Periodic Pipe Dreams and Affine Permutons     permutations

A periodic pipe dream is an $n$-periodic filling of the $k\times\infty$ grid by two tiles, the cross and the elbow. Equivalently it is a $k\times n$ pipe dream drawn on a cylinder: the left and right boundaries of the $k\times n$ rectangle are identified, so a pipe leaving through the right edge re-enters on the left.

Here every cell is an independent cross with probability $p$ and an elbow with probability $1-p$. Pipes enter through the bottom boundary and leave through the top boundary. At a cross the two pipes pass through each other; at an elbow the pipe arriving from below turns right and the pipe arriving from the left turns up. Numbering the bottom and top boundary edges by $\mathbb{Z}$, the connectivity of the pipes is an affine permutation: a bijection $f\colon \mathbb{Z}\to\mathbb{Z}$ with $f(i+n)=f(i)+n$.

Every cell passes exactly one pipe to the right, so each row contributes exactly $n$ to the total rightward displacement, and therefore $$\sum_{i=1}^{n}\bigl(f(i)-i\bigr)=kn$$ for every configuration, whatever $p$ is. The height $k$ is thus the level of the affine permutation, and $\alpha=k/n$ is the mean displacement in units of $n$. The list $f(i)-i$ is the siteswap of $f$.

Reduction and the parameter $q$. At a cross tile whose two pipes have not met before, the crossing is always realized. Where the two pipes have already crossed, it is realized with probability $q$ and otherwise forced to an elbow (shaded gray). At $q=0$ this is the Demazure product in the affine symmetric group, the periodic analogue of the reduction studied in [1], and the diagram is reduced: the number of realized crossings equals the length $\ell(f)$. At $q=1$ nothing is undone and the pipe dream is the raw i.i.d. tiling. Intermediate values interpolate, in the same way as the parameter $q$ of the staircase simulation: a re-crossing is accepted with relative weight $q$.

Affine permuton. Since $f(i+n)=f(i)+n$, the points $(i,f(i))$ are invariant under translation by $(n,n)$ and descend to $n$ points on the torus $(\mathbb{Z}/n)^2$ — one in every row and every column. Rescaling by $n$ and letting $n\to\infty$ gives a measure on the torus with uniform marginals: an affine permuton.

Reduction is what makes it nontrivial. Without reduction the displacements $f(i)-i$ are sums of $k$ nearly independent mean-one contributions, so they concentrate at $k$ and the limit is the trivial rotation $y=x+\alpha$; the observed spread $\mathrm{sd}(f(i)-i)/n$ decays like $n^{-1/2}$. With reduction that ratio instead settles down to a positive constant (about $0.11$ at $p=0.3$, $0.21$ at $p=0.5$, $0.53$ at $p=0.8$, taking $\alpha=1/2$), so the displacements spread on the scale of $n$ and the permuton is genuinely two-dimensional. Compare $q=0$ with $q=1$ at $n$ large to see the two pictures.

A row consisting entirely of crosses would carry a closed loop that never turns up, and its word has no canonical starting point on the cylinder; such rows are excluded by resampling. This has probability $p^n$ and matters only for very small $n$ or $p$ very close to $1$.