Measurable scalarization of the transverse helicity fibers

Fix a Cartesian orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{C}}^3$ and $\sigma \in \{+1,-1\}$. Since $P_{\sigma }(\mathbf{n})$ has rank one, at least one of $P_{\sigma }(\mathbf{n})e_1,P_{\sigma }(\mathbf{n})e_2,P_{\sigma }(\mathbf{n})e_3$ is nonzero. Define $j_{\sigma }(\mathbf{n})$ to be the least such index. The matrix fields $P_{\sigma }$ are smooth, so each set $\{\mathbf{n}:j_{\sigma }(\mathbf{n})=j\}$ is Borel: it is the intersection of $\{\mathrm{\parallel }{P}_{\sigma }{e}_j\mathrm{\parallel }>0\}$ with the finitely many preceding zero sets. On that set define $e_{\sigma }(\mathbf{n})=P_{\sigma }(\mathbf{n})e_j\mathrm{/}\mathrm{\parallel }{P}_{\sigma }(\mathbf{n})e_j\mathrm{\parallel }$. The denominator is positive on its defining Borel set, so the piecewise formula is Borel, has unit norm, and lies in the range of $P_{\sigma }(\mathbf{n})$.

At each $\mathbf{n}$, the vectors $e_{+}(\mathbf{n}),e_{-}(\mathbf{n})$ are orthonormal because the projectors are orthogonal, and they span $P(\mathbf{n}){\mathbb{C}}^3$ because their ranges are the two rank-one summands of $P(\mathbf{n})$. For every transverse representative $f(\mathbf{k})$, the fiberwise expansion is $f(\mathbf{k})={\sum }_{\sigma =\pm 1}{e}_{\sigma }(\mathbf{n}(\mathbf{k}))(\overline{e_{\sigma }(\mathbf{n}(\mathbf{k}))}\cdot f(\mathbf{k}))$ and $\mathrm{\mid }f(\mathbf{k}){\mathrm{\mid }}^2={\sum }_{\sigma =\pm 1}\mathrm{\mid }(Tf)(\mathbf{k},\sigma ){\mathrm{\mid }}^2$. Borel measurability of $e_{\sigma }$ makes $Tf$ measurable, and integration gives $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{L^2(X)}=\mathrm{\parallel }f{\mathrm{\parallel }}_{{\mathcal{H}}_T^{\nu }}$. Conversely, for any $g\in L^2(X)$ the field $f(\mathbf{k})={\sum }_{\sigma =\pm 1}{e}_{\sigma }(\mathbf{n}(\mathbf{k}))g(\mathbf{k},\sigma )$ is measurable, transverse and square-integrable, and $Tf=g$. Hence $T$ is unitary.

On each fiber $h{e}_{\sigma }=\sigma e_{\sigma }$ and $P_{\tau }{e}_{\sigma }={\mathbf{1}}_{\{\tau =\sigma \}}{e}_{\sigma }$. Substituting the fiberwise expansion from step 2.1 gives $Thf(\mathbf{k},\sigma )=\sigma Tf(\mathbf{k},\sigma )$ and $T{P}_{\tau }f(\mathbf{k},\sigma )={\mathbf{1}}_{\{\tau =\sigma \}}Tf(\mathbf{k},\sigma )$ for every transverse field class. These are precisely the two multiplication-operator identities in the Statement.