On cap sets and the group-theoretic approach to matrix multiplication

In 2003, Cohn and Umans described a framework for proving upper bounds on the exponent $\omega$ of matrix multiplication by reducing matrix multiplication to group algebra multiplication. In 2005 Cohn, Kleinberg, Szegedy, and Umans proposed specific conjectures for how to obtain $\omega=2$ in this framework. In this note we rule out obtaining $\omega=2$ in this framework from the groups $\mathbb{F}_p^n$, using the breakthrough results of Croot, Lev, Pach, Ellenberg, and Gijswijt on cap sets. These restrictions do not however rule out abelian groups in general, let alone nonabelian groups.


Introduction
A cap set is a subset of F n 3 containing no lines; equivalently, if u, v, and w belong to the set, then u + v + w = 0 if and only if u = v = w. In a remarkable pair of recent papers [9,11], Croot, Lev, and (1.1) If the sets involved are large enough, this yields a nontrivial bound on ω, and [4] showed how a family of sufficiently large STPP constructions would imply ω = 2. This paper contains two main results. The first is a bound on the size of tricolored sum-free sets in abelian groups of bounded exponent. A tricolored sum-free set is a generalization of a three-term progression-free set (a set whose elements satisfy u + v = 2w ⇐⇒ u = v = w) in which the elements u, v, and w range over three different subsets (see Definition 3.1). The exponent of a finite group is the least common multiple of the orders of all its elements; if a finite abelian group is generated by elements of order at most m, then its exponent is at most lcm(1, 2, . . . , m) = e m(1+o(1)) . (This asymptotic formula for lcm (1, 2, . . . , m) is a variant of the prime number theorem. See, for example, Theorem 11 in [23] for a stronger bound on ψ(m) = log lcm(1, 2, . . . , m).) Theorem A. There exists an absolute constant ε = 1 2 log((2/3)2 2/3 ) = 0.02831 . . . such that if H is an abelian group generated by elements of order at most m, then every tricolored sum-free set in H has size at most 3 · |H| 1− ε m .
In the case when H is a power of a single cyclic group Z/p k Z, we obtain a sharper bound.
Theorem A . There exists an absolute constant δ = 2ε = 0.05663 . . . such that if H ∼ = (Z/qZ) n for some prime power q, then every tricolored sum-free set in H has size at most 3 · |H| 1− δ log q .
These theorems extend the Ellenberg-Gijswijt [11] result in two directions: from progression-free sets to tricolored sum-free sets 1 and from abelian groups of prime exponent to abelian groups of bounded exponent. In particular, Theorem A disproves the strong USP conjecture via [1]. Note that when H has prime exponent p, the bound in Theorem A of 3 · |H| 1− δ log p agrees with the bound for progression-free sets obtained in [11], but with a slightly worse exponent as we have stated it uniformly in p. It remains to be seen whether ε m can be improved to c log m for general abelian groups of bounded exponent. A generalization of the Ellenberg-Gijswijt result that handles progression-free sets in abelian p-groups of bounded exponent was also obtained by Petrov in [20].
In the proof of Theorem A, we study a variant of tensor rank due to Tao [22], which we call slice rank, and we show how slice rank is related to a quantitative version of the notion of instability from geometric invariant theory. We show that functions with low slice rank are unstable, and conversely prove that a quantitative bound on instability yields a bound on slice rank of tensor powers (see Section 4.2).
Our second main result generalizes [1] by showing that every STPP construction yields a large tricolored sum-free set. Together these two facts show that it is impossible to prove ω = 2 using sets satisfying the simultaneous triple product property in abelian groups of bounded exponent: Theorem B. For every ∈ N, there is an ε > 0 such that no STPP construction in any abelian group of exponent at most is large enough to yield a bound better than ω ≤ 2 + ε via the inequality (1.1).
Note that all of the current best bounds on ω via the group-theoretic approach (in [4]), as well as the current best bounds on ω which use the Coppersmith-Winograd approach [7,10,24,15], yield STPP constructions whose underlying group is (Z/mZ) n for m fixed.
However, our results do not rule out achieving ω = 2 by using STPP constructions over abelian groups. Specifically, when the group has a large cyclic factor, it indeed contains a large sum-free subset and thus the constraints analyzed here are irrelevant. Furthermore, one can use non-abelian groups or even more general objects such as association schemes [6]. Thus, our results serve to focus the search for group-theoretic constructions, and certainly do not rule them out as an approach to achieving ω = 2.

The simultaneous triple product property
Recall that the exponent of matrix multiplication is defined as ω = inf{c : the tensor rank of n × n matrix multiplication is at most O(n c ) as n → ∞}.
In 2003, Cohn and Umans [5] described a framework for proving upper bounds on ω by reducing matrix multiplication to group algebra multiplication. This reduction is carried out by means of a triple of subsets satisfying the triple product property: . Subsets A, B,C of a group G satisfy the triple product property if Such a triple of subsets realizes |A| , |B| , |C| inside the group algebra of G. (Here m, n, p denotes the matrix multiplication tensor for multiplying an m × n matrix by an n × p matrix.) From this, letting d 1 , d 2 , . . . be the character degrees of G (i.e., the dimensions of its irreducible representations), we obtain the inequality by bounding the rank of group algebra multiplication [5,Theorem 4.1]. This inequality yields an upper bound for ω when G and A, B,C are chosen appropriately.
In the later paper [4] several concrete routes to proving ω = 2 were proposed. These proposals seemingly go beyond the framework of the triple product property in various different ways; however as described in [4, §7], all of these constructions can be uniformly described using the triple product property as follows. Several independent matrix multiplications are realized via the triple product property in the group algebra of a wreath product H S m = H m S m using certain well-chosen subsets of a group H. This general formulation is captured by the simultaneous triple product property: By a perfect matching M ⊆ S × T × U we mean a subset whose projection onto each of the three factors is a bijection. In other words, given s ∈ S there exist unique t ∈ T and u ∈ U such that (s,t, u) ∈ M, and similarly for the other factors. The cardinality of the matching M is therefore equal to |S|, |T |, and |U|. Note that any 3-dimensional matching M ⊆ H 3 (not necessarily perfect, i.e., replacing "bijection" above with "injection") uniquely determines three sets S, T,U ⊆ H such that M is a 3-dimensional perfect matching on S × T ×U.
In this section, we show how to obtain a tricolored sum-free set from any STPP construction in an abelian group. This allows us to prove that Theorem A implies Theorem B; we then prove Theorem A in Section 4.

STPP constructions imply tricolored sum-free sets
We will first construct slightly weaker objects we call border tricolored sum-free sets, which are motivated by the notion of combinatorial degeneration in the theory of border rank (see Definition 15.29 in [2]). We will then use border tricolored sum-free sets to construct genuine tricolored sum-free sets.
Definition 3.2. A border tricolored sum-free set in an abelian group H is a 3-dimensional perfect matching M ⊆ S × T ×U on a triple of sets S, T,U ⊆ H together with functions α : S → Z, β : T → Z, and γ : U → Z such that s + t + u = 0 and α(s) + β (t) + γ(u) = 0 for all (s,t, u) ∈ M, while for all (s,t, u) ∈ (S × T ×U) \ M. The cardinality of a border tricolored sum-free set is the cardinality of M, and its range is the maximum of |α(s)|, |β (t)|, and |γ(u)| over s ∈ S, t ∈ T , and u ∈ U.
One of our main new contributions in this paper is the following construction: Proof. As in Definition 2.2 (but using additive notation), we will set We will first construct a matching M out of smaller matchings M i on subsets of S i , T i , and U i . This matching will not quite be a tricolored sum-free set in general, but we will define functions α, β , and γ that repair any problems with it, so that it becomes a border tricolored sum-free set.
To begin, let n i = |A i |, m i = |B i |, and p i = |C i |, and identify , respectively, via bijections α i , β i , γ i . Let r i be the most frequently occurring value in the multiset The size of M i is the number of times r i occurs in the multiset above. The number of distinct elements of the multiset is n i + m i + p i − 2; we can ignore the 2 and bound this by |A i | + |B i | + |C i |. Since r i was chosen to be the most frequent value, Each M i is a 3-dimensional perfect matching: given the first coordinate a − b, the triple product property determines a and b from a − b, and then α i (a) + β i (b) + γ i (c) = r i determines c, and the same is true for the other two coordinates. We combine these matchings by setting M = i M i . Note that the sets M i are disjoint from each other (since the sets S i are, for example), so |M| = ∑ i |M i |. All that remains is to define the functions α, β , γ so as to obtain a border tricolored sum-free set.
Let S i , T i , U i be the projections of M i onto the three factors of H 3 . In other words, and T i and U i can be expressed similarly. Let S = i S i , T = i T i , and U = i U i , so that M is a perfect matching between S, T , and U.
To complete the proof, we must analyze when s + t + u = 0 with s ∈ S, t ∈ T , and u ∈ U. First, note that the simultaneous triple product property implies that if s i + t j + u k = 0 with s i ∈ S i , t j ∈ T j , and u k ∈ U k , then i = j = k. Thus, we cannot obtain a sum of zero from S i , T j , and U k unless i = j = k, so we can analyze each matching M i individually, without worrying about how they might interact with each other. More specifically, if for each i we can find functions on S i , T i , and U i that make M i into a border tricolored sum-free set, then this yields functions on S, T , and U that make M into a border tricolored sum-free set.
Note that s determines a and b , t determines b and c , and u determines c and a since A i , B i ,C i satisfy the triple product property. Furthermore, the triple product property tells us that However, that is not enough to conclude that (s,t, u) ∈ M i , because it is not necessarily the case that α i (a) + β i (b) + γ i (c) = r i . To address this issue, we will define functions α, β , and γ on S i , T i , and U i , respectively, so that α(s) + β (t) + γ(u) = α i (a) + β i (b) + γ i (c) − r i 2 whenever s + t + u = 0. Specifically, we can take (We simply expand the square and assign each term to one of α(s), β (t), γ(u) so that α(s) depends only on a and b , β (t) depends only on b and c , and γ(u) depends only on c and a .) By construction, α(s) + β (t) + γ(u) ≥ 0 whenever s + t + u = 0, and α(s) + β (t) + γ(u) = 0 exactly when (s,t, u) ∈ M i . Thus, we have constructed a border tricolored sum-free set, as desired.
Lemma 3.4. Let H be an abelian group in which there is a border tricolored sum-free set of cardinality |M| and range t. Then for each natural number N, there exists a tricolored sum-free set in H N of cardinality at least |M| N /(2Nt + 1) 3 .
In particular, as N → ∞ there are tricolored sum-free sets in H N of cardinality |M| (1−o(1))N .
Proof. We will use the notation from Definition 3.2 for the border tricolored sum-free set in H: let M be the perfect matching on sets S, T,U ⊆ H, with functions α : S → Z, β : T → Z, and γ : for all i, and define α(s 1 , . . . , s N ) = α(s 1 ) + · · · + α(s N ) and similarly for β and γ. This construction yields a border tricolored sum-free set of cardinality |M| N and range Nt in H N .
To control the size of the tricolored sum-free sets resulting from Theorem 3.3 and Lemma 3.4, we will need the following notion. We say that an STPP construction is

Proof. Let the original STPP construction consist of n triples
Our new STPP construction will consist of subsets of H 3N , where N is a large number to be chosen later; these subsets are indexed by triples (The products here are cartesian products of sets.) It is not hard to verify that these sets satisfy the STPP in H 3N (see [4,Lemma 5.4]). The resulting STPP construction is not yet uniform, but will become so below when we restrict the choices of u, v, and w. We first argue that the STPP construction To check that this construction meets the packing bound, we observe that because the original STPP construction meets the packing bound. Expanding the left side gives We have as desired; the same also holds for ∑ u,v,w B u,v,w C u,v,w and ∑ u,v,w A u,v,w C u,v,w .
To enforce uniformity, we restrict our attention to only certain choices of u, v, and w by observing that the cardinality depends only on the distributions of u, v, w (where the distribution of u is the vector specifying the number of times each element of [n] occurs in u). The same is true for | B u,v,w | and | C u,v,w |. There are N+n−1 n−1 possible distributions, but all we need is the crude upper bound (N + 1) n from the fact that each element of [n] occurs between 0 and N times. It follows that there is at least one triple µ 1 , µ 2 , µ 3 of distributions for which where u ∼ µ 1 means u has distribution µ 1 .
Restricting to only those sets A u,v,w , B u,v,w , and C u,v,w with u ∼ µ 1 , v ∼ µ 2 , and w ∼ µ 3 thus gives a uniform STPP construction. Combining (3.1) and (3.2) we obtain which is again |H| 3N(1−o(1)) as long as N is chosen sufficiently large relative to n and |H|. The same holds for the other two conditions in the packing bound.
Choosing N and µ 1 , µ 2 , µ 3 in this way thus yields a family of uniform STPP constructions meeting the packing bound in powers of H, as desired.

Theorem A implies Theorem B
We conclude this section by proving that Theorem A implies Theorem B; we will then prove Theorem A in Section 4.
Fix ∈ N, and suppose that for each δ > 0 there is an STPP construction in a group of exponent at most that proves ω ≤ 2 + δ . Choosing a sequence with δ tending to zero, we obtain a family of STPP constructions that meets the packing bound by Lemma 2.4. Furthermore Lemma 3.5 lets us inflate these constructions to make them uniform, while still meeting the packing bound, in powers of the original groups; in particular, the exponent of all our groups is still at most .
Consider one of these uniform STPP constructions in a group H of exponent at most , thus generated by elements of order at most . By Theorem 3.3 there exists a border tricolored sum-free set in H of Since |A i |, |B i |, and |C i | are each independent of i, without loss of generality we may assume that |C i | is the largest of these, in which case Finally, Lemma 3.4 converts the border tricolored sum-free set to a genuine tricolored sum-free set, at the cost of raising H to a high power (which of course does not change the exponent of this group). This contradicts Theorem A, which states that the cardinality of any tricolored sum-free set in H N is at most 3 |H| N(1− ε ) , and thus completes the proof of Theorem B.

Tricolored sum-free sets in abelian groups of bounded exponent via rank and instability of tensors
The remainder of the paper is devoted to the proof of Theorems A and A . Our approach builds on the symmetric formulation presented by Tao [22], although an earlier version of this paper contained a more direct application of the original methods [9,11] to the F n p case. We begin here by outlining the main ideas of the proof, giving a road map for the remaining sections. In order to formulate the symmetric version, we develop the notion of slice rank (in Section 4.1), which is a weakening of tensor rank. For tricolored sum-free sets in F n p , the upper bound is then a consequence of the following three facts about slice rank: 1. Just as the ordinary matrix rank of an m × m diagonal matrix is equal to m if all the diagonal entries are nonzero, the same holds for the slice rank of an m × m × m diagonal tensor (over any field; see Lemma 4.7 or [22, Lemma 1]).

2.
A tricolored sum-free set of cardinality m in a group G implies that the multiplication tensor of the group algebra FG restricts to an m × m × m diagonal tensor with nonzero diagonal entries. It follows that the slice rank of the multiplication tensor of FG is at least the size of any tricolored sum-free set in G (see Proposition 4.8).
3. Over a field of characteristic p, the slice rank of the F n p -multiplication tensor is at most the number of vectors in [p] n with coordinate sum at most pn/3, which is at most (p − ε p ) n for some ε p > 0. By a remarkable observation [9,11,22], this upper bound on the slice rank follows almost immediately from the fact that the (x, y, z) entry of the tensor in question equals δ 0 (x + y + z), and the delta function δ 0 (x + y + z) (indeed every function of x + y + z) is expressible as an n-variate polynomial of degree (p − 1)n; see Observation 4.11.
These three statements give an upper bound of (p − ε p ) n on the cardinality of a tricolored sum-free set in F n p . Expressed a different way, the upper bound for H = F n p is |H| 1−α p for some constant α p > 0. In Section 4.4, we prove an upper bound of the same form for tricolored sum-free sets in any group of the form G = (Z/p k Z) n with p prime, using a similar argument but using binomial coefficients instead of polynomials to describe the delta function.
We then extend this bound to any abelian group H of bounded exponent by arguing that such a group must decompose as H ∼ = G × K where G ∼ = (Z/p k Z) n and |G| ≥ |H| c for a constant c > 0. Such a decomposition expresses the H-multiplication tensor D H as a tensor product D G ⊗ D K . We establish a simple but powerful property of slice rank: the slice rank of T ⊗ T is at most the slice rank of T times the side length of T (Proposition 4.2). This allows us to reduce to the (Z/p k Z) n case, because the slice rank of D H is at most |G| 1−α |K| = |H| 1−α for some constant α > 0. As before, this bound on slice rank also bounds the cardinality of tricolored sum-free sets in G.
In the presentation below, we draw connections to (in)stability of tensors, a notion coming from geometric invariant theory (GIT). This broader context seems powerful, and potentially useful beyond the results in this paper. However, the reader who is interested only in the proofs of Theorems A and B can skip all of Section 4.2 except Lemma 4.7 and Proposition 4.8 and skip Theorem 4.10 since we only need the shaper bound on slice rank stated in Proposition 4.13, coming from triangle rank.

Tensor rank and its variants
Throughout this section, X, Y , and Z will denote finite sets. A function F : X ×Y → F with values in a field F has an unambiguous rank; rank(F) is the smallest k for which we can write F(x, y) = ∑ k i=1 f i (x)g i (y), and this coincides with the rank of the |X| × |Y | matrix described by F.
A common way to define the rank of a function F : X ×Y × Z → F is tensor rank: tensor-rank(F) is the smallest k for which we can write In this paper, we make use of another notion of rank which we call slice rank: slice-rank(F) is the smallest k for which we can write (4.1) As far as we know, this notion of rank was first used by Tao in [22]; here we take this study a bit further by establishing some basic properties of slice rank (Section 4.1) and connections to GIT (Section 4.2). Since any sum ∑ i f i (x)g i (y)h i (z) automatically fits the form (4.1), we always have Similarly, directly from the definition (4.1) we immediately conclude that and tensor-rank(F) ≤ slice-rank(F) · max(|X| , |Y | , |Z|). As another example of the difference between slice and tensor rank, every function F : X × X × X → F has slice rank at most N (and this bound is sharp; see Lemma 4.7 or [22, Lemma 1]). However, tensor rank can be much larger: the tensor rank of a generic 2 F is N 3 3N−2 ≈ N 2 /3 if N = 3 and F is algebraically closed [16].
Given functions F : X ×Y × Z → F and G : X ×Y × Z → F, set X = X × X , Y = Y ×Y , and Z = Z × Z , and let F ⊗ G denote the function F ⊗ G : X ×Y × Z → F given by The inequality tensor-rank(F ⊗ G) ≤ tensor-rank(F) · tensor-rank(G) is well-known; it has the following two variants, which bound the slice rank of a tensor product: For any F, G as above, slice-rank(F ⊗ G) ≤ slice-rank(F) · tensor-rank(G) and slice-rank(F ⊗ G) ≤ slice-rank(F) · max(|X | , |Y | , |Z |).
Proof. Set k = slice-rank(F) and choose functions f i and g i for 1 ≤ i ≤ k as in (4.1). Similarly, set = tensor-rank(G) and write G(x , y , z ) = ∑ j=1 α j (x )β j (y )γ j (z ). If we then define This demonstrates that slice-rank(F ⊗ G) ≤ k · , which verifies the first claim. For the second claim, we instead define This shows that slice-rank(F ⊗ G) ≤ k · max(|X | , |Y | , |Z |), which verifies the second claim.

Unstable tensors and slice rank
In this section we relate slice rank to the notion of an unstable tensor from geometric invariant theory. We show that functions with low slice rank are unstable and prove that a quantitative bound on instability yields a bound on slice rank of tensor powers. However, geometric invariant theory only deals with algebraically closed fields F, which is why for general F we take Definition 4.3 as the definition of unstable. To make our bounds explicit we will also need to introduce a quantitative version of this condition.
Definition 4.4. The instability of a function F : X ×Y × Z → F is the supremum of all ε ≥ 0 such that there exist bases f a , g b , h c , nontrivial weights u a , v b , w c ∈ R, and coefficients r a,b,c ∈ F such that By "nontrivial weights" we mean that the weights u a , v b , and w c should not all be constant (since in that case our definition of R becomes degenerate). Note that a function F is unstable if and only if instability(F) > 0. By convention, if for some F there is no nonnegative ε satisfying the hypotheses of the definition, we define instability(F) = −∞. For readers familiar with the terminology of GIT, we remark that (assuming F is algebraically closed) the function F is semi-stable if and only if instability(F) ∈ {−∞, 0}, and F is stable if and only if instability(F) = −∞. For consistency with this terminology, we are careful to use the term "not unstable" where appropriate (which is not the same, in GIT, as "stable").

Remark 4.5 (Notes on the definition of instability).
1. Translating all the weights u a (or all the v b , or all the w c ) by a constant translates the quantity R of (4.5) by the same amount, so the condition u a + v b + w c ≤ R is invariant under translation. Therefore if we like we may assume that u avg = v avg = w avg = 0, or that u min = v min = w min = 0, without loss of generality. Similarly, scaling all of the weights u a , v b , w c by the same constant does not change the definition of instability nor the value of instability(F).
2. It does not matter whether we require the weights to lie in R, Q, or Z. Indeed, the supremum defining instability could be taken over rational ε ≥ 0 without affecting the definition. For any given rational ε, the inequalities relating the u a , v b , and w c are a system of homogeneous linear inequalities with rational coefficients, so they have rational solutions if and only if they have real solutions. We can then transform rational weights to integer weights by scaling. (In particular, this justifies our earlier claim that the Hilbert-Mumford criterion is equivalent to Definition 4.3 when over an algebraically closed field; the usual statement of the Hilbert-Mumford criterion would require integer weights with u avg = v avg = w avg = 0.) 3. The definition of instability would be the same if the f a , g b , and h c were arbitrary functions not required to form bases, as long as |A| ≤ |X|, |B| ≤ |Y |, and |C| ≤ |Z|, where A, B, C denote index sets for the f a , g b , h c , respectively. Indeed, if one function f a is a linear combination of previous functions f a with u a ≤ u a , then the terms f a ⊗ g b ⊗ h c appearing in (4.4) can be replaced with a linear combination of terms f a ⊗ g b ⊗ h c , still satisfying u a + v b + w c ≤ u a + v b + w c ≤ R. Repeating this, we may assume that the set of functions { f a } a∈A (with A ⊆ A) appearing is linearly independent. Then extend this set to a basis arbitrarily, giving |X| − |A| of these new functions the weight u avg and giving |A| − |A | of them the weights {u a } a∈A\A . The other tensor factors are handled similarly.
Recall from (4.2) that any function F : X ×Y × Z → F has slice-rank(F) ≤ min(|X| , |Y | , |Z|). It turns out that there is a close relationship between F being unstable and this inequality being strict. More precisely, the latter implies the former (Theorem 4.6), and the former implies that the latter holds for sufficiently large tensor powers when |X| = |Y | = |Z| (Theorem 4.10).
Proof. Given an arbitrary F, choose a decomposition with p + q + r = slice-rank(F). Note that this implies that the p functions f 1 , . . . , f p are linearly independent (otherwise we could find a smaller decomposition of F), so extend them to a basis f 1 , . . . , f |X| for the functions X → F. Do the same for the other factors, and define weights by Expanding α i (y, z) as a linear combination of g j (y)h k (z) and so on, the decomposition (4.6) says that F is a linear combination of functions f i · g j · h k where at least one of 1 ≤ i ≤ p, 1 ≤ j ≤ q, or 1 ≤ k ≤ r holds. From the definition of our weight functions, these are precisely the cases when u i + v j + w k ≤ −1.
On the other hand u avg = − p |X| , v avg = − q |Y | , and w avg = − r |Z| , from which it follows that Our hypothesis that slice-rank(F) < min(|X| , |Y | , |Z|) thus guarantees that these weights satisfy u avg + v avg + w avg > −1. In this case, F is a linear combination of functions f i · g j · h k where u i + v j + w k ≤ −1 < u avg + v avg + w avg , so F is unstable.
This lets us compute the slice rank of any diagonal. Say that a function F : X ×Y × Z → F is a diagonal if its support {(x, y, z) : F(x, y, z) = 0} is a perfect matching on X, Y , and Z. Note that this implies |X| = |Y | = |Z| = |support(F)|, and that we require all the diagonal entries in F (i.e., those corresponding to the perfect matching) to be nonzero.
The following lemma was introduced to great effect by Tao [22,Lemma 1]. We first reproduce Tao's elementary proof for completeness, and then include a second proof inspired by the GIT perspective. Tao's proof from [22]. Suppose that slice-rank(F) < |X|. Then there exist k < |X| and functions f i , g i such that y). (4.7) Without loss of generality suppose that j > 0, and let This vector space has dimension at least |X| − j. Let u ∈ V have maximal support, and set Σ = {x ∈ X : u(x) = 0}. Then |Σ| ≥ dimV ≥ |X| − j, since otherwise there exists nonzero r ∈ V vanishing on Σ, and the function u + r would have a larger support than u. Multiply both sides of (4.7) by u(x) and sum over x to reduce from a tensor to a matrix. As u ∈ V is orthogonal to f i for i = 1, . . . , j, the right side of (4.7) becomes which is a function of rank at most k − j < |X| − j. Since the support of F(x, y, z) is precisely given by a perfect matching, ∑ x∈X u(x)F(x, y, z) will be a function with rank equal to | support(u)| ≥ |X| − j. Thus we have a contradiction, and it follows that slice-rank(F) = |X|.
We remark that Tao's proof is inductive, and extends to higher-order tensors as well [22]. The next proof of Lemma 4.7 uses the geometric invariant theory perspective on slice rank.
Proof of Lemma 4.7. By Theorem 4.6, it suffices to prove that a diagonal is not unstable. It will be convenient to use dual bases to compute coefficients. Given bases f a , g b , h c as in Definition 4.3, consider the dual bases f a , g b , h c ; in other words, ∑ x∈X f a (x) f a (x) = 1 if a = a , and 0 otherwise, and the same holds for g b , g b (summing over Y ) and h c , h c (summing over Z). Our proof will depend on the following two observations. First, the condition that since that sum is precisely the coefficient r a,b,c .
Second, the condition (4.8) is invariant under the change of basis that replaces f a 2 by α f a 1 + f a 2 as long as u a 1 ≥ u a 2 . For if ∑ x,y,z F(x, y, z) α f a 1 (x) + f a 2 (x) g b (y)h c (z) = 0, then either ∑ x,y,z F(x, y, z) f a 1 (x)g b (y)h c (z) = 0 or ∑ x,y,z F(x, y, z) f a 2 (x)g b (y)h c (z) = 0, so either u a 1 + v b + w c or u a 2 + v b + w c is less than u avg + v avg + w avg . Thus in either case u a 2 + v b + w c is less than u avg + v avg + w avg as it is at most u a 1 + v b + w c . The same is true for the g b and h c .
We can now proceed to the proof. For the sake of contradiction, assume that F is a diagonal and unstable. Without loss of generality, assume that X = Y = Z = [n] and that the support of F is View the functions f a : [n] → F as n-dimensional vectors. Using changes of variables of the aforementioned form, we may perform Gaussian elimination until no two f a have their first nonzero entries in the same position. In that case, each element of [n] is the position of the first nonzero entry of some f a , so we may index the f a by these positions. Do the same for the g b , but for the h c , perform Gaussian elimination so that the last nonzero entries have distinct positions, and index the h c by these positions.
Because F is diagonal, and F(x, x, x) = 0. Now for each index i ∈ [n], since i is the position of the first nonzero entry of f i and g i and the last nonzero entry of h i , which implies u i + v i + w i < u avg + v avg + w avg . Summing over i, we have which is a contradiction, so F is not unstable.  Proof. When we restrict a function F : X × Y × Z → F to subsets X 0 ,Y 0 , Z 0 , we can also restrict any decomposition as in (4.1), showing that slice-rank(F| X 0 ×Y 0 ×Z 0 ) ≤ slice-rank(F).
A set M ⊆ S × T × U on subsets S, T,U ⊆ H is a tricolored sum-free set if and only if the restriction D H | S×T ×U is a diagonal; if it is, Lemma 4.7 implies that This means we can bound the size of tricolored sum-free sets in an abelian group H by bounding the slice rank of D H from above. Remark 4.9. With the notion of slice rank in hand, it is natural to wonder whether upper bounds on slice rank could be applied directly to matrix multiplication to get upper bounds on its tensor rank. Indeed, since tensor-rank( n, n, n ) ≤ n 2 · slice-rank( n, n, n ) by (4.3), upper bounds of O(n δ ) on the slice rank of n × n matrix multiplication would imply ω ≤ 2 + δ . However, no nontrivial upper bounds on ω can be achieved this way: it is known that the matrix multiplication tensor is not unstable [3, Theorem 5.2], so Theorem 4.6 implies that slice-rank( n, n, n ) = n 2 . (However, the topic of the slice rank of matrix multiplication itself is orthogonal to the main results of this paper.)

Upper bounds on slice rank
The following result gives us upper bounds on slice rank. Theorem 4.10. If F is unstable, then for any n ≥ 1, slice-rank(F ⊗n ) ≤ (|X| n + |Y | n + |Z| n )e −2n instability(F) 2 .
We remark that in the most common case when |X| = |Y | = |Z|, this bound is always nontrivial for sufficiently large n. However if the sets have unequal sizes, this only improves on the trivial bound slice-rank(F ⊗n ) ≤ min(|X| n , |Y | n , |Z| n ) if instability(F) is sufficiently large.
Given such an ε, we can choose functions f a , g b , h c , weights u a , v b , w c , and coefficients r a,b,c as in Definition 4.4 indexed by a ∈ A, b ∈ B, and c ∈ C such that with R as in (4.5). Assume without loss of generality that u min = v min = w min = 0, and let us define bounds u bd = u avg − εu max , and similarly for v bd and w bd , so that For a ∈ A n and x ∈ X n , define f a (x) = ∏ n i=1 f a i (x i ) and u a = u a 1 + · · · + u a n ∈ R; similarly define where the sum is over triples (a, b, c) of tuples a ∈ A n , b ∈ B n , c ∈ C n satisfying the condition u a i + v b i + w c i ≤ R = u bd + v bd + w bd for each i = 1, . . . , n. This condition implies u a + v b + w c ≤ nR = n(u bd + v bd + w bd ), so each such triple must satisfy at least one of the inequalities u a ≤ nu bd , v b ≤ nv bd , or w c ≤ nw bd . Therefore we can collect terms above to write This decomposition exhibits a bound on the slice rank of F ⊗n , so all we need to do is estimate how many terms appear in these sums. The number of tuples a ∈ A n satisfying u a ≤ nu bd = n(u avg − εu max ) is a classical large deviation estimate: u a is the sum of i.i.d. variables u a i taking values in [0, u max ], and we are asking how often this deviates on one side from the mean nu avg by at least εnu max . Even without knowing anything about the distribution of u a , Hoeffding's inequality [13,Theorem 1] states that this proportion is bounded by e −2nε 2 , so the number of such tuples a ∈ A n appearing in the first sum is at most |A n | e −2nε 2 = |X| n e −2nε 2 . The same bound applies to the other two sums, so we conclude that slice-rank(F ⊗n ) ≤ (|X| n + |Y | n + |Z| n )e −2nε 2 .
The bound in Theorem 4.10 can be improved if we have more information about the distribution of the weights, by sharpening the large deviation estimate in the last paragraph of the proof. (If F is not symmetric between x, y, and z, it may also be helpful to use different cutoffs for the different variables.) One case is very common, so we single it out: we define the triangle rank triangle-rank(F) to be the smallest k for which there exist functions f a , g b , h c for a, b, c ∈ {0, . . . , k − 1} such that (4.9) We remark that in contrast with the definition of instability, here we do not require that the functions f i are linearly independent or form a basis. However, it turns out we can assume this without affecting the triangle rank by the same argument as Remark 4.5(3). We will not use the following observation directly, but it provides a representative example and was central to Ellenberg-Gijswijt's results in [11].
Observation 4.11. For any function P : F p → F p with p prime, the function F(x, y, z) = P(x + y + z) has triangle rank at most p.
Proof. We can represent P as a polynomial of degree less than p. Expanding P(x + y + z) and collecting terms expresses F(x, y, z) = P(x + y + z) as a linear combination of monomials x a y b z c . Since deg P < p, each monomial that occurs satisfies a + b + c < p.
To analyze the bounds on slice rank resulting from triangle rank, we will need to bound the proportion of tuples a ∈ {0, . . . , m} n with ∑ i a i ≤ 1 3 mn. The bounds used in the proof of Theorem 4.10 would bound this proportion by e −n/18 , but in this case the rate function can be analyzed more carefully, leading to the following definition. For m > 0 and α ∈ (0, 1 2 ), let Proof of Proposition 4.12. After dividing everything by m, the first claim in the proposition states that if X 1 , . . . , X n are independent copies of a random variable X uniformly distributed on the m + 1 values { 0 m , 1 m , . . . , m m }, then Pr ∑ i X i ≤ αn ≤ e −I(m,α)n . For any θ < 0, applying the order-reversing transformation x → e θ x shows that Pr ∑ i X i ≤ αn = Pr e θ ∑ i X i ≥ e αnθ . By Markov's inequality, the latter is at most e −αnθ E e θ ∑ i X i = e −αθ E e θ X n . Setting q = e θ /m , we have E e θ X = (1 + q + · · · + q m )/(m + 1) = 1−q m+1 1−q · 1 m+1 , and this concludes the proof of the first claim. For the remaining claims, set G(t, x) = 1−e t e tx (1+1/x)(1−e tx ) , and set G(0, x) = 1 so that G is continuous at 0. By definition I(m, α) is the supremum over θ < 0 of γ(θ , m) = αθ − log G(θ , 1 m ). Note that γ(0, m) = 0; moreover, as t → 0, 1 2 ). This guarantees that there exists some θ < 0 where γ(θ , m) is positive, proving the second claim.
Furthermore, the supremum of γ(θ , m) is attained at some θ < 0, because γ(θ , m) tends to 0 as θ → 0 and −∞ as θ → −∞, while the supremum is positive. This means that to prove that I(m, α) is increasing in m, it suffices to prove for each fixed θ < 0 that γ(θ , m) is increasing in m. Equivalently, we must prove that G(t, x) is increasing in x when t < 0 and x > 0; we do this by proving that ∂ G ∂ x > 0 in this region. Set h(x) = (1 − e tx )/x. Note that h(x+1) h(x) = (1−e t(x+1) )/(x+1) (1−e tx ) 2 , and so the proposition follows from the inequality (1 − e tx ) 2 > t 2 x 2 e tx , which we can prove as follows.
Finally, as m → ∞ we have Since − log G(θ , x) is monotonically decreasing in x, αθ − log G θ , 1 m is monotonically increasing in m, so the limit in m is the same as the supremum in m and hence commutes with the supremum in θ .

Tricolored sum-free sets in abelian groups of bounded exponent
In this section, we prove Theorems A and A , by way of the following sharper bound.
Theorem 4.14. If H is an abelian group H ∼ = (Z/qZ) n ×G where q is a prime power, then every tricolored sum-free set in H has cardinality at most 3 · |H| · J(q) n .
The upper bound in Theorem 4.14 is known to be sharp up to a subexponential factor by [14, Theorem 2], conditional on a conjecture, namely [14,Conjecture 3], that was later proved independently in [18] and [19].
The proofs of these theorems depend on the following proposition.
Proposition 4.15. If q = p r is a prime power, the triangle rank of D Z/qZ over F p is at most q.
Proof. The triangle rank is invariant under any permutation of the sets X,Y, Z, so it suffices to bound the triangle rank of the function D Z/qZ (x, y, z + 1) = 1 ∈ F p if x + y + z = q − 1 ∈ Z/qZ, and 0 ∈ F p if x + y + z = q − 1 ∈ Z/qZ.
As a consequence of Lucas' theorem, for any 0 ≤ k < q and m, m ∈ N, In other words, m → m k descends to a well-defined function Z/qZ → F p . We claim that z c ∈ F p for x, y, z ∈ Z/qZ. But for any w ∈ Z/qZ, choosing a lift in {0, . . . , q − 1} shows that w q − 1 = 1 ∈ F p if w = q − 1 ∈ Z/qZ, and 0 ∈ F p if w = q − 1 ∈ Z/qZ.
This decomposition of D Z/qZ (x, y, z + 1) shows that its triangle rank is at most q.
Proof of Theorem 4.14. Let q = p r , and consider the tensor D H over F p . The decomposition H ∼ = (Z/qZ) n × G induces a decomposition D H = D (Z/qZ) n ⊗ D G . Proposition 4.2 thus implies slice-rank(D H ) ≤ slice-rank(D (Z/qZ) n ) · |G| .
The bound on tricolored sum-free sets in H then follows from Proposition 4.8.
Proof of Theorem A . If q is a prime power, Theorem 4.14 states that the size of any tricolored sum-free set in H ∼ = (Z/qZ) n is at most 3 · |H| · J(q) n . To get a uniform bound we recall that J(q) is decreasing by Proposition 4.12, so the worst case for this bound occurs when the prime power q is 2. Therefore if we set then J(q) ≤ J(2) = e −δ = q − δ log q . The size of any tricolored sum-free set in H ∼ = (Z/qZ) n is thus at most 3 · |H| · J(q) n ≤ 3 · |H| · J(2) n = 3q n e −δ n = 3 · |H| 1− δ log q .
Proof of Theorem A. Let q 1 , . . . , q r denote all the prime powers less than or equal to m. By the Chinese remainder theorem, we can write H ∼ = ∏ r i=1 (Z/q i Z) n i . Let be the index for which n is largest. Writing ∑ i n i = N, we know that n ≥ N/r. Since ∑ i n i log q i = log |H| and q i ≤ m we have N ≥ log|H| log m . Next, we claim that r < 2m log m for all m > 1. By [23, (3.2) and (3.36)] we have that r < m log m + 3m 2(log m) 2 + . This bound is at most 2m log m as long as m ≥ 242. For the remaining 1 < m < 242, one can simply check directly that the number of prime powers less than m is less than 2m log m . Combining these bounds on N and r shows that n ≥ N r ≥ log|H| 2m . By Theorem 4.14, this implies the size of any tricolored sum-free set in H ∼ = (Z/q Z) n × G is at most 3 · |H| · J(q ) n ≤ 3 · |H| · J(q ) log|H|/(2m) = 3 · |H| 1+log J(q )/(2m) .