Section 14
Refinement by the exceptional plane of a point blow-up
Fix , and write and . Then , and . In particular, . We reserve for the exceptional divisor of .
Put and . After trivializing near , define The inclusion sends a section to its homogeneous leading term of degree at . Equivalently, is the image of restriction from to . In particular, it is the image of restriction, not a complete linear system by definition. We have .
For a prime divisor over , let be the largest value of where is a basis of , and where orders mean orders of the corresponding zero divisors on . Write for its asymptotic limit and define This is the stability threshold of a multigraded series on ; it is not automatically the threshold of a complete linear system on .
Lemma 14.1 (Abban–Zhuang at a point). For every , (14.1) Moreover, putting , every divisorial valuation centred at satisfies (14.2) Consequently, if and , then for every such .
Proof. The blow-up is a plt extraction and the different on is zero. The refinement has bounded support, since the order-at- filtration of an ample section ring is linearly bounded, and it contains an ample series by the complete jets in Lemma 14.2. Apply [4, Theorem 3.3] to and . Every centre on maps to , so the second term in that theorem is precisely . This gives (14.1).
Here is also the discrepancy calculation underlying the assertion. For an -basis divisor compatible with , write Then is not a component of , and is a basis divisor of . The coefficient of in the boundary of the pullback of is . It is at most one when . Inversion of adjunction therefore gives the stated minimum, after passage to the limit. Using a basis compatible with both and retains the additional discrepancy term. This is [4, equation (3.6)]; since , it gives (14.2). If the two hypotheses in the last assertion hold, use the same inequality with . For a valuation centred at , , proving strict positivity. ◻
The last assertion concerns valuations whose centre equals . It does not, by itself, prove , since the definition of also includes positive-dimensional centres containing .
Lemma 14.2 (The initial nef interval and the complete jets). The line bundle is very ample. For every point , If lies on a curve with , the nef threshold is exactly one. If , then (14.3) Such a line passes through every point of the set in Proposition 10.2.
Proof. By Lemma 10.5, the morphism is a closed embedding in . Hence is the restriction of the very ample bundle . Choose a section nonzero at and three sections vanishing at whose linear terms are independent. Their degree- products realize every homogeneous polynomial of degree as a leading term. This proves the equality of jet spaces.
Hyperplanes through a point of an embedded variety generate its maximal ideal, so is globally generated. Thus is nef for . Its cube is , giving its volume. On the strict transform of it has degree , proving the threshold assertion. For , a section in restricts to zero on : a nonzero section of cannot vanish to order greater than . Its degree- leading term therefore vanishes at , proving (14.3).
Every fibre of has -degree zero and -degree one. The reduced support of a double conic fibre has -degree one and -degree zero. These give the required lines through points of . ◻
Thus the part of the volume integral with is . This is only part of ; it must not be used as its full value. Likewise, completeness of the first jets does not justify completeness of the higher jets.
Lemma 14.3 (Bounding point multiplicities by a smooth flag). Let be a flag of subvarieties smooth at , of codimensions . Write and . Choose regular formal parameters at such that and locally, and trivialize there. For a nonzero section , define by first taking its order along , then the order along of , and finally the order at of the remaining nonzero restriction to after division by .
Choose a basis compatible with these lexicographically ordered values, and write . Suppose that the following limits exist: They do not depend on the compatible basis: the number of its vectors with any prescribed value is the dimension of the corresponding graded quotient. If is the exceptional divisor of , then . For the flags of Section 12, the limits are respectively , and hence
Proof. If , the formal expansion of has a nonzero coefficient. Its multiplicity at is therefore at most . The two-filtration lemma [4, Lemma 3.1] supplies a basis of compatible both with multiplicity at and with the flag values. Indeed, the finitely many flag values occurring in , ordered lexicographically, define a finite filtration by subspaces. The mean multiplicity of this basis, divided by , is . Summing the preceding inequality over the basis gives Passing to the limits proves the inequality. In the flags of Section 12, the surface and curve refinements give the three limits from their definitions and the integrals of Lemma 12.1. ◻
Proposition 14.4 (An explicit estimate for the point extraction). If lies on a smooth conic fibre, then (14.4) In addition, the cubic-surface flag can be chosen to give . Thus one may take the maximum of this numerical bound and the last bound in (14.4). For comparison, if the conic fibre through is reduced and singular, then . If it is smooth and , then .
Proof. First suppose and put . Choose a smooth cubic containing , as in Lemma 10.6. Under the identification , the point is . The line has direction at , and varies on the parameter line of cubics containing . The image of is a fixed smooth plane conic through . At most one direction is its tangent direction at . Avoid it, as well as the finitely many singular cubics. This makes . Put and ; since , we have .
Lemmas 14.3 and 12.3 now give Subtracting from proves the beta bound, and (14.1) gives the last inequality in (14.4).
For this same choice of cubic surface, Lemma 12.1 gives In particular, if , this flag already proves .
For a reduced singular fibre, use a line component and Lemma 12.6. The sum is at most . For a smooth fibre outside , choose avoiding . The sum is at most . ◻
These are upper bounds for , not asserted exact values. Their proof uses no Zariski decomposition on . The surface calculations already made supply an upper bound for ordinary point multiplicity, and Abban–Zhuang refinement then passes to the actual jet series on .
Remark 14.5 (Computing the remaining series without a threefold decomposition). There is a concrete algebraic description of every required jet space. The incidence model of Lemma 10.5 is a complete intersection of bidegrees and . Its Koszul resolution gives Surjectivity onto follows from the same resolution and the vanishing of the relevant intermediate cohomology of line bundles on .
On an affine chart at , solve the two equations formally for two of the five affine coordinates, using the rank-two Jacobian. The other three are regular parameters. Evaluation of the representatives of in these parameters gives the jet map Writing , we obtain the exact formulas The degree- terms of elements of also give the actual homogeneous polynomials spanning . Thus jet computations can replace a threefold Zariski decomposition. However, finitely many computed degrees do not establish the necessary asymptotic description, and the ranks alone do not determine : the base ideals and their infinitely near behaviour matter.
Example 14.6 (The normalization on projective space). For and , the same construction gives for . Here The weighted average degree of this complete series on is three. Since , its threshold is . The point extraction therefore gives , the reverse inequality being supplied by . In Family 2.9, the base condition (14.3) prevents this completeness argument at the points of .
To complete this alternative point-extraction proof at the points in (14.4), one must still prove , or prove and separately treat positive-dimensional centres. The surface blow-up proof in Proposition 12.4 already gives the required local bound on . For nonreduced conic fibres, Section 18 instead uses a weighted curve extraction. These assertions do not follow from the calculations of this section alone. Sections 15 and 16 improve the point-extraction term by genuine threefold decompositions, and Section 17 proves in the smooth-fibre case. Positivity of alone would not imply positivity for every divisor lying above .