Can I write: $\omega_{\nu\mu}g^{\nu\rho}=\omega^\rho_{\ \ \nu}$ even though the indices $\nu$ do not follow one after the other?

or is that not important?

What is meant with degree of freedom of a group in general, and Lorentz group in particular ?

@imbAF Why would the indices have to follow each other? The rule is that repeated indices get summed over, not that the indices have to "follow each other".

I see

ACM the questions I tend to ask, for this topic have mostly to do with me trying to understand the syntax

@imbAF I would advise against calling this a "degree of freedom" to avoid confusion with the more physical idea of degrees of freedom of a system, but it's clear that in context it can only mean the dimension of the group as a Lie group, i.e. the number of generators you need to express an arbitrary element of the group (equivalently the dimension of the Lie algebra as a vector space or like a dozen other equivalent definitions :P)

So I would like to know how do you differentiate about different objects in notations

@ACuriousMind Ok just to confirm

the generators, are the ones that produce the elements of the group

So, can I say that the generators form a basis?

or is wrong to say?

they form a basis of the Lie algebra

they are not a basis of the group since the group is not a vector space so the notion of basis does not make sense

the generators of the Lorentz group ?

@ACuriousMind indeed

Ok, so I have been reading about group theory etc

So I have a couple of questions, mostly to do with notation but also a bit of understanding how things operate

For example:\

$\Lambda=exp(-\frac{i}{2}\omega_{\mu\nu}L^{\mu\nu})$

I previously discussed with Slereah about it. But how could I tell that $L^{\mu\nu}$ is a matrix and not an element of a matrix

while $\omega_{\mu\nu}$ is an element

In both cases the indices are both either up or down

That should be part of the definition of the $L$ :P

Of the generators?

You can't just write down a mess of symbols and pretend they make sense (although admittedly that's how plenty of physicists treat math), you need to define the symbols before you can use them. So before you can write down that expression for $\Lambda$, you need to define that the $\omega_{\mu\nu}$ are numbers and the $L^{\mu\nu}$ are matrices.

I see. Well it wasn't said in class

Now I know, thanks

(the "you" there is generic, I'm not faulting you personally for it :P)

ahh ok

I thought you were telling me lol

I only have 2 more questions. Master started and we started with superficial group theory

as you can tell

@ACuriousMind .

In the notes it was asked, how many parameters/degrees of freedom does the Lorentz group has (Neither of the two terms was explained)

And as a suggestion it was said to use a symmetry argument to prove the nr.

So what I did, is I considered the condition for the spacetime interval to be invariant $g=\Lmabda^T g\Lambda$

@ACuriousMind Which incidentally is messy too since a lot of people call "degree of freedom" the dimension of phase space, not configuration space (in some specific contexts) :P

and an infinitesimal transformation. Which I substituted in the above expression. And showed that $\omega^{\mu\nu}=-\omega^{\nu\mu}$

So I used antisymmetry (symmetry argument) to prove that

yes, that's what that question wanted you to do

@Mr.Feynman I know, and even more reason to not overload the term further!

Ok but I did it in a very "cheating"/mechanical way

I will let you decide whether that is what I did or not

So, because I started with an infinitesimal transformation, which is a function of $\omega_{\mu\nu}$ (if I can use the word "function of " here). I said, that because we have 6 idependent values of the $\omega$ matrix, then we have 6 LT. 3 boosts and 3 rotations. So, correct me if I am wrong

but I argued about the nr. of parameters (generators?), without even considering them, without even considering the existence of them

What do you mean 6LT? Probably you mean 6 generators of LT

So I tied the nr. of parameters (which I have no clue what they represent) with the 6 indepedent elements of $\omega$

nothing with generators

@Mr.Feynman I think I wrote it wrong. What I meant was 3 boosts and 3 rotations

I mean, these are LT, but I guess you can have even combinations of them

I mean, that physics levels of rigor this is about what you were expected to do :P

Is there perhaps a way to go from the expression of infinitesimal transformation to the expression of $\lambda=...\omega_{\mu\nu}L^{\mu\nu}$

?

@imbAF You posit it and then derive the form of the generators

You posit it because it's a Lie group and it will of course have generators; Of course you are making a choice here and it's implicit in taking the $\omega$ of the infinitesimal forms as coefficients of the generators

posit?

You impose that relation

Ah

What I mean is that the first order term is in the Lie Algebra, so it will be a combination of generators

I just need like a sketch of what group is a sub group of which group

and all this algebra/group thing

a mess

Ok, never mind the word algebra right now. Just consider it as an "infinitesimal group" (sorry ACM)

The very definitions of generators of a Lie group entails that $\Lambda=1+\frac{i}{2}\omega_{\mu\nu}L^{\mu\nu}+\mathcal{o}(\omega^2)$

The $i/2$ coefficient is there for converience, don't mind it too much

I want to see the connection here

@Mr.Feynman ahaa.. and because the Lorentz group is a lie group, therefore, writing that expression for the LT, indicates the connection

@imbAF So, the momentum operator is not here. The momentum operator appears in the context of the Poincaré group, which also includes translations in spacetime (and the 4-momentum is precisely the generator thereof). Here you only have angular momentum and boosts

So how can I show e.g the rotatin L_i can be expressed via boosts?

I mean of course I know what $\omega^\rho_{\ \ \sigma}$ is in matrix form

@Mr.Feynman How do you argue that e.g the rotation is equal to some Generator?

@imbAF With this you can compute $[L^{\mu\nu}, L^{\rho\sigma}]$

Now you have the algebra commutation relations

You can use them with the definition of $L^i$ above to derive $[L^i, L^j]$

did you all ever feel like you were not learning physics :P

And you will discover that $[L^i, L^j]=i\varepsilon^{ijk}L^k$. Does the last equation ring any bell?

It is actually comical that these things aren't taught properly

@SillyGoose "not learning physics" as in "there is not enough physics in what I'm studying" or as in "I suck damn hard and I'm not learning a thing"?

Showing properly what form of $\Lambda$ and its generators the condition $g = \Lambda^T g\Lambda$ implies is a bit involved. The most elegant demonstration of this I know is page 2 of this [caution: my handwriting from 10 years ago :P] (there is one copy mistake where it should say "spatial rotations" instead of "spatial translations"), but it is probably not at the level you're looking for

I don't know how that makes sense. The connections between groups, algebras and generators, you can't just go along and hope to understand them

@ACuriousMind Didn't peg you as a note-taking student

Mr.Feynman told me how to do what I asked, and I still have no clue as to how and why

Cuz I don't have a fundamental understanding so that if I make a mistake I know I did one

If you have no idea how the structure looks like, you can't possibly know when you are doing a mistake or not

You need confirmation from others and reasoning why, what I did was right or wrong

From someone who has a full view of the entire picture

Are you following a book?

I don't find peskin very helpful

Every respectable QFT book has a (non-respectable) discussion of the Poincaré group

@Mr.Feynman more like i feel like the courses i am taking are not introducing any new material and also my own studies have been more leading into learning specific methods, not building a coherent picture of things. i guess i associate the latter with "learning"

And I am trying to solve previous exercises on my own.

@SillyGoose I usually think "I'm not learning enough math" :P

I will upload a section and I want to have an understanding of what I should do

The solution is irrelevant to me, as long as I can't comprehend the connection

Or that's what I did before burning out and barely standing physics as I mentioned before

@Mr.Feynman I got a tablet with a stylus specifically so I could take notes and directly upload them because I'm terrible at keeping order with physical notes but I felt that I learned a lot by taking notes instead of just listening

For solving d). What is it that I need to have an understanding of

The connection, the logic

for one, it's been rather useful when I wanted to refer back to those notes later, but I think the main benefit of taking notes was simply keeping me focused on the lecture (an easy way to tell if a lecture was boring was if I started doodling random shapes :P)

@ACuriousMind what lecture was this?

I think that was a course about the Poincaré group itself (?)

Or a DG course

Yeah, need to attend one, have no time

@Mr.Feynman 1. I'm left-handed, so I love the digital writing simply because I don't have to care about smudging the ink. 2. After getting used to it, I think I was actually faster on the tablet. 3. Skill issue. ;P

@imbAF The course from which these notes are? This was a "group theory" course held by a mathematical physicist.

Nice

@ACuriousMind The best part of lectures was that I always ended up like that. Sometimes I'd catch my fellow - who'd always take notes - staring at my doodles and heavily judging me :P

It was not the "unitary representations of Poincaré group" course Mr. Feynman is thinking of :P

i'm not sure learning differential geometry will help most directly with understanding the poincaré group

But @ACuriousMind could you help me understand the logic behind what is asked on d)

@Mr.Feynman The neat thing about the digital writing is that you can erase the doodles without a trace ;P

Because I have no link between the generators and the angular momentum compoennts or boosts. So I cannot make any statements

@ACuriousMind Not skill issue! I need a lot of time to do things properly. Even in studying Japanese now, I keep starting new notebooks to focus on different parts of understanding. I'm like on an OCD level for this D:

is there a purpose to the notation used in imbAF's screenshot? This also is the notation used by Weigand...it seems horribly unnecessary

@imbAF The angular momentum and boost generators are in the same vector space as the generators, so they are linear combinations of them. The exercise just wants you to find those specific linear combinations.

And incidentally, it turns out that nitpicking is not just about physics; I've just bought a Japanese Kanji dictionary that took the poor author 10 years to write and I already reported mistake D:

vector space? But we said there is no actual vector space

@SillyGoose What specifically? The indices?

for the generators

@imbAF no, we said the group is not a vector space

the generator live in the algebra, which is a vector space

and that is?

ok

The "infinitesimal group" for all you care now

@Mr.Feynman I just made a folder "<course name>" for each source and uploaded pdfs called "Lecture X" for X the how many-th lecture in that course it was

So now I need to show that elements of this vector space, such as rotations and boost can be expressed with the generators?

then each of those is inside a folder "Yth semester"

@Mr.Feynman well (1) why to do it starting with clifford algebras and (2) why to write the coefficients $\omega$ into a "matrix" $\omega^{\mu\nu}$

@ACuriousMind Barbarian! I had to choose the color, the name format, whether to record audio or not in all lecture, what pen to use, the size and so on. Either they were all the same or I lost any interest in taking notes :P

at least it looks like it is starting from clifford algebra based on its definition of the $L^{\mu\nu}$

@SillyGoose The practical reason why the indices are useful is that you want to label the generators clearly for the algebra commutation relations

But I don't know what the relationship between the generators is with the rotations and boost

I only have the expression for $\Lambda$. And I cannot go to index notation because

@SillyGoose Concerning Clifford algebras, I don't think you need them at this point. You are just making a possible choice of the generators

how can one tell the different rotations from one another in that way

@Mr.Feynman I don't understand the problem; I took one day to experiment which pen size and style I found most readable and stuck with that for years :P

@Mr.Feynman If I write this: $L^{i}=\frac{1}{2}\varepsilon^{ijk}L^{jk}$ and you or someone else asks why, did you write that? How do you know it's true? I don't know what to tell you

So I cannot defend my answer

= No understanding

@ACuriousMind We may argue that you're more pragmatic than me, my friend. I struggle heavily even when I have to decide what I want for breakfast.

@Mr.Feynman lol, I hate having to decide what to eat. One of the main draws for me to go to the office is that there's a canteen there where I only have to choose from three options :P

@Mr.Feynman today i considered starting some study of distribution theory. i paused and decided not to :P

@imbAF The reason is that $L^{\mu\nu}$ is a rank-2 antisymmetric tensor, so the diagonal components are zero. Those equations are just a neat way to enclose the (antisymmetric) spatial part, and the mixed-time component form another three-vector

@ACuriousMind You on an office day

why is antisymmetric? I mean I showed that $\omega^{\mu\nu}$ is but not L

@SillyGoose There is really not much you want to know about distributions. You may get away with what you care in a week or so

Well, to begin with, you can see here that $L^{\mu\nu}=-L^{\nu\mu}$

@Mr.Feynman heh, fortunately I usually have only one dish that does not contain something I don't like so the choice is easy :P

@Mr.Feynman perhaps but then the integral $\int_0^\infty dt e^{-i\eta t}$ appears

although this integral i have put behind me now :P

@SillyGoose There is not much you have to do here, I told you. Just regularize adding a small imaginary part to $\eta$ so that you ensure convergence. Finally you take the limit as the imaginary part approaches zero.

Trying to understand group theory or w/e the stupid proper name for all this discussion is, is like waking up to a strangers house, and people start telling you, "Hey can you pick this thing, in that x,y,z place" And you somehow have to know where it is

but this "regularization" procedure as presented in these physics texts makes no sense to do a priori

It's precisely in performing this limit at the end (taking it outside the integral) that the distributional nature arises. Otherwise the integral would not be defined

one is literally by hand placing something into the integral to delete the diverging contribution

@SillyGoose I'm afraid regularization are just that, a way to make sense of meaningless things :P

what if you come across another divergent integral which cannot be regularized by the same procedure?

Then can you compare the two integrals?

@imbAF I agree, and I loathe the way physics texts typically do this. The only remedy is to learn it from the mathematicians :P

mathemagicians

I don't know what to do really

I don't understand

Or better

I am unfamiliar

The same way a high school student wouldn't know what to do if you would tell him about Diracs equation

@Mr.Feynman i don't have a problem with defining a new integral. I more have a problem with coming up with something arbitrary that a priori doesn't necessarily work all the time.

@imbAF Like that time I was at a friend's house and I inadvertently opened the underwear drawer

He doesn't know because he ain't aware

This is where I stand

@Mr.Feynman :'D

@Mr.Feynman I thought of a logical way as how to solve d)

so that every step, can be explained

I'm afraid I don't have enough neurons active to follow you now

ah

Yeah, I'm about to go to bed

goodnight

@SillyGoose I'll reply another time If I remember

Good night y'all

@ACuriousMind does it make a difference if the indices of the generators are both up or both down

?

have any of you seen applications of persistent homology in physics

@imbAF I don't know how to answer that question. Not because it's a bad question, but because my only honest answer is "the index position on the generators is bogus anyway". I rant about this a bit in this answer to a question by @Mr.Feynman

The notation, in the picture I uploaded, regarding the generators, what is it called?

Does it have a name?

not really

So can I write $\epsilon^{ijk}L_{jk}$?

or $$\epsilon_{ijk}L^{jk}$

or $\epsilon^{ijk}L^{jk}$,$\epsilon_{ijk}L_{jk}$

I'm afraid that depends on your conventions :P

because the way you're writing it, it's not clear what the difference between an upper and lower index even is without first establishing a convention

since these are not "Lorentz indices" where you'd raise/lower them with the Minkowski metric

Well I need to start with the expression for the generator and land at the expression that expresses boosts and rotations as a function of generators

So, I wanted, using that expression to show the commutation relation between the generators

Then, using your argument that generators and boost belong in the same space

I could use

Ah nah, it doesn't work

I just can't give solid arguments for why $L^{i}=\frac{1}{2}\varepsilon^{ijk}L^{jk}$ and $K^i=L^{0i}$