Finite WKB Maxwell residual, null rays and polarization quotient

For any complex $C^2$ one-form $a$, write $\phi =e^{i\theta \mathrm{/}ϵ}$. By [F3], ${\mathrm{\nabla }}_c\phi =(i\mathrm{/}ϵ)k_c\phi$. Applying the product rule twice gives $\mathrm{\square }(\phi a_b)=\phi (\mathrm{\square }{a}_b+\frac{2i}{ϵ}{k}^c{\mathrm{\nabla }}_c{a}_b+\frac{i}{ϵ}({\mathrm{\nabla }}_c{k}^c)a_b-\frac{1}{{ϵ}^2}{g}^{-1}(k,k)a_b)\mathrm{.}$ Indeed, the two cross terms are each $(i\mathrm{/}ϵ)k^c{\mathrm{\nabla }}_c{a}_b$, differentiating $k_c$ contributes $(i\mathrm{/}ϵ)({\mathrm{\nabla }}_c{k}^c)a_b$, and differentiating the phase twice contributes $-{ϵ}^{-2}{k}^c{k}_c{a}_b$. The Ricci term in $P$ multiplies by the same scalar $\phi$. Therefore $P(\phi a)=\phi (Pa+\frac{i}{ϵ}La-\frac{1}{{ϵ}^2}{g}^{-1}(k,k)a)\mathrm{.}$

The first-order product rule gives ${\mathrm{\nabla }}^a(A_{ϵ}{)}_a=e^{i\theta \mathrm{/}ϵ}[i{k}^a(a_0{)}_a+ϵ(i{k}^a(a_1{)}_a+{\mathrm{\nabla }}^a(a_0{)}_a)+{ϵ}^2{d}_1]\mathrm{.}$ The first two brackets vanish by [A1], so ${\mathrm{\nabla }}^a(A_{ϵ}{)}_a={ϵ}^2{e}^{i\theta \mathrm{/}ϵ}{d}_1$.

The first-order product rule gives $d{A}_{ϵ}=e^{i\theta \mathrm{/}ϵ}[ik\wedge a_0+ϵ(d{a}_0+ik\wedge a_1)+{ϵ}^{2}d{a}_1],$ which is the stated exact field expansion under the $C^3$ coordinate definition of [F2].

In chart coordinates $y^a$, the vector field defined by metric raising has components $G^a(y):=g^{ab}(y){\mathrm{\partial }}_b\theta (y)$. Since $g$ is smooth and $\theta \in C^4$, $G$ is $C^3$ and is nowhere zero because $g$ is nondegenerate and $k$ is nowhere zero. At any chart point, continuity of each of the finitely many entries of $DG$ gives a neighborhood on which all entries are bounded by one finite $M$; shrink it to a convex coordinate box $C$ contained in the chart. For any $x,y\in C$, the segment $x+t(y-x)$ lies in $C$; the chain rule makes each component of $t\mapsto G(x+t(y-x))$ differentiable, and [F9] gives $\mathrm{\mid }{G}^a(y)-G^a(x)\mathrm{\mid }\le M{\sum }_b\mathrm{\mid }{y}^b-x^b\mathrm{\mid }$. Thus $G$ is locally Lipschitz at every chart point. The autonomous field $(\lambda ,y)\mapsto G(y)$ is continuous and locally Lipschitz in its state variable; applying [F6] gives a local $C^1$ integral curve, which is $C^2$ because $G$ is $C^1$. Torsion freedom and $k=d\theta$ give ${\mathrm{\nabla }}_a{k}_b={\mathrm{\nabla }}_b{k}_a$. Hence $k^a{\mathrm{\nabla }}_a{k}_b=k^a{\mathrm{\nabla }}_b{k}_a=\frac{1}{2}{\mathrm{\nabla }}_b(g^{-1}(k,k))=0.$ Along an integral curve this is ${\mathrm{\nabla }}_{\dot{\gamma }}\dot{\gamma }=0$, while $g(\dot{\gamma },\dot{\gamma })=g^{-1}(k,k)=0$. Its parameter is therefore affine and its tangent is null. Also $(\theta \circ \gamma {)}^{\mathrm{\prime }}=d\theta (k^{\mathrm{♯}})=g^{-1}(k,k)=0$, so $\theta$ is constant along it.

At a point, choose a real covector $v\in T_p^{\ast }M$ with $g^{-1}(k,v)=-1$ and put $\mathrm{\ell }=v+\frac{1}{2}{g}^{-1}(v,v)k$. Then $g^{-1}(\mathrm{\ell },\mathrm{\ell })=0$ and $g^{-1}(k,\mathrm{\ell })=-1$ by direct expansion. Thus ${\mathrm{span}}_{\mathbb{R}}\{k,\mathrm{\ell }\}$ has one positive and one negative direction. Since the full metric has signature $(-,+,+,+)$, its orthogonal complement $S$ is a positive two-plane. It follows that $k_p^{\perp }={\mathrm{span}}_{\mathbb{C}}\{k_p\}\oplus S_{\mathbb{C}}$, so the quotient has complex dimension two and the induced Hermitian form from $g_p^{-1}(a,\bar{b})$ is positive definite. Finally, for every $\beta \in \mathbb{C}$, $k_p\wedge (a_0(p)+\beta k_p)=k_p\wedge a_0(p)$; in a basis beginning with $k_p$, this wedge vanishes exactly when $a_0(p)$ is proportional to $k_p$.

Apply step 1.1 to $a=a_0+ϵa_1$ and multiply by $ϵ$. The $g^{-1}(k,k)$ term vanishes by [A1], the $L{a}_0$ term vanishes by [A1], and the remaining coefficient $ϵ(P{a}_0+iL{a}_1)$ vanishes by [A1]. Thus $P{A}_{ϵ}={ϵ}^2{e}^{i\theta \mathrm{/}ϵ}P{a}_1$.

Along a ray, $L{a}_0=0$ gives ${\mathrm{\nabla }}_{k^{\mathrm{♯}}}{a}_0=-({\mathrm{\nabla }}_c{k}^c)a_0\mathrm{/}2$. The integrand in $\mathcal{A}$ is continuous on each compact ray segment, so [F7] gives $k^{\mathrm{♯}}(\mathcal{A})=-({\mathrm{\nabla }}_c{k}^c)\mathcal{A}\mathrm{/}2$; the exponential never vanishes. The quotient rule now gives ${\mathrm{\nabla }}_{k^{\mathrm{♯}}}(a_0\mathrm{/}\mathcal{A})=0$, and [A1] gives $g^{-1}(k,a_0\mathrm{/}\mathcal{A})=0$.

Apply [F2] to $A_{ϵ}$ using steps 2.1 and 1.2. Differentiating ${ϵ}^2{e}^{i\theta \mathrm{/}ϵ}{d}_1$ gives $e^{i\theta \mathrm{/}ϵ}(iϵk_b{d}_1+{ϵ}^2{\mathrm{\nabla }}_b{d}_1)$. Subtraction therefore gives the stated exact residual. In a chart, expanding the defined cyclic derivative of $F_{ϵ}=d{A}_{ϵ}$ cancels the six mixed second partials in pairs by [F5], so $d{F}_{ϵ}=0$.

If $K=\mathrm{\varnothing }$, both defined norms are zero. Suppose $K\mathrm{\ne }\mathrm{\varnothing }$. For each finite component, continuity of the component and of its complex modulus makes its absolute-value function continuous; [F8] therefore supplies its finite maximum, and a maximum over the finite component set also exists. From the exact residual of step 3.1, the componentwise product bound $\mathrm{\mid }{k}_b{d}_1\mathrm{\mid }\le K_0{D}_0$, the triangle inequality [F4], and $\mathrm{\mid }{e}^{i\theta \mathrm{/}ϵ}\mathrm{\mid }=1$ give $\mathrm{\parallel }{\mathrm{\nabla }}^a(F_{ϵ}{)}_{ab}{\mathrm{\parallel }}_{\mathrm{\infty },K}\le ϵK_0{D}_0+{ϵ}^2{C}_0\mathrm{.}$ Step 1.3 gives the field bound in the same way, using $B_0,B_1$. For any $z=a+ib$, $\mathrm{\mid }a{\mathrm{\mid }}^2\le a^2+b^2=\mathrm{\mid }z{\mathrm{\mid }}^2$, so $\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }$; the real connection and exterior derivative commute with real part componentwise. Therefore the same estimates hold for the real potentials and fields.