is this PSE worthy?

or would it get closed as "homework" or "pure math"

although I guess we can just use $e^{-i\epsilon J_3}$ for small $\epsilon$, use BCH and see if it gives anything useful, and then expand to lowest order

@0celo7 What about this formula doesn't work? You have $U J_3 U^{-1} = J_3 U U^{-1} + [U,J_3]U^{-1}$, no?

si

what then

that gives you $J_3$ and an additional $J_1$

hmm

@ACuriousMind so $J_3+[U,J_3]U^{-1}=-J_3$...how?

@0celo7 Evaluate that large ugly bracket! I suspect that the power series in $\pi$ that come out are like $2\sin(\pi)$ and $2\cos(\pi)$, so they'll kill the term in front of the $J_1$ and give a $-2J_3$

ohh, interesting idea...

I haven't done this though, and I'm not particularly inclined to try that :D

me neither

can we agree that's a good idea and move on?

I mean, the $\pi$ has to play a role, and the commutators in that BCH variant will produce $J_1$ and $J_3$ alternatingly, and $\sin$ and $\cos$ are "alternating parts" of the exponential

@0celo7 No objection from my side

@ACuriousMind gut

why does angular momentum have to be a half-integer in a rotationally invariant environment?

ACM: can you protect that explosion question?

Or is it too early/soon?

@KyleKanos Too soon, two days are minimum

Unless @Qmechanic protects it, of course

Weinberg is drunk

what is he talking about

@0celo7 Because the projective representations of the rotation group are labeled by half-integers.

@ACuriousMind Hmm, I thought 20k rep could do it asap.

@ACuriousMind no he means 1/2, 3/2,...

unless that's what you're saying

@KyleKanos No, I am powerless in this case

@0celo7 oO

Why not 1,2,3...?

something with Kramers degeneracy

trying to decipher this

@ACuriousMind : Done.

Fast

@Qmechanic Thanks.

@ACuriousMind Weinberg says on the previous to consider a system with an odd number of fermions + some bosons but the way he said that made it seem like a general system

@ACuriousMind the reason I know they are labeled that way is because of single-valuedness of the wavefunction...I assume there's a much better way of looking at it?

@0celo7 Representation theory of $\mathrm{SU}(2)$, of course, or rather, of the complexified Lie algbera spanned by $\{L_-,L_z,L_+\}$. How one gets that only half-integers are allowed follows e.g. from the application of the method in this answer.

I know some of those words from BBS/BLT

what is the mathematician's def. of the Verma module

@0celo7 The one on Wikipedia

...dear god

I'll stick with non group theory group theory

Fun fact: When Wigner, Weyl et al. introduced group theory into physics, other, less enlightened physicists called their work Gruppenpest - "group plague".

(Because they didn't understand it, but it nevertheless spread rapidly)

Didn't Einstein say that he didn't understand Weyl's shit

I would not be surprised

I thought it was gonna be a fadeev popov question :p

@Slereah: You're too late ;)

So late I am

a ghost

@ACuriousMind "which is non-unitary for $j\not\in\frac{1}{2}\mathbb{N}$" where is $j$ in there

lol, I mean $l$ :D

I guess I was thinking of something else there

It's a shame to bump it for one letter, but it had to be done

@ACuriousMind so why does $l\in \frac{1}{2}\mathbb{N}$ mean the rep is unitary? no ghosts?

@0celo7 The rep is unitary because the inner product never becomes negative - look at that formula for the level $n$ norm I write down and observe that half-integer $l$ makes all the lower states just zero-norm states we can divide out

This has nothing to do with ghosts

...or did you mean "negative norm state" with "ghost"?

@ACuriousMind yes, cf. BBS/BLT

the "no ghost" theorem means the string states all have nonnegative norm in 26/10 dimensions

What about 27 dimensions

26.01 dimensions

@0celo7 I disagree with the terminology of calling any negative norm state ghost. Ghosts have a precise role in the BRST procedure, but negative norm states may appear in wholly unrelated scenarios. One should not call non-BRST negative norm states "ghosts", imo.

@ACuriousMind you can disagree with the standard terminology

it's a free country

What about Slimer

Can I call him a ghost

Also what is Slimer even a ghost of

The ghost of a fat pig?

Are you continually high, @Slereah?

Do you never wonder about ghosts

is that even a valid question

WHAT IS THE DEAL WITH GHOSTS

If ghosts go through matter

Shouldn't ghosts follow geodesics

ghosts are neutrinos

@Slereah They prefer to deal black jack, I think

And end up at earth's core

Or at least oscillate around it

@0celo7 I will, and one day I will change it!

@ACuriousMind here's the plan: get the Nobel and the Fields, then write a book and dedicate it to those who found QTF unintelligible

then on one page have the first letter of every sentence spell scru u weinberg

Can you also make people stop using pi

And use tau instead

$\tau = 2 \pi$

I will enforce $\pi = 3$.

Also can you make people stop using the cross product and use the exterior product like professionals

@0celo7 Amazing idea...

@Slereah Yes!!

Can you forbid the teaching of relativistic mass

@Slereah do you want musical isomorphisms up your ass all the time

That's beyond even my power, I fear

@ACuriousMind stupid question: what does $\tilde V_0/\tilde V_{-1}$ mean and what states are in it

Relativistic mass is pretty annoying because then you have people who cannot grasp actual relativity because of it

They get stuck on relativistic mass

@0celo7 That's the quotient space of $\tilde{V}_0$ by $\tilde{V}_{-1}$, and it is one-dimensional - it only contains one state with eigenvalue 0.

(that's why I say it's the trivial spin-0 rep)

that's what I would have said but damn if I can show that's the state it contains

@ACuriousMind I understand this but I just don't know how to verify it

@0celo7 write explicitly what vectors $\tilde{V}_0$ and $\tilde{V}_{-1}\subset\tilde{V}_{0}$ are the span of. There's only one vector that spans $\tilde{V}_0$ but not $\tilde{V}_{-1}$.

see I know all of this

I guess I wanted an intuitive explanation

@0celo7 I'm afraid I got none - these Verma modules represent nothing intuitive, I think.

Ok, so $\tilde V_0$ has 0 and all of the lower states, whereas $\tilde V_{-1}$ has the -1 state and all the lower ones, so taking out that subspace leaves the scalar 0?

@0celo7 If with "0" you mean the state of weight 0 and not the zero of the vector space, then yes

@ACuriousMind $|0\rangle$

I think yes

@ACuriousMind Weinberg's QM book does it without group theory but it's satisfactory. If $j$ is the top angular momentum and $j'$ the lowest, $j-j'\in\mathbb{N}$ because you apply a the ladder operators an integer number of times. He then shows using the commutation relations that $j'(j-1)=j(j+1)$, from which $j'=-j$, so $2j\in\mathbb{N}$, from which the result follows.

How the heck did Shankar do it...

@0celo7 How do you know there is a "lowest"?

"Now, there must be a maximum and a minimum to the eigenvalues of $J_3$ that can be reached in this way, because the square of any eigenvalue of $J_3$ is necessarily less than the eigenvalue of $\mathbf{J}^2$."

deciphering

unless you know

@0celo7 Okay, accepted

@ACuriousMind looking at Shankar, that part of the argument is standard

Yeah, my ladder method is overkill for $\mathrm{SU}(2)$, but it's nice because it generalizes to e.g. the Virasoro algebra

yup

yeah shankar uses single-valuedness of the wavefunction

Weinberg's argument is better.

yours is best, but overkill

@ACuriousMind the official spin song soundcloud.com/just-red-up-2/04-yg-im-a-real-1

nothing says rotation group better

@0celo7 If you say so

@ACuriousMind and now I get what central charges have to do with project reps, in a non-group theory sense :)

@0celo7 Ah, that's just because projective representations are actually representations of central extensions, which by Whitehead's lemma are representations of the universal cover for semi-simple complex Lie algebras.

Was that sufficiently inscrutable? :D

Weinberg shows a projective rep has a central extension in the Lie algebra...good enough for me

Behind "Whitehead's lemma" there's a beautiful bit of cohomology about the first derivative of the $\mathrm{Ext}$-functor ;)

go read Straumann...

huh, if the central extension is a linear combination of structure constants we can always construct a non-projective rep by a redefinition of the generators

@ACuriousMind So if the central extension can be eliminated and the group is simply connected, we may obtain a non-projective rep?

@0celo7 ...I think yes, although I'm not sure what you mean by "eliminated".

@ACuriousMind redefine the generators = adjust the phase of the operators

@0celo7 I don't speak physics in this case

:(

what

$[t_a,t_b]=\mathrm{i}f^c{}_{ab}t_c+c_{ab}\mathbb{1}$, right?

"eliminated" means there are $\tilde t_a$ so $[\tilde t_a,\tilde t_b]=\mathrm{i}f^c{}_{ab}\tilde t_c$

is this so unrigorous you literally have no idea what I'm saying :/

I'm trying to translate this "elimination" into my framework of central extensions

Ah. Yes. got it.

You are correct.

OK, that theorem is proved in Appendix B. I tried reading that the first time around but gave up. I'll try it again this time.

@ACuriousMind out of curiosity, what did you have to figure out

@0celo7 I had to realize that vanishing of the second cohomology group does not mean that I need to pick the trivial representative of the group, and that "elimination" means showing that the given representative (which is the central charge) is indeed in the same class as the zero central charge.

and you don't have lecture notes for that

sometimes I think this explicit stuff is more confusing

@0celo7 The lecture notes don't even tell me that the element of $H^2$ that characterizes the central extension is called the central charge, but that's kinda obvious :P

I really aim to type these up

your face is kinda obvious

@ACuriousMind please do

I probably won't have time to take the class on groups.

(it requires a full year of graduate algebra and geometry)

@ACuriousMind you've never said: (+---) or (-+++)

You also might be frustrated with these notes because they don't show some things which would require far more work (such as the proof of Whitehead's second lemma)

is he a pimple?

@0celo7 Slight preference for (-+++)

@ACuriousMind so what are the prereqs

@0celo7 I think he's this one

@0celo7 Well...a bit of diffgeo, I guess (manifolds, vector fields)

Otherwise, not much

Being comfortable with abstract symbol manipulation, I guess :D

well where would one get the Whitehead lemma

@0celo7 In a course on Lie algebras/groups that has more time :P

@ACuriousMind lol

Why this lecturer only gives such small lectures puzzles me, I'd love to hear a full-blown lecture from him.

hmm, maybe I should only do 4 years at UT and do a traditional 2 year MS at Texas...

I could drive over to see Weinberg before he croaks

too many classes!

hell even within my major there's not enough time to take everything

not even close

@ACuriousMind is there a simple proof of the polar decomposition theorem?

@0celo7 For invertible matrices, I don't think the standard proof is particularly difficult.

@ACuriousMind wow I'm bad at google

I found some MO post with words I'd never heard before :(

then some PDFs with more words

What did you search for? I just typed "polar decomposition" into Wikipedia (because I knew it was there, I've read the article before)

same, but I ignored the wiki article because reasons

@ACuriousMind what exactly is the proof there...the connection to SVD?

@0celo7 Just define $P$ as it is there, and then write $A = A P^{-1} P$ and check that $A P^{-1}$ is unitary. Done.

(Also existence of square root, but that's essentially just diagonalizing)

@ACuriousMind makes sense, checking that unitary

yay

@ACuriousMind although how do I show $P$ is hermitian? I used that

I'm not good with matrices and square roots

@0celo7 Look at how the square root is defined and note that the square root of a self-adjoint operator is self-adjoint by definition

(because it is diagonal with all positive entries in some basis)

@ACuriousMind ::facepalm:: $A^\dagger A$ is hermitian...my definition is with a power series, I assume that's wrong

@0celo7 Yeah, just diagonalize $A^\dagger A$ and take the roots of the eigenvalues.

@ACuriousMind yes any hermitian matrix can be diagonalized

@ACuriousMind yup

thanks

lol, nice edits

Is intellectual thinking more accepted over intuitive thinking on physics stackexchange

@NeilGraham What do you mean?

> "$S^3/\mathbb{Z}_2$ is not simply connected" ::sweats:: I should totally know this one

I do know it!

@0celo7 We've been over that! You saw it! ;)

@ACuriousMind Ok, just to go over it again. We have two points $u, u'$ on $S^3$ and a curve $\gamma$ from $u$ to $u'$. Now on $S^3/Z_2$ $\gamma$ could also go to $-u'$ and define the "same" curve. But since they are different curves in the cover, $S^3/Z_2$ is not simply connected.

> A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

@DavidZ is that quoted text out of date?

I thought homework questions were anything where the asker is trying to get the answer of a specific problem.

(if that's wrong, blame video games)

@0celo7 Yes! (it follows if you set $u=u'$. strictly speaking)

@ACuriousMind See Weinberg doesn't talk about closed curves.

That's why I said it the way I did.

I'm investigating.

Ah, yes

@ACuriousMind Ok he defines simple connectedness as "all curves which connect two points are homotopic".

Yep, that's equivalent to "every closed curve is homotopic to a constant curve"

This seems similar to the "curve-independent integral = 0 integral over loop"

@ACuriousMind Overall, I see the website is based around asking questions with concrete answers to them, and I find that as a large barrier to the way in which I think, where I like to discuss ideas not fully known as well argue against specific concepts that are accepted by the scientific community.

@0celo7 I think the reasoning is very similar, yes

@ACuriousMind cool

@NeilGraham Well, we are not intended as a discussion forum. SE is meant to be a place for concrete questions with concrete answers. This doesn't mean that we think all discussion should be banned everywhere, but just that SE is not the place for it.

I just got an email saying my lit class got moved

now I don't have to run a mile to get to it!

@ACuriousMind spin takes on a range of values whereas helicity is +/- one value. is this because helicity is LI, but spin is not?