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 $k$.

For
$$
E_{\mu\nu}=\eta_{\mu\nu}+\xi_\mu k_\nu+k_\mu\xi_\nu,
$$
the two transversality conditions coincide and reduce to
$$
\boxed{k^2\xi_\nu+(1+k\cdot\xi)k_\nu=0}.
$$
If $k^2=0$, this says $k\cdot\xi=-1$, with arbitrary additional component transverse to $k$. If $k^2\ne0$, it forces
$$
\boxed{\xi_\mu=-\frac{k_\mu}{2k^2}},
$$
so $E_{\mu\nu}=\eta_{\mu\nu}-k_\mu k_\nu/k^2$ is the <transverse projection operator>. Once this condition removes the third-order poles, the <conformal weights> are again $(1+\alpha'k^2/4,1+\alpha'k^2/4)$; in the null case they are $(1,1)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 3 / d / Solution

For a nonnull vector $k$ in a space with a nondegenerate bilinear form, $P_{\mu\nu}=\eta_{\mu\nu}-k_\mu k_\nu/k^2$ projects onto the subspace transverse to $k$. It obeys $P_{\mu\nu}k^\nu=0$ and $P^2=P$.


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 Incoming links: Transverse projection operator