Exact helicity projectors in vacuum

Since $n\cdot n=1$, the formula $P_{T}v=v-n(n\cdot v)$ gives $n\cdot P_{T}v=0$ and $P_{T}v=v$ whenever $v\in T_k$; hence $P_T$ is idempotent with range $T_k$. Also $n\times n=0$, so $Cn=0$. Write $N_{n}v:=n\times v$. Its real matrix is skew-symmetric, so $N_n^T=-N_n$; with the Hermitian product of [F3], $(i{N}_n{)}^{\ast }=(-i)N_n^T=i{N}_n$, hence $C(k)$ is Hermitian. The vector triple product in [F2] gives $C(k{)}^{2}v=-n\times (n\times v)=v-n(n\cdot v)=P_T(k)v$. Also $P_T(k{)}^{\ast }=P_T(k)$ because $n$ is real, $P_T(k{)}^2=P_T(k)$, its range is $T_k$, and $C(k)n=0$.

The cross product with $n$ is perpendicular to $n$, so $P_{T}C=C$; also $C(n(n\cdot v))=0$, so $C{P}_T=C$. Thus $P_{T}C=C{P}_T=C$. Since $C^2=P_T$, direct multiplication gives $P_{\sigma }^2=\frac{1}{4}(P_T+2\sigma C+C^2)=\frac{1}{2}(P_T+\sigma C)=P_{\sigma },P_{+1}{P}_{-1}=\frac{1}{4}(P_T-C^2)=0,P_{+1}+P_{-1}=P_T\mathrm{.}$ Both are Hermitian by step 1.1. To verify their ranks, let $e_1,e_2$ be the first two coordinate unit vectors. They are linearly independent, so at least one, say $e_j$, is not parallel to $n$. Set $u:=(e_j-(e_j\cdot n)n)\mathrm{/}\mathrm{\mid }{e}_j-(e_j\cdot n)n\mathrm{\mid }$; its denominator is positive, and $u$ is a real unit vector perpendicular to $n$. Put $v:=n\times u$. The norm identity in [F2] gives $\mathrm{\mid }v\mathrm{\mid }=1$, and the determinant identity gives $u\cdot v=0$; the triple-product identity gives $n\times v=-u$. Thus $u,v$ span the real transverse plane. For $w_{+}:=(u+iv)\mathrm{/}\sqrt{2}$ and $w_{-}:=(u-iv)\mathrm{/}\sqrt{2}$, expansion gives $⟨w_{+},w_{+}{⟩}_H=⟨w_{-},w_{-}{⟩}_H=\frac{1}{2}(\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }v{\mathrm{\mid }}^2)=1$ and $⟨w_{+},w_{-}{⟩}_H=\frac{1}{2}(\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }v{\mathrm{\mid }}^2-2i(u\cdot v))=0$. The relations $n\times u=v$ and $n\times v=-u$ give $C{w}_{+}=w_{+}$ and $C{w}_{-}=-w_{-}$. Since some coordinate $n_j\mathrm{\ne }0$, the equation $n\cdot z=0$ uniquely determines $z_j$ from the other two complex coordinates, so $T_k$ has complex dimension two. Hence $w_{+},w_{-}$ span $T_k$, and the complementary orthogonal projectors $P_{+1},P_{-1}$ each have rank one there. The finite coordinate choice is only for this fixed $k$; no family of choices is made.

The coefficients of $P_T(k)$ and $C(k)$ are smooth functions of $k\mathrm{\ne }0$ because $n=k\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$ is smooth there. Step 2.1 shows that $P_{\sigma }(k)$ has range in $T_k$, so multiplication by it sends a smooth transverse amplitude supported in a compact set away from zero to another such amplitude. Since $U_t$ multiplies by the scalar $e^{-ic\mathrm{\mid }k\mathrm{\mid }t}$, $P_{\sigma }(k)e^{-ic\mathrm{\mid }k\mathrm{\mid }t}a(k)=e^{-ic\mathrm{\mid }k\mathrm{\mid }t}{P}_{\sigma }(k)a(k)$ pointwise, proving $M_{\sigma }{U}_t=U_t{M}_{\sigma }$.

For $R$ as in [A1], the map $R_{\ast }a$ is smooth and its support is the rotated compact set $R(K)$, which still avoids zero. Moreover $k\cdot Ra(R^{T}k)=(R^{T}k)\cdot a(R^{T}k)=0$ by $R^{T}R=I$, so $R_{\ast }a$ belongs to the amplitude class. Preservation of orientation gives $R(n\times z)=(Rn)\times (Rz)$: for every real $w$, $(Rn\times Rz)\cdot Rw=\det [Rn,Rz,Rw]=\det (R)\det [n,z,w]=(n\times z)\cdot w=(R(n\times z))\cdot Rw$, and nondegeneracy of the Euclidean inner product identifies the two vectors. Orthogonality gives $\mathrm{\mid }Rk\mathrm{\mid }=\mathrm{\mid }k\mathrm{\mid }$. Therefore $P_T(Rk)=R{P}_T(k)R^T$ and $C(Rk)=RC(k)R^T$, hence $P_{\sigma }(Rk)=R{P}_{\sigma }(k)R^T$. Substituting this identity into the definitions of $M_{\sigma }$ and $R_{\ast }$ gives $M_{\sigma }{R}_{\ast }a(k)=P_{\sigma }(k)Ra(R^{T}k)=R{P}_{\sigma }(R^{T}k)a(R^{T}k)=R_{\ast }{M}_{\sigma }a(k)$. This proves the proper-rotation commutation claim; the determinant condition is what preserves, rather than reverses, the helicity label.