groups and physics coadjoint action momentum
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We will not write down the components of the Bargmann group momentum. Let us schematically write the Bargmann group momentum as follows:
JB = { a scalar m, plus the other components of the momentum }
The coadjoint action indicates how the different components of the momentum transform. But this coadjoint action starts with the simple relation:
(63) m' = m
The coadjoint action of the Bargmann group on its momentum starts by preserving the mass, which thus emerges with a purely geometric status.
Construction of the coadjoint action of the Poincaré group on its momentum space Jp**.**
If you are already completely lost, forget it. It's normal and it will become more and more difficult as the pages go by. I no longer know, at this stage, who this is addressed to. Probably theoretical physicists or mathematicians, but probably not plumbers. But a student from a Grandes Ecoles or physics bachelor's degree, who sticks to it, can follow. It's just matrices.
It all starts with a group of (4,4) matrices which constitute the Lorentz group, whose element is L.
These are defined axiomatically from a matrix G:
(64)
according to:
(65) tL G L = G
involving the transpose of the matrix L.
The matrices L form a group.
Proof.
The neutral element is L = 1:
Let L1 and L2 be two elements of the set. Let's check that the product L1L2 belongs to the group. If this is the case:
t( L1L2 ) G L1L2 = G
But:
t( A B ) = t B t A
Therefore:
t( L1L2 ) G L1L2 = tL2 tL1 G L1L2 = tL2 ( tL1 G L1) L2 = tL2G L2
Let's calculate the inverse of the matrix L. We start from the axiomatic definition of the elements L:
tL G L = G
We multiply on the right by L-1:
tL G L L-1 = G L-1
tL G = G L-1
We multiply on the left by G:
G tL G = G G L-1
G tL G = L-1
Therefore, the inverse matrix of L is:
L-1 = G tL G
So:
(66)
the space-time vector. The matrix G comes from the Minkowski metric, which can then be written (with c = 1):
(67)
Exercise: show that the inverse matrix obeys:
(68)
We then introduce a space-time translation vector:
(69)
From which we construct the element gp of the Poincaré group:
(70)
Exercise: show that this forms a group and calculate the inverse matrix:
(71)
Below is the "tangent vector to the group, element of its "Lie algebra":
(72)
From this we will calculate the anti-action:
(73) dgp' = gp-1 x dgp x gp
For the sake of calculation convenience, we note that:
(74) G d L
is an antisymmetric matrix. Let's call it:
(75)
so:
(76)
Let us set:
(77)
From this material we will construct the anti-action:
(78) dgp' = gp-1 x dgp x gp
After all calculations, we will obtain the application:
(79)
If you want to skip this simple matrix calculation part, refer to equation (80), bottom of the page
(79a)
(79b)
from which the elements of the anti-action are:
(79c)
but:
(79d)
so:
(79e)
but GG = 1 so :
(79f)
from which we obtain the application:
(79g)
This constitutes the desired anti-action, the application:
(80)