point-set-topology

[empty grove]

and alternating means that if you exchange any 2 of the input vectors then the sign flips

again like a determinant

so determinant is an n-form on R^n

unless I messed something up really bad

I am really rusty ugh

[pearl holly]

ye okay that makes sense. Do they have connections to the exterior algebra? Like are they elements of it?

I think you said something similar to that

[empty grove]

Graphically what should be happening is that the form applied to a tuple of k many vectors is telling you the oriented volume of the parallelopiped spanned by the vectors. That's why the 2 conditions make sense

[pearl holly]

ye right I see I see

[echo elk]

Wiki has some talk/graphics of it

[empty grove]

Yeah I think they are but I am not sure about this lol let me see

[echo elk]

In case you ever come across the terms contravariant vectors ("vectors") and covariant vectors ("covectors"), covectors are 1-forms.

[empty grove]

ok for some reason I am not able to tie this together with the universal property and I am now confused about if I ever properly worked this out or not

😵‍💫

[pearl holly]

yeah okay screw the exterior algebra stuff lmao. Thank you both so much!

[pearl holly]

oh yeah this is great lmao

[echo elk]

Alrighty, good luck. I posted an old brief file I had from someone else that talks about it.

[empty grove]

ok yes k-vectors are the elements of degree k in the exterior algebra of whatever vector space and k-forms are linear functionals on the vector space of k-vectors

so k-vectors are actually complicated objects 😵‍💫 I don't have geometric intuition for the exterior algebra though

[pearl holly]

yeah okay I see, thank you so much!

[modest agate]

We say that $f$ is continuous at a point $x_0 \in X$ if for every neighborhood $V$ of $f(x_0$ there exists a neighborhood $U$ of $x_0$ such that $f(U)\subset V$ then $f$ is cont. How do I use this to show that a component function $$f:X\to Y, (a,b,c)\mapsto \left(a,b-a,\frac{c-a}{b-a}\right)$$ is continuous

[gentle osprey]

notsushY

[modest agate]

I actually have to show that $f$ is homeomorphic, I've already shown that $f$ is a bijection mapping what remains is to show that $f$ is cont and a open mapping.

I begin with claiming that $f$ is cont everywhere and take an arbitrary point $p=(x,y,z)\in \mathbb{R}^3$. We have that $$\norm{(a,b,c)-(x,y,z)}\geq |a-x|$$ By a similar argument we get $\norm{(a,b,c)-(x,y,z)}\geq |b-y|,|c-z|$, I'm struggling with finding a $\delta$ for $f$ to satisfy the cont condition $$\norm{\left(a,b-a,\frac{c-a}{b-a}\right)-\left(u,v-u,\frac{w-u}{v-u}\right)}=\dots$$

[gentle osprey]

notsushY

[bleak path]

Thank you, this really helped quite an amount

[cursive spade]

if g is differentiable does that imply there is a linear function in it's germ-class

[wise ruin]

Intuitively, why is it that if S is locally connected, then its connected components are open? I’m having a hard time proving this as the definition of local connectedness talks about connected sets in open sets but we’re proving something about open sets in connected sets.

[cursive spade]

yeah

[wise ruin]

I’m gonna pretend TTerra didn’t delete his response and still respond to it. Finding a connected open set around every point (in a connected set) is what I’m having trouble with. I’m just not sure why this special neighborhood must exist around each point.
I’m sure there is just something obvious that I’m missing

[cursive spade]

I was thinking of R^n

I am guessing if g is differentiable R^n to R^m then there is a sequence of diff functions in it's germ-class that converges to a linear function

or it might have a linear function in it's germ

I working with the idea that diff functions can be linear approximated

@gritty widget

[abstract pagoda]

think about what it means for a set to be connected

there does not exist any nontrivial clopen subsets

if there did that would correspond to a seperation of the set

[abstract pagoda]

this statement right here is justified by what?

im thinking tbe same thing

but idk what justifies it 100%

im guessing a function on a manifold is linear if its linear wrt the chart?

so $f\circ\phi^{-1}$ is linear?

[gentle osprey]

Millionaire

[abstract pagoda]

or ig linear wrt any chart?

nah sounds wrong

probably way more specific definition

[cursive spade]

Just working on R^n for now

[abstract pagoda]

oh lol

what is equivalence for germclass?

g,f are equivalent iff ?

isnt it something with restrictions

[cursive spade]

u fix a point p

then f ~ g iff f = g on some open neighborhood of p

[abstract pagoda]

g~f iff there exist p in X with neighborhood U such that f|U=g|U

yeah

intuitively from Rn->R

yeah

true

[cursive spade]

point taken

[abstract pagoda]

what does vanishing have to do with anything?

yes

?

what does this have to do with original question?

[cursive spade]

but this is true though?

[abstract pagoda]

What do you mean any such funxtion evidently does not agree with 2x|0=0?

[gritty widget]

hurb

[gentle osprey]

Todd Male (#talks Oct 24)

[cursive spade]

but can you get a sequence of functions in germ whose limit is linear

[cursive spade]

agreed

[pearl holly]

#❓how-to-get-help

[abstract pagoda]

So what is in the germ of x^2

functions that are equal at every point in some neighborhoods?

nice sassy

[gritty widget]

[abstract pagoda]

are germs closed?

what does that mean

arent germs the equivalence classes

uh

arent germs equivalence classes of functions

like are the functions closed maps?

idk what u mean

because germs aren’t topologies?

[cursive spade]

my guess is no

[gritty widget]

you're burying the original question

[abstract pagoda]

what does it mean for a germ to be closed why ask that?

you know we have threads for a reason

why sassy

[pearl holly]

Oh we have threats now?

[abstract pagoda]

yea

is U open?

you are asking if U is closed

no ur not

[cursive spade]

the germ of a function f at point p as a subset of a space of functions with the sup topology is not close

[abstract pagoda]

sup topology?

oh

sub

[cursive spade]

yeah

but is there a linear function in the limit points of the germ

and if so is it unique

when g is differentiable

my choice of topology might be wrong

choose the most natural topology

for functions

my instinct tells my a germ is not close in general nor is a germ not closed in general.

[cursive spade]

the original question though

germ of linear functions should have there limit points if we limit our selves to smooth function?

globally

the purpose this was to build some intuition of the jacobian so diff on smooth manifolds would feel more intuitive

linear approximation of diff function at a point

$f(x) = f(x_0) + A ( x - x_0) $ + small error

[gritty widget]

We have $\psi:\mathbb{P}^1\times \mathbb{P}^1\rightarrow \mathbb{P}^3$ defined by $\psi((a_1:a_2)\times (b_1:b_2))=(a_1b_1:a_1b_2:a_2b_1:a_2b_2)$.

[gentle osprey]

lime_soup

[gritty widget]

the image of the segre embedding should be

the zero set of xw-yz

but its not clear to me that this is a homogenous polynomial

oh

homogenous is just the same degree

a homogenous polynomial doesn't need to

be invariant under "scaling"

[orchid forge]

What happens if you scale all the coordinates?

(by the same scalar)

[gritty widget]

you just get

lambda^d

where d is the degree of homogenity

[pastel coral]

Question for people who knows algebraic topology: If $X$ has closed subspaces $A,B$ with $X = A\cup B$ and $X'$ has closed subspaces $A',B'$ with $X'=A'\cup B'$ such that $A$ is homotopic to $A'$ while $B$ is homotopic to $B'$. Is $X$ homotopic to $X'$?

[gentle osprey]

blackiris

[orchid forge]

yeah, that's the usual meaning of homogeneous, and it implies the zero set is fixed under scaling

[orchid forge]

no

[gritty widget]

im not sure how to make this rigours

but any way i can thing of its a no

[orchid forge]

you can provide a counterexample

[gritty widget]

yeah so you can picture

making a sphere out of two discs

or just making a disc out of two discs

the sphere and the disc are not homotopy equivalent

I'm not sure about this

consider the segre embedding again

i want to show that

(a:b)\times P^1 is sent to a line

under the segre emebedding

[pastel coral]

Yeah, I don't think it's true as well.

[gritty widget]

its obvious that this is true if you just say that P^1 is (x:y)

[orchid forge]

not sure what you mean

take p = [a : b], then {p} x P^1 --> [ax : ay : bx : by] = [x : y : (b/a)x : (b/a)y]

assuming a isn't 0

do you disagree? or not sure why they call this a line? or something else

[gritty widget]

yes this should be a line

a line the zero set of two linear equations

so i think this should be the comon zero set of someting like

f(x,y,z,w)=(b/a)x-z

and g(x,y,z,w)=(b/a)y-w

but these aren't homogenous?

[orchid forge]

It is the zero set of the homogeneous polynomial $(b/a)x_1 + (b/a)x_2 - x_3 - x_4$

[gentle osprey]

Kogasa

[orchid forge]

I was thinking about the homogeneous coordinate description as a parametrization

So it becomes clear it's parametrized by just one variable y/x

Err i guess (b/a)x_1 - x_3 and (b/a)x_2 - x_4

[gritty widget]

i think

yes

i was confused because

[orchid forge]

Yeah you were right the first time

[gritty widget]

i thought that

but i thought that wasn't homogenous because of the coefficent

but

this is degree 1

so the coefficent doesn't matter

[orchid forge]

When would the coefficient matter?

[gritty widget]

for this

[gritty widget]

this

2x^3+y^3 would not be homogenous

[orchid forge]

Surely $2(kx)^3 + (ky)^3 = k^3(2x^3 + y^3)$

[gentle osprey]

Kogasa

[gritty widget]

oh

yeah nevermind

[violet sonnet]

hello! can someone explain how we can simply make the half open intervals on the RHS to be open on both sides?

for example (1, 5] is contained in (0, 2]U(2, 6] but then [1, 5] is not contained in (0, 2)U(2, 6) , because 2 is in [1, 5] but not in (0, 2)U(2, 6)

[orchid forge]

First note they're shrinking the interval on the lhs

[violet sonnet]

I see that

[wanton marsh]

yeah that looks false

[violet sonnet]

but it says for any epsilon

oh false...

it's a proof in my lecture notes

[wanton marsh]

what is it trying to prove ?

[fair idol]

What do people mean when it's said that algebraic geometry is a generalization of linear algebra? I understand that polynomials of order 1 are linear equations so that makes sense. But what sorts of things can be generalized from linear algebra? Can we have some sort of generalized svd?

[gritty widget]

how do i make a line between two points in projective space

in affine space a line between P and Q is given by

s(t)=tP+(1-t)Q

but does this make sense in projective space?

[violet sonnet]

@orchid forge hi, i was reading that haha, why did you delete?

[orchid forge]

It was wrong

[violet sonnet]

oh lol

[orchid forge]

The intervals are disjoint

[violet sonnet]

yeah

[orchid forge]

So there's only really one way for (a,b] to fit inside a union of (c_i, d_i]

That gives the result directly

[violet sonnet]

hmmm

reading through the proof and i think it's pointless that she removed the endpoints on the RHS anyway

i genuinely don't think it's necessary or correct

[gritty widget]

what i said does not work for projective space

[gritty widget]

If i have P=(x:y:z:w) and Q=(x':y':z':w')

is a line between them just

tx+(1-t)x'

and so on in each coordinate

this seems very wrong

[empty grove]

Why does it seem wrong? feels like it should be fine as long as you ensure that the representatives are chosen such that the line doesn't pass through the origin

[gritty widget]

i don't know im just feeling sus

I am trying to show that every line on $Z(xw-yz)\subset RP^3$ is of the form $\phi(x,RP^1)$ or $\phi(RP^1,x)$

[gentle osprey]

lime_soup

[gritty widget]

where phi is the segre emebding of RP^1 times RP^1 into RP^3

Z(xw-yz) this is the image of the segre embedding

so im trying to show thatif a line between two arbitary points on this X

it must be constant in one term

If I intersect two zariski closed sets

Is the intersection just

The zero sets of all the products

Z(f,g) cap Z(h,k)=Z(fh,fk,gh,gk)

[gritty widget]

In the projective line

we always have if (a:b) in CP^1 then, a=1/b

if b is not zero

as in

(3:7) would not be in CP^1

[plain raven]

i don't think that's how you're supposed to interpret that notation

(a:b) represents the unique line which passes through both the origin and the point (a,b)

thus

(a:b)=(2a:2b)=(-a:-b) in general

[gritty widget]

shit so we can have

(3,7) in CP^1

[orchid forge]

yes

[gritty widget]

oh yes

its just we can then scale

and there are the cooridnates

(lamda:1)

and (1:mu)

with this relation

[plain raven]

yep

[gritty widget]

damn

i was trying to find the intersection points of

some things

and used this

[gritty widget]

can someone help me with this

let O_X be some sheaf

U an open set

U_i an open cover of U

suppose f, has an inverse in each O_X(U_i)

and that these inverses agree on overlaps

does it follow directly from being a sheaf

that this is an inverse for f on all of U

i know that this is enough to glue these maps together

[orchid forge]

is O_X a sheaf of functions?

[gritty widget]

yes

[orchid forge]

and f is a section of O_X over U_i?

[gritty widget]

over U yes

[orchid forge]

oh

no, consider f : R --> R, f(x) = x^2 and let the cover be the nonnegative and nonpositive reals. f restricted to either component has an inverse, that agrees on the overlap

here I guess X = R and O_X is the sheaf of smooth functions U --> R

[gritty widget]

i dont get this example

f(x)=x^2 does have an inverse on the union of the non positive and non negatives

oh sorry i thougth these mean not zero

okay well what im really trying to see is

let f be non vanishing on U

then f has an inverse

on affine open sets this is easy because you can just define the inverse

but on non affine sets

you hvae to glue the thing together

[abstract pagoda]

what are examples of immersions that

afaik immersions are functions between manifolds with injective differentials and something else

[tight agate]

the wikipedia page has examples

[orchid forge]

I always recommend people check wikipedia for examples of immersions that

[gritty widget]

$$\bR^n \to \bR^{n+k}, x \mapsto (x, 0)$$

[gentle osprey]

TTerra

[gritty widget]

that's an immersion

every immersion is also locally of this form

[orchid forge]

but is it an immersion that

[gritty widget]

true, it might not be that

[plain raven]

lmao

[bleak path]

Can I ask for some opinions on my homework?

Question

My attempt at a solution

Relevant notation:

$f: \mathbb{R}^n \to S^n,
g: D^2 \times S^1 \to ST,
D^2 = {(u, v) \in \mathbb{R}^2: ||(u, v)|| \leq 1 } \subset \mathbb{R}^2,
ST = {(x, y, z) \in \mathbb{R}^3: (2 - ||(x, y)||)^2 + z^2 \leq 1} \subset \mathbb{R}^3,
B^2 = {(u, v) \in \mathbb{R}^2: ||(u, v)|| < } \subset \mathbb{R}^2$

[gentle osprey]

wOne

[bleak path]

Sorry about the formatting, not quite sure how to get a newline in the texit bot

[bleak path]

My thoughts are that it's fairly wordy and I would like to know if there are parts I can rewrite as equations/expressions instead of just sentences, and also I feel my proof is a little hand-wavy at times but I can't think of a way to make it more concrete

[gentle osprey]

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

[gritty widget]

specifically my situation is

the base spaces are the same

oops

let me do this all again

Let $$F\rightarrow E\rightarrow B$$
and
$$F'\rightarrow E'\rightarrow B$$
be two fibre bundles.

The $E_2$ page of the spectral sequence is given by $$E^{p,q}_r(E)=H^p(B)\otimes H^q(F)$$
and
$$E^{p,q}_r(E')=H^p(B)\otimes H^q(F')$$.

There is an inclusion of $F' \rightarrow F$. Moreover this inclusion induces isomorphisms $H^q(F)\rightarrow H^q(F')$ for $q\leq n$.

I want to know that the "induced map"
$$E^{p,q}_2(E)\rightarrow E^{p,q}_2(E')$$ is an isomorphism for $q\leq n$

[gentle osprey]

lime_soup

[gritty widget]

Its clear that there is an isomorphism

since there is an isomorphism on the level of homotopy groups

but i don't see what the induced map should be

[uncut surge]

If you knew something like the cohomology of E being trivial in some degree, then this could be hit with the borel transgression theorem or so

And if the cohomology of F is generated in odd degree @gritty widget

see this https://encyclopediaofmath.org/wiki/Transgression

or dis weaker version of the theorem, conveniently posted a couple days ago https://mathoverflow.net/a/406335/126256

[gritty widget]

I think though

I am just missing something simple about

Induced maps

Like

Are we defining a map

E’ to E

If I knew the induced map was

Identity* tensor inclusion*

Where the inclusion is in the inclusion of the fibre

[abstract pagoda]

for manifolds

locally homeomorphic condition implies for charts f:U->Rn f(U)=Rn ?

or is definition of chart wrong

[kind cedar]

two definitions are confusing me.

1. a vector field is a section on a bundle.

2. a vector field maps functions to functions, but a section maps points to vectors. what is the correct map?

a) $X:M\longrightarrow TM$
b) $X:C^{\infty}M \longrightarrow C^{\infty}M$

[gentle osprey]

Necrowizard

[kind cedar]

So I can use either definition for reasoning? I was super confused about them

In what book can I find it?

I can see how, I just cant write it formally

[finite heath]

the server logo is a torus 🙂

learned about it briefly in topology class today

basis element of R^3 right

[kind cedar]

The torus as a basis? I'm curious about it, never heard before

[gritty widget]

what

[pearl holly]

coffis cup.

[abstract pagoda]

bruh

ther

[orchid forge]

Topolg

[coarse kestrel]

@vocal anchor I think it has 2 reasons: first you want to define local homeomorphism so that the space actually looks like Y at least locally, so otherwise the inclusion R to R^2 will be a local homeomorphism but the two topologies are pretty different.
Second you probably want composition of local homeomorphisms to be local homeomorphisms, which in proving it will need f(U) to be open

[vocal anchor]

I see, the example with R -> R^2 is good 👍 makes a lot of sense (and of course a useful property of the correct definition is that local homeos are open)

Thanks

[finite heath]

Teacher said it was like balls in R2 and open ball is basis element of R2 so I just assumed!

[wise ruin]

Intuitively, is first countable-ness kinda just a local version of second countable-ness?

[wise ruin]

Or is it moreso that second countability is like a uniform version of first countability

[abstract pagoda]

they spoking about basis if topology

T2 jus make with product topology S1 x S1

[abstract pagoda]

this right her

thas better intuition ive heard

[glossy pine]

so whats the goal of topology? so far it's kinda making sense but idk wtf im learning about

I thought it was like higher level formalized geometry but I ain't seein any shapes

[orchid forge]

it's like wobbly geometry

the less rigid structure of shapes and spaces, the stuff you need to talk about continuity

[glossy pine]

hmm

will it get more interesting as I progress?

bc rn it's shit like product topology

[orchid forge]

if you progress enough, yes

[glossy pine]

so I just gotta survive for a while longer lol

[plain raven]

Yeah gmod maybe check out a book which is dedicated specifically to that stuff as opposed to the technical foundations which make it all work?

the book "introduction to topological manifolds" by Lee might be good. taking a brief look at armstrong's "Basic topology" it stresses geometric stuff.

also i've heard Janich's "Topology" is excellent

A lot of books on topology are not about geometry per se, they're dedicated to developing a theory of continuity and continuous functions which is rich enough to prove results in a wide variety of situations where you would want to reason about things varying "continuously", as an assumption or a hypothesis, and to talk about the "convergence" of a sequence to a point, which is closely related (recall that a continuous function f : R-> R is exactly one which sends convergent sequences to convergent sequences)

For example, the power series \sum x^n/n! "converges" to e^x. In what sense? There are various ways to formalize this but a convenient one involves the "topology of uniform convergence."

so this is useful in analysis.

[glossy pine]

interesting

thanks for the advice

yeah bc rn this is very abstract and im not understanding the point

so I'll probably check out those books

[plain raven]

what book are you reading, munkres?

[glossy pine]

im watching a lecture series

https://youtube.com/playlist?list=PLLq_gUfXAnkl8bjQh-hGQ9u24xZP9x0dx

@plain raven

[plain raven]

ok so yeah it follows munkres.

if you want to get an idea of why the lectures are so general and seem to be so abstracted from geometry, i would check out an analysis book

[glossy pine]

alright

[plain raven]

it will show you applications of this kind of topological reasoning using continuity, continuous functions, convergence to prove things about analysis of R^n and the properties of continuous functions.

[glossy pine]

alright will do, thank you

[orchid forge]

clerk you will never believe this

they changed my algebra course mid-semester from commutative ring theory to arithmetic geometry and now they're talking about Cech cohomology of quasi-coherent sheaves, but they rescheduled it so i have to drop it

[plain raven]

they don't want you to prove Grothenieck Vanishing...

https://tenor.com/view/vibes-vibing-we-tipsy-drunk-gif-17644302

[gentle osprey]

diligentClerk

[plain raven]

https://www.youtube.com/watch?v=y6wouVsB598

[orchid forge]

Given an open subset V of Y, how can you use the restrictions to rewrite the preimage of V?

Yes, ideally you would want to show that f^{-1}(V) is open

And you know f_a^{-1}(V) is, and the same for f_b

So how does the first set relate to the others

[bleak helm]

Try to use this

Nonsense

I meant

Show that f|A^(-1)(V) union f|B^(-1)(V) = f^(-1)(V)

[vocal anchor]

well, since f|A and f|B are continuous, what Lunsasong wrote implies that f^-1 (V) is a union of open sets

I also have a question. Let $p\colon E \to B$ be an open surjection, and let $q\colon (A, \alpha) \to (K, \kappa)$ in $\mathrm{Top}/B$ be a quotient map. Is it true that $id_E \times_B q$ is also a quotient map? I don't see it 😦

[gentle osprey]

expectTheUnexpected

[pearl holly]

daim I watched this series too while I was reading Munkres. Things get a lot more fun when you start doing AT instead of point set in my opinion

[fading vale]

You need local compactness on E

[cedar pebble]

ah so it has gone from globally commutative algebra to locally commutative algebra

[vocal anchor]

Hm. So if E is for example compact, how would that help? Or do you know where I can look it up?

[fading vale]

Do you know about the exponential law

Hom(A x B, C) cong Hom(A, Hom(B, C)) when B is locally compact

U might also need hausdorff on 2 of these spaces idr

[vocal anchor]

tensor hom adjunction in Top

[fading vale]

Yes exactl

Y

[vocal anchor]

only know algebra though 😬

[fading vale]

Im pretty sure if u play around with this u can get what u want

[vocal anchor]

Well, I don't actually know that adjunction, since I don't know any topology tbh. I'll think

[fading vale]

Im not sure how non trivial this is tbh

Cant work it out myself atm

[vocal anchor]

no worries, thanks

[vast estuary]

Just to make sure I'm not missing something

[gentle osprey]

Hausdorff

[hollow harbor]

What if gamma just moves around the unit circle in the xy plane?
Where does gamma'' point? Where does gamma point?

[vast estuary]

Well, gamma is a point on the circle, so it points away from the origin, towards the perimeter of the circle

gamma' being the tangent vector, points tangential to the circle

gamma'' is the rate of change of the tangent vector, so, umm.... it points inward does it?

looks like gamma and gamma'' are antiparallel in this case @hollow harbor

[hollow harbor]

yep!

[finite heath]

So, the subspace topology on Z (integers) from R (all reals) from the standard topology on R

any open interval (a, b) in R intersected with Z is just a single integer or a set of multiple integers

and those sets are open

so because this subspace topology on Z contains all open singletons, does that mean it is just the discrete topology on R?

Isn't that a property of the discrete topology on R?

[pearl holly]

yeah so this means that the subspace topology on Z from R is discrete

I don't understand the "on R" part

[finite heath]

i meant on Z sir

yeah!

[pearl holly]

ayy lessgo

[orchid forge]

first try to prove that it is

and if you can't, then try to prove that it is not

[cedar pebble]

it's not; to see this note that the tangent space at (0,0) is too large

[shy moss]

in this $\Delta-$complex structure of the torus, is it right that $\partial(U)=c-a+b$ and $\partial(L)=c+a-b$ ?

[gentle osprey]

Or x1

[plain raven]

i think $\partial(U) = c -a-b$ and $\partial (L) = a+b-c$

[gentle osprey]

diligentClerk

[plain raven]

$\partial(U)$ and $\partial(L)$ should both be "closed loops" in the sense that like, following all the arrows head to tail should take you back where you started. no two arrows should leave from the same point or enter the same point

[gentle osprey]

diligentClerk

[plain raven]

here the sign conventions dictate the direction of the arrow. positive = the drawn direction of the arrow, negative = opposite of the drawn direction

[shy moss]

Thanks

[finite heath]

So, the standard topology on Y is technically the subspace topology on Y right?

[rancid umbra]

there is no technicality

that’s exactly what definition 3.2 is saying

[coarse kestrel]

The lemma 5.3 here in https://math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Ford.pdf says that if I is a radical ideal in a (commutative) domain R then for every x not in I we have a prime ideal P⊆R such that I⊆P and x∉P. I feel like this looks really similar to the condition for a Hausdorff/regular space. Is there a connection here?

(@coral pivot and I were reading this)

[tough imp]

This is just because the radical of an idea is the intersection of all primes containing it

I really don’t think there’s any condition with Hausdorffness

[coarse kestrel]

It just feels really similar to me

Like sort of you have a "closed set" I and a point x not in the set then you have something containing it and doesn't contain x

[tough imp]

Well the thing is that I only really defines a closed set on Spec A

The only real natural topology on A is an I-adic one

And then the Hausdorffness of that is separate from what’s going on here

If you just say that it’s something about there exists some class of sets which can separate I and elements not in I

I guess you get that by the primes

But if that’s strictly related to Hausdorff I’d say no

[ivory dragon]

radical of an idea is the intersection of all primes containing it
someone tell political scientists

[gentle osprey]

Hausdorff

[vast estuary]

How should I go about this?

I believe it is enough to show that |U| is open in R. To do this, consider a in |U|. Then a = |z_0| for some z_0 in U. As U is open, there exists r > 0 with B(z_0, r) contained in U.

What now?

[tough imp]

Just show that any point is surrounded by an open ball

So let r in |U| then r = |z| for some z in U

But then there’s some epsilon ball around z contained in U

So that (r-epsilon,r+epsilon) < |U|

You know for any point x in (r - epsilon, r + epsilon) you have a point w in B(z,r) with |w| = x

This is kinda the definition of a ball around z

[rancid umbra]

|U| is a subset of R, is it not?

formulation of the question just seems kinda odd to me, since you just need to show that |U| is open in R. not sure whats going on with the subspace business

an aside, this shows that the map || . || : R^n --> R is an open map, where || . || is a norm on R^n

[honest terrace]

No the subspace business is important here

Take for example U to be the open ball of radius 1 centered at 0 in C

|U| is [0,1[

[tough imp]

EWWWWW

[honest terrace]

it is open in R^{>= 0}

[tough imp]

[ notation

[honest terrace]

but not in R

not my fault I am french 😔

[tough imp]

Soit G un groupe finit

[rancid umbra]

is R^{>=0} the set [0, infty)

[tough imp]

Or something like that

[rancid umbra]

because if so, wtf

[honest terrace]

yes

[tough imp]

Yes

[rancid umbra]

omg

thats such bad notation

[tough imp]

??)

Wut

It’s very clear I think

[rancid umbra]

at least make it a subscript

[tough imp]

¯_(ツ)_/¯

[rancid umbra]

the whole time i was interpreting it as "we're working in R^n with some n >= 0"

[tough imp]

[honest terrace]

"fini"*, but yes

[rancid umbra]

it was very confusing

[honest terrace]

Happens

But yeah, they're precisely asking to prove that the map |.|: C -> [0,+infty) is open

[tough imp]

I read a paper “Sur la Structure des Algèbres locales Noethériennes Formellement Lisses”

I think that says “on the structure of local Noetherian formally smooth algebras”

Also idk why it chose C tho

This is true for any R^n

[honest terrace]

That sounds like a correct translation

[tough imp]

C isn’t R^2 🤡

[honest terrace]

Well yeah but I mean, exercises stating stuff in not-full generality even though when full generality isn't harder to prove isn't that uncommon

[tough imp]

It literally is like

The same argument

I think verbatim

[honest terrace]

Yes

[tough imp]

Hahahaha

I think you could copy paste it

Hahaha

Very silly

[honest terrace]

Yeah I think you could literally just s/C/R^n

[rancid umbra]

is that a latex command?

[honest terrace]

No, a linux one

[rancid umbra]

wait yea

[tough imp]

[honest terrace]

https://www.gnu.org/software/sed/manual/html_node/The-_0022s_0022-Command.html

[rancid umbra]

dang. i miss topology proofs. multivariable calc kinda blows

[tough imp]

Wrong place

Oopsie

[rancid umbra]

no no

that was the right place

[vast estuary]

Umm how do you get this

[tough imp]

Cuz in B(e,z)

For any x between 0 and e

What’s the easiest way to explain this lol

If |z| = n

Then there’s a point w in B(e,z) which has |w| = n + x

Or n - x

Like

[rancid umbra]

draw a picture lol

[tough imp]

Yeah

Just grab a line from the origin to z

Then any point which is k away from z in the direction of that line

has norm |z| + k

Like my point is for any x in (|z| - e, |z| + e)

There’s a point w in B(e,z) which satisfies |w| = x

I can give you an argument for it if you need

But I think if you just think about it

You’ll see why it’s true

Rotate the plane so z is actually the real number |z|

Then B(e,z) contains the subset (z - e, z + e) of the real line viewed as a subset of C

[rancid umbra]

just a note: need to intersect this with [0, infty)

[tough imp]

Oh sure

[rancid umbra]

oh wait Chmonkey thats kind of a cool visual

[tough imp]

You’re kinda sweeping under the rug that rotating like that is an isometry but

I mean you can just like compute this using a matrix or whatever you want if you need to generalize this to R^n

[gentle osprey]

Hausdorff

[vast estuary]

That completes it right

[rancid umbra]

yea. i think you may be able to do away with the cases if you just scale epsilon to be less than r

[tough imp]

This isn’t strictly true

Like B(r,epsilon) isn’t equal to (r-epsilon, r + epsilon)

Since the latter is a subset of R but the former is a subset of C

Oh!

Nvm

r not z

[wise sigil]

the basis is open epsilon-balls

so open sets are arbitrary unions of those balls

yeah you need to bound the metric first or else sup{d(x_a , y_a)} might not be defined

[ocean sleet]

This is what I did. Is this the correct approach?

[plain raven]

Multivariable calculus is great! All the computations where they ask you to integrate a vector field along a curve, or integrate the divergence of a vector field across a region, can get annoying, but it's fun to use stokes' theorem or green's theorem to make it all vanish to zero

I personally feel I have to enthusiastically defend multivariable calc because I think de Rham cohomology is one of the cohomology theories where we have the opportunity to gain the most geometric intuition by working through concrete examples, and that basically boils down to doing lots of calc 3 variations. If I were to try and explain sheaf cohomology to someone, as much as possible I'd try to draw on examples from calc 3.

Plus, Maxwell's equations are so beautiful, and how can you appreciate them if you don't have a good understanding of the Divergence theorem, stokes' theorem, etc!

[tough imp]

Physics

Respect for Clerk, lost

[empty grove]

unrelated but clerk can you explain sheaf cohomology to me

[plain raven]

so sad

[empty grove]

damn

[plain raven]

chmonkey i have a bachelor's degree in electrical engineering

[empty grove]

what

[plain raven]

i designed and built the electrical load for a wind turbine as my senior project

[empty grove]

😵‍💫

[plain raven]

yeah i wasn't a math major in undergrad

[empty grove]

bruh

[plain raven]

i mean i did get a double major in math but i mostly just managed to satisfy the bare minimum requirements by taking math classes adjacent to engineering - statistics, analysis that was relevant to signal processing, etc.

I also took some classes in logic because i thought more knowledge of formal systems would be useful in computer science. i can safely say that model theory and computability theory are not that useful in computer science.

[rancid umbra]

we haven’t even gotten to this part of the course yet. we’re following munkres analysis on manifolds. idk, it’s just felt kind of dull atm. nothing really exciting. i haven’t seen really why inverse and implicit function theorems are cool yet and we’re still developing riemann integration in R^n. it’s just been….eh.

i don’t know what cohomology is but i think it gets covered in the end of the course a little bit

also, in the E & M course i took, maxwells equations just looked like some random symbols on a page. i have no idea what the divergence curl and stokes theorem say yet either.

you’re giving me hope that the course will get more interesting tho

[tough imp]

Which version?

Derived functor?

If so that’s just all formal from homological algebra

[empty grove]

I was thinking something related to calc3 in some way idk why

idk what calc3 is but yeah

[tough imp]

Multi var calc

[plain raven]

inverse and implicit function theorems are pretty basic and have lots of applications so don't sweat it i'm sure you'll see applications soon enough. i think like, to me the most obvious application of the inverse function theorem is that if f : M -> N is a map of n-dimensional manifolds, and its differential is an isomorphism, then f is a local diffeomorphism onto its image

[tough imp]

So the way it relates is that like

All the coho@ology theories under the sun agree on paracompact Hausdorff

So De Rham = Sheaf okay GG

[empty grove]

coho@ology

[tough imp]

Swag

[plain raven]

yeah i mean moldilocks my favorite example, and the one in Bott-Tu, is the following

we'll consider the space R^2 - {(0,0)}

idk if you've seen this so stop me if not

[empty grove]

I haven't seen sheaf cohom yet in ag course btw

[plain raven]

you're doing cech cohomology tho right

[empty grove]

I have seen de rham on open subsets of Rn

Ye kinda lol I know the definition but he hasn't done anything with that yet

[uncut surge]

sheaf = cech but for very smol covers

ayyyyy

[plain raven]

yeah lol

[tough imp]

For good covers

They are the same

[uncut surge]

If you're working with the constant sheaf!

[tough imp]

Good = actually so nice you’ll never have it

[plain raven]

lmfao

[uncut surge]

replace "Leray" with "good" and it's universally tru

[plain raven]

ok haha

actually this is one of the few things i totally blackboxed when studying this

i have no idea what a good cover is

[tough imp]

I think you get it for Noetherian separated schemes because of the fact affine \cap affine is affine

[plain raven]

i mean bott and tu blackbox it as well

they punt it to an exercise in spivak's diff geo

[uncut surge]

oh yeah i guess that is the context that people usually think about sheaves in

[tough imp]

I sort of reinvented good covers to compute a cohomology of a constant sheaf

Lmao

I looked at it later and was like “oh huh”

[plain raven]

wait hold on i want to explain this example for moldilocks

moldi i want to give some intuition for cech cohomology.

i gave like some gigabrained explanation previously but this is much more concrete

so say you have like, R^2 - {(0,0)}.

and i give you a smooth vector field on this space.

the basic question is, does it have a potential function?

well we know that there's an easy way to rule this out in the negative case

[empty grove]

oh ye isn't this 2nd de rham coho

[plain raven]

it's necessary that the 'curl' be zero. i say curl but like, whatever the 2d analogue of curl is that comes up in greens theorem

[empty grove]

or 1st

[plain raven]

yeah i'm talking about the first de Rham cohomology group here but like

we'll discuss it from a Cech perspective

that's the only difference

so again we start by throwing out all but the curl free vector fields.

but instead of in the de Rham approach where we just mod out by the conservative ones

we'll focus instead on how the local solutions can be woven together into a global solution

Because at every point there's a nice open nbhd, say a disk or any other simply connected region, where there is a potential function, and this potential function is unique up to being shifted up and down by a scalar

so what we can do is just cover the whole space by simply connected open subsets and choose, on each open subset, a potential function for the vector field.

so this is like an element of C^0(U, F) where U is a cover by (simply connected) open subsets and F is the sheaf of smooth scalar functions.

[empty grove]

assuming stokes for simply connected?

[plain raven]

yes.

For the sake of a reasonable analysis let's say we choose some nice concrete open cover, say that like, we have a finite cover consisting of arc segments of the form $\theta_1 <\theta <\theta_2$

[gentle osprey]

diligentClerk

[plain raven]

In general our functions might not all agree pairwise and so might not glue together into a global solution

but you can always tweak them up or down by an additive constant without there being any deep changes

so the real question is whether you can shift these all up and down by some additive constant so that they agree pairwise.

and it's not hard to see that there's an easy necessary and sufficient condition to check. Since all the scalar functions you have are solutions to this problem, the differences f_i - f_j on the overlaps are all locally constant, let's just say they're real numbers, c_{ij}.

If you arrange these segments of arc in a kind of 'cyclic ordering' and take the sum of the c_ij around this ordering (here i'm speaking informally here just because i'm impatient about the details) then you can see this takes the form of like
(f_i - f_j) + (f_j - f_k) + ...

and so if you replace f_j with f_j + a, then this decreases c_ij by a, but it increases c_jk by a, and so it cancels out

therefore you can assign this numerical quantity to the open cover which is invariant under shifting any of the arguments up and down

in particular if the family of {f_i} are gluable then the c_ij should all be zero (because the functions agree on their overlaps) and so the sum of the c_ij is zero

So this gives a necessary and sufficient condition for the 0-cochain {f_i} to be "translated" into an actual globally defined potential function (i.e. a 0-cochain which is a cocycle and thus comes from a global section of the sheaf of smooth functions)

[empty grove]

but if we take a cyclic ordering on the indices then we aren't testing all c_ij's right?

or are we doing this for every ordering?

[plain raven]

no it wouldn't be all c_ijs unless maybe the cover was chosen well so that c_ij was zero or was not-meaningful on like, i,k with i < j < k because U_i doesn't meet U_k. however that said like, we're essentially trying to measure the dimension of a certain cech cohomology group of R^2 - {(0,0)} with coefficients in a certain sheaf and through the right analysis i think you could prove here that it suffices to restrict to a certain choice of some of the c_ij's (in what is essentially a 1-cocycle in the sheaf of locally constant functions) because the other c_ij's just don't really contribute anything because they're "dependent"

i'm speaking loosely but there's some kind of linear algebra you should be able to do to reduce to the case of a certain subset of the c_ijs

but to be more formal about what we're doing, what we have is the short exact sequence of sheaves

$0\to \mathcal{F}\to\mathcal{G}\to\mathcal{H}\to 0$

[gentle osprey]

diligentClerk

[plain raven]

where $\mathcal{F}$ is the sheaf of locally constant functions, $\mathcal{G}$ is the sheaf of smooth functions, and $\mathcal{H}$ is the sheaf of curl free vector fields.

[gentle osprey]

diligentClerk

[plain raven]

and we're basically just playing around with the sequence of Cech cochain complexes which is associated to this sequence of sheaves by applying the Cech cochain complex functor $C^{\bullet}(\mathfrak{U},-)$

[gentle osprey]

diligentClerk

[plain raven]

this sequence is not exact in general but it is always right exact and given any global section of $\mathcal{H}$ you can always choose a sufficiently fine open cover $\mathfrak{U}$ so that the image ${f_i}$ $f$ in $C^0(\mathfrak{U},\mathcal{H})$ has a lift along the induced map $C^0(\mathfrak{U},\mathcal{G})\to C^0(\mathfrak{U},\mathcal{F})$,

[gentle osprey]

diligentClerk

[plain raven]

i mean for us this was easy as we can just ask that the opens of $\mathfrak{U}$ were simply connected but this holds in general for a surjective map of sheaves on a paracompact hausdorff space, by a partition of unity argument

[gentle osprey]

diligentClerk

[plain raven]

not a partition of unity argument. sorry. i don't think that's what i wanted to say.

[empty grove]

I'll have to look at the definition of C(U, F) because all it is to me right now is a formal object

[plain raven]

paracompactness doesn't come up here necessarily

Ok. well your homework is to actually just think about the short exact sequence of sheaves $0\to \mathcal{F}\to\mathcal{G}\to \mathcal{H}\to 0$ in this specific case, and the sequence of complexes $0\to C^{\bullet}(\mathfrak{U},\mathcal{F}) \to C^{\bullet}(\mathfrak{U},\mathcal{G})\to C^{\bullet}(\mathfrak{U},\mathcal{H})$ for $\mathfrak{U}$ simply connected

[gentle osprey]

diligentClerk

[plain raven]

Try and translate the informal geometric "play" we were doing with like, locally solving the problem of finding a potential function and then shifting them up and down degreewise by a locally constant function, into like, computations in this sequence of chain complexes

or not computations but a diagram chase

[empty grove]

Will try it today

[plain raven]

algebraically we're proving something here. i don't know what off the top of my head but i think all this play adds up to a loose construction of a boundary map $\partial : \Gamma(X,\mathcal{H})\to H^1(\mathfrak{U}, \mathcal{F})$ and an argument that $\delta$ carries a vector field $F$ to zero iff it lies in the image of the gradient map $\Delta : \Gamma(X,\mathcal{G})\to\Gamma(X,\mathcal{H})$. And the handwaving with the cyclic sum $\sum c_{ij}$ is an informal argument that
$H^1(X,\mathcal{F})$ is one-dimensional.

[gentle osprey]

diligentClerk

[plain raven]

ok i'm done there

[gritty widget]

what is your "area"

you seem to have a pretty wide knowledge

[empty grove]

how everything relates to sheaf theory 😌

[uncut surge]

sheaf theory is breddi stronk

[gritty widget]

Let $G$ be a group, $R(G)$ the representation ring. How does a congucay class $\gamma$ in $G$ give a prime ideal $\mathfrak{p}$ in $R(G)$

[gentle osprey]

lime_soup

[wise sigil]

https://math.stackexchange.com/questions/4281980/general-geodesic-curvature-formula

Sorry for the long question

Can anyone help me with this exercise? I still havent been able to find the mistake

[gritty widget]

Can anyone help me with this step in a proof

we have that to prove the theorem for X

its enough to prove the theorem for any relatively compact open set U

but then the proof goes

by the exact sequence of the pair $(\overline{U},\overline{U}-U)$ its enough to prove the theorem for any compact space U

[gentle osprey]

lime_soup

[gritty widget]

the theorem being some property on cohomlogy groups of X

[uncut surge]

I guess if you knew something about the relative homology of the pair when U is relatively compact, you could say things

What theorem are you trying to prove?

[gritty widget]

if you is relatively compact

yikes

If $U$ is relatively compact, $\overline{U}$ is compact.
If $K^_G(\overline{U})$ and $K^_G(\overline{U},\overline{U}-U)$ were zero then the exact sequence of the pair would give use the result that $K^*_G(U)$ was zero

[gentle osprey]

lime_soup

[gritty widget]

But i don't think it follows from $K^_G(\overline{U})=0$ that $K^_G(\overline{U},\overline{U}-U)=0$

[gentle osprey]

lime_soup

[gentle osprey]

beeswax

[cursive flume]

how does the non-degeneracy of a 2-form(be it symplectic,or the metric, riemannian-pseudo riemannian) imply the fact that the musical map is a C^{infty} isomorphism between TM and T*M?

[orchid forge]

nondegeneracy translates into injectivity of the corresponding map

[cursive flume]

could you please elaborate? I don't see how injectivity follows from non-degeneracy

also:why is the map always surjective?

[orchid forge]

If you feed a 2-form a vector field (a section of TM), it becomes a 1-form (a section of T*M). This defines a map from sections of TM to sections of T*M. Nondegeneracy says that if you feed it a nonzero vector field, the corresponding 1-form is nonzero, so this map is injective

[cursive flume]

right,makes sense

why is it surjective,though?

[orchid forge]

well, at the level of (pointwise) tangent spaces, it is an injective linear map between vector spaces of the same dimension, so it is surjective

[cursive flume]

this would follow from the rank nullity theorem, right?

[orchid forge]

yes, but it would take an argument to show that the map I mentioned is surjective

[cursive flume]

this would work on the local level,but how could I say anything globally?

so far I could show that I have an isomorphism of tangent spaces TpM T*pM

using the argument you showed

[orchid forge]

you would need to pass to a local trivialization

[cursive flume]

ugh 😅 let me google it up

[orchid forge]

you might also just want to construct the inverse map explicitly

[cursive flume]

I mean I could construct it, but not sure where I'd use the non-degeneracy in the construciton

I've been given the construction

[orchid forge]

it should break at some point if you don't assume nondegeneracy

however, you should also prove in general that a bundle map (over a common space M) which induces a fiberwise isomorphism is an isomorphism

[cursive flume]

the fiberwise isomorphism is done with the argument you said

i.e. the fibers are the vector spaces

[cursive flume]

and then this would be the only thing to prove left,righT?

[orchid forge]

yes

would recommend it because it's just a useful check to make sure we're thinking about vector bundles correctly, but also you can scrutinize the construction you were given to find where it breaks

[cursive flume]

is this not a def.?

[orchid forge]

oh

well I suppose it is for you

I'm pretty sure I used an equivalent characterization and had to prove this

[cursive flume]

I'm not sure,i have a physics background, that's why asking

[orchid forge]

If that's how you defined a vector bundle isomorphism then there's nothing to prove

(except that the map is a homeomorphism)

[cursive flume]

can I do that by assuming TM=UxR^n and T*M=UxR^m?

and since the tangent spaces are finite dim m=n

and R^n is homeomorphic to R^n

[orchid forge]

yes, that's what I mean by passing to a local trivialization. for each point x in M, there is a neighborhood U so that TU is homeomorphic to U x R^n

you can't use "R^n is homeomorphic to R^n" though, you want to show that the particular map we defined is a linear isomorphism

[cursive flume]

how would that imply anything about the topological structure?

[orchid forge]

under the local trivialization, it is a continuous family of linear maps F(x) : R^n --> R^n, which is injective (hence bijective) for each x

well linear maps R^n --> R^n are continuous

so a linear isomorphism is a homeomorphism

so the map TM --> T*M we defined restricts to a map U x R^n --> U x R^n which is the identity in the first component, and we want to know that it is a homeomorphism in the second component

[cursive flume]

right,only thing i don't see yet is why is it linear

butt I think that follows from the forms themselves being linear

from the def. of map

righ?

[orchid forge]

the map is (x, v) --> (x, g_x(v)), which is not linear itself but g_x : R^n --> R^n is

a continuous family of linear isomorphisms

given a local trivialization of T*M, you also have one for T*M (x) T*M, so you can use the same coordinates to write out what the map TM --> T*M given by a 2-form is

[cursive flume]

i don't see how this is related to the proof we're trying to do

[orchid forge]

i mean to see that the map is linear fiberwise

[cursive flume]

right,if that's true,i's a homemorphism

cause it's bijective fiberwise

[orchid forge]

you need it to be a continuous family of linear maps, not just a family of linear maps, to say it's a homeomorphism

within a local trivialization,we may write a given 2-form $\omega$ as $\sum a_{ij} e^i \otimes e^j$ where $e^i$ and $e^j$ are the standard basis elements for $\mathbb{R}^n$ and $a_{ij}$ is a smooth map $U \to \mathbb R$. similarly, a vector field $v = \sum v^i e_i$, where $v^i : U \to \mathbb{R}$ is smooth. then we can explicitly write out what the map $v \mapsto \omega(v, \cdot) \in T^*M$ is

[gentle osprey]

Kogasa

[orchid forge]

it'll be some $\sum b_i e^i$, and just have to check that the coefficients $b_i$ are smooth

[gentle osprey]

Kogasa

[orchid forge]

that would mean we have a continuous map $TM \to T^*M$. it is invertible iff it is invertible fiberwise, and using the above we can write the inverse map and check that it is also smooth (by checking its coefficients in the standard basis are smooth)

[gentle osprey]

Kogasa

[cursive flume]

i'm sorry this might be dumb question, but why could the coefficienst be no tsmooht?

every 1-form can be expressed as linear comb of standard basis with smooth coefficients

righT?

[orchid forge]

yes, but this is what we're trying to show

[cursive flume]

okay,I see. I never proved this,just accepted as a matter of fact

[orchid forge]

we explicitly write down what our map $v \mapsto \omega(v, \cdot)$ is in terms of the standard basis $e^i$, and show that we actually do get a 1-form

[gentle osprey]

Kogasa

[orchid forge]

(by checking that the coefficients are smooth)

[gentle osprey]

ProphetX

[cursive flume]

and now I expand both omega and v in a basis, right?

in the standard basis of R^n

sorry,this is not right

these are just the components we use in physics

[orchid forge]

it should basically be matrix multiplication

but the indices get confusing

[orchid forge]

so in einstein notation, if $\omega = \omega_{ij} e^i e^j$ and $v = v^j e_j$, then $\omega v = \omega_{ij} v^j e^i$, the coefficients $\omega_{ij} v^j$ are sums of products of smooth functions, hence smooth

[gentle osprey]

Kogasa

[cursive flume]

thanks a lot for your patience!

[gritty widget]

yw

[cursive flume]

and the coefficienst b_i would be sum w_ij v^j

righT?

[orchid forge]

ye

[cursive flume]

:untilted:

[orchid forge]

now figure out how to write the inverse map like this and conclude you've got a homeomorphism that's a fiberwise isomorphism, hence a bundle isomorphism

[cursive flume]

$\left(b^{-1} q\right)^{a}:=\left(\omega^{-1}\right)^{am} q_m$

[gentle osprey]

ProphetX

[cursive flume]

does this seem righ for the inverse map in coordinaest?

where q is a one-form

b should be the musical isomorphism sharp, and omega the symplectic form

[orchid forge]

yeah. do you know if/why the components of $\omega^{-1}$ are smooth?

[gentle osprey]

Kogasa

[cursive flume]

nope,not sure about it

would say because the object we obtain is a vector and every vector can be expressed in a local basis with smooth coefficiens

[orchid forge]

since $\omega|_x : \mathbb R^n \to \mathbb R^n$ is invertible, $\omega : X \to \mathbb{GL}n(\mathbb R) \subset \mathbb{M}{n \times n}$

[cursive flume]

similarly to one-forms as you showed above

[gentle osprey]

Kogasa

[orchid forge]

the inverse map $(\cdot)^{-1} : GL_n(R) \to GL_n(R)$ is smooth (consider the definition in terms of an adjugate matrix)

[gentle osprey]

Kogasa

[cursive flume]

could I also argue that the inverse map is smooth because GL_{n}(R) is a Lie group?

or that would not be an okay argumen

[orchid forge]

that is true but only because the proof that GL_n(R) is a lie group involves showing this statement

but if you've already established that GL_n(R) is a lie group, go for it

[cursive flume]

may I LaTeX the proof you showed in a document, where i'll discuss the symplectic form and its applications in classical mechanics?

[orchid forge]

sure

[gritty widget]

Let X be a compact space, let X_i be a finite sub cover of X

seemingly to show that H^*(X)=0

it suffices to show that if Y is any subspace with H^*(Y)=0

then H^*(Y union X_i) is also zero

this feels like some kind of mayer viteroies thing

does anyone have an idea why this is true?

[orchid forge]

Y intersect X_i you mean?

[gritty widget]

no i do mean union

[orchid forge]

oh, just a single X_i

nevermind

[gritty widget]

is this like a slicker way of doing "induction"?

oh yeah

thats what it is

if this is true for abtriary Y

then its true for Y=X_j

[orchid forge]

The inclusions $Y \cap U_i \to Y \sqcup U_i \to Y \cup U_i$ induce an exact sequence that gives an isomorphism $H^j(U_i) \to H^j(U_i \cap Y)$

[gentle osprey]

Kogasa

[orchid forge]

so, for example setting $Y = \emptyset$ we have $H^k(U_i) = 0$ for all $i,k$. Now set $Y = U_j$ to conclude $U_i \cap U_j$, hence all finite intersections, have trivial cohomology

[gentle osprey]

Kogasa

[orchid forge]

oh i just saw your comment, i agree it is a form of induction

[gritty widget]

thank you though

sorry i kind of fell asleep

[gritty widget]

H(X cup A, A) = H(A , X cap A)

this doesn't seem to follow form exicision

[orchid forge]

do the relative meyer vietoris sequence

[solemn yarrow]

If two objects are chiral when embedded in n dimensions, do they become nonchiral when embedded in n+1 dimensions?

[tough imp]

I feel like maybe I'm missing an obvious detail but

If we have a finite cover of a space X into closed subsets X_i

and U is a set such that U\cap X_i is open in X_i for each i

is U open in X?

the book I'm reading first notes that U doesn't intersect any of the pairwise intersections X_i\cap X_j for i not equal to j

and then says it suffices to show U\cap X_i is open in X_i for each i

but I don't see why we need the first step at all

[limpid leaf]

i thought i had a smart cool simple proof for this but i think its wrong. if we have an infinite hausdorff space, prove it contains an infinite discrete subset

i thought i could just like. pick a sequence of infinitely many distinct points and take open neighbourhoods around them such that the intersection of any two neighbourhoods is empty (cause hausdorff) and then for any of these neighbourhoods, the intersection with the entire sequence is just the {x} that is inside the neighbourhood. and so each {x} is isolated

this is wrong im 99% sure because of isolated points dont work like that

but idk how to show it o/w

well i guess the reason this is wrong is because the sequence of points is not open so the intersection w/ the neighbourhood isn't necessarily open

lol

[orchid forge]

"discrete" refers to the subspace topology, so by definition each point in your set would be open

however, how do you know that you can pick infinitely many points with disjoint neighborhoods

[limpid leaf]

no sorry, i mean just that each point is not equal to another lol

[orchid forge]

it's certainly true for finitely many points, but how do you make the jump to infinitely many

for example, let $S \subset \mathbb R$ be the set $S = {0} \cup {1/n : n \in \mathbb N}$. then each pair of points of $S$ is separated by open neighborhoods (of $\mathbb R$) but $0$ is not separated from every other point of S by any open neighborhood

[gentle osprey]

Kogasa

[limpid leaf]

ah right yes

i see

not sure how to fix it

or do it correctly

[orchid forge]

well, the thing that goes wrong is that the sequence converges

(really, that it has a limit point)

you need to find a way to make a sequence with no limit points

[limpid leaf]

[true robin]

I think I have a proof: for every t in U and for all i, since U \cap X_i is open in X_i there is a W_i open in X so that W_i \cap X_i contains t and is contained in U \cap X_i. now define W to be the intersection of all W_i's where t belongs to X_i and subtract all other X_i's. W is open and W \cap X_i is a subset of W_i \cap X_i which is a subset of U \cap X_i which is a subset of U if t is in X_i or is empty otherwise. so W is a open subset of U containing t. Since we can do this for all t in U, U must be open. I can see why he assumed the first step though, the first proof I had in mind was by writing U \cap X_i as W_i \cap X_i for some open W_i and then doing stuff like subtracting the other X_j's from it and the taking union for all i. Unfortunately you'll only get U if you have the condition he gave above.

[untold ingot]

Would this be the right channel regarding first fundamental forms and isometries?

[gritty widget]

yes

[untold ingot]

So as I understand, isometric surfaces have the same first fundamental form?

[tough imp]

My idea was that U^c = \bigcup (X_i — (X_i\cap U)) and then each of those sets is closed in X_i, so closed in X, then we’re doing a finite union of closed sets

[untold ingot]

On one hand, the catenoid and helicoid are isometric and have the same first fundamental form ONLY AFTER you find the right reparameterization. But in this case of this pdf: https://www.maths.dur.ac.uk/users/pavel.tumarkin/past/spring17/DG/outline_term2_10.pdf they immediately discount the two surfaces being isometric without even considering possible reparameterisations

Pic from pdf for reference

[true robin]

yeah that works, and is way simpler too

[tough imp]

…

This is very silly lol

Maybe secretly the thing wasn’t a finite cover?

But it’s by irreducible components of what I think is a Noetherian ring…

So it should be finite…

[true robin]

if it isn't a finite cover there is no way to save the proof even assuming that condition because we can just take all the points of R or something as our cover

[tough imp]

No, it’s definitely Noetherian lol

Oh right

U won’t hit the intersections

Because they’re all empty lol

[true robin]

ye

[tough imp]

Silly matsumura

[gentle osprey]

Hausdorff

[vast estuary]

umm hints?

[vast estuary]

ok i do have an idea

Consider the following open covering of S^1 - U_1, U_2, U_3,U_4, where these are open semicircles

U_1, U_3 are right, left semicircles, U_2, U_4 are top, bottom semicircles

I will show that U_j is evenly covered by p for all j = 1,2,3,4

Take U_2 for instance

[gentle osprey]

Hausdorff

[vast estuary]

is this correct?

we can argue for U_1,U_3,U_4 similarly

[gritty widget]

how does one actually construct or figureout what the classifying space for a group is

I want to see that the classifying space of S^1 is CP^infity

[dusk heron]

We have the Hopf fibration $S^3\to S^2$, and this is a circle bundle whose Euler number is $\pm 1$ (depending on how you define the orientation). Taking quotients, we get circle bundles $L(p,q)\to S^2$, where $L(p,q)$ is a lens space. It's true that the Euler number of this bundle is $\pm p$, right?

[gentle osprey]

gustavn64

[empty grove]

Looks right. You could also take the cover to be S¹ - {1} and S¹ - {-1}, and these being evenly covered is really obvious because taking inverse image commutes with taking complements and the inverse image of a singleton is k distinct points which obviously divide the complement into k (0,1)s

[limpid leaf]

Once again asking this question

I'm not sure I know how to make a sequence in a Hausdorff space w/ no limit points

but I also don't know if I can guarantee that a subset has limit points? idk

there's some answer on SO but it uses this property that there is an open set of X whos closure isn't equal to X and claims this is a property of Hausdorff-ness but idk where that comes from

https://math.stackexchange.com/questions/533051/every-infinite-hausdorff-space-has-an-infinite-discrete-subspace

i mean idk if proving this about subspaces is different than subsets

[orchid forge]

Given two points x,y, we can find separating open neighborhoods U, V. Then X \ (U \cup V) is Hausdorff, so do it again.

[limpid leaf]

oh there is a theorem proven in class that says that sequences in Hausdorff have exactly one limit point

hm okay Kogasa

er fuck the theorem is at most one limit point

I guess if I pick a sequence w/o a limit point I'm done?

If I have a sequence with a limit point I can do something where I like. pick the limit point and any other point in the space?

and then theres some inductive construction i can give

like I think from this I could construct a sequence of U_n and a sequence of V_n such that any U_i disjoint from any V_i and so.. ? idk

[hollow harbor]

How do you make sure the result is still infinite?

(or not just empty?)

[limpid leaf]

are you asking me

[vast estuary]

Yep thank you!

[hollow harbor]

well, anyone... in order for this to work, you need to be able to keep picking points.

[limpid leaf]

I am not sure what you're saying

[hollow harbor]

(the answer is: you can't. consider {1/n} union {0}. if the one of the first two points you pick is 0, then you're already screwed: any neighborhood of 0 contains all but finitely many of the points. your proof needs to somehow dodge picking 0 in this example).

[limpid leaf]

right Kogasa mentioned this one earlier

But 0 is a limit point right?

[hollow harbor]

yes

[limpid leaf]

So if we have an inf sequence with limit points, we can always just pick the non-limit points?

[hollow harbor]

do you know such points exist?

[limpid leaf]

Well every seq of Hausdorff has at most one limit point

[hollow harbor]

Q is a sequence. every point of Q is a limit point of Q.

[hollow harbor]

every convergent sequence in a hausdorff space

[limpid leaf]

fuck

yeah

[hollow harbor]

not just every sequence

[limpid leaf]

and I can't always pikc a convergent sequence

well, can I?

[hollow harbor]

ok, well here

if you can pick a convergent sequence you're done

[limpid leaf]

yeah

[hollow harbor]

what if you can't?

[limpid leaf]

cry

[hollow harbor]

(oh also the convergent sequence can't just be the same number over and over)

like it needs to be a convergent, distinct sequence

or something

anyway

[hollow harbor]

well, if you can't pick a convergent sequence, how many limit points does your space have?

[limpid leaf]

None, I guess?

[hollow harbor]

sorry, that's a silly way to phrase that. yeah

[limpid leaf]

Okay but then we are done? right

[hollow harbor]

i should have said "can it have limit points"

[limpid leaf]

since if we have no limit points then my argument from before works

[hollow harbor]

yeah this should work! if you make it actually rigorous.

[limpid leaf]

that we can just always take open sets around each point and intersect them with the rest of the sequence and we get just the point

Okay I think I have an approximate idea for this

god bless u ryc

[hollow harbor]

yep!

no problem

i didn't know how to do the problem either until you mentioned that if we get a sequence we can just throw out the limit point.

i was just brainstorming until then

[limpid leaf]

lets see if i can solve and type if before 11am

[hollow harbor]

[orchid forge]

woops, I meant pick another point in X \ {x,y}

then that new point p has a neighborhood U_p which is disjoint from U and a neighborhood V_p which is disjoint from V, so their (finite) intersection is a neighborhood that separates p from every prior point in the sequence

after maybe shrinking U and V

[orchid forge]

But the point of all this is just to try to do what we said, come up with a sequence with no limit points (no convergent subsequences)

It's not necessarily true that every sequence has at most one limit point in a Hausdorff space, so we need to make sure there are only finitely many and then remove them

[wise ruin]

Sorry for my poor wording. Thinking of derivations as directional derivatives, does a derivation of the algebra $C_p^\infty$ map each germ to another germ whose values at points x are the directional derivative of functions in the germ in the direction (x-p)?

[gentle osprey]

justAlex

[wise ruin]

Or this might just be a terrible way of looking at derivations, and if so, I’m cool with hearing that too

[gritty widget]

can u elaborate

[wise ruin]

So the purpose of my question is that I’m trying to understand a derivation D:C_p^\infty\to\C_p^\infty. Obviously there isn’t just one derivation but i’m trying to understand what’s going on / to what D maps a germ (f,U). I know it has to have the properties associated with derivations, but I’m just having a hard time getting an intuition for such a mapping.

And I know this is a super vague question. But I’m just not sure how to go about getting a better feel for derivations

[orchid forge]

If D is supposed to be the partial derivative in the direction v, then it sends a germ (p, f(x)) to the germ (p, (derivative of f along v). A priori there could be derivations which don't act like partial derivatives. You should prove / have proved that all of them arise like this

At least, given a naive definition of tangent vector, using embeddings into R^n. The derivations of C_p^\infty is how you (can) define T_pM in general

Or if "v" is really a vector in R^n and by "partial derivative in the direction of v" you mean with respect to a particular set of charts \phi_i : R^n --> M we take D_v of the function f \circ \phi_i

Then there's something to prove again

[orchid forge]

Is this something you remember doing?

[wise ruin]

I appreciate that bit of clarification. I kinda just wanted to first boost my intuition of derivations from the set germs to itself, and that helped.

And yes, the introduction to derivations started off with directional derivatives, but I’m trying to eventually stop thinking about them in that manner because I feel like it’s just a crutch that will stunt my understanding of derivations in general.

[orchid forge]

But the point is that really all derivations of C_p^\infty are partial derivatives

So for this particular algebra it is not a bad way to think about them

[plain raven]

@gritty widget I took a few courses in classical logic (computability theory and model theory) in undergrad. I went to grad school at a place that specializes in those things and took the qualifying exam in logic. In the first year of grad school I took the standard first year sequence in algebra and got into category theory. I read the book "Categorical Logic and Type Theory" by Bart Jacobs and this had a huge influence on me. After that as I studied more category theory, I realized that I didn't want to be one of these pie in the sky people who studies abstraction for the sake of abstraction and I wanted lots and lots of concrete examples in mathematics to understand how these categorical organizing principles could be used. I think I also find the evolution of mathematical ideas through history intrinsically interesting.

For example, I read the book "Sheaves in geometry and logic" by Mac Lane and Moerdijk having no knowledge of geometry. The book solely focuses on category theory and logic. The "Geometry" in the title is only present insofar as the book studies sheaves in geometric logic. Topos theory is an important part of category theory but I felt I didn't understand it if I didn't know about its purpose, the context in which it was conceived, the problems that motivated it. It was an interesting book but I had no knowledge of geometry and I didn't feel I could get a sense of why we give a shit. It wasn't just for personal motivation, if i am working on a question I want to be able to argue to others the importance of what i'm doing, and if you can't translate these abstract ideas into concrete applications you will have trouble. So I decided to learn basic geometry and topology.

I guess in short I am a category theorist and my core interests are applications of category theory in homological algebra and type theory, but my desire to get more intuition for constructions of homological algebra has led me to try and study many different areas of math where these theories are applied. Category theory is very abstract but de rham cohomology is very rich and geometrically concrete.

Since I said I wanted to learn about topos theory, I should have studied more algebraic geometry but i was intimidated by the pathological topology of schemes and i thought they are already pretty abstract and it's hard to get intuition by starting from schemes, so i studied sheaf cohomology on manifolds first, out of Warner and Spanier. As a result my knowledge of AG is pretty lacking. And you need lots of comm alg! but I do find AG pretty neat.

If you want something more specific, for the last six months or so I've been thinking about applications of monoidal category theory to a better understanding of the Dold-Kan correspondence, and i'm trying to write down abstract categorical characterizations of constructions from homological algebra that were initially written down concretely by means of explicit constructions and formulas and seem to appear this way in most textbooks

For example, the de Rham complex of sheaves. is there a nice categorical characterization of this guy which pins it down? I don't mean up to homotopy equivalence or even up to a strong deformation retract but actually up to isomorphism, is there a categorical characterization of this complex of sheaves up to isomorphism. I think maybe this question is answered in the literature but i'm in a research group of logicians and i don't have lots of resources who I can talk to about this so I'm trying to make more connections online and transfer to a different school where i'll be able to talk to more a categorically minded advisor

It seems to be not very well known that under the right choice of definition of the Dold-Kan correspondence, it is a monoidal equivalence of categories. The standard definition, which is imho dumb and wrong, is not strong monoidal but only like, monoidal up to homotopy equivalence.

[orchid forge]

Yep

[gritty widget]

very cool

[wise panther]

Can someone explain why the maps $H_n(S^n)\to H_n(S^n, S^n-x_i)$ is an isomorphism?

[gentle osprey]

Pan

[wise panther]

The exact sequence from homology is
\begin{equation*}
H_n(S^n-x_i)\to H_n(S^n)\to H_n(S^n, S^n-x_i) \to H_{n-1}(S^n-x_i)
\end{equation*}

[gentle osprey]

Pan

[empty grove]

The first and last term in this are 0

[wise panther]

I understand why the last term is 0

[empty grove]

S^n - x is contractible

[wise panther]

But by excision, $H_n(S^n-x_i) = H_n(S^n, {x_i}) = \tilde{H}_n(S^n) = \mathbb{Z}$

[gentle osprey]

Pan

[wise panther]

^ is this not true?

[empty grove]

Doesn't excision apply on a pair

You can't apply it on S^n - x alone

[wise panther]

The pair is $(S^n, x_i)$ to $(S^n-x_i, \emptyset)$

[gentle osprey]

Pan

[empty grove]

Ah, maybe check the conditions? I think you need the thing you are cutting out to be in the interior of the second entry of the pair

[empty grove]

Right yeah, closure of {x} is not contained in interior of {x}

[wise panther]

hmm

okay

thank you

[digital wraith]

Ok so I'm struggling to understand the construction that gives a covering space for an arbitrary group action.