There are 3 actions of the Galilean group on the free particle: On the configuration space, on the phase space and on the quantum state space (wave functions). The Galilean Lie algebra is faithfully realized on the configuration space by means of vector fields, but its lifted action on Poisson algebra of functions on the phase apace and on the wave functions (by means of differential operators) is the central extension of the Galilean algebra, known as the Bargmann algebra in which the commutator of boosts and momenta is proportional to the mass. The reasoning is given in the following arguments
1) The action on the configuration space: $Q = \{x^1, x^2, x^3, t\}$:
Here the translations and the boost operators act as vector fields and their commutator is zero:
Translation: $x^i \rightarrow x^i+c^i$, generating vector $P_i = \frac{\partial}{\partial x^i}$
Boost: $x^i \rightarrow x^i+v^i t$, generating vector $G_i = t \frac{\partial}{\partial x^i}$
This is a faithful action of the Galilean group: $[P_i, G_j] = 0$.
2) The lifted Galilean action to the phase space $Q = \{x^1, x^2, x^3, p_1, p_2, p_3\}$
The meaning of lifting the action is to actually write the Lagrangian and finding the Noether charges of the above symmetry: The charges as functions on the phase space will generate the centrally extended version of the group. An application of the Noether theorem, we obtain the following expressions of the Noether charges:
Translation: $P_i = p_i$
Boost: $ G_i = P_i t - m x^i$.
The canonical Poisson brackets at $t=0$ (because the phase space is the space of initial data): $\{P_i, G_j\} = m \delta_{ij}$
The reason that the lifted action is a central extension lies in the fact that that the Poisson algebra of a manifold itself is a central extension of the space of Hamiltonian vector fields,
$$ 0\rightarrow \mathbb{R}\overset{i}{\rightarrow} C^{\infty}(M)\overset{X}{\rightarrow} \mathrm{Ham}(M)\rightarrow 0$$
Where the map $X$ generates a Hamiltonian vector field from a given
Hamiltonian:
$$X_H = \omega^{ij}\partial_{j}H$$
($\omega$ is the symplectic form. The exact sequence simply indicates that the Hamiltonian vector fields of constant functions are all zero).
Thus if the Lie algebra admits a nontrivial central extension, this extension may materialize in the Poisson brackets (the result of a
Poisson bracket may be a constant function).
3) The reason that the action is also extended is that in quantum mechanics the wave functions are sections of a line bundle over the configuration manifold. A line bundle itself is a $\mathbb{C}$ bundle over the manifold:
$$ 0\rightarrow \mathbb{C}\overset{i}{\rightarrow} \mathcal{L}\overset{\pi}{\rightarrow} M\rightarrow 0$$
thus one would expect an extension in the lifted group action. Line bundles can acquire a nontrivial phases upon a given transformation. In the case of the boosts, the Schrödinger equation is not invariant under boosts unless the wave function transformation is of the form:
$$ \psi(x) \rightarrow \psi'(x) = e^{\frac{im}{\hbar}(vx+\frac{1}{2}v^2t)}\psi(x+vt)$$
The infinitesimal boost generators:
$$\hat{G}_i = im x_i + \hbar t \frac{\partial}{\partial x_i}$$
Thus at $t=0$, we get: $[\hat{G}_i, \hat{P}_j] = -im \hbar\delta_{ij}$
Thus in summary, the Galilean group action on the free particle's configuration space is not extended, while the action on the phase space Poisson algebra and quantum line bundle is nontrivially central extended.
The classification of group actions on line bundles and central extensions may be
performed by means of Lie group and Lie algebra cohomology. A good reference
on this subject is the book by Azcárraga, and Izquierdo. This book contains a detailed treatment of the Galilean algebra cohomology. Also, there are two readable articles by van Holten: (first, second).
Group actions on line bundles (i.e. quantum mechanics) is classified by the first Lie group cohomology group, while central extensions are classified by the second Lie algebra cohomology group. The problem of finding central extensions to Lie algebras can be reduced to a manageable algebraic construction. One can form a BRST operator:
$$ Q = c^i T_i + f_{ij}^k c^i c^j b_k$$
Where $b$ abd $c$ are anticommuting conjugate variables: $\{b_i, c_j \} = \delta_{ij}$. $T_i$ are the Lie algebra generators.
It is not hard to verify that $Q^2 = 0$
If we can find a constant solution to the equation $Q \Phi = 0$ with $\Phi = \phi_{i j} c^i c^j$
which takes the following form in components, we have
$$ f_{[ij|}^k \phi_{k|l]} = 0$$
(The brackets in the indices mean that the indices $i, j, l$ are anti-symmetrized. Then the following central extension closes:
$$ [\hat{T}_i, \hat{T}_j] = i f_{ij}^{k} \hat{T}_k + \phi_{ij}\mathbf{1}$$
The second Lie algebra cohomology group of the Poincaré group is vanishing, thus it does not have a nontrivial central extension. A hint for that can be found from the fact that the relativistic free particle action is invariant under Poincaré transformations. (However, this is not a full proof because it is for a specific realization). A general theorem in Lie algebra cohomology asserts that semisimple Lie algebras have a vanishing second cohomology group. Semidirect products of vector spaces and semisimple Lie algebras have also vanishing second cohomology provided that there are no invariant two forms on the vector space. This is the case of the Poincaré group. Of course, one can prove the special case of the Poincaré group by the BRST method described above.
This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user David Bar Moshe
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