A smooth symmetric nondegenerate 2-tensor field of inertia (2,2,0)

The component matrix $\mathrm{diag}(1,1,-1,-1)$ is constant, hence $C^{\mathrm{\infty }}$, and symmetric with determinant $1\mathrm{\ne }0$; by [L1] the family $B$ is a smooth symmetric nondegenerate covariant $2$-tensor field on $M$.

In the global chart the matrix is already diagonal with two positive entries and two negative entries and no zero entry; by [L2] the inertia of each $B_p$ is therefore $(2,2,0)$, independent of any further basis choice.

Since $(2,2,0)\mathrm{\ne }(3,1,0)$, $B_p$ is not a Lorentz form by [L3], so $(M,B)$ is not a smooth Lorentzian four-manifold; hence the refuted statement is false. The same computation with the constant matrix $\mathrm{diag}(1,1,1,1)$ gives a smooth symmetric nondegenerate field of inertia $(4,0,0)$, a Riemannian metric, which also fails the Lorentzian signature condition; both witnesses show that the signature clause in the definition of a Lorentzian metric is a genuine additional requirement beyond symmetry, smoothness and nondegeneracy.