The border support rank of two-by-two matrix multiplication is seven

: We show that the border support rank of the tensor corresponding to two-by-two matrix multiplication is seven over the complex numbers. We do this by constructing two polynomials that vanish on all complex tensors with format four-by-four-by-four and border rank at most six, but that do not vanish simultaneously on any tensor with the same support as the two-by-two matrix multiplication tensor. This extends the work of Hauenstein, Ikenmeyer, and Landsberg. We also give two proofs that the support rank of the two-by-two matrix multiplication tensor is seven over any ﬁeld: one proof using a result of De Groote saying that the decomposition of this tensor is unique up to sandwiching, and another proof via the substitution method. These results answer a question asked by Cohn and Umans. Studying the border support rank of the matrix multiplication tensor is relevant for the design of matrix multiplication algorithms, because upper bounds on the border support rank of the matrix multiplication tensor lead to upper bounds on the computational complexity of matrix multiplication, via a construction of Cohn and Umans. Moreover, support rank may be interpreted as a quantum communication complexity measure.


Introduction
Multiplication of two n × n matrices over a field F is an F-bilinear map F n×n × F n×n → F n×n called the matrix multiplication map. The matrix multiplication map corresponds naturally to the following structure tensor. Let [n] be the set {1, 2, . . . , n} and let {e i j : i, j ∈ [n]} be the standard basis for the vector space F n×n of n × n matrices. Define the structure tensor of the matrix multiplication map as n, n, n := ∑ i, j,k∈ [n] e i j ⊗ e jk ⊗ e ki ∈ F n×n ⊗ F n×n ⊗ F n×n .
(Technically, this is the structure tensor of the trilinear map that computes the trace of a product of three matrices.) Let V 1 , V 2 , and V 3 be vector spaces. The tensor rank of a tensor t ∈ V 1 ⊗V 2 ⊗V 3 is the smallest number r such that t can be written as a sum of r simple tensors v 1 ⊗ v 2 ⊗ v 3 ∈ V 1 ⊗V 2 ⊗V 3 . The computational complexity of matrix multiplication is tightly related to the tensor rank of the tensor n, n, n (see e.g. [5]). Strassen showed that the tensor rank of 2, 2, 2 is at most seven over any field [23]; Hopcroft and Kerr [13] showed that the tensor rank is at least seven over the finite field F 2 , and Winograd [25] showed that the tensor rank is at least seven over any field. Over an algebraically closed field, the border rank of a tensor t ∈ V 1 ⊗V 2 ⊗V 3 is the smallest number r such that t is in the Zariski closure of all tensors of rank at most r in V 1 ⊗ V 2 ⊗ V 3 . Landsberg proved that the border rank of 2, 2, 2 is seven over the field C of complex numbers [16], and a different proof for this based on highest-weight vectors was later given by Hauenstein, Ikenmeyer and Landsberg [12].
We extend the above results. Let t ∈ V 1 ⊗V 2 ⊗V 3 be a tensor in a fixed basis, a hypermatrix. The support of t is the set of coordinates where t has a nonzero coefficient. The support rank of t is the minimal rank of a tensor with the same support as t. This has also been called s-rank [7], nondeterministic rank [9], zero-one rank [24] and minimum rank of a nonzero pattern [2] in the literature. The border support rank of t is the minimal border rank of a tensor with the same support as t. We prove the following.
Theorem 1.1. The support rank of 2, 2, 2 is seven over any field F. Theorem 1.2. The border support rank of 2, 2, 2 is seven over C. Theorem 1.1 and Theorem 1.2 answer a question of Cohn and Umans [7], that was also posed as an open problem during the Algorithms and Complexity in Algebraic Geometry programme at the Simons Institute [1]. We note that, in general, computing the tensor rank or support rank of a tensor is a computationally hard task. Namely, given a 3-tensor t and a natural number r, deciding whether the tensor rank of t is at most r is NP-complete over any finite field [11] and NP-hard over any integral domain [22]. Moreover, given a 2-tensor (that is, a matrix) A and a natural number r, deciding whether the support rank of A is at most r is NP-hard over the real numbers [3].
Previously, it was known that the border support rank of the matrix multiplication tensor n, n, n is at least 2n 2 − n [4], so in particular that the border support rank of 2, 2, 2 is at least six. This result was obtained using Young flattenings.
Studying (border) support rank is interesting for two reasons. The first reason comes from algebraic complexity theory. As mentioned above, the tensor rank of the matrix multiplication tensor is tightly related to the computational complexity of matrix multiplication. It turns out that asymptotically, the border support rank of matrix multiplication gives an upper bound on the tensor rank of matrix multiplication, as follows. The exponent of matrix multiplication ω is defined as the smallest number β such that for any ε > 0 the tensor rank of n, n, n is in O(n β +ε ). The number ω is between 2 and 2.3728639 [18] and it is a major open problem in algebraic complexity theory to decide whether ω equals 2. One can define an analogous quantity ω s for the support rank of n, n, n . One can show with Strassen's laser method that ω ≤ (3ω s − 2)/2 [7]. To show that ω = 2, it therefore suffices to show that ω s = 2. Cohn and Umans aim to obtain upper bounds on ω s by realizing the algebra of n × n matrices inside some cleverly chosen group algebra.
The second reason, which was our original motivation, comes from quantum communication complexity. Let f : X ×Y × Z → {0, 1} be a function on a product of finite sets X, Y and Z. Alice, Bob and Charlie have to compute f in the following sense. Alice receives an x ∈ X, Bob receives a y ∈ Y and Charlie receives a z ∈ Z. Moreover, the players share a so-called Greenberger-Horne-Zeilinger (GHZ) state of rank r, which is described by the tensor GHZ r = ∑ r i=1 e i ⊗ e i ⊗ e i ∈ (C r ) ⊗3 . The players apply local quantum operations. After this, each player has to output a bit such that if f (x, y, z) = 1, then with some nonzero probability all players output 1 and if f (x, y, z) = 0, then with probability zero all players output 1. The complexity of such a protocol is the logarithm of the rank r of the GHZ-state used, and the minimum complexity of all quantum protocols for f is the nondeterministic communication complexity of f . This number equals the logarithm of the support rank of the tensor with support given by f , that is ∑ x,y,z f (x, y, z) e x ⊗ e y ⊗ e z [4]. Similarly, the logarithm of the border support rank of the tensor with support given by f equals the approximate nondeterministic communication complexity of f . Since tensor rank and border rank are natural measures of entanglement, our result may also be of interest to the quantum information theory community.
Notation. For any tensor t, we will denote tensor rank by R(t), border rank by R(t), support rank by R s (t) and border support rank by R s (t).
Paper outline. This paper is structured as follows. In Section 2 we give two proofs for Theorem 1.1. In Section 3 we give a short introduction to border rank lower bounds by highest-weight vectors and then apply this theory to prove Theorem 1.2.

Support rank
We will give two proofs for Theorem 1.1. Both proofs use the following lemma that reduces the 8parameter minimization problem at hand to a 1-parameter minimization problem. Let F be a field. Let e 11 = 1 0 0 0 , e 12 = 0 1 0 0 , e 21 = 0 0 1 0 , e 22 = 0 0 0 1 be the standard basis of the space of 2×2 matrices F 2×2 over F. Let e 1 , e 2 , e 3 , e 4 be the standard basis of F 4 . We naturally identify F 2×2 with F 4 by e 11 → e 1 , e 12 → e 2 , e 21 → e 3 , e 22 → e 4 . Let GL 4 (F) ×3 act on the tensor space F 2×2 ⊗ F 2×2 ⊗ F 2×2 accordingly. Lemma 2.1 (Parameter reduction). Let t ∈ F 2×2 ⊗ F 2×2 ⊗ F 2×2 be a tensor with the same support as the matrix multiplication tensor 2, 2, 2 . There is a tensor s in the GL 4 (F) ×3 -orbit of t, with the same support as t, such that all nonzero entries of s are 1 except possibly for the coefficient of e 11 ⊗ e 11 ⊗ e 11 .
Proof. Identify the tensor 2, 2, 2 = ∑ i, j,k∈ [2] e i j ⊗ e jk ⊗ e k with the tensor e 111 + e 123 + e 231 + e 243 + e 312 + e 324 + e 432 + e 444 ∈ F 4 ⊗ F 4 ⊗ F 4 , where e i jk = e i ⊗ e j ⊗ e k . We can view this tensor as as a 4 × 4 × 4 cube filled with elements 0 and 1 from F. Let t be a tensor in F 4 ⊗ F 4 ⊗ F 4 with the same support as 2, 2, 2 , so, in 1-slices, where a, b, c, d, e, f , g, h are nonzero elements in F. Here we index the 1-slices by the first tensor leg, the rows of the slices by the second tensor leg and columns of the slices by the third tensor leg. Scaling the 1-slices of t according to diag(1/b, 1/d, Scaling the rows of t according to diag(1/e , 1, 1/g , 1), that is, applying Finally, scaling the columns of t according to diag(1/c , 1, 1, 1), i.e., applying 1 4 ⊗1 4 ⊗diag(1/c , 1, 1, 1) to t , yields a tensor of the required form.
Our first proof of Theorem 1.1 uses a corollary of a result of De Groote on the uniqueness of the decomposition of 2, 2, 2 into simple tensors.
be an element of an arbitrary optimal decomposition of 2, 2, 2 into simple tensors over F such that the rank of each v i as an element of , where A, B and C act by matrix multiplication from the left and right on F 2×2 . Definition 2.3. For any number q ∈ F, define the perturbed matrix multiplication tensor 2, 2, 2 q := 2, 2, 2 + (q − 1) e 11 ⊗ e 11 ⊗ e 11 . This is the tensor obtained from 2, 2, 2 by replacing the coefficient of e 11 ⊗ e 11 ⊗ e 11 by q.
We now give our first proof of Theorem 1.1 using the above uniqueness statement.
Proof of Theorem 1.1; uniqueness argument. As was already observed by De Groote, Theorem 2.2 gives the upper bound R( 2, 2, 2 0 ) ≤ 6 and thus R( 2, 2, 2 q ) ≤ 7 for all q ∈ F. We claim R( 2, 2, 2 q ) ≥ 7 for all nonzero q ∈ F. Suppose q is a number in F such that R( 2, 2, 2 q ) = 6. Suppose we have a decomposition 2, 2, is an optimal decomposition of 2, 2, 2 into simple tensors. Therefore, by Theorem 2.2, there exist A, B,C in GL 2 (F) such that Let f 1 , f 2 be the standard basis of F 2 . Then by taking appropriate transposes the previous equation is equivalent to We conclude that q = 0. By Theorem 2.1 we can conclude that R s ( 2, 2, 2 ) = 7.
Our second proof of Theorem 1.1 uses a method called the substitution method. Let x i j , y i j , z i j (i, j ∈ [2]) be variables. Let X,Y, Z be the corresponding 2 × 2 variable matrices. For q ∈ F, define the function [2] x i j y jk z ki + (q − 1) The tensor rank of 2, 2, 2 q is equal to the smallest number r such that f q (X,Y, Z) can be written as a sum ∑ r ρ=1 u ρ (X)v ρ (Y )w ρ (Z), where u ρ is a linear form in the x i j , similarly for v ρ and w ρ .
Proof of Theorem 1.1; substitution method. Suppose that the function f q (X,Y, Z) has rank r in the sense that it has a decomposition into a sum of r products of three linear forms as described above. If U = (u i j ) i j∈ [2] is any upper triangular matrix, then f q (X,YU −1 ,UZ) has rank at most r and by direct computation There exists an upper triangular matrix U such that the function g q (X,Y, Z) := f q (X,YU −1 ,UZ) has a decomposition in which w r (Z) is of the form z 21 + a 12 z 12 + a 22 z 22 for some a 12 , a 22 ∈ F. Apply the substitution z 21 →w(Z) := −a 12 z 12 − a 22 z 22 to (2.1) to see that ∑ j∈ [2] x 1 j y j1 z 11 + x 2 j y j1 z 12 + x 1 j y j2w (Z) + x 21 y 12 z 22 + q x 22 We can test that y 22 occurs in the obtained decomposition of (2.2) by setting x 22 , z 22 to 1 and y 21 to 0 and the other x i j , z i j to 0. We can test that y 12 occurs in the obtained decomposition of (2.2) by setting x 21 , z 22 to 1 and the other x i j , z i j to 0. Say y 22 occurs in v r−1 and y 12 occurs in v r−2 . Then, there is a substitution To clean up, setting z 22 → 0 in (2.3) shows that ∑ j∈ [2] x 1 j y j1 z 11 + x 2 j y j1 We can test that x 21 occurs in the obtained decomposition (2.4) by setting y 11 , z 12 to 1 and the other x i j , z i j to 0. Similarly, we can test that x 22 occurs in the obtained decomposition (2.4) by setting y 21 , z 12 to 1 and the other x i j , z i j to 0. Say x 21 occurs in u r−3 and x 22 occurs in u r−4 . We apply a substitution x 21 →ũ 21 (X), x 22 →ũ 22 (X) to see that ∑ j∈ [2] x 1 j y j1 z 11 +ũ 2 j (X) y j1 Apply the substitution z 12 → 0 to (2.5) to get ∑ j∈ [2] x 1 j y j1 z 11 = (2.6) which clearly has rank 2. Therefore, r ≥ 7. By Theorem 2.1 we are done.

Border support rank
In this section all vector spaces are complex vector spaces. We will review a method that was introduced in [17] to study equations for border rank and that was later used in [12] to give a proof that R( 2, 2, 2 ) ≥ 7. Then, we will use this method to show that the border support rank of 2, 2, 2 equals seven. Our Python code is included as an ancillary file with the arXiv submission. View the space ⊗ 3 C n as an affine variety, and let C[⊗ 3 C n ] be its coordinate ring. Define σ r ⊆ ⊗ 3 C n as the subset of tensors with border rank at most r, This is called the rth secant variety of the Segre variety of C n × C n × C n . The set σ r is Zariski closed in ⊗ 3 C n by definition of border rank. In other words, if we let I(σ r ) ⊆ C[⊗ 3 C n ] be the ideal of polynomials on ⊗ 3 C n that vanish identically on σ r , then Z(I(σ r )) = σ r .

Lower bounds by polynomials
By definition, if R(t) > r then there exists a polynomial in I(σ r ) that does not vanish on t. The following standard proposition says that we may in fact assume that this polynomial is homogeneous.
Proposition 3.1. Let t ∈ ⊗ 3 C n . If R(t) > r, then there exists a homogeneous polynomial f in I(σ r ) such that f (t) = 0.
Proof. We give a proof for the convenience of the reader. If f (t) = 0 for all f ∈ I(σ r ), then we have t ∈ Z(I(σ r )) = σ r , which is a contradiction. Let f be a polynomial in I(σ r ) such that f (t) = 0. Let f = ∑ d f d be the decomposition of f into homogeneous parts. There is a d such that f d (t) = 0.
Let v ∈ σ r . For any α ∈ C, define g(α) := f (αv). This is a polynomial in α. We have that g(α) = ∑ d α d f d (v). Since σ r is closed under scaling and f (v) = 0, we have g(α) = 0 for any α ∈ C, so g is the zero polynomial. Therefore, each coefficient f d (v) is 0. This argument holds for any v ∈ σ r , so f d ∈ I(σ r ) for each d.
The polynomial ring C[⊗ 3 C n ] decomposes into a direct sum of homogeneous parts C[⊗ 3 C n ] d and, by the above argument, the vanishing ideal I(σ r ) decomposes accordingly as The space ⊗ 3 C n has a natural action of G := GL ×3 n and σ r is a G-submodule. Thus C[⊗ 3 C n ] d ∼ = Sym d (⊗ 3 (C n ) * ) has a natural action of G and I(σ r ) d is a G-submodule. We will use the well-known theory of highest-weight vectors to exploit this symmetry. The theory of highest-weight vectors holds in a much more general setting than we need here. We refer to [15, III.1.5] and [10] for the general theory, and focus on a description of the theory for the group GL ×3 n . Let W be a finite-dimensional G-module. Choose a basis so that G becomes the group of triples of invertible matrices. Let T ⊆ G be the subgroup of triples of diagonal matrices. For t = (diag(a 1 , . . . , a n ), diag(b 1 , . . . , b n ), diag(c 1 , . . . , c n )) ∈ T and z = (u, v, w) ∈ (Z n ) 3 define t z := ∏ n i=1 a u i i b v i i c w i i . As a T -module, W decomposes into weight spaces, The vectors in W z are said to have weight z. Let U ⊆ G be the subgroup of triples of unipotent matrices, that is, upper triangular matrices with ones on the diagonal. A nonzero vector v ∈ W z is a highest-weight vector if u · v = v for all u ∈ U.
A finite-dimensional (rational) representation W of GL ×3 n is irreducible if and only if it has a unique highest-weight vector v, up to multiplication by a scalar, that is, [W ] U = Span C v. If W is irreducible and v is a highest-weight vector, then one has W = Span C (Gv). Moreover, two irreducible representations are isomorphic if and only if their highest-weight vectors have the same weight. We call a sequence of n nonincreasing integers a generalized partition. It turns out that the weight of a highest-weight vector is a triple of generalized partitions. For any triple of generalized partitions λ , we will denote an abstract realisation of the G-module with highest-weight λ by V λ . For any finite-dimensional G-module W , the highest-weight vectors in W of weight λ form a vector space, which we denote by For a generalized partition λ , define the dual partition λ * as the generalized partition obtained from λ by negating every entry and reversing the order. Then V λ * = V * λ , the dual module. We note that the polynomial irreducible representations are precisely the ones that are isomorphic to V λ with λ a partition.
Recall that σ r is the variety of tensor in ⊗ 3 C n of border rank at most r. Consider the isotypic decomposition of W := Sym d (⊗ 3 (C n ) * ) and I(σ r ) d under the action of GL ×3 n , where λ runs over all triples of partitions of d with at most n parts, and k(λ )V * λ denotes an isotypic component consisting of a direct sum of k(λ ) copies of the irreducible G-representation V * λ , similarly for m(λ )V * λ . Note that, although the direct sums run over triples of partitions λ , the representations W and I(σ r ) are not polynomial since we take duals. The number k(λ ) is exactly the dimension of the highest-weight vector space [W λ * ] U , and the number m(λ ) is the dimension of the highest-weight vector space [I(σ r ) λ * ] U . The following proposition extends Theorem 3.1 by saying that we may assume that the polynomial we are looking for is a highest-weight vector, if we replace t by a random point in its G-orbit. Proposition 3.2. Let t ∈ ⊗ 3 C n . If R(t) > r, then there exists a highest-weight vector f ∈ I(σ r ) and a group element g ∈ G such that f (gt) = 0.
Proof. We provide the proof for the convenience of the reader. By Theorem 3.1, there exists a homogeneous polynomial f ∈ I(σ r ) such that we have f (t) = 0. By highest-weight theory, the polynomial f can be written as a sum ∑ λ ,i g λ ,i f λ ,i , where f λ ,i is a highest-weight vector of type λ in I(σ r ) and g λ ,i ∈ G. Since f (t) = 0, there exists a λ and an i so that f λ ,i (g −1 λ ,i t) = 0.

Highest-weight vector method
The following method was first proposed in [17] to study equations for border rank and was later used in [12] to give a proof that R( 2, 2, 2 ) ≥ 7. Let t ∈ ⊗ 3 C n be a tensor for which we want to show R(t) > r.
1. Choose a degree d ∈ N. Let W be the space Sym d (⊗ 3 (C n ) * ). Choose a partition triple λ d such that the highest-weight vector space [W λ * ] U is nonzero.

Construct a basis
3. Find a linear combination f of the basis elements b 1 , . . . , b k that vanishes on all tensors of border rank at most r, that is, f ∈ [I(σ r ) λ * ] U where σ r is the variety of tensors with border rank at most r.
4. Show that f does not vanish on gt for some g ∈ G.
The above method is guaranteed to work by Theorem 3.2. Before applying the method, we will consider each step in more detail.
Step 1. Kronecker coefficient. The dimension of the space of U-invariants [(Sym d (⊗ 3 (C n ) * )) λ * ] U is the so-called Kronecker coefficient k(λ ). We pick a partition triple λ such that the number k := k(λ ) is nonzero. Algorithms for computing Kronecker coefficients have been implemented in for example Schur [26], Sage [20] and the Python package Kronecker [6].
Before going to the next step we discuss how to efficiently implement the evaluation of a polynomial represented by a pair of permutations, as was already described in [12]. Let f = φ λ πP d and let t be the The evaluation of the polynomial f at t is equal to the contraction Note that the last expression is a sum of a product of determinants. Let us study the first factor of a summand. Let ν denote the transpose of λ (1) . We have where det m (v 1 , . . . , v m ) denotes top m-by-m minor of the matrix with columns v 1 , . . . , v m . Suppose that, in our evaluation of ∑ j , we have chosen values for j 1 , . . . , j ν 1 and suppose det ν 1 (t 1 j 1 , . . . ,t 1 ) is 0. Then whatever choices we make for j ν 1 +1 , . . . , j d , the summand at hand will be zero. Recognizing this situation early is crucial.
Step 3. Construction of a vector in I(σ r ). Pick k random tensors t 1 , . . . ,t k of rank r. Evaluate each basis element b i in each random tensor t j . If the resulting matrix (b i (t j )) i, j∈[k] has a nontrivial kernel, then we find a candidate highest-weight vector f in I(σ r ). We can verify the correctness of the candidate by evaluating f at a symbolic tensor of rank r. This evaluation should be zero. The way we do this symbolic evaluation is by working in ⊗ 3 C 6 and using the straightening algorithm, see e.g. the SchurFunctors package in Macaulay2 [21]. We used multi-prolongation to split up the computation in order to save memory. We refer to [17,19] for a discussion of multi-prolongation.
Step 4. Evaluating at gt. Evaluate f at gt for a random g ∈ G. (In our case, it turns out that taking g to be the identity is good enough.)
Proof of Theorem 1.2. The upper bound follows from Theorem 1.1, so it remains to prove the lower bound. Let σ 6 be the variety of tensors in ⊗ 3 C 4 of border rank at most 6. We will apply the method described above to the tensor 2, 2, 2 q , see Theorem 2.3.
We have thus found two highest-weight vectors such that f 19 ( 2, 2, 2 q ) = α q(3q + 2) and f 20 ( 2, 2, 2 q ) = β q 2 (q + 1), where α and β are nonzero constants. The only simultaneous root of these polynomials occurs at q = 0. This means that for any nonzero q, the point 2, 2, 2 q is not contained in σ 6 . From Theorem 2.1 we conclude that the border rank of any tensor with the same support as 2, 2, 2 is at least seven, which proves the theorem.