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A 27 x 27 x 27 counterexample to Comon's conjecture
Benjamin Lovitz
TL;DR
The paper addresses whether tensor rank and symmetric rank agree for higher-order symmetric tensors. Using Shitov’s symmetric adjoin framework, it proves a special case of Conjecture 6 and constructs a small counterexample with a 27-dimensional tensor whose entries are rational.
Problem
For symmetric tensors of order greater than two, it remains a question whether tensor rank equals symmetric rank.
Source: §1 · p. 1Method
The paper uses Shitov’s symmetric adjoin framework and proves a special case of Conjecture 6 when the minimum symmetric rank in A is 2.
Source: §1.1 · pp. 2–3, §3 · p. 5Results
The explicit tensor has complex tensor rank 55 and complex symmetric rank 56.
Source: §4 · p. 12Takeaways & Limitations
The construction yields a small counterexample to Comon’s conjecture using a symmetric tensor with rational entries and applicability over Q, R, and C.
Source: §1 · p. 2, §1.1 · p. 3Whether tensor rank and symmetric rank can differ by more than one remains open.
Source: §1 · p. 2
Abstract
from arXiv · showhide
We report an explicit construction of a 27 x 27 x 27 symmetric tensor with rational entries that has tensor rank 55 over the rational numbers and symmetric tensor rank 56 over the complex numbers, providing a small counterexample to Comon's conjecture over the rational, real, and complex numbers. The construction follows the framework of symmetric adjoins introduced by Shitov. The primary technical contribution of this work is to prove a special case of Conjecture 6 appearing in Shitov's seminal 2018 work.
1 Introduction
The introduction frames Comon’s conjecture as the question of whether tensor rank and symmetric tensor rank agree for higher-order symmetric tensors. It presents a degree-3 construction in dimension 27 and identifies a special case of Shitov’s Conjecture 6 that the paper proves.
1 Introduction
- Comon’s conjecture asks whether tensor rank equals symmetric rank for symmetric tensors of order greater than two.Source: §1 · p. 1
- A degree-3 construction in dimension 27 addresses the large-dimension drawback of earlier counterexamples, using a symmetric tensor with rational entries and considering Q, R, and C.Source: §1 · p. 2
- Earlier work produced counterexamples in several orders and fields, but the known examples discussed here required dimension greater than 100.Source: §1 · pp. 1–2
- Whether tensor rank and symmetric rank can differ by more than one remains open.Source: §1 · p. 2
1.1 Methods and technical contribution
- The paper proves the illustrated special case of Shitov’s Conjecture 6, where the minimum symmetric rank in A is 2, obtaining a small counterexample.Source: §1.1 · p. 3
2 Preliminaries
The preliminaries establish the notation for symmetric tensors, their ranks, and the symmetric product used in the paper. They also set out the symmetric adjoin construction and its associated affine space.
2.1 Tensors and the symmetric product
- The paper works with symmetric tensors as tensors invariant under index permutations and uses projection onto the symmetric subspace to form symmetric products.Source: §2.1 · p. 3
2.2 Rank and flattenings
- Tensor rank is the least number of product tensors summing to a tensor, while symmetric rank uses symmetric rank-one summands.Source: §2.2 · p. 4
- The preliminaries define tensor flattenings and products of subspaces, including the notation LH for the span of symmetric products.Source: §2.2 · p. 4
2.3 Shitov’s symmetric adjoin construction
- The symmetric adjoin setup starts from T0 and a subspace K with a basis of symmetric rank-one matrices, then defines an associated symmetric tensor on V⊕Z and an affine space A(T0,K).Source: §2.3 · pp. 4–5
3 A special case of a conjecture of Shitov
The section proves a special case of Shitov’s Conjecture 6: under the stated exclusion of zero and symmetric rank-one tensors from A, the symmetrically adjoined tensor has symmetric rank at least 3m+2. It also identifies conditions under which the rank bounds are attained.
3 A special case of a conjecture of Shitov
- The section proves Shitov’s conjectured symmetric-rank formula over C in the special case minP∈A∩S3(V) Rs,C(P) = 2.The theorem also establishes a special case of Wang and Seigal’s analogous lower-bound conjecture over the real numbers [22, Conjecture 3.17].Source: §3 · p. 5
- RF(T) ≥ 3m + 1 and Rs,F(T) ≥ 3m + 2 when A contains neither zero nor a symmetric rank-one tensor.The proof of the symmetric-rank lower bound also holds when the matrices Mν do not have rank one.Source: §3 · pp. 5, 10
- If A contains a non-symmetric rank-one tensor, the tensor-rank lower bound is attained; if A contains a symmetric rank-two tensor, the symmetric-rank lower bound is attained.In the latter case, the proof also gives Rs,F(T) ≤ 3m + 2.Source: §3 · pp. 5, 10
- The lower bound Rs,F(T) ≥ 3m + 2 also holds when the basis matrices Mν of K are not rank one.The section notes that Shitov found a counterexample to the corresponding complex lower bound when the minimum symmetric rank in A is 3 [18, Example 3].Source: §3 · p. 5
- Assuming a decomposition with at most 3m + 1 terms, the substitution argument leaves a non-symmetric rank-one tensor P in A, then uses the structure of its factor span and K to rule out the decomposition.The proof restricts K to the square-symmetric matrices on the span W of P’s factors: the intersection is either zero or Fx², and in the latter case exactly one factor is proportional to x.Source: §3 · pp. 6, 9
4 The explicit construction
The construction engineers an affine space containing a nonsymmetric rank-one tensor but neither zero nor a symmetric rank-one tensor, then uses it to obtain a one-unit rank gap. The resulting symmetric tensor has tensor rank 55 and symmetric rank 56, with the same equalities over Q and R.
4 The explicit construction
- The construction uses three linearly independent rational vectors and 18 linearly independent rational symmetric rank-one matrices to define an affine space A.A contains a nonsymmetric rank-one tensor and a symmetric rank-two tensor, but neither zero nor a symmetric rank-one tensor.Source: §4 · pp. 11–12
- R_C(T)=3m+1=55, while R_s,C(T)=3m+2=56; rational matching decompositions give the same equalities over Q and R.The construction also gives a rational symmetric decomposition of length 56.Source: §4 · pp. 12–13
4.1 The construction in detail
- Theorem 4.2 establishes that A contains neither zero nor a symmetric rank-one tensor.Its proof uses three linear maps whose transformed coordinate products are fixed at −2 on A, then rules out the possible forms of a vector whose cube lies in A.Source: §4.1 · pp. 13, 15
4.1.1 Lemmas for the proof of Theorem 4.2
- The proof lemmas classify possible vectors h with h^3 in A into sums of multiples of the designated vectors a_i or w_irs.The maps R1, R2, and R3 then contradict the required coordinate-product condition in each case.Source: §4.1 · p. 15, §4.1.2 · p. 18
4.1.2 Proof of Theorem 4.2
- Three projections in the proof of Theorem 4.2 yield the fixed coordinate-product condition used to exclude zero and symmetric rank-one tensors from A.The proof applies the maps to T0 and to the adjoined terms separately: the former contributes −2, while the latter contributes zero.Source: §4.1 · p. 13, §4.1.2 · pp. 17–18
A Proof of Lemma 3.4
The proof constructs three successive linear maps that fix V and annihilate selected terms in a symmetric-adjoin decomposition. This reduces the rank of a tensor in the affine space by at least 3m, yielding the rank relation for the adjoined tensor.
A Proof of Lemma 3.4
- Three successive maps fixing V annihilate terms indexed by three pairwise disjoint sets, each of size m, reducing the resulting tensor's rank to at most N−3m.The maps are constructed by successively choosing sets whose corresponding components span Z.Source: A Proof of Lemma 3.4 · pp. 18–19
- Applying the three maps to the symmetric-adjoin decomposition produces a tensor in A, giving the upper bound R(T)≤R(P)+3m for a minimum-rank P in A.The mapped tensor lies in A because each map fixes V.Source: A Proof of Lemma 3.4 · p. 20
- The proof also establishes that the tensor terms remaining after projection are simple because each Mν has rank one.This supports the rank bound obtained from the projected decomposition.Source: A Proof of Lemma 3.4 · p. 20