6:00 PM
@TedShifrin oops, $\exp$ on the $n$-by-$n$ matrices

@JoeShmo: You mentioned eigenvalues. What does $A^4=A^2$ tell you about an eigenvalue $\lambda$?

@BalarkaSen Nothing Personal, Kid.

So reread my comment, Leaky.

how do I know that it is defined?

well, the only thing I managed to extract is that p_(A^2)(lambda) = p_(A^4)(lambda)
p being charpoly

6:01 PM
Write down a proof that $e^A$ makes sense for all square matrices.

3 mins ago, by Leaky Nun
@TedShifrin do I use Jordan Normal Form to prove that $\exp$ is total?

@JoeShmo: Suppose $Av=\lambda v$ for some nonzero $v$. Proceed.

brb

@Narcissusjewel He said, in Liam Neeson voice

@Leaky: No normal forms needed whatsoever.

@TedShifrin hmm... I don't know how to prove it without the normal form

What's the definition?

$\displaystyle \sum_{n=0}^\infty \frac1{n!}A^n$

OK, prove that converges (absolutely) for all $A$.

@Narcissusjewel lmao this is the most beautiful video I have seen in a while
NANI????

6:04 PM
@TedShifrin as I said, I see a proof with Jordan Normal Form, but not without

@Leaky: A better question, however, is to figure out the image of the exponential. Field may matter.
@Leaky: Learn about $\|A\|$.

@Narcissusjewel if you have time please check out "MAC OR PC" by Filthy Frank
I think it might be the greatest edutational video in youtube

@TedShifrin nice!

(Leaky, I covered this stuff in my videos.) :P

@TedShifrin if lambda eigenvalue of A then lambda^4 = lambda^2

6:07 PM
@JoeShmo: Yes, and so?

... now how to know that the norm is well-defined?
without the normal form...?

sorry, not iff, i only have one direction

hi @Mr.Xcoder

@Leaky: What's the definition?

hi @LeakyNun
How is it going

6:08 PM
@TedShifrin $\sup \{\|Ax\| : \|x\| = 1\}$

Right. Linear maps are continuous. What's your point?
(We're in finite dimensions.)

hmm, ok

Q: If we have a weird (irregular)-shaped function, can a definite integral only be approximated or can it actually be precisely determined (by a human)
(newbie question)

@Mr.Xcoder if it can be done by a computer, then it can be done by a human

true...

6:12 PM
@Mr.Xcoder, ask a more specific question

a human is merely a less efficient computer

which is subjected to mistake

@BalarkaSen Now I want a mac.

@TedShifrin i fail to draw conclusions

@Narcissusjewel lmao

6:13 PM
$\lambda^4-\lambda^2 = \lambda^2(\lambda^2-1)$?

@TedShifrin so?
oh

So what are the possible eigenvalues?

well no i knew that
0 +-1

OK ... Now, what are the possible sizes of Jordan blocks? This is where your minimal polynomial comment becomes relevant.

what i dont see is why the minimal poly must divide this expression
let me dig, one sec

6:15 PM
OK.
(Definition of minimal polynomial is ... ?)

poly of minimal degree dividing charpoly

No.

ah! what is it?
you're a great debugger @TedShifrin

LOL ... is that a backhanded compliment? :P

no i hope not, i'm a software engineer by day. i can surely appreciate a good debugger...

6:18 PM
Ah.
Wait. So it's a forehanded insult?
Anyhow: How about this. The minimal polynomial is the polynomial $m(x)$ with the property that it divides every polynomial $f(x)$ satisfying $f(A)=0$ ?

gotcha
ok
i see now
the definition i gave is kinda of silly anyway

So what do you know about the possible sizes of Jordan blocks for the various eigenvalues?

by that definition m(x) = 1 for any matrix A

Yeah, you had garbage.

@TedShifrin so $\exp$ isn't injective?

6:21 PM
@Leaky: Sometimes it is; sometimes it isn't.

on the whole it isn't
on GL it is
?

I warned you that field might matter. So stop and work.

if we have "solvable" Lie algebras, can we connect it to something Galois?

I don't know why Galois. It ties into solvability of groups.

but solvability of Lie algebras is defined akin to solvability of groups

6:29 PM
right ... but why should this have anything to do with Galois theory?

because Galois theory is also related to solvability

Hey everyone!

Which (finite) groups can be realised as the Galois group of an extension of $\Bbb Q$?
Hi @Dami

hi Demonark

6:31 PM
Yo @Dami

@AlessandroCodenotti that's an open problem, right?
but I think most can

How's it going?

@LeakyNun I don't know, I wouldn't ask otherwise!

Hi Demonark

Hey @Daminark
@Alessandro This is called the inverse Galois problem
Currently no known groups provide an example

6:33 PM
Huh

@BalarkaSen

@BalarkaSen ah I see

@TedShifrin so the size of each jordan block is 1, and A is diagonalizable?

Not quite right, @JoeShmo.
What's the minimal polynomial?

6:39 PM
well actually 0 appears twice

Right.
So you can have Jordan blocks for 0 with size up to ... ?

Right.

very nice. thank you

Oh so I've been wondering recently, how exactly does analytic continuation work? I've heard about it all the time as how you extend domains of functions, but like, what's the mechanism by which you do so?

6:41 PM
So basically the functioning principle is identity theorem.
Take a holomorphic function $f$ on a disk $D$ around $0$
If it's not too wild, you can find a point $p$ on the boundary of the disk such that $f$ can be defined on the disk $D_1$ around $p$
i.e., the Taylor series of $f$ is going to define a $g$ on $D_1$ agree withing $f$ on $D_1 \cap D$
So you can patch it up and extend it to $D \cup D_1$.
You do the same thing with $D_1$, etc to extend $f$ disk-by-disk along a path on the complex plane
What goes wrong for some functions is that say you do this along a loop $\gamma$ from the origin back to itself
Then you'd return to $D$ with a different function $g$
For example, you can try to do this with $\sqrt{z}$ defined on a disk around $1$, and let the loop wind around the origin back to $1$. You'd return to $-\sqrt{z}$.
Yup, that works!

No evidence of the sniping

lmao
Well it wasn't a snipe
You brought up log, which is pretty good
The resolution for this is this

@TedShifrin actually do we know what the minimal polynomial looks like?

hi @TedShifrin

we know it divides lambda^4 - lambda^2

6:50 PM
hey you want to hear about the intuition of Hartog's theorem @Daminark ?

@Adeek do tell, though also tell the actual theorem because I don't recognize that offhand
@Ted you may like this, I intended to send it earlier but forgot
2

@Daminark :D

If you have a subset $U \subset V$ for which $f : U \rightarrow C^n$ is analytic
for specific U and V
this isn't Generalization of Riemannian contuition
So $U \subset V$ is open for open V. Then any function which is analytic on U can be extended to U.
the way it works geometrically is as follows so you consider the annulus V - U and slice it each chunk of it use Cauchy integral formula to extend each slice and we do it in uniform continous manner
and so that is how you do it
@Daminark category theory is important, but people sometime just keep generalizing everything for the sake of generalizing and don't even have a grasp about what they are actually studying in the end.

@Daminark You look at the full subdomain $\Omega$ of $\Bbb C$ where $f$ can be analytically continued by paths from the original point, say, $p$. Then, construct an object as follows: Take all the paths $\gamma$ starting at $p$, and patch together the disks of radius of convergence $D_1, \cdots, D_n$ covering $\gamma$ according to whether they intersect (this is just quotient space $\bigsqcup D_i/\sim$). Do this for all paths starting at $p$.
Then you'll end up constructing a Riemann surface $X$ where $f$ is actually a well-defined function $f : X \to \Bbb C$.
Sorry, I took ages to find a way to write that construction

abstraction is useful, but abstraction for the sake of abstraction is useless

6:57 PM
This is what it really looks like, in the process of constructing $X$

oh yeah that is good @BalarkaSen

sup ppl

Hey @Eric

Yo @Eric!

hi eric
Bist sie von deutschland @EricSilva ?

7:01 PM
@Adeek I don't have a particular view on abstraction for what it's worth. Some people care about it as a means to an end of understanding more concrete things, others seem to prefer to think about abstract things than concrete things. I'm just one of the "let people do what they want" types

i dont speak this

sorry your name sounds german
I can speak german little bit

Silva doesn't look German

Silva is a portuguese/spanish name

oh

7:04 PM
And @Balarka so basically the idea would be that if you try to build a function along a circle and fail, the Riemann surface construction would simply not bring that path back to its starting point, right?

@Daminark Yup!!

Just too verify with you guys
when people say functions on open set they mean normally it goes to $\mathbb{K}$ not $\mathbb{K}^n$ ?

Function means map to $k$, yeah

okay

3 hours ago, by Ted Shifrin
@Leaky: Because the Lie algebra is the tangent space at the identity.
so $\Lambda^2 (\Bbb R^{n*})$ is the tangent space of $SO(n)$ at the identity

7:16 PM
Literally everywhere I go I see Dolan Dark comments, regardless of how old or how classic it is. What the actual flick
@Daminark This guy has been haunting youtube like a meme ghost
And oh god this is beautiful: youtube.com/watch?v=6IAaBf1JOfQ

anyone here from imperial college?

@BalarkaSen this is gold and nothing less

@LeakyNun @quallenjäger

@LeakyNun You're at imperial right?

7:26 PM
yes

@Daminark This is my favorite one so far

@LeakyNun : @quallenjäger just asked :P

Is there a professor Richard Martin?

in Pure Math

7:28 PM
Richard Thomas

I found him also, but he should be in Mathematics department. But I cannot find a richard martin there.

Don't know any Richard Martin in maths

Ok thanks

no problem
27 mins ago, by Leaky Nun
so $\Lambda^2 (\Bbb R^{n*})$ is the tangent space of $SO(n)$ at the identity
So everything in $\Lambda^2(\Bbb R^{n*})$ is a derivative of $SO(n)$ at the identity
Skew-symmetric matrices, in particular

7:40 PM
xxxx-xx
any skew-symmetric matrix $S$ defines a quadratic form $S(x,x) = a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2$
now an element $f$ of $SO(n)$ defines a map from $\Bbb R^n \to \Bbb R^n$
I'm lost in the definitions
@AlessandroCodenotti would you help me?

@LeakyNun Whats your problem?

3 hours ago, by Leaky Nun
@TedShifrin interesting. How to see that the Lie group and the algebra have the same dimension?
3 hours ago, by Ted Shifrin
@Leaky: Because the Lie algebra is the tangent space at the identity.
@quallenjäger figuring this out

Tangent space has the same dimension as the manifold itself right?

yes

Then it should be clear that lie algebra and lie group have the same dimension
Lie algebra is just the tangent space at the identity of the Lie group

7:50 PM
@quallenjÃ¤ger yes, I'm figuring this out
for $SO(n)$ and the skew-symmetric matrices in particular

You mean why lie algebra is the tangent space at the identity?

yes

That's basically definition

@BalarkaSen yes, so I'm unfolding definitions now
in particular, the tangent space seems to be the space of all the derivatives

I see your problem
So suppose your lie algebra $so(n)$ is given by skew-symmetric matrices

7:53 PM
Technically speaking I guess the lie algebra is more than the tangent space of the group at 1
You have the Lie bracket structure

then the Lie group $SO(N)$ can be constructed as the one parameter family with exp. You can work this out by taking $so(N)$ as your generator.
if you derive the lie group at origin, it gives again your lie algebra

Uh, that's a rather sloppy way to say it
What you want to say is each vector on the Lie algebra gives a one-parameter subgroup of SO(n)
By matrix exponentiating

He will know what I mean

Hey again everyone

Sure, I don't care. I am just objecting to math said sloppily.
Hi @Perturbative

7:57 PM
@BalarkaSen Yes, I am too lazy to type:D

@LeakyNun I know nothing about that Lie stuff

@AlessandroCodenotti you must be... Lie-ing

Fair enough, that's me on a daily basis :P

Ffs @LeakyNun, that was as low a blow as you can make

hi @MatheinBoulomenos

7:59 PM
hi @LeakyNun

Hey @MatheinBoulomenos

@BalarkaSen How do you mean that?

Hi @Perturbative

That the Lie algebra is more than the tangent space?

@quallenjäger It's having the Lie bracket, like I said

8:00 PM
@MatheinBoulomenos let's consider the category of logical propositions with arrows given by implications

Hi @Mathei

Hi @Alessandro

@BalarkaSen Ah Ok, sorry I missed the second line. I got your point.

@LeakyNun okay

@MatheinBoulomenos it seems that there are only two objects up to isomorphism

8:01 PM
That's why it is a algebra and not only a vector space.

@Daminark this is awesome

Notaton wise, if $M$ is a smooth manifold, $p \in M$ and $v_p \in T_p(M)$, $v_p(f)$ means the tangent vector $v_p$ acting on $f$, and it equals the directional derivative in the direction $v$ of $f$ at $p$, which notation-ally is $D_vf(p)$ correct?

Yep

any true proposition is a terminal object, while any false proposition is an initial object

Ahh great thanks Balarka, when Ted said $Df(p)(v)$ I got confused by the notation because Lee was using $D_vf(p)$

8:04 PM
@LeakyNun yeah, this is true

so the product is the NAND gate

@Perturbative differential geometry is the study of objects invariant under change of notation
4

and the coproduct is dually the NOR gate

lmao @MatheinBoulomenos

@Perturbative In Ted's notation $Df(p)$ is the Jacobian matrix of $f$ at $p$, and $Df(p)v$ is then that matrix eating the vector $v \in T_p M$ which is the same as the directional derivative $D_v f(p)$.
Remember this from multivariable calculus?

8:05 PM
@BalarkaSen multivariable calculus?
:P

calculus?

@Mathein yeah I love it

@LeakyNun I think that product is simply AND and coproduct is OR, I don't see where the negation comes from

@MatheinBoulomenos oops, lol

@LeakyNun given that it has only two objects up to isomorphim, not that much

8:07 PM
@Daminark Did you see the sequence of puns I cc'd you earlier?

@BalarkaSen "Matrix eating the vector" is just standard matrix multiplication right? If it is then I think I should remember it

Nope, I'll look

@Perturbative Yes

Cool cool I know what you're talking bout then

@LeakyNun I think if you have "a logic" in the sense that you have some rules of inference for well-formed sentences (I don't want to make this super precise right now, I'm not an expert on this), then you can consider a category with morphisms = implications even without assigning actual truth-values to the sentences
that should be more interesting

8:09 PM
what do you mean?

@Balarka amazing

Did you see all three?

so $\phi \to \psi$ if $\vdash \phi \to \psi$?

@LeakyNun pretty much

you're literally asking for trouble :P

8:10 PM
search for cc and you should find them

I never thought that much about logic tbh

Yeah I did
I got notifications so I just checked them all

@LeakyNun this is what my category book writes on logic "We can associate to every mathematical theory in the sense of mathematical logic a category. Objects are theorems of the theory. A morphism $A\to B$ is a proof of $A \to B$. Composition is the composition of proofs"

and then?

there's nothing more on logic in the book

8:17 PM
ok

But there's a whole field of categorical logic, so there's definitely more to say about this

Homotopy type theory is like that but with higher categories

I also know that sometimes people talk about the "internal logic" of a category, but I don't know exactly what this means

2-morphisms between morphisms are like equivalence of two proofs
Also similar to homotopy between paths

weeee homotopies ftw

8:19 PM
@MatheinBoulomenos and perturbative differential geometry is the study of objects invariant under infinitesimal changes of notation

...

@Semiclassical I'm not sure if I would call the differences between the various notations in dg "infinitesimal" ..

So what domain is the study of abuse of notation?

physics.
5

8:22 PM
Oof

I do physics, so I can make that joke :P

Lolol, fair

I don't do physics but I can still make that joke
Physics is like bad math with badder notation bro

I refuse to believe a subject can have worse notation than bad math
Remember my shtick about "Let :thinking: be a compact Lie group"?

My thermodynamics lecturer used to think 'proof by contradiction' is called 'proof by falsification'
Needless to say the proof he wrote down wasn't even a proof at all

8:27 PM
Hmmm, random question for (algebraic) topology people
It's pretty known that a path-connected, locally path-connected, semilocally simply connected space has a universal cover

My favorite proof is where you say you'll prove something by contraction, give a direct proof, and be like "Our statement is true, contradicts our assumption that this was false"

But what do we know about a space if we just know that it has a universal cover?

@TastyRomeo What's your definition of a universal cover?

lol @Dami

I'd say a simply connected cover?

8:29 PM
Actually no I've seen even better than that in discrete
Someone did an induction proof where the inductive step was just the proof

Hello, who know this lemma on topology: let $X$ be a Hausdorff space, let $(x_n)$ be a sequence and $(x_{n_k})\subset (x_n)$ and $u\in X$, if there exists a sub sequence $(u_{n_{k_l}})\subset (u_{n_k}), u_{n_{k_l}}\to u$ then $u_n\to u$ ?

@BalarkaSen loops quotient homotopy?

@BalarkaSen it's the cover that covers all covers, duh

@MatheinBoulomenos a cover with UMP :P

@MatheinBoulomenos Yes, you and Tasty gave different definitions

8:31 PM
Covers the universe

That's why I was asking

Just read the damn name

In any case I'm pretty sure that covering maps being local homeomorphism is the key to Tasty's answer
If his total space is already simply connected
Uh
Uhhhhhh

@BalarkaSen big smoke

@MatheinBoulomenos So here's an example where your definition and Tasty's definition disagrees in these oddball contexts
Consider a Hawaiian earring and take the cone on it

8:36 PM
inb4 hawaiian earring
damnit
sniped

lmao
epic
So, take wedge of two of these things along the bad point

@TastyRomeo welcome to the headshot club

This is very much not simply connected, 'cuz consider the earring wedge earring that it contains in the base of the two wedged copies

is the earring compact?

There's a loop which alternates between the two copies of the Hawaiian earring
And you can never contract that in your life
@Daminark That was a 360 noscope man
Or, in legal terms described by Epic Games, "a full rotation followed by a shot in the head"

8:39 PM
@TastyRomeo if a space is covered by a simply connected cover, then it has to be semilocally simply connected: for any point in the base space, you have an open set that is evenly covered, so if you have loop in that open set, you can lift it to the universal cover and there it will be entirely contained in one leaf, so it is a loop in the universal cover which can be contraced to a point in the cover as it is simpylyconnected, you can just push down the homotopy.

@BalarkaSen I mean maybe we can find a cadaver?

this works for locally path connected spaces, path connected is not necessary

@TastyRomeo @MatheinBoulomenos But now, this is it's own universal cover
It's the only possible cover of itself, because of the two contractible wedged copies

yeah, makes sense, nice example

For a smooth manifold $M$ of dimension $n$ and a point $p \in M$ do y'all normally think of the tangent space $T_pM$ as the image of a parameterization $\varphi : U \subseteq \mathbb{R}^n \to M$ for some open set $U$ in $\mathbb{R}^n$ or as the set of all derivations of $C^{\infty}(M)$ at $p$?

8:43 PM
But yeah if you assume the total space is simply connected this should be the argument Mathei describes.

Or do you like to think of them as something freakier like derivations of the space of germs or velocity vectors of curves?

Yeah, I figured that would hold... I was hoping there would be something more, though :/

@Perturbative depends on what I'm trying to do with the tangent space

@Perturbative I think of it as a tangent space :P
Embed $M$ in large dimensions, take a tangent plane man

Depends if I'm feeling Riemannian or symplectic

8:45 PM
lol @BalarkaSen

(Not entirely true; I think of it as image of derivative of a chart map)

I think of it as the dual space of $I_p/I_p^2$, where $I_p$ is the ideal of $C^\infty(M)$ consisting of the functions vanishing at $p$

Ahh I see @BalarkaSen, same way as in the diff topology books