point-set-topology

[fading vale]

and if it has the HEP for that it has the HEP for everything else

[empty grove]

Interesting

Is this in Dieck?

[fading vale]

ya

chapter 5

4 and 5 are pretty pog

[empty grove]

Gonna read that once I finish Hatcher 2 months to go probably lmao

[fading vale]

its fun!

this is my first real exposure to homotopy theory

like properly not just pi_1

[empty grove]

Ye I enjoyed the first 2 chapters

But definitions too abstract

[sleek thicket]

hi moth! I was going to join the discussion but I realized my brain is fried

spent several hours doing comm alg with chmonkey

[fading vale]

valid

[sleek thicket]

we even worked together irl for part of it!!!

[fading vale]

i wanna do more comm alg but chapter 5 of atiyah macdonald sucks

[sleek thicket]

my class posted a homework problem when I thought it was done for the quarter

homeworkwise

so I was scrambling to try and finish it

and it's a bit of a doozy

[fading vale]

yea i skipped the last and this hw for diff top

its on like local lefschetz stuff and my brain is rly fried

[sleek thicket]

i just dropped my cs clsas (because I stopped doing homework)

[fading vale]

but we are doing differential forms stuff which i think i will do better

[sleek thicket]

i can relate

[fading vale]

i think im gonna miss the end of the class cause AP exams

[sleek thicket]

rip

if you want a fun commutative algebra problem to do

warning it is very hard

[fading vale]

also do u think profs are all unavailable for the summer lmao i want to email one and ask if he or anyone he knows would be willing to do an expository thingy but im a bit worried its too late

[sleek thicket]

idk, might as well ask

[fading vale]

i havent seen noetherian stuff yet at all monkaS

ya no harm

im going to tmrw

[sleek thicket]

r i p

it probably won't hurt their impression of you or anything

[fading vale]

ya

i want to do something on the relationship between galois theory and covering theory and maybe reformulating the fund. theorem of galois theory in terms of the way dieck does the classification of covers

i talked to buncho about it a bit and it seems like itd be nice cause itd tie together the topology im doing with also the comm alg

but idk if i have the prereqs

[sleek thicket]

ooh yeah

I was thinking of doing that at chicago this summer

[fading vale]

:O

[sleek thicket]

yeah!!

https://www.amazon.com/Fundamental-Cambridge-Studies-Advanced-Mathematics/dp/0521888506

[fading vale]

ahhhh

[sleek thicket]

but I have a lot of potential topics

so idk

kind of want to learn hodge stuff too

[tough imp]

Chmonkey

[fading vale]

i think ive talked about this before but dieck makes this very neat equivalence of the category of covers over B with the category of functors from the groupoid of B into Set

[sleek thicket]

chmoneky

[tough imp]

[sleek thicket]

yup

i talked to you about it

the first treatment I saw of covering spaces did it too

covering groupoid stuff

[tough imp]

Topology and groupoids?

[sleek thicket]

yup

[tough imp]

I remember hearing that book is kinda a meme on like MSE or MO

Something about it being oddly recommended a lot

[sleek thicket]

topology and groupoids?

[tough imp]

Or something like that

Yeah

[sleek thicket]

oh yeah absolutely

the dude shills so hard

yo ucan't miss it

[tough imp]

Oh lmao hahahaha

[sleek thicket]

like the author

it's hilarious

hello buncho spheres

[tough imp]

I don’t even know who this is without checking the picture

Maybe it’s the Buncho

[shut moat]

What makes covering spaces interesting in general? I saw the proof that the fundamental group of S^1 is isomorphic to Z but apparently in general fundamental groups are awful to compute

[sleek thicket]

i think it's approx

yeah

[shut moat]

and yea I'm approx

[sleek thicket]

simply connected spaces are very very nice

you can shrink anything to a point

[tough imp]

Froge

[sleek thicket]

there's geometric implications of being simply connected too

if you accept that simply connected is a strong property that you can get a lot out of

[tough imp]

I was gonna talk AG reasons connected is geometric

But not that useful IMO

[sleek thicket]

then the universal covering space gives you a simply connected space very closely connected to the original space

[tough imp]

Haha

[sleek thicket]

but that's just the universal one

more generally there's a correspondence between covering spaces and subgroups of the fundamental group

[bright acorn]

Hm Sham

[sleek thicket]

hello

👋

[bright acorn]

I was thinking about reading a bit of K-theory when I get the time

[sleek thicket]

ooh

nice

i was also thinking that

[bright acorn]

Is Atiyah a good book on that?

I mean

He sort of

[sleek thicket]

no idea

i haven't learned k theory

i just know the general story

[bright acorn]

Lmao

Yeah

But it seems like a classical book

He also is sort of one of the creators of the theory it seems

He has made some work on that

That was fairly important

So it must be a good read

I was thinking about going for Hatcher's book on that

But it seems that he didn't finish it yet

[fading vale]

i was talking to buncho about it and i was talking about how we kind of have a reversal of directions: with covers we have a coslice category because we have covering spaces over B so p: E -> B and in the galois case we have i: E/F -> K/F and he brought up that if you take spec the directions reverse and i was like

wtf

spec based now

[sleek thicket]

hahaha

spec is very based

yeah like

C is like a covering space of R

two sheeted

even though there's only one point in both

you have this nice map Spec C -> Spec R

[fading vale]

k theory is chapter 6 of fomenko and fuch

[sleek thicket]

go fomenko yourself

[fading vale]

shamrock i need you to convince me to not buy another AT book.

i own 2 already

[sleek thicket]

don't do it

tada

[fading vale]

i have a problem

[sleek thicket]

convinced

[fading vale]

that was not convincing!! your convincing skills are abysmal

[sleek thicket]

my gradescope keyboard shortcuts are broken

[fading vale]

you dumped charisma

[sleek thicket]

moth, if you buy another AT book I will poison you to death

roll for intimidation

[fading vale]

☘️

you cant poison me you will be trapped in seattle and i will be safe in the comfort of my own home

[sleek thicket]

fuck why did we give students 23 problems

this sucks

[fading vale]

just TA things?

[sleek thicket]

i was supposed to do this by yesterday night

but ive lost my willpower

[lunar gazelle]

hey guys, I'm trying to understand why the point at infty for vertical lines x = c is (0, 1, 0)...

following the process that is described in the text, letting z = 0, we obtain that x = 0, and y is undetermined.

thus the solutions are (0, y, 0). we can divide through by y since homogeneous coordinates never simultaneously vanish to obtain (0, 1, 0)

is this a correct way of reasoning about this?

[cerulean oriole]

What is "the map (1)"?

The affine -> projective thing at the top of the image?

[lunar gazelle]

this is how the book defines the map from R2 to projective plane

@cerulean oriole

[cerulean oriole]

Hmm

OK so verify that if (x,y) lies on x = c then (x, y, 1) lies on x = cz

In other words, using the map (1) to view R^2 as a subset of P^2R, we have that the affine line x=c is a subset of the projective line x = cz.
I assume "Proposition 4" tells you that the only other points of the projective line are those where z = 0.

So what points on x = cz satisfy z = 0? Clearly, those where x = z = 0 i.e. (0, y, 0) for some y. But because the coordinates are homogenous, all of those coordinates correspond to the same one point---(0, 1, 0).

So (0, 1, 0) is the only point on the projective line not in R^2 i.e. the point at infinity.

[stray bane]

I want to show that for a function $f: X\rightarrow Y$, and for a collection of subsets $\mathcal{A}$ which covers $X$ where for each $A\in \mathcal{A}$ the restriction of $f$ to $A$ is continuous. Now if we assume that for each $x\in X$ there is a neighborhood $U$ of $x$ such that $A\cap U$ is nonempty for a finite number of $A\in\mathcal{A}$, then $f$ is continuous.

I know this holds if every $A$ is open, however i am not sure how the proof changes now that they are closed.

[gentle osprey]

Frozaken

[stray bane]

My initial issue i see is that every neighborhood is open and every subset A is closed, thus we cannot say much about their intersection

[pearl holly]

So this is the exercise that I am working with: Let X be an ordered set in the order topology. Show that the closure of (a, b) is a subset of [a, b]. Under what conditions does equality hold? The closure of (a, b) is obviously a subset of [a, b] by definition since [a, b] is closed. But I don't understand how the equality doesn't always hold. If x is in [a, b] then a <= x <= b. All basis elements containing x are of the form (c, d) where c < x < d. This means that the intersection between (a, b) and (c, d) will always be non-empty, so by a theorem in my book, x must be in the closure of (a, b). There do I go wrong in this argument?

[cerulean oriole]

The intersection of (a,b) and (c,d) is (max(a,c), min(b,d)). Now c < x <= b, so c < b and similarly a < d. Thus max(a,c) < min(b,d).
But how do you know that the open interval (max(a,c), min(b,d)) is non-empty? ||Counterexample:|| ||suppose there is a 'gap' before b in the ordered set.||

[pearl holly]

So perhaps this holds if X has the same order type as the real numbers? Because then I get rid of that gap?

[empty grove]

Try to formalize that, you'll see what the exact condition is

[0,1] has different order type but has this property

[pearl holly]

Okay but really quick: if there's a "gap" before b in the ordered set, how can the intersection be empty? I was thinking that x could lie in the end point of the gap but then (c, d) would still contain a small portion of (a, b)

[empty grove]

I don't think you need to consider basis elements

(a,b) itself is a basis element

Also try to think of the "gap" as not a distance but more a property of order, because the space (0,1) union {2} seems to have a gap but is homeomorphic to (0,1]

(under order topology)

[pearl holly]

Okay thank you so much! (I am guessing that X need to be Hausdorff because Munkres has been going over it lmao)

[empty grove]

|| order topology is always Hausdorff xD ||

[pearl holly]

wait what??

Oh yeah that's right

Of course it is...

[cerulean oriole]

x at the end point of the gap will not be a problem, but x = b will.

[pearl holly]

Ohh okay I see now. Thank you so much!

[fading vale]

$\begin{tikzcd}
A \times I \ar[dd, "j \times id"] & A \ar[d, "j"] \ar[l, "i^A"'] \ar[r, "f"] & B \ar[d, "J"] \ar[r, "i^B"] & B \times I \ar[dd, "h"] \ar[ldd, "J \times id"'] \
& X \ar[r, "F"] \ar[ld, "i^X"'] & Y \ar[d, "i^Y"] \ar[rd, "\varphi"] \
X \times I \ar[rr, "F \times id"] \ar[rrr, bend right, dashed, "K"] & & Y \times I \ar[r, dashed, "H"] & Z
\end{tikzcd}$

tikzcd moment

[gentle osprey]

Moth In Shambles

[sweet wing]

lmfao

wait what is this for

ohhh

i think i got a nicer tikz

[fading vale]

its missing Y^I and X^I though!

[sweet wing]

and then later moving through homotopy theres this nice diagram

[fading vale]

cheating!

[sweet wing]

hu has in?

[fading vale]

for change of basis by a cofibration?

wait

[sweet wing]

yea rmb

[fading vale]

change of cobase by a cofibration

[sweet wing]

Hom(XxI,Z) = Hom(X,Z^I)

[fading vale]

Yeah but X x I (or X^I) and Y x I (or Y^I) arent in your diagram : p

[sweet wing]

wait you dont actly need them tho

you can define a cofibration as like

[fading vale]

well this is specifically for his proof which is taking the product of the pushout under I

[sweet wing]

lifting from h to H

ahh lul

[fading vale]

and then using HEP for j to induce a htpy from X to Z and then pushout to induce a htpy from Y to Z

[sweet wing]

if you just replace everything with exponential objects

your proof gets uglier

[pearl holly]

What is this?

[sweet wing]

but diagram gets nicer

[fading vale]

change of cobase by a cofibration

[sweet wing]

i.e. pushout of a cofib is a cofib

[fading vale]

[pearl holly]

So what is a cofib?

[fading vale]

have you seen homotopy

[sweet wing]

intuitively, homotopies lift

[pearl holly]

What is a homotopy?

[sweet wing]

i need to add quiver to my texit lmao

[pearl holly]

Lmao this is going to be a looonnggg chain

[fading vale]

i think at this point ud just have to read a textbook

[sweet wing]

basically for every map h, a map H exists such that this diagram commutes

[fading vale]

explaining it all would just be information overload and you wouldnt have any understanding of why we care by the end

[sweet wing]

tbh yea

[fading vale]

but they r based

[pearl holly]

Yeah I get that. But is this algebraic topology?

[fading vale]

ya

[sweet wing]

the tldr of why we care about cofib is in algebraic topology, we typically study things modulo homotopy

but the category of topological spaces modulo homotopy sucks

[fading vale]

so we make it better

[sweet wing]

cofib/fib helps give the category like less sucky nature

so you get a model category

[pearl holly]

Oh man... The algebraic part starts at page 318... I am on page 100

[fading vale]

i guess it doesnt so much change the category as it does provide alternatives to the usual defn of some things

which it turns out have the properties we actually want

Munkres moment

[sweet wing]

lol use a diff book for AT at least

munkres is only useful until like metricization

cuz that is kinda the main goal of munkres iirc?

is to build towards the metricization theorems

[pearl holly]

Well my goal is to understand the coffee cup-donut thing

That's all I care about

[sweet wing]

you know what is a homeomorphism rite

show they are homeomorphic

tada

but the usual animation is like an ambient isotopy

and you only care about ambient isotopy for knots in manifold stuff

[pearl holly]

But is like homeomorhisms and homotpy and blablabla an algebraic thing?

[sweet wing]

huh how late did munkres intro this lol

A and B are homeomorphic if there exists some continuous bijective map f:A\to B such that f^{-1}:B\to A is continuous as well

homotopy youll see in alg top

[pearl holly]

Okay I am almost at the continuous function section, just got to do the exercises first

[sweet wing]

ahhh

icicci

wait what were the first 100 pages lmao

oh crap munkres spent that long on set theory

i skipped all of it

[pearl holly]

It was first really formal math (in my opinion) and then open sets, closed sets, limit points and so on

yeah the exercises were like show that -(-1) is 1

[sweet wing]

nah its just zfcscrewing

lol

[pearl holly]

which is actually a good exercise in my opinion

[sweet wing]

idk how munkres consructed numbers lul

lemme see

[pearl holly]

But how much should I learn to start learning algebraic topology? What are the prerequisites?

Well he assumes that there's a set called the real numbers with the basic field axioms and order axioms and then he derives the integers from them

[sweet wing]

ohhh

cute

anyways

any intro alg top book typically just requires you to know say like

lemme see munkres

probably until like page 200 ish

chapter 5 is useful but you can just trust the theorem if you're lazy

as long as you have seen groups you're good to go

most alg top intro would introduce like

free groups generator stuff and like eventually cats and stuff

[pearl holly]

Okay I have learned some basic algebra like groups before so I just have to read asap to understand the homotopy things and all the cool stuff lmao

[sweet wing]

honestly

you just need the definition

[pearl holly]

I will trust you on this one

[sweet wing]

and maybe like what subgorups/normal subgoups are

and thats reallly it

[pearl holly]

yeah I have worked with them before

Okay that's good

[sweet wing]

ye then you're suited to start alg top once you understand what is like path connectedness or smt lol

[pearl holly]

Okay good. Thank you so much!

[sweet wing]

oh yea one thing munkres glosses over quotient spaces but they will be super important lmao

[pearl holly]

Okay I will make sure to read closely then!

[gritty widget]

Why can't operator T:X->X ,with T being isometry X inf dim, be compact?

Like I know that unit ball isn't compact in X

and isometry preserves distance

but like, is this the same argument as showing unit ball in X isn't compact?

(first thought was cts operators take noncompact sets into noncompact sets, but thats probably not true?)

(yep not true)

closure T(B) is comapct

T(B) = B?

yeah like I wasn't sure about this although felt like thatxd

but I guess yeah

[sleek thicket]

I don't see why T(B) = B here?

[gritty widget]

hence, question mark

because i don't feel like justifying that

[sleek thicket]

lol

[gritty widget]

but it'd work

if it were true

[sleek thicket]

ledog, is X just an arbitrary normed space? or is it like, a banach or hilbert space maybe?

[gritty widget]

l_p

[sleek thicket]

oh neat

[gritty widget]

Ive seen an argument online: "unit ball B is not relatively compact and is bounded. So T(B) is not relatively compact." but I don't get it.

[sleek thicket]

i just don't see why it would be true slim

[gritty widget]

T(B) \subseteq B and T^{-1}(B) \subseteq B, the latter of which implies B \subseteq T(B). isometry need not be invertible.

same, feels like it could be proper subset

[sleek thicket]

oh wait is T bijective or just injective?

[gritty widget]

but like -1 goes to 1 or sth

[sleek thicket]

yeah

I don't think it needs to be invertible

[gritty widget]

this is why i dislike math

[sleek thicket]

lol

it was very annoying

[gritty widget]

oh ok ye its injective

[sleek thicket]

switching from RG to functional analysis

[gritty widget]

wait so injective implies TB onto B

[sleek thicket]

I still don't see why this is true

let T be a shift forward by one cooridnate

T(x1,x2,...) = (0, x1, x2, ...)

this is an isometry l^p -> l^p

[gritty widget]

That's the problem I'm trying to solve actually.

I could do some spectrum shit but I noticed its an isometry

[sleek thicket]

so T(X) is some closed subspace I think

[gritty widget]

so idk

[sleek thicket]

like

is T a homeomorphism X -> T(X)?

er

it's an isometry whatever

look my point is T(X) is complete okay

and complete => T(X) is a closed subspace of X

But I think the unit ball of T(X) should be T(B)

so we'd end up with a banach space T(X) whose unit ball has compact closure

and that's an oopsy

I think this is a proof but i probably didn't explain it very well, so let me start over

let $Y = T(X)$

[gentle osprey]

shamrock

[sleek thicket]

then $Y$ is a normed vector space, thought of as a subspace of the codomain $X$

[gentle osprey]

shamrock

[sleek thicket]

and $T$ gives an isometric isomorphism $X \to Y$

[gentle osprey]

shamrock

[sleek thicket]

since $X$ is complete, the same is true of $Y$

[gentle osprey]

shamrock

[honest narwhal]

Oh what we doing here?

[sleek thicket]

a complete subspace of a metric space is closed, so $Y$ is closed in $X$

[gentle osprey]

shamrock

[sleek thicket]

the unit ball of $Y$ is $T(B)$

[gentle osprey]

shamrock

[gritty widget]

is that always true?

[sleek thicket]

and the closure of $T(B)$ in $Y$ is the closure of $T(B)$ in $X$, since $Y$ is closed. in particular it is compact

[gentle osprey]

shamrock

[sleek thicket]

I'm using the fact that T is an isometry

so if y is a point in the unit ball of Y then y = T(x) for some x, and the fact that T is an isometry forces |x| = |y| < 1

[honest narwhal]

Wait so are you guys allowing isometries to not be surjective?

[sleek thicket]

yes

anyways the point is, T(X) is an infinite dimensional banach space whose unit ball is precompact

contradiction

[gritty widget]

.

[sleek thicket]

dammit i forgot the scheme question I was going to ask here

@gritty widget, does my proof (outline) make sense to you? I'm happy to clarify

[gritty widget]

im trying to understand give me some time im slow

[sleek thicket]

it's no problem, just wanted to check in

[gritty widget]

wait so I get why it's a contradiction, but what were you assuming that was not true that lead to that contradiction?

T(B)=/=B?

[sleek thicket]

That such a T exists at all

sorry that T was compact

[sleek thicket]

the last sentence here is using that T is compact

so you can rephrase it as "if T were compact, then T(B) have compact closure, so Y would be an infinite dimensional banach space with precompact unit ball, a contradiction"

[gritty widget]

okaaay I think I get it, thanks.

who tf sullied

[sleek thicket]

me, by accident

look,,

they're adjacent

[gritty widget]

You too much.

Or 👍 too much.

[sleek thicket]

👍

[strong heron]

Both of these are polygon representations of the Klein bottle, right?

But why are these giving different homologies then? I'm getting a bit confused here.

Need help!

[fading vale]

How are you computing the homology here?

Simplicially?

[strong heron]

Using the chain complex given by these cell structures.

[fading vale]

Oh cellular

[strong heron]

It's H_1 that is the problem. In one case, we get a Z/3 summand while a Z/2 in the other.

That can be seen just by abelianising the fundamental group also.

But then these spaces must be different. And I can't see why.

[sleek thicket]

I don't think the left thing is the klein bottle

[fading vale]

Uh yeah I am not sure how you are identifying the sides here but

[sleek thicket]

Why do you think it is the klein bottle?

[strong heron]

It's Euler characteristic is 1-2+1, right? And it's non-orientable.

Ah, it has boundary components, is it?

[sleek thicket]

Hmm I agree with the euler char computation

I don't see how it could have boundary

[strong heron]

Then what's it?

[sleek thicket]

I'm not sure

I'm thinking about it

Hmm

[empty grove]

The lettering is wrong probably

[sleek thicket]

I agree with your reasoning whysee

[empty grove]

Because 3 a's and 3b's

[sleek thicket]

About why it's the klein bottle

It's nonorientable and has euler char 0

[strong heron]

I've been given this problem like this only. 😦

[empty grove]

hmm idk then I probably haven't seen this notation before

[sleek thicket]

Maybe it's something very strange, like it's not a manifold

What does a neighborhood of the vertex look like?

[strong heron]

Yeah, that seems so.

That's what I was trying to do now.

[sleek thicket]

Yeah I think there are too many directions

Near the vertex

So the problem probably has a typo

[empty grove]

The right representation also seems wrong unless you're pasting a's to b's

Because in Klein you identify opposite points on boundary of square

[sleek thicket]

The right one works

Think of it as a presentation of P^2 # P^2

[empty grove]

#?

[sleek thicket]

Well, I know it's correct because I just checked a textbook

Now I'm trying to figure out why it actually works geometrically

[sleek thicket]

Connected sum

[empty grove]

[sleek thicket]

I definitely understood this 1.5 years ago lol

[strong heron]

Yeah, right one is pretty standard

[sleek thicket]

Why is this P^2 # P^2...

So like, if you think about folding the a edge

You end up getting something sort of twisty

You glue points near the bottom left to points near the top left

Yes!

Cut along a diagonal

ty slim

Right right right

The aa thing will glue to be P^2

And then you have a hole cut out

which is the thing you glue along to get a connected sum

[empty grove]

I see

[sleek thicket]

So like the first thing to do to understand this would be to think about why <a|aa> is P^2

[empty grove]

Oh I can visualise it with the cut diagonal somewhat neat

[sleek thicket]

This is P^2

[sleek thicket]

You're like, gluing antipodal points on the boundary of a disk

Ty whysee!

The cut out diagonal corresponds to the ball you remove when you do the connected sum

[empty grove]

Idk what connected sum is

[sleek thicket]

Ah okay

It's a really cool thing you can do with manifolds

Take two manifolds M, M'

Cut out small disks

The boundaries of these new things are spheres

So you can glue M\ball to M'\ball along the sphere

[empty grove]

Way I'm thinking is when you cut along diagonal and identify a and b you get 2 mobius strips and then identifying diagonal is pasting their edges

[empty grove]

At some base point?

ohh

[sleek thicket]

Yeah!

And it's independent of the disks in dim 2

I think in higher dimensions you need to be careful about orientation or something

Don't remember off the top of my head

(small here means precompact and contained in some larger coordinate ball btw)

[empty grove]

We do it because edge sum won't be a, manifold?

Wedge*

[sleek thicket]

Yeah, that's one way to think about it

They aren't homotopy equivalent though

I thought they were when I first learned about the connected sum and got really confused

You can't like, shrink down the hole you've made to a point

[empty grove]

Right

ty

[sleek thicket]

np!

Oh and the sphere is a unit

The n sphere for n dim manifolds

you get the same space back

[empty grove]

oh that's epic

[sleek thicket]

and any compact connected surface is either the sphere, a connected sum of torii, or a connected sum of projective planes

This is another reason to care about the connected sum

if you want to classify or break up manifolds it comes up a lot

[empty grove]

Nice

[strong heron]

So that left guy is not a manifold right?

[strong heron]

In this picture

[empty grove]

Yeah if you take a point on an edge, its neighbourhood looks like a plane with another half plane attached (or 3 half planes), like the xy plane in R^3 union the yz plane but with only y positive.

[sleek thicket]

Pretend you are an ant at the vertex

[empty grove]

im thinking of cutting the hexagon like a pizza and then identifying the 'a' edges is like putting 3 of those slices together with the crust edge identified

🍕

[empty grove]

I dont get why the cyclic orientation of edges doesnt give a delta complex structure

It seems to me that all 3 conditions are satisfied

[gentle osprey]

Moldilocks

[bright acorn]

What books would you guys recommend as an introduction to homotopy theory, in particular spectral sequences?

I intend on giving it a try after finishing Hatcher

[tight agate]

May's concise course is good

for homotopy theory

for spectral sequences, there's vakil's notes, mcleary, weibel, bott and tu, gelfand and manin

hatcher

[thin bramble]

Can anyone help me with 2 and 3?

@gritty widget

[gritty widget]

i can't even open this file type on my phone

so vacuously i can solve it

[thin bramble]

Jeez, does no one know transformation geometry

Very last undergrad assignment

Worst math class ever. Is a good class but lack of outside resources

[sleek thicket]

i do, i simply choose to let you suffer

[thin bramble]

my only option is chegg

[sleek thicket]

(i do not)

[gritty widget]

have you heard of math stack exchange?

[thin bramble]

Jeez, does no one know transformation geometry. Stockexchange won't help me with this

[gritty widget]

stockexchange??

[sleek thicket]

its like bitcoin

[thin bramble]

sorry stack

[gritty widget]

anyways the chance of finding someone on MSE who knows transformation geometry is much higher than here

[thin bramble]

What about the cat? @digital peak

[sleek thicket]

@gritty widget might know?

or @sleek thicket

[thin bramble]

I've posted on StackExchange

What if Silmeveaus knows but doesn't want to help? 🤔

[sleek thicket]

is it worth burning my karma to tag tterra on your mse question

hmm

ah it looks like it's not possible anyways

[thin bramble]

So I found this on chegg

How do I make this into my words or something

[coral pivot]

what? are we really cheating at this level?

[vocal wharf]

just understand it first

[thin bramble]

What level are you in?

[sleek thicket]

https://tenor.com/view/vegeta-dragon-ball-z-unlimited-power-over9000-power-level-gif-12316102

[coral pivot]

in my experience when people get to this level, they are people who actually care about the subject and such and would never cheat. So i am surprised to see people cheat at high levels like this, i expect this from highschoolers, shame on you

[sleek thicket]

(this is my level)

john, stop being a fucking cop

[thin bramble]

I'm a college senior

never took a geometry course

This is my first ever geometry course and is the worst experience ever

Is a great class but it lacks outside resources. You don't see this in analysis or algebra

[coral pivot]

well I am tired of seeing people cheat(i am in bio so lol basically half my peers), and i do not like the comparision to cops lol

[thin bramble]

Oof, wait until john goes to college and half of the enigeering students look up on slader or chegg. Just to pass calc 1,2,3 and diff eq

[coral pivot]

I am aware, why do you think i abhor your behaviour so much

[thin bramble]

I wonder if john considers helping "cheating" @coral pivot what you think? So I can't have a tutor? Since that is consider cheating? lol

is going to office hours cheating?

damn

[sleek thicket]

lol don't try to bullshit this

using chegg is not the same as going to office hours

cheating doesn't really matter but own it

like 99% of the traffic on this server is cheating

[strong heron]

Yeah, I understand both of your arguments, @empty grove and @sleek thicket. But I'm wondering how to make this precise.

[sleek thicket]

lmao, I think calling my "pretend you are an ant" counts as an argument

what precisely are you trying to show? what was the problem given to you?

[strong heron]

Actually, I think the problem given to me in fact makes this precise

So I was required to compute cup products on this CW structure

With Z_3 coefficients

And I get that ab, aa, and bb are all 0

Which cannot happen if this were a manifold

[sleek thicket]

ah sure, did it tell you these present the same space or something?

I guess I'm still stuck on why you need to know <a,b|a^3 b^3> isn't the klein bottle/a manifold

[strong heron]

No. That was just my doubt. Because I felt it was (looks like at first glance) a Klein bottle and I thought I'd use Klein bottle's homology to save computation. But then we discovered that this is not anywhere close to a Klein bottle.

[sleek thicket]

ahhh okay, my bad

[strong heron]

All good

This was interesting

And kind of eye-opening

[strong heron]

No that's why I said that the problem was not wrong, there was a gap in my understanding somewhere..

[pearl holly]

So Munkres want's me to critize the following "proof" that $\overline{\bigcup A_{\alpha}} \subset \bigcup \overline{A}{\alpha}$: If ${A\alpha}$ is a collection of sets in $X$ and if $x \in \overline{\bigcup A_{\alpha}}$, then every neighbourhood $U$ of $x$ intersects $\bigcup A_{\alpha}$. Thus $U$ must intersect some $A_{\alpha}$, so that $x$ must belong to the closure of some $A_\alpha$. Therefore, $x \in \bigcup \overline{A}_\alpha$. I really can't find a flaw in this proof. What's funny about this is that there was a earlier exercise that wanted me to prove the reverse inclusion and I did it using this exact method but in "reverse" But I think that the reverse argument works, right?

[gentle osprey]

older sister

[pearl holly]

Oops, forgot that you can't have "emojis" in latex so now it says something about flonshed:629 lmao

[cerulean oriole]

This argument does have a flaw

In this sentence:
||Thus U must intersect some A_a, so that x must belong to the closure of some A_a.||

[pearl holly]

Okay before I look at the answer, if I were to show the reverse inclusion would the "reverse" argument work?

[cerulean oriole]

I thought by "reverse argument" you meant the above argument showing the reverse inclusion?

Oh reverse argument being for the inclusion not in the above message

Yes that works

[pearl holly]

Okay great! I will try to find the flaw! One more thing which might be hard to answer: how do you get better at solving these exercises? Is there any way I can get quicker at them?

Literally every time I get stuck at a problem you guys solve it in like 10 seconds. How do I get that superpower?

[cerulean oriole]

I wish I was more familiar with this server's emojis sigh

[cerulean oriole]

😓 I don't really have any clue
If this particular kind of "find a flaw" exercise, maybe look at each step really carefully, or try to find counterexamples for that step, or something?

[sharp yoke]

If you want a counter example i think you can just take each A_alpha to be a singleton set of a rational number and X to be R

[cerulean oriole]

As for how I "solved" this, I remember working out these inclusions before.

[pearl holly]

Oh okay! I will try to find the flaw and if I fail I will try to use your censored message and use wizards answer. Thank you so much!

[tough imp]

Often times some of us have just straight up done the exercise before

[pearl holly]

So perhaps the flaw lies in the fact that some A_a can be a singleton only containing x? Because if that's the case then x is not a limit point and hence can't be in the closure of A?

[tough imp]

Do you have an example of where it fails?

That will probably give you a good way to see where the proof falls through

[pearl holly]

So my take was wrong?

[tough imp]

Are you trying to work off of urahairywizard’s example?

Or just the general issue

[pearl holly]

No I have tried to ignore it, I will read it later if I can't figure it out. I just looked at a special case where A only is a singleton

[empty grove]

It fails in that case but it's not the only case where that inclusion fails

[tough imp]

I’d just grab an example it fails that makes geometric sense

Think about A_n = (0,1 - 1/n)

Compute both sets in that case, and see where it goes wrong and see if you can identify what’s going on

[empty grove]

Another thing is try to justify the given proof in your own words, sometimes if you're thinking of how you'll explain something to someone else you can find flaws in it easier

Just have a conversation in your head with someone who keeps asking why 🤡

[empty grove]

To make it precise, for a point on the edge you can find a neighborhood in the quotient which is homeomorphic to the xy plane and a half plane. You can do this pretty explicitly if needed by using the universal property of quotients. Union of the xy plane with a half plane cannot be homeomorphic to R^2 because if you remove a point from both, you get spaces that deformation retract to a firgure eight and a circle respectively, which you can easily prove to be non homeomorphic

[empty grove]

Reposting in case anyone can help

[pearl holly]

Ohhh so the flaw lies in the choice of U. If I pick neighbourhood, say U_1, of x then it could intersect A_1. But then a different U doesn't need to intersect the same A_1 so I don't know that x lies in the closure of some A_i, right?

[drifting sundial]

yes, that's correct

[pearl holly]

ayyyy thank you so much! (Sorry Moldi for disturbing your repost)

[drifting sundial]

there's no particular reason why some particular A_a has to be shared by every U

[empty grove]

np lol

[strong heron]

Well done, @pearl holly! Now you should be able to write down a counterexample.

[strong heron]

Well, I don't think you should say union of plane and half plane. Although I understand what you mean. Should perhaps say it's a "notebook with three sheets".

[empty grove]

Yeah I was struggling with the name lol

[strong heron]

I can't say I understand this very well either. But if you take the cyclic orientation, then the boundary formula of a simplex gets disturbed. You can see this for a triangle, for instance.

In other words, boundary formula is not cyclic.

[empty grove]

the boundary [a,b,c] = [a,b] - [a,c] + [b,c] formula?

[strong heron]

Yes yes

[empty grove]

Is that relevant to the definition? Hatcher defines that a bit later

[strong heron]

Yeah, true. But that does say that cyclic orientations aren't allowed.

Just with those three conditions, I don't think I understand it either, like I said.

[empty grove]

I see

Guess I'll read a bit further then come back and see if I can figure it out

[strong heron]

Actually, never really looked at those 3 conditions with great attention. Should do perhaps.

Yes, sure

Can anyone give me a hint on how to prove M×N is orientable if M and N are? I'm looking for an algebraic-topologic argument. Trying to think about orientation double cover of the product.

[cursive flume]

I've read that the definition of a lie group can be made stricter than it is usually given in books,however I am struggling to understand it,could someone help ?

apparently the smoothness of multiplication and the fact that G is a group implies that the inverse(which exists by group axioms) is smooth

[cerulean oriole]

owo reposting

me repost too 🥺
(Additionally assume U is a neighbourhood of E as mentioned below that message)

[strong heron]

This is true. My professor mentioned this but I haven't gotten around to thinking about it yet.

[cerulean oriole]

Hmm IDK about smoothness
But if multiplication is continuous
Then the preimage of {e} under it is closed (assuming Hausdorff which you should have for a manifold)
So {(g, g^-1) | g in G} is closed
Does that show that the inverse is continuous at least? Maybe not ...

[tough imp]

You can find a proof by Daniel Litt

He wrote a proof of this you can find on his website, it’s not very complicated, I think it’s < 1 page

[cursive flume]

i'll check it out,thanks!

[strong heron]

Should be an application of implicit/inverse function theorems, I think

[strong heron]

Thanks! I'll check it out, too.

[empty grove]

Can you use the fact that if a set S is contained in a ball B, and theres a countable collection of disjoint balls in S, then the total volume of those discount balls will be less than or equal to that of B

[cerulean oriole]

Yes

[empty grove]

Then given an open set S contained in a ball of volume V, you can find a cover of S with countably many balls in S whose total volume is less than V+epsilon for any given epsilon > 0 (the only difference from before being that S is covered now)

Assuming this is proved, you can solve the problem

Because you take a countable cover by balls of E of total volume less than epsilon, then intersect each ball with interior of U, but now they are open sets, so you cover them by balls contained in them, covering the nth open set with an error of at most epsilon/2^n

[cerulean oriole]

Giving total 2 epsilon

[empty grove]

Yeah

[cerulean oriole]

[cerulean oriole]

Why is this true though

[empty grove]

Yeah so transfinite recursion

[cerulean oriole]

That escalated quickly

[empty grove]

Pick a point in S and define B_1 to be the maximal radius ball that can be contained in S centred at that point. A maximal radius will exist because whenever you have a collection of "too large" radius values, their infimum also doesn't work because S complement is closed

[cerulean oriole]

Wait doesn't this make U interior measure-0?

[empty grove]

No they may not cover U interior

It's a cover of E

Intersected with U interior

[cerulean oriole]

But you're applying this with S = interior of U, right?

Oh

So (each ball) cap U_interior is covered

[empty grove]

Yes

[cerulean oriole]

But that doesn't mean U_interior is
Got it

[empty grove]

Yep

So now given an ordinal n, assume B_i has been defined for all i<n

If the collection of B_i closures covers S then terminate

If it doesn't, pick a point thats not in the closure of any B_i, and define B_n to be the maximal radius ball centred at this point and not intersecting the closure of the union of B_i's

And contained in S

So the idea is to almost cover S with balls, except you'd have points left out at all the boundaries of the balls

And all the B_i's are obviously disjoint (transfinite induction on their definition)

And contained in S, which is an open ball intersected with U interior

Now this is going to be a countable collection

Because each B_i contains a rational point

So you must terminate before this collection becomes uncountable

[empty grove]

And by this, the total volume of this whole collection is less than that original open ball

[cerulean oriole]

OK so you have countable collection of disjoint balls within the open set S, with volume at most V
And the closure of the balls covers S?

[empty grove]

Yes

[cerulean oriole]

From there expand radius by constant factor c such that (1 + c)^n V < V + epsilon

Wow

[empty grove]

Yess

[cerulean oriole]

Demm

[empty grove]

And you have countably many S

[cerulean oriole]

At least I don't feel bad about not coming up with this

[empty grove]

So make epsilon smaller with each S

So that the sum of those increases is less than any given positive quantity

[cursive flume]

this might be trivial but given that the multiplication is smooth,how can I conclude that the left multiplication is smooth?

one is a map $G \times G \to G$,the other is a map $G \to G$

[gentle osprey]

ProphetX

[cerulean oriole]

Thanks

[cerulean oriole]

TPT h -> gh is smooth
Consider G -> G x G -> G
h -> (g, h) -> gh
Both smooth
First one being smooth doesn't depend on the group structure or anything

[empty grove]

sniped

[cursive flume]

why is h->(g,h) smooth?

[cerulean oriole]

Smooth := for some chart around p and f(p) it's smooth in coordinates right?

[cursive flume]

if that is smooth,then i can use the theorem of composition of maps being smooth

[cerulean oriole]

Take some coordinates around h0 and g, let h = (x1,...,xn) for h near h0 and g0 = (y1,...,yn) in those coordinates
Then in coordinates the map is
(x1, ..., xn) -> (y1, ..., yn, x1, ..., xn)
which is definitely smooth

Or IG if you have the universal property of product then
G -> G: h -> g is constant hence smooth
G -> G: h -> h is identity hence smooth
So G -> G x G : h -> (g, h) is smooth

[cursive flume]

this is another proof right?

[cerulean oriole]

Yes, two proofs

[cerulean oriole]

of this

[cursive flume]

thanks now it makes sense

for the right translation I would need to prove that the map h->(h,g) is smooth right?

in a similar fashion

choosing charts

then conclude that h->(h,g)->(hg) is smooth cause composition of smooth is smooth

this might be very yikes,but I got another idea of a shorter proof(don't laugh if it's very incorrect please lol)

isn't the map h->(g,h) smooth by the definition of product atlas?

i.e. G->GxG is smooth,because we equipped GxG with product atlas

[cerulean oriole]

IG so?

[cerulean oriole]

If I'm not wrong this is the "by definition" part fleshed out

[cursive flume]

does this seem ok?

it might be trivial but want to be sure -tried writing it up

[uncut surge]

That's sooo much detail but yeah this is pretty good; if you want to go full hyper-correct, you should also check that L_{g^-1} L_g is the identity

[cursive flume]

right good point

i'll do that too

I could use a previous proposition for that

[gritty widget]

lie groups

[cursive flume]

can someone help me understand this please?

i understand it up to the point where it says 'inverse function theorem'

I tried looking up what it says,but I do not see how it is related at all to this proof

[gritty widget]

the inverse function theorem just says that if the differential of a smooth map at p is bijective, then the function itself is a diffeomorphism in a neighbourhood of p. so the inverse of the function, if it exists, is smooth at p

if the function is bijective then you can do this everywhere and get smoothness of the inverse

something like that

[cursive flume]

why does this hold in my case?

cause the identity map and left and right translation are bijective?

[gritty widget]

ya

[cursive flume]

okay, thanks!

is the proof of IFT tedious? do you have a reference for it?

wikipedia and some books state it totally differently

and only in R^n

[gritty widget]

there are many different versions

[cursive flume]

I checked in Lee and it also has different version than what you stated

but what you stated is the one I need

[gritty widget]

what's lee's version?

[cursive flume]

it has 'connected' extra

[gritty widget]

oh

well it doesn't hurt to assume connectedness

no loss of generality i guess

[cursive flume]

yes but his proof is very tedious

[gritty widget]

lol

[cursive flume]

uses theorems from metric spaces and what not

[gritty widget]

yeah the theorem is a bit tricky to prove

[cursive flume]

is there no 'simpler way' to prove this?

ah ok

[gritty widget]

not that i know of

it's not a particularly "easy" result

the trade off is that it's very powerful

[marble socket]

(isn't contraction mapping to prove implicit function first... easy enough?)

(even though the theorem may be hard, the idea is pretty neat, which makes it pretty easy to understand)

[cursive flume]

that's what lee does,but I have 0 metric spaces background,I only know how to define what a metric space is

then he uses some other inequalities

[gritty widget]

welcome to anal(ysis)

[cursive flume]

does the IFT hold aswell if the neighborhoods are not connected?

can I state it in my notes without the connected condition?

since not any neighbourhood is connected is why I am asking

or?

[gritty widget]

yes you do not need connectedness

[marble socket]

you can just shrink the nbhd until its connected...

[gritty widget]

any book on differentiable manifolds should have its own version of the inverse function theorem

pick the one you need lmao

[cursive flume]

having that f is a local diffeomorphism and globally bijective why is it a global diffeomorphism?

what does a local diffeomorphism mean? does it mean it is globally smooth and locally invertible?

or how is a local diffeomorphism defined?

so why does the IFT help me here if I know that f is invertible already?

it should give me global smoothness

[gritty widget]

f is a local diffeomorphism at p if there's a neighbourhood U of p and V of f(p) such that f restricts to a diffeomorphism of U onto V

(definition)

[gritty widget]

the point of using the inverse function theorem here is that it guarantees the inverse of f is smooth

[cursive flume]

but it only says it smooth at p

how can I conclude it is smooth globally?

[gritty widget]

do it at all p

lol

[cursive flume]

ohhhh lol

[gritty widget]

this implies that if f is invertible, then f^{-1} is a diffeomorphism of V onto U

[cursive flume]

since here it is bijective for all p

because each L,G is a diffeomorphism for all a,b

right?

so I apply first that $L_{a},R_{b}$ are diffeomorphisms for all a,b, in particular,they are bijective, therefore for all $a,b$ I have a bijective map between tangent spaces, therefore I can apply at all points $a,b$ the IFT

right?

[gentle osprey]

ProphetX

[cursive flume]

and then I can conclude that this is a global diffeomorphism

[gritty widget]

yeah, IFT implies that the function f's inverse is smooth near f(a, b), and since (a, b) was arbitrary, the function's inverse is smooth everywhere

sounds right

[pearl holly]

Something's wrong here. The closure of $\bigcup A_\alpha$ is defined to be the intersections of all closed sets that contain $\bigcup A_\alpha$. We know that $\bigcup A_\alpha \subset \bigcup \overline{A}\alpha$ and that $\bigcup \overline{A}\alpha$ is closed. This means that $\bigcup \overline{A}\alpha$ is one of the sets that I take the intersections of. This means that $\bigcup \overline{A}\alpha$ is finer than the closure of $\bigcup A_\alpha$, which is wrong. Where do I go wrong?

[gentle osprey]

older sister

[bleak helm]

Why is $\bigcup \overline{A_\alpha}$ closed?

[gentle osprey]

Lunasong the Supergay

[pearl holly]

Because the closure of a set is closed and the union of closed sets in closed, right?

[bleak helm]

Is the union of closed sets closed?

[gritty widget]

Only finite.

[pearl holly]

Oh, so if the union if infinite then it doesn't hold?

[bleak helm]

Not necessarily

[marble socket]

try to find an example

[pearl holly]

So maybe like the union of all possible closed intervals of the real line?

[gritty widget]

Nope, that would give a closed set.

[marble socket]

that would be all of R which is closed.

you want to union to be not closed.

[bleak helm]

Try to write || an open interval as a union of closed intervals ||

[marble socket]

what if ||singletons are closed and unions of closed was closed?||

[bleak helm]

Or that is even better

[pearl holly]

So like the union of all possible 1/n where n is a real number?

nevermind

[marble socket]

yep that works actually

(ofc n != 0)

[pearl holly]

Yeah that's what I thought

[bleak helm]

1/n is a number tho, not a set (go away set theorists)

[pearl holly]

Well if I just put it inside a set like this {1/n} or something similar

[bleak helm]

Yep

[pearl holly]

Okay but my argument above works only for FINITE unions right?

[bleak helm]

Yes

[gritty widget]

Also, you can look at [1/n, 1-1/n].

What Luna suggested probably in the hint before.

[bleak helm]

where n is a natural number, not a real number, presumably

This would show it also doesn't work for countable unions

[marble socket]

natural bigger than 0 😶

[gritty widget]

No, n is a natural number different than 0

[marble socket]

[bleak helm]

Alll natural numbers are different than 0 :)

[gritty widget]

sniped

[marble socket]

[gritty widget]

Blocked

[pearl holly]

except 0

[bleak helm]

This channel is unbased. I'm leaving.

[pearl holly]

Okay but thank you so much! I was really confused for a moment but you guys helped me a alot, thank you!

[bleak helm]

np np

Now you know

$0 \not\in \bN$

[gentle osprey]

Lunasong the Supergay

[marble socket]

[gritty widget]

Luna $\not \in$ friends list

[gentle osprey]

Ledog

[bleak helm]

[manic wave]

Looking at some MIT Topology courses for free from the web. I think the standard n-simplex is pretty neat. From what I read, I don't think it's too complicated to understand it.

[honest terrace]

and that's why nobody likes you

[bleak helm]

LE BULLYING FRANCAIS

[strong heron]

Reposting... need help

[sleek thicket]

What definition of orientation are you working with?

Consistent choice of generator of H_n(M, M{x})? Or do you have a smooth structure?

If so my instinct is to look at the kunneth formula

hmm this isn't quite right though is it...

[empty grove]

@cerulean oriole my proof from yesterday had too many mistakes 🤡

[gentle osprey]

Moldibee

Moldibee

[empty grove]

Actually this still has an issue 🤡 I don't know if the boundary of $\bigcup_i B_i$ is contained in $S$ (and in fact it won't be, except in some trivial cases probably)

Also I keep making the same mistake, covering the boundary of each ball is not enough as the boundary of the union may have extra points

This can be made to work if in the original set up, instead of U being a compact neighborhood of S, it's a neighborhood of S closure, but otherwise I'm a bit stuck

[cerulean oriole]

Sure why not

IDK if U compact is needed either as it was used for a different reason already

It just happened that whenever this fact was used, U was compact

I mean, at this point I might as well add the full context.

So it's a proof that a C^1 function f : W sub R^n -> R^n takes measure-0 sets E sub W to measure-0 sets f(E). By writing as a countable union of bounded sets, we can assume that E and W are bounded.
Then take a countable cover of W by open sets Ui such that closure(Ui) subset W. It's sufficient to show that f(E cap Ui) is measure-0 for every i, so assume E subset U_i.
Thus we have E subset U subset closure(U) subset W, all bounded, U, W open.

|Df| must attain a maximum and so be bounded on closure(U); hence |f(x) - f(y)| <= M |x - y| for some M, provided x, y in closure(U). Thus a ball subset closure(U) can grow in volume by atmost M^n.

So it comes down to showing that
If E measure-0 has a compact neighbourhood closure(U) then there are covers of E by balls subset closure(U) with arbitrarily small total volume.

[cerulean oriole]

The only explanation for how to do that was to "shrink the balls", which is either wrong or I don't understand at all.

[cerulean oriole]

I can probably make this happen. In the context, we have Ui (open) subset closure(Ui) (compact) subset W. By T4, we can add an open Vi in between so that
E cap Ui subset Ui subset closure(Ui) subset Vi subset closure(Vi) subset W
Then Vi covers W, and closure(E cap Ui) subset Vi subset closure(Vi) so you can take U := closure(Vi) as the compact neighbourhood of closure of S := (E cap Ui).

[empty grove]

How do you get that open cover by U_i? If E were closed you could use normality but otherwise I can't see how to ensure that their closures are in W

[empty grove]

And this certainly isn't trivial lol I found something related

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

The infinite version seems like it could be used

phew found exactly this as well

Just googled the statement of the theorem you sent

https://math.stackexchange.com/questions/696104/smooth-image-of-a-null-set-has-measure-zero

Last paragraph is using vitali's lemma

Too sleepy to bother reading the argument tho

[cerulean oriole]

For each x in W (open), find B(x, r) subset W, then take U_x = B(x, r/2). So U_x covers W and closure(U_x) subset W.
Use second-countability (specifically the consequence that it's a Lindelof space) to get a countable subcover.

[empty grove]

Right, makes sense

[strong heron]

Yes. In the topological setting only. No smooth structure.

[sleek thicket]

Yeah my instinct is to look at the (relative) kunneth formula

But I haven't really thought about it

[modest nexus]

Hi. How can one prove that the solid hiperboloid is a manifold with boundary?

[summer jolt]

Hi, I'm looking at the solution of this solution involving the killing equation

Should there be a minus sign in the second equality?

[chrome dew]

no I don't think so, because the covariant derivative of something with a covariant index puts a negative sign on the christoffel symbol

moving it over would make it positive

[gritty widget]

This seems to imply that {(x,y) \in R² : x=0 or y=0} is a 1-manifold, but that doesn't sound right

[uncut surge]

How would you cover the crossing point (so the origin) with an open set U that's homeomorphic to an open subset of R?

@gritty widget

also does anyone know what the "adjoint gruop" of an abstract Lie algebra is supposed to be? just the standard integrating simply connected one, or something like the inner automorphism group?

[gritty widget]

Cover the whole thing with {(x,0)} and {(0,y)}. Both of them are homeo to R which is open.

[uncut surge]

These sets aren't open in the subspace topology though (inherited from R^2), since there's no open set in R^2 whose intersection with the cross yields just a coordinate axis

Every open set in this cross which contains the origin must reach into all four "axes"

[gritty widget]

What if instead use a topology generated by open intervals of each axis?

[uncut surge]

Hmmm, maybe; I guess that probably still is Hausdorff

[gritty widget]

it is but hmm kinda weird, cuz every manifold can be embedded in R^n for some n iirc

[uncut surge]

Oh yeah sure but that n can be very big, so that's not a contradiction

[gritty widget]

ik just doesn't seem intuitively right

[uncut surge]

I think it's a topological 1-manifold by that definition, but it's for example not connected since you can write it as the union of the x-axis and (y-axis minus origin), which are both open

what a weird thing

In fact, maybe it's even homeomorphic to the disjoint union of a line and two rays?

[gritty widget]

maybe

[uncut surge]

Or rather, four rays and the origin? jesus

idk

[gritty widget]

then wouldn't be manifold

line and two rays i think makes sense

yea im pretty sure thats it

cuz then the rays don't care about the origin, only the line does

but then again it would have an automorphism by considering a different line which the other space doesn't seem to so idk

[uncut surge]

At least its decomposition into connected components equals the 4 rays and the origin

But I guess that doesn't mean it's homeomorphic to that...? I don't know, I'm not used to these kinds of examples lmao

I imagine that this manifold sucks in some subtle way, like not being paracompact, and hence escapes the usual classifications

[gritty widget]

maybe not cuz both axis are open by themselves while in a line disjoint 2 rays you'd get 2 rays and a single point. but actually maybe yes cuz that point alone is open cuz intersection of both axis

so maybe not what am i saying

[uncut surge]

Oh, I think I'm starting to see a problem. So when you generate your topology like that, and you try to, say, construct a homeomorphism from (-1,1) in the real numbers to the line segment (-1,1) x {0} in the cross, the identity will not be a homeomorphism

Because by generating your topology like that, you can easily show that the origin is by itself an open set (its the intersection of the two open sets given by the coordinate axes), so a homoemorphism would have to map {0} into an open subset of R -- but of course there are no open singletons in R

So even if you generate your topology like that, you still cannot create an open neighbourhood of the origin that maps homeomorphically into R

So this space is, after all, not a topological 1-manifold, even with the topology you gave it -- the best you could do is understand it as a disjoint union of a 0-manifold (the origin) and four 1-manifolds (the rays)

Lemme know if that makes sense to you! Really fun question @gritty widget

[gritty widget]

yeah that makes sense thank

@gritty widget Can I dm you?

[cursive flume]

is there an 'easy way' to see that SU(3) is simply connected without 'fibrations' or 'long exact sequences'? everything I find online uses those words and looking them up leads me to a loophole of definitions. or is this a difficult thing to show and I should just accept it for the time being?

[pearl holly]

So this is the exercise that I am working with: Show the T_1 axiom is equivalent to the condition that for each pair of points of X (say x_1, x_2), each has a neighbourhood (say U_1, U_2) not containing the other. What does the author mean by "each has a neighbourhood not containing the other", does he mean that the neighbourhoods are not comparable?

Okay thank you so much!

[rugged swan]

If X is a scheme, saying that a function vanishes at all points of X means that we take a f in Gamma(X,O_X) and f_x = 0 for all x ?
In this case doesn't it imply that f = 0 since O_X is a sheaf ? Am I crazy ? The book I'm reading wants us to show only that there's a n such that f^n = 0

Am I crazy ? This is general sheaf theory, if a global section s of F a sheaf on X is zero on every stalks, that one can define an open cover (V_x) for each x in X where s|Vx = 0, then s = 0 because F is a sheaf

[cerulean oriole]

I think it means x2 not in U1 and x1 not in U2
But OTOH, AFAIK that is the definition …

[pearl holly]

Okay thank you so much!

[strong heron]

I can do part (a). What does (b) want me to do? I feel I have to give covering spaces as quotients of the universal cover in (b), right? Not sure how to go about doing that though. Hints?

[strong heron]

That worked out, thanks. Although it isn't a very illuminating proof in the sense that it uses heavy technology. Would be happy to see something more concrete in terms of getting actual generators and verifying compatibility conditions and so on, another time maybe.

[sleek thicket]

Well the kunneth formula involves an explicit isomorphism, right ?

like you should be able to write down the generator of H_(n+m)(M×N, M×N\{(p, q)}) or whatever

I think consistency conditions should follow from some kind of naturality of the kunneth formula

[strong heron]

Oh, I didn't use that one actually. I just expressed the top homology of M \times N in terms of homologies of M and N. That Kunneth formula gives out that the top homology is Z.

[sleek thicket]

Oh

[strong heron]

Works well with the converse also.

[sleek thicket]

Wait sorry what exactly did you do?

I don't think you can just compute like H_(n+m)(M×N)

I was thinking of using kunneth locally

(that's why I clarified I meant the relative version)

[strong heron]

Well why can't you? M, N are given to be manifolds. They will have some CW structure. Use Kunneth for that.

[sleek thicket]

Sorry let me clarify

Orientable isn't the same as top homology being Z

[sleek thicket]

Which it seems like you were saying here

Take M = \R, N = \R

Then H_2(M×N) = 0

Gah I keep writing superscripts by accident

You need compactness in addition to orientability

[strong heron]

Hmm, you're right. My bad, I assumed that.

[sleek thicket]

Let me pull up Hatcher and think through my proof sketch in more detail

so from the top

let $M, N$ be oriented manifolds of dimensions $m, n$

[gentle osprey]

shamrock

[sleek thicket]

we define a pointwise orientation on $M \times N$

[gentle osprey]

shamrock

[sleek thicket]

let $(p, q) \in M \times N$ be arbitrary

[gentle osprey]

shamrock

[sleek thicket]

consider this relative kunneth map from Hatcher

with $R = \Z$, $X = M$, $A = M \setminus {p}$, $Y = N$, $B = N \setminus {q}$

[gentle osprey]

shamrock

[sleek thicket]

Since $H_i(X, A) \cong H_i(\R^m \setminus {0})$ is always flat (it is either $\Z$ or $0$) the tor groups in the short exact sequence always vanish

[gentle osprey]

shamrock

[sleek thicket]

and in fact, this holds if we take $A$ to be $M$ minus a regular coordinate ball around $p$ instead of $M$ minus $p$, this will be important later

[gentle osprey]

shamrock

[sleek thicket]

so we get this isomorphism $H_{m}(X, A) \otimes_\Z H_{n}(Y, B) \to H_{m+n}(X \times Y, (A \times Y) \cup (X \times B))$

[gentle osprey]

shamrock

[sleek thicket]

and this is defined without reference to a generator

observe that $(A \times Y) \cup (X \times B) = (M \times N) \setminus {(p,q)}$

[gentle osprey]

shamrock

[sleek thicket]

so if $o^M_p$ is the generator for $H_m(M, M \setminus {p})$ determined by the orientation on $M$ and $o^N_q$ is the generator for $H_n(N, N \setminus {q})$ determined by the orientation on $N$ we can push $o^M_p \otimes o^N_q$ along this kunneth map (which is an isomorphism) to get a generator for $H_{m+n}(M\times N, (M \times N) \setminus {(p,q)})$

[gentle osprey]

shamrock

[sleek thicket]

doing this over all points p, q gives a pointwise orientation of M\times N

to show local consistency, cover $M$ and $N$ by appropriate neighborhoods ${U_i}, {V_j}$ and look at $H_{n+m}(M \times N, (M \times N) \setminus (U_i \times V_j))$

[gentle osprey]

shamrock

[sleek thicket]

the same kunneth map will give an isomorphism from this to $H_m(M, M \setminus U_i) \otimes_\Z H_n(N, N \setminus V_j)$, and then by thinking about the specific construction of the kunneth map (some kind of naturality thing) you can show consistency of the orientaiton from the choice of neighborhoods Ui, Vj

[gentle osprey]

shamrock

[sleek thicket]

@strong heron okay here's what I was thinking when I said kunneth but written out in more detail

lmk if you need me to clarify anything

[strong heron]

Okay, thanks. I'll have to go through this line of proof slowly. I'm having a tough time with understanding orientability.

[sleek thicket]

definitely, it's a weird concept

have you seen it in the smooth world before?

it makes more sense there imo

[strong heron]

Yeah, it makes more sense with charts and stuff.

[sleek thicket]

Hmm I wonder if you could define it via charts still...

so you'd need to know what it means for a map to be orientation preserving

without reference to the derivative

but maybe you could frame that in terms of this whole local cohomology stuff

anyways yes I agree that it's confusing

I think the idea of homology = orientation makes sense for S^1 at least

the two generators are either a clockwise or a counterclockwise loop

and similarly for higher spheres, I guess

but then patching it all together locally is weird

[sleek thicket]

okay time for scheming

because I have a french exam I should be studying for and don't want to

let $X, Y$ be integral schemes and $f : X \to Y$ be a scheme map with the following properties

[gentle osprey]

shamrock

[sleek thicket]

$f$ is dominant, i.e. $\overline{f(X)} = Y$

[gentle osprey]

shamrock

[sleek thicket]

$f$ is finite type, i.e. there's an affine covering ${V_i}$ of the codomain $Y$ such that each preimage $f^{-1}(V_i)$ has a finite covering $U^i_j$ by affines where the ring map dual to $U^i_j \to U_i$ is finite type

[gentle osprey]

shamrock

[sleek thicket]

$f$ is generically finite, meaning that if $\eta$ is the generic point of $Y$ then $f^{-1}(\eta)$ is a finite set

[gentle osprey]

shamrock

[sleek thicket]

I need to show that there's a nonempty open subset V of Y such that f^{-1}(V) -> V is a finite map

first step: we can replace $Y$ with any nonempty affine open, so wlog $Y = \mathrm{Spec} A$ for a ring $A$

[gentle osprey]

shamrock

[sleek thicket]

the finite type assumption is then much simpler. there's a finite affine cover $X = U_1 \cup \ldots \cup U_n$ of the domain such that the maps $A = \Gamma(Y, \mathcal{O}_Y) \to \Gamma(X, \mathcal{O}_X) \to \Gamma(U_i, \mathcal{O}_X)$ are all finite type

[gentle osprey]

shamrock

[sleek thicket]

some useful examples to keep in mind for this problem are X = A^n \ {0}, Y = A^n. you can't just pass to an affine open of Y, the map still isn't finite in this case

so the hint says to show K(X) is a finite extension of K(Y), I did that

I'm trying to reduce to the affine case

i.e., wlog X affine as well as Y

[tough imp]

Hint time?

Also, didn’t we already say we can reduce to the affine case?

[sleek thicket]

yah and I can't make it work

[tough imp]

Remember you had a finite cover of opens

[sleek thicket]

i think you were wrong

okay so we can find finitely many $V_1, \ldots, V_n$ in $Y$ where $f^{-1}(V_i) \cap U_i \to V_i$ is finite

[gentle osprey]

shamrock

[sleek thicket]

wlog the Vi are even distinguished opens of Y

this is what the affine case gives us, yeah?

[tough imp]

Yeah

[sleek thicket]

then you said look at like $V = V_1 \cap \ldots \cap V_n$, we can assume this is a distinguished open too

[gentle osprey]

shamrock

[sleek thicket]

but I don't see why $f^{-1}(V)$ is even affine

[gentle osprey]

shamrock

[sleek thicket]

so we can say like $f^{-1}(V) = \left(\bigcap_{i=1}^n f^{-1}(V_i)\right) \cap \left(\bigcup_{j=1}^n U_j\right)$

[gentle osprey]

shamrock

[sleek thicket]

but if we like, distribute, we won't even end up with the union of f^-1(Vi) cap Ui. even if we did, that's only a union of affines in X

anyways yeah I don't see how this reduction would work

oh but @tough imp i did give up on my dream of factoring through Spec O_X(X)

[tough imp]

Rip

[sleek thicket]

I realized the map Spec O_X(X) -> Y might not be finite type

I don't have a counterexample

but it seems like you could have a finite type map which isn't finite type on the rings of global functions