Lemma: Pure-trace symbol on null pairs

For null $k_1,k_2$,

and it satisfies the Bianchi relation $(k_1+k_2{)}_{\mu }{\mathsf{q}}^{\mu \nu }=-(k_1\cdot k_2)(k_1+k_2{)}^{\nu }$. For the counter-propagating pair ${k_1}^{\mu }=(1,\mathbf{n})$, ${k_2}^{\mu }=(1,-\mathbf{n})$, $|\mathbf{n}|=1$, one has $k_1+k_2=2\mathrm{d}t$ as covectors, $k_1\cdot k_2=2$, and

Proof. Insert ${\psi }_a=e^{i{k}_a\cdot y}$ into (eq:J3trace): ${\partial }^{\mu }{\psi }_1{\partial }^{\nu }{\psi }_2=-{k_1}^{\mu }{k_2}^{\nu }{e}^{i(k_1+k_2)\cdot y}$, ${\psi }_1{\partial }^{\mu }{\partial }^{\nu }{\psi }_2=-{k_2}^{\mu }{k_2}^{\nu }{e}^{i(k_1+k_2)\cdot y}$, $\square {\psi }_a=0$. This is (eq:qtrace). Contracting with $(k_1+k_2{)}_{\mu }$ and using ${k_a}^2=0$, so that $(k_1+k_2)\cdot k_a=k_1\cdot k_2=:\sigma$, gives $-\frac{3}{2}\sigma (k_1+k_2{)}^{\nu }-\sigma (k_1+k_2{)}^{\nu }+\frac{3}{2}\sigma (k_1+k_2{)}^{\nu }=-\sigma (k_1+k_2{)}^{\nu }$. Independently, (eq:bianchi) polarized reads ${\partial }_{\mu }{J_3}^{\mu \nu }({\gamma }_1,{\gamma }_2)=-\frac{1}{2}[{\Gamma }_1{}_{\mu \rho }^{\nu }[{\gamma }_1]A^{\mu \rho }{\gamma }_2+(1\leftrightarrow 2)]$; for ${\gamma }_a={\psi }_a\eta$ with $\square {\psi }_a=0$, $A{\gamma }_a=-2\partial \partial {\psi }_a$ and ${\Gamma }_1{}_{\mu \rho }^{\nu }[{\psi }_1]A^{\mu \rho }{\gamma }_2=-2{\partial }_{\mu }{\psi }_1{\partial }^{\mu }{\partial }^{\nu }{\psi }_2$, which on plane waves is $2i\sigma {k_2}^{\nu }{e}^{i(k_1+k_2)\cdot y}$; the right side becomes $-i\sigma (k_1+k_2{)}^{\nu }{e}^{i(k_1+k_2)\cdot y}$ and the left side $i(k_1+k_2{)}_{\mu }{\mathsf{q}}^{\mu \nu }{e}^{i(k_1+k_2)\cdot y}$. The two computations agree. For the counter-propagating pair, ${k_1}^0{k_2}^i+{k_2}^0{k_1}^i=0$, ${k_1}^0{k_1}^i+{k_2}^0{k_2}^i=0$, ${k_1}^i{k}_2^j+{k_2}^i{k}_1^j=-2n^i{n}^j$, ${k_1}^i{k}_1^j+{k_2}^i{k}_2^j=2n^i{n}^j$, and (eq:qcounter) follows.