 Solution

For $E_{\mu \nu }={\eta }_{\mu \nu }$, transversality would require $k^{\mu }{\eta }_{\mu \nu }=k_{\nu }=0$. Hence the operator has a third-order stress-tensor pole and is not a primary operator for any nonzero $k$.

Forthe two transversality conditions coincide and reduce toIf $k^2=0$, this says $k\cdot \xi =-1$, with arbitrary additional component transverse to $k$. If $k^2\ne 0$, it forcesso $E_{\mu \nu }={\eta }_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ is the transverse projection operator. Once this condition removes the third-order poles, the conformal weights are again $(1+{\alpha }^{\prime }{k}^2/4,1+{\alpha }^{\prime }{k}^2/4)$; in the null case they are $(1,1)$.