6
We will not write down the components of the Bargmann group's momentum. Schematically, let us write the Bargmann group's momentum as follows:
J B = { 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 get more and more difficult as the pages go by. I no longer know, at this stage, who this is addressed to. Probably to theoretical physicists or mathematicians, but probably not to plumbers. But a student from a Grande Ecole or a physics bachelor's degree, who holds on, can follow. It's just matrices.
It all starts with a group of (4,4) matrices that constitute the Lorentz group, whose element is L.
These are defined axiomatically from a matrix G:
(64)
according to:
(65)
t L 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 L1 L2 belongs to the group. If this is the case:
t (L1 L2) G L1 L2 = G
But:
t (AB) = tB tA
Therefore:
t (L1 L2) G L1 L2 = tL2 tL1 G L1 L2 = tL2 (tL1 G L1) L2 = tL2 G L2
Then 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)
dg p' = g p -1 x dg p x g p
For the sake of calculation convenience, we note that:
(74)
G dL 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)
dg p' = g p -1 x dg p x g p
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)
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