Maximal Multipoint Seshadri Constants in Higher Dimensions

Proof. If W\ne0, choose the smallest exponent occurring as the least exponent of an element of W, and normalize one such element to have coefficient 1 there. Continue in the kernel of that coefficient functional, which has codimension one. After \dim W steps this gives a basis with strictly increasing least exponents. In any nonzero linear combination, the first basis element with a nonzero coefficient determines the least term. This proves the assertion about the whole set of possible exponents.

For the product assertion, write a=\nu(f) and b=\nu(g). Every exponent of f is at least a, and every exponent of g is at least b. Additivity of the order implies that any pair other than (a,b) has sum strictly greater than a+b. The coefficient of x^{a+b} in the product is therefore the product of the two nonzero least coefficients.□

Proof. The cardinality assertion follows from Lemma 2.2 in each label. Given two exponents in S'_k and S'_l, choose functions witnessing them. Their nonzero product belongs to the summed degree and label, and its least exponent is the sum of the two least exponents. Expanding products of arbitrary linear combinations therefore gives \M(S'_k)\M(S'_l)\subset\M(S'_{k+l}). The full new monomial spaces thus retain the multiplication property needed for the next operation; no simultaneous choice of lifts to earlier spaces is required.□

Proof. First omit the unchanged factor U^{\gamma_j}. On a bounded polydisc in Z, choose s_0>0 small enough that the power series converges absolutely on a slightly larger polydisc after scaling by s_0^\lambda. In the normalized expansion every exponent \lambda\cdot(\alpha-e_j) is nonnegative, and every exponent other than the least one is strictly positive. For 0<s\leq s_0, the terms are dominated by the summable series obtained at s_0. Dominated convergence gives locally uniform convergence to Z^{e_j}. Applying Cauchy’s estimates on slightly larger polydiscs gives convergence of derivatives. The unchanged Laurent monomials and their derivatives are holomorphic near every compact subset of the torus, so multiplying by them preserves these conclusions.□

Proof. Choose coordinates and frames near the points. The entries of the jet matrix are the finitely many required Taylor coefficients. A maximal minor that is nonzero in the limit remains nonzero for small s. Coordinate changes and frame changes induce isomorphisms of truncated local rings or their rank-one modules. Rescaling an individual function rescales a matrix column by a nonzero constant.□

Proof. The lattice map is bijective and additive. The torus automorphism

has pullback \phi_B^*(x^\alpha)=x^{B\alpha}. Its inverse is the monomial map associated with B^{-1}\in\operatorname{GL}_n(\Z), so the jet assertion follows by change of coordinates.□

Proof. Use the coordinate u=z-1 near z=1. Lemma 2.2 identifies the possible orders in a row with a basis of distinct leading orders. If the input sets are graded, the product of two row spaces lies in the summed degree and remaining exponent. Lemma 2.3 therefore proves both cardinality preservation and the graded support property.

Multiplying a nonzero row element by the unit z^{-a_-} gives a polynomial of degree at most a_+-a_-. Its order at 1 is unchanged and cannot exceed that degree. This proves the row bound, including for Laurent exponents.

For a fixed degree, choose in every nonempty row a basis with distinct orders and leading coefficients 1 in u. Substitute z=1+sZ, normalize each basis element by s to its order, and carry along the unchanged monomial in the other variables. Lemmas 2.4 and 2.5 apply to all rows simultaneously. At a fixed finite torus tuple, the substitution has nonzero values and is injective for sufficiently small s>0. It gives a local coordinate change, so jet separation transfers to the original space at the image tuple.

Finally, a row contained in an interval slice of \kappa Q has exponent span no greater than the length of that slice. Its compressed row is therefore contained in the interval beginning at zero with that length. For \kappa>0, slices and their lengths scale by \kappa. When \kappa=0, the only possible pair exponent is zero and compression leaves it unchanged.□

Proof. Choose general sections \sigma_1,\ldots,\sigma_{n-1} of DL so that their successive zero loci form a smooth integral flag

Such a flag exists by Bertini’s theorem for very ample hyperplane sections. Choose p\in C and a local frame e of L near p. The functions z_i=\sigma_i/e^D for 1\leq i<n have independent differentials at p; complete them by a function z_n to local analytic coordinates centered at p.

Represent every section of kL by its germ in the frame e^k. These representations are compatible with products in different degrees. They are injective, since a section on an integral variety vanishing on an open germ vanishes identically. Order Taylor exponents lexicographically, starting with the exponent of z_1, and let S_k be the set of possible least exponents of nonzero section germs. Lemma 2.2 gives

We next bound these exponents by global vanishing orders. Let \alpha=(\alpha_1,\ldots,\alpha_n)\in S_k be the least exponent of a section s. Put s_0=s. For 1\leq i<n, suppose inductively that the residual section on X_{i-1} has least exponent (\alpha_i,\ldots,\alpha_n). Its local order along X_i is \alpha_i. A restriction to the integral divisor X_i that vanishes near p vanishes identically. The Cartier divisor exact sequence therefore allows division by \sigma_i|_{X_{i-1}}, one factor at a time, exactly \alpha_i times. Define

The exact local order ensures that s_i is nonzero. Its germ is obtained by dividing by z_i^{\alpha_i} and setting z_i=0, so its least exponent is (\alpha_{i+1},\ldots,\alpha_n). This completes the induction. At i=n-1 we obtain a nonzero section on C of

Its vanishing order at p is \alpha_n. Since \deg(L|_C)=D^{n-1}V, the degree bounds that order and yields

Equivalently,

which proves (3.2). The volume formula and the Hilbert polynomial asymptotic for an ample line bundle give (3.3).

It remains to transfer jets from the least monomials to the sections. Fix k. If S_k is empty, the asserted implication has a false premise and there is nothing to prove. Otherwise, choose a basis with distinct least exponents. A single positive weight vector exposes all these least terms, even though the series may have infinite support. Indeed, put \lambda_n=1 and choose the preceding weights recursively so that

If \beta is lexicographically later than \alpha and i is their first differing coordinate, then

Here we used that all Taylor exponents are nonnegative.

Now take a torus tuple at which \M(S_k) separates the specified jets. The substitution z_i=s^{\lambda_i}Z_i sends this tuple into the chosen coordinate chart for sufficiently small s>0, without identifying any points. Lemma 2.4 gives the least monomials as limits of normalized section germs, and Lemma 2.5 gives jet separation by the original sections at the image tuple in X.□

Proof. Fix the target w. We first use a node to replace the triangular bound \Delta(A,B) by a quadrilateral of the same area. A diagonal compression then produces a triangle with a variable shape. We choose that shape and a primitive lattice direction together so that two further compressions give exactly \Delta(w,H/w).

Throughout, write \gamma=(\gamma_3,\ldots,\gamma_n) for an exponent in the unchanged coordinates. For a label occurring in degree k, the pair exponents with this label lie in

For n=2 the label is empty and \kappa=k. Each operation below acts on these two-coordinate spaces, carrying the common monomial with exponent \gamma along unchanged. In particular, a planar bound by \kappa K will hold simultaneously for every label. If \kappa=0, the only possible pair exponent is zero and it stays zero throughout; hence the geometric calculations may be made with \kappa>0.

The node and its exponent bound. In pair variables x,y, consider

This polynomial is irreducible, since 1+x is not a square in \C(x). Its curve has the parametrization

which covers the part where x\ne0. Near the origin choose the analytic square root with value 1 and set

These are local coordinates and g=\xi\eta.

For now fix a real parameter t>0, and order the Taylor exponents (\alpha,\beta) in \xi,\eta by lexicographic comparison of (\alpha+t\beta,\beta), and denote the least exponent by \nu_t. This is a multiplicative well-order: only finitely many nonnegative integer exponents have weight below any given bound. The node gives two controls on these exponents: its two branches bound the least weight after factors of g have been removed, while each extracted factor translates the least exponent by (1,1).

Let f\ne0 be supported in \kappa\Delta(A,B), and write f=g^j f_0, where j\ge0 and g does not divide f_0. The weighted polynomial degree with weights (1/A,1/B) is additive on nonzero products, since highest-weight forms multiply nontrivially. As g has weighted degree d, the weighted degree of f_0 is at most \kappa-jd, which is nonnegative. Under (4.5), its restriction is a nonzero polynomial in v: otherwise f_0 would vanish on the irreducible curve g=0. Define

The monomial inequality

shows that the restriction polynomial has degree at most

Let \mu be the weight of \nu_t(f_0). At v=1, the branch is \eta=0 and \xi vanishes simply. At v=-1, the branch is \xi=0 and \eta vanishes simply. Indeed, the chosen square root becomes v near 1 and -v near -1, and the nonzero branch coordinate in either case is 2v(v^2-1). Every nonzero pure \xi term in f_0 has exponent at least \mu, and every nonzero pure \eta term has exponent at least \mu/t. The restriction polynomial therefore has orders at least \mu and \mu/t at these two distinct points. Comparing their sum with (4.8) gives

To express this bound as a triangle, define its intercepts

For \nu_t(f_0)=(\alpha,\beta), inequality (4.9) is exactly \alpha/a+\beta/b\le\kappa-jd. Hence \nu_t(f_0)\in(\kappa-jd)\Delta(a,b).

The extracted factors account for the rest of the degree bound. Set c=1/d, so their contribution is \nu_t(g^j)=(j,j)=jd(c,c). The coefficients \kappa-jd and jd sum to \kappa, and therefore

Thus the axis triangle comes from the two branch orders, and the diagonal point comes from powers of the node equation.

From this point on, restrict the free parameter to t>T, where

The inequality

shows that Q is a quadrilateral with the vertices in the stated order. Its area is

The identity W=dH is what makes this area equal to that of the input triangle \Delta(A,B). The parameter t remains free to change the shape of Q for the later lattice-direction choice. For each label, replace its pair support by the possible values of \nu_t on its polynomial span. The product of the spaces in degrees k,l with labels \gamma,\gamma' lies in the space of degree k+l with label \gamma+\gamma', by the original support inclusion. Lemmas 2.2 and 2.3 preserve cardinality and the graded inclusion, and (4.11) supplies the bound \kappa Q.

We also need the jet implication for this change. At each fixed degree, choose a basis with distinct least exponents in each nonempty label space. One sufficiently small e>0 makes weights (1,t+e) select all these least terms strictly. To see this, choose M larger than all the selected (1,t)-weights. Among the finitely many exponents of weight at most M, a small positive perturbation preserves strict comparisons and realizes the tie-break by the second exponent. Make e smaller still so that all selected weights remain below M; exponents originally above M cannot interfere, because a positive perturbation only increases their weights. Lemmas 2.4 and 2.5 apply under (\xi,\eta)=(sX,s^{t+e}Y), with the other variables unchanged. Their inverse images are torus points. In fact, for each of the finitely many test points with X,Y\ne0, one has \eta/\xi\to0, while

Thus x,y\ne0 for all sufficiently small positive s. The local coordinate map is injective on a common neighborhood, so it also preserves distinctness of the test points.

From the quadrilateral to a triangle. Apply the integral shear

and then compress in the second coordinate. The transformed quadrilateral has vertices (0,0),(a,0),(0,c),(-b,b). For first coordinate q in [-b,0], its lower boundary is -q and its upper boundary is c+(c-b)q/b; their difference is c(1+q/b). For q\in[0,a], the vertical slice length is c(1-q/a). Compression therefore gives the triangle

Figure 1 displays these slice lengths geometrically. Lemmas 2.6 and 2.7 show that the exponent sets remain graded, retain their cardinalities, admit backward jet transfer, and are bounded, label by label, by \kappa P.

The area is now correct, but we have not yet obtained the prescribed intercept w. We choose a primitive integral direction whose transverse range is exactly H/w. In the resulting coordinates, the longest horizontal slice will have length w; compression will turn that length into the horizontal intercept.

Choosing the primitive direction. Put R=H/w and a_0=WT/(1+T). As t ranges over (T,\infty), the intercept a ranges over (a_0,W). Choose j to be the largest power of two not exceeding

Then

The open interval

has length

It therefore contains an odd integer i. This is the point of choosing j to be a power of two: the vector u=(i,j) is automatically primitive. Choose the unique t>T for which

The interval (4.16) ensures that this value lies in (a_0,W).

At the three vertices (-b,0),(a,0),(0,c) of P, the functional z\mapsto\det(u,z) takes values jb,-ja,i/d, respectively. They are distinct and ordered as

Their range is exactly i/d+ja=R. This choice used only inequalities between real numbers; in particular, no rationality of the intercepts was required. We now fix this t and this primitive vector for all degrees.

The two final compressions. Complete u to an integral basis (u,u') with determinant 1, and use the coefficients in that basis as new exponent coordinates. The second coordinate of a vector z is \det(u,z). Thus the transformed triangle still has area H/2, and its three vertex heights, say q_0<q_1<q_2, satisfy q_2-q_0=R.

Its horizontal slice length is zero at q_0, increases linearly up to q_1, and decreases linearly to zero at q_2. If the maximum slice length is h_{\max}, integration of this tent gives area Rh_{\max}/2. Hence h_{\max}=H/R=w. Compressing in the first coordinate therefore gives

Its vertical slice at first coordinate h\in[0,w] has length R(1-h/w). A final compression in the second coordinate produces \Delta(w,R), as illustrated in Figure 2.

All these changes satisfy Lemmas 2.6 and 2.7. They commute with scaling of the bounding shapes, so for the label \gamma the final pair exponents lie in \kappa\Delta(w,R). Substituting (4.3) gives (4.1). Cardinality, the graded inclusion, and the backward jet implication hold at every stage, proving all the assertions.

For clarity, the sets in every degree are defined by the fixed exact order \nu_t, the fixed integral maps, and exact orders of vanishing at 1. The perturbation e and the small scaling parameters only witness jet-rank implications at a specified degree. They may depend on that degree and on the test tuple without changing the sets or their graded inclusion.□

Proof. The lattice-point estimate for a full-dimensional compact convex set gives

Consequently the sets of missing exponents

satisfy \#H_k=o(k^n). This estimate alone would allow individual missing points deep inside kP. We use additivity to exclude them.

We may assume that K is nonempty. Choose \delta>0 so that every closed ball of radius \delta centered in K is contained in P. For a large integer k, put k_1=\lfloor k/2\rfloor and k_2=k-k_1, and fix y\in kK\cap\Z^n. Write z=y/k. Every integer vector y_1 satisfying

gives a decomposition y=y_1+y_2 with y_i\in k_iP\cap\Z^n. Indeed, both y_1/k_1 and y_2/k_2 are within distance \delta of z. The ball in this display contains at least c k^n lattice points for all sufficiently large k, where c>0 is independent of y.

At most \#H_{k_1} choices have y_1\notin F_{k_1}, and at most \#H_{k_2} choices have y_2\notin F_{k_2}, because y_1\mapsto y-y_1 is injective. The total number excluded is o(k^n). Thus some decomposition has y_i\in F_{k_i}, and additivity gives y\in F_k. All constants and degree thresholds used here are uniform over y\in kK\cap\Z^n.□

Proof. Write the coordinates as (x',Y), with n-1 coordinates in x'. At q_i, an arbitrary jet through degree m has a unique expression

For each \alpha, the Chinese remainder theorem provides a polynomial f_\alpha(Y) such that

The polynomial

has all the prescribed jets. Every monomial (x')^\beta Y^b occurring in its \alpha-summand satisfies \beta\le\alpha componentwise and hence

Thus f\in\M(E_{r,m}).□

Proof. Choose \theta/w<\lambda<1. There is a vector a with strictly positive coordinates such that a+\lambda P\subset\operatorname{int}P: it suffices to require

Choose a_k\in k^{-1}\Z^n with a_k\to a. For all sufficiently large k, the translated simplices a_k+\lambda P lie in a single compact subset K of \operatorname{int}P. Lemma 5.1 therefore gives

Consequently \M(F_k) contains the monomial space for k\lambda P\cap\Z^n, multiplied by the monomial with exponent ka_k. Multiplication by this monomial is an isomorphism on each jet algebra at a torus point, since the monomial is a unit there.

Put m=\lfloor k\theta\rfloor. For all sufficiently large k, m+1\le k\lambda w, and hence

Lemma 5.2 supplies jet surjectivity for this subspace. The monomial translation preserves surjectivity and proves the claim.□

Proof. Choose a point p_i\in C. The local ring R=\mathcal O_{C,p_i} is a one-dimensional Noetherian local domain. Its maximal ideal \frakm_R satisfies

Otherwise Nakayama’s lemma would give \frakm_R^m=0, contrary to \dim R=1. The quotient map from the ambient local ring to R lifts a nonzero element of this quotient to a jet on X with no terms of degree less than m. Prescribe this jet at p_i and the zero jet at every other marked point. By surjectivity, a section s\in H^0(X,A) has these jets. It vanishes to order at least m at every marked point, and its restriction to C is nonzero.

We recall the local intersection bound for this restriction. In a one-dimensional Noetherian local domain (R,\frakm_R), let 0\ne f\in\frakm_R^m. For every integer q\ge1, multiplication by powers of f gives

Since (f^q)\subset\frakm_R^{mq}, the Hilbert–Samuel formula yields

where e(R) is the ordinary multiplicity. Dividing by q and letting q\to\infty gives

Apply (6.5) to the local equations of s|_C at the marked points on C. The zero scheme of this nonzero section is an effective Cartier divisor on the integral projective curve, and its total length is \deg(A|_C)=A\cdot C. Summing the local contributions proves the assertion.□