The Classification of Kinematical Algebras

Contents

The generators and the hypotheses
Step one: the hypotheses reduce the algebra to five constants
Step two: the Jacobi identity eliminates two of the five
Step three: rescaling reduces the constants to their signs
Step four: the enumeration
Two corollaries the enumeration makes visible

This appendix proves Theorem 18.90 (Chapter 18): under isotropy, the two discrete automorphisms, and noncompactness of the boosts, there are exactly eleven kinematical Lie algebras, organized as displayed in Table 18.1. The result is Bacry and Lévy-Leblond's [Bacry:1968]; the derivation below is self-contained, and it is written out here because what it yields is not only a count but the parametrization that makes the whole family intelligible — three constants, two Jacobi constraints, and a lattice of limits that turns out to be exactly the contraction lattice.

The generators and the hypotheses

A kinematical algebra $\mathfrak{k}$ is a real ten-dimensional Lie algebra spanned by

\begin{equation}\tag{A10.1} J_{i}\ \text{(rotations)}\ec\quad K_{i}\ \text{(boosts)}\ec\quad P_{i}\ \text{(space translations)}\ec\quad H\ \text{(time translation)}\ec \end{equation}

with $i=1,2,3$, subject to the three hypotheses of Theorem 18.90. Written out, they are the following.

(H1) Isotropy.

The $J_{i}$ close into $\mathfrak{so}(3)$, under which $K_{i}$ and $P_{i}$ are vectors and $H$ is a scalar:

\begin{equation}\tag{A10.2} \comm{J_{i}}{J_{j}}=\epsilon_{ijk}J_{k}\ec\quad \comm{J_{i}}{K_{j}}=\epsilon_{ijk}K_{k}\ec\quad \comm{J_{i}}{P_{j}}=\epsilon_{ijk}P_{k}\ec\quad \comm{J_{i}}{H}=0\ep \end{equation}

(H2) Parity and time reversal are automorphisms.

The linear maps

\begin{align} \Pi&:\quad (J_{i},K_{i},P_{i},H)\longmapsto(J_{i},-K_{i},-P_{i},H)\ec \tag{A10.3}\\ \Theta&:\quad (J_{i},K_{i},P_{i},H)\longmapsto(J_{i},-K_{i},P_{i},-H) \tag{A10.4} \end{align}

are automorphisms of $\mathfrak{k}$. These are the algebra-level statements of $\vect{x}\mapsto-\vect{x}$ and $t\mapsto-t$: a boost velocity reverses under either, a position reverses only under the first, and the generator of translation in time reverses only under the second.

(H3) Noncompact boosts.

Each $K_{i}$ generates a one-parameter subgroup isomorphic to $\R$ and not to a circle. Only the sign of one structure constant will turn out to be at stake, and (H3) is used exactly twice, in the last two cases of Step four: the enumeration.

Step one: the hypotheses reduce the algebra to five constants

Lemma A10.1 (The general kinematical bracket).

Hypotheses (H1) and (H2) force the brackets not already fixed by Equation (A10.2) to take the form

\begin{align} \comm{K_{i}}{K_{j}}&=\alpha\,\epsilon_{ijk}J_{k}\ec& \comm{P_{i}}{P_{j}}&=\beta\,\epsilon_{ijk}J_{k}\ec \tag{A10.5}\\ \comm{K_{i}}{P_{j}}&=\gamma\,\delta_{ij}H\ec& \comm{K_{i}}{H}&=\mu\,P_{i}\ec \tag{A10.6}\\ \comm{P_{i}}{H}&=\nu\,K_{i} \tag{A10.7} \end{align}

for five real constants $\alpha,\beta,\gamma,\mu,\nu$. Rests on Equations (A10.1), (A10.2), (A10.3) and (A10.4).

Proof.

Derives Lemma A10.1. Each unknown bracket is an $\mathfrak{so}(3)$-covariant expression in the generators, so it may be expanded on the covariant objects available: the scalar $H$, the vectors $J_{k},K_{k},P_{k}$, and nothing else. Under (H2) each generator carries a definite pair of signs, and a bracket carries the product of the signs of its two entries:

	$J_{i}$	$K_{i}$	$P_{i}$	$H$
parity $\Pi$	$+$	$-$	$-$	$+$
time reversal $\Theta$	$+$	$-$	$+$	$-$

The four cases are then immediate.

The bracket $\comm{K_{i}}{K_{j}}$. It is antisymmetric in $ij$, hence a vector, hence of the form $\epsilon_{ijk}V_{k}$ with $V\in\set{J,K,P}$. Its $\Pi$-sign is $(-)(-)=+$, which excludes $K$ and $P$; only $J$ survives. Its $\Theta$-sign is $(-)(-)=+$, and $J$ is $\Theta$-even, so no further restriction arises. This gives the first of Equation (A10.5). The bracket $\comm{P_{i}}{P_{j}}$ is identical in structure, with $\Pi$-sign $+$ and $\Theta$-sign $(+)(+)=+$, giving the second.

The bracket $\comm{K_{i}}{P_{j}}$. As a rank-two tensor it decomposes into a trace, an antisymmetric part and a symmetric traceless part, of spins $0$, $1$ and $2$. There is no spin-$2$ generator in Equation (A10.1), so the symmetric traceless part vanishes identically. The trace part is $\gamma\,\delta_{ij}H$ and the antisymmetric part is $\delta\,\epsilon_{ijk}V_{k}$. The $\Pi$-sign is $(-)(-)=+$, which leaves $H$ and $J$; the $\Theta$-sign is $(-)(+)=-$, and $H$ is $\Theta$-odd while $J$ is $\Theta$-even, so the $J$ term is forbidden and $\delta=0$. This is the first of Equation (A10.6), and it is worth pausing on: the reason a kinematical algebra cannot have $\comm{K_{i}}{P_{j}}\propto\epsilon_{ijk}J_{k}$ is time reversal alone.

The brackets with $H$. $\comm{K_{i}}{H}$ is a vector of $\Pi$-sign $(-)(+)=-$, excluding $J$, and of $\Theta$-sign $(-)(-)=+$, excluding $K$ (which is $\Theta$-odd); only $P$ survives. $\comm{P_{i}}{H}$ is a vector of $\Pi$-sign $-$, again excluding $J$, and of $\Theta$-sign $(+)(-)=-$, excluding $P$; only $K$ survives. These are the second of Equation (A10.6) and Equation (A10.7).

∎

Step two: the Jacobi identity eliminates two of the five

Lemma A10.2 (The Jacobi constraints).

The brackets of Lemma A10.1 satisfy the Jacobi identity if and only if

\begin{equation}\tag{A10.8} \boxed{\alpha=-\gamma\mu\ec\qquad \beta=\gamma\nu}\ep \end{equation}

A kinematical algebra is therefore determined by the three constants $(\gamma,\mu,\nu)$ alone. Rests on Lemma A10.1 and Equation (A10.2).

Proof.

Derives Lemma A10.2. Every triple containing a $J$ is satisfied automatically, because Equations (A10.5), (A10.6) and (A10.7) were constructed to be $\mathfrak{so}(3)$-covariant. Of the remaining triples, those built from three copies of one vector generator are also automatic: for $(K_{i},K_{j},K_{l})$ the cyclic sum is

\begin{equation*} \alpha\left(\epsilon_{ijk}\epsilon_{klm}+\epsilon_{jlk}\epsilon_{kim} +\epsilon_{lik}\epsilon_{kjm}\right)K_{m}=0 \end{equation*}

by three applications of $\epsilon_{kab}\epsilon_{kcd}=\delta_{ac}\delta_{bd}-\delta_{ad}\delta_{bc}$, the six resulting Kronecker terms cancelling in pairs; the triple $(P_{i},P_{j},P_{l})$ is the same computation with $\alpha\to\beta$ and $K\to P$. Four independent triples remain.

The triple $(K_{i},K_{j},P_{l})$. Using $\comm{J_{k}}{P_{l}}=\epsilon_{klm}P_{m}$ and then the same $\epsilon$-identity,

\begin{align*} \comm{\comm{K_{i}}{K_{j}}}{P_{l}} &=\alpha\,\epsilon_{ijk}\epsilon_{klm}P_{m} =\alpha\left(\delta_{il}P_{j}-\delta_{jl}P_{i}\right)\ec\\ \comm{\comm{K_{j}}{P_{l}}}{K_{i}} &=\gamma\,\delta_{jl}\comm{H}{K_{i}}=-\gamma\mu\,\delta_{jl}P_{i}\ec\\ \comm{\comm{P_{l}}{K_{i}}}{K_{j}} &=-\gamma\,\delta_{il}\comm{H}{K_{j}}=+\gamma\mu\,\delta_{il}P_{j}\ep \end{align*}

The sum is $\left(\alpha+\gamma\mu\right)\left(\delta_{il}P_{j}-\delta_{jl}P_{i}\right)$, and the two tensors $\delta_{il}P_{j}$ and $\delta_{jl}P_{i}$ are linearly independent, so $\alpha=-\gamma\mu$.

The triple $(P_{i},P_{j},K_{l})$. Identically,

\begin{align*} \comm{\comm{P_{i}}{P_{j}}}{K_{l}} &=\beta\left(\delta_{il}K_{j}-\delta_{jl}K_{i}\right)\ec\\ \comm{\comm{P_{j}}{K_{l}}}{P_{i}} &=-\gamma\,\delta_{jl}\comm{H}{P_{i}}=+\gamma\nu\,\delta_{jl}K_{i}\ec\\ \comm{\comm{K_{l}}{P_{i}}}{P_{j}} &=\gamma\,\delta_{il}\comm{H}{P_{j}}=-\gamma\nu\,\delta_{il}K_{j}\ec \end{align*}

whose sum is $\left(\beta-\gamma\nu\right)\left(\delta_{il}K_{j}-\delta_{jl}K_{i}\right)$, so $\beta=\gamma\nu$.

The triples containing $H$. For $(K_{i},K_{j},H)$ the three terms are $\alpha\epsilon_{ijk}\comm{J_{k}}{H}=0$ by Equation (A10.2), together with $\mu\comm{P_{j}}{K_{i}}-\mu\comm{P_{i}}{K_{j}} =\mu\gamma\delta_{ij}H-\mu\gamma\delta_{ij}H=0$; the identity holds with no condition. The triple $(P_{i},P_{j},H)$ is the same with $\mu\to\nu$. For $(K_{i},P_{j},H)$,

\begin{equation*} \gamma\,\delta_{ij}\comm{H}{H}+\nu\comm{K_{j}}{K_{i}} -\mu\comm{P_{i}}{P_{j}} =-\left(\alpha\nu+\beta\mu\right)\epsilon_{ijk}J_{k}\ec \end{equation*}

so $\alpha\nu+\beta\mu=0$ — which is implied by Equation (A10.8), since $-\gamma\mu\nu+\gamma\nu\mu=0$. No further condition arises, and the two constraints of Equation (A10.8) are the whole content of the Jacobi identity.

∎

Step three: rescaling reduces the constants to their signs

Two kinematical algebras describe the same kinematics if they differ only by a choice of units for boosts, lengths and times. The corresponding transformations

\begin{equation}\tag{A10.9} K_{i}\longmapsto\lambda K_{i}\ec\qquad P_{i}\longmapsto\rho P_{i}\ec\qquad H\longmapsto\sigma H\ec\qquad \lambda\rho\sigma\neq0\ec \end{equation}

leave Equation (A10.2) and both automorphisms of (H2) intact, and they exhaust that freedom: $J$ is fixed by the normalization of Equation (A10.2).

Lemma A10.3 (Normal form).

Under Equation (A10.9) the three constants transform as

\begin{equation}\tag{A10.10} \gamma\longmapsto\frac{\lambda\rho}{\sigma}\,\gamma\ec\qquad \mu\longmapsto\frac{\lambda\sigma}{\rho}\,\mu\ec\qquad \nu\longmapsto\frac{\rho\sigma}{\lambda}\,\nu\ep \end{equation}

Consequently:

which of $\gamma,\mu,\nu$ vanish is invariant;
every nonvanishing constant can be scaled to $\pm1$, and all of them simultaneously; and
the signs of the surviving constants are determined only up to a simultaneous flip of all three, so that the invariants are the pairwise products $\sgn(\gamma\mu)$, $\sgn(\gamma\nu)$ and $\sgn(\mu\nu)$, of which two are independent.

Rests on Lemma A10.2 and Equation (A10.9).

Proof.

Derives Lemma A10.3. Equation (A10.10) follows by substituting Equation (A10.9) into Equations (A10.6) and (A10.7) and re-expressing the right-hand sides in the new generators; claim (1) is immediate from it, each constant being multiplied by a nonzero number. For (2), take logarithms of the absolute values: the exponents of $(\lambda,\rho,\sigma)$ form the matrix

\begin{equation*} \begin{pmatrix} 1 & 1 & -1\\ 1 & -1 & 1\\ -1 & 1 & 1 \end{pmatrix}\ec\qquad\text{of determinant }-4\neq0\ec \end{equation*}

so the system can be solved for $\log\abs{\lambda},\log\abs{\rho},\log\abs{\sigma}$ whatever shifts are required; when only some constants are nonzero the corresponding subsystem is solvable a fortiori, the matrix having full rank on any subset of its rows. For (3), write $s_{\lambda},s_{\rho},s_{\sigma}$ for the signs of the scale factors. Each of the three coefficients in Equation (A10.10) has the same sign, namely $s=s_{\lambda}s_{\rho}s_{\sigma}$, because each is a product of all three scale factors with one of them inverted and inversion does not change a sign. So all three constants are multiplied by one common $s\in\set{\pm1}$, which may be chosen freely. The pairwise products are unchanged by that flip, and their own product is $\left(\gamma\mu\nu\right)^{2}>0$, so only two of them are independent.

∎

Step four: the enumeration

By Lemma A10.3(1) the classification splits into the eight cases labelled by which of $(\gamma,\mu,\nu)$ vanish; within each case, Lemma A10.3(2) and (3) reduce the surviving constants to a sign pattern modulo one simultaneous flip. Writing $\ast$ for a nonzero constant, the count is as follows.

[$(0,0,0)$ — one algebra.] Every bracket among $K,P,H$ vanishes, and with them $\alpha$ and $\beta$. This is the static algebra.
[$(\ast,0,0)$ — one algebra.] $\alpha=\beta=0$, and the single surviving constant has a sign that the flip removes. Here $\comm{K_{i}}{P_{j}}=\delta_{ij}H$ with abelian boosts: the Carroll algebra.
[$(0,\ast,0)$ — one algebra.] Likewise one algebra, $\comm{K_{i}}{H}=P_{i}$ with every other new bracket zero: the Galilei algebra. Note that $\gamma=0$ here, so $\comm{K_{i}}{P_{j}}=0$ — the mass of Equation (29.15) is not in this algebra, which is exactly why it can only enter as a central extension.
[$(0,0,\ast)$ — one algebra.] The para-Galilei algebra, $\comm{P_{i}}{H}=K_{i}$.
[$(0,\ast,\ast)$ — two algebras.] $\alpha=\beta=0$ and the invariant is $\sgn(\mu\nu)$: the Newton–Hooke algebras $\mathrm{NH}_{+}$ and $\mathrm{NH}_{-}$, whose difference is exponential against oscillatory behaviour of the free motion.
[$(\ast,0,\ast)$ — two algebras.] $\alpha=0$ but $\beta=\gamma\nu\neq0$, and the invariant is $\sgn\beta$. The set $\set{J_{i},P_{i}}$ closes into $\mathfrak{so}(3,1)$ for $\beta<0$ and into $\mathfrak{so}(4)$ for $\beta>0$, while $\set{K_{i},H}$ is an abelian ideal in both cases. These are para-Poincaré and the inhomogeneous $\SO(4)$. The boosts are abelian here, so (H3) is satisfied and neither is excluded.
[$(\ast,\ast,0)$ — one algebra.] $\beta=0$ but $\alpha=-\gamma\mu\neq0$, and the invariant is $\sgn\alpha$; now $\set{J_{i},K_{i}}$ closes into $\mathfrak{so}(3,1)$ for $\alpha<0$ and into $\mathfrak{so}(4)$ for $\alpha>0$. In the second case the boosts generate a compact subgroup, so (H3) excludes it and only $\alpha<0$ survives: the Poincaré algebra $\mathfrak{t}(4)\,\bar{\oplus}\,\mathfrak{so}(3,1)$ of Equation (18.99).
[$(\ast,\ast,\ast)$ — two algebras.] Two independent sign invariants give four candidates; (H3) again forces $\alpha=-\gamma\mu<0$, halving them to two, distinguished by $\sgn(\gamma\nu)$. These are the two simple algebras of Equations (18.109) and (18.110), $\mathfrak{so}(4,1)$ and $\mathfrak{so}(3,2)$: de Sitter and anti-de Sitter.

Adding the cases gives

\begin{equation}\tag{A10.11} 1+1+1+1+2+2+1+2=11\ec \end{equation}

which proves Theorem 18.90. The assignment of the traditional names to the classes is Bacry and Lévy-Leblond's [Bacry:1968]; what is established above is the count, the parametrization, and the structure of each class.

Two corollaries the enumeration makes visible

Corollary A10.4 (The eight vanishing patterns are the contraction cube).

Each of the three constants can be sent to zero by an Inönü–Wigner contraction, in the sense of Definition 18.83, that leaves the other two fixed:

\begin{align} \nu\to0&:\quad P_{i}\mapsto\varepsilon P_{i}\ec\ H\mapsto\varepsilon H &&\text{(the flat-space limit)}\ec\tag{A10.12}\\ \gamma\to0&:\quad K_{i}\mapsto\varepsilon K_{i}\ec\ P_{i}\mapsto\varepsilon P_{i} &&\text{(the nonrelativistic limit }c\to\infty)\ec \tag{A10.13}\\ \mu\to0&:\quad K_{i}\mapsto\varepsilon K_{i}\ec\ H\mapsto\varepsilon H &&\text{(the ultrarelativistic limit }c\to0)\ec \tag{A10.14} \end{align}

in each case with $\varepsilon\to0$. The three contractions commute, so from an algebra with all three constants nonzero they generate exactly the eight vanishing patterns of Step four: the enumeration. The kinematical cube is therefore the lattice of subsets of $\set{\gamma,\mu,\nu}$: its eight vertices are the eight cases, not eight of the eleven algebras. Three of the eleven — one de Sitter, one Newton–Hooke, and the inhomogeneous $\SO(4)$ — are opposite-sign partners of a vertex rather than vertices of their own. Rests on Lemma A10.3 and Definition 18.83.

Proof.

Derives Corollary A10.4. Apply Equation (A10.10) to each substitution. For Equation (A10.12), $\lambda=1$ and $\rho=\sigma=\varepsilon$ give $\gamma\mapsto\gamma$, $\mu\mapsto\mu$, $\nu\mapsto\varepsilon^{2}\nu$. For Equation (A10.13), $\lambda=\rho=\varepsilon$ and $\sigma=1$ give $\gamma\mapsto\varepsilon^{2}\gamma$ with $\mu$ and $\nu$ fixed. For Equation (A10.14), $\lambda=\sigma=\varepsilon$ and $\rho=1$ give $\mu\mapsto\varepsilon^{2}\mu$ with $\gamma$ and $\nu$ fixed. In each case no structure constant diverges, so the limit exists and is a contraction. Commutativity is clear, the three substitutions acting as commuting diagonal matrices on $(\lambda,\rho,\sigma)$.

∎

Corollary A10.5 (The ``para'' algebras are exchanges, not new algebras).

The linear map $\Sigma$ exchanging boosts and space translations,

\begin{equation}\tag{A10.15} \Sigma:\quad K_{i}\longmapsto P_{i}\ec\qquad P_{i}\longmapsto K_{i}\ec\qquad J_{i}\longmapsto J_{i}\ec\qquad H\longmapsto H\ec \end{equation}

carries the algebra with constants $(\gamma,\mu,\nu)$ to the algebra with constants $(-\gamma,\nu,\mu)$, and so exchanges the cases $(\ast,\ast,0)$ and $(\ast,0,\ast)$, and the cases $(0,\ast,0)$ and $(0,0,\ast)$. It therefore identifies, as abstract Lie algebras,

\begin{equation}\tag{A10.16} \text{Poincar\'e}\cong\text{para-Poincar\'e}\ec\qquad \text{Galilei}\cong\text{para-Galilei}\ec \end{equation}

and identifies the inhomogeneous $\SO(4)$ with the compact-boost algebra that (H3) excluded from the case $(\ast,\ast,0)$. Rests on Lemmas A10.1 and A10.2.

Proof.

Derives Corollary A10.5. Under $\Sigma$ the bracket $\comm{K_{i}}{P_{j}}=\gamma\delta_{ij}H$ becomes $\comm{P_{i}}{K_{j}}=\gamma\delta_{ij}H$, that is $\comm{K_{j}}{P_{i}}=-\gamma\delta_{ij}H$, so $\gamma\mapsto-\gamma$; and $\comm{K_{i}}{H}=\mu P_{i}$ becomes $\comm{P_{i}}{H}=\mu K_{i}$, so $\nu\mapsto\mu$ and, symmetrically, $\mu\mapsto\nu$. The constrained constants follow from Equation (A10.8): $\alpha'=-\gamma'\mu'=\gamma\nu =\beta$ and $\beta'=\gamma'\nu'=-\gamma\mu=\alpha$, so $\Sigma$ exchanges the two quadratic constants as well. Applying this to Poincaré, which is $(\gamma,\mu,0)$ with $\alpha<0$, gives $(-\gamma,0,\mu)$ with $\beta'=\alpha<0$ — the case $(\ast,0,\ast)$ with negative $\beta$, which is para-Poincaré. The same substitution applied to $(0,\mu,0)$ gives $(0,0,\mu)$, and applied to the excluded $\alpha>0$ member of $(\ast,\ast,0)$ gives the $\beta>0$ member of $(\ast,0,\ast)$, the inhomogeneous $\SO(4)$.

∎

Remark A10.6 (Why the count is eleven, and what it is a count of).

Corollary A10.5 is the reason the eleven of Theorem 18.90 must be counted as eleven kinematics and not as eleven isomorphism classes of Lie algebras. The map $\Sigma$ is an isomorphism of Lie algebras, but it is not an equivalence of kinematics: it fails to commute with the time reversal Equation (A10.4), since it sends a $\Theta$-odd generator to a $\Theta$-even one, and it exchanges the physical roles of a boost and a translation. Two of the eleven are therefore abstractly the same algebra as another member of the list while describing a different physics, and Table 18.1 lists them separately for that reason. A statement of the theorem phrased as “eleven algebras up to isomorphism” is wrong, and the error is an easy one to make.

Appendix A10 discharges the proof obligation of Theorem 18.90. Its yield for the rest of the book is Equation (A10.8) and the three contractions of Corollary A10.4. The second of them, Equation (A10.13), is the $c\to\infty$ limit carrying the Poincaré algebra to the Galilei algebra, and the fact that it sets $\gamma=0$ — so that $\comm{K_{i}}{P_{j}}$ vanishes identically in the contracted algebra — is precisely what leaves room for the mass to reappear there as the central charge of Proposition 29.14.