point-set-topology

[abstract pagoda]

i doubt they will laff in my face unless im dog piss student

[pearl holly]

lmao max in advanced mathematics vc just to say homo-toe-pee

[abstract pagoda]

but what types of research do undergrads do with topology in general?

lol

homo toe pi

[marsh forge]

none

[abstract pagoda]

hmm ok

[marsh forge]

i mean to be honest

[abstract pagoda]

ill just ask them

[marsh forge]

you don't know enough

[abstract pagoda]

i dont

[marsh forge]

to actually help a professor

[swift fjord]

P much ye

[marsh forge]

like just bottom line

but!

[abstract pagoda]

!

[marsh forge]

you can still do reading courses with them and stuff

[pearl holly]

homo-toe-pee is advanced

[marsh forge]

and ask them to teach you stuff they are interested in

[swift fjord]

Profs can find smth an undergrad can help with a lot of the time

[marsh forge]

and eventually maybe they will give you a problem to think about

[swift fjord]

And learn in the process

[abstract pagoda]

o ok

[swift fjord]

That's the point of research exposure week at my school

Well maybe not a lot of the time but some of the tirm

[abstract pagoda]

cuz i feel like it is necessary for me to compete with other students at my school

[marsh forge]

reading courses are probably the best thing you can do with a professor

[novel acorn]

Cou-ho-mo-laa-gee is as well

[abstract pagoda]

reading courses seems like fun

[marsh forge]

this is worringly close to a toxic mentality imo

[abstract pagoda]

its like this server

[reef shore]

hoe MAW luh gee

[marsh forge]

good coldi

[reef shore]

[swift fjord]

It's a mentality some schools actively encourage sadly

[marsh forge]

I think being worried about being "competitive" is bad and will make your relationships with classmates bad

[novel acorn]

Don't even
Thats a shit mentality and it'll only bring you down

[abstract pagoda]

what is the toxic part? competing? Im not only doing it because of competition, I also want the experience

[pearl holly]

Max pronouncing homotopy: me:

[marsh forge]

you should worry about self enrichment

and improving yourself as much as you can

but it should never be about being "better"

like

[abstract pagoda]

ive always tried to be competitive with learning with friends

[marsh forge]

competitive math majors are the types who don't study together or who won't work on problem sets together or who don't help struggling classmates

[abstract pagoda]

ive seen that sometimes

[marsh forge]

ive seen many

[abstract pagoda]

i ask for help often

[marsh forge]

yeah thats good

[abstract pagoda]

sometimes some classmates dont want to join a group chat

[marsh forge]

those people are lame

[abstract pagoda]

and esp corona there was little participation

[novel acorn]

No but too much emphasis on the competition part can make you either a shit person or very anxious and self hateful since you'll never be good enough

[marsh forge]

i think you should work hard and learn a lot and enjoy math as a social activity to do with others

that is the best mentality imo

[abstract pagoda]

i cant get rid of selfhateful sadly it manifest into regret, but i still enjoy learning math so theres that

[novel acorn]

Math is my escapism vibes ngl
same

[abstract pagoda]

same here

osu! math and anime

[marsh forge]

okay this is not good

[novel acorn]

Holy shit another Osu enjoyer

[marsh forge]

You might want to talk to someone if mathematics is seriously harming your self esteem or confidence

[abstract pagoda]

im not sure how detrimental it is, because i know the past is the past and there is nothing i can do about it, but i still feel regretful. its one of those things that i think will pass with time

im just rly anxious about going to school again

because if i dont do my classes to near perfection imma feel rel shit

[marsh forge]

I think if these things are left to themselves they can really badly manifest in burnout or anxiety issues

yeah

that is

a very unhealthy mentality

[pearl holly]

this was me before and now I don't play osu

[novel acorn]

OK so to be real I have self hate issues too but thankfully math helps me with that
You have to try and seek help since you'll see that it really makes a difference

[marsh forge]

Yeah I've done a good amount of therapy in my life, and as I am entering grad school I think I will start up again

Good mental health is an incredibly pro-active goal

And it pays huge dividends to care

[novel acorn]

Imagine not playing Osu
I haven't played in 3 weeks I can already feel the adrenaline pumping through my arteries

[marsh forge]

#topology-and-mental-health

best channel

[abstract pagoda]

ig this is very off topic anyways, but therapy still stigmatized in my family and its not that i think it wont help, its just that i dont feel like spending the time and hopefully when i get older i just forget these feels like how i forget my past feelings most my life

[novel acorn]

The thing is that
They very likely won't go away

[marsh forge]

this channel is allowed to be off topic as the moderators and the top/geo people have a stalemate

Anyway

I think stigmatizing therapy is like

very very bad

its a great thing

and like

[novel acorn]

Stigmatising mental health in general is shit

[marsh forge]

obviously you can't change how your family feels

but you should really try to internalize it as a legitimate medical practice similar to any other kind of doctor

because it is

And I promise promise promise promise

that like

1 hr of therapy a week

will save you hours and hours of wasted time

and improve your efficiency

and happiness

and just overall have huge returns

(obviouslt therapy can be expensive and hard to come by so i don't want to imply that not doing it is bad, but not doing it for those reasons is bad)

Waiting until you are older to be mentally healthy is, in my opinion, a very sad way to live life

YOLO

[novel acorn]

And I'd say unrealistic
Since depression and stuff like that won't just go away because you grew old enough to wish them away
They require a lot of work to eliminate

[novel acorn]

[marsh forge]

your brain doesn't have an immune system

so unlike a flu or something

mental illness doesn't just go away over time

in some ways, the biggest part of therapy is training you how to sort of "create" an immune system of strategies and thought pattern recognition

[abstract pagoda]

ik it seems meme like but i truly dont care about how i feel when im working out or osu or learning/doing math or watching anime. Its only in times where im not doing those things when i have a feeling that i should be doing one of those.

it seems unhealthy though because when im watching movie with family or hanging out with friends i want to do that stuff in back of my mind regardless if im having fun

[novel acorn]

Is it like an urge that you're being unproductive

[abstract pagoda]

maybe, but it puts my mind to rest when im doing that stuff

i still have this big thing about productivity though, i kinda have random urges to want to make something

[novel acorn]

Well like feeling the urge to do stuff sometimes isn't bad
If it's like constantly in the back of your mind when you're doing anything else then I'd say it's bad
Especially if you start blaming yourself for being lazy

[abstract pagoda]

tbh if i go to therapy i predict the conversation will end up on my fear of becoming mentally diseased or homeless so thats why i dont wanna go, those are like my biggest fears in life

[marsh forge]

Well, therapy isn't about bashing you over the head with your biggest fears

[abstract pagoda]

i think my problems root from that though

nothing i can do about them though

i mean i have thought about it for a long time and mental disease is unavoidable so there is nothing i can do

and homelessness just makes me want to work

[marsh forge]

I mean

there's normally a pretty big gap

between working super hard

and not being homeless

[abstract pagoda]

yea i know personally

[marsh forge]

unless you are already at risk for homelessness

[abstract pagoda]

but homelessness really sucks and my safety nets are only sorta there

and if i have to go to a safety net it is by definition a failure on my goals

[marsh forge]

I think you are trying to convince yourself that there is nothing therapy could help you with

and I can tell you that seeing these thought patterns

I disagree

[abstract pagoda]

i think it would help me, but i dont know how permanent

i dont wanna do recurring visits either

[marsh forge]

You have to do recurring visits

for awhile

[abstract pagoda]

i feel like the last straw is when i see that im developing mental health issues

[marsh forge]

eventually you might be fine without them

You are currently developing mental health issues

no offense

[abstract pagoda]

🤔

but how

[novel acorn]

It's 1h a week usually so you'll visit your therapist like a bunch of times

[paper wedge]

is this topology discussion

[marsh forge]

ill repeat

[novel acorn]

#mental-health

[abstract pagoda]

ill go if my school offers free sessions

[marsh forge]

this channel is free from moderator oversight

and we can do what we want

So has always been the way of #point-set-topology

[swift fjord]

You should go to therapy BEFORE your problems exacerbate

An ounce of prevention is worth a pound of cure

[paper wedge]

yea i am just messing idc haha but i will interrupt with a topology question: what are filters

[novel acorn]

What country are you in if you're comfortable saying
Since I'm curious if the fact that your family discourages therapy is cultural

[marsh forge]

idk of a single country where that isnt a thing

maybe some euro countries?

americans certainly hate therapy

in general

[abstract pagoda]

family from uganda+caribbean but its irrelevent anyways

im in US

and second gen

[marsh forge]

okay

i have to go surf

but i will say

[novel acorn]

Well I'm in Euro and everybody has always told me yo go to therapy here so I dunno lol

[marsh forge]

you should take this seriously nine9s

thats awesome irony

[abstract pagoda]

ill consider its severity

[novel acorn]

You have literally nothing to lose by going

[abstract pagoda]

i have my dignity

going feels like i lost

[novel acorn]

Lost to who lmao

[abstract pagoda]

myself xd

[novel acorn]

This isn't a competition with the world

It's like saying I've failed in life when you break a bone

[abstract pagoda]

lol no

i failed in life if i get soft-stuck in homelessness or in trying to reach a big goal, or if i develop a debilitating mental illness

debilitating mental illness might be hereditary though

i have a history of potentially trash mental health

[paper wedge]

i think you just have to think straight and be aware man if you go to a professional things will only get better

[paper wedge]

you can fight it man

[abstract pagoda]

i got scarlet fever twice each lasting about 2-3 days, and i know what psychosis feels like. also finals week i usually have these thoughts that are incoherent right before i fall asleep, but thats rly it

psychosis sucks big time and if i notice any episodes when im not sick im probably gonna go to doctor+therapist or something, so thats my last straw

[abstract pagoda]

i hope lol, everyone in my family over 70 has dementia

some as young as 40

[paper wedge]

you can be healthy

i am not a professional but

[novel acorn]

Dementia isn't a mental illness in the usual sense I'd say

[paper wedge]

these medical things can be improven by lifestyle and diet and so son

on*

nothing is out of reach

you can do it

[abstract pagoda]

most of my family is healthy though

and im p fit

[paper wedge]

thats p good

take care of like the food u put in and stuff

what u do

[abstract pagoda]

like what else?

[paper wedge]

go to a professional

things will just get better

dont stress too much from exams man

[abstract pagoda]

i dont conciously stress

[novel acorn]

Depression and similar
Dementia is literally your brain degrading
While those others are either caused by chemical imbalances or they're of psychological origin

[paper wedge]

yea the point is

shits chemical and medical

u can fix it

aslong as you are actually aware

and willing to improve man things will get better

[abstract pagoda]

inshaallah ill just gameify my mental health and if i lose ill go to doctor, im stubborn when it comes to seeking help with these types of things

hopefully i dont get the usual suspects of depression or anxiety disorders

[paper wedge]

why

why do u say words like lose and "just"

just be aware that going to a professional will literally make you better

its good u know ur stubborn but do something about it to improve

[abstract pagoda]

well im not deep in the shits of mental health yet so being better isnt a big priority

and i dont want to be obsessed with improving my mental health/don't want to think about it too much since i mostly have no control over where it goes unless i am introspective.

[swift fjord]

Sorry nine9s but you're not an objective judge of your own mental health

No one is

[abstract pagoda]

):

[swift fjord]

And it's not a guarantee that you'll feel it declining

It's always better to take preventative measures

I'll say it again, an ounce of prevention is worth a pound of cure

[abstract pagoda]

qq

[abstract pagoda]

you are completely right

i can't really argue with what you are saying either

[swift fjord]

This is the kind of stuff you could talk to a therapist about too

[paper wedge]

go to a professional

things WILL improve

[drowsy yew]

does anyone know how to integrate with a manifold as bounds

like

$\int_M w$

i mean int

[gentle osprey]

Muffin Man

[abstract pagoda]

i feel like the goal of going is achieved once my two biggest fears go away, but they are likely to stay around until i die. i also cope with not going because millions of people for centuries didnt have mental health and everyone dies with mental issues. all in all i probably wont go unless the situation of the symptoms of disease become apparent. im also aware that i might not be aware of this decline, but i hope others around me will atleast tell me. so yes i am stubborn

[drowsy yew]

how to compute this <@&286206848099549185>

[gritty widget]

stokes bruh

[hollow harbor]

partition of unity subordinate countably many charts, pull back nth partition function * w along each chart, integrate as usual in R^n, sum.

[frosty sundial]

stokes will only work if either M or omega is d of something

[hollow harbor]

(in other words, you don't)

[gritty widget]

ye xd

[frosty sundial]

maybe you're lucky and can parametrize M with one chart

[hollow harbor]

yeah

or there could be some symmetry working to your advantage

[drowsy yew]

umm

[hollow harbor]

honestly probably a lot easier to just split your manifold into some closed pieces which are each diffeo to a subset of R^n and so that any two pieces only intersect on their boundaries

i don't think you really need to do the nasty partition of unity thing even though that's the definition

then pull back on an open neighborhood of the closed piece, but then integrate only on the closed piece.

so if M were a sphere, I would take like the bottom hemisphere and the top hemisphere (both closed). since they share only an equator, this bit won't matter if we're integrating a 2-form.

[frosty sundial]

why not just use spherical coordinates

then you only have one integral

[hollow harbor]

sure

you can remove two points

that's fine too

[frosty sundial]

i guess the point is: there are lots of ways

[hollow harbor]

i'm assuming you can always "cut up" a compact manifold along (n-1)-manifolds and get something which is a subset of R^n

maybe you just triangulate first, cut out all the simplices, and reassemble them along a spanning tree or something

but sometimes this is going to be annoying

[sacred laurel]

Where can I find a full and detailed proof of Hopf Fibration, please ?

[novel acorn]

Wikipedia has a section on how it's constructed

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

Also don't curse me for using Wikipedia OK

[raw sedge]

wikipedia is based

[shy moss]

are there vector spaces or modules in algebraic topology?

[abstract pagoda]

both right?

over field over ring

qq

u said kyfak byna

[grave maple]

I thought alg. top. dealt with groups and groupiods.

[shy moss]

which are?

y only know groups in at

[gentle osprey]

Or x1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[shy moss]

functors from Top to R-mod

with $R \neq Z,$ because this is just a group

[gentle osprey]

Or x1

[bronze lake]

In knot theory, the generalized Alexander invariant is an ideal in the module $\mathbb{Z}[t,t^{-1}]$

[gentle osprey]

cgodfrey

[wheat gulch]

Is there a name for the two opposing point the Villarceau circle intersect at? https://en.wikipedia.org/wiki/Villarceau_circles

[plain raven]

algebraic topology assigns a variety of algebraic gadgets in order to understand them better. The fundamental group and higher homotopy groups are groups (or groupoids if you modify the definition slightly) the homology of a space is a graded module, the cohomology of a space is a graded ring. rational homotopy theory assigns them vector spaces over the rational numbers. and so on.

[cedar pebble]

sure, one answer to what homotopy theory is about is studying (nice) topological spaces up to weak homotopy equivalence. One way to formalize this is that you associate to each (nice) topological space its "fundamental infinity groupoid," and then one proves that two (nice) topological spaces are weakly homotopy equivalent if and only if their fundamental infinite groupoids are isomorphic in the appropriate sense

while infinity groupoids as usually defined are a very algebraic structure, they can be pretty unwieldy. This is why people often consider weaker invariants, like various cohomology theories that capture less about a space but are more computable.

[quartz edge]

if i got a surface with one of the principal curvatures k1 constant everywhere
and i choose a principal frame e1 e2 so e1 points in the direction of k1's curvature
how tf can i "pull back" to an "e1-curve" so that the e1 curve is a geodesic?

i.e. the principal frame does not twist tangentially as i walk along e1

is this literally just like

warping the surface i'm on

[novel acorn]

Antipodal maybe?

[gritty widget]

what book is this

[reef shore]

Hatcher

[pearl holly]

Looking back at this, I don't think I get it. Like I know that an element $\sigma: \Delta^n \to X$ in $C_n$ (which is defined to be the free abelian group with basis th set of singular n-simplicies in $X$ in my book) has a path connected image so it must lie in one of the $C_n(X_\alpha)$. But what I don't get is that $C_n$ equals the direct sum of $C_n(X_\alpha)$. Doesn't the direct sum $C_n(X_\alpha)$ contain tuples and not just elements? If that is the case, then right hand side contains tuples while the left hand side doesn't so there can't be an equality, right?

[gentle osprey]

Tokidoki ✓

[reef shore]

Each simplex lies in a single component, so a sum of simplices is a sum of simplices that each lie in a single component

So this says that H_n(X) is a sum of H_n(X_alpha)

Do you follow so far?

[pearl holly]

wait this is what we are trying to prove, right?

[reef shore]

Direct sum is stronger than sum

[pearl holly]

oh direct sum

what is that lmao

no

[reef shore]

[pearl holly]

I mean just sum

[reef shore]

Wait

You are asking what a sum of abelian groups is?

[pearl holly]

well my question really boils down to how a group can be equal to a direct sum of its subgroups because the direct sum of stuff is just a "cartesian product" right?

[reef shore]

They agree for abelian groups, and finite sums and products

For infinite they don't agree

We say that A is a direct sum of subgroups B and C if every element of A can be written as b+c, b in B, c in C, uniquely

Kinda like a basis for a vector space, once you fix a basis, a vector space is a direct sum of the subspaces generated by each basis element

More generally if you have a collection B_i of subgroups of A then A is a direct sum of the B_i iff you can write every element of A as ∑ᵢ bᵢ, each bᵢ in B_i, and only finitely many of them non zero

[pearl holly]

oh lmao. I was under the impression that $A \oplus B$ is a cartesian product of $A$ and $B$. So like $(a, b) \in A \oplus B$

[gentle osprey]

Tokidoki ✓

[reef shore]

It is, they are isomorphic

[pearl holly]

But that the operation is different

oh

[reef shore]

For finitely many things

And if your groups are abelian

For non abelian I don't think direct sum is a thing, there you do free products

(direct sum is coproduct in Ab)

So if you have every element written uniquely as b+c, then you can identify it with (b,c)

That is the isomorphism

With what you called the Cartesian product

But the Cartesian product is the word for set products only I think

There are many group structures you can put on a Cartesian product of 2 groups

The one that makes it into the product group is the direct product

[pearl holly]

Right okay. So every element of A direct sum B is just a element of the form a+b?

[reef shore]

Yes, uniquely

As in it can't be written as some other a'+b'

[pearl holly]

ohhh okay I see lmao

[reef shore]

If it can then we just call it a sum

Instead of direct sum

[pearl holly]

yeah okay then I get it lmao. Thank you so much! I really needed this explanation. I was reading wikipedia about this and I thought that it was just a normal cartesian product lmao

[reef shore]

[swift fjord]

Direct sum is also called a weak direct product (iirc), it just means that only finitely many entries of any element are nonzero

Kind of like the product topology

[pearl holly]

Right so you could say that the direct sum is a product topology and that the direct product is the box topology?

Just to make an analogy

Nothing formal lmao

[swift fjord]

Yea p much

Also technically the homology of the path components aren't subgroups

Since the range of the simplices is different

Though there is a canonical inclusion ofc (induced by the inclusion of the path component)

[pearl holly]

yeah that's right. Because a element in C_n(X) goes from Delta^n to X and not from Delta^n to the path component. So therefore we have to restrict the range, right?

And restricting the range is kind of doing the same as just taking the canonical inclusion?

[swift fjord]

Exactly

Nono

[pearl holly]

oops

[swift fjord]

The other way around

You take the simplex with restricted range

Then that composed with the inclusion is a simplex in X

So that's the inclusion

[pearl holly]

ah yeah okay I see

Thank you so much!

[swift fjord]

Np

[umbral finch]

basic question here: I can't see why in the topological space (R, O), where the sets in the topology are either the empty set or contain 0, satisfies closure of singleton {0} is the whole space R (i.e. why singleton 0 is dense in this space).

[swift fjord]

Recall the closure can be defined as the smallest (w.r.t. inclusion) closed set containing (in this case {0}). A closed set is defined as the complement of a set in O, so the only closed sets are R, the empty set or the complement of a set containing 0, that is, the only closed set containing 0 is R

So in particular that is the smallest closed set containing 0

You can also think about it like this: can you show that every point in R-{0} is a limit point of {0}? (By the definition of a limit point)

[umbral finch]

Are all closed sets always the complement of the open sets in the topology?

[gritty widget]

ye

[umbral finch]

thanks for both explanations xD

[gritty widget]

because if C is closed then X/C is open and X/C is in the topology homie

[gritty widget]

say it adrian

[umbral finch]

clarification: in topology, "openness" is a local property w.r.t to a chosen topology? so the open sets are precisely those in the topology?

lol

[gritty widget]

open sets are elements of the topology ye

so one set may be open in one topology but not in another one

[umbral finch]

perfect, now it starts making sense xD

[gritty widget]

where are u learning frfom?

[umbral finch]

im not 😦 reading an appendix in order to learn measure theory

it's cohn's "measure theory"

great book, but im trying the approach of proving just the results I need later on instead of studying topology separately.

[gritty widget]

why do you want to learn measure theory

[umbral finch]

i liked analysis this year and im taking measure theory 1 month from now.

[gritty widget]

when are you taking topology?

[umbral finch]

it depends, it's not mandatory at my uni. I might take it next year (at another uni) or just self-study xD

I see it appears in measure theory and functional analysis, both of which im taking this year, so I guess self-study this month the very basics?

when did you take topology?

[gritty widget]

3rd semester

I think its odd to take those kinda advanced analysis classes without doing any point set topology before

[umbral finch]

maybe less odd given it's "applied maths" and not "maths/pure maths"?

[gritty widget]

applied measure theory

shouldve said probability course

[umbral finch]

i have structured my courses differently, since I am considering pure over applied, in general.

[umbral finch]

from the little probability theory we covered, cant say I found it very interesting. so hmm maybe geometric measure theory?

[gritty widget]

In particular stuff like functional analysis you will want to talk about eg weak topology etc, this will require a topology background

[umbral finch]

any rough idea how much to cover from Munkres (or other sources) to properly learn funct anal and measure theory?

or should I simply stick to trying to prove everything in the appendix from the measure theory book and if unable, work through corresponding chapter(s) from Munkres

[gritty widget]

to be fair my functional analysis didn't really use topology (especially weak ones were never covered)

But yeah in rudin you can see theres a lot of that

I don't think there is much topology that you need in an intro measure theory class, spaces will generally have a norm

But I guess in terms of munkres the good stuff to know is all the point set chapters except maybe the last few (metrization, etc. Tychonoff theorem is sometimes used but I feel like it's not that often. I can't remember the chapters off the top of my head)

[umbral finch]

question about proof:
when computing the set of limit points, is it usually enough to show only one inclusion instead of both? (for instance, here the direct and inverse inclusions are very similar)

or moreover, can you sometimes work with equivalence?

[swift fjord]

Usually you'll already have one inclusion. Like here you wanna shoe your entire space is the closure of {0}, so obviously {0}' is contained in R, so you just gotta show the other inclusion

I can't really say generally

It depends

[umbral finch]

I see. I was thinking it might usually be the case that both inclusions are of the same nature/idea, but it seems not. Thanks!

[novel acorn]

Tbh I don't get either this or the proof after it

Actually this one isn't even that bad

[swift fjord]

@pearl holly @novel acorn try rotman's proof if you're still unsure

4.13

[novel acorn]

Mmm thanks
But the next one fucks with my head a bit more

[swift fjord]

What is it?

[novel acorn]

I don't see how ker(epsilon) induces the isomorphism

[swift fjord]

Lemme read through it

But

Rotman's proof is very clear

If you want

[swift fjord]

1st iso. Theorem

[novel acorn]

Ah
Those that I can't remember lol
I'll read those two again

Ah OK now it makes sense

Can't believe I still don't know those two off the top of my head ffs

Wait

There are three

Brain dumb

[upper basalt]

Why we use that dotted line ,can someone please tell me?

[pearl holly]

It stands for "exists"

So basically "given those morphisms in the diagram, there exist a unique tilde phi such that the diagram commutes"

[gritty widget]

so does it stand for there exists a unique?

[reef shore]

Usually up to context

Often used in universal property statements where it would be unique but sometimes for existence of lifts etc

[pearl holly]

yeah listen to those guys instead lmao, they are more cat brained than me

[gritty widget]

is cat brained good in this server

how deep does the cat worship go

[ivory dragon]

on one hand, mniip is cat brained

but at the same time

[reef shore]

Channels moved around

[ivory dragon]

i think this is accurate

at a certain point you become too into cats

and you face a sharp decline

[gritty widget]

the hype cycle

[ivory dragon]

because people think you either
(1) program
(2) follow urs on twitter

[upper basalt]

Why soo?

[reef shore]

Why are you scribbling on discord

[ivory dragon]

but if you go all the way then youre actually a legit expert

[reef shore]

DiligentClerk

[ivory dragon]

like mniip being one of the 10 people on earth actually paid to write haskell

[ivory dragon]

asserting dominance

[pearl holly]

how do I find shit in the appendix lmao

[ivory dragon]

your mere existence is something for me to scribble over

i can simply erase you all

[pearl holly]

I'm looking for §2.A but there's no §2.A here wtf

[reef shore]

[reef shore]

Does not seem accurate though

At some point it dips into the negatives

When you do shit like

When teaching induction to HSers

[gritty widget]

or regurgitating monad as endofunctor etc.

[reef shore]

Is there any other way to view a monad

[gritty widget]

well it's usually programmers tryna seem cool

[reef shore]

Monoid in ([C,C], ∘, 1)

Ah lol

[gritty widget]

to most haskell people it's just an obstacle between them and IO

[reef shore]

I still don't fully understand how cat theory plays into haskell

[gritty widget]

not much lol

[reef shore]

I know that List and maybe are monads

But I heard it's used to abstract away some IO stuff

[gritty widget]

It's useful because it lets you have pure functions working with impure ones

[reef shore]

Lol channels back where they were

[gritty widget]

[reef shore]

A couple days ago chill was in misc momentarily

Some mod is messing around

[reef shore]

Will look it up

[gritty widget]

I'm trying to google while doing this, but the main thing is the type signature of bind

[reef shore]

Don't know what bind is

[gritty widget]

uhh

[ivory dragon]

fixed

i think this is accurate

[gritty widget]

im at oop

for sure

[reef shore]

[pearl holly]

lmao I am at "maybe I should read Riehl" right now

[reef shore]

Toki fall from grace arc

[viral atlas]

Where's the "I shit on category theory because that sounds cool" stage

[pearl holly]

Or maybe like a little bit over the "wtf do the arrows mean"

[bronze lake]

Did I miss the part where I can shit on prolog and sing the praises of haskell

[reef shore]

Toki you are well past really liking AT

[pearl holly]

but I still don't know a lot of cat theory that's the thing

[gritty widget]

yeah I read category theory for programmers, i'm basically a CT guru

[reef shore]

Good time to fall

[bronze lake]

Category theory is just an endofunctor between programming languages and math

[gritty widget]

what's the category

[reef shore]

Also toki don't read May lol

It will be difficult

[pearl holly]

yeah it does seem difficult lmao

[reef shore]

Like everything needed is there, but needs extra familiarity

[pearl holly]

ah yeah I see

[reef shore]

Preferably

[pearl holly]

I might read chapter 2 tho about the van Kampen but with cat theory language

[reef shore]

oh ye

[pearl holly]

If I manage to understand that lmao

[reef shore]

Ye I just read that today lmao

He does it slightly differently from dieck

More generality

Hatcher only does the group version, Dieck only does it for cover with 2 open sets

May does it for groupoids and groups with an infinite cover

With every finite intersection path connected

[pearl holly]

yeah I might check it out because there's subsections that talk about cat theory in that chapter so I might learn something there too

So there's the groupoid version of van Kampen there too?

[reef shore]

Yeah

Group version is an "easy" consequence of the groupoid version

Depending on how familiar you are with manipulating colimits

[pearl holly]

I see I see

[reef shore]

But May definitely seems better than Dieck lmao

Dieck is fun if you already know both AT and CT

Not fun when learning for the first time

[novel acorn]

Why must all AT books have some catch to them

[reef shore]

May good

Assuming you know cat thy

[novel acorn]

You said may hard

[reef shore]

If you don't know cat

[novel acorn]

I know
Some
(maybe a should read a bit more)

[reef shore]

[novel acorn]

I know
What a functor is

[reef shore]

Limits and colimits?

[novel acorn]

No

[reef shore]

Probably difficult then

[novel acorn]

I'll have to read then

[reef shore]

All shall be cat brained

https://tenor.com/view/kitten-genius-read-book-study-funny-gif-12040978

[hollow harbor]

Coldilocks unironically recommending May

[reef shore]

Only to those who know cats

And I've read 20 pages so far it seemed good

[hollow harbor]

I watched the first 10 minutes of this movie. It has a great plot!

[reef shore]

That is very rude rice, you shouldn't look down on other's preferences

😌

Lol it was recommended to me by others

Yesterday

[hollow harbor]

Yikes.

[reef shore]

And irony and toki were here

So they probably know I just started it

Maybe

[hollow harbor]

So it was recommended by one of this channel's resident cranks?

[novel acorn]

Yes you were shilling here yesterday and toki accidentally said he's interested (I know he isn't he's Hatcher kin)

[reef shore]

Was recommended by nobody

I was inviting toki to read together

[hollow harbor]

[novel acorn]

And he accidentally wrote yes instead of no

[reef shore]

Nobody is a user who frequents this channel

[hollow harbor]

Oh yes

I know nobody

[reef shore]

Damn, sounds lonely

I can be your friend if you need ryc

[hollow harbor]

Potentially trustworthy source of information for a category brain

I don't need friends, they disappoint me

[reef shore]

[hollow harbor]

I am the queen of the universe.

[reef shore]

[hollow harbor]

Honey, you've got a big storm comin'.

[reef shore]

Is this from some reality show

[gritty widget]

it's from a set of audition tapes

internet classic

[bronze lake]

I read so much, why can't I just fully understand AT? I just finished my 5th read-through of harry potter last week

[pearl holly]

So Hatcher states that a chain map $f_#$ induces a homomorphism $f_: H_n(X) \to H_n(Y)$ between two homology groups since it $f_#$ maps cycles to cycles and boundaries to boundaries. But how is it actually defined? Like $f_(b* im(\partial_{n+1})) = f_#(b) * f_#(im(\partial_{n+1}))$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

where * "denotes" the coset

That would make sense right?

[shy moss]

yes

$f#(b) * f#(im(\partial{n+1}))$ is an element of $H_n(Y)$ because $f#$ maps cycles to cycles and boundaries to boundaries

[gentle osprey]

Or x1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[pearl holly]

yeah okay I see, I just needed that to be clarified lmao. Thank you so much!

[novel acorn]

https://pi.math.cornell.edu/~hatcher/AT/typography.html

He gives a tutorial on how to make Tex documents look like his book lmao

[pearl holly]

yooo lmao that's kind of cool

but tbh I don't like the font. I would prefer just a standard one

and the theorems etc. don't look that nice in my opinion too

maybe this is the standard font idk

[novel acorn]

I dunno if it's distributive tho (also it annoys me immensely that he doesn't explain how it works)
I'll look online gimme a sec since I'm literally on the same page as you

[pearl holly]

yooo same page gang 😎

But what I mean is that it's defined that way

I just define it that way because then it works

Like the chain map induces exactly that map

[novel acorn]

https://math.stackexchange.com/questions/1776786/induced-homomorphism-in-homology

[pearl holly]

yeah okay so if I denote a coset by [g] then f_* = [f_# g]

which is what I wrote but with better notation I think

wait why is it a homomorphism?

wait the chain map is a homomorphism by definition, right?

[novel acorn]

No?
He never says

Or does he

AAAAAAAAAAAAAAAAAAAAAAAAAAA

[pearl holly]

no he doesn't

[novel acorn]

Homotopy is easier wah

[pearl holly]

But I'm thinking about the part where he says "then extending f_# linearly via..."

Like is that the definition of f_#?

Because then it is automatically a homomorphism

at page 111 at the very top

[reef shore]

Maps on the homologues are automatically homomorphisms yeah

[pearl holly]

yeah okay I see, great! Thank you dudles so much!

"Unable to be yeeten"

[gentle osprey]

Nobody

[grave maple]

Suppose X is a regular convergence space with two points x,y such that u(x) converges to y, where u(x) is the principal ultrafilter generated by x. Is it true that u(y) converges to x?

Hmm... the intersection of u(x) and u(y) also converges to y. And since u(y) contains this intersection, it also converges to y.

I guess I did not need regularity.

[plain raven]

homology groups are abelian so the standard notation for the group law is an addition rather than a multiplication

[plain raven]

a chain map should be a homomorphism as part of its definition, definitely. if you like, it's a homomorphism of graded Abelian groups (of degree 0) commuting with the boundary operator

i'd never talk about a map between complexes that wasn't additiive lol

[pearl holly]

yeah okay I see. Thank you so much!

[flint cove]

doesn't „map“ always implicitly mean „arrow in whatever is the obvious category“? 😉

[eternal nimbus]

dumbest question. Barely starting with projective geometry, F has not been defined, so, uh, what is it?

[gritty widget]

probably a field. can you give more context?

[eternal nimbus]

like, he just defines homogeneous coordinates and boom, F outta nowhere

that's the pont, F is never defined before. We are given V a vector space, P(V) the projective space, then homogenous coordinates, we define U0....but never has F been talked about

[gritty widget]

what book?

[eternal nimbus]

https://people.maths.ox.ac.uk/hitchin/files/LectureNotes/Projective_geometry/Chapter_1_Projective_geometry.pdf

it's literally page 6, and pages 1 to 4 are historical introduction xD

it's great for morale to stumble so close to the beginning xD

[gritty widget]

page 5

[eternal nimbus]

ah jesus fucking christ

thanks and sorry, i shouldn't do math with vaccine fever maybe

All my other curses use K, maybe that trip me over. I'll go do some penitance helping people in pre university or something like that

[gritty widget]

vaccine fever

[eternal nimbus]

:p not terrible bad, but not fun, thankfully is been passing quite fast, came this morning and it's already fading away

[bronze lake]

My first course on Alg Geo used k for a field and I hated using a lowercase letter for it

[gritty widget]

it makes me feel dirty, but in a good kind of way

[unreal stratus]

i've often seen k for a field and K for an algebraic closure of k tbf

[bronze lake]

I just hate using a lowercase letter for an algebraic structure as a whole, not just an element of one

[gritty widget]

lie algebras in shambles

[fading vale]

k makes sense

k for base field, K for function field

it comes from korper

[plain raven]

imagine trying to justify your notational choices by appeal to language and historical origin rather than just asserting that whatever notation you learned first is correct

for the record Hartshorne uses k.

this is the same with naming conventions. I know a guy who learned algebra out of Gallian as an undergrad and spent like the first two months of undergrad insisting that rings shouldn't have units

[marsh forge]

The correct notation for a field is $\mathbb{F}_p$

[gentle osprey]

MaxJ #MiuArmy

[bronze lake]

Wtf rings with units are so nice

[gritty widget]

But rings without unit are somehow not uncommon

This is a weirdness of mathematics that makes me sad

[ivory dragon]

yet another reason algebra is the superior field

[bronze lake]

I like my rings commutative, my ideals maximal, and my elements invertible

[swift fjord]

I don't understand how defining sigma just on those line segments extends it continuously on the whole simplex

the dot above the 2 simplex means boundary btw

and sigma is defined with barycentric coordinates

like what happens inside the shape abdce_1

oh wait

[reef shore]

The union of those line segments is the whole simplex

[swift fjord]

those lines are parametrised by t

[reef shore]

As t varies

[swift fjord]

yea I realised now

[reef shore]

[swift fjord]

do these lines never meet?

wait

why do we need 2 lines actually

why isn't one parametrised line enough

[reef shore]

They're all parallel

You could parameterize them at once

[swift fjord]

oh wait

[reef shore]

But they are parametrizating them using the endpoints

[swift fjord]

yea but one always touches the bottom face and one only touches the right face

yea

[reef shore]

Ye

[swift fjord]

they'll only meet on e_1

alright gotcha

why is it easy to see that it's continuous though, you can't use the gluing lemma here

[reef shore]

You can say that the restriction to the segment connecting e_0 and e_2 is continuous

Then this is projection followed by that restriction

Projection be continuous

[swift fjord]

what's the projection

[reef shore]

Projection of the plane on e_0 e_2

Orthogonally

So all those line segments map to their left end point

[swift fjord]

ah alright

yea so that's continuous

[reef shore]

[swift fjord]

I'm still a bit iffy about it but I guess it's not too important

ty

[reef shore]

[swift fjord]

btw you can apply this construction for any 3 paths where the endpoints agree right?

[reef shore]

No you use the fact that the third thing is f*g here

Because otherwise you won't have the map constant on those t-parameterized lines

[coral pivot]

(lmao did a double take when i read f*g)

[reef shore]

[swift fjord]

right

I should have said

any 2 paths whose endpoints agree + their multiplication

[swift fjord]

lmao

[reef shore]

Ye

[swift fjord]

ye my b

[swift fjord]

I wanna say here that the first fact is true since the 2-simplex is homeomorphic to the 2-disk (And there's a homeomorphism that preserves the boundary), but I don't see why the fact that say $\sigma i \vert_{S^1}$ is nullhomotopic (Where i is the homeomorphism) implies that $\sigma\vert_{\partial\triangle^2}$ is nullhomotopic

[gentle osprey]

k The Spring Constant

[swift fjord]

I guess you might not need this

I mean technically the theorem can be reformulated for any convex set, right?

[empty grove]

For any set X and CX, the cone over X

The 2 simplex is homeomorphic to the cone over its boundary

[swift fjord]

right but that raises the question again of why if it's homeomorphic does the nullhomotopy carry over

that doesn't feel trivial to me

[empty grove]

You don't need to worry about S^1 with this

It's just a more general statement

But yeah the nullhomotopy would carry over

[swift fjord]

oh wait I think I see why

[empty grove]

Because you have a homeomorphism of the simplex with the disk that restricts to a homeomorphism of their boundaries

[fading vale]

every homeomorphism induces a homotopy equivalence

thats sufficient

[swift fjord]

yea

[fading vale]

What are you trying to prove with this

[swift fjord]

this is a remark before hurewicz's theorem

just the remark, that the singular simplex restricted to its boundary is nullhomotopic

[fading vale]

isnt that sigma circ i nullhomotopic exactly what you're trying to prove

[swift fjord]

no

that sigma itself is

[fading vale]

On the boundary, no?

[swift fjord]

yea

[fading vale]

by definition thats just sigma circ i

thats what restriction is

[swift fjord]

but i'm using the fact that any map on the unit circle that can be extended to the disc is nullhomotopic

i is not a restriction

I wrote it about

it's a homeomorphism between the unit disc and the 2-simplex

that preserves the boundary

[fading vale]

Ah I see

i is not the inclusion

[swift fjord]

yeye

bad notation on my part

it's all good tho I got it now

[fading vale]

Yeah

Btw just in general precomposing with a nullhomotopy is a nullhomotopy

[swift fjord]

right

I just didn't realise that this is the case here haha

[fading vale]

[swift fjord]

In the definition of polygon, i'm assuming that it's supposed to be
$$\sigma_i(e_1)=\sigma_{i+1}(e_0)$$
Right?

[gentle osprey]

k The Spring Constant

[swift fjord]

Mod k+1 ofc

[empty grove]

would make sense yeah

[swift fjord]

HYPE

[shy moss]

what book is that?

[swift fjord]

Rotman

[shy moss]

snake's lemma proof is on the movie It's my turn

[coral pivot]

Snake.

[raw sedge]

I'm not 100% sure if this is the right channel, but suppose I have a polynomial with real coefficients of the form $ax^2 + bxy + cy^2 + dx + ey + h$, is there in general an affine change of coordinates (transforming $(x,y)$ into $(a'x + b'x + e', c'x + d'x + h')$ with $a'd' - b'c'$ nonzero, thus acting on polynomials $\mathbb{R}[x,y]$) that transforms it in into a polynomial of the form $Ax^2 + By^2 + C$ ?

[gentle osprey]

Intel

[tawdry widget]

yes,there always exists such linear isomorphism

[raw sedge]

what's a nice way to see that?

are you sure that it suffices to require it to be linear rather than affine?

[tawdry widget]

Real symmetric matrices are diagonalizable using orthogonal matrices

So your statement holds even for n variables not just 2 variables

[raw sedge]

it's been a while since I looked at linear operators on inner product spaces, and I never liked looking at linear transformations as matrices

so this explanation is difficult to follow for me

i'm gonna try to think of it while taking a shower which I need to take anyway, and then i'll come back

[tawdry widget]

Classical results of quadratic form,not hard to catch up

[regal sonnet]

can anyone here pls guide me too Riemannian Geometry/manifold/space/curvature, I'm just beginner(and 13 y/o)?

not much like, I studying it as a sub-field for Theoretical Physics & Mathematical Physics!

[raw sedge]

do you know real analysis and linear algebra?

[regal sonnet]

I know like linear algebra, I think?

[raw sedge]

I don't think you have enough mathematical maturity to learn Riemannian geometry.

since you're 13, it would be extraordinary if you did

do you know how to write proofs?

[regal sonnet]

of what?

[raw sedge]

of mathematical statements

[regal sonnet]

yeah it is, But, it is necessary for learning Topological Black holes.

[regal sonnet]

no.

[marsh forge]

have you been spending time on the nLab

[raw sedge]

then that's a good place to start learning higher math

[regal sonnet]

IDK nlab?

[raw sedge]

there are books like "how to prove it" and "the art of problem solving" that would be a good introduction

[regal sonnet]

ok

[marsh forge]

You would need (at a minimum) a background in calculus, analysis, and point-set topology

[marsh forge]

this is the right answer, just making sure

[raw sedge]

ignore what he said about nlab, that was a joke

[marsh forge]

(it wasn't a joke but I am glad the answer was no)

[obtuse meteor]

Lel

[regal sonnet]

ok, yeah, some ppl suggested me calc. previously!

[regal sonnet]

ok

[raw sedge]

learning Riemannian geometry not knowing calculus is like learning calculus not knowing addition of natural numbers

[gritty widget]

all you need to know how to do in riemannian geometry for physics is how to move indices around

[swift fjord]

Ok not exactly top (well kind of) but I know i'll get some answer here. Let $L$ be a lattice in $\mathbb R^n$. Call a subspace $V \leq \mathbb R^n$ $L$-Rational if $V\cap L$ is a lattice in $V$. Does anyone know a sufficient and necessary condition for a subspace to be $L$-rational (Cuz isomorphisms are continuous we can WLOG assume that $L=\mathbb Z^n$)

[gentle osprey]

k The Spring Constant

[raw sedge]

my initial intuition is that V intersect L is isomorphic to Z^n for some n which is at most dimV, and V is L-rational precisely when n = dimV

[swift fjord]

thing is, you could have n independent basis vectors in V intersect L that do not span the lattice V intersect L

you can have a sublattice of the same dimension as the original lattice

[swift fjord]

Consider the sublattice $span_{\mathbb Z}(2e_1,2e_2)$ of $\mathbb Z^2$

[gentle osprey]

k The Spring Constant

[raw sedge]

but wouldn't its containment in a subspace be equivalent to containing the original lattice?

[swift fjord]

Ok I reread what you said, and yes we are requiring lattices to have full rank so it'd have to be isomorphic to Z^dimV

but that doesn't really help me as it's not useful information

Why rational and not whole?

I'm also trying to understand if and when you can extend a rational subspace to a rational subspace of higher dimension

yea ig

[swift fjord]

@marsh forge You lied to me

Literally in the next chapter, the definition of chain homotopy

generalises this to Comp

lmao

[swift fjord]

I think he just.proved the snake lemma in Comp but didn't call it that

He did use the term connecting homomorphism tho

Actually idk

Probably

Just in terms of chain maps and homology groups

But I think it's equivalent

[raw sedge]

hmm, I think I thought of an interesting topological problem.

So, fix some Hausdorff topological ring. For now, let's use the real numbers.

Then for a topological space X, we have the ring Hom(X, R).

Given any subset A of this ring, define V(A) = {x in X | f(x) = 0 for all f in A}.

It is easy to see this must be a close set. (it's the intersection of V({x}) for each x in A, and each of those is the pullback of {0} under a continuous map, and R is Hausdorff)

but actually, those sets obey the axioms for closed sets, which gives you a "weakening" of X with respect to R

so the questions are, how does this weakening work? what spaces, for examples, are not strictly weakened by the real numbers?

[fading vale]

Im not really sure exactly what you are asking

[raw sedge]

It's an open-ended question.

[fading vale]

you're putting the topology on X whose closed sets are preimages of the V(A)?

[raw sedge]

the closed sets are the V(A)'s

yes

[fading vale]

Oh i see

When X is compact ||the maximal spectra of this is just X again||

[raw sedge]

Yeah, that's relates to the ||Stone-Cech compactification||.

[fading vale]

I dont think you need stone cech to prove this

[raw sedge]

Yeah, you don't need it, but it relates to it.

It's kind of the other way around.

You need the fact you mentioned as a motivation for Stone-Cech.

When X is compact, the maximal ideals correspond to points.

Then when it's not, points still give you maximal ideals, but there are more maximal ideals.

And you can get a compactification by adding those ideals as new points.

You sure? Correspond both ways?

Cause when I say correspond I don't mean there's an injection points -> maximal ideals

I mean there's a bijection points -> maximal ideals

[winged badger]

Can anyone help? I can't make any progress on this. Let $B_i$ be a sequence of open $n$-balls with radii $r_i$, such that $\sum r_i=\infty$ and $\lim_{i \rightarrow \infty} r_i=0$. Prove that for every n-dimensional box Q, there is a finite disjoint sequence of balls $B'{i_1},B'{i_2},\dots,B'{i_j}$, all contained in Q, such that $B'{i_k}$ is a translation of $B_{i_k}$, and
[\sum_{k=1}^j m(B'_{i_k})>0.99m(Q)]
where $m$ is the Lebesgue measure.

[gentle osprey]

Phorphyrion

[swift fjord]

@winged badger Just FYI might be worth to post in #advanced-analysis

like additionally

you don't have to delete it

[winged badger]

Thanks for the tip

[pearl holly]

@novel acorn sorry for the ping you but I'm stuck at the proof of theorem 2.16 and I'm wondering if you already read it and understood it because we are kind of at the same place in the book.

I'm talking about Hatcher obviously

[novel acorn]

I'll look into which one it is exactly in a bit

Oh I haven't gotten to that one yet
I just started with exact sequences

[pearl holly]

oh well frick

I can't really post the paragraph that I have trouble with too because Hatcher "invests" notation in like the last few pages so I would have to post like multiple pages

yeah okay I will give it a shot!

First pic

Second pic

In the second pic where Hatcher writes ""A homology class in Ker j_* is represented by a cycle b...." how does Hatcher know that b is a cycle?

[pearl holly]

so on the fourth row after "Proof:" he writes this

Yeah but in this case we are looking at ker(j_*)/im(i_*) and I don't see why dell b = 0

Hatcher uses the fact that dell b = 0 later on in the same "section" of text

and I assume that cycle in this case means that the boundary of it is zero

But H_n(B) is defined to be ker(j_*)/im(i_*) right?

So where does dell come from?

No wait

It's not

no way

lmao

okay yeah I see. I got confused and thought that H_n(B) = ker(j_*)/im(i_*) but this is not the case lmao. Thank you so much!

[tawdry widget]

Basically at the beginning you should learn differential geometry,then Riemann metric. And in terms of differential geometry I know some easy text book such as an introduction to manifolds written by Loring W. Tu

[flint cove]

I'm sorry but that was just pure poetry 😄

[regal sonnet]

thanks for the suggestion BTW!

[pearl holly]

yeah lmao. I have realized that I really need to take at least 10 minute breaks when I study

[pearl holly]

How do I manage to come to this conclusion

[cloud plank]

hi

i dont understand coordinate rings

for example

$R[x,y]/(xy-1)$

[gentle osprey]

pewdssssssss

[cloud plank]

can we still evalute functions in here at values not in I(xy-1)

[empty grove]

You mean V(xy-1)?

[cloud plank]

sorry what does V mean

[empty grove]

You can't, because then evaluation is not well defined. For example x and 1/y are both the same in that ring, but their evaluations are clearly not the same everywhere

[cloud plank]

V is zero set of xy-1?

[empty grove]

Set of common zeroes of an ideal. I(some variety) would be the ideal of all things which are 0 on that variety

Yeah

Unless your book is using I(ideal) for set of common zeroes

[cloud plank]

no

so functions from solutions of (xy-1) to R only

?

[empty grove]

Yeah, it is the ring of polynomial functions of that type. Polynomial function meaning a function which can be written as a polynomial, but then 2 polynomials may define the same function

So for example in the above case, x and 1/y define the same function even though they are different polynomials

That's why you quotient by all polynomials that evaluate to 0. You are just saying "treat all polynomials that are the same as functions as the same thing"

[cloud plank]

yas

okayy

[pearl holly]

real topology moment

[empty grove]

geometry moment

[languid jewel]

Heyo does anyone know anything about dual polyhedra?

[cedar pebble]

sure what about them

[grave maple]

If X is a regular top. space with two elements x, y, and the principal ultrafilter u(x) of x converges to y, does that imply that u(y) converges to x?

[plain raven]

what is your definition of regular

https://topospaces.subwiki.org/wiki/Regular_space ?

[grave maple]

I'm using a filter definition: X is regular if whenever a filter F converges to x, then filter cl F converges to x. Here, cl F is the filter obtained from F by applying the closure operator cl to all its elements.

In any case, it shouldn't matter.

[grave maple]

OK, I think I see how things play out if instead of considering u(x), we consider constant sequences. For example, suppose the sequence x, x, x, ... converges to y. Then this means every open set containing y contains x. We want to show that the sequence y, y, y, ... converges to x, i.e. we need to prove that every open set containing of x also contains y. We can do this via contradiction: If there is an open set U containing x but not containing y, then by regularity I can find open sets separating x and X - U. But the open set containing X - U contains y, so it must contain x.

I guess with filters, the proof goes like this: Since u(x) -> x, y, then also cl u(x) -> x, y. By inspection, cl u(x) contains {x, y}, so it is the filter generated by {x, y}. And since u(y) is finer than cl u(x), u(y) shares the same limits as cl u(x).

Proving with filters is so slick.

[flint cove]

wait at which point did you use regularity in this proof

[grave maple]

The cl u(x) -> x, y part uses regularity.

[flint cove]

(don't have too much experience with filters I'm afraid)

A filter converges to a point x iff every nbhd of x lies in some filter element, right?

[grave maple]

Yes. Or to put it another way, F converges to x if F is finer than then neighborhood filter of x.

[flint cove]

Wait if it is finer than N(x)
But that would be the other way around, i.e. every neighborhood would have to contain a filter element, right?

[grave maple]

"finer" means its a superset.

[flint cove]

Ah, I had a question but I just confused filters and filter bases again 🙂

[grave maple]

every neighborhood would have to contain a filter element,
This is correct, actually.

Given that my filter proof did not rely on neighborhoods, it works if X is just a regular convergence space.

[flint cove]

Okay, so the clue is that we need regularity to show for every nbhdh N that „N contains an Element U in u(x) ⇒ N contains cl(U) for an U in u(x)“?

[grave maple]

For me the clue was the cl u(x) will contain {x,y}.

[swift fjord]

Quick sanity check, so the function $f$ induces a commutative diagram of chain complexes ($S_(A)$ is regarded here as a subcomplex of $S_(X)$)
$$\begin{tikzcd}
0 & {S_(A)} & {S_(X)} & {S_(X)/S_(A)} & 0 \
0 & {S_(B)} & {S_(Y)} & {S_(Y)/S_(B)} & 0
\arrow[from=1-1, to=1-2]
\arrow[hook, from=1-2, to=1-3]
\arrow[two heads, from=1-3, to=1-4]
\arrow[hook, from=2-2, to=2-3]
\arrow[two heads, from=2-3, to=2-4]
\arrow[from=2-1, to=2-2]
\arrow[from=2-4, to=2-5]
\arrow["{{f\vert_A}{#}}"', from=1-2, to=2-2]
\arrow["{f{#}}"', from=1-3, to=2-3]
\arrow["{\overline{f_{#}}}"', from=1-4, to=2-4]
\end{tikzcd}$$
Where we've seen that the rows are exact (Inclusion is injective, projection onto the quotient is surjective and their kernel and image agree), and $f_{#}$ is the induced chain map, and $\overline{f_{#}}$ is the chain map induced by the quotient, where for every $n$ we have $$\overline{f_{#}}(z + S_n(A)) = fz+S_n(B)$$
And is well defined since we have $f_{#}(S_(A)) \subseteq (S_(B))$ (Where inclusion of complexes is degree-wise of course), and by the naturality of the connecting homomorphism this descends to a commutative diagram of homology groups as in the image, where the rows are exact because of exact triangle (As each row of the original diagram is exact).

I'm being overly pedantic on purpose just to make sure i'm not missing any basic details here.

[gentle osprey]

Deception of Maximilian

[swift fjord]

Oh I forgot to say why the diagram commutes actually

Let $i,j$ be the inclusions from $S_(A), S_(B)$ to $S_(X),S_(Y)$ respectively, and $p,q$ be the respective projections onto the quotient. For the first square, we have that
$$j \circ {f\vert_A}{#}(z) = j(fz) = fz$$ (Again regarding $S(B)$ as a subcomplex of $S_(Y)$, althrough technically it is only isomorphic to a subcomplex of $S_*(Y)$. If I wanted to be extremely pedantic i'd keep the $j$ and compare the singular complexes attained as functions in the end to see that they are the same function with the same range and domain aswell).

OTOH, we have
$$f_{#}\circ i(z) = fz$$
So clearly this square commutes. For the other square we have
$$\begin{gathered}
q\circ f_{#}(z) = q(fz) = fz + S_n(B) \
\overline{f_{#}}\circ p(z) = \overline{f_{#}}(z + S_n(A)) = fz + S_n(B)
\end{gathered}$$

[gentle osprey]

Deception of Maximilian

[swift fjord]

Where Since every element of the chain complexes is free abelian and generated by singular simplices, it's sufficient to check commutativity on the generators

So regard z as a singular simplex

similarly the definition I gave for f-hat is only on the basis elements but extending by linearity gives the same thing as the induced map from the chain map

I know it's long but it's all really basic stuff

it's just a lot of stuff all at once

[jagged ocean]

not really sure what channel Lie theory should go in, but I figured I'd try here. Given a Lie group $G$ with lie algebra $\mathfrak{g}$, $x \in \mathfrak{g}$ induces a flow $\phi_t$ on $G$ and we can consider the curve $\gamma : \mathbb{R} \to G$ given by $t \mapsto \phi_t(e)$, where $e \in G$ is the identity element. Why is it true that $\gamma$ is a group homomorphism? It isn't obvious to me that $\gamma(t)\gamma(s) = \gamma(t + s)$. Certainly $\phi_t \circ \phi_s = \phi_{t +s}$, but it's not totally clear to me how the above equality follows from this fact.

[gentle osprey]

craft

[gritty widget]

$$\gamma(t+s) = \phi_{t+s}(e) = \phi_s\circ\phi_t(e) = \phi_s(\gamma(t)).$$ now show that, for each fixed $t$, the curves $s\mapsto \gamma(t)\gamma(s)$ and $s\mapsto \phi_s(\gamma(t))$ are equal. (use uniqueness of integral curves)

[gentle osprey]

Howl of the Lone Wolf

[gritty widget]

assuming by "induces a flow" you mean the flow of the left-invariant vector field on G given by pushing x around - just to be sure

@jagged ocean

[lean marten]

I believe you can show gamma is equal to $t\mapsto Exp(tX)$?

[gentle osprey]

Oatman

[lean marten]

Oh but with lowercase X

So x

[swift fjord]

Hn(f)=Hn(g) doesn't imply that f and g are chain homotopic right?

[jagged ocean]

Yes, exactly: and ah I see we’re invoking an ODE uniqueness result, I was wondering if there was a more direct way of showing it, but this makes sense I think

[gritty widget]

i don't know of a more direct proof (ie not invoking any uniqueness results) but i'd be interested to see one. for matrix groups you can directly compute using the matrix exponential series formula, but not all lie groups are matrix groups...

[fading vale]

for all n?

[swift fjord]

Yea

[fading vale]

I think it might

[swift fjord]

I thought it might but idk if I can prove it

[fading vale]

I think its only for groups

and idk how elementary the proof is

[swift fjord]

Wait maybe I can

Wdym only for groups

[fading vale]

Like I think this is a general property of taking homology of a chain complex of groups

if your chains are valued in a ring and so the chain groups form modules idk if it necessarily holds

https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence MO overflow post if you want to confirm

[swift fjord]

Ah ok

Ok yea this is beyond me at the moment

Too many terms I don't know

[fading vale]

its pure hom alg

[swift fjord]

Seems like none of the solutions are about constructing the chain homotopy explicitly either

Wonder if that's possible

Probably not

[plain raven]

freeness is important

[gusty wing]

I’m having a bit of trouble understanding what exactly a genus is

I saw it defined as the number of “holes” in a shape, but that definition started to break down when I looked at some examples of objects like cube frames

[plain raven]

luckily in your book you'll primarily deal with chain complexes of free Abelian groups, so it's fine, but in general you can have a map of chain complexes f : C -> C' that induces an isomorphism on all homology groups but fails to have a homotopy inverse

this weaker condition is called a quasi-isomorphism

[bronze lake]

another way to think of it is the maximum number of cuts along simple closed curves you can do while keeping the surface connected

[gusty wing]

Ok ok thanks guys

I just started learning it yesterday so I wanna make sure I’m not going to start off with a rocky foundation

Apparently this is genus 5

Ye

I’m just trying to come up with a more concrete definition for genus

I think I’m starting to see it though

Thanks for the help!

[gilded coyote]

Hello, can someone help me with this one?, i need to find an set X, a collection M ⊂ 2 ^X, such that M is not a basis for the topology generated by M

[hollow harbor]

Subbase moment

What if two things in M intersect to give something not in M? Can you use this idea to produce a situation where you have an open set which is not the union of any elements of M?

[grave maple]

If it's not a basis, how is M producing a topology?

And which topology?

[swift fjord]

I think by 'topology generated by a collection of sets' they mean 'smallest topology containing that collection'

In that case it doesn't even have to be a subbasis

[grave maple]

Oh right. My bad.

Let X = {0,1} and pick M = {{0}}.

[reef shore]

If you define empty intersection to be the whole set, you can define every collection of subsets to be a subbasis

[pearl holly]

it's so sad to come to this channel and see that slim has a black name

[reef shore]

How am I supposed to get DN'd now

[swift fjord]

why hasn't he been unmuted yet

[pearl holly]

slim the DN master really do be suffering from success

:dj khaled emote:

[reef shore]

Did he get muted for DNing someone "important"

[swift fjord]

metal I think

[hazy nexus]

I feel like taking $M = \R$ and $X = { (n - \varepsilon,, n+\varepsilon ) \mid n \in \N, \varepsilon > 0 }$ might work?

[gentle osprey]

Tormeson

[hazy nexus]

Proof: just look at it

[feral copper]

Hi! Probably a dumb question: in an open book of M³, why do we care that we have a fibration M─L→S¹ ? Like, the page thingy is pretty natural, but why is it important that we ask for a fibration and not a mere map?

I can't think of a non-example

(like, something that's not a fibration and we wouldn't want to call it an open book)

[pearl holly]

Sorry to interrupt the unanswered question but I'm kind of confused with what Hatcher writes here. He writes "define $S: C_n(X) \to C_n(X)$ by setting $S \sigma = \sigma_# S \Delta^n$ for a singular n-simplex $\sigma: \Delta^n \to X$. (Here, $\Delta^n$ is the standard n-simplex and $\sigma_#$ is the induced chain map from $\sigma$). I don't get what $S \Delta^n$ is supposed to mean here. I assume that this is composition, but how are you supposed to compose a simplex with $S$? If it means $S(\Delta^n)$ then this is also weird since $S$ goes from $C_n(X)$ and $\Delta^n$ is not in that set. So what does this mean?

[gentle osprey]

Tokidoki ✓

[reef shore]

Context?

What you said makes sense yeah

Also I think you might overtake me in a couple days

[pearl holly]

this is taken from the barycentric subdivision of general chains, page 122

[reef shore]

Yeah you will overtake me in 5 mins

Send screenshot lol

I stopped right after excision

[pearl holly]

ouff okay wait a moment

Okay so Hatcher writes this in the bottom

There's not a lot of context here so I just send the whole page lmao

[reef shore]

Oh I think the S on the right is the S defined previously

Inthe previous sections he defines S for the standrad simplex

Time for notation abuse