Example: a product of Sturmian systems

Reader card. This is the running example for Parts I and II. It can be read immediately after the proof overview. The example makes the coinvariant equation, transfer functions, zero curl, and product trace explicit; the general chapters explain why the same steps work without these simplifying identities.

This example makes every ingredient visible: a minimal height direction, an irrational clopen frequency, explicit coinvariant transports, and a mixed trace.

The transfer functions in this example have zero curl directly. The passage from that integer calculation to vanishing of the Barlak corner \(K_1\)-defect uses the PV/Barlak comparison proved in Technical Engine A.

The dynamical system

Let \((X_j,S_j)\), for \(j=0,1,2\), be Sturmian subshifts of slopes \(\alpha_j\). Choose the slopes so that the diagonal product map

\[ T_h=S_0\times S_1\times S_2 \]

is minimal. A convenient sufficient choice is that the associated rotation vector is totally irrational; equivalently, choose the factors so that the diagonal product rotation is minimal. Put

\[ \Sigma=X_0\times X_1\times X_2, \qquad T_1=S_0\times 1\times1, \qquad T_2=1\times S_1\times1. \]

The three maps commute. Their exponent vectors

\[ h=(1,1,1),\qquad e_1=(1,0,0),\qquad e_2=(0,1,0) \]

form a unimodular basis of the original product action.

flowchart LR
    X0["Sturmian factor X_0"]
    X1["Sturmian factor X_1"]
    X2["Sturmian factor X_2"]
    S["Sigma = X_0 x X_1 x X_2"]
    H["height: T_h = (S_0,S_1,S_2)"]
    A["transverse: T_1 = (S_0,1,1)"]
    B["transverse: T_2 = (1,S_1,1)"]
    X0 --> S
    X1 --> S
    X2 --> S
    S --> H
    S --> A
    S --> B

The warning here is important: a product of minimal systems need not be minimal. The theorem requires minimality of the chosen single height map, so it must be checked rather than inferred factor by factor.

An explicit coinvariant class

Let \(C_a\subset X_0\) be the cylinder in which the symbol at the origin is \(a\), and define

\[ g=1_{C_a\times X_1\times X_2}. \]

Because \(g\) depends only on the first coordinate,

\[ \alpha_1(g)=\alpha_h(g), \qquad \alpha_2(g)=g. \]

Thus the transverse invariance equations are solved by

\[ h_1=g, \qquad h_2=0, \]

up to the covariance sign convention. Consequently

\[ (\alpha_i-1)g=(\alpha_h-1)h_i, \]

so \([g]\in C(\Sigma,\mathbb Z)_{T_h}\) is fixed by both transverse generators. The pairwise curvature is visibly zero:

\[ \kappa_{12}=(\alpha_2-1)h_1-(\alpha_1-1)h_2=0. \]

The general proof produces the same conclusion without needing these particularly simple transfer functions.

The trace

For the Fibonacci substitution

\[ a\mapsto ab, \qquad b\mapsto a, \]

the frequency of \(a\) is \(\varphi^{-1}\), where \(\varphi=(1+\sqrt5)/2\). Hence the invariant product measure satisfies

\[ \mu(g)=\frac1\varphi. \]

Assume first that the parameter is normalized so that \(0<\theta=\Theta_{12}<1\). Let \(E_\theta\) be the corresponding actual basic Rieffel module for the transverse two-torus. With the usual positive orientation,

\[ \tau([E_\theta])=\theta, \]

and the mixed module has

\[ \tau_\mu([\mathcal E_{\theta,g}]) =\frac{\theta}{\varphi}. \]

For a parameter outside this normalized positive range, the same formula describes the virtual \(K_0\)-class of trace \(\theta\), and the mixed output is a graded difference. Alternatively, choose an actual positive Rieffel class of trace \(m+n\theta>0\); its mixed trace is \((m+n\theta)/\varphi\). Thus no actual projective module with nonpositive trace is being asserted.

This number is the product of a pattern frequency and a magnetic Rieffel trace. It is exactly the kind of mixed gap label that is invisible if one only lists the two factors separately.