$$J(t)=(d\exp_p)_{tv}(tw)$$

That looks terribad. What does that mean?

well, I've got to get out do Carmo to check where $v$ and $w$ live

k bye

the problem with these things is that you have to consider the double tangent space

because you want to look at curves in $T_pM$ and their tangents

Right

$v\in T_pM$ and $w\in T_v(T_pM)$

So we know that $t\mapsto\exp_p tv$ is a geodesic

Ah, we're exponentiating from the double tangent space to the tangent bundle?

But what happens if we vary $v$ in $T_pM$? We end up with a "parametrized surface" $$f(t,s)=\exp_p tv(s)$$

where $s\mapsto v(s)$ is a smooth curve in $T_pM$

@BalarkaSen Don't think so. It's $d\exp_p$, which is not an exponential map

Oh, right. Differential of the exponential map.

Got it.

So if we vary $s$, we wiggle the geodesic $t\mapsto f(t,s)$, right?

Mhm

We then take the derivative of $s\mapsto f(t,s)$ ($t$ fixed), and then set $s=0$

that's the ur-jacobi field

So that measures the rate of change of the family of geodesics as $s$ varies

So if we write $J(t)=\partial_s f(t,s)$, then $J$ satisfies the Jacobi equation

That makes sense; so essentially moving a geodesic along a Jacobi field gives geodesics all-time

Ah, no, the converse is not true

Hmm

I don't have a counterexample, but I know for sure that just because you have a Jacobi field variation, you don't have "perturbed" geodesic anywhere

Gotcha. I suppose that makes sense.

Well, I gotta sleep. Talk to you in the morning

That is to say, all such variations are Jacobi fields, but not all Jacobi fields give variations

I think there's a counterexample on Wiki

bye

@Slereah you up?

← celebrating new lassaiz fair chat policy

@vzn It's laissez-faire and may carry unwanted connotations depending on who you talk to :P

@ACuriousMind Getting into Stone's theorem. I take it all the usual $U(t)$'s in QM are strongly continuous?

Sure, they're defined to be since we usually start from the Hamiltonian.

wanted to respond to DSs comments

@ACuriousMind How would one show that?

Sigh

Does that Hamiltonian need any conditions?

@0celo7 Show what?

@ACuriousMind that $e^{-i Ht/\hbar}$ is strongly cont.

You need to show the Hamiltonian is self-adjoint, then Stone's theorem yields a strongly continuous $U(t)$.

Hmm?

It looks to me like Stone starts with $U(t)$ then gives you the self-adjoint generator

@0celo7 It's much stronger, it gives a bijection between strongly continous one-parameter groups and self-adjoint operators.

← still working on building french vocabulary to more than )( words :|

@ACuriousMind Ah. He only calls one direction "Stone's theorem"

The direction we actually use in physics is a mere proposition

Ok, I should stop skipping ahead

@ACuriousMind It's interesting that strongly continuous is not defined with respect to the operator norm

Doesn't that make it...weaker than operator norm continuity?

@0celo7 It is! Deciding which norms are the "relevant" ones is a very subtle issue

And I wouldn't claim I fully understand it

@ACuriousMind Ok, I have three FA books that say Stone's theorem is specifically what I said earlier

i.e. only one direction of what you said

yuggib is probably the guy you want to talk about for such subtleties

@0celo7 Does it matter?

he doesn't come around anymore!

@ACuriousMind probably not

@0celo7 He's still available in the auto-completion for pings. I'm sure if you ask an interesting question he'll answer.

@ACuriousMind The group $t\mapsto U_a(t)$, $a\in\Bbb R^n$ corresponding to translation $x\mapsto x+ta$ is not continuous in operator norm!

@0celo7 Is that surprising, given that the generator is unbounded?

I'm not sure what the generator has to do with it to be honest

@ja72 Hello. Could you please have a look at my query which I posted as a comment on your answer here (physics.stackexchange.com/questions/310147/…).

@0celo7 It's the infinitesimal version of the group (yes, I'm being a physicist here :) ). And the infinitesimal version of continuity strikes me to be boundedness.

@ACuriousMind Do you want to know why it's discontinuous? (Maybe you will understand where I'm coming from)

@0celo7 Sure, elaborate

@ACuriousMind (From Hall, not me.) Take $\epsilon>0$ and $a\ne 0$. Let $\psi\in L^2(\Bbb R^n)$ have support in a small ball around the origin. If $a$ is large enough, $\epsilon$ small enough, or the support small enough, then it's possible that $U_a(\epsilon)\psi$ and $\psi$ have disjoint supports

So they're $L^2$-orthogonal

Then $$||U_a(\epsilon)\psi-U_a(0)\psi||=||U_a(\epsilon)\psi-\psi||$$

he has that equal to $\sqrt 2||\psi||$ because the supports are disjoint

@ACuriousMind does that make sense?

So $||U_a(\epsilon)-U_a(0)||\ge\sqrt 2$ for all $\epsilon>0$

@0celo7 Yes.

So I don't see where $-i\nabla$ being unbounded on $\mathscr D(\Bbb R^n)$ or whatever domain has to do with this

@bolbteppa ok

I submitted the thing

let me get out Waldo

@0celo7 If the operator is not unbounded, how are you gonna argue the supports are disjoint?

@ACuriousMind Hmm? It follows from geometry.

If the support is small, then moving it $a\epsilon$ away makes it disjoint from its former self

Anyone here know about equivalence relations? My algebra book is confusing me a bit in it's explanation

in particular the concept of a poset

@BernardoMeurer What do you need?

My god, posets are PhD level

what algebra book does posets?

Elements of Algebra and Algebraic Computing

I need to understand what a poset is

I think I got what an equivalence relation is

I did not understand a partition very well

what exactly about posets do you not understand

well, I did, it's pretty simple, but I'm not sure why they are meaningful

the definition

it's just that lots of things are posets :P

N with division is a poset

I'm not asking for a poset, I'm asking what makes a poset

what is a poset

a partial order

I don't get how a PO is different from an equivalence relation

A poset is a set with a reflexive, anti-symmetric, transitive relation defined on it. N with division is a poset because n | n, for all n in N, n |m & m | n -> n = m, and n |m, m |p ---> n | p.

@0celo7 Right.

Bow how is that not an equivalence relation?

Oh

Ah

An equivalence relation is symmetric, not anti-symmetric

@ACuriousMind ???

I see

Am I missing something obvious?

@bolbteppa I wasn't reading that right

it makes more sense now

@0celo7 I think the primed indices are going to explain the non-minus

@ACuriousMind I see that it's an exercise in Hall to show that $U(t)$ is continuous in the operator topology if and only if the generator is bounded

But it has three parts and uses the spectral theorem

so it's probably hard to show

@0celo7 Strongly continuous means that the function $\lvert\lvert U_a(\epsilon)(v)\rvert\rvert$ is continuous for all $v\in H$ and all $a$, right?

it means that $\lim_{s\to t}||U(s)v-U(t)v||=0$

is that the same as what you're writing

Ugh, I remember why I don't like functional analysis :)

absolute value of difference of norms is bounded above by norm of difference?

So I think what I wrote is stronger.

You're probably asking good questions but I don't have the patience or time right now to dig into them.

Patience? What are you up to

@bolbteppa I think I understand what's happening in the Minkowski case

there you just check that you reproduce the Minkowski metric

but in the curved case I don't see how you're guaranteed to get the metric

But I got a $-1$ in the Minkowski case above, definitely did something wrong

Minkowski metric has a -1...

@0celo7 I need to gain at least a passing understanding of things like the Ray-Singer torsion, affine Dynkin diagrams and thier relation to Lie groups and certain blowups in the sense of algebraic geometry. My capability of getting into things unrelated to that that I don't already know is somewhat limited right now.

@ACuriousMind I should learn algebraic geometry, but it's so far removed from what I want to do

Serious question, can we say "troll" now?

I'm not up to date on the status quo

@SirCumference I am NOT a troll.

@0celo7 wtf

Ugh, I hate reading

hey

@0celo7 Why is it badly behaved classically?

@Slereah finite time blowup

@Slereah Ok I shouldn't have read Hall. He has an example where the Heisenberg uncertainty principle is violated

turns out it's not really true for general operators without severe domain restrictions

Apparently it's still true for X and P, but still not on all of $L^2$

Does it matter if it's true for X and P?

Those are really the only two we care about

it's true, but not on all of $L^2$!

Is it true on all of the common domain of both X and P?

yes

but that's hard to prove

the usual physics proof only works on $D(PX)\cap D(XP)$

Well I mean

getting $D(X)\cap D(P)$ is harder

I'm not going to read the proof

how do you prove anything if it's not defined on it

@Slereah $\Delta A$ and $\Delta B$ are defined on $D(A)\cap D(B)$

but you can only get the usual uncertainty principle $$(\Delta A)^2(\Delta B)^2\ge\frac{1}{4}|\langle [A,B]\rangle|^2$$ on $D(AB)\cap D(BA)$

because that's where the commutator is defined

But you said that it worked okay on both of those domains

So where is the problem

@Slereah the one for $X$ and $P$ works fine in all cases

but the general one doesn't

who knows if the general one is used in QFT

Well sure, but I don't think physics really rely on any other ones?

I mean

Maybe $[L_a, L_b]$

And I guess $[f(x), f(p)]$ is important too

But unless it's a physical observable it's not that much of a problem

@Slereah Currently reading about Weyl quantization

it's not very interesting

Bochner integration though

Integration of operators

Part of my thesis was on weyl quantization

There is a neat link between that and path integral quantization

Hall references Glimm and Jeffrey

@Slereah by god, have you heard of the Groenewald no-go theorem?

I don't think so

which no go theorem is it

@Slereah

So... No 4th order polynomials of $x$ and $p$ will generally obey the commutation?

yes

third order

Quite bad indeed

f and g are third order

Although I don't think there's any real observable you can do with $> 2$

the Poisson bracket is 4th order

Although who knows

Maybe there are experiments with that

@0celo7 you have all you need to be the worst know it all student in a QM class, btw

I'm taking the QM redpill

if Alex Jones knew QM, this would be Info Wars

Have you yet brought up Norton's Dome in a classical mechanics class

I haven't taken a CM class

what kind of engineer are you

oh, I took engineering CM

I wonder what are all the worst questions you can ask to professors of various fields

I know that the twin paradox on the torus is a fun one

o.o

proof?

arxiv.org/abs/gr-qc/0101014

a good question to ask a SR professor

> Suppose $\alpha:\Bbb R\to\mathscr B(\mathscr H)$ is a differentiable map such that $$\frac{d\alpha}{dt}=\alpha(t)A$$ for some fixed $A\in\mathscr B(\mathscr H)$. Then $\alpha(t)=\alpha(0)e^{tA}$ for all $t$.

Picardy Lindyhop applies to operator valued ODE?

@Slereah halp

I dunno lol

Time to put the analysis cap on

Well, if $\beta(t)$ also satisfies the ODE, then $\beta'(t)-\alpha(t)A=0$.

Hey guys, before I've heard that AdS-CFT is in some ways a mean field theory of gravity

is that accurate?

or, well, I guess you'd have that $\frac{d}{dt}(\beta-\alpha)=0$

Jesus, how does one prove that $f'=0$ implies $f$ is constant

@0celo7 try integrating...

@Skyler there are functions that are not integrals of their derivatives

also I'm working on a Banach space and don't want to have to use the Bochner integral

@0celo7 this problem?

Mean value theorem works

If it's not constant, then you can find a secant with nonzero slope

This works in the Banach space case too

@0celo7 gosh for a second i got all excited in my head i thought someone was address my question, but then i realized you typed mean value, not mean field

I can see why ultrahyperbolic spacetimes are really badly behaved

They don't even have globally hyperbolic neighbourhoods

@Slereah So crazily enough, Picardy Lindyhop does work in that special case!

@Slereah whats the difference between hyperbolic and ultra

@Slereah is this it?

Ultrahyperbolic is when the metric is $(p,q)$ and $p > 1$

More than one time dimension

Actually, I think I just proved Picardy-Lindyhop (uniqueness) for all linear ODE on Banach spaces

@Slereah why tho?

@Skyler why not

@0celo7 :appluase:

@Slereah when you say more than 1 time dimension does it capture some dynamic i may not be thinking of

Some theories tried using more than one time dimensions

Related to weird interpretations of QM

I don't think it ever goes far because they are pretty badly behaved spacetimes

@Slereah To be honest I remember having this pontification during highschool with a few friends

without the formalism of course =P

Well the bad behaviour is pretty easy to see

Pick a plane spanned by two timelike vectors

Draw a circle in that plane

That circle is everywhere timelike

And you can contract it to arbitrary size

There are closed timelike curves through every point of spacetime and they don't even admit globally hyperbolic neighbourhoods

Ho boy: physics.stackexchange.com/questions/310468/…

it's pretty long