1 Introduction
In the present paper, we study purely algebraic notions of growth rate and entropy for an endomorphism of a finitely generated group.
Let π be a finitely generated group with a system S={s1,…,sn} of generators.
Let ϕ:π→π be an endomorphism.
For any γ∈π, let L(γ,S) be the length of the shortest word in the letters S∪S-1 which represents γ.
Then the growth rate of ϕ is defined [2] to be
GR(ϕ):=sup{lim supk→∞L(ϕk(γ),S)1/k∣γ∈π}.
For each k>0, we put
Lk(ϕ,S):=max{L(ϕk(si),S)∣i=1,…,n}.
It is known for example from [2, Proposition 1] that
GR(ϕ)=limk→∞Lk(ϕ,S)1/k=infk{Lk(ϕ,S)1/k},
and the algebraic entropy of ϕ is by definition halg(ϕ):=logGR(ϕ).
The growth rate and hence the algebraic entropy of ϕ are well-defined, i.e., independent of the choice of a set of generators [7, p. 114].
It is immediate from the definition that the growth rate and the algebraic entropy for an endomorphism of a group are invariants of conjugacy of group endomorphisms.
Furthermore, for any inner automorphism τγ0 by γ0, we have GR(τγ0ϕ)=GR(ϕ) and halg(τγ0ϕ)=halg(ϕ) ([7, Proposition 3.1.10]).
Consider a continuous map f on a compact connected manifold M, and consider a homomorphism ϕ induced by f of the group of covering transformations on the universal cover of M.
Then the topological entropy htop(f) is defined.
We refer to [7] for background.
Several authors, among them R. Bowen in [2] and A. Katok in [8], have proved that the topological entropy htop(f) of f is at least as large as the algebraic entropy halg(ϕ)=halg(f) of ϕ or f.
The problem of determining the growth rate of a group endomorphism, initiated by R. Bowen in [2], is now an area of active research (see detailed description in [5, 10] and references therein).
For known properties of the growth of automorphisms of free groups, we refer to [1, 12, 13].
The purpose of this paper is first to study the growth rate of an endomorphism on a finitely generated nilpotent group.
In [10, Theorem 1.2], it was proven that the growth rate of an automorphism of a finitely generated nilpotent group is equal to the growth rate of the induced automorphism on its abelianization.
Our main result is a generalization of this result of [10] from automorphisms to endomorphisms, using completely different arguments.
In Section 2, we recall some known results about the growth rate of a group endomorphism, sometimes correcting them.
In Section 3, we refine the calculation in [2] of the growth rate for an endomorphism of a finitely generated torsion-free nilpotent group and prove that the growth rate is an algebraic integer.
Let π be a finitely generated torsion-free nilpotent group, and let G be its Malcev completion.
Let ϕ be an endomorphism of π.
Then ϕ extends uniquely to a Lie group homomorphism D of G, called the Malcev completion of ϕ.
We call its differential D* the linearization of ϕ.
The main results are the following.
Theorem \ref{GR-nilp}.
Let ϕ:π→π be an endomorphism on a finitely generated torsion-free nilpotent group π.
Let G be the Malcev completion of π.
Then the linearization D*:G→G of ϕ can be expressed as a lower triangular block matrix with diagonal blocks {Dj} so that
GR(ϕ)=maxj≥1{sp(Dj)1/j}.
In particular, GR(ϕ) is an algebraic integer.
Theorem \ref{Ko type}.
Let ϕ:π→π be an endomorphism on a finitely generated torsion-free nilpotent group π with Malcev completion D.
Then
GR(ϕ)=GR(ϕab),
where ϕab:π/[π,π]→π/[π,π] be the endomorphism induced by ϕ.
Hence we have GR(ϕ)=sp(D1)≤sp(D*).
2 Preliminaries
We shall assume in this article that all groups are finitely generated unless otherwise specified.
For a given endomorphism ϕ:π→π, if π′ is a ϕ-invariant subgroup of π, we denote by ϕ′=ϕ|π′ the restriction of ϕ to π′.
If, in addition, π′ is a normal subgroup, we denote by ϕ^ the endomorphism on π/π′ induced by ϕ.
Then the following are known; see for example [2, 5].
GR(ϕ)≤max{GR(ϕ′),GR(ϕ^)}.
Let ϕ:ℤn→ℤn be an endomorphism yielding an integer matrix D.
Then we have GR(ϕ)=sp(D), the maximum of the absolute values of the eigenvalues of D.
Let S′ be a finite set of generators for π′, and let S^ be a finite set of generators for the quotient group π/π′.
Then it is possible to extend S′ to a finite set S of generators for π so that S is projected onto S^ under the projection π→π/π′.
For any γ∈π′, it is true that L(γ,S′)≥L(γ,S).
Consider the concentric balls B(n)={γ∈π∣L(γ,S)≤n} for all n>0, and the distortion function of π′ in π which is defined as
Δπ′π(n):=max{L(γ,S′)∣γ∈π′∩B(n)}.
The notion of distortion of a subgroup was first introduced by M. Gromov in [6].
We refer to [3] for our discussion.
For two functions f,g:ℕ→ℕ, we say that f≼g if there exists c>0 such that such that f(n)≤cg(cn) for all n>0.
We say that two functions are equivalent, written f≈g, if f≼g and g≼f.
The subgroup π′ of π is undistorted if Δπ′π(n)≈n.
The following facts about distortion can be found in [3].
If π′ is infinite, then it is true that n≼Δπ′π(n).
If [π:π′]<∞, then π′ is undistorted in π.
Assume Δπ′π(n)≼n.
By definition, there exists c>0 such that Δπ′π(n)≤c2n for all n>0.
For any γ∈π′, let n=L(γ,S).
Then
L(γ,S′)≤Δπ′π(n)≤c2n=c2L(γ,S).
Thus L(γ,S)≤c2L(γ,S) for all γ∈π′.
This inequality implies that, for all k>0,
Lk(ϕ′,S′)=max{L(ϕ′k(γi),S′)∣γi∈S′}≤c2max{L(ϕ′k(γi),S)∣γi∈S′}≤c2Lk(ϕ,S),
and so GR(ϕ′)≤GR(ϕ).
Consequently, we have the following lemma.
Lemma 1 ([5, Corollary 3.1]).
Let ϕ be an endomorphism of π.
If π′ is a ϕ-invariant undistorted subgroup in π, then GR(ϕ′)≤GR(ϕ); hence if, in addition, π′ is a normal subgroup of π, then GR(ϕ)=max{GR(ϕ′),GR(ϕ^)}.
Proof.
Since π′ is undistorted in π, we have from the definition that Δπ′π(n)≼n.
Now the proof follows from the above observation.
∎
Remark 2.
Note also the following.
If π′ is of finite index in π, then π′ is undistorted, and hence GR(ϕ′)≤GR(ϕ).
Example 4 shows that the inequality can be strict.
Thus [2, Proposition 1 (3)] (see also [5, Theorem 3.1]) is not correct.
If GR(ϕ)<GR(ϕ′), then π′ is distorted, and π′ is not of finite index in π.
Lemma 3.
Let ϕ be an endomorphism of π.
If GR(ϕ)<1, then GR(ϕ)=0 and ϕ is an eventually trivial endomorphism, and vice versa.
Proof.
Let ρ=GR(ϕ), and let ϵ=1-ρ>0.
Since limm→∞Lm(ϕ,S)1/m=ρ, there exists N>0 such that, for all m≥N, we have Lm(ϕ,S)1/m-ρ<ϵ;
Lm(ϕ,S)1/m<1⟹Lm(ϕ,S)<1⟹Lm(ϕ,S)=0
because Lm(ϕ,S) is a nonnegative integer.
This implies that ρ=0 and the endomorphism ϕN is trivial or ϕ is eventually trivial.
The converse is obvious.
∎
Example 4.
Let π=ℤ×ℤ2 with generators α and β such that β2=1.
Consider an endomorphism ϕ of π defined by ϕ(α)=1 and ϕ(β)=β.
Observing that
Ln(ϕ,S)=max{L(ϕn(α),S),L(ϕn(β),S)}=max{L(1,S),L(β,S)}=max{0,1}=1,
we have GR(ϕ)=1.
Similarly, we have GR(ϕ|ℤ)=0 and GR(ϕ|ℤ2)=1.
Notice further that ℤ2 is a distorted subgroup of π because Δℤ2π(n)=1 for all n.
Lemma 5.
Let ϕ be an endomorphism of π.
If π′ is a
ϕ
-invariant finite subgroup of
π
, then GR(ϕ′)≤GR(ϕ).
If, in addition, π′ is a normal subgroup of
π
, then
GR(ϕ)=max{GR(ϕ′),GR(ϕ^)},
and GR(ϕ)=GR(ϕ^) if and only if ϕ′ is eventually trivial or ϕ^ is not eventually trivial.
Proof.
If the ϕ-invariant subgroup π′ of π is finite, then we can show easily that GR(ϕ′) is either 0 or 1 by taking a system of generators S′=π′ for π′.
We will show that GR(ϕ′)≤GR(ϕ).
We may assume that GR(ϕ′)=1.
This implies that there is an element x∈π′ such that ϕ′n(x)≠1 for all n>0.
Considering any system of generators for π which contains x, we can see right away that GR(ϕ)≥1=GR(ϕ′).
Assume that π′ is normal in π.
If GR(ϕ′)=0, then clearly GR(ϕ)=GR(ϕ^).
On the other hand, if GR(ϕ′)=1, then GR(ϕ)=GR(ϕ^) if and only if GR(ϕ^)≥1 if and only if ϕ^ is not eventually trivial by Lemma 3.
∎
Remark 6.
However, the above lemma is not true when π′ is infinite; see Example 7.
Note further that if GR(ϕ)<GR(ϕ′), then π′ is infinite.
The following is a well-known example about subgroup distortion.
Example 7.
Let π be the Baumslag–Solitar group
B(1,n):=〈a,b∣a-1ba=bn〉,n>1.
Then S={a,b} is a generating set for π.
Let π′=〈b〉, and let S′={b}.
We observe that the subgroup π′ of π is distorted.
In fact, since bnk=a-kbak for all k>0, we have that L(bnk,S′)=nk and L(bnk,S)=2k+1.
If ϕ is an endomorphism of π given by ϕ(b)=bn and ϕ(a)=a, then we can see that GR(ϕ′)=n and GR(ϕ)=1.
Example 4 shows that [2, Proposition 1 (3)] is not correct in general, but it is almost true in the sense of Theorem 8.
By modifying the argument of the proof of [5, Theorem 3.1], we have the following theorem.
Theorem 8.
Let ϕ be an endomorphism of π, and let π′ be a ϕ-invariant, finite-index subgroup of π.
If ϕ′ is not an eventually trivial endomorphism, then GR(ϕ)=GR(ϕ′).
If ϕ′ is an eventually trivial endomorphism of π′, then
GR(ϕ′)=0 𝑎𝑛𝑑 GR(ϕ)=0𝑜𝑟 1.
Moreover, GR(ϕ)=0 if and only if
ϕ
is an eventually trivial endomorphism of
π.
Consequently, the equality GR(ϕ)=GR(ϕ′) holds except only for the case when ϕ′ is eventually trivial and ϕ is not eventually trivial.
If this is the case, then GR(ϕ′)=0 and GR(ϕ)=1.
Proof.
Let S′={γ1,…,γt} be a set of generators of π′.
Let u=[π:π′].
Then we have π=δ1π′∪⋯∪δuπ′ so that S={γ1,…,γt,δ1,…,δu} generates π.
For any j=1,…,u, there exists a unique kj such that ϕ(δj)∈δkjπ′.
We denote
p=max1≤j≤u{L(wj,S′)∣ϕ(δj)=δkjwj∈δkjπ′}.
Assume p=0.
Then ϕ(δj)=δkj for all j=1,…,u.
For each j=1,…,u, we write ϕm(δj)=δjm.
Hence it follows that L(ϕm(δj),S)=0 or 1 according to whether δjm=1 or δjm≠1.
Suppose that there is N>0 such that ϕN(δj)=1 for all j=1,…,u and hence L(ϕm(δj),S)=0 for all m≥N.
Since π′ is undistorted in π, there exists some c>0 such that
L(γ,S′)≤c2⋅L(γ,S) for allγ∈π′.
It is clear that
L(γ,S)≤L(γ,S′) for allγ∈π′.
Thus
Lm(ϕ′,S′)≤c2⋅Lm(ϕ,S),
Lm(ϕ,S)=max{L(ϕm(γi),S)}≤Lm(ϕ′,S′).
This implies that GR(ϕ′)=GR(ϕ).
Suppose on the contrary, for any m>0, there is some j such that ϕm(δj)≠1.
Then max{L(ϕm(δj),S)}=1.
Hence
Lm(ϕ,S)=max{L(ϕm(γi),S),L(ϕm(δj),S)}=max{L(ϕm(γi),S),1}≤max{Lm(ϕ′,S′),1}.
This implies that GR(ϕ′)≤GR(ϕ)≤max{GR(ϕ′),1}.
Since ϕ′ is not eventually trivial, Lemma 3 implies that GR(ϕ′)≥1, and hence GR(ϕ)=GR(ϕ′).
Next we assume that p≥1.
For each j=1,…,u, we write ϕ(δj)=δj1w1 for some j1 and w1∈π′.
Then
ϕm(δj)=δjmwmϕ(wm-1)⋯ϕm-1(w1),
and thus
L(ϕm(δj),S)≤1+p+pL1(ϕ′,S′)+pL2(ϕ′,S′)+⋯+pLm-1(ϕ′,S′).
Let L=GR(ϕ′).
By the assumption of our proposition, L≥1.
Let ϵ>0 be given.
Since
limm→∞Lm(ϕ′,S′)1/m=L,
there is some N>0 such that if m>N, then Lm(ϕ′,S′)<(L+ϵ)m.
Choose q1,…,qN>0 such that Li(ϕ′,S′)<qi(L+ϵ)i for i=1,…,N.
Put
q=max{q1,…,qN,1}≥1.
Then Lm(ϕ′,S′)<q(L+ϵ)m for all m≥1.
Hence we have
L(ϕm(δj),S)≤1+p+pq(L+ϵ)+pq(L+ϵ)2+⋯+pq(L+ϵ)m-1≤1+pq(L+ϵ)m-1(L+ϵ)-1.
Since pq≠0, this implies that
limm→∞maxj{L(ϕm(δj),S)}m≤L+ϵ.
Since π′ is undistorted in π, there exists some c>0 such that
Lm(ϕ′,S′)=maxi{L(ϕ′m(γi),S′)}≤c2⋅maxi,j{L(ϕ′m(γi),S),L(ϕm(δj),S)}=c2Lm(ϕ,S),
and hence we obtain
L=GR(ϕ′)≤GR(ϕ)=limm→∞Lm(ϕ,S)m≤L+ϵ
for all ϵ>0.
Consequently, GR(ϕ)=GR(ϕ′).
Suppose that ϕ′ is an eventually trivial endomorphism of π′.
Then it is clear that GR(ϕ′)=0.
Consider a set S={γ1,…,γt,δ1,…,δu} of generators for π as above.
For any m>0, we observe that ϕm(γi)=1 and ϕm(δj)=δjmwm for some jm∈{1,…,u} and wm in a finite subset of π′.
This implies that the sequence {Lm(ϕ,S)} is bounded.
Because Lm(ϕ,S)=0 or ≥1, it follows that GR(ϕ)=0 or 1 respectively.
When GR(ϕ)=0, Lemma 3 says that ϕ is an eventually trivial endomorphism.
Next we consider the case when GR(ϕ)=1.
From the definition, we can choose N>0 so that, for m≥N,
we have 1/2m<Lm(ϕ,S), which implies that Lm(ϕ,S)≥1 because Lm(ϕ,S) is an integer.
Therefore, for each m≥N, we can choose γ∈S such that ϕm(γ)≠1.
This shows that ϕ is not eventually trivial even though ϕ′ is eventually trivial.
∎
Before leaving this section, we observe the following elementary fact.
Proposition 9.
Let ϕ be an endomorphism of π with a finite set S of generators.
Let
GRi(ϕ)=limk→∞L(ϕk(si),S)1/k
for each si∈S.
Then GR(ϕ)=max{GRi(ϕ)∣si∈S}.
Proof.
Since L(ϕk(si),S)≤Lk(ϕ,S), it follows that GRi(ϕ)≤GR(ϕ).
Assume GRi(ϕ)<GR(ϕ) for all si∈S.
Thus there exists K>0 such that if k≥K and si∈S, then L(ϕk(si),S)1/k<GR(ϕ).
Because S is finite, it follows that Lk(ϕ,S)1/k<GR(ϕ) for all k≥K.
However, since
limk→∞Lk(ϕ,S)1/k=limk≥KLk(ϕ,S)1/k=infk≥KLk(ϕ,S)1/k,
we obtain a contradiction: GR(ϕ)=infk≥KLk(ϕ,S)1/k<GR(ϕ).
∎
3 Finitely generated nilpotent groups
Consider the lower central series of a finitely generated group π,
π=π1⊃π2⊃⋯,
where πj=[π,πj-1] is the j-fold commutator subgroup γj(π) of π.
The endomorphism ϕ:π→π induces endomorphisms
ϕj:πj→πj,ϕ^j:π/πj→π/πj,ϕ¯j:πj/πj+1→πj/πj+1.
Then it is known from [2] that GR(ϕ)≥GR(ϕ¯j)1/j for all j≥1.
The group π is called nilpotent if πj=1 for some j.
When πc≠1 but πc+1=1, we say that it is c-step.
Lemma 1 ([2, Proposition 2]).
If π is c-step nilpotent, then
GR(ϕ)=max{GR(ϕ^c),GR(ϕc)1/c}.
If π is nilpotent, then
GR(ϕ)=maxj≥1{GR(ϕ¯j)1/j}.
Recall, for example from [10, Proposition 3.1], that a finitely generated nilpotent group π is virtually torsion-free.
Thus there exists a finite-index, torsion-free, normal subgroup Γ of π.
Following the proof of [11, Lemma 3.1], we can see that there exists a fully invariant subgroup Λ⊂Γ of π which is of finite index.
Therefore, any endomorphism ϕ:π→π restricts to an endomorphism ϕ′:Λ→Λ.
By Theorem 8, we may consider only the case when ϕ′ is not eventually trivial, and hence we may assume that GR(ϕ)=GR(ϕ′).
Consequently, for the computation of GR(ϕ), we may assume that π is a finitely generated torsion-free nilpotent group.
Consider the lower central series of a finitely generated torsion-free c-step nilpotent group π,
π=π1,πj+1=[π,πj],πc≠1 and πc+1=1.
For each j=1,…,c, we consider the isolator of πj in π,
πj=πjπ:={x∈π∣xk∈πjfor somek≥1}.
Then it is known that πj is a characteristic subgroup of π with [πj:πj] finite.
Furthermore, πj/πj is precisely the set of all torsion elements in the nilpotent group π/πj, and πj/πj+1≅ℤkj for some integer kj>0.
Hence we obtain the adapted central series [4, p. 3]
π=π1⊃π2⊃⋯⊃πc⊃πc+1=1.
The following lemma plays a crucial role in our study of growth rates for endomorphisms of finitely generated nilpotent groups.
Lemma 2 ([15, Lemma 3.7]).
Let π be a finitely generated c-step nilpotent group with lower central series π=π1⊃π2⊃⋯⊃πc⊃πc+1=1.
Then there are finite sets Tj={τj1,…,τjkj}⊂πj such that
if pj:πj→πj/πj+1 denotes the projection, then pj(Tj) is an independent set of generators for the finitely generated abelian group πj/πj+1,
if j>1, then every τjr is of the form [τ1i,τj-1,ℓ],
Let G be the Malcev completion of a finitely generated torsion-free nilpotent group π, and let ϕ be an endomorphism of π.
Then ϕ extends uniquely to a Lie group homomorphism D of G, called the Malcev completion of ϕ.
We call its differential D* the linearization of ϕ.
Theorem 3.
Let ϕ:π→π be an endomorphism on a finitely generated torsion-free nilpotent group π.
Let G be the Malcev completion of π.
Then the linearization D*:G→G of ϕ can be expressed as a lower triangular block matrix with diagonal blocks {Dj} so that
GR(ϕ)=maxj≥1{sp(Dj)1/j}.
In particular, GR(ϕ) is an algebraic integer.
Proof.
Let π be a finitely generated torsion-free c-step nilpotent group with the adapted central series
π=π1⊃π2⊃⋯⊃πc⊃πc+1=1.
Let qj:πj→πj/πj+1 denote the projection.
We choose {T1,…,Tc} as in Lemma 2.
Since π2 is a fully invariant, finite-index subgroup of π2, it induces a short exact sequence
1→π2/π2→π1/π2→π1/π2=π1/π2→1.
Since π2/π2 is finite, it follows that π1/π2≅ℤk1 can be regarded as the free part of the finitely generated abelian group π1/π2.
Hence we can choose S1⊂T1 such that p1(S1) is an independent set of free generators of π1/π2 and p1(T1-S1) is an independent set of torsion generators of π1/π2.
Next we consider the short exact sequence
1→π3/π3→π2/π3→π2/π3→1.
Since π2/π3⊂π2/π3, we obtain the following commutative diagram between exact sequences:
1→π3/π3→π2/π3→π2/π3≅ℤk2→1↑↑↑1→(π2∩π3)/π3→π2/π3→π2/(π2∩π3)=π2⋅π3/π3→1.
where all vertical maps are inclusions of finite index.
So we can choose S2⊂T2 such that p2(S2) is an independent set of free generators of the free abelian group (π2⋅π3)/π3 and p2(T2-S2) is an independent set of torsion generators of π2/π3.
Note that S2⊂π2⊂π2.
Because the right-most vertical inclusion is of finite index, we can choose 𝕊2⊂π2 such that q2(𝕊2) is an independent set of free generators of π2/π3, and for each σ2∈𝕊2, there are unique ℓ2≥1 and unique τ2*∈S2 such that σ2ℓ2=τ2* modulo π3.
We remark also that #S2=#𝕊2.
Continuing in this way, we obtain {S1,…,Sc}⊂{T1,…,Tc} such that
pj(Sj) is an independent set of free generators of πj/πj+1,
pj(Tj-Sj) is an independent set of torsion generators of πj/πj+1,
qj(𝕊j) is an independent set of free generators of πj/πj+1,
for each σj∈𝕊j⊂πj, there exist unique ℓj≥1 and τj*∈Sj such that
σjℓj=τj*modπj+1.
The adapted central series of π allows us to choose a preferred basis𝐚 of π; we can choose 𝐚 to be {𝕊1,…,𝕊c} so that it generates π and π can be embedded as a lattice of a connected, simply connected nilpotent Lie group G, the Malcev completion of π.
Its Lie algebra 𝔊 has a linear basis log𝐚={log𝕊1,…,log𝕊c}.
From σjℓj=τj*modπj+1, we have
(B)ℓjlog(σj)=log(σjℓj)=log(τj*)modγj+1(𝔊).
This implies that {logS1,…,logSc} is also a linear basis of 𝔊.
Let ϕ:π→π be an endomorphism.
Then ϕ induces endomorphisms
ϕj:πj→πj,ϕ^j:π/πj→π/πj,ϕ¯j:πj/πj+1→πj/πj+1
and
φj:πj→πj,φ^j:π/πj→π/πj,φ¯j:πj/πj+1→πj/πj+1.
Moreover, any endomorphism ϕ on π extends uniquely to a Lie group endomorphism D on G, the Malcev completion of ϕ.
With respect to the preferred basis log𝐚 of the Lie algebra 𝔊 of G, we can express the linearization D* of ϕ as a lower triangular block matrix; each diagonal block Dj is an integer matrix representing the endomorphism φ¯j:πj/πj+1≅ℤkj→πj/πj+1≅ℤkj.
For details, we refer to [9] for example.
When the new basis {logS1,…,logSc} is used instead of log𝐚, the integer entries of block matrices Dj will be changed to rational entries because of identities (B), but the eigenvalues of Dj will be unchanged.
This means that, whenever the eigenvalues of D* are concerned, we may assume that πj/πj+1 is torsion-free, or πj=πj.
Consequently, we may assume that GR(ϕ¯j)=GR(φ¯j).
Since πj/πj+1≅ℤkj, by taking the tensor product with ℝ, it is known that GR(φ¯j)=sp(Dj).
Thus GR(ϕ¯j)=sp(Dj).
Now the theorem follows from Lemma 1.
∎
Remark 4.
From Theorem 3, it follows that the growth rate of any endomorphism on a finitely generated torsion-free nilpotent group is an algebraic integer.
The question of determining groups for which the growth rate of a group endomorphism is an algebraic number was raised by R. Bowen in [2, p. 27].
Example 5.
Let Nil be the 3-dimensional Heisenberg group.
That is,
Nil={[1xz01y001]|x,y,z∈ℝ}.
Consider the subgroups Γk, k∈ℕ, of Nil,
Γk={[1nℓk01m001]|m,n,ℓ∈ℤ}.
These are lattices of Nil, and every lattice of Nil is isomorphic to some Γk.
Let
a1=[100011001],a2=[110010001],a3=[101k010001].
Then S={a1,a2,a3} is a generating set of Γk satisfying
[a1,a2]=a3-k,[a1,a3]=[a2,a3]=1,
and in fact,
[1nℓk01m001]=a1ma2na3ℓ.
Let
π=Γk=〈a1,a2,a3∣[a1,a2]=a3-k,[a1,a3]=[a2,a3]=1〉.
Let π′=〈a3〉 and S′={a3}.
Since (a3-k)n2=[a1n,a2n], we have
L((a3-k)n2,S′)=kn2 and L((a3-k)n2,S)=4n.
Hence
L((a3-k)n2,S′)>L((a3-k)n2,S)
for all n with n>4/k.
It follows that π′ is distorted.
Consider any endomorphism ϕ:π→π.
Then ϕ must be of the form
ϕ(a1)=a1m11a2m21a3p,ϕ(a2)=a1m12a2m22a3q,ϕ(a3)=a3m11m22-m12m21.
We will compute GR(ϕ).
The lower central series of π is π=π1⊃π2=〈a3k〉, and its adapted central series is π=π1⊃π2=〈a3〉.
We observe that
T1={a1,a2,a3} and T2={a3k}
are sets satisfying the conditions of Lemma 2.
Then we can see that
S1={a1,a2}⊂T1,S2={a3k}⊂T2 and 𝕊1={a1,a2},𝕊2={a3}.
Furthermore, {𝕊1,𝕊2}={a1,a2,a3} is a preferred basis for π.
The linearization of ϕ with respect to this preferred basis has two integer blocks D1 and D2, where
D1=[m11m12m21m22],D2=[m11m22-m12m21]=[det(D1)].
By Theorem 3, we have GR(ϕ)=max{sp(D1),sp(D2)1/2}.
Let μ,ν be the eigenvalues of D1.
Then
GR(ϕ)=max{|μ|,|ν|,|μν|}=max{|μ|,|ν|}=sp(D1).
In fact, we show in Theorem 7 that it is always the case that GR(ϕ)=sp(D1).
We consider another example in which we obtain much information about linearizations of endomorphisms, and then we obtain an idea of proving the next result, Theorem 7.
Example 6.
Consider a 2-step torsion-free nilpotent group π generated by
τ1,τ2,τ3,σ12,σ13
satisfying the relations
[τ1,τ2]=σ12,[τ1,τ3]=σ13,[τ2,τ3]=σ12mσ13n,[τi,σjk]=[σ12,σ13]=1.
Since π2=〈σ12,σ13〉≅ℤ2 and π/π2=〈τ¯1,τ¯2,τ¯3〉≅ℤ3, it follows that the set {T1,T2}={τ1,τ2,τ3,σ12,σ13} satisfies the conditions of Lemma 2 and forms a preferred basis of our group π.
Let ϕ be an endomorphism of π.
A direct computation shows that if
ϕ(τi)=τ1d1iτ2d2iτ3d3imodπ2,
i.e., if the first block of the linearization of ϕ is
D1=[d11d12d13d21d22d23d31d32d33],
then, with σ23=[τ2,τ3], we have
ϕ(σ12)=σ12M33σ13M23σ23M13,
ϕ(σ13)=σ12M32σ13M22σ23M12,
ϕ(σ23)=σ12M31σ13M21σ23M11,
where Mij denote the (i,j)-minor of D.
These yield a matrix
K=[M33M32M31M23M22M21M13M12M11]=⋀2(D1),
the second exterior power of D1.
On the other hand, since σ23=σ12mσ13n, we have
(3.1)ϕ(σ12)=σ12M33σ13M23σ23M13=σ12M33+mM13σ13M23+nM13,
(3.2)ϕ(σ13)=σ12M32σ13M22σ23M12=σ12M32+mM12σ13M22+nM12,
(3.3)ϕ(σ23)=ϕ(σ12)mϕ(σ13)n=σ12M31σ13M21σ23M11=σ12M31+mM11σ13M21+nM11.
From (3.1) and (3.2), the second block of the linearization of ϕ is
D2=[M33+mM13M32+mM12M23+nM13M22+nM12].
Plugging (3.1) and (3.2) into (3.3), we have
(3.4){[M31M21M11]=m[M33M23M13]+n[M32M22M12]when(m,n)≠(0,0),[M31M21]=[00]whenm=n=0.
When (m,n)≠(0,0), because of (3.4), K is column equivalent to the matrix K′ with the zero third column, and then, by doing some row operations on K′, we can see that K′ is row equivalent to the matrix K′′, where
Thus the second block D2 of the linearization D* is a block submatrix of K′′.
This is obtained by removing the row and column of K′′ that are determined by (3.3) or by the relation [τ2,τ3]=σ12mσ13n.
Note also that K, K′ and K′′ have the same eigenvalues which are 0 and the eigenvalues of D2.
When (m,n)=(0,0), because of (3.4), we have
Thus D2 of D* is a block submatrix of K, and K has M11 and the eigenvalues of D2 as its eigenvalues.
On the other hand, if μ1,μ2,μ3 are the eigenvalues of D1, as K=⋀2(D1), the eigenvalues of K are μiμj (i<j).
Consequently, we have
sp(D1)=maxi=1,2,3{|μi|}≥maxi≠j{|μiμj|}=sp(K)≥sp(D2)1/2.
This proves that GR(ϕ)=sp(D1).
The following result was proved in [10] when ϕ is an automorphism using the intrinsic polynomial structure of nilpotent groups.
We will now improve [10, Theorem 1.2] from automorphisms to endomorphisms by using completely different arguments.
Theorem 7.
Let ϕ:π→π be an endomorphism on a finitely generated torsion-free nilpotent group π with Malcev completion D.
Then
GR(ϕ)=GR(ϕab),
where ϕab:π/[π,π]→π/[π,π] be the endomorphism induced by ϕ.
Hence we have GR(ϕ)=sp(D1)≤sp(D*).
Proof.
Let π be c-step and choose a family of finite sets {T1,…,Tc} satisfying the conditions of Lemma 2.
As was observed in the proof of Theorem 3, we can choose {S1,…,Sc} such that each Sj⊂Tj⊂πj projects onto free generators of πj/πj+1 and a preferred basis {𝕊1,…,𝕊c} of π so that each block matrix Dj of the linearization D* of ϕ which is determined by log𝕊j may be assumed to be determined by logSj.
Indeed, for each j with 1≤j≤c, we write Sj={τj1,…,τjkj}⊂Tj; then, if j>1, every τjr is of the form [τ1i,τj-1,ℓ].
For 1≤j≤c, if
ϕ(τjℓ)=τj1d1ℓj⋯τjkjdkjℓjmoduloπj+1,
then the jth block of the linearization D* of ϕ is
Dj=[d11j⋯d1kjj⋮⋮dk11j⋯dkjkjj].
In order to compare first the eigenvalues of D1 with those of D2, we use the following new notation: D1=[dij1]=[dij], σij=[τ1i,τ1j] for all 1≤i<j≤k1.
Then σij=τ2,ℓ±1∈S2 for some ℓ, or σij is a word of elements in S2±1 modulo π3 (see the presentation of π in Example 6).
Let S={σij∣1≤i<j≤k1}; then we may assume that S2⊂S.
Further, S2 differs from S except possibly by σij’s, words of elements in S2±1 modulo π3 (note in Example 6 that S2={σ12,σ13} and S={σ12,σ13,σ12mσ13n}).
Now we can express ϕ(σij) as follows:
(P)ϕ(σij)=σ12M1,2i,jσ13M1,3i,j⋯σ1k1M1,k1i,j⋯σk1-1,k1Mk1-1,k1i,jmoduloπ3
for some integers Mp,qi,j.
We denote by K the (k12)×(k12) matrix [Mp,qi,j],
K=[M1,21,2M1,21,3⋯M1,2k1-1,k1M1,31,2M1,31,3⋯M1,3k1-1,k1⋮⋮⋮Mk1-1,k11,2Mk1-1,k11,3⋯Mk1-1,k1k1-1,k1].
We will refer to the column vector (M1,2i,j,M1,3i,j,…,M1,k1i,j,…,Mk1-1,k1i,j)t of K as the (i,j)-column of K.
Note the following.
For any σij∈S, Mp,qi,j is unique for which σpq∈S2.
If σij∈S-S2, then σij is a word w of elements in S2±1 modulo π3.
If w≠1 modulo π3, the (i,j)-column of K is an integer combination of (p,q)-columns of K corresponding to the elements σpq appearing in the word w.
If w≡1, then Mp,qi,j=0 for which σpq∈S2.
The right-hand side of the expression (P) can be rewritten in terms of only the elements of S2 using the words σij≡w(σpq).
This yields the second block D2.
Since σij=[τ1i,τ1j], taking ϕ on both sides, we have (see [5, Lemma 4.1] or [14, p. 93, Lemma 4.1])
σ12M1,2i,jσ13M1,3i,j⋯σ1k1M1,k1i,j⋯σk1-1,k1Mk1-1,k1i,j=[τ11d1,i⋯τ1k1dk1,i,τ11d1,j⋯τ1k1dk1,j]=∏p∏q[τ1pdp,i,τ1qdq,j]=∏p∏q[τ1p,τ1q]dp,idq,j=∏1≤p<q≤k1σpqdp,idq,j-dp,jdq,imoduloπ3.
This shows that K is the second exterior power of D1, i.e., K=⋀2(D1).
Hence, if μi (1≤i≤k1) are the eigenvalues of the matrix D1, then μiμj(i<j) are the eigenvalues of K.
From part (ii) of the above remarks, we see that K is column equivalent to the matrix K′ with zero (i,j)-column for which σij=w(σpq)≠1 modulo π3.
We rearrange the elements of S so that S=S2∪(S-S2)=S2∪S21∪S22, where S21={σij∈S-S2∣σij=1} and S22={σij∈S-S2∣σij≠1}.
Rearranging S to S2∪(S-S2), we have
The effect of part (iii) on K and hence on K′ is doing some row operations using the (i,j)-rows in the last block of K′ for which σij=w(σpq)≠1 modulo π3.
By rearranging S further to S2∪S21∪S22, we have
The middle block column is determined by the fact that if σij≡w≡1, then Mp,qi,j=0 for which σpq∈S2.
Consequently, the second block D2 of D* is a block submatrix of K′′ which is obtained by removing the rows and columns associated to S-S2.
Note also that K,K′ and K′′ have the same eigenvalues which contain the eigenvalues of D2.
This observation shows that
sp(D1)=max{|μi|}≥max{μiμj}=sp(K)1/2≥sp(D2)1/2.
For the next inductive step, we recall that every element of S3(⊂T3) is of the form [τ1ℓ,σij], where i<j.
Taking ϕ, we have
ϕ([τ1ℓ,σij])=[∏rτ1rdr,ℓ,∏1≤p<q≤k1σpqdp,idq,j-dp,jdq,i]=∏r∏1≤p<q≤k1[τ1r,σpq]dr,ℓ(dp,idq,j-dp,jdq,i)moduloπ4.
This expression is unique except possibly for the exponents of the elements
[τ1r,σpq]=1moduloπ4.
This produces the matrix K=⊗D1⋀2D1.
First if [τ1r,σpq]=w(S3)≠1 modulo π4, by doing some column operations and then by doing some row operations, we obtain a matrix K′′, which can be regarded as a lower triangular block matrix.
Finally, we remove the columns and rows from K′′ which are associated with the elements [τ1r,σpq]=w(S3) modulo π4.
This gives rise to the third block D3 of D*.
Hence sp(D3)≤sp(D1)3.
Continuing in this way, we may assume that the jth block Dj of D* is obtained from (⊗j-2D1)⊗⋀2D1 so that
sp(D1)≥sp(Dj)1/j.
Consequently, GR(ϕ)=max{sp(Dj)1/j}=sp(D1)=GR(ϕab)≤sp(D*).
∎
