@Mathphile are you sure $x^{/\infty/}$ even makes sense

@s.harp Sure, anything which is isotropic. $\Bbb{RP}^n$ or $\Bbb{CP}^n$ as well.

indeed you can get similar bounds on the dimension for any geometry by considering their model space, which is a coset space $G/P$

here's how i made sense out of it

(for Riemannian geometry $G=O(n)\ltimes \Bbb R^n$, $P=\Bbb R^n$) the dimension of the symmetry group is bounded by the dimension of $G$

I'm saying I don't think infinite tetration makes sense

unless it's $0$ or $+\infty$

or something

infinte tetration does make sense

is $|x|>1$?

infinite tetration is x^x^x.....

you have a sequence

$x_1 = x, x_n = x^{x_{n-1}}$

that sequence certainly doesn't converge if $|x| > 1$

I'm not sure it converges for $0 < |x| < 1$ either

it converges from e^(-1) to e^(1/e)

oh

if you say so

it is also equal to $\frac{-W \log (x)}{\log x}$

@s.harp Correction: $P=O(n)$

where $W(x)$ is lambert W function

@GFauxPas en.wikipedia.org/wiki/Lambert_W_function

I see how that's the solution if it converges, but I don't know how to prove that that covnerges

but let's go with it

okay so the infinite tetration of x is equal to the infinite pentation of x right?

I don't know, my head isn't in the right space for this right now :(

still trying to wrap my head around this problem and I'm considering giving up

or at least trying later

well we can convert infinite tetration to expoention form and it becomes x^x^x^x.....

I think those dots are obscuring a lot of important detail

what do you mean obscuring?

their meaning is unclear and may not mean the same thing in the first and second line

it's bad notation, it's better to think of it as a sequence

$x_{n+1} = x^{x_n}, x_0 = x$

no dots

okay

not the least because ^ isn't associative

@Semiclassical buddy are you around

@s.harp And also I mean it's kind of obvious from the Killing equations that the solution space is at most $n(n+1)/2$ dimensional. In an orthogonal frame, it's just $\partial_i X_j + \partial_j X_i = 0$ for all $i, j$. These are a total of $n(n+1)/2$ independent equations :P

It's been a long time since I visited the chat. Very disappointed to see it still doesn't support latex!

You can probably make this formal by turning this system of linear homogeneous PDEs into a single $n(n+1)/2$-degree ODE.

@Mathphile actually, you should probably start by worrying about how to define pentation for non-integer values in the first place

Anyway this would prove $\dim T_{\text{id}} \text{Isom}(M)$ is at most $n(n+1)/2$, which would prove, assuming Myers-Steenrod, the result. So I wasn't entirely off on my first attempt.

@GFauxPas $|\mathrm{around}\rangle+|\mathrm{absent}\rangle$

pffft physics notations

I need your help friend

I'm trying to visualize this problem and don't know what it looks like. If $\dot{\mathbf x} = \mathbf f(t,\mathbf x(t))$ what's the graphical interpretation of $\langle \mathbf x, \mathbf f \rangle < 0$?

or $\langle \mathbf x | \mathbf f \rangle < 0$

Equally well, that’s $\langle x,\dot x\rangle $

yes

If that were merely multiplication, ie $x\dot{x}$

but I don't know what that looks like. The angle between something and something is obtuse

does $x\dot{x}$ make you think of anything?

@Semiclassical don't you need 1/sqrt(2)

(Think elementary differential calculus)

@LeakyNun baaaah

chain rule

@Semiclassical :P

wait no

half of the product rule

Sure, or just the power rule in the case where x is a function of time

$D_t (x(t)^2)$

up to a scalar

2 is inside the parentheses

whatever, I'm just looking at the sign

oh thats what you mean

meant that

Need 1/2 noob

$\dot{\frac{1}{2} x^2}$

checkmate physicists

your notations are crap

haha

Anyways

In the present case, that’s not going to work quite as written

@BalarkaSen can you also tell that the maximal dimension of the group of conformal automorphisms is $(n+2)(n+1)/2$ ?

I don't even know what it means geometrically in the scalar-valued case

But what’s the natural analogue of x^2 for vector x?

$\langle x, x \rangle$

Right. So the natural thing to then consider is the time derivative of that

$\langle \mathbf x \, | \, \mathbf x \rangle$

well that's obviously $\dot{\langle \mathbf x \, | \, \mathbf x \rangle}$

:P

Okay. And?

What happens when you differentiate a scalar product?

okay and I just looked it up and it's $D_t\langle x, y \rangle = \dot x y + x \dot y$

wait

needs more dots

no

Just double checking, a circle drawn on the hyperbolic plane looks like a normal circle but the bulk of it's disk is redistributed, right?

Needs more brackets

$\langle \dot x , y \rangle + \langle x , \dot y \rangle$

Right. So when x=y?

so $D_t\langle x, x \rangle = 2 \langle \dot x , x \rangle$

@s.harp That's a good question. Let me think.

Right. So what does your inequality tell you?

$\Vert x \Vert$ is decreasing over time?

Right.

wow that's interesting

@Balarka if I may give you a cryptic hint that will not help you: The conformal group is $O(n+1,1)$

Yeah, it’s cute

The immediate problem is that the conformal automorphism group can be drastically smaller at times. Eg, for any surface of genus $g \geq 2$, it's a finite group.

cool :) so the next part is

@GFauxPas congrats, you found a Lyapunov function

oh right we talked about those

That still allows a lot of outcomes: x could fall in directly, it could spiral in, etc

@BalarkaSen conformation group?

@s.harp Oh, this is an exercise. I thought you were asking if my argument using the tangent space generalizes.

Damn autocorrect

But you’re always getting closer to the origin with time

@Semiclassical falling in directly is just spiraling in with 0 angular velocity

@Semiclassical also derivative negative doesn't mean tend to 0...

step 2, suppose $\langle x, \dot x \rangle < 0$ for $||x|| \ge R > 0$, show that $\dot x = f(t,x)$ has a solution for all $t > t_0$. So I made some progress with this

@LeakyNun ah, you’re right

I wanted to show all solutions were bounded by $||x|| \le R$ and so you can use bolzano weierstrass

@BalarkaSen well its not really an exercise, more a reference to how these questions are basically all the same question in a certain setting where one can achieve a general answer rather easily. here the setting of cartan geometry ncatlab.org/nlab/show/Cartan+geometry

is that the right approach?

but you might actually be right @Semiclassical because 1D differential equations are nice

idk

let me look at my notes about lyapunov stuff

@LeakyNun this isn’t 1d tho

even for 2D you can invoke Poincare Bendixson

Vector x

do we have any counterexample?

@s.harp I vaguely know the formalism of $G$-manifolds but I am not super impressed by that style of thinking :)

x(t)=1+1/t?

@GFauxPas isn't there a characterisation for the endpoint of the maximal solution

Velocity is negative for any positive time but position doesn’t converge to zero

@Semiclassical oh our DE isn't autonomous

do we have an autonomous example?

if there is a maximal solution, the endpoint is either infinity at a finite point or infinity at infinity

but I need to show there is a maximal solution

@Semiclassical I think it actually works for autonomous?

@BalarkaSen dunno, I never really got into the theory but I find its very easy to think about questions involving Killing fields and automorphism groups in this language

He didn’t assume autonomous either I think

and if the maximal solution is continuous then the endpoint has to be positive infinity

okay so I have in my notes

theorem: if $\mu$ is a solution to $x' = f(x)$ with $\langle x, f(x) \rangle < 0$ for all $x$ then $\mu(t) \to 0$

if a Lyapunov function exists, then the equilibrium point is stable. If a strict lyapunov fuction exists, then it's also asumptotically stable

proof: |x| is an L. function, sublevel sets are compact, blah blah

who says there's an equilibrium point though

@GFauxPas the compactness of sublevel sets

we must have f(0)=0

(so the condition needs to be changed to $\langle x,f(x) \rangle < 0$ for all $x \ne 0$

btw "blah blah" = "Lasalle's invariance principle"

ooh

@s.harp Is the conformal automorphism group a Lie group?

@BalarkaSen yes

can I say this

on each sublevel set, all flows are heading towards the origin

I mean then I'm sure you can do a completely analogous argument as mine above; the tangent space at the identity is the space of vector fields whose flows preserve $J$

wait no that doesnt work

@GFauxPas sublevel set is compact and invariant so omega must be nonempty

who's omega and what has he done with my son

@s.harp Oh, I might have misunderstood what conformal means. Do you mean that $f : M \to M$ is conformal if $f^* g = e^\lambda g$? In that sense?

I thought you were asking about the biholomorphic automorphism group, hence my comment about surfaces above.

@BalarkaSen yes, where $\lambda : M\to \Bbb R$ is a smooth function

also i forgot, $\dim(M)>2$

OK, then I agree that noting that the conformal group is $O(n+1, 1)$ is the most natural way.

Given a point $x$ and a frame $\xi$ in $M$, $f$ can just scale each vector in $\xi$

Anyway now that we're in this diversion, I am wondering if there's an inequality on the biholomorphic automorphism group of a complex $n$-manifold.

the local Killing algebras should be infinte dimensional, should they not?

No absolutely not holomorphicity is a far stronger restriction

Generically they should be 0 dimensional

At least that's what happens in surfaces eg

I mean the stalk of local killing fields

so nothing global

I don't know what abstract nonsense you are talking about but the space of vector fields which integrate to a 1-parameter family of biholomorphic automorphism should be obviously very small dimensional

@s.harp Also, a whole section is usually determined by it's value on the stalk in the holomorphic category, so I don't see why that makes a difference

a local killing field is a killing field that is only defined on an open subset

Ok. Then obviously it's finite dimensional in the holomorphic category :)

@Ryan Aha the expert arrives

oh no

what am I supposed to know

but isn't $e^{i t\ f(x)}$ a flow of something bilholomorphic (for small enough $t$)?

hm nevermind $f$ itself needs to be locally biholomorphic

@BalarkaSen you offered to answer my question and then never did!

Maybe I'm not really sure anymore, @s.harp. $\text{Aut}(\Bbb C^2)$ has elements of the form $(z, w) \mapsto (z, w + f(z))$ for every holomorphic $f$.

is $L^2([0,1], [0,1])$ sequentially compact?

@BalarkaSen that should make it infinite dimensional

Yeah, by flowing along those, like you said.

Strange.

it is like symplectomorphisms. I think that locally you have very many symplectomorphisms between two small open sets (ie this "group" is or the local killing fields are infinite dimensional), but globally you may have less

Interesting.

this why "symplectic geometry is not geometry", and "complex manifold" is not a geometric structure if you believe in the power of blah blah

Weird but makes sense

I don't know anything about $\text{Symp}(M, \omega)$. It's generally infinite dimensional, right?

can someone help me understand the hermitian operator

@BalarkaSen I think so, I dont know anything either

it might even always be so

@Ultradark who is the hermitian operator?

You are hermitian operator.

gasp

(I am not an operator, I am a human being!)

I am the hermitian operator

you are a woman, I am a machine

get out

oof

yup

yeah, "characteristic function" is a dumb name

how do you Fourier transform a distribution?

@Semiclassical well it characterizes the distribution completely

(fourier inversion formula)

@BalarkaSen true dat

it's just a rather bland name for it

But, the concept itself isn't really too weird. You're taking a Fourier transform of the probability distribution function

which is formally equivalent to $E[e^{-i X \zeta}]$

btw I just borrowed Körner

like 3 days ago

Yeah if you think of it like that it's quite natural. My impression is that probabilists, however, think of it as a fix for the moment generating function

it's really suitable for me

@BalarkaSen might depend on the probabilist you're speaking to, of course

you shouldnt speak to probablists

main place I've seen it is in the context of the central limit theorem

@Semiclassical The real probabilists, i.e., large deviation theorists

:3

Yeah it immediately proves CLT

Convergence in char. function implies convergence in distribution

yeah, and it also is handy in establishing inequalities in that vein

since if you can turn it into a question about integrals you can do saddle-point stuff

aha

that's common to the moment-generating function idea tho

@BalarkaSen but not every char. func. have a taylor expansion?

there's some weird cases, yeah

it's not actually weird?

@LeakyNun You use a second order mean value theorem.

if it had a taylor expansion then all moments would exist

which is good enough

what if it isn't differentiable

wait I need to recall what CLT says

It's finite variance dude

so finite mean and finite variance

It's not trivial to prove that finite variance implies differentiability, of course. Finite mean implies continuity is a corollary of Levy continuity

which implies char. func. differentiable for some reason?

waaaat

usual example of things going bad is the Cauchy distribution i.e. $f_X(x) = \frac{1}{\pi}\frac{1}{1+x^2}$

I have much to learn

@Semiclassical yeah if you take sample average you are still Cauchy distributed! :P

How does one determine the convergence of $\int_1^\infty \sin{(e^x)} dx$? I'm clueless to what I could compare $\sin{(e^x)}$ with. Any hint is appreciated!

i.e., Cauchy is "symmetric 1-stable"

notably, $E[e^{-i X \zeta}]=e^{-|\zeta|}$

If you have Pareto-type tails of your random variable, decaying like $x^{-\alpha}$ (so might not have finite variance), then there's still a version of central limit theorem which says the scaled limits converge to stable $\alpha$-symmetric distributions

@Semiclassical Which is interesting, as that's $2$ times the pdf of double exponential

yeah

o shit @RyanUnger is back in chatland

@BalarkaSen Well, Levy continuity is the opposite direction. You just need dominated convergence theorem

@Semiclassical Actually the easiest way to prove that characteristic function of Cauchy is double exponential (upto scale) is to prove characteristic function of double exponential is Cauchy (upto scale) and Fourier invert :3

otherwise you have to do complex analysis which ugh

what's wrong with complex analysis

it's so beautiful

tru i just dont know much

aaaah I realized I can't use lyapunov functions to answer the question

we're only given $\langle x, \dot x \rangle < 0$ for $||x|| \ge R > 0$ which is not a neighborhood of the origin

ding dang it

gonna do the other questions and then go back to it later @Semiclassical

What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even

@LeakyNun

hey, does anyone know an "exotic" topological space, that's not very well known?:p

The dogbone

empty set

Let $G \cong \varprojlim G_\alpha$, where the $G_\alpha$ are finite discrete groups, then $\pi_\alpha : G \to G_\alpha$ is continuous because you can just send $g_\alpha \in G_\alpha$ to like.. $\lbrace g_\alpha \rbrace \times \prod \varnothing$... ?

(under $\pi^{-1}_\alpha$)

Ah wait that's just in the product, I want something $\cap G$

the projection maps are restrictions of actual projection maps from the product which are continuous there by definition of product topology

Ah right, that's better

hahaha

howdy @ÍgjøgnumMeg, a @Balarka, @GFauxPas .... hey, it's @Sha !!

Hey @Ted !

@TedShifrin heya Ted:D

it's a wild @Ted

Yo @Ted!

Guess it's really lively in here today.

Somewhat

@TedShifrin What do you think of this? :P

Oops clicked on the nerd chat

Eject thyself

:0

@s.harp: Geometry refers to some structure that's more refined, like a (Riemannian or hermitian), or a projective structure or something else. But geometry makes things much more rigid. You can't move things around with infinitely many degrees of freedom; in fact, for spaces that aren't homogeneous to some extent, you might not move things much at all.

*removes Demonark *

@Balarka: Interesting. I remember thinking about (and, I think, writing a question for my algebra book) about how many ways you can write sums of 2 squares as sums of 2 squares.

There's a nice book that talks about representing primes as sums of 2 squares!

Lots of books talk about that.

Even not nice ones.

(or a quadratic form)