Theorem: Necessary compatibility and generic empty fibres

Every member of ${{\mathfrak{C}}^{[2]}}_{{\varphi }_0}$ satisfies

If $g$ is nonconstant, choose $x_0$ with $\mathrm{d}g(x_0)\ne 0$. In the real smooth free-solution space

the set

is open dense. Every corresponding coupled-jet fibre is empty.

For $m=0$, the particular input ${\varphi }_0=t$ violates (eq:bianchi) wherever $\mathrm{d}g\ne 0$.

Proof. The exact Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. Expanding its density version to second order on $G_{\mathrm{lin}}[h]=0$ yields

Taking the divergence of (eq:metric-expansion) and using $P_m{\varphi }_0=0$ proves (eq:bianchi). Equivalently, it is the order-zero switch condition in (eq:noether). Thus this condition follows from the exact Noether identity, not from gauge fixing.

The pointwise expression is a continuous quadratic polynomial in the first scalar jet. It is not identically zero on ${\mathrm{Sol}}_m$: choose a free solution $v$ with $v(x_0)=0$ and ${\partial }_{\mu }v(x_0)={{\delta }_{\mu }}^0$. Such a solution follows from smooth Cauchy data, or from $\sin (m(t-t_0))/m$ when $m>0$, with the massless limit $t-t_0$. Its mixed stress has diagonal entries $1/2,-1/2,-1/2,-1/2$, so contraction with the nonzero covector $\mathrm{d}g(x_0)$ is nonzero. For any $u$, at least one component of the expression at $u+\lambda v$ is a polynomial in $\lambda$ with nonzero quadratic coefficient. Its zeros have empty interior. Arbitrarily small $\lambda$ give a violation. Continuity gives openness.

For $u=t$,

and the same diagonal calculation proves the final claim.