Section 3
Elementary estimates for refined series
Lemma 3.1 (Complete series on a rational curve). Let a space of sections on have a fixed effective rational divisor and complete moving part . Its dimension is , and its mean order at a point is .
Proof. The complete moving series has a basis of orders at . Restoring the fixed factor adds to every order. ◻
Lemma 3.2 (Toric alpha invariant). Let be well formed, , and , where are the coordinate divisors and . Then . For and , the local alpha invariant of is at least on .
Proof. Take with and clear denominators. A generic one-parameter subgroup of the torus degenerates its defining polynomial to a monomial. The resulting divisor is , where and . If , then for every . The toric pair is log canonical, so lower semicontinuity of log canonical thresholds gives the same conclusion for . This proves the lower bound. Taking gives the reverse inequality at the generic point of .
On , sections of are polynomials in of weighted degree at most , for weights . At a point of , translate to put the point at the origin. This preserves the degree bound and the boundary. Monomial degeneration there gives threshold at least . Away from , both coordinates may be translated and the same argument without boundary gives at least . This proves the local assertion. The semicontinuity argument applies to the germ at the fixed origin; translations do not increase weighted degree. ◻
Lemma 3.3 (Residual-degree estimate). Let be a projective klt surface, let be ample and -Cartier, and suppose . Let be a prime curve. Assume that sections of have class and that their orders along are integral. Set and suppose the limit exists. If , then for every divisorial valuation whose centre is not contained in . The same conclusion holds for a chosen valuation if the local alpha bound holds at some point of its centre.
Proof. Choose a basis in every compatible with and , using Lemma 2.3. A basis section with order along has divisor , where and . Ampleness implies that the residual coefficient is nonnegative; if it is zero, . Since the centre of is not contained in , one has . The global or local alpha bound gives . Summing and dividing by proves the claim in the limit. Compatibility with computes the left side, while compatibility with fixes the multiplicities . ◻