point-set-topology

[shut moat]

are there any good resources that go deeper into orientation? this shit is really cool

like, if a manifold is defined as the zero set of a C^1 map, then you're guaranteed an orientation on it, so you can't define a non orientable manifold that way 🤯

theorems like that ig

and/or intuition

[sleek thicket]

This is because the gradient is a normal vector, right ?

[rugged swan]

how do you prove that every curves have the cofinite topology ? @tight agate

[sleek thicket]

Anyways my recommendation as always is ISM lol

[shut moat]

yeah

[shut moat]

[sleek thicket]

hatcher also gives another perspective on orientation

In terms of the (co)homology

both good

[shut moat]

oh perfect, I've been meaning to give hatcher a look at some point

to see what this homology stuff was all about

[sleek thicket]

It's good!

[tight agate]

well, you do need to ignore the generic point, but it is quasicompact, so cover it with finitely many affines and show that each affine part has the cofinite top

over each affine part it should reduce to polynomials have finitely many zeros (1-dim)

[shut moat]

thanks!

[sleek thicket]

@obtuse meteor that person is so annoying

I have had similar interactions

(this is about a rando on Twitter not any of you)

[rugged swan]

yeah ok it should gives a one-dimension function ring, that is every non zero prime ideals are maximal then the ring is factorial. And if you take f irreductible, I(V(f)) = (f) maximal, and V(f) is only one point bc maximal ideals are in one-to-one correspondence with points of your curve. @tight agate

[sleek thicket]

I will assume it's you now ultra

why would you say this thing

duxk you

fiber bundles confuse me

apparently you can give $\mathbb{S}^1 \times \mathbb{S}^1$ the structure of a nontrivial fiber bundle over $\mathbb{S}^1$ with structure group ${-1,1}$

with the projection onto the first factor

[gentle osprey]

(T*(Terra), -dτ)

Shamrock -> E -> B

[sleek thicket]

@shut moat reveal to me your mysterious ways mx circle

[shut moat]

sorry for asking so many questions, but i just wanted to make sure my intuition is correct:

if you have a smooth 1- manifold in R^n, you can define an orientation on it with a nonvanishing continuous tangent vector field via the orientation induced by its work form.
If you have a smooth 2-manifold in R^3, you can put an orientation on it with a continuous transverse vector field via the orientation induced by its flux.

For the latter, this only works because R^3 has the special property that 3-2 = 1 (big bren maff), so you can identify an orientation on the surface with a vector via this poincare duality thing I guess. But this is equivalent to picking out a pair of continuous tangent vector fields that form a basis at each point, which is the direct analog of what you do to the 1-manifold, and this is the version that generalizes more readily to 2-manifolds embedded in arbitrary dimensional spaces, right?

also what is that horrifying symbol :o_o:

looks like that s you find on desks in middle school

[marsh forge]

why do u use mathbb sham

[sleek thicket]

its correct

[marsh forge]

in what sense

[sleek thicket]

in the sense that its correct

[marsh forge]

afaik its pretty outdated

i've not seen it at all in any post-1950 literature lol

[sleek thicket]

ie retro

im gonna make it happen

[shut moat]

tbf it looks nice lol

[sleek thicket]

we're briging it back in 2021

[marsh forge]

hot take

\mathbb{S} should only be fore symmetric groups

they don't deserve to use normal S

[sleek thicket]

wrong

[marsh forge]

ironically computing stuff w spheres ends up being something to do w k-theory related to symmetric groups

so truly notation leads to Truth

[sleek thicket]

another reason I might use this notation: my book does

[gritty widget]

Is algebraic geometry more fitting here or in abstract algebra

[sleek thicket]

also sorry @shut moat but im not big brained enough for your question so it's getting buried

[shut moat]

npnp

[sleek thicket]

I don't know what a work form is lmao

or what the flux of a surface is

[shut moat]

oh it's just like the dual of the vector field

$F_x dx + F_y dy + F_z dz$

[gentle osprey]

~S^1

[sleek thicket]

ah okay

wait so like

is your question "how does a normal vector give an orientation"

it looked like more than that

actually hm it's a little weird that it works for 1d curves in any dimension, I guess it's because the tangent space is 1d

so the top exterior power is just the cotangent bundle

smth like that lol

[shut moat]

ig like it was like "why do you use a tangent vector field for a curve but a normal vector field for a surface (in R^3)"
And I attempted to answer it with "R^3 is nice bcz dual of a 2-form is a 1-form ~ vector"

so they're actually equivalent

[sleek thicket]

ah yeah so like

here's how I would think about it

a tangent vector for a curve trivializes the tangent bundle

so you just get an orientation

does that make sense?

[shut moat]

I'm not sure what you mean since idk much about bundle stuff sry. A trivialization is a when you pick a neighborhood where the bundle looks like a product bundle or something right

[sleek thicket]

ah no worries

yeah so like

if you have a global basis for the tangent space

varying smoothly

you get an isomorphism $TM \cong M \times \R^k$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

yeah?

[shut moat]

I think yeah

[sleek thicket]

so a tangent vector to a curve is such a basis

if $X$ is a nonzero tangent vector to $C$, I automatically get a map $M \times \R \to TM$ by $(p, t) \mapsto t X_p$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

and this is an isomorphism

so our orientation is just "the bundle is trivial, i.e. globally looks like R, so we can define an orientation"

or to put in another way, just declare the direction X points in to be the positive direction at each point

so for curves something different is going on imo

the generalization of the curve thing is that if $M$ is an $n$-manifold and you have global vector fields $E_1,\ldots,E_n$ on $M$, and for any $p \in M$ the vectors $E_1|_p, \ldots, E_n|_p$ are a basis, then you get an orientation for $M$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

just declare (E_1|_p,...,E_n|_p) to be positively oriented at each point

does that make sense?

[shut moat]

yeah! This is sortof what I was thinking

[sleek thicket]

nice!

the normal vector thing is different

[shut moat]

and then normals are the special case where cross product exists, is how i tried to think of it

[sleek thicket]

hmm, I disagree

so for example, the orientation on $\mathbb{S}^2$ comes from the outward pointing normal vector, but there is no such global frame for $\mathbb{S}^2$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

(frame means global basis of vector fields like above)

[shut moat]

oh dang

[sleek thicket]

do you know (of) the hairy ball theorem?

[shut moat]

oh yeah I do

no nonvanishing continuous tangent vector field on the 2-sphere

[sleek thicket]

on spheres of even dimensions

there is one on the 3-sphere actually

and all odd dimensional spheres

but anyways

so if $(X,Y)$ was a frame for $\mathbb{S}^2$, we would have $(X_p, Y_p)$ linearly independent at each point $p$, an in particular both $X_p$ and $Y_p$ would be nonzero for each $p$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

contradicting the hairy ball theorem

does that make sense?

[shut moat]

yeah

hm my intuition was def wrong then

[sleek thicket]

so imo the normal vector thing comes from the hodge star operator

[gritty widget]

speaking of the hairy ball theorem

[sleek thicket]

which is the duality between 1 forms and 3-1 = 2 forms you mentioned

ooh

interesting

idk anything about killing vector fields

[gritty widget]

they kill the metric w/ lie derivative

[shut moat]

mathematicians are violent smh

[gritty widget]

so the geometry is preserved along their trajectories

[sleek thicket]

so $@shut moat$ say you have a riemannian manifold $(M, g)$

oops

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

this means that g is an inner product on the tangent spaces of M, smoothly varying and all

[gritty widget]

(here's the proof of the previous picture, btw)

[sleek thicket]

let $n = \dim M$. then for $0 \leq k \leq n$ we have this funny "hodge star operator" $\star : \Lambda^k(T^* M) \to \Lambda^{n-k}(T^* M)$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

you should check if hubbard defines it

I think it's really cool

anyways, here is how you can think of the normal vector giving an orientation abstractly

an orientation is equivalent to choosing a nonzero top dimensional form

oh also pretend M = R^n here lol, don't worry about general riemannian manifolds

[shut moat]

rip only a passing mention saying it's related to "perpendicularity"

[sleek thicket]

so if $M'$ is a hypersurface in $M$ (ie an $(n-1)$-dimensional submanifold) and we have a global choice of normal vector $N : M' \to TM$, we can use the metric to get a covector $\omega : M' \to T^* M$ satisfying $\omega_p(v) = g_p(N_p, v)$. Then $\ast \omega : M' \to \Lambda^{n-1} T^* M$ gives an $(n-1)$-form, and we can pull back along the inclusion $\iota : M' \to M$ to get nonvanishing $(n-1)$-form $\iota^* \circ \star\omega : M' \to \Lambda^{n-1} T^* M'$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

i do not think of the orientation coming from a normal vector this way though

i just think of it as like, declare $(v_1,\ldots,v_n) \in T_p M'$ to be oriented if $(N_p, v_1,\ldots,v_n)$ is oriented

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

¯_(ツ)_/¯

but you can look at it abstractly via the musical isomorphism and the hodge star like I did above

once again, this should be in ISM (i don't think it's spelled out but if you read the section on orientations and on riemannian manifolds this will make sense)

actually i should probably check

[shut moat]

what's the point of pulling back by the inclusion map? isn't that basically the identity

[sleek thicket]

have you seen the equivalence between an orientation and a nonvanishing top degree form?

[shut moat]

yeah

[sleek thicket]

okay cool

it's just because we want a form on M'

but what we have is a section of the restricted bundle $(\Lambda^{n-1} T^* M)|_{M'}$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

this is not the same as $\Lambda^{n-1}T^* M'$

[gentle osprey]

Shamrock -> E -> B

[shut moat]

oh ic

[sleek thicket]

i feel like you always ask simple questions and then I give super overcomplicated answers lol

[gritty widget]

idk if hodge star is in ISM but it's definitely in the exercises of sec 2 of IRM

[sleek thicket]

sorry

yeah I remember seeing it in IRM

[shut moat]

np at all, this is rly helpful!

[sleek thicket]

oh wait it's definitely in ISM

he talks about like classical stokes/greens/divergence

for surfaces in 3-manifolds

[shut moat]

I always try and link stuff I see in hubbard to smooth manifold language in lee or whatever because hubbard uses nonstandard terminology so frequently lol

[gritty widget]

page 438!

[sleek thicket]

lol

[gritty widget]

i wish i had a copy of ism

[shut moat]

same

[sleek thicket]

i dont

because i have a copy

:smug:

[gritty widget]

gib

is there a book better than ISM?

[sleek thicket]

but yeah @shut moat your intution that we were relying on 3-1=2 is right

in general you want a hypersurface

and a normal vector

or a frame

[tight agate]

Kobayashi and Nomizu?

[shut moat]

vaguely related but Milnor's topology from the differentiable viewpoint is apparently insanely good

[gritty widget]

you can read the entire thing in a day

[sleek thicket]

but i think the 2 examples you pointed out (curves vs surfaces in R^3) are distinct, one is about parallelizability (having a global frame) and the other is about this duality between normal vectors and top degree forms

[gritty widget]

get working buddy

[tight agate]

Milnor's morse theory book is really good

[sleek thicket]

im not your buddy pal

[tight agate]

Oh wait how could I forget

Griffiths and Harris

[shut moat]

glad someone latexed the morse theory book

the old typesetting was horrifying

[sleek thicket]

oh yeah ~S^1 just to be super clear, the top degree form you end up defining is $\omega(E_1,\ldots,E_n) = g(N, E_1,\ldots,E_n)$

[tight agate]

you might as well get used to it

[sleek thicket]

wait this is wrong

[tight agate]

a lot of books used that font

[sleek thicket]

wtf am I sayin

[shut moat]

mood

[sleek thicket]

I think it's like, a determinant with entries the inner products

or smth

i revoke everything I have ever said

back to bundles

when is a torus not a torus

[tight agate]

morse theory, characteristic classes, infinite loop spaces, curves on an algebraic surface, ...

all the old orange books

[shut moat]

I think wiki defined it as like $\eta \wedge \star \zeta = g(\eta, \zeta)\omega$ where $\omega$ is volume form

[gentle osprey]

~S^1

[sleek thicket]

this is true

but i didn't want to go into it

because I would need to explain the volume form and the inner product on Lambda^k T^* M

g(eta, zeta) doesn't exactly make sense

if you think of g as an inner product on TM

but there is a way to extend the inner product to the tensor/form bundles

[shut moat]

makes sense

I noticed something was weird about that because I see ppl talking about GR writing $\det(g)dx_1 \wedge \text{whatever}$ for volumes

[gentle osprey]

~S^1

[shut moat]

but then when I got to volumes on manifolds it was sqrt(det(D\gamma T D\gamma))

so I'm guessing this is somehow how you generalize the inner product if you treat D\gamma as a bunch of tensored 1-forms or something

[obtuse meteor]

hmm

topology question

if I have two bases for a space, say U_i and V_j are both bases for X

and U_i is countable

can I find a countable basis V_{j_1}, V_{j_2}, ... of X?

should be true

[sleek thicket]

I believe so

[obtuse meteor]

bc I can just pick one V_j for each U_i

I am dummy

[sleek thicket]

yeah

[obtuse meteor]

wait no

[sleek thicket]

Makes sense to me

[obtuse meteor]

not quite

hrm

I am dummy

[wanton marsh]

do they have to generate the same topology ?

[sleek thicket]

oh yeah you need to cover them

So here's my thinking

Argue that if you have such a U then any countable cover of an open has a finite subcover

[obtuse meteor]

inceresting

[sleek thicket]

Oh wait no this doesn't do what I want sorry

[obtuse meteor]

is that true?

[sleek thicket]

but it is true

[obtuse meteor]

wait doesn't that work then?

[sleek thicket]

Second countable => lindelof

Oh if it does then neat lol

[obtuse meteor]

bc then you can cover each U by finitely many Vs

[sleek thicket]

oh right

Lol

Big brain Brendan

[obtuse meteor]

ok thanks for the lemma :P

[sleek thicket]

I constantly do this

Where I prove it in my head

Then forget my proof

And confuse myself

[obtuse meteor]

the context was like

[marsh forge]

hm

[sleek thicket]

,iamnot studying

[gentle osprey]

Removed the studying! role from you.

You already have the selfroles studying!, do you want to remove them? (y(es)/n(o))
(Tip: use ,iamnot to remove roles without this prompt.)

[obtuse meteor]

I know the collection of all coordinate balls of a space form a basis and I want to use second-countability to yeet it to a countable basis of coordinate balls

and I didn't want to do that with a bunch of mess

:P

[marsh forge]

is it true that every space with a basis of cardinality k is k-compact for k sufficiently large? is it just true for countable?

[sleek thicket]

Right

should be true I think

[gentle osprey]

Session timed out waiting for user response.

Jan 24, 2021, 11:18 PM

[sleek thicket]

well actually we're using something stronger than countably compact here right?

[obtuse meteor]

👍 I'll prove the second countable -> that property about covers of opens and see what happens

[marsh forge]

i dont think so

[obtuse meteor]

I don't think so nah

[marsh forge]

oh wait

yeah we are

[obtuse meteor]

how so?

[marsh forge]

lol sorry im watching a civ stream

[sleek thicket]

Np lol

[marsh forge]

covers and basis aren't equivalent

it might end up being so

[sleek thicket]

We want that every open subset is countably compact

[marsh forge]

but i dont think its a priori obvious

[wanton marsh]

R with the usual topology has a countable basis but it is not true that every open is compact

unless I am misunderstanding something

[obtuse meteor]

hmm

[sleek thicket]

I don't understand the relevance

compact isn't equivalent to k compact for any cardinal k

[obtuse meteor]

is every open countably compact?

[sleek thicket]

and also we're not talking about being compact, we're talking about countably compact

Yes Faye, by second countability

[obtuse meteor]

yeah it should be

I just need to work through it

[wanton marsh]

Argue that if you have such a U then any countable cover of an open has a finite subcover

so you want countable instead of finite there ?

[sleek thicket]

Oh I think maybe we've been misusing the term countably compact

[gritty widget]

ah thank fuck my second letter got submitted fuck

[sleek thicket]

I read this and assumed max meant "any cover has a subcover of cardinality k"

Due to context

[marsh forge]

this is what i meant yeah

[sleek thicket]

How do you check tterra?

[gritty widget]

the fields institute sent me an email

just now

[sleek thicket]

oh fuck

I don't think I've gotten an email

[gritty widget]

"we have recieved a reference letter from prof (...)"

email them right now, shamrock.

[sleek thicket]

Wikipedia disagrees lol https://en.m.wikipedia.org/wiki/Countably_compact_space

[obtuse meteor]

oh weird

[sleek thicket]

Wait what?

[obtuse meteor]

hmmm

ok so

[sleek thicket]

"every countable open has a finite subcover" versus "every cover has a countably subcover"

Not the same

There are countable open covers of R with no finite subcover

But R is second countable and so lindelof

ah okay gotcha

[obtuse meteor]

I want to use a pigeonhole argument but don't know how to phrase it formally

[sleek thicket]

@gritty widget wait so fields emailed you?

[obtuse meteor]

which is an L

[sleek thicket]

Wtfff

[gritty widget]

yeah, i can show you what the email looks like

if you want

[sleek thicket]

One of my letter writers told me they submitted

[obtuse meteor]

yeah me too

[sleek thicket]

Please do

I am concerned now

[gritty widget]

okay, one moment

just need to crop out dox

[obtuse meteor]

dming you

[sleek thicket]

AHHHHHH

[gritty widget]

if you didn't get one of these i would email the fields and ask

like

right now lol

[sleek thicket]

Doing it

[gritty widget]

@obtuse meteor

[obtuse meteor]

I got one from one prof

not the other one

waiting a bit longer

bc the prof I haven't gotten it from said she knew about it and is working on it / will get it done

so we should be good

just gonna keep an eye on my inbox

[gritty widget]

good luck

[shut moat]

gl

[sleek thicket]

I'm so fucking anxious now

What the fuck

[gritty widget]

[sleek thicket]

I literally would have no way of knowing this if I didn't shitpost on discord

[gritty widget]

[sleek thicket]

Jack emailed me and confirmed he submitted my letter to the fields institute

like

I have it in writing

[gritty widget]

ref from lee
chad

[sleek thicket]

lol

Assuming it exists???

It's actually a little awkward to have my letters be from Lee and Beardsley

I'm working on my u mich app

And I'm saying I want to work with these like, algebraic geometers and algebraic topologists

but Lee is like the opposite side of math

diff geo pde chad

[obtuse meteor]

this is absolute chad energy

[sleek thicket]

[gritty widget]

[sleek thicket]

oh my

[gritty widget]

now that i am done worrying about my letter i will check my algebra homework, due at midnight

[sleek thicket]

Time to refresh my email for the next 8.5 hours

[cloud owl]

idk wtf you ppl are talking about but gl

[gritty widget]

sham, cohomo, and i are all applying to reus

[sleek thicket]

Summer undergrad research program at u toronto

[gritty widget]

in particular

we all applied to toronto one

[shut moat]

The actual Jack Lee? damn

[sleek thicket]

Yes lol

[cloud owl]

ah

[sleek thicket]

I'm taking my 5th class with him this quarter

And he seems to like me

Although we think about math very differently

[cloud owl]

oh wow, jack lee? the film director? or jack lee the songwriter?

:P

[sleek thicket]

I asked a question on Wednesday

And he was lkke

"this sounds right but I usually try to understand math by making it into geometry and you try to understand it by making it into algebra so I don't understand the point of the question"

Or something like that

[gritty widget]

lmao

[sleek thicket]

I wanted to know if a certain construction with bundles could be thought of as like a tensor product on G-sets

[gritty widget]

gay sets

[sleek thicket]

gay sets

[gritty widget]

i do not wish to check my algebra homework

enjoy your gay sets

[sleek thicket]

gay sets!

I'm not enjoying them

When is a torus not a torus still haunts my brain

[gritty widget]

a torus is a lie group isomorphic to a product of S^1 w itself

[sleek thicket]

w o a h

I'm supposed to make S^1 × S^ 1 a nontrivial fiber bundle

[shut moat]

god i can't wait to learn lie theory, it looks so interesting

[sleek thicket]

With structure group {-1,1}

It's so incredibly based circle

I think lie groups are the most interesting thing I've learned

In diff top/geo

[gritty widget]

lie group actions

lie groups are incredibly nice

apparently my symplectic geometry course is going to focus a lot on lie group actions

[shut moat]

does your symplectic geometry class talk much about the classical mechanics connections

[gritty widget]

so far, not really

[sleek thicket]

hnng I should be able to glue

[gritty widget]

lecture outline

1. definition of symplectic form, darboux's theorem about all symplectic 2n-manifolds being locally symplectomorphic
2. proving some lemmas behind aforementioned theorem
3. reviewing lie groups, introduction to symplectic and hamiltonian v fields
4. idk i missed this one, still waiting for the recording lol

[sleek thicket]

lol

[cloud owl]

symplectomorphic
it would be genuinely impossible for me to tell whether this is an actual term

what even is life

[shut moat]

mniip flexes it a lot in the physics server

[gritty widget]

so who wants to play with differential forms

[cloud owl]

uhhhhh think i'll pass

[shut moat]

still better than index juggling at least

[gritty widget]

cmon index juggling isn't that bad

excerpt from my solutions to lee's irm

[shut moat]

[gritty widget]

this one's not that bad tbh

lol

[sleek thicket]

Ahhh I think I might understand

[shut moat]

I remember the [physicist] derivation of the geodesic equation wasn't very pretty

[sleek thicket]

it's like

Squaring

But branch cuts

[gritty widget]

[physicist]

the geodesic equation is like

clean though

fuck

[shut moat]

I'm pretty sure a bunch of the manipulations here are wrong too

[gritty widget]

that's uh

a strange looking geodesic equation

[shut moat]

oh it goes through multiple pages

[gritty widget]

this is what i think of when i think "geodesic equation"

[shut moat]

didn't want to screenshot all of em

https://www.youtube.com/watch?v=PmmU6i0SMic

[gritty widget]

this is your brain on physics

my algebra homework is going to take like 20 minutes max to check over and i STILL don't want to do it 😡

[sleek thicket]

My idea didn't pan out and it's tterra's fault

[gritty widget]

[sleek thicket]

curse you

[gritty widget]

[gritty widget]

off topic but hamilton slaps so much

Can someone help me understand what the stalk of the sheaf of regular fuctions is

[tough imp]

Germs

[gritty widget]

So we have a variety X and its sheaf of regular functions O_X

yeah okay abstractly its the germs

but lets say explicitly

what is is it for X=A^1

[tough imp]

Pretty sure you can just do

<U,f> where U is an open set f a regular function on U

[gritty widget]

For the affine line the sheaf of regular functions is just k[x]

[tough imp]

And <U,f> = <V,g> if f and g agree on some open W < U\cap V

[gritty widget]

yeah

[tough imp]

All of these sets nbds of the point

[gritty widget]

but I'd like to actually callculate this for some examples

[tough imp]

Lol gl

[gritty widget]

so it should be a local ring

[tough imp]

Idk maybe it’s obvious I’d have to think of schemes

But in general for schemes at least they just look like a localization at a prime ideal

[sleek thicket]

Isn't it just localization?

like

Look at Hartshorne chapter 1

[tough imp]

A localization of the affine coordinate ring?

[sleek thicket]

Take an affine open around your point, take the coordinate ring, localize at the corresponding prime

That's what it is to my memory

You can see this by thinking of it as a colimit

Take the cofinal system of distinguished opens in that affine

regular functions on a distinguished open are elements of the localization, right ?

[gritty widget]

ah yes

[sleek thicket]

It's the same as for schemes

[gritty widget]

its just the localisation at the ideal coresponding to your point

I'm still not working with schemes yet

[sleek thicket]

Ah sorry that was for chmonkey's benefit

He doesn't understand anything about varieties

[tough imp]

I’m a chimp I go

Scheme cool ooo ooo aaa aaa

And then I get pwned when someone asks about A^1

🥴

[gritty widget]

So in this case then, the maximal ideal at a point p is (x-p)

so then O_x,p is the localisation of k[x] at (x-p)

I always get mixed up with localisations

[sleek thicket]

Yup

[gritty widget]

when we say a localisation at (x-a)

you really mean taking the locaisation with S={stuff that is not in (x-a)}

[tough imp]

Localize wrt the complement

[sleek thicket]

Yeah, it's confusing notation

[tough imp]

It’s just the most useful localization IMO

And it’s useful to be able to say “if a statement holds at all localizations at primes...”

[gritty widget]

so then k[x] localised at (x-a) consists of f/g where

[tough imp]

g isn’t divided by x - a

[gritty widget]

f can be anything

[tough imp]

Aka g(a)≠0

[gritty widget]

ah yes okay

g isn't divided by x-a when f and g are coprime

we can write and f and g in a stupid way so that x-a does divide g

[tough imp]

Sure haha

Well actually no

[gritty widget]

sorry just double checking

oh

[tough imp]

x-a isn’t in the complement

So a factor of x-a shouldn’t exist in the bottom

Like... symbolically SURE

I guess

[gritty widget]

oh so if you are being really formal and thinking of f/g as (f,g)

[tough imp]

But as an element of the ring no

Yeah g could never have a factor of x-a

[gritty widget]

perfect that makes sense

is there a good way to keep localisations straight in your head

[tough imp]

Practice

Lmfao

You just

[gritty widget]

i have zero geometric intuition for this

[tough imp]

Oh

Practice

L o l

You just get a feel for it

IMO

But I’m also algebrain and geometry is non existent

So it took me a while maybe some ppl find descriptions of what’s going on motivating but not for me

[gritty widget]

and the other thing

if we have a ring R, and localise at p

everything not in p becomes a unit

[tough imp]

Yup

So frankly

I don’t know the varieties side of this

Lmao

But for schemes if you look at what happens on Spec

It makes some sense I think

You could try to think about how Spec A and Spec A_p interact

[gritty widget]

I am a simple man

[tough imp]

And then just tell yourself

[gritty widget]

i see schemes

[tough imp]

“Spec is a space”

[gritty widget]

i smile and nod

I still don't even have geometric intuition for the spectrum of a ring

[tough imp]

Welcome to the club

[gritty widget]

someone draws this picture from the little red book

seemingly its a great picture

[tough imp]

Hahaha that thing is so non-helpful

Spec Z[x]

[gritty widget]

im just like

why the fuck is an ideal a squigly dot

but some other ideal is this curve running through stuff

[tough imp]

Oh lol

[gritty widget]

how does this help you think about your ring

[tough imp]

That’s describing what the closure of that ideal gets

It’s just encoding the idea that a non-maximal ideal

Has ideals above it

[gritty widget]

can i just double check one more think about localisations and units

we just need that (u/v) is a unit in k[x] localised at x-a

[tough imp]

Yup

[gritty widget]

since u and v are both non zero at a, they are units in the localisation

[tough imp]

Yup

[gritty widget]

or even just the inverse of u/v is v/u is in the localisation

[tough imp]

Right but you’d need to know that v/u exists or u/v really, even

But that’s coming from non-zero at a

Aka both are units

[gritty widget]

yeah thats g

ty king

do you have any idea for this?

I thought the whole point of affine space was that you were free to do translations?

[tough imp]

Uhhh I dunno I guess you have to show that the problem is invariant under translation

Here’s a stupid as fuck statement which wouldn’t be

The line x = 3 (in A^2) contains P

Okay WLOG by translation P = (0,0)

It’s false

[gritty widget]

so we just need to see that the translation does affect the derivative

or

doesn't make a non zero derivative zero

[tough imp]

Right

But you’d also need to know that the conclusion is invariant under translation

Since even if you can show that you maintain the hypotheses

If you can’t go back and show the conclusion of the non-translated problem it’s pointless

[gritty widget]

So basically if P is (a,b), I have the translation phi(x,y)=(x-a,y-b). Then I show that if y-y(P) is uniformizer

[sleek thicket]

Let U be S^1 minus a point

And V the same but for a different point

We have fiber bundles U × S^1, V × S^1

Glue them along the iso τ : (U cap V) × S^1 -> (U cap V) × S^1 which is τ(x, z) = (x, z) for x in the first component of the intersection and τ(x, z) = (x, - z) for x in the second component

This gives us some fiber bundle E -> S^1 with model fiber S^1 and structure group {1,-1}

what does E look like?

[tight agate]

probably klein bottle

[sleek thicket]

Oh lol that would make sense

If I pull it back along the squaring map S^1 -> S^1 I bet I get the torus

so yes, seems plausible

[tight agate]

yeah it's the klein bottle, if you disconnect the circle you have a tube, and you're identifying one end of the tube to the other end 'from behind'

[sleek thicket]

well that's annoying

[gritty widget]

from behind

[sleek thicket]

oh my

my next problem is to exhibit the klein bottle as a fiber bundle over S^1 with structure group {-1,1} and fiber S^1

but I am still stuck on this problem

🙃

Also ttrwea they haven't emailed me back

The fields people

[tight agate]

what is the problem?

[gritty widget]

monka fucking S

good luck

[sleek thicket]

yeah

I'm gonna be mad

[gritty widget]

just fucking PRAY your letters got in

[sleek thicket]

Sinc emy letter writers confirmed

yeha lol

[gritty widget]

that's so scary

i hope it goes well for you

.

[sleek thicket]

Yeah, me too :/

[sleek thicket]

it's a little hard to state because I think Lee has a different definition of fiber bundle than usual

it's not just "locally homeomorphic over the base to a product space"

The structure group is packaged into the definition

And so the problem is to show S^1 × S^1 -> S^1 can be given a nontrivial fiber bundle structure

With structure group {1,-1}

[tight agate]

There arent too many options

[sleek thicket]

Yeah

[tight agate]

the notation is a bit confusing tho

what do you mean by S1 X S1

just as a set?

or are you saying that the torus is somehow a nontrivial S1 bundle over S1

[sleek thicket]

it is somehow a nontrivial S^1 bundle over S^1

[tight agate]

what does trivial bundle mean?

[sleek thicket]

the question is "what does fiber bundle mean"

basically like, define a locally trivial bundle over $B$ with fiber bundle $F$ to mean a space $E$ with a continuous $\pi : E \to B$ such that there's a cover of $B$ by sets $U$ on which the spaces $\pi^{-1}(U)$ and $U \times F$ are homeomorphic over $U$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

so this is what people usually mean by fiber bundle

now say you have some topological group $G$, called the structure group, which acts faithfully and continuous on $F$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

if you have two local trivializations $\varphi : \pi^{-1}(U) \to U \times F$ and $\psi : \pi^{-1}(V) \to V \times F$, we say they're $G$-compatible if there's a continuous function $\tau : U \cap V \to G$ such that $(\psi\circ \varphi^{-1})(x, f) = (x, \tau(x) f)$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

an atlas for $E$ is a covering by $G$-compatible local trivializations

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

then a "fiber bundle with structure group $G$ and model fiber $F$ over $B$" is a locally trivial bundle $\pi : E \to F$ equipped with a maximal $G$-compatible atlas

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

a morphism $E \to E'$ is what you'd expect, i.e. it's a continuous map over $M$ such that the local representation with respect to trivializations in the atlases for $E$ and $E'$ is like acting by some continuous function $U \to G$

[gentle osprey]

Shamrock -> E -> B

[sleek thicket]

and a trivial bundle is one which has a global trivialization (in its atlas), or equivalently one which is isomorphic to a trivial bundle

[tight agate]

okay, say we have a morphism $f: S^0 \rightarrow G$. Then you get an S1 bundle over S1 by taking the pushout of $i_1: S^0 \times S^1 \rightarrow I_1$ given by $(x,v) \mapsto (x,v)$ and $i_2: S^0 \times S^1 \rightarrow I_1$ given by $(x,v) \mapsto (x, f(x)v)$.

[gentle osprey]

Brofibration

[tight agate]

let $S^0 = {p_1, p_2}$

[gentle osprey]

Brofibration

[sleek thicket]

what is I_1?

[tight agate]

I_j = [0,1]

the interval

sorry

[sleek thicket]

How is (x, v) a point of I_1?

if x in S^0 and v in S^1

[tight agate]

I've viewing S^0 as the boundary of both of those intervals

[sleek thicket]

bdrt? boundary?

[tight agate]

wtf

boundary*

yes

[sleek thicket]

I still don't understand the definition i1(x, v) = (x, v)

like, right hand side isn't in I1?

[tight agate]

oh

[sleek thicket]

oh I1 \times S^1

[tight agate]

it should be I_1 X S1

[sleek thicket]

okay you're just gluing two trivial bundles yeah

[tight agate]

yes

[sleek thicket]

this is what I was trying to do above

[tight agate]

yes

[sleek thicket]

with f sending one point to 1 and the other to -1

[tight agate]

so your options are all the maps p_0, p_1 to +1, -1

yeah both to -1 might be the torus

yeah both to -1 might be the torus

[sleek thicket]

right exactly

and the other is the klein bottle

so...?

both to -1 is compatible with the global trivialization of the torus

since it's locally given by multiplication by -1

if you take U to be S^1 minus a point and V to be S^1 minus another point, and define trivializations of S^1 x S^1 on U, V by phi(x, v) = (x, v) and psi(x, v) = (x,-v), the transition function will be the one you're talking about

and these trivializations are compatible with the global trivialization of the torus

[tight agate]

ig the confusing bit is when you're working with based stuff, two of those maps dont turn up

so the trivial one is the torus

[sleek thicket]

I'm not sure what you mean by based stuff

[tight agate]

and the nontrivial one is not

w/ basepoint

[sleek thicket]

i got that part lol

are you talking about based maps U cap V -> {-1,1}?

with a basepoint in like U?

[tight agate]

based maps S^0 to structure group

[sleek thicket]

sure

[tight agate]

the reason for this is people often interpret sphere bundles over other spheres are elements of homotopy groups of the structure group

and homotopy groups of the structure group are homotopy classes of based maps from some sphere to the structure group

so there should only be "2" S1 bundles over S1 with that structure group

as it's 0th homotopy group has order 2

in general bundles with structure group G over S_n correspond to elements of pi_n-1(G)

[sleek thicket]

I'm gonna think about that other bundle

the one corresponding to a nonconstant map S^0 -> G

no longer convinced it's the klein bottle

I should be able to compute the fundamental group

actually homology is probably easier because the intersection is disconnected, it'll distinguish them either way

[tight agate]

one way to study the total space is open up all the fibers into intervals

and the base S1 into an interval

you get a rectangle

and you can see how the sides are being identified

and you know which identifications give you a torus, a klein bottle, projective plane etc.

and using this you can also immediately see that the projective plane is not an S1 bundle over S1

using the fibration long exact sequence for homotopy groups is another way, but you will have to show that one of the extensions is nontrivial

[fading vale]

im not sure i understand this proof exactly

specifically the part about why h_0 and h_1 being functors means that changing the decomposition of the interval doesnt do anything

i mean that makes intuitive sense to me because you can combine the h_0 bits and combine the h_1 bits since theyrefunctors

but im not sure how youd prove it formally

[pastel linden]

groupoids

[sleek thicket]

Moth, I think that's all it's saying

like

Say you refine the partition

Then you'll split up the wi as compositions

All happening within one block of the partition

So on that wi you'll either get h0 or h1

And you can apply functorality to split up on the wi and get a new composition

also this is the proof of van kampen in my head, it's the first one I saw

[obtuse meteor]

based topology and groupoids

(not based, but meme)

[gritty widget]

For affine space, we have that the open sets D(f) form a basis. I'd like to show that on projective space D(f) where f is homogenous form a basis

I'm struggling a little bit because the topology on P^n is defined in terms of closed sets

[sleek thicket]

you should be able to reduce to the affine case, I think

well okay this is overcomplicating things sorry

[gritty widget]

so we give P^n the quoitent topology from A^n

[sleek thicket]

Sure

So what are the closed sets?

Also not A^n

A^(n+1)\{0}

[gritty widget]

oh yes

thats the important part that i think im getting tripped up on

so Y is closed in P^n if the inverse image of Y is closed in A^(n+1){0} if the inverse image of Y union {0} is closed in A^{n+1}{0}

[sleek thicket]

sure

And what are the closed sets of A^(n+1)\{0}?

[gritty widget]

they are the closed sets of A^{n+1} intersected with A^{n+1}{0}

so any closed set that does not contain 0, and any closed set containing zero we can remove 0

but a closed set in A^{n+1) is Z(f)

hmm

maybe it is true in general that if B is basis for X, and q:X to X/~ is a quoitent map then q(B) is a basis

[gritty widget]

okay this didn't work out

So now I am trying to to show directly that D(f) is a basis.

We have the intersection propert since D(f) intersect D(g) is D(fg) but its not clear to me that the D(f)'s cover projective space

well they do but I can't see why

[tepid quarry]

Sorry for the inconvenience. I want to start learning about higher dimensional shapes. I know high school geometry, Calculus till calculus III, linear algebra, very basic topology.. Can anyone recommend some books I can read?

[elder yew]

Munkres point set topology

Is a good starting point

[tight agate]

this is true whenever the quotient map is open

and the quotient map is open in this case

[tepid quarry]

Ok. Does it talk about polytopes?

[elder yew]

I'm not sure. I don't know anything about polytopes. But you should learn more about point set topology

[tepid quarry]

I see. Alright.

[gritty widget]

Are you still in high school?

Are you interested in learning about high school circle theorems but for a 4 dimensional circle kind of thing?

[tight agate]

4-dimensional circle

S3 is hard enough lmao

[sleek thicket]

if anyone remembers the problem I was stuck on last week about the tautological line bundle vs tangent bundle

It turns out the problem was false as stated

the diffeomorphism given in the statement is orientation reversing

🙃

[gritty widget]

lol

rip

[gritty widget]

yeah I was offering "bad" examples of stuff

[dusk heron]

Through the double cover $\operatorname{SU}(2)\to\operatorname{SO}(3)$, the round metric on $\operatorname{SU}(2)\cong S^3$ induces a Riemannian metric on $\operatorname{SO}(3)$. Is this the only "natural" metric to give to $\operatorname{SO}(3)$? Does it arise in a more direct way, without reference to $\operatorname{SU}(2)$?

[gritty widget]

i assume the person is a high school student who just wants to see some "sexy" maths

[gentle osprey]

gustavn64

[sleek thicket]

iirc the metric on SU(2) is also called the berger metric (it comes with a parameter but we're setting that to 1) and is bi invariant. So the descended metric on SO(3) should also be bi invariant

This doesn't uniquely specify it but it makes it seem a little more intrinsic

[gritty widget]

lol burger metric

[sleek thicket]

Eg the geodesics at the identity are the one parameter subgroups

lol burger metric

[tough imp]

lol burger metric

[sleek thicket]

Exercise 3.10 in introduction to riemannian manifolds by Lee

🍔

so I guess maybe the moral of my answer is that the connection coming from this metric is determined by the lie group structure

[obtuse meteor]

borgar

[cursive flume]

How can one prove this?

[sleek thicket]

why do you need connectedness here?

oh I didn't know the correct definition of simple lie group, my bad

[cursive flume]

Lorentz is not connected

[sleek thicket]

I thought it was just "no nontrivial normal closed subgroups"

[cursive flume]

But somewhy phys books and wiki states this

[cursive flume]

This

Idk why and how to prove

[sleek thicket]

so the image of such a representation is going to be precompact, since U(n) is compact

[cursive flume]

Maybe i didnt mention but i want to represent it on hilbert spaces

So unitary reps of lorentz on hilbert spaces

Idk if this helps/makes things more difficult

[cursive flume]

Can you link the paper pls

Ty

[summer jolt]

Hey guys, I have a question about the definition of tensor fields. I was told the usual definition that states a tensor field (r,s) is a multilinear map over smooth functions that eats r covector fields and s vector fields to give a smooth function. But then I was told that the torsion tensor is a (1,2) tensor because it eats 2 vectors and gives out a vector. I'm struggling to see why the two definition are equal.

[gritty widget]

you can identify vectors with operators on covectors

more specifically

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

there is a proposition in lee that touches on this, let me post it

this is stated fiberwise, but i'm sure you can find the analogous result for tensor fields

so the torsion isn't really a tensor field, but it's convenient to use this identification and say it is

(unless you want to start talking about shit like vector-valued forms)

[summer jolt]

@gritty widget I see. So essentially there is an associated map that is a tensor field?

Also is your $w$ an arbitrary convector field?

[gentle osprey]

snypehype46

(T*(Terra), -dτ)

[gritty widget]

to make it even more precise

given the torsion "tensor" $\tau \colon \mathfrak{X}(M)^2\to\mathfrak{X}(M)$, define a $(1,2)$-tensor field $\tilde{\tau}$ by $$\tilde{\tau}(\omega,X,Y) = \omega(\tau(X, Y)),$$ where $\omega$ is an arbitrary covector field and $X,Y$ are arbitrary vector fields

[gentle osprey]

(T*(Terra), -dτ)

[gritty widget]

and yeah

this "tau tilde" as i've written it is basically the same thing as tau

you should do the linear-algebraic exercise in the image i posted

it's pretty straightforward

[obtuse meteor]

linear algebra is good for the soul

like vegetables

[gritty widget]

agreed

[river granite]

yet both can be yucky sometimes

[pastel linden]

pin this

[pastel linden]

with this

[gritty widget]

[river granite]

https://images-ext-2.discordapp.net/external/haUc0h69ouDjhpNbgudyblotNaoQa4fRkqJhaBxeqio/https/media.discordapp.net/attachments/268882317391429632/799769127567753226/ezgif-7-c7d6b8d6ef40.gif

[sleek thicket]

it was funny

let's all remember

[gritty widget]

holy fuck the estonia picture

why is that so fucking funny

[sleek thicket]

That article was amazing

Did you see the original article?

I'm searching

Iirc it was posted to r/badmathematics

https://blog.usejournal.com/monoids-to-groupoids-492c35105113

[gritty widget]

[sleek thicket]

This channel will never reach the level of meme pins of #category-theory

[gritty widget]

:)) you guys either need to read the full question - or you guys need to expand your notion of what mathematics is. Or both :)) Mathematics among other things is the study of pure structure. All this human activities, our institutions, everything - is structured - our mind is structure, our understanding, is the way it is because it optimizes something. Any life form is a solution to such a mathematical setup expressed as our environment. Nothing is random. A tree is not random shape. It's a solution to an optimization problem. In the tree example, it optimizes the amount of energy it captures minimizing the amount of energy / matter it uses, since if it does that well enough it can grow even bigger. And once a balance is reached then that's a tree capped and we call that a tree species. Is very hard to beat it even if you simulate it, you basically end'up with the exact same structure as nature. Now if that analogy holds - then our institutions and such - are a mathematical structure which work under the same rules.
When it comes to organizing civilization - which is basically what my question is about - then we need to find the principles underneath it all. Trouble is im not smart enough to figure this one out by myself, and also we dont have time, and i also can't find a discord channel dedicated to software architecture. Because that's also very close to such an endeavor. I don't think im in the wrong place. You guys need to flip your frame and see things from this other angle. This is why im talking about having a principled design here. If i just go and make some random setup, and later things need to be changed - which most likely will - then what? Is gonna be a complete mess. Lots of people very angry. Reorganizing everything while thousands of people are already using it. So is one of this - get it wright or get it pretty right form the beginning types of problems or you are screwed. I rely need to not make the situation even crappier then it is.

pin it

[sleek thicket]

No

You can't just steal pins

It's worth reading the whole thing

It's so funny

[obtuse meteor]

staring at CW complex problems

is going to make me tear my eyes out

[ivory dragon]

reminder that it defined a categorical quiver as, paraphrasing, a "bunch of arrows in a graph that converge to something useful"

they heard the word "quiver" and know "arrows" are a thing in category thery

voila

[sleek thicket]

Omegalul

[fading vale]

ummm is that necessarily true though

since the w_i in our first partition could be totally different from our w_j in the second

though i guess since you can combine the w_j and divide to get the w_i it doesnt really matter

thats probably it

but then u get problems cuz its not like u can just combine everything since you have two functors h_0 and h_1 rather than just one

[sleek thicket]

Aren't we refining a decomposition?

So we'll split up one of the intervals in the domain of the definition of the wi

In my brain we show that you can define this thing on some partition and then that it's invariant under refinement, and since any two partitions have a common refinement that's enough

[shut moat]

I thought this was a joke but then I went back and found the original post oh my god lmao

[uncut surge]

oh my he had a discord server too

the original thing was posted on april 1st so there was still some doubt but that post was completely sincere

[obtuse meteor]

hmm

I think I'm realizing what's like

frustrating about cw complexes

there's these methods of induction that are difficult to phrase as a lemma or theorem

the one coming to mind to me is like

to show a set S is closed in a CW complex X you can pull things back to characteristic maps for X^n and show it's closed, and it's fine to assume that S is closed in X^(n - 1) when you do that checking.

And this is valid bc if you have both of the above properties S is closed in X^n

but I don't want to state that as a lemma bc there's also a lemma that if you're closed under pull back for all characterstic maps then you're closed in X

and these are essentially the same to show / prove

maybe should exploit X^n is a CW complex

I am being annoyed at this bc it is hard but grrr

[rugged swan]

I don't get what's wrong, to show S is closed you'll have to prove that S cap Xn is closed in every Xn, to do that you can use induction

[obtuse meteor]

what's wrong is that I want to use that at each stage it's enough to show pull backs under characteristic maps are closed

this is mostly annoying because there's a lot of data

[gritty widget]

how to prove the poincare group is not simply connected?

[fading vale]

i keep vanishing cuz im falling asleep lol sorry

but i dont think its necessarily refining

i think it could just be any other partition of the interval such that w([t_i, t_i+1]) is in X_0 or X_1

or rather their interiors

any two intervals having a common refinement would work but i think thats somewhat harder to prove because the refinement also needs to have a gamma: {0, ..., m+1} -> {0, 1} and showing that any two intervals have a refinement with this property seems kinda hard

[sleek thicket]

You just union all the partition points

and that's a common refinement

[fading vale]

oh wait lmao yea

[sleek thicket]

Yeah

[fading vale]

nvm im dumb

[sleek thicket]

It's okay

But I'm 90% sure this is what they're doing

It's the standard proof

[fading vale]

ok cool that makes sense

also this is a way more intuitive way to think about van kampen then the normal one

just saying groupoid preserves pushouts

[sleek thicket]

I love it!

Well not all pushouts

I don't think that's true

Brown's book talks about this

What kinds of pushouts it preserves later on

[fading vale]

neat

er

i wonder if theres a specific name for this kind of thing 🤔 i guess you can just say it sends the coproduct to the coproduct

or its compatible with coproducts

[sleek thicket]

Wym?

[fading vale]

uh like

idk if theres a more formal notion of this but when we take the pushout of X_0, X_1, and X_01 we get X, which is basically the coproduct of X_0 and X_1 with X_01 in each identified

[sleek thicket]

Yeah so there's two ways to interpret this

[fading vale]

omg fuck

[sleek thicket]

If yr category is cocomplete then pushouts are coequalizer of coproducts

[gentle osprey]

Moth | not male

[sleek thicket]

Coequalizers can be thought of as quotienting/collapsing

It sounds like you just want to say it preserves pushouts

because that's what both of these are

[fading vale]

but u said it doesnt always right?

[sleek thicket]

Pushouts are coproducts with stuff identified

Right, we need to have two open subsets and pushout along their intersection

[fading vale]

mhm

idk if theres a distinct name for that

[sleek thicket]

Okay sure the thing in your book is a little more general sorry

But interiors covering is just like

Replace by interior lol

[fading vale]

yea

[sleek thicket]

I guess it's a little more general

[fading vale]

hahahahaha

[sleek thicket]

You saw nothing

anyways

[fading vale]

literally orwells nightmare

[sleek thicket]

It's still very specific

Everything is coming from a big space

Like the maps in the top left of the pushout are injective, for one

[fading vale]

mhm

[sleek thicket]

In fact the square is both a pushout and a pullback

The point being it's not true that Π1 preserves all pushouts

I'm trying to remember/find an example

[fading vale]

even so to me it for sure makes it more intuitive to think of this as like the coproduct (coequalized?) of spaces fulfilling the assumptions in Top becomes the coproduct in Grp

vs the normal proof which is kinda just like

"lets show that directly"

[sleek thicket]

Yeah, I agree!

This is how I like to think of it too

Well in Grpd and not grp for me

[fading vale]

mhm

[sleek thicket]

I'm just pointing out that this is a little too strong

For the statement of svkt

[fading vale]

right

i think i got confused cuz

[sleek thicket]

Also "coequalized coproduct" means "pushout"

[fading vale]

he literally says "thus (2.6.1) says that Grpd preserves pushouts"

[sleek thicket]

They're the same if all colimits exist

hmmm

That's odd

[fading vale]

i assume he means in this specific case and its just like

bad writing

lol

[tight agate]

If you have a cover of the space by path-connected open sets, closed under finite intersections, then consider them as a category

where morphisms are inclusions

You have a functor from the category

by sending each open set to the fundamental groupoid

[fading vale]

that checks out i think

[tight agate]

and van kampen says the fundamental groupoid of the space is the colimit of the diagram

[fading vale]

but still i think its a bit sloppy to just say that \prod preserves pushouts in general lol

its not that big a deal tho

[tight agate]

are there operations on real/complex k-theory that are not generated by the adams operations?

[tight agate]

nvm

[gritty widget]

[violet sonnet]

hi im struggling to understand the argument here

He's saying , since B is a basis for T', a set U in T' is a union of sets in B, ok i get this
then he says; this implies U is in T
is it true that if B is a basis for T, then any union of sets in B is in B
like what if T = {(0, 1), (2, 3), (4, 5), (0, 1)U(4, 5)}, then B = {(0, 1), (2, 3), (4, 5)} is a valid basis. But (0, 1)U(2, 3) is not in T

[cloud owl]

that's not a topology?

topologies are closed under like intersection and finite union, or finite intersection and union, i forget which

[violet sonnet]

i see, i forgot this crucial point, thanks

[tough imp]

also no

a basis is not closed under union

or even under intersection

[cloud owl]

but it's that if (0, 1)U(2, 3) is not in T, T isn't a topology, right?

am i being a fool?

oh right you were replying to them

i'm blind

[marsh yarrow]

Hi guys, can someone help me understand the steps for getting the surface area of one face of a Reuleaux tetrahedron (spherical tetrahedron)?
To be specific I am talking about the steps used in this article:
http://www.math.unl.edu/~bharbourne1/ST/sphericaltetrahedron.html

Many people don't know what a Reuleaux tetrahedron is, so here's a brief definition of the shape: the intersection of four balls of radius s centered at the vertices of a regular tetrahedron with side length s. The spherical surface of the ball centered on each vertex passes through the other three vertices, which also form vertices of the Reuleaux tetrahedron. Thus the center of each ball is on the surfaces of the other three balls.

I understand, so far, that the entire spherical cap which one face of the spherical tetrahedron rests on, can be calculated using 2*pi*(r^2)*(1-costheta), and theta equals 60 degrees so the area of the spherical cap is pi*r^2. Then it goes on to calculate what fraction of that area contains the face, which is the dihedral angle of a tetrahedron divided by 2pi multiplied by the total spherical cap area to get a fractional area. After this, I get kinda lost. The article says it uses the spherical excess formula which is saying that a curved triangle constructed from three points of a sphere has a surface area equal to the sum of its interior angles subtracted by pi. However, the article not only sums the interior angles and subtract pi from it, but it also multiplies it by r^2. Why does he do that?

Also I completely don't understand this step "so we get ((arccos(1/3)/2)-(3arccos(1/3)-π ))r^2 for the area of one wedge and hence 3 times that plus the area of T for the area of T', which again gives r^2(2π - (9/2)arccos(1/3)) for the area of T', which is just one of the four faces of the spherical tetrahedron." Why does he subtract the surface area that he got from the spherical excess formula from the fractional spherical cap area? And this makes me even more confused because what area was the spherical excess formula used for calculating then? Sorry for this lengthy question, but I would appreciate any help via pings!

Sorry again scrap that entire last comment, I understand why he subtracted the surface area from the se formula from the fractional sc area now, and so I also know what he used the se formula for now too. So I only have one question really, why does he multiply the sum of the interior angles - pi by r^2 to find the surface area of T (labelled in the diagram below)? (please read everything above first!)

[tidal cedar]

Yeah your basis set basically generates the topology by taking all the unions

[alpine rain]

Hello I have a question, is there topological property (or metric space property) which imply you can exhaustively "search" so that you could find actual witness for any predicate, given that you could find out if any given element satisfy the predicate or not?

[marsh forge]

i think you would also need to place restrictions on the predicate

for this to be possible

[alpine rain]

Restrictions?

What kind of hmm

[marsh forge]

Well, there are questions you can ask that take place in very nice topological spaces

that probably wouldn't be searchable

in any topological sense

[alpine rain]

Hmm I see, I mean do you know what kind of restriction I should put to the predicate?

[marsh forge]

this isn't something I've thought about before

@dim meadow is there some kind of computability answer that comes to mind

[dim meadow]

Yeah this sort of thing is considered I think

But you wouldn't be able to do any predicate, just the sigma_1 ones

[alpine rain]

Hmm what is sigma_1

Not sigma type related right

[dim meadow]

There are notions of computability for metric spaces that work well. You essentially need a dense set which you can list out

Sigma_1 formulas have only one existential quantifier

[alpine rain]

Oh, I see. Peculiar one I guess

[dim meadow]

They are of the form \exists x \psi(x) where \psi(x) is computable

[marsh forge]

it ends up being a very natural classification

for a lot of reasons

[dim meadow]

I actually don't know too much about this, I know one of the people I am in contact with (Johanna Franklin) did work on computable metric spaces at one point

[alpine rain]

psi(x) computable? So like, it is decidable proposition?

[dim meadow]

I can look at her publications and probably find a nice definition for what a computable metric space should be

[marsh forge]

oh i wrote about those

but other than what i wrote down i dont have a reference in mind lol

[alpine rain]

For context, some compsci person bragged about exhaustive search which is topologically generalizable

Apparently they thought compactness is enough