Proof of Proposition V.1.2

(c) Conjugatingtto each term ofa^t₁^′,t^′∈T, cost zero relations. Then we basically estimate the cost of the following equatioin

a^t₁^′t=a^t₁^′t.

By the result of Lemma V.3.3, since Tr(t^′t)⊂B_(degt+degt′), then the cost is bounded by(2K)^{(degt+degt′)}.Notice deg(t^′)6Q, then the total cost is at most

(mP)(2K)^{(degt+degt′)}6(mP)(2K)^(Q+degt).

Here we use the fact Tr(t)⊂B_deg(t)since we order elements inZT degree lexicographically.

Recall thatG_∞is a factor group ofH_∞ asGis a factor group ofG_∞. Denote the epimorphism fromH_∞toG_∞ induced by identity on generating set asψ, and then we have the following homomorphism chain:

H_∞ ^ψ G_∞ ^ϕ G.

Thus kerψ=hhR₂iiH∞,ker(ϕ◦ψ) =hhR₂∪R₃iiH∞.They are all normal subgroups inH_∞as well as submodules in hhAiiH∞. H_∞contains a free module structure while each ofGandG_∞contain a factor module of it. Eventually we will convert the word problem to a membership problem of a submodule inhhAiiH∞.

Note thatGis a factor group ofG_∞and with the given presentation Lemma V.3.1, Lemma V.3.2, Lemma V.3.3 and Lemma V.4.1 all hold forG.

Since Dehn function is a quasi-isometric invariant then it enough for us to prove Proposition V.1.2 using the presentation (V.8).

Proof of Proposition V.1.2. We start with a wordw∈Gsuch that|w|=n,w=G1. WLOG we may assume

w=u₁b₁u₂b₂. . .usb_su_s+1

whereu_i∈F=F(T),b_i∈A^±1ands+∑^s+1_i=1|ui|=n. Letv_i= (u1. . .ui)⁻¹fori=1, . . . ,sandν=u₁u₂. . .u_s+1. Then we have

w=w₁:=b^v₁¹b^v₂². . .b^v_s^sν.

The equality holds in the free group generated byA ∪T thus the cost of relations convertingwtow1is 0. Since s+∑^s+1_i=1|ui|=n, in particular, we have thats6n. Moreover|vi|=∑ⁱ_j=1|uj|6nhence Tr(θ(vi))⊂B_n,i=1,2, . . . ,n.

Next sincew₁=G1,π(w1) =π(v_s+1) =1. By Lemma V.3.1,

ν=

s^′ i=s+1

∏

b^v_iⁱ

wheres^′−s6|ν|²6n²,bi∈A^±1, and Tr(θ(vi))⊂Bn,i=s+1, . . . ,s^′. By Lemma V.3.1, the cost of convertingνto the right hand side is bounded by|ν|²6n².

Thus we let

w₂:=

s^′

∏

i=1

b^v_iⁱ,s^′6n²+n,Tr(θ(vi))⊂B_n,i=1, . . . ,s^′. And the cost of convertingw₂tow₁is bounded byn².

Next, note that allv_i’s are words inF. With the help of Lemma V.3.3, we are able to organizev_ito its image in ¯F.

More precisely, we let

w₃:=

s^′

∏

i=1

b^v_i^¯i,s^′6n²+n,kθ(¯v_i)k6n.

Also followed by Lemma V.3.3,w₂=Gw₃. Let us estimate the cost of convertingw₂tow₃. To transformw₂tow₃, we need apply Lemma V.3.3 to eachb^v_iⁱ once. Since Area(b^v_iⁱb^−¯_i ^vi)6(2K)ⁿwhich provided by Tr(θ(vi))⊂Bn, each transformation costs(2K)ⁿrelations. We have in totals^′6n+n²many conjugates to convert therefore the cost is bounded by(n²+n)(2K)ⁿ.

Now letw₄be the ordered form ofw₃, which in fact is also the ordered form ofw, i.e.

w₃=Gw₄:=

m

∏

i=1

a_i^µi

whereµi are ordered under≺. By the discussion in Section V.2, we obtain the ordered form just by rearranging all conjugates ofA^±1. Note that becausekθ(v¯_i)k6n for alli, it cost at mostK²ⁿrelations to commute any two consecutive conjugatesb^v_i^¯i andb^v¯_j^j by Lemma V.3.2. To sorts^′conjugates we need commutes^′2times. Therefore the number of relations need to commutew₃tow₄is bounded above thats^′2K²ⁿ6(n²+n)²K²ⁿ.

The only thing remains is to compute the area ofw₄. Recall thatw₄ can be regarded as an element in a free T-module generated bya₁, . . . ,a_m. w₄=G1 implies that eitherw₄=H∞ 1 or it lies in the submodule generated by

R₄={f₁,f₂, . . . ,f_l}which is the Gr¨obner basis of the submodule generated byR₂∪R₃. Ifw=H_∞ 1 thenµi=/0 for

alli=1, . . . ,m. In this case Area(w4) =0. Thus

Area(w)6n²+ (n²+n)(2K)ⁿ+ (n²+n)²K²ⁿ.

We are done with this case.

Now let us consider the casew∈ hhR₄iiH∞\ {1}. LetKbe a constant large enough to satisfy both Corollary IV.4.2 and Lemma V.3.2. As an element in theT-module, degw46n sincekθ(¯v_i)k6nfor alli. Also recall that for an elementα∈ZT,|α|is defined to be thel₁-norm of it regarded as a finite suppported function fromT toZ. Thus|w4| represents the number of conjugates inw₄which iss^′. Then by Corollary IV.4.2 we have

w4=H_∞ l

∏

i=1

f_i^αi,fi=a₁^µi1a^µ₂ⁱ². . .a^µ_m^im∈R₄,deg(f_i^αi)6n,

l i=1

∑

|αi|6s^′K^n2k6(n²+n)K^n2k.

whereµi=∑^l_j=1αjµjiinZT. Note that f_i^αi is the product consisting of exactly|αi|many relators. In conclusion we have

Area(

l

∏

i=1

f_i^αi)6

l

∑

i=1

|αi|6(n²+n)K^n2k.

Last, let us estimate the cost of converting∏^l_i=1f_i^αi tow₄. This process consists of two different steps: 1. converting all f_i^αi’s to their ordered form; 2. adding thelterms of orderedf_i^αi.

To transformf_i^αi to its ordered form, we write

αi=

∑

u∈suppαi

αi(u)u.

Let us denoteP=max^l_i=1|f_i|,Q=max^l_i=1deg(f_i). Then

f_i^αi=f_i^{∑u∈suppαiαi(u)u}=

∏

suppαi

f_i^αi(u)u=

∏

suppαi

f_i,u=a^µ₁^′i1a^µ₂^i2′ . . .a_m^µim′ =OF(fi), (V.9)

wheref_i,uis the ordered form off_i^αi(u)uandu^′_{i j}=αiµi jhence OF(f_i)is the ordered form off_i^αi. The first two equalities above hold in the free groupF(A ∪T)thus the cost is 0. In the third equality, applying Lemma V.4.1 (b) and (c), the cost of convertingf_i^αi(u)uto f_i,uis bounded bym|f_i|(2K)^{k(degfi+degu)}+ (|αi(u)| −1)m²|f_i|²K^{2(degfi+degu)}. Here we first conjugateuto f_ithen add|αi(u)|terms of f_i^u. Because degu6degαi,∑_u∈suppαi|αi(u)|=|αi|,|suppαi|6|αi|.

Consequently the cost of the third equality of (V.9) is bounded by

u∈suppα

∑

i

(m|f_i|(2K)^{k(degfi+degu)}+ (|αi(u)| −1)m²|f_i|²K^{2(degfi+degu)}) 6

∑

u∈suppα_i

(m|f_i|(2K)^{k(degfi+degαi)}+ (|αi(u)| −1)m²|f_i|²K^{2(degfi+degαi)})

=|suppαi|m|f_i|(2K)^{k(degfi+degαi)}+

∑

u∈suppαi

(|αi(u)| −1)m²|f_i|²K^{2(degfi+degαi)}

=|suppαi|m|fi|(2K)^{k(degfi+degαi)}+ (|αi| − |suppαi|)m²|fi|²K^{2(degfi+degαi)} 6|αi|m|f_i|(2K)^{k(degfi+degαi)}+|αi|m²|f_i|²K^{2(degfi+degαi)}

=|αi|(m|f_i|(2K)^{k(degfi+degαi)}+m²|f_i|²K^{2(degfi+degαi)}) 6|αi|(mP(2K)^kn+m²P²K²ⁿ).

The last inequality is obtained by the condition deg(f_i^αi)6n, i.e degf_i+degαi6n.

The forth equality of (V.9) is adding allf_i,u’s up. Since

degf_i,u6degf_i^αi 6n,|f_i,u|6|αi(u)||f_i|6|αi||f_i|6|αi|P,

by Lemma V.4.1 (a), the cost of adding|suppαi|terms of f_i,uis bounded by(|suppαi| −1)m²(|αi|P)²K²ⁿ. Here we use the fact that the size of the addition of any step is bounded by|f_i^αi|. Therefore the total number of relations we need to convert each f_i^αito its order form f_i^′is bounded by

|αi|(mP(2K)^kn+m²P²K²ⁿ) + (|suppαi| −1)m²(|αi|P)²K²ⁿ 6|αi|(mP(2K)^kn+ (1+|αi|²)m²P²K²ⁿ).

In general, the cost of converting all f_i^αi’s to their order forms is bounded by

l i=1

∑

|αi|(mP(2K)^kn+ (1+|αi|²)m²P²K²ⁿ)

=(mP(2K)^kn+m²P²K²ⁿ)(

l

∑

i

|α_i|) +m²P²K²ⁿ

l i=1

∑

(|α_i|³)

6(mP(2K)^kn+m²P²K²ⁿ)(n²+n)K^n2k+m²P²K²ⁿ(n²+n)³K^3n2k.

The next step, as described above, is to add all OF(f_i)up. We have that|OF(f_i)|6|αi||f_i|and deg OF(f_i)6n for alli=1,2, . . . ,l.Moreover, the size of any partial product∑^l_i=1^′ OF(f_i),16l^′6lis controlled by the following inequalities:

|

l^′ i=1

∑

OF(f_i)|6

l^′ i=1

∑

|αi||f_i|6P

l^′ i=1

∑

|αi|6P(n²+n)K^n2k,deg(

l^′ i=1

∑

αif_i)6n.

This is similar to add f_i,u’s. By Lemma V.4.1 (a), the cost of the(|l| −1)additions is bounded by

(l−1)m²(P(n²+n)K^n2k)²K²ⁿ6(l−1)m²(n²+n)²P²K^2n2k+2n.

Now we need to verify the process of those steps above indeed resultw₄. This is provided by the factµi=∑^l_j=1αjµji=

∑^l_j=1µ_{i j}^′ and eventually following Lemma V.4.1 we have

l

∏

i=1

f_i^αi=

l

∏

i=1

f_i^′=

l

∏

i=1

(a^µ₁^i1′a^µ₂^i2′ . . .a^µ_m^im′ ) =

m

∏

j=1

a^∑

l j=1µ_{i j}^′

j =a^µ₁¹a^µ₂². . .a^µ_m^m=w₄.

By our estimation, the cost of the first equality is bounded by(mP(2K)^kn+m²P²K²ⁿ)(n²+n)K^n2k+m²P²Q²ⁿ(n²+ n)³K^3n2k and the cost of the third equality is bounded by(l−1)m²(n²+n)²P²K^2n2k+2n.Other equalities hold in the free group hence no cost. Therefore

Area(w4) =(n²+n)K^n2k+ (mP(2K)^kn+m²P²K²ⁿ)(n²+n)K^n2k+m²P²K²ⁿ(n²+n)³K^3n2k + (l−1)m²(n²+n)²P²K^2n2k+2n.

Now we choose a constantC>Klarge enough such that

(n²+n)K^n2k+ (mP(2K)^kn+m²P²K²ⁿ)(n²+n)K^n2k+m²P²K²ⁿ(n²+n)³K^3n2k + (l−1)m²(n²+n)²P²K^2n2k+2n

6C^n2k

It is clear that suchCexists, for example we can chooseCto be 4m²P²QK. Note thatP,Qonly depends onf₁,f₂, . . . ,f_l, henceR₄, and so doesK. ThereforeCis independent ofw.

In conclusion, we start withw=G1 of length at mostn. By converting it four times, we end up with a wordw₄, of which area is bounded byC^n2k. Thus

w ⁰ w₁ ⁶ⁿ² w₂ ⁶⁽ⁿ⁺ⁿ w₃ w₄.

2)(2K)ⁿ 6(n+n²)²K²ⁿ

Summing up all the cost fromw₁tow₄and with the factC>K, we conclude that the area ofwis bounded above by

Area(w)6C^n2k+ (n+n²)²C²ⁿ+ (n+n²)(2C)ⁿ+n². This completes the proof.

Chapter VI

Relative Dehn Functions of Finitely Generated Metabelian Groups