point-set-topology

[bitter yoke]

I think you need more assumptions because I don't think this is true in all topologies

[ivory dragon]

uh, the problem is that that statement is false

consider the trivial/indiscrete topology on a set X

then the only open sets are X and the empty set

so X - {a} cannot be open for any a (unless X only has one element)

are you referring specifically to the euclidean/standard topology on R^n?

[bitter yoke]

This is true if your space is hausdorff since finite sets are closed in hausdorff

[ivory dragon]

^

[fathom cave]

okk so i worked out the transformation and that term in the parantheses transformed as $Z^{i'}_{j'}=Z^i_jJ^{i'}i J^j{j'}$ which i should have seen

[gentle osprey]

Zero0

[thorny flare]

Yeah it was just for R but I figured it out

[gritty widget]

Can someone help me compute the hodge start of $sin\theta d\theta \wedge d\phi$?

[gentle osprey]

lime_soup

[gritty widget]

so i know its going to be a one form and it will be f dr

but in trying to do this calcuation i need to know what the metric on a sphere in spherical coordinates is

[sleek thicket]

the metric in spherical coordinates is dr^2 + r^2 dθ^2, you can compute this by pulling back the form dx^2 + dy^2 + dz^2 along the spherical coordinates map U -> S^2

So an orthonormal frame would be
\begin{align*}
X &= \frac{\partial}{\partial r} \
Y &= r^{-2}\frac{\partial}{\partial \theta}
\end{align*}

[gentle osprey]

Shamrock (not a furry)

[sleek thicket]

oh wait sorry this is wrong

I've confused myself, this is the metric on R^2\{0}. Sorry lime_soup

I think it should be g = dr^2 + r^2 dθ^2 + r^2 sin(θ)^2 dφ^2. You should check that though

[gritty widget]

Okay

[sleek thicket]

sorry just to check, you want spherical coordinates on (an open subset of) R^3}{0} right?

Not on S^2?

[gritty widget]

Oh sorry yes

[sleek thicket]

So if my metric computation was right then an orthonormal frame would be
\begin{align*}
X &= \frac{\partial}{\partial r} \
Y &= r^{-2}\frac{\partial}{\partial \theta} \
Z &= r^{-2}\cos(\theta)^{-2}\frac{\partial}{\partial \phi}
\end{align*}
and you can use the definition of the hodge star in terms of an orthonormal basis to compute it for your form

I think it'll be sin(θ) dr but still not sure

[gentle osprey]

Shamrock (not a furry)

[gritty widget]

Now does dr have a “singularity” at the origin

As in is dr not defined properly at 0

[sweet wing]

yea it isnt afaik

r ranges from is [0,\infty) only

[gritty widget]

Surely you mean (0,\infty)

[sweet wing]

r can take on 0 but dr doesnt make sende

[gritty widget]

Oh sorry yes

[sleek thicket]

Oops

Ah yeah sorry

I see exactly what I got wrong

I didn't convert dθ and dφ into my basis X, Y, Z

If you do that you'll get the answer here (probably)

[gritty widget]

One more think

Thing

How do you calculate the divergence of a differential form

I’m used to calculating divergence of a vector field

And I know differential forms are related to vector fields

[cold vine]

Isn't a cone of any space X always semilocally 1-connected since cones are contractible?

(I guess sl1c and semi-locally simply connected are the same thing)

[uncut surge]

Yeah "0-connected" and "1-connected" are the fancy big boy names for "path connected" and "simply connected"

[cold vine]

Ok thanks!

Any hint on how to show that in a cone of the Hawaiian earring any nbd of (0,0,0) contains a non contractible loop`?

[uncut surge]

Can that be true? If I take the whole cone as the neighbourhood, that should be contractible and thus simply connected like you said earlier

My topology brain is rusty but idk

Maybe you're rather looking to prove that (0,0,0) has some neighbourhood in which not all loops can be contracted

[cold vine]

Ah that is correct. I was thinking that the Nbd would be H x [0,1/2)

but I'm not sure if this works since I think union of all half circles containing (0,0,0) cross the interval might be open here

[patent mural]

I'm new to affine geometry, just had the first lecture, but i've received a list of problems. Can someone give me some intructions on how to solve, at least how to start?

Let a, b, and c tree distinct points in vectorial space V(dim V >4) and
p = a+<d1, d2> a plan.
Determine a linear 4-dimensional variety A witch contains the points a, b, c and is parallel with p.

[shut moat]

the keyword you're looking for is the exterior derivative

[cold vine]

Is it true in general that a path from x to -x pushes down to the generator of fundamental group instead of 0 when we have covering from sphere S^n to projective space RP^n ?

In this https://math.stackexchange.com/questions/567434/on-double-covering-of-projective-plane-and-map-preserving-antipodal-points

[gentle osprey]

slimvesus

[cold vine]

Ah I get this mostly

[marsh forge]

As explcitly phrased

i think the answer is technically no

because it might be some multiple of the generator

you want like a nice straight path from x to -x

[cold vine]

The only thing I'm not seeing is why would some antipodal points mapping to a trivial loop lead to all of them being mepped to trivial loop

[marsh forge]

Oh wait I think you're correct because we are mixing paths and loops, and I think your claim about antipodals is also correct, my b

[cold vine]

Thanks so much! ^^ I mostly get it, I'll think some more to fully absorb it

[marsh forge]

Ok this is a bit geometric for me to spend much time on, but if you take S^2, do a partial projection so that you get D^2 (basically, fix a meridian, and quotient the hemispheres), and look at what happens to paths its a lot easier to see their homotopy class

And I think that this works out nicely

[thin bramble]

So I'm having a hard time trying to find an example of an isometry which is not a linear map; 2) a linear map which is not an isometry

[marsh forge]

For paths that take place entirely in one hemisphere this works immediatley, and for more comlicated paths you need to think about what happens as hemispheres are crossed, but you should just see the paths enter the center of the disk on one of the two sides and then push the paths up to the border

Killerwhale: What have you tried

I think the second is probably you overthinking

and the first is you forgetting a very strict fact about linear maps

@thin bramble

[thin bramble]

they aren't isometries?

[marsh forge]

well they arent all isometries

some are

What is an isometry?

[thin bramble]

A distance that preserves a pair of points

A transformation which preserves the distance between any pair of points is an isometry.

T(U+V) = T(U)+T(V) is this what you talking about simple description?

ohh that was the simple description

So will this work ? translation by (0,0)

[marsh forge]

does that preserve distances?

[pastel linden]

I think a linear map is an isometry iff it is inner product preserving, so you can probably cook up an example in an inner product space pretty easily

[gentle osprey]

slimvesus

[sleek thicket]

No

Err

The divergence of a form isn't well defined as far as I know

But you can talk about the divergence of a vector field on a riemannian manifold

(and you get a smooth function)

My first thought would be slime means to take the divergence of the vector field paired with their form

[marsh forge]

i think its easier than this if you want to preserve distances lol

youre overthinking

[shut moat]

yeah that's what I was referring to

[sleek thicket]

That's not the exterior derivative though?

Like it's a 0 form and not a 2 form

[shut moat]

like $d(F_z dx\wedge dy - F_y dx\wedge dz + F_x dx\wedge dy) = \text{div}(\mathbf{F})d^3\mathbf{x}$

[gentle osprey]

~S^1

[shut moat]

I hope I got the signs right lmao

[sleek thicket]

oh sure that looks plausible

[shut moat]

so yeah it's not a literal divergence but it falls out of it right

[sleek thicket]

yeah I just meant you're not taking the exterior derivative of the form

[gritty widget]

physicist spotted

[sleek thicket]

lol

[shut moat]

[shut moat]

why do you need a metric?

[sleek thicket]

well, what's your general definition of the divergence of a vector field?

[shut moat]

hm, well it'd be the coefficient of the exterior derivative of det[V, . ,.] which is a top dim form

i think

[sleek thicket]

Right so the point is we need a distinguished top dim form

[shut moat]

ooh, so this'd be replaced with the canonical volume form

makes sense

[sleek thicket]

Exactly!

So for notation, given a $(p+1)$-form $\omega$ and a vector field $X$, let $X\lrcorner \omega$ be the $p$-form $(X \lrcorner \omega)(Y_1,\ldots,Y_p) = \omega(X, Y_1,\ldots, Y_p) $

[gentle osprey]

Shamrock (not a furry)

[sleek thicket]

does this makes sense? It's called interior multiplication

[shut moat]

yeah

$d(X\rlcorner \omega) = \text{div}(X) \omega$?

[sleek thicket]

the divergence of a vector field $X$ is then specified by the property that $d(X \lrcorner dV_g) = (\div X) dV_g$

[gentle osprey]

~S^1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Shamrock (not a furry)

[sleek thicket]

Haha almost

[shut moat]

oops

lmao

[sleek thicket]

Oh ω is the volume form?

Then yeah!

[shut moat]

[sleek thicket]

Some care must be taken

The volume form only exists on an oriented riemannian manifold

but a general riemannian manifold does not come with an orientation

However it turns out that this definition of div X is well defined independent of the orientation, since swapping the orientation flips the sign of dV_g and dV_g shows up once on each side

So this is well defined for all orientable riemannian manifolds

Does that make sense S^1?

[shut moat]

yeah!

Could you extend it to any orientable manifold that has a volume form? Ik symplectic manifolds or whatever have volume forms attached to them

i don't see why you need a metric specifically

[sleek thicket]

Well hold on a sec lol

My next point was going to be we can extend to all riemannian manifold

[shut moat]

oh shit

[sleek thicket]

On any orientable open submanifold there's this function which is unique specified by some global equation

Right?

in a subset diffeomorphic to R^n, we can define the divergence as a local function

[shut moat]

I'm not sure what you're referring to

[sleek thicket]

Does that make more sense?

locally any manifold is orientable

[shut moat]

what's this unique function specified by a global equation?

locally any mfld is orientable --> local divergence
yeah this makes sense

[sleek thicket]

div(X)

It's uniquely determined at each point by this equation d(X _| dV) = div(X) dV

[shut moat]

oh yeah, cool

[sleek thicket]

Sorry I'm actually wrong

it's not global

[shut moat]

oop

[sleek thicket]

The whole point is we can't define dV globally

The important thing though is we have some kind of uniqueness

On an orientable riemannian manifold, for any choice of riemannian volume form we have this equality

yeah?

[shut moat]

yeah

[sleek thicket]

Okay cool

So what this implies is that if we have two orientable open subsets U, V of (M, g) the functions div(X|U) and div(X|V) agree on the overlap U \cap V

So we can use the gluing lemma to find a global smooth function div X

[shut moat]

oooh

doesn't gluing lemma require that U and V are closed

oh wait no that's pasting lemma

is that different lmao

[sleek thicket]

idk lol I just said gluing lemma to make sure you understand

I think it's only for continuous functions too

You certainly can't glue smooth functions on closed subsets, eg the absolute value isn't smooth

the point is if you have an open cover {Ui} and smooth function fi : Ui -> R such that fi(x) = fj(x) for x in Ui cap Uj there is a unique global smooth function f such that f(x) = fi(x) when x in Ui

because the collection of smooth functions forms a sheaf

[thin bramble]

A linear map L: V --> V is an isometry if
<Lv, Lw> = <v,w> like this?

[shut moat]

ah here we are

that was neat, ty sham

[thin bramble]

what is that suppose to tell me?

[cold vine]

I'm trying to show that if S is the circle S^1 and f:S->S is continuous map then assuming (f(z))^n = z where n>1 leads to contradiction. So we know z^n is a covering and now f is a lift of identity map since the triangle S -f-> S -z^n-> S , S-id-> S commutes. Is this a good start and where do I go from here?

[sleek thicket]

makes sense to me

So you know what the degree of a continuous map is?

[cold vine]

Yeah I've used it a bit

[sleek thicket]

Then that's my hint, try to look at degrees

[cold vine]

ok cheers ^^

[sleek thicket]

welp, it was nice following my bundle class

RG flashbacks

[gritty widget]

[sleek thicket]

i literally have no idea what's going on

[obtuse meteor]

Oop

[sleek thicket]

I saw this slide and my brain shut off

[gritty widget]

some funky connection stuff

[sleek thicket]

yeah we're generally talking about connections on vector bundles

[tidal wadi]

Jost's Geometric Analysis and Riemannian Geometry book is a good reference btw

His treatment of connections is quite nice in my opinion

If you want to challenge yourself, do what I did and learnt it via Kobayashi--Nomizu 😉

By the way, those notes just say that the covariant derivative is given by the flat Euclidean part (name the exterior derivative d) and the non-flat Euclidean part (which has curvature) given by the matrix varphi

The exterior derivative is a flat connection on the trivial bundle

[gritty widget]

good taste in books

[shut moat]

woah that looks really good

[tidal wadi]

By the way, the exterior derivative features in my video series on Euler's number

Although, I don't explicitly ever use the exterior derivative, you should try to find where it is lurking in the background: https://www.youtube.com/watch?v=rbmUqseGOOM

#notsoshamelessplug

Can someone grant me permission to post these videos in the resources page?

I can't post there for some reason

[obtuse meteor]

you really only need to post this in one channel

[tidal wadi]

ah ok, sorry about that

[gritty widget]

geometry

[tight agate]

[gritty widget]

[tight agate]

catwiggle

[gritty widget]

[fathom cave]

lmao

can someone explain what am i missing? why i cant do the thing on the right side

https://cdn.discordapp.com/emojis/667391296452952084.gif?v=1

oo lol

[ivory dragon]

what "thing" are you referring to, sorry?

[fathom cave]

the step i did right before going to the last step making it 0

[ivory dragon]

i dont really get what youre doing; are you trying to differentiate both sides wrt x'^k?

A presumably isnt necessarily a constant here

[fathom cave]

I am good with the first term in 3.42 but I dont understand why the second term cant be 0

$A^j\frac{\partial}{\partial x^p}\bigg(\frac{\partial x^{i'}}{\partial x^j}\bigg)\frac{\partial x^p}{\partial x^{'k}}$ which can be written as
$\frac{\partial^2 x^{i'}}{\partial x^j\partial{x^{'k}}}=0$?

oof

[gentle osprey]

Zero0

[gritty widget]

huh isn't this just the product rule

oh

[fathom cave]

ye

[gritty widget]

you're asking why the second derivative of the coordinate function vanishes?

let me tex something up

[fathom cave]

no i am asking why the second term is not vanishing

book

[gritty widget]

using primes on coordinates like this should be a felony

can you show us what the book says

i.e. the text below (3.42)

[fathom cave]

the pic i posted above and

shit i cut that part

[gritty widget]

can you show us what the book says
i.e. the text below (3.42)

[fathom cave]

ok

[gritty widget]

well it explains why the second term in (3.42) might not be zero, when the x' coordinates are not linear functions of the x coordinates

you do have to be a bit careful since in the second term there's an implicit sum

[fathom cave]

buut it should always be 0, should not it?

[gritty widget]

i don't understand how you went from the first thing to the second

[ivory dragon]

did you cancel the partial x^ps?

you cant do that

[gritty widget]

the second thing will always vanish, yes, because the derivative of any x' coordinate with respect to another is always 1 or 0, constant. but this rearrangement is not justified, as nami is probably typing

[ivory dragon]

you cant cancel partial derivatives like you (usually) can single-variable derivatives

the multivariate chain rule isnt quite that "nice"

[fathom cave]

i cant do chain rule with partial derivatives?

fuck i accidentletly deleted all the texits thing i was writtng

[gritty widget]

ctrl z

[fathom cave]

its gone

[nimble jolt]

alt f4

[fathom cave]

lol

[gritty widget]

put a magnet on your hard drive, the magnetic power will pull the text back to life

[tough imp]

Maybe your pc is thirsty. Try pouring water on it?

[fathom cave]

how is that different from this $\frac{\partial V}{\partial x^{j'}}=\frac{\partial V}{\partial x^{j}}\frac{\partial x^j}{\partial x^{j'}}$?

[gentle osprey]

Zero0

[tough imp]

That’s not how function composition works

I’m assuming you’re saying that V is a function of the x^j

And then x^j and x^j’ are different variables in that?

Correct?

[fathom cave]

xj and xj' are different coordinates

[tough imp]

Right, so this doesn’t make sense because that’s not how function composition works right?

This would assert that x^j is a function which takes in x^j’ in one coordinate

[fathom cave]

yes

xj is a function of xj'

[tough imp]

??

[nimble jolt]

https://tenor.com/view/stare-cat-confused-confused-look-serious-gif-16884211

[versed pivot]

Your statement and use of the chain rule is fine
The problem is when you commuted the partial derivatives corresponding to different coordinate systems

Keep the d/dx^k' on the outside, then you can fiddle with the product rule to get it into the form the book wrote

[tough imp]

Maybe I’m just a peanut brain

I will go back to swinging from vines

And playing with schemes

[fathom cave]

can we show me excatly where i commuted Seoin? i am not seeing it

[versed pivot]

This

On the left you have d/dx^j of x^i'

On the right you have moved the d/dx^j to the outside

Specifically on the left you have A^j d/dx^k' (dx^i' / dx^j) by the chain rule, and I'm assuming that from here you swapped the d/dx^k' and the d/dx^j

[fathom cave]

o yes i cant do that?

[versed pivot]

You can only commute partial derivatives from the same coordinate system

[fathom cave]

oooo

thank you. that makes sense
do you have anything that i can read that goes on detail about it?

[versed pivot]

Think about cartesian and polar coordinates for example
If you go a little bit in the x direction, and then a little bit in the theta direction, and then back the same amount in the negative x direction, and then back the same amount in the negative theta direction... you aren't going to end up back where you started

In contrast to if you go a little bit in the x direction, then a little bit in the y direction, then back in the x direction, then back in the y direction

[fathom cave]

ye

[versed pivot]

It just doesn't work if the other variables are not being held constant

https://math.stackexchange.com/questions/52882/interchangeable-derivatives

[fathom cave]

ooo nice that helps. i have literally spent 2 days why tht thing was not a tensor
thank you

also i have another question if u dont mind

see this says the expression in the parantheses in tensor because its a component of a covariant vector

this is the section it is refrencing

but the component V^i is tensor only when V^iZ_i is an invariant right?

[versed pivot]

Right the component functions of a vector need to transform in the opposite way as the d/dx^j (the latter being determined by the chain rule)

[fathom cave]

I will come for ur help later sometime after i have poured few more time into it.

basically, this is the reasoning why V^i is a tensor but i am not sold with that big thing in the parantheses being a tensor when the term in the right side is not invariant

[drowsy trout]

Hey, I am trying to learn étale cohomology and I have a question about inverse image of sheaves on sites.
I'm learning from Milne, here's the context :
We have two sites (C'/X')_E' and (C/X)_E, and a continuous morphism pi from X' to X, which defines a morphism of sites.
We consider a sheaf P on (C/X)_E and we define the pull back of P using a limit.
Now, the following pictures gives us a more concrete way to understand the constructions for some cases where finite inverse limits exist:

But I'm not sure I understand what type of morphism g is.

As a concrete example, take $X$ a scheme and $\pi = Id$ the identity on $X$.
We get a morphism of sites $X_{et} \rightarrow X_{zar}$.
Now, consider a zariski-sheaf $P$ over $X$.
$\pi^P(P)$ is a sheaf over $X_{et}$.
Taking $U'$ an étale $X$-scheme, a representent of $\Gamma(U',\pi^P(P))$ is a couple $g : U' \rightarrow U$ and $s \in P(U)$.
So $U$ is an object of $X_{zar}$, equivalently it is an open subset of $X$.
Does the definition ask that $g$ be étale ? Or just a scheme morphism over $\pi$ ?

[gentle osprey]

Dagnyr

[fading vale]

so the thing in the second paragraph defines a map from the path components of the fiber of b to the path components of the fiber of c as stated

apparently this map is G-equivariant when G is a left-principal or right-principal covering

which is """"immediate from the definition""""

except its not immediate for me so if anyone could explain why thatd be very cool

[tight agate]

Let $x \in F_b$ and $g \in G$. Let $V_x : I \rightarrow E$ be the lift starting from $x$. Let $V_{gx}$ be the lift starting from $gx$. Then $g \circ V_x$ is a path from $gx$ to $gV_x(1)$.

[gentle osprey]

Brofibration

[tight agate]

So you get a path from $V_{gx}(1)$ to $g(V_x(1))$

[gentle osprey]

Brofibration

[tight agate]

by doing one of them in reverse

so they are in the same path component

[fading vale]

@tight agate isnt this a path in E though

and not the fiber of c

actually nvm

[marsh forge]

group acts on fibers

[fading vale]

uh yea

[marsh forge]

so you stay in the fiber

[fading vale]

uh isnt the path brofibration said given by going from V_gx(1) to gx and then to gV_x(1)

[marsh forge]

im on mobile i can reread

[fading vale]

but also i think u can use like

[tight agate]

yeah my path doesnt stay in the fiber

[fading vale]

path lifting htpy conditions here or something

like if two paths starting at the same point E are htpic iff their images in B are homotopic

or something

[tough imp]

Have you tried looking at the definition to see why it’s immediate?

[marsh forge]

sorry what is the complaint

@fading vale

[fading vale]

i get it now

[marsh forge]

ah okay

[fading vale]

it follows from uniqueness of path lifting xd

[marsh forge]

i dont think it matters

yes

[tough imp]

Moth what book is this again

[fading vale]

tom dieck

[tough imp]

What’s the moth approved opinion on it

Good

Bad

[fading vale]

its pretty good

[tough imp]

Moth book review time

[fading vale]

i like it

[marsh forge]

to answer this question its immediate that gV_x is a lift from the definition

[fading vale]

its super category brained though

[marsh forge]

i think thats what dieck meant

[fading vale]

yea

[tough imp]

So you mean that

You now understand what an equalizer is

[fading vale]

no :egg_hank:

[tough imp]

Bromoment

Super category brained

Wat is equalizer

Smh my head

[marsh forge]

moth finish dieck then read more concise with me

[fading vale]

i will max

concise sounds pog

[tough imp]

I was told by Max L

To read concise

[marsh forge]

im dicking around w ravenel in the meantime

[tough imp]

For learning AT

[marsh forge]

concise is a good book

[tough imp]

When I didn’t know any

Sham said

“No, don’t do this”

[fading vale]

tom dieck is hard but there are parts of it thats really nice

[marsh forge]

i think concise is fine w sufficient mathrity

maturity

[tough imp]

No

Mathrity

Math maturity

[marsh forge]

also it helps to read like

[tough imp]

[marsh forge]

riehl first

[tough imp]

For?

[marsh forge]

bc peters intro to categories is a meme

[tough imp]

Knowing cat theory?

[marsh forge]

concise

[tough imp]

I don’t think that should be a problem for me

[marsh forge]

like just a few chs of riehl

[tough imp]

Lol

[marsh forge]

yeah for sure

i meant like

[tough imp]

In general

[marsh forge]

for a young person

[tough imp]

Ah

[fading vale]

like it constructs a fiber sequence $\pi_1(F_b, x) \to \pi_1(E, x) \to \pi_1(B, b) \to \pi_0(F_b, x) \to \pi_0(E, x) \to \pi_0(B, b)$

[marsh forge]

who hasnt endured ag

[tough imp]

Fair

[gentle osprey]

Moth | not male

[tough imp]

Joe mama is a fiber sequence

[marsh forge]

ravenel is so cursed

i got thru like

1 page per hour

yesterday

[tough imp]

Nice

I spent 2 hours on

One sentence

[fading vale]

and it does universal cover stuff by showing how path connected base space or path connected total space or simply connected total space gives u more and more stuff in the sequence as trivial

[marsh forge]

homotopy groups of spheres go brrrr

[tough imp]

In Goertz and Wedhorn

My peanut brain cannot comprehend why I can give a sheaf an O_X-module structure locally

Cuz of gluing

[marsh forge]

the longer u take to read either the more or the less you are understanding material

[tough imp]

But Hartshorne says the same thing

[fading vale]

ur bible

[tough imp]

Hartshorne is not my bible

It’s like

[marsh forge]

your penance

[tough imp]

It’s like a human leather skin bound book

Full of spells

[fading vale]

its ur zohar

[tough imp]

Too dangerous, but the folly of man leads them to open it anyway

[fading vale]

much like kabbalah magic it too, is not actually useful

[tough imp]

-_-

[fading vale]

[tough imp]

You know what is useful is

Vector calculus

[fading vale]

is it tho

[tough imp]

Yes

[fading vale]

[tough imp]

Also how come it seems

No one reads Bott & Tu

[fading vale]

i had to tell these geologists i want to do diff geo

xd

we'll see how that goes

[tough imp]

Why are you talking to geologists

[fading vale]

summer program thingy

[tough imp]

And you lied to them

[fading vale]

noooooooooooooooooooo

its not a lie

i read a paper on it and everything it was cool

https://structuralgeology.sites.stanford.edu/sites/g/files/sbiybj10271/f/mynatt_et_al.pdf

[tough imp]

Did you know that

A recent study found out that children who do too many memes without doing calculus have a 90% chance to end up as life-long memers

[fading vale]

ok

first of all

ive done calculus

fuck u

second of all

fuck u

😡

[tough imp]

Moth ur gonna end up ugct’ing

Unless you do vector calc

[fading vale]

thats literally not even true!

[tough imp]

No it is

[fading vale]

im not a category memer

[tough imp]

You will become one

[fading vale]

i dont even care about pure category mem estuff at all

[tough imp]

Unless you have vector calc to ground you

[fading vale]

only applications to AT

[tough imp]

Chmonkey

[fading vale]

many such cases

[tough imp]

Many such cases

https://tenor.com/view/cat-meme-funny-gif-5754572

Live footage of moth

[fading vale]

yay

i finished a tom dieck section

[tidal cedar]

algenomicon

[tidal cedar]

AT is just category theory where every once in a while we check if an arrow is continuous

[fading vale]

no

lies

seethe!

[gritty widget]

[marsh forge]

no they are all trivially continuous

i havent checked continuity in a long time

[sleek thicket]

there was a nontrivial argument that a map was continuous on my last bundles hw

I will not forgive my professor

[gritty widget]

all maps are continuous, and smooth if you're on a smooth manifold

[obtuse meteor]

@sleek thicket since you're not working (twitter lol) I am going to complain at you about topology work

http://www.math.lsa.umich.edu/~jchw/2021Math592Material/Homework5-Math592-W2021.pdf

[sleek thicket]

I was going to ask lol

(also I'm watching someone general exam !)

[obtuse meteor]

Casually just do homotopy lifting, sheets, universal cover, and galois correspondence in one homework

[sleek thicket]

(learning about spectra!)

[obtuse meteor]

oh neat!

idk what someone general exam means

[sleek thicket]

Holy shit this is so long

[obtuse meteor]

you can ignore the warm-up qs

[sleek thicket]

It's like the thing you do before you're a PhD candidate I think

[obtuse meteor]

ahhh I see

[sleek thicket]

A turning point in your PhD ig

[obtuse meteor]

yeah for sure

when question 3 has parts (a)-(k)

🧠

[sleek thicket]

God Faye this is ass

20 page homework here we go

[obtuse meteor]

we're at 12 pages currently and I haven't done any of Problem 4, parts (f)-(k) on Problem 3, the last part of Problem (2), or (d)-(f) on problem 1

it's due tomorrow night

this is a one-week homework????

Jenny wtf

[sleek thicket]

Holy fuck

uhhh

Good luck lmao

[obtuse meteor]

What drops were made for my friend

it's actually so ass

[sleek thicket]

here take this

[obtuse meteor]

Like am I not wrong that this is two homeworks crammed into one???

[sleek thicket]

You're not wrong

This is absurdly long

[obtuse meteor]

at least we're like allowed to read hatcher for some of this

but still

I also haven't started any of my manifolds homework that's due Sunday

but I think that's going to be a weekend problem from here on out

due to spicy alg top

[sleek thicket]

oh no

sorry Faye

This sounds really hard

if you need help with manifolds or AT stuff due to time pressure or w/e feel free to ping me :C

[obtuse meteor]

¯_(ツ)_/¯

I'll vibe

hard classes are what I wanted out of this semester. And I'm surviving

and doing pretty well on homeworks / exams overall

[sleek thicket]

nice!

[obtuse meteor]

and with grading and keeping everything together

it's just like

[sleek thicket]

yeah, I've had quarters like that

I'm going to be TAing a grad course next quarter which is very very weird... I feel like I shouldn't be allowed to do this

[obtuse meteor]

yeah that's a bit unreal

[tight agate]

which topic?

[sleek thicket]

Programming language theory

[obtuse meteor]

wow lmao

curveball

[sleek thicket]

yeah I mean I did actually start out in this stuff

and TA it pretty frequently

[sleek thicket]

@gritty widget the milnor paper actually says any left invariant metric is complete!

The proof isn't so hard

A small metric ball around the identity will be diffeomorphic to the unit ball in R^n, so compact. By homogeneity any ball of radius ε is compact. Any cauchy sequence eventually lies in such a ball, so converges

[gritty widget]

interesting

nice proof

[sleek thicket]

This paper seems very approachable so far

[gritty widget]

well that's complete as a metric space. does that give complete in the sense that all e.g. one-parameter subgroups exist for all time?

if you're working with a bi-invariant metric that follows from hopf rinow

but what about here

i'll eventually sit down and read it, i don't have a lot of free time

[sleek thicket]

why do we need bi invariance for hopf rinow?

oh one paramarer subgroups

Not geodesics

[gritty widget]

since for a bi invariant metric the geodesics at the identity coincide with 1 param subgroups

[sleek thicket]

All one parameter subgroups exist for all time

Always

[gritty widget]

fug

[sleek thicket]

left invariant vector fields are complete

lol

[gritty widget]

oof

got me

[sleek thicket]

[gritty widget]

look i haven't had much time to do lie group stuff lately

[sleek thicket]

It's okay haha

but this is still cool

[gritty widget]

analysis midterm next weeek

[sleek thicket]

I just like lie groups

[gritty widget]

who doesnt

[sleek thicket]

ooh spooky

My analysis course didn't have a midterm this quarter

It was nice

I just had a French oral exam this morning though and it was terrifying

[gritty widget]

horrifying

oh yeah the proof that left invariant v fields are complete is super easy too

analysis brain-rot

your brain on epsilon-N-delta's

[sleek thicket]

rip

I think it's like

translate

can extend a little

tada

[gritty widget]

yeah you extend it a little and apply uniqueness

i remember doing it a while ago

[sleek thicket]

A lot of the basic vector field stuff is locked in a weird place in my brain

like I know it

[gritty widget]

how do you feel about lie derivatives

[sleek thicket]

But I strongly associate it with the start of the pandemic

I feel a lot better about them than I used to

I think I have a better sense for them now

Cool objects

I need to liepill myself more

lie algebra classification stuff, lie group reps, lie groupoids and their connection to foliations

[quartz edge]

the first fundamental form is a differential 2-form, isn't it?

how then do you write it out as a wedge product

given coordinates u, v

would it just be

[gentle osprey]

gristle

[quartz edge]

or rather just the first part

im confused

[gritty widget]

"form" in the term "first fundamental form" refers to merely a bilinear form, not a differential form

what does it mean to say that a bilinear form is invariant under an adjoint action of SU(2)

I understand if we had some map f:\mathfrak{su}(2) to R, we would say it was invariant under the adjoint action if f(x)=f(ad(g)*x) for all g in SU(2)

but when we have a bilinear map im not sure what require

is it f(ad(g)x,ad(g)y), or f(ad(g)x,ad(h)y)?

[quartz edge]

o

[nimble jolt]

I would assume the former.

[quartz edge]

@gritty widget my mistake was in assuming that it acting on tangent vectors made it a diff form

[nimble jolt]

The latter seems pretty nuts if you say fix h and vary g arbitrarily.

[fading vale]

do physicists care about the conway knot

sadge

wait nvm

i no longer need you conway knot.

you are dead to me

[gritty widget]

physicists dont really use knot theory

[frigid patrol]

Inb4 string theory

[gritty widget]

i said physicists

[fathom cave]

let me know if this transformation does not work this way
$$\frac{\partial V^{i'}}{\partial Z^{j'}}+\Gamma^{i'}{j'}{m'}V^{m'}=\Bigg(\frac{\partial V^i}{\partial V^j}+\Gamma^i_j_mV^m\Bigg)\frac{\partial Z^{i'}}{\partial Z^i}\frac{\partial Z^j}{\partial Z^{j'}}$$

[gentle osprey]

Zero0
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fathom cave]

@versed pivot

[shut moat]

I think there was some cool stuff with knotted currents in classical electrodynamics

[fathom cave]

we never know imao what we might end up using 😂

[shut moat]

there we go

@fading vale it's pretty niche but it's cool

[fading vale]

neat

i went with this pretty cool paper on this weird other knot thing

https://science.sciencemag.org/content/367/6473/71

i literally just had to write this short essay on a recent discovery and i had no idea what id write about if not math

lmfao

[gritty widget]

there's some stuff that comes up

but usually it's more niche

there's also knots in polymers which can be classed as either chemistry or physics

but in general knot theory isn't used much

[tidal wadi]

Be careful with broad claims: You are not aware of the possibly vast applications. I've repeatedly found myself baffled by the number of applications a field of initially seeming abstract nonsense has had in the real world

[chrome dew]

eh, fairly often it flows the opposite direction, people claiming things to have some grand applications but really being overhyped

[gritty widget]

usually <pet math concept/theory> in physics is hyped as groundbreaking and will revolutionize physics

gets a ton of articles and people think it's super relevant

https://tenor.com/view/oh-no-top-gear-jeremy-clarkson-no-one-cares-gif-18925814

[shadow ermine]

Quick question about stereographic productions π: S² -> Σ between the unit sphere and the Riemann sphere C ∪ {∞}

I'm given a set of points on the unit sphere, A = {(x, y, z) ∈ S² : x + y + z = 1}
And I'm asked to find the image under stereographic projection π(A)

Now, I'll first note that (0, 0, 1) is in A, and π((0, 0, 1)) = ∞, so let's look at A\{(0,0,1)}
Picking (ξ, η, ζ) in A\{(0,0,1)}
I know there's a point z = x + iy ∈ C where
ξ = 2x/(x² + y² + 1)
η = 2y/(x² + y² + 1)
ζ = (x² + y² - 1)/(x² + y² + 1)
Now, using the fact ξ + η + ζ = 1
<=> 2x/(x² + y² + 1) + 2y/(x² + y² + 1) + (x² + y² - 1)/(x² + y² + 1) = 1
<=> 2x + 2y + x² + y² - 1 = x² + y² + 1
<=> x + y = 1

So I conclude that the image π(A) = { x + iy ∈ C : x + y = 1 } ∪ {∞}
My question is, is this adequate? Or would I need to do a more explicit "both ways set inclusion"?

[shadow ermine]

<@&286206848099549185>

[shadow ermine]

Came back to the problem, and just gave a little more care justifying the set equality π(A) = { x + iy ∈ C : x + y = 1 } ∪ {∞}

no need for help now

[gritty widget]

[fathom cave]

https://cdn.discordapp.com/emojis/771287728305471528.gif?v=1

https://cdn.discordapp.com/emojis/771287728305471528.gif?v=1

[gilded lance]

Question. I'm a physics grad student studying gravitational wave sources. There's a comprehensive physics way of doing what I'm doing, but I feel like it seems outdated and brutish, but I don't exactly know what to be looking for in terminology used by the mathematicians. In short, I have spinning, compressible spheroid. For a given ratio of rotational to gravitational energies, this shape becomes unstable, and it morphs into an ellipsoid. There are greater complexities, but that's the basic idea. Anyone know of any maths texts or work that would contain ideas analogous to this process of sphere -> ellipsoid due to rotation?

[marsh forge]

My best guess is that this would fall under dynamics as far as mathematicians are concerned

But its also kind of an odd form of dynamics so I am not sure

[lyric wadi]

Is singular cohomology a fibre bundle?
So for H^n(X;R), we have R\rightarrowX\rightarrow(sometotalspace)

I guess it might just be a bundle?

[marsh forge]

Sorry I'm not sure I am understanding

What is the map R->X here

how have you topologized R?

If you want R->X->B to be a fibration you want R to be some kind of subspace of X

which doesnt seem very natural

[lyric wadi]

oh R is just some sort of abelian group

[marsh forge]

No i know

but I'm asking why you think a nice map R->X should exist?

R needs a topology for this map to be continuous at all

you can always give it a discerete or trivial topology

[lyric wadi]

yeah, I've not 100% thought it through tbh, I just noticed that the association of an element group to every point of a space is fairly similar to how vector bundles are defined

[marsh forge]

If you want a fun preview of material you probably won't learn for awhile

[lyric wadi]

but since the groups used in the dualisation to cohomology aren't necessarily vector spaces, they would have to be fibers or something

[marsh forge]

While cohomology has nothing really to do with maps

R->X

There is a space called K(R,n) for all n

such that cohomology classes in H^n(X,R)

are the same thing as homotopy classes of maps

X->K(R,n)

Here I am thinking of R only as an abelian group

[lyric wadi]

is that K-theory

[marsh forge]

no hahaha

similar notation

but no

These are "Eilenberg Mac Lane spaces"

or EM Spaces for short

[lyric wadi]

oh

[marsh forge]

ironically K-Theory does get a sort of generalized version of an eilenberg maclane space

but this is even more advanced

[lyric wadi]

ok, let me try state the reasoning a little more clearly
if X is some space and G is an abelian group, then a (0-)cochain is a function which associates an element of a G to a point of X. If we take all the cochains at defined on a single point, then we can form a projection, p, from G to a point in X by using all of their inverses.
This next bit I'm not 100% sure on, but the requirement for local triviality seems to follow from how n-chains relate to 0-chains.
so we have a projection p:B->X which seems to satisfy the local triviality condition
ofc, it might be that I'm just misunderstanding the definition of a fibre bundle and that there's some issue with how the group G gets topologised or it's possible that I might even be missing some other necessary condition for fibre bundles

in fact, I'd put money on it

actually, they'd be 0-cochains of single points

[marsh forge]

Sorry I don't see that first construction yet

can you explain explicitly what f(g) is for f:G->X

I am not certain about your claims for local triviality. I also don't know what map you are calling a fiber bundle

Is it G->X? X->something?

[lyric wadi]

it's taking all the functions from a point x in X to G and then inverting them to get a function f(g)=x for all g in G (the first construction that is)

[marsh forge]

Can you be more explicit

All the functions?

The elements of C^0(X,G) are homomorphisms C_0(X) to G

[lyric wadi]

yes, so within each C_0(X), there are all the points of X?

[marsh forge]

Yes

Its a basis

[lyric wadi]

so for each point x in X and for each element of g in G, there is a homomorphism f(x)=g?

[marsh forge]

You can find a homomorphism for which that is true, but it is not necessarily unique

[lyric wadi]

hmm, I might just be finding a fibre bundle-like structure in singular cohomology then

rather than the cohomology forming a fibre bundle of sorts

[marsh forge]

I think you should be really careful about like

mixing maps G->X with topological stuff

If G has a discerete topology all such maps are continuous

but X looks nothing like G locally

in general

like the total space in a fiber bundle is suppose to look locally like Gx(something)

[lyric wadi]

Makes sense

[marsh forge]

but G is a terrible topology

That said there are lots of algebraic analogies that show up in topology!

so i dont mean to act like it was a bad idea

its just that the details in topology can be hard and if you find yourself handwaving you gotta be really careful

[lyric wadi]

Oh, I didn't take it as such

If anything, i was just wondering why I couldn't find any reference to the idea

So thanks for getting me to think it through :)

[marsh forge]

np

[median glade]

If you take any group, is there always some (pointed) top space with that group as fundamental group?

Are all groups infinitely presented?

Well, some groups aren't finitely presented, which is sort of what you described

[cloud owl]

if infinitely presented means what i think it means, a presentation is just a list of all the elements at worst?

under the operation

y/n?

[gritty widget]

Does the set of generators have to be minimal?

[median glade]

I just meant, not finite

[cloud owl]

i mean for finite groups can you really present them infinitely without duplicating elements, which you're technically not supposed to do for sets iirc

[median glade]

Nothing

I had just forgotten how to make every group presented

Yeah thanks :)

Another question: are infinity groupoids equivalent to spaces?

Or can two spaces that aren't homeomorphic share an inf groupoid of homotopies?

[cloud owl]

oh right, yeah, you can have all the relations in the world

[sleek thicket]

idk why you would overcomplicate things like this

I would simply use milnor's infinite join construction on the group with the discrete topology to get a classifying space

presentations? Who needs em

[tight agate]

you don't even need to construct it

BG exists by Brown rep.

and then use fibration LES

[obtuse meteor]

idk why you would complicate this, the delooping of G is an infty-groupoid (aka a space) that stops at the 1-skeleton, and has pi1 G :)

[tight agate]

this is almost the same thing as what Sham said tho right?

except you're using the nerve construction instead of Milnor's

[obtuse meteor]

shhhhhhhhh

[wicked matrix]

okay

i am learning about continuity in Calculus

is that the same type of continuity that they talk about in topology?

[ivory dragon]

"continuity in calculus" is a specific case of metric space continuity

which is an equivalent condition to general topological continuity on metric spaces.

(considering the course was just called "calculus", i think i can guess how general it is)

impossible

[tough imp]

If you have a geometric vector bundle (in the context of schemes) X over Y, this is the data of a morphism f:X -> Y and some cover {U_i} of Y along with isomorphisms f^{-1}(U_i) -> A^n_{U_i} such that the transition functions are linear

To take a geometric vector bundle and get a locally free sheaf, we let F(U) be the collection of sections of the map f on the open set U, so the collection {g:U -> X: f\circ g = id}, and suitably locally this is identified with O_U^n so that this sheaf is locally free

One issue I get though is that I don't see how I'm supposed to glue these together, I can't figure out a global way to define an O_Y-module structure on F, and I don't see how I can glue the local module structures together

Every source I look at just says this is local, which sure, once you get a global O_Y-module structure you can check that it's locally free of rank n well... locally, but I don't see how you can even get that module structure to begin with

This has been gnawing on my neurons for two days now

[tough imp]

NVM I did it

[dim radish]

asking for a friend: is there any point in defining a structure like a topological space, but instead of having arbitrary union of open sets be open, to only have finite union of open sets is open

cool

:D

[viral atlas]

Was this supposed to be a pun?

[quasi forum]

Okay, so I really need help with 3i). I believe I know the answer (the relationship is a subset), but I am having a really hard time figuring out how to formalize my intuition here.
Any help would be really great

Oh?

Okay, so there are our definitions. Ignore the top bit

Yea. We are really only dealing with metric spaces right now

Did you also notice the max metric assunption?

That statement means nothing to me

Let me write it out real quick and let me get back to you.

This is what I have so far

I don't see how this helps

Yea, I altered that

I can agree to that

Wait, why does this tell us x is in cl(a)

Ohhh, okay. And then the other order is simply just reversing the idea

Alright. Let me get a shot at 3ii) before I ask for help again. Something tells me we have to use a different approach for it

So because the interior does not contain all the limit points, this is not an approach we can use, right?

Closure is the superset of the interior if that is what you mean

Tbh, I am not so sure

Ohhh, that one

Hold on. Is that 2nd part right?

I'm not sure how that implies (x,y) is in cl(AxB)

Oh. So should I actually make (x_n,y_n) in AxB

Gotcha! That makes sense now. Did that present an issue in the first problem?

I think the answer is yes

Okay got it! This makes a lot more sense now. Thank you for this? I was definitely not thinking about this approach, but it's really cool!

So with the closure interior relation, are you also suggesting they are equivalent again?

So correct me if I am wrong, but is A^CxB^C=(AxB)^C?

If so, we are in business

Oh. Then we are not in business

I'm not sure this approach will work then

@limpid vault I need to take a food break (haven't eaten yet). Will you be around in about an hour? This has been very helpful

Okay no worries. Thank you very much for the help

[long coyote]

$A,B$ are connected subsets of $X$, if $\bar{A}\cap B\neq\phi$, then $\bar{A}\cup B$ is connected

[gentle osprey]

亜城木 夢叶

[long coyote]

closure of union is union of closure, this is key i believe

[obtuse meteor]

manifolds homework actually ass

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

Anyone wanna compute

@sleek thicket gonna complain at you again about my classes

granted I left this till like

the day before it was due

so everything is my fault

but also

also there's this which while it's not conceptually difficult, we did it last year in Linear Algebra and I really don't wanna write up the proofs of these facts

https://estrogen.fun/i/o5fx.png

and also like

it's not manifolds

why care about it

Angry woman noise

[sleek thicket]

Oh this stuff doesn't matter at all, it's so nasty

You check it once and then like

tada, associative

[obtuse meteor]

exactly

why assign this

as homework

when you could just like

say

"read Munkres"

?????

[sleek thicket]

Lol

[obtuse meteor]

idk it's just like

very frustrating

[sleek thicket]

No I agree

[obtuse meteor]

at least there's a problem about Lie groups too on here

it's basically just a "verify all these things are examples of Lie groups"

but that's a lot cooler than like

this shit

https://estrogen.fun/i/o7jy.png

[marsh forge]

i had a prof who would like

include a ton of problems that you didnt have to turn in

and was like

[obtuse meteor]

part (iv) is just like. Chase definitions and extend smooth maps properly

[marsh forge]

if you fail the midterm bc you can't prove X is a lie group

thats on you

[obtuse meteor]

lol

[marsh forge]

i like that idea

[obtuse meteor]

yeah

My alg top professor

has warm up problems

[marsh forge]

like people should practice easy problems exactly as much as they need to but doing extra basic stuff is painful

[obtuse meteor]

I never do them formerly / write them up

but I look at them

idk it's just like

[marsh forge]

its good to make sure defns are in working memory

before starting

[obtuse meteor]

I feel like why assign problems like this

it doesn't help me understand any of the concepts

it just makes me like

verify facts

and want to tear my eyes out bc of formalism

it's not as bad as the measure theory IBL

which is just "here's a bunch of basic properties of the Lebesgue integral that are like "verify this is <= this and linearity works and everything" that just relies on tracing back definitions through the last 4 worksheets about simple functions"

[marsh forge]

isnt that every analysis problem

i mean yeah thats bad

[gritty widget]

Hey guys, just a bit confused. is the topological space a topology within a topology?

or am i just way off?

elaborate

sounds off

[marsh forge]

A topological space is a set X with a topology T where T is a subset of the powerset of X subject to a bunch of axioms

Normally we just refer to the topological space as "X" where the topology T is obvious

or we specify T informally

[sleek thicket]

Does anyone know how to draw out a simplicial set like this in latex?

idk how to stack the arrows correctly

(would just use this picture but I prefer the one where you interleave the arrows)

hello @gritty widget feel free to post your question here

[gritty widget]

lol its ok

[marsh forge]

i think stackrel could work

not sure

[shut moat]

$$\begin{matrix}\xrightarrow{\quad d_0\quad} \ \xrightarrow{\quad d_1 \quad} &&\\quad \dots\quad&&\end{matrix}$$

[gentle osprey]

~S^1

[shut moat]

it's a bit hacked together but I think that technically works lol

[sleek thicket]

hmm

Ty for the suggestion

[sleek thicket]

okay, this looks passable

ty @shut moat!

[shut moat]

np!

[topaz shuttle]

Hi can someone help me with this

Can i use a 2x2 matrix to show isometry here?

[gritty widget]

presumably not since n might not equal 2

[topaz shuttle]

how do I do that

d(x,y) = d(f(x), f(y))

[marsh forge]

yes

you know what f is explicitly

and d

so just substitute

[flint cove]

Does anybody know off the top of their head how the classifying space construction G ↦ BG (i.e. the representing object of the „principal G-Bundles over -“ functor) gives rise to a functor G→H ↦ BG→BH?

I mean I would guess that EG→BG would somehow get an induced principal H-action, but I don't see how

[cedar pebble]

@sleek thicket

$\begin{tikzcd}
X_0 \arrow[r,"s_0" description] & X_1 \arrow[l,shift right=3,"d_0" description] \arrow[l,shift left=3,"d_1" description] \arrow[r,shift left=3,"s_0" description] \arrow[r,shift right=3,"s_1" description] & X_2 \arrow[l,shift right=6,"d_0" description] \arrow[l,"d_1" description] \arrow[l,shift left=6,"d_2" description] \arrow[r,shift left=6,"s_0" description] \arrow[r,"s_1" description] \arrow[r,shift right=6,"s_2" description] & \hdots \arrow[l,shift right=9,"d_0" description] \arrow[l,shift right=3,"d_1" description] \arrow[l,shift left=3,"d_2" description] \arrow[l,shift left=9,"d_3" description]
\end{tikzcd}$

[gentle osprey]

nGroupoid

[cedar pebble]

if you want the most compact possible

(this renders runny on TeXbot but it looks normal in a document)

[sleek thicket]

ty

[cedar pebble]

$\begin{tikzcd}
X_0 \arrow[r,"s_0" description] & X_1 \arrow[l,shift right=3,"d_0" description] \arrow[l,shift left=3,"d_1" description] \arrow[r,shift left=3,"s_0" description] \arrow[r,shift right=3,"s_1" description] & \hdots \arrow[l,shift right=6,"d_0" description] \arrow[l,"d_1" description] \arrow[l,shift left=6,"d_2" description]
\end{tikzcd}$

the more truncated version

[gentle osprey]

nGroupoid

[cedar pebble]

there's a nice macro you can write that does all the arrows automatically but I would have to dig it up from somewhere

[sleek thicket]

this is a good question and I don't know off the top of my head

I think I would try to use an explicit construction of EG and BG like milnor's join or smth

I don't see how to get it abstractly

oh wait maybe I do

[flint cove]

Hm, okay, that might be a bit far off from what I know.
My whole background on this is that I know BO(n) and BU(n) from real and complex VBs via this grassmannian construction

[cedar pebble]

if you want this at the level of these infinite dimensional spaces that represent BG then this is a little annoying to write down explicitly

[flint cove]

At some point someone used that the embedding U(1)^n → U(n) lifts to BU(1)^n → BU(n)

I actually only want to understand where that comes from

Oh wait I just need to read one sentence further

[sleek thicket]

lol

[cedar pebble]

well so these maps say BU(1)^n → BU(n) really come from the maps Gr_1(C^k)^n->Gr_n(C^{nk}) sending subspaces to direct sums

and then you pass to the colimit

[flint cove]

Wait, don't you mean Gr_1(ℂ^k)^n?

[cedar pebble]

yes 😛

[tight agate]

in this case, as the map is injective, you get a free action of BU(1)^n on EBU(n)

[cedar pebble]

So more generally like Gr_n(C^k)->Gr_n+1(C^k+1) passes to BU(n)->BU(n+1) in the colimit

and Gr_n_1(C^k_1)xGr_n_2(C^k_2)->Gr_n_1+n_2(C^k_1+k_2) passes to BU(n_1)xBU(n_2)->BU(n_1+n_2) in the colimit

[sleek thicket]

oh maybe you can yoneda it?

[tight agate]

be very careful when you do this

[sleek thicket]

like, using the map to turn H-bundles into G-bundles

I was concerned about the action no longer being effective

not sure how to fix that

[tight agate]

maps into these things correspond to principal bundles when the domain is nice

you want something like compact

[sleek thicket]

ah this isn't how I've seen it presented

[cedar pebble]

I guess the way you can see the map U(n)->BU(n) is like

[sleek thicket]

I've seen it as a condition on the bundles

"numerable"

[tight agate]

you will need to write BG as a limit

[cedar pebble]

take the U(n) principle bundle U(n)->V_n(C^k)->Gr_n(C^k)

[sleek thicket]

which is preserved under pullback

[cedar pebble]

so you get a map U(n)->Gr_n(C^k)

[tight agate]

and look at the milnor sequence

[cedar pebble]

pass to the colimit

[tight agate]

and pray that the lim^1 term vanishes

[cedar pebble]

oh lord

[sleek thicket]

I don't see the concern?

BG represents the functor sending X to the set of iso classes of numerable principal G-bundles on X

(on the homotopy category)

[tight agate]

but X needs to be a finite CW-complex

[sleek thicket]

no, you can set it up as a condition on the bundle

[tight agate]

[sleek thicket]

https://ncatlab.org/nlab/show/numerable+bundle

this is how it's done in Lee's book and I'm suddenly very appreciative of it hahaha

[tight agate]

so no restrictions on the space at all?

[cedar pebble]

nlab has a decent coverage of this (don't worry, a lot of pages on nlab are written in a sane way now) https://ncatlab.org/nlab/show/classifying+space

[sleek thicket]

nope

[cedar pebble]

also https://ncatlab.org/nlab/show/Chern+class

[sleek thicket]

it is kind of tedious proving all this numerable stuff

and like, not making assumptions on the base

but it does allow you to yoneda directly

[tight agate]

hmm

I'll read through it

this is pretty cool tho

[sleek thicket]

yeah!

[tight agate]

because if you prove existence of BG via brown rep

[sleek thicket]

also you don't need either compact or finite cw

[tight agate]

then you can't use yoneda directly

[sleek thicket]

just paracompact (so it works for any cw complex or manifold)

[tight agate]

because the representing object comes from a larger cat

[cedar pebble]

I really like the proof in the nlab Chern class article

since you really only need to understand BU(1) for this

[tight agate]

and this is the source of stuff like phantom maps and shit

[cedar pebble]

and then the rest is formal

[sleek thicket]

hey lux, is this discussion making any sense? sorry if we've been distracting you

[flint cove]

Huh, lot to unpack here.
I mean we can ignore what the assumptions on the base space are, let's just say paracompact hausdorff.
Perhaps something stronger if we still don't get a countable set of trivializations subordinate to whatever we have.

I'm just assuming that whatever (homotopy) cat T we work on, we have a functor Vec_{ℂ,n}: T→Set mapping base spaces to iso classes of bundles over that base space, and that this functor is represented by BU(n).

[flint cove]

Sure, just need to unpack stuff.

I haven't completely internalized the Gr-construction as well so it might take some time

[sleek thicket]

sure, we're not actually using a specific construction here

but you can trace it to get an explicit map

[tight agate]

yo lux in your case, as the map is an inclusion H --> G, you get a free action of H on EG

so you have EG/H

and then you get a map from EG/H ---> BG by quotienting out by G

ig you need H normal lol rip

for this to work

no you don't, I was being stupid

[sleek thicket]

is this what you're thinking of brofib?

this kind of result

[tight agate]

yes

[sleek thicket]

(the lie group condition is because G -> G/H might not even be a locally trivial bundle for H a closed subgroup of G if G is an arbitrary topological group)

[tight agate]

you probably just need the inclusion to be a cofib or something

this is why things can get messed up if H = {0} and G is the p-adics or something

I did not manage to prove it tho

[sleek thicket]

this lemma?

[tight agate]

no

[sleek thicket]

it's a little complicated

oh okay

[tight agate]

that (Segal's model) EG ---> BG is not locally trivial when G = Z_p

[sleek thicket]

the counterexample I was thinking of was (S^1)^infty -> (S1)^infty/{1,-1}^infty not being a locally trivial bundle (^infty meaning countable product)

[flint cove]

okay, so in this case, U(1)^n acts freely on all the grassmannians and by the limit on EU(n), and (hopefully) in a fiber-preserving way, so the quotient is actually a bundle EU(n)/U(1)^n… but wait, this would be a bundle over BU(n). I want a bundle over BU(1)^n.

[sleek thicket]

oh no this makes me sad

[flint cove]

Okay, so let me backtrack a bit. Just on representability grounds it would suffice to turn this embedding U(1)^n → U(n) into some induced bundle over BU(1)^n.

[tight agate]

lux look at the result shamrock posted

[tight agate]

it works whenever the inclusion of the identity gives you an NDR pair

[tight agate]

look at EU(n) ---> EU(n)/U(1)^n

it has fiber U(1)^n

[flint cove]

Ohhh, right
I somehow completely forgot how quotients work, lol.

sorry about that

[tight agate]

and the corollary shamrock posted shows that that bundle is a universal H-bundle

and you get a map to BG by "finish quotienting"

[flint cove]

May I ask where that is from?

[tight agate]

Might be Lee's bundles book

(which hasn't been published yet )

only Lee and his apostles (his bundles class) have access to it