Re: unitary irreps of Galilei group

whopkins_at_csd.uwm.edu
Date: 01/30/05 

Date: Sun, 30 Jan 2005 19:01:00 +0000 (UTC)

Arnold Neumaier wrote:
> Alfred Einstead wrote:
> > Here's a side-by-side comparison of the two Lie algebras:
> > Galilei (with the central extension by mass m included):
> > [J_a,J_b] = e^c_{ab} J_c
> > [J_a,K_b] = e^c_{ab} K_c
> > [J_a,P_b] = e^c_{ab} P_c
> > [K_a,P_b] = m delta_{ab}
> > [K_a,E] = P_a
> > All other commutators are 0.
> >
> > Poincare' looks like:
> > [J_a,J_b] = e^c_{ab} J_c
> > [J_a,K_b] = e^c_{ab} K_c
> > [J_a,P_b] = e^c_{ab} P_c
> > [K_a,P_b] = E/c^2 delta_{ab}
> > [K_a,E] = P_a
> > [K_a,K_b] = -e^c_{ab} J_c/c^2
> > All other commutators are 0.
> But this E diverges for c to infinity, while there is an E
> in the Galilei formulas, too. So something must still
> be incorrect. What about using H=mc^2+E in place of E in
> the quantum case?

Actually I just came onto that idea, as well. In general it's best to
consider the E generator, not as the total energy, but only as the
KINETIC energy. To make the distinction, call the kinetic energy T and
the total energy E. Then one can write down a general Lie algebra that
accomodates the central extensions of all the above:

[J_a,J_b] = e^c_{ab} J_c
[J_a,K_b] = e^c_{ab} K_c
[J_a,P_b] = e^c_{ab} P_c
[K_a,P_b] = M delta_{ab}
[K_a,T] = P_a
[K_a,K_b] = -A e^c_{ab} J_c
[P_a,P_b] = [P_a,T] = [T,T] = 0
where
M = m + A T
is the total mass; and the 3 cases for A are:
Poincare': A = (1/c)^2 > 0
Galilei: A = 0
Euclidean: A = -(1/r)^2 < 0.

> > [K_a,P_b] = (m+E/c^2) delta_{ab}
> > [K_a,E] = P_a
> and the limit is apparent. But now one has a central extension
> of Poincare!?

For all the cases, except Galilei, the constant m may be reabsorbed
into a redefinition: E = M/A = T + m/A.

For Poincare' this, of course, gives you E = M c^2.

This generalizes further, if you consider the case where the 3 groups
are actually on a hypersphere or hyperbolic geometry. Then you also
get non-zero commutators for the P's and T's:

[P_a,P_b] = LA e^c_{ab} J_c
[P_a,E] = L K_a.

As is the case for Galilei and Poincare', you also get 2 invariants in
the general case:
p^2 - 2MT + AT^2 + L(A J^2 - K^2)
|ML + PxK|^2 - A(J.P)^2 - LA(J.K)^2.
(Check for correct factors and signs, I'm doing this extemporaneously
from memory).

One can proceed to write down the Lie-Poisson bracket (which
technically is defined over the space of C^{infinity} functions over
T*L*, the cotangent space of the dual L* of the Lie algebra L), which
in ordinary 3-D vector notation becomes:

{f,g} = J.(fJ x gJ - A fK x gK + LA fP x gP
+ K.(fJ x gK + fK x gJ + L(fP gT - fT gP)
+ P.(fJ x gP + fP x gJ + fK gT - fT gK)
+ M (fK.gP - fP.gK)
where
df = fJ.dJ + fK.dK + fP.dP + fT dT
dg = gJ.dJ + gK.dK + gP.dP + gT dT
(the differentials dJ_a, dK_a, dP_a, dT residing in T*L* = L).

An invariant f is a function which gives you 0 Poisson bracket with all
g's. This yields a condition for each differential coefficient of g:
gJ: J x fJ + K x fK + P x fP = 0
gK: -AJ x fK + K x fJ - P fT - M fP = 0
gP: LAJ x fP - LK fT + P x fJ + M fK = 0
gT: LK.fP + P.fK = 0.
Parcelling out the cases (where different combinations of the
coefficients give you 0 multipliers) you get the full range of
possibilities for invariants. The two invariants above are for the
general case, but reduce to trivial constants in the specialized cases.

If M is non-zero, then the equation for gT becomes redundant, because
of the quadratic invariant. If M is 0, then its Poisson brackets are
all also 0, which gives you restrictions on the possibilities for (L,A)
and (J,K,P) that in turn yields some of the special cases. For
non-zero M, you'll find that the quantity (M^2 + L (AJ)^2) becomes of
special significance. If 0, the equations for gK and gP do not yield
unique solutions for fP and fK; otherwise they do. The first condition
(gJ) essentially implies that the invariant must be scalar polynomials
[of a constant degree] whose monomials are formed out of the vectors
(J,K,P) by dot products, triple product; and multiplied by powers of E
and M. So, adopting that ansatz, merely solving for the gP and gK
equations is almost enough to get the most general form of the
differential for f; which in turn gives you the function f.