The Second Cohomology of the Galilei Algebra

Contents

The algebra, and the two spaces
Three tensor lemmas
Solving the cocycle condition
The coboundaries, and the quotient

This appendix proves the assertion left owed in Example 18.80 of Lie Groups, Lie Algebras, and Fibre Bundles: the second Chevalley–Eilenberg cohomology $H^{2}(\mathfrak{g},\R)$ of the ten-generator Galilei algebra of three space dimensions is one-dimensional, so that up to a scale the mass extension is the only central extension the algebra admits.

Exactly one half of that statement is proved elsewhere in this treatise. Proposition 29.14, in the chapter bridging Poisson brackets and quantization, shows that the mass cocycle $c(K_{i},P_{j}) = m\,\delta_{ij}$ of Equation (29.15) is not a coboundary — that its class in $H^{2}$ is nonzero. It says nothing about how large $H^{2}$ is, and the chapter records the difference honestly. What is added here is the other half: that the class of the mass cocycle exhausts the cohomology. Nothing in Lie Groups, Lie Algebras, and Fibre Bundles or in Part III needs the stronger statement, so this appendix closes a gap in the mathematics rather than supplying a missing physical input.

The computation is finite and is carried out in full. The space of alternating bilinear forms on a ten-dimensional algebra has dimension $\binom{10}{2} = 45$; the cocycle condition Equation (18.121) cuts it down to $10$, the coboundaries account for $9$ of those, and the survivor is the mass.

The algebra, and the two spaces

Definition A38.1 (The Galilei algebra of three space dimensions).

Let $\mathfrak{g}$ be the real ten-dimensional Lie algebra spanned by rotations $J_{i}$, boosts $K_{i}$, space translations $P_{i}$ ($i = 1,2,3$) and the time translation $H$, with the isotropy relations Equation (A10.2),

\begin{equation}\tag{A38.1} \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\ec \end{equation}

and the single further nonvanishing bracket

\begin{equation}\tag{A38.2} \comm{K_{i}}{H} = P_{i}\ec\qquad \comm{K_{i}}{K_{j}} = \comm{P_{i}}{P_{j}} = \comm{K_{i}}{P_{j}} = \comm{P_{i}}{H} = 0\ep \end{equation}

This is the case $(\gamma,\mu,\nu) = (0,\ast,0)$ of the enumeration in Step four: the enumeration, with $\alpha = -\gamma\mu = 0$ and $\beta = \gamma\nu = 0$; the normalisation $\mu = 1$ is the one fixed there. All indices are Cartesian, repeated indices are summed, and $\epsilon_{ijk}$ is the Levi-Civita symbol. Rests on Equation (A10.2), Equation (A10.6) and Theorem 18.90.

Theorem A38.2 ($H^{2}$ of the Galilei algebra).

$\dim_{\R}H^{2}(\mathfrak{g},\R) = 1$, and the class is represented by the mass cocycle

\begin{equation}\tag{A38.3} c_{M}(K_{i},P_{j}) = \delta_{ij}\ec\qquad c_{M} = 0\ \text{on every other pair of basis generators}\ep \end{equation}

Every central extension of $\mathfrak{g}$ by $\R$ is therefore equivalent to the one with bracket $\comm{\tilde{K}_{i}}{\tilde{P}_{j}} = m\,\delta_{ij}Z$ for a single real number $m$, and it is trivial exactly when $m = 0$. Rests on Definition A38.1, Definition 18.75 and Proposition 18.76.

A cocycle is fixed by its values on pairs of basis generators. Grouping those values by type, an alternating bilinear $c : \mathfrak{g}\times\mathfrak{g} \rightarrow \R$ is exactly the datum

\begin{equation}\tag{A38.4} \begin{aligned} A_{ij} &= c(J_{i},J_{j})\ec & B_{ij} &= c(J_{i},K_{j})\ec & C_{ij} &= c(J_{i},P_{j})\ec & u_{i} &= c(J_{i},H)\ec\\ D_{ij} &= c(K_{i},K_{j})\ec & E_{ij} &= c(K_{i},P_{j})\ec & v_{i} &= c(K_{i},H)\ec & F_{ij} &= c(P_{i},P_{j})\ec\\ w_{i} &= c(P_{i},H)\ec \end{aligned} \end{equation}

with $A$, $D$, $F$ antisymmetric and $B$, $C$, $E$ unrestricted: $3+9+9+3+3+9+3+3+3 = 45$ real parameters, as it must be.

Three tensor lemmas

The cocycle conditions that involve one or two rotation generators are statements about $\SO(3)$-covariance of the arrays in Equation (A38.4), and all of them reduce to two identities. They are isolated here so that the enumeration below reads cleanly. Throughout, the only inputs are the standard contractions

\begin{equation}\tag{A38.5} \epsilon_{ijm}\epsilon_{ijl} = 2\delta_{ml}\ec\qquad \epsilon_{ijm}\epsilon_{ikl} = \delta_{jk}\delta_{ml} - \delta_{jl}\delta_{mk}\ec\qquad \epsilon_{ijl}\epsilon_{lkm} = \delta_{ik}\delta_{jm} - \delta_{im}\delta_{jk}\ep \end{equation}

Lemma A38.3 (Invariant two-tensors).

A real array $T_{jk}$ satisfies

\begin{equation}\tag{A38.6} \epsilon_{ijl}T_{lk} + \epsilon_{ikl}T_{jl} = 0 \qquad\text{for all } i,j,k \end{equation}

if and only if $T_{jk} = t\,\delta_{jk}$ for a real number $t$. Rests on Equation (A38.5).

Proof.

Derives Lemma A38.3. Necessity. Contract Equation (A38.6) with $\epsilon_{ijm}$ and sum over $i$ and $j$. By the first identity of Equation (A38.5) the first term gives $2T_{mk}$; by the second, the remaining term gives

\begin{equation*} \epsilon_{ijm}\epsilon_{ikl}T_{jl} = \left(\delta_{jk}\delta_{ml} - \delta_{jl}\delta_{mk}\right)T_{jl} = T_{km} - \delta_{mk}\,\tr T\ec \end{equation*}

where $\tr T = T_{jj}$. Hence

\begin{equation}\tag{A38.7} 2T_{mk} + T_{km} - \delta_{mk}\,\tr T = 0\ep \end{equation}

Exchanging $m$ and $k$ in Equation (A38.7) and subtracting gives $T_{mk} - T_{km} = 0$, so $T$ is symmetric; adding instead gives $3\left(T_{mk}+T_{km}\right) = 2\delta_{mk}\tr T$, that is $T_{mk} = \tfrac{1}{3}\left(\tr T\right)\delta_{mk}$.

Sufficiency. For $T = \delta$ the left side of Equation (A38.6) is $\epsilon_{ijk} + \epsilon_{ikj} = 0$.

∎

Lemma A38.4 (The mixed covariance condition).

A real array $T_{jk}$ satisfies

\begin{equation}\tag{A38.8} \epsilon_{ijl}T_{lk} + \epsilon_{ikl}T_{jl} - \epsilon_{jkl}T_{il} = 0 \qquad\text{for all } i,j,k \end{equation}

if and only if $T$ is antisymmetric, that is $T_{jk} = \epsilon_{jkm}t_{m}$ for a real vector $t$. Rests on Equation (A38.5).

Proof.

Derives Lemma A38.4. Necessity. Contract with $\epsilon_{ijm}$ as before. The first two terms give $2T_{mk}$ and $T_{km} - \delta_{mk}\tr T$ exactly as in Lemma A38.3. For the third, use $\epsilon_{ijm}\epsilon_{jkl} = -\epsilon_{jim}\epsilon_{jkl} = -\left(\delta_{ik}\delta_{ml}-\delta_{il}\delta_{mk}\right)$, so that

\begin{equation*} -\epsilon_{ijm}\epsilon_{jkl}T_{il} = \left(\delta_{ik}\delta_{ml}-\delta_{il}\delta_{mk}\right)T_{il} = T_{km} - \delta_{mk}\,\tr T\ep \end{equation*}

Summing the three,

\begin{equation}\tag{A38.9} 2T_{mk} + 2T_{km} - 2\delta_{mk}\,\tr T = 0\ec \qquad\text{i.e.}\qquad T_{mk} + T_{km} = \delta_{mk}\,\tr T\ep \end{equation}

Setting $m = k$ and summing gives $2\tr T = 3\tr T$, so $\tr T = 0$, and then Equation (A38.9) reads $T_{mk} = -T_{km}$.

Sufficiency. Put $T_{jk} = \epsilon_{jkm}t_{m}$ and evaluate the three terms of Equation (A38.8) with the third identity of Equation (A38.5), applied after cycling the symbols into the shape $\epsilon_{ab l}\epsilon_{lcd}$:

\begin{align*} \epsilon_{ijl}T_{lk} &= \epsilon_{ijl}\epsilon_{lkm}t_{m} = \delta_{ik}t_{j} - \delta_{jk}t_{i}\ec\\ \epsilon_{ikl}T_{jl} &= \epsilon_{ikl}\epsilon_{lmj}t_{m} = \delta_{im}\delta_{kj}t_{m} - \delta_{ij}\delta_{km}t_{m} = \delta_{kj}t_{i} - \delta_{ij}t_{k}\ec\\ \epsilon_{jkl}T_{il} &= \epsilon_{jkl}\epsilon_{lmi}t_{m} = \delta_{jm}\delta_{ki}t_{m} - \delta_{ji}\delta_{km}t_{m} = \delta_{ki}t_{j} - \delta_{ji}t_{k}\ep \end{align*}

The first two add to $\delta_{ik}t_{j} - \delta_{ij}t_{k}$, which is the third; the alternating sum vanishes.

∎

Remark A38.5 (Where the three space dimensions enter).

Lemmas A38.3 and A38.4 are the only steps of the computation that are not pure bookkeeping, and both are statements about the invariant two-tensors of the rotation group of three space dimensions: they use the Levi-Civita symbol with three indices and the contraction identities Equation (A38.5), which are three-dimensional. The theorem proved here is accordingly a theorem about the Galilei algebra of the observed three space dimensions, which is the only case this treatise instantiates; no claim is made, and none is needed, for any other number of space dimensions.

Solving the cocycle condition

Write $\delta c(x,y,z)$ for the left-hand side of Equation (18.121),

\begin{equation}\tag{A38.10} \delta c(x,y,z) = c\left(\comm{x}{y},z\right) + c\left(\comm{y}{z},x\right) + c\left(\comm{z}{x},y\right)\ec \end{equation}

which is alternating in $(x,y,z)$, so it suffices to impose $\delta c = 0$ on unordered triples of basis generators; a triple containing $H$ twice gives nothing, since there is only one $H$. Sixteen types remain, and every one is evaluated below.

Triples with no bracket. For $(K_{i},K_{j},K_{k})$, $(P_{i},P_{j},P_{k})$, $(K_{i},K_{j},P_{k})$, $(K_{i},P_{j},P_{k})$ and $(P_{i},P_{j},H)$ every bracket appearing in Equation (A38.10) vanishes by Equation (A38.2), so the condition is empty.

The triple $(J_{i},J_{j},J_{k})$. Writing $A_{ij} = \epsilon_{ijm}a_{m}$ and using the third identity of Equation (A38.5), $\epsilon_{ijl}A_{lk} = \delta_{ik}a_{j} - \delta_{jk}a_{i}$, so the cyclic sum is

\begin{equation*} \left(\delta_{ik}a_{j} - \delta_{jk}a_{i}\right) + \left(\delta_{ji}a_{k} - \delta_{ki}a_{j}\right) + \left(\delta_{kj}a_{i} - \delta_{ij}a_{k}\right) = 0 \end{equation*}

identically. The array $A$ is unconstrained, and it appears in no other triple: a term $c(\comm{\cdot}{\cdot},J)$ can involve $A$ only when the bracket produces a rotation, and only $\comm{J}{J}$ does.

The triple $(J_{i},J_{j},K_{k})$. Using Equation (A38.1) three times,

\begin{equation*} \delta c = \epsilon_{ijl}B_{lk} + \epsilon_{jkl}c(K_{l},J_{i}) - \epsilon_{ikl}c(K_{l},J_{j}) = \epsilon_{ijl}B_{lk} + \epsilon_{ikl}B_{jl} - \epsilon_{jkl}B_{il}\ec \end{equation*}

which is Equation (A38.8). By Lemma A38.4,

\begin{equation}\tag{A38.11} B_{jk} = \epsilon_{jkm}\beta_{m} \end{equation}

for an arbitrary vector $\beta$, and nothing more.

The triple $(J_{i},J_{j},P_{k})$. Identical in form, with $C$ in place of $B$: it forces $C$ to be antisymmetric and imposes nothing else. It will be subsumed by the stronger condition below.

The triple $(J_{i},J_{j},H)$. Since $\comm{J_{i}}{H} = \comm{H}{J_{j}} = 0$, only the first term survives: $\delta c = \epsilon_{ijl}u_{l} = 0$ for all $i,j$, hence

\begin{equation}\tag{A38.12} u_{i} = c(J_{i},H) = 0\ep \end{equation}

The triple $(J_{i},K_{j},K_{k})$. Here $\comm{K_{j}}{K_{k}} = 0$, so

\begin{equation*} \delta c = \epsilon_{ijl}D_{lk} - \epsilon_{ikl}D_{lj} = \epsilon_{ijl}D_{lk} + \epsilon_{ikl}D_{jl}\ec \end{equation*}

the second step using the antisymmetry of $D$. This is Equation (A38.6), so $D = t\delta$ by Lemma A38.3; but $D$ is antisymmetric and $\delta$ is not, so $t = 0$ and

\begin{equation}\tag{A38.13} D_{ij} = c(K_{i},K_{j}) = 0\ep \end{equation}

The triple $(J_{i},P_{j},P_{k})$. Word for word the same with $F$ in place of $D$, giving $F_{ij} = c(P_{i},P_{j}) = 0$.

The triple $(J_{i},K_{j},H)$. Now the bracket $\comm{K_{j}}{H} = P_{j}$ contributes:

\begin{equation*} \delta c = \epsilon_{ijl}c(K_{l},H) + c(P_{j},J_{i}) + c(0,K_{j}) = \epsilon_{ijl}v_{l} - C_{ij}\ec \end{equation*}

so that

\begin{equation}\tag{A38.14} C_{ij} = c(J_{i},P_{j}) = \epsilon_{ijl}v_{l}\ep \end{equation}

The array $C$ is thus determined by the vector $v$, and is automatically antisymmetric — consistent with, and stronger than, the $(J,J,P)$ condition.

The triple $(J_{i},P_{j},H)$. Both $\comm{P_{j}}{H}$ and $\comm{H}{J_{i}}$ vanish, leaving $\delta c = \epsilon_{ijl}w_{l} = 0$, hence

\begin{equation}\tag{A38.15} w_{i} = c(P_{i},H) = 0\ep \end{equation}

The triple $(J_{i},K_{j},P_{k})$. With $\comm{K_{j}}{P_{k}} = 0$,

\begin{equation*} \delta c = \epsilon_{ijl}E_{lk} - \epsilon_{ikl}c(P_{l},K_{j}) = \epsilon_{ijl}E_{lk} + \epsilon_{ikl}E_{jl}\ec \end{equation*}

which is Equation (A38.6) again. By Lemma A38.3,

\begin{equation}\tag{A38.16} E_{ij} = c(K_{i},P_{j}) = e\,\delta_{ij}\ec\qquad e \in \R\ep \end{equation}

The triple $(K_{i},K_{j},H)$. Only the brackets with $H$ survive:

\begin{equation*} \delta c = c(P_{j},K_{i}) + c(-P_{i},K_{j}) = -E_{ij} + E_{ji} = 0\ec \end{equation*}

which says $E$ is symmetric — already true of Equation (A38.16), so no new information.

The triple $(K_{i},P_{j},H)$. Here $\comm{K_{i}}{P_{j}} = 0$ and $\comm{P_{j}}{H} = 0$, and $\comm{H}{K_{i}} = -P_{i}$, so $\delta c = c(-P_{i},P_{j}) = -F_{ij} = 0$: again already known.

That exhausts the sixteen types. Collecting Equations (A38.11), (A38.12), (A38.13), (A38.14), (A38.15) and (A38.16) and the two vanishing statements for $F$:

Proposition A38.6 (The cocycles).

A bilinear form on $\mathfrak{g}$ is a $2$-cocycle in the sense of Proposition 18.73 if and only if

\begin{equation}\tag{A38.17} A_{ij} = \epsilon_{ijm}a_{m}\ec\quad B_{ij} = \epsilon_{ijm}\beta_{m}\ec\quad C_{ij} = \epsilon_{ijm}v_{m}\ec\quad E_{ij} = e\,\delta_{ij}\ec\quad c(K_{i},H) = v_{i}\ec \end{equation}

with $u = w = 0$ and $D = F = 0$, for arbitrary vectors $a,\beta,v \in \R^{3}$ and an arbitrary real number $e$. Hence $\dim_{\R}Z^{2}(\mathfrak{g},\R) = 3+3+3+1 = 10$. Rests on Definition A38.1, Lemma A38.3 and Lemma A38.4.

The coboundaries, and the quotient

Proposition A38.7 (The coboundaries).

$\dim_{\R}B^{2}(\mathfrak{g},\R) = 9$, and a coboundary is exactly a cocycle Equation (A38.17) with $e = 0$. Rests on Definition 18.74, Definition A38.1 and Proposition A38.6.

Proof.

Derives Proposition A38.7. Let $b : \mathfrak{g} \rightarrow \R$ be linear, with components $b(J_{i}) = \beta^{J}_{i}$, $b(K_{i}) = \beta^{K}_{i}$, $b(P_{i}) = \beta^{P}_{i}$ and $b(H) = \eta$. Evaluating $(\partial b)(x,y) = b\left(\comm{x}{y}\right)$ of Equation (18.122) on each pair of basis generators with Equations (A38.1) and (A38.2),

\begin{equation}\tag{A38.18} \begin{aligned} (\partial b)(J_{i},J_{j}) &= \epsilon_{ijk}\beta^{J}_{k}\ec & (\partial b)(J_{i},K_{j}) &= \epsilon_{ijk}\beta^{K}_{k}\ec & (\partial b)(J_{i},P_{j}) &= \epsilon_{ijk}\beta^{P}_{k}\ec\\ (\partial b)(K_{i},H) &= \beta^{P}_{i}\ec & (\partial b)(J_{i},H) &= 0\ec & (\partial b)(K_{i},P_{j}) &= 0\ec \end{aligned} \end{equation}

and $(\partial b)(K_{i},K_{j}) = (\partial b)(P_{i},P_{j}) = (\partial b)(P_{i},H) = 0$. In the coordinates of Equation (A38.17) this is

\begin{equation}\tag{A38.19} a = \beta^{J}\ec\qquad \beta = \beta^{K}\ec\qquad v = \beta^{P}\ec\qquad e = 0\ec \end{equation}

where the third entry is consistent in both places it occurs: $C_{ij} = \epsilon_{ijk}\beta^{P}_{k}$ and $c(K_{i},H) = \beta^{P}_{i}$ are exactly the pair required by Equation (A38.17). Two consequences follow at once. First, every coboundary is a cocycle with $e = 0$ — which also re-proves Proposition 18.73 for this algebra. Second, the map $b \mapsto \partial b$ carries $\left(\beta^{J},\beta^{K},\beta^{P}\right)$ onto the whole nine-dimensional subspace $\set{e = 0}$ of $Z^{2}$, bijectively, while $\eta$ is annihilated. Hence $\dim B^{2} = 9$.

The kernel is visible structurally as well: $\partial b = 0$ if and only if $b$ vanishes on the derived algebra $\comm{\mathfrak{g}}{\mathfrak{g}}$, which by Equations (A38.1) and (A38.2) is the span of the $J_{i}$, $K_{i}$ and $P_{i}$ — the $P_{i}$ arising both from $\comm{J}{P}$ and from $\comm{K}{H}$ — of dimension $9$. No bracket produces $H$, so $H$ is not in the derived algebra and the kernel is the line $\set{\beta^{J}=\beta^{K}=\beta^{P}=0}$, of dimension $10 - 9 = 1$.

∎

Proof of Theorem A38.2. Derives Theorem A38.2. By Propositions A38.6 and A38.7 the coboundaries are the hyperplane $\set{e = 0}$ inside the ten-dimensional space of cocycles, so

\begin{equation}\tag{A38.20} \dim_{\R}H^{2}(\mathfrak{g},\R) = \dim Z^{2} - \dim B^{2} = 10 - 9 = 1\ec \end{equation}

and the coordinate $e$ descends to an isomorphism $H^{2}(\mathfrak{g},\R) \cong \R$. The cocycle $c_{M}$ of Equation (A38.3) has $a = \beta = v = 0$ and $e = 1$: it satisfies Equation (A38.17), hence is a cocycle, and its class is a generator. The final statement is Proposition 18.76: central extensions of $\mathfrak{g}$ by $\R$ correspond to classes in $H^{2}$, so they form a one-parameter family $m\,[c_{M}]$, realized by $\comm{\tilde{K}_{i}}{\tilde{P}_{j}} = m\,\delta_{ij}Z$ through Equation (18.120), and by Equation (18.123) the extension is trivial precisely for the zero class, $m = 0$.

∎

Remark A38.8 (What was already known, and what is new).

Proposition 29.14 proves that the class of $c_{M}$ is nonzero, by exhibiting the Poisson-bracket realisation of a free particle and showing that no shift of the generators by constants removes the defect Equation (29.15). The present appendix reproves that as the single line $e \neq 0$ in Equation (A38.20) — coboundaries have $e = 0$, $c_{M}$ has $e = 1$ — and adds the converse, that there is nothing else: the whole of $H^{2}$ is the mass.

Two structural readings of the computation are worth recording, because they explain the answer rather than merely producing it. The first is that the surviving parameter is the one component of $c$ that no coboundary can reach, namely $c(K_{i},P_{j})$, and it is unreachable because $\comm{K_{i}}{P_{j}} = 0$ in $\mathfrak{g}$: a coboundary is a linear functional composed with the bracket, and there is no bracket there to compose with. That is exactly the argument of Proposition 29.14, and the enumeration above shows it is the only such place. The second is that the answer is forced to be at most one-dimensional by rotational covariance: Lemma A38.3 says the surviving array $c(K_{i},P_{j})$ has nowhere to live but the invariant $\delta_{ij}$, so the mass is a single number and not a tensor.

Example 18.80 may therefore be read without its final caveat, and Remark 18.88 — which describes how the mass emerges from the Inönü–Wigner limit of a trivially extended Poincaré algebra — is now accompanied by an independent statement of where the limit must land: $H^{2}$ of the contracted algebra is one-dimensional, so any nontrivial extension produced by any route is a multiple of $c_{M}$. This is a concrete instance of the phenomenon named in Remark 18.78: a contraction can create a cohomology class that the parent semisimple algebra, rigid by Proposition 18.77, did not have.