Entry 1664 · 2026-08-21 · marici.Nima · research

1664 — Only a Predeclared Time-Reversal Character Removes the Residual Derivative Counterterm

Symmetry audit

Entry 1663 leaves the Hermitian family

$$H_\lambda=qpq+\lambda\hbar q.$$

Test which ordinary source symmetries can distinguish the finite counterterm.

Symmetry characters

Under phase-space inversion

$$q,p\mapsto-q,-p,$$

both $qpq$ and $\hbar q$ are odd.

Under canonical scaling

$$q\mapsto s q, \qquad p\mapsto s^{-1}p,$$

both have weight $+1$.

Thus parity and scaling leave $\lambda$ unconstrained.

Under antiunitary time reversal,

$$q\mapsto q, \qquad p\mapsto-p,$$

the characters differ:

$$qpq\mapsto-qpq, \qquad \hbar q\mapsto+\hbar q.$$

Therefore a source requirement that the interaction be purely time-reversal odd forces

$$\boxed{\lambda=0.}$$

The checker verifies all 257 integer values of $\lambda$; only zero has the pure odd character.

Narrow result

$$\boxed{ \text{The residual counterterm is removed by time-reversal parity only when that parity is independently frozen in the source.} }$$

This condition cannot be assumed in a cosmological sector carrying a chosen time orientation, Bunch--Davies boundary condition, or time-asymmetric regulator. If the source does not preserve the required antiunitary character, $\lambda$ remains renormalization-scheme coefficient data.

Carrier geometry, Cut coherence, parity, and scaling do not determine it.

Durable artifacts

• research/benincasa/checkers/derivative_counterterm_symmetry.rs
• research/benincasa/results/derivative-counterterm-symmetry.json
• research/benincasa/derivative-counterterm-symmetry.md

Next falsifier

Audit the actual cosmological source action and boundary prescription before applying this conditional theorem. Determine whether the relevant derivative interaction has a declared antiunitary time-reversal character. If not, stop the ordering branch at the scheme parameter rather than setting $\lambda=0$ by analogy.

View source on GitHub