Example. Reflection positivity without a quotient time shift [fdip-007Q]

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On time sites \(\mathbb Z/6\), let \(\phi _t\) be independent centered real Gaussians with variances \[(v_0,v_1,v_2,v_3,v_4,v_5)=(1,1,2,3,3,2).\] Use the link reflection \(\theta (t)=1-t\) and positive half \(X_+=\{1,2,3\}\). Since \(v_{\theta t}=v_t\), the product law is reflection invariant. The halves are independent and their laws agree under reflection, so for every positive-half square-integrable \(F\), \[B_\mu (F,F)=|\mathbb E_\mu F|^2\ge 0.\] This proves reflection positivity directly, without a translation-invariance assumption.

Define a one-step shift of a coordinate expression by replacing each \(\phi _t\) by \(\phi _{t+1}\). On expressions whose sites remain in \(X_+\), this at least gives another positive-half function. In particular, \[F=\phi _1^2-1,\qquad \tau F=\phi _2^2-1\] both belong to the positive-time \(L^2\) domain. Their pairings are \[B_\mu (F,F)=0,\qquad B_\mu (\tau F,\tau F)=1.\] Indeed their expectations are \(v_1-1=0\) and \(v_2-1=1\). The centering constant is part of the shifted expression and does not change from \(1\) to \(v_2\).

Thus a zero vector in the reflection quotient would be sent to a nonzero one. The shift cannot descend even on this pair of positive-half polynomials. A transfer operator requires additional structure; reflection positivity alone does not construct it.