Finite-dimensional Lorentz normalization cocycles split

In ${\mathfrak{l}}_{\mathbb{C}}$ set $L_i=(J_i+i{K}_i)\mathrm{/}2$ and $R_i=(J_i-i{K}_i)\mathrm{/}2$. Expanding the stated brackets gives $[L_i,L_j]={ϵ}_{ijk}{L}_k$, $[R_i,R_j]={ϵ}_{ijk}{R}_k$, and $[L_i,R_j]=0$. The real span $\mathfrak{k}={\mathrm{span}}_{\mathbb{R}}\{L_i,R_i\}$ has compact Lie-algebra bracket: with ${\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, ${\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$, ${\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, the basis $-i{\sigma }_i\mathrm{/}2$ of $\mathfrak{s}\mathfrak{u}(2)$ has the same ${ϵ}_{ijk}$ brackets. The two summands commute, and $\mathfrak{k}{\otimes }_{\mathbb{R}}\mathbb{C}={\mathfrak{l}}_{\mathbb{C}}$ because $J_i=L_i+R_i$ and $K_i=(L_i-R_i)\mathrm{/}i$.

Let $G=\mathrm{S}\mathrm{U}(2)\times \mathrm{S}\mathrm{U}(2)$, whose Lie algebra is $\mathfrak{k}$. The matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2=1$ parametrize $\mathrm{S}\mathrm{U}(2)=\{M\in M_2(\mathbb{C}):M^{\ast }M=I,\det M=1\}$ by $(\mathrm{Re}a,\mathrm{Im}a,\mathrm{Re}b,\mathrm{Im}b)\in S^3\subset {\mathbb{R}}^4$, so $G$ is compact. To prove simple connectivity in the form needed here, take any piecewise-smooth based loop in $S^3$. Uniform continuity gives a finite subdivision on which each subarc lies in an open hemisphere. If $\delta (t)$ is the short great-circle arc between its endpoints, then $H(s,t)=((1-s)\gamma (t)+s\delta (t))\mathrm{/}\mathrm{\parallel }(1-s)\gamma (t)+s\delta (t)\mathrm{\parallel }$ is defined throughout that hemisphere, fixes the endpoints, and homotopes the subarc to $\delta$. Each great-circle arc lies in the intersection of $S^3$ with a proper two-dimensional linear subspace $P_j\subset {\mathbb{R}}^4$. A finite union of proper linear subspaces cannot cover ${\mathbb{R}}^4$: induct on their number; after choosing $x$ outside $P_1,\dots ,P_{n-1}$, if $x\mathrm{\notin }{P}_n$ we are done, and otherwise choose $y\mathrm{\notin }{P}_n$; the line $x+ty$ meets each $P_j$ in at most one parameter value, so a real $t$ outside the finite forbidden set gives a point outside all $P_j$. Thus the resulting polygonal loop omits some $q\in S^3$. Stereographic projection identifies $S^3\setminus \{q\}$ with ${\mathbb{R}}^3$; straight-line contraction there to the basepoint fixes the loop endpoints and is piecewise smooth. Hence every piecewise-smooth based loop contracts through piecewise-smooth based loops. Applying this factorwise proves the required property for $G$.

Complexify $\rho$ and $c$ and, on $W=V\oplus \mathbb{C}$, define $\tilde{\rho }(X)(v,z)=(\rho (X)v+zc(X),0)$ for $X\in \mathfrak{k}$. The representation identity for $\rho$ and the cocycle identity for $c$ give $[\tilde{\rho }(X),\tilde{\rho }(Y)](v,z)=([\rho (X),\rho (Y)]v+z(\rho (X)c(Y)-\rho (Y)c(X)),0)=\tilde{\rho }([X,Y])(v,z)$, so $\tilde{\rho }$ is a finite-dimensional Lie-algebra representation.

Reparameterize each smooth path segment to $[0,1]$. For a piecewise-smooth path $\gamma$ put $a(t)=\gamma (t{)}^{-1}\dot{\gamma }(t)\in \mathfrak{k}$ and $A(t)=\tilde{\rho }(a(t))$. Solve $U^{\mathrm{\prime }}=UA$, $U(0)=I$, separately on each segment and concatenate. On a segment with $\mathrm{\parallel }A\mathrm{\parallel }\le M$, the Picard series has $n$th term ${\int }_{0\le t_1\le \cdots \le t_n\le t}A(t_1)\cdots A(t_n)d{t}_1\cdots d{t}_n$, bounded by $M^n\mathrm{/}n!$; it converges uniformly and termwise integration verifies the integral equation. If $D$ is the difference of two solutions with $D(0)=0$, then $\mathrm{\parallel }D(t)\mathrm{\parallel }\le M{\int }_0^t\mathrm{\parallel }D(r)\mathrm{\parallel }dr$; iterating gives $\mathrm{\parallel }D(t)\mathrm{\parallel }\le \sup \mathrm{\parallel }D\mathrm{\parallel }{M}^n{t}^n\mathrm{/}n!\to 0$, proving uniqueness. Solving $V^{\mathrm{\prime }}=-AV$, $V(0)=I$, by the same series gives $(UV{)}^{\mathrm{\prime }}=0$ and $UV=I$, so each transport matrix is invertible. Concatenating paths multiplies their transports in order, since on the second segment the solution with initial value $U_1$ is $U_1$ times the solution with initial value $I$. For a piecewise-smooth endpoint-fixed homotopy $\mathrm{\Gamma }(s,t)$, the series terms and their $s$-derivatives are uniformly bounded by $M^n\mathrm{/}n!$ and $nL{M}^{n-1}\mathrm{/}n!$, respectively, where $L=\sup \mathrm{\parallel }{\mathrm{\partial }}_{s}A\mathrm{\parallel }$ and we enlarge $M$ so $M\ge 1$. Both series converge uniformly; integrating the derivative series in $s$ shows the limit $U$ is differentiable in $s$. Put $a={\mathrm{\Gamma }}^{-1}{\mathrm{\partial }}_t\mathrm{\Gamma }$ and $b={\mathrm{\Gamma }}^{-1}{\mathrm{\partial }}_s\mathrm{\Gamma }$. Direct differentiation yields ${\mathrm{\partial }}_{s}a-{\mathrm{\partial }}_{t}b=[a,b]$. The matrix-valued form $\theta =g^{-1}dg$ satisfies $d\theta =-\theta \wedge \theta$, by differentiating $g^{-1}g=I$; equivalently $d\theta +\frac{1}{2}[\theta ,\theta ]=0$. Applying the representation from step 2.1 gives $A_s-B_t=[A,B]$, where $B=\tilde{\rho }(b)$. For $K=U_s-UB$, differentiation now gives $K_t=KA+U(BA+A_s-AB-B_t)=KA$, and $K(s,0)=0$ because the initial endpoint is fixed. The uniqueness just proved gives $K=0$; at the fixed terminal endpoint $B(s,1)=0$, so $U_s(s,1)=0$. Thus transport is invariant under these homotopies. If ${\gamma }_1,{\gamma }_2$ run from $e$ to the same $g$, the loop ${\gamma }_1$ followed by the reverse of ${\gamma }_2$ contracts by step 1.2, so its transport is $I$; the analogous loop using ${\gamma }_2$ shows the two transports agree. Hence transport is path-independent. Locally near $g_0$, choose paths from $e$ to $g_0^{-1}g$ varying smoothly with $g$ in a coordinate ball; the same uniform-series bounds imply continuous dependence on $g$, so $U$ is continuous. Finally, left-translating a path from $e$ to $h$ and concatenating it after a path from $e$ to $g$ gives $U(gh)=U(g)U(h)$, since left translation preserves ${\gamma }^{-1}\dot{\gamma }$. Thus $U$ is a continuous group representation.

The block form of $\tilde{\rho }$ makes $U(g)=\left(\begin{array}{cc}R(g)& b(g)\\ 0& 1\end{array}\right)$, and its group law gives $b(hg)=R(h)b(g)+b(h)$. Use the matrix parametrization of $S^3$ from step 1.2. For an explicit normalized round-volume integral, let $\chi (t)=0$ for $t\le 1\mathrm{/}3$ and $\chi (t)=e^{-1\mathrm{/}(t-1\mathrm{/}3)}$ for $t>1\mathrm{/}3$. For $x=(x_1,\dots ,x_4)\in S^3$, put $D(x)={\sum }_{j=1}^4(\chi (x_j)+\chi (-x_j))$ and ${\psi }_{j,\pm }(x)=\chi (\pm x_j)\mathrm{/}D(x)$. Since some $\mathrm{\mid }{x}_j\mathrm{\mid }\ge 1\mathrm{/}2$, $D(x)>0$; these eight smooth nonnegative functions sum to $1$ and have supports compactly contained in $U_{j,\pm }=\{\pm x_j>1\mathrm{/}4\}$. In the chart given by the other three coordinates $y$, the sphere is $x_j=\pm \sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$, the metric matrix is $G=I+y{y}^{\mathsf{T}}\mathrm{/}(1-\mathrm{\mid }y{\mathrm{\mid }}^2)$, and therefore $\sqrt{\det G}=(1-\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$. By [L1], integrate each complex chart integrand and each finite-dimensional $V$-valued function coordinatewise in any basis. Define $I(f)={\sum }_{j,\pm }{\int }_{{\mathbb{R}}^3}{\psi }_{j,\pm }(x(y))f(x(y))(1-\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy,$ extending each integrand by zero outside its chart. Each real and imaginary part is continuous with compact support inside the chart, so [F1]–[F2] make each coordinate integral finite. The chart weights are nonnegative, and their partition-of-unity sum gives $I(1)$; applying the triangle inequality to the complex Riemann sums and passing to their limits gives $\mathrm{\mid }I(h)\mathrm{\mid }\le {\sup }_{S^3}\mathrm{\mid }h\mathrm{\mid }I(1)$. The number $I(1)$ is finite and positive: at $x=(1,0,0,0)$ one partition function is positive on a neighborhood, and on a small coordinate cube there the integrand has a positive lower bound. Thus $\nu (h):=I(h)\mathrm{/}I(1)$ is complex-linear, normalized, positive on nonnegative real functions, and satisfies $\mathrm{\mid }\nu (h)\mathrm{\mid }\le {\sup }_{S^3}\mathrm{\mid }h\mathrm{\mid }$. Its chart definition is independent of a finite subordinate chart partition: insert the second partition into each summand; each product has compact support in the chart overlap, and the metric identity $G(x)=DT(x{)}^{\mathsf{T}}G(Tx)DT(x)$ gives $\sqrt{\det G(x)}=\mathrm{\mid }\det DT(x)\mathrm{\mid }\sqrt{\det G(Tx)}$. The real and imaginary chart integrands, extended by zero, are continuous and compactly supported in the overlap image, and the transition is a smooth diffeomorphism of open subsets of ${\mathbb{R}}^3$. Apply the Riemann formula [F3] to these parts separately to transfer each compactly supported integral; summing the finite terms gives equality. For $q=(a,b)\in S^3$ and $x=(c,d)\in {\mathbb{C}}^2\cong {\mathbb{R}}^4$, matrix left multiplication induces $L_q(c,d)=(ac-b\bar{d},ad+b\bar{c})$. This map is real-linear and $\mathrm{\mid }ac-b\bar{d}{\mathrm{\mid }}^2+\mathrm{\mid }ad+b\bar{c}{\mathrm{\mid }}^2=(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2)(\mathrm{\mid }c{\mathrm{\mid }}^2+\mathrm{\mid }d{\mathrm{\mid }}^2)=\mathrm{\mid }c{\mathrm{\mid }}^2+\mathrm{\mid }d{\mathrm{\mid }}^2$, since the two cross terms cancel. Its inverse is left multiplication by the inverse matrix, so $L_q$ is orthogonal and an isometry of the induced metric; using the transported charts and partition in the same identity gives $\nu (h\circ L_q)=\nu (h)$. Extend $\nu$ to continuous $V$-valued maps by the same coordinatewise rule and call this extension ${\nu }_V$. Define the product functional on continuous $F:G\to V$ by $\mu (F)={\nu }_V(q\mapsto {\nu }_V(r\mapsto F(q,r)))$. For continuity, fix $q_0\in S^3$ and $\epsilon >0$. For each $r\in S^3$, continuity of the finitely many coordinate functions of $F$ at $(q_0,r)$ gives neighborhoods in both factors on which their values differ from $F(q_0,r)$ by less than $\epsilon \mathrm{/}3$. Compactness of the second $S^3$ selects finitely many such neighborhoods; intersecting their first-factor neighborhoods gives $U\ni q_0$ with ${\sup }_r\mathrm{\mid }{F}_j(q,r)-F_j(q_0,r)\mathrm{\mid }<2\epsilon \mathrm{/}3$ for each coordinate $j$ and $q\in U$. The scalar bound for $\nu$ therefore makes each coordinate of $q\mapsto {\nu }_V(r\mapsto F(q,r))$ continuous. The product functional is linear, normalized, and left invariant by applying the factor identity twice. Since $b$ is continuous on compact $G$ and $V$ is finite-dimensional, $\bar{v}:=\mu (b)$ exists coordinatewise. Left invariance and the cocycle formula give $R(h)\bar{v}+b(h)=\mu (g\mapsto b(hg))=\bar{v}$ for every $h\in G$.

Differentiate the fixed-point identity of step 4.1 at $h=\exp (tX)$ and $t=0$. The path equation defining $U$ gives $d{R}_e(X)=\rho (X)$ and $d{b}_e(X)=c(X)$: these are the upper-left and upper-right blocks of $\tilde{\rho }(X)$ in step 2.1. Hence $\rho (X)\bar{v}+c(X)=0$ for $X\in \mathfrak{k}$. Set $v=-\bar{v}$. The identity $c(X)=\rho (X)v$ extends complex-linearly from $\mathfrak{k}$ to ${\mathfrak{l}}_{\mathbb{C}}$ by step 1.1, and hence holds for every original $X\in \mathfrak{l}$.