= Solution
First differentiate the unit normalization $X^\alpha X_\alpha=-1$. <Metric compatibility> gives
$$
B_{\alpha\beta}X^\alpha
=X^\alpha\nabla_\beta X_\alpha
=\frac12\nabla_\beta(X^\alpha X_\alpha)=0.
$$
The <timelike geodesic vector field> equation gives the other contraction,
$$
B_{\alpha\beta}X^\beta
=X^\beta\nabla_\beta X_\alpha=0.
$$
For the <spatial projector of a timelike congruence>,
$$
h_{\alpha\beta}X^\alpha
=X_\beta+(X_\alpha X^\alpha)X_\beta=0,
$$
and symmetry also gives $h_{\alpha\beta}X^\beta=0$. Thus
$$
\boxed{B_{\alpha\beta}X^\alpha=B_{\alpha\beta}X^\beta
=h_{\alpha\beta}X^\alpha=h_{\alpha\beta}X^\beta=0.}
$$
Splitting the <deformation tensor of a timelike geodesic congruence> into symmetric and antisymmetric parts and substituting the definitions gives
$$
B_{\alpha\beta}
=B_{(\alpha\beta)}+B_{[\alpha\beta]}
=\frac13\Theta h_{\alpha\beta}+\sigma_{\alpha\beta}+\omega_{\alpha\beta}.
$$
Moreover, the orthogonality of $B_{\alpha\beta}$ implies
$$
\Theta=B^{\alpha\beta}h_{\alpha\beta}
=B^\alpha{}_\alpha,
$$
while in four dimensions
$$
g^{\alpha\beta}h_{\alpha\beta}
=4+X^\alpha X_\alpha=3.
$$
It follows that
$$
g^{\alpha\beta}\sigma_{\alpha\beta}
=B^\alpha{}_\alpha-\frac13\Theta\,g^{\alpha\beta}h_{\alpha\beta}=0.
$$
Hence
$$
\boxed{B_{\alpha\beta}=\frac13\Theta h_{\alpha\beta}
+\sigma_{\alpha\beta}+\omega_{\alpha\beta},
\qquad g^{\alpha\beta}\sigma_{\alpha\beta}=0,}
$$
which is the <kinematic decomposition of a timelike geodesic congruence>.
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