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8:00 PM
@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
the infinite tetration of x is equal to the infinite pentation of x right?
 
(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
 
8:02 PM
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
 
8:04 PM
it is also equal to $\frac{-W \log (x)}{\log x}$
 
@s.harp Correction: $P=O(n)$
 
where $W(x)$ is 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
 
8:12 PM
well we can convert infinite tetration to expoention form and it becomes x^x^x^x.....
when we convert infinite pentation to exponentiation it also is of the form of 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
 
8:27 PM
@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 $|\mathrm{around}\rangle+|\mathrm{absent}\rangle$
2
 
8:33 PM
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?
 
8:38 PM
@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
 
8:40 PM
$D_t (x(t)^2)$
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
 
8:43 PM
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
 
8:44 PM
$\langle \mathbf x \, | \, \mathbf x \rangle$
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
 
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
 
8:47 PM
$\langle \dot x , y \rangle + \langle x , \dot y \rangle$
 
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?
 
8:49 PM
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
 
8:50 PM
@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
 
8:51 PM
@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?
 
8:52 PM
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?
 
8:53 PM
@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
 
8:55 PM
@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
 
8:57 PM
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"
 
the hypothesis was that $\langle x, \dot x \rangle < 0$ for $ ||x|| \ge R > 0$
 
ooh
 
@s.harp Is the conformal automorphism group a Lie group?
 
9:02 PM
@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
 
9:07 PM
@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
 
9:23 PM
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
 
9:29 PM
@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
 
9:35 PM
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.
 
9:47 PM
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
 
9:50 PM
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
 
9:54 PM
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
 
9:56 PM
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?
 
9:58 PM
@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
 
10:00 PM
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"
 
10:02 PM
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
 
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
 
10:12 PM
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
 
10:28 PM
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
 
10:34 PM
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!
 
10:47 PM
Guess it's really lively in here today.
 
Somewhat
 
@Ted can I ask you a "blah blah" question? Why is it important for a geometric notion on manifold to have a finite dimensional automorphism group? (The phrase motivating this question being: "If it has an infinite dimensional automorphism group its topology and not geometry")
What does "geometry" even refer to I guess
 
@TedShifrin What do you think of this? :P
 
Oops clicked on the nerd chat
 
Eject thyself
 
10:55 PM
: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)
 

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