Positive harmonic functions on the Heisenberg group II Tome 8 (2021), p. 973-1003.
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POSITIVE HARMONIC FUNCTIONS ON THE HEISENBERG GROUP II
by Yves Benoist
Abstract. — We describe the extremal positive harmonic functions for finitely supported mea- sures on the discrete Heisenberg group: they are proportional either to characters or to translates of induced from characters.
Résumé(Fonctions harmoniques positives sur le groupe de Heisenberg II)
Nous décrivons les fonctions harmoniques positives extrémales pour les mesures à support fini sur le groupe de Heisenberg discret : elles sont proportionnelles à des caractères ou à des translatées d’induites de caractères.
Contents
1. Introduction. . . 973
2. Notation and preliminary results. . . 978
3. Induced harmonic functions. . . 980
4. Z-Invariance of harmonic functions. . . 986
5. Existence of induced harmonic functions. . . 993
References. . . .1003
1. Introduction
In this paper, we present the classification of the positive harmonic functions on the discrete Heisenberg groupG=H3(Z).
1.1. Positive harmonic functions. — Letµ=P
s∈Sµsδsbe a positive measure onG with finite supportS⊂G. We recall that a functionhonGis said to beµ-harmonic if it satisfies the equality h= Pµh, where Pµh(g) := P
s∈S µsh(sg) for all g in G.
We want to describe the coneHµ+ of positiveµ-harmonic functions honG. By the Choquet theorem, it is enough to describe its extremal rays.
Mathematical subject classification (2020). — 31C35, 60B15, 60G50, 60J50.
Keywords. — Harmonic function, Martin boundary, random walk, nilpotent group.
The main aim of this paper is to prove that the extremal positive µ-harmonic functions on G are proportional either to a character of G or to a translate of a function which is induced from a character of an abelian subgroup (Theorem 1.1).
The special case whereµis thesouthwest measurewas handled in the introductory paper [2]. This case was striking because the classical partition functionh(x, y, z) :=
py(z)with
py(z) :=number of partition ofzbyy non-negative integers
occurs as one of these extremal positive harmonic functions. This partition function py(z)is the simplest instance of a “harmonic function induced from the character of an abelian subgroup” that we will introduce in this paper.
1.2. Construction of harmonic functions. — The simplest examples ofµ-harmonic functions areµ-harmonic characters. Those are the charactersχ:G→R>0such that P
s∈Sµsχ(s) = 1. Such a functionh=χis an extremal positiveµ-harmonic function onGwhich is invariant by the centerZ ofG, see Lemma 2.1.
We now introduce another construction of extremal positiveµ-harmonic functions by inducing harmonic characters. Let S0 ⊂S be a maximal abelian subset and G0
be the subgroup of G generated by S0. Denote by µ0 := P
s∈G0µsδs the measure restriction of µ to G0. Let χ0 be a µ0-harmonic character of G0. We extend χ0
as a function on G, still denoted χ0, which is 0 outside G0. This function χ0 is µ-subharmonic, so that the sequencePµnχ0 is increasing. We set
hG0,χ0 = lim
n→∞Pµnχ0. We will tell exactly for which pairs(G0, χ0)the functionhG
0,χ0 is finite, in Lemma 3.8 and in Propositions 5.1, 5.4 and 5.5. When it is finite, the function hG
0,χ0 is an extremal positiveµ-harmonic function onG, see Lemma 3.1. We will callhG0,χ0 the harmonic function on Ginduced from theµ0-harmonic characterχ0 ofG0.
Forg inG, we denote byρg:g0 7→g0g the right translation byg onG. Whenever a functionhisµ-harmonic, the functionhg:=h◦ρg is alsoµ-harmonic.
1.3. Main results. — Our main theorem tells us that conversely these three con- structions are the only possible ones.
Theorem1.1. — Let G=H3(Z)be the discrete Heisenberg group andµ be a positive measure onGwhose supportS is finite and generates the group G. Then every extre- mal positiveµ-harmonic functionhonGis proportional either to a characterχofG or to a translate hG0,χ0 ◦ρg of a function induced from a harmonic character of an abelian subgroup.
Remark1.2
– Of course the case whereµ(G) = 1is the major case. However, even when dealing with a probability measure µ, the induction process forces us to work with positive measuresµ0 which are not probability measures.
(1) (2b) (2b)
Figure1.1. In Case (1) and in Case (2b) of Theorem 5.10, no har- monic function is induced from a character of an abelian subgroupG0.
(2a) (3a) (3a)
Figure1.2. In Case (2a), exactly two harmonic functions are induced from a character of G0=Gµ0 and no other. In Case (3a), only one harmonic function is induced from a character ofG0=Gµ0 and one or infinitely many are induced from a character ofG1=Gµ1.
(3b) (3b)
(3b) (3b)
Figure1.3. In case (3b), infinitely many harmonic functions are in- duced from a character of G0 =Gµ0 and one or infinitely many are induced from a character ofG1=Gµ
1.
– Theorem 1.1 can not be extended to all nilpotent groupsG. Indeed, the conclu- sion of Theorem 1.1 is not always valid for a probability measureµon the nilpotent groupGof rank 4 with cyclic center. See Section 5.5.
Theorem 1.1 has been announced in [2]. It will be proved in Section 4. Indeed it is a direct consequence of Propositions 4.8 and 4.10. We will give a more precise description of the extremal positiveµ-harmonic functionshin Theorem 5.10. In particular, we will say exactly when and how many of these new examples occur. This is illustrated in the schematic Figures 1.1, 1.2 and 1.3. In these figures, we have drawn various cases of semigroup G+µ generated by S that are described in Theorem 5.10. Note that the support of a positive µ-harmonic function h is invariant by the opposite semigroup, i.e., by the semigroup generated by S−1. In particular when G+µ = G, a positive harmonic functionhis either identically zero or vanishes nowhere. Here are two corollaries of Theorem 5.10 that we will prove in Section 5.4. The first corollary tells us that these new examples always vanish somewhere.
Corollary1.3. — Same notation. Lethbe an extremal positiveµ-harmonic function onG which does not vanish. Thenhis a character ofG.
The second corollary tells us exactly when no new example occurs. We denote byG+µ the semigroup generated by S.
Corollary1.4. — Same notation with µ(G) = 1. The following are equivalent:
(i) Every extremal positive µ-harmonic function honGis a character of G.
(ii) G+µ contains two non-central elements whose product is in Zr{0}.
1.4. Previous results. — The study of harmonic functions on groups has a very long history. I will just point out the part of it which is relevant for our purposes.
As a general motivation, let us recall that the boundedµ-harmonic functions on a group G are described thanks to bounded functions on the Poisson boundary of (G, µ). They are used to study random walks on G-spaces. The extremal positive µ-harmonic functions onG are related to the Martin boundary of(G, µ). They are used to study more precisely the behavior of these random walks, see [1], [9], [13]
or [15].
1.4.1. Abelian groups. — This part of the history begins with the Choquet–Deny theorem in [5]:
Let Gbe a finitely generated abelian group andµbe a positive finite measure onG whose support generatesGas a group. Then every extremal positiveµ-harmonic func- tionhonGis proportional to a character.
Indeed the proof of this theorem is very short: one notices that the harmonicity equation (2.1) is a decomposition ofh as a sum of positive harmonic functions and hence all the terms in this sum are proportional toh.
1.4.2. Bounded harmonic functions. — The Choquet–Deny theorem has been ex- tended to nilpotent groups when µ has mass 1 and h is bounded. This is due to Dynkin and Maljutov in [7]:
Let Gbe a finitely generated nilpotent group and µbe a probability measure on G whose support generatesGas a group. Then every boundedµ-harmonic function onG is constant.
1.4.3. WhenSgeneratesGas a semigroup. — The Choquet–Deny theorem has also been extended to nilpotent groups forhunbounded under an extra assumption. This is due to Margulis in [12]:
Let G be a finitely generated nilpotent group and µ be a positive measure on G whose support generates Gas a semigroup. Then every extremal positiveµ-harmonic function onGis proportional to a character.
1.4.4. The Heisenberg group. — The main significance of our Theorem 1.1 is that even though the Choquet–Deny theorem can not be extended to finitely generated nilpotent groups without this extra assumption, for the Heisenberg group one can describe all the positive harmonic functions. Note that, because of Margulis theorem, most of our paper will deal with a positive measure whose support generatesGas a group but does not necessarily generateGas a semigroup.
Many recent works focus on the random walks on the discrete Heisenberg groupG as in [3], [6] and [8], or on nilpotent groups as in [4] and [10], or on the geometry of words in Gas in [11] and [14]. We mention these related results even though we will not use them.
1.5. Strategy of proof. — We now explain the strategy of proof of Theorem 1.1 and the organization of the paper.
In Section 2, we recall well-known facts on positive harmonic functions and nota- tions for the discrete Heisenberg group G and its positive measures µ with a finite supportS.
In Section 3, we begin the proof of Theorem 1.1. Whenhis an extremalµ-harmonic function onG, we merely focus on the equalityh(g) =Pµnh(g), where the right-hand side is written as a weighted sum of values h( ˙wg) for words w of length n in S, as in Equation (2.2). In Lemmas 3.1 and 3.2, we check that when the contribution in this sum of the words w whose letters are in a proper subgroup of G, is not negligible, then h is an “induced harmonic function”. In Lemma 3.10, we prove a useful generalization: we alloww to be a concatenation ofk subwords whose letters are in a proper subgroup with k >1 fixed. The proofs are very general and do not assumeGto be nilpotent.
In Section 4, we assume that “his not induced”, and we want to prove that his invariant by the centerZofG. The main idea is to construct a symmetric relationRn
among the words inSn such that two related wordswandw0 have same weight and their image w˙ and w˙0 in G differ by a non-trivial element z of Z. A key point is to be able to compare the number of words related to w and the number of words related tow0, see Lemma 4.4. This allows us to prove thathis proportional to one of its translate hz, see Proposition 4.3. The last step is to prove that h is indeed equal to its translate hz. This is done in Propositions 4.8 and 4.10. The key point there, Lemma 4.11 is based on a counting argument that again involves the partition function. This finishes the proof of Theorem 1.1.
In Section 5, we give a complete classification of the extremalµ-harmonic functions that are “induced from a character”, see Theorem 5.10. Their existence is an important new feature of this article. The proof of this classification in Propositions 5.1, 5.4 and 5.5 uses a transience property for random walks onZsimilar to the large deviation inequality, see Lemma 5.3.
In the last Section 5.5, we explain how to construct, for a rank4nilpotent group, new extremal positiveµ-harmonic functions that are not induced.
2. Notation and preliminary results
We introduce in this section notations that will be used all over this article.
2.1. The cone ofµ-harmonic functions. — We first recall classical facts on positive µ-harmonic functions.
LetGbe a finitely generated group andµbe a positive measure with finite support S ⊂ G. We denote by G+µ the subsemigroup of G generated by S and by Gµ the subgroup ofGgenerated byS.
A positive function h : G → [0,∞[ is said to be µ-harmonic if it satisfies the equality
(2.1) h=Pµh, where Pµh:g7−→X
s∈S
µsh(sg).
A non-zero positiveµ-harmonic function is said to beextremal orµ-extremal if every smaller positiveµ-harmonic functionh06his a multiple ofh.
A functionhis said to be µ-superharmonic, respectivelyµ-subharmonic, if it sat- isfies the inequalityh>Pµh, respectivelyh6Pµh.
We will often write thenth power of the operatorPµ under the form
(2.2) Pµnh(g) = X
w∈Sn
µwh( ˙wg),
where, for a word w = s1. . . sn ∈ Sn of length `w = n, the constant µw > 0 is the product µw := µs
1· · ·µs
n > 0 and where the element w˙ ∈ G is the product
˙
w:=s1· · ·sn inG.
LetHµ+ be the convex cone of positiveµ-harmonic functions honGandE be a Borel set of extremal µ-harmonic functions containing exactly one function in each extremal ray ofHµ+. We endowHµ+with the topology of the pointwise convergence.
When G+µ = G the cone Hµ+ has a compact basis, this means that there exists a compact subset ofHµ+that meets all rays ofHµ+. In general, the coneHµ+might not have a compact basis but it iswell-capped, this means that it is a union of closed convex subconesHµ,i+ with compact basis such thatHµ+rHµ,i+ is also convex. This coneHµ+
is alsoreticulated, this means that every two positiveµ-harmonic functionsh1andh2
admit a maximalµ-harmonic lower boundhmand also a minimalµ-harmonic upper
boundhM. Indeed one has hm= lim
n→∞Pµn(min(h1, h2))>0 and hM = lim
n→∞Pµn(max(h1, h2))6h1+h2<∞.
By the Choquet theorem, it is enough to describe the extremal rays of this coneHµ+. Indeed, sinceHµ+ is well-capped, this theorem tells us thatevery positive µ-harmonic function h can be written as an integral of non-proportional extremal µ-harmonic functions: h=R
Efdα(f), for a positive measure αon the setE. SinceHµ+is reticulated, this theorem also tells us thatsuch a measureαis unique.
In this paper acharacterwill always mean a multiplicative morphismχ:G7→R>0. A characterχisµ-harmonic if and only if it satisfies the equationP
s∈Sµsχ(s) = 1.
2.2. Harmonic characters. — We discuss here harmonic characters on nilpotent groups.
Let Gbe a nilpotent finitely generated group and µ be a positive finite measure onGwith finite support generatingG.
Lemma2.1. — Every µ-harmonic character of Gis an extremal positiveµ-harmonic function.
Proof of Lemma2.1. — Letχbe aµ-harmonic character such that χ=h0+h00 with both h0 andh00 positive and µ-harmonic. We want to prove that the function eh0 :=
χ−1h0 is constant. We notice that the measureµe:=χµonGis a probability measure and the functioneh0is a boundedµ-harmonic function. Therefore by Dynkin–Maljutove theorem, see Section 1.4, the functioneh0 is constant.
2.3. The Heisenberg group. — We gather here notation that we will use in this article for the discrete Heisenberg group.
Recall that the discrete Heisenberg groupG:=H3(Z)is the setZ3of triples seen as matrices(x, y, z) :=
1x z 0 1y 0 0 1
.It is endowed with the product (2.3) (x, y, z) (x0, y0, z0) = (x+x0, y+y0, z+z0+xy0).
We will denote by 0 := (0,0,0) the identity element of G, and by z0 the generator z0:= (0,0,1)of the centerZ ofG.
For two elements g = (x, y, z), g0 = (x0, y0, z0) of G, we will denote by cg,g0 the integer cg,g0 :=xy0−yx0 so that
(2.4) gg0g−1g0−1=z0cg,g0.
LetG:=G/Z 'Z2 be the abelianization ofGthat we embed in the real vector spaceV :=G⊗ZR'R2.
Letµbe a positive measure onGwith finite supportS. We denote byµthe image ofµin Gand byS its support.
We denote byVµ the vector subspace ofV generated byS and byVµ+the smallest convex cone of V containingS. Note that, when Gµ =G, one always hasVµ = V, and, whenG+µ =G, one always hasVµ+=V.
The description of Hµ+, when Gµ =G will heavily depend on the shape of Vµ+. We will often distinguish the three cases:
(2.5) Vµ+ = the plane, a half-plane, or a properly convex cone.
3. Induced harmonic functions
In this section we present general facts onµ-harmonic functions on a finitely gen- erated group G. These facts will be particularly useful when G is the Heisenberg group.
3.1. Construction of induced harmonic functions. — The following lemma gives us a method to constructµ-harmonic functions starting from a harmonic function for a smaller measureµ0. This lemma will be mainly useful whenµ0is the restriction ofµ to a proper subgroupG0.
LetGbe a finitely generated group andµandµ0 be positive measures onGwith finite support such thatµ0< µ, i.e., such thatµ1:=µ−µ0is also a positive measure.
Lemma3.1. — Let h0 be a positiveµ0-harmonic function onGsuch that the function h:= supn>1Pµnh0 is finite.
(i) Then one hash= limn→∞Pµnh0 andhis a positiveµ-harmonic function.
(ii) One can recover h0 from hash0= limn→∞Pµn
0h.
(iii) Moreover whenh0 isµ0-extremal then hisµ-extremal too.
When it is finite, the function h will be called induced from the harmonic func- tionh0.
Proof of Lemma3.1
(i) We first notice that, sinceh0=Pµ
0h06Pµh0, the sequencePµnh0is increasing.
Hence, when this sequence is bounded it converges to aµ-harmonic function.
(ii) Sinceh=Pµh>Pµ0h, the sequencePµn
0his decreasing. SincePµn
0h>Pµn
0h0=h0, this sequencePµn
0hconverges to aµ0-harmonic functionh00:= limn→∞Pµn
0hsuch that h00>h0.
We want to prove that the function h000 := h00−h0 is zero. Since h0 6 h00 6 h, one has Pµnh0 6 Pµnh00 6 h. Therefore one also has limn→∞Pµnh00 = h and hence limn→∞Pµnh000 = 0. Since h000 is µ0-harmonic, this last sequence is increasing and hence one hash000 = 0.
(iii) Assume now thath0isµ0-extremal and assume thathis the sum of two positive µ-harmonic functionsh=h0+h00. We want to prove thathandh0 are proportional.
The functions h00 = limn→∞Pµn
0h0 and h000 = limn→∞Pµn
0h00 are µ0-harmonic and, by (ii), they give a decompositionh0=h00+h000.
Therefore, one hash00=λ0h0andh000 =λ00h0 for positive constantsλ0 andλ00 with λ0+λ00= 1. One has the inequalities
h0> lim
n→∞Pµnh00=λ0h and h00> lim
n→∞Pµnh000 =λ00h.
Since h=h0+h00, these inequalities are equalities: one has h0 =λ0handh00=λ00h.
This proves that the functionhisµ-extremal.
3.2. Recognizing induced harmonic functions. — The following lemma is a converse of Lemma 3.1. It tells us how to recognize aµ-harmonic function that is induced from a µ0-harmonic function.
Let Gbe a finitely generated group and µ0 < µ be positive measures onG with finite support.
Lemma3.2. — Let h be a positiveµ-harmonic function onG such that the function h0:= infn>1Pµn
0his non-zero.
(i) Then one hash0= limn→∞Pµn
0handh0 is a positiveµ0-harmonic function.
(ii) One has the inequality h>limn→∞Pµnh0.
(iii) Moreover whenhisµ-extremal, one has the equalityh= limn→∞Pµnh0andh0
isµ0-extremal too.
In particular, when his µ-extremal, h0 is supported by a translate Gµ0g of the subgroupGµ0.
Proof of Lemma3.2. — The argument is very similar to that of Lemma 3.1.
(i) Since the functionhis positive andµ-harmonic, the sequencePµn
0his positive and decreasing. Hence it has a limith0 which isµ0-harmonic.
(ii) By assumption, this limith0is non-zero. By construction, one has the inequality h>h0. Sincehisµ-harmonic, the sequencePµnh0is bounded byhand, by Lemma 3.1, the limit h0:= limn→∞Pµnh0 exists, isµ-harmonic and is bounded by h.
(iii) Assume now thath isµ-extremal. Then one has h0 =λ0hfor some constant λ0>0. Again by Lemma 3.1, one also has
(3.1) h0= lim
n→∞Pµn
0h0 =λ0 lim
n→∞Pµn
0h=λ0h0. Therefore one hasλ0= 1.
It remains to check thath0isµ0-extremal. Assume thath0=h00+h000 with bothh00 andh000 positiveµ0-harmonic. The limith00:= limn→∞Pµnh000 is aµ-harmonic function bounded by h. Hence one has h00 =λ00hand by the same computation as (3.1), one getsh000 =λ00h0. This proves thath0 is extremal.
The following definition relies on the previous lemmas:
Definition3.3. — A µ-harmonic functionhonGis said to beinduced from a sub- groupG0 if
(3.2) lim
n→∞Pµn
0h6= 0, whereµ0 is the restriction ofµto G0.
By Lemma 3.2, when h is µ-extremal this limit (3.2) is equal to h0◦g, where g is inGandh0 is an extremalµ0-harmonic function supported onG0. Therefore one hash=hG
0,h0 ◦ρg, wherehG
0,h0 := limn→∞Pµnh0. In this casethe functionh is a translate of the harmonic function induced from h0. Equivalently, the function h is induced fromh0◦ρg.
Definition 3.4. — A µ-harmonic function is said to be induced, if there exists a subgroup G0 of infinite index inG such thath is induced from G0. It is said to be non-induced otherwise.
Remark3.5. — The reason why we require in this definitionG0to have infinite index will be explained in Lemma 4.1. A posteriori, for an extremal positive µ-harmonic function h on the Heisenberg group G with Gµ = G, this requirement is not so useful. Indeed, by Corollary 3.6, the characters of G are not induced from proper finite index subgroups. Moreover, by Definition 3.3, if his induced from an infinite index subgroup G0, it is also induced from all the finite index subgroup of G that containG0.
Corollary3.6. — Let Gbe a finitely generated group andµa positive measure onG with finite support such thatGµ=G. Aµ-harmonic characterχofGis never induced from a proper subgroup G0⊂G.
Proof. — SinceGµ =G, the restrictionµ0 ofµ to G0 satisfiesµ0 < µ. Sinceχ is a character, one has Pµ
0χ =αχ with some constant α > 0. Since Pµχ =χ, one has α < 1. Therefore, one has limn→∞Pµn
0χ = 0, and the µ-harmonic function χ is not
induced fromG0.
3.3. Double induction. — The following lemma tells us that two successive induc- tions of a positive harmonic function is equivalent to a direct induction. LetG be a finitely generated group.
Lemma3.7. — Letµ0< µ00< µbe positive measures onGwith finite support. Leth0
be a positiveµ0-harmonic function onG. The following are equivalent:
(i) the function h:= limn→∞Pµnh0 is finite.
(ii) the functions h00:= limn→∞Pµn0 0
h0 andh0 := limn→∞Pµnh00 are finite.
In this case, the two inducedµ harmonic functions are equal h=h0.
Proof of Lemma3.7
(i) ⇒(ii) Since h0 6 h, one has the inequalities Pµn0
0h0 6Pµn0
0h6 Pµnh =h and h006h. Therefore, one also has the inequalitiesPµnh006Pµnh=handh06h.
(ii)⇒(i) Sinceh06h00, one hasPµnh06Pµnh00 andh6h0.
3.4. Induction of characters. — We give now a few conditions that have to be satisfied in order for the induction of a harmonic character to be a finite function.
LetGbe a finitely generated group andµbe a positive measure onGwith finite supportSsuch thatG=Gµ. We writeµ=µ0+µ1as a sum of two positive measures and setS0:= suppµ0andG0:=Gµ0. Letχ0 be aµ0-harmonic character ofG0that we extend by0 as a function onG. We denote by
ZG(G0) :={g∈G|gg0=g0g for allg0 inG0} the centralizer of G0 inG, and by
NG(G0, χ0) :={g∈G|gg0g−1∈G0andχ0(gg0g−1) =χ0(g0)for allg0 in G0} the normalizer of(G0, χ0)inG.
Lemma3.8. — If the induced µ-harmonic function hG0,χ0 is finite, then:
(i) The measure µ0 is the restriction ofµ toG0 andS0=S∩G0. (ii) The subgroup G0 has infinite index inG.
(iii) One hasG+µ
1 ∩G0=∅. (iv) One has G+µ
1 ∩ZG(G0) =∅. (v) One has G+µ
1∩NG(G0, χ0) =∅. Remark3.9
– In particular, the supportsS0ofµ0 andS1ofµ1are disjoint and the semigroup G+µ
1 does not meet the centerZ ofG.
– Note also that if one wants hG
0,χ0 to be µ-extremal, the group G0 must be generated byS0. Indeed if this is not the case, the µ0-harmonic characterχ0 is not µ0-extremal and, by Lemma 3.2, the functionhG0,χ0 is notµ-extremal.
– The above conditions are not the only necessary conditions, as we will see in Section 5.
Proof of Lemma3.8
(i) This is equivalent toµ1(G0) = 0which follows from (iii).
(ii) This follows from (iii). Indeed pick an element s1 in the support ofµ1, if the index were finite, there would exist a positive powersd1 belonging toG0.
(iii) This follows from (v) becauseG0⊂NG(G0, χ0). (iv) This follows from (v) becauseZG(G0)⊂NG(G0, χ0).
(v) This point is the main content of Lemma 3.8. We proceed by contraposition.
Let S1 be the support of µ1 and w1 = s1. . . s` ∈ S1`, with ` > 1 be a word such thatw˙1 belongs toNG(G0, χ0).
The proof relies on a cautious analysis of the words that occur in Equality (2.2).
We recall the notationµ1,w1 :=µ1,s1· · ·µ1,s` >0. We will denotePw1 for the operator of left translation by w˙1 := s1· · ·s` ∈ G; it is given by Pw1h(g) = h( ˙w1g) for all
functionhonGand allgin G. One computes Pµn+`χ0( ˙w−11 )> X
16i6n
µ1,w1Pµi
0Pw1Pµn−i
0 χ0( ˙w−11 )
= X
16i6n
µ1,w1Pµi
0Pw1χ0( ˙w1−1) because χ0 isµ0-harmonic
= X
16i6n
µ1,w1
X
w0∈S0i
µ0,w0χ0( ˙w1w˙0w˙1−1) by definition of Pµ0
= X
16i6n
µ1,w
1
X
w0∈S0i
µ0,w
0χ0( ˙w0) becausew˙1 normalizesχ0
= X
16i6n
µ1,w
1χ0(0) =nµ1,w
1 because χ0 isµ0-harmonic.
This goes to infinity withn, and the induced function is not finite.
3.5. Negligible trajectories. — We now discuss a lemma on non-induced extre- mal positive µ-harmonic functions. This lemma will be useful for the proof of the Z-semiinvariance of these functions on the Heisenberg group.
LetGbe a finitely generated group andµbe a positive measure onGwith finite supportS generating G.
For every word w=s1. . . sn ∈Sn, we define kw >0 to be the smallest integer k for which we can write w=w0. . . wk as a concatenation of strongly non-generating subwordswj. Strongly non-generating means that there exists an infinite index sub- groupGj ofGcontaining all the letterssioccurring in the subwordwj. The following lemma tells us that the words withkw bounded are negligible in the sum (2.2) for a non-inducedµ-harmonic function.
Lemma 3.10. — Let h be a non-induced positive µ-harmonic function on G Then, for all k>0, andg in G, the partial sums
(3.3) In,k(g) := X
w∈Sn kw6k
µwh( ˙wg)
converge to 0when n→ ∞.
Proof of Lemma3.10. — Fixg inG. Forwin Sn we introduce the maximalstrongly non-generatingsuffixσofw. Suffix means that one can writew=w0σ. We denote by S0,w the set of letters ofσ and by `0,w the length of σ. Since there are only finitely many subsets S0 of S, we can write In,k(g) as a finite sum In,k(g) = P
In,k,S0(g), where In,k,S
0(g) involves the words w for which S0,w = S0. Here this finite sum is indexed by the subsets S0 of S that generates an infinite index subgroup of G.
We argue by induction onk.
First assumek= 0. — For suchS0⊂S one has In,0,S0(g)6 X
w0∈Sn
0
µw0h( ˙w0g) =Pµn
0h(g),
where µ0 is the restriction of µ to S0. By Definitions 3.3 and 3.4, since h is non- induced and sinceS0generates an infinite index subgroup ofG, the sequencePµn
0h(g) converges to0 whenn→ ∞, and the claim (3.3) is true fork= 0.
Now assumek>1. — Fixε0>0. Sincehis non-induced, as above, we can choose`0 such that, for any subsetS0ofS that generates an infinite index subgroup ofG, one has Pµ`0
0h(g) 6 ε0, where µ0 is the restriction of µ to S0. We decompose the sum In,k,S0(g)as a sum of two terms
In,k,S0(g) =In,k,S0
0,`0(g) +In,k,S00
0,`0(g), whereIn,k,S0
0,`0(g)involves the wordswfor which`0,w>`0 andIn,k,S00
0,`0(g)involves the wordswfor which`0,w< `0.
BoundingIn0. One computes, using theµ-harmonicity ofh, In,k,S0
0,`0(g)6 X
w0∈S0`0
µw0
X
w1∈Sn−`0
µw1h( ˙w1w˙0g)
6 X
w0∈S0`0
µw0h( ˙w0g)6ε0.
BoundingIn00. One decomposesIn,k,S00
0,`0 as a finite sum In,k,S00
0,`0(g) =X
σ
µσIn,k,σ00 (g) over the finitely many wordsσof length` < `0, where
In,k,σ00 (g)6 X
w0∈Sn−`, kw06k−1
µw0h( ˙w0σg)˙
6In−`,k−1( ˙σg).
Therefore by the induction hypothesis one haslimn→∞In,k,σ00 (g) = 0. Sinceε0can be chosen arbitrarily small, one deduces thatlimn→∞In,k(g) = 0.
3.6. WhenG+µ meets the center. — There is a simple case where the semiinvariance ofµ-harmonic functions is easy to prove, namely whenG+µ meets the center.
LetGbe a finitely generated group,Zbe the center ofGandµbe a finite positive measure onG.
Lemma3.11. — Assume that an elementz of Z belongs to the semigroup G+µ. Then, for every extremal positiveµ-harmonic functionhonGthere exists a constantq >0 such that hz=qh.
We recall thathzis the function g7→h(gz).
Proof of Lemma3.11. — This is a slight generalization of the Choquet–Deny theorem.
Letn>1 be an integer such thatz is in the support of µ∗n. The equality h=Pµnh is of the form h = αhz +h0, where α > 0 and h0 is a positive function. Since the functionhz is alsoµ-harmonic, the extremality of himplies that hz is proportional
toh.
4. Z-Invariance of harmonic functions In all this section we keep the following notation:
(4.1)
Gis a finite index subgroup inH3(Z),Z is the center of G, µis a positive measure with finite support S such that Gµ=G, his a positiveµ-harmonic function onG.
In this section we will mainly focus on non-induced µ-harmonic functions (see Defi- nitions 3.3 and 3.4) and we will prove that they areZ-invariant.
We begin by a lemma that explain our choices in Definition 3.4.
Lemma 4.1. — The positive µ-harmonic function h is non-induced if and only if limn→∞Pµn0h= 0,for all restriction µ0 of µto an abelian subsetS0 of S.
Proof. — By Definition 3.4, “hnon-induced” means that his non-induced from an infinite index subgroup G0 of G. Note that the subgroups G0 of infinite index in G are exactly the abelian subgroups of G. Indeed any two non-commuting elements of
H3(Z)generate a finite index subgroup ofH3(Z).
Remark 4.2. — A finite index subgroup G of H3(Z) is not always isomorphic to H3(Z), but it contains a finite index subgroup that is isomorphic toH3(Z). Extending our theorem 1.1 to these groupsGwould be straightforward but not so interesting.
The main reason we want to work with this slightly larger class of groupGin this section is that, in the “proof by induction” of Proposition 4.10, we need to apply the
“induction hypothesis” to a finite index subgroup ofG.
4.1. Semiinvariance of harmonic functions. — In this section we prove that h is semiinvariant by one central element. The proofs below are self-contained. They are inspired by the more intuitive proofs for the south-west measure in [2] that rely on Young diagrams.
Proposition4.3. — Keep notation (4.1) and assume that his µ-extremal and non- induced. Then there exist z6= 0 inZ andq >0 such thathz=qh.
Proof of Proposition4.3. — By Lemma 3.11, we can assumeS∩Z=∅. Forn>2, we introduce a symmetric relation on Sn given by
Rn:={(w, w0)∈Sn×Sn|w=w0ss0w00andw0=w0s0sw00, where
w0∈Si, w00 ∈Sn−i−2, s∈S, s0 ∈S withss06=s0s}.
This means thatwandw0are obtained from one another by switching two consecutive non-commuting letters. For a wordw∈Sn we let
kw =the number of pairs of consecutive non-commuting letters inw.
SinceGis the Heisenberg group H3(Z)and sinceS∩Z =∅, this number kw is the same as the one occurring in Lemma 3.10. Indeed, there exists a unique partition S = S0∪ · · · ∪S` of S such that two elements s, s0 of S commute if and only if
they belong to the sameSi. To go on the proof of Proposition 4.3, we will need the following two lemmas.
We setp0:= max
s,s0∈S|cs,s0|, where the integers cs,s0 are defined in (2.4).
Lemma4.4. — For(w, w0)∈Rn, one has
(i) w˙ = ˙w0z0p for some integer pwith0<|p|6p0, (ii) µw0 =µw,
(iii) |kw0−kw|62.
Proof of Lemma4.4
(i) This follows from the equalityss0=s0s zc0s,s0. (ii) The same letters occur inwandw0.
(iii) The pairs of adjacent letters inwandw0 are the same except for at most two
of them.
Lemma4.5. — Forg in G, one hash(g)6P
0<|p|6p0h(z0pg).
Proof of Lemma4.5. — Replacing h by its translatehg, we can assume that g = 0.
We want to prove that the following difference is non-positive:
D:=h(0)− X
0<|p|6p0
h(z0p)60.
Using notations (2.2), we computeD as D= X
w∈Sn
µwh( ˙w)− X
0<|p|6p0
X
w0∈Sn
µw0h( ˙w0zp0).
We fixε0>0andk0>2 + 2ε−10 . By Lemma 3.10, one can find an integern>1such that the first sum limited at the trajectoriesw for whichkw< k0 is bounded by ε0. Using the fact that, forwinSn, the fiber
{(w, w0)|w0∈Sn, (w, w0)∈Rn} of the mapsRn7→Sn; (w, w0)7→whas cardinality kw, one gets
D6ε0+ X
(w,w0)∈Rn, kw>k0
µw kw
h( ˙w)−µw0 kw0
P
0<|p|6p0h( ˙w0zp0) .
By Lemma 4.4, the element w˙ is equal to at least one of those w˙0zp0, therefore one gets
D6ε0+ X
(w,w0)∈Rn, kw>k0
µw
kw0−kw kwkw0
h( ˙w).
By Lemma 4.4, one has|kw0−kw|62, and2/kw0 62/(k0−2)6ε0, and D6ε0+ε0 X
(w,w0)∈Rn
µw kw
h( ˙w).
Using again thatkwis the cardinality of the fiber and using the harmonicity ofh, one gets
D6ε0+ε0 X
w∈Sn
µwh( ˙w) =ε0+ε0h(0).
Sinceε0 can be chosen arbitrarily small, this givesD60as expected.
End of proof of Proposition4.3. — Lemma 4.5 tells us that there exists a finite subset F ⊂Zr{0} and a positiveµ-harmonic function h0 such that
X
z∈F
hz=h+h0.
Since the coneHµ+ is well-capped and reticulated, both the functionh0 and the sum P
z∈Fhz admit a unique desintegration in µ-extremal functions (see Section 2.1).
Hence, since all the positiveµ-harmonic functionshandhzareµ-extremal, the func- tionhhas to be proportional to one of these translateshz. Remark 4.6. — We now want to deduce from the semi-invariance of h proved in Proposition 4.3, theZ-invariance ofh. This is not a general fact. Indeed, the harmonic functionhin Case (3b) of Theorem 5.10 can beZ-semiinvariant but is notZ-invariant.
Hence, we have to use once more the assumption thathis not induced. One technical difficulty comes from the fact that, whenG+µ 6=G, the coneHµ+ often does not have a compact basis. This prevents us from using the same arguments as in [12].
4.2. z-invariance and Z-invariance. — We first notice that in order to prove the Z-invariance of a positive µ-harmonic function h on the Heisenberg group G, it is enough to check that it is invariant under one non trivial element ofZ.
Lemma4.7. — Keep notation (4.1)and assume that there existsz6= 0inZ such that hz=h. ThenhisZ-invariant. In particular, if hisµ-extremal, it is proportional to aµ-harmonic character ofG.
Note that in this lemma the positiveµ-harmonic function his not assumed to be µ-extremal.
Proof of Lemma4.7. — We writez=z0p. We can assume thatpis the smallest positive integer for whichhz=h. We can also assume thathis extremal in the convex cone
Hµ,z+ :={positive, µ-harmonic andz-invariant functions onG}.
Therefore the functions hzi
0, for i = 1, . . . , p, are non-proportional functions which are extremal in this cone, and the functionf :=hz0 +· · ·+hzp
0 isµ-harmonic and Z-invariant.
We claim that f is extremal among the µ-harmonic functions on G/Z. Indeed, assume that one can writef =f0+f00 with bothf0 andf00 positive,µ-harmonic and Z-invariant. We argue as in the proof of Proposition 4.3 with the well-capped and reticulated cone Hµ,z+. Both the function f0 and f admit a unique desintegration in extremal functions in this cone (see Section 2.1). Hence, since all the functionshzare
extremal in this cone, one must have f0 =P
16i6pλihzi
0 for some constants λi >0.
Since f0 is z0-invariant, all these constants are equal to some λ > 0 and one has f0=λf. This proves thatf is extremal among theµ-harmonic functions onG/Z.
Since G/Z is abelian, by the Choquet–Deny theorem, this function f is a µ-har- monic character of G. Therefore, by Lemma 2.1, this function f is µ-extremal and
one hasp= 1. This means thathisZ-invariant.
4.3. Z-invariance whenVµ+ contains a line. — In this section, we finish the proof of our main Theorem 1.1 when the coneVµ+ is the plane or a half-plane, see (2.5).
Proposition4.8. — Keep notation (4.1), assume that the cone Vµ+ contains a line and that theµ-harmonic function his not induced. Then hisZ-invariant.
Proof of Proposition4.8. — We can assume thath isµ-extremal and apply Proposi- tion 4.3. Then our claim follows from the following slightly stronger Proposition 4.9.
This stronger version will also be useful in Section 5.
Proposition 4.9. — Keep notation (4.1) and assume that the cone Vµ+ contains a line. Assume also that there exists z6= 0 inZ andq >0 such thathz=qh. Then the function hisZ-invariant.
Proof of Proposition4.9. — According to Lemma 4.7, it is enough to prove thatq= 1.
Replacing h by a multiple of a suitable translate, we can assume that h(0) = 1.
Replacing z by its inverse if necessary, we can also assume that q > 1. Since the cone Vµ+ contains a line, there exists two words w0 in Sn0 and w00 in Sn00 whose product is in the center:
˙
w0w˙00=za for some ainZ.
Since the coneVµ+ is not a line, there exists also a wordw1 inSn1 such that (4.2) w˙0w˙1w˙−10 w˙1−1=zb for someb>1.
Note that one might have to switchw0 andw00to ensure that b>1.
Assume, for a contradiction, that q6= 1, so thatq > 1. Choose an integer` >1 such thatC:=µw0µw0
0qa+b`>1. Notice the equality, for allk>1, (4.3) w˙k0w˙`1w˙00kw˙−`1 =zak+b`k.
Note that both Equations (4.2) and (4.3) rely on the bilinear formula (2.4) for the commutators in the Heisenberg group. Now we can compute withn:=kn0+`n1+kn00,
h( ˙w1−`) =Pµnh( ˙w1−`)
>µkw
0µ`w
1µkw0 0
h( ˙wk0,w˙`1w˙00kw˙−`1 )
>µkw
0µ`w
1µkw0 0
qak+b`k=µ`w
1Ck.
SinceC >1and since this inequality is valid for all integerk>1 one gets a contra-
diction. This proves thatq= 1.
4.4. Z-invariance whenVµ+contains no line. — In this section, we finish the proof of our main Theorem 1.1 when the coneVµ+ is properly convex, see (2.5).
Proposition4.10. — Keep notation (4.1), assume thatVµ+ contains no line and that the µ-harmonic function his not induced. ThenhisZ-invariant.
Beginning of proof of Proposition4.10. — The proof is by induction on the cardinality of the supportS ofµ, simultaneously for all the finite index subgroupsG ofH3(Z).
We will use the induction hypothesis inside the proof of Lemma 4.12. We begin the proof by a few reduction steps.
First step. — We can assume thathisµ-extremal. Indeed by Definitions 3.3 and 3.4, almost all theµ-extremal µ-harmonic positive functionsf that occur in the desinte- grationh=R
Efdα(f)of hare non-induced. In this case, by Proposition 4.3, there exist z =z0p 6= 0 in Z and q >0 such that hz =qh. According to Lemma 4.7, it is enough to prove that q= 1.
(a) We can assumez=z0.Because we can replacehby the functionf :=q0−1hz
0+
· · ·+q0−phzp
0, where q0 >0 is chosen so that q0p =q. This function f is µ-harmonic andZ-semiinvariant. It might not beµ-extremal, but this property will not be used in the argument below.
(b) We can assume S ∩Z = ∅. Indeed, by (a), if µZ is the restriction of µ to the center, one has Pµ
Zh= λh for a constant0 6λ < 1. But then the function h is harmonic for the measure (1−λ)−1(µ−µZ). It might not be extremal for this measure, but, as we just said, this is not important.
(c) We can assumeh(0) = 1.Because we can replacehby a multiple of a suitable translate.
(d) We can assume q <1.Because we can replace the generatorz0 by its inverse.
We are now looking for a contradiction.
Second step. — We now can enter the key argument of the proof. Since the coneVµ+ is properly convex and sinceS∩Z =∅, we can find a partition of the support of µ in two non-empty subsets
(4.4) S=S1∪S2,
such that
(4.5) cs1,s2>1 for alls1 inS1 ands2 inS2.
The partition (4.4) is given by a suitable decomposition Vµ+ = V1+∪V2+ of the properly convex cone Vµ+ in two cones V1+ and V2+ of disjoint interior so that the inequalities (4.5) will follow from the bilinear formula (2.4) for the commutators in the Heisenberg group.
We will use the decomposition µ = µ1 +µ2, where µ1 := 1S1µ and where µ2:=1S2µ. The proof again starts with the equality (2.2) which tells us that, for
