= Solution
Treat the printed $e^\mu$ as an <orthonormal coframe>, with frame metric $\eta=\operatorname{diag}(-1,1,1,1)$. Work on a patch $z\ne0$ and $F=1-\alpha z^3>0$, so the given real coframe exists. Define $f=\sqrt F$ and $A=zf'-f$. The <exterior derivatives> are
$$
de^0=A e^3\wedge e^0,\qquad
de^1=-f e^3\wedge e^1,\qquad
de^2=-f e^3\wedge e^2,\qquad de^3=0.
$$
The <connection 1-forms> satisfying <Cartan's first structure equation> and <metric compatibility> are
$$
\boxed{\omega^0{}_3=\omega^3{}_0=Ae^0,\qquad
\omega^1{}_3=-fe^1,\quad\omega^3{}_1=fe^1,\qquad
\omega^2{}_3=-fe^2,\quad\omega^3{}_2=fe^2.}
$$
All other forms vanish. With a Lorentzian frame, it is the lowered forms $\omega_{ab}=\eta_{ac}\omega^c{}_b$ that are antisymmetric; the two mixed time–space forms are equal. These displayed forms solve the torsion-free structure equation, and uniqueness of the <Levi-Civita connection> identifies them as the required connection.
Apply <Cartan's second structure equation>. For example,
$$
\Theta^0{}_3=d(Ae^0)=-(zfA'+A^2)e^0\wedge e^3,
\qquad
\Theta^0{}_1=\omega^0{}_3\wedge\omega^3{}_1=Af e^0\wedge e^1.
$$
The remaining derivatives give $\Theta^1{}_3=Af e^1\wedge e^3$ and $\Theta^1{}_2=-f^2e^1\wedge e^2$, with analogous forms for index 2. Put $q=\alpha z^3$ and $B=1+q/2$. From $f^2=1-q$,
$$
Af=zf f'-f^2=-B,\qquad zfA'+A^2=F.
$$
Therefore all six independent <curvature 2-forms> are
$$
\boxed{\begin{aligned}
\Theta^0{}_1&=-B e^0\wedge e^1,&\Theta^0{}_2&=-B e^0\wedge e^2,\\
\Theta^0{}_3&=-F e^0\wedge e^3,&\Theta^1{}_2&=-F e^1\wedge e^2,\\
\Theta^1{}_3&=-B e^1\wedge e^3,&\Theta^2{}_3&=-B e^2\wedge e^3.
\end{aligned}}
$$
The other six are fixed by $\Theta^b{}_a=-\eta_{aa}\eta_{bb}\Theta^a{}_b$ for $a\ne b$, with no sum: equal for time–space pairs and opposite for spatial pairs. All diagonal forms are zero. The <Lorentzian connection-form antisymmetry> is essential to these signs.
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