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bias(ϕ)= Ex∈A[k] χ(ϕ(x)) where χ : Fq → T is any standardized character such as χ(x)= e((tr x)/p), where tr : Fq → Fp is the absolute trace. Bias in this context was introduced by Gowers and Wolf in [GW] and studied extensively by several authors. The term analytic rank is used for the quantity −1 AR(ϕ) = logq(bias(ϕ) ). The partition rank PR(ϕ) of ϕ is the minimal number r such that ϕ is the sum of r multilinear maps of the form

ϕ1(xI )ϕ2(xIc ), where I ⊆ [k] and 0 < |I| < k. Multilinear maps of bounded partition rank are uniformly biased: in fact, AR(ϕ) ≤ PR(ϕ). This is one of several nice lemmas appearing in a paper by Lovett [L]. The main theorem in this field is a converse: uniformly biased multilinear maps have bounded partition rank, or in other words partition rank is bounded in terms of analytic rank.

Theorem 1.1. Suppose A1,...,Ak are finite Fq-vector spaces and ϕ : A[k] → Fq is multilinear over Fq. If bias(ϕ) ≥ ǫ > 0 then ϕ has partition rank Oǫ,k,q(1). In other words, PR(ϕ) ≤ F (k, AR(ϕ), q) for some F . This result was essentially proved by Green and Tao [GT]. Subsequent research has focused on quantitative aspects (which were initially poor). The dependence on q was removed by Bhowmick and Lovett in [BL] (so PR(ϕ) ≤ F (k, AR(ϕ))). In breakthrough work Janzer [J] and Mili´cevi´c[M] independently proved that PR(ϕ) ≤ Dk Ck(AR(ϕ) + 1) for constants Ck,Dk. Recently, Cohen and Moshkovitz [CM] k−1 proved that PR(ϕ) ≤ (2 + 1) AR(ϕ) + 1 provided that q > q0(k, AR(ϕ)). In this note we consider biased multilinear maps of arbitrary abelian groups. We will use Theorem 1.1 as a block box (and only the prime field case) and we will deduce an analogous structure theorem for arbitrary abelian groups. To state this structure theorem we need a suitable notion of partition rank. For q a prime power, 2 let mq : (Z/qZ) → T be the defined by m(x, y) = xy/q mod 1. We say that ϕ : A[k] → T factors through mq if it has the form

ϕ(x)= mq(ϕ1(xI ), ϕ2(xIc )) for some I ⊆ [k] with 0 < |I| < k. Here ϕ1 : AI → Z/qZ and ϕ2 : AIc → Z/qZ must both be multilinear.

Theorem 1.2. Suppose A1,...,Ak are finite abelian groups and ϕ : A[k] → T is multilinear. If bias(ϕ) ≥ ǫ> 0 then ϕ is the sum of Oǫ,k(1) multilinear maps each of which factors through mq for some prime power q ≤ Oǫ,k(1). It is easy to see that Theorem 1.1 and Theorem 1.2 agree in the case of vector spaces over a prime finite field, but neither result is more general than the other. (A common generalization would consider R-modules for some commutative R; Theorem 1.1 would be the case R = Fq and Theorem 1.2 would be the case R = Z.) The following corollary follows from the connection mentioned already between bias and multilinear kernels. It states that a multilinear function F with P(F = 0) BIASEDMULTILINEARMAPSOFABELIANGROUPS 3 bounded away from zero must be the sum of boundedly many functions each of which “crushes” a nontrivial subset of the variables before mapping to the range. Here cod(g) denotes the codomain of the function g.

Corollary 1.3. Suppose A1,...,Ak−1,B are finite abelian groups and F : A[k−1] → B is a multilinear map such that P(F = 0) ≥ ǫ> 0. Then there is an expression

F (x)= GI (gI (xI ), x[k−1]\I ), ∅6=IX⊆[k−1] where for each I the functions gI and GI are multilinear maps

gI : AI → cod(gI ),

GI : cod(gI ) × A[k−1]\I → B, and | cod(gI )|≤ Oǫ,k(1). In the last section we give an application of Theorem 1.2 to the set all possible biases. Let Φk be the set of all multilinear maps ϕ : A1 ×···× Ak → T (for any finite abelian groups A1,...,Ak) and let

Bk = {bias(ϕ): ϕ ∈ Φk}.

We show that Bk is a small subset of [0, 1] in various senses; for example, all its limit points are algebraic. Elsewhere we will give an application to probabilistically nilpotent finite groups.

2. Basic properties of bias Several nice lemmas about bias over vector spaces were proved by Lovett in [L]. We need analogues in the category of abelian groups. In this section we will sometimes consider functions ϕ : A[k] → T which are not multilinear, but we still define bias(ϕ) by (1). If ϕ : A[k] → T is a map we write c ϕxI : AI → T for the map obtained by restriction:

c ϕxI (xI )= ϕ(x). Obviously,

(2) bias(ϕ)= ExI bias(ϕxI ).

c In particular, consider (2) in the case I = {i} . If ϕ(x) is linear in xi then bias(ϕxI ) is 1 or 0 according to whether ϕxI ≡ 0, so

(3) bias(ϕ)= PxI (ϕxI ≡ 0). This shows that, while in general bias(ϕ) is complex-valued, bias(ϕ) ∈ [0, 1] pro- vided that ϕ is linear in at least one of its arguments.

Lemma 2.1. Assume ϕ : A[k] → T is multilinear. For each i ∈ [k],

bias(ϕ) ≥ 1 − (1 − 1/|Aj|). Yj6=i If ϕ is nontrivial then

bias(ϕ) ≤ 1 − (1 − 1/pj), Yj6=i where pj is the smallest prime divisor of |Aj |. 4 SEAN EBERHARD

c Proof. Let I = {i} . If xj = 0 for any j ∈ I then ϕxI ≡ 0, so from (3) we have

1 − bias(ϕ)= P(ϕxI 6≡ 0) ≤ P(xj 6=0)= (1 − 1/|Aj |). jY∈I Yj6=i The second inequality is proved by induction. The case k = 1 is clear, because if ϕ is linear and nontrivial then bias(ϕ) = 0, and the right-hand side is also 0. Let k> 1. By (2) with I = {j} we have

1 − bias(ϕ)= Exj (1 − bias(ϕxj )).

Let Bj = {xj ∈ Aj : ϕxj ≡ 0}. By induction if xj ∈/ Bj then

′ 1 − bias(ϕxj ) ≥ (1 − 1/pj ). ′ j Y6=i,j Thus 1 − bias(ϕ) ≥ (1 − |Bj |/|Aj |) (1 − 1/pj′ ). ′ j Y6=i,j But Bj is a subgroup of Aj , and proper since ϕ is nontrivial, so |Bj |/|Aj | ≤ 1/pj. This completes the induction. 

Lemma 2.2. For I ⊆ [k], let ϕI : AI → T be |I|-linear. Let ϕ : A[k] → T be the function

ϕ(x)= ϕI (xI ). IX⊆[k] Suppose J ⊆ [k] is any subset such that ϕI =0 for I ) J. Then

| bias(ϕ)|≤ bias(ϕJ ). Proof. Let i ∈ [k]. Write ϕ = ϕi + ϕi′ , where

ϕi = ϕI , I⊆X[k],i∈I ϕi′ = ϕI . I⊆X[k],i/∈I Note that ϕi is linear in xi and ϕi′ is indepenent of xi, so

′ bias(ϕ)= Ex{i}c (e(ϕi )Exi e(ϕi)) Hence by the triangle inequality

| bias(ϕ)|≤ Ex{i}c |Exi e(ϕi)| .

But since ϕi is linear in xi, Exi e(ϕi) ≥ 0, so we may drop the absolute value signs. Hence

| bias(ϕ)|≤ ExI\i Exi e(ϕi(x)) = bias(ϕi). Repeat for every i ∈ J. 

Lemma 2.3. Let ϕ, ψ : A[k] → T be multilinear. Then bias(ϕ + ψ) ≥ bias(ϕ) bias(ψ).

Proof. Let x, y ∈ A[k] be independent. Then

bias(ϕ) bias(ψ)= Ex,ye(ϕ(x)+ ψ(y)) = Ex,ye(ϕ(x)+ ψ(x + y)). We may expand

ψ(x + y)= ψI (xI ,yIc ), IX⊆[k] BIASEDMULTILINEARMAPSOFABELIANGROUPS 5 where ψI : AI × AIc → T is again k-linear. Note that ψ[k] = ψ(x). For each fixed y ∈ A[k], the previous lemma with J = [k] implies that

|Exe(ϕ(x)+ ψ(x + y))|≤ Exe(ϕ(x)+ ψ(x)) = bias(ϕ + ψ). Hence the lemma follows from the triangle inequality. 

Suppose ϕ : A[k] → T is k-linear and ψ : B[l] → T is l-linear. We say ϕ factors through ψ if there is an l-partition

[k]= I1 ∪···∪ Il (I1,...,Il 6= ∅) and for each j ∈ [l] an |Ij |-linear map

ϕj : AIj → Bj such that ϕ factors as

ϕ = ψ(ϕI1 ,...,ϕIl ), i.e., (ϕ ) ∼ Ij j ψ ϕ : A[k] = AIj −−−−→ B[l] −−→ T. j Y Lemma 2.4. Suppose ϕ : A[k] → T and ψ : B[l] → T are multilinear and ϕ factors through ψ. Then bias(ϕ) ≥ bias(ψ).

Proof. Suppose [k]= I1 ∪···∪ Il and

ϕ = ψ(ϕI1 ,...,ϕIl ).

For b ∈ B[l], let

ϕb = ψ(ϕI1 + b1,...,ϕIl + bl). Clearly

Eb bias(ϕb) = bias(ψ).

By Lemma 2.2, for every b ∈ B[l] we have

| bias(ϕb)|≤ bias(ϕ0) = bias(ϕ). Hence bias(ψ) ≤ bias(ϕ). 

Corollary 2.5. Suppose ϕ : A[k] → T and ψ : B[l] → T are multilinear and ϕ factors through ψ. Then bias(ϕ) ≥ 1 − j6=i(1 − 1/|Bj|) for each i ∈ [l]. Proof. Combine the previous lemma withQ Lemma 2.1. 

Recall that mq denotes the map Z/q×Z/qZ → T defined by m(x, y)= xy/q mod 1, where q is a prime power.

Corollary 2.6. Suppose ϕ : A[k] → T is the sum of maps ϕ1,...,ϕr and ϕi factors through mqi . Then −1 −1 bias(ϕ) ≥ q1 ··· qr . Proof. Combine the previous corollary with Lemma 2.3. 

′ Corollary 2.7. Suppose Ai ≤ Ai for each i. Let ϕ : A[k] → T be multilinear and ′ ′ ′ let ϕ be the restriction of ϕ to A[k]. Then bias(ϕ ) ≥ bias(ϕ). Proof. Since ϕ′ factors through ϕ, we may apply Lemma 2.4 to (ϕ′, ϕ). 

The following lemma, which gives a little more information than the second part of Lemma 2.1, illustrates the utility of the last corollary. 6 SEAN EBERHARD

Lemma 2.8. Suppose ϕ : A[k] → T is multilinear and suppose the image of ϕ has an element of order q = pn for some prime p. Then bias(ϕ) ≤ bias(ψ), where ψ : (Z/qZ)k → T is defined by

ψ(x1,...,xk)= x1 ··· xk/q (mod 1). In particular, for k ≥ 2, bias(ϕ) ≤ (n + 1)k−2/q. n ′ Proof. Suppose ϕ(a1,...,ak) has order q = p . Let Ai = haii for each i. Then

bias(ϕ) ≤ bias(ϕ| ′ ). A[k]

Hence without loss of generality Ai is generated by ai for each i. Note that ϕqai =0 k for each i, so ϕ factors through i=1haii/hqaii. Hence we may assume that qai =0 n−1 for each i. Also note that p ai 6= 0 for each i, since Q n−1 n−1 ϕ(...,ai−1,p ai,ai+1,... )= p ϕ(a1,...,ak) 6=0.

Hence each ai has order q, so ϕ is equivalent to the map ψ in the statement of the lemma. For the second statement we may assume ϕ = ψ. Let vp denote the p-adic valuation. From (3),

bias(ϕ)= P(x1 ··· xk−1 = 0)

= P(vp(x1)+ ··· + vp(xk−1) ≥ n)

≤ P(vp(xi) ≥ vi for each i < k) v +···+v =n 1 Xk−1 = p−v1−···−vk−1 v +···+v =n 1 Xk−1 = M/q, where M is the number of solutions to v1 + ··· + vk−1 = n in nonnegative integers. Clearly M ≤ (n + 1)k−2. 

3. The structure theorem In this section we prove Theorem 1.2. Throughout we consider k to be fixed and dependence on k will not be explicit. We start with some simple reductions. Suppose ϕ : A[k] → T is multilinear and bias(ϕ) ≥ ǫ. We may decompose each Ai into its p-primary parts: (p) Ai = Ai . p M Then ϕ decomposes as

ϕ = ϕp, p X where ϕp factors as (p) (p) A[k] → A1 ×···× Ak → T. The bias of ϕ correspondingly decomposes as

bias(ϕ)= bias(ϕp). p Y k−1 k−1 By Lemma 2.1, bias(ϕp) ≤ 1 − (1 − 1/p) ≤ 1 − 1/2 whenever ϕp is nontrivial, so since bias(ϕ) ≥ ǫ it follows that there are at most log ǫ−1/ log(1 − 1/2k−1)−1 primes such that ϕp 6≡ 0. Hence it suffices to consider the p-primary parts separately, so without loss of generality A1,...,Ak are p-groups. BIASEDMULTILINEARMAPSOFABELIANGROUPS 7

Let Ki = {ai ∈ Ai : ϕai ≡ 0}. By replacing Ai with Ai/Ki we may assume

Ki = 0, so ϕ is nondegenerate in the sense that ϕai 6≡ 0 for each nonzero ai ∈ Ai. n Then if Ai has an element of order q = p , the image of ϕ also contains an element of order q, so by Lemma 2.8, k−2 k−2 bias(ϕp) ≤ (n + 1) /q ≤ (log2 q + 1) /q. This implies that q is bounded in terms of ǫ (and k). Hence we may assume that A1,...,Ak have bounded exponent (and in particular p is bounded). We will now deduce Theorem 1.2 from Theorem 1.1 using the following elemen- tary lemma about extending multilinear maps. For any abelian p-group, let A[p] denote the subgroup {x ∈ A : px =0} and let pA denote the subgroup {px : x ∈ A}.

Lemma 3.1. Let A1,...,Ak be finite abelian p-groups and q a power of p.

(1) (domain enlargement) Suppose ϕ : pA1 × A[2,k] → Z/qZ is a multilinear map. Assume ϕ(x)=0 whenever pxi = 0 for any i > 1. Then ϕ factors through pA1 × i∈[2,k] Ai/Ai[p], and there is a multilinear map ψ : A[k] → Z/(pq)Z extending ϕ in the sense that the following diagram commutes: Q

pA1 × i∈[2,k] Ai/Ai[p] Q ϕ . pA1 × A[2,k] Z/qZ

×p ψ A[k] Z/(pq)Z

(2) (range enlargement) Suppose ϕ : A[k] → Z/qZ is a multilinear map such that ϕ(x)=0 whenever pxi = 0 for any i ∈ [k]. Then ϕ factors through i∈[k] Ai/Ai[p], and there is a multilinear map ψ : A[k] → Z/(pq)Z such that ψ mod q = ϕ, as in the following commutative diagram: Q

i∈[k] Ai/Ai[p] Q ϕ . A[k] Z/qZ ψ mod q Z/(pq)Z

(3) (extension of rank-one maps) Suppose ϕ : pA1 ×A[2,k] → T is a multilinear map such that ϕ(x)=0 whenever pxi = 0 for any i > 1, and factoring through mq. Then ϕ extends to a multilinear map ψ : A[k] → T factoring through mpq, as in the following commutative diagram:

pA1 × i∈[2,k] Ai/Ai[p] Q ϕ 2 mq . pA1 × A[2,k] (Z/qZ) T

mpq

ψ 2 A[k] (Z/(pq)Z)

Proof. By the structure theorem for finite abelian groups, we may write

Ai = hei1i⊕···⊕heidi i (1 ≤ i ≤ k), 8 SEAN EBERHARD where each eij has order some power of p. (1) Write

ϕj1···jk = ϕ(pe1j1 ,e2j2 ,...,ekjk ) ∈ Z/qZ.

We may identify Z/qZ with p(Z/(pq)Z). For each j1,...,jk, let ψj1···jk ∈ Z/(pq)Z be an arbitrary solution to

pψj1···jk = ϕj1···jk . n Suppose i > 1 and eiji has order p . By the assumption that ϕ factors through n−1 n pA1 × i∈[2,k] Ai/Ai[p], we must have p ϕj1···jk = 0. Hence p ψj1···jk = 0. Similarly, if e has order pn, then Q 1j1 n n−1 n p ψj1···jk = p ϕj1···jk = ϕ(p e1j1 ,e2j2 ,...,ekjk )=0.

Hence we may define ψ : A[k] → Z/(pq)Z by linearly extending the definition

ψ(e1j1 ,...,ekjk )= ψj1···jk .

The maps ψ|pA1×A[2,k] and pϕ agree on generators, so they are the same. (2) Write

ϕj1 ···jk = ϕ(e1j1 ,e2j2 ,...,ekjk ) ∈ Z/qZ.

For each j1,...,jk, let ψj1···jk ∈ Z/(pq)Z be an arbitrary solution to

ψj1···jk ≡ ϕj1···jk (mod q). n n−1 Suppose p is the order of one of the generators eiji . Then p ϕj1 ···jk = 0 (mod q), n so p ψj1···jk = 0 (mod pq). Hence we may define ψ : A[k] → Z/(pq)Z by linearly extending the definition

ψ(e1j1 ,...,ekjk )= ψj1···jk . Then ψ mod q and ϕ agree on generators, so they are the same. (3) By assumption ϕ = mq(ϕ1, ϕ2) where ϕ1 is defined on pA1 × AI for some I ( [2, k] and ϕ2 is defined on A[2,k]\I . By (1), ϕ1 may be extended to ψ1 :

A1 × AI → Z/(pq)Z in such a way that ψ1|pA1×AI = pϕ1. By (2), ϕ2 may be extended to ψ2 : A[2,k]\I → Z/(pq)Z in such a way that ψ2 mod q = ϕ2. Let ψ = mpq(ψ1, ψ2). Then for x ∈ pA1 × A[2,k] we have

ψ(x)= ψ1(x)ψ2(x)/(pq)mod1= ϕ1(x)ϕ2(x)/q mod 1 = ϕ(x), so ψ extends ϕ. 

We can now prove Theorem 1.2.

ni Proof of Theorem 1.2. Let the exponent of Ai be p . We will prove the theorem by induction on n1 + ··· + nk. If ni = 1 for each i then each Ai is elementary abelian, hence a over Fp. In this case the result follows from Theorem 1.1. Hence assume n1 > 1 without loss of generality. Then pA1 and A1/pA1 both have smaller exponent than A1. ′ The restriction ϕ1 of ϕ to pA1 × A[2,k] factors through a map ϕ1 defined on pA1 × i∈[2,k] Ai/Ai[p], as in the following diagram: Q pA1 × A[2,k] pA1 × i∈[2,k] Ai/Ai[p]

ϕ1 ′ . Q ϕ1 ϕ A[k] T

′ By Corollary 2.7, bias(ϕ1) = bias(ϕ1) ≥ ǫ, so by induction ′ ′ ′ ϕ1 = ψ1 + ··· + ψr BIASEDMULTILINEARMAPSOFABELIANGROUPS 9 for r = Oǫ(1) and some ′ ′ ψ1,...,ψr : pA1 × Ai/Ai[p] → T i∈Y[2,k] ′ ′ where each ψi factors through mqi for some qi ≤ Oǫ(1). By Lemma 3.1, each ψi extends to a multilinear map ψi : A[k] → T factoring through mpqi . Let

ϕ2 = ϕ − (ψ1 + ··· + ψr).

Then ϕ2 also has bias bounded away from zero (by Lemma 2.3 and Corollary 2.6), and ϕ2 factors through (A1/pA1) × A[2,k]. By induction again there are multilinear maps ψr+1,...,ψr2 : (A1/pA1) × A[2,k] → T, where r2 = O(1), such that each ψi factors through mq for some bounded q, and such that

ϕ2 = ψr+1 + ··· ψr2 . Hence

ϕ = ψ1 + ··· + ψr2 , and the induction is complete.  We end by deducing Corollary 1.3.

Proof of Corollary 1.3. Let Ak = B and let ϕ(x) = xk(F (x1,...,xk−1)). Then ϕ is multilinear and bias(ϕ) = P(f = 0) ≥ ǫ > 0, so Theorem 1.2 implies that ϕ is the sum of Oǫ(1) multilinear functionsb of the form

ψ(x)= mq(ϕ1(xI ), ϕ2(xIc )), c where 0 < |I| < k and q ≤ Oǫ(1). Assume without loss of generality that k ∈ I . ∼ Since ψ is linear in xk and Ak = B, there is some b = b(x[k−1]) ∈ B such that

ψ(x)= xk(b(x[k−1])). b Moreover, b must be multilinear in x[k−1], and b can depend on xI only through ϕ1(xI ). Hence ψ has the form

ψ(x)= xk(G(g(xI ), x[k−1]\I )), where g = ϕ1. This implies that F is the sum of Oǫ(1) maps of the form G(g(xI ), x[k−1]\I ). Now fix I and suppose the terms of the form G(g(xI ), x[k−1]\I ) are

Gj (gj (xI ), x[k−1]\I ) (1 ≤ j ≤ n).

Let g = (g1,...,gn), and note that | cod(g)| is still bounded and Gj (gj (xI ), x[k−1]\I )= ′ ′ Gj (g(xI ), x[k−1]\I ), where Gj is the composite of Gj and the projection πj . Let ′ ′ G = G1 + ··· + Gn. Then n

Gj (gj (xI ), x[k−1]\I )= G(g(xI ), x[k−1]\I ). j=1 X Hence F has the claimed form. 

4. Possible biases

Let Φk be the set of all multilinear maps ϕ : A[k] → T for any finite abelian groups A1,...,Ak. In this section we probe the set of all possible biases

Bk = {bias(ϕ): ϕ ∈ Φk}. For example,

B1 = {0, 1}, B2 = {1/n : n ≥ 1}.

By (3), Bk ⊆ [0, 1]. We will show that Theorem 1.2 implies that Bk is a small subset of [0, 1] in various senses. The results of this section are analogous to those 10 SEAN EBERHARD proved in [E] essentially for B3 and for the set of commuting probabilities of finite groups. A slight generalization is useful. Call ϕ : A[k] → T multiaffine if ϕ(x) is affine- linear in each variable xi separately. Equivalently, ϕ is multiaffine if and only if it can be written

ϕ = ϕI , IX⊆[k] where ϕI is a multilinear map AI → T. The greatest d = |I| such that ϕI is nontrivial is the degree of ϕ. LetΦk,d be the set of all multiaffine maps ϕ : A[k] → T (for any A1,...,Ak) of degree at most d and zero constant term, and let

Bk,d = {bias(ϕ): ϕ ∈ Φk,d}.

Thus Bk,d is a subset of the unit disc D = {z ∈ C : |z|≤ 1}.

Example 4.1 (Gauss sums). Let A1 = A2 = A3 = Fp, where p is an odd prime, and let

ϕ(x,y,z) = (xy + xz + yz)/p mod 1 (x,y,z ∈ Fp)

Thus ϕ ∈ Φ3,2. Then −2 bias(ϕ)= Ex,y,z∈Fp e((xy + xz + yz)/p)= p G(p), where G(p) is the Gauss sum G(p)= e(x2/p). F xX∈ p It was proved by Gauss that G(p)= p1/2 or ip1/2 according to whether p is 1 or 3 ±(p−1)2/4 −3/2 mod 4. Hence B3,2 includes i p for every odd prime p. If X is a subset of a compact metric space, let X′ be the set of limit points of X. We iterate this operation transfinitely according to the following rules: X(0) = X, X(α+1) = (X(α))′ (α an ordinal), X(λ) = X(α) (λ a nonzero limit ordinal). α<λ\ If X is countable then there is a countable ordinal β such that X(β) = ∅. The smallest such β must be a successor ordinal, say α + 1, and this ordinal α is called the Cantor–Bendixson rank of X and denoted CB(X). It is the unique ordinal such that X(α) is finite and nonempty.1

Theorem 4.2. The set Bk,d has the following smallness properties.

(1) All limit points of Bk,d are algebraic (in fact, rational linear combinations of roots of unity). d−2 (2) CB(Bk,d) ∈{1,ω,...,ω } for d ≥ 2.

In particular, Item (1) shows that Bk,d is countable, so in particular it is nowhere dense and measure-zero.

Proof. For ǫ> 0 let

Bk,d,ǫ = Bk,d ∩{z ∈ D : |z|≥ ǫ}.

1More often β itself is called the Cantor–Bendixson rank, but the present usage is a common variant for countable sets. BIASED MULTILINEAR MAPSOF ABELIAN GROUPS 11

Let ϕ ∈ Φk,d,ǫ, so ϕ : A[k] → T is a multiaffine map of degree d and zero constant term defined on some finite abelian groups A1,...,Ak and | bias(ϕ)|≥ ǫ. Write

ϕ = ϕI , IX⊆[k] where ϕI : AI → T is multilinear. Let ϕd be the sum of the terms with |I| = d. By Lemma 2.2, for each I ⊆ [k] of size d we have

bias(ϕI ) ≥ | bias(ϕ)|≥ ǫ, so by Theorem 1.2 there is an expression for ϕI as the sum of Oǫ(1) maps factoring through mq for some bounded prime power q. Hence there is an expression for ϕd of the form r L R ϕd = mqi (ϕi , ϕi ), i=1 LX R where r = Oǫ(1) and, for each i, ϕi and ϕi (1 ≤ i ≤ r) are multilinear maps with disjoint sets of variables and codomain Z/qiZ, where qi = Oǫ(1). Hence L L R R e(ϕd)= F (ϕ1 ,...,ϕr , ϕ1 ,...,ϕr ), where F (x1,...,xr,y1,...,y1)= e(x1y1/q1 + ··· + xryr/qr). r 2 1/2 Let Γ=( i=1 Z/qiZ) and let Q = q1 ··· qr = |Γ| . The Fourier expansion of F is r Q 1 F (x ,...,x ,y ,...,y )= e (−u v + u x + v y )/q . 1 r 1 s Q i i i i i i i i=1 ! (u,vX)∈Γ X Hence r 1 e(ϕ )= e (−u v + u ϕL + v ϕR)/q d Q i i i i i i i i=1 ! (u,vX)∈Γ X and 1 bias(ϕ)= e(−u v /q −···− u v /q ) bias(ψ ), Q 1 1 1 r r r u,v (u,vX)∈Γ where r L R ψu,v = (uiϕi + viϕi )/qi + (ϕ − ϕd). i=1 X Note that ψu,v is multi-affine of degree at most d−1 with zero constant term. Since |Γ|≤ Oǫ(1), we have proved that

(4) Bk,d,ǫ ⊆ Oǫ(1)HOǫ(1)Bk,d−1, where HX is the (finite) set of all complex numbers of the form ζ/m where ζ is a root of unity of order at most X and m is a positive integer of size at most X. Here we are also using the standard notation ST = {st : s ∈ S,t ∈ T } for the product of sets S and T and nS = S + ··· + S for the sum of n copies of a set S. We will now use induction on d to prove Items (1) and (2). As the base of the induction, note that an affine linear map with zero constant term is just a linear map, so Bk,1 = B1 = {0, 1}. Hence, for d =2, (4) shows that Bk,2,ǫ is finite for all ′ ǫ> 0, so Bk,2 ⊆{0}. Hence we may assume d> 2. Let z be a limit point of Bk,d. Then either z =0 or z is a limit point of Bk,d,ǫ for ǫ = |z|/2. In the latter case (4) shows that

z ∈ Oǫ(1)HOǫ(1)Bk,d−1. By induction on d this shows that z is a rational linear combination of roots of unity. This proves Item (1). 12 SEAN EBERHARD

Let α = CB(Bk,d). Then it follows from (4) that

(αOǫ(1)) Bk,d,ǫ = ∅. Since this holds for all ǫ> 0, it follows that (αω) Bk,d ⊆{0}. Hence CB(Bk,d) ≤ CB(Bk,d−1)ω, d−2 and this proves CB(Bk,d) ≤ ω by induction. Finally, since Bk,d is closed under γ multiplication, the argument of [E, Lemma 5.1] shows that CB(Bk,d)= ω for some d−2 ordinal γ, so CB(Bk,d) ∈{1,ω,...,ω }. 

Since Bk ⊆ Bk,k, the theorem applies in particular to Bk ⊆ [0, 1] and shows

(1) all limit points of Bk are algebraic, k−2 (2) CB(Bk) ∈{1,ω,...,ω }.

It is very likely that all limit points of Bk are actually rational, but a slightly more sophisticated analysis is required to prove this. As to the Cantor–Bendixson rank, basic examples demonstrate that CB(Bk) > 1 for k> 2, so

CB(B1)=0, CB(B2)=1, CB(B3)= ω, but the value of CB(Bk) for k > 3 is unclear. Relatedly, it is plausible that Bk is well-ordered by the reverse of the natural order, i.e., for every x ∈ Bk there is some ǫ> 0 such that (x − ǫ, x) ∩ Bk = ∅; if this is the case it follows that Bk has order type ωCB(Bk ). We leave these questions as open problems. Problem 4.3.

(1) Prove that all limit points of Bk are rational. (2) Determine the Cantor–Bendixson rank of Bk. In particular, determine 2 whether CB(B4) is ω or ω . (3) Determine whether Bk is reverse-well-ordered. References

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