Composition theorem for majority

In this section using Theorem 14 we give a composition theorem forF₂-sketching of the composedM aj3function. Unlike in the deterministic case for which the composition theorem is easy to show (see Lemma 53) in the randomized case composition results require more work.

IDefinition 27 (Composition). For f:Fⁿ₂ →F₂ and g: F^m₂ → F₂ their compositionf ◦ g:F^mn₂ →F2 is defined as:

(f◦g)(x) =f(g(x₁, . . . , x_m), g(x_m+1, . . . , x_2m), . . . , g(x_m(n−1)+1, . . . , x_mn)).

Consider the recursive majority functionM aj₃^◦k ≡M aj3◦M aj3◦ · · · ◦M aj3 where the composition is takenktimes.

ITheorem 28. For any d≤n,k= log₃nandξ > ^4d_n it holds that dimξ M aj₃^◦k

> d.

First, we show a slighthly stronger result for standard subspaces and then extend this result to arbitrary subspaces with a loss of a constant factor. Fix any setS⊆[n] of variables. We associate this set with a collection of standard unit vectors corresponding to these variables.

Hence in this notation∅ corresponds to the all-zero vector.

ILemma 29. For any standard subspace whose basis consists of singletons from the set S⊆[n]it holds that:

X

Z∈span(S)

M aj\₃^◦k(Z)²

≤ |S|

n

Proof. The Fourier expansion ofM aj3is given as M aj3(x1, x2, x3) =1

2(x1+x2+x3−x1x2x3).

Fori∈ {1,2,3} letNi={(i−1)n/3 + 1, . . . , in/3}. LetSi=S∩Ni. Letαi be defined as:

αi= X

Z∈span(Si)

M aj\^◦k−1₃ (Z) ²

.

Then we have:

X

Z∈span(S)

M aj\₃^◦k(Z)²

=

3

X

i=1

X

Z∈span(Si)

M aj\₃^◦k(Z)² + X

Z∈span(S)−∪³_i=1span(S_i)

M aj\₃^◦k(Z)² .

For each Si we have X

Z∈span(Si)

M aj\₃^◦k(Z)²

= 1 4

X

Z∈span(Si)

M aj\₃^◦k−1(Z) 2

= αi

4. Moreover, for eachZ ∈span(S)− ∪³_i=1span(S_i) we have:

M aj\₃^◦k(Z) =

(−¹₂M aj\₃^◦k−1(Z₁)M aj\^◦k−1₃ (Z₂)M aj\₃^◦k−1(Z₃) ifZ∈ ׳_i=1(span(S_i)\ ∅)

0 otherwise.

Thus, we have:

X

Z∈(span(S1)\∅)×(span(S2)\∅)×(span(S3)\∅)

M aj\₃^◦k(Z)²

= X

Z∈(span(S1)\∅)×(span(S2)\∅)×(span(S3)\∅)

1 4

M aj\^◦k−1₃ (Z1) ²

·

M aj\₃^◦k−1(Z2) ²

·

M aj\₃^◦k−1(Z₃) 2

=1 4





X

Z∈(span(S1)\∅)

M aj\₃^◦k−1(Z₁) ²



·





X

Z∈(span(S2)\∅)

M aj\^◦k−1₃ (Z₂) ²



·





X

Z∈(span(S3)\∅)

M aj\₃^◦k−1(Z3) ²





=1

4α1α2α3.

where the last equality holds sinceM aj\^◦k−1₃ (∅) = 0. Putting this together we have:

X

Z∈span(S)

M aj\₃^◦k(Z)²

= 1

4(α1+α2+α3+α1α2α3)

≤1 4

α1+α2+α3+1

3(α1+α2+α3)

=1

3(α1+α2+α3).

Applying this argument recursively to each αi fork−1 times we have:

X

Z∈span(S)

M aj\₃^◦k(Z)²

≤ 1 3^k

3^k

X

i=1

γ_i,

whereγi = 1 ifi∈S and 0 otherwise. Thus,P

Z∈span(S)

M aj\₃^◦k(Z)²

≤^|S|_n . J

To extend the argument to arbitrary linear subspaces we show that any such subspace has less Fourier weight than a collection of three carefully chosen standard subspaces. First we show how to construct such subspaces in Lemma 30.

For a linear subspace L ≤Fⁿ₂ we denote the set of all vectors in L of odd Hamming weight asO(L) and refer to it as theodd set ofL. For two vectorsv1, v2∈Fⁿ₂ we say that v1 dominatesv2if the set of non-zero coordinates ofv1 is a (not necessarily proper) subset of the set of non-zero coordinates ofv2. For two sets of vectorsS1, S2⊆Fⁿ₂ we say thatS1

dominates S₂(denoted as S₁≺S₂) if there is a matching M betweenS₁ andS₂ of size|S2| such that for each (v1∈S1, v2∈S2)∈M the vectorv1dominates v2.

ILemma 30(Standard subspace domination lemma). For any linear subspaceL≤Fⁿ₂ of dimension dthere exist three standard linear subspaces S1, S2, S3≤Fⁿ₂ such that:

O(L)≺ O(S₁)∪ O(S₂)∪ O(S₃),

anddim(S₁) =d−1,dim(S₂) =d,dim(S₃) = 2d.

Proof. Let A∈ F^d×n₂ be the matrix with rows corresponding to the basis in L. We will assume thatAis normalized in a way described below. First, we apply Gaussian elimination to ensure thatA= (I, M) where I is a d×didentity matrix. If all rows of A have even Hamming weight then the lemma holds trivially sinceO(L) =∅. By reordering rows and columns of A we can always assume that for some k ≥1 the firstk rows of A have odd Hamming weight and the lastd−k have even Hamming weight. Finally, we add the first column to each of the lastd−krows, which makes all rows have odd Hamming weight. This results inAof the following form:

A=

























1 0· · ·0 0· · ·0 a

0^...

I _k−1 0 M ₁

0

1^...

0 I _d−k M ₂

1

























We use the following notation for submatrices: A[i1, j1;i2, j2] refers to the submatrix of A with rows betweeni₁ andj₁ and columns betweeni₂ andj₂ inclusive. We denote to the first row byv, the submatrixA[2, k; 1, n] asAand the submatrixA[k+ 1, d; 1, n] asB. Each x∈ O(L) can be represented asP

i∈SAi where the setS is of odd size and the sum is over Fⁿ₂. We consider the following three cases corresponding to different types of the setS.

Case 1. S⊆rows(A)∪rows(B). This corresponds to all odd size linear combinations of the rows ofAthat don’t include the first row. Clearly, the set of such vectors is dominated byO(S1) whereS1 is the standard subspace corresponding to the span of the rows of the submatrixA[2, d; 2, d].

Case 2. Scontains the first row,|S∩rows(A)|and|S∩rows(B)|are even. All such linear combinations have their first coordinate equal 1. Hence, they are dominated by a standard subspace corresponding to span of the rows thed×didentity matrix, which we refer to as S2.

Case 3. S contains the first row, |S∩rows(A)|and|S∩rows(B)|are odd. All such linear combinations have their first coordinate equal 0. This implies that the Hamming weight of the firstdcoordinates of such linear combinations is even and hence the other coordinates cannot be all equal to 0. Consider the submatrixM =A[1, d;d+ 1, n] corresponding to the lastn−dcolumns of A. Since the rank of this matrix is at mostdby running Gaussian elimination onM we can construct a matrixM⁰ containing as rows the basis for the row space ofM of the following form:

M⁰=

It M1

0 0

wheret =rank(M). This implies that any non-trivial linear combination of the rows of M contains 1 in one of the first t coordinates. We can reorder the columns ofA in such a way that theset coordinates have indices fromd+ 1 tod+t. Note that now the set of vectors spanned by the rows of the (d+t)×(d+t) identity matrixI_d+t dominates the set of linear combinations we are interested in. Indeed, each such linear combination has even Hamming weight in the firstdcoordinates and has at least one coordinate equal to 1 in the set{d+ 1, . . . , d+t}. This gives a vector of odd Hamming weight that dominates such linear combination. Since this mapping is injective we have a matching. We denote the standard linear subspace constructed this way byS₃ and clearlydim(S₃)≤2d. J The following proposition shows that the spectrum of theM aj₃^◦k is monotone decreasing under inclusion if restricted to odd size sets only:

IProposition 31. For any two setsZ1⊆Z2 of odd size it holds that:

M aj\^◦k₃ (Z1) ≥

M aj\^◦k₃ (Z2) .

Proof. The proof is by induction onk. Consider the Fourier expansion ofM aj3(x1, x2, x3) =

1

2(x1+x2+x3−x1x2x3). The casek= 1 holds since all Fourier coefficients have absolute value 1/2. SinceM aj₃^◦k =M aj₃◦(M aj₃^◦k−1) all Fourier coefficients ofM aj₃^◦k result from substituting either a linear or a cubic term in the Fourier expansion by the multilinear expansions ofM aj^◦k−1₃ . This leads to four cases.

Case 1. Z1and Z2 both arise from linear terms. In this case ifZ1 andZ2 aren’t disjoint then they arise from the same linear term and thus satisfy the statement by the inductive hypothesis.

Case 2. IfZ1 arises from a cubic term and Z2 from the linear term then it can’t be the case thatZ1⊆Z2 sinceZ2contains some variables not present inZ1.

Case 3. IfZ₁andZ₂both arise from the cubic term then we have (Z₁∩N_i)⊆(Z₂∩N_i) for eachi. By the inductive hypothesis we then have

M aj\^◦k−1₃ (Z₁∩N_i)

≥

M aj\₃^◦k−1(Z₂∩N_i) . Since for j = 1,2 we haveM aj\₃^◦k(Zj) = −¹₂Q

iM aj\₃^◦k−1(Zj∩Ni) the desired inequality follows.

Case 4. IfZ1arises from the linear term and Z2 from the cubic term then w.l.o.g. assume thatZ₁ arises from the x₁ term. Note thatZ₁⊆(Z₂∩N₁) sinceZ₁∩(N₂∪N₃) =∅. By the inductive hypothesis applied toZ1 andZ2∩N1 the desired inequality holds. J

We can now complete the proof of Theorem 28

Proof of Theorem 28. By combining Proposition 31 and Lemma 29 we have that any setT of vectors that is dominated byO(S) for some standard subspaceSsatisfiesP

S∈TM aj\^◦k₃ (S)²≤

dim(S)

n . By the standard subspace domination lemma (Lemma 30) any subspaceL≤Fⁿ₂ of dimensiondhasO(L) dominated by a union of three standard subspaces of dimension 2d,d andd−1 respectively. Thus, we haveP

S∈O(L)M aj\₃^◦k(S)²≤ ^2d_n +^d_n+^d−1_n ≤ ^4d_n. J We have the following corollary of Theorem 28 that proves Theorem 5.

ICorollary 32. For any ∈[0,¹₂],ξ >4² andk= log₃nit holds that:

D^lin,U₁₋^√_ξ

2

(M aj₃^◦k)> ²n, D^→,U1−ξ 6

(M aj^◦k₃ ⁺)>²n 6 .

Proof. Fixd=²n. For this choice ofdTheorem 28 implies that forξ >4² it holds tha t dimξ M aj₃^◦k

> d. The first part follows from Part 2 of Theorem 14. The second part is by

Part 3 of Theorem 14. J

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