Theorem: Exact FR order stratification

On the positive-amplitude simplex with two phase parameters:

1. The order-one region is open and dense.

2. The order-two region is a smooth codimension-two locus. It is real-algebraic with nonvanishing inequalities when expressed in real and imaginary amplitude coordinates. Its intersection with $\alpha =\beta =0$ is exactly $x=y=z=1/\sqrt{3}$.

3. There is no infinite-order point: the three nonempty question coefficients cannot all vanish.

Proof. Set

Order two requires

Since $x,y,z>0$, these equations require $A,B,C$ to be nonzero and have the same sign. Conversely, on that open phase domain the unique positive normalized solution is

Thus the order-two set is the smooth graph of amplitudes over an open two-dimensional phase domain, inside the four-dimensional parameter space. Its codimension is two. At zero phases, (eq:frgraph) gives the symmetric point.

For the algebraic description, write the amplitudes as $x$, $u+iv$, and $w+it$, with $x>0$. The two equations become

with normalization $x^2+u^2+v^2+w^2+t^2=1$ and $w\ne 0$. These are polynomial equations and inequalities. In angular coordinates the same locus is real-analytic, rather than polynomial in the angles themselves.

If all three coefficients vanished, $\cos \beta =0$ would follow. Then $\cos \alpha =0$. Both phases would be odd multiples of $\pi /2$, so $|\cos (\alpha -\beta )|=1$. The second coefficient would be nonzero, a contradiction.

Finally, the order-two locus has empty interior, while the nonvanishing of at least one singleton coefficient defines an open set. The explicit graph description and the absence of an infinite-order point establish density of that open order-one set.