Zero Divisors and L

New York J. Math. 7(2001) 49–58.

Zero Divisors and L

( G ), II

Peter A. Linnell and Michael J. Puls

Abstract. LetGbe a discrete group, letp ≥1, and letL^p(G) denote the Banach space{

g∈Gagg |

g∈G|ag|^p <∞}. The following problem will be studied: Given 0=α∈CGand 0=β∈ L^p(G), isα∗β= 0? We will concentrate on the caseGis a free abelian or free group.

Contents

1. Introduction 49

2. Statement of main results 50

3. A characterization ofp-zero divisors 51

4. A key proposition 52

5. Proofs of Theorems 2.1, 2.2, and 2.3 53

6. Free groups andp-zero divisors 53

References 57

1. Introduction

LetGbe a discrete group and letf be a complex-valued function onG. We may representf as a formal sum

g∈Ga_ggwhere a_g ∈Candf(g) =a_g. ThusL^∞(G) will consist of all formal sums

g∈Gagg such that sup_g∈G|ag| <∞, C₀(G) will consist of those formal sums for which the set{g| |ag|> } is finite for all >0, and forp≥1,L^p(G) will consist of those formal sums for which

g∈G|ag|^p<∞. Then we have the following inclusions:

CG⊆L^p(G)⊆C₀(G)⊆L^∞(G).

Forα=

g∈Gagg∈L¹(G) andβ=

g∈Gbgg∈L^p(G), we define a multiplication L¹(G)×L^p(G)→L^p(G) by

α∗β =

g,h

agbhgh=

g∈G

h∈G

agh−1bh

g.

(1.1)

Received May 17, 2001.

Mathematics Subject Classification. Primary: 43A15; Secondary: 43A25, 42B99.

Key words and phrases. zero divisor, free group, Fourier transform, radial function, free abelian group.

2001 State University of New Yorkc ISSN 1076-9803/01

49

In this paper we consider the following:

Problem 1.1. Let G be a torsion free group and let1 ≤p≤ ∞. If 0 =α∈CG and0=β ∈L^p(G), isα∗β= 0?

Some results on this problem are given in [7, 8]. In this sequel we shall obtain new results for the casesG =Z^d, the free abelian group of rank d, and G=F_k, the free group of rankk.

Part of this work was carried out while the first author was at the Sonder- forschungsbereich in M¨unster. He would like to thank Wolfgang L¨uck for organizing his visit to M¨unster, and the Sonderforschungsbereich for financial support.

2. Statement of main results

Let 0=α∈L¹(G) and let 1≤p∈R. We shall say thatαis ap-zero divisor if there existsβ ∈L^p(G)\0 such that α∗β = 0. Ifα∗β= 0 for all β ∈C₀(G)\0, then we say thatαis auniform nonzero divisor.

Let 2 ≤ d ∈ Z. It was shown in [8] that there are p-zero divisors in CZ^d for p > _d²₋₁^d . In this paper we shall show that this is the best possible by proving:

Theorem 2.1. Let 2 ≤ d ∈ Z, 1 ≤ p ∈ R, let 0 = α∈ CZ^d, and let 0 = β ∈ L^p(Z^d). Ifp≤ _d²₋₁^d , thenα∗β = 0.

LetT^d denote thed-torus which, except in Section 4, we will view as the cube [−π, π]^d in R^d with opposite faces identified, and let p: [−π, π]^d →T^d denote the natural surjection. Forn∈Z^d andt∈T^d, letn·tindicate the dot product, which is well defined modulo 2π. Ifα=

n∈Z^dann∈L¹(Z^d), then fort∈T^d its Fourier transform ˆα:T^d→Cis defined by

ˆ

α(t) =

n∈Z^d

a_ne^−i(n·t).

Let Z(α) = {t ∈ T^d | α(t) = 0ˆ }. We say that M is a hyperplane in T^d if there exists a hyperplane N in R^d such that M =p([−π, π]^d∩N). We will prove the following theorem, which is an improvement over [8, Theorem 1].

Theorem 2.2. Let α∈CZ^d. Then αis a uniform nonzero divisor if and only if Z(α)is contained in a finite union of hyperplanes in T^d.

Let V = p

(−π, π)^d

, let α ∈ L¹(Z^d), let E = Z(α)∩V, and let U be an open subset of (−π, π)^d−1. Let φ:U → (−π, π) be a smooth map, and suppose {p(x, φ(x))|x∈U} ⊆E. If the Hessian matrix

∂²φ

∂xi∂xj

ofφhas constant rankd−1−ν onU where 0≤ν ≤d−1, then we say thatφhas constant relative nullityν. We shall say thatZ(α) hasconstant relative nullityν if every localizationφofE has constant relative nullityν [6, p. 64]. We shall prove:

Theorem 2.3. Let α∈CZ^d, let1≤p∈R, and let2≤d∈Z. Suppose thatZ(α) is a smooth(d−1)-dimensional submanifold ofT^d with constant relative nullityν such that 0≤ν≤d−2. Thenαis ap-zero divisor if and only ifp > ²⁽_d−1−^d−ν_ν⁾.

Fork∈Z≥0, letFk denote the free group onk generators. It was proven in [7]

that ifα∈ CFk\0 and β ∈L²(Fk)\0, thenα∗β = 0. We will give an explicit example to show that ifk ≥2, then this result cannot be extended to L^p(Fk) for any p >2. This is a bit surprising in view of Theorem 2.1. We will conclude this paper with some results aboutp-zero divisors for the free group case.

3. A characterization of p -zero divisors

Let G be a group, not necessarily discrete, and let L^p(G) be the space of p- integrable functions on G with respect to Haar measure. Let y ∈ G and letf ∈ L^p(G). The right translate off byywill be denoted byfy, wherefy(x) =f(xy⁻¹).

DefineT^p[f] to be the closure inL^p(G) of all linear combinations of right translates of f. A common problem is to determine whenT^p[f] = L^p(G); see [3, 4, 11] for background.

Now suppose that G is also discrete. Given 1 ≤ p∈ R, we shall always let q denote the conjugate index ofp. Thus if p >1, then ¹_p+¹_q = 1, and ifp= 1 then q=∞. Sometimes we shall require p=∞, and thenq= 1. Letα=

g∈Ga_gg∈ L^p(G),β=

g∈Gbgg∈L^q(G), and define a map·,·:L^p(G)×L^q(G)→Cby α, β=

g∈G

agbg.

Fixh∈L^q(G). Then·, his a continuous linear functional onL^p(G) and ifp=∞, then every continuous linear functional onL^p(G) is of this form. We shall use the notationβfor

g∈Gbgg⁻¹,βfor

g∈Gbgg, andβ^∗for

g∈Gbgg⁻¹. Also the same formula in Equation (1.1) gives a multiplicationL^p(G)×L^q(G)→L^∞(G). Then we have the following elementary lemma, which roughly says thatα∗β= 0 if and only if all the translates ofαare perpendicular toβ.

Lemma 3.1. Let 1 ≤p∈R orp=∞, let α∈L^p(G), and letβ ∈L^q(G). Then α∗β= 0 if and only if(α) y, β= 0 for ally∈G.

Proof. Writeα=

g∈Ga_ggandβ =

g∈Gb_gg. Then α∗β =

y∈G

(

g∈G

a_yg−1b_g)y and(α) y, β=

g∈Ga_yg−1bg. The result follows.

The following proposition, which is a generalization of [8, Lemma 1], character- izesp-zero divisors in terms of their right translates (the statement of [8, Lemma 1]

should have the additional condition thatp= 1).

Proposition 3.2. Let α∈L¹(G)and let1< p∈Rorp=∞. Thenαis a p-zero divisor if and only if T^q[α] =L^q(G).

Proof. The Hahn-Banach theorem tells us thatT^q[α] =L^q(G) if and only if there exists a nonzero continuous linear functional on L^q(G) which vanishes on T^q[α].

The result now follows from Lemma 3.1.

Remark 3.3. If p = 1 in the above Proposition 3.2, we would need to replace L^q(G) withC₀(G), and T^q[α] with the closure in C₀(G) of all linear combinations of right translates ofα.

4. A key proposition

In this section we prove a proposition that will enable us to prove Theorems 2.1, 2.2 and 2.3.

Let 1≤p∈ R, let y ∈ R^d and let f ∈L^p(R^d). We shall use additive notation for the group operation in R^d; thus the right translate off byy is now given by f_y =f(x−y). We say thatf has linearly independent translates if and only if for alla₁, . . . , am∈C, not all zero, and for all distincty₁, . . . , ym∈R^d,

m i=1

a_if_yi= 0.

For the rest of this section we shall viewT^d as the unit cube [0,1]^d with opposite faces identified. LetL^p(T^d×Z^d) denote the space of functions onT^d×Z^d which satisfy

t∈T^d

m∈Z^d

|f(t, m)|^pdt <∞. Then forα=

n∈Z^dann∈CZ^d andf ∈L^p(T^d×Z^d), we defineαf ∈L^p(T^d×Z^d) by

(αf)(t, m) =

n∈Z^d

a_nf(t, m−n), and this yields an action ofCZ^d onL^p(T^d×Z^d).

Lemma 4.1. Let α∈CZ^d. Then there exists β ∈L^p(Z^d)\0 such thatα∗β = 0 if and only if there exists f ∈L^p(T^d×Z^d)\0 such thatαf = 0.

Proof. Let β ∈ L^p(Z^d)\0 such that α∗β = 0 and define a nonzero function f ∈L^p(T^d×Z^d) byf(t, m) =β(m). Forn∈Z^d, set bn=β(n). Then

(αf)(t, m) =

n∈Z^d

a_nf(t, m−n) =

n∈Z^d

a_nβ(m−n)

=

n∈Z^d

a_nb_m−n= (α∗β)(m) = 0.

(4.1)

Conversely suppose there exists f ∈ L^p(T^d×Z^d)\0 such that αf = 0. This means that (αf)(t, n) = 0 for all n, for allt except on a set T₁ ⊂T^d of measure zero. Also

n∈Z^d|f(t, n)|^p<∞for alltexcept on a setT₂⊂T^d of measure zero.

Since f = 0, we may chooses∈ T^d\(T₁∪T₂) such that f(s, n)= 0 for some n.

Now defineβ(n) =f(s, n). Then β ∈L^p(Z^d)\0 and the calculation in Equation

(4.1) shows thatα∗β= 0.

Forα=

n∈Z^da_nn∈CZ^d andf ∈L^p(R^d), we defineαf ∈L^p(R^d) by (αf)(x) =

n∈Z^d

a_nf(x−n).

Ifα= 0 andαf = 0, then there is a dependency among the right translates off, i.e.,f does not have linearly independent translates. We are now ready to prove:

Proposition 4.2. Let α ∈ CZ^d. Then α is a p-zero divisor if and only if there existsf ∈L^p(R^d)\0 such thatαf = 0.

Proof. Define a Banach space isomorphismζ:L^p(R^d)→L^p(T^d×Z^d) by the for- mula (ζf)(t, n) =f(t+n) forf ∈L^p(R^d). We want to show that this isomorphism commutes with the action of CZ^d. Clearly it will be sufficient to show that ζ commutes with the action ofZ^d. Ifm∈Z^d, then

m(ζf)

(t, n) = (ζf)(t, n−m) =f(t+n−m)

= (mf)(t+n) = (ζ(mf))(t, n).

Thus the action ofCZ^dcommutes withζ. We deduce that forα∈CZ^d, there exists f ∈ L^p(R^d)\0 such that αf = 0 if and only if there exists f ∈ L^p(T^d×Z^d)\0 such thatαf= 0. The proposition now follows from Lemma 4.1.

Remark 4.3. ReplacingL^p(R^d) by C₀(R^d) in the above arguments, we can also show thatαis a uniform nonzero divisor if and only ifαf = 0 for allf ∈C₀(R^d)\0.

5. Proofs of Theorems 2.1, 2.2, and 2.3

The proof of Theorem 2.1 is obtained by combining [11, Theorem 3] with Propo- sition 4.2. The proof of Theorem 2.2 is obtained by combining [3, Theorem 2.12]

with Remark 4.3.

Before we prove Theorem 2.3, we will need to define the notion of aq-thin set.

See [4] for more information on this and other concepts used in this paragraph.

Let Gbe a locally compact abelian group and let X be its character group. Let β ∈ L^∞(G) and let ˆβ indicate the generalized Fourier transform of β. The key reason for using the generalized Fourier transform is that for α∈L¹(G), we have α∗β= ˆαβˆwhich tells us thatα∗β= 0 if and only if supp ˆβ ⊆Z(α). LetE ⊆X. We shall say thatE isq-thin ifβ∈C₀(G)∩L^p(G) and supp ˆβ ⊆E impliesβ = 0.

Recall thatpis the conjugate index of q. The result of Edwards [4, Theorem 2.2]

says that ifα∈L¹(Z^d) andZ(α) isq-thin, thenT^q[α] =L^q(G). Here ourqis used in place of Edwards’sp, and ourpis used in place of hisp.

We are now ready to prove Theorem 2.3. SupposeZ(α) satisfies the hypothesis of the theorem. Let β ∈ L^p(Z^d)\0 such that α∗β = 0 and p ≤ ²⁽_d−1−^d−ν_ν⁾. Since

2(d−ν)

d−1−ν >1 and increasingpretains the propertyβ∈L^p(Z^d), we may assume that p >1. Thenα∗β= 0 and using Proposition 3.2, we see thatT^q[α]=L^q(Z^d). But [4, Theorem 2.2] tells us thatZ(α) is notq-thin, and this contradicts [6, Theorem 1].

Conversely, let T be a smooth, nonzero mass density on Z(α) vanishing near the boundary ofZ(α). Using [6, Theorem 3], we can constructβ ∈L^p(R^d)\0 for p > ²⁽_d−1−^d−ν_ν⁾ such that ˆβ =T. Then supp ˆβ ⊆Z(α), that isαβ = 0. An application of Proposition 4.2 completes the proof of Theorem 2.3.

6. Free groups and p -zero divisors

Throughout this section, 2≤k∈Z. In [7] it was proven that if 0 =α∈CFk, thenαis not a 2-zero divisor. In this section we will give explicit examples to show that this result cannot be extended toL^p(Fk) for any p >2. We will conclude this section by giving sufficient conditions for elements ofL¹_r(Fk), the radial functions ofL¹(Fk) as defined below, to bep-zero divisors.

Any element xof Fk has a unique expression as a finite product of generators and their inverses, which does not contain any two adjacent factorsww⁻¹orw⁻¹w.

The number of factors inxis called thelength ofxand is denoted by|x|.

A function in L^∞(F_k) will be called radial if its value depends only on |x|. Let E_n = {x ∈ F_k | |x| = n}, and let e_n indicate the cardinality of E_n. Then e_n = 2k(2k−1)ⁿ⁻¹ for n ≥ 1, and e₀ = 1. Let χ_n denote the characteristic function ofE_n, so as an element ofCF_k we haveχ_n =

|x|=nx. Then every radial function has the form _∞

n=0anχn where an ∈ C. Let L^p_r(Fk) denote the radial functions contained inL^p(Fk) and let (CFk)rdenote the radial functions contained inCF_k. ThenL^p_r(F_k) is the closure of (CF_k)_rin L^p(F_k). Letω=√

2k−1. It was shown in [5, chapter 3] that

χ₁∗χ₁=χ₂+ 2k∗χ₀

χ₁∗χn=χn+1+ω²χn−1, n≥2,

henceL¹_r(F_k) is a commutative algebra which is generated byχ₀ andχ₁. Later we will need the following elementary result.

Lemma 6.1. Let x, y∈Fk with|x|=|y|, and let0≤m, n∈Z. Then χm∗x, χn=χm∗y, χn.

Proof. We haveχ_m∗x, χ_n=x, χ^∗_m∗χ_n=x, χ_m∗χ_n. By the above remarks, χ_m∗χ_n is a sum of elements of the form χ_r. Therefore we need only prove that x, χ_r=y, χ_r. But

x, χ_r=

1 if|x|=r 0 if|x| =r,

and the result follows.

Letαbe a complex-valued function onF_k. Set an(α) = 1

e_n

x∈E_n

α(x)

and denote byP(α) the radial function_∞

n=0an(α)χn.

Lemma 6.2. Let 1 ≤ p∈ R or p= ∞, letα ∈ L¹_r(Fk), and let β ∈ L^p(Fk). If α∗β= 0, then α∗P(β) = 0.

Proof. Let f, h ∈ CFk. It was shown in [9, Lemma 6.1] that P(f)∗ P(h) = P(P(f)∗h). Write β =

g∈F_kbgg. If p =∞ and 0 ≤a₁, . . . , an ∈ R, then by Jensen’s inequality [10, p. 189] applied to the functionx^p forx >0,

a₁+· · ·+an

n

p

≤ a^p₁+· · ·+a^p_n

n ,

consequently

P(β)^pp= ∞ n=0

e_n 1

en

|g|=n

b_g

p

≤

g∈F_k

|b_g|^p=β^pp.

Therefore P is a continuous map from L^p(Fk) into L^p_r(Fk) forp =∞. It is also continuous forp=∞. The lemma follows because the mapL¹(G)×L^p(G)→L^p(G) is continuous; specificallyα∗βp≤ α1βp.

Forn∈Z_≥0, define polynomialsPn by

P₀(z) = 1, P₁(z) =z, P₂(z) =z²−2k and P_n+1(z) =zP_n(z)−ω²P_n−1(z) for n≥2.

Letα=_∞

n=0anχn ∈L¹_r(Fk). In [9], Pytlik shows the following.

1. X ={x+iy∈C|(₂^x_k)²+ (_2k^y₋₂)²≤1} is the spectrum ofL¹_r(F_k).

2. The Gelfand transform ofαis given by ˆα(z) =_∞

n=0a_nP_n(z) forz∈X. LetZ(α) ={z ∈X |α(z) = 0ˆ }. Forz ∈X we defineφz ∈L^∞_r (Fk), the space of continuous linear functionals onL¹_r(Fk) [1, p. 34], by

φ_z= ∞ n=0

Pn(z) e_n χ_n. We can now state:

Lemma 6.3. Let α ∈ L¹_r(Fk) and let z ∈ X. Then α∗φz = 0 if and only if z∈Z(α).

Proof. Letβ∈L¹_r(Fk) and writeβ=_∞

m=0bmχm. Then β, φz=

m,n

bmPn(z)

en χm, χn

=

n

b_nP_n(z) = ˆβ(z).

Applying this in the case β =α∗χ_n, we obtain α∗χ_n, φ_z= ˆα(z)P_n(z). Using Lemma 6.1, we deduce that ify∈F_k and|y|=n, thenα∗y, φ_z= ˆα(z)P_n(z)/e_n. Sinceα=α, the result now follows from Lemma 3.1.

If α∈L¹_r(Fk), we shall say thatα∗χn is a radial translate ofα. We then set T R¹[α] equal to the closure in L¹_r(Fk) of the set of linear combinations of radial translates ofα.

Proposition 6.4. Let α∈L¹_r(Fk). Then α∗β = 0for all β ∈L^∞(Fk)\0 if and only ifZ(α) =∅.

Proof. Ifz∈Z(α), thenφz∈L^∞(Fk)\0 andα∗φz= 0 by Lemma 6.3.

Conversely suppose there exists β ∈ L^∞(Fk)\0 such that α∗β = 0. Then β(y) = 0 for some y ∈ F_k, so replacing β with β ∗y⁻¹, we may assume that P(β)= 0. If γ =β, then α∗γ = 0 andP(γ)= 0. Using Lemma 6.2 we see that α∗P(γ) = 0, and we deduce from Lemma 3.1 thatαy, P(γ)= 0 for ally∈Fk. It follows thatα∗χ_n, P(γ)= 0 for alln∈Z≥0, consequentlyT R¹[α]=L¹_r(F_k). Let J be a maximal ideal inL¹_r(F_k) which contains T R¹[α]. By Gelfand theory there existsz∈X such thatJ ={δ∈L¹_r(Fk)|δ(z) = 0ˆ }, soz∈Z(γ).

We can now state:

Example 6.5. Let k≥2. Thenχ₁ is ap-zero divisor for allp >2.

Proof. Since 0 ∈ Z(χ₁), we see from Lemma 6.3 that χ₁∗φ₀ = 0. Of course φ₀= 0. We now prove the stronger statement thatφ₀∈L^p(Fk) for allp >2. We have

φ₀= ∞ n=0

Pn(0) e_n χn =

∞ n=0

(−1)ⁿ (2k−1)ⁿχ₂n. Therefore

g∈F_k

|φ₀(g)|^p= 1 + ∞ n=1

e_2n

(2k−1)^pn = 1 + ∞ n=1

2k(2k−1)²ⁿ⁻¹ (2k−1)^pn

= 1 + 2k 2k−1

∞ n=1

1 (2k−1)^n(p−2)

and the result follows.

We can use the above result to prove that the nonsymmetric sum of generators in Fk is ap-zero divisor for allp >2 in the casek is even andk >2. Specifically we have

Example 6.6. Let k >3 and let {x₁, . . . , xk} be a set of generators for Fk. If k is even, thenx₁+· · ·+x_k is ap-zero divisor for allp >2.

To establish this, we need some results about free groups.

Lemma 6.7. Let 0 < n ∈ Z and let F be the free group on x₁, . . . , xn. Then no nontrivial word in the2n−1 elementsx²₁, . . . , x²_n, x₁x₂, x₂x₃, . . . , xn−1xn is the identity; in particular these 2n−1 elements generate a free group of rank 2n−1.

Proof. The result is clearly true if n = 1, so we may suppose that n > 1.

We shall use induction on n, so assume that the result is true with n−1 in place of n. Let T denote the Cayley graph of F with respect to the generators x₁, . . . , x_n. Thus the vertices of T are the elements ofF, andf, g ∈F are joined by an edge if and only if f = gx^±1_i for some i. Suppose a nontrivial word in x²₁, . . . , x²_n, x₁x₂, x₂x₃, . . . , x_n−1x_n is the identity, and choose such a word w with shortest possible length.

Note thatwmust involvex²₁, becauseF is the free product of the group generated byx₂, . . . , xnand the group generated byx₁x₂. By conjugating and taking inverses if necessary, we may assume without loss of generality thatwbegins withx²₁.

Writew=w₁. . . wm, where w₁ =x²₁, and each of the wi are one of the above 2n−1 elements. Let us consider the path whose (2i+ 1)th vertex isw₁. . . wi. Note thatw= 1, butw₁. . . wi= 1 for 0< i < m.

Observe that the path of length 2 from x²₁ to x²₁w₂ cannot go through x₁ (just go through the 4n−2 possibilities forw₂, noting thatw₂=x⁻²₁ ). Now remove the edge joining x₁ and x²₁. SinceT is a tree [2, I.8.2 Theorem], the resulting graph will become two trees; one component T₁ containing 1 and the other component T₂ containingx²₁. Since the length 2 path fromx²₁to x²₁w₂ did not go throughx₁, for i ≥ 1 the path w₁w₂. . . wi remains in T₂ at least until it passes through x²₁ again. Also the path must pass throughx²₁ again in order to get back to 1. Since the pathsw₁. . . wiall have even length (all thewiare words of length 2), it follows thatw₁. . . wl=x²₁for somel∈Z, where 2≤l < m. We deduce thatw₂. . . wl= 1, which contradicts the minimality of the length ofw.

Corollary 6.8. Let n∈ Z≥1 and let F be the free group on x₁, . . . , xn. Then no nontrivial word in the2n−1 elementsx²₁, . . . , x²_n, x⁻¹₁ x₂, x⁻¹₂ x₃, . . . , x⁻¹_n−1x_n is the identity; in particular these 2n−1 elements generate a free group of rank 2n−1.

Proof. This follows immediately from Lemma 6.7: replacexixi+1withx⁻²_i xixi+1

for alli < n.

Corollary 6.9. Let n∈ Z≥1 and let F be the free group on x₁, . . . , xn, w. Then the elementswx₁, wx⁻¹₁ , . . . , wx_n, wx⁻¹_n generate a free subgroup of rank2n.

Proof. The above elements generate the subgroup generated by x²₁, . . . , x²_n, x⁻¹₁ x₂, x⁻¹₂ x₃, . . . , x⁻¹_n−1xn, wx₁.

The result follows from Corollary 6.8.

Proof of Example 6.6. LetG=Fk and letF be the free group ony₁, . . . , yk, w.

By Corollary 6.9 there is a monomorphismθ:G→F determined by the formula θ(x₁) =wy₁, θ(x₂) =wy⁻¹₁ , . . . , θ(xk) =wy_k/⁻¹₂.

Note that θ induces a Banach space monomorphism L^p(G) → L^p(F). Set α = wy₁+wy₁⁻¹+· · ·+wy_k/2+wy_k/⁻¹₂. Since y₁+y⁻¹₁ +· · ·+y_k/2+y⁻¹_k/2 is a p-zero divisor by Example 6.5, we see that α is a p-zero divisor, say α∗β = 0 where 0=β ∈L^p(F). WriteF =

t∈Tθ(G)t whereT is a right transversal forθ(G) in F. Thenβ=

t∈Tβtt whereβt∈L^p(θ(G)) for allt. Alsoα∗βt= 0 for allt and βs= 0 for some s∈T. Defineγ∈L^p(G) byθ(γ) =βs. Then 0=γ∈L^p(G) and (x₁+· · ·+xk)∗γ= 0 as required.

We conclude with some information on the existence ofp-zero divisors inL¹_r(F_k).

Let α ∈ L¹_r(F_k) and define p(α) as follows. If Z(α)∩(−2k,2k) = ∅, then set p(α) =∞. IfZ(α)∩(−2k,2k)=∅, then setm(α) = min{|t| |t∈Z(α)∩(−2k,2k)}. If m(α)∈[0,2ω], then set p(α) = 2. Finally ifm(α)∈(2ω,2k), then letp(α) be the positive root of the following equation inp:

m(α) =√ 2k−1

(2k−1)^12−1p+ (2k−1)^1p−12 . We can now state:

Proposition 6.10. Let α∈L¹_r(Fk). Thenαis ap-zero divisor for allp > p(α).

Proof. Lett∈(−2k,2k) such thatm(α) =|t| and supposep > p(α). Sinceφ_t is a positive definite function by [9, Lemma 6.1], we can apply [1, Theorem 2(a)] to deduce thatφt∈L^p_r(Fk). By Lemma 6.3 α∗φt= 0 and the result is proven.

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Math, VPI, Blacksburg, VA 24061–0123

[email protected] http://www.math.vt.edu/people/linnell/

New Jersey City University, Jersey City, NJ 07305–1597 [email protected] http://ellserver3.njcu.edu/math/puls/Puls.htm This paper is available via http://nyjm.albany.edu:8000/j/2001/7-4.html.