 Solution

For $v,w\in k^{\perp }$, define $\mathcal{A}([v],[w])=A_{ab}{v}^a{w}^b$. 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 $V=k^{\perp }/⟨k⟩$. After imposing $A^a{}_a=0$, the remaining gauge parameters obey $k\cdot B=0$, 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 $k\cdot \ell =-1$ and orthonormal spacelike vectors $e_1,e_2$ representing a basis of $V$. Since $A(k,\ell )=0$,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.