point-set-topology

[hollow harbor]

so that would be like V'

and then i want to think about the complement of V'

since it's an open set, the complement is a closed set

hmm

[limpid leaf]

hm

[hollow harbor]

i don't know of that helps

oh

you get neighborhoods of each

ok

so i have x, and i have cl V

and i get U'' around x, and V' around cl V

right?

[limpid leaf]

yeah

thats where we are now

they have empty intersect yeah

[hollow harbor]

ok so V' complement contains U''

since V' and U'' are disjoint

[limpid leaf]

hm

yeah

[hollow harbor]

this is easier to see if you draw it

[limpid leaf]

yes

[hollow harbor]

it's like we have a point and a blob, there are circles around both, and if you look at everything outside the circle around the blob, that includes everything inside the circle around the point.

(as well as everything outside both circles)

[limpid leaf]

yes right

this is fact i am familiar with about empty intersections as well

[hollow harbor]

alright but V' complement is closed

[limpid leaf]

right

[hollow harbor]

and we have a subset of a closed set

so what can you say about the closure of that subset?

[limpid leaf]

must be the set, right?

er. contained in

[hollow harbor]

yep!

if you have A inside B, and B is closed, then closure(A) is also inside B.

so now we can boost this to saying closure(U'') is inside V' complement.

[limpid leaf]

yes

[hollow harbor]

(if you want to think about this intuitively: we have 2 disjoint open sets, and if i take the closure of one of them, it can't contain points from the other. if it did, then whatever point we added couldn't have possibly been in the interior of its open set)

[limpid leaf]

right

yeah i have a picture in my head for this part

[hollow harbor]

ok, now we know closure(U'') is disjoint from V'.

but cl(V) was inside V'.

[limpid leaf]

right, but this doesn't prove the result unless we can show cl U subset cl U''?

[digital wraith]

Are two arbitrary simply connected spaces not always homotopy equivalent?

Like am I just dumb

[hollow harbor]

don't you just need to show that there are some open neighborhoods where this is true?

it doesn't matter if it's true for U

[limpid leaf]

maybe im misreading the question

ok if thats the case then I solved this yesterday i think

[hollow harbor]

"we can find a pair of open neighborhoods U and V"

[limpid leaf]

right

our pair is U'' and V'

[hollow harbor]

you just need to find 2 neighborhoods

they don't have to be the ones you get from regularity or whatever

i mean

idk

like

[limpid leaf]

i figured it was those sets

[hollow harbor]

this could very easily not be true for certain pairs of open neighborhoods

(for example, (-1, 0) and (0, 1) in R)

[hollow harbor]

well that's good

[limpid leaf]

right

not good that i misraed question

[hollow harbor]

are they connected too?

[limpid leaf]

okay well this still was helpful thx ryc

[hollow harbor]

also like, it's not true anyway

S^2 and S^3 are not homotopy equivalent

but they are both simply connected

[digital wraith]

Simply connected yes

[hollow harbor]

pretty sure simply connected does not preclude connected

i mean

maybe it does

i've never seen that as part of the definition, just "every path contracts to a point"

[digital wraith]

Topologists name thing well challenge

I mean path connected implies connected right

[hollow harbor]

sure, but just because every path contracts to a point doesn't mean any two points are connected to a path

[digital wraith]

Oh right yeah

[hollow harbor]

also what i mean to say is every loop contracts to a point

[digital wraith]

No I see

[hollow harbor]

anyway S^2 and S^3 are a counterexample to what you say (why are these homotopy inequivalent? i don't think there's an easy way to see it besides computing homology)

[digital wraith]

So any two simply connected spaces with the same number of path components should he homotply equivalent

[hollow harbor]

no

[digital wraith]

Or wait I'm dumb

[hollow harbor]

it's just not true

[digital wraith]

Actually just stupid

[hollow harbor]

i mean it's easy to think this

once you realize it's not true though

it starts to make the whole higher homotopy thing sound a lot less silly

[digital wraith]

S^n is just the obvious counterexample

[hollow harbor]

it's definitely not obvious that S^n and S^m are not homotopy equivalent unless you've done some work

[digital wraith]

Is it not obvious?

I think that means my intuition is bad

[hollow harbor]

you can't visualize S^3?

it's just R^3 but every direction takes you to the same point at infinity...

anyway this doesn't tell me why they're homotopy inequivalent

i can only see it by thinking about how spheres contract to a point on S^3

but not always on S^2

sorry i mean S^2s

like

you can kind of convince yourself that pi_2(S^3) = 0

wat

the point is that S^2s dont contract on S^2

so pi_2(S^2) != pi_2(S^3)

this is something you can see sort of

if you have the right picture of S^3 in mind

yeah

ok for S^3

pick a point not on the S^2

that you've embedded

and then stereographically project to R^3 about that point

and now contract the S^2 to the origin

and undo the projection

that's how you see pi_2(S^3) = 0

it's still a 2-sphere

how is a loop contractible?

i don't mean contractible as a topological space

i mean homotopic to 0

clearly because no one is reading this shit

it's fine

[wise sigil]

real line and S^2 should be easier to visualize

because S^2 isnt contractible

[hollow harbor]

true, true

much nicer

[coral pawn]

Can someone verify this?

[mossy ermine]

this is not true

[limpid leaf]

[mossy ermine]

unless regular means regular hausdorf

[limpid leaf]

Yes

[mossy ermine]

ok then its true

[limpid leaf]

xd

[mossy ermine]

but fuck ur definition of regular

[empty grove]

Regular usually includes T_1

[limpid leaf]

i think we just use Munkres?

^ this is our lecture notes, not munkres

[empty grove]

If you don't include T_1 then you don't get T_(n+1) stronger than T_n

[mossy ermine]

definitions are not supposed to make sense

[empty grove]

Did they say call a point and a set disjoint

[mossy ermine]

also imagine having closed points

[empty grove]

Sounds like a cope ngl

[limpid leaf]

u get what he means Moldi

[mossy ermine]

nooooooooo my primes are points and must be closed

[limpid leaf]

closed points is good

i think

idk i dont know what im doing really

[empty grove]

[limpid leaf]

just finished the assignment tho 😎

[coral pivot]

My book is defining totally disconnected groups to be groups who have a neighborhood basis of open subgroups around the identity. I cannot figure out how this is equivalent to the usual definition (no nontrivial connected sets). any hints?

[strong siren]

Any feedback on this proof?

[empty grove]

Open subgroups are closed too, so you have a neighbourhood basis of clopens. By taking all translates you get a basis of clopens. This along with Hausdorff gives the standard definition (without hausdorff the above definition allows the indiscrete topology to be totally disconnected I think?)

Replace hausdorff with T_1

[coral pivot]

Right makes sense (yeah its assumed all groups are haussdorf, which is equivalent to T_1 for groups)

what about the converse? thats what i was stuck thinking about

[honest narwhal]

@coral pivot so you want no nontrivial connected sets => basis of open subgroups around the identity?

[coral pivot]

yeah

[rancid umbra]

confused on the definition of f. d is a function from X^2 to R, since it is a metric. how are you taking in a subset of X as an argument of d?

[empty grove]

A is fixed

[strong siren]

$f(x)=d(x,A)$ relates the distance of the point $x$ to the set $A$. So $d(x,A)=inf(d(x,a)|a\in A)$.

[gentle osprey]

SelahW

[strong siren]

And yes, A is fixed

[rancid umbra]

yea no i got that. is there really no problem with just writing d(x, A) and d(x,y)?? i feel like it should be d’(x,A) or something

since this is measuring distance from a point to a set

should be notated differently imo, but ig that’s just me

[empty grove]

It doesn't matter because the arguments disambiguate it anyway

[rancid umbra]

ehhhh gross

like, if i were a computer, and i had two local variables named d, i would be confused.

[empty grove]

You should probably say that the inverse triangle inequality is happening in R rather than in X

[empty grove]

Good thing you aren't

And not really, overloading is a thing programming languages have too

[strong siren]

So replace $x,y$ with $a,b\in\mathbb{R}$?

[gentle osprey]

SelahW

[strong siren]

For thm. 1

[empty grove]

The variable names are fine, but just add the in R part

[strong siren]

Ah okay

[vocal anchor]

If
\begin{tikzcd}
X
\arrow[r, shift left=1, "f"] \arrow[r, swap, shift right=1, "g"]
& Y
\arrow[r, "h"] & Q
\end{tikzcd}
is a coequalizer and $f, g$ are open surjections, why must $h$ also be an open surjection?
It's somehow supposed to be obvious, but it's really not 🙈
please anyone? 😦

[gentle osprey]

expectTheUnexpected

[empty grove]

Coequalisers in Top are surjective, so you only need to prove that it is open. Given an open set U in Y, you can saturate it with respect to the equivalence relation f(x) ~ g(x) by taking
U ∪ g(f⁻¹(U)) ∪ f(g⁻¹(U))
And this saturation is open
An open set with an open saturation wrt ~ maps to an open set under the quotient map defined by ~

Saturated wrt ~ means union of equivalence classes of ~

The saturation should be that because the saturation of a singleton is that and both image and pre image operations on subsets commute with unions

Is there some categorical characterisation of open maps in Top?

[vocal anchor]

there's this, but I don't understand it haha https://ncatlab.org/nlab/show/open+morphism

[vocal anchor]

dang, never had proper topology, so I don't know what saturated means. brb learning

[empty grove]

Another definition of U being saturated wrt ~ is U being an inverse image of the (set) quotient map by ~

So the quotient topology can be interpreted as "the open sets are exactly the images of saturated open sets"

A set and its saturation will have the same image under the quotient map because you're just adding all the things that map to the same image as something already there

So saturated wrt a function (ie wrt to equivalence relation defined by mapping to the same image under it) is a set such that adding anything enlarges the image

This is the language munkres uses to define quotient topology and I quite like it

[vocal anchor]

So just that I get this right. You take the open set U in Y. Because h coequalizes f, g, we have that h(U) = (h o f)(g^-1(U)) = (h o g)(f^-1(U)).
I don't see why this means U' = U \cup f( g^-1(U)) \cup g( f^-1(U)) is a saturation w.r.t. h,
Then because h is quotient, it maps saturated sets to open sets, hence h(U) = h(U') is open, and h is open.

[empty grove]

Oh you're right, that may not be the saturation

Saturation will be an even larger union

Any sequence of going back to X then coming back to Y will be there

They will all be in the saturation because the equivalence relation is f(x) ~ g(x) so for any x the image of any of its preimages of it must be there

And the saturation will be in this because this is saturated

Being saturated is easy enough to check because the equivalence relation is the equivalence closure of f(x) ~ g(x) and we are just taking the equivalence closure in this process pretty much

[empty grove]

Saturated open to open

[vocal anchor]

Oops

What are some conditions such that we can find an open saturation of a given open set? I suspect that these are exactly satisfied in this scenario

(they have to be lel)

[empty grove]

The equivalence relation can be very different so idk if there's a general thing

But if you are coequalising stuff then maybe

Like we can say from f and g being open continuous that the saturation of U is open

But it may be open in other cases

It's an infinite union

[vocal anchor]

right, but then the existence is not so clear yet?

[empty grove]

Existence of what?

[vocal anchor]

wait

confused

[empty grove]

same

🥴

[vocal anchor]

I did it!

confusing moldi that is

[empty grove]

[vocal anchor]

since I'm new to saturation: a subset U \subset B is saturated w.r.t. h iff U = h^-1(h(U)), right?

[empty grove]

Yep

[vocal anchor]

So this is the idea, right? If that's correct, then at least I know what things to check now.

[empty grove]

Yep, though I don't see why the first equation about U is needed

[vocal anchor]

Yeah your right

my left

you're right*

my bad*

[empty grove]

[vocal anchor]

Just thought of that, now I feel accomplished 🙏

[empty grove]

But then I don't see where we use surjectivity of f and g

[vocal anchor]

Oh

the line is maybe not unnecessary, since it's basically saying: this holds for all open sets, since f,g are onto

just that it's not saying it explicitly

[empty grove]

Yeah but that fact doesn't seem to be used later

[vocal anchor]

I mean, we need to show that all opens get mapped to opens, but w/e. Those are just notes for me anyway

[empty grove]

Right, but doesn't the argument work even without the image of inverse image under f being itself

[vocal anchor]

Well, one might use it here: h(U) = (hf)(f^-1(U)) = (hg)(f^-1(U) ?

[empty grove]

Hmm right, though it feels to me like this is true without surjectivity if you look at the element level (and view h as the quotient map)

And anyway you don't need equality

You need the left side to contain the right side

And image of preimage of U under f is contained in U

[vocal anchor]

so, is the statement then that the coequalizer of two open maps is an open surjection?

[empty grove]

Seems like it to me

[vocal anchor]

https://tenor.com/view/patrick-star-spongebob-squarepants-wow-mindblown-gif-13169052

[empty grove]

I might very easily be wrong lol

yo https://mathoverflow.net/questions/117174/category-of-topological-spaces-with-open-or-closed-maps

same proof without surjectivity

Todd's answer

[vocal anchor]

oh man, almost same formula as me, I'm trimbling

very good, thanks 🙂

Actually, I dont understand why this formula holds. I mean, I can believe that the right hand side is contained in the LHS by some induction argument, but I don't really understand why that's everything there is in h^-1 h (U).
Intuitively it seems right, but my intuition is a heaping pile of dung, so...

[empty grove]

Suppose y is in the left set. then there is some x in U such that x ~ y. This means that there is a sequence of elements a_1,...,a_n of X such that f(a_1) = x and g(a_n) = y and f(a_i+1) = g(a_i) (or exchange f and g in this)

That is how you construct the transitive closure of a relation

or equivalence closure in this case I guess

reflexive symmetric closures are trivial anyway

[vocal anchor]

Ok, turns out I don't understand coequalizers in Set

wtf

[empty grove]

F

[vocal anchor]

brb enrolling in bachelors again

[vocal anchor]

right, so if x~y, then there must be some "chain" of f's and g's connecting them, i.e. they must be in the image of f or g - because elements not in the image of f or g have equivalence class with just 1 element? wut

[empty grove]

the statements are both true on their own

elements outside are not equivalent to anything else

and the chain thing

I don't get what you mean by the because

[vocal anchor]

I guess im just big dumb

nvm

[empty grove]

[empty grove]

derivations are functions from the stalk to R, so composition doesn't make sense. Is this some other operation?

right, makes sense

thank

[cursive flume]

what’s the difference between level set,fibre of a point of a real valued fn and the preimage of that point?

3 notions but they seem equivalent to me

[empty grove]

level set is apparently a special case of a fibre

fiber is a shorter name for preimage of a point

level set is a fiber of a function R^n → R in particular, according to wikipedia

[gentle osprey]

james_ash_.

[empty grove]

What is G_alpha?

"[0,1/2] can't be covered by finite subcovers." For this I would first say "there is an open cover of [0,1/2] such that..."

You say intersection of the U_n is {x} for some real number x, are you using Cantor intersection theorem here? That requires compactness. But here you could just say that the intersection is {0} by inspection

@wispy veldt

[little hemlock]

"are you using cantor intersection theorem." I think its common to have this result for closed-bounded intervals in R without mentioning compactness

[wispy veldt]

doesnt seem like compactness is required only closed bdd intervals and diameter coverging to 0 (which i forgot to mention)

[empty grove]

You'd be using sequential compactness to prove it though? At least that's the proof i know

[little hemlock]

yea

[empty grove]

[little hemlock]

but i mean, the idea is this is heine borel in R.

u start with a notion like "closed and bounded intervals" and get the topological def of compactness.

[empty grove]

right

The rest of the proof still doesn't make sense to me though

[little hemlock]

yea james, ur not supposed to let Un = [0, 1/n], but instead you should continue the process of splitting up non-compact subintervals into more non-compact subintervals

[tough imp]

Also

You can just explicitly compute that the intersection of the Un is {0}

So you didn’t have to pull out the nested interval theorem to make an argument anyway

[empty grove]

I already said that

[tough imp]

Wowza

I just got ’d

[empty grove]

What's the point, in the end chmonkey wins anyway

[wispy veldt]

so what i did i was always taking the bottom part of the interval and made a sequence instead and i guess since they intersect at 0 we can use that as an open cover? cause i was both of the intervals wont be covered but then i realized we can cover 1 half by finite subcovers but the other can still be impossible to do so
so i made this adjustment with that in mind

[gentle osprey]

james_ash_.

[wispy veldt]

does that maybe makes more sense or am i still missing the point?

[tough imp]

I think you want to rewrite this to be more clear, it’s a bit hard to tell exactly what you’re doing and the assumptions blah blah but

My understanding is [un,vn] is always not compact

Or specifically there’s no finite sub cover of O which covers it

Which you get because if all of them had a finite subcover you can just union the finite ones for each one

Hoohee finite subcover of O

Then you used the nested interval thm to grab a point x in each [un,vn]

Then grab an open set U in O with x in U

This contains some open ball B_e(x)

But now because the diameter of [un,vn] goes to 0

There’s some n so that [un,vn] < B_e(x) < U in O

So wait [un,vn] is covered by a finite subcover of O (in fact it’s a single open)

Is that correct?

[wispy veldt]

yeah that sums it up well
ile try to write it more clearly i kinda rushed this your right

[tough imp]

I think the proof works then

It’s just a little unclear how you’re taking the un,vn

And then the explanation of why it’s contained in your N_r(x) is kind of lacking

But the idea is good I think

Just need to polish it up

[wispy veldt]

ye fair enough i can see the issues in my proof
thanks for the help

[tough imp]

[empty grove]

hoohee

[dusk heron]

Why does that suffice? (I know that $SO(n)$ is connected)

[gentle osprey]

gustavn64

[dusk heron]

Perhaps it has something to do with the fact that $\lambda$ is a Lie group homomorphism, and is therefore an open map.

[gentle osprey]

gustavn64

[orchid forge]

You could also try lifting a path from the quotient Spin(n) / ker lambda

Using the fact that a quotient map of Lie groups is a covering map, cover the path (compact) with finitely many evenly covered neighborhoods. Lift the path in each one, by picking one of the two sheets arbitrarily. Now for two consecutive neighborhoods, pick a point x in the intersection, and connect the sheets with the path x --> -x you found.

[gritty widget]

Not specifically topology and geometry related but I think you lot will give the best answer

When you are late king the Mayer vitreous sequence

Or such similar things

Do you use tikzcd

Or just a standard display math with \rightarrows

The arrows seem too long when using tikzcd

[dusk heron]

You mean that it's a covering map on each connected component of Spin(n), right? So since ker lambda has two elements, this means that either we have two coverings with one sheet each, or one covering with two sheets. If it were two coverings with one sheet each, every loop based at the identity in SO(n) would lift to a loop. But since we found a non-loop in Spin(n) which maps onto a loop in SO(n), it has to be the case that we have one covering with two sheets.

Does that sound about correct?

[cursive spade]

how should we think about wedge products?

Or how do you think about it?

[orchid forge]

Xymatrix

Strongly recommend over tikzcd

[orchid forge]

A wedge of vectors is intuitively the oriented area/volume formed by their convex hull

[cursive spade]

what about a differential forms

1 or k-form

[orchid forge]

At face value, just dual (multi)vectors at each point, but they vary smoothly as you move around the manifold, so they have an extra structure. Intuitively a k-form is precisely the kind of thing you can integrate over a k-dimensional manifold.

[orchid forge]

It's a covering map of the quotient space. Locally, each point [x] is evenly covered by 2 sheets, one containing x and the other -x

[orchid forge]

If a wedge of vectors represents an oriented convex hull, it becomes natural to express "how does a linear map distort volumes." You apply it to each vector in a wedge, and you get a new wedge. If the new wedge is a positive multiple of the old one, then the linear map acts on volumes by scaling.

A smooth map f is locally approximated by a linear transformation df : T_pX --> T_qY. So you can ask, how does it (locally) distort volumes? Form wedge products of tangent vectors (so the exterior algebra of T_pX), and look at how df acts on wedges.

Assume f : R^n -> R^m for now. Then df is a linear map at each point, that varies smoothly. The induced map on the exterior algebra should also be smooth. Fix a basis for T_pR^n and T_qR^m, say dx_1, ... dx_n and dy_1, ... dy_m, and you get a basis for the exterior algebra, given by wedges of these things. We can characterize df's action on wedges by writing down what it does to this basis. The induced map sends a basic oriented volume (a wedge of dx_i's) to a sum of wedges of dy_i's, weighted by coefficients which vary smoothly. This kind of thing, we call a differential form. (Technically we really want to do this on the cotangent space instead of the tangent space.)

So, in the same way wedges let us measure how linear maps distort space, differential forms let us measure how smooth maps locally distort space

where k-forms measure the distortion of various (oriented) k-dimensional spans

Now you play with this idea and see that you can make sense of integration on manifolds

And you get the slightly more mature intuition of "the things you can integrate over k-submanifolds"

[gritty widget]

Thanks I’ll check this out

[gritty widget]

excision gives that H(X,A)=H(X-A,\emptyset)=H(X-A)

but there are details you have to be careful about

when does thsi go wrong?

[pearl holly]

Honestly H(X, A) = H(X-A) sounds convincing to me on a intuitive level. Is that wrong?

[tough imp]

Something something A not closed?

I’m a chmonkey so just ignore me. Chmonkeys don’t know AT, only sheaf cohomology

[pearl holly]

yeah like excision does only work when like the closure of B is inside the interior of A or something like that

wait how do you think of it intuitivly

oh yeah I see

wait isn't that exctly excision tho

or is there something else to it

[pearl holly]

because this is how I think of intuitibly lmao

like you just ignore A

ye that's what I said

but I used B instead

B gang

[gritty widget]

technical obstacles

thank you Toki

[tough imp]

Hey

I said that!

Something something A not closed

😤

I am the AT master

[gritty widget]

chmonkey

thank you

both for having a name that is nice to say

and for your answer

chmonkey

[orchid forge]

question about algebraic topology: is it pronounced "chuh-monkey" or more like "shmonkey"

[rancid umbra]

i’ve been saying it like ch-maunky

[gritty widget]

chmunky

[pearl holly]

chmonkey

[gritty widget]

I've been ~actively saying aloud~ chowe monkey

chowe as in

[pearl holly]

I've been saying c h monkey

c h as in ch I guess

idk

lmao

[gritty widget]

sorry

choe

as in

type choe into google translate and use the english pronunciation

[pearl holly]

but honestly

[gritty widget]

joe monkey

[pearl holly]

what if everyone just said

[gritty widget]

joe monkey

[pearl holly]

LMAO

yeee

joe monkey

[plain raven]

i don't know if you got a good answer to this question because fuck reading the backlog lmao

The suspension is a functor, meaning that for two spaces $X, Y$, if $f : X\to Y$ is continuous, then there is an induced map $S(f)$ between the suspensions, $S(X)\to S(Y)$. It's not hard to work out what $S(f)$ should be, and how it should be defined.
You can prove that if $f,g$ are homotopic maps, then so are $S(f),S(g)$ homotopic as maps $S(X)\to S(Y)$.
A famous theorem of Freudenthal - the Freudenthal Suspension Theorem - is as follows

[gentle osprey]

diligentClerk

[plain raven]

Here the suspension map is the map $f\mapsto S(f)$.

[gentle osprey]

diligentClerk

[pearl holly]

hmm ye okay I see

but like when talking about suspensions and stuff, do you talk about reduced suspesion or just "regular" suspeinsion?

because I've mostly seen the reduced suspension in play

[plain raven]

The answer to that is: figure out what base category you're working in, the category of all spaces and the category of pointed spaces and basepoint preserving maps

Now, if $Y$ is $k$-connected for some $k$, then the suspension should be $k+1$ connected unless i'm badly screwing up here, and if $X$ is $k$-dimensional then $S(X)$ should be $k+1$ dimensional. So if we keep taking iterated suspensions, over and over again, then $Y$ becomes $n$ connected, $n+1$ connected, $n+2$ connected... and so on, and so the $2n$ in the question becomes $2n+2, 2n+4,\dots$

[gentle osprey]

diligentClerk

[plain raven]

So the right hand side of the inequality $dim X \leq 2n$ grows by $2$ every time we take the suspension of $Y$, and $\dim X$ grows by $1$ every time we take the suspension of $X$.

[gentle osprey]

diligentClerk

[plain raven]

It follows that if $X$ is finite dimensional and $Y$ is $0$ connected, then eventually, for some $k$, $S^k(X)$ and $S^k(Y)$ should satisfy the hypotheses of the theorem.

[gentle osprey]

diligentClerk

[plain raven]

Thus, for large enough $k$, for all $n\geq k$, $[S^n(X), S^n(Y)]\cong [S^{n+1}(X), S^{n+1}(Y)]$.

[gentle osprey]

diligentClerk

[plain raven]

Does this make sense?

[pearl holly]

okay wait let me read everything

[plain raven]

This is what is meant by "stability." The maps between these spaces eventually settle down and become regularized after we take iterated suspensions enough times. One way to think about this is that there is a "stable homotopy category", where the objects are spaces $X,Y$, .... and the maps between them are homotopy classes of maps $S^nX\to S^nY$ for very very large $n$, large enough that the set $[S^n(X),S^n(Y)]$ is constant as a function of $n$.

[gentle osprey]

diligentClerk

[plain raven]

i don't think this definition is totally sufficient but it's close enough. The point is that like, because this set $[S^nX,S^nY]$ is eventually constant, perhaps it's better for us to study that constant set in some situations rather than $[X,Y]$. The theorem suggests that maybe this is a category worth studying.

[gentle osprey]

diligentClerk

[plain raven]

Stable homotopy theory is the study of spaces and maps between them after the suspension functor has been applied to them enough times that the hom sets stop changing.

Roughly speaking.

So, knowing that an operation on spaces plays nicely with suspension is important, because if something is unchanged by the suspension operation then you can study it as part of stable homotopy theory. This whole idea of "take the suspension however many times and then look at the maps between spaces" only works when you're talking about operations which are eventually constant under the behavior of this operation

For example, the homotopy groups $\pi_{k+n}(S^\ell+n)$ should eventually stop changing for large enough $n$. Asking that a cohomology operation is stable is basically asking that we are able to study it from the POV of stable homotopy theory.

[gentle osprey]

diligentClerk

[pearl holly]

yeee okay I see

man daim

thank you so much for all the answer I got, including this

[plain raven]

well the way hatcher says is it is that when A is a nice subset of X, H(X,A) = H(X/A). so this is a better way to look at it, i think.

but it is useful to think of it this way in some cases of relative homology, like for example if A is a subspace of X and you want to study 'the homology of A as embedded in X' then a natural thing to look at is like, the homology of X but we forget everything that happens outside A, i.e. H(X,X-A). which is what you said

i guess if it's reduced homology it's basically the same thing, because the base point is "nothing"

[pearl holly]

yeeee okay I start to see this

I will read this whole convo tomorrow tho because I'm super tiered to today

like I barely know what I'm typing in now

[plain raven]

goodnight!

😄

[pearl holly]

yeah goodnight diligentClerk! 💤

[charred tartan]

Can someone explain to me how the exterior algebra definition for determinant gives Cauchy Binet formula

[old violet]

Are open balls necessarily connected in a metric space?

[rancid umbra]

discrete metric broski

[finite heath]

Hello

I am learning about the quotient topology

but then they dive into this concept of partition(ing) and I reckon I am quite confused, even after trying to look it up online.

What are they trying to say here

[finite heath]

@rancid umbra can I call you?

[rancid umbra]

like in a voice chat?

[finite heath]

yeah

u dont gotta talk

just hard to articulate my confusion

[rancid umbra]

sure

[finite heath]

also late af

a, b, c, d, e, f, g, h, i, f

[rancid umbra]

X = {a, b, c, d, e, f, g, h, i}

X* = {{a,b} , {c,h,i}, {e}, {f,g,d}}

A = {a,b}, B = {c,h,i} , C = {e}, D = {f,g,d}

X** = {{a}, {b}, {c} ,..., {f}}

Y* = {{a,b}, {c,h,i}} is a subset of X*

[finite heath]

I* = {{0, 1}} U {(0, 1)}

[loud scarab]

When we want to show that two spaces are homeomorphic, we need to come up with a suitable homeomorphism, my question is, what do you do to come up with one? Is it:

i) By geometry reasoning such as the polar representations or stereographic projection? i.e trying to find meaningful equation and then define a map based on that equation

ii) By trial and error

iii) By other means that I do not know? Sometimes in textbooks authors come up with the weirdest kinds of maps that I am so desperate to know how do they even come up with one

[rancid umbra]

theres really no set of rules or algorithm to do this

[loud scarab]

I know, but I am struggling with defining maps on my own, I'm not looking for rules, but rather hints

[empty grove]

Visual intuition lol assuming you don't have access to universal properties

The maps in intro top can usually be explained with some visualisation, do you have specific examples of ones that feel random?

[loud scarab]

To find the connection between geometrical objects in intro topology is not that hard, but I have to be careful with the map I choose based on intuition, because it requires bijectivity, continuity and continuous inverse, so I have to use the map in such a way to satisfy all of this

In other words, I have to choose my open sets well, and I have to be aware that my map that I extracted from a geometric equation, is going to be bijective

I just want to ask, when mathematicians encounter these problems, does it come natural to them and define the map in a matter of minutes or could it take more than an hour to define a simple intuitive map

[dusk heron]

So you are using a definition of covering which allows a covering space to be disconnected?

[orchid forge]

why would you require a covering space to be connected?

[dusk heron]

I don't know, this is what I have learned in the books I have read about it. But I suppose most theorems hold anyway.

[orchid forge]

are you sure?

[dusk heron]

Let me check

This is from John Lee's introduction to topological manifolds, at least

[orchid forge]

weird

[dusk heron]

Anyway, suppose that we drop that requirement, then sure, I agree that Spin(n) -> SO(n) is a covering map. Given the fact that its kernel is {1,-1}, and the fact that there exists a path from 1 to -1 in Spin(n), how exactly does the argument go?

[orchid forge]

i said how it goes earlier

[dusk heron]

You just say to try to lift a path, if I understood correctly? I'm not sure how that implies that Spin(n) is connected

[orchid forge]

i said more right after

[dusk heron]

"It's a covering map of the quotient space. Locally, each point [x] is evenly covered by 2 sheets, one containing x and the other -x"? I don't see anything else you wrote related to this

I mean, this path from 1 to -1 in Spin(n) shows that there is a loop which lifts to such a path, i.e. the fundamental group of SO(n) acts transitively on each fiber

But why does this imply that Spin(n) is connected?

[orchid forge]

Using the fact that a quotient map of Lie groups is a covering map, cover the path (compact) with finitely many evenly covered neighborhoods. Lift the path in each one, by picking one of the two sheets arbitrarily. Now for two consecutive neighborhoods, pick a point x in the intersection, and connect the sheets with the path x --> -x you found.

[dusk heron]

Lifting the path? It is already in the cover, not in the base space. Not sure I understand that argument. However, I think I understand how one can do it. Given any points x,y of Spin(n), we can take a path between their images in SO(n) and lift this path to Spin(n), to a path starting at x. This produces a path in Spin(n) which starts at x and ends at y or -y. If it ends at -y, we can use the path from 1 to -1 to produce a path from -y to y. Thus x and y lie in the same path-component.

[orchid forge]

I'm saying to lift a path in the base space

the quotient should automatically be connected

[dusk heron]

So your argument seems like the same argument as mine, except that you use the path x --> -x while lifting the path, but I first lift to a connected path and then use the path x --> -x at the end if needed, if I understand correctly.

[orchid forge]

you're assuming my entire argument

"lift this to a path in Spin(n)"

how do you know this is possible?

[dusk heron]

Because Spin(n) is a covering space of SO(n)?

Let's assume that we already know that paths in the base space can be lifted to paths in the covering space.

So what we could do in your argument is that instead of picking one of the sheets arbitrarily, we pick the sheet where the last lifted subpath ended. Then at the end, we use the path x --> -x if needed

[orchid forge]

Sure

[proud coral]

It depends, something it's easy and sometimes it's very hard

[feral copper]

Hi! In the paper https://link.springer.com/article/10.1007/BF01078120 (which you can read on sci-hub by c/p the url above), there is the following which I don't quite understand:
F(x0,x1,x2) is homogeneous of degree 2k with real coefficients. We choose the sign of F so that the points corresponding the the non-orientable component of the complement of {F=0} in RP² is negative.

We now consider the equation z²=F(x0,x1,x2), where x0,x1,x2 are complex variables not all simultaneously zero. This equation gives a compact complex algebraic surface Y embedded in 3-dim complex space E of the 1-dim vector fibration over the complex projective plane P':E→CP²={[x0:x1:x2]} whose sections are homogeneous functions of degree k in the variables x0,x1,x2. From the real point of view, surface Y is a 4-dim compact smooth orientable connected manifold w/o boundary.
I don't quite understand the construction of Y... What is the meaning behind that equation z²=F(x0,x1,x2)? What is that vector bundle P':E→CP²? Does he define the vector bundle by means of all sections on open subsets of CP²? (which I don't understand because we have global sections here...)

[feral copper]

(who buys legal copies of 1971 springer articles anyways? 😈)

Actually, reading through the rest of §2, is E defined as this?

$$E={(z,[x_0:x_1:x_2])\in{\bf C}\times{\bf CP}^2\mid z^2=F(x_0,x_1,x_2)}$$

[gentle osprey]

Matplotlib

[feral copper]

(which doesnt make sense because F(x0,x1,x2) is only defined up to a scalar)

[wanton marsh]

it takes place in the line bundle O(k)

F(x0,x1,x2) becomes a section in that line bundle

well rather

F is a section of O(2k)

[feral copper]

Yeah okay this I agree

[wanton marsh]

and we look at the square roots in sections of O(k)

[feral copper]

Okay, but how is this formally defined then?

[wanton marsh]

well I guess you can cover P²(C) with 3 affine subsets

say U0 is the one where x0 doesn't vanish

so U0 = {[1 : x1 : x2] | x1,x2 in C} in an affine plane

[feral copper]

Yup

Same for U1, U2

[wanton marsh]

the line bundle O(k) will be uuh U0 x C

and a homogeneous polynomial P of degree k corresponds to the section {([1 : x1 : x2], P(1,x1,x2)}

or rather

[feral copper]

On U0, and then on U1 and U2 and you glue

[wanton marsh]

{([x0 : x1 : x2], P(x0,x1,x2)/x0^k)}

wait I'm getting confused lol

[feral copper]

I mean, okay I agree that F is a global section of O(2k)

[wanton marsh]

or also {([x0 : x1 : x2], P(1,x1/x0,x2/x0))}

[feral copper]

But my question was rather: how to make sense of this equation z²=F?

[wanton marsh]

well then it becomes the set of points

of the form {([1:x1:x2], z) ; where z² = P(1,x1,x2)}

on U0

[feral copper]

So E is defined on each three affine charts U0, U1 and U2 then?

[wanton marsh]

yeah and they also glue together to make E = O(k)

[feral copper]

Okay okay, I think I can work out the details now, thank you very much 🙂

That was helpful, I actually didn't think of O(2k) in the first place ^^”

[wanton marsh]

and Y is compact because it is a branched 2-covering of P²(C)

basically the chart I picked for O(k) above U0 takes something called P that pretends to be a homogeneous polynomial of degree k and asks what is the value of P/x0^k

[feral copper]

Actually, to put it formally, do you agree with: you have a natural map O(k)→O(2k) by squaring the sections
Then, Y is the surface of square roots of F in O(k), that is, you pre-image F under this map?

[wanton marsh]

yeah

I'm not completely sure of what algebraicity means for subsets of O(k)

[vocal anchor]

Is this true? I would think so, but don't see it immediately

So the first half is obviously just a question in Set

sh*t, also, A = X obviously

[vocal anchor]

So, I'm doing things in Set now, and I only need to show one final thing. Namely, if a subset of $E \times_B X$ is $E \times_B f$ -saturated, then it is in particular of the form $U \times V$. I claim that it is actually given by $(E \times f^{-1}(f(V))) \cap E \times_B X$, but I can't see it.

[gentle osprey]

expectTheUnexpected

[vast estuary]

What is the meaning of a "formal" linear combination?

[vocal anchor]

element in free Z-module

[vast estuary]

I haven't learnt modules yet, should I do that first?

[feral copper]

It's elements of the form a1[x1]+...+an[Xn]
with [xi] formal elements labelled the same as the xi's

And ai are integers

Basically, the same as a vector space whose basis is the set of xi, but coefficients are integers

[vocal anchor]

What Matplotlib said. And at least for me, "formal" refers to the fact that you don't interpret the operation "+" in any meaningful way (yet)

[feral copper]

Yup, [x1]+[x2] is nothing more than that, it's not related to [x1 * x2] for some operation *

[vast estuary]

but this is confusing

[feral copper]

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

Nothing, it's not a map

It's a formal element

[gentle osprey]

Hausdorff

[vast estuary]

Oh so it's just there but has no meaning?

[feral copper]

Yes and no, but it doesn't correspond to a map

[vast estuary]

Should I read all of this?

[feral copper]

I was just giving the reference for the name, but you don't need it 🙂

Here, the notation is confusing, you should write elements as sums of [\sigma_i] and not \sigma_i

And so the \sigma_i are actual maps, and the [\sigma_i] make the basis of your group

[vast estuary]

What is [sigma_i]?

sigma_i is a map, obviously

See, I might be lacking algebraic background here

[vocal anchor]

[sigma_i] is just a symbol

[feral copper]

Well, that's the confusing part

It's something, it's an element of your free group

It just happens to be labelled the same as \sigma_i

[vast estuary]

Free group generated by?

[vocal anchor]

Maybe think of it as a way of keeping track of data. Have you learned about "homologous cycles" in Complex Analysis?

[feral copper]

Generated by the \sigma_i means that your group has a basis whose elements are symbols [\sigma_i] ine 1:1 correspondance with the actual maps \sigma_i

[vast estuary]

I haven't

[vast estuary]

Oh interesting

I think I'm getting it now

[feral copper]

It just means "take a group with basis in 1:1 correspodance with elements of your given set"

You might as well label them the same because of that correspondance

[vast estuary]

Yep, okay

[feral copper]

For instance the free abelian group over {a,b,c} is Z³, where each element is of the form (x,y,z), but also written as x[a]+y[b]+z[c] with [a]=(1,0,0), [b]=(0,1,0) [c]=(0,0,1)

[vast estuary]

By the way, what algebraic background would you suggest I know for sure at this point?

[feral copper]

So that ([a],[b],[c]) is your preferred basis

[vast estuary]

My instructor has started talking about singular homology

[feral copper]

For singular homology? Group theory, quotient groups etc

[vast estuary]

I'm familiar with that

What about modules? and free abelian groups?

I know free groups

[feral copper]

Would make you a bit more comfty but that's not absolutely mandatory, it's almost the same as vector spaces so I'd say be fine with linear algebra and you'll work things out

It depends if you only talk about (co)homology over Z or any ring

If you do talk about homology over any ring, know what a tensor product is

[vocal anchor]

Given a subset $S \subset X$ of a topological space and a map $f \colon X \to Y$, can we always find a smallest \emph{open} $f$-saturated set containing S? So, it's like a closure, but in general if would be an infinite intersection, so not necessarily open, right?

[gentle osprey]

expectTheUnexpected

[empty grove]

Yeah not necessarily. Consider f: [0,1] → {a,b} by 1/2 mapping to a, everything else mapping to b. Set S = {1/2}

really you can make any subset f-saturated by taking an appropriate f into {a,b} discrete

so your question turns into "does every subset have a smallest open set containing it?"

I guess we would call that operation open closure if it worked lmao

[vocal anchor]

lel

fugg

[vast estuary]

Alright, thanks!

Why is the stuff in blue true?

Am I missing something obvious?

[wanton marsh]

that follows more or less immediately from the definition of free abelian group

note that it's about group morphisms from Sn(X) into other abelian groups

not into all groups

[feral copper]

It's just saying that f:S_n(X)→G is uniquely determined by its image on all generators [\sigma_i]

Just the same as transpositions generate the symmetric group, the elements [\sigma_i] generate S_n(X) and so you only need to define f on such elements to have a well-define group morphism

[empty grove]

Except in this case since it's free abelian you can also map the generators arbitrarily

Hausdorff I believe you have enough background to start thinking in terms of universal properties

[feral copper]

Now the true question: what is the universal property of Hausdorff spaces?

[feral copper]

Seems like nlab has got you covered, as usual: https://ncatlab.org/nlab/show/Hausdorff+space

[remote marsh]

What’s an easy way to show that identifying the boundary of a disk gives a sphere? I.e D^n/S^(n-1) = S^n

I have visual intuition how to construct an explicit homeomorphism but I was curious if there is a purely algebraic way to do it

[pearl holly]

This is a really dumb question but I don't understand why f can be extended over X^n union sigma iff fg_\sigma is null homotopic. Can't you just take an f such that fg_\sigma is not null homotpic and then extend f to X^n union \sigma by just sending \sigma to a single point?

\sigma grindset 😎

[fading vale]

why exactly are maps with image in A homotopic to the identity in pi_n(X, A, x0)?

I guess a softer question related to this is that it feels odd to me that pi_n(X, A, x0) should be "smaller" than pi_n(X) so to speak. like i think the idea behind it is that we're treating A rather than x_0 like the basepoint by treating maps contained in A rather than x0 to be in the class of the identity

But unless im mistaken pi_n(X, A, x0) is homotopy classes of maps f: (D^n, S^n-1) -> (X, A) rel A

And an arbitrary map g(D^n, S^n-1) -> (X, A) with g(D^n) subseteq A shouldnt be necessarily homotopic to the constant map rel A?

since its not necessarily homotopic to the constant map in general?

[swift fjord]

I feel like i'm losing my mind, why is this a valid triangulation of a rectangle. I thought triangulations are supposed to be homeomorphisms, so how isn't the line through the middle getting in the way. Some points which were interior points aren't anymore

Moving this to a thread

[gritty widget]

The exact sequence of the pair is this long exact sequence

and it is long

can we instead just have a short exact sequence of rings

$H^(X,A)\rightarrow H^(X)\rightarrow H^(A)\rightarrow H^(X,A)$

[gentle osprey]

lime_soup

[pulsar lynx]

Perhaps dumb question:

is it true that discrete spaces can never be compact, since the open cover {{x} | x in X} has no finite subcover?

Oh yeah that makes sense

Any finite space is compact right? Since its power set is finite so every cover is finite

Law of excluded middle?

Isn’t that always given in these contexts

Can topology even like, work, without that

Ah

I mean that’s pretty interesting, I thought lem was like everywhere in math

Like how sets are everywhere yknow

[pulsar lynx]

That’s cool, thanks

[vocal anchor]

Is a pullback $X \times_Z Y$ automatically an open subset of $X \times Y$?

[gentle osprey]

expectTheUnexpected

[tough imp]

No

I don’t think so at least, I’m pretty sure that this is sometimes closed

Like if X = Y then the pullback is basically the place that your two maps agree

Or even easier, take Z a point

Then X x_Z X = diagonal of X inside X x X

And this isn’t open, it’s closed iff X is Hausdorff

[vocal anchor]

I see
Thnak

But I wanted it to be true :(

[tough imp]

:(

[vocal anchor]

If $fg$ and $g$ are quotient maps, then $f$ is also a quotient map. The proof is: $f^{-1}(U) open$ iff $g^{-1}( f^{-1}(U)) = (fg)^{-1}(U)$ open iff $u$ open, and surjectivity is the general categorical fact that if $fg$ is an epimorphism then $f$ is an epimorphism.

[gentle osprey]

expectTheUnexpected

[vocal anchor]

Since quotient maps are exactly the extremal epimorphisms in Top, I was wondering if the statement "fg , g extremal epis => f extremal epi" is true in general.

Wait, this might be better for #category-theory

Oh, so actually $f, g$ continuous and $fg$ a quotient map is enough to conclude that $f$ is quotient.

[gentle osprey]

expectTheUnexpected

[crimson path]

If I have for n>=1 shown S^{n-1} is a strong deformation retract of R^n \ {0},
how does this follow next: for n >= 3 that R^n \ {0} is simply connected?

Hm alright I'm sure that is true. I just don't see how it is a corollary

For context, it is from Lee (p 201, topological manifolds):

[gentle osprey]

Nobody

[crimson path]

I know a retract of a simply connected space is itself simply connected.
Hm, not yet. Thus far we use the fact that S^1 has a non-trivial fundamental group

Oh wait, they do. Alright then it is straightforward. Since for n >=3, S^{n-1} is simply connected, so is R^n \ {0}.
thanks for the help!

[pearl holly]

I’m still stuck on this

So the red part is Y, the blue part is the image of f and the yellow/green part is g. Let’s say that the white thing is a hole. Then fg is not nullhomotopic but can’t I still just attach the n+1 cell over that hole with no problems?

What am I missing?

[pearl holly]

Okay Max helped me lessgoo

[quasi forum]

Correct me if I am wrong, but I am pretty sure the answer for (a) is jack squat.

Hold on, dont answer yet.

If you take an open cover in Tau, it is also an open cover in Tau'. Since Tau' is compact, there is a finite subcover, which is also in Tau

Tau' compact implies Tau is compact

[vocal anchor]

if by "tau' compact" you mean "x compact wrt tau'", then yes

[quasi forum]

Yes, my bad.

But the other direction wont cut it because there may be an open cover in Tau' that is not in Tau, so we don't have any information for that

Ooo, Hausdorff is the opposite

(X,Tau) Hausdorff means (X,Tau') Hausdorff

Discrete vs Indiscrete topologies is always a good starting point

Shoot, I always get them mixed up

[vocal anchor]

discrete means powerset

[quasi forum]

Shoot, I always get them mixed up

But the Discrete Topology is not compact if X has infinitely many elements

Because you can take the open cover to be {x} for all x in X

What do you mean by collect along the top/bottom?

[vocal anchor]

and comparable topologies will be in the same "string"

[quasi forum]

So I think the contrapositive is a good way to try and prove (b)

Actually, I'm not sure with this one ...

Why can't Tau be strictly finer than Tau'?

[gentle osprey]

Nobody

[quasi forum]

Clearly the goal is to prove it is a bijection

Why do I know it is a bijection?

[gentle osprey]

Nobody

[quasi forum]

Oh, duh

[pearl holly]

wtf is the infinite-dimensional sphere?

[gentle osprey]

Nobody

[pearl holly]

oh no

not the colimit

okay so you basically take every S^n and all the inclusions and then draw a big diagram and add S^\infty and arrows between S^\infty so that everything commutes?

[gentle osprey]

Nobody

[pearl holly]

yeah nvm I need to read up on the colimit

apparently it has a CW complex structure

yeah okay I see. Is there a way to maybe visualize all this?

I assume it will be hard

but like, is it always the case that the colimit of cw complexes is itself a cw complex?

or is this a special case?

yeah okay I see. Do you maybe know where I could quickly read up on this?

okay I will give it a try. Thank you so much!

[obtuse meteor]

So

I’m trying to prove that if X -f-> Y is an iso on homology and X, Y are simply connected then f is a weak equivalence

Using the hints from class I’m stuck at one particular point

Namely I need to show that H_q(F f) = 0 using the serre spectral sequence (F f being the homotopy fiber)

I’ve gotten that like the bottom row of the serre spectral sequence stabilizes on the second page by using edge map shenanigans

But idk where to go from there

[dusk heron]

Not sure if this fits here or in #groups-rings-fields, but I will post it here. What is the intuition behind the Lie algebra of a Lie group? I know that as a vector space it represents the tangent space at the identity. But what does the Lie bracket represent here, intuitively? I mean, if you take a basis {e_1, ..., e_n} for the Lie algebra, then you can view them as a finite set of infinitesimal generators, in the same sense as a finitely generated group has a finite set of generators. If we look at the case finitely presented instead, we have a finite set of relations between the generators. Can the Lie bracket be viewed as an analogue of this, giving "infinitesimal relations" between these infinitesimal generators?

If I'm not mistaken, the Lie algebra determined the Lie group if you assume the latter to be simply connected.

[vocal anchor]

Do you know the exponential map (from the lie algebra of a lie group to the lie group)?

[fair idol]

I hear the phrase" the lie algebra is a linearization of a nonlinear lie group" but idk what the heck that means lmao

[gritty widget]

but that's defined for all t?

[clever badge]

i am getting tripped up with this question on the tautological one form

i dont get how they want me to define "α*θ=α"

because it seems that these are elements of two different cotangent spaces

(this is assessed work so im looking for general understanding instead of direct answers)

im confused since we have α being an element and a map at the same time

this is what i have come up with

we can sort of "promote" a tangent vector of M to a tangent vector of T*M such that its sort of seen as an equivalence?

but honestly i have no idea what im doing

<@&286206848099549185>

[quartz edge]

i'm having a hard time with proving chain rule on smooth maps (milnor)

i've got f : M -> N and g : N -> P

f g both smooth, x is in M and f(x) = y

i'd like to show then that d(g o f)_x = dg_y o df_x

i can't see anything useful i can get from a comm diagram because there's no guarantee that dg or df are injective (how would you even use such a diagram anyway?)

i thought about trying to exploit the continuity of f and g by moving to the limit definition of the derivative

but i can't seem to wrangle anything from that either. I'm now considering using a taylor series expansion of f and the remainder theorem to get something workable, which ought to do it, but i have a feeling there's another approach the author wants me to get

[gritty widget]

a $1$-form $\alpha$ is a section $\alpha\colon M \to T^M$, so it induces a pullback $\alpha^\colon\Omega^1(T^*M)\to\Omega^1(M)$. since $\theta\in\Omega^1(T^M)$, $\alpha^\theta\in\Omega^1(M)$.

[gentle osprey]

TTerra

[gritty widget]

so \alpha^*\theta = \alpha makes sense

equality of 1-forms

your flow is correct and you have correctly deduced that X is complete

[clever badge]

i have not seen "induced pullbacks on one-forms" so im still a little confused

[gritty widget]

let $F,G$ be extensions of $f,g$ in neighborhoods of $x,y$, respectively. then $d(g\circ f)_x = d(G \circ F)_x$ by the definition in milnor, since $G\circ F$ extends $g\circ f$. the ordinary multivariable calculus chain rule will give you that this equals $dG_y \circ dF_x$. again, by the definition in milnor, $dG_y = dg_y$ and $dF_x = df_x$, so you may conclude $d(g\circ f)_x = dg_y\circ df_x$

[gentle osprey]

TTerra

[quartz edge]

ah shit

that clears up 2 things i was unsure about

well

[gritty widget]

if $f\colon M \to N$ is a smooth map between manifolds, and $\omega\in\Omega^1(N)$, then define $f^\omega \in \Omega^1(M)$ by $$(f^\omega)m(v) = \omega{f(m)}(df_m(v)), \qquad v \in T_mM$$

[quartz edge]

oh, F and G are extensions to open sets in the embedding space.

[gentle osprey]

TTerra

[quartz edge]

vry insightful

i think however the taylor series thing works too albeit much more ugly

[gritty widget]

taylor series
functions on manifolds

[quartz edge]

yep

[gritty widget]

[quartz edge]

multivariable taylor series

[clever badge]

i see, its just a standard pullback applied to a one form, that makes sense

the problem here is that we havent assumed yet that $\theta\in \Omega^1(T^*M)$

[gritty widget]

so prove it

write it in coordinates

[gentle osprey]

aritmos

[clever badge]

that would be part b)

[gritty widget]

ok, well, the pullback is still defined even if your form isn't smooth

[clever badge]

and therefore i assume that there is some way to do it without coordinates

[gentle osprey]

TTerra

[gritty widget]

so use this definition to prove part (a), and then go prove in part (b) that the thing is actually smooth

[clever badge]

im trying to see if we have seen that in the class, but as far as im aware we have not and therefore would assume that is not the case

again, i am just given

[gritty widget]

ok, sure. and then you have to prove that α^*θ = α for one forms α. here, α^*θ is defined by the formula i just wrote

don't worry about whether or not α^*θ makes sense if you haven't shown θ is smooth yet. there's literally only one thing α^*θ could mean

[clever badge]

my main problem is justifying the formula you wrote since we havent touched it

[gritty widget]

it's the definition of the pullback of a form

how have you not seen it if you're doing an exercise involving it?

or maybe you've seen a different definition of pullback. if so, what was it?

[clever badge]

oh wait

found it

the difference in notation is tripping me up a bit but it should be the same thing

[gritty widget]

the notation makes me want to cry

but yes

[clever badge]

:,)

[gritty widget]

this is what you're looking for

it's the same thing i wrote, just without the argument v

[clever badge]

but wait dont i have the opposite? this is the pull back of a one form via F, we have the pull back of θ (whatever it is for now) via a one form

[gritty widget]

a (smooth) one form $\alpha$ on $M$ is a (smooth) map $\alpha\colon M \to T^*M$, so the pullback of a form on $T^*M$ still makes sense

[gentle osprey]

TTerra

[gritty widget]

it's a little confusing

you're pulling back the tautological 1-form (a 1-form on T^*M) by a 1-form considered as a map M -> T^*M

[clever badge]

yeah my bad my point was that we still havent shown that θ is a one form

i just have that weird equation.

sorry if what im saying makes no sense

im not following the class very well

(i main phys and thought this math fit in well esp later with relativity but man does the class go fast and its so algebraic that it gets hard to follow)

[gritty widget]

i think i get what you're saying now

you could prove it's a 1-form before doing part (a) if that helps

$\theta$ still makes sense as a map $\theta\colon T^M \to T^(T^*M)$ so the formula for the pullback still makes sense, but i can understand why you'd be uncomfortable doing that if you've only defined the pullback for smooth forms, and haven't shown $\theta$ is smooth yet

[gentle osprey]

TTerra

[gritty widget]

showing it's actually a 1-form should have been the first part of the problem tbh

[gritty widget]

DG is just hard

i was barely following my first class in it

now i've taken like 3 of them and it's my favorite subject

[clever badge]

it just seems like all the math majors get it

and so ive kinda shut off

because i dont wanna ask stupid questions in lectures

while everyone else is asking about how to extend what we just learnt and stuff

just leading to a negative feedback loop

[gritty widget]

put them in their place by asking the prof questions about how it relates to physics

i can guarantee you you'll make some of the pmath majors confused

[clever badge]

well the course then turns into semi-physics

sem1 we do differentiable manifolds

sem2 we do geometry of general relativity

(technically two separate courses but for the prof its one smooth yearlong class)

[gritty widget]

sounds fun

in my experience math majors like to ask complicated sounding questions of no actual substance just to flex what they know. don't worry about it too much

[clever badge]

anyways let me have a go at showing that theta is a one-form, we have them defined as (smooth) sections

i took an intro course to diff geometry last year and so i still have the normie definition in my head (as just things that kill tangent vectors to R^n)

ok i see how by its definition, theta is a one form on T^*M (albeit we skipped smoothness)

now ill try plugging it into the lemma

idk if the q subscripts (your (p) evaluation) are in the right places but at least now all the stuff makes sense

[gritty widget]

the very first thing should be $(\alpha^\theta)(q)$ instead of $\alpha_q^\theta$, but other than that, it looks good

[gentle osprey]

TTerra

[clever badge]

youre right the q should be on the outside

[gritty widget]

but other than that it's perfect

[clever badge]

alright two more parts to go

next is just local coords, which ill have to look at the notes to figure out (doesnt seem like it should be too hard)

and the last part is that for a diffeomorphism F:M->M, (T*F)*θ=θ

thanks for the help

oh wait

i used the formula without showing that θ was smooth, you said that it worked regardless right?

[gritty widget]

yes

the pullback of a form still makes sense even if it's not smooth

[clever badge]

alright i guess ill just say that and hope theyre cool with it

[gritty widget]

they should be

they're the ones that put the pullback question before the showing it's smooth question

[gritty widget]

[haughty anvil]

I wanted to be sure that this answer my partner and I have are right here. We just need to do c and d.

This is what he sent me for his answer. He fell asleep now but I just wanted to be sure it was all right.

Namely I'm confused why and how he made the vertices 2a - 2d.

[rancid umbra]

wrong geometry channel lol

[haughty anvil]

Oh. Which one would it be in?

[empty grove]

#geometry-and-trigonometry

[haughty anvil]

This is pre university? But it's an advanced geometry class.

[empty grove]

oh idk then

[haughty anvil]

Here's the full question if it makes it any better.

[rancid umbra]

uhh, mp of AB is (1/2,0)

[haughty anvil]

Where are you getting that?

[rancid umbra]

why is he doing all of this extra stuff

the mp formula

[haughty anvil]

We aren't trying to find the mp.

That first picture isn't the actual question, it's just one I found that also had it and a visual representation.

[haughty anvil]

This is the actual question.

1a is the same as in the first image, just c and d.

[rancid umbra]

ah wait, so you can do this just by scaling

like, a is just some vector pointing in the direction of the lower left vertex, so that when you scale it up by a factor of two, that vector lands on the lower left vertex. same for all the other points

then the vector pointing to the midpoint m is (2a + 2b)/2 = a + b

this was probably done for notational convenience

[tough imp]

This is from a while back but I absolutely abhor telling people wrong things. The pullback X x_Z X where Z is the 1 point space is not the diagonal like I had claimed, it’s the entire product X x X lol. When Z = X and the maps X -> Z are identity, THEN X x_X X is the diagonal lmfao

[coral pawn]

For c, why can't I define the flow to be $\theta_{(x,y)}(t)=(x+ty,y=4xt)$?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

Why do I have to do the weird stuff with the differential equations?

[uncut surge]

The derivative vector field associated to your flow needs to equal your vector field at every point; I think yours only does so at t = 0?

[coral pawn]

Oh I thought it was only supposed to agree at t = 0

[uncut surge]

For example, starting at the point $(x,y)$ and $t = 1$, your flow ends up at the point $(x + y, y - 4x)$, where the vector field you are given has the shape $((y - 4x) \partial_x, 4(x + y) \partial_y)$, but the flow you define has tangent vector field $(y \partial_x, x \partial_y)$

[gentle osprey]

Lartomato

[uncut surge]

Then you'd basically just define a linear approximation for your vector field at t = 0, but the flow is really supposed to give you the curve that you get when walking (or flowing wink wink nudge nudge) along your vector field

[coral pawn]

Got it

Thanks

[uncut surge]

naisu

[atomic ember]

Hello everyone. Assuming T and S are linear maps what are some consequences that follow from ||Tv|| = ||Sv|| where v is a any element of a vector space V. Thank you for reading.

I meant norm of Tv = norm of Sv. Also let’s assume S and T are operators on V

[empty grove]

||Tv|| = ||Sv|| for all v or some v?

[bleak helm]

For v = 0

[empty grove]

I'm guessing that S and T would differ by composition with an isometry then (assuming for all v)

[quasi forum]

So question: what does a homeomorphism tell us here?

I'm not sure I am following how that helps

[empty grove]

Because take a basis of the image of S, choose a preimage of it under S, take image of that under T and this should be an isometry from im S to im T and then you should be able to extend this to the whole space as long as dimension and codimension are equal?

[empty grove]

If the identity set map is a homeomorphism then the open sets in both topologies are exactly the same

[quasi forum]

Why is that exactly?

[empty grove]

Homeomorphism tells you that image and pre image of open set is open

[quasi forum]

Oh duh!

Well, we proved a seemingly niche theorem a while ago: If a continuous bijection maps a compact space onto a hausdorff space, then the function is a homeomorphism

Turns out this niche theorem is gonna come in handy :p

[empty grove]

It's an important theorem

[quasi forum]

Well, "seemingly niche." It seemed like a very specific claim when we first did it

I'm not sure what maximally compact and minimally hausdorff means?

Oh wow! That's quite the powerful claim

[empty grove]

The 2nd exercise is asking you to prove pretty much exactly that

[quasi forum]

Yeah, Nobody was very helpful. I was just having a hard time processing everything yesterday.

[quasi forum]

How so? I am not sure I see why this is?

Oh yeah, I understood that bit. I guess I am having a hard time understanding why coarser topologies can't be hausdorff, and finer ones can't be compact

[atomic ember]

Thank you for the responses. In conclusion it seems that there needs to be more specifications to have more interesting consequences.

[empty grove]

If you have compact hausdorff X, you make it finer, you get hausdorff space Y. Suppose this were compact as well, but then this will contradict problem 2 because you have comparable unequal compact hausdorff topologies on the same set

Similar reasoning works for coarser topologies not being hausdorff

[quasi forum]

Hmm, alright.
I got another question. I need to induce R with the cofinite topology and prove every subspace is compact.
What is so special about R here?

[empty grove]

And we use the fact that topologies finer than hausdorff topologies are hausdorff, and topologies coarser than compact topologies are compact

[empty grove]

Nothing

[quasi forum]

Phew, that's what I was thinking. I just had to make sure

Okay, so let me share my idea for this one.
If we take a cover of a set in R, then the complement of this union is the intersection of the complements. Since there is at least one set that has one of these elements as a complement, we can just choose a set that has an element in its complement: and we just do this for all of the elements in the complement.
That is a finite number of sets, producing a finite subcover

But I am actually not sure if this actually the job (Perhaps all the sets I chose are all missing an element in my subspace)

Oh wait, I know what to do!

If we just considered one set in the cover, then there are finitely many holes in the subspace (possibly zero), then for each missing element, we choose a set that contains that element. This is a finite number of sets, giving a finite subcover

[empty grove]

Yep

[quasi forum]

Perfect. For part b) I am pretty sure the answer is no, but I can't figure out a nice counter example

[bleak helm]

@quasi forum

Nvm

Big crank moment

||R - {1/n, 1/(n+1), ...}|| are the open sets you should consider

[empty grove]

Loon star

[bleak helm]

[finite heath]

🐐

[quasi forum]

Yep, already found it :)

[empty grove]

The big crank moment?

[summer terrace]

What is Topology exactly a study of?

And how does it relate to things like geometrical shapes or 'knots'?

[summer terrace]

Ooga booga

Okay, I think I understand. @wooden falcon

[clever badge]

im not sure how to show that the tautological one-form $\theta\in\Omega^1(T^*M)$ relative to local coordinates $(x^1,...,x^n,p_1,...,p_n)$ for $T^*M$ is of the form $p_i dx^i$

[gentle osprey]

aritmos

[clever badge]

in the past we've found the local coordinates for maps between manifolds F:M->N but im not sure how to adapt that reasoning to what we have here

[clever badge]

<@&286206848099549185>

[gritty widget]

it's going to be of the form $a_i dx^i + b^j dp_j$ for some smooth functions $a_i, b^j$, and here, $$a_i(q) = \theta_q\left(\left.\frac{\partial}{\partial x^i}\right|_q\right)$$ and $$b^j(q) = \theta_q\left(\left.\frac{\partial}{\partial p_j}\right|_q\right)$$

[gentle osprey]

TTerra

[gritty widget]

now use the definition of the tautological 1-form to compute these

oh, i guess the notation might be misleading. these are functions on (a chart of) T*M, so the q here should be a covector

[clever badge]

hmm i still dont really know how to proceed

i just think im missing conceptual understanding for the tangent and cotangent bundle stuff

(your definition with the a_i and b^j did make sense)

its just the inserting the tautological nature that is confusing

[quasi forum]

Yep. I chose to look at the sets $U_k=\R - {\frac{1}{n}: n\neq k}$. Not only does $\bigcup_{k\in \N} U_k$ cover $[0,1]$, but removing any of them would mean it will no longer cover it.

[gentle osprey]

dackid

[empty grove]

Where's the crankery

[quasi forum]

I think this is the big crank moment (whatever that is) because I found this before anyone suggested anything.

[empty grove]

Crank moment is like the opposite

[quasi forum]

Oh, well then I didn't have that

I had a big brain moment

[empty grove]

It's the stuff you find in the pinned messages of this channel

[empty grove]

Indeed

[bleak helm]

I had a wrong suggestion (big crank moment) which I deleted before my right suggestion

[quasi forum]

Ohhh, I get it now

[gritty widget]

you should have pi(x, p) = x in your local coordinates, so it's just a matter of working through the definition of the tautological 1 form and seeing what pops out

it might help to recall what p_i does to a covector as well

[clever badge]

does it just pop out the "i-th coordinate of the covector"?

[gritty widget]

pretty much

[clever badge]

alpha -> alpha_i

alright ill take a look at it after i finish this lecture

we're covering vector bundles now

[gritty widget]

if you're still stuck, you can ping me and ill see if i can help

[clever badge]

alright perfect

thanks for all the help

[fading vale]

Ok let p: E -> I^n be a fibration

if you dont know a fibration just means that there exists some F such that every x in I^n has some neighborhood U with p^{-1}(U) = U x F

I fibration p: E -> B is trivial if E is homeomorphic to B x F itself

anyway I need to find a way to subdivide the cube I^n up into N^n subcubes of length 1/N small enough that the restriction to each subcube is trivial

My idea was to take every point x in I^n, pick an open U around it, and then use compactness to find some U_1, ..., U_k covering I^n

and then take some N smaller than the diameter of each

But im still not confident that this actually works because it doesnt imply that each subcube is in fact contained in some U_i

Its a bit rude to post your question when someone else already has

[kind cedar]

Oh, I'm deeply sorry. What is the correct protocol in this situation? I wait until your question is answered or no one answers it for too long?

[fading vale]

Yeah pretty much

[kind cedar]

I'm not used to Discord netiquette

[fading vale]

No problem

[kind cedar]

I'll delete it and repost later then

[fading vale]

anyway moldi do you have any ideass

This seems like it shouldnt be too hard at least from how its presented in the book but i cant think of a nice way to do it

[empty grove]

I didn't even bother reading because you usually ask advanced stuff

[fading vale]

This is basically a point set question

Sad!

[empty grove]

[fading vale]

I dont really have intuition for why this should even be correct, an arbitrary fibration might be trivial over subsets U_1, ..., U_k and its entirely possible that you cant fill them with cubes

[empty grove]

Feels like lebsegue number lemma

Rather than choosing the size of the cube to be less than all diameters you'd want it to be a lebesgue number of the given covering

[fading vale]

Oh

I didnt know that existed

Well that makes this immediate kek

[empty grove]

[fading vale]

i should file that away

and remember it

@kind cedar my question is resolved now if you want to post yours

[kind cedar]

thanks

I was asked to prove that the vector bundle $T\mathbb{S}$ is diffeomorph to $\mathbb{S}^1\times\mathbb{R}$. I managed to prove that in a chart $U_i$ I have a map from TS to SxR, but my professor said I could prove it "globally" instead of locally. I don't know how exactly I should do that. I managed to construct a chart transition map, but what do I do with it?

[gentle osprey]

Necrowizard

[gritty widget]

to prove it globally you'll want to come up with a diffeomorphism

[pearl holly]

do you guys think it's okay if I start a thread here to ask a question?

[gritty widget]

that's what they're for

[pearl holly]

How is this a homotopy?

[kind cedar]

A bijective smooth map? I did it on a chart, isn't it enough?

[obtuse meteor]

can someone help me actually compute a geometric realization?
for some reason I can't seem to verify for myself that geometric realization of horns is like
geometrically what it should be

[gentle osprey]

Nobody

[clever badge]

@gritty widget finally finished my other work and can get back to manis

by the "definition of the tautological one-form" should i be focusing on what they gave me at the start or what i showed in the first part

i wish i had some other example of how to find local coordinates based on some criterion