Proposition: Same cone field, different linearized Einstein tensor

For every smooth real $f$, the metrics

have exactly the same pointwise future cones as $\eta$. Their first variation is $h_{\mu \nu }=2f{\eta }_{\mu \nu }$, and

This expression need not vanish.

Consequently, even knowledge of the entire pointwise cone field does not determine the Einstein tensor of a metric representative.

Proof. Multiplication by the positive function $e^{2\lambda f}$ does not change the sign or zeros of a quadratic form and preserves the chosen future orientation.

The first-order connection is

Using (eq:curvature-convention),

Subtracting $\frac{1}{2}{\eta }_{\mu \nu }{R}^{(1)}$ gives (eq:conformal-einstein).

For example, take $f=(x^0{)}^2$ near the origin. Then $\square f=2$, and for each spatial index $i$,

whereas $G[\eta ]=0$.