x) = p(x)(x-a) where |a| > 1, then q/|q| has the same degree as p/|p|

I could buy that this is an outline of a proof strategy

I don't know offhand how to execute it/if it will have any different idea than just doing it in one shot thanks to FTA

But I could see it working

[obtuse meteor]

yeah

now to actually

work the details

[honest narwhal]

I think the second point is what's relevant tbh

[obtuse meteor]

which seem

hard

proving the degrees add seems pretty hard (tm)

but maybe I am missing something clear

[honest narwhal]

I mean start from 1

And just say okay

[marsh forge]

my first strategy is to prove that you can chop off roots outside the circle, prove it for z^n, and then somehow reduce the degree n general case with all roots inside the ball to z^n

[honest narwhal]

If I multiply p by (z-a)

Then either |a| < 1 and the degree is +1

Or |a| > 1 and nothing happens

I think either that or it's really just better to do it in one go

[marsh forge]

i despite functions C->C bc i cant visualize them but i feel like I should be able to

despise*

i really think this problem wants you to reduce to z^n

[honest narwhal]

Oh it 100% does

[obtuse meteor]

well in both my and your strategies we're doing that

we're just doing it in different ways

and you're doing it "all at once" I think

[marsh forge]

rue

true

[honest narwhal]

The point is

(z-at)

[marsh forge]

gl i guess

[honest narwhal]

Becomes a homotopy from (z-a) to z

And it doesn't cut through the unit circle

At least I think that's the point lmfao

[obtuse meteor]

ye

that makes things nice and removes the local degree

just have to prove the lemma for |a| > 1

thank you for talking with me <3

[honest narwhal]

Sure thing fam

[nimble jolt]

@honest narwhal you should be able to use the same idea for the roots outside the circle, but with (tz-a) instead. this lets you do it all at once to get a homotopy from p(x) to cx^k where k is the number of roots inside the unit circle.

[honest narwhal]

Yup seems good

Also @nimble jolt you wanna read this paper with me?

https://www.jstor.org/stable/2118522?seq=1

[nimble jolt]

Lol probably don't have time to read that paper properly but will talk to you about it at some point. I know the non-arithmetic side quite well.

Oh also I should say with my previous answer, we are using FTA implicitly when we factorise p(t) into linear factors. You can avoid this by simply factorising out the roots that exist inside the unit circle, and then the remaining factor has all roots outside the unit circle (although we do not need the assumption that it has as many roots as its degree). Then the same homotopy q(x) -> q(tx) still works for this factor.

[frigid patrol]

Dami teach this paper to me

[fading vale]

@pulsar thunder you've seen metrics right

[pulsar thunder]

yea, but only very briefly

[fading vale]

have you seen the definition of a topological space

[pulsar thunder]

that is i "know" what a metric is

[pulsar thunder]

well yea

[fading vale]

a set + a subset of its power set?

thats closed under union and finite intersection?

[pulsar thunder]

i havent seen this definition, no

[fading vale]

hurb

You have not seen the definition of a topological space then

[pulsar thunder]

oh

[fading vale]

https://mathworld.wolfram.com/TopologicalSpace.html

[pulsar thunder]

but i do get what this means

[fading vale]

in the same way metrics let us "distinguish points" by taking d(x, y) and defining open balls around x and y

open sets let us "distinguish points" by taking open sets around x and y that dont intersect

[pulsar thunder]

so, basically we take disjoint sets around x and y?

[fading vale]

like this

yes but those sets have to be open

thats basically the underlying spacial meaning of the topology

in metric spaces "close" points are ones that are very close to each other, i.e d(x, y) is small

as d(x, y) gets bigger and points get further away from each other

we can make more and more disjoint open balls around x and y

e.g if d(x, y) = 1 then our balls must be of radius less than 0.5 around x and y to separate them

but if d(x, y) = 100 they have to be of radius less than 50

we generalize by getting rid of the metric and retaining the notion of "separating by open sets"

points that are separated by many open sets are far away

if theyre separated by only a few then theyre closer

if they cant be separated at all then they are topologically indistinguishable

no topological data will let you tell one from the other

[pulsar thunder]

so if they are topologically indistinguishable, they are kinda same point?

[pulsar thunder]

ohk

[fading vale]

for the purposes of topology more or less

obviously theyre set theoretically different

[pulsar thunder]

okk, that makes a ton of sense now

[fading vale]

anyway i gtg

[pulsar thunder]

oh, ok

also a lot!

[fading vale]

max typing and stop typing is so ominous

i feel like i was about to be reamed for bad pedagogy

Even worse

[hollow harbor]

pedagogy in shambles

[marsh forge]

no u did good

@fading vale

@frosty sundial ull appreciate this too

[gritty widget]

compact mode

[frosty sundial]

hahahaha

it's ok moth pedagogy, like math, is something you can improve on over time :^)

[hollow harbor]

moth's explanation was fine

[pulsar thunder]

yea, i understood it just fine

[fading vale]