I am trying to understand how one obtains the Galilean algebra from the Poincare algebra, specifically through the method of central extension. I'm doing this by imposing that the generators of the Poincare group scale with the velocity in certain ways, and then taking the small velocity limit. Also I was told I should make the definition
$$H = M+W,\tag{1}$$
where $H$ is the Hamiltonian, and $M$ will turn out to be a central element of the algebra. Now I have been able to fine many of the commutation relations and they make sense (i.e. match the Galilean algebra) but I don't really understand how to show $M$ commutes with everything. Also, is the physical interpretation that $M$ is the potential energy and $W$ is the kinetic energy?
Homework on Hamiltonian and Inonu-Wigner Group Contraction: Lie Algebra and Galilean Relativity
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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).
Here is how I understand Weinberg's discussion:
First of all: following Weinberg's words precisely, he only says that "Such a set of commutation relations is known as a Lie algebra". He motivates the commutation relations by considering a representation $U(T)$ of a Lie group $T$, but he does not clarify the relationship between these notions. The commonly accepted terminology has to be learned from elsewhere; I'll try to expand on it here.
Note that mathematicians distinguish between "a Lie algebra" and "the Lie algebra associated to a Lie group".
Any algebra of operators that fulfills (2.2.22) is "a" Lie algebra.
But for each Lie group $T$, there is "the" "most general" Lie algebra $\mathfrak{t}$ that fulfills (2.2.22). It arises from the structure constants $f^a_{bc}$ that Weinberg mentions. Note that the structure constants do not depend on the representation on some Hilbert space, the only depend on the Lie group $T$.
To find this "most general" Lie algebra associated to a Lie group, you can look at a very special vector space, namely the tangent space at the identity element $1\in T$. The ajdoint representation of the group on this space will give rise to a Lie bracket $[·,·]$ on that vector space and turn it into said "most general" Lie algebra.
As for your particular questions:
- Weinberg only the mentions "a Lie algebra", though it comes from a group. It's "the" lie algebra of the group $U(T)$.
- In general, every such Lie algebra is a homomorphic image of "the" Lie algebra associated to $T$. These Lie algebras are different from each other; for instance, the vector spaces may have different dimensions.
Weinberg invokes Wigners theorem for the following reason: a priori, the symmetry group $T$ acts on rays. Remember that the vector $|ψ\rangle$ represents the same physical state as the vector $ξ|ψ\rangle$ which arise from multiplication with an arbitrary complex number of mangitude $|ξ|=1$. A ray is the set of all vectors that arise in this fashion; they all represent the same physical state.
Now, the symmetry maps physical states to physical states, i.e. sets of vectors to sets of vectors. But each symmetry operation is allowed to permutate vectors within the set, after all, they are indistinguishable physically. It is not clear at all that a mapping $T_1$ on rays can be recast as a mapping $U(T_1)$ that acts on individual vectors and is linear. Neither is it clear that $U(T_1T_2) = U(T_1)U(T_2)$, because these operations may map the rays to each other, but they might permute the vectors within a ray rather differently, which would manifest in a phase $U(T_1T_2)|ψ\rangle = ξ·U(T_1)U(T_2)|ψ\rangle$.
We want the symmetry to act on a Hilbert space, though, because this allows us use the familiar expression for the commutator of two operators $[X,Y] = XY-YX$. After all, it involves the addition (subtraction) of two operators, something that is not available for symmetries merely acting on rays.
Other than that, Hilbert spaces are completely unnecessary for defining Lie algebras or Lie groups.
The question of how to find all representations of a Lie group or of a Lie algebra is beyond the scope of this answer. It leads to topics such as semisimple Lie algebras and their classification.
Mathematicians obtain the Lie algebra associated to a Lie group by considering the tangent space at the identity element, as mentioned above. The Lie group doesn't need to be connected for this to be well-defined. (But it is only useful for studying the part of the group that is connected to the identity.)
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In physical, non-covariant language, (WP conventions), the Poincaré algebra presents as $$ [P_0,P_i]=0,\\ [P_i,P_j]=0, \\ [J_i,P_0] = 0 ~, \\ [K_i,P_k] =- i \delta_{ik} P_0 ~,\\ [K_i,P_0] = -i P_i ~,\\ [K_m,K_n] = -i \epsilon_{mnk} J_k ~, \\ [J_m,P_n] = i \epsilon_{mnk} P_k ~,\\ [J_m,K_n] = i \epsilon_{mnk} K_k ~, \\ [J_m,J_n] = i \epsilon_{mnk} J_k ~, $$ where one of the Casimir invariants is $P_\mu P^\mu= P_0^2-\vec{P}^2$.
Now redefine the boosts and $P^0$ up and down by the speed of light $c$, so $K^i\equiv c C^i$ and $P_0\equiv \frac {1}{c} E$.
The Wigner-İnönü contraction $c\to\infty$ (slowness!) results in the naive Galilean Lie algebra G(3), $$[E,P_i]=0, \\ [P_i,P_j]=0, \\ [L_i,E]=0 , \\ [C_i,P_j]= 0 ~,\\ [C_i,E]=i P_i , \\ [C_i,C_j]=0, \\ [L_{m},L_{n}]=i \epsilon _{ink} L_{k} ,\\ [L_{m},P_k]=i \epsilon _{mkj}P_j , \\ [L_{m},C_k]=i \epsilon _{mkj}C_j . $$
In effect, the boost has lost its time-translation piece and is but space translations proportional to the time, Galilean boosts, $C^i$; and the timelike momentum is a plain time-translation oblivious of c, namely a "hamiltonian", E. The spacelike $P_i$ are generators of translations as before (momentum operators), and $L^i$ stand for generators of space rotations, having merely changed name from J, to banish any inapposite thoughts of spin.
Observe how this limit has trivialized several right-hand sides to 0. In fact the 10D regular representation (matrix of structure constants, Gilmore p 220) is a very sparse matrix, indeed. It amounts to extreme structural violence.
Note the quadratic invariants $P^iP^i$ and $L^iL^i$.
The Bargmann central extension algebra is the above, but now with $[C_i,P_j]=iM\delta_{ij} $ instead of the above trivial relation ($E/c^2\to M$ as $c\to \infty$), where the central charge M is an invariant, as the name implies, easy to check consistency of. The quadratic momentum invariant now morphs into a new invariant, $ME-P^2/2$, the mass-shell invariant, and since M is invariant, $E-\frac{P^2}{2M}$ is invariant as well, the potential energy.
E is like the Hamiltonian, but it is not an algebra invariant, as it fails to commute with the Galilean boosts. It is merely a time invariant, i.e. it commutes with itself--pfffft....