The h Bar

user218912

7:00 PM

is it $\frac{1}{2}(d-2)$

user218912

in mass units

user218912

where d is the dimension of the lagrangian density?

Balarka Sen

@Danu I think there's also a symplectic structure involved.

(I should mention that I don't know crap about Kahler manifolds)

user218912

hi @ACuriousMind I'm having an issue with dimensional analysis can you please help?

0celo7

@BalarkaSen Correct.

user218912

7:10 PM

@ACuriousMind my prof didn't teach dimensional analysis in qft during the lectures but one of the problems requires it.

0celo7

You're supposed to know that from high school.

Like completing the square.

user218912

I know high school dimensional analysis

user218912

but idk dimensional analysis in mass units

0celo7

This is exactly the same.

user218912

$\hbar = c = 1$

user218912

7:11 PM

it's the same?

0celo7

@IceLord Then go to regular units, convert to mass units at the end.

ACuriousMind

@IceLord I don't really know what there is to teach except "The units must match up", but what is your question?

Balarka Sen

I don't like to keep track of units.

It gets complicated often.

0celo7

@BalarkaSen Suppose I have some group acting smoothly on a manifold

user218912

i'm learning this from a book right now and it says because the lagrangian is $\mathcal{L} = -\frac{1}{2} \partial^\mu \phi \partial_\mu \phi$ the units of $\phi$ are $\frac{1}{2}(d-2)$

0celo7

7:13 PM

And I consider the subgroup that leaves a point fixed. Is this always a compact subgroup?

user218912

and $[\mathcal{L}] = d$

ACuriousMind

@IceLord "the units of $\phi$"? Are you talking about its mass dimension?

user218912

yes

user218912

sorry bad wording

Balarka Sen

@0celo7 Eh, under what topology of the group?

ACuriousMind

7:16 PM

@IceLord So? What's your problem?

user218912

@ACuriousMind how do you derive it?

user218912

I've never seen this before

0celo7

@BalarkaSen Compact open. I should mention the group is a subgroup of the diffeomorphism group.

ACuriousMind

@IceLord Well, do you know why the Lagrangian has mass dimension $d$?

user218912

yes

Balarka Sen

7:17 PM

Hm. I am inclined to believe this is true, but I am not certain yet. Let me think for a bit.

user218912

because the action is dimensionless? @ACuriousMind

ACuriousMind

Okay, do you know what a derivative does to the mass dimension of a field?

user218912

nope

ACuriousMind

Well, think about it! The definition of the derivative contains all information you need for this.

user218912

the definition with the limit?

Balarka Sen

7:18 PM

@0celo7 So the subgroup of diffeomorphisms of $\Bbb R^n$ fixing the origin is compact in $\text{Diff}(\Bbb R^n)$?

ACuriousMind

Yes, the standard definition as a limit of a difference quotient.

user218912

okay. let me think.

user218912

alright so the derivative has dimensions of mass

Balarka Sen

I mean, that can't be right because take a translation of $\Bbb R^n$ translating the origin to $(0, 0, \cdots, \epsilon)$ say. You should be able to uniformly approximate that by diffeomorphisms fixing the origin (take diffeomorphisms which fix balls of smaller and smaller radius around the origin, but translates by $\epsilon$ outside a slightly bigger ball).

I am not yet convinced this is uniform though. Let's see.

0celo7

@BalarkaSen Is the diffeomorphism group even a finite-dim Lie group?

Balarka Sen

7:23 PM

No, of course not.

0celo7

the parent group is Lie

Balarka Sen

Well, you never said that.

ACuriousMind

@IceLord Yes, correct

0celo7

My bad.

ACuriousMind

Now you should see how the $\frac{1}{2}(d-2)$ came about, and also that no further knowledge of "dimensional analysis" was needed.

Danu

7:26 PM

@BalarkaSen You're right lel

Balarka Sen

@Danu Ah, ok.

Danu

But it comes for free

Balarka Sen

Oops.

Danu

Because if you have the inner product and compatible almost copmlex structure

then $\omega=\langle I \cdot,\cdot\rangle$ is a 1,1 form which I think gives the symplectic form

Yeah, exactly

It comes for free so the book I have never mentions it

ACuriousMind

Arrrrrrgh, I finally figured out what the weird chirping noise is I've been hearing for the last few hours

Danu

7:27 PM

@ACuriousMind sooooo?

ACuriousMind

Damn smoke detector is running out of battery

And I have no clue how to make it shut up till I can get someone to replace the battery -.-

0celo7

Get someone?

@ACuriousMind is there PhD level dimensional analysis?

ACuriousMind

@0celo7 Well, I neither have a replacement battery nor do I know what kind of battery I need nor could I get any today since it's Sunday

Danu

@ACuriousMind Hehehe, I had that happen

I put so much tape on it that I didn't hear the noise anymore :)

ACuriousMind

lol

Balarka Sen

7:31 PM

@0celo7 Well, then, stabilizer of $p \in M$ under the action of $G$ is cut out of $G$ by the equation $gp = p$. So $G_p$ is preimage of $p$ by the map $G \to M$ given by $g \mapsto gp$. Preimage of compact set is compact.

user218912

@ACuriousMind when I solve for $[\phi]$ I get $-\frac{1}{2}$ ._.

ACuriousMind

@IceLord How did you solve for that?

Balarka Sen

@Danu I believe ya. Unfortunately I only pretend to know what a symplectic form is.

Danu

@DanielSank You... also have this experience?

0celo7

@ACuriousMind Sunday? Aren't Germans all atheists now?

Danu

7:31 PM

@BalarkaSen The definition is easy

@0celo7 I wish they weren't so religious here

Religious holidays are the fucking worst

ACuriousMind

@0celo7 That doesn't mean shops would be open

Danu

NO FOOD :(

0celo7

@BalarkaSen Preimage of compact set is compact?

Are group actions proper?

Danu

No

But usual ones are

Balarka Sen

Oops, $G$ ain't compact.

For noncompact Lie groups there should be counterexamples, then. Not that I can immediately come up with one.

0celo7

7:33 PM

I'll have to read the original paper then.

user218912

@ACuriousMind I did something wrong but

user218912

can I show you my reasoning?

ACuriousMind

I don't know, can you? :P

user218912

I'm probably approaching this wrong and being excessively retarded so please don't laugh at me.

Balarka Sen

@0celo7 This time I fell in my own trap because subconsciously my manifolds are compact :P

If $G$ is compact however, then that's trivial, so... oh well.

user218912

7:36 PM

@ACuriousMind $[S] = 0$ so $[d^dx] = -d$ therefore $[\mathcal{L}] = d$ because $S = \int \mathcal{L}\, d^dx$

ACuriousMind

Yeah, I thought we had already established that

user218912

yes and from that

user218912

I did

user218912

$\mathcal{L} = -\frac{1}{2}\partial^\mu \phi \partial_\mu \phi$

user218912

so

user218912

7:38 PM

$d = -\frac{1}{2} \, 1 \, [\phi] \, 1 [\phi]$

user218912

is that wrong?

ACuriousMind

I do not understand what happened to your r.h.s.

user218912

I just replaced the terms by their dimension

DanielSank

Hi, everybody.

ACuriousMind

Uh

That's not how it works

user218912

7:39 PM

:o

user218912

okay so that's my issue

user218912

idk how it works

user218912

can you explain please? @ACuriousMind

DanielSank

@IceLord You're not retarded. However, it's pretty clear that more math experience would help you quite a lot. Isn't it frustrating how we have to learn math while learning physics?

user218912

@DanielSank yes I'm learning both at the same time.

DanielSank

7:40 PM

It seems so many people learn math "on the job" as they do physics.

ACuriousMind

@IceLord Well, "replacing the terms by their dimensions" is the correct idea. Unfortunately, you didn't replace $\partial_\mu \phi$ by its dimension.

Balarka Sen

One of the joys of learning math is that it's strictly speaking independent from any other branches of science. Although learning some physics isn't bad for getting intuition about things.

user218912

@ACuriousMind :o

user218912

what did i do then?

ACuriousMind

When you said that "the dimension of the derivative is 1", what I thought you meant (because that's the correct meaning) was "taking a derivative raises the dimension by one". So $[\partial_\mu\phi] = [\phi] + 1$.

user218912

7:42 PM

oooh

user218912

okay :P

user218912

so it works out then

user218912

I switched from exponents to regular numbers

Balarka Sen

@Danu Anything interesting?

user218912

that's why I messed up @ACuriousMind

ACuriousMind

7:45 PM

@IceLord I don't know what that means but I'm happy you've found your error

user218912

thanks

user218912

I know how mass dimension works now i think

DanielSank

@IceLord Have you noticed that ACM and I frequently can't tell what you're talking about because you're using words to try to describe equations?

Might want to reconsider that.

user218912

I'll work on it.

DanielSank

That's all a person can do :)

Danu

7:48 PM

@BalarkaSen I took a break---had dinner n stuff

DanielSank

@ACuriousMind download QFT into my head, please.

@Danu Whadya eat?

ACuriousMind

@DanielSank Alrighty, need to make room for it first, though

Danu

@DanielSank Basic stuff---pasta with oil & garlic

Some fresh tomatos (the small ones) on top

ACuriousMind

You don't need all those octopus facts, do you?

Danu

I'm experimenting with lemon stuff

Lemon peel and juice

Balarka Sen

7:48 PM

Don't find pasta interesting.

Danu

Me neither, typically

DanielSank

@ACuriousMind Sure do. I can delete real analysis though...

Danu

That's why I'm trying this lemon stuff.

Balarka Sen

Good idea.

DanielSank

@Danu Sounds good.

ACuriousMind

7:50 PM

@BalarkaSen I usually want to find my food "delicious" rather than "interesting" ;)

Danu

Putting a bunch of lemon peel in the water has surprisingly little effect

Balarka Sen

@ACuriousMind Can't imagine what I eat being not delicious but interesting.

DanielSank

Is there an easy way to drop images into chat?

ACuriousMind

@DanielSank Depends on what you mean by "easy". Clicking the "upload" button doesn't strike me as hard, though

DanielSank

@ACuriousMind smartass

Thanks.

Made that.

It's delicious.

1.5 kg chocolate. Twelve eggs.

user218912

7:53 PM

so that's the chocolate cake you kept talking about making?

Balarka Sen

Dear god.

user218912

how big is it?

DanielSank

@IceLord yes

@IceLord Two layers. Diameter is... I dunno... cake size.

ACuriousMind

@BalarkaSen Oh, I've had things I'd describe as "interesting" but certainly not "delicious".

Balarka Sen

Octopus legs?

DanielSank

7:55 PM

@ACuriousMind I gotchu: gfycat.com/GroundedGracefulGrouper

@BalarkaSen :(

ACuriousMind

@DanielSank lol

Balarka Sen

That's an... odd recipe.

@DanielSank Oh, forgot you were an octopus-fan. Sorry about that.

DanielSank

@BalarkaSen np

user218912

@ACuriousMind when I try to show my work I still mess up because idk what to do with the $-\frac{1}{2}$

user218912

how do I treat it?

0celo7

8:00 PM

@ACuriousMind Nonsense

If you replace each term by the units, it must be consistent

ACuriousMind

19 mins ago, by ACuriousMind

@IceLord Well, "replacing the terms by their dimensions" is the correct idea. Unfortunately, you didn't replace $\partial_\mu \phi$ by its dimension.

Read the entire conversation before commenting :P

0celo7

Hmm.

DanielSank

@ACuriousMind yes

Please.

ACuriousMind

@IceLord Well, is the dimension of $\frac{1}{2}\phi$ any different from that of $\phi$?

user218912

ah

user218912

8:03 PM

ok it's not so I remove the 1/2?

ACuriousMind

"remove"?

user218912

uhm

user218912

then what do I do?

user218912

it works if I just remove it

user218912

ignore it i mean

ACuriousMind

8:05 PM

I don't know what exactly you mean by "remove" or "ignore"

user218912

what is the correct way to say it?

ACuriousMind

Are you averse to just writing $[\frac{1}{2} X]=[X]$ instead of trying to find an awkward word for that step?

user218912

what I'm saying is that i'll just write

user218912

xD

user218912

@ACuriousMind alright then

user218912

8:06 PM

thanks

user218912

wow it says to find it using the equal time commutation relations

user218912

same deal though right? @ACuriousMind

ACuriousMind

what says to find what?

user218912

the question asks to find the dimensions of the canonical scalar field using the equal time commutation relations for it.

user218912

i'll try it out

user218912

8:10 PM

using dimensions

user218912

okay I have no idea how to do this

user218912

I'll do it later. time to work on analysis

user218912

@ACuriousMind is the schrodinger equation the most fundamental thing in QM?

user218912

or is $[x, p] = i\hbar$ more fundamental?

user218912

8:25 PM

where does $\hbar$ come from?

ACuriousMind

I don't know what "most fundamental" means.

Danu

WHAT IS LIFE

Haddaway - What Is Love [Official]

►

Sanya

@ACuriousMind I'd guess "what do we need for an axiomatic construction" or sth

ACuriousMind

Both of the things you mention are basically postulates

user218912

okay but why is there an $\hbar$

ACuriousMind

8:26 PM

@Danu lol, you were faster because my bookmarked link for that is deleted :D

user218912

@ACuriousMind ignore everything I said

user218912

back to the problem I'm working on, can you give me a small hint how to find the dimension of the scalar field using the ETCR for the fields?

user218912

because it's not obvious to me

ACuriousMind

@IceLord Uh, there's not really a reason for that.

@IceLord Find the dimension of the delta function first.

Danu

@ACuriousMind :D

user218912

8:28 PM

@ACuriousMind okay

user218912

@ACuriousMind thanks I think I know what to do now

Danu

@ACuriousMind This is so tricky ^^

ACuriousMind

@Danu It is? :O

Danu

@ACuriousMind It was for me, when I learned about it. Just "obviously dimensionless" is my kneejerk reaction.

ACuriousMind

@Danu Yeah, that's my "intuitive" reaction, too. Just goes to show one can't trust intuition :P

Bass

8:45 PM

If you think of $\delta$ as a very sharp rectangle, i.e. $\delta(x)=\frac{1}{2\epsilon}$ for $|x|<\epsilon$, it seems pretty intuitive :P

Sanya

physics.stackexchange.com/questions/280948/… does this qualify as "weird personal theory" or not?

ACuriousMind

@Sanya Sigh...kpv's been at this for a while.

It's not really a personal theory though, it's just that they analyzed some data and believe that there are patterns in them. Too bad that they don't do any statistical analysis to back up the claim those patterns are in any way significant.

Sanya

@ACuriousMind I mean, I've got nothing to say against the stuff he writes in his post here; basically "I want to have a detailed statistical analysis of Bell Inequality tests" - but if you look at the paper he linked - well, it's very badly presented and doesn't involve you to really go through it AND he actually makes some more dramatic claims in the paper

ACuriousMind

I know, I've looked at the paper before.

Sanya

@ACuriousMind ah ok - well, I mean, in the first place, it would be very weird if no one had done decent statistical analysis in Bell tests anyway :|

ACuriousMind

8:58 PM

@Sanya Well, the question is - what kind of analysis would one run? If one doesn't go in with a particular idea of what pattern to find, what is one looking for?

Sanya

@ACuriousMind tbh I've never gotten into statistics as deep as I should have, maybe because I in the end never did too much experimental stuff - but to me, the only obvious tests would be to test how likely it is that the data would have been produced even though our hypothesis is wrong and well ... other things not, I'd guess, I see your point

0celo7

9:28 PM

@BalarkaSen How do you get to the point where you assume all manifolds are compact?

@ACuriousMind I have a really important question for you.

@ACuriousMind What does $L_X\nabla$ mean?

Seems like some physicist thing

ACuriousMind

@0celo7 Context! I have no idea what you're talking about

2

0celo7

@ACuriousMind $L$ - Lie derivative, $\nabla$ - LC.

$X$ - smooth vf.

ACuriousMind

Then it's the operator that first takes the covariant and then Lie derivative, no?

0celo7

No, because I have the claim: $L_X(\nabla)=0$ iff $X$ is Killing.

But it's certainly not true in general that $L_X(\nabla T)=0$ if $X$ is Killing.

ACuriousMind

@0celo7 Well, I can't magically divine what whatever you're reading is talking about, either

0celo7

9:41 PM

If this were representation theory I bet you could!

Wtf $(L_X\nabla)_UV=(L_X\nabla)(U,V)$??

What does that even mean

$=L_X(\nabla_UV)-\nabla_{L_XU}V-\nabla_U(L_XV)$

have you seen such sorcery before @ACuriousMind

ACuriousMind

no

0celo7

I don't think moderators should be allowed to delete/edit posts past the 2 min limit unless there's a rule violation.

Sanya

@0celo7 I have seen \nabla used as the symbol for a connection :o

0celo7

Why ":o"

I'm using it as a connection here.

Sanya

but then you know what your symbols mean

0celo7

9:47 PM

I'm not sure what $L_X(\nabla)$ means even if I know what all the symbols involved are.

I think ACM knows

I just have to extract it

ACuriousMind

wtf

0celo7

what is wrong?

ACuriousMind

I have clearly stated I don't know. Why do you believe otherwise?

0celo7

Oh I believe you.

Doesn't mean it's not in there somewhere.

You could have been walking past some geometers who talked about it.

@ACuriousMind Correct me if I'm wrong please, but $(Z,Y)\mapsto \nabla_ZY$ is not tensorial.

does that even make sense?

Yeah that's not a $(1,2)$ tensor.

So what is $L_X\nabla$ even?

$L_X$ is defined on tensors.

Ok, it's a formal product rule.

We take the object $\nabla_UV$, then differentiate it: $$L_X(\nabla_UV)=(L_X\nabla)(U,V)+\nabla_{L_XU}V+\nabla_UL_XV$$

then rearrange.

seems useless

Balarka Sen

10:06 PM

@0celo7 Not sure what you mean.

0celo7

@ACuriousMind I'm sorry I gave you shit for not knowing why the isometry group is Lie. The proof is horrible :/

ACuriousMind

I suspected that :P

0celo7

@ACuriousMind Petersen has a "full" proof if you do a 9 part functional analysis exercise first

ACuriousMind

Heh

0celo7

@ACuriousMind How much time have you spent with the compact-open topology, Arzela-Ascoli, etc.?

ACuriousMind

10:10 PM

None with the compact-open topology, some with A-A

Danu

@ACuriousMind Do you know a bit of complex algebraic geometry?

Perhaps familiar with the "Kodaira dimension" of a connected complex manifold [see math chat if interested]?

ACuriousMind

@Danu Not really, and the only thing I know of the Kodaira dimension is its name

Danu

@ACuriousMind I'm not asking for much more :P

Just a little bit of motivation behind its definition

0celo7

@ACuriousMind Did you prove A-A in your FA course?

ACuriousMind

@0celo7 yes

@Danu I mean, I literally only know its name. I don't know what its definition or its use is :D

Danu

10:17 PM

@ACuriousMind It's related to the algebraic dimension... Somethingsomething.

0celo7

@ACuriousMind Ok, my analysis book proves it for functions on $\Bbb R$, then claims (I think) that the idea is the same for $\Bbb R^n$, or at least that the generalization is easy.

Balarka Sen

This is why complex varieties are better. Dimension is easy enough.

0celo7

because when the author references it in his geometric analysis book he writes the pages of his analysis book where he proves it on $\Bbb R$

@ACuriousMind Can I see the proof you learned?

Danu

@BalarkaSen Dimension is no problem either way.

Algebraic dimension is something different

As is Kodaira dimension

The normal definition of dimension is not so bad (just take a regular point of your variety and look at the dimension of the manifold).

Balarka Sen

Sure.

ACuriousMind

10:20 PM

@0celo7 Here

Danu

Anyways, I've got a huge headache "rising"... So I guess I better go to bed.

Bye, guys.

0celo7

@ACuriousMind wtf you did PDEs in that course?

Don't tell me you don't know anything about PDEs

Balarka Sen

The tangent space definition isn't really the right version of the defn anyway.

Transcendence degree of the field of rational (or as you would say, meromorphic) functions over it is.

0celo7

@ACuriousMind What is $\mathscr L$?

Balarka Sen

Bye, @Danu.

I should head to bed too.

0celo7

10:22 PM

Spitting bars, again!

ACuriousMind

@0celo7 continuous linear operators

0celo7

Proof: Omitted?

You skipped the inverse function theorem?

Balarka Sen

That's standard.

ACuriousMind

@0celo7 Ain't nobody got time for that!

0celo7

@BalarkaSen What's standard?

Balarka Sen

10:25 PM

Inverse function theorem.

0celo7

So? Most things in math are "standard"

Balarka Sen

No.

0celo7

@BalarkaSen So one shouldn't prove the transversality homotopy theorem in a diff top class because it's standard?

Balarka Sen

I didn't say that.

Just that people would probably do calculus before functional analysis.

0celo7

Inverse function theorem is not...oh forget it.

@ACuriousMind Um, has anyone told you your Ys are backwards?

ACuriousMind

10:28 PM

@BalarkaSen This inverse function theorem is a variant for general Banach spaces. We didn't do that in my analysis class.

Balarka Sen

'Tis the exact same proof tho

ACuriousMind

@0celo7 Yes, multiple times :D

0celo7

@BalarkaSen Do you actually remember the proof of it?

Balarka Sen

sure

0celo7

Don't ever give that "bad memory" BS.

Balarka Sen

10:29 PM

All you need is a normed vector space where contraction mapping holds.

aka a complete normed space

aka Banach space

same proof.

@0celo7 I don't know if there's anything to memorize about it. It's a straightforward application of contraction mapping theorem.

0celo7

Do you mean the Banach fixed point theorem?

Balarka Sen

Those are not distinct theorems.

0celo7

I recall having to prove the implicit function theorem first, and that taking a couple pages.

Balarka Sen

I have never heard of proving implicit function theorem before inverse function theorem.

0celo7

cf. Bredon.

Balarka Sen

10:34 PM

I'll look at it. Weird.

0celo7

cf. Jost

Balarka Sen

For me it's always the implicit function theorem the proof idea of which I forget. One has to come up with an appropriate function to use inverse function theorem but that requires a few manipulations.

0celo7

@BalarkaSen I remembered a geometry book you might be interested in

I heard about it from T. Tao

amazon.com/Differential-Geometry-Graduate-Studies-Mathematics/…

Balarka Sen

Hmm ok.

0celo7

It's super short.

180 pages.

Balarka Sen

10:42 PM

Nice.

Pretty quick-and-dirty.

0celo7

@BalarkaSen I'm going to the library to print something and get a book. I'll pick up this too and take a look.

Balarka Sen

alright. thanks.

I want to read something on differential forms where you actually do something nontrivial with them except introduce them as "fun, elementary, and beginning de Rham theory". Bott-Tu is too technical, I'd prefer something more geometric and non-homotopical.

Don't want to do algebraic topology with forms.

0celo7

Bott & Tu is very technical

I will attack that book again next summer.

Balarka Sen

so, any ideas what to read?

0celo7

@BalarkaSen Honestly...GR.

Diff forms are incredibly powerful in GR

Balarka Sen

10:47 PM

Eh. Maybe not now.

I'll ask Ted or Mike.

0celo7

Kobayashi & Nomizu use forms a lot but they put B-T to shame in terms of technicality.

Balarka Sen

Ted's a fan of diff geom done with forms, so hopefully he'll tell something interesting. Mike was also saying a thing or two about sympletic and contact topology after smooth manifolds.

0celo7

Has Ted recommended Chern's book?

Balarka Sen

I haven't talked to Ted about this, so no.

do you know anything about foliation geometry?

0celo7

Just the basics, i.e. what's in Lee and other Lee.

Balarka Sen

10:51 PM

what does it do

oh nevermind found an MO post on that

0celo7