Embeddings of Wreath Product of Abelian Groups

The class of examples we investigate in this section was introduced by Baumslag in 1973 [4]. LetAbe a free abelian group of finite rank freely generated by {a1,a₂, . . . ,a_r}. Furthermore letT be a finitely generated abelian group with basis {t1,t2, . . . ,tk, . . . ,t_l}, wheret₁, . . . ,t_k are of infinite order andt_k+1, . . . ,t_l are respectively of finite order m_k+1, . . . ,m_l. Finally letF={f₁,f₂, . . . ,f_k}be a set of element f_ifromZT, where each f_iis of the form

f_i=1+c_i,1t_i+c_i,2t_i²+· · ·+c_i,di−1t_i^di−1+t_i^di,d_i>1,c_i,j∈Z.

Now let us define a groupW_F corresponds toF. The generating set is the following

X={a1,a₂, . . . ,a_r,t1,t₂, . . . ,t_l,u₁, . . . ,u_k},

wherer,k,lare the same integers as above.

The defining relations ofW_F are of four kinds. First we have the power relations

t_i^mi =1,i=k+1, . . .,l.

Next we have the commutativity relations

























[ui,u_j] =1, 16i,j6k;

[ti,t_j] =1, 16i,j6l;

[ti,u_j] =1, 16i6l,16j6k;

[ai,a_j] =1, 16i,j6r.

Thirdly we have the commutativity relations for the conjugates of the generatorsa_i:

[a^u_i,a^w_j] =1,16i,j6r,

whereu,w∈ {t₁^α1t₂^α2. . .t_l^αl |06αi6d_ifori=1, . . . ,k,06αi<m_ifori=k+1, . . . ,l}.Finally we have relations defining the action ofu_jona_i:

a^u_i^j=a_i^fj,16i6r,16 j6k.

It is not hard to show thatWFis metabelian [4]. Moreover, Baumslag showed the following:

Proposition VII.2.1(Baumslag [4]). Given a free abelian groupAof finite rank and a finite generated abelian group T, there existsFsuch thatA≀T֒→W_F.

In particular, ifr=k=land we let f_i=1+tifor alli,W_Fcontains a copy of the free metabelian group of rankr.

We claim that

Proposition VII.2.2. Ifk>0,WFhas exponential Dehn function.

Note that wheni=j=k=1, f₁=1+t₁,W_F is the Baumslag groupΓ=ha,s,t|[a,a^t] =1,[s,t] =1,a^s=aa^ti.

The exponential Dehn function of this special case is proved in [21].

We need a few lemmas before we prove Proposition VII.2.2. First, let us denote the abelian groups generated by {t1,t₂, . . . .tl}and{u1,u2, . . . ,u_k}byT andUrespectively.

Lemma VII.2.3. Let M be a free(U×T)-module with basis e1, . . . ,er. Let S be the submodule of M generated by {(ui−f_i)ej|16i6k,16j6r}. If h=h₁e₁+h₂e₂+· · ·+h_re_r∈S such that h_i∈ZT for all i. Then h=0.

Proof. Ifk=1, thenh∈Smeans there existsα1,α2, . . . ,αr∈Z(U×T)such that

h=α1(u1−f₁)e1+α2(u1−f₁)e2+· · ·+αr(u1−f₁)er.

Sinceh=h₁e₁+h₂e₂+· · ·+h_re_r, thenh_i=αi(u1−f₁). Note thath_i∈ZT. It follows thatαi(u1−f₁)does not have

any term involvesu₁. Supposeαi6=0 for somei. Because f₁∈ZT, deg_u1(αiu₁)>deg_u1(αif₁). Thusαi(u1−f₁)has at least one term containsu₁, that leads to a contradiction.

If the statement ofk=nhas been proved, fork=n+1, we have

h=

r

∑

i=1 n+1

∑

j=1

αi,j(uj−f_j)ei.

We choose an integerNlarge enough such thatu^N₁α_i,jdoes not have any negative power ofu1for alli,j. Then

u^N₁h=

r

∑

i=1 n+1

∑

j=1

f₁^Nα_i,j(uj−f_j)ei=:

r

∑

i=1 n+1

∑

j=1

β_i,j(uj−f_j)ei,

whereβ_i,j=u^N₁α_i,j. We regardβ_i,j(u1)as a polynomial ofu₁. Replacingu₁by f₁, we have

f₁^Nh=

r

∑

i=1 n+1

∑

j=2

βi,j(f_i)(uj−f_j)ei.

Note that f₁^Nhi∈ZT fori=1, . . . ,r, then by the inductive assumption, f₁^Nhi=0 for alli. Since f₁=1+c_1,1t₁+ c_1,2t₁²+· · ·+c_i,d1−1t₁^d1−1+t₁^d1andt₁has infinite order, then f₁is not a zero divisor inZ(U×T). Thush_i=0 for alli.

Thereforeh=0. The induction finishes the proof.

It follows that ifa^h₁¹a^h₂². . .a^h_r^r=WF 1 such thath_i∈ZT for alli, thenh_i=0 as an element inZ(U×T)for everyi.

To convert it to 1, we only need those metabelian relations to commute all the conjugates ofa_i’s. By Theorem VI.1.1, it will cost at most exponentially many relations with respect to the length of the word to kill the word.

Next, let w=WF 1 and consider the minimal van Kampen diagram∆overW_F. There are two types of relations containui: (1) commutative relations[ui,uj] =1,[ui,ts] =1,j6=i,16s6l; (2) action relationsa^u_jⁱ =a^f_jⁱ,16 j6r.

Those cells, in the van Kampen diagram, form aui-band.

Figure VII.1: an example of au₁-bands

We have some properties foru_i-bands in a van Kampen diagram overW_F.

Lemma VII.2.4. (i) The top (or bottom) path of a u_i-band is a word w that all t_s,uj for s,j6=i are in the same orientation, i.e. the exponents of each letter t_s,u_j’s are either all 1 or all−1. In particular,

w=WF a^h₁¹a^h₂². . .a^h_r^rt₁^α1. . .t_l^αlu^β₁¹. . .u^β_k^k,

where h_i∈Z(U×T),sgn(αi) =sgn(βj)for all i,j, andαs(orβj) is equal to the number of times of t_s(resp. u_j) appears in w for s,j6=i.

(ii) u_i-bands do not intersect each other. In particular, a u_i-band does not self-intersect.

(iii) If i6=j, a u_i-band intersects a u_j-band at most one time.

Proof. (i) By the definition of au_i, all letterst_s,uj,s,j6=iof the top (or bottom) path must share the same direction.

The second half of the statement can be proved basically the same way as we did for the ordered form (See??).

(ii) Because there is nou_ion the top or the bottom path of au_i-band, twou_i-bands cannot intersect each other.

(iii) Ifi6=jand au_i-band intersects au_j-band. Since the van Kampen diagram is a planer graph, by comparing the orientation, it is impossible for au_i-band to intersect au_j-band twice (or more).

Last, we have

Lemma VII.2.5. Let f(t) =t^d+c_d−1t^d−1+· · ·+c₁t+1∈Z[t],c_i∈Z,d>0. Then there existsα >1such that

|(f(t))ⁿ|>αⁿfor all n.

Proof. We denote that(f(t))ⁿ=∑^ndi=0c_n,itⁱ.

Consider the corresponding holomorphic functiong(z) =z^d+c_1,d−1z^1,d−1+· · ·+c_1,1z+1. If∃z0,|z0|=1 such that|g(z0)|>1, we have

|g(z0)|ⁿ=|(g(z0))ⁿ|=|

nd

∑

i=0

c_n,izⁱ₀|6

nd

∑

i=0

|cn,i|=|fⁿ|.

Then we are done.

Now suppose|g(z)|61 for all|z|=1. Then by Cauchy’s integral formula we have

1=g(0) = 1 2πi

Z

|z|=1

g(z) z dz.

Take modulus on both sides:

1=| 1 2πi

Z

|z|=1

g(z)

z dz|6 1 2π

Z 2π

θ=0|g(e^iθ)|dθ61.

Therefore|g(z)|=1 for|z|=1 almost everywhere. Letz=e^iθ. We have

g(e^iθ) = (

d

∑

j=0

c_1,jcos(iθ)) +i(

d

∑

j=0

c_1,jsin(jθ)).

Then

(

d j=0

∑

c_1,jcos(iθ))²+ (

d

∑

j=0

c_1,jsin(jθ))²=

d h=0

∑

c²_1,j+2

∑

j<k

c_1,jc_1,kcos((k−j)θ) =1,

holds for allθ. But cos((k−j)θ)is a polynomial with respect to cosθ, i.e. cos((k−j)θ) =T_k−j(cosθ), whereT_m(x) is them-th Chebyshev polynomial. The leading term ofTm(x)is 2^m−1x^m. Thus

d h=0

∑

c²_1,j+2

∑

j<k

c_1,jc_1,kT_k−j(cosθ) =1,∀θ.

Note the leading term of the left-hand side is 2^d−1cos^dθ. That leads a contradiction since the equation above has at mostdsolutions for cosθ.

Proof of Proposition VII.2.2. First, we show that the lower bound is exponential. Consider the wordw= [a^u₁ⁿ¹,a₁].w is of length 2n+4 andw=WF 1. Let∆be a minimal Van-Kampen diagram with boundary labelw. By comparing the orientation,u₁-bands starting at the top left of∆will end at either bottom left or top right. By Lemma VII.2.4, u₁-bands do not intersect each other, then we can suppose at least half of theu₁-bands starting at the top left end at the top right. See in Figure 4, the shaded areas areu1-bands.

Figure VII.2:u₁-bands in∆

We first claim that there are no cells containingts,uj,s,j>1 on eachu₁-band. We denote the top and bottom path of thei-thu₁-band from the top byγ_i^topandγ_i^bot, wherei=1,2, . . . ,m,m>ⁿ₂. Assuming au_i-band intersect one of the u1-band, again by Lemma VII.2.4 (ii), (iii), it can neither intersect au1-band twice nor intersect itself. Thus it has to end all the way to the boundary of∆. A contradiction.

Then if there exists a cell containingt_sfors>1 in the top mostu₁-band, then by Lemma VII.2.4 (i),γ₁^topis a word a^h₁¹a^h₂². . .a^h_r^rt₁^α1. . .t_l^αl. Thusγ₁^topanda₁form a cycleγ. We have

a^h₁¹⁺¹a^h₂². . .a^h_r^rt₁^α1. . .t_l^αl=1,αi6=0.

It leads to a contradiction since the image of the left hand side inU×T is not trivial. And by definition of au₁-band, ifγ_i^topdoes not have anyts,s>1, neither doesγ_i^bot. Next consider two consecutiveu₁-bands. Ifγ_i^botdoes not havets, then by the same argument, neither doesγ_i+1^top. Therefore the claim is true.

Denote the words ofγ_i^topandγ_i^botbyw^top_i andw^bot_i respectively. Such words only consist ofa_i’s andt₁. Note that w^bot_i =w^top_i+1fori=1, . . . ,m−1. Sincew^bot₁ =a⁻₁^f1, by the same discussion above,w^top_i =a^−f

i−1 1

1 ,w^bot_i =a^−f

i 1

1 (See in Figure 4). Next we focus on the number ofa₁in eachw^bot_i , which is at least|f₁ⁱ|. By Lemma VII.2.5, there exists α>1 such that|f₁ⁱ|>αⁱ. Therefore, the number ofa₁inw^top_m is at leastα^m−1. Sincem>ⁿ₂, the number of cells in the m-thu₁-band is at leastα^[n2]. Thus the area of[a^u₁ⁿ¹,a1]is at leastα^[n2]. It follows that the lower bound is exponential.

For the upper bound, as Theorem VI.1.1 suggests, all we need is to consider how to solve the membership problem of the submoduleSwhereSis generated by{(ui−f_i)ej|16i6k,16j6r}. Supposew=1 with|w|6n, thewhas a ordered form as

w=W_F a^g₁¹a^g₂². . .a^g_r^r,gi∈Z(U×T).

And the cost of convertingwto its ordered form is exponential with respect tonas we showed in Section V.4. Also note that deg(gi),|gi|6nfor alli. WLOG, we assume that the all exponents ofu_i’s are positive. The corresponding module element ofwis

g₁e₁+g₂e₂+· · ·+g_re_r.

For each termt₁^α1t₂^α2. . .t_l^αlu^β₁¹u^β₂². . .u^β_k^k,αi∈Z,βi>0, we replaceu_ibyu_i−f_i+f_i. Then we convertt₁^α1t₂^α2. . .t_l^αlu^β₁¹u^β₂². . .u^β_k^k to a form

k i=1

∑

ηi(ui−f_i) +τ,ηi∈Z(U×T),τ∈ZT.

If|α₁|+· · ·+|α_l|+|β₁|+· · ·+|β_k|<n, then deg(ηi),deg(τ)<Dn,|ηi|,|τ|<Dⁿ, whereD=max{d1, . . . ,d_k,|f₁|, . . . ,|f_k|}.

Therefore, replacinguibyui−fi+fiin every term ofw, we have

g₁e₁+g₂e₂+· · ·+g_re_r=

r

∑

i=1 k

∑

i=1

µi,j(uj−f_j)ei+ρ,µi,j∈Z(U×T),ρ∈ZT.

Since w=1, then ρ lies in the submodule S. By Lemma VII.2.3, ρ =0. Also note that deg(µi,j),deg(ρ)<

Dn,|µi,j|,|ρ|<nDⁿ. It follows from Lemma V.4.1 that all module computations in the process cost exponentially many relations with respect ton. And it also cost at exponentially many relations to convertρto 0. Therefore

w=W_F r

∏

i=1 k

∏

j=1

a^µ_i^i,j(uj−fj),

and the cost of converting is exponential with respect ton. And the area of the right hand side is bounded by∑i,j|µi,j|6 rknDⁿ. The upper bound is exponential.

Theorem VII.1.4 follows immediately from Proposition VII.2.2.

Proof. LetAbe a free abelian group of finite rank andT be a finitely generated abelian group. If the torsion-free rank ofT is greater than 0, by Proposition VII.2.2,A≀T embeds intoWF, which has exponential Dehn function. If the torsion-free rank is 0,A≀T can be first embedded intoA≀(T×Z). Then the problem is reduced to the first case.

Chapter VIII

Further Discussions