这是一份 Imperial帝国理工大学 PHYS97026作业代写的成功案例
Within this framework (as well as in chapter three) we need to know that $\psi_{A B C D}$ has two scalar invariants:
$$
\begin{gathered}
I \equiv \psi_{A B C D} \psi^{A B C D} \
J \equiv \psi_{A B}^{C D} \psi_{C D}^{E F} \psi_{E F} A B
\end{gathered}
$$
Type-II space-times are such that $I^{3}=6 J^{2}$, while in type-III space-times $I=I=$
0 . Moreover, type-D space-times are characterized by the condition
$$
\psi_{P Q R\left(A \psi_{B C}\right.}^{P Q} \psi_{D E F)}^{R}=0
$$
while in type-N space-times
$$
\psi_{(A B}{ }^{E F} \psi_{C D) E F}=0
$$
PHYS97026 COURSE NOTES ：
$$
\left[\nabla^{a} \nabla^{d}+\nabla^{a} \omega^{d}-\omega^{a} \omega^{d}\right] C_{a b c d}=0
$$
We now re-express $\nabla^{a t} \omega^{d}$
$$
\nabla^{a} \omega^{d}=\omega^{a} \omega^{d}+\frac{1}{8} T g^{a d}-\frac{1}{2} R^{a d}
$$$$
\left[\nabla^{a} \nabla^{d}-\frac{1}{2} R^{a d}\right] C_{a b c d}=0
$$
This calculation only proves that the vanishing of the Bach tensor, defined as
$$
B_{b c} \equiv \nabla^{a} \nabla^{d} C_{a b c d}-\frac{1}{2} R^{a d} C_{a b e d}
$$