The h Bar

8:00 PM

What math/physics is not tedious?

Judging from how everyone just keeps writing down "the bosonic terms" and refers to that original paper for the derivation and the full action, I think at least for the 11d SUGRA there isn't

Ben Niehoff

but there are a handful of "parent" SUGRAs from which all others can be obtained by dimensional reduction

0celouvsky

$\mathscr{GL}(H)$

ACuriousMind

It's easy to figure out which fields are there, it's hard to actually get the action

bolbteppa

Dim Red is another equivalent way apparently

Ben Niehoff

8:00 PM

so once you get those once the tedious way, then you can get everything

ACuriousMind

@BenNiehoff Sure, but you still have to get the action of those parents

Ben Niehoff

yes

0celouvsky

what is dimensional reduction?

ACuriousMind

It's where you reduce dimensions

Ben Niehoff

haha

bolbteppa

8:01 PM

If you set up the first order formulation of a super particle
$$S = \int dt \frac{1}{2}(\dot{x}^{\mu} \dot{x}_{\mu} - i \chi^{\mu} \dot{\chi}_{\mu})$$
by incorporating the constraints
$$p^{\mu} p_{\mu} = 0, \chi^{\mu} p_{\mu} = 0$$
as a constrained Lagrangian
$$S = \int dt ( \dot{x}^{\mu} p_{\mu} - \frac{e}{2}p^{\mu}p_{\mu} - \frac{i}{2} \chi_{\mu} \dot{\chi}^{\mu} - \frac{ik}{2} \psi \chi^{\mu} p_{\mu}),$$
eliminate the momentum from this via it's equations of motion
$$\frac{\partial L}{\partial p_{\mu}} = 0,$$

But you can do it with this Noether thing too

Ben Niehoff

I can never quite remember the distinction between dimensional reduction and KK reduction

ACuriousMind

That's literally what you do - you go from a d-dimensional theory to a d-k dimensional thoery, I'm not being flippant :P

@BenNiehoff dim. red. is a bit more general, in KK reduction you do expansions in harmonic forms in analogy to taking the Fourier series on the circle in the original KK example. I think.

That is, I think KK reduction is a special case of a dimensional reduction, but there are methods in some cases that are not KK

Ben Niehoff

hmm, ok, but now I'm not seeing what could be more general than that

ACuriousMind

I'm having trouble to come up with an example myself :P

Ben Niehoff

say I'm reducing IIB on S^5...you're saying there's something more general than keeping the infinite tower of KK modes?

0celouvsky

8:04 PM

topos theory is more general

Ben Niehoff

or maybe the notion of KK modes only applies when those modes are small

especially for metric modes

Ben Niehoff

8:18 PM

@0celouvsky There are some people who jokingly refer to the opposite of dimensional reduction as "oxidation" :P

bolbteppa

Freedman gives an example of dimensional reduction, $0 = (\Box_{D+1} - m^2)\psi = (\Box_D + \partial_y^2 - m^2)\psi(x^{\mu},y) = (\Box_D + \partial_y^2 - m^2)\sum_{k=-\infty}^{\infty} e^{\frac{iky}{L}}\psi_k(x^{\mu}) = [\Box - \{(\frac{k}{L})^2+m^2 \}]\psi_k$ this is saying?

KK infinite towers, massive scalars

Ben Niehoff

@bolbteppa So in that case, you're reducing on a circle, which is the easiest case

ACuriousMind

@BenNiehoff lol, I like that

0celouvsky

Ugh I have a proof but it requires the closed graph theorem

I don't know if I'm allowed to use that

Ben Niehoff

you expand your fields in a Fourier series on that circle, and you'll find you get an infinite tower of massive fields in the lower-dimensional theory

@ACuriousMind Read Clifford Johnson's book, he's one of those people ;)

0celouvsky

8:22 PM

@ACuriousMind my google fu is growing more powerful

Ben Niehoff

soon you'll be throwing hadoukens!

0celouvsky

I...don't think he actually proves the first claim.

Ben Niehoff

so I'm beginning to look at applying to real jobs, which means I need to write something about my "pedagogical vision"

0celouvsky

Banach spaces for all

Ben Niehoff

that will go over great teaching undergrad E&M for pre-med students

"OK, so, first you start with a Banach space..."

0celouvsky

8:28 PM

I mean, you do need some function spaces to solve Maxwell's equations

Ben Niehoff

you don't need them, if you're a physicist...

you just need Dirac delta!

0celouvsky

So apparently there is a way to make sense of $\delta(f(x))$ in distribution theory

Ben Niehoff

but that's ok, undergrads start with the integral form of Maxwell's equations

0celouvsky

I don't know what it is or if the physicist formula works for it

Ben Niehoff

which is the only true form, anyway!

bolbteppa

8:29 PM

Just set up the EM action and let them do it themselves

ACuriousMind

@BenNiehoff $\mathrm{d}{\star}F = 0$ or bust :P

bolbteppa

Talk about using digital technology to enhance learning experience yada yada

Ben Niehoff

ah yes, time to leverage some technological solutions in order to enhance our synergy while surfing the information superhighway

bolbteppa

So the Noether thing is basically motivated by taking an action invariant under some symmetry, modifying your symmetry then asking what modified action is invariant under this symmetry I guess

Ben Niehoff

kind of, but for SUGRA it's not so ad-hoc

like I think at each step there is a definite thing you need to add that you can work out

so it's not just arbitrarily changing things

0celouvsky

8:37 PM

Ayy, this proof is impossible

bolbteppa

Ok will just go for it

Another buzzword

Non linear realisation

Ever come across this stuff?

Ben Niehoff

yes, but I can't remember what it means

for SUSY, anyway

bolbteppa

Nonlinear realization

In mathematical physics, nonlinear realization of a Lie group G possessing a Cartan subgroup H is a particular induced representation of G. In fact, it is a representation of a Lie algebra g {\displaystyle {\mathfrak {g}}} of G in a neighborhood of its origin. A nonlinear realization, when restricted to the subgroup H reduces to a linear representation. A nonlinear realization technique is part and parcel of many field theories with spontaneous symmetry breaking, e.g., chiral models, chiral symmetry breaking...

ACuriousMind

It's what happens in SSB

The original "broken" symmetry group is realized non-linearly on the fields

bolbteppa

Ah wow

So basically, if you make supersymmetry local, and you use dim red or this Noether thing to find the action, you get superg with gravitons as goldstone bosons or something?

0celouvsky

8:46 PM

@ACuriousMind I need some words of wisdom for this proof

I can't find $\epsilon$ small enough :(

$0^+$ works

too bad it's $0$

Ben Niehoff

Ain't no mountain high enough, ain't no $\epsilon$ low enough...

6

0celouvsky

9:07 PM

@ACuriousMind Does $A\subset B$ imply $A\cap B^\bot=\{0\}$?

I'm in linear algebra hell

It should because $B\cap B^\bot=\{0\}$.

ACuriousMind

@0celouvsky "Let it be."

@0celouvsky yup

0celouvsky

@ACuriousMind You want me to give up?

ACuriousMind

Nah, I'm just quoting the Beatles becaues I could think of actual words of wisdom :p

0celouvsky

Because I think I have the proof. Let $N=\ker F$, where $F$ is Fredholm. Then $N$ is finite dimensional, and $H\ominus N$ is a Hilbert space. Also $F|_N$ is Fredholm and quasi-invertible since its range is closed. So I can perturb it by some $T$ with $||T||<\epsilon$ since "Fredholm" and "invertible" are open conditions. This implies $\ker(F+T)\subset \ker F$.

The only question is if I can extend $T$ off of $H\ominus N$ legitimately

just set it to zero on $N$, surely that works.

Oh, I just need that $(F+T)|_N$ is injective.

@ACuriousMind Is "injective" an open condition in Hilbert spaces? It totally should be.

ACuriousMind

I don't know what an "open condition" is

0celouvsky

9:15 PM

@ACuriousMind $\{T\in\mathscr B(H):T\text{ is injective}\}$ is open.

dmckee

@BernardoMeurer Not at all. I skimmed a technical description once when I was wondering if it was worth hacking a PSone, but that is as close as I've gotten.

ACuriousMind

@0celouvsky Ah, yes, my intuition agrees with yours that that is probably open

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