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).
The multiplication by a phase of the wave function commutes with the
action of the Galilean group.
It is always possible to add a generator, commuting with a Lie algebra generators, to form a Larger Lie algebra. In this case, the larger Lie algebra is called a central extension of the former. The origin of the name is that the added generator (or generators) commute with all the Lie algebra generators, thus they belong to the Lie algebra's center.
For example, The Heisenberg-Weyl algebra:
$$[x, p] = i \hbar \mathbb{1}$$
is a central extension of the two dimensional translation algebra
$\mathbb{R}^2$:
$$[x, p] =0$$
If the central element does not appear in any commutator of the original
Lie algebra generators, the central extension is trivial (all the
$b_{ij}$s are zero) and the algebra is just a direct sum of the two algebras. This is the case in the specific example of the extension described in Ballentine's book. However, this is not the case in the Heisenberg-Weyl algebra, where the commutator of $x$ and $p$ produces the central element. In this case, the central extension is not trivial.
However, this is not the whole story yet. Ballentine is preparing the
backgound for the description of a nontrivial central extension of the Galilean group:
It turns out that the Galilean algebra does not close in neither classical or quantum mechanics without a nontrivial central extension. The Poisson brackets in classical mechanics and the commutator in quantum mechanics of boosts and momenta is not trivial and has the form:
$$[G_i, P_j] = m \delta_{ij}$$
where: $m$ is the particle's mass. Note that the corresponding commutator in the Galilean algebra is vanishing. This result is due to V. Bargmann.
In the classical case, it can be seen quite easily. The Poisson brackets of the Noether charges computed from the free particle Lagrangian corresponding to the boosts and the momenta just satisfy the above relation and they do not Poisson-commute as in the unextended Galilean group algebra.
Finally, let me note that the central element is always represented by a unit matrix in an irreducible representation and a representation of the centrally extended algebra is called a ray representation of the original algebra.
Best Answer
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.