point-set-topology

[gentle osprey]

goblin shamrock

[sleek thicket]

this is a homeomorphism

take $g(x) = - x$

[gentle osprey]

goblin shamrock

[sleek thicket]

then for $x \geq 0$ we have $f(g(f^{-1}(x))) = f(g(x^{1/3})) = f(-x^{1/3}) = - 2 x$

[gentle osprey]

goblin shamrock

[sleek thicket]

and for $x \leq 0$ we have $f(g(f^{-1}(x))) = f(g(2^{-1/3} x^{1/3})) = f(-2^{-1/3} x^{1/3}) = - \frac{x}{2}$

[gentle osprey]

goblin shamrock

[sleek thicket]

and so $(f \circ g \circ f^{-1})(x) = \begin{cases} -2x &\text{if } x \geq 0 \ -x/2 &\text{if } x \leq 0\end{cases}$

[gentle osprey]

goblin shamrock

[sleek thicket]

which is not smooth!

@silver umbra

holy shit i cannot hold a straight thought this was hard

we can even simplify with $f(x) = \begin{cases}
x &\text{if } x \geq 0 \ 2x &\text{if } x \leq 0\end{cases}$ lol

[gentle osprey]

goblin shamrock

[bright acorn]

Ok so

I was trying to prove something about Grassmanians over a finite dimensional vector space

And proving somehow that it is a functor would be really helpful

But idk exactly how to construct such a functor

More specifically

I would want a functor Gr(k, *) : R-Vec -> Smooth for each natural number k

which sends a vector space V to its k-th grassmanian

[sleek thicket]

yooo I fucking love grassmannians

[bright acorn]

and that induces a vector space homomorphism into a smooth math between manifolds

[sleek thicket]

it's not clear to me that you can do thus

Like independent of the rank of the linear map

bc L : V -> W might send a k dim subspace to an m < k dim subspace

In fact it might act differently depending on the subspace

Like projecting R^2 onto R

[bright acorn]

that's exactly the problem I was having

[sleek thicket]

You could restrict to injections in the domain category

Or so some kind of disjoint union of grassmannians

Although I don't think it would be continuous if you did that

[bright acorn]

I was searching this up

And it seems that we can somehow construct such a functor in the case of schemes

[sleek thicket]

hmm

[bright acorn]

In this case

[sleek thicket]

What's the reference?

like where did you see that claimed and could you link it

[bright acorn]

https://mathoverflow.net/questions/6732/what-functor-does-grassmannian-represent#:~:text=The Grassmannian G(r%2Cn,vector%20bundle%5D%2C%20mod%20isomorphism.

[sleek thicket]

Right so this is talking about a different kind of functor

If X is an object of a category C, the functor it represents is F(Y) = Hom(Y, X), F : C^op -> Set

It's common to ask what functor a scheme represents, e.g. Spec Z[x] represents the global sections functor

So this post is about what Hom_Sch(X, Gr) is for a scheme X

[bright acorn]

Oh

Well, sorry

[sleek thicket]

no you're fine!

there's no reason you should know this

[bright acorn]

Ok, I should try a different strategy then.

[sleek thicket]

Representable functors are important because of the yoneda lemma, which tells us that (1) sending an object to its represented functor is an embedding of categories C -> [C^op, Set] and (2) a functor C^op -> Set is determined by the maps from representable functors into it (there's a specific, easy formula relating them, it's not some abstract thing)

fun fact my Twitter username is @grassmannian

So feel free to ask me grassmannian questions and I will try to help as much as possible

[bright acorn]

I was trying to prove that Gr(k, n) is diffeomorphic to Gr(n-k, n) and my idea was to use the fact that there is a one to one correspondence between k-th dimensional subspaces of R^n and n-k dimensional subspaces of R^n.

This correspondence being made by the double dual

[sleek thicket]

That's exactly the right idea

er wait is it

Gr(k, n) and Gr(k, n-k)?

Not Gr(k, n) and Gr(n-k, n)?

[bright acorn]

yup

[bright acorn]

no, I mean it should be this

I made a mistake back there lmao

[sleek thicket]

Yeah so you have the right idea

Although I don't think it's the double dual

anyways, not important

So this correspondence is literally a bijection between Gr(k, n) and Gr(k, n-k)

Right?

[bright acorn]

Yes

[sleek thicket]

So you just need to check that it's a smooth function

[bright acorn]

Now I would haev to prove it is smooth

yeah

[sleek thicket]

(the inverse is also of this form so you don't need to check the inverse is smooth)

yee so like

I mean this is maybe an annoying suggestion

But check it in coordinates

How did you define the grassmannian?

[bright acorn]

Oh

[sleek thicket]

Like the topology and smooth structure i mean

[bright acorn]

We defined the topology this way, let $W \in \text{Gr}(k,n)$, then define $U_{W} = { W' \in \text{Gr}(k,n) , \vert , W' \cap W^{\perp} = 0 }$. Then I had to prove that I can define a bijection $\psi_{W} : \text{Hom}(W, W^{\perp}) \rightarrow U_{W}$ for each $W \in \text{Gr}(k,n)$ and the inverse of this bijection gives an atlas with open sets of $\mathbb{R}^{k(n-k)}$

[sleek thicket]

Ah yeah that's about what I expected

[gentle osprey]

MisterSystem

[sleek thicket]

So yeah check it on these atlases is my first instinct

[bright acorn]

aight

thanks!

Ok, sincerely I am have much trouble constructing this map psi_W

Is it possible to do without a choice of basis?

[sleek thicket]

Isn't it supposed to be to Hom(W, W')? Or something?

And W' has dim n-k maybe?

[bright acorn]

No, it really is Hom(W,W^perp) on my exercise sheet

[sleek thicket]

That's what I said

You wrote Hom(W, W^perp)

[bright acorn]

I meant perp really lmao

[sleek thicket]

I feel like the dimensions aren't right here

Er wait is this right?

Sorry yeah I think this is reasonable

[bright acorn]

I don't know really

And I can't do the rest of the exercises if I don't know if this is stated correctly

like

For instance

End(W,W^perp) makes no sense

so I corrected that

ofc that's just a silly mistake

[sleek thicket]

Yeah I noticed that

I'll check a book in a sec

[bright acorn]

but maybe there's some other stuff that makes more difference

I hope with the possible corrections this is all fine

It would be kind of nice to see a clean construction of the topology of Gr(k,n)

[sleek thicket]

Yeah this just doesn't seem right lmao

Er maybe it is

Okay so I'll tell you the map at least

Say you have a map T : W -> W^perp

Then you can associate this to the space { v + T v : v in W }

If you think of V as W × W^perp then this sends T to its graph

I was thrown off by the fact that we used W, W^perp instead of W and any other complementary space U

This is a special case of the construction I'm used to but I guess it suffices to cover the space

[bright acorn]

What is the construction you are used to btw?

[sleek thicket]

This, but with Hom(P, Q) for any P, Q subspaces of X which intersect trivially and have dim P = k, dim Q = n-k

[tropic flicker]

hello

can i ask a question or are you guys still discussing something/

[bright acorn]

Alright, I will write down the proof for both exercises now.

I think this should be it

[bright acorn]

Oh sure, you can ask.

[tropic flicker]

i'm sorry if it is a stupid question, but how can i obtain the y1 and y2?

[sleek thicket]

Don't they explain it at the end?

Formula (2.9)?

[tropic flicker]

the author says check it for yourself

[silver umbra]

haha ty for the help

[sleek thicket]

Oh so you want to know how to do the computation which gives that map

you're trying to find the intersection of a plane and a line in R^3

right?

So find a general form for the line (eg parameterize) and figure out when that satisfies the equation of the plane

[tawdry widget]

(0,0,1)+t(x^1,x^2,(x^3-1)) is contained in the plane x^3=-1

Therefore t=-2/(x^3-1)

[tropic flicker]

i'm sorry but how did you formalize that?

[sleek thicket]

do you know how to parameterize a line between two points in R^3?

[tropic flicker]

apparently no

sorry

[sleek thicket]

sure, so the idea is to think about the displacement vector between them

Say the points are p, q

Then p-q represents the vector which starts at q and points to p

if you keep repeating this vector over and over you stay on the line

[tawdry widget]

The image of (x^1,x^2,x^3) is the intersection of the plane x^3=-1 and the line containing (x^1,x^2,x^3) and (0,0,1) therefore this image can be written as (0,0,1)+t(x^1,x^2,(x^3-1))

[sleek thicket]

So in fact the line is given by γ(t) = q + t(p-q)

Start at q and follow this displacement vector p-q "t-many times" (this intuition works better if you pretend t is a whole number)

Does that make sense @tropic flicker?

[tropic flicker]

tbh i don't understand what this equation means, is it the equation of the line tangent to the sphere and intersecting the plane x3 = -1?

[sleek thicket]

You can't define the line by a single equation

[tawdry widget]

Not tangent, read my context

[sleek thicket]

A single equation will cut out a surface

and yeah as cogwheels says it won't be tangent

you want the line which goes through both the north pole and an arbitrary point on the rest of the sphere

I'm telling you how to parameterize a line through two points

In your case, p would be the north pole and q would be the other given point

do you know what a parameterization is?

[tropic flicker]

i'm sorry, but i don't understand what is meant by parameterize in this context

is there a book that i can use to help me solve this? i didn't get a good calculus course at my uni, so my math is pretty non-existant

[sleek thicket]

you want to find a formula which takes in a single number t and produces a point on the line, and which hits all the points on the line exactly once

Probably Stewart's calculus talks about lines through two points in the calc 3 section

But this is really just geometry

[tropic flicker]

so this is not calculus?

[sleek thicket]

Not imo

I think a lot of students would see this in a calc 3 class, but others would see it in precalc

[tropic flicker]

doesn't that mean it produces a line?

[sleek thicket]

that's the idea

This will let you figure out the point where the line hits the plane

Because you have a way to write down every point of the line

[tropic flicker]

aha so this is parameterization, i parametrize the line in terms of t, such that for every value of t i get a point on the line

[sleek thicket]

yes!

[tropic flicker]

and if i specify the two end points i can get the value of t that tells me the coordinates at which the line intersects the plane

[sleek thicket]

And you get all of the points of the line in this way, and you never repeat a point

[sleek thicket]

exactly!

And so what I was explaining was how to get that parameterization

[tropic flicker]

aha i understand now, thx alot , sorry for my shit math knowledge

[sleek thicket]

np

Stereographic projection is really cool geometrically

[tropic flicker]

so that is what it is called?

[sleek thicket]

Yup!

[tropic flicker]

where can i find more details on this topic?

[bright acorn]

Let
$$
\forall k \in \mathbb{N} , \text{Gr}(k,n) = { W \subset \mathbb{R}^{n} , \vert , W , \text{is a subspace of } , \mathbb{R}^{n}, , \text{dim} W = k }
$$
Define for each $W \in \text{Gr}(k,n)$ the set
$$
U_{W} := {W' \in \text{Gr}(k,n) , \vert , W' \cap W^{\perp} = 0}
$$
Then, define
$$
\begin{align*}
\psi_{W}& : \text{Hom}(W,W^{\perp}) \rightarrow U_{W} \
T& \mapsto \text{graph}(T) = {v + T(v) , \vert , v \in W}
\end{align*}
$$
We have that $\psi_{W}$ is well defined since since for each $T \in \text{Hom}(W, W^{\perp})$ we have that $\text{graph}(T)$ is a subspace of $\mathbb{R}^{n}$ and $\text{dim} (\graph(T)) = k$, moreover if $v \in W$ and $v + T(v) \in \text{graph}(T) \cap W^{\perp}$, then $\forall w \in W$ we have $\langle v + T(v), w \rangle = 0 \iff \langle v, w \rangle + \langle T(v), w \rangle = 0$.
\
\
Since $T(v) \in W^{\perp}$, we have that $\langle v,w \rangle = 0$ for all $w \in W$ which implies that $v \in W^{\perp}$, but since $W \cap W^{\perp} = 0$ it must be the case that $v = 0$ and so $\text{graph}(T) \cap W^{\perp} = 0$.
\
\
Now, notice that since $V = W \oplus W^{\perp}$ we have that if $W' \in U_{W}$, then $\forall w' \in W$ there are unique $w_{1} \in W$ and $w_{2} \in W^{\perp}$ with $w' = w_{1} + w_{2}$.
\
\
Suppose now that for $w'' \in W'$ there is also $w_{3} \in W^{\perp}$ for which $w'' = w_{1} + w_{3}$. Then, $w' - w'' = w_{2} - w_{3}$. Since $W' \cap W^{\perp} = 0$, it must be the case then that $w' = w''$ and $w_{2} = w_{3}$. This means that $\forall w \in W, \exists ! w' \in W'$ and $\exists ! T_{W'}(w) \in W^{\perp}$ where $w' = w + T_{W'}(w)$.
\
\
With this in mind, construct
$$
\varphi_{W}& : U_{W} \rightarrow \text{Hom}(W,W^{\perp})
$$
Where $ W' \mapsto T_{W'}$ and $T_{W'}(w)$ is given as previously discussed.
\
\
Then, it is easy to easy that $\psi_{W}$ and $\varphi_{W}$ are inverses and $\psi_{W}$ is a bijection.

[gentle osprey]

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

[bright acorn]

Nice

I did the details

[sleek thicket]

pog!

[amber karma]

hi folks, as a disclaimer, I'm coming from a physics background, so please excuse what may be a silly question.

if a system has order-parameter manifold described by homotopy group π₂ [S²] ≅ ℤ, does this imply, necessarily, that we can gauge the system with a U(1) field?

[tough imp]

@dull goblet

You know of topological vector spaces already which are a special case of a topological group

Think of R^n for R the reals

[dull goblet]

ok

[tough imp]

If you take the addition function where you just add vectors together, this is continuous in the Euclidean topology

So that’s a good basis to base some ideas about topological groups from

[dull goblet]

ohhh

[tough imp]

So to describe what’s going on in that image

The point is if you have a nbd of an element say g

Then the function “subtraction by g”

Is an automorphism as a topological space

Cuz it’s inverse “addition by g” is also continuous

So any nbd of g corresponds to a nbd of 0 by subtracting g

And conversely any nbd of 0 corresponds to a nbd of g by adding g and this is bijective

So it suffices to look at nbds of 0 for a lot of stuff to understand the topology

It’s like the following, the Euclidean topology on R^n is generated by open balls around points

But an open ball around a point x

Is the same thing as an open ball around the origin just like… shifted

When you add or subtract x it’s kinda like you’re just drawing the axes at a different point

[empty grove]

[tough imp]

Hi Toki 👋

[pearl holly]

Ay hello!

[tough imp]

I only now noticed you and moldilocks are #verified

✅

[empty grove]

✓

How's Hatcher going toki

[pearl holly]

Well I haven't read that much tbh, I'm on page 191 now. I will read some more later tho

[tough imp]

haven’t read much

page 191

[empty grove]

[tough imp]

Is Hatcher a quick read?

For the books I read 191 pages is like

Considerable

[empty grove]

Chmonkey what do you mean do you not read at least 300 pages of each book before deciding whether to finish it or not?

[tough imp]

Smh Moldi

[empty grove]

[tough imp]

The books I read I’d be done in 300 pages!

[pearl holly]

I meant that I haven't read a lot for the last couple of days

[empty grove]

Chmonkey reads nursery rhymes confirmed

[tough imp]

It’s true

[pearl holly]

Moldi where you at tho?

[tough imp]

What year are you in Moldi?

Inb4 2021

[empty grove]

I started 3.1

MSc 1st year 2021

[tough imp]

Oh are you in Wurope

Europe*

[empty grove]

I'm in Wndia

[tough imp]

Oh I see

Is it usual to do a Masters before PhD in Wndia as well?

[empty grove]

Everywhere except the US I think loo

[tough imp]

I’m unsure if the straight to PhD thing is an exclusively American thing

[empty grove]

l

[tough imp]

Lol

[pearl holly]

whatcu waiting on then slim?

[tough imp]

Toki is a first year right?

[empty grove]

waiting on busy

[pearl holly]

no. Read.

Now.

[empty grove]

He'll get to it once he is busy

[tough imp]

Rip

What’s chapter 3 about

[empty grove]

Comohogoly

[tough imp]

Ohhh

[pearl holly]

Comohogoly.

[tough imp]

Just learn sheaf cohomology

Use Hartshorne

Oh

Ewwww

What are you doing that requires that

O

I wonder if people have computed homotopy groups of schemes

That seems mega trash

[cedar pebble]

kinda

[tough imp]

Sheaf cohomotopy

[cedar pebble]

I mean the way to do it is with something called etale homotopy theory

[tough imp]

Okay I don’t care already

[cedar pebble]

[tough imp]

Etale homotopy theory

[cedar pebble]

a lot of examples you run into in nature are the etale homotopy version of K(π,1)'s though

which is nice

only nontrivial homotopy group is π_1 (other than π_0)

for example moduli spaces of curves are K(π,1)'s in this sense

[tough imp]

Eilenberg Maclane space yeah?

[cedar pebble]

yea

[tough imp]

BG is an example?

For 1

[cedar pebble]

yea that too

[tough imp]

Swag

I’m so goddamn good

Universal…

Something

Classifies principal G-bundles

[cedar pebble]

yea

[tough imp]

Oky nerd

You can make it via geometric realization

So it’s related to simplicial stuff

Idk if that’s actually enough to say it’s related

But you can do it that way lol

[cedar pebble]

yea there's some simplicial realization involving the bar construction for EG->BG

there's also some geometric realizations when G is a Lie group related to Steifel manifolds or whatever

[tough imp]

Wtf is EG?

[cedar pebble]

EG->BG is the universal principal G-bundle

EG is contractable

[tough imp]

Ohhhh

Okay so

Is it the universal object

[cedar pebble]

yea

[tough imp]

Makes sense

Idk why it gets an E tho

¯_(ツ)_/¯

Dude

https://alex-youcis.github.io/torsors.pdf

This made so much sense

Based Alex, nice name

[cedar pebble]

A ‘brief’ discussion of torsors
31 pages

[tough imp]

I only read section 1

Since I was short in time

But the intro explanation was like

:sippy:

Now I get it

[cedar pebble]

lol at student 20 minutes before an exam asking if we can meet and discuss material

[tough imp]

Bruh

The answer

no

[cedar pebble]

actually the better answer is to just not respond to the email

[tough imp]

What can u possibly ask in 20 minutes to change ur results on the exam

[hollow harbor]

[tough imp]

Like there’s no time to stew on it

And if it’s just a formula you can get that from the book

Oh right

[cedar pebble]

I literally met with like half of them on zoom yesterday and announced to the class that I am free to meet

[tough imp]

🤦‍♂️

[hollow harbor]

Respond 5 minutes after exam starts with "sure, let's meet!"
Then email after the exam saying "shoot, seems like my email failed to send. Sorry."

[tough imp]

Did you remember to tell them not to do it 20 mins before tho

[tough imp]

Actually psychopath moment

Just say “sorry I didn’t see”

Ryc is just an empath slim

[cedar pebble]

respond 5 minutes after exam "no"

[hollow harbor]

Damn chmonkey, lying that you didn't see it? Heartless.

[tough imp]

Words that are new (maybe) that I hate

Empath

Adulting

There are other examples that I will remember as they come up

Adulting

[hollow harbor]

Manifesting

[tough imp]

Pretending it’s an accomplishment to do mundane tasks humans have done for centuries

[cedar pebble]

I'm an empath (I assume that my guesses as to people's emotions are always correct)

[hollow harbor]

Middle america conservatives be like "Berkeley is a joke school, they have an adulting class!"

[tough imp]

Does Berkeley actually?

[hollow harbor]

We don't have an adulting class

[cedar pebble]

adulting classes are what trade school is for

[hollow harbor]

We have an adulting student-run "seminar"y thing

Which no one cares about

[tough imp]

Berkeley must be nuked from orbit

[hollow harbor]

[tough imp]

:O

[cedar pebble]

alright time to punish my students with this exam

[tough imp]

Owned

[hollow harbor]

[tough imp]

nG power trip

[hollow harbor]

Going to berkeley

Rezakhanlou? Who tf else even

[shy moss]

I think here i need to apply van Kampen theorem, but how can i choose a base-point which is in every X_n?

[pearl holly]

I don't think you need to pick a base point "immediately" here

[shy moss]

so how can i apply van kampen?

[pearl holly]

I would use induction first

[marsh forge]

As a side note though, you absolutely can choose such a basepoint if you want to

[tawdry widget]

You don’t

You just need to directly show that X is path-connected which is obvious

After that any path f from I to X for each point a from I whose image is contained in X_i you can choose an open interval of a such that the image of this interval is contained in X_i

[marsh forge]

You absolutely can use Van Kampen, fwiw

I am not sure which method is faster if you are new to van kampen

[reef shore]

isn't this basically the proof of van kampen in Hatcher

[reef shore]

and actually I am now not so sure about this are you saying that they all must intersect? I thought this should be by induction but I am getting stuck

[tawdry widget]

Those open intervals cover I and I is compact. the image of each open interval alone with two line segments connecting to the base point is contained in a convex set.

[reef shore]

seems like a counterexample to all of them intersecting

ah no the middle vertical is there in all the intersections

[tawdry widget]

This is not a counterexample

The intersection of / \ and - is empty

[reef shore]

ye ty

[shy moss]

in the case n=0, X is simply connected because every loop in X is homotopic to constant loop via linear homotopy

[reef shore]

Then X is empty anyway

[shy moss]

yes

[shy moss]

for the inductive step suposse that if X is the union of n convex sets such that the interseccion of three of them is non-empty then X is simply connected. Then suppose that X' is the union of n+1 convex sets such that the interseccion of three of them is non-empty, then X'=XUX_{n+1}, by the van Kampen theorem pi_1(X')=pi_1(X) * pi_1(X_{n+1})/N =e*e/N which is just the trival group

[reef shore]

You do have to show that X intersection X_{n+1} is path connected

to apply van kampen

[shy moss]

Suppose $X \cap X_{n+1}$ is non-empty. Let $a,b \in X \cap X_{n+1},$ then $a,b \in X$ and $a,b \in X_{n+1}$ this implies that the segment of line which joins a and b is in X and in $X_{n+1}$, then the segment of line which joins a and b is in $X \cap X_{n+1}$. Hence $X \cap X_{n+1}$ is path connected

[gentle osprey]

Or x1

[reef shore]

Are you assuming that X_n+1 is convex?

[shy moss]

yes

[reef shore]

oh wait lol

I thought it was the union of previous X_i's nvm

I meant X

Line joining a and b is in X, why?

[coarse kestrel]

wooow

[tough imp]

OWNED OWNED OWNED

[shy moss]

Suppose $X \cap X_{n+1}$ is non-empty. Let $a,b \in X \cap X_{n+1},$ then $a,b \in X$ and $a,b \in X_{n+1}$ , this implies that $a$ and $b$ are in a convex set (because X is a union of convex sets), say, $a \in X_i$ and $b \in X_j$, then there is a point $c$ which is in $X_i ,X_j$ and $X_{n+1}$ (because the intersection of any three $X's$ is non-empty), there is a line wich joins $a$ and $c$ and then $c$ and $b$ and this line is also in $X_{n+1}$ because $c$ is in $X_{n+1}$ and $X_{n+1}$ is convex

Hence $X \cap X_{n+1}$ is path-connected

[gentle osprey]

Or x1

Or x1

[empty grove]

Perfect

[pearl holly]

wait, does "dualizing" a short exact sequence preserve the exactness?

[marsh forge]

No

Hom is in general left exact

But you can do better depending on what group is the receiving one

[pearl holly]

oh lmao, I was trying to prove that and got stuck

So Hatcher writes this: the dual of a split short exact sequence is a split short exact sequence and so I thought that it does preserve exactness, so there must be something with the splitting thing I guess

[marsh forge]

Yes split is sufficient to know it is maintained

(In fact any additive functor is exact wrt split SESs)

[empty grove]

Split anything is usually preserved by functors

Because the conditions for being a split short exact sequence are just equations

And (additive*) functors preserve equations that only involve composition and addition

[pearl holly]

ahh yeah okay I see lmao

hmm what does it mean for a functor to be additive?

[marsh forge]

It’s just some axioms u can google

[empty grove]

Preserves addition of morphisms

[pearl holly]

Okay great, thank you both so much!

[plain raven]

yeah it's all about the splitting.

this is why we often do algebraic topology over abelian groups or more generally over a PID - because subgroups of free modules are free

and if most of the chain complexes you're working with are complexes of free modules

well, every surjection onto a free module splits.

so lots of your short exact sequences of chain complexes are split, and thus plays nicely with dualization

[pearl holly]

yee okay that makes sense. Thank you so much!

[plain raven]

😦

[shy moss]

if i have a quotien space $X/\sim$ and i know what is $\pi_1(X)$, how can i calculate $\pi_1(X/\sim)$?

[gentle osprey]

Or x1

[sleek thicket]

you will need more information

[shy moss]

what more i need to know?

[sleek thicket]

It's specific to X and ~

there are formulas for some very nice quotients

like if you know this quotient is the quotient by a nice group action

but in the general case you get pretty much nothing

is any space the quotient of a contractible space?

that feels vibes?

oh probably take like a disjoint union of a bunch of points and lines

and 2 cells I guess

you can squash anything that doesn't matter to a point

but otherwise you can make any K(G,1) for G generated by a number of generators and relations < the number of lines and 2 cells

[sleek thicket]

do you have a specific example in mind?

also general question, is every space a quotient of a contractible space?

maybe every CW complex?

[trail tiger]

Say I have a collection of compact sets with nonempty interiors. Loosely speaking in their union they form polytopes in R^d.

I'm shooting around for a quick, computable characterization of those compact sets who have segments of their boundaries that are also boundaries for their union. The naive approach is "for all points on the boundary of the set, draw an arbitrary ball around it, check if the ball is contained in the union of all sets" but that's not very computable at all.

[obtuse meteor]

feels like this should be true by taking cone and then identifying each slice with X x {0} and then the tip with the basepoint

kinda like just a squish kinda thing

like you squish the tip of the cone to the basepoint

and then flatten everything else out

probably true

not gonna thing about it any more than that 👍

[marsh forge]

You at least need the condition that X is connected for this to work

But ofc this can be modified to like unions of cones on connected components

I don’t think it works anyway though as this should not be continuous on say S^1 and it’s cone

For any choice of basepoint you can approach the top of the cone from the antipode of the basepoint and break the limit defn

[shy moss]

the proyective plane S^2/~

[sleek thicket]

makes sense

oh max is right

lol

[sleek thicket]

this actually is a case where you're quotienting out by a nice group action

[marsh forge]

I’m actually stumped on your Q sham

[sleek thicket]

how you would go about computing pi1 of this space depends on what techniques you know for computing a fundamental group. If you know the seifert van kampen theorem try to use that, if you know about covering spaces try to use those

[sleek thicket]

yeah it seems hard

[marsh forge]

I don’t think any tools of AT can really approach it bc like

Well the reason this all came up is that quotients are bad

[sleek thicket]

yeah haha

so like you can assume simply connected

bc i said cw complex

and so you can pass to the universal cover

but that is as much progress as I've made

how do you unquotient a space with all homotopy groups < k zero to get a space with all homotopy groups < k+1 zero

what about the sphere

like

S^2

oh this one is easy right

just cut it in half

and then glue back together

this should give a surjection disk -> sphere

Or just like disk/boundary = sphere lol

So can we somehow generalize from spheres to cw complexes?

Inductively like

Say you want to glue on a sphere

To a space X

If you can unfurl X into a contractible space

Er wait I guess you're gluing a ball anyways

[sleek thicket]

@marsh forge if you're still interested, I think this construction works: https://twitter.com/NikhilBukowski/status/1437533179217420290?s=19

Nikhil Bukowski (@NikhilBukowski)

@grassmannian @tormeson Take a graph whose vertices are closed disks corresponding to cells, with edges between any two cells that intersect. Pick a spanning tree (your CW complex is connected and thus so is your graph). For any edge in the tree, add an interval and attach the endpoints to points…

Twitter · Sep 13, 2021, 9:46 PM

(and also this is untrue in general bc path connected ≠ connected)

oh

Well that's obviously impossible

for cardinality reasons

[silver umbra]

does it suffice to prove continuity for basis elements

[sleek thicket]

yes

because preimages commute with unions

[silver umbra]

ah ok

[bright acorn]

Does anyone here have some knowledge of manifold surgery?

And some references?

[tough imp]

https://youtu.be/KNHgeykDXFw

[marsh forge]

All the surgery theory i know is baby stuff used for AT

[wary furnace]

Might be a trivial question, but how many holes does a Clebsch surface have? And how do I find it algebraically?
I know it has to do something with fundamental groups but I'm still rather new to algebraic topology

[wary furnace]

4 would be my guess too but I wanna be certain about it lol

[cedar pebble]

The Clebsch surface, as a real algebraic surface, is diffeomorphic to the connected sum W_7=(RP^2)^#7. This has homology H_0(W_7,Z)=Z and H_1(W_7,Z)=Z^6xZ/2Z, and H_n(W_7,Z)=0 for n>1. So this has 6 holes.

There are 5 isotopy classes of real smooth cubic surfaces in P^3, diffeomorphic to either W_7, W_5, W_3, W_1, or W_1 disjoint union S^2, with 27, 15, 7, 3, and 3 real lines contained in them respectively. Since the 27 lines on the Clebsch surface are all real, this determines the isotopy class.

[rancid umbra]

wot? i swear it has 4 holes. 1 on top, 3 on bottom. if it’s six then i just don’t know what a hole is fr 😓

[cedar pebble]

the 6 holes should be fairly visible in the picture

there's three on top, three on bottom

if not holes, think of loops you can draw here

[rancid umbra]

so that’s not one big hole at the top?

[cedar pebble]

I could be mistaken, there could be a hole at infinity that I'm missing and then the holes up top only contribute two

but there are three obvious loops you can draw up top and I don't think these can be deformed into one another

[rancid umbra]

but i only see one hole on top. i see three openings to the same hole on top

somebody explain to me what a hole is lmao

[cedar pebble]

in this case we're thinking of 1-dimensional holes, so they correspond to the rank of H_1 (alternatively, the rank of π_1)

[rancid umbra]

like i’m 5

[cedar pebble]

no

[rancid umbra]

huh?

what did i do wrong lol

[cedar pebble]

[rancid umbra]

oh r/woosh

it’s a reddit sub with a bunch of posts of ppl missing the joke

i missed the joke

[cedar pebble]

yes I know, stop talking like you're on reddit

[rancid umbra]

welp. this is awkward…

[cedar pebble]

idk if there is a good explanation for holes that is precise and also understandable

[rancid umbra]

aw darn

[cedar pebble]

I guess one nice way to explain it is like

the number of n-dimensional holes is the maximum number of n-dimensional curves you can remove while the space remains connected

or at least this is a definition, it's maybe not clear why this is the "correct" definition of n-dimensional holes

[gritty widget]

for an algebraic variety we can define the tangent space in the following sense

The tangent space of X at p is defined to be all v such that

for all f in the ideal generated by X, the map lambda to f(p+v lambda) has order greater than 2

(here by variety i only mean a closed subset in affine space)

i see that this definition is supposed to relate to some kind of directional derivative at p

but i don’t see it

[dim radish]

Can someone give an example of a open connected set which is not path connected

Ive proven that open connected => path connected in R^n, and i would like a counterexample or a characterization of the spaces where this implication holds

[pearl holly]

I'm looking back at this thing here and tbh I'm still confused I think. If I have a "dualized" sequence 0←A ← B ← C ← 0 that is obtained from a split short exact sequence, then how do I know that this dualized sequence is exact at B?

I know that this will be a chain complex but I can't prove the other inclusion between the kernel and the image in B

[wary furnace]

wait what was W_7?

[hollow harbor]

(RP^2)^#7

Also wait I'm having trouble seeing where the cycle is whose double cover is homologous to 0

[wary furnace]

by ^#7 you mean that it's RP^2 connect-summed 7 times?

[hollow harbor]

Yeah, that should be right

[wary furnace]

ah gotcha

damn I'm sometimes so blind

[hollow harbor]

I still have no idea why this thing is diffeo to that connect sum lol

So I'm no less blind than you

[tough imp]

So here’s a lemma you can prove

In an Abelian category (or just do it in R-mod or Ab if you like) if you have a split exact sequence then the middle term is the direct sum of the two outside ones

Then your dualized complex looks like
0 <- A <- A (+) C <- C <- 0

And well now it should be obvious why it’s exact yeah?

In fact you can show that left split <==> right split <==> this direct sum thing

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

[pearl holly]

yee right I know that. So since A (+) B would correspond to B* (+) A(*) in the dualized sequence, you get the sequence that you wrote down. But why is this exact?

[tough imp]

Oh so that’s because splitness is preserved via duals

Uhhh let me think what’s the easiest way to see that lol

Well okay so

You’re applying like

Hom(-,k)?

Is that right?

[pearl holly]

yee right

[tough imp]

Right so Hom commutes with direct sums

So

Hmmm okay this is like “here’s a fact”

But what ends up going on is like

If you trace out the maps you get that you have

0 <- B* <- B* (+) A* <- A* <- 0

And the maps here are like

The “canonical” inclusion of A* into this direct sum

And then the “canonical” projection

So like let’s see how the map A* -> A* (+) B* is formed

Well before that you actually need to look at how (A (+) B)* = A* (+) B*

So the exact nature of this morphism is that given f:A (+) B -> k you get a pair (g,h) where g is defined to be f•i_A where i_A is the canonical inclusion of A into A (+) B

And similarly h = f•i_B

So does that make sense?

[pearl holly]

ye okay but what is k here?

[tough imp]

Just whatever the base field is

Your * functor is Hom(-,k) yeah?

[pearl holly]

oh yeah right lmao

yeah okay then it does make sense

[tough imp]

Okay so now

Given f:A -> k we get this map A (+) B -> k

That’s the map A* -> A* (+) B*

So a priori we send f to f•pi_A

Where pi_A is like the projection A (+) B -> A

Now this is a map from A (+) B -> k so an element of (A (+) B)*

But now we apply the explicit iso we found earlier

So this corresponds to (f•pi_A•i_A, g•pi_A•i_B)

Okay… but f•pi_A•i_A is literally just f

And g•pi_A•i_B is 0

Maybe it’ll take a sec to verify but you should see that pi_A•i_A is identity

And pi_A•i_B = 0

[pearl holly]

yeah okay I see that

[tough imp]

Sweet so our map is

[wise sigil]

Topologist's sine curve as a space in itself is open, connected but not path connected.

One sufficient condition off the top of my head is if your space is locally path connected:
Connected & locally path connected set is path connected
Open subset of locally path connected space is locally path connected

[tough imp]

f maps to (f,0)

So this is the canonical inclusion of A* into A* (+) B* right?

[pearl holly]

yeah right

[tough imp]

So now verify in a similar way that the map A* (+) B* -> B*

Is given by (f,g) maps to g

Then it should be obvious it’s exact yeah?

[pearl holly]

oohhh yeah right okay

[tough imp]

Note you’ll need the inverse of the iso (A (+) B)* -> A* (+) B* we came up with, but this is just given by like (f,g) maps to f•pi_A + g•pi_B

This is cuz the map in our sequence a priori is actually a map (A (+) B)* -> B*

But we want a map from A* (+) B*

So we need to like precompose with the iso A* (+) B* -> (A (+) B)*

[pearl holly]

yeah okay I see I see. I didn't even try to "split" the sequence and tbh I didn't know what I was doing. But now I see it, thank you so much!

[tough imp]

This is really valuable so keep it in mind

Split exactness is preserved under all additive functors I think…

I’m not 100% sure on that one but it’s true for tensor product

And that tells you a lot of nice stuffs

[pearl holly]

yeah I think that max and moldi said this yesterday

but I don't even know what an additive functor is

[tough imp]

Oh it means like

F(A (+) B) = F(A) (+) F(B)

So it commutes with direct sums

[pearl holly]

ohh okay I see

[tough imp]

It’s like how we have that iso A* (+) B*

With (A (+) B)*

For Hom the fact they commute is well…

Basically the definition of (+) lol

It’s a coproduct so maps from A (+) B correspond to a map from A and a map from B

Which is exactly what the iso represents

[pearl holly]

yeee right I see

[tough imp]

Swag

[gritty widget]

Is anyone good with charactersitic classes?

[gritty widget]

sorry yes

Is there any way to relate an euler class to products of chern classes

I know that the euler class is the top chern class

[cedar pebble]

sort of, the Euler class is the top Chern class c_n and the Euler class of the determinant line bundle is the Chern class c_1

but there are obstructions to e.g. expressing the Chern class c_2 as an Euler class of some plane bundle

[shy moss]

is it better van kampen theorem for fundamental grupoids than vkt for fundamental group?

[cedar pebble]

there is a Van Kampen theorem for fundamental groupoids, yes

[hazy nexus]

Depends, are you Ronnie Brown?

[shy moss]

no

[cedar pebble]

it's certainly a better theorem and you can use it in more cases than you can use the Van Kampen theorem for fundamental groups

[shy moss]

if i can prove something using van kampen theorem for fundamental group, then is it possible to probe it using van kampen theorem for grupoid?

[cedar pebble]

yes, but maybe not conversely

for example I don't think you can easily compute π_1(S^1) using the Van Kampen theorem for the fundamental group

but it's definitely possible to do this using the groupoid version of the theorem

[bright acorn]

I am trying to compute some Cech cohomology groups and I am a bit stuck on this one

[sleek thicket]

I can't read, sorry for the ping @dim radish

[bright acorn]

I have a surface X endowed with the cofinite topology, and I consider C the sheaf of locally constant functions from X to C. Given that X is compact, I want to show that the Cech cohomology groups H^i(X,C) all vanish of for i > 0

[dim radish]

lol np

[bright acorn]

I am a bit stuck

Because of the generality

And all that I have in mind is try to calculate H^i(U,C) for U a finite cover of X and show all these are 0

but because of how general X and the covering can be

idk how I could approach this

[shy moss]

what is a cubical space?

[obtuse meteor]

me: thinks up a question
me: Imma ask the big math server
next second: this question is one route to the universal coefficient theorem isn't it duh

relevant question: homology answers how far off sequences are from being exact. What measures how far off homology with coefficients in A is from homology with coefficients in Z tensored with A (aka how do you measure the lack of commutativity of some functors)

torsion noise

[marsh forge]

Is there still a question

[obtuse meteor]

no

this is like, very close to what Tor does bc that measures failure to have exactness of tensor

[marsh forge]

It’s not just very close it is exactly Tor

[obtuse meteor]

ye

that's cool

[shy moss]

derived functors measure exactness of a functor

[marsh forge]

I do think it’s a bit mysterious that the Tor obstruction draws only from lower homology tho

Or at least not what I’d expect using only intuition

[pearl holly]

Okay it's getting late and I'm starting to feel tired so this might not be a good idea but I really need to ask. Hatcher wants me to prove that if A → B →C →0 is exact, then dualizing by applying Hom(-, G) yields an exact sequence A* ← B* ← C* ← 0. I know that this will be a chain complex and that the kernel of the map B* ← C* is 0. What I'm struggling to show now is that the kernel of f*: A* → B* is a subset of the image of g*: C* → B*. So if h is in this kernel then hf = 0. I must show that f can be written as ug for some u. I don't know how to go about this and I don't know how to construct such a u. Any hints?

I might just think about this tomorrow tbh

[bright acorn]

Oh, I know this result from here:

I am a bit busy now doing hw

But I will try to think about the construction and so on

Because I know this result

And maybe later on tell you something

[pearl holly]

yeah okay great! Thank you so much!

[shy moss]

you can also see corolary 1.6.9 from Weibel's book

[pearl holly]

yeah I will check that out if I really get stuck, thank you so much!

But now it's time

[marsh forge]

@pearl holly from now on you can only ask questions when you aren’t tired lol

Learn to rest

[pearl holly]

I'm sorry lmao

But sometimes I get stuck on something and then it becomes really frustrating and it feels like I really need to understand so I just go on and on until I become tired and ask in this server lmao

[marsh forge]

Yes

That is a bad habit

If anything you should stop doing math awhile before you get very tired

Unless you have a deadline

You are worse at math when you are frustrated and you get more frustrated when you are tired

[pearl holly]

yeah that's actually true

I will keep this in mind, thank you!

[gritty widget]

thanks (sorry am so slow at replying)

i’m not sure what i need then

i’m actually working in an equivariant setting

and i’m trying to get euler class for the normal bundle of some fixed point set

and the text says that

since we are on the fixed point set the action lifted to the normal bundle splits into however man non trivial representations

which is fine

and each irreducible represents is just some character j

and then it’s just stated that the euler class is the product of these j

i thought it might be something like

in this setting we can relate characters to homology in degree 2

and then we could construct the euler class from products of classes in degree two

[alpine bolt]

Do I use rank or IVT for this part?

I know since the matrix associated with f is injective, its rank is equal to m

[lunar yoke]

for the m >= n you know that Df(a) is an injective linear map, so it just follows by elementary linear algebra like you said

but i think you get that the rank of Df(a) is n, not m

since we start in R^n

[alpine bolt]

I'm going to post a similar problem

My professor wants me to use chain rule for this problem, but I don't think that's necessary. Wouldn't it suffice to use knowledge of matrices?

[lunar yoke]

how would you go about that?

I think you need the chain rule

[alpine bolt]

since we are given that the matrix is injective

[lunar yoke]

no

note that F is injective, not Df(a)

[alpine bolt]

F can be represented as a matrix tho

[lunar yoke]

no

not necessarily

its just an injective C^1 mapping

does not have to be linear

like (x,y) -> (x^3,y) if n = m = 2

[alpine bolt]

Wait

that means Df(a) in the first photo I sent may not be linear so I can't even use linear algebra for that problem lmao

[lunar yoke]

I think you need to review the definition of the total derivative

[unreal stratus]

Df(a) is linear by its very definition

[alpine bolt]

yeah youre right

derivative and integrate are vector spaces

i remember now

[unreal stratus]

wdym by saying 'integrate' is a vector space?

[alpine bolt]

ass i messed up wording a bit

[unreal stratus]

I assume you mean the operations of integration and differentiation are linear?

[alpine bolt]

yeah sorry

[unreal stratus]

That's not quite what we mean when we talk about Df(a) being linear though

differentiation is linear (as a map from functions to functions) but Df(a) is linear as a map between the underlying spaces (as in R^n and R^m)

[alpine bolt]

but since Df(a) is linear we can represent it as a matrix. In this case m x n since it starts @ R^n

[unreal stratus]

yes

[alpine bolt]

if f is injective can we say that df is also injective?

[lunar yoke]

no

take identity on R with constant derivative

[alpine bolt]

f(x)=x? Just making sure

[lunar yoke]

yep

[unreal stratus]

In that case df is injective though

It is the identity matrix/map after all

[lunar yoke]

ah right i see

i forgot the identification

whats not injective is the map x -> f'(x)

[unreal stratus]

Yeah

[lunar yoke]

wait no i think that was correct. df(x) is the identity matrix on R, identified with the scalar 1, for any x in R. But hence the map df : R -> L(R, R) = R, x -> df(x) = 1 is constant

generally Df for any linear f should be constant, which agrees with the second derivative being 0 then

(in that case Df(x) = f for all x)

[unreal stratus]

by Df(x) do you mean the derivative evaluated at x

[lunar yoke]

yeah

[unreal stratus]

That's what i mean, that is the identity and its not injective

as a map from x to Df(x) sure its not injective

[lunar yoke]

yeah by df i mean x -> df(x)

[alpine bolt]

\

This excerpt implies that there does exist a matrix for f even if it's not linear.

[lunar yoke]

yeah we just need f to be differentiable

then its derivative is by definition a linear map

but f does not have to be linear

the "rank of f" is just notation, it does not imply that f is linear with associated rank, if you mean that

[alpine bolt]

The "rank of f" means there exists a matrix representation of f, which can only be done if f is linear?

[lunar yoke]

no

its just notation

namely the rank of f at p is defined to be the rank of the linear map Df(p)

you could also call it some other word instead of rank, i.e. "the size of f at p is defined to be the rank of the linear map Df(p)"

its just a word in this case

[alpine bolt]

thank you that resolved my confusion haha. at the end of the day it really is about notation. Almost all my frustrations are about notation...

[lunar yoke]

well the notation can be really bothersome sometimes, especially in differential geometry and related topics

[alpine bolt]

is there anything I can use relating the injectivity of f to the rank of df? You can say yes or no would prefer not to give it out directly. I've tried looking up some related topics and found nothing

Using the chain rule, I was able to get the product of two matrices which is <= rank of either factor

actually wait i think i figured somehting out lol

[lunar yoke]

with what other map are you using the chain rule?

I dont think you can relate injectivity of f and rank of Df in general, but thats just my guess

note also that "rank of Df" is actually a function here, since Df(x) is a linear map, but x -> Df(x) need not be linear

so "rank of Df" would be "x -> Rank(Df(x))"

thats why in the definition they furthermore specify the "at p", which gets you a single number

[alpine bolt]

then it would be Df(p) where p is a point?

[lunar yoke]

Df(p) is still a linear map; the derivative of f at p, the best linear approximation to f at p

you can see it in your picture as well

Df(p) is the linear map $v \mapsto \frac{d}{dt}\bigg|_{t=0}f(p + tv)$

[gentle osprey]

Phil P

[alpine bolt]

All I need to do is use the fact that f is injective somewhere

I think

[lunar yoke]

well, but do you know that f^{-1} is differentiable?

[alpine bolt]

yes

both f & f^-1 are C^1

[lunar yoke]

ah ok then yes that works out

you have a C^1-Diffeomorphism

[bright acorn]

@sleek thicket Sorry for the ping, do you have some knowledge or Riemann Surfaces?

[sleek thicket]

not really no

do you have a problem on riemann surfaces you want help with? or just like asking about the general idea

[bright acorn]

I wanted a hint of how to start the proof of a problem yeah

Like, if we have a Riemann surface X and a weil divisor D of X, we can associate to it a line bundle O_X(D)

This then induces a map f : X -> Pic(X)

that sends a point p of X

to O_X(p)

And I want to prove this map is injective

the hint is to use Riemann Hurwitz

But I don't see how

btw

X is compact and has genus g > 0

[tough imp]

He won’t know this sort of thing

[bright acorn]

that's ok!

[long hornet]

Hi, I want to learn homology from Munkres' book. Now he starts with simplicial homology, and directly proves that it's invariant, then introduces singular homology and proves that it's isomorphic to simplicial homology. The latter route seems obviously better, so why does he present the former route too? Is it of independent interest?

[tough imp]

Try to compute something explicitly with singular homology

You’ll find it’s incredibly hard

[long hornet]

But I can prove that they are isomorphic first, and then just compute with simplicial homology, right?

[tough imp]

I mean you have to introduce simplicial homology to show they’re isomorphic right?

[long hornet]

Oh, yes. I mean why should I learn anything more than how to compute the homology of simplicial complexes (that is, not prove invariance directly).

[tough imp]

¯_(ツ)_/¯

I dunno enough to speak on that, I thought you were just asking why he introduces two homologies and then bothers to show they’re isomorphic lol

[long hornet]

Oh, I understand why. Thanks!

[marsh forge]

I’m not sure I understand the question

Simplicial homology is only useful for computations for sure

But it is also much easier to get a geometric intuition for homology with it

Eventually you never reference singular really

And CW is usually strictly superior imo

[long hornet]

The book proceeds as follows, roughly:

1. Introduce simplicial homology, gives some tools for computation.
2. Uses the simplicial approximation theorem to show that it is indeed an invariant.
3. Introduce singular homology.
4. Prove that it's isomorphic to simplicial homology.

So why shouldn't I just skip 2?

[long hornet]

In what sense?

[marsh forge]

easier to compute

you can indeed skip two if you like

in many books you can skip material or even reorder it

[long hornet]

Yes, and he said so.

But I thought maybe it will be useful somehow

Sufficiently useful to warrant the effort*

[marsh forge]

I almost always listen to the author when they something can be skipped, especially for a first pass

[long hornet]

Fair enough

I am just not sure what's the best thing to do

[marsh forge]

ive never thought about that material ever

[plain raven]

hatcher's construction of CW homology is so tedious and difficult to follow
later on I was reading a paper on spectral sequences and they showed that CW homology just pops right out of the spectral sequence for singular homology given by filtering the complex along its skeleton and it blew my mind

almost made me think that spectral sequences should be introduced like, way, way earlier in alg top/homological algebra, if only for that result

[marsh forge]

I think its fine to just like, construct it and assert that it is the same as singular

give the defn of the boundary map

and refer students to where to find the construction

(i def don't think we should introduce spectral sequences before like early grad/late late undergrad material tho)

Maybe a second course in AT in grad school

[empty grove]

Where can I read about spectral sequences

[plain raven]

hahaha yeah i didn't mean like.... junior year of undergrad haha

[marsh forge]

no source on SS is good imo

[plain raven]

later than that

[marsh forge]

There are some lecture notes scattered about

hatcher has a thing on them

i didnt like it

the issue here

pedagogically

is one should absolutely ignore the proof of existence for most spectral sequences

until after you already know how to use them

it is completely irrelevant do actually doing computations

My honest advice would be to like find a source you like

and message here if (when) you get confused

[plain raven]

Weibel chapter 5 is a decent intro because he starts off with some black boxed applications to show you how to use it
Godement's book on sheaf cohomology is an all time fav of mine, lots of applications of spectral sequences in there
Chapter 15 of Switzer's book on algebraic topology is dedicated to (applications of) spectral sequences, and it's pretty good

[empty grove]

Weibel homoalg?

[marsh forge]

I honestly think you should just start w the Serre Spectral sequence

before seeing it from a non-topologicla pov

[plain raven]

absolutely do not spend time trying to learn the necessary and sufficient conditions for the gadgets to converge

stick to first quadrant spectral sequences only

assume everything converges and works out nicely

[plain raven]

yes

[empty grove]

Thanks, I'll take a look

[marsh forge]

anyway i love spectral sequences so im always happy to talk about them if you ping

[plain raven]

someone was telling me (Maybe @honest narwhal ?) that they had a homological algebra professor who just explained everything in the class in terms of spectral sequences.

Five lemma? easy spectral sequence argument.
Snake lemma? apply the obvious spectral sequence argument

[marsh forge]

lolwhat

[plain raven]

i also a fan of spectral sequences
when i do a calculation with them successfully and get a useful/interesting result it makes me feel like a god

i think i've said this before but you know how there's a Cech-de Rham spectral sequence?
i figured out a Cech-Cech spectral sequence lol

like, if you take two open covers U and V it shows you how the associated Cech complexes relate to the cech complex determined by the cover consisting of pairwise intersections U_i\cap V_j

and if you fix U and let V vary across like, arbitrary refinements of U, this gives you a nice tool for figuring out when the cech cohomology on the open cover U agrees with the colimiting cohomology

[honest narwhal]

Rohit Nagpal @marsh forge and @plain raven

[bright acorn]

Does the one point compaction of a topological space have some interesting universal property?

More specifically

I have a map f : C -> X

And I want to lift it to a map f : P^1 -> X

I am thinking if maybe f has some properties

I could do this lift

More specifically, I want to know if f : C -> X is a holomorphic map between Riemann Surfaces then, under suitable conditions, there exists a f* : P^1 -> X holomorphic with ι : C -> P^1 being the inclusion and f*(ι(x)) = f(x) for every x in C.

[long hornet]

Call a sequence divergent if every compact set contains only finitely many terms of the sequence. Then I think that's equivalent to this sequence converging in X*? (X* is the one point compactification)

Maybe then the maps that can be lifted are those which interact with divergent sequence in a nice way

[tight agate]

gelfand and manin

weibel is also alright

[empty grove]

Will check

[tight agate]

more than reading bout them , it might be better to work out the details on your own

like the spectral sequence of a filtered complex

imagine you have a chain complex

[empty grove]

Will try once I know what they are

[tight agate]

and a 2-step filtration

[empty grove]

You've lost me

[tight agate]

so essentially subcomplexes C_0 \subset C_1 \subset C

choose better notation

but try to relate the cohomology of C

and the cohomology of the quotient complexes you get from the filtration

it is worth working this out in all gory detail

and then try to generalize for longer filtrations

some books introduce spectral sequences through exact couples

and I feel this is a horrible way to do it

it hides everything

moldi try doing it for stupid filtrations

like if you have a complex C

then you get subcomplexes C_i where you set everything above (or below) i to be 0

C is clearly the union of these

[empty grove]

Wait so a filtration is literally just ascending chain of subobjects?

[tight agate]

yeah

[empty grove]

[tight agate]

like a filtered module

yeah in hindsight that doesnt help lol

[empty grove]

[tight agate]

but yeah read gelfand and manin

[empty grove]

Will try it, thank

[tight agate]

but work out the details on your own

there are exercises which will make you do it the exact couple way

there's another exercise which makes you find a spectral sequence associated to a postnikov tower (they explain what this is)

great book

if youre familiar with the stuff in Weibel, the Hochschild-Serre SS is a good example to work with

or if you know sheaf cohomology, then leray

[empty grove]

Haven't read Weibel past page 10

[tight agate]

oh learn group cohomology

it's very cool

[empty grove]

Will do

[tight agate]

homological algebra is cool

it's a miracle that it works

like it's pure wizardry

[empty grove]

[tight agate]

[true garden]

Hello. When is the union of locally connected spaces a locally connected space?

[reef shore]

union inside some ambient space or disjoint union?

[true garden]

Arbitrary union

My question is unclear?

[reef shore]

Yeah, like are these spaces subspaces of some larger space containing them?

[true garden]

So, when talking about the union of connected spaces, why do we not ask such questions?

[reef shore]

we do

union alone doesn't make sense

because you don't say what the topology is then

union is a set theoretic operation

When we say disjoint union it usually refers to topological sum/coproduct which is a top space operation

[true garden]

For example, we have the following theorem:

The union of connected spaces with a point in common is a connected space.

[gritty widget]

Does anyone have reference for classifying spaces of finite groups?

[reef shore]

Yes this is happening in a larger space

that's why you can talk about points in common

So for locally connected spaces, I imagine openness of each space should be enough

[true garden]

Why?

They should not have a point in common?

[reef shore]

because then given x in the union and an open set U, x is contained in one of those spaces X, and by openness of X, U cap X is open, and by local connectedness of X, U cap X contains a connected neighbourhood of x (open in X) but open in an open subspace → open in the whole space

[reef shore]

not needed

[gritty widget]

I think you need to try draw some pictures to think about this

just take two intervals in R

that don't have any points in common

open intervals

and its obvious that you dont need a point in common

but locally connected is a weird property, you should google for or check out counter examples in topology, for the comb space

you can be connected by not locally connected

[true garden]

Ok. How about the product of them?

[gritty widget]

these are good questions

but they are good for you to check them out yourself

when you think of a question like this try draw some examples yourself

and ask your lecturer

[true garden]

I do not have any lecturer.

[empty grove]

You can try solving these on your own, they are direct applications of definitions

[true garden]

The product of them is locally connected?

[empty grove]

Yeah

[true garden]

Why?

[empty grove]

Take a point in the product and an open neighborhood

You have to show that this contains a connected neighborhood of the point

Try

[true garden]

I think the answer is not true in general.

You may not be able to find a connected neighborhood.

Some people may not be smart enough to answer math questions independently.

However, thank you all for your helps.

[long hornet]

Is there a simple criterion for when a compact subset A of R^n with nonempty interior is homeomorphic to the n-ball D^n? For example, I know that being convex is enough.

When n = 2, that's the Jordan curve theorem, I guess? So maybe there is none

Yeah

I just checked. For n = 2, a criterion follows from the Jordan-Schönflies theorem: We already want A to be connected and has boundary homeomorphic to S^1, and the theorem says the interior is homeomorphic to the open disc.

[gritty widget]

Also, I hope this didn’t come across cheeky

Just that the value in these small questions is figuring out the correct statement yourself

You’ll probably never use the statement “a union of open locally connected spaces is locally connected”

But trying to figure out the precise statement makes you check things like, when does taking unions even make sense topologically

Can anyone help me with this? If G is a finite group then H^i(BG:Z) are torsion

[tight agate]

Do you know the fadeev-shapiro lemma

restriction (to the trivial subgroup) composed with corestriction is multiplication by the order of the group

it is also the zero map (as it factors through 0)

so every element of H^i(BG, Z) is killed by |G|

[gritty widget]

no, im just reading in a paper and its stated that the borel cohomology of finite groups is torsion

[tight agate]

that's the argument

https://en.wikipedia.org/wiki/Shapiro's_lemma

[gritty widget]

okay thanks, ill read this and maybe ask you a question after parsing it if thats okay?

[tight agate]

it's a pretty neat result

[gritty widget]

okay lmao i need to ask a question early

I thought BG need not be a group?

[tight agate]

the singular cohomology of BG is the group cohomology of G

[gritty widget]

oh

so for finite groups

the G-equivariant cohomology of point, is the same as the group cohomology for some representation N?

[tight agate]

the representation is the trivial rep

Z with the trivial G action

[gritty widget]

because we are taking coefficents in Z in the equivariant case

[tight agate]

yes

group cohomology was pretty much invented to compute the singular cohomology of BG

and the singular cohomology of BG (w/ Z coeffs) is the equivariant cohomology of a point

[gritty widget]

okay so now i need to see why Hi(G:N) is trivial where N is this trivial representation

[tight agate]

it is not

[gritty widget]

oh its

we take the trivil subgroup

[tight agate]

it can be nontrivial

yes

[gritty widget]

H(0,N) is trivial

[tight agate]

yes

[gritty widget]

and then this lemma related H(G;N^G) to this

oh sorry sorry i am gone very wrong

we are trying to show its torsion

not trivial lmao