point-set-topology

[gritty widget]

hmm i suppose this is sufficently slick

modulo being careful about the Ext(H_0, R)

[kind cedar]

let $f:M\longrightarrow \mathbb{R}$ smooth and non-singular. Show there is a field $X\in\mathfrak{X}(M)$ such that $X(f)=1$

[gentle osprey]

Necrowizard

[kind cedar]

I honestly have no idea what to do here. Any hints?

[gritty widget]

one way to do it is to pick a riemannian metric on M and consider something like the gradient of f

[kind cedar]

I haven't learned gradient in this class yet.

[gritty widget]

you could do it locally (in a coordinate chart) and then patch the vector fields together with a partition of unity

[kind cedar]

Professor told me something about it but I honestly didn't understand 😦

[gritty widget]

what didnt you get

[kind cedar]

what is the meaning of "Patching the fields together" and why do I need a partition of unity for that

But at this moment I'm guessing even my attempts to do it locally will be frustrated

[gritty widget]

if you're doing it locally then you can use the gradient on R^n

that is

you can assume that M = R^n first, prove it there, and then work on "patching them together"

as for what that means

[gentle osprey]

TTerra

[gritty widget]

actually by phi_alpha X_alpha in the summand i mean the vector field on M which is equal to phi_alpha X_alpha on U_alpha and 0 outside

this is a common abuse of notation

if you've seen partitions of unity then you must have seen a similar construction

[kind cedar]

Okay, I'll work on probing it locally first then

[kind cedar]

somewhat

[gritty widget]

i should remark

the two answers i gave are basically the same

that riemannian metrics exist on M uses partitions of unity

and the proof is: cover by charts, in each chart bring over the metric on R^n, and patch together with pou

here you're covering by charts, using the metric on R^n to construct such a vector field, and patching together

[kind cedar]

thank you

[plain raven]

Does your book give you rigorous definitions of these concepts you can work with? I would appeal to definitions here

What is meant by an "edge"

i would also advise like, coming up with a nice explicit description of the set of points in the convex hull as well as a notational system by which they can be uniquely specified

|| barycentric coordinates ||

[silver umbra]

how do u show that a topological space with an isolated point isn't a manifold

or that a topological space with a self intersection isnt a manifold

[tight agate]

a point is a manifold

unless your definition does not allow that

[silver umbra]

but like lets say the object in question

is the disjoint union of an isolated point and a curve

in R2, say

how do i show that that can't be a manifold

[tight agate]

that is a manifold

what is your definition of a manifold

[silver umbra]

okay my object here is the set of (x,y) satisfying x^2(x-1) - y^2

in R2

i want to show that it isn't an embedded submanifold of R2

because it has an isolated point at (0,0)

but im not sure how to do that rigorously

@tight agate

[tight agate]

again, what is your definition of a manifold

[silver umbra]

? the standard definition

hausdorff, second countable, locally euclidean

but then embedded submanifold has its own definition too

[tight agate]

this space is hausdorff, second countable and locally euclidean

[silver umbra]

where we have to consider an embedding from one smooth mfd to another, etc

oh yeah i sort of misspoke

i meant to ask how to show this object is not an embedded submfd of R2

bc of its isolated point at (0,0)

is this more clear?

[tight agate]

what is your definition of an embedded submanifold

[silver umbra]

an embedded submanifold is a submanifold such that the inclusion map to the ambient mfd is an embedding

an embedding between manifolds is a smooth immersion such that it's a homeomorphism onto its image

is this sufficient?

[tight agate]

doesnt your space satisfy all of those conditions?

oh good, tterra is here

[gritty widget]

"locally euclidean" could be hiding something like "all components have the same dimension"

since sometimes people say "manifold of dimension n" and require in "locally euclidean" that the space is locally homeo to R^n, not just some euclidean space

(this is directed at pdk)

[silver umbra]

in my case though

so is the set in question an embedded submanifold?

[tight agate]

if your definition does not include the condition that all components have the same dimension, then yes

[silver umbra]

wdym by components

[tight agate]

connected components

[silver umbra]

are my definitions just like

not clear enough or smth

bc the problem is explicitly telling me to show

that it isn't an embedded submfd of R2

[tight agate]

if that is the case, then yes your definitions are incomplete/not clear

idk

tterra probably knows

[gritty widget]

just admit you can't solve the problem brofib

[silver umbra]

is this enough? im rly just going off my book here (lee)

[gritty widget]

lee
all the components must have the same dimension

so your space having a point (dim 0) and a curve (dim 1) disqualifies it from being a manifold

[silver umbra]

does lee define manifolds that way?

[gritty widget]

yes, go look at the first chapter

[silver umbra]

thm 1.2?

lol

[gritty widget]

no

the definition of a manifold

though invariance of dimension is what ensures that dimension is well-defined (at least for the components of your space) when you take the latter definition of locally euclidean i mentioned

[silver umbra]

not sure how the definition of a manifold here implies that all connected components must have the same dimensino

[gritty widget]

right before theorem 1.2

[silver umbra]

i see, ty

[gritty widget]

many differential topology/geometry questions today

i am pleased

[kind cedar]

Seriously, I don't get it. It seems utterly like running in circles. If {B_i} is already a basis for M, why do I need to set the B_p,alpha's and declare it a basis that is, as far as I understand, composed of the very same elements?

I understand that there are many things in diff. manifolds that seem to be obvious although they are not. This is one of the confusing ones for me. Help...

[gritty widget]

the B_{p, alpha}'s are coordinate open sets because they're contained in coordinate opens U_alpha

(by restricting the coordinate map defined on U_alpha to B_{p, alpha})

[kind cedar]

Ok, I kinda understand the reason for this

Let me rethink a bit

[gentle osprey]

fajitas

[gritty widget]

maybe try measuring the angle from a point not in U

[fair idol]

That's a good idea. Maybe an inner product?

[gritty widget]

[gusty wagon]

is there a precise definition for “reformulation” in math?

i read earlier that “if Y is compact then pi:XxY->Y is a closed map” is a reformulation of the tube lemma

i understand how to prove one using the other

[gritty widget]

synonymous with "equivalent"

[gusty wagon]

okay but

they’re both just true statements

obviously they’re equivalent

every true statement ever is equivalent

does that mean every true statement is just a reformulation of every other true statement?

?_?

the other day i asked someone whether the two parts of the FTOC were equivalent

and he said “well they’re both true so yes”

so i said “okay but is it okay to say that you can prove one using the other and only elementary methods”?

and he said “define elementary” and i was like )=

like am i going crazy here what am i missing

[gritty widget]

this doesn't really fit in this channel, you'll have better luck in something like #proofs-and-logic

[plain raven]

try checking out the wikipedia page on reverse mathematics. there have been a variety of specific formal systems proposed that would constitute "elementary methods" in the context of what you're saying

[gusty wagon]

yeah i moved over there sorry my bad

[plain raven]

you may get different answers for different systems

if you want to talk about reverse mathematics ask in #foundations

[gusty wagon]

thank u

[fair idol]

I saw an exercise in John lee that says a vector field can be defined as the partial derivative with respect to an angle function. How exactly is a partial derivative a vector field?

[gritty widget]

derivations are vector fields

[fair idol]

Maybe it's mapping a point p on the sphere to the tangent vector of the associated level set containing p

[plain raven]

fajitas imo you have it backwards

rather than ask "how is a partial derivative a vector field" the right question to ask is like "What is a vector"

this question unfortunately doesn't have a good answer because like, generally there's no good notion of what a thing is. What is a natural number? What is a real number? We are able to give definitions of a natural number in terms of set theory and we are able to define real numbers as cauchy sequences of rationals. but that's not really saying that a real number is a cauchy sequence of rationals, we just do that because it's a technical construction that works for us, and the resulting model of the real numbers has lots of good properties we expect of the real numbers

ok so

scratch the question "what is a vector"

bad question

a more sophisticated question is "How can we precisely define a vector so that all the things we're used to doing with intuitive notions of vectors can be done with this formal model"

and that requires us to sit down and think about what we use vectors for in math. there are different things we want to use them for - for example we like to integrate vector fields along curves - but one of the most important aspects of vectors is that if you have a scalar function f and a tangent vector v, you can talk about the directional derivative of f in the direction v

and so like, for the sake of having a precise definition of a vector, you could define a vector to literally just be a differential operator defined at a point. this is commonly what is done.

so speaking literally, partial / partial x is a vector immediately by appeal to the definition of a vector

[fair idol]

Yeah I can imagine a partial as a vector in a vector space.

John lee defines a vector field as a mapping from a manifold into its tangent bundle. I was wondering what tangent vector the partial derivative operator will map a point on a manifold to.

[plain raven]

well

what is a tangent bundle

it's a bundle of tangent spaces

what is a tangent space

it's a vector space associated to a point

so again the answer comes back to what a vector is

you can't really define the tangent bundle afaik without explaining what the tangent space is at a point

and you can't explain what the tangent space is at a point without explaining what a tangent vector is

nvm i'm just gonna look at lee

oh wait

are you talking about Lee's topological manifolds

or his book on smooth manifolds

[fair idol]

Smooth manifolds

[plain raven]

ok what exercise is this

[fair idol]

Is was example 8.4 of chapter 8.

[plain raven]

ok. well i think if you're reading chapter 8 and you have questions about something which was introduced and systematically developed in chapter 2 you should try reading chapter 2!

[fair idol]

I was wondering about vector fields more so than angle functions (the stuff from chapter two).

Forget about angle functions for a month, even though I brought them up.

How does a partial derivative define a vector field?

[plain raven]

sorry i meant chapter 3

not chapter 2

i think you'll be able to answer your questions if you read the chapter on the tangent bundle and tangent vectors

[fair idol]

Thanks a lot

[wanton marsh]

well in this case it converges to some integer

if you really can't find it I think you can use that x -> 10x+9 is continuous and so the limit x must satisfy x = 10x+9

[pearl holly]

I would never commit such a sin I’m still doing AT, turns out that the cohomology ring of the n-torus is the “n-th exterior algebra”. So if the exterior algebra somehow is connected to some cross product in linear algebra then it makes me wonder if there’s some “hidden” geometric intuition behind why the torus has that cohomology ring

[coral pivot]

derham cohomology

[winged badger]

I'm having trouble with exercise 16.4.O in Vakil's FOAG. It's as follows. If $X_1, X_2$ are universally closed $A$-schemes, where $A$ is a ring, then the gluing of $X_1, X_2$ along some isomorphic closed subschemes is also universally closed over $A$. Could someone please help?

[gentle osprey]

Porphyrion

[pearl holly]

So the de Rham cohomology explains this?

[empty grove]

It uses the exterior algebra 😌

There you work with smooth manifolds. You consider smooth functions from the manifold to the degree n part of the exterior algebra over R and these are called differential n-forms. You can integrate differential n-forms over an n-manifold. You can also differentiate a differential n form to get a differential n+1 form (this operation is called exterior derivative). Differential n-forms form a vector space over R and along with the exterior derivatives this forms a cochain complex and you can take cohomologies and the dimension of the nth cohomology tells you the number of n dimensional holes in the manifold

There's a theorem that says that this counts the same holes as the singular stuff

Aleph 0 has a really cool video on this

On some intuition

[pearl holly]

Oh man

That’s some good stuff

Ye okay I see, I will need to look up more of this later lmao this seems really cool. Thank you so much!

[empty grove]

You can watch aleph 0's video no prereqs

[pearl holly]

Oof yeee I can

[gritty widget]

any good recommendation for spectral sequences

that isn’t hatchers chapter 5

[empty grove]

I was recommended Gelfand Manin by Brofibration and Weibel by Clerk 😌

I am reading Gelfand Manin but haven't gotten to that part yet

You can go straight to ch3 in Gelfand Manin

The book is called Methods of Homological Algebra

Weibel I haven't looked at at all so idk

Brofib also gave me some exercises to figure out myself before looking at the book

Maybe you should try too I'll link it wait

here

[gritty widget]

okay thanks

also thanks to brofibration

And completely unrelated

But there seem to be many papers

What is…

Is there any good way to find these

Googling for what is… math

Does not give the result

And another question related to the original question

Hmm nevermind I’ll read more first

[winged badger]

does anybody know?

is ag allowed in this channel?

[tough imp]

Can you try using the valuative criterion for universal closedness?

I haven’t ever done this before but that would be my first attempt

[winged badger]

Well, those types of criteria require finite-typeness of morphisms... I don't think it would work generally.

[tough imp]

It only requires quasi-compactness of the morphism

But I think this might be a hiccup in the setup when you consider arbitrary schemes

:/

I think one thing to maybe consider

Is it true that the disjoint union is still universally closed?

If so, I think that the disjoint union of X1 and X2 surjects onto the gluing of X1 and X2 along their closed subscheme

Then this should hopefully give you a map over A

Once you base change the universal closedness of X1 disjoint union X2 should give you closedness

Namely, you take a closed set S in the gluing of X1 and X2

Pull this back to X1 disjoint union X2

Then when you push this into X1 glue X2 you recover S because the map is surjective (surjectivity is preserved under base change)

Then pushing S into A is the same as pushing the preimage of S inside X1 disjoint union X2 into A via your universally closed map

So you know the image of S is closed

[gritty widget]

is there a nice “program” for drawing spectral sequences

[orchid forge]

spooky xymatrix maybe

[plain raven]

i would recommend chapter 5 of weibel
Spectral sequences are hard
study the "Blackboxed" examples in the first section carefully to get as much intuition as you can without diving into the technical details. i really like the stuff about the serre spectral sequence
Don't study convergence conditions or necessary or sufficient conditions for convergence at first, you can push them way off.
Learn the spectral sequence associated to a filtered chain complex and the spectral sequence associated to a double complex / bicomplex. Then stop. Don't learn any more kinds of spectral sequences after that, most examples I know of are examples of these two kinds
Make your life simpler. Assume that all filtrations start at 0 and their union is the whole complex. Assume all chain complexes are concentrated in nonnegative degree.
Weibel gives two proofs of the theorem that Tor and Ext are "Balanced". One is in chapter 2 before spectral sequences are introduced, and the other is in chapter 5 using spectral sequences. These two proofs have essentially the same mathematical content so it might be helpful to look at those intuition.

The place i got really familiar with spectral sequences was Godement's book on sheaf cohomology. it's in french and i read it with lots of help from a dictionary and google translate. but it's a great book!

I would look online to see if you can find youtube videos. spectral sequences are a very dynamic and interactive thing, like a lot of diagram chasing related things you either have to do it yourself or watch someone else do it, it's very difficult to understand by reading it

a good exercise is to use spectral sequences to prove that a Cartan-Eilenberg resolution of a complex (regarded as a total complex) has the same homology groups as the original complex, i.e. the augmentation is a quasi-isomorphism

another cool exercise: let X be a cw complex, and let C(X) be its chain complex of singular simplices. C(X) is naturally filtered by the subgroups C(X_n) associated to its n-skeleta. Play with the spectral sequence associated to this filtration and show that the concept of CW homology naturally pops out.

[uncut surge]

@gritty widget in latex there's a bunch of options for drawing https://tex.stackexchange.com/questions/1085/what-is-the-best-way-to-typeset-spectral-sequences

[plain raven]

I think another book that develops spectral sequences in some depth is bott-tu. spectral sequences are a very natural tool for comparing two cohomology theories to show they give the same results. I think bott-tu gives what they call a "tic-tac-toe" argument to prove that Cech cohomology and de Rham cohomology agree for manifolds, which is a standard application of spectral sequences

nice

[uncut surge]

https://q.uiver.app/ and for simple diagrams this is a fun little website, might extend to some smol spectral sequences

also i coincidentally wrote a little blurb as well to somebody on where to learn spectral sequences, just this week

i'll just copypaste it: "Chapter 8 in "Hilton, Stammbach - A Course in Homological Algebra" gives a very rigorous way of defining what spectral sequences generally are, where they come from and what they're good for, but in very high level of abstractness; if you're not very well-versed in the basics of homological algebra & category theory, it's easy to get lost ;
There's also this paper: https://arxiv.org/pdf/1702.00666.pdf which is very down to earth and the rigor only comes at the very end, but, in my opinion, it might be simplifying things a bit too much.
And then there's the famous "User's Guide to Spectral Sequences", here https://people.math.rochester.edu/faculty/doug/otherpapers/McCleary-UGSS.pdf which is somewhere in between the two other sources. It's somehow approachable, and somewhat rigorous, too. Didn't really click with me, but it might be nice"

[gritty widget]

thanks to both of you!

[gritty widget]

i'll think about this and get back to youi

but i am prety slow so it might be a while

[plain raven]

np!

[crimson path]

I have a compact hausdorff space X and a closed continuous map $f: X \rightarrow X$
I have a sequence of nested, non-empty, closed and compact sets
$C_1 = X; C_2 = f(X); C_3 = f(f(X)); ...$

I want to show $f\left( \bigcap_{n=1}^\infty C_n \right) = \bigcap_{n=1}^\infty C_n$.
I have shown one inclusion namely that the left-hand side is a subset of the right-hand side.

To show the other inclusion: I take $y \in \cap C_n $. That implies there is a sequence
$(x_n)$ such that $x_n \in C_n$ for each $n$ and furthermore $y = f(x_n)$.

So we have a set $A = {x_n | n \in \mathbb{N}}$.
If this set is finite then there is a constant subsequence $(t,t,t,...)$ in $(x_n)$, so $y = f(t)$ for some $t \in \cap C_n$.

But if A is infinite, then there is a limit point x of A in X.
I have managed to show $x \in \cap C_n$, but don't see how to show $y = f(x)$.

[orchid forge]

I have also only seen filtration types in the wild and double complexes in books (just realized this message was pretty old when I replied, sorry, not deleting because I already pinged )

[gentle osprey]

hardisc

[proud coral]

Is Godement's book still good if you mostly want to do arithmetic geometry ?

I'm asking because the book was written before the invention of schemes

[orchid forge]

Sequential characterization of continuity?

[crimson path]

@orchid forge I know x is a limit point but I dont know if X is a first countable space. Is it still possible to construct a sequence converging to the limit point?

[orchid forge]

Nevermind that, but actually I'm not sure exactly what sequence you're talking about.

[plain raven]

godement's book is not really about algebraic geometry. it's more general than that. it does talk about algebraic varieties but also about like, paracompact hausdorff spaces, etc. as examples it deals with de Rham cohomology, etc. It develops spectral sequences relating various cohomology theories.
Ultimately tho, everything it says about Cech cohomology and flasque sheaf cohomology is applicable to schemes. It does contain some results on the relation between ordinary dimension of an algebraic variety and cohomological dimension.
Also the first 100-150 pages in the book are a general introduction to homological algebra. which is probably valuable for anyone.
For arithmetic geometry i suspect you'll have to learn etale cohomology which is definitely not covered in Godement's book. I don't know anything about etale cohomology so I can't comment much beyond that. but certainly for everything you would see in Chapter 3 of Hartshorne, the stuff in Godement is applicable. there's a lot of overlap.

[crimson path]

@orchid forge
Well $x$ is a limit point of $A = {x_n | n \in \mathbb{N}$. Here $x_n \in C_n$ for each $n$, and $y = f(x_n)$ for each $n$.

From here I would like to show $y = f(x)$.
The space X is hausdorff but I unfortunately don't necessarily have first countability.
Otherwise I could have constructed a subsequence of $(x_n)$ converging to $x$.

[plain raven]

FWIW Godement's flasque sheaf cohomology computes the same theory as grothendieck's sheaf cohomology with injective resolutions.

[gentle osprey]

hardisc

[proud coral]

Alright, thanks for the detailled answer !

I'll need to learn about homological algebra anyway

[rich cliff]

If X is finite, isn't the resulting set just the empty set? Or are finite sets also considered countable?

yeah I see why it has to be true for the above to be a topology

but

finite sets can't be put into 1-1 correspondence with N

isn't that kinda bad

oh I thought it was the standard one

I guess not

I skipped the first chapter

yeah

I still like the "at most countable" terminology better

but I should get use to this

[orchid forge]

Sorry, I'm not sure if it's my fault but I have no idea how this sequence is well defined. If x is in C_n, then f(x) is in every C_(n+1) = f(C_n), and f^(-1)(x) is a closed set contained in C_(n-1) (taking C_0 = X). So if x is in each C_n, then it is the image of something contained in each C_(n-1)

[crimson path]

@orchid forge

Hmm, well how I defined the sequence is I take $y \in \cap_{n=1}^\infty C_n = \cap_{n=1}^\infty C_{n+1} = \cap_{n=1}^\infty f(C_n)$.

So there is an $x_n \in C_n$ such that $f(x_n) = y$. This gives us the sequence.

[gentle osprey]

hardisc

[crimson path]

$\cap_{n=1}^\infty C_n = \cap_{n=1}^\infty C_{n+1}$ holds owing to the nested property of the sets $C_n$

[gentle osprey]

hardisc

[kind cedar]

Back to this same question, yesterda @gritty widget helped me with some hints and I'm still lost. I (kinda) understood the part of patching solutions together but I still don't exactly know how to prove that the solution exists locally.

If I write $X(f) = X^{i}\frac{\partial{f}}{\partial{x^i}}=1$, can I just isolate the $X^i$s?

The function f is non-singular, but that doesn't imply it's derivative can't be zero.

[gentle osprey]

Necrowizard

[gritty widget]

you can try choosing some X^i's that make this true

as a hint, try computing (grad f)f when M = R^n

[kind cedar]

Isnt' the gradient of f like $\nabla f = \frac{\partial{f}}{\partial{x^i}}\cdot \hat{e}_{i}$ ?

[gritty widget]

yes

[kind cedar]

the versor is is just the del/del x^i isn't it?

I don't see how this is equal to 1

[gentle osprey]

Necrowizard

[gritty widget]

what does it equal

(grad f)f

[kind cedar]

$(\nabla f)f = (\frac{\partial{f}}{\partial{x^i}}\cdot \hat{e}_{i})f = (\frac{\partial{f}}{\partial{x^i}}\cdot \frac{\partial{f}}{\partial{x^i}})f = (\frac{\partial{f}}{\partial{x^i}})^{2}$ ??

[gritty widget]

please write in the sums

[gentle osprey]

Necrowizard

[kind cedar]

oh, I was using Einstein summation notation

sorry

ah, I can't. the indexes are not up and down

[gritty widget]

yeah lol

anyways i know what you mean

the end result is |grad f|^2

this is non-zero since we assumed f is non singular

so if you divide both sides by that you have your X

[orchid forge]

Oh I see. Then it's simple to conclude, {y} is closed since X is hausdorff, f^(-1)(y) is a closed set, and x_n is contained in f^(-1)(y), so its limit point x is too

[kind cedar]

ok, let me try that

got that, thanks

[crimson path]

@orchid forge oh lord it was that simple, how did I not see that. big big thanks mate

[gritty widget]

I heard something like this

does anyone know what the precise statement is

If a space X can be covered by n contratible open sets

the homology groups of X cannot have more than n generators

[uncut surge]

Hmmm, sounds like some Cech homology fun; the statement that I'm more familiar with and that is easy to prove is that "If a space $X$ has a good cover, i.e. an open cover $\mathcal{U}$ with $n$ elements so that all nonempty intersections of elements in $\mathcal{U}$ are contractible, then $H^k(X) = 0$ for $k > n$"

[gentle osprey]

Lartomato

[uncut surge]

I haven't heard of a statement like yours, but it sounds interesting

[orchid forge]

Yeah I'm more familiar with the second statement as well, I would guess it is a relatively simple consequence of the lebesgue covering theorem but I'm not sure

[tight agate]

there's a very similar exercise in hartshorne

oh nvm it was about cohomological dimension

but the idea is the same

[orchid forge]

Larto is right though, the k-th cech cochain group is the free group on nCk <= n generators (one for each k-fold intersection of your n sets), so the cohomology is going to be a quotient of that

The precise statement would then be whatever you need to say your cohomology theory is equivalent to Cech

At least, that would be a sufficient condition

Actually sorry, this isn't the cech cohomology if your cover isn't a "good cover"

[gritty widget]

ah this is it

[orchid forge]

It's just a cohomology that's associated to that cover

[gritty widget]

if X is covered by n contractible open sents

n-fold cup products are zero

its this generaliseation of suspension destroying cup products

[obtuse meteor]

I'm calculating pi_m(O) and struggling to see why pi_1(O) is not zero

so like

the action O(k - 1) -> O(k) -> S^{2k - 1} gives you a fibration sequence

wait

I retract my statements

wait no I don't

and in particular in this fibration sequence we have

pi_2(S^{2k - 1}) -> pi_1(O(k - 1)) -> pi_1(O(k)) -> pi_1(S^{2k - 1})

if k >= 2 then 2k - 1 > 2

so this gives an iso

but then pi_1(O(1)) = 0

and we must cry

someone explain to me why I'm dumb :)

does it have to do with O(k) having two path components?

[obtuse meteor]

I believe that it does bc basepiints noise

In fiber sequences

[gritty widget]

can anyone direct me to a source that will tell me what a homotopy equivalence of kan complexes is

[obtuse meteor]

Now I have 5 of the 8 groups. Pi1,pi2,pi3 to go

[plain raven]

"Simplicial objects in algebraic topology" Peter may, more combinatorial definitions
"Simplicial homotopy theory" Goerss, Jardine, more categorical/abstract definitions

[wise ruin]

The union of two closed discs in R^2 is obviously disconnected. But what two disjoint non-empty open sets comprise this union?

[orchid forge]

open (in the subspace topology)

since connectedness is a property of a topological space, not of a "space, subspace" pair

does that help?

[wise ruin]

Yes it does, thank you so much!

[marble socket]

is there a really really really cool definition/generalization of a manifold that isn't like saying glue copies of R^n a bunch of times? Definitions of manifolds and schemes are weird... they feel like a very-simple-minded generalization of R^n and Spec A.

[tough imp]

Can you define a manifold purely in terms of functors the way you can a scheme?

You can define a scheme as special types of functors from CRing to Set

[gritty widget]

maybe nestruev's book has what you're looking for

i know there's a funky definition of manifold in there

[marble socket]

haven't checked all the details... but this feels a lot like the sheaf-theory definition. a locally ringed space which is locally isomorphic to the locally ringed space of R^n with smooth functions.

[tawny jewel]

Hey, if I remove a point from the unit circle in R^2 I still get an open set in R^2 right?

[lunar yoke]

if by circle you mean the filled-in open disk, i.e. {x in R^2 with |x| < 1}, then yes

[tawny jewel]

Yep

Thanks

[pearl holly]

Okay so I know that the cohomology ring help us distinguish spaces with the same (co)homology groups, for example the torus and S^1 v S^1 v S^2. But how does this work geometrically? From my perspective it almost seems like a coincidence that the cohomology ring has this "property", you first define the cup product and it just turns out that induced maps are ring homomorphisms that "respect" the cup product, but how would I guess before the theorem that induced maps are ring homomorphisms that the cohomomology ring would have this "property"? Like what does the cohomology ring/the cup product measure? I kind of understand that the cup product counts the number of intersections with each simplex if you triangulate your space, but what does this geometrically mean in terms of the torus and S^1 v S^1 v S^2?

[orchid forge]

Well, when you write e.g. H^2 = R + R, you're forgetting what the generators are and any geometric/algebraic properties they have. The cup product just recovers a bit of that. Geometrically I don't have much intuition besides knowing that cup products are poincare dual to intersections. Maybe it would help to explicitly figure out what generates the cohomology in both spaces you wrote down, and see if you can interpret what the cup product is doing

(remember the cup product is defined at the level of cocycles so it's not necessarily misleading to pick representatives of cohomology classes here)

[gritty widget]

How seriously should I take the statement

Chain complexes are generalizations of abelian groups

My goal is to understand how

Spectra are mutual generalizations of spaces and abelian groups/chain complexes

[empty grove]

The category of chain complexes over a module is an abelian category for which the prototype is the category of abelian groups

[gritty widget]

But I feel like there is probably interesting stuff already at this level that I don’t know

[empty grove]

I don't know if that counts as a generalization though

[gritty widget]

When you say

AbGrps are the prototype category for abelian categories

Aside from being the main first example

Are you also hinting that we can make some statement like

Abelian categories are enriched over abelian groups?

[empty grove]

lol I was just saying this

[gritty widget]

Oh

Cause I know

Sets are kind of the prototypical category

And categories are enriched over sets

Modulo

Being precise

[empty grove]

I guess that is true

[gritty widget]

And in the infinity category case

The infinity category of spaces, which is the infinity category of Kan complexes

Is the pro typical example

And in a way infinite categories are enriched over Kan complexes

I know maybe this should go in abstract algebra or category theory

But top x geom gang just gives the best answers

[empty grove]

happy smiley dog ftw

[gritty widget]

Oh shit

Abelian categories are enriched over Abelian groups

Yessss

We getting an intuition

[empty grove]

😎

[pearl holly]

ye okay I see. So it's all about the generators and the properties they have? So if two spaces have different generators that have different properties in some sense then the cup product captures that and so the cohomology rings will be different?

[orchid forge]

yeah, the cup product structure tells you how n+m forms are built out of n- and m-forms

[gritty widget]

One thing I find helpful is

[orchid forge]

still do the example though, i think it will be helpful

[gritty widget]

Arbitrary top spaces can be as bad as you like

[pearl holly]

ye I will think about those later

[gritty widget]

Working at the cochain leave is very hard

[pearl holly]

what are m forms?

[orchid forge]

uhh elements of H^m is all i meant

[pearl holly]

oh okay

[pearl holly]

I don't really get it tbh

okay sure, working at the cochain level might be hard, but does the cup product help with it? Is that what you mean?

[gritty widget]

Oh sorry I forgot to finish that thought

If you want a “geometric” intuition

Look up stuff like

What does the cup product look like for manifolds

Kogasa mentioned about the cup product being related to poincare duals of intersection forms

Getting a good understanding of the cup product in this situation can help

[pearl holly]

ooh okay I see I see. Thank you both so much!

[abstract pagoda]

I think I need help

So I want pushforward of transsition map

But the problem is idk how to do it

maybe im being a baby nvm

i hve been struggling with this problem for longer than necessary

[orchid forge]

yeah i don't think there's much that anyone can help with, you have to just write down the definitions and go through the calculation carefully

[abstract pagoda]

ok

so ive done jt

and its cancer

so much cancer I might cry

ill write up my proof and show in around 3 hrs

which field has worst notation?

im guessing geometry

[orchid forge]

the computations are bad but at the end it should be a pretty simple formula

there's just a lot of indices

[plain raven]

this is "a thing" but i don't remember the details. maybe a really categorical book on diff geo would have the answer. something like "Natural operations in differential geometry"

[plain raven]

yeah this is because hatcher's definition of the cup product is shit lmao

this is what i was complaining about the other day

but there are much more natural ways to motivate the cup product

one is in de rham cohomology where it's just the wedge product of differential forms, which is a very basic operation.

another is using stuff from diff top, the intersection of transversal submanifolds is somehow related to the cup product (please forgive me if that sentence isn't grammatically correct, i don't know any diff top)

even in singular theory there are better ways to motivate it than just giving that bullshit formula hatcher gives

i was going to write something up on this the other day, sorry i never got around to this

but like, speaking simply

the geometric intuition i have for this stuff

and i have like, most of the formal details worked out, but not all

but

there's a binary operation called the join on topological spaces $X\ast Y$. The join is defined as $(X\amalg X\times I\times Y\amalg Y)$ modded out by the equivalence relation $\sim$ defined by requiring that $x\sim (x,0,y)$ and $(x,1,y)\sim y$ for all $x,y$. basically this is the operation that takes two spaces $X, Y$ and creates a new space out of them by freely drawing lines from each point in $X$ to each point in $Y$.

[gentle osprey]

diligentClerk

[plain raven]

this binary operation on spaces $\ast$ is functorial, in that given maps $f : X\to X', g: Y\to Y'$ there's an induced map $f\ast g: X\ast Y\to X'\ast Y'$

[gentle osprey]

diligentClerk

[plain raven]

Hatcher defines the join in Chapter 0 if you want to see a reference for this stuff

anyway this is a very natural operation for simplicial stuff because $\Delta^p\ast \Delta^q\cong \Delta^{p+q+1}$

[gentle osprey]

diligentClerk

[plain raven]

this is easy to work out

more generally if $X,Y$ are two simplicial complexes (or for you, delta complexes, because you're studying hatcher) the join of $X\ast Y$ is again a simplicial complex (delta complex) in a natural way. no subdivision etc is required like for the cartesian product, it's just a corollary of the fact that the join of two simplices is again a simplex.

[gentle osprey]

diligentClerk

[plain raven]

moreover, and this is pretty cool, if $X, Y$ are delta complexes, and $C(X), C(Y)$ denote their associated chain complexes (derived from the triangulation, then the chain complex $C(X\otimes Y)$ of the join is easy to calculate using the complexes $C(X)$ and $C(Y)$ - it's exactly the total tensor product $C(X)\otimes C(Y)$, up to a dimension shift (which i'm going to be a bit handwavy about here because the details are pretty complicated)

[gentle osprey]

diligentClerk

[plain raven]

here by the total tensor product I mean $(C(X)\otimes C(Y))n = \bigoplus{p+q=n}C(X)_p\otimes C(Y)_q$

[gentle osprey]

diligentClerk

[plain raven]

with the usual differential

Ok. So, topologically, there's a natural map $X\times Y\to (X\ast Y)^I$. It's easy to describe what this is: it sends the pair $(x,y)$ to the canonical path from $x$ to $y$ in $X\ast Y$, remember $X\ast Y$ is constructed exactly by freely drawing lines from points in $X$ to points in $Y$.

[gentle osprey]

diligentClerk

[plain raven]

And then of course $(X\ast Y)^I$ is naturally homotopy equivalent to $X\ast Y$, although not by a unique choice of natural homotopy equivalence. You have a map $(X\ast Y)^I\to X\ast Y$ given by "evaluate the path at $0$" and $X\ast Y\to (X\ast Y)^I$ given by "send $p$ to the constant path at $p$" and these two are naturally homotopy equivalent, so $X\ast Y$ is a deformation retract of $(X\ast Y)^I$.

[gentle osprey]

diligentClerk

[plain raven]

In the strong sense, I guess.

Grr. I'm worried on the edge of saying something nonsensical so i'm trying to be careful.

Here's an exercise you can do that will hopefully give you a sense of the point i'm trying to make

Let's work in the category of pointed topological spaces. If $(X,x_0)$ and $(Y,y_0)$ are given, i assume you know what is meant by the wedge product $\land$ and the reduced suspension $S$. I define the reduced join of $(X,x_0)$ and $(Y,y_0)$ as the join $X\ast Y$ modded out by the equivalence relation given by asking that $(x,t,y_0)\sim y_0$ for all $x,t$, and $x_0\sim (x_0,t,y)$ for all $t,y$.

You can prove that $S(X\land Y)\cong X\ast Y$ where all operations are interpreted in the reduced sense. This shows that the join and smash product are kind of the same up to a degree shift, a dimension shift, because we interpret the suspension operation as shifting all the 'holes' in the space up by one degree. conversely the smash product is like a downshift of the join

[gentle osprey]

diligentClerk

[plain raven]

a 'delooping'

ok. so there's some nice relationship between the cartesian product (or in the pointed case the smash product) and the join, in that they agree up to a dimension shift.

We have already said that if $C(X)$ and $C(Y)$ are the chain complexes associated to delta complexes $X$ and $Y$, then the chain complex of their join $C(X\ast Y)$ is just the total tensor product $C(X)\otimes C(Y)$ shifted up by one. Or to say this another way, $C(X)\otimes C(Y)$ is a kind of "delooping" or downshifting in the category of chain complexes which is given by taking the "join" of the two chain complexes and shifting everything down by one dimension.

[gentle osprey]

diligentClerk

[plain raven]

On the other hand, how should the operation of forming these chain complexes play with the cartesian product? In the singular case at least, the singular $n$ simplices $\Delta^n\to X\times Y$ are in bijection with the set of ordered pairs $(\sigma, \tau)$ where $\sigma : \Delta^n\to X$ is a singular $n$ simplex in $X$ and $\tau$ is a singular $n$ simplex in $Y$. This follows from the universal property of the Cartesian product of spaces. So if $C(X)$ is the chain complex of singular simplices in $X$ and $C(Y)$ is the chain complex of singular simplices in $Y$, so that $C(X)_n$ is free on $n$ simplices in $X$ and $C(Y)_n$ is free on singular $n$ simplices in $Y$, then $C(X\times Y)_n \cong C(X)_n\cong C(Y)_n$. (There's a theorem about abelian groups which says that the free abelian group on the cartesian product $A\times B$ is isomorphic to the tensor product of the free Abelian groups on $A,B$ respectively. this isn't hard to show.)

[gentle osprey]

diligentClerk

[plain raven]

I'll denote the chain complex $C(X)_n\otimes C(Y)_n$ by $C(X)\times C(Y)$. of course it's not really the categorical product in the category of chain complexes, which could be confusing, but normally we'd use the direct sum notation for that, and i don't need to talk about the direct sum here anyway. We view $C(X)\times C(Y)$ as algebraically representing the cartesian product of the two spaces.

[gentle osprey]

diligentClerk

[plain raven]

So, from the homeomorphism $S(X\land Y)\cong X\ast Y$ that we talked about earlier, we might hope that in the category of chain complexes we can get the "Cartesian product" $C(X)\times C(Y)$ to agree with the "delooping of the join" $C(X)\otimes C(Y)$. Indeed this is the case. These two complexes are naturally homotopy equivalent! This is called the Eilenberg-Zilber theorem.

[gentle osprey]

diligentClerk

[plain raven]

It's a great theorem. the actual formulas are a bit of a combinatorial nightmare but it is possible to get geometric intuition for what they say, I promise!

i won't go into that right now. But the point is, this Eilenberg-Zilber theorem $C(X)\times C(Y)\simeq C(X)\otimes C(Y)$ is really really fundamental. It's the basis for all the product structures in singular homology and it comes from this subtle relationship between the Cartesian product and the join.

[gentle osprey]

diligentClerk

[plain raven]

Now, one of the important things about the homology functor from chain complexes to graded groups (i.e. sending $C_\bullet$ to $H_\bullet(C) =\bigoplus_n H_n(C)$ is that it actually plays nicely with the total tensor product in that there's a natural transformation $H_\bullet(C) \otimes H_\bullet(C')\to H_\bullet(C \otimes C')$. A pretentious way of saying this is that the homology functor is 'lax monoidal'. The famous Kunneth theorem gives necessary and conditions for this map to be an isomorphism using homological algebra to describe the quotient. similarly in cohomology, say with coefficients in modules $G,G'$, there is a natural map $H^\bullet(C,G)\otimes H^\bullet(C',G')\to H^\bullet(C\otimes C',G\otimes G')$.

[gentle osprey]

diligentClerk

[plain raven]

It's not hard to work out what these natural transformations are on your own. Take an educated guess!

Now. If we combine the Eilenberg-Zilber theorem i described a minute ago with this natural transformation, what do we get? Let $X,Y$ be spaces, we'll look at singular homology.
We have natural maps
$H_\bullet(C(X))\otimes H_\bullet(C(Y))\to H_{\bullet}(C(X)\otimes C(Y))$.
But of course $H_\bullet(C(X))$ is just $H_\bullet(X)$, and $H_\bullet(C(Y)=H_\bullet(Y)$. And by the eilenberg-zilber theorem there is a natural homotopy equivalence $C(X)\otimes C(Y)\cong C(X\times Y)$, so $H_\bullet(C(X)\otimes C(Y))\cong H_\bullet(X\times Y).$ Thus we get a natural map
$H_\bullet(X)\otimes H_\bullet(Y)\to H_\bullet(X\times Y)$.

[gentle osprey]

diligentClerk

[plain raven]

This map is called the cross product in homology.

It is the most fundamental product structure in singular homology.

There is also a cross product in cohomology

$H^\bullet(X;G)\otimes H^\bullet(Y;G')\to H^\bullet(X\times Y;G\otimes G')$.
We can derive this almost exactly the same way as the other one, with minor changes, so i won't explain that in detail.

[gentle osprey]

diligentClerk

[plain raven]

The most important case of the cross product is when we take $G =G'=R$ for some ring, $R$. Then we can use the multiplication map $R\otimes R\to R$ to give a map $H^\bullet(X\times Y;R\otimes R)\to H^\bullet(X\times Y;R)$. It's obvious how to do this i think.

[gentle osprey]

diligentClerk

[plain raven]

Thus combining these two we get a map $H^{\bullet}(X;R)\otimes H^{\bullet}(Y;R)\to H^\bullet(X\times Y;R)$. This i think is also sometimes called the cross product map in cohomology.

[gentle osprey]

diligentClerk

[plain raven]

Ok. So now we arrive at the cup product, at the very end. Set $X=Y$ and look at the diagonal map $\delta : X\to X\times X$. Then because cohomology is a contravariant functor, there's an induced map $\delta^{\ast} : H^\bullet(X\times X;R)\to H^\bullet(X;R)$.

[gentle osprey]

diligentClerk

[plain raven]

If we combine this with the cross product from earlier we get a map $H^\bullet(X;R)\otimes H^\bullet(X;R)\to H^\bullet(X;R)$.

[gentle osprey]

diligentClerk

[plain raven]

That's the cup product.

So I think we should understand the cup product in terms of the cross product and not the other way around, as hatcher does.

seems a bit bizarre to me honestly.

This is the exposition Spanier gives in his book and this was kind of a revelation to me after Spanier's attempt

he doesn't give much geometric intuition but at least the order of development makes more sense

Anyway i'll stop there but talk to me another time about if you want some more intuition for the formulas in the eilenberg-zilber map and where they come from

we also have the alexander whitney map

i've been working pretty hard on my own to try and get these formulas to just fall out of the sky naturally by appeal to general categorical principles

i've gotten some of them

ok.

done there

peace

🥤✖️

i didn't see it as i'm on dark background.

thanks for pointing it out.

[gritty widget]

:pettheclerk:

[plain raven]

is this like

"time to drink"

[orchid forge]

I think he wants you to talk about the categorical properties of the topological space "cup with a straw" and how it relates to this cartesian product map

[plain raven]

oh i'm an idiot

[orchid forge]

(i don't actually think that)

[plain raven]

cup product

i thought it meant "drinking times"

[orchid forge]

oh true that is what that is

[plain raven]

tteppa now that it's like 24 hours later what do you have to say about warner's book on smooth manifolds and lie groups

[gritty widget]

i never had a moment to look at it or read the original question i got pinged for

that night i went and got wasted with my friends

and i spent today hiking

[plain raven]

nice

[long hornet]

Wikipedia says the homology of spheres can be computed just from the axioms. That seems interesting - is there a simple characterization of such spaces? And am I right in thinking they are precisely the ones whose homology can be computed without reasoning about what the maps that show up between homology groups do?

Also, if a space is assigned the same homology groups by every homology theory, can we compute them from the axioms?

(By the "axioms" I meant Eilenberg-Steenrod axioms with the dimension axiom)

[orchid forge]

the axioms are enough to give you the Mayer-Vietoris sequence, which you can use to do a bunch of simple examples

i dunno if there is an exact characterization of spaces whose (co)homology is determined by the axioms though

[abstract pagoda]

is mater vietoris sequence the LES made with relative homology?

like LES of (Sn,dSn)

or am i thinking something not it

yea thats not it nvm

[orchid forge]

re: this, I think you can relatively easily show that the axioms are sufficient for a space with a sufficiently nice cover

because of this fact

[long hornet]

I think even then we need to reason about maps somehow?

[orchid forge]

well yes

inclusion maps in particular

but inclusion maps are mentioned in the axioms, so you can still say some stuff

intuitively it should be possible for any space that is built in a nice way out of other spaces whose homology is computable from the axioms

[long hornet]

Reasonable

[orchid forge]

if you don't include the dimension axiom though

it might get weird

you still have MV but one fewer building block, in this analogy

[long hornet]

We wouldn't know the homology of, say, D^n?

[orchid forge]

yeah I don't think so

[long hornet]

I think this should follow from the completeness theorem in logic, but sadly I don't even know its precise statement. If it is true, then maybe we can use the result that the axioms determine the simplicial homology for all triangulable spaces, or something, and in this case my question is kinda trivial because a whole theory is "built" from the axioms.

[orchid forge]

if you replace the homotopy axiom with weak homotopy then the question definitely does become trivial

[long hornet]

What does it say?

[orchid forge]

if weak homotopy equivalences induce isomorphisms in homology then, since any space is weakly homotopy equivalent to a CW complex, we can just compute the cellular homology of that, which i think is the unique homology theory satisfying the axioms for cw complexes

[long hornet]

Aha

So the axioms are "too" strong in this sense

[orchid forge]

i'm not sure

[long hornet]

Most "identities" in specific groups cannot be deduced from group theory axioms

I mean there are many groups but only one homology (when we restrict to nice spaces)

[orchid forge]

there is a paper by Milnor that shows that there is (up to isomorphism) only one additive (co)homology theory on the category of pairs (X,A) where both X and A have the homotopy type of a CW complex

still using the dimension axiom though

[long hornet]

Do we get a consistent theory by picking an arbitrary sequence of groups, and saying these are the homology groups of the one-point space?

I mean replacing the dimension axiom

[orchid forge]

I was wondering the same thing, there's all kinds of generalized homology theories but I dunno what you can get by fixing the homology of a point

[gritty widget]

the first derived things you meet are derived functors in the form of Tor and Ext

and the as far as i understand

these are best thought of as obstructions to tensor and him being exact

i assume we can do something similar for any other functor possibly modulo small conditions like being a functor in abelian categories

can someone tell me

what the big picture of derived algebraic geometry is

why do you do it

feel free to use terms i won’t understand

i am more so lookin for an answer that i can stew on and eventually learn where thins are going

[haughty anvil]

I'm a bit confused as for what to do here. I'm on 1c and I don't think I really understand vanishing lines very well.

This is what I have so far.

[frigid patrol]

nonorientable Riemann surface

[haughty anvil]

Huh?

<@&286206848099549185>

[pearl holly]

Yo omg I didn’t even see that you wrote all of that! I will need to take a careful read through it later on because this is good stuff. Thank you so much!

[sonic hill]

I don’t claim to know all of derived algebraic geometry but the “why do you do it” is the same reason as you listed above; if for example you are considering an intersection theory, there are cases of intersection multiplicities being less than you want because of Tor groups which are “invisible” unless you take the entire derived tensor product into account

[pearl holly]

I also just very quickly want to ask if $\alpha \smile \beta[v_0, \cdots, v_n] = \alpha \smile \beta [v_{\sigma(0)}, \cdots, v_{\sigma(n)}]$ where $\sigma \in S_n$, so is the cup product independent of how I choose to "represent" my simplex? Because I know that if you permute some $v_i$'s in $[v_0, \cdots v_n]$ then the orientation might change, so does the cup product respect that? If this holds up to sign change, then how do I know how I'm supposed to "label" a simplex when I'm calculating the cup product?

[gentle osprey]

Tokidoki ✓

[pearl holly]

so let's say that I have this torus and I want to look at $\cup: H^1 \times H^1 \to H^2$. I know that $H^1(T^2; \mathbb{Z}) = \text{Hom}(H_1(T^2), \mathbb{Z})$. The boundary of the sum of the simplicies in the picture above is zero so I want to look at that. $a$ and $b$ are the generators of $H_1(T^2)$ and so the dual basis for $\text{Hom}(H_1(T^1), \mathbb{Z})$ is $\alpha$ that sends $a$ to 1 and everything else to 0 and the same goes for $\beta$. Let $\alpha$ be a representative of $\alpha$ in $H^1(T^2)$ and the same for $\beta$. I want to have $\alpha \partial = \beta \partial = 0$ and this is true since I have drawn the $\alpha$ and $\beta$ with orientations (which looking at it now, does it really matter what orientation $\alpha$ and $\beta$ has? It seems like no matter what I choose, $\alpha \partial$ will always be zero). Now the thing is that I have $\alpha \smile \beta [1, 3, 0] = 1$ but $\alpha \smile \beta[0, 3, 1] = 0$ even though $[1, 3, 0]$ is the "same simplex" as $[0, 3, 1]$. So how do I know which one to pick?

[gentle osprey]

Tokidoki ✓

[dim radish]

i dont need any hints, i just dont understand what is being asked

what does it mean for A-tilde, p to be the covering space corresponding to the kernel of that map

proving that A tilde is a covering space isnt difficult

[gritty widget]

i’ve heard this example before of bezouts theorem and trying to find the right way to count multiplicity

i suppose i more so want to know

what objects do we derive

[gritty widget]

Is it possible that in some metric space, we can never attain the minimum length among all the curves connecting two points so only the infimum of the length is obtained?

[gentle osprey]

weilam06

[gritty widget]

Oops it's arbitrary metric space

My bad

I think it's correct by considering R^2 with origin removed

[gritty widget]

My concern is that if we pick (-1,0) and (1,0), let says, then yeah; but what if I take two points on R^2 which its shortest line segment does not pass through the origin?

[hollow harbor]

so you want this to be true for all points?

[gritty widget]

Yeap

[sonic hill]

Ideally we derive schemes and stacks. Derived tensor product of rings becomes derived fiber product of schemes (apparently this still not a fully developed theory though)

All of the stuff like spectral schemes are basically attempts to make a rigorous theory of that

[tough imp]

Tfw serre intersection formula

[sonic hill]

[orchid forge]

It's talking about the galois correspondence between covering spaces and subgroups of pi_1

[gritty widget]

okay thanks

[pearl holly]

I think I'm slowly realizing how shitty Hatcher's examples of the cup product are

[gritty widget]

My definition of a variety is a ringed space $(X,\mathcal{O}_X)$ such that there is an open cover $U_i$ of $X$ such that each $(U_i,\mathcal{O}_X|_U)$ is isomorphic to a closed subset of affine space. I want to show that if $f$ does not vanish on some open set $U$ then $f$ is a unit in $\mathcal{O}_X|_U$

[gentle osprey]

lime_soup

[gritty widget]

this is obvious if U is an affine open set

[pearl holly]

yoo you lost the light blue name

[gritty widget]

but dont' see how to do this for an arbitrary open set

[true robin]

Cover U by affine sets, then for each V_i in the cover we have a local inverse g_i now notice that g_i and g_j have to be equal on any intersection because on V_i \cap V_j fg_i=1=fg_j. So that means that the g_i satisfies the sheaf condition so there exists g on U st g restricted to V_i=g_i and g is the inverse

[long hornet]

Maybe take R^2 - (Q x Q)

[gritty widget]

thank you!

the g_i g_j intersectin was what i was missing

[gritty widget]

Ohh then the denseness of Q in R showing that the shortest path is not attainable right

[true robin]

No problem, also this should hold in general for locally ringed spaces, because the germ of f at each point x is not in the maximal ideal of the stalk so it has an inverse in that ring so it will have an inverse on some open set containing x, now we can use the logic I gave to finish off

[gritty widget]

thanks!

[orchid forge]

If you have only learned cup product from hatcher, i strongly recommend you look into the de Rham formulation for geometric intuition

[pearl holly]

ye I really need to do that

[orchid forge]

read (the beginning of) bott & tu maybe

[pearl holly]

there's so many details in these examples that Hatcher skips

[orchid forge]

it's an awesome book to read

[pearl holly]

He just claims a bunch of stuff like "oh it's an easy calculation that blablabla" when in reality you can't even compute that thing because that thing is dependent on how you label the vertices and he hasn't done that

okay roger that! Thank you so much!

[gritty widget]

I'm trying to under the segre embedding

sorry i keep accidentally hitting enter ill type this all and then post it

[modest nexus]

Hi! I have the following question: prove that if F -> E -> B is a locally trivial fibration, where the base space B is contractible, comapact and T2, then this is a trivial fibration, that is E = B x F. Can you give me some hint regarding the solution?

[orchid forge]

I could be wrong but I don't think there's a particular trick here. Lifting a nullhomotopy of $B$ lets you identify all these copies of $F$ in the local trivialization: given $(b,v) \in B \times F$, the nullhomotopy gives a path $b \to b_0$, which lifts to a path $U \times F \cong p^{-1}(b) \to p^{-1}(b_0) \cong V \times F$, and you associate $(b,v)$ with the other endpoint of the path ending at $(b_0, v)$

[gentle osprey]

Kogasa

[orchid forge]

err those should be $b \times F$ and $b_0 \times F$, not $U \times F$ and $V \times F$

[gentle osprey]

Kogasa

[orchid forge]

does this make sense at all? i can draw a picture if not

maybe a simpler way to say it would be, if B collapses onto b then you can collapse a local trivialization U x F onto b x F in a unique, continuous way (by some lift). Each e in E lies in a fiber over p(e), and collapses onto (b,v) for some v in F, and we identify e ~ (p(e), v)

then you can either prove this is an isomorphism directly or use the universal property of cartesian products

[quartz edge]

i'm trying to compute the derivative of the curvature of a curve and i keep getting the following (in terms of the unit tangent vector T in the frenet frame)

[gentle osprey]

geometroid

[quartz edge]

this appears to be zero

curvature is defined as:

[gentle osprey]

geometroid

[quartz edge]

what am i doing wrong here?

i figured you could just like

chain rule that sucker

[hollow harbor]

How exactly are you chain ruling the sucker

[quartz edge]

,,\kappa' = \frac{d|T'|}{dT'} T''

[hollow harbor]

Oh I see

[orchid forge]

why is it 0?

[gentle osprey]

geometroid

[hollow harbor]

Its zero because the curve is unit speed

[quartz edge]

T'' has a (3, 1) matrix representation

yes

it is

[orchid forge]

oh that is true

[quartz edge]

but that seems weird, why would the curvature derivative be zero?

implies curvature is constant

[hollow harbor]

Well

Wait

No

I dont think that is 0

[quartz edge]

unit speed =/=> constant curvature

[hollow harbor]

Sorry

T dot T' should be 0

[quartz edge]

oh

you're righct

[hollow harbor]

But not T' dot T''

[quartz edge]

because T is constant length

you're right, thanks

[hollow harbor]

Yeah I always screw that up lol

[quartz edge]

that resolves my problem, thx u

[hollow harbor]

Cause T itself is the derivative of a thing

[quartz edge]

yeah

fixed length so the tangent space is a plane

[orchid forge]

this is still not true

[quartz edge]

it is

[hollow harbor]

It is if |T|^2 is constant.

[quartz edge]

^

T is defined as a unit vector

[hollow harbor]

(which, if 1, is what it means to be unit speed)

[quartz edge]

thanks u saved my night

[hollow harbor]

I think the best way to remember this is that T' when normalized is N, the normal vector.

But then T'' is not the binormal

[quartz edge]

ok so this finally yields an expression for torsion

[gentle osprey]

geometroid

[quartz edge]

in terms of just derivatives of the unit tangent vector

[quartz edge]

i needed it

to do uhh

to compute the variation of an integral over torsion

https://tenor.com/view/get-real-get-fake-meme-airplane-when-the-gif-20448665

ah shit it turns out theres another difficulty hiding right after this one

turns out my night is stunted again

[orchid forge]

simply overcome it

[quartz edge]

well

it's too much symbolic computation to overcome

i need a CAS

[gritty widget]

[hollow harbor]

It's giving cope

[sullen patio]

This is for my analysis class but i thought posting it here might be more appropriate

All we discussed about dense subsets is that their closures r equal to the parent set

So the image of the closure of E is equal to the image of X, any hints on what to do next?

[empty grove]

That statement is equivalent to "f(E) closure contains f(E closure)", and you can prove this by looking at sequences converging to points in the closure (I'm assuming you're working with metric spaces)

[sullen patio]

Yes we r

f(E) closure contains all the limit points of f(E) right?

How do we relate that to f(E closure)

[empty grove]

Take a point of the latter and show that the point must be in the former

[sullen patio]

So i said let f(x) be in f(E closure), so x is in E closure

Ultimately we want to show f(E) closure is equal to f(X) dont we?

[empty grove]

Yes, when you take closure inside the subspace f(X)

So it suffices to show only one inclusion

As in, you have some U ⊂ A ⊂ B, and you want to show that U is dense in A, ie closure of U in A is A, and this is equivalent to showing that closure of U in B contains A

Because then everything in A is a limit point of U

[empty grove]

If f(x) is in f(U), you can't be sure that x is in U

But you can start by saying
Let y ∈ f(E closure), then there is some x in E closure such that y = f(x)

[gritty widget]

there was a brief mention made in an algebra lecture that if X is compact and hausdorff, and if C(X) is the ring of all continuous functions from X to C, then all maximal ideals of C(X) consist of the functions that evaluate to 0 at a given particular point x of X. I can easily see that each such ideal is maximal, but I can't see why all maximal ideals have this form, and where the properties of X being compact and hausdorff come in at all

it was just meant to be a fun little cute example 😭😭

[empty grove]

If m is a maximal ideal such that the elements of m have a common zero, then there must be at most 1 common zero, otherwise you have a larger proper ideal containing it. Now suppose there are no common zeroes. Then for each point in X there's a function f_x in m which is non zero at x therefore in a neighborhood U_x of x. By taking powers of f_x you can assume that the real part of f_x is positive on x (hence on small enough U_x). Take a finite subcover of this cover and the sum over (the x in the finite subcover) of f_x and this is non zero everywhere hence a unit

[trail tiger]

if i just wrote $\int \alpha\otimes\mathrm{d}\beta$ where it's understood that $\otimes$ is a tensor product, is this even well-defined? you need antisymmetry in the tensor algebra to use geometric integration, aka the wedge product, so I'm puzzled at friz and hairer using this in their papers/books

[gentle osprey]

teafortwo

[trail tiger]

the idea is for the regular euclidean case to collapse to lebesgue intengration of alpha wrt the induced measure of dbeta

[tight agate]

the exterior algebra is a quotient of the tensor algebra

so it could just mean the corresponding equivalence class in the exterior algebra

[trail tiger]

so im taking this integral modulo that class? i guess i should check that preserves representation invariance

[gritty widget]

ohhhh ok that's so nice, I assume somehow X being hausdorff was used secretly, but I didn't catch it and maybe my mind jumped assumptions. what step fails if X is non hausdorff?

[empty grove]

I don't think hausdorff is needed for this part, it's needed to say that x ↦ m_x is injective

[gritty widget]

oh right

actually I'm not seeing why the sum over the finite subcover is nonzero everywhere

each f_x is nonzero on U_x

but when you add them

[empty grove]

Oh you're right that doesn't work

[gritty widget]

they can create zeros

I have an idea

[empty grove]

oof yeah, this works when you have real valued functions and you take squares

[gritty widget]

you can form a function algebra generated by the finite set {f_x}, so like polynomials in those functions

but I'm forgetting what neat properties those have

like if that ring satisfies certain properties then I know it's dense in C(X)

so you could just make it close to some unit

[empty grove]

oh yeah that could work

[gritty widget]

and therefore make it a unit

[empty grove]

How would you show that this algebra separates points?

[gritty widget]

that's what I'm thinking of rn

[orchid forge]

If $f_x \in m$, then $\bar{f_x}$ is continuous and $|f_x|^2 = f_x \cdot \bar{f_x} \in m$ right?

[gentle osprey]

Kogasa

[gritty widget]

I was thinking of saying like maybe it doesn't

but if it doesn't, then your functions must already be nice

in some way

[empty grove]

Perfect

I was trying to work with weaker condition of being closed under multiplication 😵‍💫

[spring relic]

Good evening guys, I'm kinda stuck on how to go about this problem.
So some background first.

Two metrics, d1 and d2 on the same set X are called strongly equivalent if there are constants
0 < c < C such that
c d1(x, y) ≤ d2(x, y) ≤ C d1(x, y)
for all x,y in X

Prove that the following metrics are strongly equivalent (R^k)

1. Taxicab (Norm 1)
2. Standard Euclidean (Norm 2)
3. Max Metric (Norm inf)

[empty grove]

I think this might actually work lol

You can change X to X/~ where ~ identifies all points that aren't separated and quotients of compact spaces are compact

So this algebra is a dense subalgebra of C(X/~)

[gritty widget]

yes yes yes

so the other points are filled in because they are defined as the ones that aren't separared

it seems silly to separate points if all we are looking for is a function with no roots 😭

but this is my only way of using the results I know 🤞🤞

[empty grove]

Lol yeah this is nice way to do it

Though kogasa's solution is much simpler

[gritty widget]

my other option would just be modifying the Weierstrass approximation theorem for a weaker condition... and no thanks

[gritty widget]

sorry!!!! I didn't see this

that's wonderful

that's wonderful

now I'm 🤦‍♀️

i was literally like "hmm if these were real we could just square them"

"too bad there's nothing like that for complex numbers" 💀

good work team

[gritty widget]

I don't remember sending this twice 🤔🤔

[spring relic]

appreciate if anyone has any pointers on how to get started.

[empty grove]

Try sketching what an open ball looks like in each metric, and the equivalence condition talks about fitting scaled versions of the open balls in different metrics inside one another (after you fix an x)

You can guess what c and C would be by looking at those sketches and then formally prove the inequalities

[tough imp]

Taxicab and standard are kinda easy

It’s basically drawing a triangle

Like, just consider points on the unit circle

Since you can just scale stuff

[spring relic]

ohh, ok, so for 1) and 2) I see that the c = 1 and C = 2/sqrt(2) does that work?

[tough imp]

Right

2/sqrt(2)

Has a umm

More standard name tho

Hahaha

[trail tiger]

c and C should depend only on the dimensions

i wonder what we can say about strongly equivalent metrics that are dimension-independent. can they only be scaled versions of the other?

[gritty widget]

hahahaha

[spring relic]

right sqrt(2)

[tough imp]

Right

For the infinity norm

Well

was it true in general that taxicab >= standard >= infinity

I think so by just

Looking at the unit circle again

So I think strongly equivalent is transitive?

Like… pretty easily?

[empty grove]

Infinity balls are squares

[tough imp]

So I think it boils down to the infinity and standard

And for this one…

I remember being presented this proof on day 1 of analysis

Freshman year

But I don’t remember the idea

Isn’t it something stupid like uhh

Oh well

Isn’t the infinity norm minimized at like

45 degree angles?

If that makes sense

[trail tiger]

ez to prove by showing two sided inequality

[tough imp]

It’s when the two coordinates have the same absolute value

[trail tiger]

[tough imp]

(Again on the unit circle)

[empty grove]

Infinity balls are actually squares lol like sides parallel to the axes

[tough imp]

Oh right

What if this isn’t R^2

Well whatever, find a point where infinity norm is minimized

[orchid forge]

you maximize the infinity norm by just moving a single coordinate

[tough imp]

Wrt the unit ball

[orchid forge]

minimize it by making them all as close as possible

[tough imp]

Well you wanna minimize it

Yeah

You should be able to minimize it and get a constant

[empty grove]

bonk

[tough imp]

Just work on the unit ball (in Euclidean, lol)

And everything else will scale out the same way

[spring relic]

ok, I'll try

[tough imp]

Hopefully this was actually helpful

[empty grove]

It would have been if you didn't nitpick dimensions

[tough imp]

[spring relic]

I'm just stuck on showing the inequalities. I assume it goes something like the triangle inequality proof

[tough imp]

I’ll let moldi handle this

[empty grove]

Oh shit the problem was in R^k

Chmonkey was right

Chmonkey wins again

It won't be √2 in higher dimensions

Will be √k

[spring relic]

ye

[empty grove]

Now maybe you'll have more success

[gritty widget]

btw @empty grove I just realized, even though we are over this, all ideals separate points, it's necessary for them to absorb multiplication.

[empty grove]

The 0 ideal doesn't

But yeah as long as the ideal doesn't vanish at a point that makes sense

Urysohn right?

[gritty widget]

urysohn what?

[empty grove]

Wait how are you proving this

[tough imp]

So to prove the inequality, do it based on the norm of the point wrt the euclidean one.

When you have a point on the sphere of radius r in euclidean

You just need to prove that the taxicab length is at most sqrt(k)r

You can do this by say, I think am-gm

Maybe?

If not you can use calculus… if you have access to that

[spring relic]

I think I just have the definition of the metric to work with

so the metric function

[tough imp]

What is gonna be the best way to do this then

[empty grove]

Fix an x, prove the inequality for all y, choosing c and C to be independent of x, so that the inequality is shown for all x as well

And without loss of generality you can do this for x=0

[tough imp]

Oh well

In my head I’ve already done this kekw

I’m just struggling to come up with a nice proof that this bound works for points on the sphere

Without calculus

I actually don’t think am-gm works?

[empty grove]

AM-GM works right?

Oh

[tough imp]

Does it?

[empty grove]

Let me have a think

[tough imp]

It seems like you’re bounding the arithmetic mean by the wrong thing

[empty grove]

True

It's the AM RMS inequality or something

[tough imp]

TFW useless w/o calculus

[empty grove]

[orchid forge]

draw a triangle and do induction

[tough imp]

Okay well

THATS WHAT I WAS THINMING

But it’s messy

So my shitty idea was

[spring relic]

^ yeah

[tough imp]

Fix M

[empty grove]

Thinming 🥸

[tough imp]

Write out k things in the form

x_i = M/k + epsilon_i

And uh

I guess the sum of epsilon_i = 0

[spring relic]

So my general proof struct, give c and C. Fix arb. x,y in R^k, try and show inequality holds

[tough imp]

Then expand out what the Euclidean norm of

(x_1,…,x_k) is

Like the point is I want to show this is less than the norm of

(M/k,M/k,…,M/k)

In terms of the differences epsilon_i

Or something

Lmfao

Idk if this works though lmfao

Like well

We can ignore the sqrt

So it comes down to showing that

(M/k + epsilon_1)^2 + … + (M/k + epsilon_k)^2 <= k (M/k)^2

And like on the left hand side

We get k(M/k)^2

Along with like

2M/k(sum epsilon_i)

= 0 by assumption

And then the sum of epsilon_i^2

So we need to show that sum epsilon_i^2 <= 0

When these are numbers chosen so that they sum to 0

Does this… work?

¯_(ツ)_/¯

Wait this has to be bullshit

I messed up somewhere

These are all nonnegative cuz square

[spring relic]

I'm just gonna do induction

[tough imp]

Friggin

Chmonkey moment

[spring relic]

for the inequality

[spring relic]

monkey magic

[tough imp]

Oh wait I’m dumb

It’s cuz

My inequality is supposed to be backwards

Lmfao

I want to show that

For a fixed number for the sum of the coordinates

The Euclidean norm is GREATER when the coordinates aren’t all M/k

And so it should be showing sum epsilon_i^2 >= 0

But this is… obvious

So it should be like this

Fix x

If the taxicab norm is M

Call the Euclidean norm uhhh

N

I swear this should work

I am just drawing a blank on order to do it hurb

Oh RIGHT

Okay sorry

So N(x) >= N((M/k,…,M/k))

And note our taxicab norm is M

By assumption

So it suffices to show that for the constant C

That CN((M/k,…,M/k)) >= Taxicab norm(x) = M

But also the point (M/k,…,M/k) has taxicab norm M

[empty grove]

If you have radius r, then you're trying to compare distance in standard norm (which is √k times RMS of coordinates) and distance in taxi cab (which is k times the AM of coordinates). RMS ≥ AM gives √k standard distance ≥ taxi cab distance and taxi cab distance ≥ standard should be just Pythagoras

[tough imp]

So we can basically just show that CN(M/k,…,M/k) = CM/sqrt(k) = M

Wtf is the RMS AM inequality bruh

[empty grove]

RMS ≥ AM ≥ GM ≥ HM

I know because I just looked it up

[tough imp]

Idk what RMS is

Anyway my proof

Actually ducking works

[empty grove]

Lul it's root mean square

[tough imp]

It’s mega pepega

[empty grove]

RMS(a, b) = √(AM(a²,b²))

So literally root of mean of squares

[tough imp]

Hurbed

[empty grove]

It's annoying to prove RMS ≥ AM and I'm sure you're just trying to do that ultimately

But it should be fine to assume this inequality

RMS is also called QM (quadratic mean apparently)

And I think RMS ≥ AM might be just Cauchy Schwartz

[spring relic]

Found this online

Seems legit

[bleak path]

Hey, does anyone know an explicit example of a homeomorphism between S1 and D2 to the torus? Like an example of a function

[vocal anchor]

Do you mean from S1 x D2 to the solid torus?

[bleak path]

Yes, the solid torus, sorry I wasn't clear on that

[vocal anchor]

For me, an explicit homeomorphism would be the identity, but what is the solid torus for you?

[bleak path]

The filled in donut?

Oh I see where I've gotten mixed up

I mean from the circle S1 and the solid disk D2 to the solid torus which is the product space of S1 and D2

[vocal anchor]

what does "and" mean in "from the circle S1 and the solid disk D2"?

[bleak path]

So I know that the solid torus is the product space of the circle and the disk. I'm asking if there is an example of f(s, d) where s is in S1 and d is in D2 and the image of f is the solid torus

[vocal anchor]

But then aren't you asking for a function f: S1 x D2 -> S1 x D2?

I mean, the solid torus, as you said, is S1 x S2. And you are asking for a function that takes two inputs, namely from S1 and D2, and spits out something in S1 x D2. Why not simply take the identity?

[bleak path]

So I guess I've been skipping a step somewhere - I have subspaces D2 and ST defined explicitly, and I'm supposed to construct a homeomorphism between D2 x S1 to ST. I thought by looking at an example or two I could have an intuition of how to do it.

$D^2 = {(x, y) \in \mathbb{R}^2: ||(x, y)|| \leq 1} \subset \mathbb{R}^2$

[vocal anchor]

subspaces of what?

[gentle osprey]

wOne

[vocal anchor]

Do you view ST as a subset of R^3?

[bleak path]

I can? I'm not sure if I have to

$ST = {(x, y, z) \in \mathbb{R}3: (2 - ||(x, y)||)^2 + z^2 \leq 1} \subset \mathbb{R}^3$

[empty grove]

you just said \subset R^3

[bleak path]

So I was given these 2 expressions for D2 and ST, and I'm supposed to find a homeomorphism g: D2 x S1 -> ST

[bleak path]

I did, and admittedly I was using them interchangably because I don't know of a difference between them now

difference between viewing the space ST and viewing ST as a subspace of R3

[gentle osprey]

wOne

[empty grove]

saying "space ST" doesn't tell you what the topology is, only the underlying set, unless you say that you are inducing the subspace topology from R3

I mean all you did was give ST as a set, you have to describe the topology somehow, and saying that it is a subspace is probably the easiest way to say it

otherwise you haven't given enough info

[bleak path]

Sorry, a bit confused now - so you're saying that by me describing ST as a subspace of R3, I have given enough information to describe it?

[empty grove]

yes because you have told us that it has the subspace topology

without that we don't know the topology

[empty grove]

like if you want to describe the space ST separately you have to describe the topology in some other way

[bleak path]

So I suppose this earlier definition of ST described it as a subspace of R3 then

Then in that case, is there a way to come up with a homeomorphism from my definition of D2 plus an arbitrary S1 to my definition of ST?

[empty grove]

Plus doesn't make sense, you are asking for an isomorphism S¹ x D² → ST

[bleak path]

Okay, then given that you seem to know what I'm asking for, do you have an example of such an isomorphism?

[empty grove]

I'll tell you the isomorphism going backwards. So ST has a map onto the circle, which is just "ignore the thickness of ST", and it has a map onto the disk, which you get by taking each disk cross-section of ST and mapping each point in this cross section to the corresponding point in the actual disk

This is scuffed

But this gives you maps from ST to S¹ and to D² and put them together in a pair to get a map to the product

Like imagine S¹ x D² as a bunch of disks glued together in a circle, that's exactly what ST is

[pearl holly]

yo can I ask a quick question?
** **
👉 👈

okay lmfao so what is the difference between a k-vector and a k-form?

I'm watching some video online but they don't really say the definitions

[empty grove]

k-form in the sense of? element of exterior product of copies of k?

[pearl holly]

ye I think so

the guy in the video just says "volume"

[echo elk]

Mathematically, they're dual to one another. Lemme see if I can find the bells and bongs interpretation.

[pearl holly]

oh yeah, so a k-form "measures" a k-vector right?

[empty grove]

let me just review this before I say something wrong

[echo elk]

I'd say that's good.

[empty grove]

wait what does a k-vector mean? k is a natural number and you mean k-tuples of vectors?

[echo elk]

Misner Thorne and Wheeler's gravitation talk about it pages 115-120, I really shouldn't screenshot all of that lol.. ||You can find it online for free though, shh||

[pearl holly]

I honestly don't know, the video keeps saying "k-vectors" and "k-forms" without really defining them

but I guess that k-vectors are just vectors with k components

[empty grove]

So a k-form is an alternating multilinear function from product of k copies of R^n to R (replace R with any field and this works)

[echo elk]

So k-vectors aren't used that much (and I don't know if the wedge product is defined for them), but again they're ideally dual to the k-forms.

[echo elk]

Isn't that just a k-dimensional vector? Not the same in this context, heh.

[pearl holly]

ye it is I guess but I don't really know what else this guy could be talking about lmao

[empty grove]

multilinear means that if any of the k input vectors is a sum of a scalar multiple then you can break the form applied to the vector linearly

like a determinant