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).
Yes. Both universal covers and central extensions incurred during quantization come from the same fundamental concept:
Projective representations
If $\mathcal{H}$ is our Hilbert space of states, then distinct physical states are not vectors $\psi\in\mathcal{H}$, but rays, since multiplication by a complex number does not change the expectation values given by the rule
$$ \langle A\rangle_\psi = \frac{\langle \psi \vert A \vert \psi \rangle}{\langle \psi \vert \psi \rangle}$$
nor the transition probabilities
$$ P(\lvert \psi \rangle \to \lvert \phi \rangle) = \frac{\lvert \langle \psi \vert \phi \rangle\rvert^2}{\langle \phi \vert \phi \rangle\langle \psi \vert \psi \rangle}$$
The proper space to consider, where every element of the space is indeed a distinct physical state, is the projective Hilbert space
$$ \mathrm{P}\mathcal{H} := \mathcal{H} /\sim$$
$$ \lvert \psi \rangle \sim \lvert \phi \rangle :\Leftrightarrow \exists c\in\mathbb{C}: \lvert \psi \rangle = c\lvert\phi\rangle$$
which is just a fancy way to write that every complex ray has been shrunk to a point. By Wigner's theorem, every symmetry should have some, not necessarily unique, unitary representation $\rho : G \to \mathrm{U}(\mathcal{H})$. Since it has to descend to a well-defined ray transformation, the action of the symmetry is given by a group homomorphism into the projective unitary group $G \to \mathrm{PU}(\mathcal{H})$, which sits in an exact sequence
$$ 1 \to \mathrm{U}(1) \to \mathrm{U}(\mathcal{H}) \to \mathrm{PU}(\mathcal{H}) \to 1$$
where $\mathrm{U}(1)$ represents the "group of phases" that is divided out when passing to the projective space. It is already important to notice that this means $\mathrm{U}(\mathcal{H})$ is a central extension of $\mathrm{PU}(\mathcal{H})$ by $\mathrm{U}(1)$.
To classify all possible quantumly allowed representations of a symmetry group $G$, we need to understand the allowed Lie group homomorphisms $\sigma : G\to\mathrm{PU}(\mathcal{H})$. Since linear representations are nicer to work with than these weird projective things, we will look at
Classifying projective representations by unitary linear representations
For any $g\in G$, choose a representative $\Sigma(g)\in\mathrm{U}(\mathcal{H})$ for every $\sigma(g)\in\mathrm{PU}(\mathcal{H})$. This choice is highly non-unique, and is essentially responsible for how the central extension appears. Now, since for any $g,h\in G$ we have $\sigma(g)\sigma(h) = \sigma(gh)$, the choices of representatives must fulfill
$$ \Sigma(g)\Sigma(h) = C(g,h)\Sigma(gh)$$
for some $C : G\times G\to\mathrm{U}(1)$. Applying associativity to $\Sigma(g)\Sigma(h)\Sigma(k)$ gives the consistency requirement
$$ C(g,hk)C(h,k) = C(g,h)C(gh,k)\tag{1}$$
which is also called the cocycle identity. For any other choice $\Sigma'$, we must have
$$ \Sigma'(g) = f(g)\Sigma(g) $$
for some $f : G \to \mathrm{U}(1)$. $\Sigma'$ has an associated $C'$, and so we get
$$ C'(g,h)\Sigma'(gh) = \Sigma'(g)\Sigma'(h) = f(g)f(h)C(g,h)f(gh)^{-1}\Sigma'(gh)$$
which yields the consistency requirement
$$ C'(g,h)f(gh) = f(g)f(h)C(g,h)\tag{2}$$
Therefore, projective representations are classified giving the choice of unitary representatives $\Sigma$, but those that are related by $(2)$ give the same projective representation. Formally, the set
$$ H^2(G,\mathrm{U}(1)) := \{C : G\times G \to \mathrm{U}(1)\mid C \text{ fulfills } (1)\} / \sim$$
$$ C \sim C' :\Leftrightarrow \exists f : (2) \text{ holds }$$
classifies the projective representations of $G$. We want to use it to construct a unitary representation of something that classifies the projective representation:
Define the semi-direct product $G_C := G \ltimes_C \mathrm{U}(1)$ for any representative $C$ of an element in $H^2(G,\mathrm{U}(1)$ by endowing the Cartesion product $G \times \mathrm{U}(1)$ with the multiplication
$$ (g,\alpha)\cdot(h,\beta) := (gh,\alpha\beta C(g,h))$$
One may check that it is a central extension, i.e. the image of $\mathrm{U}(1)\to G \ltimes_C\mathrm{U}(1)$ is in the center of $G_C$, and
$$ 1 \to \mathrm{U}(1) \to G_C \to G \to 1$$
is exact. For any projective representation $\sigma$, fix $\Sigma,C$ and define the linear representation
$$ \sigma_C : G_C \to \mathrm{U}(\mathcal{H}), (g,\alpha) \mapsto \alpha\Sigma(g)$$
Conversely, every unitary representation $\rho$ of some $G_C$ gives a pair $\Sigma,C$ by $\Sigma(g) = \alpha^{-1}\rho(g,\alpha)$.
Therefore, projective representations are in bijection to linear representations of central extensions.
On the level of the Lie algebras, we have $\mathfrak{u}(\mathcal{H}) = \mathfrak{pu}(\mathcal{H})\oplus\mathbb{R}$, where the basis element $\mathrm{i}$ of $\mathbb{R}$ generates multiples of the identity $\mathrm{e}^{\mathrm{i}\phi}\mathrm{Id}$. We omit the $\mathrm{Id}$ in the following, whenever a real number is added to an element of the Lie algebra, it is implied to be multiplied by it.
Repeating the arguments above for the Lie algebras, we get that the projective representation $\sigma : G \to \mathrm{PU}(\mathcal{H})$ induces a representation of the Lie algebra $\phi : \mathfrak{g}\to\mathfrak{pu}(\mathcal{H})$. A choice of representatives $\Phi$ in $\mathfrak{u}(H)$ classifies such a projective representation together with an element $\theta$ in
$$ H^2(\mathfrak{g},\mathbb{R}) := \{\theta : \mathfrak{g}\times\mathfrak{g} \to \mathbb{R}\mid \text{ fulfills } (1') \text{ and } \theta(u,v) = -\theta(v,u)\} / \sim$$
$$ \theta \sim \theta' :\Leftrightarrow \exists (b : \mathfrak{g}\to\mathbb{R}) :\theta'(u,v) = \theta(u,v) + b([u,v])$$
with consistency condition
$$ \theta([u,v],w) + \theta ([w,u],v) + \theta([v,w],u) = 0 \tag{1'}$$
that $\theta$ respects the Jacobi identity, essentially.
Thus, a projective representation of $\mathfrak{g}$ is classified by $\Phi$ together with a $\theta\in H^2(\mathfrak{g},\mathbb{R})$. Here, the central extension is defined by $\mathfrak{g}_\theta := \mathfrak{g}\oplus\mathbb{R}$ with Lie bracket
$$ [u\oplus y,v\oplus z] = [u,v]\oplus\theta(u,v)$$
and we get a linear representation of it into $\mathfrak{u}(\mathcal{H})$ by
$$ \phi_\theta(u\oplus z) := \Phi(u) + a$$
Again, we obtain a bijection between projective representations of $\mathfrak{g}$ and those of its central extensions $\mathfrak{g}_\theta$.
Universal covers, central charges
We are finally in the position to decide which representations of $G$ we must allow quantumly. We distinguish three cases:
There are no non-trivial central extensions of either $\mathfrak{g}$ or $G$. In this case, all projective representations of $G$ are already given by the linear representations of $G$. This is the case for e.g. $\mathrm{SU}(n)$.
There are no non-trivial central extensions of $\mathfrak{g}$, but there are discrete central extensions of $G$ by $\mathbb{Z}_n$ instead of $\mathrm{U}(1)$. Those evidently also descend to projective representations of $G$. Central extensions of Lie groups by discrete groups are just covering groups of them, because the universal cover $\overline{G}$ gives the group $G$ as the quotient $\overline{G}/\Gamma$ by a discrete central subgroup $\Gamma$ isomorphic to the fundamental group of the covered group. Thus we get that all projective representations of $G$ are given by linear representations of the universal cover. No central charges occur. This is the case for e.g. $\mathrm{SO}(n)$.
There are non-trivial central extensions of $\mathfrak{g}$, and consequently also of $G$. If the element $\theta\in H^2(\mathfrak{g},\mathbb{R})$ is not zero, there is a central charge - the generator of the $\oplus\mathbb{R}$ in $\mathfrak{g}_\theta$, or equivalently the conserved charge belonging to the central subgroup $\mathrm{U}(1)\subset G_C$. This happens for the Witt algebra, where inequivalent $\theta(L_m,L_n) = \frac{c}{12}(m^3 - m)\delta_{m,-n}$ are classified by real numbers $c\in \mathbb{R}$.
Best Answer
The boosts and translations do not commute in neither relativistic or nor relativistic systems, please see for example the case of the Poincare group.
Since $K_i = M_{0i}$ , $(i=1,2,3)$, we get for the Poincare group: $[K_i, P_j] =i (\eta_{0j}P_i-\eta_{ij}P_0) = -i \delta_{ij}P_0$
Now, The Galilean group can be obtained from the Poincare group by means of Wigner-Inonu contraction, see for example this link.
In the non relativistic limit the rest momentum is dominated by the mass term thus under the contraction the above relation becomes:
$[K_i, P_j] =-i \delta_{ij}M$
The above reasoning shows that the parameter $M$ is the mass.
Considered as an Lie algebra element M commutes with all other generators and cannot be removed by a smooth redefinition of the generators, this is the reason that it is called a central extension, see for example chapter 3 of Ballentine's book, where several representations of the Galilean group are constructed.
The consequence is that in each irreducible representation of The Galilean group $M$ must be represented by a scalar (the particle's mass), and two representations with different masses are unitarily inequivalent.
The formula of the product of a translation and a boost can be directly obtained the cation of the Baker–Campbell–Hausdorff formula.
This formula means that the wavefunctions acquires a phase factor in a transformed inertial frame (in other words, the wavefunction representation is projective). In quantum mechanics a global phase is not measurable, thus no physical consequence is caused
However, Representations corresponding to different masses fall into different superselection sectors, this means that linear superpositions of wavefunctions of particles with different masses are unphysical, because in this case each component would acquire a different phase which contradicts experiment because differential phases are observable in quantum mechanics.
It is worthwhile to mention that this is exactly the example which was originally analyzed by V. Bargmann in his paper "On unitary ray representations of continuous groups", Ann. of Math.,59 (1954), 1–46.
Update
This update contains the answers to the questions appearing in Aruns comments:
The noncommutativity of the momentum and boost operators is quantum mechanical. The commutators are written in $\hbar= 1$ units, working in general units, the boost-momentum commutator has the form:
$[K_i, P_j] =i \hbar \delta_{ij}M$
The noncommutativity is observable only in the action of the operators on the wave functions. Let me elaborate this point for the case a free particle considered in Ballentine's book.
First, the phase space (i.e., the manifold of initial data) is $\mathbb{R}^6$ covered with the coordinates $\{ \mathbf{q}, \mathbf{v} \}$, which upon quantization satisfy the commutation relations:
$[ q_i, v_j] = \frac{i}{M} \delta_{ij}$
Consider the operators:
The finite translation operator $\mathcal{D}_{q_0} =\exp(i \frac{ M\mathbf{v}. \mathbf{q_0}}{\hbar})$.
The finite boost operator $\mathcal{B}_{v_0} = \exp(i \frac{\mathbf{K}. \mathbf{v_0}}{\hbar}) = \exp(i \frac{M \mathbf{q}. \mathbf{v_0}}{\hbar})$
Consider a wave function on the phase space $\psi(\mathbf{q})$. It is not difficult to see that:
$\mathcal{B}_{v_0} \mathcal{D}_{q_0} \psi(\mathbf{q})= \exp(\frac{i M \mathbf{v_0}.\mathbf{q_0}}{\hbar}) \mathcal{D}_{q_0} \mathcal{B}_{v_0} \psi(\mathbf{q})$
The fourth relativistic momentum components is given by
$ P_0 = \sqrt{M^2 c^4 + p^2 c^2}$
In the non-relativistic limit $c \rightarrow \infty$, the first term dominates and we get $ P_0 \approx M c^2$. Now, you absorb the factor $c^2$ into the definition of the generators, (or work in units c=1). This is essentially the Wigner-Inonu contraction of the Poincare group to the Galilean group.
I'll give you here the standard argument of the uniatry inequivallence between the quantization of two free particles with different masses $ M_1 \ne M_2$. Let us use the subscript 1 and 2 for the individual particle operators (Each acting on a different Hilbert space ). If the two representations are unitarily equivalent means that there is an isometry $U$ such that
$\mathbf{K_2} = U \mathbf{K_1}U^{*} $ and $\mathbf{P_2} = U \mathbf{P_1}U^{*} $, thus we obtain $[\mathbf{K_2}, \mathbf{P_2}] = U [\mathbf{K_1}, \mathbf{P_1}]U^{*}$ , which means $M_1 = M_2$, a contradiction.
Unitary inequivalence is similar to representations of $SU(2)$ having different $J$, consquently having different dimensions. In the finite dimensional case, it is obvious that we cannot find an isometry between different dimensional Hilbert spaces. One might think that all infinite dimensional representations of the Galilean are unitarily equivalent which is not the case.