The h Bar

Emilio Pisanty

19:08

@JohnRennie it's hardly plowing

I found it riveting, to be honest

it really paints a good picture of the huge web of tensions that was so tightly spring-loaded in early 1914

and it makes a really compelling case of the importance of facts like that Russia did not have separate mobilization plans against Austria and Germany, to the point that it was logistically impossible for them to mobilize against Austria but not Germany

even if they tried

which they did

well, the Tsar tried, much to the displeasure of the military general staff

but more importantly it shows how you end up building this huge web of deadliness without realizing it, piece by piece, with individual steps that look reasonable to the people doing them

kinda like we're doing now

Ben Niehoff

\mathcal is great for saying "here's some kinda abstract thing that I don't quite understand myself"

AccidentalFourierTransform

$$\mathcal{LOVE}$$

Emilio Pisanty

@BenNiehoff in which role it is second only to \mathfrak, of course

Ben Niehoff

or it says "oops, ran out of letters"

\mathfrak strikes me as a bit more concrete than \mathcal

since we use it only for Lie algebras

and when we run out of letters

Emilio Pisanty

@BenNiehoff there's no such thing as "ran out of letters"

Ben Niehoff

19:15

haha

Emilio Pisanty

there is "did not plan correctly for how many symbols would come up in this thing", though

Ben Niehoff

there are only so many accented letters you can write before it starts to look ridiculous

Emilio Pisanty

@BenNiehoff that's why God gave us subscripts

Ben Niehoff

oh, no, when you use indices belonging to a bunch of different vector spaces, you run out of letters right quick

Emilio Pisanty

@BenNiehoff number your vector spaces and then you don't need different letters for vectors from different spaces

Ben Niehoff

19:18

no, I mean indices, for index notation

Emilio Pisanty

@BenNiehoff subscripts on your indices

Ben Niehoff

ew

Emilio Pisanty

$T_{i_1\cdots i_n}^{j_1\cdots j_n}$

Ben Niehoff

yeah, that works for that, fine

but now say you have indices belonging to 6 different vector spaces

Emilio Pisanty

$T_{i_1\cdots i_{n_2}}^{j_1\cdots j_{n_1}}$

@BenNiehoff example?

Ben Niehoff

19:20

well

say I want to do an uplift from 5d gauged SUGRA into 10d IIB

Emilio Pisanty

@BenNiehoff no idea what any of that means, but sure

Ben Niehoff

well, that's at least two distinct index sets you need already

Emilio Pisanty

@BenNiehoff no, but how does this look like in practice

symbols-wise

Ben Niehoff

if you take type IIB supergravity and reduce it on S^5, you get a 5d supergravity theory which is "gauged"

0celou7

Latin upper & lower, Greek upper & lower, then Hebrew

Ben Niehoff

19:22

hold on, that's not even the half of it

0celou7

cyrillic

Ben Niehoff

in the 5d gauged SUGRA, there are 42 scalars that sit in an SL(6)/SO(6) coset inside of E_6(6)

Emilio Pisanty

Until you show me examples, imma stick with indices can totes solve it

0celou7

@BenNiehoff what is $E_6$?

Ben Niehoff

so, you need E_6(6) indices

the Lie group E_6

E_6(6) is a different real form with 6 non-compact directions (I think)

0celou7

19:23

@BenNiehoff what's the definition?

Ben Niehoff

it's faster to link you to a paper, hold on

sorry, it's an old paper, and I think it's behind a paywall

inspirehep.net/record/219727

0celou7

@yuggib I'm having trouble with Plancherel's theorem. I understand the proof in e.g. Conway where he extends $\mathscr F:\mathscr S\to L^2$ using the bounded linear operator theorem. Yosida defines the extension via TF of distributions $\mathscr S'$. He needs the estimate $||\hat f||\le ||f|$, and says it follows from the estimate $\int f\hat \phi \le ||f||||\phi||$ for $\phi\in\mathscr S$ using density of $\mathscr S$ in $L^2$.

Ben Niehoff

but anyway, to write the E_6 Lie algebra, they use an SL(6) basis, so there are branching rules and they involve gamma matrices

so you have SL(6) indices and Spin(6) indices

I think

0celou7

@yuggib What I think actually works is to take $\phi\in\mathscr S$ with $||f-\phi||<\epsilon$ and $||\phi||=||f||$. This works because $\mathscr S\cap B_R$ is dense in $B_R$, where $B_R$ is the $R$-sphere in $L^2$. Actually that doesn't work. I wanted to do $||\hat f||\le ||\hat f-\hat\phi||+||\hat\phi||=||\hat f-\hat \phi||+||f||$, but I don't have an estimate for $||\hat f-\hat \phi||$.

Ben Niehoff

actually, USp(8) is in there somehow, now I'm actually forgetting!

the point is, there are a shit-ton of different portions of the alphabet that have to be used for different things

and "running out of alphabets" is not to uncommon of an occurence!

AccidentalFourierTransform

19:32

what is the beta function of the alphabet?

bolbteppa

Since the beta function is an inverse binomial coefficient in disguise, $C$?

Ben Niehoff

ah, now I remember!

there are 42 scalars (from the 5d perspective) that live inside $E_{6(6)}/USp(8)$

so, you need to construct that coset

and it's a pain

($USp(8)$ is the maximal compact subgroup of $E_{6(6)}$, of course!)

bolbteppa

How do people know this crazy lie stuff

Ben Niehoff

I have no idea

from a bunch of ancient papers in the 80s, I guess

but how did they know?

bolbteppa

This is my favorite example of lie theory

motls.blogspot.co.uk/2007/11/…

"Unfortunately, Garrett Lisi, painting himself as an E8 expert, already doesn't know that E8 has a SU(5) x SU(5) subgroup. It's kind of amazing to be ignorant about these elementary facts for a person who is pictured as an expert."

Ben Niehoff

19:40

lol, Lubos Motl...

bolbteppa

I don't blame him, because the subgroups are crazy, but it's still hilarious

Only realised why Engel's theorem even arises in Lie theory like 2 days ago :(

0celou7

@BenNiehoff one of the great minds in string theory, no?

Ben Niehoff

I think he did a handful of moderately important things before becoming a bit of a pariah

0celou7

A pariah?

Ben Niehoff

I think he had a permanent position at Harvard? And then either left or was made to leave

I don't actually know, because nobody talks about it

all I know is his personality which shines through his blog :)

0celou7

19:45

I like his personality

A rational thinker in science which is shockingly rare

Ben Niehoff

oh, he's certainly rational and brilliant

no question there

just...abrasive

0celou7

I'm being slightly facetious

Ben Niehoff

alas, I literally know no facts about his situation!

this summer I'm looking forward to meeting one of the other prominent physics bloggers, Sabine Hossenfelder

0celou7

@BenNiehoff same

I hate not knowing, it keeps me up at night

Ben Niehoff

and knowing is half the battle!

bolbteppa

19:49

Show her this and ask her to write a proper response to it motls.blogspot.ie/2004/10/…

Ben Niehoff

that's from 2004, I'm sure she already knows about it

I don't think she's very pro-LQG, aside from just generically thinking we should keep our options open

0celou7

lumo has been shitposting for over a decade :O

Ben Niehoff

I'm actually very little aware of what she does exactly

but I think it's a variety of quantum gravity things

as for me, I'm pretty anti-LQG

I got to meet Rovelli once, he was invited to give a math seminar on LQG

bolbteppa

As of now, that article has convinced me to ignore it as a subject, and any time I look into it I see people basically validating that article, so I stay away

Ben Niehoff

I went in with an open mind, but after seeing him present it, it seemed like it was full of holes :shrug:

0celou7

19:54

You've met every famous physicist of the past half century

That's a little crazy

Ben Niehoff

maybe it means the wrong ones are getting famous

none of these people has a Nobel

Witten has a Field's medal, I guess

0celou7

Because they haven't done anything real lol

I'm going to name my next cat Witten

Ben Niehoff

oh, and then there's that Milner prize, invented by some Russian oligarch more or less specifically to give out to string theorists because he likes string theory

0celou7

Either that or Sobolev

Emilio Pisanty

^ that's some joyful use of notation there

Ben Niehoff

19:56

ye gods

what does it mean?

Emilio Pisanty

that's the Jacobi identity

Ben Niehoff

are you sure? don't you need a third thing in there?

Emilio Pisanty

you define ${:}f{:}$ as the operator that takes $({:}f{:})(g) = [f,g]$

Ben Niehoff

oh

that is totally not how I've ever seen $:f:$ used before

0celou7

Lol

What the hell is that notation for

@EmilioPisanty that's sad

Emilio Pisanty

19:58

some monkey business with $[]$ and $\{\}$, but whatever

@0celou7 well, it lets you write things like $e^{\alpha \, {:}f{:}}$

Ben Niehoff

usually it means some sort of canonical ordering

say f is a function of x and p (position and momentum)

AccidentalFourierTransform

whats wrong with $e^{\alpha\{f,\cdot\}}$

Emilio Pisanty

toddsatogata.net/2011-USPAS/lieAlgebras.pdf there it is if you're that angry about it =P

Ben Niehoff

quantum mechanically, x and p get promoted to operators, and now you have ordering ambiguities if you expand f in some series in x and p

Emilio Pisanty

@AccidentalFourierTransform dunno, ask Todd

Ben Niehoff

19:59

:...: tells you that the operators inside go in a standard order

0celou7

@EmilioPisanty is that even convergent :P

Emilio Pisanty

but if it puts ${:}{=}{:}$ on the page then I'm on board

dammit, LaTeX, behave

@0celou7 yes

proof by because we're physicists

0celou7

You're going to say that every time I raise the mathematician flag, aren't you

Emilio Pisanty

@0celou7 probably, yeah

though in this specific example asking about the "convergence" of $e^{\alpha \, {:}f{:}}$ is kinda pointless because it's not really defined as a series (which you probably already know), so there

0celou7

@EmilioPisanty I don't really know how you want to define it.

Functional calculus?

Emilio Pisanty

20:04

@0celou7 no, its action on $g$ is the solution of the differential equation $g' = \{f,g\}$, roughly speaking

0celou7

Ohhh, I wouldn't have guessed that.

Seems like a physicist thing, is that differential equation even solvable?

Ben Niehoff

it's a first order ODE

0celou7

It's a PDE assuming that's a Poisson bracket

Emilio Pisanty

@0celou7 it's this kinda thing

Exponential map (Riemannian geometry)

In Riemannian geometry, an exponential map is a map from a subset of a tangent space TpM of a Riemannian manifold (or pseudo-Riemannian manifold) M to M itself. The (pseudo) Riemannian metric determines a canonical affine connection, and the exponential map of the (pseudo) Riemannian manifold is given by the exponential map of this connection. == Definition == Let M be a differentiable manifold and p a point of M. An affine connection on M allows one to define the notion of a geodesic through the point p. Let v ∈ TpM be a tangent vector to the manifold at p. Then there is a unique geodesi...

I think

0celou7

Doubtful.

Ben Niehoff

20:06

true...the prime is differentiation w.r.t. time, which is a parameter in Hamiltonian mechanics

Emilio Pisanty

pretty standard object

0celou7

What's the manifold?

What's the metric?

Ben Niehoff

the manifold is phase space

0celou7

What are the geodesics?

Ben Niehoff

I don't think we care whether phase space has a metric, although we do care that it has a Poisson structure

0celou7

20:07

@BenNiehoff he's trying to exponentiate an operator on functions

bolbteppa

In physics you treat them like the weak topology, whatever works you just use that

0celou7

It's not an exponential map in the riemannian sense

@EmilioPisanty it's a physicist expression, I'm happy with that

I don't expect it to be correct or well defined

Emilio Pisanty

@0celou7 You've got so much joyful geometric analytical mechanics in front of you

I take it you haven't gone through VI Arnold's mechanics

0celou7

@EmilioPisanty what?

I have, what about it?

bolbteppa

Arnold is just Landau where he tries to define the domains properly

and some frills

which are hard

Plug of Arnold's teacher:

Ben Niehoff

20:12

I can't remember whether the Arnold book I have his mechanics or just differential geometry

and I'm too lazy to get up and check

0celou7

Maybe there's a way of understanding it in terms of Hamiltonian vector fields. Not sure

Ben Niehoff

yes there is

Emilio Pisanty

@0celou7 of course there is

Ben Niehoff

f is a moment map

I think

0celou7

Well, what is it?

Ben Niehoff

20:13

in $g' = \{f,g\}$, f is the "Hamiltonian" that generates a flow

this is also in Goldstein, it's pretty standard

0celou7

If you think I don't know that, then I might as well stop asking

I'm asking why that PDE should have a solution

Ben Niehoff

it's not a PDE

well, kinda

0celou7

Is it not? You've got a time derivative on one side and other derivatives on the other

Maybe you can flow along f and get a solution. I don't remember how exactly to do that

Emilio Pisanty

@0celou7 not really

you're solving for the flow, not for $g$

i.e. you have some manifold $M$ and you're solving for $\Phi: M\times \mathbb R\to M$

its derivative at $t=0$ is $X= \frac{d\Phi}{dt}$

0celou7

I don't need a lecture on geometry

Emilio Pisanty

20:19

you want this to be equal to the hamiltonian under the symplectic identification

so you're solving for $d\Phi/dt$, i.e. not a PDE

0celou7

@EmilioPisanty ok it make it sound like you were trying to solve for g

Maybe you can do that with phi, I'd have to check Arnold

Emilio Pisanty

@0celou7 yeah, well, solving for the flow is messy

0celou7

I'm not going to argue there

Emilio Pisanty

and you're mostly interested on the effect on observables $g:M\to \mathbb R$

which handily enough is probably enough information to reconstruct $\Phi$ if you really want to

0celou7

Ok, I read Arnold, we can talk about something else

Emilio Pisanty

20:21

(I think)

0celou7

Like atoms

Or Fourier transforms

Emilio Pisanty

what about 'em?

Ben Niehoff

better not talk about Fourier transforms

0celou7

@EmilioPisanty I'm working through the details of the Plancherel theorem and I need an estimate

@BenNiehoff why not?

Emilio Pisanty

OK, question

consider the operator $$\mathbf S = \sum_{\mathbf k} \frac{\hbar \mathbf k}{k}(a_{\mathbf k,+}^\dagger a_{\mathbf k,+}^\phantom{\dagger} - a_{\mathbf k,-}^\dagger a_{\mathbf k,-}^\phantom{\dagger})$$

paper claims that

> for an arbitrary unit vector $\mathbf u$, the operator $R(\alpha) = \exp(-i\alpha \mathbf u \cdot \mathbf S/\hbar)$ rotates the polarization of each $\mathbf k$-component of the field around its wave-vector $\mathbf k$ over an angle $\alpha \cos(\mathbf u\cdot \mathbf k/k)$

surely that's a typo and the angle in question is $\alpha \times \mathbf u{\cdot} \mathbf k/k$, right?

0celou7

20:44

strange, why would $\cos$ of something be the angle

So I want to show that $||\hat f||\le ||f||$. I can take a sequence $(\phi_n)\subset\mathscr S$ with $\phi_n\to f$ in $L^2$. I claim that $\hat f=\lim \hat \phi_n$.

Emilio Pisanty

@0celou7 well, there's a cos in the inner product, right?

but that would make the angle $\alpha\cos(\cos(\theta))$ for $\theta$ the angle between $\mathbf u$ and $\mathbf k$

0celou7

For $\psi\in\mathscr D$, I have $(\hat f,\psi)=(f,\hat\psi)=(\lim \phi_n,\hat\psi)=\lim (\phi_n,\hat\psi)=\lim (\hat\phi_n,\psi)$

I guess I don't know that $\lim \hat\phi_n$ exists :(

Emilio Pisanty

it's a shame, too, it's a really important paper

tandfonline.com/doi/abs/10.1080/09500349414550911

I don't know what any of those symbols mean off the top of my head

$\mathscr S$ probably

$\mathscr D$ no idea

0celou7

@EmilioPisanty that's $a\cdot b=||a||||b||\cos\theta$

the angle isn't inside of the cosine

Emilio Pisanty

@0celou7 no, the angle itself is $\alpha\,(\mathbf u\cdot \hat{ \mathbf k})$

0celou7

20:51

@EmilioPisanty $\mathscr S$ are the rapidly decreasing functions, $\mathscr D$ are the test functions

Emilio Pisanty

because geometry

0celou7

@EmilioPisanty why that $\times$?

Emilio Pisanty

just to indicate a product

AccidentalFourierTransform

the angle should be $\boldsymbol u$-linear, so yeah, the $\cos$ makes no sense

0celou7

Maybe I should show that $\hat\phi_n$ is a Cauchy sequence

@EmilioPisanty Ahhhh

One uses the fact that norms in $L^2$ can be computed as operator norms on the dual, which is $L^2$

And that $\mathscr S$ is dense in $L^2$, so one can compute $L^2$ norms using $\mathscr S$ functions

$$||\hat f||=\sup_{||\phi||=1,\phi\in\mathscr S}\int \hat f\phi\, dx$$

0celou7

21:17

@EmilioPisanty you'd love Yosida's notation

$\mathfrak D\subset\mathfrak S\subset\mathfrak O\subset\mathfrak E$

$\subset \mathfrak C$

heather

21:33

@HDE226868 no need to apologize - I'm honestly just glad I didn't unload my sarcastic response =)

Slereah

21:50

that is too much Fraktur

peterh

I am thinking on tachyons. With positive real mass. They would live in a Universe with 3 timelike and 1 spacelike dimensions

They would interact with ordinary matter only on nondeterministic ways. It would solve the problem of the tachyonic antitelephone.

Physics Meta

22:23

0

Q: Under which conditions are static links displayed by their title?

Whenever I link a related/relevant question in a comment, I use the "share" static link of the question or answer I want to refer to. I've seen some comments where the link is replaced by the question title, but I'm not able to reproduce that behavior (without doing the [URL](link description) in...

bolbteppa

22:55

Another amazing thing about Kolmogorov - the open mapping theorem doesn't appear, he only states and proves the bounded inverse theorem as a natural part of inverting equations, proof becomes obvious

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