lin-0078

5.6 Two counterexamples that fix the interpretation

Example 5.9 Residual without a latent variable

On \(\mathbb R^3\), let

\[ u=\partial _x, \qquad v=\partial _y+x\partial _z. \]

Then \([u,v]=\partial _z\), which is not in \(\operatorname {span}\{ u,v\} \) near the origin. The visible distribution is not involutive, yet the example contains no hidden random variable. A nonzero residual can arise because the declared visible controls are not closed under order-sensitive composition.

Example 5.10 Latent structure with commuting visible fields

Suppose latent variation changes a distributional family, while the two available visible interventions act as independent translations \(u=\partial _x\) and \(v=\partial _y\). Then \([u,v]=0\) despite the latent structure. A zero residual shows closure of the tested visible span; it does not show absence of hidden causes.

Boundary

A nonzero bracket residual is a typed failure of visible closure, not by itself a certificate of latent confounding. A zero residual certifies involutivity only for the declared protocol and stratum; it does not certify causal completeness.