galilean-relativitygroup-theorypoincare-symmetryspecial-relativity
What is the relationship between the Galilean group and the Poincaré group?
Are they siblings within the Lie group? Or does the Poincaré group contain the Galilean group as a subgroup?
I'm not so much interested in the Galilean group being the limit of the Poincaré group for c -> inf.
It is important to distinguish between three group actions that are named "Galilean":
-The Galilean transformation group of the Eucledian space (as an automorphism group).
-The Galilean transformation group of the classical phase space (whose Lie algebra constitute a Lie subalgebra of the Poisson algebra of the phase space). This is the classical action.
-The Galilean transformations of the wavefunctions (which are infinite dimensional irreducible representations). This is the quantum action.
Only the first group action is free from the central extension. Both classical and quantum actions include the central extension (which is sometimes called the Bargmann group). Thus, the central extension is not purely quantum mechanical, however, it is true that most textbooks describe the central extension for the quantum case. I'll explain first the quantum case, then I'll return to the classical case and compare oth cases to the Poincare group.
In quantum mechanics, a wavefunction in general is not a function on the configuration manifold, but rather a section of a complex line bundle over the phase space. In general the lift of a symmetry (an automorphism of the phase space) is an automorphism of the line bundle which is therefore a $\mathbb{C}$ extension of the automorphism of the base space. In the case of a unitary symmetry, this will be a $U(1)$ extension. Sometimes, this extension is trivial as in the case of the Poincare group. Now, the central extensions of a Lie group $G$ are classified by the group cohomology group $H^2(G, U(1))$. In general, it is not trivial to compute these cohomology groups, but the case of the Galilean and Poincare groups can be heuristically understood as follows:
The application of the Galilean group action $\dot{q} \rightarrow \dot{q}+v$ to the non-relativistic action of a free particle: $S = \int_{t_1}^{t_2}\frac{m }{2}\dot{q}^2dt$, produces a total derivative leading to $S \rightarrow S + \frac{m}{2}v^2(t_2-t_1) + mvq(t_2) - mvq(t_1)$: Now Since the propagator $G(t_1, t_2)$ transforms as $ exp(\frac{iS}{\hbar})$ and the inner product $\psi(t_1)^{\dagger} G(t_1, t_2) \psi(t_2)$ must be invariant, we get that the wavefunction must transform as:
$\psi(t,q) \rightarrow exp(\frac{i m}{2\hbar}(v^2 t+2vq) \psi(t,q) $
Now, no application of a smooth canonical transformation can romove the total derivative from the transformation law of the action, this is the indication that the central extension is non-trivial.
The case of the Poincare group is trivial. The relativistic free particle action is invariant under the action of the Poincare group, thus the transformation of the wavefunction doen not acquire additional phases and the group extension is trivial.
Classically, the phase space is $T^{*}R^3$ and the action of the boosts on the momenta is given by: $p \rightarrow p + mv$, thus the generators of the boosts must have the form $K = mvq$, then the action is easily obtained using the Poisson brackets{q, p} = 1, and the Poisson bracket of a Boost and a translation is non-trivial {K, p} = m.
The reason that the Lie algebra action acquires the central extension in the classical case is that the action is Hamiltonian, thus realized by Hamiltonian vector fields and vector fields do not commute in general.
The Iwasawa decomposition of the Lorentz group provides the answer to your second question:
$SO^{+}(3,1) = SO(3) A N$ where $A$ is generated by the Boost $M_{01}$ and $N$ is the Abelian group generated by $M_{0j}+M_{1j}$, $j>1$. Now both subgroups $A$ and $N$ are homeomorphic as manifolds to $R$ and $R^2$ respectively.
To your third question: The limiting process which produces the Galilean group from the Poincare group is called the Wingne-Inonu contraction. This contraction produces the non-relativistic limit. Its relation to quantum mechanics is that there is a notion of contarction of Lie groups unitary representations, however not a trivial one.
Update
In classical mechanics, observables are expressed as functions on the phase space. see for example chapter 3 of Ballentine's book for the explicit classical realization of the generators of the Galilean group.
This is a case where the full geometric quantization recepie can be carried out. See the following two articles for a review. (The full proof appears in page 95 of the second article. The technical computations are more readable in pages 8-9 of the first article).
The central extensions appear in the process of prequantization.
First please notice that the Hamiltonian vector fields $X_f$ corresponding to the Galilean Lie algebra generators close to the non-centrally extended algebra, (because, the hamiltonian vector field of constant functions vanish).
However, the prequantized operators
$\hat{f} = f - i\hbar(X_f - \frac{i}{\hbar}i_{X_f} \theta)$, ($\theta$ is a symplectic potential whose exterior derivative equals the symplectic form) close to the centrally extended algebra because their action is isomorphic to the action of the Poisson algebra.
The prequantized operators are used as operators over the Hilbert space of the square integrble polarized sections, thus they provide a quantum realization of the centrallly extended Lie algebra.
Regarding your second question, the Wingne-Inonu contraction acts on the level of the abstract Lie algebra and not for its specific realizations. A given realization is termed "Quantum", if it refers to a realization on a Hilbert space (in contrast to realization by means of Poisson brackets, which is the classical one).
In general, there are two classic ways to represent the Poincaré group. The first one comes from its definition. They are the "rigid" motions for a 4D Minkowski spacetime. Identifying space-time with $\mathbb R^4$ equipped with the metric: $$\eta = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{pmatrix} \tag{1} $$ you can view the Poincaré transformations as the affine transformation $f: x\in\mathbb R^4 \to\Lambda x+u$ with $u$ a constant fourvector and $\Lambda$ the analogue of a rotation matrix satisfying: $$ \Lambda \eta \Lambda^T = \eta $$ ie a linear transformation preserving the metric, more commonly known as a Lorentz transformation.
Translations are just the case when $u=0$ and Lorentz transformations about the origin when $\Lambda=I_4$. Within the Lorentz transformations, you typically distinguish rotations and boosts. This is done by fixing a time coordinate (note that the distinction is observer dependent). In our case, let's fix time to be the first coordinate $x^0$, with corresponding basis vector $e_0$.
Rotations are the subgroup of the Lorentz group that fix $e_0$ ie don't mess with time. You can show that you can write: $$ \Lambda = \begin{pmatrix} 1 & 0 \\ 0 & R \end{pmatrix} \tag{2} $$ with $R$ a rotation matrix ie a $3\times 3$ satisfying $RR^T = I_3$.
For boosts, you can't rigorously talk about a representation as they do not form a subgroup of the Lorentz group (this is the origin of the famous Thomas precession). This subset can be parametrised by a $3$ vector $v$ with corresponding Lorentz factor $\gamma = (1-v^Tv)^{-1/2}$ by: $$ \Lambda = \begin{pmatrix} \gamma & \gamma v^T \\ \gamma v & I_3+\frac{-1+\gamma}{v^T v}v v^T \end{pmatrix} \tag{3} $$
Note that there is another convenient way of representing the Lorentz group as a matrix subgroup. It is the mathematical trick of homogenising The idea is to add an extra dimension. You can view them as $5\times 5$ the matrices of the form: $$ \begin{pmatrix} 1 & 0 \\ u & \Lambda \end{pmatrix} \tag{4} $$ To prove the equivalence with the previous one, you can easily show it has the same action on the affine subspace $\begin{pmatrix}1 \\ x\end{pmatrix}$ with $x\in\mathbb R^4$. In a similar way, you can view the Galilean transform more explicitly as a degenerate Lorentz group by viewing the group of boosts and rotations as a matrix subgroup. In fact, this representation shows that it is in some sense a degenerate version of a bigger group, which leads to AdS and dS spaces.
Hope this helps
Best Answer
It is assumed you have appreciated Inönü, E.; Wigner, E. P. (1953), "On the Contraction of Groups and Their Representations" Proc. Natl. Acad. Sci. 39 (6): 510–24, and the super-helpful Gilmore text in Group contraction.
Very crudely, the Poincaré Lie algebra, $$ [J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad [J_i, P_0] = 0 ~, \\ [K_i,P_k] = i \delta_{ik} P_0 ~, \\ [P_0,P_i]=0 \qquad [P_i,P_j]=0 \qquad [K_i, P_0] = -i P_i ~, \\ [J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad [J_m,K_n] = i \epsilon_{mnk} K_k ~, \\ [K_m,K_n] = -i \epsilon_{mnk} J_k ~, $$ given relabelings $E=-cP_0$ and $K_i=cC_i$ contracts upon $c\to \infty$ to the Galilean algebra, $$ [J_m,P_n] = i \epsilon_{mnk} P_k ~, \qquad [J_i,E]=0 \\ [C_i,P_j]= 0,~\\ [E,P_i]=0, \qquad [P_i,P_j]=0, \qquad [C_i,E]=i P_i \\ [J_m,J_n] = i \epsilon_{mnk} J_k ~, \qquad [J_m,C_n] = i \epsilon_{mnk} C_k ~, \\ [C_i,C_j]=0 . $$
There are a few subtleties and wrinkles, extensions, to be sure, which I gather you are not focussing on, but, crudely, the third and the last commutation relations trivialized/collapsed. (There is more, but I am oversimplifying...). This collapse/amputation is the Lie algebraic manifestation of a group contraction.