Abstract. Singular-vector span defects beyond boundary format are governed
by multiplication in Artinian reductions of Segre rings. We prove that the first
Lefschetz map of every Segre ring has maximal rank in every characteristic, and use
it to classify all defects one step beyond boundary format. The same comparison
shows that a generic tensor of any format lies in the span of its singular rank-one
tensors. Integral multiplication for two factors and rigidity of products of chains
give the rank theorem; the rigidity argument also yields full spans at arbitrary
excess, both for sufficiently many factors and for sufficiently large dimensions with
the number of factors fixed.
1. Introduction
The singular rank-one tensors attached to a generic tensor form a finite set, but their span
can be smaller than the critical space cut out by the singular-vector equations. Over ,
Draisma, Ottaviani and Tocino [1] proved equality when the projective factor dimensions
satisfy the triangle inequalities. Immediately beyond that range, Teixeira Turatti and
Ventura [4] conjectured the exceptional formats and their defects. We prove this
classification in every characteristic by identifying the defect with a multiplication kernel
in a Segre ring. The identification also explains why the tensor itself remains in the span,
even when the span is defective, and gives full-span ranges farther beyond boundary
format.
Let be an algebraically closed field. We use the coordinate bilinear forms . A tuple is
singular for if contraction against all slots except is proportional to , for every .
Write for the singular-tuple scheme in the Segre embedding. The critical space
is cut out by these equations before restricting to rank-one tensors; thus and
.
One step beyond boundary format, the multiplication kernel is the kernel of the first
Lefschetz map, so the classification rests on a maximal-rank theorem for every Segre ring.
For a tuple of positive integers, put
This is the homogeneous coordinate ring of the Segre embedding of .
Theorem 1.1.Let be a general linear Artinian reduction, and let be general.
Multiplication
has maximal rank. For and , its only nonzero kernel dimensions, up to
transposition, are
For , the only nonzero kernel dimensions, up to permutation, are
For two factors with , one has ; for , one has .
Thus every exceptional defect is forced by the positive difference . For a tensor with
smaller projective factor dimensions , write . Boundary format has last vector-space
dimension ; one step beyond it has last dimension . The graph-span formula in Section 2
identifies the defect of in with the kernel in Theorem 1.1.
Corollary 1.2.Assume either , or and . Put . For a generic tensor of format ,
the singular tuples span , with precisely the following exceptions
Permuting factors gives the same conclusions.
The remaining three-factor tensor family has defect , because its Artinian reduction
has and . This accounts for all formats with at least three factors one step beyond
boundary format. For matrices, and the defect is zero.
The graph-span formula applies to every last dimension at least . It also shows that the
additional linear equations vanish on itself. Together with the range satisfying the
triangle inequalities, this gives the following consequence without a restriction on
format.
Theorem 1.3.A generic tensor of every format is a linear combination of its
singular rank-one tensors.
A cohomology gap in the Koszul resolution shows that the linear equations on the
singular-tuple scheme are precisely those detected by multiplication in the Segre ring. We
obtain this gap by applying the critical-space argument of Draisma, Ottaviani and
Tocino to the graph of the last contraction map with a multiquadratic twist.
Writing the contractions directly keeps this part of the proof valid in characteristic
two.
The rank calculation uses two forms of the same staircase degeneration. For two Segre
factors, an integral shelling proof of the descent basis of Oeding, Raicu and Sam [2]
reduces multiplication to two unimodular starting cases and an induction on the last
column. With more factors, degree-two multiplication becomes an infinitesimal rigidity
matrix; coning and gluing propagate rigidity through products of chains, while successive
completions of one hexagon supply the small starting boxes. Both arguments work without
exceptional characteristics.
The rigidity argument also controls larger excess. If the last dimension is , Section 6
proves full span when , and when for . For each fixed , it gives explicit sufficient lower
bounds on the smaller dimensions. The same span–multiplication comparison therefore
treats both the exact first-excess classification and uniform ranges farther beyond
boundary format.
2. The graph-span formula
Let , , , and . Write a tensor in slices , using the bilinear identifications. Define
Degree-two evaluations span , so is spanned by , for . Thus is the linear span of the
graph of the last contraction map. The singular tuples are cut out on this graph by the
remaining critical equations. The following proposition says that these equations account
for every linear relation on that finite scheme.
Proposition 2.1.Suppose the slices have no common zero and is finite and reduced.
Then
In particular, . The hypotheses hold generically whenever .
We use the following characteristic-independent cohomology facts
The Euler sequence and its exterior powers give the first two lines; the degree- Koszul
complex gives the additional vanishing of global sections, and Serre duality gives the last
line. See [3, Section 50.11].
Proof of Proposition 2.1. Base-point-freeness makes the last singular-vector
equation the graph , scheme-theoretically. On the graph, the remaining equations are
the zero scheme of
in the rank- bundle
Here contracts all slots except , and the is in position . Finiteness of makes this
section regular.
Put . For , the summands of are
A factor with nonzero degree-zero cohomology would require , which is impossible.
Every nonzero factor cohomology has degree at least , and Künneth gives
The twisted Koszul resolution yields
Pairing the section with gives , exactly the image of a small-mode critical equation
under . Write for the span of those equations. The displayed isomorphism shows that is
injective on , whose dimension is .
A linear form on tensor space restricts to the graph as its image under . It vanishes on
precisely when that image lies in . Hence
The last-mode critical equations already belong to , since their images are . Taking
annihilators gives . Subtracting the independent equations from gives the dimension
formula in (1).
The functional defined by
satisfies , hence . Thus . Genericity follows from the next lemma. □
Lemma 2.2.The singular-tuple scheme of a generic tensor is finite and reduced.
A general family of multilinear forms on is base-point-free.
Proof.Write , , and . At , the linear conditions on span
If no is isotropic, the summands are independent and . If exactly are isotropic,
their only overlaps are the common tensor line, giving . The corresponding stratum
has dimension , so its singular incidence has dimension . In characteristic two the
reduced isotropic locus is a hyperplane; the same calculation applies. By projectivity,
the union of these incidences has closed proper image.
Over the nonisotropic open, the incidence is a smooth vector bundle of rank ,
hence has dimension . Its projection is separably dominant; at and the tuple of
coordinate points, the local singular section is , with derivative . Remove the isotropic
image and the closure of the critical image. The remaining fibers are projective,
zero-dimensional and ètale, hence finite and reduced.
At a fixed point of , vanishing of arbitrary slices imposes independent conditions.
The base-point incidence has dimension and closed image when . □
2.1. The defect as Koszul homology. Suppose the slices are independent
and span . Their last-mode critical equations form the Koszul subspace . The
small-mode equations are independent of this subspace, since their images under
are independent. Comparing the dimension of with the graph-span formula
therefore leaves exactly the multiplication relations modulo the Koszul relations
For , choose parameters of dimension , put , and let generate . Resolve first by the
Koszul complex on , which is exact because the Segre ring is Cohen–Macaulay. The
remaining complex is multiplication by on . Its homological-degree-one part in internal
degree two is precisely the map , giving
Canonical-module duality also identifies this kernel with the dual of the cokernel of ,
recovering the cohomological map of [4, Corollary 4.5].
2.2. All tensor formats. To prove Theorem 1.3, omit one-dimensional factors;
order-one tensors are immediate. If one factor has dimension at least one plus the sum of
the other projective dimensions, apply Proposition 2.1. Otherwise for every . On the full
product, the singular section belongs to
For , the summands of are , with . For , a degree-zero factor requires . All other factors
then have negative twist, and nonzero total cohomology has degree . If no factor has
degree-zero cohomology, the total degree is again at least . Thus the analogue of (2) holds.
Lemma 2.2 and the Koszul resolution identify with all critical equations. Hence in this
range. Since , Theorem 1.3 follows.
3. Two Segre factors
Put
At a rank-one point , vanishing of all would give
Thus the are parameters, and Cohen–Macaulayness makes them a regular
sequence.
Lemma 3.1.Over every field, bases of the first two graded pieces are
In particular, and .
Proof.In the staircase degeneration, variables of rank have the same weight, so the
rank-sum parameters degenerate to the same sums in the order complex. A maximal
chain is a lattice path, encoded by a word with letters and letters . Order these
words lexicographically, with . This is a shelling, whose restriction face consists of
the vertices at descents .
Here is why the restriction faces give an integral basis after quotienting by the
rank sums. When a chain is attached, its new faces are exactly those containing its
restriction face. The corresponding relative module is the product of the restriction
variables times the polynomial ring on that chain. There is one chain vertex of
each rank, so the rank-sum parameters restrict to those individual variables, with
coefficient matrix the identity. The successive quotients of the shelling filtration are
therefore free rank-one modules over the parameter ring, generated by the restriction
monomials. The filtration splits as a parameter-module filtration. Elimination along
the shelling has diagonal coefficients one and works over the integers.
A descent occurs at with and ; two descents have both coordinates strictly
increasing. The restriction monomials in degrees one and two are therefore exactly
those in the statement.
The parameters remain regular in the flat family. The monomials are therefore
a basis near the initial fiber, and torus rescaling identifies all nonzero fibers while
scaling the parameters and the monomials. This proves the assertion in every
characteristic, and recovers the relevant degrees of [2, Theorem 5.2]. □
Lemma 3.2.If , , and some linear form acts injectively from degree one to degree
two in , then the same is true in .
Proof.Setting the last column to zero gives , with kernel satisfying . Multiplication
by is injective on . For , the products are distinct basis monomials; for , the minors
and the relations , give
a further basis monomial. The open set of forms injective on meets the inverse
image of the open set of injective forms in the smaller algebra. Any form in the
intersection is injective on . □
The two necessary starting points can be read from the dimensions; for ,
Up to transposition, the minimal pairs where injectivity is possible are and . The
column lemma will propagate these two cases, while their quotients will handle the smaller
targets.
In , multiplication by is an isomorphism. Write the source coefficients in the order
as . In the target order
the image has coefficients
They recover the source, successively in the order , using only additions and sign
changes.
In , write , , and for , and take . The defining relations give
If lies in its kernel, the rows for give successively , , and . Thus . The other rows give
Hence and all coefficients vanish. Thus multiplication is invertible over every field; its
integral determinant is .
Lemma 3.2, in both directions, gives injectivity for and for , , together with their
transposes. The quotient transfers surjectivity to , where the kernel dimensions are .
Nonvanishing of the required minors is open in the parameter coefficients, so this proves
the two-factor part of Theorem 1.1.
4. Multiplication and rigidity
Because degree-two multiplication in a Stanley–Reisner ring depends only on vertices and
edges, its relations can be read from a graph. The square coefficients determine one
component at each vertex; eliminating those components leaves exactly the equations of
infinitesimal rigidity. Let have vertices and graph . For a general placement , write for
the rank of the rigidity matrix, whose row at has in the columns, in the columns, and
zero elsewhere. For rigidity in dimension , assume that the graph has at least vertices, so a general placement affinely spans . The graph is generically -rigid if , and generically stress-free if its edge
rows are independent.
Lemma 4.1.Let , and let be a general -dimensional subspace. Then
If is Cohen–Macaulay of dimension , is a general linear Artinian reduction, and is
general, then
Thus rigidity gives injectivity, and stress-freeness gives surjectivity.
Proof.Rescale variables so that has basis and , . A vector in the multiplication kernel
has coefficients with . The square coefficient gives ; the edge coefficient becomes
Thus the kernel has dimension . Quotienting by the injective Koszul subspace
proves (5). For an Artinian reduction, use the same Koszul identification as in (4).
If is the number of edges, then
so stress-freeness gives surjectivity. No division by two occurs. □
Trivial infinitesimal motions are , where is alternating; and . Their dimension on an
affinely spanning placement in is , also in characteristic two.
Lemma 4.2.Over , the join of an -rigid graph and an -rigid graph is -rigid. In
particular, coning raises rigidity dimension by one. Coning preserves stress-freeness
as well, and is an equivalence for rigidity.
Two -rigid graphs with at least common vertices have -rigid union.
Adding a vertex with at least neighbors preserves -rigidity; adding one with at
most neighbors preserves stress-freeness in dimension . A one-extension, replacing by a new vertex joined to and other vertices, preserves both properties in dimension
.
Proof.For the join, place the frameworks in the skew affine subspaces and . Choose
an affine anchor basis in each part; together they form an ambient affine basis.
Internal rigidity implies all pairwise edge constraints within each part, since those
constraints involve only the tangential velocity components. The cross edges therefore
make the anchor vertices a rigid simplex. Subtract its ambient rigid motion. Internal
rigidity now kills the tangential components at every vertex, and the cross edges to
the opposite anchors kill the remaining normal components.
For the converse cone assertion, put the apex at zero and generic base positions
at . After fixing the apex velocity, the cone edges determine the last coordinates.
Substituting in the remaining coordinates turns each base-edge equation into the
original one multiplied by . Thus, for a base with vertices,
This also proves the stress assertion. The gluing assertion follows because rigid
motions agreeing on an affine basis agree everywhere.
The vertex-addition assertions follow from independence of the incident directions.
For a one-extension, place the new vertex on the old edge, away from its endpoints,
with the other directions independent modulo that line. Eliminating its velocity
from the two collinear edge equations recovers the old edge constraint. The remaining
directions then determine its whole velocity. Dually, equilibrium at the new vertex
kills the transverse stresses and combines the two remaining stresses into a stress
on the old edge. This proves the required rank at one placement, hence at a general
placement. □
4.1. The staircase degeneration. Put and . Let join distinct comparable elements.
The Segre straightening relations are
Give weight . If , , and , then the first monomial has weight more than the second.
Chain monomials form a basis by sorting the indices in each factor. Thus the initial ring is
the Stanley–Reisner ring of the order complex of . Its maximal chains have a lexicographic
shelling, so it is Cohen–Macaulay of dimension .
A maximal multiplication rank in this initial ring passes to the Segre ring;
parameter regularity and nonvanishing of a minor are open in the flat family.
For the first Lefschetz map, the Artinian Hilbert function is constant on the
parameter locus. For arbitrary , the universal rank bound from is the same on every
fiber. Consequently the graph statements below give the desired multiplication
ranks.
5. Products of chains
Proposition 5.1.For and , the graph is generically -rigid except at , , and their
permutations. At these three tuples it is generically stress-free in dimension .
It is enough to construct a few rigid cores and then lengthen the chains. Boolean
products supply the cores with at least four factors, and two three-chain boxes account
for the remaining tuples; we begin with the propagation step shared by both
constructions.
5.1. Increasing a chain length. Suppose is rigid in dimension , where is the sum of
the other chain lengths. After increasing by one, take the lower old box coned by the new
maximum, and the translated upper old box coned by the global minimum. They are rigid
in dimension , cover every vertex, and share vertices. Gluing applies exactly when
For , the other product has at least two positive chain lengths, so and the inequality
holds.
5.2. Boolean products.
Lemma 5.2.For , the graph of the barycentric subdivision of the boundary of an
-simplex is generically -rigid over every field.
Proof.Start with and perform stellar subdivisions of faces in decreasing dimension.
Subdivision of a face with at least three vertices preserves the old graph and adds
a vertex with at least neighbors. For an edge, its link in the current triangulated
-sphere has at least vertices. The subdivision contains a one-extension using the
endpoints and any link vertices. Additional edges preserve rigidity, and subdividing
a vertex merely renames it. Apply Lemma 4.2. □
Let be the comparability graph of the nonempty proper subsets of . It is the graph in
Lemma 5.2 with . The full Boolean graph is its double cone, hence is -rigid for . Increasing
chain lengths handles every tuple with at least four factors.
5.3. Three chains. Every sorted positive triple outside the three exceptions
dominates or . We construct these two bases from one sequence of minimally
rigid completions. Write for the proper-part graph of , and name its vertices
A minimally -rigid graph is both -rigid and stress-free. Coning, adding a vertex on
exactly neighbors, and one-extension preserve minimal rigidity in the appropriate
dimension.
The temporary edges in these completions explain both sides of the classification.
Deleting them proves stress-freeness for the exceptional boxes, whereas consuming them in
successive one-extensions produces the two rigid starting boxes. We keep track of the edges
so that the same construction proves both assertions.
The graph is the hexagon with cyclic order . Add the triangle on the alternate vertices
. Starting with this triangle, each remaining vertex has two neighbors, so the completion is
minimally -rigid.
Figure 1: The hexagon . The three added edges form the central triangle of its
minimally rigid completion.
Cone this completion by . Add by a -dimensional one-extension removing , with
neighbors . The edge is temporary.
Add on , then add by a one-extension removing , with neighbors . The result is the
minimally -rigid completion
Cone it by . A one-extension at , on , removes . The symmetric extension at
, on , removes . Finally add on . This gives the minimally -rigid completion
Deleting the added edges shows that are stress-free in dimensions , respectively.
To obtain , cone the last completion by . Add by a -dimensional one-extension
removing the remaining temporary edge . Choose any six of its seven old neighbors that
contain the two endpoints. Vertex also has seven old neighbors, and has the five
neighbors . Add these two vertices and restore any omitted edges. Every edge now belongs
to , proving its -rigidity.
For the proper-part graph of , return to the completed and cone it by . Write and for
the new vertices. The three added edges are removed by one-extensions at , in that order
The two further edges may be restored. Add on its four old neighbors , then on its
five old neighbors . The resulting graph is precisely the proper-part graph of , and is
-rigid. Double coning restores the full graphs; increasing chain lengths now proves
Proposition 5.1.
5.4. The first Lefschetz map. Write
The first Artinian Hilbert values are and . At the exceptional triples they are , , and .
Stress-freeness gives surjectivity and kernel dimensions ; all other tuples give injectivity.
The staircase degeneration proves Theorem 1.1.
The parameter and maximal-rank loci pull back to nonempty open loci in tensor space,
because a basis of is an arbitrary array of slices. Intersect with the open from Lemma 2.2
and apply (4). This proves Corollary 1.2.
6. Arbitrary excess
The graph comparison is not confined to the first Lefschetz map. If with , equations (3)
and (5) identify the additional relations with the failure of rigidity in dimension . Thus
the staircase degeneration shows that -rigidity of gives full singular-vector span for a
general tensor of format
There are two useful regimes. The number of factors may grow with the excess while
their dimensions remain small, or the number of factors may stay fixed while their
dimensions grow. The constructions in both regimes contain at least vertices, so the slices
can be independent.
Theorem 6.1.Assume for all . The generic singular tuples span whenever . They
also span if and for some integer . In particular, suffices when .
Both bounds come from the same family of interval cores in the Boolean lattice. Fix
ordered block sizes and a partition . Put and take the core
It is the join of and the clique on the intermediate prefixes.
Two partitions related by swapping one element between adjacent blocks of sizes have
exactly
common core vertices. Indeed, in the changed window the common intervals are
the lower interval of size and the upper interval of size ; the other intervals
are unchanged. Subtract the repeated endpoints and the two global endpoints.
Adjacent-block swaps connect all partitions of the prescribed block sizes. The cores cover
every proper nonempty subset; order its elements first, then cut at the prescribed
sizes.
Proof of the bound . Take and . Each is -rigid by Lemma 5.2. Joining these graphs
gives rigidity dimension ; the intermediate prefix vertices are cones, raising it to .
This explains the choice of blocks. For there is one core. For , when every , the
overlap (7) is . Increasing a block size increases the overlap by more than it increases
the required dimension. Thus all cores glue, giving -rigidity of , and double coning
gives -rigidity of the full Boolean graph.
To increase a chain length, use (6). The other positive lengths give
so propagation holds. This proves -rigidity for every positive tuple in the stated
range. □
The same cores give a sharper bound if we use the complete joins between their interval
interiors. With three blocks of size , each interior has vertices. The following
tripartite calculation converts that number of vertices into a much larger rigidity
dimension.
Lemma 6.2.If , the complete tripartite graph is generically -rigid over every field.
Every placement in which each part affinely spans is infinitesimally rigid.
Proof.Let have rows for the three parts. Let have rows . All three position matrices
have full column rank. The first two parts satisfy
Applying left annihilators of gives for a square matrix , and hence . Similarly .
The remaining edges give , so and , , .
The velocity encoding also gives for every row of . Outside characteristic two,
skew-symmetry already makes the diagonal zero. In characteristic two, this last
identity is . Choose an invertible square submatrix of ; its entrywise square is
invertible, so again every . Thus is alternating in every characteristic. Splitting its
last row and column shows that all velocities are one common translation plus one
common alternating linear motion. □
Proof of the three-block bound. First take and three blocks of size . The three
interval interiors have vertices each, are completely joined, and have two universal
prefix vertices. Put . The hypothesis gives , so Lemma 6.2 and two cones make every
core -rigid. Adjacent cores intersect in vertices. The preceding connectivity and
coverage argument proves rigidity of .
To add a Boolean factor, cover by the subgraphs of subsets containing a fixed
element. Each is a cone over , and two intersect in vertices. The required inequality
holds at and persists thereafter. Thus is -rigid for every .
For arbitrary positive chain lengths,
The first inequality is minimized when every other length is one. It implies (6),
so double coning and coordinate increments finish the proof.
Finally take . For , one has , , and ; hence . □
6.1. A fixed number of large factors. The same use of joins permits a fixed number
of chains, provided they are sufficiently long. We arrange diagonal boxes whose
interiors are pairwise joined, using the intervening boxes to supply additional
rigidity dimensions. Enlarging the first and last boxes ensures that every outside
vertex has enough neighbors to attach to this core. For fixed , put if and if
.
Theorem 6.3.Let , , and choose . Put . If
then the generic singular-vector span is whenever every .
Proof.In , take consecutive diagonal boxes with side lengths
By Proposition 5.1, the proper interval graph for a box of side is -rigid. These
graphs are disjoint and pairwise joined. Together with the diagonal prefix vertices
they form a core rigid in dimension , by the join and cone assertions of Lemma 4.2.
An outside vertex has at least
neighbors in the first and last boxes. These two sets are disjoint. Since , at least
one factor in each paired coordinate equals , and the other is at least one. Decreasing
the smaller factors to one gives a sum , minimized at the two integers nearest .
Thus (8) permits every outside vertex to be added to the rigid core.
Finally . In every coordinate increment, and , so . Equation (6) extends the result
to all larger tuples. □
For three smaller factors, the criterion is ; the bound suffices. For fixed , put and .
Taking gives the sufficient bound . Indeed, with , the inequality makes the left
side of (8) at least its right side. Thus the thresholds are for and for fixed
.
Taking also gives the uniform sufficient bound for . Substitution in (8) proves this for
, and the inequality persists when increases by two. For , taking improves this bound to
.
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