Lorenz auxiliary subsidiary null quotient

The map $Q:\mathcal{K}\to L^2({\mathbb{R}}^3)$ defined by $Qϵ=-{ϵ}^0+n\cdot \mathit{ϵ}$ is bounded because $\mathrm{\mid }Qϵ{\mathrm{\mid }}^2\le 2(\mathrm{\mid }{ϵ}^0{\mathrm{\mid }}^2+\mathrm{\mid }\mathit{ϵ}{\mathrm{\mid }}^2)$ pointwise. Thus ${\mathcal{K}}_0=\ker Q$ is closed. For $ϵ\in {\mathcal{K}}_0$, $n\cdot \mathit{ϵ}={ϵ}^0$, so $n\cdot Tϵ=0$ and $Tϵ\in {\mathcal{H}}_T$ by [F1]. Also $\mathrm{\parallel }Tϵ{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathit{ϵ}{\mathrm{\parallel }}_2+\mathrm{\parallel }{ϵ}^0{\mathrm{\parallel }}_2$, proving boundedness.

Write $\mathit{ϵ}=Tϵ+n{ϵ}^0$ and $\mathit{\delta }=T\delta +n{\delta }^0$. The cross terms in ${\mathit{ϵ}}^{†}\mathit{\delta }$ vanish because each $T$ vector is orthogonal to $n$, while the remaining $n$ term is $\overline{{ϵ}^0}{\delta }^0$ and cancels the time component $-\overline{{ϵ}^0}{\delta }^0$ of $\eta$. Hence $[ϵ,\delta {]}_{\eta }=⟨Tϵ,T\delta {⟩}_2$. Given any $u\in {\mathcal{H}}_T$, $ϵ=(0,u)$ belongs to ${\mathcal{K}}_0$ and $Tϵ=u$, so $T$ is onto.

The equation $Tϵ=0$ is equivalent to $\mathit{ϵ}=n{ϵ}^0$, so $ϵ=b(1,n)$ with $b={ϵ}^0\in L^2$; each such vector is in ${\mathcal{K}}_0$. Step 2.1 shows these vectors pair to zero with every member of ${\mathcal{K}}_0$. Conversely, if $ϵ$ pairs to zero with all of ${\mathcal{K}}_0$, choose $\delta =(0,Tϵ)$, which lies in ${\mathcal{K}}_0$ by [F1]; then $0=[ϵ,\delta {]}_{\eta }=\mathrm{\parallel }Tϵ{\mathrm{\parallel }}_2^2$, so $Tϵ=0$. Thus the radical is exactly $\mathcal{N}$.

The induced map ${\mathcal{K}}_0\mathrm{/}\mathcal{N}\to {\mathcal{H}}_T$, $[ϵ]\mapsto Tϵ$, is well-defined, injective by the kernel calculation, onto by step 2.1, and preserves the inner product by that same step. It is therefore an isometry onto the complete Hilbert space ${\mathcal{H}}_T$, proving positivity and completeness of the quotient.