Connections Between Dehn Functions and Relative Dehn Functions

the usual presentationBS(1,2) =ha,t|a^t=a²i^S₂. But one can prove that the relative Dehn function ofBS(1,2)isn instead of the usual Dehn function 2ⁿ[15]. In general, it is difficult to compute the relative Dehn function of a finitely generated metabelian group. We will list some known examples in Section VI.4.

So what is the connection between the relative Dehn function and Dehn function? On the surface, in the relative presentation, we make all metabelian relations cost 0, which should result in a significant reduce in the cost in Lemmas like Lemma V.3.2, Lemma V.4.1. In the next section, we will estimate how much cost we reduce by introducing the relative presentation.

Then for every element f inS, there existsα1,α2, . . . ,αl∈Rsuch that

f =α1f₁+α2f₂+· · ·+α_lf_l.

We denote byAreadA(f)the minimal possible∑^l_i=1|αi|. Thenthe Dehn functionof theR-moduleAis defined to be

δˆA(n) =max{AreadA(f)| kfk6n}.

As expected, the Dehn function of a module is also independent from the choice of the finite presentation [15].

Remark. Now we have three different types of Dehn functions in this thesis: the Dehn function, the relative Dehn function and the Dehn function of a module. They are similar and we distinguish them by the notation. We denote by δ_G(n),Area(w)the Dehn function ofGand the area of a wordw; ˜δ_G(n),Area(w)g the relative Dehn function and the relative area of a wordw, ˆδA(n),Area(d f)the Dehn function of the moduleAand the area of a module elementf.

Let k=rk(G)be the minimal torsion-free rank of an abelian groupT such that there exists an abelian normal subgroupAinGsatisfyingG/A∼=T.

First, ifk>0 we notice that the problem can be reduced in the same way as Proposition V.1.2 does. Because for a finitely presented metabelian groupGthere exists a subgroupG₀of finite index such thatG₀is an extension of an abelian group by a free abelian group of rankk. Most importantly, by Corollary II.1.3 and Proposition VI.2.2, their Dehn functions are equivalent as well as their relative Dehn functions. Therefore from now on, we assume thatGis an extension of an abelian groupAby a free abelian groupT. The projection ofGontoT is denoted byπ:G→T.

LetT ={t1, . . . ,t_k} ⊂Gsuch that{π(t1), . . . ,π(tk)}forms a basis forT andA be a finite subset ofGsuch that it contains all commutatorsa_{i j}= [ti,tj] for 16i< j6kand generates theT-moduleA. ThenA ∪T is a finite generating set for the groupG.

Recall thatGhas a finite presentation as follows

G=ha1,a₂, . . . ,a_m,t1,t2, . . . ,tk|R₁∪R₂∪R₃∪R₄i,

where

R₁={[ti,t_j] =a_{i j}|16i<j6k};

R₂={[a,b^u] =1|a,b∈A,u∈F,¯ kθ(u)k<R};

R₃={

∏

u∈F¯

(a^λ(u)ˆ)^u=a|a∈A,λ ∈Λ∩C(A)} ∪ {

∏

u∈F¯

(a^λ(u)ˆ)^u−1=a|a∈A,λ∈Λ∩C(A^∗)};

andR₄is the finite set that generates kerϕas a normal subgroup. All the notations are the same as in Section V.5.

Since we are dealing with relative Dehn function, we can reduce amount of redundant relations inR₂. We set R^′

2={[a,b] =1,[a,b^t] =1|a,b∈A,t∈T}. Then we have Lemma VI.3.2. R^′

2generates all commutative relations[a,b^u] =1,a,b∈A,u∈F(T)in the presentation relative to the variety of metabelian groups. Moreover, the relative area of[a,b^u]is bounded by4|u| −3.

Proof. Suppose the result is proved for|u|6n, i.e.,[a,b^u] =1 can be written as a product of conjugates of words in R^′

2and metabelian relations. For metabelian relations, we mean those relations make commutators commute to each other. Note that the relative area of any metabelian relations is 0.

Now for that case|u|=n+1, letu=vt,|v|=n,t∈ {t1,t2, . . . ,tk}. By metabelian relations, we have that

1= [a⁻¹a^t,b^−tb^u].

Sincea⁻¹a^t= [a,t]andb^−tb^u= [b,v]^t. Then by inductive assumption, we are able to use relations like[a,b^w] =1 whena,b∈A,|w|6n. In particular,[a,b^v] =1.

And notice that

1= [a⁻¹a^t,b^−tb^u] = a^−ta

commute|{z}

b^−ub^t a⁻¹a^t

| {z }

commute

b^−tb^u

=a a^−tb^−u

| {z }

commute

b^ta^t

|{z}

commute

a⁻¹b^−tb^u

=ab^−ua^−ta^t b^ta⁻¹

| {z }

commute

b^−tb^u

=ab^−ua⁻¹b^u.

This shows that[a,b^u]can be generated byR^′

2and metabelian relations. Let us count the cost. In the computation

above we use[a,b^v] =1 once (notice that[a^t,b^u] = [a,b^v]^t),[a,a^t] =1 twice,[a,b] =1 once, and[a,b^t]once. Therefore,

g

Area([a,b^u])6Area([a,g b^v]) +464(|v|+1)−3=4(n+1)−3.

This completes the proof.

The lemma allows us to replaceR₂byR^′

2in the relative presentation. And we immediately get the relative version of Lemma V.3.2.

Lemma VI.3.3. Let u be a reduced word in F(T) andu be the unique word in T representing u in the form of¯ t₁^m1t₂^m2. . .t_k^mk. Then we have

Area(ag ^−ua^u¯)64|u|²+2|u|.

Proof. The only difference of this proof to the proof of Lemma V.3.2 is that now it only costs 4|u| −3 to commute conjugates of elements inA every time.

Thus in the relative sense, we save a lot of cost due to the fact we assume metabelianness is free of charge.

Our solution for the membership problem of a submodule has a better control for the degree ofαif_i than for

∑^l_i=1|αi|, resulting enormous upper bound for the area. When we considerα1,α2, . . . ,αl that minimizes∑^l_i=1|αi|, one trade-off is that we lose control of the degree ofαi, sort of. The following lemma shows that in this case even though the degree ofαicannot be linearly controlled just by degf but can still be linearly controlled by both degfand

∑^l_i=1|αi|.

Lemma VI.3.4. There exists a constant C such that for every f∈S where S is a T -submodule generated by{f₁,f₂, . . . ,f_l}, assume that there existsα1,α2, . . . ,αl∈ZT such that

f =α₁f₁+α₂f₂+· · ·+α_lf_l,

and∑^l_i=1|αi|is minimized, thendegαif_i6degf+C∑^l_i=1|αi|for all i.

Proof. We denote by∆(f)the difference in the degree of the highest term and lowest term. Let

C:=max{∆(f₁),∆(f₂), . . . ,∆(f_l)}.

In addition we lets=∑^l_i=1|αi|.

We rewrite the sumα1f₁+α2f₂+· · ·+αlf_lto the form

f =u₁f_i1+u₂f_i2+· · ·+u_sf_is,

wherei_j∈ {1,2, . . . ,l}andu_jis an element in∪^l_i=1supp_αi, such that deg(ujf_ij)>deg(uj+1f_ij+1)forj=1,2, . . . ,s−1.

Letgnbe the partial sumgn=∑ⁿ_j=1ujfi_j. Assume that deg(u1fi₁)>degf, otherwise there is nothing to prove.

Every term of degree greater than degf will be cancelled. Since we assume that deg(ujf_ij)>deg(u_j+1f_ij+1), then deggn>degg_n+1when degg_n>degf. We claim that degg_n+1>degg1−Cnwhenever degg_n>degf. If the claim is not true, letn₀be least number such that

deg(

n0

j=1

∑

u_jf_ij)>deg(u1f_i1)−C(n0−1),deg(

n0+1

∑

j=1

u_jf_ij)6deg(u1f_i1)−Cn₀.

Note that∆(ujf_ij)6Cfor all j=1,2, . . . ,s. The least degree among terms ing_n0 is greater than deg(un0f_in

0)− C. Since deg(gn0)>deg(gn0+1), then deg(un0+1f_n0+1) =deg(gn)>deg(u1f_i1)−C(n0−1). Moreover, because deg(un0f_n0)>deg(un0+1f_n0+1), the least degree among terms ing_n0 is greater than deg(u1f_i1)−Cn₀. Therefore if g_n0+16=0, the least degree among terms ing_n0+1is also greater than deg(u1fi₁)−Cn0. It follows thatg_n0+1=0. We have

f =

s j=n

∑

0+2

u_jf_ij. It is a contradiction to the minimality ofs=∑^li=1|αi|.

Letn₁be the largest number that degg_n1>degf. By the claim, we have that degg_n1+1>deg(g1)−Cn1. Straight from the definition ofn₁, deggn₁+16degf. Combining those two inequalities, we finally have

degf >deg(g1)−Cn1.

Sincen₁<s=∑^l_i=1|αi|and degg1=maxi{degαif_i}, the lemma is proved.

Next, we focus on theT-moduleA. It is not hard to see thatAis the quotient of the freeT-module generated by A by the submodule generated byR₃∪R₄. We then replaceR₃∪R₄by the Gr¨obner basisR^′

3={f₁,f₂, . . . ,f_l}for the same submodule. Therefore we finally have the relative presentation ofGwe want:

G=ha1,a₂, . . . ,a_m,t1,t2, . . . ,t_k|R₁∪R^′

2∪R^′

3iS₂.

We letMbe the freeT-module generated byA andSbe the submodule generated byR^′

3overT. So thatA∼=M/S.

Then we have a connection between the relative Dehn function ofGand the Dehn function of the submoduleS.

Lemma VI.3.5. Let G be a finitely generated metabelian group and A is defined as above, then

δˆA(n)4δ˜G(n)4max{δˆ_A³(n³),n⁶}.

Proof. Now letwbe a word of lengthnsuch thatw=G1. We then estimate the cost of converting it to the ordered form. The process is exactly the same as in the proof of Proposition V.1.2. We replace the cost by the cost in relative presentation by Lemma VI.3.2 and Lemma VI.3.3. It is not hard to compute that it costs at mostn²+ (4n−3)(n²+ n) + (4n−3)²(n²+n)²to convertwto its ordered formw^′:=∏^m_i=1a^µ_iⁱwhere∑ⁿ_i=1|µi|6n²,degµi6nand|w^′|62n³. Sincew^′lies in the normal subgroup generated byR^′

3, then there existsα1,α2, . . . ,αlsuch that w^′=

l

∏

i=1

f_i^αi,

l i=1

∑

|αi|6δˆA(2n³).

The relative area of the left hand side is less than∑^l_i=1|αi|. Then we just repeat the same process of the proof of Proposition V.1.2, and compute the cost of addingf_i^αi up tow^′. By Lemma VI.3.4, deg(αif_i)6n+CδˆA(2n³)for ev- eryiand some constantC. It follows that, by Lemma VI.3.3, conjugatingαito f_icosts at most|αi||f_i|(4 deg²(αif_i) + 2 deg(αifi)). Last, we rearrange∑^l_i=1|αi||fi| terms whose degree are at most n+Cδˆ_A(2n³), which costs at most max{δˆ_A³(n³),n²δˆ_A(n³)}up to equivalence. Thus the relative area ofw^′is asymptotically bounded by max{δˆ_A³(n³),n²δˆ_A(n³)}

up to equivalence. And hence the relative area ofwis bounded by max{δ_A³(n³),n⁶}. Thus the right inequality is proved.

For the left inequality in the statement, let∏^m_i=1a^µ_iⁱ be a word of ordered form such that it realizesδA(n). The length of the word, by definition, is bounded byn. We claim that the relative area of∏^m_i=1a_i^µi is greater thanδA(n). If not, by the definition of the relative area, we have that

m

∏

i=1

a_i^µi=

s

∏

j=i

r_i^hi,r_i∈R^′±1

1 ∪R^′±

2 ∪R^′±

3 ,h_i∈M_m+k,

wheres=Area(∏^m_i=1a_i^µi)<δS(n). If we only keep all relations fromR^′

3and combine the same relations together, we will get∏^l_i=1f_i^αi and∑^l_i=1|αi|6s<δA(n). Since canceling relations like[ti,t_j] =a_{i j},[a,b^t] =1 and commuting f_i^hj’s do not change the value of left hand side as an element in freeT-module generated by basis{a1,a₂, . . . ,am}.

Therefore we eventually get

m i=1

∑

µia_i=

l

∑

j=1

αjf_j,

l i=1

∑

|αi|<δA(n).

It leads to a contradiction.

We have

Theorem VI.3.6. Let G be a finitely generated metabelian group. Let k=rk(G), the minimal torsion-free rank of an abelian group T such that there exists an abelian normal subgroup A in G satisfying G/A∼=T .

Then the relative Dehn function of G is asymptotically bounded above by (1) n²if k=0;

(2) 2^n2k if k>0.

Proof. LetGbe a finitely generated metabelian group. Ifk=0,Ghas a finitely generated abelian subgroup of finite index. Then the relative Dehn function is asymptotically bounded byn²by Theorem II.4.2.

Ifk>0, similarly, we can reduce the case to that Gis an extension of a moduleA by a free abelian groupT such that the torsion-free rank ofT isk. Then a wordw=G1 with |w|6n can be converted to its ordered form w^′:=∏^m_i=1a^µ_iⁱwhere|w|62n³,deg(w)6n,∑^m_i=1|µi|6n². Then by Corollary IV.4.2, there exists a wordw^′′such that

w^′=Gw^′′,Area(w^′′)62^n2k. The theorem follows immediately.

Finally, we have all the ingredients to prove Theorem VI.3.1.

Proof of Theorem VI.3.1. The left inequality is obvious since the finite presentation of Gis also the relative finite presentation ofG.

Ifk=0,Ghas a finitely generated abelian subgroup of finite index. The result follows immediately.

Ifk>0, letwbe a word of lengthnandw=G1. Then there existsα1,α2, . . . ,αl

w=

l

∏

i=1

f_i^αi,

l i=1

∑

|αi|6δˆA(2n³),degαi6n+CδˆA(2n³). (VI.1)

According to the proof of Proposition V.1.2, adding the left hand side of (VI.1) costs at most max{δˆ_A³(2n³),2ⁿ}up to equivalence. All other steps of converting cost at most exponential with respect ton. Then by the left inequality in Lemma VI.3.5, Area(w)6max{δ˜_G³(n³),2ⁿ}. Therefore the theorem is proved.