For a foolish definition of quaternion analytic --- The difference quotient definition included --- the only quaternion analytic functions are $f(x) = xz$

Iirc the correct definition which gives a novel theory is one based on the CR equations explicitly

the only ones with the naive definition are the "affine linear" ones, according to that paper

where affine linear is in quotations cause it isn't actually linear, but it looks like that

the equation you want is $\frac{\partial f}{\partial t}+i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}=0$, it seems

Ahsure

Ah sure you can also add a constant. :P

This looks like the paper I remember

Can someone explain the reason behind the statement:

If $\log_m (x/y) = \log_m (x)/ \log_m (y)$, then $x\in (0, 1) \cup (m^2, \infty)$

the paper looks cool

though it's not what Balarka needs, alas

What does he need

some quaternionic open mapping theorem

why isn't $df \in \Bbb H$ the right defn

i missed the convo

@MikeMiller $\Bbb{HP}^1$ is a hypercomplex manifold, and a polynomial should be a quaternionic analytic map $f : \Bbb{HP}^1 \to \Bbb{HP}^1$. If you have open mapping theorem you have proved quaternionic FTA

Can you make this "proof" into a proof

it's not the right defn cause the paper uses a different one, dunno

That should be the same as the "Cauchy Riemann" definition; it's just different than the difference quotient definition

yeah i thought it should be the right thing

i dont care about diff quotient

is it?

$df \in \Bbb H$ is an elliptic PDE it should be right

I thought the PDE looks different

how man it's just a 4x4 Cauchy Riemann thing

@Thorgott garbage reason

there must be better reasons

is this equivalent to the PDE you got?

I didnt check

looks like CR to me

yeah

it's like $\partial f/\partial \overline{q} = 0$

it should be the same as $df \in \Bbb H$

it looks like CR, but I dunno whether it looks like CR in the same way the other thing looks like CR

What does that mean

just check if they're the same man pretty sure they are

its a calculation

i'll do it when im bored

when you write out $df\in\mathbb{H}$, you get a PDE that's reminiscent of CR, but idk whether it's actually equivalent to the PDE above

the PDE above is $\partial f/\partial \overline{q} = 0$

CR in complex is the same as $\partial f/\partial \overline{z} = 0$ is the same as $df \in \Bbb C$

yes

so they should b same

Y'all boring

if you write $q=v+jw$ and $f=g+jh$, the above is equivalent to $\frac{\partial g}{\partial\overline{v}}=\frac{\partial h}{\partial\overline{w}}$ and $\frac{\partial g}{\partial w}=-\frac{\partial h}{\partial v}$

That says $df \in \Bbb H$ right

because $(z, w|-\bar{w}, \bar{z})$

Yeah, should be

nice, so the definition actually wasn't wrong

Cool so your reason was not only garbage it was not a reason at all!

owned

ye

let's see this paper

i want open mapping

but q-analytic is the same as r-analytic

so no hope for open mapping with these methods, no?

Number theoretic interjection

If $K^{\operatorname{ab}} \subsetneq K^{\text{ur}}E_\sigma$ then $\operatorname{Gal}(K^{\text{ur}}E_\sigma/K^{\text{ab}}) \neq 1$ and $\operatorname{Gal}(K^{\text{ur}}E_\sigma/E_\sigma)/\operatorname{Gal}(K^{\text{ur}}E_\sigma/K^{\text{ab}}) \cong \operatorname{Gal}(K^{\text{ab}}/E_\sigma)$, but this implies $\widehat{\Bbb Z}$ is isomorphic to a proper quotient of itself, which is false since $\widehat{\Bbb Z}$ is Hopfian, so $\operatorname{Gal}(K^{\text{ur}}E_\sigma/K^{\text{ab}}) = 1$

what is r-analytic man

real-analytic

R^4 -> R^4 projection to R^3 is R-analytic

clearly df is not in H

so i dont understand

wdym

yeah, it's q-analytic, but it's not q-differentiable, I think

so my remark misses the point

y u confus me bro

cause i am confused myself

q-differentiable they define as the difference quotient crap

nobody cares about that

everything that way is linear or smth

well, not linear, but looks linear

multiplication by a scalar, that is

plus constant

so what I think is the case is that q-analytic doesn't imply q-regular

i dont know these terms man

where I use regular as in the paper, i.e. df in H

hm is that what they call em

yeah

i see they call the question Cauchy-Riemann-Fueter equations

check prop.3

ah nice

it's equivalent to the weird differential forms defnm

yeah $f(q)dq$ should be closed as a quaternionic diff form

thats what i was trying to say earlier

arxiv.org/abs/0802.3861

the fudge is Cullen-regular

that's stronger?

bizarre definition

along every quaternionic line something happens

Ah I see what's happening

Cullen-regularity means a function on $\Bbb H$, when restricted to a complex subspace of $\Bbb H$, is complex differentiable

Slice by slice stuff

These have power series, Fueter-regular guys don't

Nuts man why do people develop these kind of theories

I'll ask an MSE question soon

ah, that sounds nice

surely that's satisfied for polynomials, so you get what you want?

yeah I think so

I'll let experts answer in MSE to be 100% sure because I don't understand the subtleties and so on

I'll let you know when I post a q

Thanks for the references/engagement/interest

what's $\mathbb{Z}/k$?

Yes, you have a sum of non-negative numbers. That sum is gonna be 0 iff all the summands are zero.

A different, but equivalent way of seeing this is to note that, since all the summands are non-negative, $0=\sum_{i=0}^na_i16^i\ge a_j16^j\ge0$, implying $a_j16^j=0$ and hence $a_j=0$ for all $j$

With just base-10 arithmetic and some modular arithmetic people will think you are a genius

indeed, that's a stronger statement

try reducing modulo powers of 16

@S.D. Try to think in base 10 why this is true, and see if that argument also works in base 16

that is indeed the crucial point

Bingo

I’m unable to see how $z=cr $ gives us the right circular cone (z and r are in cylindrical coordinates).

For example, if I place the base of my right circular cone in $xy$ plane and let its axis coincide with the $z-$ axis. Then, at $z=0$ we have $r=R$ (R is radius of the base). But according to that formula at $z=0$ $r$ is also zero.

And at $z=h$ (h is the height of the cone) we should have $r=0$ but we get $r= h/c$

OKAY! If that equation is meant to be an inverted cone, means the apex lies at $(0,0,0)$ and base is at $z=h$ then it seems reasonable. But again, at $z=h$ we have $r= h/c$ but we want $r=R$ at $z=h$.

How can I integrate $$\int_{r=0}^{R} \int_{z=0}^{h}\int_{\phi=0}^{2\pi} zr ~dr ~d\phi dz $$

Given that: $z= cr $

How would I manage $dz$ and the limits of $z$?

Is there a standard name for a function between metric spaces that sends bounded sets to bounded sets? I want to call it a bounded function, but that's already a function with bounded image

bounded²

idk man

non-escalating

no-blow-uppy

^

@Thorgott I'll go with "a map $f\colon X\to Y$ is called chill if..."

"Blowing up in finite time" is now known as "harshing the vibes"

loool

@AlessandroCodenotti Call them maps of bounded type

bounded-preserving

Turns out that what I wanted was actually a bit stronger, I want $d_X(x,y)\leq r\implies d_Y(f(x),f(y))\leq G(r)d_X(x,y)$ for some $G:[0,\infty)\to[0,\infty)$, which is called coarsely uniform

yeah i was wondering why you would need weaker than coarsely uniform

At first I though that bounded to bounded was enough, but then I realized that I want to be able to say that if I have many bounded sets with the same bound, then their images are also uniformly bounded

yeah, it does seem that just preserving boundedness is quite weak

Are you doing some sort of coarse geometry?

Alessandro is the physical manifestation of coarse geometry

@TobiasKildetoft Yes, I'm finishing to type up my thesis, which is pretty much on coarse geometry (asymptotic dimension in particular)

Neat

(not that I actually know anything about coarse geometry)

I know next to nothing about coarse geometry in the sense of coarse structures, controlled subsets and so on, I only work with honest metric spaces

Is black book of maths sufficient for jee advanced preparation ?

Never heard of it. But I recommend watching Black Books (not as preparation though, just because I like that series)

@TobiasKildetoft any recommendations , if you like to give for book. I need to strengthen my topic of function

I am not sure I understand what that topic entails

@AlessandroCodenotti I would say that this is an "everywhere bounded" function maybe

Not that you need the notion

@Balarka what's a current and how do I integrate over it

It's something you can integrate over, and you integrate by doing so

This is an entirely formal and correct answer

It's like a distribution but on the space of forms

Hey that's what I said

I know!

what you said is obviously better

@MikeMiller Isn't it true that wild mild hypothesis you can realize a current as a fuzzy manifold

I dunno there are things called varifolds

Did they change the stackexchange beep?

like a measure on (open set in R^n) x Grass(n, m) or whatever

@BalarkaSen

I have my thing muted

Unmute it

You mute

Just once

The beep changed

@BalarkaSen Fine, let's see

@BalarkaSen

No it's the same horrendous thunk

It's different for me

weird

in which sense can the graph of a function be a current

the graph of a function is a submanifold man

you can integrate on a submanifold

^

any submanifold naturally gives a current, define $[M](\omega) = \int_M \omega$ :)

no, it isn't a submanifold in my context, but I don't know why it isn't either

what kind of function do you have

a vector field and I know you want to tell me that the graph of that is a perfectly fine submanifold, but I have yet to figure out the reason why I'm supposed to disagree with that

what are you talking about? a vector field on what

i mean the function needs to be regular enough

a manifold

it takes values in the sphere bundle though

(manifold's Riemannian)

Your vector field is probably just something like bounded variation instead of smooth or whatever

graph of a good enough function can't be too wild in general

ah, I think I figured out what I have to figure out

apparently we want to count boundary components of the graph with multiplicity, whatever that means

Lipschitz is enough

BV is probably also good

yeah, I think I'll just give up on understanding this

I think functions with rectifiable graphs are the same as BV functions

why are you taking graphs of vector fields man

doing that proves Chern-Gauß-Bonnet

he actually has finitely many vector fields, which he calls nodes, and has homotopies between them, which he calls edges

lol

the category of vector fields

i wish

Can someone ping me? I'm curios about the discussion above on the sound being different

what's the right proof of CGB actually I never figured that out. If $\Omega$ is the curvature $2$-form, the coefficients of $\det(I - t\Omega)$ computes various char classes, right?

@AlessandroCodenotti

I don't even understand 10% of this proof, hell if I know what the right one is

Hm I think it's the usual sound

I was never interested enough in CGB to read a proof

Ping me again then

@MikeMiller

Oh

wow, you guys are making me curious

Looks like I just had my audio set low

I thought they made it quieter

lol

lmao

@MikeMiller

@MikeMiller more spam

Blocked

lmao

knew you'd crank the sound up

I don't really understand the Pfaffian well

man, we're proving Poincaré-Hopf and Chern-Gauß-Bonnet simultaneously and it's super weird

its the square root of determinant right

GB is corollary of PH so its not that surprising actually

I never learnt CGB

the integral is equal to the sum of indices of a vector field with isolated singularities, thus proving independence and then you just pick a nice vector field

I don't even know what the Pfaffian is tbh

Huh? What's the statement of CGB for you then

To me it's integral of Pfaffian is Euler characteristic

integral of Lipschitz-Killing-form is Euler characteristic

oly shit whats the Lipschitz Killing form dude

I would understand this shit better if I had a good answer to that question

just tell me the defn

It's probably the Pfaffian thing lol

how does he know the Pfaffian without knowing the curvature tensor

thats nuts

it's $\lambda_0=\frac{\operatorname{vol}(B^n)}{2^n\pi^n}\sum_{\pi\in S_n}\operatorname{sgn}(\pi)\Omega_{\pi_1}^{\pi_2}\wedge...\wedge\Omega_{\pi_{n-1}}^{\pi_n}$, where the $\Omega$ are curvature forms

Oh yeah thats the Pfaffian

Lmfao

Blocked

i c

I should specify 0-th Lipschitz-Killing-Form, because you can define a $\lambda_k$ for any $k$ of the same parity as $n$

@Thorgott yo dude look if $A$ is skew-symmetric matrix $\det(A)$ is actually square of some polynomial on the entries of $A$

$\det(A)$ is always a polynomial in the entries of $A$, why do you care about it being the square of one

I dunno cool fact. Here's a clever way to prove it; consider the $2$-form $\sum_{i < j} a_{ij} dx_i \wedge dx_j$. Call this $\omega$

This is a nondegenerate $2$-form on $\Bbb R^{2n}$, so $\omega^n \neq 0$.

$\omega^n$ is what?

ok, this works

yeah just $n$-th wedge power. I mean $A$ is $2n\times 2n$ because for odd dimensional guys the determinant vanishes

right

So $\omega^n = (stuff) dx_1 \wedge \cdots \wedge dx_{2n}$

I was just confused for the first version cause wedging a 2-form with itself n times won't yield something nonzero on R^n

ye

yeah i messed up

Hi all

Do you see why $(stuff)^2 = \det(A)$ (maybe I'm missing a factor of $n!$)

I was wondering; Given a monotonically increasing function $f(x)$ and the standard normal CDF $\Phi$, can it then be stated for sure that $\Phi(f(x)) \rightarrow 1$ as $x\rightarrow +\infty$?

or are there corner cases where this would not hold

should that be clear for abstract reasons or do I have to do a computation

I mean there's no reason to believe f(x) goes to infinity

what if $f(x)=\int^x_0 g(y) dy$ where $g(y)$ is a strictly positive function

@Thorgott you dont need to compute no

then it will always go to infinity as x goes to infinity right

Again why should that go to infinity

or am i missing something

Take g(y) = 1/(1+y^2), so its integral is f(x) = arctan(x)

hmm yeah I was scared of such g(y) formulations where they get infinitely close to zero

ok that tells me all I need to know then, thanks for the quick example

If g(y) is larger than some positive constant c then this is true

Since then f(x) >= cx

Which goes to infinity

that makes sense

@Thorgott if you care about invariants, this is critical. This is the Pfaffian and leads to the Euler class of oriented vector bundles.

okay I'd have to impose such a condition then

thanks alot @MikeMiller

Hi @MikeM, a @Balarka

The Pfaffian overlord has arrived

time to flee

Or flea

I care about invariants but not CGB (:

"care" is a strong word though

i feel like a flea, whence i flee

the fleeing flea is my rap name

awful rap

I meant just algebra, @MikeM. After all, this was Thor.

what did Thor's family say after they learn of his algebraic fetish

"Odiner"

Ah, in the sense of invariant theory?

??

"Oh dear"

"Odin"

Come on

These are awful "jokes" not even 1/10

Go to sleep

these are rather Loki jokes

agree?

hi Ted

@Balarka I'm afreyjd you're coming across as a little crazy

@Balarka I think there's something off about your $\omega$. If you wedge it $n$ times, you get terms involving $2n$ $1$-forms, but they only involve $dx_1$ up to $dx_n$, so that will be $0$.

$A$ is a $2n\times 2n$ matrix, the indices vary as $1 \leq i < j \leq 2n$

i dont understand $dx_1$ up to $dx_n$

$\omega$ is just the skew symmetric bilinear form given by $A$ if you want to think like that

$\omega(u, v) = u^T A v$

oh, ofc, I was still thinking $A$ is $n\times n$

mb

@MikeMiller fine I'll go read Norse theory

from Mjolnir's book

lool

top puns

thank u

the haters gonna hate

ok, I believe this works out computationall, but I don't see the conceptual reason

Hi @Alessandro

Why is the sum $\sum\limits_{A} (-1)^(\vert A \setminus B\vert)$ over all $B \subset A \subset C$ for some sets $A, B, C$ equal to zero when $B \neq C$?

I assume that's supposed to be $(-1)^{|A\setminus B|}$ as summand?

Yes

Hi! Do you think $\langle f_i(A)\rangle^{n-1}_{i=0}$ is the correct way to write the group that's generated by the images of $f_i$?

I mean the group that's generated with all of them together

@Taufi: Does your $\subset$ mean $\subsetneq$ or $\subseteq$?

The notation is only $\subset$.

I suggest you think about the $B=\emptyset$ case first to figure out what's going on

Well, if it means $\subsetneq$ the result is false, so it must not mean that.

Yes, you want to think about $|C\backslash B|$.

It immediately reduces to the case $B=\emptyset$.

Thank you, I will think about it.

I remember there was a nice combinatorial argument for this stuff, but I forgot it..

It's immediate from the binomial theorem, or yeah there's probably a combinatorial argument for that.

The presence of sets there is unnecessarily confusing, @Taufi. Think about what it means in terms of numbers. Suppose $|C\backslash B| = n$. What is this sum?

The presence of sets stems from its connection to the inclusion-exclusion principle. In that context I encountered it.

OK. That makes sense. My suggestion is to write the sum in terms of the $n$ I defined.

Okay, I will think about that. Thank you!

Sure.

You should be able to spot a Gaussian

You should also feel comfortable with a uniform distribution

I'd say the rest depends on what you're trying to do

I don't think anyone really cares whether you can call the Irwin-Hall distribution by its name or spot it from its PDF when you have the necessary conceptual background in order to understand it when you first encounter it

Maybe also know the assumptions that need to be satisfied in order to apply the common ones accurately to a real life problem.

oh, I've been ignoring discrete distributions

you should know Bernoulli/Binomial/Geometric/Hypergeometric distributions too, of course

i.e. something like "when can I approximate what I'm modeling with a poison distribution"

gotcha

Is there a secret, unifying mathematical description of the modeling applicability question, analogous to how the twelvefold way uses various types of functions between labeled and unlabeled domains and codomains to describe counting without all the physical scenarios?

@TedShifrin Hello professor .