point-set-topology

[uncut surge]

oooh i love it when snappy slogans are true

tyvm

[obtuse meteor]

https://estrogen.fun/i/vbt3.png

[lost estuary]

Hello, I have a question for you my friends

Is the projection onto the unit sphere infinitely differentiable ?

the $p(x) = \underset{a \in S^{n-1}}{\text{argmin}}|x - a|$ projection

[gentle osprey]

Jean-Jérôme

[lost estuary]

if you omit the origin of course

[obtuse meteor]

definitely

oh wait one sec

what's argmin ?

not familiar with this type of projection

familiar more often with x |-> x/||x||

[feral bay]

argument for which the function is minimal would make sense?

[quartz edge]

it's the same thing

as the normalization projection

& yea its smooth

[obtuse meteor]

yeah the normalization projection is smooth simply because besides a square root it's a rational function with nonzero denominator

and sqrt is smooth when you're > 0 so

[lost estuary]

oh god you are right

but now my more serious question

if now I have a $\beta$ smooth manifold, is the projection mentioned above still at least $\beta$ smooth

[gentle osprey]

Jean-Jérôme

[lost estuary]

ok by $\beta$ smooth I mean $\beta$ times differentiable

[gentle osprey]

Jean-Jérôme

[lost estuary]

sorry I am pretty unfamiliar with differential geometry

[gritty widget]

what does the projection you wrote even mean on an arbitrary manifold

do you have a notion of distance in mind?

are you projecting to some submanifold?

(give context)

[lost estuary]

ok

[lost estuary]

I have a function $f \colon [-1, 1]^d \mapsto R$ that is $\beta$ times differentiable, say $\max_{[-1, 1]} f(x) =1$, that this maximum is attained at $x=0$ and suppose that $f$ vanished on the border of the hypercube. Now I'd like to take a function $\lambda_\alpha(x) = min_{R_+} {\lambda, f(\lambda x) = \alpha} $ and now I would like to consider the manifold defined by $M = {x\lambda_\alpha(x), x \in [-1, 1]}$ for some $\alpha \in (0, 1)$ and I would like to use the projection defined above for some stuff, but I would like it to be $\beta$ differentiable

[gentle osprey]

Jean-Jérôme

[lost estuary]

The distance here is simply a norm

Do I have to explicit what is $argmin$ ?

[gentle osprey]

Jean-Jérôme

[lost estuary]

Ok I rewrite it here because something was wrong before

I have a function $f \colon [-1, 1]^d \mapsto R$ that is $\beta$ times differentiable, say $\max{[-1, 1]} f(x) =1$, that this maximum is attained at $x=0$ and suppose that $f$ vanished on the border of the hypercube. Now I'd like to take a function $\lambda\alpha(x) = min_{R+} {\lambda, f(\lambda x) = \alpha} $ and now I would like to consider the manifold defined by $M = {x\lambda_\alpha(x), x \in [-1, 1]^d}$ for some $\alpha \in (0, 1)$ and I would like to use the projection defined above for some stuff, but I would like it to be $\beta$ differentiable

[gentle osprey]

Jean-Jérôme

[quartz edge]

it's at least beta differentiable

[gentle osprey]

Buncho Grist

[quartz edge]

like say you're projecting f(x) = |x|+1 (i.e. (x, |x|+1)) onto the unit circle

f'(0) doesn't exist but dp(f(x))/dx does

and in fact p(x) is C^infty on f

[rugged swan]

Hi, I'm actually reading the divisor section of Harthshorne's book and I'm confused, why does the local ring O_eta,X - of a prime divisor Y where eta is its generic point - is a discrete valuation ring ?

[tough imp]

You should be using the thing that like a umm, dim 1 integrally closed domain local Noetherian ring is a DVR

Or something like that

Just look up the equivalent conditions for a DVR and IIRC it satisfies one of them pretty obviously, it definitely uses the integrally closed domain thing I think

@rugged swan

Oh, also you need to use the following fact I think, if eta is the generic point of a codim n irreducible closed set, then dim O_X,eta = n

This tells you that O_X,eta is dimension 1, which then is integrally closed

[tough imp]

For this fact, I remember proving this for some exercise in the past, it was probably like II.3.20??

[lost estuary]

Thanks, is there a way to prove it without doing any calculations ? like is there some results that I could combine together ?

[lost estuary]

yes thanks 🙂

[rugged swan]

ah it's bc it's dim 1 and notherian hence it's principal

[tough imp]

Yeah

[rugged swan]

wait

[tough imp]

Anyway, I think I have a proof of the thing about the codim and the dimension of the local ring

[rugged swan]

you're saying that dim O_X,eta = n and then 1 wtf xD

[tough imp]

Sorry, n was for the general result

If {eta}-closure is codim n

The local ring is dim n

But you’re looking at codim 1 stuff for divisors

So in this case n = 1

Also I have a proof for that fact if you want one

[rugged swan]

btw I still don't understand how you did all the exercices of the Hartshorne lmao, its explanation are dark af and without a good understanding of things I can't do any of the exercises

[tough imp]

Lots of hard work

And a lot of frustration

[rugged swan]

I've just figured out myself why a divisor was named a divisor but wtf, why it's not written anywhere that it generalize the valuations on Z

[tough imp]

Yeah, idk lol

I think it briefly mentions this a bit later sort of?

Like it talks about how the class group is the same as the one in NT

If you have a UFD or whatever the fuck, I don’t remember haha

[rugged swan]

yes

it seems legit

[tough imp]

Anyway, to prove the thing about codim and dimension of local ring:
The dimension of a space is the supremum of dimensions of its open sets in an open cover

So this lets you WLOG to Z being an irreducible closed set in an affine by intersecting with an affine cover

Then it just looks like V(p) for a prime ideal p, and codim V(p) is the same as ht p = dim A_p

And A_p will be the local ring at p

(p is the same thing as eta, it just looks like a prime ideal now cuz you’re in an affine)

Oh fuck, wait hnggggggg

We’re dealing with codim, and codim and dim aren’t related nicely unless you’re in a variety

[rugged swan]

yeah but here we are in a notherian separated scheme

[tough imp]

I think it’s true that codim(Z,X) is the same as inf{U\cap Z, U) for you U in an open cover

[tough imp]

This isn’t enough 😟

You need to be ft integral over a field to get it

[rugged swan]

yes ok

[tough imp]

For example Spec R[x] for a DVR R won’t have codim(Z,X( + dim Z equal to dim X for all Z

[rugged swan]

in hartshorne it's defined as the supremum of lenght of such chains Z<=Z1<=...<=Zn-1<=X

somth like that

[tough imp]

Yup

Anyway, I think what I typed about the inf is true

I think on some affine open, you need to have the right codim

And then you know the result on an affine by the argument I gave

[rugged swan]

cursed af

yeah I think I will manage to do it myself thx, you gave me the essential argument

[tough imp]

Okay, I think I have it. Actually the codimension is preserved upon intersecting

This is because there’s an order preserving bijection between irreducible closed sets of U < X, and irreducible closed sets of X which touch U

So if Z\cap U is nonempty, then given a chain like Z < Z_1 <... < Z_n < X maximal (this says Z is codim n) we know that all Z_i intersect U, so we get this strictly increasing chain
Z\cap U < Z_1 \cap U <... < Z_n\cap U

So codim(Z\cap U, U) >= codim(Z,X) but conversely a chain inside of U gives rise to the same length chain in X

By taking closures

So codim(Z,X) >= codim(Z\cap U, U)

[lost estuary]

Hum sorry to interrupt was my question in the wrong topic ? if yes I apologize for that 🙂

[tough imp]

At first it looks like analysis maybe, but I think it was about smooth manifolds? If it was about manifolds then this is the right place.

This channel houses topology, differential geometry, differential topology, algebraic topology, and algebraic geometry

[lost estuary]

yeah I kinda hesitated on this one, I think it's about smooth manifolds

[tough imp]

So sometimes things look wildly different

If it’s at that cusp I think people don’t really mind. If it was about a smooth manifold then turns into just real analysis I think ppl don’t care

[lost estuary]

Thanks for clarifying that, I'm a probability/statistics guy

[tough imp]

It’s just if someone says “how can I show this is C^infinity” or something ppl say you should maybe switch channels

I find myself asking the same questions sometimes since AG is sometimes practically pure algebra :)

[lost estuary]

Oh okok but so if I have a function mapping onto a manifold it's kinda ok to ask whether it is C^k

I guess

[tough imp]

I think as soon as a manifold gets introduced it’s fair game for this channel

[lost estuary]

haha ok

[hard wind]

Not sure what part c) means lol, if someone could clarify I would appreciate it

Mostly the word "carried"

[sharp yoke]

why are these 2 covering spaces for S^1 v S^1 distinct

is the map sending the outer a loops to the outer b loops and the inner b loop to the inner a loop not an isomorphism

yea actually i guess that wouldn't be continuous

actually yes it would ?

idk any help lol

oh it would be continuous but you wouldn't have p_1=p_2f so its not an isomorphism

i think i got it

[sharp yoke]

I'm trying to do question 5 on page 86 of hatcher

Construct a connected graph X and maps $f,g: X \rightarrow X$ such that $fg=id$ but f and g do not induce isomorphisms on $\pi_1$

[gentle osprey]

『Urahairywizard』

[obtuse meteor]

A good way to think about this problem @sharp yoke

is with covering space theory

namely consider S^1 v S^1

this has free group F_2 for pi_1

you can factor F_2 -> F_3 -> F_2 where the composition is an iso

and you can embed F_3 in F_2

and I claim there's a way to induce these by continuous maps

[sharp yoke]

what would be the continuous map between S^1 v S^1 and itself that induces the homomorphism from F_2 -> F_3

[obtuse meteor]

that one's easy

wrap the first loop around the first loop then the second loop then back through the first loop in the opposite order

wrap the second loop to the first loop twice then the second loop then back through the first loop twice

[sharp yoke]

so ur sending the loop a to aba^-1

and the loop b to a^2ba^-2

this might be a stupid question but if ur trying to map F_2 into F_3 should you also have something mapping to the third generator of F_3?

[limpid leaf]

hi topologists what does this double arrow mean

for the projection map

thanks ultra

why does this person find it necessary to write that

this is the proof of tychonoffs btw

[gritty widget]

i've seen it to mean "fiber bundle" as well

[limpid leaf]

i hope it doesnt mean vector bundle

[gritty widget]

im just trying to spook you jesse

[limpid leaf]

this proof is already spooky enough

oh

I thought hookarrow had something to do with exact sequences

not that I know what an exact sequence is

[pseudo crane]

mathematicians use a hook arrow when they r feeling extra stylish

[shut moat]

I thought it was just for inclusions

[sharp yoke]

is it even possible to come up with another function that composes with this function to become the identity?

it seems to me that to "reverse" what you did for the function you defined you'd have to send points in X to at least 2 different other points in the image

[sharp yoke]

The hint i got from the TA said to try to emulate the behavior of g(n)=n+1 with an appropriately chosen infinite graph so i don't think considering S^1 v S^1 was the right approach

[rugged swan]

hi, does anyone have a good book on how to do the parallel between sheaves reasoning and bundle ?

[obtuse meteor]

ah sorry, the problem you stated was very similar to a problem I had, where we were required to have f_* g_* = Id

I think for what you said, it would be good to consider the infinite wedge of circles

[gritty widget]

Can a 2D manifold embedded in R^4 disconnect it?

[gritty widget]

Is my question not clear?

[flint cove]

no I find it interesting. I would think no, but have no idea how to prove it.

[gritty widget]

Well, can you prove that a simple curve can't disconnect R^3?

[flint cove]

Not rigorously.
My reasoning would be something like if we look at the embedded curve C locally, we get a diffeomorphism of U sub ℝ³ to ℝ× ℝ² where C\cap U lands in the first component, giving us that the complement U×ℝ²\{0} is at least on this region path-connected
So it's not able to „split up the space locally“
But that might be the completely wrong approach here.

[strong heron]

And what maps are you suggesting?

[strong heron]

Do you mean a translation? Can you explain the hint a bit more?

[gentle osprey]

ayylmao

[fading vale]

this probably belongs in #geometry-and-trigonometry or maybe #multivariable-calculus since you probably have to take an integral to get 2

you are welcome!

[obtuse meteor]

@sharp yoke You can take the maps to be like. Label your circles a_1, a_2, .... Map a_n homeomorphically to a_{n + 1} for the first map, and map a_{n + 1} to a_n as well as a_1 to a_1 for the second map

[thin jewel]

ELI5 what is k-theory?

And is hard to learn

[gritty widget]

In mathematics, K-theory is, roughly speaking, the study of a ring generated by vector bundles over a topological space or scheme. In algebraic topology, it is a cohomology theory known as topological K-theory. In algebra and algebraic geometry, it is referred to as algebraic K-theory. It is also a fundamental tool in the field of operator algebras. It can be seen as the study of certain kinds of invariants of large matrices.[1]

[thin jewel]

👀

Ok but why do we want to use it?

Bonus points if anyone can relate the answer to linear Algebraic Groups

[gritty widget]

In high energy physics, K-theory and in particular twisted K-theory have appeared in Type II string theory where it has been conjectured that they classify D-branes, Ramond–Ramond field strengths and also certain spinors on generalized complex manifolds. In condensed matter physics K-theory has been used to classify topological insulators, superconductors and stable Fermi surfaces. For more details, see K-theory (physics).

[pseudo crane]

what a good writer

you sound so professional

[tight agate]

if you know anything about vector bundles, you should be able to read hatcher's book on it

or if you know some algebra (modules) you could learn it in theory

but it may not make too much sense

[limpid leaf]

Why is "The topology on X is generated by sets of the form V = ..." true?

(this is Tychnoff's Theorem)

I think I'm being really silly here

yep nevermind I figured it out thanks everyone

[honest terrace]

that was fast

[limpid leaf]

the projection map inverse giving a topology for X confused me for some reason

[honest terrace]

that's like literally the definition of the product topology, right ?

[limpid leaf]

Yes

lmao

or it's very related I'm not 100% clear on it

[sharp yoke]

yea i was thinking about this as well

[thin jewel]

At the store

[thin jewel]

Ok thanks - my advisor has a paper that makes heavy use of it and was wondering what I'd need to get through to make sense of it

Lol

[tough imp]

bdobba

[thin jewel]

Yeah?

[tough imp]

bdobba

[tight agate]

depends on the kind of K-theory

[thin jewel]

Equivariant K-theory

Whatever that is

Great thanks

This looks cool

[tight agate]

ultra they could have meant equivariant top k theory

oh ok

some of my homework from last quarter makes more sense now

coflasque tori should be related to coflasque lattices right

yes

[thin jewel]

Yeah i hate it

[tight agate]

bdobba was ultra right

was this the equivariant k-theory that you were looking for?

[thin jewel]

Yeah I think so

[tight agate]

petthecat

[thin jewel]

Well the notes are making more sense than I thought they would so that's good

[tight agate]

noice

[honest terrace]

I mean we call squares "balls", can you do dumber than that ?

slim agrees with himself

[marble socket]

It's weird if you don't. :p

[thin jewel]

Clopen is a pretty stupid word as well though

[honest terrace]

Well epistemic logic is weird

[tight agate]

[honest terrace]

[tight agate]

is there an atiyah-segal completion theorem for the algebraic case

K for klasse

[tight agate]

lmao there is one

https://en.wikipedia.org/wiki/Atiyah–Segal_completion_theorem

is there an atiyah-hirzebruch ss

that's the real question

krothendieck

[honest terrace]

obvsly stands for KKaspersky-antivirus

[tight agate]

hurwitz thm

atiyah singer index thm

that should be a big enough application

also, rep theory

the AHSS specializes to a spectral sequence that lets you compute the rep ring from group cohomology

wanwan

do the top k theory class

[thin jewel]

I feel like I missed out on really understanding homology by not doing Algebraic Topology

There is a class offered but I don't really think it's going to help with my project

Yeah I've been working through Weibel and found it ok

😥

Algebra is basically all I study

Well plus some Algebraic Geometry and combinatorics

[gritty widget]

Oh, why is that?

[tight agate]

what kind of AG

[gritty widget]

(I haven't studied really any alg top)

[thin jewel]

Linear Algebraic Groups

There's some funky interaction between Linear Algebraic Groups and combinatorics

Which is what my project will be on

[gritty widget]

I thought homological algebra is just about arrows

[tight agate]

homalg is a miracle

[thin jewel]

Homological Algebra lets you study hard things like modules without going insane

[tight agate]

homalg = "SES gives you LES"

[thin jewel]

Liebeck

[tight agate]

isnt fulton and harris standard?

[gritty widget]

There should be a book like homology for analytic minded people, I would maybe read that

[tight agate]

miranda

[thin jewel]

Yeah F&H is the gold standard

[tight agate]

huybrechts

[thin jewel]

But it's hard

[tight agate]

any complex geometry book

[thin jewel]

Probably Lie Algebras if you're into analytic things

[tight agate]

(im talking to 8da btw)

[thin jewel]

But finite dim rep theory is fairly chill

[gritty widget]

"miranda homology" didn't give any hits on google

[thin jewel]

Whereas rep theory of lie algebras is much harder imo

[tight agate]

miranda algebraic curves and riemann surfaces

[thin jewel]

Great book

[gritty widget]

Ah

[tight agate]

meh

the book is fine

but it does cech cohomology

[thin jewel]

That's what got me into AG

[tight agate]

Him

who is whom

[thin jewel]

Or just learn Hopf Algebras to get big brained

[tight agate]

you can probably blaze through basic stuff on reps of finite groups in a week

there isnt too much

definition

maschke

characters

that's about it

[thin jewel]

You've seen Hopf Algebras basically everywhere if you've been doing mathematics

[tight agate]

maybe some stuff on splitting would be cool

[thin jewel]

Lol

[tight agate]

bdobba are you an undergrad or a grad student?

[thin jewel]

grad student

[tight agate]

cool

[thin jewel]

you?

[tight agate]

undergrad

no

I do not know a source

dummit?

[thin jewel]

well you know a ton of stuff for an undergrad lol

I feel like I didn't make the most of my undergrad time and should have learnt a bunch more

suffering for it now

[tight agate]

lots of examples

[thin jewel]

Yeah it really is great through

[honest terrace]

[tight agate]

voldemort

[thin jewel]

yeah you're probably right

[honest terrace]

Yeah voldemort is a really great mathematician

everyone knows that

[thin jewel]

is he though

I bet he knows loads of category theory but can't do a double integral

[tight agate]

enough?

[honest terrace]

that's exactly what being a great mathematician means

the things he's telling you about

[tight agate]

all of rep theory ?

[honest terrace]

the same sense

[tight agate]

maybe in a formal sense

but actually?

[thin jewel]

yeah "same" is a loaded term in mathematics

[honest terrace]

nah it always means the same thing

[tight agate]

what's it

the rep?

what is a

by group algebra do you mean k G

[thin jewel]

k[G]?

[tight agate]

yes

basis given by elts of G

[thin jewel]

yeah, and multiplication given by foil expansion

[tight agate]

multiplication given by mult on G

what is a

[thin jewel]

TIL you can have a norm in a group algebra

[tight agate]

there is the augmentation map

which is mult

[thin jewel]

I'm guessing this doesn't work for finite fields (having a norm)

this is a better algebra discussion than the algebra channel

lol

are you a grad student Ultra?

cool, what year?

me too

[tight agate]

this is easy, just read about cool shit

[thin jewel]

Try to prove the theorems yourself

After you've seen them

@limpid vault Do you know what you're going to specialise in yet for your project?

[tight agate]

ultracrank /s

hermitian sex operator

wut

[viral atlas]

[tight agate]

.pin

[thin jewel]

Doxing is not cool

[tight agate]

but boxing is

unless youre an alligator

[thin jewel]

Agreed

[tight agate]

are kangaroos actually good at boxing

[thin jewel]

Probably

They can beat the shit out of you regardless

[tight agate]

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

lmao

muay thai-d the crap out of the other one

https://tenor.com/view/animals-judging-judgemental-kangaroo-gif-4246612

[thin jewel]

It would basically rip your face off

No it would

[tight agate]

age doxxed

[thin jewel]

I'm also old

[viral atlas]

[tight agate]

all of them are slim

[viral atlas]

Ultra didn't say the hexagonal numbers have to be unique

So 6+6+6+6+1+1=26 still works

[tight agate]

are you the only one born on that day

probably not a child

[gritty widget]

hermitian sex operator

[tight agate]

they watched that kangaroo thing

[viral atlas]

Child capable of remembering in 2003

I'd say at least 6 by 2003

[tight agate]

very spiritual

[gritty widget]

i wonder if this author is polish

[tight agate]

spanish

spaniard

espanol

espania

[gritty widget]

This means that the pinal gland remains closed for such bindings and the belly glows. An alike Polish marriage
is one where λ ∼ µ which opens the pinal gland of both when the respective
partners consume it. Such marriages give peace as the third eye sees everything
but they evoque disgust or jealousy with others. So such a Polish marriage is
energetically the least favourable one an really puts an extremely strong binding
energy between both partners. In case both partners are extremely white, they
have a very strong cosy home but outside they need to individualize where the
woman becomes mostly much more black to evoque the impression of neutrality. Another possiblity would be to engage in quantum contracts at that point
which makes them look at other partners. For Poles who value beauty, social
interactions with other couples are a difficult exercise. However, such couples
are individually extremely energetic whereas the opposite Polish marriage are
a strain. Indeed, the opening of the eye and its associated energetic cost is far
outnumbered by the energetic gain due to white-white interactions. The Belgian
couples do have a poor understanding of one and another, this could improve
by looking for a suitable mixture of both the belly and astral gland operator C2.

[tight agate]

lmao

[gritty widget]

Woman, in the same vein, correspond to imaginary
momenta in the male psychic Fourier transformation which reflects itself all the
time into human behaviour.

They are extremists which go with full
force whereas the white male is much more modest. Indeed, white woman destroy everything which stands in their way whereas white men are much better
negotiators

"Indeed, white woman destroy everything which stands in their way "

For this very reason, black men often use white females against
white dominant (but not complete white) males, by concquering their minds for
another goal with money and hereby causing for a spiritual discrepancy breaking the heart. The pure white male cannot be destroyed in this way and literally
slices a white female’s throat if she were to oppose him because that testifies
of evil behaviour.

this is

wow

it just keep delivering

[tight agate]

mods?

[gritty widget]

slim how did u find this

[tight agate]

slim is the author

[gritty widget]

yeah bend over

[tight agate]

equivariant pinal gland theory

someone just disproved RH on vixra

#math-discussion message

on the same day too lmao

[tough imp]

I forget which DVD, but I remember watching some DVD and at the start before the movie or w/e there was an ad for kangaroo Jack

I saw it many times

Speaking of which, why the fuck are there ads on the DVD of a movie I bought?????

[thin jewel]

capitalism

[tough imp]

bdobba

[thin jewel]

hi

why do you keep just typing my name

it's kind of creepy

[tight agate]

bdobba

[bleak helm]

bdobba

[tight agate]

bdanobs

chcreepy

[tough imp]

I’m just doing a play on me saying “Chmonkey”

And bdobba is fun to say, so I like typing it haha

[uncut surge]

what the fuck

[gritty widget]

🙌 🙌 🙌 🙌 🙌 🙌 🙌 #chill message
🙌 🙌 🙌 🙌 🙌 🙌 👏 👏 👏 👏 👏
#point-set-topology IS OFFICIALLY A WEEB CHANNEL @boreal osprey @distant phoenix @strong sandal @nimble pebble @modern osprey 🥂 🥂 🥂 🥂

it's official! 👏

[honest terrace]

YAY

🥂

[bleak helm]

mutes channel

[bleak helm]

Homophobia happened

[uncut surge]

on the one hand maybe it should bother me that the channel for my favourite math stuff is just memes

on the other hand, i love memes

not a fan of the term "homo posting" but i will gladly indulge in some homo posting regardless

[bleak helm]

I don't have a problem with the term, but I will cancel Yohan for it anyway if you'd like

[gritty widget]

homomorphism

This can be the "everything" channel 🤗

[bleak helm]

I don't think anyone was confused about that, but thanks for explaining anyway

[gritty widget]

Is anyone going to the pcmi summer school this year?

[gritty widget]

Reposting an unanswered question: can a 2D manifold embedded in R^4 disconnect it?

[gritty widget]

Do you mean R^4-M is not connected

[gritty widget]

Yes, I mean exactly that, @gritty widget

Like a circle disconnects R2 but not R3. Does something similar happen in higher dimensions?

I thought it was a well-known thing... Hmm...

[long coyote]

Let $[\alpha], [\beta]\in\pi_1(X,x_0,x_1)$. Show that the composition $\Phi([\alpha])^{−1}\comp\Phi([\beta])$ corresponds to a conjugation in $\pi_1(X,x_0) $ and conclude that $\Phi([\beta]) = \Phi([\alpha])\comp c[\gamma]$ for some $[\gamma]\in\pi_1(X,x_0)$. $\Phi :\pi_1(X,x_0,x_1)\to I(x_0,x_1)$. Here $I(x0,x1)$ stands for the collection of groupisomorphisms from $\pi_1(X,x0)$ onto $\pi_1(X,x1)$

[gentle osprey]

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

[long coyote]

how do you present $\Phi([\alpha])$, is that a curve

[gentle osprey]

亜城木 夢叶

[uncut surge]

@gritty widget Apparently there's a standard result in Differential Topology, called the transversality theorem, from which you can deduce things like that, see Theorem 1.1.3 and Theorem 1.1.4 in these lecture notes: https://ivv5hpp.uni-muenster.de/u/jeber_02/skripten/bordism-skript.pdf

At least if you're familiar with homotopy groups, this will answer your question (note that in the inequalities at the end, there should be an i rather than a j)

Cus then, choose M = R^4, N = some two-dimensional manifold, and g the embedding of N in M. Then, if i < 4 - 2 - 1 = 1, the map of homotopy groups are isomorphic by the theorem

So, when i = 0 this tells you that M \ N is just as connected as M

This also neatly shows that this generalizes: This holds for any m-manifold in which you embed an m-2 manifold

(i'm being a bit more detailled than necessary because i've not learnt about this statement before so i think this is really cool)

[fading vale]

im not sure this holds when i = 0

or rather

[gentle osprey]

Moth In Shambles

Moth In Shambles

[fading vale]

@uncut surge

[uncut surge]

oh nooo i forgot how homotopy groups work if they depend on a basepoint

but wait, isn't \pi_0(M,x) basically just the space of path components, pretty much independently of x

since it's the set of homotopy classes of maps S^0 to M, mapping one of the points of S^0 to x; so basically one homotopy class for every path component, depending on which path component the second point maps to, so I think the reasoning above holds

@fading vale uwu owo

[fading vale]

i think ur right

do u have a link to the proof of this thm

[uncut surge]

not immediately but i'll google around

[gritty widget]

A bullshit attempt at the same problem: write Rn=MUAUB, A and B are the connected components of Rn minus M. A and B are probably open. Then A is an n dimensional manifold, and boundary of A is probably equal to M, and it is probably n-1 dimensional

[fading vale]

catthink

[tough imp]

probably, probably, probablyf

I'm optimistic

[empty grove]

$\pi_0(X,x)$ is the set of path components of $X$ with the component of $x$ as the base point, so $\pi_0$ becomes a functor from Top* to Set*

[gentle osprey]

Moldilocks

[empty grove]

The functor from Top to Set which gives you path components without base points also carries the same name, but if there is a base point in the topological space then there will be a base point in the set of components too

[gentle osprey]

bdobba

bdobba

[gritty widget]

Does anyone know of any good upcoming algebraic topology summer schools/workshops/conferences

[thin jewel]

ofc

I'm an idiot

I just take $ e \times Y$ and $e \times X$

[gentle osprey]

bdobba

[thin jewel]

I've got to say I'm struggling more with Linear Algebraic Groups than anything I've studied in a while

I think not having a background in Lie Algebras is one factor

yeah

Like I want something that just deals with them using maps between their (Hopf) Algebras the same way you do in Algebraic Geometry with maps between coordinate algebras but that doesn't seem to be what people do

[worthy canopy]

how do i find the average distance from the origin to the points on a disk centered at (a,b) and radius r

[wanton marsh]

looks like a painful integration

[tough imp]

Shouldn’t it just be |(a,b)|?

This is what my gut says

Maybe my gut is wrong tho

[worthy canopy]

thats just the average coordinate. distance is absolute and always positive. so you cant make 2 points cancel each other's influence

[tough imp]

Yeah but like

Okay you can WLOG (a,b) = (0,1)

Or (1,0) for example

By just scaling and rotating

Does that make it easier?

[worthy canopy]

just take a,b = 0,0. the answer becomes 2r/3

[tough imp]

Oof

True

Well

Rip in pepperoni

Wait doesn’t the answer become

Just r?

Like every point is r away in that case

But it still definitely drives home the point that my gut was wrong

😭😭

[gritty widget]

having a smooth brain moment

why is tau(alpha)=alpha

ah damn

everytime, i get stuck for so long, the minute i go to ask someone its obvious

[rugged swan]

yes lmao

[gritty widget]

i think i need a cat

[rugged swan]

get some sleep

🤪

[gritty widget]

@ gomez

oh wait i don't understand it

sigma_a(a)=a because sigma_a is reflection in the plane but then sigma only leaves phi invariant it doesn't have to be the identity

[tight agate]

what is sigma_a ?

[distant phoenix]

-a

[tight agate]

I think it's supposed to be a to -a

ye

[gritty widget]

ah yes

ty

[gritty widget]

i should see this advice

@tight agate may I ask I saw in the previous messages you mentioned pcmi?

are you going to that this year?

[tight agate]

did not apply

im doing an reu

are you going?

oh wait also which one?

@gritty widget

[gritty widget]

The motiovic week

[tight agate]

nice!

[gritty widget]

gg on the reu though

[tight agate]

should be fun

[gritty widget]

europe is not great for reu style stuff

do you know what kind of stuff you are doing?

[tight agate]

in the reu?

[gritty widget]

yes

[tight agate]

dunno yet

I do want to learn something in the bloch-kato direction

so it might be on motivic stuff

but yeah idk how much freedom you get on topic choice

[gritty widget]

what is bloch-kato stuff?\

[tight agate]

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

[gritty widget]

swanky

[tight agate]

there's that

another idea was to read through thomason-trobaugh

or deligne hodge

or perverse sheaves

there's too much shit to do

[gritty widget]

chew coffee beans

saves time on drinking coffee

good buzz none the less

actually

be carefull if you chew coffee beans

they can be very yummy

but you get wayy more caffeine from eating the whole beans than drinking coffee (which in hindsight is obvious)

but i thought you would get less since when you are brewing tea you need heat to extra the caffeine

[tight agate]

[gritty widget]

so i was throwing these back like a bag of pocporn

[tight agate]

[gritty widget]

and i'd never really find myself getting a buzz off coffee, just drink it for the taste

but by golly

my heart was pumping

[tight agate]

just read the abstracts for the summer school

looks dope

[gritty widget]

looks sick as shit

I also did not realise it it was princeton

[tight agate]

I wonder if there's a way to sneak in

[gritty widget]

yeah this looks a good summer school to apply to that actually accepts non americans

i will try take good notes and share them with you

[tight agate]

thanks!

are you an undergrad or grad student?

[gritty widget]

in europe so im in a masters

[tight agate]

ah

[gritty widget]

you

[tight agate]

undergrad

[gritty widget]

gg

sounding large for an undergrad

[tight agate]

do you intend on getting into motivic homotopy stuff?

[gritty widget]

I really like algebraic topology

I enjoyed algebraic geometry

but was not a big fan of the tools

[tight agate]

algebra?

[gritty widget]

I don't know how to describe it

the stuff you get to look at is cool

but i prefered the "flavour" of algebraic topology

i did really enjoy commutative algebra

[tight agate]

interesting

[gritty widget]

but the motivic homotopy stuff does seem sick

do some nice alg top

get to see some of the mad spaces from alg geom

[tight agate]

yea idk anything about it

my algebra prof keeps talking about it

[gritty widget]

this is funny

to do this exercise i need that and automorphism of a finite set is idempotent

which is not too hard to see but I don't think ive ever noticed that before

[charred tartan]

Can someone help me understand the theoretical motivation for sidh

[pine ingot]

Georgia Tech Summer School, Summer Trisectors Meeting, but those are definitely more geometric topology in flavor.

[honest terrace]

what's sidh ?

[charred tartan]

Supersingular isogeny diffie helman

[thin jewel]

Do you know about elliptic curves?

It looks like it's one of those crypto alpgorithms that uses the group law but idk much about cryptography

I think Silverman has a nice book for undergrads on EC

[wanton timber]

Do you have a specific question in mind? Depending on your background, there might be a lot to learn to feel like you really understand what is going on.

[wise cedar]

are we talking about animu now?

[gritty widget]

Thanks

[strong heron]

I want to calculate the homology of the torus using Mayer-Vietoris on the above decomposition.

What would be the map from $H_1(A \cap B) \to H_1(A) \oplus H_1(B)$?

[gentle osprey]

whysee

[empty grove]

You know the map $C_1(A \cap B) \to C_1(A) \oplus C_1(B)$ right?

[gentle osprey]

Moldilocks

[empty grove]

You just look at the map this induces on the equivalence classes

[strong heron]

Actually I think my confusion lies here only. What are singular simplices in A \cap B? Like, the image should be connected, but A \cap B is not connected.

[empty grove]

which image should be connected?

[strong heron]

I mean, my simplex should lie in one of the components of A \cap B

[empty grove]

yep

each simplex would lie in a single component

[strong heron]

I think I see something now

So, a cycle in each of the path components lies in A and B as generating cycles of A and B respectively

So, the map Z+Z \to Z+Z is something like (m,n) --> (m+n,m+n)?

[empty grove]

The exact isomorphism would depend on the isomorphism you have to Z+Z for each homology

like it could be (m,n) --> (m+n, -m-n)

[strong heron]

But this doesn't fit in the MV for calculating H_1, it seems

Sorry it does

[thin jewel]

How are the Dehn–Sommerville equations related to Poincaré duality? I feel like they should have a close relationship but I can't quite figure it out

Like the symmetry of the h-vector looks just like the dimensions of a homology group

But maybe that's just a coincidence - I'm not a topologist

[strong heron]

Thanks, Moldilocks!

[fading vale]

@thin jewel from wikipedia lol:

https://en.wikipedia.org/wiki/Dehn–Sommerville_equations

[thin jewel]

cool thanks

feel like explaining it to me?

[fading vale]

I super don't know anything about this stuff haha http://math.mit.edu/~rstan/pubs/pubfiles/69.pdf but this is the paper being referred to i believe

[worthy canopy]

how do i find the average distance from the origin to the points on a disk centered at (a,b) and radius r

[thin jewel]

Didn't you ask this the other day?

[tough imp]

Didn’t get an answer I think

[gritty widget]

What do you get when you try to calculate the integral?

[fading vale]

i dont understand the part about showing that subset (i called it K) is open

it says to use coordinates to make X look locally like R^n-1 in R^n but i dont see how that helps

[empty grove]

Say x is in K, you need to show a nbhd of x is in K. Take a neighborhood U of x such that U intersection X in U looks like R^(n-1) in R^n. As x is in K, there is some t in U - X st z connects to t in X complement. Then z will then connect to every point in the component of t in R^n - R^(n-1). Then U intersection X is in K because for every point in this, and a neighborhood of the point, its intersection with U would be non empty so will contain a point connecting to z

[fading vale]

hmm im not totally sure the last bit follows. even if U intersects the nbhd we dont necessarily have that the y in U which connects to z is contained in the nbhd right?

we just have that U contains a point connecting to z, not that every point of U connects

unless im misunderstanding ur proof :c

[empty grove]

That's where you use the fact that there's no single y, rather everything in the component of y will connect to z

So in the given picture, the half of U which is outside X can all be connected to z

[fading vale]

that requires that R^n - X has (at most) 2 components, right

because uh

[empty grove]

Not R^n - X

[fading vale]

U - X?

[empty grove]

R^n - R^(n-1)

[fading vale]

oh

yeah

[empty grove]

Yes

[fading vale]

ah okay i see

[empty grove]

But identified via homeomorphism

[fading vale]

we need to add on the restriction that U is connected at first then right

or i guess just let it be like a very small ball about x

[empty grove]

We choose U such that such an identification is possible

We don't need any other conditions on U

[fading vale]

are you identifying U with R^n itself?

[empty grove]

Yep

[fading vale]

and U cap X with R^n-1?

[empty grove]

Yes

I thought that's what the hint is

Idk how to prove the hint tho

[fading vale]

wdym

[empty grove]

Like I'm using a single homeomorphism from U to R^n, which restricts to a homeomorphism between U cap X and R^(n-1)

R^(n-1) viewed as subspace of R^n in natural way

[empty grove]

Isn't this what that hint means?

[fading vale]

are u saying u dont know how to show that X can be locally identified w/ R^{n-1}

[empty grove]

Not exactly that

[empty grove]

We need to show that there is a neighborhood of x in R^n, U such that there exists a homeomorphism such that...

[fading vale]

oh i see

[empty grove]

How are you interpreting that hint?

[fading vale]

as that haha, i just wasnt sure what u meant when u said u didnt know how to prove the hint

well

i was thinking of it as U being a neighborhood of 0 in R^n but like

same thing functionally

[empty grove]

Ah ok

But assuming the hint I think the argument works

[fathom cave]

ugh im not sure what an open subspace is.

[feral bay]

it's open with respect to the topology on Y

[inner fiber]

Hint: Any open subset of X is also open if viewed as a subset of Y in the respective topology.

That follows more or less directly from the definition of the subspace topology on X

[fathom cave]

X is an open set of Y with topology induced by T_1 right @feral bay

[feral bay]

τ₁ is the topology on Y, and (X, τ) is an open subspace if X is an open set of Y with τ₁ (aka X ∈ τ₁)

it will always be open wrt to the subspace topology induced by τ₁, since X is open in X

[fathom cave]

i meant X \in T_1 and T is topology induced on X by T_1

[feral bay]

aaaa

yea

that's right

[cursive glade]

if the interior product exists, why dont we have a de rham homology?

wait i think i misread

[empty grove]

Inner product

[cursive glade]

nah, i thought it acted like the reverse differential but its just contracts an n form

[empty grove]

[cursive glade]

[gritty widget]

well, it's square is still zero, so you can look at ker/im

[obtuse meteor]

Hmmm

Interior product you need a fixed tangent vector no?

I am not like

Fully clear on this part of my manifolds class lol

[gritty widget]

homology parametrized by space of vector fields

yes

[obtuse meteor]

Hmm

Gives me like

Sheaf cohomology vibes

Where you take coefficients parameterized by a sheaf

Hmmm

Is there an easy example calculation of this

With a nice like vector field

Someone calculate vector homology of a 3-sphere with a non vanishing vector field go

Oh I think it’s just stupid

Nvm only stupid in the degrees it should be stupid

Which is a good sign

👍

Hmm

Maybe this is a bad property though

For a nonvanishing vector field

It has the property that

H_n = 0

Bc take your n-form omega and your vector field X

Extend X to a basis X, Y_1, …, Y_{n -1} locally

Omega(that stuff) is zero if and only if omega is zero

So your homology is always zero in your dimension

But perhaps it’s less stupid in low degree (obviously it’s also trivial in high degree)

Bc like in low degree you’d have to be zero in all the combinations to be zero, and some of those might not involve X

@gritty widget am I speaking sense?

[cedar pebble]

@obtuse meteor

[obtuse meteor]

Take a tangent vector field on a manifold M

[cedar pebble]

mhm

[obtuse meteor]

This defines a chain complex

On the forms

Via contraction with the vector field

What does homology of that complex look like

[cedar pebble]

ahh okay I see what you're asking

if the vector field never vanishes then the homology of this complex is trivial

[obtuse meteor]

:o

[cedar pebble]

if the vector field does vanish then uhhh

[obtuse meteor]

I got zero in degree n and up if it never vanished

[cedar pebble]

you can have some nonvanishing homology groups but they are typically infinite dimensional

so in this sense the theory is kind of a pain

[obtuse meteor]

Dunno how to get nonzero anywhere

[obtuse meteor]

Holy fuck

That’s

Got a reference text?

[cedar pebble]

Not off the top of my head

[obtuse meteor]

Oof

[cedar pebble]

let me search around hold on

https://mathoverflow.net/questions/184699/take-contraction-wrt-a-vector-field-twice-and-define-kernel-mod-image-does-that

there's some good answers here

oh hey the last answer is really nice

if a compact Kähler manifold admits a holomorphic vector field 𝑣 with nonempty zero locus 𝑍(𝑣), then the Hodge numbers vanish for |𝑝−𝑞|>dim𝑍(𝑣). The key idea is to analyze the complex of sheaves …Ω^𝑝→Ω^𝑝−1→… which is really a Koszul complex.

huh I didn't think about the connection with Koszul complexes this is cool

[gentle osprey]

bdobba

[obtuse meteor]

Oh this is cool

[agile shadow]

Hi

How can I prove the cosimplicial identities in the ordinal number category?

[cold vine]

Hmmm, any idea on showing that pi_4(U(3)) is finite? I'm quite stuck

[gritty widget]

@cold vine did you get that question out?

[cold vine]

still stuck on it

[gritty widget]

okay I don't know the answer, and it seems like there could be a slicker answer

but

can you get pi_4(SU(3)) instead?

[cold vine]

I haven't done anything with SU(n) so I think it should at least be doable without it. But if it's easier that way 🤔

[gritty widget]

what things do you know homotopy groups of

i'm really not sure how do do this question

I just know that pi_4(SU(3)) is 0

and impretty sure U(n) and SU(n) have the same higher homotopy groups

and for calculatin gthe homotopy groups of SU(N) i think you should have a nice fibration in terms of S^(2n-1) and SU(n-1)

maybe this helps?

i was more so interested in seeing the answer myself

if you get it out please @ or dm me

[cold vine]

Yeah. I know the groups pi_i(U_(n)) when i=1,2,3, n>=2. Yeah I'm trying something with the fibrations of U(n)

I'll message you if I get something ^^ proll continuing tomorrow and letting it rest now

But I'll check the SU fibrations out if I get nothing to work. Thanks!

[gritty widget]

no worries

[frigid river]

Could someone tell me how that bases stuff works? If U were an element of the Basis of X, then A would be open by definition, right? But how do I know that U is an element of the Basis of X? Is it because U is open?

[empty grove]

Use the openness of U to get a possibly smaller basis element containing x

U itself may not be basic

[red garden]

or maybe just take the union

A is just {x,y,z,....}

[frigid river]

Ohh, thanks. So I can just say A must be open, since it can be written as the union of open subsets, right?

Does the stuff I learnt about openness with metric spaces also apply to topological spaces?

[empty grove]

The union of open sets is open in topological spaces, it's one of the axioms of a topology

Not everything about openness in metric spaces generalizes, like metric spaces are Hausdorff but topological spaces may not be

[frigid river]

I think it follows from this, right?

[empty grove]

Yeah

It is exactly that

[frigid river]

Great. Thank you.

[cold vine]

I have A0->A1->A2->A3->... cofibrations of closed subspaces of X and for A= union An, where C in A is closed iff C \cap An is closed in An for all n. Now I'm trying to show that A0->A is cofibration

I can extend the homotopy of a restriction to A0 of f:A->Y to any degree since I have that An is closed in An+1 and each inclusion is cofibration but I'm having trouble on figuring out how to show that it can be extended to the union A

[cold vine]

So I guess now I have HEP for any n and the inclusion An to A. Maybe I'm missing some thrm which helps with extending to A=union of An, I don't think I can do it naturally. Or maybe it somehow follows from the given restriction that C closed in A if and only if A\cap An closed in An for all n but I don't see that helping

[light plinth]

hello guys, im going to try to give a good explanation of my question !
I have a theory on a possible solution to my master thesis that is about genetic algorithms (no need to understand what that is to understand my question, but geometry and topology is one of the main advanced components that are deeply rooted into this topic)...
in order to achieve and test my hypothesis I need to be able to, in any n-dimensional sphere, calculate the cartesian coordinates of each point in the line that unites the center of that sphere with its boundary with a certain angle... my objective is create a spherical scan (assuming im not doing only this line im exemplifying but all possible lines in the 360 degrees possible) that possesses every point in that line

[light plinth]

think of it like you're finding all the points that are inside that red line... if I want to check all the space that is in that line I need to get all of the points of that line (assume r = 100, I have to get the cartesian coordinates of every unit "1,2,3...till 100" --> assuming the smaller distance is one unit)
then I will change the angle (assume the smaller unit is 1degree) and if I do that for all the 45 lines i will get each and every point on that quadrant cartesian coordinates)

this is easy to imagine in 2d... but harder to imagine in n-d, because an angle is just a relation between 2 axis and for example in 32d I will have 32 axis

and I will get a lot of data points but that's not a problem at all, i just need to be able to calculate them in a sequenced way

if someone is interested in helping me you can DM me and i would be ok with discussing this even if it was not in depth

[cold vine]

Isn't it correct that SO(n) and O(n) have the same homotopy groups since O(n) is just two disjoint copies of SO(n), i.e. it has path component of identity matrices and path component of the inverses? Also I have that O(n)/O(n-1) = S^n-1 and I think it follows by the above logic that SO(n)/SO(n-1)=S^n-1

[cold vine]

So can I not just say straight up that since SO(3)=RP3 we have also pi_1(O(3))=pi_1(SO(3))=pi_1(RP3)=Z/2Z

[flint cove]

I may be missing some basics here, but since when does it make sense to talk about the fundamental group of a non-path-connected space without choosing a basepoint?
Or do we just implicitly take *=id for lie groups

[cold vine]

From what I understand the path components are homeomorphic since is is of det 1 and the other det -1

[flint cove]

I mean yes, but that doesn't apply to all spaces obv

[cold vine]

yeah

[flint cove]

otherwise I'd say that rescaling the matrices of positive determinant to det=1 gives a deformation retract to SO(n), right?

(wrt the obvious inclusion SO→O+)

[cold vine]

Yeah that would make sense to me

[honest narwhal]

Yeah for Lie groups usually you implicitly say connected component of the identity

[flint cove]

But the argumentation seems correct, assuming that SO(3) ≅ ℝP³

[honest narwhal]

Yeah taking the appropriate basepoint this is correct

[flint cove]

(ah yes, of course it is the projective space, just got the right mental image)

[honest narwhal]

@flint cove if you haven't seen this proof before it's a cute exercise

Lol sniped

[cold vine]

nice!

I just know it has something to do with Sp(1) acting on H but I don't know the details 😄

[honest narwhal]

There's an easy proof

[flint cove]

I mean I know the construction SO(3) ≅ (3-ball with antipodals identified) by using „rotation around $foo-axis around angle 2π $blah“, but I jusf forgot how the right hand side was just ℝP³

[cold vine]

Hmm I'm trying to calculate pi3(O(4)) so I got the LES -> Z/2Z -> Z-> pi3(SO(4)->Z->0 but this doesn't seem to lead to anything

from the fibration SO(n)->SO(n+1)->S^n

ah nevermind that just leads to 0 -> Z -> pi3(SO(4)) -> Z -> 0 which splits and implies pi4(O(4) = Z^2

had it incorrectly first

[cold vine]

Hmmm how do I show that this is injective

[empty grove]

If 2 maps from the n-cell to X were homotopic in the wedge sum via H, you could change that into a homotopy entirely in X by mapping everything in H^(-1)(Y) to the base point

Idk what the second map is

or that last space for that matter

[gritty widget]

any symplectic kings able to help me out

I want to show that for a presymplectic manifold (M,\omega), TM^\omega is an involutive distribution

what's TM^\omega?

is a sub bundle of TM

and it is

at every point p defined to be, since T_pM is symplectic vector space, the "orthogonal" compliment of T_pM

won't that just be {0}

ohh maybe i have that wrong

oh oops i missed the "pre" part of presymplectic. that means not necessarily non degenerate, right?

no i yes

yes*

oh i think i have it, this is clever

$d\omega(X,Y,Z)=d(\omega(X,Y))(Z) + ... + \omega([X,Y],Z) +...

i think is true for a genral form

the since its presymplectic its closed so the left is zero

and if we take X,Y in TM^omega then all the other stuff vanises exept

\oemga([X,Y],Z)

i think i see

the formula for d\omega is a little unwieldy but that sounds like it works

i was thinking you could use i_[X, Y] = [L_X, i_Y] along with cartan's formula, but it's probably the same proof

ya, your proof works, nice

[gentle osprey]

狼人

[gritty widget]

is sharmrock still here?

no

[obtuse meteor]

Symplectic kings

[gritty widget]

@gritty widget shamrock left (more accurately, asked to be banned) because they were spending too much time here

[tough imp]

They’re gonna be back soon

I think just a couple days

IIRC he said he’d rejoin once he moves back to WA which is in a few days

[little dew]

anyone know why a manifold is called a manifold?

[gritty widget]

"manifold" means "many and various" (google definition) and a manifold is built up from many and various coordinate charts

[little dew]

ic that makes some sense, tyty

[gritty widget]

i actually don't know if that's the historical reason

[tough imp]

Gaussw

Gauss used the word manifolden or some shit

sorry

I mean Riemann

I think it was around when he was developing the ideas of Riemann surfaces

[gritty widget]

so if (M, \omega) is presymplectic then lime_soup shows that the distribution TM^\omega is involutive. so it determines a foliation of M. what could be said about this foliation?

reminds me of the foliation theorem for poisson manifolds

[empty grove]

Manifolds are called that because whenever you think of them you think of sheet of paper with many folds

[gritty widget]

like, what could be said about e.g. restricting the presymplectic form to each leaf

hm

if only shamrock were here to confirm my suspicions

in certain situations the space of leaves of this foliation is a symplectic manifold

👀

neat

tfw missed the lecture on this so it's all new

[cedar pebble]

@gritty widget someone said you're good at complex geometry

[gritty widget]

im not

[cedar pebble]

fuck

[gritty widget]

who said that

[cedar pebble]

brofibration

[gritty widget]

but

he's literally taken a full course in it

[cedar pebble]

anyways here's a fun challenge: let X be a smooth connected variety over C (or say the complement of a simple normal crossings divisor in a compact Kahler manifold), let E be a unipotent C-vector bundle with flat connection on X. Show that E admits a canonical real analytic trivialization.

I can do this very very indirectly

but I'm trying to do it directly and it's like

it also sounds totally false when you first look at it!!

[cold vine]

So inclusions X-> X V Y and Y -> X V Y should induce this to be a isomorphism, when X p-connected and Y q-connected for n< p+q. So I guess this is Blakers-Massey thrm but I don't know how to show this. Any hints?

[red garden]

problem 1?

no

i can try proving this

well ig i see a map right away

sending x --> F_x

where F_x = f(x)

f is in N*

is this right?

but why does this not work for inf

i still dont see why

inf dim --> somethings just dont work

okay so

suppose X is finite

C(X) is finite ?

i really dk lmao

yea lmao

soo umm

yes

oohhhh

X is finite here means

finite as a vec space

?

okay okay

so if X is finite

let [e_i] be some basis

i meant for C(X)

so those are just continuous functions iirc

so umm

C(X)** would be functions defined on C(X)*

C(X)* would be functions defined on C(X) umm

okay

so

suppose X is finite

suppse C(X) is inf dim

B its basis

thats not easy is it

and compact

lmao i didnt see that

its stated in the problem

hahaa okay

yea

soo

if X is finite hausdorff

[little hemlock]

Am I being dumb? The set of functions X to R is a finite dim space when X is finite. So C(X) is a sub space

[red garden]

yea

if X is finite

and hausdorff

then the space can be like

thought of just discrete elements

[little hemlock]

Of The set of functions X to R

[red garden]

so

if like X = {a1,a2,....,an}

C(X) is R^n by f_x --> x ( i hope )?

so C(X) has dim n

where n is card X

an element in C(X)

some element in X

X is discrete

[nimble jolt]

what are you asking/trying to prove mo2men

[red garden]

can i say

the basis for C(X) are the maps coming from tietzse

as X is hausdorff and compact its normal

[little hemlock]

I think ur overthinking this mo

[red garden]

yea im confused

[little hemlock]

Start with the set of functions X to R. This is a vector space. Do u see why?

[red garden]

yes

[little hemlock]

X is finite, so I claim u can find a finite basis

[red garden]

how

[little hemlock]

Well, maybe not for C(X) w/o some effort, but the larger space is easier

[red garden]

finally mo2men figured out something

lmao

[little hemlock]

Yea, I don’t think you even need the hausdorff condition in principle

[red garden]

delta?

those delta shitty functions

in lin algebra

yea

omg that was so hard for me

tysm

sending x --> F_x
[2:26 AM]
where F_x = f(x)
[2:27 AM]
f is in N*?

[red garden]

using basis

that its finite dim

im jusst gonna like

linear ocmbinate any element

and then take the function

right?

[empty grove]

Yeah the S(a,b) form a basis

They didn't say that those are the only sets that are open so not not true

[gentle osprey]

slimevesus

[red garden]

rank nullity

[gentle osprey]

مرزا