Section 34

Monomial degenerations and exact moment certificates

Part VI · Family 2.12: intersections of three bilinear divisorsPDF page 129

We use the word pivot monomial for the leading monomial of a row-echelon basis of a polynomial subspace in a fixed monomial order. The set of pivot monomials is independent of the row-echelon basis. Its cardinality is the dimension of the subspace. Leading monomials multiply, so the pivot sets of an algebra form a semigroup.

Lemma 34.1 (Bounding all valuations by pivot moments). Let $T$ be $\mathbb P^2_{[t:a:b]}$ or $\mathbb P(2,1,1)_{[t:a:b]}$ , and let $W_{m,j}$ be a series of homogeneous, respectively weighted homogeneous, polynomials with bounded support containing an ample series. Put $N_m=\sum_j\dim W_{m,j}$ . Suppose that the sums of the three pivot exponents, divided by $mN_m$ , converge to $\mu_t,\mu_a,\mu_b$ . Then $\delta(T;W)\ge\frac1{\max\{\mu_t,\mu_a,\mu_b\}}.$ The corresponding finite-level bound holds with the finite-level means $\mu_{m,t},\mu_{m,a},\mu_{m,b}$ for every basis divisor of $W_m$ .

Proof. Fix $m$ and a divisorial valuation $v$ on $T$ . Choose an integral diagonal one-parameter subgroup with distinct monomial weights realizing the given lexicographic order on the finite degree range in question. On each $W_{m,j}$ choose a basis adapted simultaneously to the $v$ -filtration and the induced weight filtration, using Lemma 2.3. The weight leaves have dimension at most one. The initial monomials of this basis are consequently exactly the pivot monomials.

The resulting $m$ -basis divisor $D_m$ has $v(D_m)=S_m(W;v)$ . Under the one-parameter subgroup it degenerates on the fixed surface to $D_{m,0}=\mu_{m,t}(t=0)+\mu_{m,a}(a=0)+\mu_{m,b}(b=0).$ The coordinate boundary is log canonical when each coefficient is at most one. For the weighted plane this follows as well on its finite quotient charts; the quotient is unramified in codimension one. Thus $\mathop{\mathrm{lct}}(T;D_{m,0})=1/\max_i\mu_{m,i}$ . Lower semicontinuity of log canonical thresholds [11], with all nonzero fibres obtained by automorphisms of $T$ , gives $A_T(v)\ge\frac{S_m(W;v)}{\max_i\mu_{m,i}}.$ An arbitrary basis has order at most $S_m(W;v)$ for each $v$ , so the same bound holds for every basis divisor. Passing to the limit proves the assertion. The one-parameter subgroup may depend on $m$ . We split exactly two filtrations on each fixed homogeneous piece; invariance of $W$ under this subgroup is not needed. ◻

Lemma 34.2 (A full-volume certificate). Let $A=\bigoplus_{m\ge0}A_m$ be a graded algebra with $\dim A_m=(10/3)m^3+O(m^2)$ . Suppose a pivot filtration with one-dimensional leaves assigns to $A_m$ a set of $\dim A_m$ distinct vectors in a fixed full-rank lattice $\Lambda\subset\mathbb R^3$ , and that every such vector has norm $O(m)$ , uniformly in $m$ . Assume products of nonzero homogeneous elements are nonzero and their pivot vectors add. Suppose finitely many certified vectors $(m_i,v_i)\in\mathbb Z_{>0}\times\Lambda$ generate a semigroup whose group is $\mathbb Z\times\Lambda$ . Set $P=\operatorname{conv}\{v_i/m_i\}\subset\mathbb R^3.$ Normalize Lebesgue measure so that a fundamental parallelepiped of $\Lambda$ has volume one. If $\mathop{\mathrm{vol}}_\Lambda(P)=10/3$ , then the asymptotic pivot means of the whole algebra are the coordinates of $\frac{1}{\mathop{\mathrm{vol}}_\Lambda(P)}\int_P v\,d\mathop{\mathrm{vol}}_\Lambda.$ In particular, exponent coordinates may be negative; a proper exponent sublattice is allowed when its lattice-normalized volume is used.

Proof. Write $\Gamma$ for the generated semigroup and $\overline\Gamma$ for the points of $\mathbb Z\times\Lambda$ in its rational cone. The saturation is a finite $\Gamma$ -module. Hence there is $\eta\in\Gamma$ with $\eta+\overline\Gamma\subset\Gamma$ : choose module representatives, write each as a difference of two elements of $\Gamma$ , and add their negative parts. The lattice points lost from a degree- $m$ slice by this fixed translation lie in bounded-width boundary layers and number $O(m^2)$ . Ehrhart counting in $\Lambda$ therefore gives $\#\Gamma_m=\mathop{\mathrm{vol}}_\Lambda(P)m^3+O(m^2),\qquad \sum_{v\in\Gamma_m}v =m^4\int_Pv\,d\mathop{\mathrm{vol}}_\Lambda+O(m^3).$ All these vectors occur in the full pivot set by multiplicativity. Equality of the leading Hilbert coefficients leaves only $O(m^2)$ additional pivots. Their norms are $O(m)$ , so their contribution to the vector sum is $O(m^3)$ , including when some coordinates are negative. Divide by $m\dim A_m$ . ◻

Remark 34.3. The lemma requires the full group $\mathbb Z\times\Lambda$ , not merely full real span. In the finite certificates in this Family 2.12 part, $\Lambda=\mathbb Z^3$ and the degree-one vectors include $(1,0,0,0)$ and its translates by the three exponent unit vectors, which prove this condition. Later applications specify their value lattice separately. We prove enough leading terms to account for the whole asymptotic dimension; we do not assert that the finite list generates the entire initial algebra.

For clarity, all the moment calculations below use ordinary Euclidean volume in the exponent lattice. If $c$ is an interior point and $(p_i,p_j,p_k)$ a triangle in a boundary triangulation, the contribution of its cone to volume and first moment is $\label{f212:eq:tetra} V_{ijk}=\frac{|\det(p_i-c,p_j-c,p_k-c)|}{6},\qquad M_{ijk}=\frac{V_{ijk}}4(c+p_i+p_j+p_k).$ Thus the rational vertex and facet data in the lemmas specify their integrations explicitly. For completeness, the boundary triangulation can be recovered using only rational arithmetic. For every triple of affinely independent listed points, form its plane and retain it exactly when all listed points lie in one of its closed half-spaces. Identify repeated planes. Every facet occurs, since a two-dimensional facet contains three affinely independent vertices. In each retained plane take the convex polygon of the points on it and triangulate that polygon. The mean of all the listed points is an interior point of their full-dimensional convex hull, so coning these triangles to that mean gives tetrahedra with disjoint interiors and union the hull. Formula (34.1) then computes both volume and first moment exactly. This is a finite exhaustive procedure, not a floating-point test of selected facets. If the exponent lattice is a proper sublattice $\Lambda$ , divide both Euclidean volume and first moment by its covolume before applying Lemma 34.2; the centroid is unchanged.