The mass-shell weighted transverse space matches the Lebesgue transverse space

The multiplier $Wf(\mathbf{k})=f(\mathbf{k})\mathrm{/}\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}$ is measurable on $M$, and $\mathrm{\parallel }Wf{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}^2={\int }_M\frac{\mathrm{\mid }f(\mathbf{k}){\mathrm{\mid }}^2}{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}{d}^3\mathbf{k}=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(d\nu )}^2\mathrm{.}$ The same identity for inner products shows that $W$ is an isometry with the second-variable-linear convention of [F2].

For any $u\in {\mathcal{H}}_T^0$, set $f(\mathbf{k})=\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}u(\mathbf{k})$. Then $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(d\nu )}^2={\int }_{M}2\mathrm{\mid }\mathbf{k}\mathrm{\mid }\mathrm{\mid }u(\mathbf{k}){\mathrm{\mid }}^2\frac{d^3\mathbf{k}}{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}^2,$ so $f\in L^2(d\nu ;{\mathbb{C}}^3)$, $Wf=u$, and $\mathbf{k}\cdot f(\mathbf{k})=\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}\mathbf{k}\cdot u(\mathbf{k})=0$ almost everywhere. Conversely, multiplication by the positive finite scalar $(2\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-1\mathrm{/}2}$ preserves the transverse condition. Lebesgue measure and $\nu$ have the same null sets on $M$: on every annulus $1\mathrm{/}m\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le m$ their density ratio is bounded above and below by positive constants, and these annuli cover $M$. Thus $W$ maps ${\mathcal{H}}_T^{\nu }$ onto ${\mathcal{H}}_T^0$. The isometry and surjectivity transfer completeness from the Hilbert space [F1] to ${\mathcal{H}}_T^{\nu }$, so $W$ is unitary.