point-set-topology

[sleek thicket]

So I'm not touching that

[clever badge]

fellow physicist taking an intro to geometry course here.

I have an assignment question dealing with a regular surface

the solution i have found to the problem includes a euclidean motion of said surface and fixing a point of interest to the origin such that the normal vector to the surface points in the z axis.

I have then taken it as a fact that i can describe the surface in some neighborhood around the origin as a graph (x,y,g(x,y))

given that this is a course from the math department i assume they wouldnt let this slide without a proof. How would i go about showing this?

[snow gull]

I have no idea of how to solve this, can someone please help?

[clever badge]

<@&286206848099549185>

[gritty widget]

Sounds like some inverse function theorem kind of thing

[snow gull]

I think it's related to this one, if I can show one homeomorphism from S(omega) \ {some point} to S'(Omega) then it extends to a homeomorphism of their one point compactifications

[gritty widget]

What's S_Omega?

[snow gull]

minimal uncountable well ordered set

[hard wind]

Am I able to apply Theorem 22.2 of Munkres to this problem?

I am not entirely sure.

[gritty widget]

I was gonna say you can write that you don't believe in choice so such a space doesn't necessarily exist, but sadly I guess they exist even without choice

Sorry, I am of no help

[hard wind]

All good

I'm just not sure if the setup of the commutative diagram implies the condition on "g" in Theorem 22.2

I have no experience in category theory so I am a bit unprepared for this

I don't need answers for the problem, just need to know if the theorem is able to be applied if anyone knows :)

[high hill]

This is hwk, so don't want answer --- just a point in the right direction as I'm completely stuck https://i.imgur.com/M4AVRYb.png

In our course [V, W] := (V . del) W - (W . del) V

This is what I have but doubt it helps (not 100% sure I did it right as well)

https://media.discordapp.net/attachments/496784996510466057/774429741465600070/IMG_20201107_002402.jpg?width=902&height=677

I think maybe I'm supposed to use this, but don't see how I'd find the flows https://i.imgur.com/oRiTTl8.png

I also notice the expressions for the ith components for V and W are similar to the pushforward defn

https://media.discordapp.net/attachments/496784996510466057/774430419600670740/unknown.png

Pointer appreciated 🙂

[gritty widget]

lie brackets

[gritty widget]

maybe you can just use the standard coordinate expression for the lie bracket here? $$ [V, W] = c^k\frac{\partial}{\partial x^k} $$ where $$ c^k = V^i \frac{\partial W^k}{\partial x^i} - W^i \frac{\partial V^i}{\partial x^i} $$ (einstein notation)

[gentle osprey]

TTerra:

[gritty widget]

and then push some derivatives around and hope it works lol

@high hill

like it'll suck but it should work out

what is this from?

[high hill]

uni yr 3 hwk

[gritty widget]

i mean like

what book

[high hill]

not book - lecturer sets Qs

[gritty widget]

ah i see

[high hill]

ty -- ill look again

[gritty widget]

there's probably a better way of doing it that isn't just pushing through derivatives like i suggested

but like

it absolutely should work out

[high hill]

My working --- is it not wrong?

(although maybe not helpful)

Rusty on this stuff

I think what I wrote is pushing thru derivatives as u suggest (or meant to be)

[hard wind]

Is it possible for f and g to be not quotient maps, but f o g be a quotient map?

[gritty widget]

when i said pushing through derivatives i guess it basically amounts to what you were doing lol

[elder yew]

So I recently saw a problem that said

Prove fundamental theorem of algebra, assuming Stokes Theorem is Valid

Hrm

Any thoughts?

[gritty widget]

use stokes' theorem to go through all the complex analysis stuff needed to prove liouville and then do it from that

"it suffices to prove liouville's theorem, for which it suffices to prove cauchy's integral formula, which is trivial"

[elder yew]

Lol

[modern holly]

At this point I'm more surprised at results which you can't use to prove the fundamental theorem of algebra

[gritty widget]

now that i think about it ive never actually seen an algebraic proof of the fundamental theorem of algebra

[clever badge]

can one always locally represent a regular curve implicitly?

[clever badge]

<@&286206848099549185>

example that i found online from a differential geometry midterm:

is it always fine to say that we can implicitly express the surface in some neighborhood of a point within it?

[errant rapids]

Problem:

Let $\mathbf{x_1}, \mathbf{x_2}, ,...$ be a sequence of the points of the product space $\Pi X_{\alpha}$. Show that this sequence converges to the point $\mathbf{x}$ if the sequence $\pi_{\alpha} (\mathbf{x_1}), , \pi_{\alpha}  (\mathbf{x_2}), ,...$ converges to $\pi_{\alpha} (\mathbf{x})$ for each $\alpha$.

[gentle osprey]

gelearndeutsch:

[errant rapids]

What's the mistake in the following counterexample:

$\pi_{\alpha} (\mathbf{x_n})=\begin{cases}  1 & \alpha \geq n \ 0 & \alpha < n \end{cases}$

for each $\alpha$, the sequence $\pi_{\alpha} (\mathbf{x_1}), , \pi_{\alpha} (\mathbf{x_2}), ,...$ converges to $0$. But $\nexists , N\in \mathbb{N}$ s.t. $\forall n>N$, $\mathbf{x_n} \in (\frac{-1}{2}, \frac{1}{2})\times (\frac{-1}{2}, \frac{1}{2}) \times ....$

[gentle osprey]

gelearndeutsch:

[errant rapids]

I'm new to general topology. so please give a hint

lemme think

[wicked mirage]

$\pi_{\alpha}$ is the projection on the $\alpha$-th coordinate, is it not? You are redefining it with a seemingly unrelated function, completely missing the point

[gentle osprey]

welperooni:

[errant rapids]

@wicked mirage Yes $\pi_{\alpha}$ is there projection. I defined only $\mathbf{x_n}$. The definition of $\pi_{\alpha}$ is the same. I didn't redefine it

[gentle osprey]

gelearndeutsch:

[errant rapids]

ok I'll think along the openness

[errant rapids]

yes the product is not open because all but finitely many coordinates should be $\mathbb{R}$. Thanks

[gentle osprey]

gelearndeutsch:

[clever badge]

@gritty widget the way we have defined for a surface to be regular is if for each point in the surface there exists a regular local surface. Where a regular local surface is a surface given by a smooth injective map some D to E^3 with a continuous inverse, where the tangent vectors (which we call x_u and x_v) are always linearly independent

so for the point of the regular surface im interested in i know there must be some regular local surface

then the surface is described locally by some parametrization (u,v) where the functions are smooth

but i am not aware of any theorem that states that we can go from parametrization to implicit

[nimble cipher]

How do I show that the topological dimension of the Cantor set is zero?

clopen in the Cantor set?

I was thinking of building an open cover then get a refinement with order less than or equal to 1.

[clever badge]

doesnt the implicit function theorem require that you go down in dimensions?

from R^n+m to R^m

and here our surface is defined as a map R^2 to R^3

so idk how i would apply the implicit function theorem

[meager python]

Why can we assume e_i are those points? The explanation should probalby not be under advanced but something more elementary.

[sleek thicket]

suppose you know it for this triple

then for any $p_1,p_2,p_3$ and $q_1,q_2,q_3$ you can find a group element $g$ such that $g \cdot e_i = p_i$ for $i=1,2,3$ and a group element $h$ such that $h \cdot e_i = q_i$ for $i=1,2,3$. Then $hg^{-1} \cdot p_i = h \cdot (g^{-1} \cdot p_i) = h \cdot e_i = q_i$

[gentle osprey]

shamrock:

[sleek thicket]

it's like how if you want to show that a group action is transitive it suffices to show that you can go from one fixed single element to any other

actually it's exactly the same, if $G$ acts on $X$ then this action is triply transitive iff the action $g \cdot (x,y,z) = (g\cdot x, g \cdot y, g \cdot z)$ of $G$ on $X \times X \times X$ is transitive

[gentle osprey]

shamrock:

[meager python]

oh ofc... I was thinking about coordinate change

thanks!

[nimble cipher]

Hi. How can I create an open cover of the Cantor set and a refinement of it such that the refinement has an order less than or equal to 1

[nimble cipher]

a the order of a subset of the power set is the maximum number of sets in the collection that contains an element of the set

[sleek thicket]

So order 1 means pairwise disjoint?

So by compactness your cover is finite

and actually a finite cover by intervals

right?

I mean yeah, but the argument has to work for finite covers anyways

And it's easier to work with finite covers

Right?

Oh, I thought it was for all covers

They were talking about topological dimension earlier

Which I think would need all covers

Only one cover is very easy lol

why?

I thought order 1 = pairwise disjoint?

So for an arbitrary cover

We can refine it to a finite cover by intervals

Why is the symmetric difference of open sets open?

Anyways

Finitely many open intervals

we can truncate the cantor set construction to finitely many steps maybe?

Like

hmm

[gritty widget]

The complement of Cantor set in $\mathbb{R}$ is dense in $\mathbb{R}$, so if you have an open covering of the Cantor set $C$ then you can find a refinement of it covering $C$ with open intervals whose endpoints are non-Cantor points. Taking a finite subcover of this latter, since $C$ is compact, we have $C \subset I_1 \cup I_2 \cup \cdots I_n$ for some refinement ${ I_j }_j$ with open intervals with endpoints not in C. Then $C \subset I_1 \cup (I_2 \setminus \overline{I_1}) \cup (I_3 \setminus (\overline{I_1 \cup I_2})) \cup \cdots$ as the desired refinement (the closing of the finite unions of these intervals only adds non-cantor points, so still covers C).

@nimble cipher

[gentle osprey]

ℓ-adic:

[gritty widget]

You can do this with any nonwhere dense closed set C by using sigma-compactness of R to reduce to countable covering with open intervals whose endpoints are not in C then taking out closures of finite unions of intervals successively.

[sleek thicket]

nice!

[nimble cipher]

Thank you @gritty widget and everyone else who answered!!

[cursive flume]

idk if it fits here,but could someone help?

[sweet wing]

so suppose there is no convergent subsequence right

at every point, you can find a open set that only contains finitely many other points

[cursive flume]

why can I find at every point?

convergence= every nbhd of x contains all x_n except finitely many

[sweet wing]

wait nvm ignore that uh

but similar reasoning: so like for every point in the topological space, you can find a open set with finitely many points in the sequence in it

cuz we are assuming there is no convergent subsequence

this gives a open cover of the whole space by taking a union, meaning there exists a finite subcover, but the sequence is infinite (notice the contradiction?)

[cursive flume]

i might be knowing convergence wrong

my prof gave me the def $\lim_{n \to \infty} x_n=x$ iff every nbhd ofx contains all x_n except finitely many

[gentle osprey]

ProphetX:
Compile Error! Click the reaction for details. (You may edit your message)

[sweet wing]

yup

[cursive flume]

how does assuming there is no convergent susbseq lead to this

[sweet wing]

which is equivalent to you cannot find a nbhd that has finitely many points of the sequence

x is limit point = no nbhd with finitely many points in seq

so
x is not a limit point = exists a nbhd ...

[cursive flume]

negation of that would be there exists a nbhd of x which contains all x_n except infinitely many?

sry in advance for dumb q,i'm not used to rigorous maths,physicist lad here

[sweet wing]

x is limit point -> for all nbhd, there are infinitely ...

x not limit point -> exists one nbhd, there is only finitely many

Maybe more abstractly it is like

$$\forall U:\phi(U)$$
and in this case $U$ takes on all nbhd and $\phi(U)$ is true if $U$ contains infinitely many points in the sequence

The negation is

$$\exists U:\neg\phi(U)$$

[gentle osprey]

ariana:

[cursive flume]

wait. x is a limit pint=every neighbourhood of x contains infinitely many x_n except finitely many?

[sweet wing]

yup

ah missed the finitely many part in that yh

[cursive flume]

ok so I have this. negation of this is: there exists a neiighbourhood of x containing finitely many x_n except infinitely many. neighbourhood=open set => i can find an open set with finitely many points in the sequence. i don't understand your last argument tho.

this gives a open cover of the whole space by taking a union, meaning there exists a finite subcover, but the sequence is infinite (notice the contradiction?
this

compact=every open cover has a finite subcover

right?

also where have we used first countability, why was it needed at all?

[sweet wing]

contains finitely many already implies it misses infinitely many so you can drop that part out

compact=every open cover has a finite subcover
yup

[cursive flume]

i have one open set. cause one neighbourhood exists. how can I take union of one set?

[sweet wing]

hm good qn about first countability 🤔

[cursive flume]

i can't follow your last argument

[sweet wing]

oh so you define such a open set for every point and take a union of all of them

cuz every point is included it covers X

[cursive flume]

suppose the sequence is convergent. it converges to a point. suppose it is not<=> it converges everywherE?

😮

[sweet wing]

uh if it isn’t convergent then it doesnt converge anywhere

[cursive flume]

then how can you do that with every point i'm confused

[sweet wing]

we assume it doesnt converge first right

[cursive flume]

yes

[sweet wing]

so at every point $x$ we can have a open set $x\in U_x$ such that $U_x$ only has finitely many points in the sequence

[gentle osprey]

ariana:

[cursive flume]

yes that's my question

if it converges to a point(convergent)

if it converges to every point(Divergent?)

i can't see how you get to every point

so: a sequence converges to a point iff every nbhd of x contains infinitely many x_n except finitely many

a sequence does not converge=(??) to every point there exists a nbhd of x contains finitely many x_n except infinitely many?

[sweet wing]

maybe let me just give a explicit example

Consider the sequence $x_n=n$ in $\mbb R$

[gentle osprey]

ariana:

[sweet wing]

at the point $2.5$, we can choose the open set $(2.3,2.7)$

[gentle osprey]

ariana:

[sweet wing]

at the point $3$ we can choose like $(2.5,9.5)$, it only has finitely many terms in the sequence contained in it

[gentle osprey]

ariana:

[cursive flume]

ok. so we found an open cover, which has a finite subcover. this does not imply compactness

every open cover has to have a finite subcover,or?

[sweet wing]

we are assuming X is compact

[cursive flume]

then where is the contradiction?

[sweet wing]

so you have your finite subcover

[cursive flume]

we have an open cover with finite subcover(this is true cause compact)

yes

[sweet wing]

it covers the whole space

but your sequence is infinite

and every open set has finitely many terms

(note: seems like cuz some more technical details we do need first countability in the U_x exists part:
suppose no convergent subsequence but you cannot find U_x, then we can construct a convergent subsequence by considering a bunch of open sets V_n, V_n\in V_{n+1} and choose a point of the sequence in each V_n, counterexample to where you cant do this is like stone cech of integers which honestly fake topological space)

[sleek thicket]

Let $\varphi : \R^n \setminus {0} \to \R^n \setminus {0}$ be a diffeomorphism. Is $\varphi$ the restriction of a diffeomorphism $\R^n \to \R^n$ fixing the origin?

[gentle osprey]

shamrock:

[sleek thicket]

No lol take n = 2 and look at z |-> 1/z

or generally diffeomorphisms of S^n which interchange the north and south pole

hello mx moth

[fading vale]

hi

[sleek thicket]

do you have a topology inquiry

[fading vale]

im trying to figure out how many times the topology role has actually been pung

[sleek thicket]

Hmm you should try pinging it and asking

[fading vale]

[sleek thicket]

I only remember one time, maybe 2?

[fading vale]

yea i dont think the advanced roles get pinged very often

[sleek thicket]

wait

I'm smart and never make mistakes

So new question, can any diffeomorphism of S^n \{N,-N} be extended to a diffeomorphism of S^n?

Where N is the north pole

[fading vale]

fixing the north pole and its antipode?

[sleek thicket]

Not necessarily

[fading vale]

oh is there supposed to be a set minus there

[sleek thicket]

The 1/z example says this can't happen

Oh lol

[fading vale]

lol

i was kinda thonking

[sleek thicket]

discord ate my backslash

[fading vale]

yea it does that

[sleek thicket]

But yeah possibly swapping the two points

[fading vale]

this osunds like

the kind of problem that could be answered in the book im supposed to be reading

[sleek thicket]

ooh

[fading vale]

its like

applying the borsuk ulam theorem to combinatorial stuffs

but i assume results in it would be helpful

[sleek thicket]

I was trying to show that Iso(S^n) = O(n+1)

And like, you can extend an isometry of S^n to a diffeomorphism of R^(n+1)\{0} pretty easily (just apply it on the sphere of radius r centered at the origin for each r)

And it's gotta fix the origin

[fading vale]

thats pretty cool

[sleek thicket]

But I was having trouble showing smoothness of the origin

anyways if I could show this extension is an isometry of R^n I think I have it

Since Iso(R^n) is the group generated by O(n) and translations

and things fixing the origin are just O(n)

[fading vale]

i guess that sounds believable

i dont know anything about this stuff so i couldnt really say

[sleek thicket]

yr fine

oh wait I think I'm overcomplicating things a little

I don't actually need to show smoothness or anything, any map V -> V fixing the origin and preserving distances is an isometry (where V is a fd inner product space)

[gritty widget]

did someone say pinging my role

maybe i can post my riemann geo homework here and have someone do it

[sleek thicket]

I mean id look at it just out of curiosity

[gritty widget]

ill post when i get to my computer

no answers pls

[fading vale]

@sleek thicket do u know anything about like

problems re: partitioning regions of the circle

w/ a cap

[sleek thicket]

Not really, sorry

[fading vale]

hnsdnfsldf

u know hat its ok

ill just bullshit it

[gritty widget]

that's the only problem that isn't right from do carmo shamrock

the rest are just random qs from various sections of do carmo's rg

this was one of them, pretty nice question

[sleek thicket]

What's a jacobi field?

[sonic hill]

I wonder if anyone can answer this confusion I have:

1. The hairy ball theorem says that the tangent bundle TS^2 of S^2 has no everywhere nonvanishing section.
2. Therefore, TS^2 has no trivial line subbundle.
3. Every real line bundle on S^2 is trivial (because H^1(S^2, Z/2Z)=0).
4. Conclusion: TS^2 has no line subbundles.
   Is everything I said right? I never thought about whether it makes sense for a vector bundle to have no line subbundles...

[sleek thicket]

I don't see a reason why we'd expect all bundles to have a line subbundle tbh

Like, that's saying something big and global about the bundle imo

[gritty widget]

@sleek thicket a jacobi field is a type of vector field along a geodesic that solves a certain second order ODE called the jacobi equation. they arise naturally when you study stuff like the exponential map and when you want to study "how radial geodesics spread"

[sleek thicket]

Oh okay

[gritty widget]

i think do carmo's chapter 5 does a good job on motivating them and deriving the jacobi eqn

[sleek thicket]

I think my course is a little behind yours right now

[gritty widget]

mine's kinda speedrunning do carmo lol

and do carmo is a lot less comprehensive than lee i think

i think my prof really wants to get to the comparison geometry stuff

[fading vale]

what class is this

just diff geo?

[gritty widget]

i guess

itd be called riemannian geometry since "differential geometry" is usually reserved for either a curves and surfaces class, or for an intro to manifolds class (like out of lee ism)

it's correct to call it differential geometry

[fading vale]

interesting

[gritty widget]

nice topology role btw

one of us 😌

[fading vale]

oh i was given it

yes

i dunno that much but i learn every time i come here so i figured it would be nice to have

in case any cool discussions happen

hmmm

is there a name for the theorem that any two regions on the 2-sphere can be bisected by a cap

i see a not that hard way to prove it but im interested if there r any generalizations

(may have found something for the latter will add it to my list of things to read)

[elder yew]

what do you mean bisected by a cap?

The way I've seen cap used is to "cap off a disk"

[fading vale]

its like

the intersection of a plane and the sphere i mean

lol

so like if u were to embed S^2 into R^3 then ud have some hyperplane of R^3 which bisects any two regions on the sphere

and this is proven (apparently)

[elder yew]

Ahh I see what you're saying

[fading vale]

ya

idk i have a lot of time ill figure out more things later

[tight agate]

I think it's called the ham sandwich theorem

[sonic hill]

Got another topology question:

Consider ${(x,y,z)\in\bC^3\mid x^2+y^2+z^2=1}$. Now looking at it algebraically, there is a rational parametrization of this complex surface, making its projective closure in $\bP_\bC^3$ isomorphic (and therefore homeomorphic) to $\bP_\bC^2$.

On the other hand, it's ``well known" that the surface is diffeomorphic to the total space of tangent bundle of $S^2$. So in conclusion, the total space of $TS^2$ embeds as a dense open subset of $\bP_\bC^2$?

[gentle osprey]

Icy001:

[little hemlock]

S is a tetrahedron thought of as a oriented 3-simplex. I don't understand why C_3(S), the group of oriented 3-chains of S is 0, and not the free abelian group on one generator?

My understanding is that the tetrahedron is a single oriented 3-simplex, so C_3(S) should have a single generator?

[tight agate]

I think they're only looking at the boundary of the tetrahedron, no interior

"surface"

[little hemlock]

okay yea, i think i get it

[gritty widget]

So I'm trying to learn category theory and I'm finding I need learn a companion math to apply category theory too. I've studied an Intro to Topology book that focuses on Point Set Topology, basically getting one familiar with open and closed sets with metric spaces and such, should I now be ready to learn algebraic topology, or do I need more point set or general topology, just wondering topology kind of scares me as a subject, but I love the ideas of continuity and deforming shapes and learning their properties.

?

[frigid patrol]

https://topology.pubpub.org/

@gritty widget this is the book for you

[gritty widget]

ok, thanks!

can it be approached from an upper undergraduate level too?

[tight agate]

Peter May's book is also great

[gritty widget]

is that A Concise Course in Algebraic Topology?

[tight agate]

yup

a lot of people like Hatcher too

you don't need too much point-set top to start off with algebraic topology

as long as you're comfortable with the notion of a topological space, compactness, connectedness, products, coproducts and quotients you should be ready

[gritty widget]

ok

[tough imp]

A concise course is so fucking hard

I do not recommend it

Brofibration seems to be a mega genius

[tight agate]

no, it's because mathacka said they want to apply category theory

which is sorta the point of view that may takes

[tough imp]

I mean it's very categorical sure... but it's so ludicrously hard

[tight agate]

I agree May is hard, but there's so much great stuff in there

especially chapter 9 onwards

sorry chapter 6

oh, Tom Dieck is also pretty categorical

maybe read the fundamental groups and van kampen part from hatcher and then move onto may?

[gritty widget]

ok

[tight agate]

it depends on what you're interested in, Hatcher is very geometric

[gritty widget]

the topology book from @frigid patrol then?

I do like geometric presentations of material

[frigid patrol]

It's used as a first semester graduate course

There are no prereq for the book that I'm aware of

[gritty widget]

great, I love self contained math books!

are homtopies talked about in any of those books?

[frigid patrol]

Yes

If you learn about fundamental group, you'll learn about homotopies and covering spaces

[gritty widget]

So if I understand correctly homotopies have something to do with paths between spaces much like functors in CT

ok

AT?

oh, duh

I don't mean to ask a dumb question, but I'm not seeing yet significant connection between Topology and Geometry in anything I've read yet, is that on purpose on introductory material?

I guess I've thought of Geometry as kind of an extension of Analytic Geometry.

Something, again, I didn't see in point set stuff

Excuse my nievity

Yes

and I've peeked at differential geometry, definately something I want to study later

I do lean towards applied math and differential can be applied to many real world problems, but I don't want to downplay the significance of pure math in the process, I need more of those.

yeah, like abstract algebra for groups and such

I don't want to get to many opinions going, but is N J Wildberger someone I should be learning topology from?

I've sensed that he's a bit out of the norm

ok, suspected so, he seems to deny the reals, but other then that ok

also I have Principles of Topology by Croom is that considered a good book?

ok, yes I watched part of that lecture, looks like something worth watching more of.

[sweet wing]

honestly i feel like it's better to just spend a month or two on point set fundamentals then moving on with algebraic topology

[tight agate]

If X and Y are surfaces, and if $p_2: X \times Y \rightarrow Y$ is the projection map is there a way to identify $H^0(Y, \mathscr{O_Y})^*$ and the image of the pullback map $p_2 *: Pic(Y) \rightarrow Pic(X \times Y)$?

[gentle osprey]

Brofibration:

[tight agate]

X is a reduced projective scheme

[gentle osprey]

Brofibration:

[tight agate]

or just a projective variety

[tight agate]

Oh lol nvm I was being stupid

invertible global sections are precisely global sections of O*

which is the same data as a line bundle

[cold vine]

I'm supposed to prove that Hn(X,A) is isomorphic to Hn(X,{a}) when I know A in X, {a} in A, Hn(A,{a})=0 for all n so I suspect I at least inspect long exact sequences ...-> Hn(X,A)-> Hn(A) -> Hn(X) -> Hn(X,A) ->... and the same for Hn(X,{a}) and injections between them but the assumption Hn(A,{a}) does not arise here. And if I expand Hn(A,{a}) I dont see how that helps eiter.

I know I'm supposed to use a ladder with two long sequences but I'm kinda lost

[gritty widget]

Write down the LES for the pairs (X, {a}) and (X,A) and apply the H_* to (X,{a}) -> (X,A) and diagram chase (or 5-lemma if you feel lazy).

The point is that the square H_n(X, {a}) -> H_n(X,A) (on the row) and H_{n-1}({a}) -> H_{n-1}(A) (on the bottom) and the vertical maps as the boundary maps H_n(X, {a}) -> H_{n-1}({a}) and H_n(X,A) -> H_{n-1}(A) commute, and that the map H_{n-1}({a}) -> H_{n-1}(A) is an isomorphism because if fits into another long exact sequence with a snip 0 = H_n(A, {a}) -> H_{n-1}({a}) -> H_{n-1}(A) -> H_{n-1}(A, {a}) = 0

so that's where you use H_n(A,{a}) = 0

So I'm using 3 different long exact sequences here

[cold vine]

ahh alright I'll think about this. I got a hint that you could do it with two LES but is this by using the 5-lemma?

[gritty widget]

Yea you can certainly use 5-lemma

[cold vine]

alrught, thanks! I'll try to figure it out

[gritty widget]

np

[quiet pilot]

I'm having some trouble showing that V(J) is a subset of V(I(p(V(I))))

Given an f that is 0 on p(V(I)) I want to show it is also 0 on V(J)

I'm thinking I might need to use the nullstellensatz somewhere, but I'm not really getting anything

[gritty widget]

This is actually a consequence of the standard proposition that if $A \hookrightarrow B$ is an injection of rings then $\textnormal{Spec }B \to \textnormal{Spec }A$ has image that is dense in $\textnormal{Spec }A$.

[gentle osprey]

ℓ-adic:

[gritty widget]

In your case, apply with $B = k[X_1, \ldots, X_n, Y_1, \ldots, Y_m]/I$ and $A = k[X_1, \ldots, X_n]/J$.

[gentle osprey]

ℓ-adic:

[gritty widget]

To check this, just take a nonempty basic open set of the form $D(f)$ (so $f$ is not nilpotent) in $\textnormal{Spec }A$ and note that the image of $f$ in $B$ under the injection $A \hookrightarrow B$ is still a non-nilpotent, so the inverse image of $D(f)$ under the natural map $\textnormal{Spec B} \hookrightarrow \textnormal{Spec }A$ is nonempty (since the inverse image is, explicitly, $D(\overline{f})$ where $\overline{f}$ is the image of $f$ under $A \to B$).

[gentle osprey]

ℓ-adic:

[gritty widget]

Also in your case, the condition $f$ is not "nilpotent" is the same as the condition that $f \neq 0$ at all (classical) points (since nilradical = jacobson radical for f.g. k-algebras and k is algebraically closed so we know the "points" are the maximal ideals) so you can just check on points.

[gentle osprey]

ℓ-adic:

[gritty widget]

So that's connecting to a general standard fact. It'd look like this in your particular situation: We know that $p(V(I)) \subset V(J)$, so to show that the closure of the former gives you the latter, you need only show that the former is dense in $V(J)$. To that end, take an open basic set $D(f)$ so that $D(f) \cap V(J) \neq \emptyset$, i.e. that $f$ doesn't become nilpotent in the ring $k[X_1, \ldots, X_n, Y_1, \ldots, Y_m]/J$ (by Hilbert-Nullstellensatz, if you will). Since $k[X_1, \ldots, X_n, Y_1, \ldots, Y_m]/J$ injects naturally into $k[X_1, \ldots, X_n]/I$, $f$ remains non-nilpotent in that latter ring, i.e. there's a maximal ideal $m$ of the latter that does not contain $f$, i.e. corresponding to a point $x \in V(I) \cap p^{-1}(D(f))$. So $p(x) \in D(f) \cap p(V(I))$, so $D(f) \cap p(V(I))$ is nonempty, so $p(V(I))$ is dense in $V(J)$ so the closures have to equal.

[gentle osprey]

ℓ-adic:

[gritty widget]

Anyway the point is that p(V(I)) is dense in V(J). So check this in however manner you want.

Btw replace the hookrightarrows for Spec B, Spec A with just arrrows (Spec B -> Spec A), since Spec B -> Spec A is often not an injection (if A -> B is an injection).

[quiet pilot]

Thanks for the answer!

[quiet pilot]

I was trying to prove the contrapositive of what you did. But doing it this way makes the usage of the nullstellensatz seem very natural

[cursive glade]

If w(x, v2, v1) (where v1= v2) is a 2 form on R^2, then it has to be 0 no?

[gritty widget]

i've never seen the notation w(x, v_2, v_1) for a differential form, can you elaborate?

is it like x is a point of R^2, and then w(x,) takes in two tangent vectors v_2, v_1 at x to get a real number w(x, v_2, v_1)? if so then yes, it will be zero, because dx \wedge dy is alternating

(something something every 2-form on R^2 is a function times dx \wedge dy)

[gritty widget]

There is a recent result that all smooth jordan cubes have an inscribed rectangle. Can someone ELI5 (and in particular ELI-didnt read the paper, and ELI-dont know symplectic manifolds) why the proof won't work for general jordan curves?

[cursive glade]

@gritty widget Sorry, the prof defined it as function w:D×R^2×R^2 -> R , where D (the x argument) is a subset of R^2 . I was thrown off by the x part too, but it might indeed be interpeted as a point on the manifold, where the manifold just a subset of R^2

[gritty widget]

that's quite the weird way to define a differential form lol. either way, it still holds on a manifold

can write the form in coordinates and then use the same reasoning basically

[cursive glade]

Yeah, problem is that he lets it equal 1, so it must be a typo

[gritty widget]

[cursive glade]

Kinda questioning my sanity for the last couple of hours 😂

[gritty widget]

can you post what they wrote?

[cursive glade]

Q 3

[gritty widget]

,rotate

[cursive glade]

Friends pic sorry

[gritty widget]

wait, i may have been wrong

let me check something

seems sus

[cursive glade]

This is where its defined

[gritty widget]

you didn't get the whole definition of a 2-form in the picture

[cursive glade]

It just says anti symmetric bilinear homomorphism like the rest

[gritty widget]

pls show

[cursive glade]

Friend sent it to me, i dont have it on me handy

[gritty widget]

then omega(x, (1,0), (1,0)) must absolutely be zero

your instructor made a typo

[cursive glade]

I dunno

[gritty widget]

this is an immediate consequence of antisymmetry

[cursive glade]

The antisymmetry is sufficient for for it to be 0

Yeha

[gritty widget]

omega(x, v, v) = -omega(x, v, v), pretending that you swapped the two arguments

this is R, so omega(x,v,v) = 0, done

[cursive glade]

Thanks for your help anyways

[fading vale]

@cold vine this is a late reply but i figure its worth mentioning that there is a relative homology LES for triples as well

let Z subset Y subset X then the sequence you get is H_n(Y, Z) -> H_n(X, Z) -> H_n(X, Y) -> H_n-1(Y, Z) and so on

letting X = X, Y = A, Z = {a} solves your problem immediately lol

^_^

[gritty widget]

Yea I was thinking of mentioning that but a proof of that is basically the same thing as I suggsted: "braiding" of 3 different LES if you have three spaces braided X < Y < Z

[fading vale]

mhm

[gritty widget]

There's a nice diagram in Bredon's book as I recall (used that book in my alg top class)

[fading vale]

lemme check the proof in hatcher its been a while so im not sure i remember it

nice

my AT progression has been really wack with basically like

going thru a shitton of different sources and getting a little further each time

but at the moment i am on the hatcher bend and probably will just finish it

lol hatcher literally just handwaves it hes like

"just take the SES of relative chain complexes :^)"

[gritty widget]

k I found the proof of this triple sequence in pg 188 of Bredon's Top & Geometry.

[fading vale]

ty!

[gritty widget]

Pretty pictures. NP! 🙂

[fading vale]

i have bredon on my shelf i think if (when tbh) i pass back through AT i will go for it

for a different perspective

this is hatcher lmfao

[gritty widget]

My alg top professor was all about having to learn AT together with interactions with smoothie manifolds so he made us read notes of his own and Bredon's. He seemed to dislike Hatcher with some sort of passion.

[fading vale]

yeah thats fair

i think both perspectives are valid

pedagogically bredon is probably better i feel? but hatchers exposition is >>>>>>

[cold vine]

Ah thats interesting :D Something with triples did cross my mind when thinking through the problem but we havent at least yet seen thwm

Interesting to know

[fading vale]

AT actually has a lot of good textbooks i feel weirdly

hatcher spanier concise tam-dieck bredon

lurie

[gritty widget]

Switzer's AT is also highly recommended after a first course in AT

[fading vale]

neat

hi shamrock : )

[sleek thicket]

hi moth :)

I'm gonna call them smoothie manifolds from now own

[fading vale]

delicious

[gritty widget]

Ha ha 🙂 Yum

[sleek thicket]

smoothie manifold, creammanian manifold

[fading vale]

man i have no idea where im gonna go topology wise once i finish hatcher

there are like a billion options

[gritty widget]

creamy manifold

[sleek thicket]

diff top

[fading vale]

and they are all exciting

[cold vine]

We are working with Rotman. Do you guys know spesifically what book is good to complement with that. I have a couple I took from the library but they cover the material in such a different way

[sleek thicket]

Enmanifold yourself

[fading vale]

valid shamrock

[gritty widget]

Oh I used one of Rotman's book (his homological algebra book, not the AT one).

[fading vale]

hatchers exposition is like top tier

[gritty widget]

feels like im the only person who posts in this channel that knows no AT

[fading vale]

bredon covers more material i think (at least not including appendices)

and does more manifold stuff

[sleek thicket]

well it is a topology channel lol

[fading vale]

shamrock maybe i will do lee

[sleek thicket]

I don't know AT very well

I am a big Lee fan!!

[fading vale]

[gritty widget]

He started doing injective modules and "essential" modules early on in the book when I was quite new to algebra in sophomore and I was confused for a bit.

[sleek thicket]

Im gonna be a beta tester for the hot 5th book by Lee

[fading vale]

wait thats so cool!!

[gritty widget]

what's the 4th

[sleek thicket]

Bundle stuff

@gritty widget its Euclidean geometry

[gritty widget]

By that time I had not seen injective modules in the "wild" so-to-speak. My first encounter with injective things in the wild were..... in Ch 3 of Hartshorne when one builds injective sheaves...

[sleek thicket]

@fading vale yeah we're the unreleased textbook for our bundles course with him next quarter :DDD

[fading vale]

oh does lee teach at UW??

man im so jealous of uni students

[sleek thicket]

@gritty widget https://sites.math.washington.edu/~lee/Books/AG/

[fading vale]

you guys get to talk to so many cool people

[sleek thicket]

yeah lol I namedrop him constantly

He is a cool dude

He's retiring this year though

[fading vale]

aw

[sleek thicket]

Well he might actually be retired this year

I'm not sure

He's still teaching infrequently I think

But it's either at the start or end of this year

[fading vale]

but yeah manifold stuff seems really neat so i definitely wanna do that at some point

but also i need to finish AM

[sleek thicket]

I will probably still be able to get a rec letter from him which is good

[fading vale]

...and rudin...

[sleek thicket]

AM is good

[fading vale]

: |

[sleek thicket]

Rudin is also good

[fading vale]

AM is very fun

i do not have a lot of self discipline so i have been avoiding learning basic analysis by jumping around algebra and topology land

but realistically that is a Bad Idea

[sleek thicket]

Yeah I mean

I kind of did that last year

But to a less extreme degree

[fading vale]

i feel like

idk

[sleek thicket]

self discipline is hard

[fading vale]

analysis is prolly cool and i just have to power thru the very early bits of rudin

[sleek thicket]

Analysis is very cool

[gritty widget]

AM has some excellent exercises

[fading vale]

and i will like it

[sleek thicket]

And it would be sad if you never saw the really cool stuff

[fading vale]

i will force myself to do rudin after hatcher and AM lol

yeah the AM exercises are great

TTerra : (

dont chmonkey me!!!

[gritty widget]

[sleek thicket]

im chmonkey reacting because he did Hartshorne before munkres

[fading vale]

i never did munkres

[sleek thicket]

he took a scheme course before an intro topology course lol

[fading vale]

i did ch 1 of bredon

based

[sleek thicket]

Okay well whatever

okay ty for the conversation moth I need to do my French homework now

Got a midterm on Thursday

[fading vale]

good luck sham : )

i have tmrw off

so im chilling

[gritty widget]

your next assignment is to read all of lee's trilogy. exercises due to my dms tomorrow

[fading vale]

monkaS

[gritty widget]

ultra's gone so im taking his place

[fading vale]

b-but i am too busy for mathematics until december

i hate the mit primes app its so mean

[tough imp]

I did do Hartshorne before munkres hurb

[fading vale]

i just want to learn cohomology instead i am being cyberbullied by probability memes

[tough imp]

thanks to... who again?

[sleek thicket]

uhh

Sorry I gtg

French

[fading vale]

shamrock go study > : (

[sleek thicket]

And shame

Yes

I have a good reason not to answer chmonkey's question

[fading vale]

you and chmonkey can bully each other tmrw

[cold vine]

Just browsing through Hatcher here. This book seems amazing. So much explanation and clarity.

[fading vale]

yeah its really good for intuition

the downside is that its kinda idk fluffy and it doesnt necessarily cover as much as bredon but i think as a supplement to a course its very solid

floaty is such a cute emote

[cold vine]

Yapp seems great on the side of a main book. Ill def read up on stuff from here when something isnt clear.

[sleek thicket]

The "CW" in CW-complex stands for Constantine Whitehead, because the person who defined them was named John Henry Constantine Whitehead

[fading vale]

thank u sham

cool shamrock facts

[gritty widget]

ty sham

[sleek thicket]

Whitehead died from an asymptomatic heart attack at 55 :C

This one is true but sad

[fading vale]

i get pinged every time any msg is sent in this channel

[sleek thicket]

Nice

hi moth

[fading vale]

hi shamrock

[gritty widget]

hi sloth

[fading vale]

hi TTerra

[sleek thicket]

hi ttera

Moth do you want to hear a cool riemannian geometry fact?

[gritty widget]

i do

[sleek thicket]

okay well that's nice thank u for ur input

[fading vale]

maybe

[gritty widget]

[sleek thicket]

curvature makes no goddamn sense

[fading vale]

shamrock story time

[sleek thicket]

Fun fact :)

[fading vale]

also shamrock i want u to see this video https://twitter.com/i/status/1324149180139524100 i cannot stop thinking about it

j🦃ey (@joeba_fettt)

Gays walking into their conservative relatives’ house for Thanksgiving after Biden wins https://t.co/go2W1SWNim

Retweets

1022

Likes

7058

▶ Play video

Twitter · Nov 5, 2020, 12:39 AM

[sleek thicket]

Shamrock story time was secretly shamrock whining time

[fading vale]

u do not have to keep it a secret

[sleek thicket]

I love that video

It's great

What is that from??

[fading vale]

degrassi but the music is edited in

[sleek thicket]

I figured lol

"She has a crush on him and he's pretending to be gay so she'll stop liking him."

King shit imo

[fading vale]

please

i do not need to pretend

this strat works automatically for me

passive buff

[sleek thicket]

I have some important tweets to share with you

https://twitter.com/grassmannian/status/1325965817431748610?s=19

Brendan (@grassmannian)

Biden will be the first twink president

Twitter · Nov 10, 2020, 12:57 AM

https://twitter.com/grassmannian/status/1326096178736295937?s=19

Brendan (@grassmannian)

Biden will be the second twink president

Twitter · Nov 10, 2020, 9:35 AM

[fading vale]

followed

but also ur so wrong

[sleek thicket]

https://twitter.com/grassmannian/status/1326398070112874503?s=19

Brendan (@grassmannian)

Biden will be the third twink president

Twitter · Nov 11, 2020, 5:35 AM

[fading vale]

like half of us presidents were twinks

[sleek thicket]

lol

[fading vale]

keep counting!!

[sleek thicket]

I will!!!

[fading vale]

look at this video

[sleek thicket]

My tweets are all implicitly in a disjunctiom

Okay except

[fading vale]

if you tell me you play league i will leave this server

[sleek thicket]

I have met gay gamers

No lol

I'm just saying

They're fucking insufferable too

my friend group is very gay and big into tabletop and good lord

[fading vale]

yes but its more funny if we make fairly universal personality flaws revolve around heterosexuality to cope with homophobia

[sleek thicket]

counterpoint: yes all gamers

AGAB

[fading vale]

[sleek thicket]

Err wait that's assigned gamer at birth isn't it

anyways

Topology

K(Gamer, 1)

[sleek thicket]

@gritty widget did you like my riemannian geometry fact?

[gritty widget]

yes

[fading vale]

shaaaaamrock

[gritty widget]

curvature is difficult

[fading vale]

what does lee cover

and what comes next

[sleek thicket]

umm

Idk

It covers a lot

[fading vale]

woke

is all of it worht doing

[sleek thicket]

I didn't do all of it

I didn't do the last two chapters

But I liked all the material that was in there

And I intend to learn about symplectic manifolds at some point

[fading vale]

cool

how long did it take u to cover this material

[sleek thicket]

Like half a year

[fading vale]

woke

[sleek thicket]

2 quarters

[fading vale]

h m mmmm

maybe once i do rudin i will do this

ok now i am going to sleep cuz its 3 am

[sleek thicket]

night

[fading vale]

good night

[elder yew]

Anyone familiar with tangles?

What the hell is a tangle intuitively

[sleek thicket]

Oh lol I avoided understanding these this summer

It was right at the edge of the stuff I was doing and I needed to understand it if I wanted to fix my project, so instead I didn't and ran out the clock

[elder yew]

haha

I'm updating a previous project to incorporate more advanced techniques

and stuffing it into my PhD apps, so I'm trying to get through this paper

and summarize the techniques

[sleek thicket]

nice nice nice

[elder yew]

https://arxiv.org/pdf/1705.08490.pdf

This is what I'm reading

There's like a page and a half of definitions

[sleek thicket]

Oh wow that's rough

I just took a look

[elder yew]

I'm going through all of them right now

I think I understand most of them, but it's like

why do you write it this way

[sleek thicket]

if it makes you feel any better, for the short period of time where I understood anything about tangles it was all super duper categorical

So I wouldn't have been able to help you very much anyways :P

[elder yew]

It's fine. I've heard the referee comments on my profs papers

where the ref is like

"This is all very rigorous, and to the point - but this is hard to read"

[gentle osprey]

Divya Ranjan:
Compile Error! Click the reaction for details. (You may edit your message)

[sleek thicket]

could you be a little more specific lol

[feral dragon]

well you can categorize me as a Bourbakian

Divya Ranjan:
Compile Error! Click the :errors: reaction for details. (You may edit your message)
@gentle osprey admins delete this one, isn't required

[sweet wing]

bourbaki

[feral dragon]

I need to fix the LaTeX and would send again this is the theorem to prove though https://cdn.discordapp.com/attachments/762590388858126358/775748802299691068/2020-11-10_21h18_27.png

[gentle osprey]

Divya Ranjan:

Divya Ranjan:

Divya Ranjan:

[feral dragon]

Okay sorry for messing up this place but this was just a damn thing, i'mma stop learning Differential Geometry if this proof turns out to be flawed

[gritty widget]

yeah just use the chain rule to show that the differential of the chart is an isomorphism. this works i guess, but it's really long

ill more carefully read it later

also i found the physicist using a and b for indices

[exotic badger]

can someone help me out? i dont seem to understand how this equivalence relation ~ is defined

[cursive glade]

@exotic badger You're then left with a set of equivalence classes X/A = {[a], [x1_], [x_2], ....}, where [a] represents all the points in A (they all go to one point in X/A since they're equivalent) and the [x_n] are just regular points in X, they're equivalence classes with 1 element.

For instance i can take the interval X = [0, 2xpi] with the equivalence relation 0 ~ 2xpi, then X/~ is the unit circle, with all points except 0 and 2pi being themselves and 0 and 2pi being the same point

[exotic badger]

im workign thorugh your answer. thanks for the help!

[cursive glade]

These are called quotient spaces, im sure you can find a better explanation online @exotic badger

[exotic badger]

is this here the same idea?

i read that u basically "glue" them together?

found the pic on wikipedia

[fading vale]

yes

@exotic badger have you done quotient groups before

[exotic badger]

thanks! yes i had that last year in an algebra course

[fading vale]

so think about how in Z/2Z

{0, 2, 4, ...} all get identified to a point

imagine plotting this on a number line

and then basically collapsing them all to a point

[gritty widget]

now tell me what R / Z is

[fading vale]

it is the same idea when working with a topological space

[exotic badger]

and this point is Z/2Z?

[fading vale]

does this image help?

im trying to give the intuition behind why quotienting corresponds to collapsing things in a space

quotienting by Z2 glues all the even integers together into one point, and then glues all the odd integers together into one point

[exotic badger]

yes it helps a bit

[fading vale]

the S^1 case is in fact the same idea yea

just like how for Z/2Z we can identify all the even points together and identify all the odd points together

[exotic badger]

so Z/2Z = { 0 + 2Z , 1 + 2Z}?

[fading vale]

w/ the interval we can identify {0, 1} together to basically glue the ends to each other

yeah cone

fair warning topological quotients and group quotients do NOT always agree

so for example we can consider Z as a group and quotient by 2Z to get cosets

but in the topological case there is no additive structure right

so we dont get cosets

in the case of X/A we just collapse A because there are no group operations to preserve

[exotic badger]

this idenitfy process is basically the equivlance relation ~?

[fading vale]

yes!

: )

the analogy to groups works pretty well mostly you just have to keep in mind that in the topological case our quotients are really much simpler

[cursive glade]

Z/2Z as just spaces is just a 2 two element space. Its doesn't have to be {0 + 2Z ...

[exotic badger]

ok im slowly making progress, thanks for all of your help

[fading vale]

its a little confusing at first but you'll get used to it lol

[exotic badger]

its defintely the hardest math course since i started 2 years ago lol

[gritty widget]

quotient spaces

i didn't really "get" them until like a full semester after taking topology

much fun

[sleek thicket]

you just glue things together

And then everything's terrible

And the topology makes no sense

ez

[fading vale]

idk why it started making sense to me

i think the problem is that

the way point set is taught does not really convey much geometric intuition tbh

[sleek thicket]

I am still not convinced that topological spaces are actually geometric

Like I think of them as a general foundation for building up things which are somehow geometric

[fading vale]

kinda agree

[sleek thicket]

Like manifolds or cw complexes or schemes

[fading vale]

ya

[sleek thicket]

But you need extra data

Yee

[fading vale]

point set is just a weird field

and a weird subject

[gritty widget]

just slap a riemannian metric on your manifold and you have geometry

[fading vale]

ew metrics

[tough imp]

I am unconvinced algebraic geometry is geometry

[fading vale]

lame!

[sleek thicket]

No!!

Moth riemannian metrics are based

[fading vale]

no distance only

loop

i remain unconvinced

[sleek thicket]

smh

[fading vale]

[gritty widget]

imagine not even having TM and T*M be isomorphic

cringe!

[sleek thicket]

I mean they are always isomorphic

[gritty widget]

ree

yes

[sleek thicket]

Just pick a riemannian metric :^)

[gritty widget]

you know what i meant to say

[sleek thicket]

Yeah lol

[gritty widget]

[sleek thicket]

is this new memes

Am I missing out

[fading vale]

catThhhhh

[gritty widget]

[sleek thicket]

I'm being bullied (presumably)

[fading vale]

itso k shamrock you can convince me its cool in like ummmm

sometime next year

[sleek thicket]

Yes

[gritty widget]

im not bullying you shamrock

[fading vale]

does lee need material from rudin

or can i do them side by side

[sleek thicket]

Everyone always bullies the shamrock

[fading vale]

wtf

ive never bullied shamrock

[sleek thicket]

How's your multivariate calculus?

Like

Inverse/implicit fonction theorems

Are the workhorse of diff top

[fading vale]

nonexistence pepelaugh

[sleek thicket]

Then yes you need Rudin

[fading vale]

how much of lee can i get thru without doing it

bc if i do them side by side

[sleek thicket]

Like a chapter maybe

If that

[fading vale]

oh sad

ok then i will just finish rudin first

[gritty widget]

doing partitions of unity without learning series stuff

[fading vale]

STOP

dont mention partitions of unity in my presence

im still traumatized from bredons treatment of them

[sleek thicket]

Um

You're gonna have a bad time

With Lee

Partitions of unity are like

Breathing

Or differentiating

[gritty widget]

excuse me

[sleek thicket]

You just do it

All the time

[gritty widget]

partitions of unity are extremely useful

[sleek thicket]

Forever

Okay but to be fair

I didn't understand them until I read ISM

[gritty widget]

imagine not being able to integrate on manifolds

[sleek thicket]

But you must release your fear

Also moth like, the basic stuff about tangent vectors involves some Taylor series stuff

[sleek thicket]

In multiple variables

[fading vale]

if you say this shit isnt hideous you're lying

[sleek thicket]

Yeah that's how it's done in ITM lol

But you never think about the proof

You just use them

[fading vale]

[sleek thicket]

Like

They're a useful thing to know exists

[tough imp]

Lol

Imagine thinking about the actual construction

[sleek thicket]

I don't even remember what paracompact means most of the time, I just think of it as "the thing you need for partitions of unity"

[fading vale]

bredons organization sucks ass because he does this shit and then doesnt use it at all for ther est of the chapter

[sleek thicket]

Oof

[fading vale]

its really dumb

he should have just put it in when it was actually needed

[sleek thicket]

anyways moth, you agree that riemannian metrics are good right? The way you prove that every manifold admits a riemannian metric is by doing it in coordinates and using a partition of unity

To integrate you pick a cover by charts/partition of unity and integrate over the cover elements

[tough imp]

Wdym every manifolds has a riemann Ian metric

[sleek thicket]

They're based okay

[tough imp]

Don’t u have to be smooth

[gritty widget]

all manifolds are smooth

[sleek thicket]

By manifold I mean smooth manifold

[tough imp]

To even be a riemannian manifold.

Wut

Smh sham

[fading vale]

i have never agreed to such a thing

[tough imp]

Yeah you did

[fading vale]

wtf

slander

[tough imp]

Sloth has dementia

[sleek thicket]

Yeah chmonkey? What do you mean when you say ring again?

[tough imp]

Yeah but noncommutative rings actually don’t exist

[sleek thicket]

*me interrupting chmonkey doing AG* doesn't that ring have to be commutative?

[tough imp]

Also when I say ideal I mean prime

[sleek thicket]

Yeah same for non smooth manifolds

[tough imp]

Okay shamrock

Fair

[sleek thicket]

It's a little different because a smooth manifold is a structure and not a property

[fading vale]

lol one of the problems on our pset was "find a finite noncommutative ring" a while ago

i think

[sleek thicket]

Ooh

Matrices

[fading vale]

yea

[sleek thicket]

Done

[fading vale]

thats what i did

its not hard lmao

[sleek thicket]

Wait better one

Which is based

[fading vale]

i think it wasnt just finite

[sleek thicket]

Group ring over a finite field

[fading vale]

there must have been some other condition

ill find

[sleek thicket]

Partitions of unity are like the chad gluing lemma

imo

[fading vale]

nvm there were no extra conditions

it was literally just

"find a finite noncommutative ring"

lol

[sleek thicket]

tfw

anyways sloth, I would not do ISM without seeing multivariate calc/analysis at a lower level

[fading vale]

yes i will just

rudin

[sleek thicket]

Like I did implicit/inverse function theorem and not super rigorous stokes/greens theorem and I was fine

The little bit about measure zero sets can be picked up on the fly

[gritty widget]

just do spivak calc on manifolds over a weekend

[fading vale]

chad yes

[fading vale]

why would i know everything

hurb

[sleek thicket]

I have ananalysis exam on Friday

I was worried until I saw this message

ty for the advice

[gritty widget]

bananalysis

🍌

[sleek thicket]

a nanalysis

I think that's what happened to orange

Like

It used to be norange

And then people said a norange

And it morphed into an orange

[gritty widget]

[fading vale]

lol nerds

what class is it for slim

yuck

[sleek thicket]

sick

[fading vale]

elementary probability is not

and thats what im being forced to do

it sucks

hurb

[sleek thicket]

My exam is for measure theory

[fading vale]

primes pset bad

very very very bad

[gritty widget]

start each possible solution with "assume CH" or "assume not CH"

[sleek thicket]

I'm hoping one problem will be "apply DCT"

[stable lance]

Does anyone know how to prove this, or a reference to look up? Here X is a manifold, A,B closed and H_q is the singular homology

[fading vale]

is the bar reduced cohomology

[stable lance]

yes
that one I know . My question was why the singular homology preserved by taking the limit

[fading vale]

https://math.stackexchange.com/questions/401302/does-taking-the-direct-limit-of-chain-complexes-commute-with-taking-homology

@stable lance Does this post help?

https://math.stackexchange.com/questions/121122/why-do-direct-limits-preserve-exactness here is another post in the case of modules

hopefully it helps

[stable lance]

unfortunately not
because the way homology is defined on those links don't take the topology on account, they're more algebraic exercises. It's quite different here

but thank you for searching those

[fading vale]

uh

hm

im not totally sure what you mean by that

oh i see

yeah sorry

¯_(ツ)_/¯

[stable lance]

it's ok 😂

I think we have to use those cobounded neighborhoods somewhow, and maybe the fact that X is a manifold

[fading vale]

gl

[stable lance]

ty

[gritty widget]

@stable lance : $A$ is locally compact Hausdorff, so you're going to have $B = \cap_N N$ as $N$ ranges over all closed and cobounded neighborhoods in $A$ (this is just another way of saying that in a LCH space an open set is a union of precompact open neighborhoods). So you have $X - B = \cup_N (X - N)$ (same index range for $N$). Now the point is that a cycle in $\Delta_q(X - B, X - A)$ is represented by sums of continuous functions (with coefficients in $G$) from a compact simplex $\Delta_q$, say $\sigma: \Delta_q \to X - B = \cup_N (X - N)$. $\Delta_q \subset \cup_N \sigma^{-1}(X - N)$ then, so by compactness of the simplex the image of $\sigma$ must land in some single $X - N$. That is to say, the singlar complex $\Delta_q(X - B, X-A)$ is actually equal to the directed limit of $\Delta_q(X - N, X - A)$ over such $N$.

[gentle osprey]

ℓ-adic:

[gritty widget]

The rest follows by the "algebraic" exercise.

(say by a G-linear combination of a bunch of those sigmas, but finite in number, so you can still take a single N closed and cobounded in A with the images of these sigmas representing a cycle in Delta_q(X-B,X-A) all landing in X - N)

[gritty widget]

btw A is LCH and A - B is open in A, so A - B = \cup_{V \in K} V where each V in K is open and precompact in A. So then B = \cap_{V in K} (A - V), so taking N = A - V (for V in K) does the job : A - V is closed in A hence closed in X and is cobounded. It's also a neighborhood of B becaause A - N <= closure of (A - N) <= A - B, so N >= A - closure of (A - N) >= B, and A - closure of (A - N) is open in A.

[meager python]

F is a qcoh over a scheme X. Is H^i(X, F) all the same depending on how you look at the global sections functor as over Ab, QCoh or O_X-mod?

[little hemlock]

Would anyone happen to have a hint on how to show that the following is a metric on $\bR^2$?
$$\rho(x,y) = \frac{|x-y|_2}{(1 + |x|_2^2)^{1/2}(1 + |y|_2^2)^{1/2}} $$
I have a hint that says that the bijection from $\bR^2$ to the open unit disk given by $x \mapsto x/(1 + |x|_2^2)^{1/2}$ might help, but i've staring at this forever, and im not seeing how to proceed

[gentle osprey]

kxrider:

[fading vale]

jesus

[bleak helm]

Is your hint: Ask Jesus? Lol

[little hemlock]

believe me, ive already tried

[fading vale]

no this is just gross

[woeful oasis]

This is much like the Lorentz metric on $\bR^2$: given two metrics on the two 1-dimensional subspaces, you can form a metric from either by Pythagorean theorem (this would be putting the two metrics together in a "straight" way) and you would end up with the square root of the sum of squares of the metrics.

However, in this problem, you're investigating more of a "hyperbolic" direct sum of metric spaces.

[gentle osprey]

Apopheniac:

[woeful oasis]

Also, the only tricky part is proving the triangle inequality.

I think it helps to think of it as some combination of metrics, because then you can keep track of what's what.

[uncut surge]

Aha, I see now where it's coming from. But it might be more helpful as a hint to namedrop "stereographic projection".

Hint: Consider the map $F: \mathbb{R}^2 \to \mathbb{R}^3, x \mapsto \left( \frac{x}{1 + |x|_2^2}, \frac{1}
{1 + |x|2^2} \right)$ and show that $|F(x) - F(y)|{\mathbb{R}^3} = d(x,y)$.

[gentle osprey]

Lartomato:

[uncut surge]

uuh, on closer inspection, i think the last argument of F needs to be 1/(1 + |x|2^2)

[stable lance]

@gritty widget thank you very much.
So we have $\bigcup_N X-N = X-B$ and for each $N$ there's the inclusion $i_N:C(X-N)\to C(X-B)$ (here $C(U)$ is the set of the singular simplexes in $U$). A simplex $\sigma\in X-B$ is a simplex in $\bigcup_N (X-N)$ so it's a simplex in some $C(X-N')$ because compactness. Therefore the map $i:lim_N C(X-N)\to C(X-B)$ is surjective. And $i$ is injective because each $i_N$ is injective. $i$ also commutes with the boundary $\partial$ operator, so $i_*:lim_N H_q(X-N)\to H_q(X-B)$ is an isomorphism.
It looks about right to me

[gentle osprey]

gb214:

[fading vale]

[uncut surge]

nvm i'm fine and this is the wrong channel anyway

[ivory dragon]

[karmic smelt]

Actually I already asked the question here, but I still did not understand

[ivory dragon]

what dont you understand about the answer?

[gritty widget]

I'm a little confused at what some simple examples of sets would look like, a discrete space with a metric? What would that look like?

[gentle osprey]

mathacka:

[fading vale]

uh

what do you mean "examples of sets"

in the topology?

[gritty widget]

example of a discrete space with a metric in set notation

what does a simple one look like as a set?

the topology as a set

[fading vale]

do you mean

the discrete metric

"a discrete space" is just a set with the discrete topology on it

[gritty widget]

yeah, I'm trying to see a simple example of one as a set

in the integers

[fading vale]

ok idk why you keep saying "as a set" here lol theres nothing else the topology can be

[gritty widget]

ok

[fading vale]

but you are probably looking for the discrete metric

given by d(x, y) = 1 if x neq y and 0 if x = y

the resulting topology is always just the topology where all subsets are open

the discrete topology

[gritty widget]

so if have elements {0} and {1} then you have d({0},{1}) = 1

the the set looks like {{0}, {1}, (0, 1, 1)}?

[fading vale]

uhhh the metric acts on elements of the space not on subsets

🤔

what do you mean (0, 1, 1)

[gritty widget]

yes

the ordered pair that is a subset of X cross X?

d(0,1) -> Z?

[fading vale]

there is no ordered pair involved here

i dont understand what you mean lol

the set of open sets w/ the discrete topology on Z

is just P(Z)