One special case Fuh [15, Theorem 6.1] concerned is whenm>2,m=n+1. In this case, we have that a=
[an,t]. Sincea itself is a commutator, it follows that the relative area of words like[atk,a] is at most 4 instead of
linearly depending onk. Therefore we can improve the result in [15, Theorem 6.1] by the following corollary of Proposition VI.4.4.
Corollary VI.4.5. The metabelianized Baumslag-Solitar groupBS(n,˜ m) =ha,t|(an)t=amiS2,m>2,m=n+1has at most quadratic relative Dehn function.
The lamplighter groups are another interesting class of infinite presented metabelian groups with a simple module structure. We have
Proposition VI.4.6. The lamplighter groups have at most cubic relative Dehn function.
Proof. Consider the lamplighter groupLmwith the standard presentation.
Lm=ha,t|am=1,[a,atn] =1,n∈Ni.
By Lemma VI.3.3, we have a finite relative presentation as the following
Lm=ha,t|am=1,[a,at] =1iS2.
The rest of the proof is the same as the proof of Proposition VI.4.4. The only difference is that the submodule is generated by{m}.
This slightly improves the estimation in [15, Theorem B2].
We choose the following relative presentation ofL2:
L2=ha,t|a2=1,[a,at] =1iS2.
For the upper bound, consider a wordw∈L2that represents the identity. Thuswhas the form
w=tn1atn2an3. . .tn2katn2k+1,wheren2,n3, . . . ,n2k6=0.
Suppose the length ofwisn, combining the fact thatw=1, we have
2k+
2k+1 i=1
∑
|ni|=n,
2k+1 i=1
∑
ni=0.
Insertingtt−1ort−1t, we can rewritewas the following form:
w=at−n1at−(n1+n2). . .at−(n1+n2+···+n2k).
Thuswrepresents an element in⊕i∈ZZ2, where theati is the generator of thei-th copy ofZ2. Sincew=1, then every element in the set{−n1,−(n1+n2), . . . ,−(n1+n2+· · ·+n2k)}occurs even many times in the sequence−n1,−(n1+ n2), . . . ,−(n1+n2+· · ·+n2k). Our goal is to gather the conjugates ofaof the same exponents together at a linear cost with respect ton.
Sincea−1=a, we notice that
atsatl = (aatl−s)ts= [a,tl−s]ts,l,s∈Z.
Thus any consecutive pair of two conjugates ofais a commutator. It follows that any such pair commutes with any other pair of this form without any cost inside the variety of metabelian groups.
For convenience, letmi=∑2ki=1−ni. We now turn the problem of estimating the relative area ofwto a problem of cancelling numbers in a sequence and estimate the cost. Consider a sequence of number
m1,m2, . . . ,m2k.
The goal is to cancel all the pairs of the same value. We have three operations allowed:
(i) Cancel two consecutive numbers without any cost.
(ii) Commute a pair of consecutive numbers with another pair of consecutive numbers next to it without any cost.
(iii) Commute two consecutive numbersc,dwith a cost of|c−d|.
Applying all three operations to the original sequence many times, the result might seems chaotic. To analyze the process, for a sequence of numbers, we define theι(mi)be the position ofmiin the sequence. At the beginning, ι(mi) =i. Then we defineσ(mi,mj) =|ι(mi)−ι(mj)| mod 2. Soσ(mi,mj) =0 ifmiandmjare even positions apart andσ(mi,mj) =1 ifmiandmjare odd positions apart. We notice that
(a) operations from (i) and (ii) do not changeσ(mi,mj);
(b) ifmjis next tomj, applying the operation (iii) to commutemiandmjwill change all values ofσ(mi,ml),σ(mj,ml) forl6=i,j.
From the above observation, we have that
(1) ifσ(mi,mj) =0 andi<j,mjcan be moved to the position next tomijust using operations from (ii).
(2) ifσ(mi,mj) =0,i<jandmi=mj, thenmiandmjcan be cancelled using just operations from (i) and (ii).
(3) formi,mj,mlsuch thatmi=mj,σ(mi,mj) =1,σ(mi,ml) =0, we can cancelmi,mjwith the cost of|mi−ml|.
(1) can be achieved by commuting two consecutive pairs of numbers. (2) is a direct consequence of (1). Let us show how to achieve (3). By (1), we can moveml next tomi. Then by using operation (ii), the pairmiml (ormlmi) can be moved to the position next tomj, resulting the form ofmimlmj ormjmlmi. Finally, we commutemiandmj using operation (iii) at a cost of|mi−mj|and cancelmimj.
Now we are ready to estimate the cost to cancel the sequencem1,m2, . . . ,m2k to the empty sequence. By(2), we can assume that we have already cancelled all the pairsmi,mj whereσ(mi,mj) =0 using operations (i) and (ii).
This step costs nothing and does not change anyσ(mi,mj)formi,mj remaining in the resulting sequence. Let the remaining elements after cancellations bemi(1),mi(2), . . . ,mi(4s) for some 2s6kandi(1)<i(2)<· · ·<i(4s). The remaining sequence satisfies the following properties:
(a) σ(mi(s),mi(l)) =i(s)−i(l) mod 2, (b) ifmi(s)=mi(l)thenσ(mi(s),mi(l)) =1,
(c) σ(mi(s),mi(l)) =σ(mi(s′),mi(l′))formi(s)=mi(s′),mi(l)=mi(l′).
Here the property (a) is true because in the original sequenceι(mi(s)) =i(s)and we only use operation (i) and (ii) which do not changeσ(mi(s),mi(l)). (b) and (c) follow from the definition ofσand the remaining sequence.
Figure VI.1: the corresponding graph of the sequence 2,3,5,3,5,8,2,8
We define the weighted graphΓ0associated withmi(1),mi(2), . . . ,mi(2s))where the vertex set is{mi(1),mi(2), . . . ,mi(4s)} and there is an edge with weight|mi(s)−mi(l)|connectsmi(s),mi(l)ifσ(mi(s),i(l)) =0. Note that this graph is invariant under operations (ii) and may have multi-edge.
By (3), we are allowed to cancelmi(s),mi(l) at a cost of |mi(s)−mi(j)| for some mi(j) that σ(mi(s),mi(j)) =0.
After the cancellation, since we use operation (iii) once,σ(mi(j),mi(j′))change to 0 for somemi(k′)thatmi(j)=mi(j′). Therefore we can then cancelmi(j),mi(j′)without any cost. In summary, we have
(4) formi(s)6=mi(j)thatσ(mi(s),mi(j)) =0, we can cancel a pair of numbermi(s)and a pair of numbermi(j)at a cost of|mi(s)−mi(j)|whereσ(mi,mj)remains the same for numbers that have not been cancelled.
(4) will delete an edge of(mi(s),mi(j))in the graph. If no edges connectingmi(s)andmi(j), we delete the two vertices mi(s),mi(j). The cost is the weight of that edge. LetCbe a cancellation ofΓ0whereC consists of an ordered sequence of edges inΓ0, where we cancel the edge by the order of the sequence. Thus the total cost of a cancellationC to the empty graph is just a sum of the weight of edges inC. Every cancellation can be associated with a pathpC where the path passes through the sequence of edges inC in the same order.
Now we delete edges in the following way. We first delete one edge(mi(1),mi(2))sinceσ(mi(1),mi(2)) =0. We let the resulted graph to beΓ1.
Figure VI.2: two different cancellationsC,C′and their correspondingpC,pC′. The total cost ofC is 3 and the total cost ofC′is 8.
Inductively,Γi+1is obtained by deleting an edge(mi(s),mi(j))wherei,jare the smallest numbers remained inΓi. Γswill be an empty graph since every time we delete four numbers in the sequence.
Let us estimate the cost fromΓ0toΓs. Since every time we cancel pairs of numbers based on the order of the original sequencem1,m2, . . . ,m2k(always cancel the first two numbers remained). The cost is bounded by
2k−1
∑
i=1|mi+1−mi|=
2k i=2
∑
|ni|<n.
Let inequality can also be realized by the following interpretation: the sequencemi(1),mi(2), . . . ,mi(4s)defines a pathp inΓ0(sinceσ(mi(i),mi(j+1)) =0) thatp(j) =mi(j), the weight of the pathpis bounded bynby the definition ofmi.p happens to be the path associated with this cancellation. It follows that the cost of the cancellation is bounded by the total weight ofp. Thus the total cost is bounded byn.
By Lemma VI.3.2, the total cost of converting
atm1atm2. . .atm2k
to 0 is bounded by 4n−3. We finish the proof.