Conjecture on symmetry of complements in the finite k-chain

In fact, the entire stationary distribution has an explicit formula. Write $\lambda^*$ for the complement defined by

$$l_i(\lambda^*)=k-i-l_i(\lambda),$$

and let

$$C(\lambda)=\frac{d_\lambda^{(k)}}{|\lambda|!}.$$

Then, for every $k\ge2$,

$$\boxed{\displaystyle \pi(\lambda)= \frac{C(\lambda)C(\lambda^*)} {\sum_{\nu\in\mathcal R_k}C(\nu)C(\nu^*)}.}$$

In particular, $\pi(\lambda)=\pi(\lambda^*)$.

Let $\operatorname{ex}$ denote exponential specialization, defined by $\operatorname{ex}(h_j)=1/j!$, or equivalently $\operatorname{ex}(p_1)=1$ and $\operatorname{ex}(p_j)=0$ for $j>1$. The strong $k$-dimension satisfies

$$\operatorname{ex}(s_\lambda^{(k)})=C(\lambda).$$

For the $k$-rectangle $R_i=(i^{\,k-i+1})$, put

$$\gamma_i=\operatorname{ex}(s_{R_i})>0.$$

The rectangle identity stated in the source,

$$s_{R_i}s_\mu^{(k)}=s_{\mu\cup R_i}^{(k)},$$

therefore gives

$$C(\mu\cup R_i)=\gamma_i C(\mu). \tag{1}$$

Each directed transition $\lambda\to\mu$ comes from a weak cover $\lambda\to_k\Lambda$, where either $\Lambda=\mu$ or $\Lambda=\mu\cup R_i$. Its transition probability is

$$P(\lambda,\mu) =\frac{d_\Lambda^{(k)}}{(|\lambda|+1)d_\lambda^{(k)}} =\frac{C(\Lambda)}{C(\lambda)} =\gamma(\lambda,\mu)\frac{C(\mu)}{C(\lambda)}, \tag{2}$$

where $\gamma(\lambda,\mu)=1$ if no rectangle is removed, and $\gamma(\lambda,\mu)=\gamma_i$ if $R_i$ is removed.

To identify the reverse edge, use the cyclic particle construction from the source. Starting with the fixed largest label $k+1$, insert $k,k-1,\ldots,1$; the insertion offset for label $i$ is precisely $l_i(\lambda)$. Reversing cyclic orientation replaces that offset by

$$k-i-l_i(\lambda)=l_i(\lambda^*).$$

Thus complement is spatial reversal. An allowed swap moving a smaller label past a larger one consequently satisfies

$$\lambda\longrightarrow\mu \quad\Longleftrightarrow\quad \mu^*\longrightarrow\lambda^*. \tag{3}$$

The labels being swapped do not change. By Corollary 3.7 of the source, a rectangle is removed exactly when the crossed labels are consecutive; in that case its type is determined by those same labels. Hence both edges in (3) remove the same rectangle, or neither removes one:

$$\gamma(\lambda,\mu)=\gamma(\mu^*,\lambda^*). \tag{4}$$

Set $W(\lambda)=C(\lambda)C(\lambda^*)$. Equations (2) and (4) give skew detailed balance:

$$\begin{aligned} W(\lambda)P(\lambda,\mu) &=\gamma(\lambda,\mu)C(\mu)C(\lambda^*)\\ &=W(\mu)P(\mu^*,\lambda^*). \end{aligned}$$

Summing over $\lambda$, and using the fact that complement permutes the finite state space, yields

$$\sum_\lambda W(\lambda)P(\lambda,\mu) =W(\mu)\sum_\lambda P(\mu^*,\lambda^*) =W(\mu).$$

Thus $W$ is stationary. Irreducibility gives uniqueness, and normalization proves the boxed formula. Since $W(\lambda)=W(\lambda^*)$, the conjectured symmetry follows for every $k$.

Source: S. Linusson and A. Özdemir, The $k$-Plancherel Measure and a Finite Markov Chain, arXiv:2512.24346, Sections 3.3–3.5, Conjecture 1.