Run the Robinson–Schensted–Knuth algorithm on a long random word, then insert one more letter z. A new box appears somewhere on the boundary of the Young diagram. Where? A law of large numbers pins it, to leading order, at one specific point of the Logan–Shepp–Vershik–Kerov curve Ω. This paper is about what happens at the next order: the fluctuations around that point, which live on the scale n1/4, are conjecturally Gaussian, and are squeezed onto the tangent line of Ω.

The three panels below run genuine Schensted row insertion — no precomputed data, no fitted curves. Colours are the paper's own.

1Which letter is responsible for which box?

Insert n i.i.d. uniform values w1,…,wn and paint every box of the resulting diagram with the colour band of the letter that created it. Small letters build boxes near the Oy axis, large letters near the Ox axis: the value of a letter already decides, to first order, the ray along which its box will land. (Figure 4 of the paper, live.)

w ∈ [0, ¼] (¼, ½] (½, ¾] (¾, 1] — Ω  ·  ● natural parametrization at z = 0, ¼, ½, ¾, 1

Push n up and the rays sharpen. Schensted insertion costs Θ(n3/2) elementary bumps — about 0.48√n per letter — so the slider carries a time estimate and the diagram is built in the background; you can stop it. Beyond n ≈ 105 a box is smaller than one screen pixel, so each pixel is painted with the colour band of the mean of the letters responsible for the boxes it covers. At n = 106 the four regions are separated by visibly straight rays and the boundary sits on Ω — the law of large numbers, drawn.

2The natural parametrization of Ω

The map z ↦ (RSKcos z, RSKsin z) = (FSC−1(z), Ω(FSC−1(z))) is the RSK analogue of the angle parametrization of the circle: the ray through the point cuts off a curvilinear triangle whose area is proportional to z. The whole region between Ω and the two axes has area 2, so the shaded piece has area 2z.

u₀ = RSKcos z
v₀ = RSKsin z
tangent slope Ω′(u₀)
σ²u₀ = (π/3)√(4−u₀²)

3The fluctuations — Conjecture 1.5, live

Fix z. Insert n i.i.d. uniform letters, then insert z itself and record the position (xn, yn) of the new box. Plot n1/4[((xnyn)/√n, (xn+yn)/√n) − (u₀, v₀)], once per independent trial. The conjecture: this converges to a centred, degenerate Gaussian carried by the line v = Ω′(u₀)·u, whose u-marginal is N(0, σ²u₀) with σ²u₀ = (π/3)√(4−u₀²).

trials done
0
predicted σu₀
empirical sd of u
empirical mean of u

Left: the rescaled cloud, with the conjectured supporting line (red) and the axes of the blue (u, v) system of Figures 5–6. Right: the histogram of the u-coordinate against the conjectured density N(0, σ²u₀). The scaling is n1/4, so at moderate n the empirical spread only approaches the conjectured value slowly — the paper's own figures use n = 100 and n = 104.

You will notice that the cloud sits slightly above the red line. That is a finite-size effect, not a defect of the conjecture: a box is a unit square, and the formula above labels it by (xn, yn) rather than by its centre, which displaces v by exactly n−1/4 — 0.178 at n = 103, 0.100 at n = 104, and 0 in the limit. Tick box centres and watch the cloud settle onto the line.