Exact massless scalar on a conformally flat external metric

Put $u_a={\mathrm{\Omega }}^{-1}{\mathrm{\partial }}_a\mathrm{\Omega }$. Since ${\eta }_{ab}$ is constant, [F14] gives ${\mathrm{\Gamma }}^a{}_{bc}(\eta )=0$. Coordinate fields commute, so the curvature definition [F2] gives $R({\mathrm{\partial }}_c,{\mathrm{\partial }}_d){\mathrm{\partial }}_b=0$ for all indices; every curvature component and its Ricci and scalar contractions therefore vanish, in particular $S_{\eta }=0$. Thus for the base metric $\eta$, $\mathrm{\nabla }=\mathrm{\partial }$, $\theta ={\eta }^{ab}{\mathrm{\partial }}_a{u}_b$, and $s={\eta }^{ab}{u}_a{u}_b$ in [F3]. Applying that conformal formula gives $R_g=-6{\mathrm{\Omega }}^{-2}({\eta }^{ab}{\mathrm{\partial }}_a{u}_b+{\eta }^{ab}{u}_a{u}_b)$. Since ${\mathrm{\partial }}_a{u}_b={\mathrm{\Omega }}^{-1}{\mathrm{\partial }}_a{\mathrm{\partial }}_b\mathrm{\Omega }-{\mathrm{\Omega }}^{-2}({\mathrm{\partial }}_a\mathrm{\Omega })({\mathrm{\partial }}_b\mathrm{\Omega })$ and $u_a{u}_b={\mathrm{\Omega }}^{-2}({\mathrm{\partial }}_a\mathrm{\Omega })({\mathrm{\partial }}_b\mathrm{\Omega })$, the gradient-square terms cancel, giving $R_g=-6{\mathrm{\Omega }}^{-3}{\mathrm{\square }}_{\eta }\mathrm{\Omega }$.

Fix compact $K$ and let $f\in {\mathcal{D}}_K$. The product rule and boundedness on $K$ of each $x^{\alpha }{\mathrm{\partial }}^{\gamma }({\mathrm{\Omega }}^3)$ give, for each $\alpha ,\beta$, $p_{\alpha \beta }({\mathrm{\Omega }}^{3}f)\le C_{K,\alpha ,\beta }{p}_{\mathrm{\mid }\beta \mathrm{\mid }}(f)$; multiplication preserves support. Thus $f\mapsto {\mathrm{\Omega }}^{3}f$ is continuous ${\mathcal{D}}_K\to \mathcal{S}$ by [F11,F12]. For arbitrary $\psi \in \mathcal{S}$, [F7,F16] identify $K_0\psi$ as the restriction of ${\mathcal{F}}_{+}\psi$ to $p=(\mathrm{\mid }\mathbf{k}\mathrm{\mid },\mathbf{k})$, where ${\mathcal{F}}_{+}$ is the standard continuous Schwartz Fourier transform followed by a fixed invertible linear change of frequency variables. This pullback is continuous on Schwartz seminorms by the chain rule and fixed linear coordinate bounds. The weight $(1+\mathrm{\mid }p\mathrm{\mid }{)}^2$ is bounded by a finite sum of coordinate monomials of degree at most two, so $q_2(F):={\sup }_{p\in {\mathbb{R}}^4}(1+\mathrm{\mid }p\mathrm{\mid }{)}^2\mathrm{\mid }F(p)\mathrm{\mid }$ is controlled by finitely many Schwartz seminorms from [F11]. Continuity of the Fourier map [F8] therefore gives $q_2({\mathcal{F}}_{+}\psi )\le C{\sum }_{(\alpha ,\beta )\in E}{p}_{\alpha \beta }(\psi )$ for some finite set $E$ and finite $C$. For $r=\mathrm{\mid }\mathbf{k}\mathrm{\mid }$, the dispersion in [F17] and mass-shell measure in [F20] yield $\mathrm{\parallel }{K}_0\psi {\mathrm{\parallel }}_{H_0}^2\le \frac{q_2({\mathcal{F}}_{+}\psi {)}^2}{2(2\pi {)}^3}4\pi {\int }_0^{\mathrm{\infty }}r(1+\sqrt{2}r{)}^{-4}dr<\mathrm{\infty };$ the radial integrand is $O(r)$ at zero and $O(r^{-3})$ at infinity. This proves continuity $K_0:\mathcal{S}\to H_0$, so [F13] and the fixed-support multiplier estimates prove continuity $f\mapsto K_0({\mathrm{\Omega }}^{3}f)$ from $\mathcal{D}$ to $H_0$. For $h\in H_0$, Cauchy--Schwarz gives $\mathrm{\parallel }{C}_h\mathrm{\parallel }\le \mathrm{\parallel }h\mathrm{\parallel }$. [F9] bounds its tensor extension by the same norm, and by [F19], $P_{s,n+1}$ is the averaging operator $P_{+}$ of [F10], so it is orthogonal and has norm at most one. The product tensor pairing in [F15] gives $\mathrm{\parallel }h\otimes {\mathrm{\Psi }}_n\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }{\mathrm{\Psi }}_n\mathrm{\parallel }$. Its sector formulas [F6] therefore imply $\mathrm{\parallel }{a}_n(h){\mathrm{\Psi }}_n\mathrm{\parallel }\le \sqrt{n}\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }{\mathrm{\Psi }}_n\mathrm{\parallel }$ and $\mathrm{\parallel }{a}_n^{†}(h){\mathrm{\Psi }}_n\mathrm{\parallel }\le \sqrt{n+1}\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }{\mathrm{\Psi }}_n\mathrm{\parallel }$. By [F18], each $\mathrm{\Psi }\in {\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ has sectors only through some $N$; for either creation or annihilation, different input sectors give orthogonal output sectors. Summing squared sector norms thus bounds each full operator by its largest sector bound, and the triangle inequality between creation and annihilation gives $\mathrm{\parallel }{\mathrm{\Phi }}_0({\mathrm{\Omega }}^{3}f)\mathrm{\Psi }\mathrm{\parallel }\le (\sqrt{N}\mathrm{\parallel }{K}_0({\mathrm{\Omega }}^3\bar{f})\mathrm{\parallel }+\sqrt{N+1}\mathrm{\parallel }{K}_0({\mathrm{\Omega }}^{3}f)\mathrm{\parallel })\mathrm{\parallel }\mathrm{\Psi }\mathrm{\parallel }\mathrm{.}$ Both shell-vector terms depend continuously on $f$ (complex conjugation preserves the fixed-support seminorms). For each $\chi ,\mathrm{\Psi }\in {\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$, Cauchy--Schwarz now makes $f\mapsto ⟨\chi ,{\mathrm{\Phi }}_g(f)\mathrm{\Psi }{⟩}_2$ continuous; it is complex-linear by [F4] and the second-variable-linear pairing. Hence ${\mathrm{\Phi }}_g$ is the stated weak operator-valued distribution.

For every real compact test $f$, ${\mathrm{\Omega }}^{3}f$ is real, so [F4] makes ${\mathrm{\Phi }}_g(f)={\mathrm{\Phi }}_0({\mathrm{\Omega }}^{3}f)$ symmetric on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$. Each generator preserves ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$, so finite products and linear combinations do too; repeated use of symmetry gives $⟨\chi ,B\psi {⟩}_2=⟨B^{\ast }\chi ,\psi {⟩}_2$ for every polynomial $B$ and $\chi ,\psi \in {\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$. Define ${\omega }_g(B):=⟨{\varpi }_0,B{\varpi }_0{⟩}_2$. Then ${\omega }_g(B^{\ast }B)=⟨B{\varpi }_0,B{\varpi }_0{⟩}_2=\mathrm{\parallel }B{\varpi }_0{\mathrm{\parallel }}^2\ge 0$, and ${\omega }_g(1)=1$.

At every point $g(v,v)={\mathrm{\Omega }}^2\eta (v,v)$, so the future causal cones for $g$ and $\eta$ coincide and use the direction selected by ${\mathrm{\partial }}_0$. Define causal reachability by piecewise $C^1$ curves whose tangent is future-causal on each smooth piece, allowing constant curves; such a curve is admissible for $g$ exactly when it is admissible for $\eta$. In $(-+++)$ coordinates a future causal tangent satisfies ${\dot{x}}^0\ge \mathrm{\mid }\dot{\mathbf{x}}\mathrm{\mid }$; integrating along each piece and using the triangle inequality gives $\mathrm{\mid }\mathrm{\Delta }\mathbf{x}\mathrm{\mid }\le \mathrm{\Delta }{x}^0$ for a future causal curve. Conversely, if $\mathrm{\Delta }{x}^0\ge \mathrm{\mid }\mathrm{\Delta }\mathbf{x}\mathrm{\mid }$, the straight segment has future causal tangent (including the constant segment when the endpoints agree). Reversing endpoints handles the past direction. Thus two points are causally related exactly when their coordinate difference has nonpositive $\eta$-squared norm, for both $g$ and $\eta$; positive-$\eta$-squared pairs are spacelike for both.

Matrix inversion and $\det ({\mathrm{\Omega }}^2\eta )=-{\mathrm{\Omega }}^8$ give $g^{ab}={\mathrm{\Omega }}^{-2}{\eta }^{ab}$ and the density ${\mathrm{\Omega }}^4{d}^{4}x$. Thus ${\mathrm{\square }}_{g}h={\mathrm{\Omega }}^{-4}{\mathrm{\partial }}_a({\mathrm{\Omega }}^2{\mathrm{\partial }}^{a}h)$, where ${\mathrm{\partial }}^a={\eta }^{ab}{\mathrm{\partial }}_b$. For $h={\mathrm{\Omega }}^{-1}u$, the differentiated numerator is ${\mathrm{\partial }}_a(\mathrm{\Omega }{\mathrm{\partial }}^{a}u-u{\mathrm{\partial }}^a\mathrm{\Omega })=\mathrm{\Omega }{\mathrm{\square }}_{\eta }u-u{\mathrm{\square }}_{\eta }\mathrm{\Omega }$, because the two cross-gradient terms cancel. Hence ${\mathrm{\square }}_g({\mathrm{\Omega }}^{-1}u)={\mathrm{\Omega }}^{-3}{\mathrm{\square }}_{\eta }u-{\mathrm{\Omega }}^{-4}u{\mathrm{\square }}_{\eta }\mathrm{\Omega }$. By step 1.1, $-R_{g}h\mathrm{/}6=+{\mathrm{\Omega }}^{-4}u{\mathrm{\square }}_{\eta }\mathrm{\Omega }$, so the curvature term cancels the second term and $P_g({\mathrm{\Omega }}^{-1}u)={\mathrm{\Omega }}^{-3}{\mathrm{\square }}_{\eta }u$.

Write $f=f_R+i{f}_I$ and $h=h_R+i{h}_I$ with real compact tests. Their supports and the supports of ${\mathrm{\Omega }}^3{f}_R,{\mathrm{\Omega }}^3{f}_I,{\mathrm{\Omega }}^3{h}_R,{\mathrm{\Omega }}^3{h}_I$ lie in the original spacelike-separated support sets by step 1.2; step 1.4 confirms that these pairs are spacelike for $g$ as well as $\eta$. Bilinearity of the commutator and the real-test locality of [F5] on each of the four pairs therefore give $[{\mathrm{\Phi }}_g(f),{\mathrm{\Phi }}_g(h)]=0$ on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$.

For $F\in C_c^{\mathrm{\infty }}({\mathbb{R}}^8;\mathbb{C})$ define $\mathcal{G}F(p):={\int }_{{\mathbb{R}}^4}\text{\hspace{0.17em}⁣}{\int }_{{\mathbb{R}}^4}F(x,y)e^{-ip\circ x}{e}^{ip\circ y}{d}^{4}x{d}^{4}y\mathrm{.}$ This is the standard eight-dimensional Fourier transform of $F$ restricted to the fixed injective linear frequency map $p\mapsto ((p^0,-\mathbf{p}),(-p^0,\mathbf{p}))\mathrm{/}(2\pi )$, determined by the phase $p\circ x$; the chain rule and fixed coordinate bounds make this a continuous map $\mathcal{S}({\mathbb{R}}^8)\to \mathcal{S}({\mathbb{R}}^4)$ by [F8]. Let $q_4(G):={\sup }_p(1+\mathrm{\mid }p\mathrm{\mid }{)}^4\mathrm{\mid }G(p)\mathrm{\mid }$. The weight is bounded by finitely many degree-at-most-four coordinate monomials, so [F11] and Fourier continuity give $q_4(\mathcal{G}F)\le C{\sum }_{(\alpha ,\beta )\in E}{p}_{\alpha \beta }(F)$ for some finite set $E$ and constant $C$. For $p=(\mathrm{\mid }\mathbf{k}\mathrm{\mid },\mathbf{k})$, $\mathrm{\mid }p\mathrm{\mid }=\sqrt{2}\mathrm{\mid }\mathbf{k}\mathrm{\mid }$. Thus [F17,F20] imply ${\int }_{H_0^{+}}\mathrm{\mid }\mathcal{G}F(p)\mathrm{\mid }d{\nu }_0(p)\le \frac{q_4(\mathcal{G}F)}{2(2\pi {)}^3}4\pi {\int }_0^{\mathrm{\infty }}r(1+\sqrt{2}r{)}^{-4}dr<\mathrm{\infty };$ the radial integrand is $O(r)$ at zero and $O(r^{-3})$ at infinity. Define $W_0(F):={\int }_{H_0^{+}}\mathcal{G}F(p)d{\nu }_0(p)$. On each fixed compact support, every Schwartz seminorm of $F$ in the bound is controlled by a fixed-support derivative seminorm from [F12], since each coordinate weight is bounded there. The bound and the fixed-support criterion [F13] therefore make this a continuous distribution on ${\mathbb{R}}^8$. For $F(x,y)=f(x)h(y)$, its pairing is $W_0(f,h)={\int }_{H_0^{+}}\overline{K_0\stackrel{‾}{f}(p)}{K}_{0}h(p)d{\nu }_0(p)=⟨K_0\bar{f},K_{0}h{⟩}_2,$ which is the vacuum covariance in [F4]. Since $\mathrm{\Omega }$ is smooth and positive, define the curved-volume pairing of $W_g$ on $F$ by $W_g(F):=W_0({\mathrm{\Omega }}^3(x){\mathrm{\Omega }}^3(y)F(x,y))$. This is a distribution, and relative to $d{\mathrm{v}\mathrm{o}\mathrm{l}}_g(x)d{\mathrm{v}\mathrm{o}\mathrm{l}}_g(y)$ its scalar kernel is $\mathrm{\Omega }(x{)}^{-1}\mathrm{\Omega }(y{)}^{-1}{W}_0(x,y)$ because $d{\mathrm{v}\mathrm{o}\mathrm{l}}_g={\mathrm{\Omega }}^4{d}^{4}x$. On product tests it equals $W_0({\mathrm{\Omega }}^{3}f,{\mathrm{\Omega }}^{3}h)$, the covariance of ${\mathrm{\Phi }}_g$ by [F4]; step 1.2 confirms the pulled-back field matrix coefficients are continuous on the same test space.

For compact $f,h$, integrate inside a finite coordinate box containing both supports. Using ${\mathrm{\square }}_{g}h={\mathrm{\Omega }}^{-4}{\mathrm{\partial }}_a({\mathrm{\Omega }}^2{\mathrm{\partial }}^{a}h)$ and vanishing boundary terms gives $\int \bar{f}{\mathrm{\square }}_{g}hd{\mathrm{v}\mathrm{o}\mathrm{l}}_g=-\int {\mathrm{\Omega }}^2{\eta }^{ab}({\mathrm{\partial }}_a\bar{f})({\mathrm{\partial }}_{b}h)d^{4}x=\int \overline{{\mathrm{\square }}_{g}f}hd{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. Since $R_g$ is real by step 1.1, the multiplication term $R_g\mathrm{/}6$ is symmetric too, so $P_g$ is formally symmetric.

Substituting $u=\mathrm{\Omega }f$ in step 2.1 gives ${\mathrm{\Omega }}^3{P}_{g}f={\mathrm{\square }}_{\eta }(\mathrm{\Omega }f)$, with $\mathrm{\Omega }f\in \mathcal{D}$. For $\psi \in \mathcal{D}$ and a massless shell point $p=(\mathrm{\mid }\mathbf{k}\mathrm{\mid },\mathbf{k})$, two integrations by parts in the compact support give $K_0({\mathrm{\square }}_{\eta }\psi )(p)=\int \psi (x)(-{\mathrm{\partial }}_0^2+\mathrm{\Delta })e^{i(\mathrm{\mid }\mathbf{k}\mathrm{\mid }{x}^0-\mathbf{k}\cdot \mathbf{x})}{d}^{4}x=({\omega }_0(\mathbf{k}{)}^2-\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2)K_0\psi (p)=0,$ and the same holds for $\bar{\psi }$ because ${\mathrm{\square }}_{\eta }$ has real coefficients. The field formula [F4] therefore gives ${\mathrm{\Phi }}_0({\mathrm{\square }}_{\eta }\psi )=a(K_0{\mathrm{\square }}_{\eta }\bar{\psi })+a^{†}(K_0{\mathrm{\square }}_{\eta }\psi )=0$. Taking $\psi =\mathrm{\Omega }f$ proves ${\mathrm{\Phi }}_g(P_{g}f)=0$. The operator $P_g$ preserves compact smooth tests because its coefficients are smooth.