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A stationary directed cycle is not similar to its discriminant

Statement refuted

For every finite row-stochastic kernel with a strictly positive stationary distribution, the discriminant equals the stationary similarity transform $diag (\sqrt{π}) P diag (1 / \sqrt{π})$ , even when detailed balance fails.

Facts & Assumptions

Given: A finite row-stochastic kernel, a stationary probability vector, and the discriminant defined by Szegedy row-preparation map and discriminant.

[F1]

Finite row-stochastic kernel: each row of $P$ is nonnegative and sums to one.

[F2]

Invariant and stationary distribution for a Markov kernel: $π$ is stationary exactly when $π P = π$ .

[F3]

Reversible measure and detailed balance: reversibility requires $π_{x} P_{x y} = π_{y} P_{y x}$ for every pair.

[F4]

The Szegedy discriminant is self-adjoint and contractive: $D_{x y} = \sqrt{P_{x y} P_{y x}}$ .

Counterexample

Take $X = {0, 1, 2}$ with indices modulo $3$ and $P_{x y} = 1$ exactly when $y = x + 1 (m o d 3)$ , with all other entries zero. Let $π = (1 / 3, 1 / 3, 1 / 3)$ .

1.1F1F2given

Every row of $P$ has one entry equal to $1$ , and each column also has one such entry, so $P$ is row-stochastic and $π P = π$ .

2.1F3step 1.1algebra

Detailed balance fails at $(x, y) = (0, 1)$ because $π_{0} P_{01} = 1 / 3$ but $π_{1} P_{10} = 0$ . Thus the chain is stationary but not reversible.

3.1F1F2F4step 1.1algebra∎

No pair of distinct states has transitions in both directions, and all diagonal entries are zero, so $P_{x y} P_{y x} = 0$ for every $x, y$ and [F4] gives $D = 0$ . Since the stationary vector is uniform, its diagonal similarity transform is $diag (\sqrt{π}) P diag (1 / \sqrt{π}) = P \neq 0$ . The claimed identity fails.

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, §§18.2–18.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, §8.2 (standard reference, not scraped)