• One can define the Poincare group as the group of isometries of the Minkowski space. Is its Lie algebra given either by the equations 2.4.12 to 2.4.14 (as also given in this page - https://en.wikipedia.org/wiki/Poincar%C3%A9_group) or equations 2.4.18 to 2.4.24 of Weinberg's volume 1 of his QFT books?

    What confuses me is that in deriving the commutation relations between $J^{\mu \nu}$ and $P^\mu$ he did use quantum theoretic arguments about the Hilbert space operator $U$ but I guess there is nothing quantum about the Lie algebra he derives in the aforementioned equations. Is that right?

  • This quantum confusion steepens when one looks at the $K_i$ (..relativistic boost along the $i^{th}$ spatial direction..), $P_j$ (..linear momentum along the $j^{th}$ spatial direction..) commutator being non-zero. This is justified by saying that the exponential action of the boosts and the translations on the Hilbert space states do not commute and that is being reflected here. (..they pick up an extra phase proportional to the mass and the dot product of the boost velocity and the displacement vector..)

But if the afore mentioned equations are really the Lie algebra of the isometry group of the Minkowski spacetime then in the Galilean limit shouldn't they be instead reflecting the fact that Galilean boosts and translations when acting on the spacetime coordinates do infact commute? But the $K_i$ and $P_j$ commutation continues to be non-zero even when the Galilean limit is taken on page 62.

This makes me strongly suspicious that the equations 2.4.12 to 2.4.14 are not the Lie algebra of the isometry group of the Minkowski spacetime but are the Lie algebra of the group whose elements are $U(\Lambda, a)$ (..using Weinberg's notation..) …right?

  • So is the "low velocity" limit taken on page 62 recovering non-relativistic quantum theory ? (and not Newtonian physics)

  • On page 89 of the same book he derives the topology of the inhomogeneous Lorentz group as being $R^4 \times R^3 \times S^3/Z_2$. Since this is a connected manifold, I guess that by the term ``inhomogeneous Lorentz group" he is meaning only the proper orthochronous component of the full relativistic symmetry group. right?

  • I can't see how the above topology matches with the semidirect product structure for posibly the same thing as given on this Wikipedia page -http://en.wikipedia.org/wiki/Poincaré_group?

When people talk of the Poincare group is the full symmetry group of relativity is what is being referred to or is it just its proper orthochronous component (and not the other 3 components) ?

  • I am familiar with the notion of "central charge" as in the "first" term on the RHS of the TT OPE of CFT… what also has the interpretation as the zero-point energy when doing the plane-minus-point<->cylinder conformal transformation.

In this light it is not clear to me as to what is meant when one says that one can add in the "mass" as a central charge to the Galilean group... an extra generator which commutes with all the rest so that with this "central extension" the free particles will lie in the unitary representations of the Galilean group rather than the projective representations before the extension.

I would be grateful if someone can shed light on this issue and help reconcile the two "different" notions of central charge.

  • When one takes a "low velocity" limit of the Poincare algebra to get the Galilean algebra then is one taking just a non-reltivistic limit or is one also taking a non-quantum limit?

(... I guess this will depend on my first query about whether what Weinberg calls as the Poincare algebra in the quoted equations has any quantum effect encoded in it {as it seems to be!} or is it just the Lie algebra of the isometry group of the Minkowski spacetime...)

  • $\begingroup$ I started reading Weinberg and I've asked myself many of the same questions! I was wondering about your second bullet point too, Moshe provided a quantum explanation when he answered my question about it, but now I too am wondering if that means there is something "quantum" about the Poincare algebra. I believe you are correct about the third bullet point, and I too would love to see more discussion on central charges. $\endgroup$ Jul 16, 2011 at 23:25
  • 3
    $\begingroup$ Could you perhaps exchange the equation numbers to the equations themselves please? $\endgroup$
    – qftme
    Jul 17, 2011 at 0:10
  • $\begingroup$ @Arun I had "studied" these issues long back. It was in fact your questions and comments which made me rethink about these issues and discover these "blind spots" in my reading of Weinberg's book. Let us see if the confusions get clarified this time round. $\endgroup$
    – Student
    Jul 17, 2011 at 13:26
  • $\begingroup$ @qftme Well..these are lot of sets of equations. It would be very complicated to write them all over again. I have reference the first set of commutation relations to a Wikipedia page and I guess people can look up on google books for the second set…if they need the details of it. The second set is merely a rewriting of the first by splitting time and space. $\endgroup$
    – Student
    Jul 17, 2011 at 13:36

3 Answers 3


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.


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).

  • $\begingroup$ Thanks for this awesome answer. Can you give some reference or this concept that you alluded to that - the wavefunction is a section of a complex line bundle over the phase phase and hence symmetries of the phase space lift to it with a extension. This is totally new to me! $\endgroup$
    – Student
    Jul 18, 2011 at 5:02
  • $\begingroup$ So when you derive the relation $\{K,p\}=m$ are you thinking of the Galilean group to be acting on the classical phase space (..your second meaning?..). So you mean that the algebra gotten by Weinberg on page 62 of his volume 1 of QFT books is the Lie algebra of the group of transformations of the classical phase space? This connects to your last comment - I am not being clear as to how do you decide whether any quantum effect is left in the algebra that remains after the Wigner-Inonu contraction. Kindly explain about these. $\endgroup$
    – Student
    Jul 18, 2011 at 5:07
  • $\begingroup$ May be you can give me some expository reference regarding these. $\endgroup$
    – Student
    Jul 18, 2011 at 5:14
  • $\begingroup$ @Anirbit I added an update which contains some clarifications $\endgroup$ Jul 18, 2011 at 15:37
  • $\begingroup$ Thanks for the details. I thought my questions were naive. But your answer seems to take me to the whole new terrain of Geometric Quantization! This is exciting. It will take me some time to understand what you are saying. All this is totally new to me. I hope to come back to you after some more studies. $\endgroup$
    – Student
    Jul 18, 2011 at 21:12

You are still asking way too many questions at once. So again, consider splitting them next time. I will address only the topology part here.

As topological spaces we have $$SO(3) = {\mathbb R \mathbb P}^3 = S^3 /\sim ,$$ $$SO(4) = S^3 \times S^3 / \sim, $$ $$SO^+(1,3) = {\mathbb R}^3 \times S^3 / \sim $$ (in all of these cases $\sim$ is an identification of the antipodal points on the relevant spheres). One can get these results by noting that the double cover of $SO(3)$ is $SU(2)$, double cover of $SO(4)$ is $SU(2) \times SU(2)$ and the double cover of $SO^+(1,3)$ is $SL(2, {\mathbb C})$. (Note that the above factorizations can be also written as $X/({\mathbb Z} / 2{\mathbb Z})$ which is a factorization of a space by the action of the group on the space; in this case the action is sending a point to its antipodal point).

Adding translations only means using semidirect product on the level of groups, or direct product on the level of topological spaces. In total, we have that the connected component of the Poincaré group is ${\mathbb R}^4 \times {\mathbb R}^3 \times S^3 / \sim$. As for the wikipedia page, I am not sure about your confusion. It defines Poincaré group as ${\mathbb R}^4 \rtimes O(1,3)$ and topologically this is disjoint union of four copies of ${\mathbb R}^4 \times {\mathbb R}^3 \times S^3 / \sim$.

  • $\begingroup$ Thanks for your answer. Can you elaborate on this statement of yours - `Adding translations only means using semidirect product on the level of groups, or direct product on the level of topological spaces" … this seems unfamiliar to me. Can you also elaborate as to why you see that the semi-direct product of $\mathbb{R}^4$ and $O(1,3)$ is $4$ disjoint copies of $\mathbb{R}^4 \times \mathbb{R}^3 \times S^3/~$. This is not obvious to me! $\endgroup$
    – Student
    Jul 18, 2011 at 5:13
  • 2
    $\begingroup$ @Anirbit: $O(1,3)$ containts four disjoint components itself (determined by whether the determinant of the matrix is positive or negative and whether the transformation is ortochronous or not). Extending the group by ${\mathbb R}^4$ doesn't really change anything about this picture: you just multiply each component separately because as a set (or here rather a space) $H \ltimes G = H \times G$, they differ only by the group operation but the topology doesn't see this. E.g., adding translations into the game doesn't help you to find a continuous deformation of identity to a reflection, right? $\endgroup$
    – Marek
    Jul 18, 2011 at 10:18

The question of the non-relativistic limit is obscured by the fact that the non-relativistic kinematic symmetry group, as we now understand it, has 11 dimensions, with the inclusion of the eleventh generator - the central charge - which means that we're actually talking about the Bargmann group, not the Galilei group, while the kinematic symmetry group for relativity has only 10 dimensions, because historically the symmetry transforms were devised before we had our current understanding of the situation in non-relativistic theory.

The formulation of Relativistic symmetry was never retro-updated to match this updated view of non-relativistic symmetry, so as to retain the validity of the correspondence limit. The correspondence limit was broken by the revision of non-relativistic theory.

In order to have consistency with the correspondence limit, you also need an 11th generator for the Poincaré group, and adding in the trivial central extension will also happen to clarify a great number of other issues.

Everything centers around the issue of the mass and energy. I make a distinction between the kinetic energy used in non-relativistic theory, denoting it $H$, versus the "total" energy used in relativity, denoting it $E$; while also using the "moving mass" $M$ interchangeably, where $E = M c^2$. The total energy $E$ can be decomposed into the kinetic part $H$ plus an invariant part $μc^2$ proportional to a mass invariant $μ$. That's the additional generator. The use of $M$ is preferable, since it has a meaningful non-relativistic limit, by virtue of the relation $M = μ + H/c^2$, while $E$ does not.

In Weinberg's treatment (in section 2.4), he actually does make a similar distinction - his $W$ corresponds to my $H$, and his $H$ to my $E$, and his $M$ to my $μ c^2$, so you could write $E = μ c^2 + H$, in place of his $H = M + W$. So, he sneaked in the 11th generator too, but tried to hand-way his way out of explicitly calling it out as such.

This is a slight generalization of the Poincaré symmetry of Relativity that reduces to Relativity - at least for Tardyons - by identifying $μ$ with the rest mass $m$. In place of the invariant $m^2 c^2 + P^2 - E^2/c^2$, where $𝐏$ is the momentum 3-vector, you have the quadratic invariant $λ = P^2 - 2MH + H^2/c^2$, whose non-relativistic limit is just $P^2 - 2MH$. For the linear invariant $μ = M - H/c^2$, the non-relativistic limit is just $μ = M$. The restriction of the generalized Poincaré symmetry to Relativity then corresponds to setting $λ = 0$.

Otherwise, if you allow for a non-zero rest value for $H$, as an invariant $U$, then (for Tardyons) in the rest state (where $P^2 = 0$ and $M = m$, the rest mass), you would have $λ = -2mU + U^2/c^2$ and $μ = m - U/c^2$. In the non-relativistic limit the quadratic invariant yields the relation $P^2 - 2MH = -2mU$, with $M = m = μ$ all coinciding, and the decomposition $H = P^2/2m + U$: the kinetic energy for the center of mass motion plus the internal energy. So, the generalization allows you extra room to add in an internal energy. In the general case, the rest mass would only be defined by $$m^2 = μ^2 - \frac{λ}{c^2} = M^2 - \frac{P^2}{c^2},$$ and only for Tardyons, where $M^2 > P^2/c^2$. It is an abuse of terminology to use the term "rest mass" with Luxons or Tachyons, when they have no rest frame! At best, you can catch Luxons on the margin by pretending $m = 0$; when actually all you're really saying is that $P = M c$, or equivalently: $E = P c$. But, there's no such thing as a "rest mass" for Luxons, since they can never be at rest.

Once you do all of that, then the situation with the $[K,P]$ brackets becomes clear. In indexed form, they would be written as $$[K_i, P_j] = iħ δ_{ij} M = iħ δ_{ij} \left(μ + \frac{H}{c^2}\right) = iħ δ_{ij} \frac{E}{c^2},$$ and its non-relativistic limit would be $[K_i, P_j] = iħ δ_{ij} μ$, as expected, with $μ$ as the central charge.

As to what the transforms correspond to geometrically, you're not going to find any answers in 4-dimensional geometry at all. You can already see that, non-relativistically, by the fact that the mass, energy and momentum transform together as a 5-vector, not as a 4-vector. So, for consistency, you also need a 5th coordinate to serve as conjugate to the mass, with the invariant 1-form (from which the operator correspondence is ultimately drawn) written down as: $$𝐏·d𝐫 - H dt + μ du, \hspace 1em 𝐫 = (x, y, z).$$ The action under an infinitesimal boost by $𝞄$ is then given by $$ΔH = -𝞄·𝐏, \hspace 1em Δ𝐏 = -𝞄M, \hspace 1em ΔM = -α𝞄·𝐏,$$ with the non-relativistic case ($α = 0$) and relativistic case ($α = 1/c^2 > 0$) being distinguished from one another by the deformation parameter $α$.

The (trivial) central extension of the Poincaré group is a deformation of the centrally-extended Galilei group, which is called the Bargmann group.

As a footnote: it's possible to embed these two groups into a much larger family of kinematic symmetry groups (that includes, for instance the [anti-]de Sitter groups), as a three parameter family of deformations of the "static group" ... and to provide a uniform 5-dimensional geometric representation applies across the board to the central extensions of them all.

The coordinate transforms which make the above one-form invariant are given in infinitesimal form as $$Δ𝐫 = -𝞄t, \hspace 1em Δt = -α𝞄·𝐫, \hspace 1em Δu = 𝞄·𝐫,$$ and yield the following two geometric invariants: $$|d𝐫|^2 + 2 dt du + α du^2, \hspace 1em ds ≡ dt + α du.$$ Minkowski geometry is recovered by setting the quadratic invariant to 0 and identifying $s$ as the proper time. So, both the relativistic and non-relativistic geometries are contained in this geometry as light cones. The ambient (4+1)-dimensional geometry, in the case $α = 0$ is called the Bargmann geometry. It is still (4+1)-dimensional, for all other values of $α$, the only difference being on the direction and nature of the invariant $ds$ within this geometry.

If you integrate the infinitesimal transforms to finite form, taking boosts in the $x$ direction, the resulting transforms are $$x → \frac{x - vt}{\sqrt{1 - α v^2}}, \hspace 1em y → y, \hspace 1em z → z, \hspace 1em t → \frac{t - αvx}{\sqrt{1 - α v^2}},\\ \hspace 1em u → u + \frac{v x}{\sqrt{1 - α v^2}} - \frac{v^2}{1 + \sqrt{1 - α v^2}} \frac{t}{\sqrt{1 - α v^2}}.$$ The coordinate $u = c^2 (s - t)$ is just the time-dilation, itself, ramped up by an order of $c^2$. Ramping it up gives it a non-zero and non-trivial non-relativistic limit, in contrast to $s - t$ which goes to zero. So, the extra coordinate is a non-relativistic vestige of time-dilation.

In the non-relativistic limit, this transform becomes: $$x → x - vt, \hspace 1em y → y, \hspace 1em z → z, \hspace 1em t → t,\hspace 1em u → u + v x - \frac{v^2}{2} t.$$

So, now do the translation and boosts. Translation first, then boost: $$x → x + a → x + a - vt, \hspace 1em y → y → y, \hspace 1em z → z → z, \hspace 1em t → t → t, \\ u → u → u + v (x + a) - \frac{v^2}{2} t.$$ Boosts first, then translation: $$x → x - vt → x - vt + a, \hspace 1em y → y → y, \hspace 1em z → z → z, \hspace 1em t → t → t, \\ u → u + v x - \frac{v^2}{2} t → u + v x - \frac{v^2}{2} t.$$ As you can see: they no longer commute. If you do the first, followed by the inverse of the second, the net result is a translation in $u$; and - as noted above - the mass invariant $μ$ has $u$ as its conjugate, so it corresponds to $u$ translation.

If you look carefully at Weinberg's treatment, you'll be able to spot where he slipped in vestiges of the $u$ coordinate. It sneaked into his treatment (at least in vestigial form), even beneath his own guard.

Here's a hint. The non-relativistic Schrödinger equation has a differential operator that is the operator version of $\hat{H} - \hat{P}^2/2m$. It should actually be written as $\hat{P}^2 - 2\hat{M}\hat{H}$, with an extra operator eigenvalue equation for $\hat{μ} = m$, if restricting attention to Tardyons. The solutions, when applying this to $ψ$, when expressed as a function of $u$, are given by $ψ(u) = e^{imu/ħ} ψ(0)$. If you assume that the operator $\hat{μ} = -iħ ∂/∂u$ indexes "superselection" sectors (as Weinberg did); i.e. that it commutes with all observables, then that actually expresses the $u$-independence of the observables: a kind of gauge symmetry.


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