Maximal Seshadri constants on arbitrary polarized surfaces

Proof. A nonzero N\times N minor of the limiting jet matrix remains nonzero for all sufficiently small s, since each of its entries converges. The final assertion follows from the Cauchy integral formula on small polydiscs around the finitely many points.□

Proof. The statement is immediate if F=0, so write N=\dim F>0.

Choose a basis with distinct initial terms. Among the initial exponents of nonzero elements of F, choose the least one, e_1. Every element of F has zero coefficient at every earlier exponent. The coefficient at e_1 is a nonzero linear functional on F. Choose f_1 on which it equals 1, and continue inside its codimension-one kernel. After N steps this gives a basis f_1,\ldots,f_N with strictly increasing initial exponents e_i=(\alpha_i,\beta_i) and initial coefficients 1. In a nonzero linear combination of the basis vectors, the first vector with nonzero coefficient supplies its uncancelled initial term. Consequently E_t(F)=\{e_1,\ldots,e_N\}.

Separate the initial terms by a single positive weight. Set

Choose a small number \eta>0. For \eta sufficiently small, all the weights \alpha_i+(t+\eta)\beta_i remain below M. Every term of old weight at least M has new weight at least M, since its second exponent is nonnegative, so it cannot become a least term.

Only finitely many exponent pairs have old weight below M. For a noninitial term of f_i in this finite set, either its old weight is strictly greater than that of e_i, or its old weight is equal and its second exponent is strictly greater. In the first case the strict inequality persists for all sufficiently small \eta>0. In the second case increasing t by \eta makes the new weight strictly greater. One common choice of \eta>0 therefore makes e_i the unique term of least weight (1,t') in every f_i, where t'=t+\eta. Put

Dilate the convergent series. Choose representatives of the finitely many basis germs on a common polydisc, and define, for positive real s,

These functions converge locally uniformly on \C^2 to X^{\alpha_i}Z^{\beta_i}, in the sense that they are defined on any fixed compact set for all sufficiently small s. To verify the convergence, write f_i(x,z)=\sum_{\alpha,\beta}c_{i,\alpha,\beta}x^\alpha z^\beta. Fix R>0 and choose 0<s_0<1 so that the radii (s_0R,s_0^{t'}R) lie strictly inside a common polydisc of absolute Taylor convergence. Every supported exponent satisfies \alpha+t'\beta-\lambda_i\ge0, with equality only at e_i. For 0<s\le s_0, the sum of uniform majorants for the rescaled terms on |X|,|Z|\le R is

Each noninitial term tends uniformly to zero. The summable majorant therefore proves the asserted uniform convergence. Applying this argument on a slightly larger polydisc and using the Cauchy integral formula gives convergence of all derivatives of any fixed finite order on the smaller polydisc.

Transfer the jet test. At the given points q_j, the limiting monomials have an injective jet test. Lemma 2.1 shows that the functions g_{1,s},\ldots,g_{N,s} have an injective jet test at the same points for every sufficiently small positive s. The map

is invertible for each such s. Its images p_j=D_s(q_j) are distinct, lie in the common coordinate neighborhood, and approach the origin. Pullback by D_s preserves orders of vanishing at corresponding points. The nonzero scalar factors s^{-\lambda_i} change only the basis. Hence the original space F has an injective order-m jet test at p_1,\ldots,p_r.□

Proof. Choose an integer e>0 such that eL is very ample, embed S in \PP^N using this line bundle, and choose any p\in S. Take homogeneous coordinates with p=[1:0:\cdots:0]. The restrictions of X_1,\ldots,X_N generate \mathcal I_p\otimes\mathcal O_S(eL): away from p at least one is nonzero, and near p their ratios with X_0 generate the maximal ideal because the embedding is a closed immersion.

Let \pi:X=\operatorname{Bl}_pS\to S be the blow-up, with exceptional curve E. The generators just described give a surjection onto the twisted Rees algebra

Taking relative \operatorname{Proj} gives a closed immersion X\hookrightarrow S\times\PP^{N-1}, the closure of the graph of projection from p. The second factor pulls \mathcal O(1) back to \mathcal O_X(e\pi^*L-E); see also [19, Tag 01OF]. After embedding the first factor in \PP^N, the very ample bundle \mathcal O(1,1) therefore restricts to

In particular, \mathcal A^{\otimes2}=\mathcal O_X(4e\pi^*L-2E) is very ample.

The point blow-up X is a smooth integral projective surface. By Bertini, a general divisor \widetilde V in this last linear system is smooth and integral [19, Tags 0FD6 and 0G4F]. We may also require that it meet E transversely at two distinct points. Indeed, \mathcal A|_E\simeq\mathcal O_{\PP^1}(1), and the restrictions of its global sections generate this bundle, hence span its full two-dimensional space of sections. Products of these sections show that

is surjective. Having two distinct zeros on E is consequently a nonempty open condition, which can be imposed along with the Bertini conditions.

Set d=4e and V=\pi(\widetilde V). Pushing down the divisor class shows that V\in|dL|, and V is integral because \widetilde V is. The blow-down is an isomorphism away from E, so V is smooth away from p. In a local blow-up chart the map has the form (a_0,u_0)\mapsto(a_0,a_0u_0) with E=\{a_0=0\}. A curve transverse to E has a_0 as a local parameter, and its image is a smooth branch. The two different points of \widetilde V\cap E give different tangent directions. Thus the two image branches are smooth and transverse at p.□

Proof. First suppose that the divisor D of s does not contain V, and write f(x,z) for its local function. Neither f(x,0) nor f(0,z) vanishes identically. To see this, D\cap V is a proper closed subset of the integral algebraic curve V, hence is finite. An identically vanishing branch restriction would put an entire analytic branch in this finite set. This argument uses the global integrality of V without asserting that its nodal analytic germ is irreducible.

Let \ell_x=\ord_0f(x,0) and \ell_z=\ord_0f(0,z), and let v be the weight of \nu_t(f). The pure terms defining these two finite orders have weights \ell_x and t\ell_z, respectively. Therefore \ell_x\ge v and \ell_z\ge v/t. Intersection multiplicity is additive over the two smooth branches, so

The analytic computation gives the algebraic local intersection number: both are the colength of (f,xz) in the common completed local ring \C[[x,z]]. The final inequality in (3.4) follows from nonnegativity of all the other local intersection numbers. Consequently v\le ka, and

since a=tb.

For a general nonzero s, let j be the coefficient of V in its divisor. Dividing by the defining section of this effective Cartier divisor gives

where the effective divisor R of \rho does not contain V. Intersecting with the ample line bundle gives nH=L\cdot R\ge0, so n\ge0. If n=0, then R=0 by ampleness and \rho is a nonzero constant, since S is integral and projective. If n>0, the preceding argument gives \nu_t(\rho)\in n\Delta_t.

The order on exponents is compatible with addition, so initial exponents add under multiplication. A unit has initial exponent (0,0). Equation (3.1) therefore yields

When n>0, this implies

because the two coefficients are nonnegative and sum to one. When n=0, the normalized exponent equals (c,c) directly.

Finally, restriction to the analytic chart is injective on global sections, by the identity theorem. Apply the finite coefficient elimination in Proposition 2.2 to this finite-dimensional space of germs. It gives a basis with distinct initial exponents, each of which satisfies the bound just proved.□

Proof. The strict inequality defining t is equivalent to

Thus (c,c) lies strictly beyond the segment joining (a,0) to (0,b), on the side away from the origin. In the coordinates (q,\beta)=(\alpha-\beta,\beta), the four vertices of Q_t, in cyclic order, are

For 0\le q\le a, the slice runs from \beta=0 to \beta=c(1-q/a). For -b\le q\le0, it runs from

Subtracting the endpoints gives the stated formula. The region below this tent is precisely P_t, as illustrated in Figure 1.□

Proof. Use the torus coordinates

A monomial x^\alpha z^\beta becomes u^qv^\beta, where q=\alpha-\beta. For every nonempty integer slice q of kQ_t, write

Convexity of the slice implies that all integers between these endpoints occur. Hence the slice has the basis

Indeed, expanding (v-1)^h gives a triangular change of basis from the consecutive powers of v, with diagonal entries equal to one. By Lemma 3.3,

Consequently every pair (q,h) in (3.6) belongs to kP_t\cap\Z^2. These pairs are distinct, both within each slice and between different slices.

Let (u_i,y_i), 1\le i\le r, be distinct torus points giving an injective jet test for \mathcal M(kP_t), now written in the variables (u,y). For a positive parameter s, substitute v=1+sy into (3.6) and divide its member indexed by h by s^h. The resulting functions are

The convergence is locally uniform, with every spatial derivative, on fixed neighborhoods of the finitely many test points. To see this, choose bounded neighborhoods on which u stays away from zero and then take s small enough that |sy|<1/2 there. The factors (1+sy)^{\beta_q^-} are holomorphic and converge uniformly to one on slightly larger neighborhoods, so the derivative assertion follows from the Cauchy estimates. The argument also applies to negative Laurent powers, since 1+sy stays away from zero.

The distinct limiting monomials in (3.7) span a subspace of \mathcal M(kP_t), so their jet matrix has full column rank. Lemma 2.1 gives the same conclusion for the functions on the left for every sufficiently small positive s. These functions form a rescaled pullback of a basis of \mathcal M(kQ_t).

For such an s, the corresponding points in the original torus are

They lie in \T because u_i\ne0 and 1+sy_i\ne0. They are distinct: equality of two images forces equality first of their u coordinates and then of their y coordinates. The coordinate map

has Jacobian determinant s/u\ne0, and therefore preserves orders of vanishing at these points. The full-column-rank conclusion is thus exactly the required injectivity for \mathcal M(kQ_t).□

Proof. If H^0(S,kL)=0, the conclusion is immediate. Otherwise, Proposition 3.4 gives such a jet test for \mathcal M(kQ_t). Trivialize L in the fixed coordinate chart at the node and regard the sections of kL as holomorphic germs there. This identification is injective: a global section vanishing on a nonempty open subset of the integral surface vanishes identically. Proposition 3.2 gives a basis whose distinct initial monomials lie in kQ_t. Their span therefore has injective jets at the torus tuple just constructed. Applying Proposition 2.2 gives injective jets for the section germs at distinct points in the chart, and hence for H^0(S,kL) on S.□

Proof. For a height y in the vertical projection of \Delta, let \ell(y) be the horizontal length of the slice of \Delta at that height. Write y_0\leq y_1\leq y_2 for the three vertex heights; thus y_2-y_0=\rho. If these heights are distinct, \ell increases linearly from zero at y_0 to its maximum at y_1, then decreases linearly to zero at y_2. If two vertex heights coincide, \Delta has a horizontal side and there is just one linear segment. In either case, the area is one half the vertical range times the maximum slice length. The maximum is therefore \rho.

More precisely, for 0\leq e\leq\rho, the heights with \ell(y)\geq e form an interval of length \rho-e (Figure 2). When the three heights are distinct, its endpoints are

The same formulas apply when two adjacent heights coincide, including the cases e=0 and e=\rho.

Write F=\sum_q f_q(U)Z^q, with each displayed row f_q nonzero, and let e be the least order of these rows at U=1. Dividing every row by (U-1)^e gives

Indeed, at least one d_q(1) is nonzero, and distinct powers of Z are linearly independent. All d_q remain Laurent polynomials, including when some exponents are negative.

The horizontal exponent span of the row f_q=(U-1)^e d_q is exactly e plus that of d_q: the extreme coefficients of the product are nonzero. Thus \ell(q)\geq e for every height occurring in D. There is no cancellation between distinct heights, since the multiplying factor depends only on U. It follows that e\leq\rho and that the heights occurring in D span an interval of length at most \rho-e.

On setting U=1, some rows may disappear, but no new height appears. The nonzero Laurent polynomial D(1,Z) therefore has exponent span at most \rho-e. Multiplication by a power of Z, which is a unit at Z=1, turns it into an ordinary polynomial of degree at most \rho-e. Consequently

The first inequality holds because restriction cannot introduce a term of smaller total degree into the local Taylor series. Orders add under products of nonzero holomorphic germs, so \ord_{(1,1)}F\leq\rho. All exponent spans and orders used here are integers; no integrality of \rho, the vertices, or the translating vector is required.□

Proof. Complete u to an integer basis (u,v_0) with \det(u,v_0)=1. Put v=rv_0 and

This sublattice has index r. For \eta=(\eta_1,\eta_2)\in\Z^2, write \chi_\eta(x,z)=x^{\eta_1}z^{\eta_2} for the corresponding character on \T, and define

The characters X=\chi_u and Y=\chi_{v_0} are torus coordinates: their exponent matrix is unimodular, so its inverse also has integer entries. In these coordinates,

Over \C, this consists of exactly r distinct points.

Suppose, toward a contradiction, that a nonzero f\in\mathcal M(kP) has order at least m at every point of G. For g\in G, let (\tau_gf)(p)=f(gp). Translation by g permutes G and is locally invertible, so every \tau_gf has the same prescribed vanishing. For a coset \eta+M, its character projection is

This is exactly the sum of the monomials of f with exponents in \eta+M. Indeed, if \xi-\eta=n u+\ell v_0, then \chi_{\xi-\eta} restricts to \zeta^\ell on G. The sum over the rth roots of unity gives

Thus the displayed projection keeps precisely the required coset. Each f_\eta still has order at least m at every point of G, and its support remains inside kP. Since the components sum to f, choose one that is nonzero.

Multiplication by the Laurent monomial \chi_{-\eta} preserves local orders and shifts this component’s support into M. There is therefore a nonzero Laurent polynomial F such that

The map (x,z)\mapsto(U,Z) is locally invertible at the identity: in coordinates (X,Y) it is (X,Y)\mapsto(X,Y^r), whose derivative at (1,1) is \operatorname{diag}(1,r). The complex inverse function theorem therefore makes it a biholomorphism on sufficiently small analytic neighborhoods. In particular,

Only local invertibility is used; the map need not be injective on the whole torus.

Let N be the matrix with columns u,v. The exponent support of F lies in the real triangle

Since \det N=r, this triangle has area k^2H/(2r). The second coordinate of N^{-1}\xi is \det(u,\xi)/r, so its vertical range is k\sqrt{rH}/r=k\sqrt{H/r}. Translation by \eta changes neither calculation. Lemma 4.1, with \rho=k\sqrt{H/r}, now gives \ord_{(1,1)}F\leq k\sqrt{H/r}<m, the required contradiction.□

Proof. Put W=dH and a_0=WT/(1+T). As t ranges over (T,\infty), a=Wt/(1+t) ranges continuously and bijectively over (a_0,W). For R=\sqrt{rH} sufficiently large, choose j to be the largest power of two at most R/(2W). Then

The open interval

has length

which exceeds 2 for every sufficiently large r. Hence I_r contains an odd integer i. Its lower endpoint is positive, so i>0; as j is a power of two, the vector u=(i,j) is primitive.

The strict inequalities defining I_r imply

Choose the unique t>T for which a=a_*. Since b=W-a, we have

The determinant functional has the three vertex values

Its minimum is therefore -ja and its maximum is i/d, giving the required range R exactly. All lower bounds on r in this construction depend only on the fixed data d,H,T.□

Proof. First fix r and choose t and u by Lemma 4.3. The area of P_t is H/2, so Proposition 4.2 produces G with the stated property simultaneously for all k,m. Thus r_0, and then t,G, are chosen before these two jet parameters.□

Proof. Let q_1,q_2:S\times S\to S be the projections and let \mathcal I_\Delta be the ideal of the diagonal. The bundle of jets of orders less than m is

This is the algebraic principal-parts construction [19, Tags 0G3P and 0638]; it is locally free of rank \binom{m+1}{2}. Indeed, the diagonal is a regular embedding with conormal bundle \Omega_S^1; the successive quotients of the displayed sheaf, pushed forward to S, are L^{\otimes k}\otimes\operatorname{Sym}^j\Omega_S^1 for 0\le j<m. They are locally free of ranks j+1. The thickened diagonal is finite over S, so these pushforwards are exact and give the asserted filtration. The fiber at p is (L^{\otimes k})_p/\mathfrak m_p^m(L^{\otimes k})_p.

Writing \operatorname{pr}_i:U_r\to S for the coordinate projections, consider the algebraic bundle map

Its fiber kernel consists exactly of sections vanishing to order at least m at every point. If n=h^0(S,L^{\otimes k})>0, failure of injectivity is cut out locally by the n\times n minors. If n=0, the map is injective everywhere. In either case the injective locus is Zariski open, and a single injective tuple makes it nonempty. The analytic and algebraic finite jets agree at complex points: in local coordinates both are the same Taylor polynomial modulo terms of total degree at least m. Thus a tuple obtained by the analytic transfers of the preceding sections is a valid witness here.

The variety U_r is smooth and irreducible. A proper algebraic closed subset has empty interior in its complex manifold of points. For a countable collection of nonempty injective loci, their complements are therefore closed nowhere-dense subsets of U_r(\C). This locally compact Hausdorff space is a Baire space, so their common complement is nonempty, indeed dense. By construction it is the complement of a countable union of proper Zariski-closed subsets of U_r, as required for very general tuples.□

Proof. Choose a positive integer q such that qD is Cartier and let k run through positive multiples of q. By Serre duality [19, Tag 0FVV, Lemma 48.27.1(6)–(7)],

For sufficiently large k,

A divisor with a nonzero section is linearly equivalent to an effective divisor, which has nonnegative intersection with the nef class A. Hence the displayed h^2 vanishes for all such k. Surface Riemann–Roch [19, Tag 02UO] now gives

for sufficiently large k, since D^2>0.□

Proof. Choose a Cartier divisor representing L and retain the notation L for it. Write A=\pi^*L and E=\sum_iE_i. The class A is nef and

If D_0 were not nef, there would be an integral curve C\subset Y with D_0\cdot C<0. Since D_0\cdot E_i=w>0, this curve is not exceptional, and A\cdot C=L\cdot\pi_*C>0. For sufficiently small positive rational \delta, both

hold. Fix such a \delta. By continuity, there is a rational number s>w, sufficiently close to w, such that

Here A\cdot D=H-\delta A\cdot C is independent of s.

Choose a sufficiently large positive integer k divisible by the denominators of both s and \delta. Lemma 5.2 supplies an effective divisor F\sim kD. Since k\delta and m=ks are positive integers,

is effective. The blow-up identity [19, Tag 0AGP, Lemma 54.3.4(4)]

and the projection formula give a nonzero section of L^{\otimes k} vanishing to order at least m at every p_i. To recall the local content of this identity, sections of a pullback line bundle descend across the missing points because S is normal, and the exceptional order of a local equation on the blow-up of a smooth point is its ordinary multiplicity. Thus the condition at each E_i is precisely membership in \mathcal I_{p_i}^{\,m}. But m=ks>kw, contradicting the hypothesis. Therefore D_0 is nef.□

Proof. Put H=L^2, let \pi:Y\to S be the blow-up, and write A=\pi^*L and E=\sum_iE_i. Suppose A-\lambda E is nef for some \lambda\ge0. Fix 0<\alpha<\sqrt{H/r} and set m_k=\lceil k\alpha\rceil. The proof of Lemma 5.2, applied on S with A=D=L, gives

The dimension of the target of the order-m_k jet evaluation at the r points is

Since r\alpha^2<H, evaluation has a nonzero kernel for all sufficiently large k. A section in that kernel gives an effective divisor F_k\sim kA-m_kE on Y. Nefness implies

Letting k\to\infty gives \lambda\le H/(r\alpha), and then letting \alpha\uparrow\sqrt{H/r} gives \lambda\le\sqrt{H/r}. Taking the supremum over the nef coefficients \lambda proves the claim.□

Proof of Theorem 1.1. Choose the nodal divisor and parameters in Section 3. Corollary 4.4 supplies an integer r_0, depending only on these choices and hence on (S,L), with the following property. For every integer r\ge r_0, there is a parameter t>T and an r-point torus tuple such that the jet test on \mathcal M(kP_t) is injective for every pair of positive integers k,m with m>k\sqrt{H/r}.

Fix any such integer r. For each of these pairs, Corollary 3.5 gives an injective jet test on H^0(S,L^{\otimes k}) at some tuple of r distinct points of S. The tuples may depend on k,m. Lemma 5.1 makes all these countably many tests hold simultaneously at very general tuples. Proposition 5.3 proves that the class \pi^*L-\sqrt{H/r}\sum_iE_i is nef at every such tuple. Combining this lower bound for the Seshadri constant with Lemma 5.4 proves the required equality. The integer r was arbitrary subject to r\ge r_0, and r_0 was fixed before k,m were chosen. This proves the assertion for every sufficiently large integer r.□

Proof. Descend the projective equations and the finite gluing data for L to a subfield F\subset K finitely generated over \mathbb Q. This gives a smooth geometrically integral projective surface S_0/F and an ample line bundle L_0 whose base change is (S,L); these properties descend through field extension, and the finite-data descent and ampleness criteria are recorded in [19, Tags 01ZM and 0D3C]. The field F is countable and embeds into \C. Fix such an embedding and choose r_0 from Theorem 1.1 for (S_{0,\C},L_{0,\C}). The intersection number H=L_0^2=L^2 is unchanged by field extension.

Fix r\ge r_0 and put w=\sqrt{H/r}. For every pair of positive integers a,m with m>aw, form on U_{r,0}=U_r(S_0) the evaluation map

using the principal-parts construction in the proof of Lemma 5.1. Its noninjectivity locus B_{a,m} is closed, cut out by maximal minors, and is empty if the source space is zero. Principal parts and these minors commute with field extension, as does H^0 by flat base change [19, Tag 02KH]. Each B_{a,m} is proper over \C: at a tuple where A-wE is nef, with A=\pi^*L_{0,\C} and E=\sum_iE_i, a section in the evaluation kernel would give an effective divisor of class aA-mE, whose intersection with A-wE is aH-rmw<0. Thus B_{a,m} is proper over F and remains proper over K. The single threshold r_0 works for all these pairs.

Let Z_r=\bigcup_{a,m:\,m>aw}B_{a,m,K}. This is a countable union of proper closed subsets. Its complement contains a K-point: the function field F(U_{r,0}) has transcendence degree 2r over F and embeds over F into K. Indeed, choose 2r algebraically independent elements of K over the countable field F and extend the resulting embedding of a rational function field across the finite algebraic extension, using that K is algebraically closed. The resulting K-point maps to the generic point of U_{r,0} and therefore avoids all the B_{a,m}.

At every tuple outside Z_r, all the homogeneous jet tests are injective. The proofs of Lemma 5.2, Proposition 5.3, and Lemma 5.4 use only surface Riemann–Roch, Serre duality, and the blow-up ideal identity, and are valid over K. They give nefness of \pi^*L-w\sum_iE_i and the matching upper bound, respectively.□