point-set-topology

[marsh forge]

not sure i understand the question

[rotund thicket]

well i guess im thinking about it in terms of equivalence classes because that's all i know

like if 3 points of a triangle lying on a circle were equal

and then the quotient space of that

[marsh forge]

If you aren’t familiar w homotopy this question will be a problem

in general you have to be careful with quotients

as they can really mess up homotopy types

Our goal is to compute pi_1 which is only a homotopy invariant

not a quotient space invariant

[rotund thicket]

ok, thanks

[flint flicker]

Thanks again for the help

[marsh forge]

the transition from what i drew to a wedge of circles is a little tricky

you basically just have to slide stuff around

[obtuse meteor]

another thing that's like good to see I think

is when you do the unfilled in version

you get a wedge of five circles, which should have pi_1 the free group on five generators

if you glue on an open square to make the filled in version

this "kills" one of your generators

by providing a relation on them

namely it'll be like if you have generators a, b, c, d, e

then you'll have something like abcd is trivial

(ofc this depends on what generators you pick in particular, but this is the idea)

[flint flicker]

Okay thank you very much very helpfull

[marsh forge]

ive just realized theres actually an alternate way to do the filled in square lol

take all the little line segments

we can slide their endpoints along any part of the square we want

[obtuse meteor]

ah yeah

[marsh forge]

so just slide them all to one corner

[obtuse meteor]

get it to be wedge of a disk with

[marsh forge]

and then contract the square

[obtuse meteor]

mhm

[marsh forge]

exercise

generalize this to the un filled in version

[obtuse meteor]

sliding argument go brrrrrrrrr

[pulsar lynx]

having trouble understanding condition (ii)

[gentle osprey]

slimvesus

[bright acorn]

I am pretty sure that most of the use of this construction comes from this property, and I was thinking if this is indeed actually a universal property of the space $M_{f}$ which defines it at least up to homotopy. Is that true?

[gentle osprey]

MisterSystem

[bright acorn]

I am trying to explicit construct a homotopy equivalence between two spaces that satisfy simultaneously this property.

But I don't know if even what I am trying to do is true.

[pulsar lynx]

I see. I’m confused bc I’m thinking “okay you have a point X on V1. Then you map that to V2 through phi. Then you put that point into f (where f is in the ideal of V2), but doesn’t that just give you 0?”

Even still, what’s the point of that condition

[tight agate]

@bright acorn the mapping cylinder does satisfy a universal property

it's the homotopy pushout of f and the identity map on X

just embedding + homotopy equivalent to Y doesn't characterize it

the map from X to the mapping cylinder is a cofibration, which is nicer than a general embedding

[bright acorn]

I still don't have a good notion of what a fibration or the dual notion of a cofibration are tho

are these notions in category theory?

[tight agate]

model categories, yes

but yeah, look up the homotopy extension property

[bright acorn]

Could you give a brief description of what it is tho? Does the name comes from like fiber bundles or something? At least historically.

[bright acorn]

I am still at the very beginning of algebraic topology

Soon I will take a look at it

It's just that the prof mentioned this property

[tight agate]

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

it isnt hard to read

[bright acorn]

And I thought if you could somehow make it into a universal property

because they are usually easier to work with

rather than explicit constructions

[bright acorn]

aight

I will take a look

thanks

[sleek thicket]

Yes

Fiber bundles have this thing called the homotopy lifting property

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

It turns out this property is a good analogue of being a fiber bundle for homotopy theory

Since being a fiber bundle is very much not homotopy invariant

I think every fibration is homotopy equivalent to a fiber bundle though

[bright acorn]

Hmm

I didn't know this question of mine would have such interesting connections to other things

That's kinda nice

I will take a look at these articles

But I am pretty sure Hatcher covers these, right?

On chapter 1

[marsh forge]

is this true?

[sleek thicket]

I remember looking into it and finding a paper

[marsh forge]

im not doubting it

[sleek thicket]

Let me search

[marsh forge]

im just surprised ive never heard it

[sleek thicket]

https://www.jstor.org/stable/2046409?seq=1

True if you restrict the base and total spaces to be cw complexes

[marsh forge]

as any reasonable person should anyway

nice!

[hazy siren]

ok this is probably a silly basic doubt but in the metric space of Real numbers with the discrete metric, how do i prove that the subset [0,1] is closed. Can i use the direct definition that every closed interval in R is a closed set?

[bright acorn]

Try to prove that the topology induced by the discrete metric is equivalent to the discrete topology

in particular

[hazy siren]

can i say that since every subset in discrete is closed and open, so hence [0,1] is closed?

[bright acorn]

every subspace of such a space would be closed

because its complement would trivially be open

[marsh forge]

if you can prove the trivial metric is homeomorphic

then yes

there is probably an easier way

or at least a more direct one

[hazy siren]

what is that?

[marsh forge]

What is your definition of closed

[bright acorn]

hmmm

Yeah

Prolly

this is the easier way I thought tho

but we should stick to the definition of closed sets he knows

[marsh forge]

the proof i have in mind is effectively the same thing without the word discrete or cross topology comparison

[hazy siren]

for me right now we prove closed only by proving that the complement is oopen

[bright acorn]

Nice

[marsh forge]

Okay lets start there

[bright acorn]

Try proving that every singleton set is open

[marsh forge]

what are the basic open sets of a metric space

[marsh forge]

please stop giving it away hahaha

[bright acorn]

oook I am sorry 😦

[marsh forge]

nbd

[hazy siren]

wait i thought singleton is closed?

[marsh forge]

It will help if you answer my question

[fading vale]

open and closed are not necessarily mutually exclusive properties

[marsh forge]

not in every space

but they happen to be here—its not relevant to us though

(there is no direct proof based on closedness of points bc [0,1] is the uncountable union of points)

anyway

what is the standard basis for a metric space?

what do open sets look like?

[hazy siren]

where the boundary points are not included?

im sorry my defs are not great

[marsh forge]

You should review this hopefully

your notes should have this

or textbook

i dont want to just tell you bc knowing this off the top of your head is essential

[hazy siren]

okay yes, i will review

[marsh forge]

you can just look up this specific defn for now

[fading vale]

hmm if the channel is open for now then

[marsh forge]

[bright acorn]

it's actually closed tho

or both

[fading vale]

let A = k[t1, ..., tn], k an infinite field, and let P a prime ideal and consider the variety V defined by P. atiyah macdonald says you can identify the localization of A at P, A_P, with the ring of rational functions defined on almost all points of V

so i understand that for any g notin P (so f/g in A_P) you have some x in V with g(x) non-zero

but the "almost all" part is pretty confusing to me

or maybe im just not totally clear on what that term means

usually its all but finitely many i think but cant V be finite?

[marsh forge]

Can you post the passage

[fading vale]

uh yes

[tight agate]

the set on which it is not defined has lower codim

[fading vale]

uh no dimension stuff has been introduced but that sounds right from the diff top ive seen

i guess

[tight agate]

almost all = dense open set here ig

[fading vale]

is there a reason that g is non-zero on almost all points of V?

[tight agate]

oh I think I see what they mean

all but finitely many points in Spec

not k-points on the variety

[fading vale]

wait sorry wdym

[tight agate]

an irreducible subvariety is a point in Spec K[...]/p

i.e a prime ideal in K[...]/p

[fading vale]

uh i think i kind of vaguely recall that

[tight agate]

eh this is probably not the right way to look at it

[fading vale]

something about a correspondance between points on the variety and maximal ideals in k[...]/I where I is generated by the variety

[tight agate]

almost all = dense open set

im pretty sure that's what they mean

[fading vale]

that makes sense i think

i guess i just dont see why f/g would be defined on almost all points in V?

[tight agate]

in this sense?

[fading vale]

yeah

[marsh forge]

g shouldnt share too many 0s with V i think

[tight agate]

show that the vanishing subscheme has lower dim

[fading vale]

um

[tight agate]

oh wait you didnt talk about dim

[fading vale]

this is from AM chapter 3

lol

[tight agate]

the complement of vanishing set is open

you are on a variety

any nonempty open set is dense

in the zariski top

[fading vale]

im not sure i really get what u mean by zariski top on the variety

is this using the nullstellensatz or whatever where

points correspond to maximal ideals in the quotient of k[...]

[tight agate]

what topology are you working with?

[fading vale]

honestly i dunno. i mean V is a subset of k[...] right?

wait

k^n

lol

[tight agate]

A-M does the zariski top pretty early on

[fading vale]

yeah

[tight agate]

the first or second chapter

[fading vale]

im just not super clear on the details of how that relates to varieties

[tight agate]

k[...]/I is a ring

the variety is spec k[...]/I

and the topology is the zariski top

[fading vale]

okay right

so we say that that complement of the points that g vanishes on is open and non-empty and thus dense in the zariski top?

[tight agate]

that is a true statement

im not sure if that's what they mean in the exercise

probably

[fading vale]

idk what else it could be

varieties were discussed for like two exercises in the very end of ch1

[tight agate]

it certainly isnt true cardinality-wise

[fading vale]

and the "varieties are secretly Max Spec(k[...]/I)" is like the only thing from that that wouldbe relevant here

so i guess?

yeah thats what tripped me up xd

[tight agate]

not max spec

spec

[fading vale]

uh

AM says Max i think

[tight agate]

they're working with the classical definition

[fading vale]

anyway thank you :^)

i think i see now

[tight agate]

weird exercise

[fading vale]

its not even xd

its an example

[tight agate]

oh

the main point is

"p" is the generic point of the variety

nbhds of p are dense open subsets

anything that happens at p holds almost everywhere

[fading vale]

interesting

AG seems really neat i wanna learn more

[tight agate]

this might make a bit more sense once you start working with the structure sheaf

[marsh forge]

it also probably will make more sense once it is formalized

in general if an example or expository statement is not like

obvious or even rigorous

its often best to keep it in mind and move toward the rigorous parts of the text

[fading vale]

i mean i guess but AM is so terse that like

[marsh forge]

and come back later to make sure you understand it

[fading vale]

if you move on from a few paragraphs you're missing a substantial chunk of the chapter

[marsh forge]

haha yeah learning what you can skip is a skill

[fading vale]

god the list of exercises is nearly as long as the actual exposition in this chapter

its less than 8 pages and it has 30

[hazy siren]

so, if the channel is open again

[marsh forge]

go for it

[fading vale]

yeah i'm done :^) sorry for hijacking it while you were rereading haha

[hazy siren]

can i show that the subset [0,1] in discrete metric in R is closed by saying its complement is open using the neighbourhood ball definition?

[hazy siren]

np

[marsh forge]

Yes

That is what I had in mind

(you need to use unions of open balls but yes)

[hazy siren]

yes makes sense, thanks

and i can generalise my radius of the ball to be half of the distance between the point in the complement and the set A?

will that be correct?

[marsh forge]

you can choose any distance you want less than 1

[hazy siren]

oh yes because its discrete, of course

[marsh forge]

B(x,\epsilon)={x}

if \epsilon < 1

[hazy siren]

cool thank you again

[marsh forge]

np hopefully my refusal to just tell you the stuff wasn't too annoying

[hazy siren]

no it helped, i really should know my basics clearly

[pulsar lynx]

Well when you do f(phi(X)), for any X in V1, isn’t it always equal to 0? Why would he say that it’s now in I(V1)

[bright acorn]

So I was thinking here

This is sort of a conceptual soft question

But like

When you have an oriantable manifold

You can defined integration of differential k-forms over it

How is this definition of integration

Related to other definitions of integration in measure theory?

[sleek thicket]

Definition of differentiation or of integration?

[bright acorn]

Integration

If you have a smooth oriantable manifold

It makes sense to talk about integrating differential forms

And I was thinking

[sleek thicket]

Sorry, I just wanted to make sure

[bright acorn]

If it is also related to measure theory

Oh

Silly mistake right there

I am sorry

[sleek thicket]

I think if you fix a volume form on the manifold then you should be able to define a measure from it, but I've never actually worked out the details

By integrating characteristic functions, I mean

hello

[bright acorn]

Hi TTerra

[gritty widget]

seconding what shamrock is saying

[sleek thicket]

One specific connection which might be of interest is the Haar measure on a compact lie group

Vs the Haar form

[bright acorn]

Hmm

[bright acorn]

That might actually be a neat exercise

[sleek thicket]

in both cases there's a unique-up-to-scaling measure/form whcih is translation invariant

And the definitions of integration you get are the same

Oh nice! That makes sense

[bright acorn]

Haven't formally studied functional analysis

So I don't really know the details of this theorem

But I might look up into it

[sleek thicket]

Given a locally compact hausdorff space, you have an associated set of continuous compactly supported functions

Right?

Real valued

[bright acorn]

Yup

[sleek thicket]

This has the structure of a real vector space, and there's a norm on it given by ||f|| = sup_x |f(x)|

have I screwed up yet ultra?

ah whoops thanks

[bright acorn]

Oh, so the usual operator norm

[sleek thicket]

Hm, but I do not want to define that

[sleek thicket]

No, that's something different

I guess it's not so bad

I've heard this called the sup or uniform norm

Also sometimes the infinity norm

Nah, I said locally compact

I think you're right

Wait

Isn't functions vanishing at infinity the completion of functions of compact support?

I'm checking folland now to make sure I don't say the wrong thing

because in that case, shouldn't the duals should be the same?

Okay I have gotten off track

Let's just do the compact case

So I don't get things wrong

MS, say you have a compact hausdorff space X

In particular something like a compact manifold

Then C(X) is a metric space

[bright acorn]

So you want to fix a compact Hausdorff space and look at the subspace of C(X,[0,1]) given by of all functions with compact support.

[sleek thicket]

Not [0,1], all of R

[bright acorn]

Ok

[sleek thicket]

And since I'm working with a compact space all functions are compactly supported

I may or may not have screwed up when trying to state it more generally and I want to not lie

So focus on X compact

[pulsar lynx]

Yeah no I get what he’s trying to say, I just don’t understand how it works out. f o phi (x) is always 0, is what I’m thinking, so why it be in I(V1)

[bright acorn]

Oh

That's why it's easier to state

[sleek thicket]

Yup haha

[bright acorn]

Makes sense, since the image of a compact set by a continuous function is always compact

What would be the problem if X was locally compact tho?

[sleek thicket]

Not quite. The support is a closed subset of the domain, so it's automatically compact if the domain is compact

Part of the problem is that we're going to talk about integrating functions

And you can't integrate arbitrary functions against a measure, you need them to be nice in some way

Continuous on a compact space gives you that they're bounded and borel measurable

Okay so

Okay so if you have a (borel) measure μ on X, you get a "functional" I : C(X) -> R given by integrating a function against μ

So $I(f) = \int_X f , \mathrm{d} \mu$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

What makes this a "functional" is that's it's linear and continuous (remember, C(X) is a metric space)

Does that make sense?

@bright acorn did my statement about getting a functional associated to a measure make sense?

integrating functions is linear and respect uniform convergence

Ah wait I did screw up, I want this measure μ to be finite. Otherwise eg you might not be able to integrate

[bright acorn]

Yeah

[bright acorn]

Yup

[sleek thicket]

So this defines a map from finite borel measures into the dual space C(X)^* of C(X), ie the space of all continuous linear maps C(X) -> R

yeah?

[bright acorn]

Let me write this map down

[sleek thicket]

oh goddammit I screwed up again. Technical topological difficulty: restrict to the case where X is second countable. This still loses no generality in the case of manifolds

[bright acorn]

So, suppose that $\Omega$ is the set of all finite borel measures induced by a borel sigma algebra on X.
\
\
Then $F : \Omega \rightarrow C(x)^{\ast}$ where $\mu \mapsto I : C(x) \rightarrow \mathbb{R}$ and $I(f) = \int_{X} f d\mu$, for every $f \in C(X)$.

[sleek thicket]

Well, what I wanted originally was just manifolds :P

[bright acorn]

One sec

So \Mu is not a thing apparently

[sleek thicket]

Lmao

[bright acorn]

:/

[sleek thicket]

Ah true. But (I am actually checking folland now) it would've worked if I'd done compactly supported functions to start!

[gentle osprey]

MisterSystem

[sleek thicket]

It's okay haha

[bright acorn]

This would be the map

You were talking to, right?

[sleek thicket]

Integrating differential forms is nicer in the compact case anyway

Yup, this is the map

[bright acorn]

So like

F(mu)(f) = integral of f over X dμ

[sleek thicket]

Yup

[bright acorn]

Nice nice

[sleek thicket]

So this map is actually injective

[bright acorn]

Mind if I interrupt?

[sleek thicket]

Our weird topological condition implies that if two finite borel measures agree on all open sets they are equal (i will not prove this)

Yeah sure

[bright acorn]

What book is this one by folland you are talking about?

[sleek thicket]

Real Analysis
Modern Techniques and Their Applications

I like it

[bright acorn]

Reading the preface

It seems like a nice book on measure theory and functional analysis

[sleek thicket]

It's where I learned both of those from

Okay so this map is injective, the idea is that it suffices to check that the measures agree on open sets, and then for a fixed open set you can approximate inside by compact sets and use urysohn's lemma

To sort of approximate the characteristic function of U

[gritty widget]

i'm going to do the opposite and say that if you wish to read a book on measure theory you should not look at the sixth chapter of pugh

[sleek thicket]

This is where the topological nicety condition ("second countable") comes in

Anyways don't worry, about it, F is injective

[bright acorn]

I've read a bit of Bogachev's first book on the subject

[sleek thicket]

So what's the image of this function F?

[bright acorn]

Oh

Let me think about this for a second

[sleek thicket]

Yup

[bright acorn]

F maps finite borel measures into a linear funtional of the form $\int_{X} \cdot , d\mu$

So ok

Hmmm

[sleek thicket]

Yup

First question: can F be surjective?

[gentle osprey]

MisterSystem

[bright acorn]

This is like

Asking if for every linear funtional from C(X) to IR

There exists a certain finite measure mu

[sleek thicket]

Yup

[bright acorn]

Such that this linear functional is nothing more than integrating a continuous function over mu

I have no idea seriously

[sleek thicket]

Hahaha

[bright acorn]

And finding a counterexample seems hard

[sleek thicket]

Well the riesz representation theorem says, pretty much, yeah!

[bright acorn]

Wtf

Ok you got me there

[sleek thicket]

However there's an easy way to see this isn't exactly true

In the way I've set this up

If I = F(μ) then I(f) >= 0 for any function f such that f >= 0 everywhere

Yeah?

[bright acorn]

Yup

[sleek thicket]

Well, what if I consider -I(μ)

Where μ is just some random measure

Does this have the same property?

||but I'm working with real vector spaces ||

||and yeah i just didn't want to have to define signed measures||

||I am lazy||

[bright acorn]

-F(μ), right?

Nah

[sleek thicket]

Oh yeah lol

[bright acorn]

It couldn't be roughly because of what you said

[sleek thicket]

Sorry -F(μ)

[bright acorn]

Take any non negative valued funtion

Then

In order for there to exist a certain ν such that -F(μ) would be integral over X in dν

We would have that for for a non negative valued function

Integral over X f dν would be positive

But since (-F(μ))(f)=-(F(μ)(f)) and F(μ) is always positive

Then (-F(μ))(f) would be negative

[sleek thicket]

Yup, that's the idea!

[bright acorn]

Which would be a contradiction

[sleek thicket]

Minor nitpick, we need to start with a measure μ which isn't just 0

And we need to take f to be like constant so that the integral is actually positive

But basically yeah

[bright acorn]

Oh yeah

[sleek thicket]

But yeah, you see why F isn't exactly surjective

[bright acorn]

Hm hm

[sleek thicket]

So there's two ways we can fix this

The first is to consider only positive functionals

(continuous also)

[sleek thicket]

Positive is this condition on I

In the way we've set things up, the image of F is exactly the positive functionals

Does that make sense?

[bright acorn]

Yup

I am still keeping up I guess haha

Most of the details I will work out by myself tomorrow tho

But I think that I am getting the main ideas

Ok, so we consider a compact second countable hausdorff space X and the space of all real valued functions with compact support (which are actually all of them)

And we construct a certain map

Which brings you from the finite borel measures induced by a certain borel sigma algebra on X

Into linear functions of the space of continuous real valued functions defined on X

This linear funtional is actually injective

And to force this thing to be surjective we have to take a little bit more of care of

[sleek thicket]

Sorry I had irl stuff

But I am back

[bright acorn]

Np

[sleek thicket]

Yes, and the way you've said it here you don't actually need X compact, just locally compact hausdorff

Which is nice for when we return to manifolds

Ultra was concerned earlier but both them and I agree it works for X just locally compact now

[bright acorn]

Oh

Ok then

[sleek thicket]

You do need compactly supported functions in this case though

(and still second countable)

I agree with your summary so far

[bright acorn]

So we are working with functions of compact support defined on a locally second countable hausdorff space

This seems almost like a manifold lmao

[sleek thicket]

Locally compact

yeah haha

There's a more general statement

I'm just trying to simplify things

I want to avoid defining radon measures

And second countable does that

Okay so the image of this map F is exactly the positive continuous functional

🤯

and, even better, any (continuous) functional can be written as the difference of two positive ones

[bright acorn]

I think that I have seen this covered in real analysis actually tho

You just take like

[sleek thicket]

ah okay, nice!

[bright acorn]

If f is a continuous linear functional

Take g=max{f,0} and h=max{0,-f} which are both positive

Proving continuity of these is trickier I guess

[sleek thicket]

Those won't be linear

[bright acorn]

Oh

[sleek thicket]

I think you're thinking of eg functions X -> R, not functionals C(X) -> R

(apologies if not)

[bright acorn]

True

So yeah

You are right

[sleek thicket]

but yeah, the other solution is to generalize our notion of "measure"

which is what ultra was scolding me for not doing earlier

if you allow your measure to take on negative values then you can eg add or subtract or scale measures by real numbers, and M(X) = {finite borel signed measure on X} forms a vector space

and in fact it has a norm $|\mu| = |\mu|(X)$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

don't worry about what |mu| is if you haven't seen it

but then the theorem gets even better, not only if F injective and surjective it's also an isometry!

[bright acorn]

|μ| would be like

Ok I have no idea

[sleek thicket]

hahaha

[bright acorn]

But like

[sleek thicket]

look at a maximal subset P of X on which mu(E) >= 0 for each subset E of P

[bright acorn]

A measure is a function so like

[sleek thicket]

do the same to get a subset N of X but for negative

[bright acorn]

Can't you just apply the sup norm here?

[sleek thicket]

$|\mu|(E) = \mu(E \cap P) - \mu(E \cap N)$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

nah, that's not quite right

you want to account for both the positive and negative parts together

[bright acorn]

I see

[sleek thicket]

anyways

if we return to manifolds

say you have an smooth oriented $n$-manifold $M$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

and a top dimensional differential form $\omega \in \Omega^n(M)$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

you then get a functional on continuous compactly supported functions on $M$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

namely $f \mapsto \int_M f \omega$

[gentle osprey]

Shomrack (not a furry)

[sleek thicket]

and then this must be associated to some signed measure(!)

(equivalently, there are two positive measures where this is the difference of their integrals)

[bright acorn]

Omg

[sleek thicket]

so we get a linear map from n-forms into signed measures

🤯

[bright acorn]

Ok this is nice

The main thing we used here

[sleek thicket]

yeah haha I had never thought about using riesz rep here

I just sort of figured it worked out lmao

ty ultra for pointing me in this direction

oh also this map should be injective but don't quote me on that

yeah I think injectivity follows because knowing the integral of omega against any function lets you take the average value of each coordinate of omega at a given point once inside some fixed chart

and the limit of average values of omega will equal the value of omega by continuity

[bright acorn]

Was the riez rep theorem, right? From which we deduced that a certain map from the space of finite borel sigma algebras to the space of linear functionals on C(X) with compact support is surjective.

[sleek thicket]

Yup!

That's exactly it

[bright acorn]

So I will take a look into it

[sleek thicket]

yeah, it's a cool theorem

[bright acorn]

Looks a very nice application of something I haven't even properly studied yet lmao

But when I study functional analysis more seriously

I will have this example in mind

And fill out the details

Proving that map F is injective tho follows from what?

[sleek thicket]

this is where the second countability comes in

the right way to do this, in more generality, is to let you work with measures where you can approximate outside by open sets/inside by compact sets

and so if two measures have the same integral against every function, use something like urysohn on increasingly large compact sets to say they have the same value on any open set

it's a little hairy

[bright acorn]

Nice

Thanks for having the time to explain it

Really appreciate

[sleek thicket]

np!

I am procrastinating on french homework due in 2.5 hours

[bright acorn]

lmao

procrastinating on math tho...

is that a good thing?

[sleek thicket]

i procrastinate via math

less so on math

[gritty widget]

@sleek thicket i think that it would be a good idea to do your french homework

[sleek thicket]

[gritty widget]

look

i got a fair amount of my analysis hw done today

u can do french

[sleek thicket]

i hope you enjoyed my integration thing

I will now go do french

,iam studying but it locks me out of this channel too

[gentle osprey]

No selfroles matching studying but it locks me out of this channel too.
See ,selfroles --list for the list of valid selfroles.

Apr 10, 2021, 4:41 AM

[bright acorn]

Good luck

Also

oh nvm 😳

TTerra

If that map F is actually injective

I was thinking of the converse statement

Let mu be a finite borel measure on an orientable n dimensional manifold

then there exists a differential form

[sleek thicket]

it will not be integration against a differential form, consider evaluation at a fixed point

okay goodbye for real sorry I am cringe

[bright acorn]

SORRY

[sleek thicket]

Although

hmmm

No

I cannot think about this

Sorry

Oh my fucking god I have to write about how my experience as a student changed during the covid 19 pandemic

I want to not do this

«C'est mal» fin

[gritty widget]

why would you make a student write about that

ew

[sleek thicket]

I don't like it

[gritty widget]

that's uhh

look i know language courses are usually starved of actual rich topics to write about

but could they have at least picked something that most people definitely dont wanna think about

[sleek thicket]

The present tense/during corona part turned out to be really easy to write

I just said everything was shit

oh maybe I should include the word merde somewhere

I talked about how I teach but I don't see students during class

[long hornet]

R^2 is not homeomorphic to R^3 because, by deleting a point in the former, we get a space that is not simply connected. Can we generalize this to prove that R^n isn't homeomorphic to R^(n + 1)? I think by removing a certain copy of R^(n - 1) of R^n, we get a space that is not simply connected.

(I don't know any algebraic topology)

[sweet wing]

yes (look at π_n / H_n (nth homotopy/homology group)

or jus like

space not contractible

hence done

simply connected isn't sufficient

(show that R^n-* is not contractible, not super trivial but doable)

oh lol

right oops

🤦‍♀️

iirc theres some way to define dimension that may work in munkres but i havent rlly read it in too much detal as well

[long hornet]

I can imagine R^3 - * and R^4 - * aren't easy to distinguish using elementary methods.

But the thing is that R^3 - line seems easier, to me. For example, a circle around the (now nonexistent) line isn't contractible, right?

[sweet wing]

any circles in R^3-* is contractible unfortunately

[long hornet]

But its image under a homeomorphism should be

[long hornet]

Yes, that's why I considered R^3 - line.

[sleek thicket]

It's true that R^n \ k plane is homotopy equivalent to R^(n-k) \ a point

By projecting down

[sweet wing]

there's this purely nonalgtop way to do in munkres

[sleek thicket]

But I'm not sure this is true for arbitrary subsets homeomorphic to a k plane

And so the homeomorphism R^n ≈ R^m might distort your line in a weird way

[long hornet]

Is it easy to show that the dimension is indeed an invariant?

[sweet wing]

by definition it kinda is?

[long hornet]

Is it easy to compute the dimension of R^n, then?

Yeah

[sleek thicket]

I think the answers are yes and no

Respectively

[long hornet]

Oh, true..

[long hornet]

Makes sense, because otherwise any reference to the invariance of dimension would be quoting it.

[sweet wing]

isnt trivial but doable without AT

[long hornet]

I see

[sleek thicket]

Yeah just to reiterate what ari said earlier

[sweet wing]

in some sense elementary in regards to the prerequesite

but nonelementary in like proofsense

[sleek thicket]

This same argument works but replacing "connected" with "nonvanishing nth homology group"

(homotopy also works but is more involved)

[sweet wing]

yea

[long hornet]

I see

Well, is there a reason to expect the R^n =/= R^(n + 1) case to be easier?

[sleek thicket]

Than n, m ?

Easier than what?

[long hornet]

Yes.

[sleek thicket]

I don't see why, but maybe

[long hornet]

My only argument was the R^n - plane thing, but as you said it doesn't work..

[sleek thicket]

I wonder if there's an algebraic proof

Looking at the algebras of real valued functions

Probably hard

[sweet wing]

probably exists but quite tricky

[long hornet]

What are some potential invariants one could use?

[sleek thicket]

¯\_(ツ)_/¯

I mean my first thought is the coordinates

But they don't really generate it in any meaningful way

[sweet wing]

or just cry and this

[long hornet]

lol

Which proof do you think is more satisfactory?

[sweet wing]

AT

[long hornet]

The one based on algebraic topology or dimension theory?

[sweet wing]

You can actually use it elsewhere

like you can easily throw AT onto many sufficiently nice spaces

[sleek thicket]

I looked it up and it seems like the most straightforward proof that dim R^n = n uses brouwers fixed point theorem

[sweet wing]

hmm interesting

[sleek thicket]

At which point you might as well just use the homology proof

[long hornet]

True lol

How about other definitions of dimension?

[sleek thicket]

There's the algebraic geometry one ("combinatorial dimension")

This is pretty neat here in that dim R^n = 0 for all n

Not particularly helpful

[sweet wing]

lmao ikr

theres the hausdorff dimension as well

but it's pretty disgusting

[long hornet]

Hausdorff dimension, for example

[sleek thicket]

Yes, I agree

I'm not suggesting you can get it for free

Just wondering if there's a nice alternate proof

[sweet wing]

but idt it's a topological invariant

[sleek thicket]

This is very neat

[long hornet]

Oh..

[sweet wing]

looking at it's definition itself convinces me it isnt as well

[sleek thicket]

oh

We saw this the other night

Hahaha

For those who missed this discussion https://en.m.wikipedia.org/wiki/Osgood_curve

[sweet wing]

urgh

cursed

[long hornet]

So, it's homeomorphic to the circle?

[sleek thicket]

To the interval

I mean you can get a circle version too

Nope

Maybe she saw my course eval from last quarter

I said harsh words

It was a bad class

(this is why I'm not actually taking GMT)

[long hornet]

That seems interesting, could you reply to a message of the discussion?

[sleek thicket]

If you search Osgood curve you should find it in #advanced-analysis

Essentially we were curious how regular an Osgood curve could be

It can't be rectifiable

But what about differentiable ae?

That is, a continuous injection I -> R^2 whose image has positive area

[long hornet]

Seems interesting. Although I don't have tools to attack it at moment, I would think of the cases "differentiable at a cofinite/cocountable set" first.

[sleek thicket]

I'm actually not sure how to handle the differentiable everywhere case

If you're not C^1 then the curve might not be rectifiable

[long hornet]

C1 means f' is continuous?

[sleek thicket]

Yup

[long hornet]

Weird!

But yeah I see why

[sleek thicket]

If the derivative is bounded then you're rectifiable

[long hornet]

Not only that, but my length is an effectively computable constant.

[sleek thicket]

What do you mean?

[long hornet]

It's a joke lol

[long hornet]

I see..

[lyric wadi]

So, in "the wild world of 4-manifolds" chapter 3, Scorpan claims that any 2-homology class of a simply-connected 4-manifold may be represented by images of maps f:S^2->M thanks to Hurewicz's theorem. He then claims that these images may be perturbed to be immersed spheres and that any double points which obstruct embedding may be removed by increasing the genus before going on to list a few examples.
So, my questions are, "why can these spheres always be perturbed to be immersed? Is it the consequence of the whitney immersion theorem?
and "why can these double-points be eliminated in general? Is it just the whitney trick?"

[sweet wing]

sounds a lot like whitney trick

[lyric wadi]

yeah, my problem with it is that I thought that the whitney trick failed (aside from freedman's work on casson handles) for surfaces in 4-manifolds

but that might well be a misunderstanding

[pulsar lynx]

Ohhhh I get it now haha, thank you lol

[lyric wadi]

Hmm, unless it is just removing the points and keeping the remaining map which is now an embedding

And having that submanifold represent the homology class (though I still don't really like that tbh since it loses the directness in my understanding)

[obtuse meteor]

homework this week in manifolds be like "generalize the cross product to R^d in order to show (d-1)-submanifolds have smooth unit normal vector fields"

And I'm like "Professor Hani i do not know what the cross product in R^3 is"

[tight agate]

If you know how the exterior product works

and the hodge star operator

then cross product is exterior composed with hodge

[obtuse meteor]

🧠

[cedar pebble]

@obtuse meteor re: your twitter question

cohomology theories are represented by spectra. If you have the dimension axiom, this corresponds to ordinary cohomology theories, and all of these are represented by Eilenberg-MacLane spectra HA for some Abelian group A

but these spectra are particularly simple: they correspond to the embedding of the category of Abelian groups into the category of spectra

in some sense this is no more interesting than the category of Abelian groups (although to be fair there are plenty of interesting things you can ask about Eilenberg-MacLane spectra and the Eilenberg-MacLane spaces that constitute them)

[obtuse meteor]

I just looked at this lel

[cedar pebble]

aa

[obtuse meteor]

literally have https://en.wikipedia.org/wiki/Brown's_representability_theorem open

give me a sec

[cedar pebble]

yea!

[obtuse meteor]

sure okay this makes sense

[tight agate]

the proof of brown rep is pretty neat

[obtuse meteor]

I have never like seen Eilenberg Maclane spectra be constructed or the isomorphism with homology/cohomology

but this seems like a really cool thing

[tight agate]

it is

[obtuse meteor]

it also feels unintuitive that like

[cedar pebble]

well so have you seen Eilenberg-MacLane spaces before?

[obtuse meteor]

all cohomology theories are representable in this way

nop

I know what they do

never seen a construction

or seen them be used

[cedar pebble]

yea the construction is a little annoying

[obtuse meteor]

just know they're CW complexes whose homotopy groups are zero except for in one degree

[cedar pebble]

right exactly

I guess the way you can construct these is like

[obtuse meteor]

and you can specify any group you want in that degree

[cedar pebble]

keep gluing in cells so that it kills the lower homotopy groups

[obtuse meteor]

as long as abelian

ofc

in degree > 1

[cedar pebble]

it's generally not something that's nice to write down obviously

[obtuse meteor]

yeah

[cedar pebble]

but it's possible in principle

the point is that like

[obtuse meteor]

I guess like the harder question to me is

how do you kill the higher homotopy groups

[cedar pebble]

very carefully

[obtuse meteor]

lmao

bc the lower homotopy groups you can kinda just add cells above that degree

and they won't effect the lower ones

and so you can just use them to kill shit

inductively

[cedar pebble]

yea there's a few ways to construct them

[obtuse meteor]

but as for higher degrees

[cedar pebble]

one way is to represent them as simplicial Abelian groups and then take geometric realization

another way is like

[obtuse meteor]

holy fuck what

pi_{n + k}(sigma^n X) stabilizes????

wh-what

[cedar pebble]

oh yee

[obtuse meteor]

this feels like a miracle???

[cedar pebble]

yea it's wild

this is basically the starting point of stable homotopy theory

[obtuse meteor]

like I haven't worked enough with reduced suspension

to have any intuition for whether or not this is true

but

it feels like

it shouldn't be

[cedar pebble]

oh yea it's really bizarre that this happens I agree

another way you can construct K(A,n) is using something called the Moore space M(A,n)

since these have π_n(M(A,n))=A and all the lower homotopy groups vanish

but it has higher homotopy groups

so you just have to kill these

[obtuse meteor]

mhm

[cedar pebble]

anyways

[obtuse meteor]

this shit wild

[cedar pebble]

the thing that the Eilenberg-MacLane spaces have is they behave nicely under suspension and delooping

namely like

oh god I'm going to fuck it up hold on

it's either like

suspension takes K(A,n) to K(A,n+1)

[obtuse meteor]

nG I've got the career trajectory planned out

take Igor Kriz's AT class => meet Igor Kriz / Sophie Kriz and make good impressions => Get a letter from Igor Kriz and get into Uchicago bc JP May => work with Sophie Kriz for the rest of my life because she's 1000% smarter than me

😌

/s

[obtuse meteor]

huh

[cedar pebble]

ah okay I didn't fuck this up

okay so suspension takes K(A,n) to K(A,n+1)

and conversely

loops takes K(A,n) to K(A,n-1)

so the Eilenberg-MacLane spaces K(A,n) form a so called sequential spectrum: this is an infinite sequence of spaces related by suspension/loops in an invertible way like this

[obtuse meteor]

mhm

ok I need to take a bath

but like

[cedar pebble]

[obtuse meteor]

wanna keep talking about this at some point so put a pin in it <3

gah

[cedar pebble]

yea ping me if you wanna keep chatting about this I love this stuff

[obtuse meteor]

I know I shouldn't take the next algtop course next semester

but like

[cedar pebble]

don't let me hold you up lol

[obtuse meteor]

I really want to

[cedar pebble]

nah take more

[obtuse meteor]

tru

[cedar pebble]

also damn you're a bath chad

not a shower virgin

[summer jolt]

When people write a Riemannian metric like this

Is it usually understood that tensor product is symmetric?

i.e in this we would have $g = g_{ji} dx ^j \otimes dx^i$?

[gentle osprey]

snypehype

[wanton marsh]

it's some einstein notation

it means $g = \sum_{i=1}^n \sum_{j=1}^n g_{ij} dx^i \otimes dx^j$

[gentle osprey]

Zef Klop 🍃 🌿 🌻

[wanton marsh]

tensor product isn't symmetric commutative

[summer jolt]

I understand that. What I mean was that on a Riemmannian manifold if g is symmetric then $g = g_{ji} dx ^j \otimes dx^i = g_{ij} dx ^i \otimes dx^j$?

[gentle osprey]

snypehype

[summer jolt]

The reason what I'm asking was because I saw the following calculation

This can only be true if you treat the mixed tensor product terms to be the same i.e. $d\theta \otimes d\phi = d\phi \otimes d\theta$

[gentle osprey]

snypehype

[summer jolt]

I'm aware that tensor product is not in general symmetric hence my question.

[wanton marsh]

with einstein notation, $g = g_{ji} dx ^j \otimes dx^i = g_{ij} dx ^i \otimes dx^j$ is just swapping the names of the two indices and is always true. The tensor is symmetric when forall i and j, $g_{ij} = g_{ji}$, and that's not always true.

[gentle osprey]

Zef Klop 🍃 🌿 🌻

[wanton marsh]

if you do the computation, the coefficients of both $d\theta \otimes d\phi$ and $d\phi \otimes d\theta$ vanish.

[gentle osprey]

Zef Klop 🍃 🌿 🌻

[summer jolt]

@wanton marsh I see thanks. So in general if I have I have let's say $2 dx \otimes dy$ and $2 dy \otimes dx$ and I'm working with a symmetric tensor, am I correct in assuming that these two terms are the same?

[gentle osprey]

snypehype

[wanton marsh]

no

[summer jolt]

I see. Now I'm a bit confused by this.

This is from my lecture notes, where $dx^i dx^j$ is the symmetric tensor product and I was under the impression that the notation above meant the same thing

[gentle osprey]

snypehype

[shut moat]

$dx^i dx^j = \frac{1}{2}(dx^i \otimes dx^j + dx^j \otimes dx^i)$

[gentle osprey]

Buncho Spheres

[shut moat]

to see why they're the same for $g$, $g_{ij}dx^i dx^j = \frac{1}{2}\qty(g_{ij}dx^i\otimes dx^j + g_{ij}dx^j \otimes dx^i)$ and by the symmetry in the indices, $g_{ij}dx^j \otimes dx^i = g_{ji}dx^j\otimes dx^i = g_{ij}dx^i \otimes dx^j$, so $$g_{ij}dx^idx^j = \frac{1}{2}\qty(g_{ij}dx^i\otimes dx^j + g_{ij}dx^i\otimes dx^j) = g_{ij}dx^i\otimes dx^j$$

god i hate latex

[gentle osprey]

Buncho Spheres

[summer jolt]

@shut moat thanks helped a lot! Ok so we have $dx^i dx^j \neq dx^i \otimes dx^j$ but we have $g{ij}dx^idx^j = g_{ij}dx^i\otimes dx^j$

[gentle osprey]

snypehype

[shut moat]

np! 😄 and yeah

[summer jolt]

I wish people stuck to the same notation!

[wanton marsh]

what does (8.16) say ?

[summer jolt]

It's what @shut moat said #point-set-topology message

I was under the impression that tensor product symbol in this is case meant something different (symmetric tensor product)

but it turns out you were right, it's just the ordinary tensor product

[fierce rock]

Hey! Does anyone have formula sheet for shapes in 3d (analytic geometry)

[gritty widget]

[sweet wing]

honestly #geometry-and-trigonometry

[thin bramble]

Can anyone help me?

[obtuse meteor]

this should probably be in #geometry-and-trigonometry

[thin bramble]

Are you sure?

The sigma is halfturn and rho is rotation

[tidal cedar]

Yes this should be in #geometry-and-trigonometry

[viral atlas]

I'm kinda confused if it should be in #geometry-and-trigonometry

I don't think this is a part of standard school geometry

[thin bramble]

is an upper 300 level course.

[tidal cedar]

This is for geometry in an AG, manifolds, etc. sense

Oh

[thin bramble]

yea

[viral atlas]

Yeah, this seems to be a classical geometry course for undergrads

[tidal cedar]

Oh ok

[obtuse meteor]

algtop seems

very spicy

this week

http://www.math.lsa.umich.edu/~jchw/2021Math592Material/Homework12-Math592-W2021.pdf

[thin bramble]

you have a weeek to complete that homework? damn, i'm scared of grad school 😦

[obtuse meteor]

let's just sum this up shall we

1. Mayer-Vietoris Sigma_g inductively
2. Euler Characteristic up to Euler Characteristic of a cover
3. Moore Spaces
4. Classification of Surfaces stuff (but it sounds like it's a "draw and explain why this works" problem)
5. Define Homology with Coefficients and Use Universal Coefficients without proof
   6 (bonus): Fundamental classes for orientable manifolds from Hatcher

That bonus looks hyper spicy

so I might just read it

and not do it

[thin bramble]

oh

[cedar pebble]

oh that's spicy

[obtuse meteor]

https://estrogen.fun/i/g2sa.png

meanwhile

kinda funny nG how we were just talking about moore spaces

[cedar pebble]

assignment: health comes first uwu
also the assignment:

and yea that is kinda funny

[obtuse meteor]

t-that can't be that bad

@thin bramble I'm not a grad student but yeah this is a grad qualifying exam course

[cedar pebble]

can you make goals SMART?

I can't even make goals

[obtuse meteor]

they are known to be scary

[obtuse meteor]

the fact that you pointed out this one--which I expect to be the easiest one--makes me very scared

[thin bramble]

wow that is nice. I never took a math grad course as an undergrad.

[cedar pebble]

oh it's easy it's just like

tedious

[obtuse meteor]

oh

IDK how to draw that that's gonna be a pain

like wtf does that look like on a torus lol

[gritty widget]

tex it

[obtuse meteor]

no?

[cedar pebble]

UGHHHH

[obtuse meteor]

hm?

[cedar pebble]

speaking of tedious exercises

this is for my quivers course

this is Homework 13

[obtuse meteor]

but yeah the qual courses are burtal @thin bramble, so it has become tradition for me to complain about how long the algtop psets are here every sunday

quivers?

[cedar pebble]

man quivers are so cool

mind if I chat spam

[obtuse meteor]

go for it

[thin bramble]

not unless someone can help, then you can spam chat

[obtuse meteor]

idk anyone here who knows classical geometry unfortunately :(

[cedar pebble]

a quiver is a directed graph where multiple edges/loops are allowed

[obtuse meteor]

ah

pre-category

I have heard of these

[cedar pebble]

yes!

[obtuse meteor]

was wondering if it was like

not that use of quiver

[cedar pebble]

no no it's precisely that kind

given a quiver Q you can consider K-linear representations of that quiver: you assign to each vertex a K-vector space and to each directed edge a K-linear map between the corresponding vector spaces

morphisms of quiver representations are what you suspect

[obtuse meteor]

so I guess I should think of a quiver as a presentation for a category

and a K-linear representation of a quiver

is an equivalence between the category the quiver presents

and a subcategory of K-Vect

[cedar pebble]

you can complete such a quiver to a category

and then the representations are like

functors from that category to Vect_K

[obtuse meteor]

yeah

that makes sense

they have to satisfy some injectivity condition though right?

actually no

this isn't what group reps do

like 0-rep is fine

[cedar pebble]

not necessarily! What you're talking about corresponds to faithful representations

[obtuse meteor]

ok cool

[cedar pebble]

anyways

[obtuse meteor]

yeah yeah makes sense

[cedar pebble]

another way to think about this: given a quiver Q with finitely many vertices and edges, you can define an associative K-algebra KQ

[obtuse meteor]

and then morphisms of representations are functors in a slice category

[cedar pebble]

the underlying vector space is spanned by paths in Q

[obtuse meteor]

[cedar pebble]

and the multiplication is given by like

[obtuse meteor]

mmm

I don't know algebras

[cedar pebble]

either composition of paths if they compose nicely

or 0 if they don't

K-linear representations of Q are equivalently KQ-modules

[obtuse meteor]

hmmm

ok I kinda see this

you don't have composition in a quiver though

so I'm a bit confused

example?

[cedar pebble]

err sorry

KQ is spanned by paths in Q

so you have composition of those

[obtuse meteor]

ah ok I see

yes

[cedar pebble]

okay

[obtuse meteor]

and an algebra is just vector space + multiplication

where the obvious axioms work

[cedar pebble]

mhm!

[obtuse meteor]

which they do here

bc

of course they do

[cedar pebble]

right and for instance

you need finiteness because

you can take the trivial path e_v at each vertex v of Q

and the unit of this algebra is \sum_v e_v

which only makes sense when this is vertex-finite

[obtuse meteor]

ye

[cedar pebble]

okee so

[obtuse meteor]

and presumably something similar for edges?

[cedar pebble]

mhm

you can do something a little more general: you can take a two-sided ideal I of KQ and consider the quotient KQ/I

on the representation side this is like

if you think of I as being generated by some paths

you demand that the composition of the K-linear maps in the representation compose to 0 along those paths

so for example if you have a quiver Q like 1->2->3 with edges a:1->2 and b:2->3

a K-linear representation of Q (equivalently a KQ-module) is just K-vector spaces V_1, V_2, V_3 with K-linear maps V_1->V_2 and V_2->V_3

if you take the ideal I=(ab)

a KQ/I-module is the same thing but you demand that the composition V_1->V_2->V_3 is zero

good so far?

[obtuse meteor]

mmmm

probably need to do homework instead of this right now

but I would like another rant about it!

[cedar pebble]

okay let me say the punchline and then you can leave

without getting into the technical details and which adjectives you have to add

the main theorem is essentially the following:

given any associative K-algebra A

you can produce a quiver Q and an (admissible) ideal I of KQ

such that A is Morita equivalent to KQ/I: that is the category Mod_A of A-modules is equivalent to the category Mod_KQ/I of KQ/I-modules

that is to say, the representation theory of quivers completely captures the representation theory of associative K-algebras

this is great because quiver representation theory is extremely explicit

for example: you can enumerate all the simple KQ/I-modules and indecomposible projective KQ/I modules essentially for free, and in particular you can compute projective resolutions in a completely combinatorial way

[obtuse meteor]

huhhhhhhhh

this is like

[cedar pebble]

yeaaa

[obtuse meteor]

really weird

[cedar pebble]

it's really bizarre

but it's insanely powerful

like for instance

as soon as you can find the quiver/relations for an associative K-algebra

you can just like

compute ext groups algorithmically and really quickly too

which if you didn't know this would be like

impossibly hard