Non commutator of symmetries giving rise to a gauge symmetry

[binbagsss]

If there are two symmetries of a Lagrangian, perhaps they are transformations, A and B, and they don't commute $[A, B] \neq 0$. Let this act on some field, then if $(BA) ^{-1}AB$ does not return the original field, i.e. if $(BA) ^{-1}AB \neq \mathbb{1}$, then this gives a rise to a new symmetry so it will have an associated charge/ current by Noethers theorem. I feel like this is a common occurrence but I can't find any references on considering this as a symmetry/computing an associated conserved charge/current. Is there certain terminology/framework I need to be searching for? Thanks

 

[haushofer]

If there are two symmetries of a Lagrangian, perhaps they are transformations, A and B, and they don't commute $[A, B] \neq 0$. Let this act on some field, then if $(BA) ^{-1}AB$ does not return the original field, i.e. if $(BA) ^{-1}AB \neq \mathbb{1}$, then this gives a rise to a new symmetry ...

Not necessarily so. E.g. the commutator between translations and rotations gives a new translation.

In the algebra $so(3)$ one has

$[L_i, L_j] = i \epsilon_{ijk} L_k$

so

$[L_1, L_2] = i L_3$

i.e. the commutator $L_1$ and $L_2$ generate the third symmetry generated by $L_3$. But those three generators span the whole algebra, so we only need to look at the corresponding conserved charges of those generators.

 

[binbagsss]

Not necessarily so. E.g. the commutator between translations and rotations gives a new translation.

In the algebra $so(3)$ one has

$[L_i, L_j] = i \epsilon_{ijk} L_k$

so

$[L_1, L_2] = i L_3$

i.e. the commutator $L_1$ and $L_2$ generate the third symmetry generated by $L_3$. But those three generators span the whole algebra, so we only need to look at the corresponding conserved charges of those generators.

ok thanks, I will check the translation and rotation one you say. Is there a way to know whether there will be a 'new' symmetry or not from (BA)^{-1} AB from the Lie algebra of the commutators without having to verify and compute each time?

 

[haushofer]

I guess those are given by the structure constants of the Lie-algebra.

 

[binbagsss]

"Not necessarily so. E.g. the commutator between translations and rotations gives a new translation."

So here you are saying that this can be written in terms of the known translations and so the charge/current associated with the new translation would be able to be written in terms of the known charges/currents associated with the original translations?

 

[haushofer]

"Not necessarily so. E.g. the commutator between translations and rotations gives a new translation."

So here you are saying that this can be written in terms of the known translations and so the charge/current associated with the new translation would be able to be written in terms of the known charges/currents associated with the original translations?

I don't understand what you mean by "original" and "new" translations. There are only 4 of them (or D, in D spacetime dimensions if you're a string- or sugra-theorist). To every generator you can assign a Noether charge, and these Noether charges obey the same algebra as the generators.

 

[binbagsss]

I don't understand what you mean by "original" and "new" translations. There are only 4 of them (or D, in D spacetime dimensions if you're a string- or sugra-theorist). To every generator you can assign a Noether charge, and these Noether charges obey the same algebra as the generators.

"so. E.g. the commutator between translations and rotations gives a new translation." is what I meant by it. Obviously it's a simple case but I am asking for the purposes of thinking about other cases. obviusly I am aware its not a *new* translation.

 

[haushofer]

Ah yes, sorry, that can be confusing. But I still don't understand your precise question. Maybe you can provide some more context.