point-set-topology

[sweet wing]

[sweet wing]

Is K algebraically closed? it would have to be for it to be true

but ye theres a really natural parametrization

[quiet pilot]

Not sure how to proceed, I tried looking in the chart Y=1, and was trying to check if (t,1,t^3)+(s,1,s^3) was equal to (s+t,1,(s+t)^3) but it didn't really work out I think

Yeah K is algebraically closed

[sweet wing]

nonprojective case we essentially have y^2=x^3

can you find a reallly simple rational parametrization x(t)=?, y(t)=?

[quiet pilot]

Yeah t^2,t^3

[sweet wing]

yup

[quiet pilot]

It looks like I made a mistake though, I forgot to change signs

[sweet wing]

and theres a super simple way to get t from x and y

that is your isomorphism

[quiet pilot]

Thanks!

[sweet wing]

yw

[sleek thicket]

So the conversation about T^2\pt being homotopy equivalent to a wedge of two circles was a couple hours ago

But

I feel like it was missing an important visualization about the homotopy

It's goatse

I should do more topology

geometry is kind of bad, actually

[gritty widget]

embrace the love-hate relationship with rg

[sleek thicket]

lol

[gritty widget]

it can be cringe sometimes

thank god that seems like the last extreme computation ill have to do in this course

[river granite]

differential operators eww

[small obsidian]

Differential operators uwu

[sleek thicket]

lol

The E_is are even differential operators, oh no!

[gritty widget]

shit like this is why you pretend everything is C\infty

[tight agate]

boy is dg/rg rough

heard it gets better tho

[sleek thicket]

i hope lol

Kind of considering never looking at RG ever again

[gritty widget]

RG is nice modulo computations like this

where you're just mindlessly applying definitions

some things are disgusting but lead to really nice results

like the formulas for variations of energy give you bonnet myers

[tight agate]

variations of energy wtf lol

[gritty widget]

whatever they're called

like

[elder yew]

Oh brofibration, has sarkar posted or emailed a syllabus

for 236

[tight agate]

dont think so

[elder yew]

I might try to sneak into a few lectures

[gentle osprey]

TTerra:

[gritty widget]

then take derivatives

it's awful!

[elder yew]

why is the derivative 0

why tho

BUT WHY

[gritty widget]

well by take derivatives i mean

consider a family of curves

blah blah blah

it's not pretty

[tight agate]

I will stick to just using RG theorems for now

the proofs are hard af

and the definitions for a lot of things are so much simpler in the algebraic case

[elder yew]

Yeah, sometimes these things get rather complicated

[tight agate]

yeah I'll be happy if I can retain just the Hodge decomposition and chern-weil theory from my class

screw the rest

[elder yew]

are you gonna take the other 226s?

[tight agate]

I don't think they're being offered this year

[fading vale]

this is your brain on RG

[tight agate]

AG has some horrible messy stuff too

just a matter of taste ig

[fading vale]

hurb

AG was not the alternative i had in mind

[elder yew]

are you taking AG?

[tight agate]

took it last year

[elder yew]

Did totaro teach it?

[tight agate]

yup

[elder yew]

I had totaro for 131H

[tight agate]

lol

[elder yew]

That was wild

[tight agate]

wild how?

[elder yew]

His exams were fucking insane

[tight agate]

well, which G then

no G is not an option

[elder yew]

Now they look easy, but at the time it was like holy fk

[tight agate]

my 131bh final was hard

or hard back then

[elder yew]

who did you take for 131H?

[tight agate]

a lot of those feel easier after prepping for the basic

visan

[elder yew]

Oh she's good

[tight agate]

yup

[elder yew]

I had totaro, garnett, then gangbo

[tight agate]

oh you did 131c?

[fading vale]

what do you have against zero G

: (

[tight agate]

or 245?

[elder yew]

Back when it was measure, I did 131C

I know they changed it since some of my friends are there

[tight agate]

yup

[elder yew]

I did 246ABC and 226A for grad courses

[tight agate]

who taught 246?

[elder yew]

Garnett, Garnett, Terry

[tight agate]

I was considering doing 246 this quarter lol

tao is teaching it

[elder yew]

yA, his wait list is long. I preferred garnett over terry

[tight agate]

really? I think there's empty spots in his class rn

[elder yew]

My friend tried to get in and was denied

Ppl drop terry's course like flies

[tight agate]

there's a zillion people in the lecture tho

[elder yew]

ofc. he's terry!

I liked Garnett more because I understood the way garnett does and processes mathematics better

terry's too big brain for me

Also garnett had homework due every 2 weeks instead of 1 week

Which I also preferred

[tight agate]

noice

maybe ill do 246 next year

umm

nah

[elder yew]

What are you planning for next quarter?

[tight agate]

not sure yet

homological algebra

maybe topics in alg top

maybe fukaya categories

[elder yew]

You're so far in the algebra direction oh jeez

[tight agate]

maybe NT (dunno what the topic is)

or maybe some AG thing I can get going

[elder yew]

I feel like the LA students are either very heavily in the Alg. camp

or the analysis camp

[tight agate]

ye pretty much

[elder yew]

I was on the analysis side of things, obviously

[tight agate]

I'm interested in analysis tho

[elder yew]

I didn't even finish 110C, I disliked the courses so much

[tight agate]

lmao

[elder yew]

226 conflicted with 110

So I was like fk this I might enjoy RG

So I went to that instead

[tight agate]

the 210 sequence was great

[elder yew]

did you do 110H?

[tight agate]

yup

[elder yew]

Who was the 210 prof?

[tight agate]

merkurjev

[elder yew]

Oh merkurjev. ppl say he's good

[tight agate]

yup

[elder yew]

I went with Elman for 110H and I wasn't a fan

I loved Elman for 115AH/B though!

[tight agate]

I had elman for 115ah too

did not do B

[elder yew]

great course. I think that's his best one

115AH

[tight agate]

well that's the only class I've had with him, so it's automatically his best and worst one

[elder yew]

He does his 3 hour midterm, 6 hour final thing for 110

And everyone just cheated

[tight agate]

rip

[elder yew]

It was very frustrating cuz I refused to cheat

but he grades on points/rankings

[tight agate]

did he make y'all classify all groups up to some pretty high order or something on the final?

[elder yew]

Yeah, it wasn't good at all

[tight agate]

HAHAHAHAHAHAHA

up to which order

[elder yew]

I did so bad on that final

I don't even wanna remember

I wouldn't be surprised if it was like in the hundreds or so

There was one student that got a near perfect on the final

[tight agate]

how tf do you classify all up to 100 on a final

[elder yew]

Have you seen his 110H finals? They're insane

He gives at least 20 problems or so

Of varying points. He says "get to 100 points"

[tight agate]

konstantin?

[elder yew]

No it wasn't konstantin

I didn't take algebra with him. He took it a year prior

[tight agate]

oh ok

[elder yew]

It was uh

Clark

[tight agate]

ah

[elder yew]

IDK if you knew john, but he was also not a fan of algebra

[tight agate]

I do know a John

and he isnt a fan of algebra

[tough imp]

There’s no way you can classify groups up to order 100

Groups of order 64 are so numerous

https://www.amazon.com/Groups-Order-Senior-J-K-Hall/dp/B002GQQ11S
This book does groups of order 2^n up to n = 6 aka 64

And it’s gigantic and costs 500 dollars lol

[elder yew]

Dude, his fucking finals were crazy

He's put open problems on finals before

You can put it on there just to see how far students will get

[tight agate]

I've heard the classify all groups up to some large order legend from multiple sources now lol

3 sources

[tough imp]

...

[tight agate]

and Ken Ono is one of them

[tough imp]

https://users.fmi.uni-jena.de/~green/Coho_v3/64gps/index.html

Maybe this guy took his class

[tight agate]

maybe

https://youtu.be/Jliw4bpefGU?t=188

there ya go

the link takes you to the part where he mentions it

[tough imp]

But why...

I could do up to 16 maybe

With effort

And then some random bits interspersed

The main issues are when the prime decomposition has high powers

That’s when things get... messy

[tight agate]

up to 15 isnt bad at all

[tough imp]

Yeah

[tight agate]

16 idk

[tough imp]

8 and 16 are the first sort of big road blocks

I think there’s 9 of 16?

8 isn’t bad once you learn semi direct products

[tight agate]

yup

[tight agate]

there's more

14 lol

rip anyone in that class

51 groups of order 32

lmao

[tough imp]

How many abelian are there?

5?

Yeah there’s 5 abelian I think

So 9 was referring to the non abelian ones which are all rhat matter

😎

I remember 9 being related to groups of order 16

[fading vale]

classify all groups up to order 100

what the fuck?

????

thats literally easily over a thousand

it might be way way more

[sweet wing]

haha gap.finite group brrrr

[fading vale]

idek if gap has everything up to order 100

[sweet wing]

gap has 64

hence yes

[gritty widget]

You can probably grind it out if you turn your brain off and spam sylow

[elder yew]

Yeah that's basically the solution mathbath

It's like a lot of iterations

[river granite]

I had a group theory course where the final class project was classifying as much groups as we could up to order 500 (we worked in pairs, 20 orders for each pair), but we skipped most powers of 2 lmao

it was fun tbh, I wouldn't have learnt as much otherwise

though half of it was checking our answers in https://people.maths.bris.ac.uk/~matyd/GroupNames/

[tough imp]

Lmao 20 orders for each pair

If the orders were interspersed that seems not too bad

else people that got like "460-480" got mega fucked

vs "1 - 20"

[elder yew]

It's very easy to intersperse it

I'd imagine profs aren't brain dead

[uncut surge]

tfw group theory in #point-set-topology

[gritty widget]

inb4 someone starts talking about how algebra = geometry

no moth

[fading vale]

Yes moth

[gritty widget]

elaborate

[uncut surge]

g r o u p o i d s

or so idk

[fading vale]

i worded that poorly

[gritty widget]

something something

fundamental group

homotopy groups

blah blah blah

[fading vale]

but you can construct up to weak htpy equivalence a CW space with any group G as its nth homotopy group

[gritty widget]

you know it's good when the first google result is nlab

[fading vale]

why did i say fundamental space??

[fading vale]

lol

yeah

[gritty widget]

2nd is also nlab

3rd is an MO post

[fading vale]

lens spaces

[gritty widget]

[fading vale]

Ultra do I actually have to care about lens spaces

they are so ugly

something something S^infty/(Z/q) or whatever???

[tight agate]

geometra = algebra + gluing data

[elder yew]

lens spaces are so cool

[fading vale]

they are Not

K(Z/p, 1) bullshit

[elder yew]

schultens 3 manifold topology has a good intro I think

[tight agate]

Let C1 and C2 be projective curves. For any irreducible curve D \neq C1 on C1 X C2, is the intersection number with C1 positive?

[elder yew]

What have you tried? Have you looked at exampless?

[tight agate]

well it's true for C2

D = C2

[tight agate]

ah nvm it looks like there's a workaround and I dont need this

phew

[little hemlock]

Okay, so if ${X_\alpha}$ is a family of topological spaces, and ${Y_\alpha}$ with $Y_\alpha \subset X_\alpha$ for each $\alpha$, there are two ways to put a topology on $Y = \prod Y_\alpha$. You can give each $Y_\alpha$ the subspace topology and then give $Y$ the product topology, or you can give $Y$ the subspace topology from $\prod X_\alpha$ (with the product topology). Call the former $Y_a$ and the latter $Y_b$. I am trying to show that these are homeomorphic using only the universal properties. i.e. i am trying to show that $\operatorname{id}_Y$ is a homeomorphism from $Y_a \to Y_b$.

[gentle osprey]

kxrider:

[little hemlock]

To show $\operatorname{id}_Y : Y_a \to Y_b$ is continuous, it suffices to show that the composition of $\operatorname{id}Y$ with the inclusion $i: Y_b \to \prod X\alpha$ is continuous, and to show that $i\operatorname{id}Y $ is continuous, it suffices to show that the composition of $i\operatorname{id}Y $ with each projection $\pi\alpha$ is continuous, but I don't really see why \emph{just from the universal properties} that $\pi\alpha i \operatorname{id}_Y$ is continuous. I reach a similar impasse trying to show $\operatorname{id}_Y : Y_b \to Y_a$ is continuous.

[gentle osprey]

kxrider:

[sleek thicket]

Let i_α : Y_α -> X_α. You essentially want to show π_α ° i = i_α, yeah ?

I guess I don't see what you mean by via universal properties

hmm

actually sorry I think I'm too sleepy to help here, I hope you can figure it out

[gritty widget]

go to sleep

or ill post RG

[little hemlock]

its np, have a gn sham

[little hemlock]

okay i think i got it. $\pi_\alpha i \operatorname{id}Y : Y_a \to X\alpha$ is the same as $i_Y \pi_\alpha^Y$ where $\pi_\alpha^Y : Y_a \to Y_\alpha$ is the projection (which is continuous by assumption), and $i_Y : Y_\alpha \hookrightarrow X_\alpha$ is the inclusion which is also continuous by assumption.

[gentle osprey]

kxrider:

[little hemlock]

map theoretic stuff is cool but kind of messes with my head lol

[sleek scaffold]

I have a question from an exercise about complete measure spaces

If, given a measure space $(X,\mathcal{A},\mu)$, I'm trying to prove that ${A\cup N : A\in \mathcal{A}, N \text{ is a subset of some } \mathcal{A}\text{ measurable null set}}$ is a $\sigma$-algebra, how do I show that the complement $(A\cup N)^c = (A^c\cap N^c)$ is also a part of that sigma algebra?

[gentle osprey]

maple:

[sleek scaffold]

I think what I am asking boils down to showing that $(A\cup N)^c$ is $A'\cup N'$ for a different pair of sets, but I'm having trouble with that.

[tepid depot]

@sleek scaffold this probably belongs in #advanced-analysis (gonna think about it though give me a minute)

[gentle osprey]

maple:

[sleek scaffold]

OK

[tepid depot]

ah ok think i got it

it helps to draw a picture

draw A, N, and a null set M containing M

[sleek scaffold]

Should I draw with intersection between A and N?

[tepid depot]

yeah since that's fully general

[sleek scaffold]

OK

[tepid depot]

draw it in a box X

and shade in (A U N)^c

hopefully your picture looks something like that

you should be able to see that (A U N)^c is the union of a measurable set with something contained in a null set...

just stare until you see it lol

[sleek scaffold]

Oh

Yes

Thank you 🙂

[tepid depot]

no problem 🙂

[crimson imp]

Hey, people, I’m struggling to prove that $TS^n \times \mathbb{R}$ is diffeomorphic to $S^n \times \mathbb{R}^{n+1}$. Does anyone know an approach to this problem?

[gentle osprey]

pmorelli:

[tepid depot]

@crimson imp there's a straightforward map you can define taking you from $S^{n} \times \mathbb{R}^{n+1} \to TS^n \times \mathbb{R}$. have you checked if that's a diffeomorphism?

[gentle osprey]

doubledual:

[sleek thicket]

what's the map? I was trying to think of one and I couldn't

Because S^n isn't parallelizable

Oh sorry I think I see, you use the embedding of S^n into Euclidean space

[gritty widget]

[tepid depot]

let me expand a little since maybe the map isn't so straightforward

you're given a point on S^n and a vector in R^{n+1}

and you're thinking of S^n as being embedded in R^{n+1}, and the tangent plane as living in R^{n+1} too

how do you turn a point on S^{n} and a vector in R^{n+1} into a point in TS^n x R?

thinking of everything as being embedded is key

[sleek thicket]

So if you have (p, v) you can write v = w + tp for w, p orthogonal, and φ(p, v) = ((p, w), t)

I think you can compute the Jacobian of this as a map Rn+1 × Rn+1 -> Rn+1 × Rn+1 × R

[tepid depot]

yeah that's it

[sleek thicket]

lol sorry i disconnected but it sounds like this is the same thing I came up with

oh or instead of computing the Jacobian just write down an inverse

[crimson imp]

I That’s a great strategy, I didn’t think about looking for a map defined on $S^n\times \mathbb{R}^{n+1}$

[gentle osprey]

pmorelli:

[crimson imp]

I will try to write down an argument

[crimson imp]

Is this map well defined? I mean, is it possible to find other t’s and w’s that satisfie v=w+tp?

[sleek thicket]

@crimson imp yes, we're projecting onto the subspace spanned by p

The fact that w and p are orthogonal is what makes this unique

t = <v, p> and w = v - <v, p> p

[crimson imp]

Oh, think I get it

Thanks, guys 🙂

[nimble salmon]

In the highlighted section, should $T \pi_{TM}$ be simply $\pi_{TM}$? It's from Lee's Manifolds and Differential Geometry. I'm pretty sure it's a typo (many typos in that book) but I can't find it in the errata, so I might be just having a brain fart.

[gritty widget]

"a nice map"

[nimble salmon]

Yeah not sure what he means by that either

[gritty widget]

what does the T on a map mean?

e.g. T\pi

the differential?

[gentle osprey]

frittata:

[gritty widget]

no i mean like this thing

[nimble salmon]

Ahhh yes that's the tangent map

Which restricts to the differentials on the fibers of the tangent bundle

[gritty widget]

wow that's fucking cancer to look at

[gentle osprey]

TTerra:

[gritty widget]

and it doesnt make sense to say they're the same

in which case it does make sense to say they're the same

[gentle osprey]

TTerra:

[gritty widget]

as you say

although i haven't done the exercise so idk

yeah there shouldn't be any T\pi_{TM}'s

neither equation makes any sense

@nimble salmon

okay now for actually defining s...

christ

that's aids

so if $(p,v) \in T_pM$ then you can describe the elements of $T_{(p,v)}TM$ using curves: pick a curve $\alpha : (-\delta,\delta) \to TM$ with $\alpha(s)=(p(s),v(s))$, $p(0)=p$,$v(0)=v$. then you have the tangent vector $\alpha'(0) \in T_{(p,v)}TM$ which you can write like $$\alpha'(0) = ((p,v),\alpha'(0))$$ i guess?

[gentle osprey]

TTerra:

[gritty widget]

but one must ask themselves: why

[fading vale]

never ask why

asking why in math is always a bad idea

[gritty widget]

and then uhh you can look at how \pi_{TM} and T\pi act on this and try to come up with a definition of s

yes moth

especially with weird ass problems like these

wait does s = id work

https://arxiv.org/pdf/math/9611223.pdf

on page 2

this seems to be it but i don't wanna read into the details 🤢

[sleek thicket]

@gritty widget post pfp

[sleek thicket]

Cute

[gritty widget]

yeah

#competition-math message
i said something about never looking in the preuni or early uni channels and then arch did this

[tepid depot]

that's sick

[sleek thicket]

Lol

double have you seen beastars?

It's a good anime

[gritty widget]

season two in january

[tepid depot]

have not actually

[sleek thicket]

POG

I didn't know that

[gritty widget]

yes they're going to cover the manga arc

where they find out the killer

and whatnot

my favorite part of the manga

i hope they only cover that because it drops off a bit after

anyways i'm hungry so i'm gonna leave this here then go make food, rg prof told us to read it because it has our homework answers

[nimble salmon]

Sorry I was away! Yeah so it looks that the statement had a typo lol. Btw my solution is that

[gentle osprey]

frittata:

[gritty widget]

yeah i was working on the problem and i figured it was a swap type thing (i mean, your book's author calls it s for swap lol)

got lazy and googled it

[nimble salmon]

Nice, it didn't occur to me that s stands for swap

And thank you for the paper link, gonna read through it (if it doesn't burn my eyes too much)

[gritty widget]

repost

ah okay you got it

ye this talks about that map s a bit i guess

the notation looks awful in this lol

[nimble salmon]

Yeah honestly notation is what keeps me away from diff geo

Or should I say notation_s_, since each branch of math and physics use their notation for it

I think somebody once described differential geometry as the branch of mathematics which is invariant under change of notation

[gritty widget]

haha

lee (the other lee) writes that in his ism book i think

[nimble salmon]

What book is this?

[gritty widget]

https://www.springer.com/gp/book/9780387217529

note that the book you're reading is also by a lee

but it's not the same one

yours is the jeff lee

[nimble salmon]

As if diff geo couldn't get more confusing lol, I thought the two books were by the same author!

[gritty widget]

y'know while looking at your exercise i was like "why the fuck would anyone do this"

but

that paper confirms it

people actually care about shit like TTTM

[gentle osprey]

frittata:

[gritty widget]

cotangent bundle of the cotangent bundle

[fading vale]

scary

[gritty widget]

trivial

@ doubledual is isomorphic to the original space

[fading vale]

i was going to say something about how that should morally happen but i decided not to

[gritty widget]

i'll put a /s next time

this is your brain on AT

[fading vale]

hurb

it was obviously a joke

i also said "morally true"

[gritty widget]

too bad

you've already been sully'd

it's over

[fading vale]

oh yeah?

[gritty widget]

modulo honorable abuse

[fading vale]

check the message again

[gritty widget]

if moth deletes this message they're gay

[gritty widget]

@ moderators power abuse

[fading vale]

theyre gay

[sleek thicket]

I thought that you used morally true to mean true

aren't you intuit gang

[fading vale]

no

there are stages of moral truth

[sleek thicket]

I'm p sure max once said morally true false things are truer than morally false true things

[fading vale]

ok its like

you have morally true statements like "topological spaces are semilocally simply connected/path connected/locally path connected"

[gritty widget]

philosophy goes in #chill

[sleek thicket]

lol

[fading vale]

where its like a restriction of the general case

and you have morally true statements that dont correspond to reality at all but are a descriptor of what reality should be

[sleek thicket]

This fits my theory that morally true is a synonym for false

[fading vale]

its not!

[sleek thicket]

no other connotations

[fading vale]

no!

[sleek thicket]

okay so unrelated

I was doing this year's manifolds qual at my uni (for fun)

*uni

Wanted to discuss/check my solutions

But also, I used riemannian geometry on 8 and you're maybe not allowed to use that/not supposed to know it when taking the qual

Like, it's not taught in the manifolds course

[tough imp]

LOL imagine thinking “topological spaces have sort of nice property” is morally true

Your brain on 0 AG hurb

[sleek thicket]

So I was wondering if anyone had a nice way of seeing the following: if $G$ is a compact, connected Lie group then $\det(Ad(g)) = 1$ for all $g\in G$

shhmonkey

[gentle osprey]

shamrock:

[fading vale]

AG doesnt deal with topological spaces

spec isnt topology its a mutation

[tough imp]

Hurb

[sleek thicket]

my proof is that by compactness we can choose a bi-invariant metric, where $Ad(G) \subseteq O(\mathfrak{g})$

[gentle osprey]

shamrock:

[tough imp]

Topology is too weak to deal with big strong AG things hence Groth had to generalize topological spaces

[sleek thicket]

So they all have determinant ±1

[fading vale]

hurb

AG is just ugly

topology had to be made weaker to allow it into its folds

[sleek thicket]

Then by connectedness and the fact that $\det(Ad(e)) = 1$ we see it's always 1

[gentle osprey]

shamrock:

[sleek thicket]

Does anyone have a way of seeing this without using RG?

[tough imp]

“Topology couldn’t handle a wide array of things because it’s strong”
Hurb

[sleek thicket]

like on any compact lie group the adjoint representation always has det ±1

[fading vale]

Yes.

It was pure.

[sleek thicket]

Oof I can't bug ttera because he doesn't know lie groups

[fading vale]

nerd

[gritty widget]

[tough imp]

Wait wat did TTerra skip smooth manifolds to do RG?

[gritty widget]

sorry i can't hear you over my ricci curvature bounds

my ass is too curved

[sleek thicket]

No he just didn't do lie groups

[tough imp]

Oh

[sleek thicket]

In his smooth manifolds course

And this is like lie groups heavy

[gritty widget]

yeah my manifolds course did not mention lie groups once LOL

[tough imp]

Hurb

[gritty widget]

kind of a shame

they're nice

so everything i know is from talking with sham

e.g.

theyre complete

or something

[sleek thicket]

All vector fields are complete

[gritty widget]

some funny stuff about left/right/biinvariant metrics

[tough imp]

If you don’t know Lie groups you won’t know Lie algebras and then Algebraic groups won’t make sense to you

[sleek thicket]

They might not be complete as riemannian manifolds unless they have a bi invariant metric

[gritty widget]

ah i see

[tough imp]

That’s kind of a shame they would shoot you in the foot like that

[sleek thicket]

Eh

Like our manifolds course is very very long and fast and stuff

Like

It's pretty hard and maybe not the best way to teach it

ISM is more comprehensive than most books

[gritty widget]

we did not do ism 😔

out of tu's book

which does like

nothing

[sleek thicket]

Yeah I mean I don't think every class has to do ISM

[tough imp]

Do you think there’s Negi-type people who ask if there’s an orbifold-theoretic approach to all this so they can just subsume manifold theory

[sleek thicket]

We did ISM over like 6 months or so and it was insane

really really hard to keep up with

I don't know that I learned it as well as if we went slower

anyways, my question

[gritty widget]

quick shamrock tell me if S^1 x RP^2 can be given a metric of negative curvature

[tough imp]

You should repost it lol since it got buried

[sleek thicket]

ffs

On any compact lie group the adjoint representation always has det ±1

[elder yew]

It can't? Tterra?

[gritty widget]

right

it cannot

preissman's theorem

[elder yew]

w00t

wot

RP^2 is just fucky

[gritty widget]

oh wait

no

synges theorem?

idr

yeah its an orientability thing

[elder yew]

that's what I figured

[tough imp]

RP^2 = CP^1 = Riemann Sphere

[gritty widget]

nvm nope im going in circles, it is preissman. if it had a metric of negative curvature then preissman's theorem says every nontrivial abelian subgroup of pi_1 would be isomorphic to Z, not the case

[sleek thicket]

yeah, it has a C2

[gritty widget]

synge concerns positive curvature

anyways sham repost for the third time

i will go

[elder yew]

moon is having dumplings with red wine

[sleek thicket]

Sham is having cocktails

Okay so my question

On any compact lie group the adjoint representation always has det ±1

I can prove this by choosing a bi-invariant riemannian metric, but I'd like to do it more elementarily. Like, just the contents of ISM lol

[elder yew]

define the adjoint representation?

Can you make a path connectedness argument?

I've seen things like you go around on points, on each neighborhood you get a result

but you have a connected chain so the result holds

[sleek thicket]

So for any g in G, we have a conjugation action G -> G. The derivative of this is an automorphism of the lie algebra, so it's a representation of G on the lie algebra

[elder yew]

shot in the dark if something like that would work

[sleek thicket]

You can reduce to a nbhd of the identity on which the exponential map is surjective

oh hm no that's only for the case of G connected

So I'm not sure you can use path stuff

[elder yew]

Hrmm

I'm not sure what ISM methods would entail

That was my go to nuke in a few courses

[gritty widget]

anything without a metric

[sleek thicket]

Well the actual test problem is for the connected case

[elder yew]

Can you do it on each path connected component

[sleek thicket]

I just realized that the RG proof does more than I need

[elder yew]

and then try to patch together with AT?

[sleek thicket]

Okay sure assume G connected

Then the problem is to show det(Ad(g)) = 1 for all g

[elder yew]

Can you assume it's not 1 at a point

then do a nbhd argument

with complements

I'm just throwing paint at a wall

[sleek thicket]

I mean, maybe? I don't see why the set where it's 1 would be open too

[elder yew]

hoping to get a solid color

[sleek thicket]

So you can reduce to a nbhd of e onto which the exponential map is surjective

and it suffices to show det(Ad(exp(X))) = 1 for any X in the lie algebra

But I didnt see a nice way to compute Ad ° exp...

[elder yew]

Hrm I guess idk

I'd have to look at a lie theory text

Nothing stuck to my head when I took Lie Theory

Tterra any thoughts?

[gritty widget]

i'd have to recall the definitions of these things

i don't play with lie groups a lot so

i'm not comfortable/familiar with things like ad and Ad

[sleek thicket]

I did try to use ad but it didn't work

hmm okay maybe I should ask about other problems first and then this later

Okay so

Problems from qual I just did

I had a really cute solution to 5

so like

[gentle osprey]

TTerra:

[gritty widget]

blegh

too lazy to fix that

[sleek thicket]

What's that last bit?

it's the derivative of γ(1) α(t) γ(t)^-1 at time 0

[gritty widget]

$$(\gamma(1) \cdot \alpha \cdot \gamma(1)^{-1})'(0)$$

[gentle osprey]

TTerra:

[sleek thicket]

Right

[gritty widget]

like

[sleek thicket]

Yup

[gritty widget]

conjugate the curve and differentiate

but wtf is that

[sleek thicket]

But that didn't seem nice lol

anyways

Other problems

So for 5

If $γ$ is an integral curve it satisfies $γ'(t) = A γ(t)$, where $A = \begin{bmatrix} 0 & c & - b \ -c & 0 & a \ b & -a & 0\end{bmatrix}$

[gentle osprey]

shamrock:

[sleek thicket]

By the theory of linear ODEs, the solution with initial condition $γ(0) = p$ is $γ(t) = e^{At} p$

[gentle osprey]

shamrock:

[sleek thicket]

so $Θ_t$ is a linear transformation, namely $e^{A t}$

[gentle osprey]

shamrock:

[sleek thicket]

now why is this in $SO(3)$?

[gentle osprey]

shamrock:

[elder yew]

By definition

[sleek thicket]

?

[elder yew]

I am

Now drunk

Is it by definition?

[sleek thicket]

you gotta compute the exponential lol

which sucks

[elder yew]

Oh is that the

[sleek thicket]

You can diagonalize but it's ass

[elder yew]

$$ \Lambda D \Lambda ^{-1} $$

[gentle osprey]

MoonBears-C-:

[sleek thicket]

right, and D is actually complex too (to make it worse)

[elder yew]

Jesus I haven't done that since lower division

[gritty widget]

[sleek thicket]

Well you need to compute arbitrary powers of A to find the exponential

Since it's a power series

[elder yew]

That's one way

[sleek thicket]

So the only real way is to diagonalize

however!

[elder yew]

Or you can give a matrix representation

[sleek thicket]

wym?

[elder yew]

I'm in accordance with your diagonalization

Again

Drunk

[gritty widget]

https://math.stackexchange.com/questions/1949814/adjoint-representation-of-exponential-on-lie-group
maybe you can use this to compute the thing i was trying to do earlier sham

[sleek thicket]

You can see directly from the definition of $A$ that $\tr A = 0$, so $A \in \mathfrak{sl}(3)$ and it's skew symmetric $A^T + A = 0$ so $A \in \mathfrak{o}(3)$. Because $SL(3)$ and $O(3)$ are lie subgroups of $GL(\R, 3)$, we have $\exp(tA) \in SL(3) \cap O(3) = SO(3)$

[gentle osprey]

shamrock:

[sleek thicket]

isn't that sick??

You can completely avoid thinking about the flow

You just need to know the lie algebra of SO(3)

It's traceless skew symmetric matrices

Which A is

Also you can compute directly that $A (a, b, c) = 0$, so the same is true for all powers, and by using the power series definition $\exp(tA) (a, b, c) = 1$

[gentle osprey]

shamrock:

[sleek thicket]

so you can skip computing Θ at all!

Just using the matrix exponential and a tiny bit of lie theory

[elder yew]

that is slick

[sleek thicket]

Yeah lol I felt very smort

[elder yew]

How long is the computation

[sleek thicket]

Not long at all

Like

[elder yew]

I feel it would take like ~ 2 mins

[sleek thicket]

You just gotta combine terms in $aX + bY + cZ$

[gentle osprey]

shamrock:

[sleek thicket]

That's literally the only computation

but ofc I spent like 15 minutes of time trying to diagonalize

[elder yew]

A born geometer

[gritty widget]

just plug into WA

[sleek thicket]

tfw you don't have wolfram alpha acces on a qual 😔

[gritty widget]

weak

you don't have the entirety of WA remembered?

[sleek thicket]

For problem 6a I said that $\omega$ is cohomologous to $\frac{-y dx + x dy}{x^2 + y^2}$ since that's a closed but not exact form and $H^1(\R^2\setminus{0})$ so if $C$ is the circle of radius $\varepsilon$ then $\int_C \omega = \int_C \frac{-y dx + x dy}{x^2 + y^2} = 2\pi$. But we can also compute $\int_C \omega = \varepsilon \int_0^{2\pi} \cos(t) g(\gamma(t)) - \sin(t) f(\gamma(t)) dt$, so if $|f|, |g|$ are bounded by $K$ we have $\abs{\int_C \omega} \leq \varepsilon \int_0^{2\pi} B + B = 4B\pi \varepsilon$. Thus $2\pi \leq 4B\pi \varepsilon$ for all $\varepsilon > 0$, which is a contradiction

[gentle osprey]

shamrock:

[elder yew]

yA

[sleek thicket]

nice

And I already gave an explicit example for b

[elder yew]

Looks like you're ready for quals

[sleek thicket]

Nice

sick

I did algebra yesterday but it was with chmonkey so it doesn't count

[elder yew]

Just bug UW till they let you take it

[sleek thicket]

lol

[elder yew]

The Analysis ones are weird

[sleek thicket]

i mean they told me they'd let me into the masters program no question asked

I could probably get into the PhD program, they like admitting their own students

[elder yew]

Oh really?

[sleek thicket]

Yup

[elder yew]

Dude fuck UCLA, I should've gone to UW

[sleek thicket]

I know a bunch of people

[elder yew]

UCLA actively discourages their own students from going there

by not admitting them or low-balling their stipends

The UCLA Mathematics Department encourages students to change their educational institution between their Bachelor and PhD degrees to broaden their horizons. In particular, no admissions advantage is given to UCLA undergraduates.

https://ww3.math.ucla.edu/admissions/

It's on the fucking website

anyways

Yeah!

[sleek thicket]

Yeah I mean, I would like to do that

But I'm gonna apply as a safety

[elder yew]

Would you go to berkeley?

or davis?

[sleek thicket]

Yup

[elder yew]

I met a girl whose first name was berkeley-davis

You can guess where their parents went to school

[sleek thicket]

Berkeley because it's Berkeley and also my parents live here

lol

live in the bay area at least

I could Bart/drive over

so 7 is secretly easy. T_e G = T_e R^n (+) T_e GL(n) = R^n (+) GL(n)

The subspaces have the usual lie algebra structure

So you just gotta compute $[(0,A), (v, 0)] = ad((0,A)) (v, 0) = \frac{d}{dt} Ad((0,tA)) (v, 0) = \frac{d}{dt}\frac{d}{ds} (0,tA) (sv, 0) (0, t^-1A^-1) = \frac{d}{dt}\frac{d}{ds}(s t A v, 0) = (Av, 0)$

[gentle osprey]

shamrock:

[sleek thicket]

And finally I have a riemannian manifolds argument for 8, which may or may not be valid

I got this all in 1.5/4 hours so I think I would've solidly passed the qual!

Oh yeah so

[elder yew]

I'm applying to Cal too

I won't get in, but it's a shot

[sleek thicket]

Yeah

I need to make a list

😢

Okay so my original question again

[elder yew]

the adj

Hrmm

[sleek thicket]

Oh wait ttera had a proof

[gritty widget]

wat

i just wrote out the definition lol

[sleek thicket]

Not you lol

[gritty widget]

[sleek thicket]

The stack exchange proof

[gritty widget]

oh

doing qual problems
when is it time for me to take the qual pill

inb4 "now"

[sleek thicket]

So it suffices to show $ad(X) \in \mathfrak{sl}(\mathfrak{g})$

Ie it's traceless

hmm

[gentle osprey]

shamrock:

[sleek thicket]

@gritty widget I'm mostly just doing it for fun

And to test how well I understood my manifolds course

[gritty widget]

nice

seems like you understand it pretty well

i should look at some of my uni's manifolds qual problems

anyways continue ur problem

[sleek thicket]

I feel so too

So like

In the past I tried this for previous years

And it was too hard/I got scared

But after actually finishing the class I think I'm in a good spot

I didn't like prep for this

@gritty widget you should look at uw ones

I think they're really good

otoh lie groups...

Shamrock has moved on from cocktails

To white wine

[gritty widget]

https://www.math.toronto.edu/cms/graduate-program/current-students-grad/past-comprehensive-exams/

[sleek thicket]

I'll take a look

What's with the I/II ?

[gritty widget]

the grad courses here split into two semesters

e.g. topology i is differential, ii algebraic

this is reflected in the exams

except for grad complex, which is one semester

[sleek thicket]

Ahh yeah that made sense

The second seemed very differential lol

Uw has a post quals course on AT

Year long, every other year

First two quarters are like basic AT/homology/cohomology and then I think it terminates in either homotopy theory or vector bundles/applications to manifolds stuff

[sleek thicket]

Okay so uhh

Let $V$ be a finite dimensional real lie algebra

[gentle osprey]

shamrock:

[sleek thicket]

Why is $ad(X)$ traceless

[gentle osprey]

shamrock:

[sleek thicket]

I guess like, what are the eigenvalues?

We can extend $X$ to a basis $X = X_1, \ldots, X_n$ and write $[X_i, X_j] = c_{ij}^k X_k$, so $ad(X)$ has matrix $(c_{1,j}^i)$

[gentle osprey]

shamrock:

[sleek thicket]

Then I want to show $\sum_{i=1}^n c_{1i}^i = 0$

[gentle osprey]

shamrock:

[sleek thicket]

Oh wait

This isn't algebraic

I have to use compactness somehow

hmm

Can you detect when a lie algebra is the lie algebra of a compact lie group?

Real

what's the killing form?

I don't know lie algebra theory, sorry

oh I looked it up

Hm

So my original problem

Is to show Tr(ad(X)) = 0 in Lie(G) for G a compact connected lie group

Or really that det(Ad(g)) = 0 for g in G

What does unitary invariant mean?

like

I don't have a metric

Oh okay

I don't see why this implies it vanishes on ad(X)?

ad(X)Y = [X, Y]

oh I see your confusion

I'm looking at tr(ad(X))

Not tr(ad(X) Y)

I'm also not in a matrix lie algebra (I mean I can embed but I can't embed the lie group so I lose compactness...)

ad(X) is a linear map \g -> \g

And \g is a finite dimensional vector space

Yeah but I lose information that way

re:compactness of G

So I'm not sure it holds in an arbitrary lie algebra

Oh sure okay

I would like to only use things I know though lol

This was a problem on the manifolds qual at UW in fall

I solved it with riemannian geometry

But I didn't know riemannian geometry when the test was administered

And it isn't covered in the class so my proof might not be admissible

the RG proof is that a compact lie group automatically has a bi invariant metric, and Ad(g) in O(\g) wrt a bi invariant metric, so det(Ad(g)) = ±1 for all X. Using connectedness and det(Ad(e)) = 1 we get that det(Ad(X)) = 1 everywhere

[fading vale]

I'm not really sure what to do here

or what this problem is asking for exactly

like i mean

oh maybe its like

h_n is isomorphic to H_n or something

lets find out

[tight agate]

it wont be iso to Hn

in general

[fading vale]

yeah

i think i didnt realize at first that its assuming C_n(X) is a Z module

but i see now

it is isomorphic in the case G = Z

[tight agate]

yup

the other cases are pretty stupid

[fading vale]

i think just zero

[tight agate]

yup

[tough imp]

You're a Z-module moth

[fading vale]

[tough imp]

[meager python]

Since we can define grassmanians in a coordinate free way as hilbert schemes, can we do something similar for proj?

[tight agate]

are you asking if it can be defined by a universal property?

[meager python]

Or just intristically, without gluing or similar

[tight agate]

https://stacks.math.columbia.edu/tag/01O4

[meager python]

The scheme quotient way of introducing proj seems to not be coordinate free either

Thanks I’ll try to unwind that

[fading vale]

axioms here are these btw i know theyre different

so im pretty sure (3) works and im guess that this fails for (2)?

but im not rly sure how to show it

[sleek thicket]

hmm

[fading vale]

hello shamrock

also i hope you are doing ok i saw tweets : (

[sleek thicket]

I am just like

drinking and angry

fuck academia

[fading vale]

yeah : |

[sleek thicket]

angry at self but also everything else

also I lied on Twitter for clout

I only have half a bottle of tequila

[fading vale]

wdym according to your twitter you are a minor with no tequilla smh

[sleek thicket]

I should play minecraft

but

I need to set up my desktop

takes work

I'm thinking maybe you X/A acyclic

But maybe not

Or like

One of them should be acyclic

For simplicity

Like what about X = disk, A = circle

Simple example

Ahh no

If all homotopy groups are free I think it just holds

[tight agate]

you prolly want some torsion in there

[sleek thicket]

So what about a projective plane somewhere

[tight agate]

yeah

[sleek thicket]

yeah

oh so like

X = RP^2, A = S^1

Then X/A = S^2

you have Hom(H^3(A), Z) -> Hom(H^2(X/A),Z) -> Hom(H^2(X),Z) exact right

wait uhh

This must be wrong

It feels right though...

Like

Isn't that 0 -> Z -> 0?

@fading vale what do you think of this

[fading vale]

hrm

[tight agate]

the map is going from 3 to 2?

[fading vale]

oh i think this is a typo but it should be H_i not H^i

but. yeah

i guess this proves that its not exact?

[sleek thicket]

oh lol sorry idk

I am sloshed

[fading vale]

oh hm

i think actually H_2(RP^2) = Z so Hom(H^2(X), Z) = Hom(Z, Z) = Z

so

hn

counterexample might not work

[tight agate]

H2(Rp2) = 0

[sleek thicket]

sorry

[fading vale]

oh wait yeah 2 is even

im brainlet

mhm

okay

[tight agate]

okay yeah it works

[fading vale]

so then this isnt exact and it is a counterexample

nice nice nice

[tight agate]

H1(X)* is 0

H1(A)* is Z

H2(X/A)* is 0

by * I mean hom(_, Z)

so you get 0 --> Z --> 0

oh fuck no

H2(X/A)=Z

[fading vale]

yea

but we dont even need to do that i dont think

exactness breaks at Hom(H2(S^2), Z)

[sleek thicket]

hell yeah

I am good at math

yes

[fading vale]

based shamrock

ty

[sleek thicket]

the time I got the most drunk in my life and threw up and my roommates got mad I did an algebra final for the undergrad class at my school

It was so easy

I don't get it

[tight agate]

does it?

[fading vale]

mhm because its 0 -> Z -> 0

[tight agate]

H1(S1) = Z

so Z -> Z ->0

[fading vale]

uh you do
Hom(H3(S^1), Z) -> Hom(H2(S^2), Z) -> Hom(H2(RP^2), Z)

wait

fuck

fuuuuuck

[tight agate]

the map goes from 2 to 3

[fading vale]

yea

aghh

[tight agate]

not 3 to 2

[fading vale]

yeah its Hom(H_i(X, A), Z) -> Hom(H_i(X), A) -> Hom(H_i(A), Z) ->

[sleek thicket]

okay so shamrock example failed

wq

Qq

[fading vale]

Q_Q

[sleek thicket]

Wait I'm confused

Isn't the start of the sequence h^3(A) -> h^2(X/A) -> h^2(X)?

[fading vale]

cohomology goes up instead of down

[sleek thicket]

fuck lmao

okay

Time for shamrock to not do math

[fading vale]

lol

go sleep

[sleek thicket]

thank you for not being patronizing moth

Also no I'm gonna play minecraft with my brother

Because he can't be jere

[fading vale]

or play mc!

[sleek thicket]

Because of covid

[fading vale]

aw

also why would i be patronizing : c

[sleek thicket]

You aren't!! I like that!!

[fading vale]

[sleek thicket]

Also I am drunk and if I call you dude I hope you know it's because I'm drunk and not because I think you're q man

Like

I have been calling a lot of people dude rittht now

[fading vale]

haha its ok

Hello Ultraproduct

I

okay.

[uncut surge]

the well-known gender neutral pronoun "dude"

my personal replacement for that is "fam"

at least online, it replaces it perfectly

[fading vale]

hn

yeah fam is weird to say irl

[uncut surge]

that is true, and i don't put it in my mouth

[sleek thicket]

I mean this is gonna sound like bullshit

[uncut surge]

but online it's perfect

[sleek thicket]

But I do use guys and dudes in a gender neutral fashion

And it takes mental effort to not do that

[fading vale]

also H^i(S^n/S^n-1) = H^i(S^n)^2 right

i think it depends on the person

[uncut surge]

"hey do you fuck dudes"

[fading vale]

also the context

[sleek thicket]

Because I know people who would rather I not say that

Yeah moth

I try to not

[fading vale]

like

yeah what lartomato said

[sleek thicket]

okay so like

That argument always soujded sus to me

[fading vale]

in that context because you're introducing something gendered here (sex) dudes suddenly has a gendered connotation

[sleek thicket]

because words have different meanings depending on context

[fading vale]

mhm

[sleek thicket]

like always

Sorry all I meant was

I have a mental filter

This is common and good for interacting with other humands

[fading vale]

lol

[sleek thicket]

And I am bad at keeping it up when drunkv

[fading vale]

Probably not often Ultra

[sleek thicket]

@limpid vault no ultra it's just an example

[fading vale]

But if they theoretically did