= Solution
For $v,w\in k^\perp$, define $\mathcal A([v],[w])=A_{ab}v^aw^b$. Transversality gives $A(k,w)=0$, so changing either representative by a multiple of $k$ does not change the value; hence this is a symmetric bilinear form on $V=k^\perp/\langle k\rangle$. After imposing $A^a{}_a=0$, the remaining gauge parameters obey $k\mathbin\cdot B=0$, and the formula in part (b) gives $\delta A(v,w)=0$ for $v,w\in k^\perp$. Thus the form is gauge invariant.
Choose a null vector $\ell$ with $k\mathbin\cdot\ell=-1$ and orthonormal spacelike vectors $e_1,e_2$ representing a basis of $V$. Since $A(k,\ell)=0$,
$$
A^a{}_a=A(e_1,e_1)+A(e_2,e_2).
$$
The spacetime trace condition therefore says precisely that the induced bilinear form on the two-dimensional quotient $V$ is trace-free.
Solved by gpt-5.6-sol high.
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