Bounded Stein degree for log Calabi–Yau boundaries

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Bounded Stein degree for log Calabi–Yau boundaries

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You pick a boundary component S of a log Calabi–Yau pair over a field that is not algebraically closed, and you explain why the number of Galois-conjugate pieces it breaks into stays below a constant N(d,t). That constant depends only on the dimension and on how large the coefficient of S is.

What is being bounded

Over the algebraic closure, S can split into conjugate components. Their number is c(S/k) = [k_S : k], where k_S is the field of constants inside k(S). It also bounds the Stein degree dim_k H⁰(S, 𝒪_S) [0:08]. The theorem: if (X,B) is projective and lc, K_X + B ∼_ℚ 0 and H⁰(X,𝒪_X) = k, then every component with coefficient ≥ t has c(S/k) ≤ N(d,t). The same N works over every field of characteristic zero [0:44].

The worked case: curves

For d = 1, deg B = 2 − 2g, so g = 0 and deg B = 2. S is a closed point, so t·[k(S):k] ≤ 2. The point x⁴ = 2 on ℙ¹ over ℚ with coefficient ½ uses up that budget exactly. With coefficient 1/m you can take a point of degree 2m. Without a threshold t there is no bound [1:15]. Check: with coefficient 2/5 the largest degree is 5 [2:08].

Higher dimension: three branches

Run the MMP to a Mori fibre space X′ → Z and keep S as a valuation. Then ask where S sits [2:27]:

Z is a point. Bounded complements give n(K + C) ∼ 0 with n = n(d,t), so X′ is 1/n-lc and BAB puts it in a bounded family. On a bounded resolution, an lc place is fixed by a stratum plus integer weights. Galois permutes at most M strata, so every orbit has at most [F:k]·M! elements [3:00].
S horizontal. Restrict to the generic fibre, a smaller log Calabi–Yau pair over k(Z). Use c(S/k) ≤ c(S_η/k(Z)) and induction [3:50].
S vertical. S was made big, so some coefficient-one divisor E is horizontal. A relative MMP turns the prime into a pullback that meets E. Adjunction to E with induction bounds the trace J. Then at most 2/b components pass through a codimension-two locus, so c(S/k) ≤ (2/b)·c(J/k) [4:08].

Rule of thumb: a horizontal prime goes to the generic fibre. Adjunction is only needed in the vertical case [4:57]. The open direction is Birkar's conjecture for log Calabi–Yau fibrations over a positive-dimensional base [5:23].

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