Equivalence between the Markov-Doob integrability and the local move integrability for usual-spec Macdonald RSK-type dynamics

local_move markov_doob_integrability robinson_schensted_knuth

Theorem. A usual-spec RSK-type dynamics satisfies Markov-Doob integrability if and only if it satisfies the local move integrability.

Proof. Due to the weight-preserving property of RSK-type dynamics, the Markov-Doob integrability equation (0) can be written as follows

$\sum_{\lambda^{n - 1, k - 1}, w_{n, k}} R_{w_{n, k}}(\lambda^{n - 1, k - 1}, \lambda^{n - 1, k}, \lambda^{n, k - 1}, \lambda^{n, k}) {(t)_{w_{n, k}} \over (q)_{w_{n, k}}} {\psi_{\lambda^{n - 1, k} / \lambda^{n - 1, k - 1}} \phi_{\lambda^{n, k - 1} / \lambda^{n - 1, k - 1}} \over \psi_{\lambda^{n, k} / \lambda^{n, k - 1}} \phi_{\lambda^{n, k} / \lambda^{n - 1, k}}} = 1$

On the other hand the local move integrability (-1) is

$\sum_{s'} {(t)_{s'_{n k}} \over (q)_{s'_{n k}}} {M_\Theta(s') \over M_\Lambda(s)} \prob(\rho_{n, k} s' = s) = 1.\label{eq:local2}$

Since $$R_{w_{n, k}}(\lambda^{n - 1, k - 1}, \lambda^{n - 1, k}, \lambda^{n, k - 1}, \lambda^{n, k}) = \prob(\rho_{n, k} s' = s)$$, it suffices to show

${\psi_{\lambda^{n - 1, k} / \lambda^{n - 1, k - 1}} \phi_{\lambda^{n, k - 1} / \lambda^{n - 1, k - 1}} \over \psi_{\lambda^{n, k} / \lambda^{n, k - 1}} \phi_{\lambda^{n, k} / \lambda^{n - 1, k}}} = {M_\Theta (s') \over M_\Lambda(s)}. (1)$

When $$n > k$$, denote $$a, b, c, d$$ by

\begin{align} \lambda^{n - 1, k - 1} &= (s'_{n - 1, k - 1}, s'_{n - 2, k - 2}, \dots, s'_{n - k + 1, 1}) =: (d_1, d_2, \dots, d_{k - 1}) \\ \lambda^{n - 1, k} &= (s'_{n - 1, k}, s'_{n - 2, k - 1}, \dots, s'_{n - k, 1}) = (s_{n - 1, k}, s_{n - 2, k - 1}, \dots, s_{n - k, 1}) =: (b_1, b_2, \dots, b_k)\\ \lambda^{n, k - 1} &= (s'_{n, k - 1}, s'_{n - 1, k - 2}, \dots, s'_{n - k + 2, 1}) = (s_{n, k - 1}, s_{n - 1, k - 2}, \dots, s_{n - k + 2, 1}) =: (c_1, c_2, \dots, c_{k - 1}) \\ \lambda^{n, k} &= (s_{n, k}, s_{n - 1, k - 1}, \dots, s_{n - k + 1, 1}) =: (a_1, a_2, \dots, a_k). \end{align}

Due to (1)(2) in Explicit formula of Macdonald polynomials, both sides of (1) are equal to

\begin{align} &{\prod_{1 \le i < j \le k - 1} g(d_i - d_j, j - i - 1) g(d_i - d_j, j - i) \over \prod_{1 \le i < j \le k} g(a_i - a_j, j - i - 1) g(a_i - a_j, j - i)}\\ &\times{\prod_{1 \le i \le j \le k - 1}g(b_i - a_{j + 1}, j - i) g(a_i - c_j, j - i) g(c_j - a_{i + 1}, j - i) \prod_{1 \le i \le j \le k} g(a_i - b_j, j - i) \over \prod_{1 \le i \le j \le k - 2} g(d_i - c_{j + 1}, j - i) \prod_{1 \le i \le j \le k - 1} g(c_i - d_j, j - i) g(b_i - d_j, j - i) g(d_i - b_{j + 1}, j - i)}. \end{align}

When $$n = k$$, denote $$a, b, c, d$$ by

\begin{align} \lambda^{n - 1, k - 1} &= (s'_{n - 1, k - 1}, s'_{n - 2, k - 2}, \dots, s'_{1, 1}) =: (d_1, d_2, \dots, d_{k - 1}) \\ \lambda^{n - 1, k} &= (s'_{n - 1, k}, s'_{n - 2, k - 1}, \dots, s'_{1, 2}) = (s_{n - 1, k}, s_{n - 2, k - 1}, \dots, s_{n - k + 1, 2}) =: (b_1, b_2, \dots, b_{k - 1})\\ \lambda^{n, k - 1} &= (s'_{n, k - 1}, s'_{n - 1, k - 2}, \dots, s'_{2, 1}) = (s_{n, k - 1}, s_{n - 1, k - 2}, \dots, s_{2, 1}) =: (c_1, c_2, \dots, c_{k - 1}) \\ \lambda^{n, k} &= (s_{n, k}, s_{n - 1, k - 1}, \dots, s_{1, 1}) =: (a_1, a_2, \dots, a_k). \end{align}

Both sides of (1) are equal to

\begin{align} &{\prod_{1 \le i < j \le k - 1} g(d_i - d_j, j - i - 1) g(d_i - d_j, j - i) \over \prod_{1 \le i < j \le k} g(a_i - a_j, j - i - 1) g(a_i - a_j, j - i)} {\prod_{1 \le i \le k} g(a_i, k - i) \over \prod_{1 \le i \le k - 1} g(d_i, k - 1 - i)}\\ &\times{\prod_{1 \le i \le j \le k - 1}g(b_i - a_{j + 1}, j - i) g(a_i - c_j, j - i) g(c_j - a_{i + 1}, j - i) g(a_i - b_j, j - i) \over \prod_{1 \le i \le j \le k - 2} g(d_i - c_{j + 1}, j - i) g(d_i - b_{j + 1}, j - i) \prod_{1 \le i \le j \le k - 1} g(c_i - d_j, j - i) g(b_i - d_j, j - i) }. \end{align}

The case when $$n < k$$ is similar to $$n > k$$ case but symmetric. $$\square$$