 Solution

Put $u=t-z$, $\ell =du=e^0-e^3$, $p=A^{\prime }/A$, $q=B^{\prime }/B$, with primes denoting $d/du$. The exterior derivatives are $d{e}^0=d{e}^3=0$, $d{e}^1=p\ell \wedge e^1$, $d{e}^2=q\ell \wedge e^2$. For a Levi-Civita connection, the connection 1-forms obey both Cartan's first structure equation and metric compatibility. With lower indices ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$, the result isAll other lower-index forms vanish or follow from ${\omega }_{ab}=-{\omega }_{ba}$. Raising the first time index changes its sign: for example ${\omega }^0{}_1={\omega }^1{}_0=p{e}^1$, whereas ${\omega }^1{}_3=-p{e}^1$ and ${\omega }^3{}_1=p{e}^1$.

For direct verification, ${\omega }^1{}_0\wedge e^0+{\omega }^1{}_3\wedge e^3=-p\ell \wedge e^1=-d{e}^1$, and the same holds for index two. For indices zero and three each term wedges $e^1$ or $e^2$ with itself, so it is zero. These forms are metric-compatible, and uniqueness follows from the existence and uniqueness of the Levi-Civita connection. The printed uniqueness statement implicitly includes metric compatibility: the torsion equation alone permits adding nonzero forms $C^a{}_{bc}{e}^c$ with $C^a{}_{bc}=C^a{}_{cb}$, since their wedge with $e^b$ vanishes.