Arithmetic Stein-degree bounds for log Calabi–Yau pairs

1. Preserve the constants and the boundary component

Lemma 2.1 identifies $c(S/k)$ with the number of geometric components, so tracking the same divisorial valuation through birational operations preserves this degree. Lemma 4.1 supplies a $\mathbb Q$-factorial klt model with effective crepant boundary. Run the MMP for $K+B-bS\sim_{\mathbb Q}-bS$. Every contracted extremal ray is positive for $S$, and negativity prevents $S$ from being contracted. On the resulting $X_1\to Z_1$, the transform $S_1$ is relatively ample and retains its coefficient lower bound $t$.

Lemmas 2.1, 4.1–4.2 · pp. 2–3, 7–8

2. End on the generic fibre when the base has positive dimension

If $\dim Z_1>0$, relative ampleness makes $S_1$ horizontal. The generic fibre has global function field $k(Z_1)$ and is an lc log Calabi–Yau pair of dimension $r=d-\dim Z_1<d$ with threshold $t$. Induction and the comparison of constants for horizontal components give $c(S/k)\le M_{<d}(t)$. This branch needs neither a complement nor the second MMP.

Lemma 2.1; §5.2, (5.3) · pp. 2–3, 11

3. Choose a bounded-index complement when the base is a point

If $Z_1=\operatorname{Spec}k$, then $X_1$ is Fano and $-(K_{X_1}+bS_1)$ is nef. Lemma 3.1 applies bounded complements with the fixed coefficient $b$, then chooses a general member of a linear system over $k$. It gives an lc boundary $C_1\ge bS_1$ with $n(K_{X_1}+C_1)\sim0$. First choose $b=b(t)$, then $n=n(d,b)$. The next step uses this $n$ to control singularities without depending on the original boundary denominators.

Lemma 3.1; §5.2, (5.4) · pp. 4, 11 · Bir19, Theorem 1.7 · p. 4

1. Turn index discreteness into a uniform discrepancy bound

Corollary 3.3 makes the valuations with $a(v,X_1,0)<1/n$ a finite set. Their log discrepancies for the complement are nonnegative multiples of $1/n$, no larger than their original discrepancies, so they are all lc places. Extract exactly these valuations to obtain a $1/n$-lc variety $X_2$; every extracted divisor has coefficient one in $C_2$. After a $K$-MMP, $X_3$ remains $1/n$-lc, retains an lc complement of the same index, and satisfies $a(v_S,X_3,C_3)\le1-b<1$.

Corollary 3.3; §5.3, (5.5)–(5.6) · pp. 5, 11–12

2. End with the valuation-orbit bound over a point

If $\dim Z=0$, then $X_3$ is a $1/n$-lc Fano variety. Proposition 3.5 applies with $q=d$, $\epsilon=1/n$, and boundary $C_3$, giving $c(S/k)\le A(d,1/n,n)$. Its input is a valuation satisfying $a(v_S,X_3,C_3)<1$, so the strict transform of $S$ may already have been contracted. Section 4 explains this arithmetic bound.

Proposition 3.5; §5.3, (5.7) · pp. 6–7, 12

This exit uses the arithmetic proof of Proposition 3.5.

3. Recover the original valuation over a positive-dimensional base

If $\dim Z>0$, choose $\Gamma=(1-\lambda)C_3$ for sufficiently small $\lambda>0$. This is klt with relatively ample anti-log-canonical divisor. Extract only $v_S$ if it is exceptional; otherwise use the divisor already present. On the resulting model $U\to Z$ of relative Fano type, the prime $P$ satisfies $c(P/k)=c(S/k)$ and has coefficient at least $b$. The relation $n(K_U+C_U)\sim0$ is preserved.

§5.4 · p. 12 · LM26, Theorem 22.1 · p. 87

4. Use induction for a horizontal prime and adjunction for a vertical prime

If $P$ is horizontal, apply induction with threshold $b$ on the generic fibre of dimension $d-\dim Z<d$, obtaining $c(S/k)\le M_{<d}(b)$. A vertical $P$ disappears on that fibre, so the same induction cannot apply. Section 5 constructs a different horizontal coefficient-one component and transfers the intersection with $P$ to its normalization.

§5.4, (5.8); §§5.5–5.6 · pp. 12–14

For the vertical case, continue to adjunction retaining the same index.

1. Remove the polarization's descent obstruction after a bounded extension

BAB supplies a bounded very ample bundle $L$ over $\bar k$. The manuscript embeds the Picard group into a lattice of bounded rank and bounds its finite Galois image, obtaining a bounded extension $F/k$ fixing the class of $L$. Fixing a class does not finish descent of the bundle. With $r=h^0(L)$, the following determinant factor cancels the scalar two-cocycle and supplies a bounded very ample bundle $A$ on $Q_F$.

Proposition 3.5, (3.3) · pp. 6–7 · Bir21, Theorem 1.1 · p. 2

2. Bound a resolution containing the boundary support

Integrality of $n\Lambda$ and $n(K_Q+\Lambda)\sim0$ bound the degree of the boundary support with respect to $A$. Apply Lemma 3.4 to this embedding and support to obtain a resolution over $F$ and a reduced SNC divisor $H$ containing the boundary. A uniform bound $M$ on its geometric components and strata supplies the finite set needed in the group-action argument.

Lemma 3.4; Proposition 3.5, (3.4) · pp. 5–7

3. Fix each valuation by fixing components and strata

If $a(v,Q,\Lambda)<1$, then $a(v,W_{\bar k},H)$ is a nonnegative integer less than one, hence zero. Lemma 3.2 uniquely determines this valuation from its centre stratum and positive integral weights. Thus the Galois subgroup fixing every element of the finite set fixes each valuation, and its orbit has size at most $[F:k]M!\le A(q,\epsilon,n)$. There is no need to bound the total number of valuations.

Lemma 3.2; Proposition 3.5 · pp. 4–5, 7

1. Recover a horizontal coefficient-one component from earlier bigness

Since $S_1$ is ample on the Fano variety $X_1$, the divisor $\pi^*S_1$ on $X_2$ is effective and big. Its pushforward to $X_3$ remains big: the map extracts no divisors and preserves the required growth of sections. A divisor supported only in vertical components cannot restrict to a big class on the positive-dimensional generic fibre. The original $S$ has vertical centre, so a horizontal component $E_3$ comes from an extracted exceptional divisor and has coefficient one in $C_3$. Generic-fibre induction for this component gives $c(E_3/k)\le M_{<d}(1)$.

§5.5, (5.9)–(5.10) · p. 13

2. Use a relative MMP to force a nonzero intersection

On the relative Fano type model $U$, verticality of $P$ makes $-P$ relatively pseudo-effective. The big-boundary MMP of Lemma 4.3 preserves $P$ and the horizontal $E$, and basepoint freeness makes $-P'$ semiample. For its contraction $f:U'\to Z'$, the map $Z'\to Z$ is birational. Descending the canonical section gives the actual divisor identity $mP'=f^*T$, where $T\ne0$ is effective Cartier. Since $E$ is proper and horizontal, it surjects onto $Z'$; hence $P'|_{E^\nu}\ne0$. Choose a prime component $J$ of this restriction.

Lemma 4.3; §5.6 · pp. 8–9, 13

3. Preserve the index by residue and discretize positive coefficients

The effective lc pair $(U',C')$ has coefficient one along $E$, coefficient $\beta\ge b$ along $P'$, and $n(K_{U'}+C')\sim0$. Lemma 4.4 gives an effective lc different with $n(K_{E^\nu}+C_{E^\nu})\sim0$ for the same $n$. The manuscript's additional argument takes the residue of a rational $n$-canonical form over $k$ and recovers the degree-$n$ divisor identity from an identity for a sufficiently divisible power.

Equation (122.7) of Kol10 supplies the residue isomorphism used in this comparison. Equation (122.10) also assumes that $K+E$ itself is $\mathbb Q$-Cartier, so that formula alone does not settle the general case. The manuscript compares residues before and after adding $\beta P'$ at a common Cartier multiple, obtaining $C_{E^\nu}\ge\beta P'|_{E^\nu}$. Thus $J$ has positive coefficient; integrality of $nC_{E^\nu}$ makes it at least $1/n$.

Lemma 4.4, (4.2) · pp. 9–10 · Kol10, Definition 122, Proposition 123, Lemma 125 · pp. 61–64

4. Induct over the component's own constants and multiply degrees

Let $k_E$ be the constant field of $E^\nu$. Lemma 2.1 makes $E^\nu/k_E$ normal and geometrically integral with $H^0(\mathcal O_{E^\nu})=k_E$. This finite separable extension preserves the canonical divisor, so induction in dimension $d-1$ with threshold $1/n$ applies to the adjunction pair. Since $k_E\subset k(J)$, the bound for the horizontal component gives

Lemma 2.1; §5.6, (5.11) · pp. 2–3, 13–14

5. Count at codimension two and return to the original component

Over $\bar k$, the image of a component of $J$ has codimension two in $U'$. The geometric components of $P'$ form one Galois orbit, and each contains the image of a conjugate of $J$. Lemma 4.5 bounds by $2/b$ the number of components of coefficient at least $b$ through each image. It cuts down to a klt surface quotient and blows up a point on a smooth cover, bounding the boundary multiplicity by two. Consequently,

Lemma 4.5; §5.6, (5.12) · pp. 10, 14

6. Combine the four bounds and close the induction

The positive-dimensional base of the first MMP, the point base of the second, horizontal $P$, and vertical $P$ give four bounds. Choose an integer satisfying

First fix $b=b(t)$, then $n=n(d,b)$; every occurrence of $N$ on the right is already known in lower dimension. The original boundary denominators, the field, and the number of extracted divisors do not enter the dependence of the bound.

§5.7, (5.13) · p. 14