51. Why finite predictive rank need not mean finitely many positive states
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The following controlled rational example makes the boundary explicit. It is a worked adaptation of the irrational-rotation obstruction in [R12, Section 7]. Related distinctions among finite linear, positive and quantum realizations are discussed in [R14]. Only the elementary cone argument below is used.
Let \(z=(t,x,y)^\top\), \(\lambda=(1,0,0)\), \(z_0=(1,1,0)^\top\), and
There are two actions and two observations. A rotate action has branches
A probe action has branches
The rotate outcome zero is certain. The probe has probabilities \((t+x)/2\) and \((t-x)/2\) and resets every nonzero branch to \(z_0\) after normalization. Both actions conserve the mass \(t\).
Proposition 46: A rational rank-three model with no polyhedral probability-domain certificate
The Lorentz cone
is probability valid for these operators. It is the canonical all-test cone. Every closed convex invariant cone containing \(z_0\) and admitting the nonnegative mass interpretation equals \(C_{\mathrm L}\). In particular no finite normalized generator certificate exists for this exact model. Its reachable and observable predictive dimension is three.
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Proof
Rotation preserves \(t\) and \(x^2+y^2\). On \(C_{\mathrm L}\) both probe effects are nonnegative and each probe image is a nonnegative multiple of \(z_0\). Thus the displayed cone is invariant and valid.
Write the rotation angle as \(\theta\). It is not a rational multiple of \(2\pi\). Otherwise \(\zeta=e^{i\theta}\) would be a root of unity, so \(\zeta+\zeta^{-1}=6/5\) would be a rational algebraic integer. Every rational algebraic integer is an integer, by the rational-root argument for a monic integer polynomial, contradicting \(6/5\notin\mathbb Z\). The forward powers of an irrational circle rotation are dense. Compactness supplies powers approaching the identity, so their closure contains the inverse rotation and is a closed subgroup. An infinite closed subgroup has arbitrarily small positive angles. Integer multiples of such angles approximate every point on the circle, so that subgroup is the entire circle.
A state passing every test must pass a probe after any number \(k\) of rotations. Hence
Density gives \(t\geq\sqrt{x^2+y^2}\). The reverse inclusion was already proved, so the canonical cone is exactly \(C_{\mathrm L}\).
Any invariant cone with nonnegative mass lies inside this canonical cone, by Proposition 38. If it is closed, convex and contains \(z_0\), it contains the orbit \(T^kz_0\) and the closed convex hull of that orbit. On \(t=1\), this hull is the full unit disk. Conical scaling gives all of \(C_{\mathrm L}\). The unit disk has infinitely many extreme points, so it is not the convex hull of finitely many points.
Finally the reachable columns \(z_0,Tz_0,T^2z_0\) are independent. The mass row and probe rows before zero and one rotations span the dual space because the rotation has nonzero sine. Reachability and observability establish predictive dimension three. \(\square\)
This is not a failure of finite-dimensional prediction. The exact model has only three coordinates and a simple quadratic cone. It is a failure of the narrower requirement that every valid model admit a finite polyhedral domain. A certificate language limited to finitely many positive generators would exclude a legitimate low-rank signal model.
No general impossibility of all finite certificates follows. This very example has a finite quadratic certificate. The obstruction concerns polyhedral certificates, not invariant cones described by other mathematics.