Canonical nonvanishing for fourfolds via moving jets

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Canonical nonvanishing for fourfolds via moving jets

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You'll be able to explain how the fourfold nonvanishing proof gets its first pluricanonical form. It never writes the form down. It takes a hypothetical counterexample and traps the vanishing order of one determinant between a ceiling and a floor that can't both hold.

The setting and the counterexample

[0:08] For a smooth projective complex variety, K_X pseudo-effective means X is not uniruled. Nonvanishing asks for a single nonzero section of mK_X. That's much less than abundance, which asks for semiampleness. The theorem needs no condition on numerical dimension or irregularity.
[0:47] Suppose it fails. Fix a terminal minimal model Y with K nef, κ = −∞ and K⁴ = 0. Fujino's and Ambro's results rule out q(Y) > 0 and nef dimension ≤ 3. So every curve through a very general point has K·C ≥ 1/ι, and every proper subvariety through it is of general type.

Ceiling and floor

[1:35] Ceiling. The classes L_t = r_t(K + tA) have small volume (≈ ε⁴), but their curve degrees go to infinity. A section that vanishes too much at a moving point leaves a persistent base component. Differentiating in the point (Ein–Küchle–Lazarsfeld) and using general type then gives a curve of bounded degree, which is a contradiction. So ord_x(s) ≤ 32kε.
[2:34] On the two-slot bundle Z over Y×Y (dimension 9, P = L₁ + L₂ + qξ), the Seshadri constant is greater than 10, so P separates many jets.
[3:13] Floor. The finite data are reduced mod a large prime p. There, the evaluation map Φ_p has generic rank R = p⁸. On the diagonal, the Frobenius filtration (truncated powers of the cotangent bundle) pushes the rank below R/4.
[4:02] Think-first: a rank drop below R/4 forces the determinant to vanish to order ≥ 3R/4.
[4:33] The fixed movable curve caps that same order at about 80ε·R. Since 80ε < 3/4, the ceiling sits below the floor, so the counterexample cannot exist.

The reading rule

[5:13] Follow one number: the vanishing order of one section at one point. Find the estimate that caps it and the one that forces it, then check the inequality between them. The proof gives no bound on m. Getting from the first section to log abundance is the job of the companion work.

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