Motivation Introduction Bounds Applications Discussion Slice rank of tensors and its applications Wenjie Fang, LIP, ENS de Lyon Work of Terence Tao and William Sawin One Day Meeting in Discrete Structures April 14 2017, ENS de Lyon
Motivation Introduction Bounds Applications Discussion Two problems in extremal combinatorics Sunflower-free set problem Let U be a finite set with n = | U | . Three subsets A, B, C of U form a sunflower if A ∩ B = B ∩ C = C ∩ A . What is the size of the largest subset family of U that has no sunflower? Cap set problem Three vectors a, b, c ∈ F n 3 form a progression of length 3 if a + b + c = 0 . What is the cardinal of the largest cap set (set of vectors avoiding such progressions) in F n 3 ?
Motivation Introduction Bounds Applications Discussion Naslund–Sawin bound on sunflower-free set Theorem (Naslund–Sawin 2016) Let F be a sunflower-free family of { 1 , 2 , . . . , n } . Then � n � 3 · 2 − 2 / 3 � n � � e o ( n ) . |F| ≤ 3( n + 1) = k k ≤ n/ 3 Idea: A notion called slice rank , first used implicitly by Croot–Lev–Pach (2016) on progression-free sets in Z n 4 . First result that breaks 2 n e o ( n ) !
Motivation Introduction Bounds Applications Discussion A polynomial model for the sunflower-free set Let U = { 1 , 2 , . . . , n } , and v 1 , . . . , v n be the canonical base of F n 3 . For A ⊆ U , we define v A = � i ∈ A v i . Given a polynomial P ( X 1 , . . . , X n ) and a vector u = � n i =1 x i v i ∈ F n 3 , we define P ( u ) = P ( x 1 , . . . , x n ) . Proposition Let A, B, C be three sets without one set being the proper subset of another. The sets A, B, C form a sunflower or A = B = C iff P ( v A , v B , v C ) = 1 , with n � P ( X 1 , . . . , X n , Y 1 , . . . , Y n , Z 1 , . . . , Z n ) = (2 − ( X i + Y i + Z i )) . i =1 Proof: Since no set is a proper subset of the other, w.l.o.g., we only need to avoid i ∈ ( A ∩ B ) \ C , which means x i = y i = 1 , z i = 0 , which implies x i + y i + z i − 2 = 0 .
Motivation Introduction Bounds Applications Discussion Polynomial as tensor A polynomial P ( X 1 , . . . , X n , Y 1 , . . . , Y n , Z 1 , . . . , Z n ) in F 3 ⇔ A tensor T in F n 3 ⊗ F n 3 ⊗ F n 3 with T ( u, v, w ) = P ( u, v, w ) Let F be a sunflower-free family in U , and T F the sub-tensor of T with coordinates restricted to all v A with A ∈ F . Proposition T F is a diagonal tensor, that is, T F ( u, v, w ) = 1 iff u = v = w . Idea: Upper bound on “big diagonals” ⇒ upper bound on sunflower-free set. We want some notion of rank to capture the size of “big diagonals”.
Motivation Introduction Bounds Applications Discussion Slice rank of a function Let A be a finite set. A function S : A ⊗ A ⊗ A → F is a slice if it has one of the following forms: S ( u, v, w ) = f ( u ) g ( v, w ) or f ( v ) g ( u, w ) or f ( w ) g ( u, v ) . The slice rank of a function F : A ⊗ A ⊗ A → F , denoted by sr( F ) , is the minimum number of slices needed to sum to F . Property : Let T A : A ⊗ A ⊗ A → F , and T B its restriction on B ⊗ B ⊗ B with B ⊆ A . Then sr( T B ) ≤ sr( T A ) . Lemma (Special case of Tao (2016)) The slice rank of the function F ( u, v, w ) = � a ∈ A c a δ a ( u ) δ a ( v ) δ a ( w ) is the number of non-zero coefficients c a ∈ F . Proof: delayed.
Motivation Introduction Bounds Applications Discussion Slice rank of the sunflower polynomial n � P ( X, Y , Z ) = (2 − ( X i + Y i + Z i )) . i =1 For a monomial X a 1 n Y a 1 n Z a 1 1 · · · X a n · · · Y a n 1 · · · Z a n in P ( X, Y , Z ) , we 1 n have � n i =1 a i + � n i =1 b i + � n i =1 c i ≤ n . One of the total powers of X , Y and Z must be ≤ n/ 3 . � X a 1 1 · · · X a n P ( X, Y , Z ) = n P a 1 ,...,a n ( Y , Z ) + · · · . a 1 + ··· + a n ≤ n/ 3 Thus we have (since all a i ≤ 1 ) � n � � sr( P ) ≤ 3 . k k ≤ n/ 3
Motivation Introduction Bounds Applications Discussion Proof of upper bound Let F be a sunflower-free family, with F = � ℓ ≥ 0 F ℓ the partition by number of elements. Sets in F ℓ are never proper subset of each other. Let A ℓ = { v A | A ∈ F ℓ } . The function P is diagonal on A ℓ , thus |F ℓ | = sr A ℓ ( P ) ≤ sr( P ) . We thus have � n � 3 · 2 − 2 / 3 � n � � e o ( n ) . |F| ≤ 3( n + 1) = k k ≤ n/ 3
Motivation Introduction Bounds Applications Discussion New bound on cap set problem Polynomial: n � (1 − ( X i + Y i + Z i ) 2 ) . P ( X, Y , Z ) = i =1 Theorem (Ellenberg–Gijswijt (2016)) The size of a cap set in F n 3 is o (2 . 756 n ) . General result for any finite field. Kleinberg–Sawin–Speyer gave a concrete construction on a lower bound that matches within a subexponential factor.
Motivation Introduction Bounds Applications Discussion A general strategy Given a problem concerning avoiding some structure. Construct a polynomial P whose zeros are exactly on “everything 1 equal” or “things forming the structure”, which is a product of the same polynomial on different sets of variables in many cases; The function P restricted to an avoiding family F will then be 2 diagonal; Compute the slice rank of P , which is an upper bound of the size of 3 F ; Hopefully this bound will be a breakthrough, or not. 4 Can we know the power of the method?
Motivation Introduction Bounds Applications Discussion Slice rank for tensors We consider tensors in V 1 ⊗ V 2 ⊗ · · · ⊗ V k . We define in the natural way the j th tensor product � � ⊗ j : V j ⊗ V i → V i . 1 ≤ i ≤ k,i � = j 1 ≤ i ≤ k A slice is any element of the form v j ⊗ j v � = j for any j . The slice rank of a tensor T is the minimum number of slices that sum to T . Example : For V 1 (resp. V 2 , V 3 ) the space of polynomials of X i (resp. Y i , Z i ) in F , the slice rank of tensors in V 1 ⊗ V 2 ⊗ V 3 is the slice rank of polynomials. Property : Let T be a tensor in V 1 ⊗ V 2 ⊗ · · · ⊗ V k and T ′ a sub-tensor of T . Then sr( T ′ ) ≤ sr( T ) .
Motivation Introduction Bounds Applications Discussion Slice rank of a polynomial and its value tensor Let P be a polynomial in a finite field F with k sets of n variables. P is a tensor in V 1 ⊗ · · · ⊗ V k , where V i is spanned by monomials in the i th set of variable. Let T P be the value tensor of P in ( F n ) ⊗ k defined by � T P = P ( v 1 , . . . , v k ) v 1 ⊗ · · · ⊗ v k . v 1 ,...,v k ∈ F n Proposition We have sr( P ) = sr( T P ) . Proof: Equivalence on slices.
Motivation Introduction Bounds Applications Discussion Slice rank and diagonal We now consider tensors of the form V ⊗ k . Let S be a basis of V . Lemma (Special case of Tao (2016)) a ∈ S c a a ⊗ k , denoted by sr( F ) , is the The slice rank of the tensor T = � number of non-zero coefficients c a ∈ F . Proof: Again delayed. For S ⊆ V k structures to avoid ( e.g. sunflowers), suppose we have a polynomial P in F with non-zero values only on u 1 = · · · = u k or S . An avoiding family F ⊆ V gives a sub-tensor T P | F ⊗ k that is a diagonal. We thus have |F| = sr( T P | F ⊗ k ) ≤ sr( T P ) = sr( P ) . Upper bound on sr( P ) ⇒ upper bound on F .
Motivation Introduction Bounds Applications Discussion Slice rank (dual version) Let T be a tensor in V = V 1 ⊗ V 2 ⊗ · · · ⊗ V k . Let W i be the dual space of V i , with the canonical pairing �· , ·� i . Let W = W 1 ⊗ · · · ⊗ W k , and we define the pairing k � � w 1 ⊗ · · · ⊗ w k , v 1 ⊗ · · · ⊗ v k � = � w i , v i � i . i =1 Proposition We have sr( T ) ≤ r iff there are sub-spaces W T for all i such that the i co-dimensions of W T for all i sum to r , and that �· , v � is zero on i � k i =1 W T i . Proof: There must be a component that annihilates the pairing.
Motivation Introduction Bounds Applications Discussion Projections and upper bound We fix a basis S i for each V i . We define π i ( s 1 ⊗ · · · ⊗ s k ) = s i for all v i in S i . Proposition Let T be a tensor in V 1 ⊗ · · · ⊗ V k , and Γ its support w.r.t. ( S i ) 1 ≤ i ≤ k . We have k � sr( T ) ≤ min | π i (Γ i ) | . Γ=Γ 1 ∪···∪ Γ k i =1 Proof: Decompose by the vector obtained after projection: k � � T = c ∗ s 1 ⊗ · · · ⊗ s k i =1 ( s 1 ⊗···⊗ s k ) ∈ Γ i k � � = c ∗ s i ⊗ i v s i , � = i . i =1 s i ∈ π i (Γ i ) Each summand is a slice.
Motivation Introduction Bounds Applications Discussion Lower bound We suppose that, for each S i , we have a total order ≤ i . They induce a partial order on vectors s 1 ⊗ · · · ⊗ s k for s i ∈ S i . Proposition Let T be a tensor in V 1 ⊗ · · · ⊗ V k , Γ its support w.r.t. ( S i ) 1 ≤ i ≤ k , and Γ ′ the set of maximal elements in Γ . We have k � | π i (Γ ′ sr( T ) ≥ min i ) | . Γ ′ =Γ ′ 1 ∪···∪ Γ ′ k i =1 Remark : sr( T ) does not depend on basis. k of Γ ′ such that We only need to show that there is a covering Γ ′ 1 , . . . , Γ ′ sr( T ) ≥ � k i =1 | π i (Γ ′ i ) | .
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