9:15 PM
This is just beautiful

@TastyRomeo why are you feeling geometric?
We need to reel this room back to rep/number theory before it's too late

It's too late, category theory already won over the room

I'm fine with category theory/number theory/algebra

Categories are aight too
I'm just concerned with geometry
:P

I downloaded 0AD. I haven't played a strategy in a long long time

9:30 PM
For a smooth manifold $M$ with $p \in M$ and $v \in T_pM$ and $f \in C^{\infty}(M)$, $v$ 'acting' on $f$, just means the derivation $v$ taking $f$ as its input right? So $v(f) = c$ for some $c \in \mathbb{R}$ since $v : C^{\infty}(M) \to \mathbb{R}$
And notation-wise $vf = v(f)$ correct?

I love that picture. I saw it from this page called "Derived memes for spectral schemes"

b0ss meme

Hello ladies and gentlemen

Hai

9:39 PM
Hey @Astyx

Long time no see

Hi, $$\text{Let }f\in C([0,1]), g\in C^1([0,1]) \text{ such as } f\geq g \text{ and increasing } g. \\\text{Is there the existence a sequence of function in }C^1([0,1]) \text{ minus by } g \text{ and having the uniform limit }f?$$

How've you been?

Not doing maths
Which is terrible :(
But appart from that good

Hey @Astyx

9:46 PM
6

Fucking utilitarians

tbf that computations look more like GR than pure differential geometry

@Perturbative Did you make this

Sniped by Balarka
Yep I made it, but it's just a joke, I still like diff geom so far

Well it's mine now
I made it

9:53 PM
@BalarkaSen prepare to get sued

LMAO
That is legendary

@MatheinBoulomenos tbf that was the only overly complicated expression I could find on Google images

well, of course you'll end up with the physics results if you look for overly complicated expressions

you could also end up with intergalactic teichmuller theory
or whatever the name was
@Daminark "These days angel of topology and the satan of category theory fight for the soul of each individual mathematical internet chatrooms"

10:00 PM

Let $f, g: \Bbb S^1 \to X$ be two loops, and $\varphi, \psi : \Bbb S^1 \times I \to X$ be two homotopies of $f$ and $g$. Must there be a homotopy $\Bbb S^1 \times I^2 \to X$ between $\varphi$ and $\psi$? If the universal cover exists then the answer is certainly yes, because the universal cover is loops quotient homotopy and is simply connected, and the homotopy required is a path between the two points in the universal cover. What if the universal cover doesn't exist?
@MatheinBoulomenos @BalarkaSen

@MikeMiller pure gold this

cc @Daminark

10:20 PM
Now that I think about it without category theory I would've never understood why free groups, and free vector spaces and free modules all popped up everywhere
It's as if mathematicians were all trying to give me this free stuff I didn't want
Badum tss

@Perturbative but I can understand it without category theory
it's simply everything generated by those elements, without any extra assumption

@LeakyNun I knew the definition of it, just didn't know why it popped up in AT in stuff like Seifart van Kampens theorem

I see
@GFauxPas hi!

hello Mr Nun

@MatheinBoulomenos Do I have an epimorphism from $D_{2mn}$ to $D_{2m}$?

10:28 PM
what's the interpretation of a linear map which is diagonal? or diagonalizable, more generally?
what does such a map look like?

@LeakyNun not sure. I'd guess probably not in general

@GFauxPas It is similar (i.e. after changing the basis) to a map that is stretching each component

hmm

@GFauxPas In an appropriate choice of coordinates the map stretches along each axis

without rotating it you mean?

10:31 PM
but do note that the new basis does not need to have vectors that are perpendicular in the old basis

Nope, no rotation

interesting

In general the eigendirections of a linear map are the directions the map stretches along
In this case there's a basis of eigendirections
The stretch factors vary direction-to-direction though, btw

@MatheinBoulomenos do you know other sequences of groups other than $S_n$ and $D_{2n}$?

unless it's a multiple of identity or something

10:33 PM
Sure

oh, and do I have an embedding from $D_{2m}$ to $D_{2mn}$?
@KasmirKhaan goddag

@LeakyNun sure, you can look at various finite groups coming from linear algebra over finite fields
@LeakyNun yeah

I'm thinking about an extra credit problem and I'm not sure I'll get to it because i have thr egular problems first

@MatheinBoulomenos oh, nice
what do you suppose $\displaystyle \lim_{\longleftarrow} D_{2n}$ look like?
$D_{\infty}$?

it's:

10:34 PM
LEAKY :D
guten abend

let $A^{[k]}$ denote the matrix $A$ with each entry raised to the $k$th power

@KasmirKhaan was bringst du hier?

Given $A \in \operatorname{GL}_n(\mathbb C)$ , suppose that $A^k = A^{[k]}$ for all $k \ge 1$. Prove $A$ is diagonal.

@LeakyNun hunger for Learning :D

@Semiclassical already noticed that it is sufficient to consider only $k$ from $1$ through $n$, because for $k$ larger than $n$ $A^{k+i}$ can be written as a linear combination of lower powers of $A$ by Cayley-Hamilton
which suggests induction, but it also suggests proof by contradiction
but it's only extra credit so i probably wont get to it, but im sharing it because ti's interesting

10:38 PM
@LeakyNun nevermind, there is an epimorphism $D_{nm} \to D_{n}$

@MatheinBoulomenos then what does $\displaystyle \lim_{\longleftarrow} D_{2n}$ look like?

@LeakyNun it's not $D_\infty$

What is it then?

some profinite group? I don't know

interesting

10:40 PM
I think it's $\hat{\Bbb{Z}} \rtimes \mathbb{Z}/2\mathbb{Z}$

what is the involution on $\widehat {\Bbb Z}$?

$x \mapsto -x$

fair enough
@KasmirKhaan find $[\Bbb S_4,\Bbb S_4]$ :P

@MatheinBoulomenos Hello :D
Leaky
what is that notation ?

@LeakyNun but there's an embedding $D_\infty \to \displaystyle \lim_{\longleftarrow} D_{n}$ which has dense image, so it's kind of surjective

10:45 PM
@MatheinBoulomenos btw mathein , gonna repeat all abstract algebra from 0, gonna take dummit and foote book and hierstein =p

@KasmirKhaan $[A,B] = \{[a,b] \mid a \in A, b \in B\}$

during december

Hey @KasmirKhaan

and $[x,y] = xy(yx)^{-1} = xyx^{-1}y^{-1}$

that is a commutator
but what is the use of the other one ?

10:46 PM
@KasmirKhaan if you want to train your proving skills: for a group $G$ and a normal subgroup $N \trianglelefteq G$, prove that $G/N$ is abelian iff $[G,G] \subseteq N$

@MatheinBoulomenos Cute, I didn't know that!

one can show that a subgroup N is normal iff it has the commutator for any pair of elements in it

@KasmirKhaan did you?

What's the embedding?

@LeakyNun >< we done this on exam

10:47 PM
how did you do?

G/N is abelain , xN yN = xyN ,
but xyN = yx N
hence first direction

the second one i reversed the steps and inserted N in the product

G A W D
A
W
D

@Balarka as I said $\displaystyle \lim_{\longleftarrow} D_{n} \cong \hat{\Bbb{Z}} \rtimes \mathbb{Z}/2\mathbb{Z}$ and $D_\infty = \Bbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$, so basically you take the embedding $\Bbb{Z} \to \hat{\Bbb Z}$ on the first component and the identity on $\Bbb Z/ 2\Bbb Z$

10:50 PM
@BalarkaSen did you just found out that pluto is not a planet anymore? or why is that reaction ?>>

@Kasmir Click the link :P

@BalarkaSen oh my gah

@MatheinBoulomenos Hm, not entirely sure why the first isomorphism holds
Ah I see

boy band won so what

8 mins ago, by Leaky Nun
@KasmirKhaan find $[\Bbb S_4,\Bbb S_4]$ :P
@KasmirKhaan

10:51 PM
@BalarkaSen that could almost be the name of a light novel

I never Heard of those and from the page it is not news

@MatheinBoulomenos that title is bordering on being a light novel itself lmao

@LeakyNun hmm ill do it later leaky :D gonna make some coffe and study AA from 0 :D

Yeah I see why it's \hat Z semidirect Z/2. The Z/nZ's have that inverse limit on the left factor
That was dumb

@MatheinBoulomenos That could almost be a light novel

10:53 PM
It is a light to decent novel/10

@Perturbative 720 noscoped right there tbh

(only true melonheads will understand what I mean)
@KasmirKhaan It's the longest album title of all time.

@MatheinBoulomenos not sure how $\displaystyle \lim_\longleftarrow \Bbb Z/n\Bbb Z = \widehat {\Bbb Z}$ and $D_{2n} = \Bbb Z/n\Bbb Z \rtimes \Bbb Z/2\Bbb Z$ imply $\displaystyle \lim_\longleftarrow D_{2n} = \widehat {\Bbb Z} \rtimes \Bbb Z/2\Bbb Z$
Is it obvious?

@BalarkaSen haha okay >< I dont listen to Music so that is why I did not get it :D

I'm sad that nobody got my melon joke though
@KasmirKhaan You don't need to listen to music to read that it's a huge album title...
It's written in bold letters in the wikipedia page :P

10:57 PM
@LeakyNun not entirely obvious, no. I can prove it by messing around with an explicit construction of projective limits in the category of groups, but I guess that's not really pretty

@BalarkaSen i do accually, if you seen for the first time 1 apple, would you know if it was small or big ? you need at least 2 thing to be able to compare

...

@BalarkaSen melon memes/10

Aha, here's a man who appreciates my melon jokes

Since I'm here, @BalarkaSen I got a follow up from our work yesterday
Can I do induction proofs showing it's true for $n-1$?

11:00 PM
Ah, no, not I am afraid :P

A. asked me that and I didn't know

Well, unless you want to prove it for all integers, instead of just natural numbers like the examples we were discussing
So you have to do it for negative integers too

Yeah, I said no because the "direction" of the inductive step is the same as the bound of the set ($\Bbb N$) and that didn't make much sense

So you move backwards from n to n - 1

::nods::

11:02 PM
@BernardoMeurer That's a non-rigorous, but accurate explanation.

Just making sure I told her right
What's a rigorous explanation? :P

I would say that is because of the Peano axioms on $\Bbb N$.
More accurately, the "successor" function
Successor of each natural number $n$ is defined to be $S(n) = n+1$
The fundamental working principle of induction is to rig the successor function to prove a statement for all natural numbers knowing only for $n = 1$ (or $0$, whatever $\Bbb N$ starts with)

Bam, cool

There are "mathematical induction" techniques that are defined for more general sets than naturals, or the integers, however... there are things called "real induction", which is induction on the real numbers? I have no idea how it works
People in here knows more about this than me

@BalarkaSen Yes, but I'm afraid of mathematicians ever since that analysis professor touched me
mentally, not physically

11:09 PM
l m a o

> A change of variables $y = Sx$ (where $S$ is an invertible matrix) so that $\langle x, Ax \rangle = \|y\|^2$ (equivalently, $S^T AS = I$)
I fail to see how that is equivalent
@MatheinBoulomenos

@BalarkaSen You should try solving the proofs on my analysis exercise list, github.com/bemeurer/hard-calc
I think you might have a fun time
They're genuinely hard

@Bernardo Thanks! I'll bookmark these

@BalarkaSen There's a couple of typesetting errors here and there, but it should be overall understandable. I typeset all of that on a single night in rage, so it can only be so good :P

Hahah

11:14 PM
@LeakyNun I think it should be $S^TS=A$, not $S^TAS=I$

@Bernardo Hey. That's pretty good.
I'm looking at it right now

@MatheinBoulomenos I agree

This is going to be quite useful to me. Thanks

@BalarkaSen You're welcome! I can add to it if you finish them all, I have more I never got to put in there :)

I'll let you know for sure

11:16 PM
Also, if you want to be a sweetheart, open issues on the GH repository when you find problems, or just ping me on the h-bar :)

Nov 17 at 17:27, by MatheinBoulomenos
@TedShifrin how do show that $SO(n)$ is path-connected?
Nov 17 at 17:28, by Ted Shifrin
Aha, @Mathei. Let's not do Leaky's work for him.
@MatheinBoulomenos @TedShifrin I will show that there is a path from $I$ to any matrix $A \in SO(n)$:
Now, $A$ can be diagonalized over the complex numbers (follows from a simple argument by induction where you reduce the dimension of the space by one each time by taking the orthogonal complement)

Okay so I'm trying to prove the equivalence of definitions of solvability, anyone want to join in this endeavor?

nope

@Daminark sounds fun

11:19 PM
come on dude

@BalarkaSen fucking hell speak for yourself and for geometers

you single handedly make the algebra genre survive in this chat

I feel ignored

I think Balarka was talking about you

Ugh, algebra, I'm outie

11:20 PM
I was talking about @Mathei

I occasionally am willing to lower my standards and discuss other things, needs to change

ah I see

Handsome folks
what does SO stand for?

I'm going to go smoke filterless cigarettes, it's better for my health

Special orthogonal

11:21 PM
I discuss non-algebra things as well

thanks dami :)

Disgusting

Disgusted that I did not know that? tuff but fair

No, that Mathei discusses things non-algebraic. We can both work together to improve
Anyway so my definition definition of a solvable group is that the derived series terminates at the trivial group

11:23 PM
Hi @Ted

Hi @Ted

hi Mathei, Balarka â€” hey, Balarka, aren't you supposed to be on a sleep schedule?

@MatheinBoulomenos @TedShifrin Now, let $(\lambda,v)$ be an eigenpair. Then, $\|\lambda\|^2 \langle v, v \rangle = \langle \lambda v, \lambda v \rangle = \langle Av, Av \rangle = \langle v, v \rangle$, so $\|\lambda\|^2=1$. Now, pair each non-real eigenvalue with its conjugate, and on the diagonal matrix, change $\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}$ to $\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}$.

"Schedule: Don't sleep"

I slept 11 pm yesterday and woke up at 11 am
now i can't sleep
it'll alternate, i am guessing

11:24 PM
Your real matrix is wrong, Leaky. Right column is off by negative.

@TedShifrin thanks

Derived subgroups are characteristic so that just gives you a normal series whose quotient factors are abelian
Hey @Ted!

hi Demonark

@MatheinBoulomenos @TedShifrin now, since the determinant is $1$, the real eigenvalues have product $1$, so the number of $-1$ is even, so they can still be paired up and there $\theta = \pi$. Now, the path is given by changing each $R_{\theta}$ block to $R_0$ simultaneously.
@TedShifrin did you read my first message?

Or you could learn the general version of the spectral theorem.

11:26 PM
So for bounded normal operators?

Dami will probably spell out the functional analysis theorem

we're doing finite dimensional, not functional analysis

I got ... sniped

Rehi Ted

rehi Alessandro

11:27 PM
I'm not wearing my functional analysis cap just yet

@Balarka I just watched Svankmajer's Alice, you might like it, it's the Czech surrealist stop motion version of Alice in wonderland

@Daminark which other definition of solvable are you trying to prove?

@Alessandro Ohhh nice. Let me check it out and put it on my to-watch list
I actually really like the stop-motion surrealist genre.
Have you seen Frankenweenie?

Nope, should I add it to my list?

Definitely do
It's awesome and it's by Tim Burton

11:30 PM
Two of them: that there is a subnormal series with abelian quotients, and that for finite groups, you have a composition series with Z_p quotients

Ok, Svankmajer's Alice is on my list. Thanks

@Daminark fundamental theroem of finitely-generated abelian groups?

Haven't done that theorem

@LeakyNun that doesn't help much

@BalarkaSen added

11:31 PM
you guys didnt do that?? @Daminark man that's weird

@Daminark you probably just don't know the name

Alice is actually a mix of stop motion and live action, but there is a single actress in the whole movie

Demonark: That's fundamental. Well, better is to learn the theorem characterizing finitely generated modules over a PID (or ED).

@EricSilva our class was REALLY slow

I see

11:32 PM
that's weird man

But our prof decided to wait for modules

I continue to be angered by Demonark's UC curriculum.

in my HA last year we did it like halfway through

We proved that in freshman LA
for PIDs

OK, waiting for modules is OK.

11:33 PM
Lol, for a second I was surprised, like wait you're mad that we're slow? Interesting change of events

no, choices of coverage
lack of concreteness and computational skills

@Daminark the basic idea for one direction is that if $N \subset G$ is normal such that $G/N$ is abelian, then $[G,G] \subset N$
@TedShifrin that's a good thing

But yeah I mean honestly we could've condensed the class into 6-7 weeks

tbh I go to UC and I'm with Ted, our curriculum is weird

you're already on my shit list, Mathei

11:34 PM
@MatheinBoulomenos How is that a good thing?

and even lots of profs here agree that the curriculum here is incoherent

And our prof did a good bit of computation, a lot of what we did was talking about cycles and the like

well, why the hell doesn't the departmental curriculum committee do some work for a change and change things?

@Narcissus I'm just trying to annoy Ted

And we also did computation with matrices a good bit

11:35 PM
not true, Mathei

@MatheinBoulomenos That makes more sense :P.

you actually believe what you said
don't be gullible, Narcissus.

@Ted they finally added linear algebra after like a billion years of not having it

what structures do the symmetric matrices have?

what does that mean?

11:36 PM
the orthogonal matrices form a group

look at the real spectral theorem

even a Lie group

symmetric matrices form a vector space, eh?

I swear I left for 4 hours, and Leaky is still smashing questions out at light speed.
(Left = fell asleep on my desk)

@TedShifrin they are just $\Sigma^2 \Bbb R^{n*}$ right?

11:37 PM
Right.

what is $\Sigma^{2}$
is it $\text{Sym}^{2}$

@EricSilva yes

never saw someone use a $\Sigma$ for it

@TedShifrin I just think that certain matrix computations (with actual numbers) are among the most dull things I did in college. I don't like following an algorithm. I had fun computing Galois groups etc.

$\bigotimes^2 \Bbb R^2 \cong \Lambda^2 \Bbb R^2 \oplus \Sigma^2 \Bbb R^2$
@EricSilva apparently Ted did

11:39 PM
Let us drop this conversation now lest things get too heated
How about that?

i like sym cause i will never forget what it means

^^

Some things just can't be made concrete, like the homotopy category

Instead, let me post something funny
Looking through some old comments in chat, I found this from years ago: "lol, no, i mean batman vs sherlock. who would win?".
I say the person to win will be Neil deGrasse Tyson

Eric: It might remind you to symonize your floors or car.

11:40 PM

is that an actual word or is it a pun @Ted

Do you expect me to answer that?!

oh i see
simonize
i have never seen or heard this word in my short existence

Aight so we know $H_i' \le H_{I+1}$, so double prime is contained in the i+2, and so forth

@MatheinBoulomenos Surely that's an unrelated point?

11:43 PM
No one waxes (eloquent or otherwise) any more, Eric.

So we just bash out derived subgroups, eventually one is contained in the identity, and we're done

@Narcissusjewel He's making a joke on concretizability
"To be fair, you need a very high IQ to understand category theory jokes..."

That explains my distaste for category theory, Balarka.

Haha it's a Rick and Morty meme

i feel like category theory is also a meme

11:45 PM
It's like Rick and Morty so

Last one is the fact that for finite groups this is equivalent to a composition series with abelian quotients. One direction is obvious

I both like that show and hate that show

Maybe that means you're going to have a love-hate relationship with category theory
v e r y c o m p l i c a t e d

Actually the other direction should be too, if you have the right subnormal series you can just extend it to a composition series and the quotients have to remain abelian, right @Mathein?

@Daminark yeah

11:47 PM
Sick

@Balarka actually i think it's not a love hate sort of thing, I like the show but hate when people think it's smarter than it is

@EricSilva Oh yes that's pretty much the reasons why the memes were formed

yeah ik lol

Rick & Morty memes are basically ridiculing the Rick & Morty elitists
Memes are not always right-wing channels of communication, contrary to popular belief

which is all good laughs until you actually meet a rick and morty elitist and the memes become too real

11:51 PM
Thought those guys were a myth

nah dude, they actually exist
if youve ever been in a place that sells a lot of rick and morty merch u stumble on em and it gets real weird

@EricSilva Well, to be fair, you have to have a very high IQ to understand how to make a Rick and Morty meme. Without a good understanding of theoretical physics the harsh realization of the memes coming to real life and the realism of the elitism would be completely lost to the memer....
Please don't hate me for this
I had to

2 late bruv
engaging hatred

Yeah the Szechuan Sauce thing tho
d00d wtf

I think they're like all small but vocal interest groups. The incel. The Bernie-bro. The red-piller. The algebraist. They're out there, but hard to find.

11:54 PM
lmao

lololol

Kevin, it sounds like you're on a bird walk.

not an inaccurate statement tho

World is a strange place. Everyone seems to think the other person is a sheeple, and that person himself is thought as a sheeple by a person watching both, and ad infinitum to hell
It's like increasing layers of irony and memes

I think Balarka is in a pensive mood. To quote Radar from MASH: "No, I'm just thinking."

11:56 PM
No, I'm just thonking

Darn, asterisks don't work in here. Duh.

M * A * S * H

Blah ... not worth it :P

@TedShifrin I think you're going to have to tell me what a bird walk is. because the only bird walk I know is a pretty bad hiphop dance by Souljah Boi

decent show

11:58 PM
@KevinDriscoll lolol

DECENT!?

It means you go walking out in the country looking for/listening to different species of birds, Kevin. LOL.

that sounds nice

Ah ha, then definitely.

Ugh, I should have used a new thonk emoji

11:59 PM
At this very moment though, I'm listening to the call of some drunk tigers and bulldogs because 'the game' is on next door