Photon quotient Lorentz representation and positive dynamics

By [F5], $\mathrm{\Lambda }$ preserves $\eta$ and sends a future null vector $p$ to a future null vector $\mathrm{\Lambda }p$. It maps $p^{\perp }$ onto $(\mathrm{\Lambda }p{)}^{\perp }$ and $\mathbb{C}p$ onto $\mathbb{C}\mathrm{\Lambda }p$, so it induces a well-defined map on the quotient fiber. The induced form is preserved because $\eta (\mathrm{\Lambda }v,\mathrm{\Lambda }w)=\eta (v,w)$; hence the fiber map is an isometry. The map $T_p$ is independent of the representative: replacing $ϵ$ by $ϵ+cp$ changes its value by $c(\mathbf{k}-(\mathbf{k}\mathrm{/}\omega )\omega )=0$. For $ϵ\in p^{\perp }$, $\mathbf{k}\cdot T_p[ϵ]=0$; its pairing is the positive quotient pairing of [F1]. The lift $ϵ=(0,\sqrt{2\omega }u(\mathbf{k}))$ for transverse $u$ proves surjectivity, and $d\nu =d^3\mathbf{k}\mathrm{/}(2\omega )$ gives $\int \mathrm{\parallel }{T}_p[s(p)]{\mathrm{\parallel }}^{2}d\nu =\int \mathrm{\parallel }u(\mathbf{k}){\mathrm{\parallel }}^2{d}^3\mathbf{k}$. Thus the stated section identification is well-defined and unitary. Applying two Lorentz maps successively gives the representation composition law.

The component identification of [F3] uses $X={\mathbb{R}}^3\times \{1,2,3\}$ with Lebesgue measure times counting measure; $X$ is sigma-finite because the boxes $[-m,m{]}^3\times \{1,2,3\}$ have finite measure by [F8]. The finite measurable map $\mathit{\omega }(\mathbf{k},j)=(\mathrm{\mid }\mathbf{k}\mathrm{\mid },-\mathbf{k})$ lets us apply [F3]. It gives the PVM $\mathsf{E}(B)$ whose $n$-particle multiplier is ${\mathbf{1}}_{\{({\sum }_r\mathrm{\mid }{\mathbf{k}}_r\mathrm{\mid },-{\sum }_r{\mathbf{k}}_r)\in B\}}$, with vacuum multiplier ${\mathbf{1}}_B(0)$, and it gives the maximal componentwise multiplier domains. By [F7], $P_T(\mathbf{k})$ is pointwise self-adjoint and idempotent, so it commutes with each scalar multiplier; its tensor powers commute with the n-particle PVM and multiplier. The symmetric transverse sector is reducing, so restriction preserves the maximal domain, which for the energy is exactly $D(H_E)$ displayed in the Statement.

For the boost $B_n(\chi )$ in [F2], put $\beta =\tanh \chi$ and $\gamma =\cosh \chi$, so $\mathrm{\mid }\beta \mathrm{\mid }<1$ and $\gamma =(1-{\beta }^2{)}^{-1\mathrm{/}2}$. In Euclidean unit direction $n$ [F5], its spatial cone coordinates satisfy ${\mathbf{k}}^{\mathrm{\prime }}=\mathbf{k}+(\gamma -1)(n\cdot \mathbf{k})n+\gamma \beta \omega n,{\omega }^{\mathrm{\prime }}=\gamma (\omega +\beta n\cdot \mathbf{k})\mathrm{.}$ Taking $n$ as the third coordinate axis, the first two spatial coordinates are unchanged and $\frac{\mathrm{\partial }{k}_3^{\mathrm{\prime }}}{\mathrm{\partial }{k}_3}=\gamma (1+\beta \frac{k_3}{\omega })=\frac{{\omega }^{\mathrm{\prime }}}{\omega }\mathrm{.}$ Thus $\det (\mathrm{\partial }{\mathbf{k}}^{\mathrm{\prime }}\mathrm{/}\mathrm{\partial }\mathbf{k})={\omega }^{\mathrm{\prime }}\mathrm{/}\omega$. A rotation has determinant one and preserves $\omega$. By [F2], every proper orthochronous $\mathrm{\Lambda }$ is a product of these, so $d^3{\mathbf{k}}^{\mathrm{\prime }}\mathrm{/}(2{\omega }^{\mathrm{\prime }})=d^3\mathbf{k}\mathrm{/}(2\omega )$. Combining this change of variables with the fiber isometry in step 1.1 proves unitarity on the quotient-section space and verifies the stated identification with ${\mathcal{H}}_T$.

Each joint multiplier value $({\sum }_r\mathrm{\mid }{\mathbf{k}}_r\mathrm{\mid },-{\sum }_r{\mathbf{k}}_r)$ lies in ${\bar{V}}_{+}$ by the triangle inequality [F9], so the PVM vanishes outside that closed cone. Fix a nonzero ${\mathbf{k}}_0$ and the boundary point $(\mathrm{\mid }{\mathbf{k}}_0\mathrm{\mid },-{\mathbf{k}}_0)$. Choose $v\in {\mathbb{R}}^3$ not parallel to ${\mathbf{k}}_0$. By [F7], $P_T({\mathbf{k}}_0)v\mathrm{\ne }0$, and continuity gives a bounded open momentum neighborhood $V$ of ${\mathbf{k}}_0$, disjoint from zero, on which $\mathrm{\parallel }{P}_T(\mathbf{k})v\mathrm{\parallel }$ is bounded below by a positive constant. Given any open neighborhood $G$ of $(\mathrm{\mid }{\mathbf{k}}_0\mathrm{\mid },-{\mathbf{k}}_0)$, shrink $V$ so the map $\mathbf{k}\mapsto (\mathrm{\mid }\mathbf{k}\mathrm{\mid },-\mathbf{k})$ sends $V$ into $G$. Choose a ball inside $V$ and a positive finite-measure box $Q$ inside that ball by [F8]. The vector $u(\mathbf{k})={\mathbf{1}}_Q(\mathbf{k})P_T(\mathbf{k})v$ is a nonzero element of ${\mathcal{H}}_T$ supported over $G$, so $\mathsf{E}(G)u=u\mathrm{\ne }0$. The vacuum supplies the origin.

Matrix multiplication is continuous, and the Lorentz equation gives ${\mathrm{\Lambda }}^{-1}=\eta {\mathrm{\Lambda }}^T\eta$, so inversion is continuous in the stated matrix topology. For $v\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;{\mathbb{C}}^3)$ with support in $B_R$, $R\ge 1$, take the four-component representative $s_v(p)=(0,\sqrt{2\omega }{P}_T(\mathbf{k})v(\mathbf{k}))$; under the stated unitary identification its class corresponds to $u=P_{T}v\in {\mathcal{H}}_T$. Put $q={\mathrm{\Lambda }}^{-1}p$ and $a_{\mathrm{\Lambda }}(\mathbf{k})=T_p\mathrm{\Lambda }(0,\sqrt{2q^0}{P}_T(\mathbf{q})v(\mathbf{q}))$, the transverse chart representative of the transformed quotient section. If $\mathrm{\Lambda }$ is sufficiently close to $I$, uniformly for future null $p$ one has $\omega \mathrm{/}2\le q^0=\mathrm{\mid }\mathbf{q}\mathrm{\mid }\le 2\omega$ and $\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2$: use $\mathrm{\parallel }p{\mathrm{\parallel }}_2=\sqrt{2}\omega$, the displayed inverse formula and the Euclidean Cauchy–Schwarz bound for a matrix acting on a vector. Thus $a_{\mathrm{\Lambda }}$ vanishes outside $B_{2R}$, and $\mathrm{\parallel }{T}_p{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \sqrt{2}$, $\mathrm{\parallel }{P}_T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$ give $\mathrm{\mid }{a}_{\mathrm{\Lambda }}(\mathbf{k})\mathrm{\mid }\le CM\sqrt{\omega }$, where $M=\sup \mathrm{\mid }v\mathrm{\mid }$ and $C$ is independent of sufficiently small $\mathrm{\Lambda }-I$. Consequently $\mathrm{\mid }{a}_{\mathrm{\Lambda }}-a_I\mathrm{\mid }$ is bounded by a fixed constant $C_{R}M$ on the common support. Its squared $L^2(d\nu )$ contribution on $B_{\delta }$ is at most $C_R^2{M}^2\nu (B_{\delta })=C_R^2{M}^2\pi {\delta }^2$ by [F6, F10]. On the annulus $\delta \le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le 2R$, extend the explicit formula to matrices $A$ in a sufficiently small closed Euclidean matrix box about $I$ by putting $q=\eta A^T\eta p$. On that box $q^0$ and $\mathrm{\mid }\mathbf{q}\mathrm{\mid }$ remain positive, bounded away from zero on the annulus; the formula using $P_T(\mathbf{q})=I-(\mathbf{q}\mathrm{/}\mathrm{\mid }\mathbf{q}\mathrm{\mid })(\mathbf{q}\mathrm{/}\mathrm{\mid }\mathbf{q}\mathrm{\mid }{)}^T$ is therefore continuous. The box times annulus is closed bounded in ${\mathbb{R}}^{19}$, hence compact, and [F11] gives uniform continuity of this formula there. Restricting back to Lorentz matrices yields $a_{\mathrm{\Lambda }}\to a_I$ uniformly on the annulus, which has finite $\nu$-measure by [F6]. First choose $\delta$, then $\mathrm{\Lambda }-I$, to obtain $\mathrm{\parallel }{a}_{\mathrm{\Lambda }}-a_I{\mathrm{\parallel }}_{L^2(d\nu )}\to 0$; this is exactly the quotient-section norm. No fiber value at the omitted apex is used. These sections are dense by [F4, F7]: approximate any $u\in {\mathcal{H}}_T$ by smooth compactly supported $v$ and project by $P_T$. Unitarity from step 2.1 gives $\mathrm{\parallel }{U}_{\mathrm{\Lambda }}s-s\mathrm{\parallel }\le 2\mathrm{\parallel }s-s_v\mathrm{\parallel }+\mathrm{\parallel }{U}_{\mathrm{\Lambda }}{s}_v-s_v\mathrm{\parallel }$, extending continuity to every section. At any ${\mathrm{\Lambda }}_0$, the representation identity reduces continuity to $U_{{\mathrm{\Lambda }}_0^{-1}\mathrm{\Lambda }}\to I$, and ${\mathrm{\Lambda }}_0^{-1}\mathrm{\Lambda }\to I$ in the matrix topology. For an elementary $n$-tensor, telescoping and unitarity give $\mathrm{\parallel }(U_{\mathrm{\Lambda }}^{\otimes n}-I)(h_1\otimes \cdots \otimes h_n)\mathrm{\parallel }\le {\sum }_{j=1}^n\mathrm{\parallel }(U_{\mathrm{\Lambda }}-I)h_j\mathrm{\parallel }{\prod }_{l\mathrm{\ne }j}\mathrm{\parallel }{h}_l\mathrm{\parallel }\to 0$. Finite linear combinations are dense in the Hilbert tensor product; the symmetric projection is a contraction commuting with the tensor action, so continuity holds on each symmetric sector. The vacuum is fixed. Truncating an arbitrary Fock vector to finitely many sectors leaves an error bounded by twice its tail norm under these unitary maps; letting that tail tend to zero proves strong continuity on all Fock space.

Fix $(E,\mathbf{P})$ with $E>\mathrm{\mid }\mathbf{P}\mathrm{\mid }$. Choose a unit $n\perp \mathbf{P}$; if $\mathbf{P}\mathrm{\ne }0$, select a standard basis vector $w$ not parallel to $\mathbf{P}$ and normalize $w-(w\cdot \mathbf{P}\mathrm{/}\mathrm{\mid }\mathbf{P}{\mathrm{\mid }}^2)\mathbf{P}$, while if $\mathbf{P}=0$ take any standard basis vector. Put $q=\frac{1}{2}\sqrt{E^2-\mathrm{\mid }\mathbf{P}{\mathrm{\mid }}^2}$ and let ${\mathbf{k}}_1=-\mathbf{P}\mathrm{/}2+qn$, ${\mathbf{k}}_2=-\mathbf{P}\mathrm{/}2-qn$. Since $n\cdot \mathbf{P}=0$, $\mathrm{\mid }{\mathbf{k}}_i{\mathrm{\mid }}^2=\mathrm{\mid }\mathbf{P}{\mathrm{\mid }}^2\mathrm{/}4+q^2=E^2\mathrm{/}4$, so they are distinct nonzero momenta with $\mathrm{\mid }{\mathbf{k}}_1\mathrm{\mid }+\mathrm{\mid }{\mathbf{k}}_2\mathrm{\mid }=E$ and $-({\mathbf{k}}_1+{\mathbf{k}}_2)=\mathbf{P}$. Given any open neighborhood $G$ of $(E,\mathbf{P})$, continuity of the two-particle sum map gives disjoint bounded momentum neighborhoods $V_1,V_2$ of ${\mathbf{k}}_1,{\mathbf{k}}_2$ whose pairwise sums map into $G$. Choose fixed $v_i$ not parallel to ${\mathbf{k}}_i$ and shrink $V_i$ so $P_T(\mathbf{k})v_i$ never vanishes there by [F7]; choose balls inside $V_i$ and positive finite-measure boxes $Q_i$ inside those balls by [F8]. The vectors $u_i(\mathbf{k})={\mathbf{1}}_{Q_i}(\mathbf{k})P_T(\mathbf{k})v_i$ are nonzero in ${\mathcal{H}}_T$. The symmetric vector ${\mathrm{\Psi }}_2=P_{s,2}(u_1\otimes u_2)$ is nonzero because $Q_1$ and $Q_2$ are disjoint, and its two-particle multiplier values lie in $G$. Hence $\mathsf{E}(G){\mathrm{\Psi }}_2={\mathrm{\Psi }}_2\mathrm{\ne }0$. Thus every neighborhood of $(E,\mathbf{P})$ has nonzero two-particle spectral projection. Together with steps 1.2–2.2 this proves $\mathrm{supp}(\mathsf{E})={\bar{V}}_{+}$; [F3] gives the exact energy domain from step 1.2.