point-set-topology

[pastel linden]

so isn't the 1-form then dependent on our choice of $X_p \in T_p(\R^n)$?

[uncut geyser]

Im sorry
wym dependant

it can be seen as a function

[gritty widget]

i think one of the things tu means to get at here is that when you consider how tangent vectors act on (germs of) functions, you get a map T_p(R^n) x C_p^\infty(R^n) -> R. tangent vectors act on functions by fixing the first component, which you should be familiar with. however, if you fix a (germ of a ) function in the second component, you get a map that sends tangent vectors to real numbers. if you do this pointwise with a smooth function on R^n, you get the differential (a 1-form)

(this isn't in response to anything anyone said, just interjecting and giving my interpretation of the page of tu bacono linked)

[uncut geyser]

takes vectors to the dir derivative of f

[pastel linden]

sorry wait I think I understand what's been fucking me up now

the notation just writes (df)|p which hides the reliance on choice of X_p

[uncut geyser]

what is (df)|p defined as?

[pastel linden]

so I've been thinking the differential is invariant no matter what vector we choose in the tangent space

[gritty widget]

(df)_p isn't defined relative to a tangent vector X_p. (df)_p is defined as a covector, which takes a tangent vector X_p to the real number X_p(f)

[pastel linden]

OH

[gritty widget]

there's no reliance on tangent vectors in the definition of (df)_p

[uncut geyser]

it takes the tangent vector you put in

if thats what you meant

[pastel linden]

ok that's exactly what I needed

[gritty widget]

and when you do that pointwise, you get the 1-form df

[uncut geyser]

its expecting a tangent vector in its argument for it cooom and give you a dir derivative

[gritty widget]

(the thing (df)_p is called a covector because it's an element of the dual space to T_pR^n)

[pastel linden]

ok that makes a lot more sense

[uncut geyser]

df(v) = vf
if that helps with anything

makes notation less clumsy

[pastel linden]

thank you guys, this has been very helpful

diff geo notation is hell

[uncut geyser]

I only know the diff geo notation

never had manifolds other than there

[gritty widget]

don't worry tu is relatively light on the notation

except for the lie derivative section

[pastel linden]

oh god oh fuck

[gritty widget]

that's section 20 though so dw

[uncut geyser]

this book seems really fun

[pastel linden]

I'm actually really enjoying it so far

I tried learning classical diff geo first and it was terrible but this is significantly more intuitive

[gritty widget]

it's a great book but i think the one major downside is that it doesn't cover a whole lot of stuff

[uncut geyser]

always fun when they define concepts
not fun when they use them abusively and you have to recall

[gritty widget]

which could be seen as a plus as well, i guess

[uncut geyser]

I used O'neil for diff geo

said elementary, but its more like between elementary and modern

[pastel linden]

I just think modern approaches work significantly better

[uncut geyser]

theyre also significantly harder tbh

[pastel linden]

looking at it from a much more algebraic standpoint does wonders

[uncut geyser]

old approaches were like
hey look at this arrow with a direction

new approaches were like
hey look at this derivation over smooth functions in a tangent space

[pastel linden]

well in do carmo he does fucking everything in 2 and 3 space

virtually every proof involves the fucking jacobian

[uncut geyser]

the jacobian is a good tool in diff geo tho
it becomes the basis matrix for maps from Rn to Rm

[gritty widget]

tu is much more of a differential topology book than a differential geometry book

but that's me being pedantic

since do carmo does everything in R2 and R3 there is already a riemannian metric inner product

tu's introduction to manifolds doesn't even touch that kind of stuff

@pastel linden when you're going through the first 5 sections of tu, see how much of it uses the fact that Rn has a standard inner product

might be a fun thing to keep track of

[pastel linden]

given i'm on section 4 and he hasn't used it once I doubt it'll be non-zero

[gritty widget]

he mentions the cross product like once in the lie algebras section

apart from that you can forget all about those geometrical concepts from R3 and R2

[gentle osprey]

gustavn64:

[dusk heron]

I am trying to understand the following construction, where $\sigma^i$ are the standard left-invariant one-forms on $S^3\cong\operatorname{SU}(2)$. As far as I can see, this really only gives a Riemannian metric on $(0,\infty)\times S^3$. How does one go from this to a metric on $TS^2$? My guess is that we are at least supposed to quotient the $S^3$-factor by $\mathbb{Z}_2$ (by identifying antipodal points) to obtain a metric on $(0,\infty)\times(S^3/\mathbb{Z}_2)=(0,\infty)\times\mathbb{R}P^3$, since $\mathbb{R}P^3$ is diffeomorphic to the unit tangent bundle of $S^3$. This gives a metric on $TS^2$ minus the zero section, but how do we show that this metric extends smoothly across the zero section?

[gentle osprey]

gustavn64:

[sleek thicket]

Let φ : G -> H be a lie group homomorphism with dim G = dim H and H connected. If Γ = ker φ is discrete, is φ surjective?

I'm thinking that φ descends to an injection G/Γ -> H which is constant rank since it's equivariant wrt the G action on each (note that G/Γ might not be a group) and thus is an immersion, but since dim G/Γ = dim G = dim H, this means it's also a local diffeomorphism. This means it's an open map, and so im φ is an open subgroup of H. Since H is a connected topological group, it has no open subgroups, and so φ is surjective

I guess this shows more, i.e. it shows that the induced map G/Γ -> H is a (G-equivariant) diffeomorphism

[uncut surge]

the above proof looks good to me

and i now feel justified to ask my own question: Consider the subspace $A := {(x,y,z) \in S^1 \times S^1 \times S^1 : x = y \text{ or } x = z \text{ or } y = z } \subset S^1 \times S^1 \times S^1$. Is there any smart way to approach calculating its singular homology?

[gentle osprey]

Lartomato:

[uncut surge]

oh right this is the fat diagonal of the space, maybe that helps

[rugged swan]

Maybe you can write this as an union of equalizers. But idk if homology is preserved here

[uncut surge]

actually, i believe this was a silly question to ask

What I'm more interested in is relative homology of the pair $(S^1)^3, A$. That may be actually a bit simpler, since you can use that relative homology is (in this case) equal to reduced homology of the quotient space. But I'm not sure what to do with the quotient space

[gentle osprey]

Lartomato:

[uncut surge]

nvm i posted it on stackexchange instead

[rugged swan]

Yeah maybe S1^3/A is a simple spacea

[uncut surge]

I'd expect it to be something like S^3 times S^2; cause that's what the author in the book I'm reading sort of alludes to...

[west spindle]

anyone wanna tackle a hyperbolic geometry problem

[wind portal]

Is axiomatic geometry permissable?

[west spindle]

?

[wind portal]

https://en.wikipedia.org/wiki/Foundations_of_geometry#:~:text=The term axiomatic geometry can,from this point of view.

or is that stuff better placed in foundations?

[west spindle]

...

[wind portal]

We used a book at my college that taught Euclidean geometry like this. I wanted to go back and give it a shot but I didn't know if I had questions where could I post them.

You kindof just start out with a set of axioms and prove as much as you can with it. So we would start with some undefined terms and a set of axioms like what it means for lines to be collinear and stuff and then once we could prove everything we could with those axioms, we added another.

It's very formal so I didn't really know where it should be. In here or foundations.

[marsh forge]

its fine here

but its not clear if you have a question

[wind portal]

@marsh forge No questions atm pertaining to actual math. Just where to post.

[sleek thicket]

@uncut surge could you use excision?

To compute the singular homology of A

We can cover it by A1 = { (x, y, z) in A : x = y } and A2 = { (x, y, z) in A : x = z } and A3 = { (x, y, z) in A : y = z }. Fatten those up a little so their interiors cover A. Then the homology of A relative to A1 is the same as the homology of (A2 union A3) relative to (A1 intersect (A2 union A3)) = { (x, y, z) : x = y = z }

No this is too messy

[signal venture]

following a good conversation with mathemagician a few days back i proved some stuff and feel like i have a good grasp on what the image of a morphism phi:F->G of sheaves is. essentially the elements of im(phi)(U) are elements s of G(U) with the property that for some cover {V_i} of U we have that s restricted to V_i is in im(phi(V_i)). so to me this is sort of like enlarging im(phi(U)) in the smallest way possible which allows gluing to work

anyways i'm feeling really good about images now and i'm moving on to quotients of sheaves. i'm looking for some other intuition about what they look like. hartshorne defines the quotient F/G to be the sheaf associated to the presheaf U -> F(U)/G(U). with enough effort i could write out what the elements of (F/G)(U) look like, but i'm wondering if there's a more intuitive description of F/G, similar to the above scenario with the image sheaf

i imagine it's a similar process in that you want to add in the minimal amount of stuff which makes gluing in F/G work, but this time there isn't a larger object from which to pull from like in he image presehaf case

[tough imp]

I honestly just think about it in terms of that

I feel like I kinda have a feeling for what a sheafification looks like so in my head it’s just the sheafifcatuin of that presheaf haha

[sage hamlet]

So I'm trying to understand a gauge map transformations wrt a Ehressman one form. I know the first term represents local changes in the one form (vertical projection of tangent base vector via lift) but what does the second term mean? How does the Maurer Cartan form act on a vector within the base manifold? My gut tells me it has to do with how the gauge map transforms the base space (but again how does a group element act on a base manifold not the total principle bundle) and it is the lie algebra equivalent of that transformation wrt a direction. So both terms together would be the change in the lie group due to moving within the base space and the change in lie group due to the gauge map transformation.

[gentle osprey]

tolaria:

[honest narwhal]

@meager python at a glance it seems that you need to show that p^{-1}(V) is preserved by the action?

[gentle osprey]

tolaria:

tolaria:

[meager python]

But it feels like it's in different order than I want even

[uncut geyser]

sup

[coarse kestrel]

Yeah that's the counterexample I know

About path connectedness does not imply locally path connected

[uncut geyser]

ah weird
I thought local properties were made to be more general

well yes

wait

why is that space lpc?

ah

why is it not lpc

does 0,0 have some weirdness

that looks surprisingly similar to the counter example of pc<=>c

can you walk me through why no nb of 0 is pc?

cant immediatelly see

cant I just take nbs which vertically involve the wave

oh

is it

[tough imp]

I thought the standard counterexample for one of these is hawaiian earring

[uncut geyser]

ah right
looks homeomorphic to it

but why not lpc?

oh

oh truee

wait

maybe not so true

if you think about it(0,1) is hom to R

so thats not nec so intuitive

ah, but [0,1] compact

wait

oh ye, makes sense

if only there was some lindelof equiv of pc

(there may be)

fking lpc tho

not v pretty of you

I thought of any path

but maybe compact path may be a better way to think

[tough imp]

wait so

[uncut geyser]

well ye

[tough imp]

the like

[uncut geyser]

I just meant intuitive path

[tough imp]

sinusoidal thing on the right going to the point with x-value 0

Hold up

[uncut geyser]

like going from one point to another

[tough imp]

let me get my proof back from where I showed this isn't path connected

Without the loop around

I think that's the same proof for it not begin lpc right?

Yeah

[uncut geyser]

is it compact
not sure

is it bounded

totally bounded

but thats not the space

otherwise itd be lpc

same
2am

closed hausdorff are

but like

uhmm

lemme think

this is weird to think about

it should be compact

but I think the problem is the pseudopath not being

so where is the intuition mistake

for some reason I feel like not using subspace top here

[tough imp]

Okay so I'm just gonna show that the set is the sinusoidal thingy

and the line (0,y) where -1 <= y <= 1

[uncut geyser]

yes

well trivially

compact are always closed and bounded

isnt heine borel that compacts are equiv to that in Rn?

ye
but I meant that always is the case
compact is always closed and bounded

even if the rec is not always true

there is something about it being a line there

[tough imp]

So suppose you could draw a path between (2,0) and like (0,0) so f(0) = (2,0) and f(1) = (0,0)

then f(t) has to assume all x-values between 2 and 0

as you increase x from 0 to 1

Then this says that you can get t_k such that f(t_k) = (1/2k,0) for all k

and t_k are increasing

pass to a convergent subsequence of the t_k

[uncut geyser]

wait wait wait
why cant I just take the nb to be the whole set?

[tough imp]

Lpc is for all open neighborhoods

right

[uncut geyser]

no

uhmm

[tough imp]

Okay so just take a disck

such that you can't loop around

[uncut geyser]

feels diff than usual local defs

[tough imp]

then any smaller one still won't let you

you'd have to make the path through the sinusoidal stuff

But by passing to a convergent subsequence we can assume that t_k convegres to t_0

[uncut geyser]

ah
its that

right

[tough imp]

however as t runs from t_k to t_{k + 1}

f(t) assumes all y-values between -1 and 1

[uncut geyser]

this feels way too strong for a local def

damn topologists

[tough imp]

so that f(t_k) couldn't converge

[uncut geyser]

its not

its weaker than lc

[tough imp]

also

[uncut geyser]

pc

I meant

[tough imp]

The hawaiian earring is an easier example for pc but not lpc I think

[uncut geyser]

like local compactness
weaker than compactness

[tough imp]

It's pc cuz you can go from any loop to the origin

then out on any other path to any other point

but if you draw a circle away from the point all the circles meet

you can't get from one loop to another

No?

Here's a shitty drawing of it

take the open set to be the red circle

intersect it

you can't get from the inner loop to the outer one

??

Oh I think so

you're right

shrink more

lmao

wait

is it locally path connected

but not connected was it?

then...

[uncut geyser]

maybe pc and not c

[tough imp]

what was this a counterexample for

No

pc implies c

[uncut geyser]

wait

the opposite

mb

not pc and c

jfc

every local def is different

[tough imp]

Although {\displaystyle \mathbb {H} }\mathbb {H} is locally homeomorphic to {\displaystyle \mathbb {R} }\mathbb {R} at all non-origin points, {\displaystyle \mathbb {H} }\mathbb {H} is not semi-locally simply connected at {\displaystyle (0,0)}(0,0).

[uncut geyser]

Im just now realizing that

[tough imp]

so apparently... that

idek what that is

[uncut geyser]

local compactness, locally hausdorff and locally homeomorphic are completely diff than those connected ones

[tough imp]

lol

locally compact is meme

contained in a compact set

[uncut geyser]

all of those are implied from their non local counterparts

but connectedness had to be special

[tough imp]

This is because connectedness isn't a local condition I think

[uncut geyser]

isnt comp l comp, l comp paracomp and paracomp metacomp

[tough imp]

Like I have strong reasons to say this

[uncut geyser]

well
compactness also isnt
but Ig youre right
the motivation for classifying local compactness makes sense

same with local homeomorphisms

and local hausdorffism

wer that means

[tough imp]

compact is sort of a local condition

if you're a finite union of compact sets

[uncut geyser]

yeah, you can also be sigma compact and want to be loc compact

like reals

heine borel may be closely related to this

total boundness being equiv to boundness may be similar to the space being locally compact

[gritty widget]

A universal cover is just like an initial cover in some category of covering spaces, right?

up to unique isomorphism I'm sure 🙂

I think it's unique if you take into consideration the "other data"

namely the projection map

at least, that's usually how categorical constructions work

[dim meadow]

Lol

[gritty widget]

I never got that far into topology. All I know is that the circle's universal cover is the line.

and that the covering spaces are just n-fold spinny-wiggle maps

oh neat

[dim meadow]

Z acts properly discontinuously on R by \phi_n(x) =x+n. Quotienting by this action gives S^1, and shows that \pi_1(S^1) = Z

That's a nice way to think about the relationship between R, Z and S^1

There's a nice story re a space being the quotient of it's universal cover under the action of the deck transformation group

Lol

[gritty widget]

(Why is there just one overloaded Wiki page for coverings? 😦 )

[tough imp]

@gritty widget are these linear transformations from the same degree vector spaces?

If so surjective is the same as injective same as bijective

Is this like smooth manifolds stuff?

ISM?

Oh lol you said that

Well

@sleek thicket

You like smooth manifolds

[sleek thicket]

what

The functionals are the rows of the matrix dg_x, right?

The row rank of a matrix equals the column rank

So the image (= column space) has to be l-dimensional

And thus dg_x has to be surjective

@gritty widget

when I say "rows" and "matrix" I'm saying to choose an arbitrary basis, you don't really need to. I'm sure there's a direct abstract linear algebra way to see this

but yeah this is just the fact that if you have an n×m matrix with m > n and with linearly independent rows the associated linear map is surjective

[dim meadow]

The term is fullrank

[sleek thicket]

yee

[sleek thicket]

Haha no worries, I didn't really learn linear algebra until I took a course on smooth manifolds

[tough imp]

Am I being brainlet? I wanted to show that the composition of quasi-compact morphisms is quasi-compact, but I think you only need one of the morphisms to be quasi-compact lmfao

I am brainlet

thank you for coming to my ted Talk

lmfao

I can't just cover the pullback of an affine by quasi-compact sets, I need to know the pullback is quasi-compact

but if you assume both are you can cover the pullback by finitely many sets which are quasi-compact which means...

😉

[gritty widget]

we're all varying levels of brainlet

[west spindle]

hm it looks like i am just the tiniest bit stuck

this exercise from rotman

so obviously what's being hinted at is that i should construct a function $\varphi: I^2 \to I^2$ such that $\varphi(s,t) = (s,t)$ corresponds to $f(s) = g(t)$ (and derive the existence of a fixed point of $\varphi$ from BFPT, which is theorem 0.3 here)... but i'm a bit lost on how to do that exactly

[gentle osprey]

Ann:

[west spindle]

i've tried phi(s,t) = (f(s)+g(t))/2 as the first thing that popped into my mind but beyond this im a bit out of ideas lol

ping me if anyof yall have like. hints or whatever

[timber estuary]

maybe umm ... ||phi(s,t) = ((1,1)+f(s)-g(t))/2|| could work?

||then the range is definitely in I × I and also phi(0,0) = (1/2+, 1/2-) and phi(1,1) = (1/2-, 1/2+) [where k+ means larger than or equal to k, and the same for k-] so the fixed point(?) can be phi(s,t) = (1/2, 1/2) cause only when f(s) = g(t) we can have phi(s,t) = (1/2, 1/2) .. i dont know how this works but um yeah||

sorry if its not helpful

[uncut geyser]

that is a great idea
but the problem is the point would have to be (1/2,1/2)

we can guarantee a fixed point exists

not which it is

I though of the same with (s,t) initially instead of (1,1)
not sure if it is on range tho

and Id need to use the info of the points in g and f

[fervent citrus]

with I=[0, 1], you can maybe try φ(s, t) defined by
|| φ(s, t) = exp(-||f(s)-g(t)||)(s, t)||

[timber estuary]

ooh wait so fixed point means somewhere where f(P) = P?

[uncut geyser]

yes

exactly

there is somewhere I need to use the f and g being in those points hypothesis

meaning Im almost sure a,b,c and d will be somewhere

I tried normalizing f and g
but nothig came out

so something like defining
p = (f -(a,0))/(b-a) and q = (g-(0,c))/(c-d)

now p and q go from (0,0) to (1,1)

I dunno what to do with those ideas tho

[west spindle]

@fervent citrus yeah I = [0,1]

actually hm. that works

or does it?

i think you run into some problems if phi only has (0,0) as fixed point

[uncut geyser]

I dont recall
is bijective part of brouwer?

I assume it is, right?

[fervent citrus]

ah shit

[uncut geyser]

wait @west spindle is I said to be [0,1]?

[west spindle]

yeah

[uncut geyser]

thats why it didnt look like an intersection would nec happen at first. I dum

[timber estuary]

huum a can equal b, no ?

[uncut geyser]

ye

[timber estuary]

and c can equal d ?

then q isnt defined

[uncut geyser]

I assume they could

ye
those are ideas Im trying

p and q wouldnt

[coarse kestrel]

Is that continuous tho

[gritty widget]

phi(s,t)=|1-(II f(s)-g(t) II)/(10^100)|(s,t)

works

[coarse kestrel]

Is the image IxI?

[gritty widget]

wha

[coarse kestrel]

Well for brouwer to work you need the codomain and domain to be the same

[gritty widget]

now it is

[fervent citrus]

it still has the same issue of potentially having (0,0) as the only fixed point

[gritty widget]

ah

[coarse kestrel]

Also I think brouwer doesn't require the map to be bijective

Not even surjective

[gritty widget]

only continuous

[coarse kestrel]

yeah

[coral pawn]

I had a question about subsheafs

Let F be a sheaf and f_UV be the family of maps

Then F' is a subsheaf if F'(U) \in F(U) and f'_UV is the restriction of f_UV

Say we are given some U, V open sets such that V \in U. Then how do we know that f'_UV(U) will be contained in F'(V)?

[gritty widget]

sorry quick question

probably stupid but what is \eta_U here

[burnt spruce]

not sure what all the notation is , but I think there's a function $\eta_U$ for each open set U

[gentle osprey]

lorenzo:

[burnt spruce]

(is this Peter May's book?)

[gritty widget]

ya

[honest narwhal]

His hypersquiggle categories are just an immediate identifier

[burnt spruce]

at first i thought hypersquiggle was a technical term

like, hypersquiggle cohomology

[gritty widget]

what are those maps though

[honest narwhal]

Lemme pull up Concise and find out

[gritty widget]

sorry thanks

[honest narwhal]

Oh so, it's like

Given eta

You're trying to verify Pi(X) is the colimit of a diagram, so it's like, alright for each eta build eta squiggly

[tough imp]

Say we are given some U, V open sets such that V \in U. Then how do we know that f'_UV(U) will be contained in F'(V)?
@coral pawn This follows fromt he line above

Oh wait

no you also require that

F' need to be a sheaf

[gritty widget]

oh I see lmao sorry this is confusing

[honest narwhal]

Lol it's all good, the wording threw me off too

[burnt spruce]

someone should write a 'non-concise guide to peter mays concise AT'

[honest narwhal]

I've entertained for some time the possibility of writing an algebraic topology book that's pretty formal and covers a lot, a la Concise, but is longer

Except that book kinda already exists as I've found out

"Algebraic Topology" by Tammo tom Dieck

(Fwiw I don't know the subject myself, the idea was that I'd be writing it as I learn)

(So basically texing lecture notes but without lectures)

[burnt spruce]

writing a book is (ive heard) a terribly time consuming project

like, even writing a paper is terribly time consuming even when you have all the ideas down

so the general advice to grad students is to not write books ( tho some people try and are somehow successful)

i personally like the idea of learning through writing and teaching

generally feels more motivated to me

[gritty widget]

uhhhgg I forgot that a morphism of diagrams is a natural transformation

[burnt spruce]

yeah isnt that cool 🙂

[honest narwhal]

Yeah I definitely see that if the act of writing something isn't otherwise doing something for you (e.g. lecture notes) then I'd prob be worried to do so pre-tenure

[burnt spruce]

yeah. but honestly maybe for some people it can be a source of energy, not a drain, idk.

personally I find writing math really hard, but maybe being in the practice of writing a lot bc of lecture notes would help

[honest narwhal]

That's prob a good point

Anyway yeah I would recommend checking out tom Dieck lorenzo

I think once I decide to learn algebraic topology "for real this time" it'll be through either that or Bredon's Topology and Geometry

Much more detail than Concise and much more formal than Hatcher

[burnt spruce]

I ended up using Hatcher mostly

[honest narwhal]

Also ninespam in case you get stonewalled in Concise, TD is worth looking into

[burnt spruce]

and supplementing

with lecture notes and stuff

there are some really good lecture notes on bundles and homotopy / BG kind of stuff

[gritty widget]

yea ok I have a feeling that’s going to happen sooner then later

[burnt spruce]

https://www3.nd.edu/~mbehren1/18.906/prin.pdf

this was great

[honest narwhal]

Oh Behrens

I applied to Notre Dame and was considering him as a possible advisor

Though prob preferred Andy Putman

[burnt spruce]

notes are by mitchell though

this is also a good supplement: https://math.berkeley.edu/~hutching/teach/215b-2011/cup.pdf

people always talk about how cup products are dual to intersections , but this was the only place that I saw it written down in a way that I was able to digest

[honest narwhal]

Ooh I should read this for sure

Yeah same I know what's supposed to happen but not the proofs

[tough imp]

Am I being dumb?

Yes

[burnt spruce]

I am dumb too

[sleek thicket]

Yes

[burnt spruce]

😦

[sleek thicket]

Sorry I didn't mean you're dumb

Just chmonkey

[burnt spruce]

hey why does chmonkey get to be called dumb and not me 😦

[sleek thicket]

I don't want to claim things I can't back up

[burnt spruce]

fair enough

[gritty widget]

this exercise from rotman
@west spindle
how about restrict the domain to [a,1] for some a , since there clearly exists a interval [0,a] on which f(s)≠g(t) by their construction

that will resolve the (0,0) being the only fixed point problem

And instead construct phi : [a,1]^2→[a,1]^2

[west spindle]

hm yeah you're right sounds like it will

it feels a bit hacky tho

[gritty widget]

yeh

but there exists such 1>a>0

[west spindle]

i don't doubt the validity of this fix

it's the inelegance that i'm unhappy about

(also why are you writing your interval boundaries backwards lmao)

[gritty widget]

eh

Proof is Proof

[willow spear]

How does someone read through this, any advice:

http://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

[tough imp]

https://discordapp.com/channels/268882317391429632/496785752936546304/741464294377521253
@gritty widget

it happens to all of us lmao

https://discordapp.com/channels/268882317391429632/496785752936546304/741860565877784716

Also this lol

I think I agree

after algebra it's the most active I think

[uncut geyser]

thats because top kinda intuitive

notice I said kinda btw

[tough imp]

So I think I'm almost done with II.3.20 a) of Hartshorne, the problem is to assume that X is an integral scheme of finite type over a field k. Then to prove that for any closed point $P \in X$ that dim $X = \text{dim }\mathcal{O}_P$

[gentle osprey]

Chmonkey:

[tough imp]

So we can cover $X$ with finitely generated $k$-algebras, and then it follows that $P$ is a maximal ideal in whatever Spec, then for finitely generated $k$-algebras we know that ht $P +$ coht $P = \text{dim }A = \text{dim Spec }A$

[gentle osprey]

Chmonkey:

[tough imp]

Vakil says that you can show that a scheme has dimension n if and only if it admits
an open cover by affine open subsets of dimension at most n, where equality is
achieved for some affine open subset. Hint: You may find it helpful, here and later,
to show the following. For any topological space X and open subset U ⊂ X, there is
a bijection between irreducible closed subsets of U, and irreducible closed subsets
of X that meet U.

I believe this, this at least says that some Spec $A$ has dimension equal to the dimension of $X$ itself, the problem I'm having is to show that the dimension of all Specs in that cover are equal

[gentle osprey]

Chmonkey:

[tough imp]

Essentially this says that if $X$ is an integral scheme of finite type over a field that every open affine has the same dimension, now if there is a closed point in the intersection of any two open affines then I can use that closed point to get an equality of the dimensions of those two affines, but I don't see how that's possible? I thought I could look at the generic point but it actually doesn't give me equality of the dimensions I think

[gentle osprey]

Chmonkey:

[uneven stratus]

hey ppl i'm getting introduced to topology since last week, and i've got one question

is the notion of continuous functions between metric spaces the same as a continuous function definition in calculus and the ones in numerical analysis (c0, c1, cn continuity) or more general, or more loose?

or totally different?

from what i could gather, it's the same thing as c0

[small obsidian]

It is more general. A continuous function on a metric space is, of course, something you can place on any metric space.

Whereas in calculus, you're only looking at continuous functions on R or Rⁿ

Yes, if your metric space is Rⁿ, then the two concepts are the same.

[uneven stratus]

so that would also apply to spaces that are subsets of Rn, or manifolds, or things like projective space?

that's where the difference lies?

just throwing some examples

[cursive flume]

hello! I have a question, I am missing something, but idk what. somehow it seems to me that every two vector fields commute if you expand them in charts

where do I miss the point?

if you say there are two vector fields X,Y and you take their commutator [X,Y] and expand them in a chart U(x) they will look like $[X^a \frac{\partial}{\partial X^a}, Y^b \frac{\partial}{\partial X^b}]$

[gentle osprey]

ProphetX:

[cursive flume]

you can take out the component functions I think cause they are just functions and you remain with the two 'partial derivatives' which actually commute because of schwartz's rule

however it is not intuitive, that if X,Y do not commute, in a chart why would they commute..

i'm asking this question regarding Riemann tensor-and deriving its components

Riemann tensor field is defined as follows:

if you want to compute the components it turns out that the last term will vanish.. but if it vanishes then it should vanish for any arbitrary vector field WHEN YOU GO IN CHARTS

however this is not intuitive

what am I missing?

[cursive flume]

or is it true that every 2 vector fields commute locally, but not necessarely globally too?

[coral pawn]

When proving surjectivity, how do we know that such s_i exist?

[tough imp]

This is just

By definition of surjectivity on stalks

If you have a t in G(U), then t_x is in the stalk, so there has to exist some element of the stalk which maps to it

That’s saying some <s_i,U_i> such that it maps to t_x, so after restricting to the intersection this just says s_i maps to t|_{U_i}

[coral pawn]

uhhh

[tough imp]

Do you know what the maps of stalks look like?

[coral pawn]

ivory hasn't been introduced yet

[tough imp]

Whoops

[coral pawn]

Map of stalks is a map on the direct limits right?

[tough imp]

Yes but there’s an explicit formula for it

In terms of the <s,U>

Definition of elements

If you give me a few minutes I can type it all out

But rn I’m on phone so

[coral pawn]

Yeah no hurry

[tough imp]

Okay so, I will not show why everything is well-defined, you can do so if you wish

but first of all elements of $F_x$ are equivalence classes $<s,U>$ where $s \in F(U)$ and $<s,U> = <t,V>$ if and only if there exists some $W\subseteq U\cap V$ such that $s|{W} = t|{W}$, this is a bit different from what I said before cuz I forgot you need to include $W$

(Here all sets are open sets, and contain $x$

[gentle osprey]

Chmonkey:

Chmonkey:

[tough imp]

You have a group operation on this, via $<s,U>\cdot<t,V> = <s|{U\cap V}\cdot t|{U\cap V},U\cap V>$

[gentle osprey]

Chmonkey:

[tough imp]

Now for each open set $U$ containing $x$, you have a map $F(U)\to F_X$ given by mapping $s$ to $<s,U>$

[gentle osprey]

Chmonkey:

[tough imp]

Now, given a map of sheaves $\varphi:F\to G$, we get a map $\varphi_x:F_x\to G_x$ by the formula

[gentle osprey]

Chmonkey:

[tough imp]

$\varphi_x(<s,U>) = <\varphi(U)(s), U>$

[gentle osprey]

Chmonkey:

[tough imp]

Try to show this is well-defined

but this makes a commutative square with the maps $F(U)\to F_x$ and $G(U)\to G_x$ and with $\varphi$ itself

[gentle osprey]

Chmonkey:

[tough imp]

So now suppose you have surjectivity of stalks

given a $t \in G(U)$, we have $t_x = <t,U>$ in $G_X$

[gentle osprey]

Chmonkey:

[tough imp]

we then have some element of a stalk, which is something like $<s,V> \in F_x$ such that $\varphi_x(<s,V>) = <\varphi(V)(s), V> = <t,U>$

[gentle osprey]

Chmonkey:

[tough imp]

that last equality tells us that for some $W\subseteq V\cap U$, that $\varphi(V)(s)|_{W} = \varphi(W)(s|_W) = t|_W$

[gentle osprey]

Chmonkey:

[tough imp]

so let this $W = U_x$, and let $s_i = s|_W$

[gentle osprey]

Chmonkey:

[coral pawn]

I think I kinda get what you're trying to say

I was using the universal property definition of the direct limit so I'll try and translate this into that

[tough imp]

Yeah I mean

there's a time and place for both

100%

But from the universal property you can get it too

I tended to use the universal property in the exercises

but I remember specifically in showing that f_* and f^{-1} are adjoint

I ended up doing an element wise sort of construction

right at the end

since f^{-1}F is defined pretty similarly to the stalk

just you go with a colimit of open sets containing some set, not just a point

[coral pawn]

This definition confuses me a bit. Why is f# a function into fO_X but f#_x is a function into O_X,x and not fO_X,x?

[tough imp]

https://discordapp.com/channels/268882317391429632/496785752936546304/742222760507605052

Just a bump for a question I had yesterday

Still not too sure how to approach this

[west spindle]

should i do all the category theory exercises in rotman chapter 0?

i have like... a passing knowledge on the very basics of cat theory

[honest narwhal]

It covers everything it needs in exposition. But the exercises aren't too involved so tbh it's fine if you just glance at it and say

Okay I could definitely write the details down

[west spindle]

thats what i did

i just don't know if i should expend my energy (which is a rather scarce resource at the moment) actually writing down the details

[honest narwhal]

Probably not in that case. Having gone through them there's not much to it

[willow spear]

Are you asking specifically about these notes, or just how to read mathematical writing in general? @willow spear
@gritty widget both

[willow spear]

@gritty widget Any good texts for analysis?

[willow spear]

ok thanls

*thanks

[willow spear]

ok ic

how long would it take to go through an analysis book

yeah

Im mainly reading topology as a background for a reserach project

with a professor

no clue lol....i'm interested in pure math and im goign to be contact professors soon to see if i can do something relatated to topology

ok

yeah

so there are two types of analysis right?

real and complex

Yeah eventually my goal si to read through algebraic topology

oh ok

ok cool

yeah so im goign to do some analysis reading then for now

yeah true

so this the track then:

real analysis

point set topology

algebraic topology

no lol im in HS

rising junior

yeah thats what i was thinking too

ive done linear algebra

oh

no actually i learned as part of a AI/ML program

so defintely not proof priented

yeah also....you said a consice book for point set topology

any suggestions

oh lol

alright cool

also is topology a doable field to do research in

are other areas better

yeah fs

true

makes sense

so the recap

rudin would be fine then right for analysis

ok thanks...welll then ive got get to analysis grinding

thanks a lot for your help

yeah caus ei was reading through it and some of the stuff amde sense

its just the notation and stuff that trips me up

^word

oh ic

yeah hopefully

ok

[crisp sparrow]

Napkin by Evan Chen is quite interesting

yea im currently doing it now, and i like it a lot

yea that is the plan

as long as it doesnt get too complex but i think it should be ok

[willow spear]

wait should i read Napkin...is it a good start

i mena the oarts related to topology lol

[crisp sparrow]

i like it as it is mostly goes over the important and interesting parts

not too thorough

[willow spear]

ok will check out.....are there any other good youtube video for PS topology and beyond

ok np

[cursive flume]

how can one define rotations on manifolds coordinate free?

does anyone have a good resource on this?

[uncut geyser]

search for manifold isometries, probably

[tough imp]

Let $X$ be a topological space, then call $X$ irreducible if whenever $X = Y\cup Z$ can be written as the union of two closed subsets, then either $Y = X$ or $Z = X$. Now suppose that $X$ is a topological space, $U\subseteq X$ an open set, and $V\subseteq U$ an irreducible closed subset of $U$. Show that there exists some irreducible closed subset $Z\subseteq X$ such that $V = Z\cap U$

[gentle osprey]

Chmonkey:

[tough imp]

Essentially I want to show that I can extend $V$ to a set which is irreducible and closed in $X$, not just $U$, but I'm not sure how I can go about this. I know I can embed it into some irreducible closed set, but I don't see how I can do this

[gentle osprey]

Chmonkey:

[bleak helm]

Is your problem statement correct? Otherwise what prevents$ Z = V$ so that $Z \cap U = V \cap U = V$

[gentle osprey]

Lunasong:

[tough imp]

V is not closed in X

V is closed in U

[bleak helm]

Ah, can't read :(

[tough imp]

No worries

[tough imp]

I had a solution, but now I think it doesn't work unless I have some brilliant fix

Idea, let $V\subseteq U$ be an irreducible closed subset, then let $Z\subseteq X$ be minimal among closed sets such that $Z\cap U = V$ (this is the issue, why does $Z$ exist?). Suppose that $Z$ weren't irreducible, then $Z = Z_1\cup Z_2$ for proper subsets, but then $V = (Z_1\cap V)\cup (Z_2\cap V)$, since $V$ is irreducible this implies that $Z_1$ or $Z_2$ contains $V$, WLOG $Z_1$ does, but this contradicts $Z$'s minimality.

[gentle osprey]

Chmonkey:

[tough imp]

So can we show there must exist a minimal such $Z$?

[gentle osprey]

Chmonkey:

[bleak helm]

Isn't it always the case that if V is closed in U then V = U int Z for a closed set Z in X?

[tough imp]

Yes that's the definition

but how can you take a minimal one?

[bleak helm]

Oh, I see

[tough imp]

What stops there just being an infinite chain down?

Oh wait,

I can just take the intersection of an entire chain I guess?

I can do Zorn's lemma I think

[bleak helm]

Yeah

[tough imp]

Cool

sweet I'm done

[bleak helm]

Yay

[uncut geyser]

the minimal closed set would be taking the closure of V relative to X, right?

[tough imp]

Oh yeah, I think that's true

WAIT

[uncut geyser]

thats the def of closure

[tough imp]

I already know that the closure of V is irreducible in X

..............

Lol

This is like a fact I have seen proven before

☠️

Bruh moment

Lmao

You know what this isn't even the worst, I spent like 3-4 hours stuck on a problem over two days, and I knew if I can show that the intersection of two open affine sets has a closed point, that I would be done

After struggling wondering "how can I show that's true" I emailed my professor and he told me to prove that the set of closed points are dense given my conditions, so that I don't just have 1, I have a dense set of them

Then I realized the last problem I proved was showing that the closed points are dense...

[meager python]

Does anyone have mumfords abelian varieties in pdf form?

[bleak helm]

@meager python I don't, but I checked and it's on libgen

[meager python]

Can’t find a download link there 😦

[bleak helm]

@meager python Can you send documents on discord? If yes, I'll download it and dm to you

[meager python]

I think so, yes. Would appreciate it

[bleak helm]

Okay np

[dusk heron]

What's a simple way of computing the homology of $(S^2\times S^1)/\sim$, where $(x,y)\sim(-x,-y)$?

[gentle osprey]

gustavn64:

[random sequoia]

@dusk heron can we say that the given space is homeomorphic to RP2 x RP1?

[random sequoia]

Nah that's probably not even true

[cursive flume]

why is the metric defined on a smooth manifold different from the one on a set?

one can make a set a manifold, do the tangent space,bundle structures and then talk about the metric

why is the thus defined metric different from the one defined on a set?

it should be just merely a special case of that

but on a set we define the metric as follows:

and on a smooth manifold, we define the metric as follows:

well its defined on the manifold

and the manifold is a special structure of a set

[gritty widget]

is this not a riemannian metric?

[cursive flume]

im not sure,he defines it as just a metric

[gritty widget]

if it's a riemannian metric / related to a riemannian metric then it's just abuse of terminology i guess

[cursive flume]

later on he says riemannian metric is just where the sign is +++++

[gritty widget]

edited my message

that's my take

[cursive flume]

so they have nothing to do a-priori with each other?

the thus defined g should be a special case of the metric on a set

[gritty widget]

semi-related is that a riemannian metric gives you a "topological metric"

[cursive flume]

hmm yes this makes sense

but my brain somehow cant take it that we could take a bare set,equip it with a metric

or we could add more structure and just then equip it with a metric

why are these 2 not related?

it makes no sense to me

not really

i just studied this

but yes you ae right,you measure on tangent spaces

but that induces a notion of length of a curve on the manifold

first you have tangent space-velocities(as vectors), inducing metric you get speed, and integrating speed you get length

yes,at each point of each tangent field

so its an inner product field

a smooth inner product field basically

isn't every smooth manifold riemann-metric compatible?

i thought it is

like you can establish a riemann metric on any smooth mfd

ye,makes sense

[gritty widget]

aren't all second-countable manifolds metrizable by urysohn

true

i think so ultraproduct

regularity is local (or can be stated as a local thing) so you just say "locally euclidean" to finish the proof lol

(not really, but i think that you can do it this way without using any fancy topology theorems)

depends on their definition of a manifold i guess

hausdorff lets you have things like flows of vector fields

i was assuming that "manifold" meant "second countable hausdorff locally euclidean space" so there mightve been some confusion there ultraproduct

mb

prophetx's definition of a manifold (i checked the notes they are reading) does not include hausdorffness it seems

just locally euclidean

[jagged ocean]

I apologize if this is inappropriate to share our own work like this, but reading the conversation about the inclusion of the Hausdorff condition in the definition of a manifold made me want to share this paragraph I wrote as a part of a document describing/motivating each component of the definition of a smooth manifold.

I don't pretend to be very knowledgable about these things, but this kind of summarizes what I found when I was looking into the same question

[uncut geyser]

nice text

[cursive flume]

ok so I got an answer:it is different,because if you want to impose a metric as tensor field, you have to make each fibre of the tangent bundle( the tangent spaces) into metric spaces. they rae just vector spaces. if you impose a metric on each of the tangent spaces and require it to be invariant under change of charts you will get the tensor-field point of view

so the tensor-field point of view is more than simply a metric on a space

that's why they do not coincide

however constructing the tensor field structure you need to use the classic metric defintion on each tangent space

[sacred jolt]

assume this is in R^2. how can U be open to A but not to the metric space R^2?

i'm reading POMA and it's not clicking for me; remark 2.29 is just totally not doing it for me

[small obsidian]

I'm guessing the dark green boundary is contained by A

[sacred jolt]

yes

U \cup V is an open ball

[small obsidian]

Actually, what's your definition of open? Haha

What's POMA?

[sacred jolt]

baby rudin

[small obsidian]

Ah good okay

[sacred jolt]

open -> there is a minimum upper bound r on d(p, q) for any p, q in the open set such that d(p, q) < r

this is where he is doing so

i'm struggling with it

[small obsidian]

Rudin says that a set is open if every point in it is interior

And says that a point is interior if there's a neighborhood of the point contained in the set

[sacred jolt]

indeed

[small obsidian]

Also says this

[sacred jolt]

was trying to find that again, i'm picking up where i got stuck yesterday

i'll stare at this for a while and see if i can get it to click

mmh, that's what my friend was saying

is the idea here that, say, for example, i can take a half ball and go to the "closed" (better word needed) boundary somewhere near the arc and say like

[small obsidian]

Rudin is terse haha

[sacred jolt]

there's no like

max distance i can go away from this point on the boundary

and still be in the half ball

because all points aren't interior

but then, isn't U just the same relative to A?

ah i see

balls around points in U which are not interior points do not contain anything in A that isnt in U

so that's what's being looked at

[small obsidian]

We care about the metric space that A creates now. Some point on the boundary is still interior, because all points that are a "distance" r away are in A.

I say "distance" because "distance" in A is different from "distance" in R²

[sacred jolt]

why is distance in A not like distance in R^2?

ah yes

well

actually i think it should be

[small obsidian]

If I looked at a neighborhood on A's boundary, with reference to R², the neighborhood would leave A.

Instead, with reference to A itself, the neighborhood can't leave as that doesn't make sense

[sacred jolt]

neighborhoods around points in the boundary of the closure of U do not contain anything in A that isn't in U

wait

there's a lot of stuff going on at once in this construction and it's hard to really pinpoint one specific thing to focus on

OH

YES!

that's what i needed to hear

thank you

[sacred jolt]

i notice that a lot of the proofs in rudin use notions of n-balls and k-cells and whatnot; in point-set topology it seems like a lot of the time illustrations will use these really uniform set shapes and it makes me wonder: do i lose anything in doing this to visualize and work through proofs? doesn't this, working with open/closed disks and rectangles and connected regions of the real line lead to some failure to deduce some property about non-connected, not-simple sets?

is that something that will be addressed later on once i've covered the simple topics like compactness and connectedness?

can i safely use these visualizations to "cement" my intuition for these ideas, or do i need to think about sets which are not connected when i read over these basic concepts

and then later on, when we deal with disconnected sets, be able to apply that intuition from before to successfully understand what's going on

or do i need to fully memorize the theorems and corollaries symbolically in order to generalize to such cases?

i suppose not; i should just read, read, read and not fret too much about oddities until i come across one

[granite copper]

I don't know if the answer to your question can be a calming yes, but what I have to point out is that the all-important Zariski-topology in algebraic geometry is non-Hausdorff and not induced by a metric so I suppose from its properties it's quite different.

[marsh forge]

zariski topology isnt topology tho

yes use these visualizations

you dont need to remember anything ‘symbollically’ ever really

but you should know the details that make your intuition precise

[tough imp]

I don't know if the answer to your question can be a calming yes, but what I have to point out is that the all-important Zariski-topology in algebraic geometry is non-Hausdorff and not induced by a metric so I suppose from its properties it's quite different.
@granite copper I still visualize stuff in Zariski topology, but I just draw big open circles to draw open covers and stuff, and to think through some normal point set topology stuff. Just as long as you let go of any sort of distance-based geometric intuition it works for me

[meager python]

Every topological space that is the inverse limit of finite T_0 spaces is homeomorphic to a Spec R for some ring R

Or equivalently a quasicompact space with a basis that is stable under intersections and that is sober

[uncut geyser]

for the above problem, if it is about metric spaces, just notice the definition of open ball in A is different than R^2

[sacred jolt]

thank you, moxwell

[granite copper]

@meager python Can you say that in a more basic way again?

[sacred jolt]

how can a metric space be a closed subset of itself?

[marsh forge]

What do you mean

How are you defining closed here

[sacred jolt]

i'm reading rudin's theorem 2.33 in baby rudin

[marsh forge]

How are you defining closed here
@marsh forge

[sacred jolt]

is it rigorous enough to just say 'it contains all its limit points'

[marsh forge]

Sure

[sacred jolt]

ok that's straightforward

[marsh forge]

Do you see it then?

[sacred jolt]

yea, easily

thanks

[marsh forge]

np

[tough imp]

[tough imp]

I cannot make sense of the scheme-theoretic image of a scheme

how do you define it when f isn't qcqs?

[tidal forge]

I'm reading through Francis Su's "Topology Through Inquiry", and came across the following theorem:
Theorem 6.8. Let A be a closed subspace of a compact space. Then A is compact.
I'm a little confused why this was chosen as a theorem. Isn't any subspace of a compact space compact?

[tough imp]

No

It's stated as is because it's false if A is note closed

take the interval [0,1] which is compact in R, then consider the subspace (0,1), then an open cover of (0,1) by the sets (0, 1 - 1/n)

No finite subcovering will still cover (0,1)

[tidal forge]

Aaaa I see. I'll have to think about it a little more but I think it's starting to make sense. Thanks!

[cursive flume]

does the push forward map only push forward tangent vectors from a manifold to the other, or you can push forward whole fields by it? do you need additional structure to do that?

[nimble jolt]

You'd want the map to be a diffeo, otherwise you get two issues:

1. Not all points in the codomain will be in the image of the map, so how would you assign a tangent vector to these points?
2. Some points in the codomain will be hit by the map several times, this gives us multiple choices for tangent vectors at these points, and no natural way to single one out. @cursive flume

[cursive flume]

how can you 'push forward' whole fields?

which object does it,and how is it defined?

[nimble jolt]

if the map F:M->N is a diffeo, and V is a vector field on M, you can define the pushforward F_*(V) to be the vector field on N that takes a point F(p) on N and spits out DF(p)(V_p)

I.e. you just pushforward each vector comprising the vector field on M

[cursive flume]

so if i say the map is a diffeo,then this def works,right?

this doesn't contain that the map is invertible and inverse is smooth

and also says that only VECTOR FIELDS can be pushed forward, and only covector fields can be pulled back. if the map is invertible/diffeo then one can pull back vectors and push forward covectors too

is this wrong?

[nimble jolt]

it's a little subtle, read: https://en.wikipedia.org/wiki/Pushforward_(differential)#Pushforward_of_vector_fields

The subtlety comes about from the injectivity/surjectivity issue I mentioned above, and is only relevant for talking about fields. For individual vectors you can push them forward and covectors you can pull back always, and if the differential is invertible then you can pushforward and pullback both.

[cursive flume]

but if i make the map to be invertibe and the inverse be smooth, everything works for fields just as fine as it does for vectors,right?

[nimble jolt]

yep, because the obvious way of pushing forward each individual tangent vector works.

[cursive flume]

thank you! 😄

[nimble jolt]

no worries 🙂

[cursive flume]

does anyone know papers/books discussing the hawking-penrose singularity theorems from a mathematically rigorous PoV?

[mellow latch]

if an affine transformation doesn't involve translation, is it always an isomorphism?

[cursive flume]

does it hold for the connection coefficient functions,if you multiply the symmetric part by the antisymmetric part,then it is 0?

since they are not tensors

[dusk heron]

Consider a Riemannian 2-manifold with boundary $M$, and assume that $M$ is non-compact and simply connected (and therefore orientable). Furthermore, assume that we have a harmonic function $f:M\to\mathbb{R}$ which vanishes on the boundary, and is positive on the interior. We can view the interior of $M$ as a Riemann surface, and by the uniformization theorem it is conformally equivalent to either $\mathbb{C}$ or the upper half-plane $H={z\in\mathbb{C}\mid \mathrm{Im}(z)>0}$. The former case can't happen, since the complex plane has no non-constant positive harmonic functions, so $\mathrm{int}(M)$ must be conformally equivalent to $H$. Sounds correct?

[gentle osprey]

gustavn64:

[gritty widget]

Can someone help me find a clear definition of "(non-)singular curve"?

[gritty widget]

Can you clarify?

I hope this helps but I'm specifically talking about algebraic curves

Does it suffice to just say "every point has a tangent line"?

Also I'm not sure whether this is equivalent to just having a self-intersection or not but I suspect so

Thanks

[tough imp]

Over what field though?

I think over an algebraically closed field derivative 0 means self intersection... but I miiiiiiiight be wrong

[gritty widget]

well I ended up finding a clear definition after getting to the exercises lol

so gabe u got it xd

[noble dock]

Question

In 3-dimensional space, if you take an object and two perpendicular axis of rotation relative to the object, you can achieve any rotation you want.

You don’t actually need the third axis is my point

Does this generalize to higher dimensions? Does an n-dimensional object always require n-1 axis of rotation to be able to trace out all possible rotations, or is there a “cheaper” way?

[noble dock]

Actually, do objects in higher dimensions even have “axis” of rotation?

Or is rotation a 2-dimensional phenomenon, meaning an n-dimensional object has an (n-2)-dimensional object it rotates around?

Are there transformations that correspond to higher dimensional versions of rotation?

[tough imp]

I remember people talking about this

And I want to quality that my memory isn’t great and I’m not an expert

But you have to specify what you mean by rotate

But I think the definition I heard that’s the right one (someone correct me if I’m wrong) is that you fix a 1-dimensional sub space

Which is the axis of rotation so to speak

But I definitely might be wrong here so take that with a grain of salt

[chrome dew]

axis of rotation is a 3D thing, rotations really happen in a 2D plane

[uncut geyser]

arent some elements of the euclidean group rotations around hyperplanes?

I think those are called axes too, right?

[uncut geyser]

the hairy ball

[gritty widget]

oh hey my manifolds class is also doing a sort of project

[dusk heron]

Let's say we have a smooth 2-manifold (surface) with boundary, $\Sigma$, assume that $\mathrm{int}(\Sigma)$ is endowed with a complex structure (which a priori does not extend to the boundary). Now let $f=u+iv$ be a smooth complex-valued function on all of $\Sigma$, which is holomorphic on $\mathrm{int}(\Sigma)$. Assume that $u>0$ in $\mathrm{int}(\Sigma)$ and $u=0$ on $\partial\Sigma$. Can we say anything about the directional derivative of $u$ at the boundary, the derivative being taken along an outward-pointing direction? For instance, is it necessarily non-zero? Can we say anything about that same derivative of $v$ at the boundary?

[gentle osprey]

gustavn64:

[dusk heron]

I think what the first question boils down to is really: We are given a smooth surface with boundary, equipped with a Riemannian metric defined only on its interior, and given a function which is positive and harmonic on the interior, and vanishes identically on the boundary. Is the normal derivative of that function non-zero?

The Hopf lemma seems to say that the answer is yes, under the extra assumption that the metric is defined on the boundary too, but does this still hold when the metric is not defined on the boundary?

[tough imp]

I HAVE SINCE SOLVED THIS
Given a quasi-coherent sheaf $\mathcal{I}$ associated to a scheme $X$, I tried to define the closed subscheme asssociated to it as $(\text{Supp}(\mathcal{O}_X/\mathcal{I}),\iota^{-1}\mathcal{O}_X/\mathcal{I})$, where $\iota$ is the inclusion of Supp$(\mathcal{O}_X/\mathcal{I})$ into $X$. I've shown that for an affine neighborhood Spec $A \subseteq X$ that the open set Spec $A\cap\text{Supp}(\mathcal{O}_X/\mathcal{I}))$ of Supp$(\mathcal{O}_X/\mathcal{I})$ is equal to $V(\mathcal{I}(\text{Spec }A))$, which is splendid, the only issue I'm having is showing that this scheme restricted to $V(\mathcal{O}_X/\mathcal{I}(\text{Spec }A))$ is actually affine, presumably it's supposed to look like Spec $A/\mathcal{I}(\text{Spec }A)$, or really the image of the closed subscheme of Spec $A$ associated to this scheme, but I can't for the life of me compute what happens to the sheaf because of this annoying $\iota^{-1}$.

[gentle osprey]

Chmonkey:

[surreal flint]

From

#❓how-to-get-help

Rule 7

When asking for help, do not insist on getting just the answer; we are here to help you learn, not cheat. Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer.

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[modern trail]

and this is not even topology

[fading vale]

its geometry lol

unless something was deleted and you're not referring to magician's message

[ivory dragon]

the message was deleted

something about topological sortings of a digraph

[muted swift]

Let's say I have a given set of points in an arbitrary known shape(8 verticies), and I want to convert this arbitrary shape to a cube (n dimensional) via minimizing the amount distances between points are changed (nonlinearity ok). Is there a "canonical" way to do this?

In addition, this arbitrary shape is infinitely repeating, with some verticies connecting. In the case of a cube, it's an infinite amount of cubes with each cube having another cube touching all faces.

So distances to "outside" shapes need to to be considered.

[marsh forge]

It is very unclear what you mean

You have 8 points in what space? R^n?

You want an n-cube with 8 points?

What do you mean here by n-cube?

[muted swift]

A cube in n dimensions. 8 verticies for the arbitrary shape. N points. Arbitrary shape lies in R3

All arbitrary shapes have the same points inside of them. I want to convert this arbitrary shape to a cube, while changing the distances between the points (a point to all other points, including all neighboring shapes) the least.

Naively, I could just "stretch" the 8 verticies to a cube in R3, but I don't think that's the ideal solution.

I want to keep distortion to a minimum, so if I need to project to a higher dimension that's fine.

[marsh forge]

You cant use 8 points to create an arbitrary n cube

A cube in R3 has 4 verticies

Im really not following

Can you explain in a longer message and make sure to define your terms

Ie what does distortion mean

[muted swift]

In r3, since cubes have 8, and my arbitrary shape had 8, I can naively stretch space to convert the arbitrary shape to a cube.

But you're right. I can't do that for n-Cubes.

I'll type up something more detailed. Give me second.

Here's an example in R2

there is an infinite amount of these shapes, each with their own points inside of them.

In this case, lets say I want to convert this arbitrary shape to a square.

the resulting transformation might look something like this

we can look at the ratio of the side length and the distance between points

and see that this isn't a perfect transformation (im pretty sure)

my question is primarily:

Given an arbitrary shape with 8 verticies in R^3, how can I create the most perfect cubic representation in R^N?

A representation is perfect if forall points, the distance from one point all others expressed in terms of side lengths is as close to the original as possible.

All arbitrary shapes in R^3 are rhombohedrons

I'm not sure if my criteria for "best" is correct, but it's roughly there.

[sharp frost]

does anyone know what it's on about with the dot products thing?

I can show that it's closed by considering all the convergent sequences, but I don't know how to use this hint

[vocal wharf]

hint: continuous preimages of closed sets are closed

[sharp frost]

do I sum up all the dot products in a matrix or something? The dot product on its own wouldnt map a matrix to a number

[vocal wharf]

tbh the easiest way to see this is to note that a matrix A is orthogonal iff AA^T = I

and that A \mapsto AA^T is continuous

[sharp frost]

yeah I did that with a sequence $A_n\to A$

[sharp frost]

but I do want to understand this hint

[vocal wharf]

i mean, you can define some map from R^(n^2) to R

like, sum up all the dot products <v_i, v_j> for all (i, j)

then the matrix is orthogonal iff it gets mapped to n

[sharp frost]

ah yeah, and so that will map to {n} which is closed in R

[vocal wharf]

ye

[sharp frost]

ah yes I see

thanks

[vocal wharf]

but as i said the easier thing is a map R^(n^2) -> R^(n^2), A \mapsto AA^T

and then its preimage of {I}

but it's the same thing i guess 🤷

[sharp frost]

oh yeah nice thinking

the thing I was doing before was considering any convergent sequence $A_n\to A$, then the limit of $A_nA_n^T$ is $I$, so the limit $A$ is orthogonal

[sharp frost]

but I think your way is nicer

[vocal wharf]

yeah, when i want to show that some space is closed/open, the first thing i try to do is show its the preimage of some "nicer" space

[sharp frost]

I also have a scuffed proof that I feel is wrong but I don't know how to fix, if anyone wants to check

actually maybe it isnt terribly wrong

excuse the "clopen"

[unkempt bolt]

You should be careful as Int(A) isn't necessarily the same as the interior of the closure of A

[unkempt bolt]

You also shouldn't be able to prove that B is connected given only A is connected with Int(A) = Int(B). Consider for example
A = (- infinity, 0)
B = A union {1}
Inside of the reals R

[sharp frost]

hmm good points @unkempt bolt I will throw the proof out and try a different approach

when is int(A) not equal to int(clos(A))?

[unkempt bolt]

Essentially when some limit point of A has a neighborhood consisting entirely of A and itself

There examples that aren't that bad if you want to think of one

I can give you an explicit one if you want

[sharp frost]

yea I'm struggling to imagine such a case

[unkempt bolt]

Something like A = (0,1) u (1,2) works where the u is a union

Do you see why?

[sharp frost]

oh true

[unkempt bolt]

Cool

Unrelated but there's a really nice problem by Kuratowski about closures and interiors

If you want I can state it for you at some point

[sharp frost]

once I get these proofs done I might take a look

[unkempt bolt]

Yeah for sure, just wanted to mention it

[sharp frost]

$f:\mbb{R}\mapsto S^1,\ f(x)=(\sin x,\cos x)$. Am I right in thinking that $f$ is both an open and closed mapping?

[unkempt bolt]

It's locally a homeomorphism and thus an open mapping but I don't think it's a closed mapping

[coarse kestrel]

I mean if it’s locally a homeomorphism then it should be a closed mapping as well

Shouldn’t it

[unkempt bolt]

No

You could say its "locally a closed mapping"

Maybe by defining that to mean that given any point p there's a neighbourhood U of p so that f maps closed subsets of U (closed in the original topology, not subspace topology) get mapped to closed sets

But the given f is not a (globally) closed mapping

I'm kinda new to this btw, if I say something that's unclear or you want some hints just ask

[dim meadow]

I think you can take a discrete set in R that maps to a dense set in S^1

As a more concrete way of confirming this idea

[coarse kestrel]

Oh right

So like $\brc{2n\pi+1/n\middle|n\in\bN}$

[gentle osprey]

Whoever:

[unkempt bolt]

Yeah

You can also do
$$\bigcup_{k = 1}^{\infty} [2\pi k, 2\pi k + 1 - 1/k]$$

[gentle osprey]

The_Vman:

[sharp frost]

😍

thanks guys

[unkempt bolt]

No problem

[knotty pasture]

wo want to help me with an exercise?

I want to prove that there is a unique topology on X such that every $U_i$ is open and every $\phi_i:U_i\rightarrow V_i$ is an homeomorphism. In this topology, a subset $U\subset X$ is open $\Leftrightarrow$ the sets $\phi(U\cap U_i)$ are open for every i.

[gentle osprey]

Davide:

[knotty pasture]

where $X$ be any set and $\phi_i$ be a bijection onto an open subset on $\mathbb{R}^n$ and $U_i$ covering X

[unkempt bolt]

Sure I think I can help

So there are two steps basically

[gentle osprey]

Davide:

[unkempt bolt]

You need to show that your proposed topology (is a topology) and has the property you claim

[knotty pasture]

Ok

[unkempt bolt]

And that any topology with the property is actually exactly this one

[knotty pasture]

I'm sure that is a topology but i can't prove that is unique with this proprieties

[unkempt bolt]

Hmm ok

[marsh forge]

show that its both finer and coarser

than any topology with this property

that will give you the result

[knotty pasture]

Nice idea

I try

[unkempt bolt]

I think you are missing one assumption actually

You probably need to assume that for any $i$ and $j$ the set $\phi_i(U_i \cap U_j)$ is open in $\mathbb{R}^n$

[gentle osprey]

The_Vman:

[marsh forge]

where did R^n come from

Davide i think your problem is poorly stated, could you write it exactly as given

[knotty pasture]

The_Vman:
@gentle osprey yess sorry

[marsh forge]

in particular, what is V_i

[knotty pasture]

The probelm is relative to non prior topology, a way to define a smoth manifold structure to a set X without a prior topology on X

[unkempt bolt]

where $X$ be any set and $\phi_i$ be a bijection onto an open subset on $\mathbb{R}^n$ and $U_i$ covering X
These are the assumptions right? They seem fine

[knotty pasture]

Then you prove that there is a unique topology on X compatible with the no prior topology

[marsh forge]

Thats not sufficient to make the problem well defined lol

[gentle osprey]

The_Vman:

[marsh forge]

but its possible that im missing some prior info

so ill just leve it to you

i think he wants like

charts on a set X induce a topology on X and it comes with smth manifold structure for free

[knotty pasture]

I'm just describing where the problem comes from

[marsh forge]

yeah the problem just doesnt make a ton of sense as written

[knotty pasture]

i think i do first by showing you where the problem is coming from
http://people.dm.unipi.it/martelli/didattica/matematica/2018/Manifolds.pdf

in page 56

[marsh forge]

yeah okay that makes everything well defined

the question made no sense without that paragraph

anyway! my approach to the problem will still work

[knotty pasture]

I think I am convinced of the uniqueness of the topology

I try to prove that, with this topology, a subset $U\subset X$ is open $\Leftrightarrow$ the sets $\phi(U\cap U_i)$ are open for every i

[gentle osprey]

Davide:

[knotty pasture]

$\phi(U\cap U_i)=\phi(U)\cap V_i$ then $\phi(U\cap U_i)$ is open iff $\phi(U)$ is open iff U is open

[gentle osprey]

The_Vman:

Davide:

[knotty pasture]

while, if $U $ is open, then $U=\cup_j U_j$ then $\phi(U\cap U_i)=\bigcup_j \phi(U_j\cap U_i)$ that is open

[gentle osprey]

Davide:

[knotty pasture]

it's right?

[gentle osprey]

The_Vman:

[knotty pasture]

YEs

[unkempt bolt]

Ok so there are still some slight errors

[gentle osprey]

The_Vman:

[knotty pasture]

I don't undestand

[unkempt bolt]

$\phi(U\cap U_i)=\phi(U)\cap V_i$ then $\phi(U\cap U_i)$ is open iff $\phi(U)$ is open iff U is open
Here you wrote $\phi(U)$

[gentle osprey]

The_Vman:

[unkempt bolt]

I think you sort of have the right idea

[gentle osprey]

The_Vman: