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Galilean, SE(3), Poincare groups - Central Extension

+ 5 like - 0 dislike

After having learnt that the Galilean (with its central extension) with an unitary operator

$$ U = \sum_{i=1}^3\Big(\delta\theta_iL_i + \delta x_iP_i + \delta\lambda_iG_i +dtH\Big) + \delta\phi\mathbb{\hat1} = \sum_{i=1}^{10} \delta s_iK_i + \delta\phi\mathbb{\hat1} $$

This makes sure the commutation relations hold good in the group (especially for the boosts). However, in the case of Poincare group, the commutators still hold without a central extension. Similarly, is the case with SE(3) (there are no central extensions involved).

My question is why is there a necessity for central extensions in the first case, but not later ??

PS: This answer is somewhat related to the question, but am not able to get to the bottom of this thing.

This post imported from StackExchange Physics at 2014-03-24 09:16 (UCT), posted by SE-user user35952
asked Mar 20, 2014 in Theoretical Physics by user35952 (155 points) [ no revision ]
Oh ! So far from what I have seen to understand, the central extension of Poincare and SE(3) groups are trivial (the coefficient of identity in all commutators is zero) and the algebra is just a direct sum of the two algebras !!

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user user35952

2 Answers

+ 3 like - 0 dislike

The central extensions are classified by the second cohomology group: http://en.wikipedia.org/wiki/Group_extension . If this group is trivial then each central extension is semidirect (and hence in some sense trivial). In particular, this is the case for the Poincare group but not for the Galilei group.

However, if you want to take a nonrelativistic limit starting with the Poincare group, you need to introduce the nonrelativistic energy $E= cp_0 -mc^2$, which can be done only in a (trivial) 11-dimensional central extension of the Poincare group by a central generator, the mass $m$. In this form, the presentations of the Poincare group and the Galilei group can be made to look very similar.

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user Arnold Neumaier
answered Mar 21, 2014 by Arnold Neumaier (11,685 points) [ no revision ]
Thanks !! Now of one of my professors told me that the central extension is used here (in Ballentine's book), since the state vectors are used in projective representation. I am sorry, I am not getting the right statement, but this is something close to what he said !! Can you clarify on that part ??

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user user35952
@user35952: A projective representation of a group is the same thing as an ordinary representation of a corresponding central extension.

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user Arnold Neumaier
I am sorry, this is being cyclic. I neither know projective representations nor central extension with proper rigor. But, considering your statement, a question that comes up is why do we need central extension for states in QM (just because we have an additional symmetry of phase ?) ?

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user user35952
@user35952: Look the terms up in Wikipedia; they are defined rigorously, there is no cycle. - Wave fucntions $\psi$ are ambiguous; only the associated density matrix $\rho=\psi\psi^*$ contains physical information. Thus projective representations are the natural objects in QM. Expressed in terms of the symmetry group represented, it leads automatically to central extensions. In case these are nontrivial, they cannot be avoided.

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user Arnold Neumaier
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-24 09:17 (UCT), posted by SE-user David Bar Moshe
answered Mar 24, 2014 by David Bar Moshe (3,505 points) [ no revision ]

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