Nagata–Biran–Szemberg threshold via Okounkov simplex reshaping

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Nagata–Biran–Szemberg threshold via Okounkov simplex reshaping

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Take a polarized variety (X, L), a large number r of points, and turn its flag simplex into the long, thin simplex that interpolation needs. Then check that every reshaping step is allowed and see why the jets you get force ε(X, L; p) = (Lⁿ/r)^{1/n} at very general points.

The problem

The multipoint Seshadri constant never exceeds the volume bound w = (V/r)^{1/n}, with V = Lⁿ. The theorem says that in dimension n ≥ 3 there is a threshold r₀(X, L), and for every r ≥ r₀ equality holds exactly. It is not a convergence statement. [0:09]

The lower bound comes from jets. If H⁰(X, kL) surjects onto jets of degree m at the r points, then ε ≥ m/k. So you aim for jets through ⌊kθ⌋ for every θ < w, and you get them from monomials whose exponent sets fill the right simplex. [0:53]

The procedure

Start from the flag simplex P₀ = Δ(1/D,…,1/D, D^{n−1}V). Its volume is V/n!, #S_k = h⁰(X, kL), and S_k + S_l ⊂ S_{k+l}. [1:22]
Pick r so that w < ℓ/100, where ℓ is the smallest intercept of P₀.
Reshape two intercepts at a time: (A, B) ↦ (w, AB/w). The product stays the same, so the volume does too. So do the counts, additivity and the backward jet transfer. The step needs w·max(3/A, 2/B) < 1/16. Carried intercepts only grow, so checking 3w/ℓ < 1/16 covers the whole chain. [2:00]
Interior saturation: additivity fills every lattice point of kK, for K compact inside P*. Each interior point has about kⁿ splittings, and holes block only o(kⁿ) of them. [4:35]
Hermite interpolation: put the points at (1,…,1,c_i). In the last variable, the Chinese remainder theorem builds f_α with deg_Y < r(m+1−|α|). These exponents form the set E_{r,m}, which fits inside a shrunk, translated copy of kP*. [5:16]
Transfer back through each reshaping and the flag. Intersect countably many open loci to get very general tuples. The jet lemma then gives ε ≥ ⌊kθ⌋/k → w.

Worked case

For ℙ³ with O(1): V = 1 and P₀ = Δ(1,1,1), so ℓ = 1. Take r = 8·10⁶, so w = 1/200. Step 1 gives (1,1) → (1/200, 200), with 3w = 0.015 < 1/16. Step 2 gives (200,1) → (1/200, 40 000), with 2w = 0.01. Check: rw = 40 000 and the product of the intercepts is still 1. [2:47] [3:31]

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