Poincare semidirect group, dual pairing and moment components

On $GL(4,\mathbb{R})$ define $\mathrm{\Phi }(L)=L^T\eta L$ with values in the ten-dimensional space of symmetric matrices. At $\mathrm{\Phi }(L)=\eta$, its derivative is $D{\mathrm{\Phi }}_L(\mathrm{\Delta })={\mathrm{\Delta }}^T\eta L+L^T\eta \mathrm{\Delta }$. For any symmetric $S$, set $\mathrm{\Delta }=L{\eta }^{-1}S\mathrm{/}2$; using $L^T\eta L=\eta$ and $S^T=S$, each summand equals $S\mathrm{/}2$. Thus $D{\mathrm{\Phi }}_L$ is onto.

Every point of ${\mathrm{\Phi }}^{-1}(\eta )$ is regular by step 1.1, and the level is nonempty because it contains $I$. Thus [F3] makes $O(1,3)={\mathrm{\Phi }}^{-1}(\eta )$ an embedded six-dimensional submanifold of $GL(4,\mathbb{R})$. For $L\in O(1,3)$, $L{e}_0$ is timelike and $L$ is a homeomorphism of the timelike set; [F4] implies that $L$ preserves the selected future component exactly when $L{e}_0$ lies in that component. Since $e_0$ is the positive time-axis vector, this is exactly $L^0{}_0>0$. Also $\mathrm{\mid }{L}^0{}_0\mathrm{\mid }\ge 1$, and taking determinants in $L^T\eta L=\eta$ gives $(\det L{)}^2=1$. Therefore $\det L=1$ and $L^0{}_0>0$ select an open subset of $O(1,3)$. By [F1] this subset is $S{O}^{+}(1,3)$. Matrix multiplication and inversion restrict smoothly from $GL(4)$, so this subgroup is a six-dimensional Lie group.

Give $G={\mathbb{R}}^4\times S{O}^{+}(1,3)$ the product manifold structure from step 2.1 and multiplication $(a,\mathrm{\Lambda })(b,L)=(a+\mathrm{\Lambda }b,\mathrm{\Lambda }L)$. Both parenthesizations of a triple product equal $(a+\mathrm{\Lambda }b+\mathrm{\Lambda }Lc,\mathrm{\Lambda }LM)$; the identity is $(0,I)$; and direct substitution shows $(-{\mathrm{\Lambda }}^{-1}a,{\mathrm{\Lambda }}^{-1})$ is a left and right inverse. Multiplication and inversion are smooth because matrix multiplication and inversion on $S{O}^{+}(1,3)$ are smooth and the translation operations are smooth. The affine matrix map $\rho (a,\mathrm{\Lambda })=\left(\begin{array}{cc}\mathrm{\Lambda }& a\\ 0& 1\end{array}\right)$ is injective, has smooth inverse onto its image by block projection, and is a group homomorphism because matrix multiplication gives $\rho (a+\mathrm{\Lambda }b,\mathrm{\Lambda }L)$. The product manifold tangent at identity is ${\mathbb{R}}^4\times T_{I}S{O}^{+}(1,3)$. By [F5], the latter tangent space is contained in the six-dimensional space of matrices satisfying $A^T\eta +\eta A=0$; both have dimension six by step 2.1 and [F5], so they are equal. Hence $d{\rho }_I(a,A)=\left(\begin{array}{cc}A& a\\ 0& 0\end{array}\right)$. A group homomorphism carries left-invariant vector fields to related left-invariant fields: differentiating $\rho \circ L_g=L_{\rho (g)}\circ \rho$ gives this relation. Related fields' brackets are related, since applying each bracket to a smooth test function and expanding $X(Yf)-Y(Xf)$ gives the same identity after composition with $\rho$. For matrices $M,N$, the left-invariant fields on $GL(5)$ are $g\mapsto gM$ and $g\mapsto gN$; differentiating these fields gives their commutator at the identity as $MN-NM$. Thus the bracket of two pair matrices is $\left(\begin{array}{cc}[A,B]& Ab-Ba\\ 0& 0\end{array}\right)$, yielding $[(a,A),(b,B)]=(Ab-Ba,[A,B])$.

The cotangent lift preserves the covector-tangent pairing: $p^{\mathrm{\prime }}(d{x}^{\mathrm{\prime }})=({\mathrm{\Lambda }}^{-T}p)(\mathrm{\Lambda }dx)=p(dx)$. For the one-parameter element $\exp (-s(a,A))$, differentiation of $x^{\mathrm{\prime }}={\mathrm{\Lambda }}_{s}x+a_s$ gives the fundamental field $(-(a+Ax),A^{T}p)$ on $(x,p)$; its negative is the positive cotangent-lifted field $(a+Ax,-A^{T}p)$. Pairing the covector with the positive base generator gives $\mathrm{\ell }=p_b(a^b+A^b{}_c{x}^c)$.

Antisymmetry of $A^{bc}$ gives $J_{bc}{A}^{bc}=-2p_b{A}^b{}_c{x}^c$: expand $J_{bc}=x_b{p}_c-x_c{p}_b$ and interchange $b,c$ in the first term. Substitution yields $\mathrm{\ell }=p_b{a}^b-\frac{1}{2}{J}_{bc}{A}^{bc}$. The definitions of $P,J$ then give $[P^a]=\mathrm{k}\mathrm{g}\mathrm{m}\mathrm{/}\mathrm{s}$ and $[J^{ab}]=\mathrm{J}\mathrm{s}$ from $[x]=\mathrm{m}$, $[p]=\mathrm{k}\mathrm{g}\mathrm{m}\mathrm{/}\mathrm{s}$, and $[c]=\mathrm{m}\mathrm{/}\mathrm{s}$.