Proposition: Transversality leaves contact freedom

The nonzero contact distribution $C$ is symmetric in each tensor pair, symmetric under pair exchange, and satisfies

It has scaling degree eight at the origin.

Consequently, a defined linearly transverse response kernel $\Pi$ and $\Pi +aC$ have the same off-origin values and linear transversality for every $a\in \mathbb{C}$, although their local contact normalizations differ.

Proof. Constant-coefficient derivatives commute, and

This proves the transverse identities and the symmetries. The Fourier transform of $C$ is a nonzero homogeneous degree-four tensor polynomial, so $C\ne 0$. A nonzero order-four derivative combination of ${\delta }_0$ in four dimensions scales with degree $4+4=8$. The remaining statements follow by linearity and the support of $C$.