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Dehn functions of finitely presented metabelian groups

  • Wenhao Wang EMAIL logo
Published/Copyright: May 19, 2021

Abstract

In this paper, we compute an upper bound for the Dehn function of a finitely presented metabelian group. In addition, we prove that the same upper bound works for the relative Dehn function of a finitely generated metabelian group. We also show that every wreath product of a free abelian group of finite rank with a finitely generated abelian group can be embedded into a metabelian group with exponential Dehn function.

1 Introduction

The class of groups we are interested in is the class of finitely presented metabelian groups. Recall that a group is metabelian if its derived subgroup is abelian and a group is finitely presented if it is generated by a finite set subject to finitely many defining relations. For finitely presented groups, there is a geometric and combinatorial invariant called the Dehn function. The Dehn function of a finite presented group G = X R is defined to be

δ ( n ) = sup | w | n inf { k | w = i = 1 k r i f i , where r i R R - 1 , f i G } .

It was introduced by computer scientists Madlener and Otto to describe the complexity of the word problem of a group [17], and also by Gromov as a geometric invariant of finitely presented groups [12]. There are a lot of significant results about Dehn functions in the past 30 years, revealing the relationship between this geometric invariant and algebraic properties. For instance, a finitely generated group is hyperbolic if and only if it has sub-quadratic Dehn function [12, 22]. It is also well known that a finitely presented group has decidable word problem if and only if its Dehn function is bounded above by a recursive function [17]. The word problem for any finitely generated metabelian group is decidable, which follows from the fact that finitely generated metabelian groups are residually finite. Hence the Dehn function of a finitely presented metabelian group is always bounded above by a recursive function.

In this paper, we are most interested in the asymptotic behavior of Dehn functions rather than the explicit Dehn function for a particular presentation of a group. In order to compare different functions asymptotically, we define the following relation: for f , g : N N , we write f g if there exists C > 0 such that, for all 𝑛,

f ( n ) C g ( C n ) + C n + C .

And we say f g if f g and g f . One can verify that ≈ is an equivalence relation. This relation has many nice properties. One of which is that it distinguishes polynomials of different degrees, while all polynomials of the same degree are equivalent. Moreover, for exponential functions, we have a n b n for a , b > 1 and n m a n for m > 0 , a > 1 . Despite the dependence of Dehn function on finite presentations of a group, all Dehn functions of the same finitely presented group are equivalent under ≈ (see [12]).

In this paper, we show that there exists a universal upper bound for Dehn functions of all finitely presented metabelian groups (up to equivalence). For example, one choice of such an upper bound is the double exponential function 2 2 n .

First, let us give some examples of Dehn functions of finitely presented metabelian groups.

  1. The first class of examples is the class of metabelian Baumslag–Solitar groups BS ( 1 , n ) , n 2 , which has the presentation

    BS ( 1 , n ) = a , t t a t - 1 = a n

    for any n 2 . It is well known that metabelian Baumslag–Solitar groups have exponential Dehn function up to equivalence. The proof can be found in many places, for example, [13, 1].

  2. Another class of important examples is given by the Baumslag groups, which were first introduced by Baumslag in 1972 [3],

    Γ = a , s , t [ a , a t ] = 1 , [ s , t ] = 1 , a s = a a t and Γ m = Γ a m = 1 .

    Note that Γ contains a copy of Z Z , while Γ m contains a copy of the lamplighter group Z m Z . Recall the wreath product A T is defined to be the semidirect product of t T A t by 𝑇 with the conjugation action. Γ is the first example of a finitely presented group with an abelian normal subgroup of infinite rank. M. Kassabov and T. R. Riley [15] showed that Γ has an exponential Dehn function, while the Dehn function of Γ m is at most n 4 . In particular, Y. de Cornulier and R. Tessera [9] showed that Γ p has a quadratic Dehn function when 𝑝 is a prime number.

  3. The third example consists of groups that are a semidirect product of a finitely generated free abelian group and an infinite cyclic group, namely, Z n Z . Bridson and Gersten have shown that the Dehn functions of such groups are either polynomial or exponential depending on the action of ℤ on Z n (see [8]).

  4. Lattices in R n α R n - 1 , n 3 , have quadratic Dehn function [12], where α : R n - 1 GL ( n , R ) is an injective homomorphism whose image consists of all diagonal matrices with diagonal entries ( e t 1 , e t 2 , , e t n ) verifying

    t 1 + t 2 + + t n = 0 .

    Drutu extends the result for the case that a 1 t 1 + a 2 t 2 + + a n t n = 0 for any fixed vector ( a 1 , a 2 , , a n ) with at least three non-zero components [10].

  5. Let

    G = a , b , t [ a , b ] = 1 , a t = a b , b t = a b 2 .

    𝐺 is metabelian and polycyclic and is also the fundamental group of a closed, orientable fibered 3-manifold. It has been shown that 𝐺 has exponential Dehn function [5]. The lower bound can be proved using the structure of the second homology group of 𝐺.

The Dehn functions of those examples are not very large. In fact, all known cases of finitely presented metabelian groups have at most exponential Dehn functions. It is natural to ask the following question.

Question 1.1

Is the Dehn function of any finitely presented metabelian group bounded above by the exponential function?

The upper bound we obtain in this paper is slightly bigger than the exponential function. It remains unknown if there exists a metabelian group with a Dehn function that exceeds 2 n . We will talk about all the obstructions to seeking such groups in the last section. In this paper, we in fact show that it costs at most exponentially many relations, with respect to the length of the word, commuting two commutators of a finitely presented metabelian group (see Theorem 6.1). From those pieces of evidence, we find that the only hope for seeking such a group is to find a complex enough membership problem for a submodule in a free module over a group ring (see in Section 7.2).

For other varieties of solvable groups, the question has been studied extensively. It is not hard to show that Dehn functions of finitely generated abelian groups are asymptotically bounded above by n 2 . For varieties of solvable groups of derived length three or higher, we have the following result.

Theorem 1.2

Theorem 1.2 (O. Kharlampovich, A. Myasnikov, M. Sapir [16])

For every recursive function 𝑓, there is a residually finite finitely presented solvable group 𝐺 of derived length 3 with Dehn function greater than 𝑓.

Our main result is the following.

Theorem 1.3

Theorem 1.3 (Theorem 5.1)

Let 𝐺 be a finitely presented metabelian group. Let 𝑘 be 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 Dehn function of 𝐺 is asymptotically bounded above by

  1. n 2 if k = 0 ,

  2. 2 n 2 k if k > 0 .

In particular, if we let T = G ab , the abelianization of 𝐺, and we suppose 𝑇 is infinite, δ G is asymptotically bounded by 2 n 2 k , where 𝑘 is the torsion-free rank of the abelianization G ab . The proof of Theorem 1.3 can be found in Section 5.1.

This theorem immediately gives us a uniform upper bound for Dehn functions of finitely presented metabelian groups. For example, let H ( n ) : N N be any super-polynomial function; then 2 H ( n ) will be an upper bound of the Dehn function of any finitely presented metabelian group.

Since metabelian groups form a variety, we also discuss the Dehn function relative to the variety of metabelian groups (defined in Section 6.2). Since finitely generated metabelian groups satisfy the maximal condition for normal subgroups [14], a finitely generated metabelian group is always finitely presentable in the variety of metabelian groups. Therefore, the relative Dehn function exists for any finitely generated metabelian group. The relative Dehn function δ ~ G ( n ) has a close connection to the complexity of the membership problem of the submodule G over the group ring of G / G . And we can translate this connection to a connection between relative Dehn functions and Dehn functions. We prove that the Dehn function δ G ( n ) is bounded above by max { δ ~ G 3 ( n 3 ) , 2 n } (Theorem 6.4). We also improve a result in [11], as in Proposition 6.11, which gives a better estimate of the relative Dehn function in a special case. In addition, we prove that the same upper bound works for the relative case.

Theorem 1.4

Theorem 1.4 (Theorem 6.9)

Let 𝐺 be a finitely generated metabelian group. Let 𝑘 be 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. n 2 if k = 0 ,

  2. 2 n 2 k if k > 0 .

The proof of Theorem 1.4 can be found in Section 6.3.

The general method we establish in this paper provides a way to compute the Dehn function of metabelian groups. In the last section, we generalize one result in [15] and show the following.

Theorem 1.5

Every wreath product of a free abelian group of finite rank with a finitely generated abelian group can be embedded into a metabelian group with exponential Dehn function. In particular, any free metabelian group of finite rank is a subgroup of a metabelian group with exponential Dehn function.

The proof of Theorem 1.5 can be found in Section 7.1.

The structure of this paper

We will state some preliminary concepts and lemmas in Section 2. In Section 3, we prove Corollary 3.4 that allows us to solve the membership problem for some special free modules in a reasonable time. Then we shall briefly revisit the proof of the “if” part of a result of Bieri and Strebel [6] in Section 4, which characterizes finitely presented metabelian groups. In Section 5, we provide the proof of Theorem 1.3. We will talk about relative Dehn functions and the connections between them and Dehn functions in Section 6. In Section 7, we prove Theorem 1.5 and discuss the obstructions for finding metabelian groups with Dehn function greater than the exponential function.

2 Preliminaries

2.1 Notation

We denote the set of rational integers by ℤ and the set of real numbers by ℝ. The set of natural numbers is indicated by ℕ, where our convention is that 0 N . In addition, we let R + = { x R x > 0 } .

If 𝐺 is a group, we will denote by G = [ G , G ] the derived (commutator) subgroup and by G ab G / G the abelianization. For elements x , y G , n N , our conventions are x n y = y - 1 x n y , [ x , y ] = x - 1 y - 1 x y . We use a double bracket G to denote the normal closure of a set in the group 𝐺. Sometimes, we omit the subscript when there is no misunderstanding in the context. For a set 𝒳, we denote the free group generated by the set 𝒳 as F ( X ) . We also use F ( X ) to represent the set of reduced words in the alphabet X X - 1 .

In addition, for a group 𝐺 and a commutative ring 𝐾 with 1 0 , we let K G be the group ring of 𝐺 over 𝐾. An element λ K G is usually denoted as

λ = g G α g g , α g K ,

where all but finitely many α g are 0. We also regard 𝜆 as a function λ : G K with finite support, where λ ( g ) = α g .

We say a group 𝐺 is an extension of a group 𝐴 by a group 𝑇 if 𝐴 is a normal subgroup of 𝐺 and T G / T . If 𝐴 is abelian, then 𝐴 is a module over Z T and the action of 𝑇 on 𝐴 is given by conjugation. In this case, we also say that 𝐺 is an extension of a 𝑇-module 𝐴 by 𝑇.

2.2 Dehn function

Let 𝐺 be a finitely presented group in the category of all groups, i.e. G = X R , where | X | , | R | < . We denote this presentation by 𝒫. The length of a word w G is the length of the corresponding reduced word in F ( X ) . A word 𝑤 in the alphabet X X - 1 is equal to 1 if and only if it lies in the normal closure of 𝑅, i.e.

w = F ( X ) i = 1 k r i f i , where r i R R - 1 , f i F ( X ) .

The smallest possible 𝑘 is denoted by Area P ( w ) . In addition, if w 1 = G w 2 , the cost of converting w 1 to w 2 in 𝐺 with respect to the presentation 𝒫 is defined to be Area P ( w 1 w 2 - 1 ) . The Dehn function of 𝐺 with respect to the presentation 𝒫 is defined as

δ P ( n ) = sup { Area P ( w ) | w | F ( X ) n } .

It is convenient for us to talk about functions up to the equivalence relation ≈ which we defined in the introduction, because the Dehn function is independent of finite presentations of a group in terms of ≈.

Proposition 2.1

Let P , P be two finite presentations of 𝐺; then

δ P ( n ) δ P ( n ) .

We shall then denote the Dehn function of a finitely presented group 𝐺 by δ G ( n ) . Moreover, the Dehn function is a quasi-isometric invariant up to ≈.

Theorem 2.2

Theorem 2.2 ([12])

Let 𝐺 be a finitely presented group, G = X . Let 𝐻 be a finitely generated group with generating set 𝑌, | X | , | Y | < . If G , H are quasi-isometric, then 𝐻 is finitely presented and δ G , X δ H , Y .

Proposition 2.1 follows immediately from the fact that Cayley graphs of the same group over different generating sets are quasi-isometric.

Here is another useful consequence of Theorem 2.2.

Corollary 2.3

Let 𝐺 be a finitely presented group, and let H G be such that [ G : H ] < . Then 𝐻 is finitely presented and quasi-isometric to 𝐺, and hence δ H δ G .

2.3 Van Kampen diagrams

One way to visualize the area of a given word is to consider what is called a van Kampen diagram. Let G = X R be a finitely presented group and 𝑤 a reduced word which is equal to 1. Then, by the previous discussion, 𝑤 has a decomposition as follows:

(2.1) w = F ( X ) i = 1 k r i f i , where r i R R - 1 , f i F ( X ) .

For every decomposition (2.1), we can draw a diagram which consists of a bouquet of “lollipops”. Each “lollipop” corresponds to a factor r i f i , the stem of which is a path labeled by f i and the candy of which is a cycle path labeled by r i . Going counterclockwise around the “lollipop” starting and ending at the tip of the stem, we read f i - 1 r i f i . Thus the boundary of the bouquet of “lollipops” is labeled by the word which is the right-hand side of (2.1).

Note that we obtained 𝑤 from the right-hand side of (2.1) by canceling all consecutive pairs of x x - 1 or x - 1 x , x X on the boundary and removing subgraphs whose boundaries labeled by x x - 1 or x - 1 x , x X (which is a “dipole” or a sphere). In the diagram, the corresponding process is identifying two consecutive edges with the same label but different orientation on the boundary. After finitely many such reductions, we will obtain a diagram whose boundary is labeled by 𝑤.

Figure 1 
                  A bouquet of “lollipops” and its corresponding van Kampen diagram
Figure 1

A bouquet of “lollipops” and its corresponding van Kampen diagram

The resulting diagram is the van Kampen diagram of 𝑤. The edges are labeled by elements in 𝑋, and cells (i.e. the closure of a bounded connected component of the plane minus the graph) are labeled by words from R R - 1 .

For example, in the group a , b [ a , b ] = 1 , the van Kampen diagram of

[ a 2 , b ] = [ a , b ] a [ a , b ]

is depicted in Figure 2.

Figure 2 
                  Obtain the van Kampen diagram of 
                        
                           
                              
                                 [
                                 
                                    a
                                    2
                                 
                                 ,
                                 b
                                 ]
                              
                           
                           
                           [a^{2},b]
Figure 2

Obtain the van Kampen diagram of [ a 2 , b ]

The following is called the van Kampen lemma.

Lemma 2.4

If a reduced group word 𝑤 over the alphabet 𝑋 is equal to 1 in G = X R , then there exists a van Kampen diagram over the presentation of 𝐺 with boundary label 𝑤.

Conversely, let Δ be a van Kampen diagram over G = X R , where X = X - 1 and 𝑅 is closed under cyclic shifts and inverses. Let 𝑤 be the boundary of Δ. Then 𝑤 is equal in the free group F ( X ) to a word of the form u 1 r 1 u 2 r 2 u k r k u k + 1 , where

  1. each r i is from 𝑅,

  2. u 1 u 2 u k + 1 = 1 in F ( X ) ,

  3. i = 1 m + 1 | u i | 4 e , where 𝑒 is the number of edges of Δ.

In particular, 𝑤 is equal to 1 in 𝐺.

We say a van Kampen diagram is minimal if it has the minimal number of cells over all such diagrams of the same word. For a word w = G 1 , the area of 𝑤 is the same as the number of cells of a minimal van Kampen diagram. This fact is useful to estimate the lower bound of the Dehn function. We will use the van Kampen diagram to compute Dehn functions of a class of examples in the last section.

Other applications of the van Kampen diagram can be found in many books. For example, in the book [21], one can find the study of van Kampen diagrams used to construct groups with extreme properties such as infinite bounded torsion group, Tarski monsters, etc.

3 The membership problem of a submodule of a free module

3.1 A well-order on the free module

Let R := Z [ x 1 , , x k ] be a polynomial ring over ℤ. Consider a free 𝑅-module 𝑀 with basis elements e 1 , , e m . A term in 𝑀 is a product of an integer, a monomial in 𝑅 and an element from the basis. A typical term looks like a μ e i , where a Z , 𝜇 is a monomial in 𝑅. Let 𝒯 be the set of all terms in 𝑀. In addition, we will call μ e i a monomial in 𝑀 and denote by 𝒰 the set of monomials of 𝑀. The set of monomials in the polynomial ring 𝑅 in the usual sense will be denoted by 𝒳. For a term g T , we denote by C ( g ) and M ( g ) the coefficient and monomial part of 𝑔, respectively. An element in 𝑀 is a finite sum of terms. From now on, we only consider reduced elements in 𝑀, in a sense that no terms share the same monomial. We also denote by supp ( f ) the set of monomials with non-zero coefficients.

Our first goal is to put a well-order on 𝒯. To construct such an order, we put well-orders on ℤ, 𝒳 and { e 1 , , e m } separately. Then we extend this to 𝑀.

On ℤ, we define an order as follows:

0 1 2 - 1 - 2 .

Under this order, all negative numbers are larger than any positive number. Let a , b Z , where a b ; then there exist unique 𝑞 and 𝑟 such that a = q b + r , 0 < r < | b | . Note that r a , depending on whether 𝑎 is positive or negative; thus we can reduce any number to its remainder in this sense. The remainder is always positive, which is important to us. It is not hard to see that on ℤ is a well-order.

For monomials in 𝑅, we use the degree lexicographical order (also called shortlex or graded lexicographical order) which is defined with respect to

x 1 x 2 x k ,

i.e. for μ 1 = x 1 n 1 x 2 n 2 x k n k , μ 2 = x 1 m 1 x 2 m 2 x k m k ,

μ 1 μ 2 if i = 1 k | n i | < i = 1 k | m i | or i = 1 k | n i | = i = 1 k | m i | , μ 1 lex μ 2 ,

where lex is the usual lexicographical order which in defined in the following way:

x 1 n 1 x 2 n 2 x k n k lex x 1 m 1 x 2 m 2 x k m k if n i < m i for the first i where n i and m i differ .

In fact, on 𝒳 is a well-order, while lex may not be. (See [2].)

Finally, we fix an order e 1 e 2 e s . We now set on 𝑇 to be the lexicographical order based on X { e 1 , , e m } Z . For instance,

7 x 1 2 x 2 e 2 5 x 1 3 e 1 , 3 x 1 3 x 2 5 e 2 3 x 1 3 x 3 6 e 2 , 2 x 1 5 x 3 2 e 3 4 x 1 5 x 3 2 e 3 .

Since on each of 𝒳, { e 1 , , e m } , ℤ is a well-order, we obtain a well-order on 𝒯.

With the well-order , we are able to compare any two terms. Consequently, we can define the leading monomial LM ( f ) of 𝑓 to be the largest monomial among supp ( f ) . For example,

LM ( x 1 7 e 1 + 3 x 1 3 x 2 4 e 2 ) = x 1 7 e 1 , LM ( x 2 3 e 1 + ( x 2 5 x 3 2 + x 2 3 x 4 5 ) e 2 + x 2 5 x 3 2 e 3 ) = x 2 3 x 4 5 e 2 .

Next, we define the leading coefficient of 𝑓 to be the coefficient of the leading monomial, denoted by LC ( f ) . Then the leading term of 𝑓 can be defined as

LT ( f ) := LC ( f ) LM ( f ) .

We then extend to 𝑀. For g , f M , we define g f inductively as follows:

g f if LT ( g ) LT ( f ) or LT ( g ) = LT ( f ) , g - LT ( g ) f - LT ( f ) .

Since on 𝑇 is a well-order, then so is on 𝑀.

Note that is compatible with multiplication by elements from 𝒳, i.e. if g h , then μ g μ f , μ X .

One remark on is that it is Noetherian on 𝒰 as well as 𝒳, the set of monomials in 𝑀, i.e. there is no infinite descending chain of monomials. However, the corresponding statement is not true for on 𝒯 because we have an infinite descending chain for negative numbers. This issue can be avoided by what we will introduce in the next subsection: polynomial reduction.

3.2 Polynomial reduction and Gröbner bases

Now let us define the key ingredient for the application of Gröbner bases: polynomial reduction.

For two monomials μ e and μ e from 𝒰, we say μ e μ e if μ μ and e = e . Let F = { f 1 , , f l } be a finite subset of 𝑀, and let 𝑆 be the submodule generated by 𝐹. Given g , h M , we define the polynomial reduction g F h as follows: if there exists f F and a term g 0 T of 𝑔 such that LM ( f ) M ( g 0 ) , LC ( f ) C ( g 0 ) , then

g = q M ( g 0 ) LM ( f ) f + h ,

where C ( g 0 ) = q LC ( f ) + r , q , r are unique integers such that 0 < r < | LC ( f ) | .

For g F h , read “𝑔 reduces to ℎ modulo 𝐹”. If there is no such 𝑓 and g 0 , then we say that 𝑔 is irreducible modulo 𝐹.

Note that we naturally have h g if g F h . We claim that F is Noetherian, i.e. there is no infinite reduction sequence. First, note that we turn the coefficient of M ( g 0 ) of ℎ to a positive number after a reduction; then there are only finitely many possible reductions that can be applied to monomial M ( g 0 ) . Thus if we assume that g 0 is the largest term that can be reduced in 𝑔 modulo 𝐹, then after finitely many reductions, the monomial of the largest term that can be reduced is strictly less than the original one. Since is Noetherian for monomials, we only have a reduction of finite length for any given g M .

Let F * be the reflexive and transitive closure of F . Then, for each g M , there exists h M such that g F * h and ℎ is irreducible modulo 𝐹. We call ℎ a reduced form of 𝑔 modulo 𝐹. Unfortunately, the reduced form of an element in 𝑀 may not be unique. In fact, at each step of reduction, we may have multiple choices of f F that can be applied to this reduction. This yields our motivation for defining a Gröbner basis: a generating set 𝐹 such that every element in 𝑀 has a unique reduced form modulo 𝐹. In theoretical computer science, this property is called the Church–Rosser property (see [7]).

We write g S h if g - h S . It is not hard to check that S defines an equivalence relation on 𝑀. We define the normal form of 𝑔 to be the smallest element in its equivalence class with respect to . It is well-defined since is a well-order. We denote the normal form of 𝑔 by NF ( g ) .

Definition 3.1

Let 𝑀 be a free 𝑅-module of finite rank, and 𝑆 a submodule of 𝑀. A finite generating set 𝐹 of 𝑆 is called a Gröbner basis if g F * NF ( g ) for all g M .

Remark

𝑅 is a Noetherian ring; hence 𝑀 is a Noetherian module. Thus any submodule of 𝑀 is finitely generated.

Let S u = { g S LM ( g ) = u } , u U and L u = { LC ( g ) g S u } . It is not hard to see that L u is an ideal in ℤ. Thus it is generated by the smallest element in this ideal with respect to . We write h u for an element such that LC ( h u ) generates L u since ℤ is a principle ideal domain. Note that the leading coefficient of h u is always positive since, by our definition of , negative numbers are larger than positive numbers. For our purpose, we denote this by c u ; hence LT ( h u ) = c u u . Let 𝑃 be the set of all such h u which generate 𝑆 over ℤ whenever h u can be defined (since S u might be empty).

Theorem 3.2

Theorem 3.2 ([24, Proposition 10.6.3])

For any submodule of a free module of finite rank over R = Z [ x 1 , , x k ] , there exists a Gröbner basis.

Let 𝐹 be a Gröbner basis for 𝑆. From the proof of [24, Proposition 10.6.3], 𝐹 can be constructed as a finite set such that P = X F := { u f u X , f F } , which allows us to reduce g S by deleting the leading terms with elements in 𝐹. In the next subsection, we will use this specific 𝐹 for the Gröbner basis.

3.3 Division algorithm

For an element g M , 𝑔 can be written as a finite sum of distinct terms, i.e.

g = c 1 u 1 + c 2 u 2 + + c d u d ,

where c i Z , u i U and u 1 u 2 u d . We define the length of 𝑔 to be | g | := i = 1 d | c i | . And if the leading monomial of 𝑔 is x 1 n 1 x 2 n 2 x k n k e i , then we define deg ( g ) = i = 1 k n i . One immediate observation is that if g h , then deg ( g ) deg ( h ) .

Let F = { f 1 , , f l } be a Gröbner basis for a submodule 𝑆 and

g = c 1 u 1 + + c d u d S

such that deg ( g ) n , | g | p . Since g S , then g F * 0 . Thus there exists a finite sequence of reductions

g = g 0 F g 1 F g 2 F g 3 F F g r = 0 .

At each step, if we always choose to cancel the leading term of g i using the polynomial reduction (this is always possible since 𝑔 can be reduced to 0), we may assume that LM ( g 0 ) LM ( g 1 ) LM ( g 2 ) LM ( g r ) = 0 . Thus the number of steps of reduction is bounded by the number of monomials less than or equal to LM ( g ) . Recall that 𝑚 is the rank of the free module; then

r | { u U u LM ( g ) } | m G k ( n ) ,

where G k ( n ) is the growth function of a free commutative monoid with a free generating set of size 𝑘 (see [23, Example 3.7.1]). It is well known that G k ( n ) is a polynomial of degree 𝑘. In fact,

G k ( n ) = ( n + k k ) .

At the 𝑗-th step of reduction, we have g j = g j - 1 - a j μ j f i j , where a j Z , μ j X , 1 i j l and LT ( g j - 1 ) = LT ( a j μ i f i j ) . Then

| a j | LC ( g j - 1 ) | g j - 1 | .

Let C = max { | f 1 | , | f 2 | , , | f l | } . We also observe that

(3.1) | g j | | g j - 1 | + | a j | | f i j | | g j - 1 | + C | a j | .

Additionally, we have | a 1 | | g 0 | = p , and

(3.2) | a j | LC ( g j - 1 ) | g j - 1 | .

Combining (3.1) and (3.2) inductively,

| a j | | g i - 1 | | g i - 2 | + C | a j - 1 | | g i - 2 | ( 1 + C ) p ( 1 + C ) j - 1 , j 1 .

Adding all the steps up, we have

g = j = 1 r a j μ j f i j = i = 1 l α i f i , α i R .

Note that

i = 1 l | α i | = j = 1 r | a j | p ( 1 + ( 1 + C ) + ( 1 + C ) 2 + + ( 1 + C ) r - 1 ) p ( ( 1 + C ) m G k ( n ) - 1 ) C .

In general, one important consequence of Gröbner bases is the following.

Corollary 3.3

Corollary 3.3 (Division)

Suppose that 𝑀 is a free module over a polynomial ring R = Z [ x 1 , , x k ] . Let F = { f 1 , , f l } be a Gröbner basis for a submodule 𝑆. Then there exists a constant 𝐾 such that, for every g M , deg ( g ) n , | g | p , one can write

g = i = 1 l α i f i + r

with α i R , r = NF ( g ) and

deg ( α i f i ) deg ( g ) , 1 i l , i = 1 l | α i | p K n k .

Remark

This provides an algorithm to solve the membership problem for submodules of a finitely generated free module over polynomial rings with integral coefficients. Given g , f 1 , , f l , to decide if 𝑔 lies in the submodule 𝑆 generated by f 1 , , f l , we first find a Gröbner basis for 𝑆. The algorithm which finds Gröbner bases can be found in [24]. Once we have Gröbner bases in hand, we can compute the normal form of 𝑔. Lastly, we use the fact that g S if and only if NF ( g ) = 0 .

Let 𝑇 be the free abelian group of rank 𝑘 with basis t 1 , , t k . We can regard the group ring Z T as a factor ring of Z [ t 1 , t 1 - 1 , t k , t k - 1 ] , i.e.

Z T Z [ t 1 , t 1 - 1 , t k , t k - 1 ] / t 1 t 1 - 1 - 1 , , t k t k - 1 - 1 .

Then a submodule generated by a finite set 𝐹 over Z T can be identified as a submodule generated by F { t 1 t 1 - 1 - 1 , , t k t k - 1 - 1 } over Z [ t 1 , t 1 - 1 , t k , t k - 1 ] .

Therefore, we have a similar result for group rings.

Corollary 3.4

Let 𝑀 be a free module over Z T , where 𝑇 is a free abelian group of rank 𝑘. Let 𝑆 be a submodule of 𝑀. Then there exists a finite generating set F = { f 1 , , f l } and a constant 𝐾 such that, for g S with deg ( g ) n , | g | p , there exist α 1 , , α l Z T such that

g = α 1 f 1 + + α l f l , deg ( α i f i ) deg ( g ) , i = 1 l | α i | p K n 2 k .

Remark

deg ( g ) and | g | for an element g Z T are inherited from

Z [ t 1 , t 1 - 1 , t k , t k - 1 ] .

4 A characterization of finitely presented metabelian groups

4.1 A geometric lemma

Let R n be the Euclidean vector space with the usual inner product , . We denote the norm induced by this inner product by x = x , x . If r > 0 , then B r denotes the open ball of radius 𝑟, i.e. B r = { x R n x < r } .

We consider a finite collection ℱ of finite subsets L R n . We say that an element x R n can be taken from B r if either x B r or if there exists L F such that

x + L = { x + y y L } B r .

Lemma 4.1

Lemma 4.1 ([6, Lemma 1.1])

Assume that, for every 0 x R n , there is some L F such that x , y > 0 for all y L . Then there exists a radius r 0 R + and a function ε : ( r 0 , ) R + with the property that, for r > r 0 , each element of B r + ε ( r ) can be taken from B r by ℱ.

We omit the proof of this lemma but we will use the explicit choice of r 0 and ε : ( r 0 , ) R + in the proof of [6, Lemma 1.1]. Let S n - 1 be the unit sphere in R n , and consider the function f : S n - 1 R given by

f ( u ) = max L min y { u , y y L F } for u S n - 1 .

The function 𝑓 is continuous. By the assumption on ℱ, we have f ( u ) > 0 for all u S n - 1 . Since S n - 1 is compact, we can define

C = inf { f ( u ) u S n - 1 } > 0 , D = max L min y { y y L F } > 0 .

Then our choice of r 0 and 𝜀 are

r 0 = D 2 2 C , ε ( r ) = C - D 2 2 r .

4.2 A theorem by R. Bieri and R. Strebel

Let 𝑇 be a finitely generated abelian group, written multiplicatively. A (real) character of 𝑇 is a homomorphism χ : T R of 𝑇 into the additive group of the field of real numbers ℝ. Let tor T be the torsion subgroup of 𝑇. Then

T / tor T Z k R k ,

where 𝑘 is the rank of 𝑇. We fix a homomorphism θ : T R k . For every character χ : T R , there is a unique ℝ-linear map χ ¯ : R k R such that χ = χ ¯ θ . And by the Riesz representation theorem, there is a unique element x χ R k such that χ ¯ ( y ) = x χ , y for all y R k , whence χ ( t ) = x χ , θ ( t ) (see [6]). Therefore, each character 𝜒 corresponds a vector x χ in R k . Conversely, given a vector 𝑥 in R k , we can define a corresponding character by χ ( t ) = x , θ ( t ) . This will be a useful realization for characters on 𝑇.

Every character χ : T R can be extended to a “character” of the group ring χ : Z T R { + } by putting χ ( 0 ) = + and

χ ( λ ) = min { χ ( t ) t supp ( λ ) } , where  0 λ Z T .

One can check that χ ( λ μ ) χ ( λ ) + χ ( μ ) for all λ , μ Z T . Moreover, if 𝑇 is free abelian, the group ring Z T has no zero divisors. In this case, it follows that χ ( λ μ ) = χ ( λ ) + χ ( μ ) (see [6]).

For every 𝑇-module 𝐴, the centralizer C ( A ) of 𝐴 is defined to be

C ( A ) = { λ Z T λ a = a for all a A } .

If 𝐴 is a left (right) 𝑇-module, then we write A * for the right (resp. left) 𝑇-module with 𝑇-action given by a t = t - 1 a (resp. t a = a t - 1 ).

We say a 𝑇-module 𝐴 is tame if 𝐴 is finitely generated as a 𝑇-module and there is a finite subset Λ C ( A ) C ( A * ) such that, for every non-trivial character χ : Z T R , there is λ Λ with χ ( λ ) > 0 .

Robert Bieri and Ralph Strebel proved the following theorem that characterizes finitely presented metabelian groups.

Theorem 4.2

Theorem 4.2 ([6, Theorem 5.1])

Let 𝐺 be a finitely generated group, and let A G be a normal subgroup such that both 𝐴 and T = G / A are abelian. Then 𝐺 is finitely presented if and only if 𝐴 is tame as a 𝑇-module.

For our purpose, let us sketch the proof of the “if” part of this theorem more precisely.

Theorem 4.3

Theorem 4.3 ([6, Theorem 3.1])

If 𝑇 is a finitely generated abelian group and 𝐴 is a tame 𝑇-module, then every extension of 𝐴 by 𝑇 is finitely presented.

To prove Theorem 4.3, we have to introduce some preliminary concepts in order to provide a reasonable sketch. We first define ordered and semi-ordered words. Let 𝐹 be the free group freely generated by T = { t 1 , , t k } . Let F ¯ F denote the subset of all ordered words of 𝐹, i.e.

F ¯ = { t 1 m 1 t 2 m 2 t k m k m 1 , , m k Z } .

If w F , we write w ¯ for the unique word from F ¯ representing 𝑤 modulo the derived subgroup F . In addition, a word w F is said to be semi-ordered if it is of the form

w = t σ ( 1 ) m 1 t σ ( 2 ) m 2 t σ ( k ) m k ,

where 𝜎 is a permutation of the symbols { 1 , , k } .

Let θ : F R k be the homomorphism given by

θ ( t i ) = ( δ i 1 , , δ i k )

for 1 i k . For every w F , define the trace Tr w R n as follows: if

w = s 1 s 2 s m , where s j T T - 1 ,

is freely reduced, then

Tr ( w ) = { θ ( s 1 s j ) j = 0 , 1 , , m } .

Next, we define a sequence of auxiliary groups. Let 𝒜 be a finite set, and choose an assignment picking an element a i j A for every pair of integers ( i , j ) with 1 i < j k . For every r R + { + } , let H r be the group generated by the set A T with the following defining relations:

[ t i , t j ] = a i j for  1 i < j k ,
[ a , b u ] = 1 for a , b A , u F ¯ with θ ( u ) < r .

We have some useful properties for the group H r .

Proposition 4.4

If r R + , then

  1. [6, Lemma 3.2] a w ¯ = a w for every a A and every w F with Tr ( w ) B r .

  2. [6, Lemma 3.4a] For u , v F such that

    Tr ( u ) B r , Tr ( v ) B r , θ ( u v ) < r ,

    [ a , b u v ] and [ a , b u v ¯ ] are conjugate in H r for every a , b A .

  3. [6, Lemma 3.4b] Assume r > 2 k . Let u , v be semi-order words in 𝐹 such that

    θ ( u ) r 2 k , θ ( v ) r + 1 2 k , θ ( u v ) < r

    are satisfied. Then [ a , b u v ] , [ a , b u v ¯ ] are conjugate in H r for every a , b A .

For r = , H is metabelian, and it is an extension of a free abelian group by another free abelian group. In fact, by the definition of H and Proposition 4.4 (a), we have [ a u , b v ] = 1 for any u , v F , a , b A , and a u is of infinite order. Then A H is free abelian of infinite rank with basis { a u a A , u F ¯ } . By definition, H / A is generated by 𝒯. It is abelian since we include all commutators [ t i , t j ] in 𝒜. Each t i is of infinite order. It follows that H / A is also free abelian. But let us emphasize that H is infinitely related.

Now back to the proof of Theorem 4.3. We first claim that the problem can be reduced to the case when 𝑇 is a free abelian group. Let π : G T be the epimorphism, and let T 1 T be a complement of the torsion subgroup of 𝑇. Then G 1 = π - 1 ( T 1 ) has finite index in 𝐺, and G 1 is an extension of an abelian group by a finitely generated free abelian group. Note that 𝐺 is finitely presented if and only if G 1 is finitely presented. Moreover, if 𝐴 is a tame 𝑇-module, then 𝐴 is also a tame T 1 -module [6, Proposition 2.5]. Therefore, the statement of Theorem 4.3 is true for 𝐺 if and only if it is true for G 1 .

Now we assume that 𝑇 is a free abelian group of rank 𝑘, 𝐴 is a tame 𝑇-module and 𝐺 is an extension of 𝐴 by 𝑇. Denote by π : G T the epimorphism such that A ker π .

Let T = { t 1 , , t k } be a subset of 𝐺 such that { π ( t 1 ) , , π ( t k ) } forms a basis of 𝑇, and let 𝒜 be a finite subset of 𝐴 containing all commutators a i j = [ t i , t j ] for 1 i < j k and generating 𝐴 as a 𝑇-module. We write w ^ T for the image of w F under 𝜋.

Since 𝐴 is a tame 𝑇-module, there is a finite subset Λ C ( A ) C ( A * ) with the property that, for every character χ : T R , there exists λ Λ such that χ ( λ ) > 0 . Recall that F := F ( T ) and F ¯ is the set of ordered words of 𝐹. For every r ( 0 , + ] ,we define the group G r to be given by generators A T and defining relations

(4.1) [ t i , t j ] = a i j for  1 i < j k ,
(4.2) [ a , b u ] = 1 for a , b A , u F ¯ with θ ( u ) < r ,
(4.3) u F ¯ ( a λ ( u ^ ) ) u = a for a A , λ Λ C ( A ) ,
(4.4) u F ¯ ( a λ ( u ^ ) ) u - 1 = a for a A , λ Λ C ( A * ) .
In relations (4.3) and (4.4), we regard 𝜆 as a finite supported function from 𝑇 to ℤ. Hence λ ( u ^ ) is just the value of 𝜆 at u ^ .

If r + , then G r is finitely presented. If r = , although G is not finitely presented, it is metabelian once we realize G is a factor group of H . For each λ Λ , θ ( supp ( λ ) ) is a finite subset of R k , denoted by L λ . Let F = { L λ λ Λ } . As in the previous discussion, there is a one-to-one correspondence between each character χ : T R and a linear functional v χ , . Therefore, if 𝐴 is tame, ℱ is a collection of finite sets which satisfies the assumptions of Lemma 4.1.

Let

C = inf u S n - 1 max λ Λ min y L λ { u , y } , D = max λ Λ min y L λ { y } .

In addition, let R = 2 k max { D , D 2 / 2 C } . We have the following lemma.

Lemma 4.5

Lemma 4.5 ([6, Lemma 3.5])

For r [ R , ) { } , G r G R . In particular, G is finitely presented.

Since relations (4.1)–(4.4) hold in 𝐺, 𝐺 is a factor group of G . The epimorphism φ : G G is induced by the identity map on A T . By the fact that the normal subgroup of a finitely generated metabelian group is the normal closure of a finite set [14], 𝐺 is finitely presented. Thus we have finished the proof of Theorem 4.3.

In summary, given a tame 𝑇-module 𝐴, any extension of 𝐴 by 𝑇 is always a factor group of G . The group G is finitely presented, and the defining relations are given by (4.1)–(4.4) for any fixed positive real number r R .

5 Main theorem

5.1 Reduction step

Suppose that we have a finitely presented metabelian group 𝐺 with the short exact sequence 1 A G T 1 such that 𝐴 and 𝑇 are abelian. In addition, we suppose that the torsion-free rank of 𝑇 is minimized over all such short exact sequences of 𝐺 where both the normal subgroup and the quotient group are abelian. The torsion-free rank of such 𝑇 is denoted by rk ( G ) . Since 𝐺 is finitely presented, in particular, it is finitely generated. Then 𝑇 is a finitely generated abelian group, and 𝐴 is finitely generated as a 𝑇-module (see [14]).

The main goal of this chapter is to prove the following.

Theorem 5.1

Let 𝐺 be a finitely presented metabelian group. Let rk ( G ) = k , i.e. 𝑘 is 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 Dehn function of 𝐺 is asymptotically bounded above by

  1. n 2 if k = 0 ,

  2. 2 n 2 k if k > 0 .

We can apply the same technique as in Section 4.2 to reduce the problem to a simpler case. Denote by π : G T the epimorphism such that A ker π . Suppose T 1 T is the complement of the torsion subgroup of 𝑇. The preimage G 1 = π - 1 ( T 1 ) has finite index in 𝐺; then G 1 is quasi-isometric to 𝐺. It follows that δ G = δ G 1 due to Theorem 2.2. Therefore, an upper bound of δ G 1 is also an upper bound for δ G .

Next, we show that rk ( G ) = rk ( G 1 ) . If G 1 can be written as an extension of two abelian groups A 2 and T 2 , where the torsion-free rank of T 2 is strictly less than 𝑘, consider the following commutative diagram:

where f ( g , h ) := ( π ( g ) , h ) , g G 1 , h G / G 1 . By the snake lemma, there exists an exact sequence

ker π = A 2 ker f ker i = 1 coker π = 1 coker f coker i = 1 .

It follows that ker f is abelian and 𝑓 is surjective. Then 𝐺 can be represented as an extension of ker f by T 2 × G / G 1 , where the torsion-free rank is of T 2 × G / G 1 , strictly less than 𝑘. This contradicts the minimality of 𝑘. Therefore, 𝑘 is preserved when passing to G 1 .

Thus, from now on, we shall assume that 𝑇 is a free abelian group of rank 𝑘 and 𝐺 is an extension of a tame 𝑇-module 𝐴 by the free abelian group 𝑇. Also, let us assume that k > 0 , since if k = 0 , then 𝐺 has a finitely generated abelian subgroup of finite rank, which is not interesting to us.

Let T = { t 1 , , t k } G be such that { π ( t 1 ) , , π ( t k ) } forms a basis for 𝑇, and let 𝒜 be a finite subset of 𝐺 such that it contains all commutators a i j = [ t i , t j ] for 1 i < j k and generates the 𝑇-module 𝐴. Then A T is a finite generating set for the group 𝐺.

By Theorem 4.2, since 𝐺 is finitely presented, 𝐴 is a tame 𝑇-module. Thus there is a finite subset Λ C ( A ) C ( A * ) such that, for each character χ : T R , there exists λ Λ such that χ ( λ ) > 0 . Let 𝐹 be the free group generated by 𝒯 and F ¯ the set of all ordered words in 𝐹 (see Section 4.2). As in the previous section, we let θ : F R k be the homomorphism given by

θ ( t i ) = ( δ i 1 , . δ i k ) , 1 i k .

If w F , we write w ¯ for the unique word in F ¯ representing 𝑤 modulo F . In addition, we write w ~ T for the image of w F under 𝜋.

Then we are able to define a sequence of groups G r as we did in Section 4.2, but for our purpose, we will need a larger 𝑅. Let

R = 2 k max { D 2 / 2 C , D , D 2 / ( 4 k C - 4 ) } ,

where C , D are defined as before Lemma 4.5. Since R > 2 k max { D , D 2 / C } , we have G R G , and in particular, 𝐺 is a factor group of the finitely presented group G . Then we can list all the defining relations of G here:

(5.1) [ t i , t j ] = a i j for  1 i < j k ,
(5.2) [ a , b u ] = 1 for a , b A , u F ¯ with θ ( u ) < R ,
(5.3) u F ¯ ( a λ ( u ^ ) ) u = a for a A , λ Λ C ( A ) ,
(5.4) u F ¯ ( a λ ( u ^ ) ) u - 1 = a for a A , λ Λ C ( A * ) .

To simplify our notation, we write relations (5.1) and (5.2) as R 1 and relations (5.3) and (5.4) as R 2 .

Denote by φ : G G the epimorphism induced by the identity map on A T . Note that 𝜑 induces an isomorphism on G / A T . Therefore, ker φ A is abelian, where A := A G G . Let ker φ = R 3 G , where R 3 is a finite set.

Thus we obtain a finite presentation for 𝐺,

(5.5) G = A T R 1 R 2 R 3 .

With (5.5), we have the following proposition.

Proposition 5.2

If rk ( G ) > 0 , then δ G ( n ) 2 n 2 k .

Proposition 5.2 will be proved in Section 5.5.

Proof of Theorem 5.1

If k = 0 , 𝐺 has a finite index abelian subgroup. Therefore, δ G n 2 by Theorem 2.2.

If k > 0 , the result follows directly from Proposition 5.2 by passing the problem to a finite index subgroup. ∎

5.2 The ordered form of elements

For convenience, we assume that | A | = m and write A = { a 1 , , a m } .

To understand the module structure in 𝐺, we have to go back to the group H corresponding to 𝐺. As in Section 4.2, H has a presentation as follows:

H = A T [ t i , t j ] = a i j ,  1 i < j k , [ a , b u ] = 1 , a , b A , u F ( T ) .

We have that 𝐺 is an epimorphic image of H , where the epimorphism is induced by the identity map on the set A T . Let 𝑀 be the free 𝑇-module with basis 𝒜. We will show that A H is isomorphic to 𝑀.

Note that we only consider words that are fully reduced in F ( A T ) , the free group generated by A T , since each group element in A H can also be regarded as an element in the 𝑇-module 𝑀. Different words in the group might represent the same element in the module. For example, a 1 t 1 a 2 t 1 and a 2 t 1 a 2 t 1 both represent t 1 a 1 + t 1 a 2 in 𝑀. We now pick a canonical element among all words representing the same element in 𝑀. This canonical element, which we will call the ordered form of an element in A H , is defined in the following way.

Definition 5.3

Let F ¯ be the set of ordered exponent words in F ( T ) (see in Section 4.2), and let be the well-order defined in Section 3.1. Let 𝑤 be an element in A H ; then the ordered form of 𝑤 is a word OF ( w ) of the form a 1 μ 1 a 2 μ 2 a m μ m , where

  1. μ i Z T for 1 i m , and each μ i is of the form μ i = j = 1 n j c i j u i j such that c i j Z , u i j F ¯ and u i 1 u i 2 u i n i ,

  2. w = H OF ( w ) .

To check the definition is well-defined, we have to show the existence and uniqueness of the ordered form.

To show its existence, let us construct the explicit algorithm that rewrites a word w A H to a word of the ordered form.

Let π : H T be the canonical quotient map. For g A H , we have π ( g ) = 0 . It follows that the sum of exponents of each t i is 0.

Let us start with a word

w = u 1 b 1 u 2 b 2 u s b s u s + 1 A H ,

where u i F ( T ) , b i A ± 1 . Here, u 1 , u s + 1 could be empty. Then

w = b 1 u 1 - 1 b 2 ( u 1 u 2 ) - 1 b s ( u 1 u 2 u s ) - 1 u 1 u 2 u s + 1 .

The equality holds in the free group generated by A T . Note that u 1 u s + 1 is a word in 𝐹. It also has the property that the sum of exponents of each t i is 0. We then write this word as a product of conjugates of { [ t i , t j ] ± 1 , i < j } algorithmically in the following fashion: assume we have already written u 1 u s + 1 as w 1 w 2 , where w 1 is a product of conjugates of { [ t i , t j ] ± 1 , i < j } and w 2 is a word in 𝐹 such that the sum of the exponents of each t i is 0. Let t i be the letter with the smallest index among all letters in w 2 . Then w 2 can be written as w 2 t i ε w 2 ′′ , ε = ± 1 , where w 2 does not contain any t i ± 1 . Then

w 2 t i ε w 2 ′′ = [ t i ε , t j 1 ε 1 ] ( w 2 t j 1 - ε 1 ) - 1 [ t i ε , t j 2 ε 2 ] ( w 2 t j 1 - ε 1 t j 2 - ε 2 ) - 1 [ t i ε , t j l ε l ] t i ε w 2 w 2 ′′ ,

where w 2 = t j l ε l t j 1 ε 1 . Since the sum of the exponents of t i is 0, by repeating this process, we can gather all t i factors to the left, and hence they will be canceled eventually. We end up with a word w 3 w 4 , where w 3 is a product of conjugates of { [ t i , t j ] ± 1 , i < j } and w 4 is a word in 𝐹 in which the sum of the exponents of each t i is 0 and whose length is strictly less than w 2 . Thus, by repeating this algorithm, we are able to write 𝑔 as a product of conjugates of { [ t i , t j ] ± 1 , i < j } in a unique way. Now we just apply relations like [ t i , t j ] = a i j to replace all the commutators by their corresponding letters in 𝒜.

Since 𝑔 can be written as a product of conjugates of elements in 𝒜, applying commutator relations like [ a , b u ] , a , b A , u F , we are able to commute those conjugates and hence gather all conjugates which share the same base. In addition, combining with the fact a u = a u ¯ from Proposition 4.4 (a), we can write 𝑔 in the ordered form of the type g = a 1 μ 1 a 2 μ 2 a m μ m , where μ i Z T and the terms of μ i are written in order from high to low with respect to defined in Section 3.1. The result is a word satisfying all conditions of Definition 5.3. Therefore, the existence of the ordered form is shown.

The uniqueness of the ordered form can be seen from the fact that the set of words in the ordered form is isomorphic to the free 𝑇-module 𝑀. The isomorphism is given by the canonical map

a 1 μ 1 a 2 μ 2 a m μ m μ 1 a 1 + μ 2 a 2 + + μ m a m .

For 𝑤 in A G (or A G ), we define the ordered form by lifting 𝑤 to H , that is, as the ordered form of ι ( w ) , where ι : G H (resp. G H ) is the combinatorial map induced by identity on A T . Note that, by the way we define the ordered form, the ordered form of each word is unique. The ordered forms distinguish different elements in the 𝑇-module A H . In fact, two elements in A H are equal in H if and only if they have the same ordered form. One remark is that two words which are equal in 𝐺 or G may have different ordered forms, for example, a 1 and u F ¯ ( a λ ( u ^ ) ) u , λ C ( A ) .

Recall that G = A T R 1 R 2 R 3 . Note that both R 2 and R 3 are contained in the normal closure of 𝒜. From now on, we write all relators from R 2 R 3 in their ordered form.

5.3 Main lemmas

Before we embark on the proof of Proposition 5.2, we shall establish some preliminary lemmas.

Now consider an arbitrary factor group 𝐻 of G equipped with the presentation

(5.6) H = A T R 1 R 2 R ,

where ℛ is a finite subset in G . Then H G / R G . Note that if R = R 3 , then H = G , and if R = , then H = G , which are two major examples we are concerned with. We have the following lemmas for 𝐻.

Lemma 5.4

Let 𝐻 be a factor group of G equipped with presentation (5.6), and let 𝑤 be a word in ( T T - 1 ) * such that | w | = n ; then

w = H w ¯ i = 1 p b i u i ,

where p n 2 , b i A ± 1 , u i F , Tr ( θ ( u i ) ) B n . In addition, the cost of converting the left-hand side to the right-hand side is bounded by n 2 .

Proof

Since w ¯ = t 1 m i t k m k for some m 1 , , m k Z such that i = 1 k | m i | n , to move each letter in 𝑤 to the desired place, it will cost at most 𝑛 commutators of the form [ t i , t j ] , 1 i < j k . In total, we need at most n 2 such commutators. That is,

w = w ¯ i = 1 p [ t i 1 , t i 2 ] ε i u i , where u i F , p n 2 ,  1 i 1 < i 2 k , ε i { ± 1 } .

Moreover, since the length of 𝑤 is bounded by 𝑛, we have Tr ( θ ( u i ) ) n .

By applying relations in { a i j = [ a i , a j ] 1 i < j n } 𝑝 times, we immediately have

w = H w ¯ i = 1 p b i u i , Tr ( θ ( u i ) ) n .

The cost of relations is bounded by p n 2 . ∎

In particular, for w F such that π ( w ) = 1 , it costs at most n 2 relations in 𝐻 to convert it to a product of conjugates of elements in 𝒜.

Lemma 5.5

Let 𝐻 be a factor group of G equipped with presentation (5.6); then there exists a constant 𝐾 only depending on R 1 R 2 such that

Area ( [ a , b u ] ) K n for all a , b A , θ ( u ) < n .

Proof

Let F = { θ ( supp ( λ ) ) λ Λ } ; then ℱ is a finite collection of finite sets. By the choice of Λ, ℱ satisfies the assumptions of Lemma 4.1.

By Lemma 4.1, each x B r + ε ( r ) can be taken from B r by ℱ for r > R . Recall that 𝑅 is defined to be max { D , D 2 / 2 C , D 2 / ( 4 k C - 4 ) } and ε ( r ) = C - D 2 / 2 r , where C , D are purely determined by Λ; hence we have R 1 R 2 as we stated in Lemma 4.5.

According to our choice of 𝑅, we note that ε ( r ) ε ( 2 k D 2 / ( 4 k C - 4 ) ) = 1 2 k for r > R . Let K 1 be the constant which is large enough such that f ( n ) K 1 n for n R , and let K 2 be the constant

K 2 := max λ Λ { u F ¯ | λ ( u ) | } + 2 .

Since each 𝜆 has finite support, K 2 is well-defined. Now let K := max { K 1 , K 2 2 k } .

Suppose for n > R , Area ( [ a , b u ] ) K n for all a , b A , θ ( u ) < n . We then prove our lemma by induction. Let us first consider the case r = n + 1 2 k . Fix some v F ¯ satisfying θ ( v ) < r . Since ε ( n ) 1 2 k , B n + 1 2 k can be taken from B n by ℱ. Then there is λ Λ with θ ( supp ( λ v ^ ) ) B n by the definition of “taken from”.

Therefore, we have two cases depending on λ C ( A ) or C ( A * ) . Firstly, assume that λ Λ C ( A ) . Then, by applying the commutator formula

[ x , y z ] = [ x , y ] x - 1 z x [ x , z ] ,

we obtain

[ a , b v ] = G [ a , u F ¯ ( b λ ( u ^ ) ) u v ] = u F ¯ [ a , b λ ( u ^ ) u v ] h ( u ) ,

where the h ( u ) are certain elements in 𝐻 which need not concern us. Note that, in the first equality above, we apply relations in (5.3) twice to replace 𝑏 by

u F ¯ b λ ( y ) u .

Since supp ( λ ) B ¯ D , we have θ ( u ) < D < n 2 k . Additionally, θ ( v ) < n + 1 2 k and θ ( u v ) < n . It meets all assumptions of Proposition 4.4 (c). Note that 𝐻 is a factor group of H n which we defined in Section 4. Then [ a , b λ ( u ^ ) u v ] is conjugate in 𝐻 to [ a , b λ ( u ^ ) u v ¯ ] , the area of which is bounded by | λ ( u ¯ ) | K n . It follows that

Area ( [ a , b v ] ) 2 + u F ¯ Area ( [ a , b λ ( u ^ ) u v ] ) 2 + u F ¯ | λ ( u ^ ) | K n K 2 K n .

Repeating this process 2 k times, we obtain that

Area ( [ a , b v ] ) K 2 2 k K n K n + 1 for v F ¯ , θ ( v ) < n + 1 .

If λ Λ C ( A * ) , the only difference is that

[ a , b v ] = [ a v - 1 , b ] v - 1 = [ u F ¯ ( a λ ( u ^ ) ) u v - 1 , b ] v - 1 .

Similarly, we obtain that Area ( [ a , b v ] ) K 2 2 k K n K n + 1 . ∎

Furthermore, Lemma 5.5 allows us to estimate the cost of commuting two conjugates of elements in 𝒜. Since the normal closure of 𝒜 in 𝐻 is abelian, this lemma provides a tool to estimate the cost of converting words in A H , in particular, in 𝐺. Also, Lemma 5.5 reveals how much metabelianness costs in a finitely presented metabelian group. We will discuss this in Section 6.1.

Lemma 5.6

Let 𝐻 be a factor group of G equipped with presentation (5.6), and let 𝐾 be the same constant as in Lemma 5.5. Then, in 𝐻, we have

Area ( a u a - u ¯ ) ( 2 K ) n for all a A , u F , Tr ( u ) B n .

Proof

We prove this by an induction on 𝑛. Suppose for i n the result holds. Then, for the case n + 1 , we write u = u t s ± 1 ; then Tr ( u ) B n + 1 , Tr ( u ) B n ,

a u = a u t s ± 1 = ( a u ¯ ) t s ± 1 ν 1 ,

where Area ( ν 1 ) ( 2 K ) n by our inductive assumption. Writing u ¯ = t 1 m 1 t k m k , we claim that

u ¯ t s ± 1 = u ¯ j = 1 m c j α j ,

where

c j { [ t s , t l ] ± 1 1 s < l k } , α j F ¯ , m = i = s + 1 k | m i | n .

We need to be really careful here. Let us first consider the case where the exponent of t s is 1. We assume s < k ; otherwise, it is trivial. Note that if m k 0 ,

t k m k t s = t k m k - 1 t s t k [ t s , t k ] - 1 = t k m k - 2 t s t k 2 [ t s , t k ] - t k [ t s , t k ] - 1 = t k m k - 3 t s t k 3 [ t s , t k ] - t k 2 [ t s , t k ] - t k [ t s , t k ] - 1 = t s t k m k [ t s , t k ] - t k m k - 1 [ t s , t k ] - t k [ t s , t k ] - 1 .

If m k < 0 , we have

t k m k t s = t k m k + 1 t s t k - 1 [ t s , t k ] t s = t k m k + 2 t s t k - 2 [ t s , t k ] t s t k - 1 [ t s , t k ] t s = t k m k + 3 t s t k - 3 [ t s , t k ] t s t k - 2 [ t s , t k ] t s t k - 1 [ t s , t k ] t s - 1 = t s t k m k [ t s , t k ] t s t k m k + 1 [ t s , t k ] t s t k - 1 [ t s , t k ] t s .

Repeating this process, we then prove the claim for the case where the exponent of t s is 1. On the other hand, if the exponent of t s is −1, then similarly, consider, if m k 0 ,

t k m k t s - 1 = t k m k - 1 t s - 1 t k [ t s , t k ] t k = t k m k - 2 t s - 1 t k 2 [ t s , t k ] t k 2 [ t s , t k ] t s = t k m k - 3 t s - 1 t k 3 [ t s , t k ] t k 3 [ t s , t k ] t k 2 [ t s , t k ] t k = t s - 1 t k m k [ t s , t k ] t k m k - 1 [ t s , t k ] t k 2 [ t s , t k ] t k ,

and if m k < 0 ,

t k m k t s - 1 = t k m k + 1 t s - 1 t k - 1 [ t s , t k ] - 1 = t k m k + 2 t s - 1 t k - 2 [ t s , t k ] - t k - 1 [ t s , t k ] - 1 = t k m k + 3 t s - 1 t k - 3 [ t s , t k ] - t k - 2 [ t s , t k ] - t k - 1 [ t s , t k ] - 1 = t s - 1 t k m k [ t s , t k ] - t k m k + 1 [ t s , t k ] - t k - 1 [ t s , t k ] - 1 .

Again, by repeating this process, the claim is proved. Thus, by induction on 𝑘, we can move t s to the desired place.

Now we have

a u = a u ¯ t s ± 1 ν 1 = ( j = 1 m c j α j ) - 1 a u ¯ ( j = 1 m c j α j ) ν 1 .

Applying relations from { a i j = [ a i , a j ] 1 i < j n } 2 m times, we have that

a u = ( j = 1 m d j α j ) - 1 a u ¯ ( j = 1 m d j α j ) ν 2 ν 1 ,

where d j A ± 1 and Area ( ν 2 ) 2 m by our discussion.

Next, we need to commute a u ¯ and d j α j for j = 1 , , m to the left and estimate the cost. Note that [ a u ¯ , d j α j ] is conjugate to [ a , d j α j ( u ¯ ) - 1 ] . From the computation above, α j is either a tail of u ¯ or a tail of u ¯ multiplied by t s ± 1 . Therefore, ( u ¯ ) - 1 , α j , α j ( u ¯ ) - 1 satisfy the assumption of Proposition 4.4 (b). Thus

[ a , d j α j ( u ¯ ) - 1 ] is conjugate to [ a , d j α j u - 1 ¯ ] .

Since θ ( α j u - 1 ¯ ) n + 1 , the area of [ a , d j α j u - 1 ¯ ] , by Lemma 5.5, is bounded by K n + 1 .

Applying [ a , d j α j u - 1 ¯ ] to a u and d j α j for j = 1 , , m , we can commute each d j α j to the left such that it cancels with d j - α j . Then we finally have

a u = a u ¯ ν 3 ν 2 ν 1 , where Area ( ν 3 ) m K n + 1 .

In total, the cost of converting a u to a u ¯ is bounded by

Area ( ν 3 ν 2 ν 1 ) Area ( ν 3 ) + Area ( ν 2 ) + Area ( ν 1 ) ( 2 K ) n + 2 m + m K n + 1 ( 2 K ) n + 1 .

Note that we use the fact that m n and we can choose K 1 . ∎

Lemma 5.6 provides a method for us to “organize” the exponent of a conjugate. In particular, combining Lemma 5.5 and Lemma 5.6, we are able to convert any word in A H to its ordered form.

5.4 The 𝑇-module in metabelian groups

In H , A H is naturally a 𝑇-module by the conjugation action. Let

A = { a 1 , , a m } .

Each element g A H can be written in its ordered form, i.e.

g = i = 1 m a 1 λ 1 a 2 λ 2 a m λ m Z T .

For λ i , we always write it from high to low with respect to the order . Then 𝑔 can also be regarded as an element ( λ 1 , , λ m ) in the free 𝑇-module with basis a 1 , , a m . From now on, we treat an element in A H as an element in the group H as well as an element in the free 𝑇-module with basis a 1 , , a m .

Let us first state the relation of operations between the group language and module language in Table 1, where c C , t T .

Table 1

Operations in groups and modules

Group Module
( i = 1 m a i λ i ) ( i = 1 m a i λ i ) ( λ 1 , , λ m ) + ( λ 1 , , λ m )
= H i = 1 m a i λ i + λ i = ( λ 1 + λ 1 , , λ m + λ m )
( i = 1 m a i λ i ) c = H i = 1 m a i c λ i c ( λ 1 , , λ m ) = ( c λ 1 , , c λ m )
( i = 1 m a i λ i ) t = H i = 1 m a i t λ i t ( λ 1 , , λ m ) = ( t λ 1 , , t λ m )

Let 𝒳 be a subset of A H . Then the normal closure of 𝒳 in the group H coincides with the submodule generated by 𝒳 over Z T . One direction is trivial since, by the table we have above, elements in the submodule are obtained by group operations and conjugations. Conversely, let g A H ; if h Z T , then g h can be obtained by finitely many scalar products and module operations, and if h A H , then g h = g . The general case is a combination of those two cases. Thus g h must lie in the submodule generated by 𝑔. Conversely, the subgroup generated by 𝒳 coincides with the submodule generated by 𝒳 over ℤ.

Again, we consider an arbitrary factor group 𝐻 of G with the finite presentation

H = A T R 1 R 2 R ,

where ℛ is a finite subset of G . Then H G / R . We now estimate the cost of relations in the group 𝐻 to make each of the module operations above. Note that notations like deg ( λ ) and | λ | for an element g Z T are inherited from the polynomial ring Z [ t 1 , t 1 - 1 , t k , t k - 1 ] (see in Section 3).

In the following lemma, 𝐾 is the same constant as in Lemma 5.5, which only depends on R 1 R 2 .

Lemma 5.7

Let 𝐻 be a factor group of G equipped with presentation (5.6).

  1. Let

    f = i = 1 m a i λ i , g = i = 1 m a i λ i ,

    and denote

    P = max { | λ i | , | λ i | i = 1 , , m } ,
    Q = max { deg ( λ i ) , deg ( λ i ) i = 1 , , m } .
    Then the cost of relations in 𝐻 of converting

    f g = H i = 1 m a i λ i + λ i

    is at most m 2 P 2 K 2 Q , where the right-hand side is written in its ordered form.

  2. Let

    f = i = 1 m a i λ i ,

    and denote

    P = max { | λ i | i = 1 , , m } ,
    Q = max { deg ( λ i ) i = 1 , , m } .
    For c Z , the cost of relations in 𝐻 of converting f c to i = 1 m a i c λ i is at most ( | c | - 1 ) ( m ) 2 P 2 K 2 Q , where the right-hand side is written in its ordered form.

  3. Let

    f = i = 1 m a i λ i ,

    and denote

    P = max { | λ i | i = 1 , , m } ,
    Q = max { deg ( λ i ) i = 1 , , m } .
    For t T , the cost of relations in 𝐻 of converting

    ( i = 1 m a i λ i ) t = H i = 1 m a i t λ i

    is bounded by ( m P ) ( 2 K ) k ( Q + deg t ) .

Proof

(a) First, we consider a simpler case when g = a 1 λ 1 . Then it is essential to estimate the cost of converting the left-hand side to the right-hand side of

(5.7) ( i = 1 m a i λ i ) a 1 λ 1 = H ( a 1 μ 1 ) ( a 2 λ 2 a m λ m ) , μ 1 = λ 1 + λ 1 .

In order to commute a 1 λ 1 with a m λ m , , a 2 λ 2 , we apply Lemma 5.5 ( m - 1 ) -times. Each step costs at most P K 2 Q since deg ( λ i + λ 1 ) 2 Q , | λ i | , | λ 1 | P . Therefore, the cost of

( i = 1 m a i λ i ) a 1 λ 1 = ( a 1 λ 1 a 1 λ 1 ) ( a 2 λ 2 a m λ m )

is bounded by ( m - 1 ) P 2 K 2 Q . When it comes to the last step, i.e.

a 1 λ 1 a 1 λ 1 = a 1 μ 1 ,

the only thing we need to do is move each term a 1 u u F to its position corresponding to . In fact, we sort all conjugates a 1 u in order. Note that those conjugates in a 1 λ 1 and a 1 λ 1 are already in order, respectively. Thus we only need to insert each a 1 u of a 1 λ 1 into terms of a 1 λ 1 . Again, from Lemma 5.5, the cost is bounded by P 2 K 2 Q .

Therefore, the cost of (5.7) is bounded by m P 2 K 2 Q .

In general, if g = i = 1 m a i λ i , by repeating the previous process 𝑚 times, we get an upper bound m 2 P 2 K 2 Q . This completes the proof.

(b) This follows by applying (a) | c | - 1 times.

(c) Conjugating 𝑡 to each term of a 1 t , t T , costs zero relations. Then we basically estimate the cost of the following equation:

a 1 t t = a 1 t t ¯ .

By the result of Lemma 5.6, since Tr ( t t ) B ( deg t + deg t ) , the cost is bounded by ( 2 K ) ( deg t + deg t ) . Note that deg ( t ) Q ; thus the total cost is at most

( m P ) ( 2 K ) ( deg t + deg t ) ( m P ) ( 2 K ) ( Q + deg t ) .

Here, we use the fact Tr ( t ) B deg ( t ) since we order elements in Z T degree lexicographically. ∎

Recall that G is a factor group of H as 𝐺 is a factor group of G . Denote the epimorphism from H to G , induced by the identity on the generating set, by 𝜓, and then we have the following homomorphism chain:

H ψ G φ G .

Thus ker ψ = R 2 H and ker ( φ ψ ) = R 2 R 3 H . They are all normal subgroups in H as well as submodules in A H . In H , we have a free module structure, while each of 𝐺 and G contain a factor module of it. Eventually, we will convert the word problem to a membership problem of a submodule in A H .

5.5 Proof of Proposition 5.2

Now we are ready to prove Proposition 5.2. It is enough to show that, for any given word w = 1 of length 𝑛, 𝑤 can be written as a product of at most C n 2 k conjugates of relators for some constant 𝐶. Since 𝐺 is a factor group of H , w = G 1 if and only if

w ker ( φ ψ ) = R 2 R 3 H .

Note that R 2 R 3 H A H . Recall that A H has a natural module structure: it is a free 𝑇-module with basis a 1 , , a m . By the previous discussion, R 2 R 3 H coincides with the submodule generated by R 2 R 3 over Z T . Let R 4 = { f 1 , f 2 , , f l } be the Gröbner basis of the submodule generated by R 2 R 3 . We then add R 4 to our presentation (5.5), and in addition, we assume that all relators of R i , i = 2 , 3 , 4 , are written in their ordered form. Note that R 2 R 3 and R 4 generates the same submodule in A H . It implies that R 2 R 3 H = R 4 H . We obtained an alternating presentation of 𝐺 as

(5.8) G = A T R 1 R 2 R 3 R 4 .

Although R 4 is equivalent to R 2 R 3 , it is convenient to keep R 2 , R 3 in our presentation since all the estimation we have done previously is based on R 2 R 3 .

Note that 𝐺 is a factor group of G , and with the given presentation, Lemmas 5.4, 5.5, 5.6 and 5.7 all hold for 𝐺.

Since the Dehn function is a quasi-isometric invariant, then it is enough for us to prove Proposition 5.2 using presentation (5.8).

Proof of Proposition 5.2

We start with a word w G such that | w | = n , w = G 1 . Without loss of generality, we may assume

w = u 1 b 1 u 2 b 2 u s b s u s + 1 ,

where u i F = F ( T ) , b i A ± 1 and s + i = 1 s + 1 | u i | = n . Let v i = ( u 1 u i ) - 1 for i = 1 , , s and ν = u 1 u 2 u s + 1 . Then we have

w = w 1 := b 1 v 1 b 2 v 2 b s v s ν .

The equality holds in the free group generated by A T ; thus the cost of relations converting 𝑤 to w 1 is 0. Since s + i = 1 s + 1 | u i | = n , in particular, we have that s n . Moreover, | v i | = j = 1 i | u j | n ; hence Tr ( θ ( v i ) ) B n , i = 1 , 2 , , n . Next, since w 1 = G 1 , π ( w 1 ) = π ( v s + 1 ) = 1 . By Lemma 5.4,

ν = i = s + 1 s b i v i ,

where s - s | ν | 2 n 2 , b i A ± 1 and Tr ( θ ( v i ) ) B n , i = s + 1 , , s . By Lemma 5.4, the cost of converting 𝜈 to the right-hand side is bounded by | ν | 2 n 2 .

Thus we let

w 2 := i = 1 s b i v i , s n 2 + n , Tr ( θ ( v i ) ) B n , i = 1 , , s .

And the cost of converting w 2 to w 1 is bounded by n 2 .

Next, note that all v i are words in 𝐹. With the help of Lemma 5.6, we are able to organize v i to its image in F ¯ . More precisely, we let

w 3 := i = 1 s b i v ¯ i , s n 2 + n , θ ( v ¯ i ) n .

Also by Lemma 5.6, w 2 = G w 3 . Let us estimate the cost of converting w 2 to w 3 . To transform w 2 to w 3 , we need to apply Lemma 5.6 to each b i v i once. Since Area ( b i v i b i - v ¯ i ) ( 2 K ) n , which is given by Tr ( θ ( v i ) ) B n , each transformation costs ( 2 K ) n relations. We have in total s n + n 2 many conjugates to convert; therefore, the cost is bounded by ( n 2 + n ) ( 2 K ) n .

Now let w 4 be the ordered form of w 3 , which in fact is also the ordered form of 𝑤, i.e.

w 3 = G w 4 := i = 1 m a i μ i ,

where μ i are ordered under . By the discussion in Section 5.2, we obtain the ordered form just by rearranging all conjugates of A ± 1 . Note that, as θ ( v ¯ i ) n for all 𝑖, it costs at most K 2 n relations to commute any two consecutive conjugates b i v ¯ i and b j v ¯ j by Lemma 5.5. To sort s conjugates, we need to commute s 2 times. Therefore, the number of relations needed to commute w 3 to w 4 is bounded above so that s 2 K 2 n ( n 2 + n ) 2 K 2 n .

The only thing that remains is to compute the area of w 4 . Recall that w 4 can be regarded as an element in a free 𝑇-module generated by a 1 , , a m . The fact that w 4 = G 1 implies that either w 4 = H 1 or it lies in the submodule generated by R 4 = { f 1 , f 2 , , f l } which is the Gröbner basis of the submodule generated by R 2 R 3 . If w = H 1 , then μ i = for all i = 1 , , m . In this case, Area ( w 4 ) = 0 . Thus

Area ( w ) n 2 + ( n 2 + n ) ( 2 K ) n + ( n 2 + n ) 2 K 2 n .

We are done with this case.

Now let us consider the case w R 4 H { 1 } . Let 𝐾 be a constant large enough to satisfy both Corollary 3.4 and Lemma 5.5. As an element in the 𝑇-module, deg w 4 n since θ ( v ¯ i ) n for all 𝑖. Also recall that, for an element α Z T , | α | is defined to be the l 1 -norm of it regarded as a finitely supported function from 𝑇 to ℤ. Thus | w 4 | represents the number of conjugates in w 4 , which is s . Then, by Corollary 3.4, we have

w 4 = H i = 1 l f i α i , f i = a 1 μ i 1 a 2 μ i 2 a m μ i m R 4 , deg ( f i α i ) n , i = 1 l | α i | s K n 2 k ( n 2 + n ) K n 2 k ,

where μ i = j = 1 l α j μ j i in Z T . Note that f i α i is the product consisting of exactly | α i | many relators. In conclusion, we have

Area ( i = 1 l f i α i ) i = 1 l | α i | ( n 2 + n ) K n 2 k .

Last, let us estimate the cost of converting i = 1 l f i α i to w 4 . This process consists of two different steps: (1) converting all f i α i to their ordered form, (2) adding the 𝑙 terms of ordered f i α i .

To transform f i α i to its ordered form, we write

α i = u supp α i α i ( u ) u .

Let us denote P = max i = 1 l | f i | , Q = max i = 1 l deg ( f i ) . Then

(5.9) f i α i = f i u supp α i α i ( u ) u = supp α i f i α i ( u ) u = supp α i f i , u = a 1 μ i 1 a 2 μ i 2 a m μ i m = : f i ,

where f i , u is the ordered form of f i α i ( u ) u and u i j = α i μ i j ; hence f i is the ordered form of f i α i . The first two equalities above hold in the free group F ( A T ) ; thus the cost is 0. In the third equality, applying Lemma 5.7 (b) and (c), the cost of converting f i α i ( u ) u to f i , u is bounded by

m | f i | ( 2 K ) k ( deg f i + deg u ) + ( | α i ( u ) | - 1 ) m 2 | f i | 2 K 2 ( deg f i + deg u ) .

Here, we first conjugate 𝑢 to f i , then add | α i ( u ) | terms of f i u . Since deg u deg α i , we have u supp α i | α i ( u ) | = | α i | , | supp α i | | α i | . Consequently, the cost of the third equality of (5.9) is bounded by

u supp α i ( m | f i | ( 2 K ) k ( deg f i + deg u ) + ( | α i ( u ) | - 1 ) m 2 | f i | 2 K 2 ( deg f i + deg u ) ) u supp α i ( m | f i | ( 2 K ) k ( deg f i + deg α i ) + ( | α i ( u ) | - 1 ) m 2 | f i | 2 K 2 ( deg f i + deg α i ) ) = | supp α i | m | f i | ( 2 K ) k ( deg f i + deg α i ) + u supp α i ( | α i ( u ) | - 1 ) m 2 | f i | 2 K 2 ( deg f i + deg α i ) = | supp α i | m | f i | ( 2 K ) k ( deg f i + deg α i ) + ( | α i | - | supp α i | ) m 2 | f i | 2 K 2 ( deg f i + deg α i ) | α i | m | f i | ( 2 K ) k ( deg f i + deg α i ) + | α i | m 2 | f i | 2 K 2 ( deg f i + deg α i ) = | α i | ( m | f i | ( 2 K ) k ( deg f i + deg α i ) + m 2 | f i | 2 K 2 ( deg f i + deg α i ) ) | α i | ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) .

The last inequality is obtained by deg ( f i α i ) n , i.e. deg f i + deg α i n .

The fourth equality of (5.9) is adding all the f i , u up. Since

deg f i , u deg f i α i n , | f i , u | | α i ( u ) | | f i | | α i | | f i | | α i | P ,

by Lemma 5.7 (a), the cost of adding | supp α i | terms of f i , u is bounded by

( | supp α i | - 1 ) m 2 ( | α i | P ) 2 K 2 n .

Here, we use the fact that the size of the addition of any step is bounded by | f i α i | . Therefore, the total number of relations we need to convert each f i α i to its order form f i is bounded by

| α i | ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) + ( | supp α i | - 1 ) m 2 ( | α i | P ) 2 K 2 n | α i | ( m P ( 2 K ) k n + ( 1 + | α i | 2 ) m 2 P 2 K 2 n ) .

In general, the cost of converting all f i α i to their order forms is bounded by

i = 1 l | α i | ( m P ( 2 K ) k n + ( 1 + | α i | 2 ) m 2 P 2 K 2 n ) = ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) ( i l | α i | ) + m 2 P 2 K 2 n i = 1 l ( | α i | 3 ) ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) ( n 2 + n ) K n 2 k + m 2 P 2 K 2 n ( n 2 + n ) 3 K 3 n 2 k .

The next step, as described above, is to add all the f i up. We have | f i | | α i | | f i | and deg f i n for all i = 1 , 2 , , l . Moreover, the size of any partial product i = 1 l f i , 1 l l , is controlled by the following inequalities:

| i = 1 l f i | i = 1 l | α i | | f i | P i = 1 l | α i | P ( n 2 + n ) K n 2 k , deg ( i = 1 l α i f i ) n .

This is similar to adding the f i , u . By Lemma 5.7 (a), the cost of the ( | l | - 1 ) additions is bounded by

( l - 1 ) m 2 ( P ( n 2 + n ) K n 2 k ) 2 K 2 n ( l - 1 ) m 2 ( n 2 + n ) 2 P 2 K 2 n 2 k + 2 n .

Now we need to verify that the process of the above steps indeed yields the result w 4 . This is given by the fact that μ i = j = 1 l α j μ j i = j = 1 l μ i j , and eventually, following Lemma 5.7, we have

i = 1 l f i α i = i = 1 l f i = i = 1 l ( a 1 μ i 1 a 2 μ i 2 a m μ i m ) = j = 1 m a j j = 1 l μ i j = a 1 μ 1 a 2 μ 2 a m μ m = w 4 .

By our estimates, the cost of the first equality is bounded by

( m P ( 2 K ) k n + m 2 P 2 K 2 n ) ( n 2 + n ) K n 2 k + m 2 P 2 Q 2 n ( n 2 + n ) 3 K 3 n 2 k ,

and the cost of the third equality is bounded by ( l - 1 ) m 2 ( n 2 + n ) 2 P 2 K 2 n 2 k + 2 n . Other equalities hold in the free group, hence incur no cost. Therefore,

Area ( w 4 ) = ( n 2 + n ) K n 2 k + ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) ( n 2 + n ) K n 2 k + m 2 P 2 K 2 n ( n 2 + n ) 3 K 3 n 2 k + ( l - 1 ) m 2 ( n 2 + n ) 2 P 2 K 2 n 2 k + 2 n .

Now we choose a constant C > K large enough such that

( n 2 + n ) K n 2 k + ( m P ( 2 K ) k n + m 2 P 2 K 2 n ) ( n 2 + n ) K n 2 k + m 2 P 2 K 2 n ( n 2 + n ) 3 K 3 n 2 k + ( l - 1 ) m 2 ( n 2 + n ) 2 P 2 K 2 n 2 k + 2 n C n 2 k .

It is clear that such a 𝐶 exists; for example, we can choose 𝐶 to be 4 m 2 P 2 Q K . Note that 𝑃 and 𝑄 only depend on f 1 , f 2 , , f l , hence R 4 , and so does 𝐾. Therefore, 𝐶 is independent of 𝑤.

In conclusion, we start with w = G 1 of length at most 𝑛. By converting it four times, we end up with a word w 4 , whose area is bounded by C n 2 k . Thus

w w 1 n 2 w 2 ( n + n 2 ) ( 2 K ) n w 3 ( n + n 2 ) 2 K 2 n w 4 .

Summing up all the cost from w 1 to w 4 and with the fact C > K , we conclude that the area of 𝑤 is bounded above by

Area ( w ) C n 2 k + ( n + n 2 ) 2 C 2 n + ( n + n 2 ) ( 2 C ) n + n 2 .

This completes the proof. ∎

6 Relative Dehn functions for metabelian groups

6.1 The cost of metabelianness

First, we state an important consequence of Lemma 5.5. Consider a finitely presented group 𝐺 with a short exact sequence 1 A G T 1 , where A , T are abelian groups. By definition, 𝐺 is metabelian; this follows from the abelianness of 𝐴. More precisely, if we let 𝒜 be a generating set for 𝐴 and 𝒯 a set in 𝐺 whose image in 𝑇 generates 𝑇, the abelianness of 𝐴 is given by all commutative relations like [ a u , b v ] = 1 for all a , b A , u , v F ( T ) . It follows that all commutators commute. Recall Lemma 5.5, from which we immediately have the following theorem.

Theorem 6.1

The metabelianness of a finitely presented metabelian group 𝐺 costs at most exponentially many relations with respect to the length of the word, i.e. there exists a constant 𝐶 such that

Area ( [ [ x , y ] , [ z , w ] ] ) C 2 | [ [ x , y ] , [ z , w ] ] | for all x , y , z , w G .

Proof

Let 𝑘 be the minimal torsion-free rank of an abelian group 𝑇 for which there exists an abelian normal subgroup 𝐴 in 𝐺 satisfying G / A T . The projection of 𝐺 onto 𝑇 is denoted by π : G T .

If k = 0 , 𝐺 has a finitely generated abelian subgroup of finite index. Then the result follows immediately.

If k > 0 , we first consider the case where 𝑇 is free abelian. Let

T = { t 1 , , t k } G

be such that { π ( t 1 ) , , π ( t k ) } forms a basis for 𝑇, and let 𝒜 be a finite subset of 𝐺 that contains all commutators a i j = [ t i , t j ] for 1 i < j k and generates the 𝑇-module 𝐴. Then A T is a finite generating set for the group 𝐺. Recall that 𝐺 has a finite presentation as follows:

G = a 1 , a 2 , , a m , t 1 , t 2 , , t k R 1 R 2 R 3 R 4 ,

where

R 1 = { [ t i , t j ] = a i j 1 i < j k } ,
R 2 = { [ a , b u ] = 1 a , b A , u F ¯ , θ ( u ) < R } ,
R 3 = { u F ¯ ( a λ ( u ^ ) ) u = a | a A , λ Λ C ( A ) } { u F ¯ ( a λ ( u ^ ) ) u - 1 = a | a A , λ Λ C ( A * ) }
and R 4 is the finite set generating ker φ . All the notation is the same as in Section 5.1.

By Lemma 5.5, we have that there exists a constant C 1 such that

Area [ a , b u ] C 1 2 | [ a , b u ] | , a , b A , u F ( T ) .

Now let x , y , z , w be elements in 𝐺 and n = | [ [ x , y ] , [ z , w ] ] | . We use two commutator identities [ a , b c ] = [ a , c ] [ a , b ] c and [ a b , c ] = [ a , c ] b [ b , c ] to decompose [ x , y ] and [ z , w ] into products of [ a , b ] u , where a , b A A - 1 T T - 1 , u G . There are three cases to be considered.

  1. If a , b A A - 1 , [ a , b ] u = G 1 and the cost for converting [ a , b ] u to 1 is 1.

  2. If a , b T T - 1 , we have two cases. If a , b { t i , t i - 1 } for some 𝑖, then [ a , b ] u = G 1 with no cost. If a { t i , t i - 1 } and b { t j , t j - 1 } , where i j , then [ a , b ] u = c ε u , where c A A - 1 , ε { ± 1 } , | u | | u | + 1 . This is due to cases like [ t i - 1 , t j ] = [ t i , t j ] - t i - 1 . The cost of converting [ a , b ] u to c ε u is 1.

  3. If a A A - 1 , b T T - 1 (or b A A - 1 , a T T - 1 ), then we have [ a , b ] = G a a b (resp. [ a , b ] = b a - 1 b ). Thus we get [ a , b ] u = G a u a b u (resp. [ a , b ] u = b a - 1 u b u ). The cost of converting is 0.

It follows that [ x , y ] = G i = 1 l b i ε i u i , where l 2 | x | | y | , b i A , ε i { ± 1 } , u i G , | u i | | x | + | y | . The cost of converting [ x , y ] to i = 1 l b i ε i u i is bounded by | x | | y | .

Let 𝑢 be a word in 𝐺; we claim that u = w 1 w 2 i = 1 p c i v i , where w 1 F ( A ) , w 2 F ( T ) , c i A A - 1 , v i F ( T ) . The claim can be proved by always choosing to commute the left most pair of t a , where t T T - 1 , a A A - 1 . Then, for an element a A , we have

a u = F ( A T ) a w 1 w 2 i = 1 p c i v i = G a w 2 , w 1 F ( A ) , w 2 F ( T ) .

Note that | w 2 | < | u | . Similar to Lemma 5.6, there exists a constant C 2 such that the cost of the second equality in terms of relations is bounded C 2 2 | u | .

Therefore,

[ x , y ] = G i = 1 l b i ε i u i ,

where l 2 | x | | y | , b i A , ε i { ± 1 } and u i F ( T ) . The cost is bounded by 2 n 2 C 2 2 n . Consequently, [ [ x , y ] , [ z , w ] ] can be written as a product of at most 8 n 2 conjugates of elements in 𝒜 at a cost of at most 8 n 2 C 2 2 n . Lastly, converting this product to 1 costs at most ( 8 n 2 ) 2 C 1 2 n by Lemma 5.5. The theorem is proved in this case.

If 𝑇 is not free abelian, we choose T = { t 1 , t 2 , , t k , t k + 1 , t k + 2 , , t s } such that the set T 0 := { π ( t k + 1 ) , π ( t k + 2 ) , , π ( t s ) } generates a finite abelian group and { π ( t 1 ) , π ( t 2 ) , , π ( t k ) } generates Z k . Then we can write down a presentation of 𝐺 as follows:

G = a 1 , a 2 , , a m , t 1 , t 2 , , t s R 1 R 2 R 3 R 4 ,

where

R 1 = { [ t i , t j ] = a i j , t l n l = a l 1 i < j s , k + 1 l s } ,
R 2 = { [ a , b u ] = 1 a , b A , u F ¯ , θ ( u ) < R } ,
R 3 = { u F ¯ ( a λ ( u ^ ) ) u = a | a A , λ Λ C ( A ) } { u F ¯ ( a λ ( u ^ ) ) u - 1 = a | a A , λ Λ C ( A * ) }
and R 4 is the finite set generating ker φ . Note that θ : F ¯ R k kills all t l , l > k . The rest of the proof is the same as the case when 𝑇 is free abelian. ∎

6.2 The relative Dehn functions of metabelian groups

Recall that a set of groups forms a variety if it is closed under subgroups, epimorphic images and unrestricted direct products. The set of metabelian groups naturally form a variety, denoted by S 2 , since metabelian groups satisfy the identity [ [ x , y ] , [ z , w ] ] = 1 . Inside a variety, we can talk about relative free groups and relative presentations. Firstly, a metabelian group M k is free of rank 𝑘 if it satisfies the following universal property: every metabelian group generated by 𝑘 elements is an epimorphic image of M k . It is not hard to show that M k F ( k ) / F ( k ) ′′ , where F ( k ) is a free group of rank 𝑘 (in the variety of all groups).

Next, we shall discuss the relative presentations. Recall that the usual presentation of 𝐺 consists of a free group 𝐹 and a normal subgroup 𝑁 such that G F / N . For relative presentations, we shall replace the free group with the relative free group. Now let 𝐺 be a metabelian group that is generated by 𝑘 elements; then there exists an epimorphism φ : M k G , where M k is generated by X = { x 1 , x 2 , , x k } . We immediately have that G M k / ker φ . Note that ker φ is a normal subgroup of M k , so it is a normal closure of a finite set. We let R = { r 1 , r 2 , , r m } be the finite set whose normal closure is the kernel of 𝜑. Therefore, we obtain a relative presentation of 𝐺 in the variety of metabelian groups G = x 1 , x 2 , , x k r 1 , r 2 , , r m S 2 . The notation S 2 is used to indicate that the presentation is relative to the variety of metabelian groups S 2 . Here, the subscript two stands for the derived length two. We denote by 𝒫 the relative presentation X R S 2 . Note that if 𝐺 is finitely presented, then the finite presentation in the usual sense is also a relative presentation, with some redundant relations.

Let us give an example of a relative presentation of a metabelian group which is not finitely presented in the variety of all groups. The group H , which we introduced in Section 4.2, is a free metabelian group of rank 𝑘. It has two different relative presentations depending on how many generators we choose,

H = t 1 , t 2 , , t k S 2 = a i j , t 1 , t 2 , , t k a i j = [ t i , t j ] ,  1 i < j k S 2 .

Now we return to a finitely generated metabelian group 𝐺 with finite relative presentation X R S 2 . Let 𝑤 be a word in 𝐺 such that w = G 1 . Then 𝑤 lies in the normal closure of 𝑅. Thus 𝑤 can be written as

w = M k i = 1 l r i f i , where r i R R - 1 , f i M k .

The smallest possible 𝑙 is called the relative area of 𝑤, denoted by Area ~ P ( w ) . The difference between the area and the relative area is that we take the equality in different ambient groups, one in free groups and the other in free metabelian groups. Consequently, the Dehn function relative to the variety of metabelian groups with respect to the presentation 𝒫 is defined as δ ~ P ( n ) = sup { Area ~ P ( w ) | w | X n } . Here, | | X is the word length in alphabet 𝑋. Similar to the usual Dehn functions, the relative Dehn functions are also independent of the choice of finite presentations up to equivalence.

Proposition 6.2

Proposition 6.2 ([11])

Let 𝒫 and 𝒬 be finite relative presentations of the finitely generated metabelian group 𝐺. Then δ ~ P δ ~ Q .

Therefore, it is valid to denote the relative Dehn function of a finitely generated metabelian group 𝐺 by δ ~ G . One remark is that every finitely generated metabelian group is finitely presentable relative to the variety of metabelian groups. Thus the relative Dehn function can be defined for all finitely generated metabelian groups. Not also that, unlike Dehn functions, it does not make sense to talk about whether or not the relative Dehn function is a quasi-isometric invariant. This is because groups that are quasi-isometric to a finitely generated metabelian group may not even be metabelian. For example, A 5 Z and Z 60 Z are quasi-isometric, while the former is not metabelian.

But the relative Dehn function still inherits some nice properties from the Dehn function: one of which is that it is preserved under taking a finite index subgroup.

Proposition 6.3

Let 𝐺 be a finitely generated metabelian group and 𝐻 a finite index subgroup of 𝐺. Then 𝐻 is finitely generated and metabelian, and the relative Dehn functions of 𝐺 and 𝐻 are equal up to equivalence.

Proof

Since 𝐻 is a finite index subgroup of a finitely generated metabelian group, it is finitely generated and metabelian.

Let X R S 2 be a finite relative presentation of 𝐻 with X = { x 1 , x 2 , , x n } . Then we have a finite relative presentation of 𝐺 as follows:

G = X Y R 0 R 1 R 2 S 2 ,

where

Y = { y 1 , y 2 , , y m } ,
R 0 = R ,
R 1 = { y i y j = y f ( i , j ) w i , j , y i - 1 = y g ( i ) u i } , w i , j , u i ( X X - 1 ) * , f : { 1 , 2 , , m } × { 1 , 2 , , m } { 1 , 2 , , m } , g : { 1 , 2 , , m } { 1 , 2 , , m } ,
R 2 = { x l y i = y j v i , j } , v l , i ( X X - 1 ) * .
We claim that there exists a constant 𝐿 such that, for every word w = G 1 , there exists a word w such that w = w , w ( X X - 1 ) * and | w | L | w | . Moreover, it costs at most | w | relations from R 1 R 2 to convert 𝑤 to w .

If the claim is true, then we have that

Area ~ G ( w ) Area ~ H ( w ) + | w | .

It immediately implies that δ ~ G ( n ) δ ~ H ( L n ) + n . Thus δ ~ G ( n ) δ ~ H ( n ) . And the other direction δ ~ H ( n ) δ ~ G ( n ) is obvious since w H = 1 implies w G = 1 .

To prove the claim, let L = max { | w i , j | , | u i | , | v l , i | 1 i , j m ,  1 l n } . Let 𝑤 be a word such that w = G 1 . Without loss of generality, we assume that 𝑤 has the following form:

w = F ( X Y ) a 1 b 1 a 2 b 2 a k b k a k + 1 , a i ( X X - 1 ) * , b i ( Y Y - 1 ) * ,

where only a 1 , a k + 1 can be the empty word. For b k , using relations in R 1 , we have that

b k = y h ( k ) b k , h ( k ) { 1 , 2 , , m } , b k ( X X - 1 ) * ,

and | b k | L | b k | . Thus w = a 1 b 1 a 2 b 2 a k y h ( k ) b k a k + 1 , while the cost of converting is bounded by | b k | and all relations are from R 1 .

Next, we commute y h ( k ) with a k using relations from R 2 ,

a k y h ( k ) = y h ( k ) a k , a k ( X X - 1 ) * ,

and | a k | L | a k | . Substituting this in, we get w = a 1 b 1 a 2 b 2 y h ( k ) a k b k a k + 1 , while the cost of converting is bounded by | a k | and all relations are from R 2 .

Therefore, repeating the above process, we eventually have

w = y h ( 1 ) a 1 b 1 a 2 b 2 a k b k a k + 1 .

Since w = 1 , y h ( 1 ) is actually an empty word. Consequently, we have

w = a 1 b 1 a 2 b 2 a k b k a k + 1 ( X X - 1 ) * ,

and the length of the left-hand side is controlled by

| a 1 b 1 a 2 b 2 a k b k a k + 1 | i = 1 k L ( | a i | + | b i | ) + | a k + 1 | L | w | .

The cost of relations is bounded by i = 1 k ( | a i | + | b i | ) | w | , while all relations are from R 1 R 2 . The claim is proved. ∎

Let us consider one classic example: the Baumslag–Solitar group BS ( 1 , 2 ) . The relative presentation is the same as the usual presentation

BS ( 1 , 2 ) = a , t a t = a 2 S 2 .

But one can prove that the relative Dehn function of BS ( 1 , 2 ) is 𝑛 instead of the usual Dehn function 2 n (see [11]). 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 6.4.

So what is the connection between the relative Dehn function and the Dehn function? On the surface, in the relative presentation, we make all metabelian relations cost 0, which should result in a significant reduction in the cost in lemmas like Lemma 5.5, Lemma 5.7. In the next section, we will estimate how much cost we reduce by introducing the relative presentation.

6.3 Connections between Dehn functions and relative Dehn functions

The goal of the section is to prove the following theorem.

Theorem 6.4

Let 𝐺 be a finitely presented metabelian group. Then

δ ~ G ( n ) δ G ( n ) max { δ ~ G 3 ( n 3 ) , 2 n } .

Before we prove the theorem, we have to introduce the third “Dehn function” in this paper: the Dehn function of a finitely generated module. Let 𝑅 be the group ring Z T , where 𝑇 is a free abelian group with basis { t 1 , t 2 , , t k } and 𝐴 is a finitely generated 𝑅-module generated by 𝑚 elements. Let 𝑀 be a free 𝑅-module of rank 𝑚. Suppose a basis of 𝑀 is { a 1 , a 2 , , a m } . Then there exists a submodule 𝑆 of 𝑀, generated by { f 1 , f 2 , , f l } , such that A M / S .

For an element f = μ 1 a 1 + μ 2 a 2 + + μ m a m in 𝑀, we define its length, denoted by f , to be the following:

f = i = 1 l | μ i | + reach ( f ) ,

where reach ( f ) is the minimal length of a closed loop that starts at 1 and passes through all points in i = 1 l supp μ i in the Cayley graph of 𝑇. Another way to think of this length is that it is the minimal length of a word that is equal to a 1 μ 1 a 2 μ 2 a m μ m in H (see Section 4.2), i.e. the minimal length of a word among words that are a rearrangement of all conjugates of elements in 𝒜 in a 1 μ 1 a 2 μ 2 a m μ m . For example, suppose m = k = 1 ; then we have

( t 1 n + t 1 n - 1 + + t 1 + 1 ) a 1 = ( n + 1 ) + 2 n = 3 n + 1

because the minimal length of a loop passing { 1 , t , t 2 , , t n } is 2 n . Note that a 1 t 1 n + t 1 n - 1 + + t 1 + 1 = t 1 - n a 1 t 1 a 1 a 1 t 1 a 1 is a word of length 3 n + 1 .

Then, for every element 𝑓 in 𝑆, there exist α 1 , α 2 , , α l R such that

f = α 1 f 1 + α 2 f 2 + + α l f l .

We denote by Area ^ A ( f ) the minimal possible i = 1 l | α i | . Then the Dehn function of the 𝑅-module 𝐴 is defined to be

δ ^ A ( n ) = max { Area ^ A ( f ) f n } .

Remark

Now we have three different types of Dehn functions: 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 ) and Area ( w ) the Dehn function of 𝐺 and the area of a word 𝑤, by δ ~ G ( n ) and Area ~ ( w ) the relative Dehn function and the relative area of a word 𝑤, and by δ ^ A ( n ) and Area ^ ( f ) the Dehn function of the module 𝐴 and the area of a module element 𝑓.

Let k = rk ( G ) be 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 .

First, if k > 0 , we note that the problem can be reduced in the same way as Proposition 5.2 does because, for a finitely presented metabelian group 𝐺, there exists a subgroup G 0 of finite index such that G 0 is an extension of an abelian group by a free abelian group of rank 𝑘. Most importantly, by Corollary 2.3 and Proposition 6.3, their Dehn functions are equivalent as well as their relative Dehn functions. Therefore, from now on, we assume that 𝐺 is an extension of an abelian group 𝐴 by a free abelian group 𝑇. The projection of 𝐺 onto 𝑇 is denoted by π : G T .

Let T = { t 1 , , t k } G such that { π ( t 1 ) , , π ( t k ) } forms a basis for 𝑇, and let 𝒜 be a finite subset of 𝐺 that contains all commutators a i j = [ t i , t j ] for 1 i < j k and generates the 𝑇-module 𝐴. Then A T is a finite generating set for the group 𝐺.

Recall that 𝐺 has a finite presentation as follows:

G = a 1 , a 2 , , a m , t 1 , t 2 , , t k R 1 R 2 R 3 R 4 ,

where

R 1 = { [ t i , t j ] = a i j 1 i < j k } ,
R 2 = { [ a , b u ] = 1 a , b A , u F ¯ , θ ( u ) < R } ,
R 3 = { u F ¯ ( a λ ( u ^ ) ) u = a | a A , λ Λ C ( A ) } { u F ¯ ( a λ ( u ^ ) ) u - 1 = a | a A , λ Λ C ( A * ) }
and R 4 is the finite set that generates ker φ as a normal subgroup. All the notation is the same as in Section 5.5.

Since we are dealing with the relative Dehn function, we can reduce the amount of redundant relations in R 2 . We set

R 2 = { [ a , b ] = 1 , [ a , b t ] = 1 a , b A , t T } .

Lemma 6.5

R 2 generates 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 by 4 | u | - 3 .

Proof

Suppose the result is proved for | u | n , i.e. [ a , b u ] = 1 can be written as a product of conjugates of words in R 2 and metabelian relations. For metabelian relations, we mean those relations that make commutators commute to each other. Note that the relative area of any metabelian relation is 0.

Now, for the case | u | = n + 1 , let u = v t , | v | = n , t { t 1 , t 2 , , t k } . By metabelian relations, we have that 1 = [ a - 1 a t , b - t b u ] . This is because a - 1 a t = [ a , t ] and b - t b u = [ b , v ] t . Then, by the inductive assumption, we are able to use relations like [ a , b w ] = 1 when a , b A , | w | n . In particular, [ a , b v ] = 1 .

Now note that

1 = [ a - 1 a t , b - t b u ] = a - t a commute b - u b t a - 1 a t commute b - t b u = a a - t b - u commute b t a t commute a - 1 b - t b u = a b - u a - t a t b t a - 1 commute b - t b u = a b - u a - 1 b u .

This shows that [ a , b u ] can be generated by R 2 and metabelian relations. Let us count the cost. In the computation above, we use [ a , b v ] = 1 once (note that [ a t , b u ] = [ a , b v ] t ), [ a , a t ] = 1 twice, [ a , b ] = 1 once and [ a , b t ] once. Therefore,

Area ~ ( [ a , b u ] ) Area ~ ( [ a , b v ] ) + 4 4 ( | v | + 1 ) - 3 = 4 ( n + 1 ) - 3 .

This completes the proof. ∎

The lemma allows us to replace R 2 by R 2 in the relative presentation. And we immediately get the relative version of Lemma 5.5.

Lemma 6.6

Let 𝑢 be a reduced word in F ( T ) , and let u ¯ be the unique word in 𝑇 representing 𝑢 in the form of t 1 m 1 t 2 m 2 t k m k . Then we have

Area ~ ( a - u a u ¯ ) 4 | u | 2 + 2 | u | .

Proof

The only difference between this proof and the proof of Lemma 5.5 is that now it only costs 4 | u | - 3 to commute conjugates of elements in 𝒜 every time. ∎

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

Our solution for the membership problem of a submodule has a better control for the degree of α i f i than for i = 1 l | α i | , resulting in an enormous upper bound for the area. When we consider α 1 , α 2 , , α l that minimize i = 1 l | α i | , one trade-off is that we lose control of the degree of α i , sort of. The following lemma shows that, in this case, even though the degree of α i cannot be linearly controlled just by deg f , it can still be linearly controlled by both deg f and i = 1 l | α i | .

Lemma 6.7

There exists a constant 𝐶 such that, for every f S , where 𝑆 is a 𝑇-submodule generated by { f 1 , f 2 , , f l } , with α 1 , α 2 , , α l Z T such that

f = α 1 f 1 + α 2 f 2 + + α l f l ,

and i = 1 l | α i | minimized, we have deg α i f i deg f + C i = 1 l | α i | for all 𝑖.

Proof

We denote by Δ ( f ) the difference in the degree of the highest and lowest terms. Let

C := max { Δ ( f 1 ) , Δ ( f 2 ) , , Δ ( f l ) } .

In addition, we let s = i = 1 l | α i | .

We rewrite the sum α 1 f 1 + α 2 f 2 + + α l f l to the form

f = u 1 f i 1 + u 2 f i 2 + + u s f i s ,

where i j { 1 , 2 , , l } and u j is an element in i = 1 l supp α i , such that

deg ( u j f i j ) deg ( u j + 1 f i j + 1 ) for j = 1 , 2 , , s - 1 .

Let g n be the partial sum g n = j = 1 n u j f i j . Assume that deg ( u 1 f i 1 ) > deg f ; otherwise, there is nothing to prove. Every term of degree greater than deg f will be canceled. Since we assume that deg ( u j f i j ) deg ( u j + 1 f i j + 1 ) , we have that deg g n deg g n + 1 when deg g n > deg f . We claim that deg g n + 1 deg g 1 - C n whenever deg g n > deg f . If the claim is not true, let n 0 be the least number such that

deg ( j = 1 n 0 u j f i j ) deg ( u 1 f i 1 ) - C ( n 0 - 1 ) ,
deg ( j = 1 n 0 + 1 u j f i j ) deg ( u 1 f i 1 ) - C n 0 .

Note that Δ ( u j f i j ) C for all j = 1 , 2 , , s . The least degree among terms in g n 0 is greater than deg ( u n 0 f i n 0 ) - C . Since deg ( g n 0 ) > deg ( g n 0 + 1 ) , we have

deg ( u n 0 + 1 f n 0 + 1 ) = deg ( g n ) deg ( u 1 f i 1 ) - C ( n 0 - 1 ) .

Moreover, since deg ( u n 0 f n 0 ) deg ( u n 0 + 1 f n 0 + 1 ) , the least degree among terms in g n 0 is greater than deg ( u 1 f i 1 ) - C n 0 . Therefore, if g n 0 + 1 0 , the least degree among terms in g n 0 + 1 is also greater than deg ( u 1 f i 1 ) - C n 0 . It follows that g n 0 + 1 = 0 . We have

f = j = n 0 + 2 s u j f i j .

This is a contradiction to the minimality of s = i = 1 l | α i | .

Let n 1 be the largest number such that deg g n 1 > deg f . By the claim, we have that deg g n 1 + 1 deg ( g 1 ) - C n 1 . By the definition of n 1 , deg g n 1 + 1 deg f . Combining those two inequalities, we finally have deg f deg ( g 1 ) - C n 1 . Since n 1 < s = i = 1 l | α i | and deg g 1 = max i { deg α i f i } , the lemma is proved. ∎

Next, we focus on the 𝑇-module 𝐴. It is not hard to see that 𝐴 is the quotient of the free 𝑇-module generated by 𝒜 by the submodule generated by R 3 R 4 . We then replace R 3 R 4 by the Gröbner basis R 3 = { f 1 , f 2 , , f l } for the same submodule. Therefore, we finally have the relative presentation of 𝐺 we want,

G = a 1 , a 2 , , a m , t 1 , t 2 , , t k R 1 R 2 R 3 S 2 .

We let 𝑀 be the free 𝑇-module generated by 𝒜 and 𝑆 the submodule generated by R 3 over 𝑇 so that A M / S .

Then we have a connection between the relative Dehn function of 𝐺 and the Dehn function of the submodule 𝑆.

Lemma 6.8

Let 𝐺 be a finitely generated metabelian group and 𝐴 defined as above, then

δ ^ A ( n ) δ ~ G ( n ) max { δ ^ A 3 ( n 3 ) , n 6 } .

Proof

Now let 𝑤 be a word of length 𝑛 such that w = G 1 . We then estimate the cost of converting it to the ordered form. The process is exactly the same as in the proof of Proposition 5.2. We replace the cost by the cost in relative presentation by Lemma 6.5 and Lemma 6.6. It is not hard to compute that it costs at most n 2 + ( 4 n - 3 ) ( n 2 + n ) + ( 4 n - 3 ) 2 ( n 2 + n ) 2 to convert 𝑤 to its ordered form w := i = 1 m a i μ i , where i = 1 n | μ i | n 2 , deg μ i n and | w | 2 n 3 . Since w lies in the normal subgroup generated by R 3 , then there exist α 1 , α 2 , , α l such that

w = i = 1 l f i α i , i = 1 l | α i | δ ^ A ( 2 n 3 ) .

The relative area of the left-hand side is less than i = 1 l | α i | . Then we just repeat the same process as in the proof of Proposition 5.2 and compute the cost of adding f i α i up to w . By Lemma 6.7, deg ( α i f i ) n + C δ ^ A ( 2 n 3 ) for every 𝑖 and some constant 𝐶. It follows that, by Lemma 6.6, conjugating α i to f i costs at most | α i | | f i | ( 4 deg 2 ( α i f i ) + 2 deg ( α i f i ) ) . Lastly, we rearrange i = 1 l | α i | | f i | terms whose degrees are at most n + C δ ^ A ( 2 n 3 ) , which costs at most

max { δ ^ A 3 ( n 3 ) , n 2 δ ^ A ( n 3 ) }

up to equivalence. Thus the relative area of w is asymptotically bounded by max { δ ^ A 3 ( n 3 ) , n 2 δ ^ A ( n 3 ) } up to equivalence. And hence the relative area of 𝑤 is bounded by max { δ A 3 ( n 3 ) , n 6 } . Thus the right inequality is proved.

For the left inequality in the statement, let i = 1 m a i μ i be a word of ordered form that realizes δ A ( n ) . The length of the word, by definition, is bounded by 𝑛. We claim that the relative area of i = 1 m a i μ i is greater than δ A ( n ) . If not, by the definition of the relative area, we have that

i = 1 m a i μ i = j = i s r i h i , r i R 1 ± 1 R 2 ± R 3 ± , h i M m + k ,

where s = Area ( i = 1 m a i μ i ) < δ S ( n ) . If we only keep all the relations from R 3 and combine the same relations, we will get i = 1 l f i α i and i = 1 l | α i | s < δ A ( n ) . Since canceling relations like [ t i , t j ] = a i j , [ a , b t ] = 1 and commuting the f i h j do not change the value of the left-hand side as an element in the free 𝑇-module generated by basis { a 1 , a 2 , , a m } , we eventually get

i = 1 m μ i a i = j = 1 l α j f j , i = 1 l | α i | < δ A ( n ) .

This leads to a contradiction. ∎

Theorem 6.9

Let 𝐺 be a finitely generated metabelian group. Let k = r k ( 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. n 2 if k = 0 ,

  2. 2 n 2 k if k > 0 .

Proof

Let 𝐺 be a finitely generated metabelian group. If k = 0 , then 𝐺 has a finitely generated abelian subgroup of finite index. Then the relative Dehn function is asymptotically bounded by n 2 .

If k > 0 , similarly, we can reduce the case to that where 𝐺 is an extension of a module 𝐴 by a free abelian group 𝑇 such that the torsion-free rank of 𝑇 is 𝑘. Then a word w = G 1 with | w | n can be converted to its ordered form w := i = 1 m a i μ i , where | w | 2 n 3 , deg ( w ) n and i = 1 m | μ i | n 2 . Then, by Corollary 3.4, there exists a word w ′′ such that w = G w ′′ , Area ( w ′′ ) 2 n 2 k . The theorem follows immediately. ∎

Finally, we have all the ingredients to prove Theorem 6.4.

Proof of Theorem 6.4

The left inequality is obvious since the finite presentation of 𝐺 is also the relative finite presentation of 𝐺.

If k = 0 , then 𝐺 has a finitely generated abelian subgroup of finite index. The result follows immediately.

If k > 0 , let 𝑤 be a word of length 𝑛 and w = G 1 . Then there exist α 1 , α 2 , , α l such that

(6.1) w = i = 1 l f i α i , i = 1 l | α i | δ ^ A ( 2 n 3 ) , deg α i n + C δ ^ A ( 2 n 3 ) .

According to the proof of Proposition 5.2, adding the left-hand side of (6.1) costs at most max { δ ^ A 3 ( 2 n 3 ) , 2 n } up to equivalence. All other conversion steps cost at most exponential with respect to 𝑛. Then, by the left inequality in Lemma 6.8, Area ( w ) max { δ ~ G 3 ( n 3 ) , 2 n } . Therefore, the theorem is proved. ∎

6.4 Estimate relative Dehn functions

Computing the relative Dehn function is harder than computing the Dehn function. Many techniques are no longer useful for the relative case. For the variety of metabelian groups, fortunately, the structure of these groups is not complicated. The key is to understand the natural module structure of a finitely generated metabelian group.

First, let us list some known results for relative Dehn functions; they are computed by Fuh in her thesis. Note that most of them only give the upper bound of the relative Dehn function.

Theorem 6.10

Theorem 6.10 (Fuh [11])

The following statements hold.

  1. The relative Dehn function of a wreath product of two finitely generated abelian groups is polynomially bounded.

  2. The Baumslag–Solitar group BS ( 1 , 2 ) has linear Dehn function.

  3. Let G = BS ~ ( n , m ) = a , t ( a n ) t = a m S 2 , where m > 2 , m = n + 1 . Then δ ~ G ( n ) n 3 .

There is one result in [11, Theorem E], we can improve.

Proposition 6.11

Let 𝑇 be a finitely generated abelian group and 𝐴 a finitely generated 𝑇-module, and form the semidirect product G = A T . Then we have δ G ( n ) max { n 3 , δ ^ A 3 ( n 2 ) } .

Proof

It is not hard to reduce the problem to the case when 𝑇 is free abelian. Thus we just assume that 𝑇 is a finitely generated free abelian group. Suppose T = { t 1 , t 2 , , t k } is a basis of 𝑇 and A = { a 1 , a 2 , , a m } generates the module 𝐴 over Z T . Let 𝑀 be the free 𝑇-module generated by 𝒜, and let 𝑆 be a submodule of 𝑀 generated by f 1 , f 2 , , f l , where f i = j = 1 m α i , i a j for 1 i k , α i , j Z T . Then we can write down a presentation of 𝐺 as follows:

G = a 1 , a 2 , , a m , t 1 , t 2 , , t m | [ t i , t j ] = 1 ( 1 i < j k ) , [ a i , a j w ] = 1 ( 1 i < j m , w Z T ) , j = 1 m a j α i , j = 1 ( 1 i l ) .

Then, by the same discussion as in Section 6.3, we have a finite relative presentation of 𝐺,

G = a 1 , a 2 , , a m , t 1 , t 2 , , t m | [ t i , t j ] = 1 ( 1 i < j k ) , [ a i , a j ] = 1 , [ a i , a j t s ] = 1 ( 1 i < j m ,  1 s k ) , j = 1 m a j α i , j = 1 ( 1 i l ) S 2 .

Now let w = G 1 and | w | n . Since, in this case, all t i , t j commute, it is much easier than the general case. Following the same process as in the proof of Proposition 5.2, 𝑤 can be converted to its ordered form a 1 μ 1 a 2 μ 2 a m μ m , where‌

deg ( μ i ) < n , i = 1 m | μ i | n .

The cost is bounded by n 3 . Notice that the length of a 1 μ 1 a 2 μ 2 a m μ m is bounded by n 2 . Then there exist α 1 , α 2 , , α l Z T such that

a 1 μ 1 a 2 μ 2 a m μ m = i = 1 l f i α i , i = 1 l | α i | δ ^ A ( n 2 ) .

The rest of the proof is the same as the proof of Lemma 6.8, since it is just a special case of Lemma 6.8. ∎

7 Examples and further comments

7.1 Subgroups of metabelian groups with exponential Dehn function

It is time to apply our technique to some concrete examples and investigate all the obstacles preventing us from constructing a finitely presented metabelian group with Dehn function that exceeds the exponential function.

The class of examples we investigate in this section was introduced by Baumslag in 1973 [4]. Let 𝐴 be a free abelian group of finite rank freely generated by { a 1 , a 2 , , a r } . Furthermore, let 𝑇 be a finitely generated abelian group with basis { t 1 , t 2 , , t k , , t l } , where t 1 , , t k are of infinite order and t k + 1 , , t l are, respectively, of finite order m k + 1 , , m l . Finally, let F = { f 1 , f 2 , , f k } be a set of elements f i from Z T , where each f i is of the form

f i = 1 + c i , 1 t i + c i , 2 t i 2 + + c i , d i - 1 t i d i - 1 + t i d i , d i 1 , c i , j Z .

Now let us define a group W F corresponding to 𝐹. The generating set is the following:

X = { a 1 , a 2 , , a r , t 1 , t 2 , , t l , u 1 , , u k } ,

where r , k , l are the same integers as above.

The defining relations of W F are of four kinds. First, we have the power relations

t i m i = 1 , i = k + 1 , , l .

Next, we have the commutativity relations

{ [ u i , u j ] = 1 , 1 i , j k , [ t i , t j ] = 1 , 1 i , j l , [ t i , u j ] = 1 , 1 i l ,  1 j k , [ a i , a j ] = 1 , 1 i , j r .

Thirdly, we have the commutativity relations for the conjugates of the generators a i ,

[ a i u , a j w ] = 1 , 1 i , j r ,

where

u , w { t 1 α 1 t 2 α 2 t l α l 0 α i d i for i = 1 , , k , 0 α i < m i for i = k + 1 , , l } .

Finally, we have relations defining the action of u j on a i ,

a i u j = a i f j , 1 i r ,  1 j k .

It is not hard to show that W F is metabelian [4]. Moreover, Baumslag showed the following result.

Proposition 7.1

Proposition 7.1 ([4])

Given a free abelian group 𝐴 of finite rank and a finitely generated abelian group 𝑇, there exists some 𝐹 such that A T W F .

In particular, if r = k = l and we let f i = 1 + t i for all 𝑖, W F contains a copy of the free metabelian group of rank 𝑟.

We claim that the following statement holds.

Proposition 7.2

If k > 0 , then W F has exponential Dehn function.

Note that, when i = j = k = 1 and f 1 = 1 + t 1 , then W F is the Baumslag group Γ = a , s , t [ a , a t ] = 1 , [ s , t ] = 1 , a s = a a t . The exponential Dehn function of this special case is proved in [15].

We need a few lemmas before we prove Proposition 7.2. First, let us denote the abelian groups generated by { t 1 , t 2 , . t l } and { u 1 , u 2 , , u k } by 𝑇 and 𝑈 respectively.

Lemma 7.3

Let 𝑀 be a free ( U × T ) -module with basis e 1 , , e r . Let 𝑆 be the submodule of 𝑀 generated by

{ ( u i - f i ) e j 1 i k ,  1 j r } .

If h = h 1 e 1 + h 2 e 2 + + h r e r S such that h i Z T for all 𝑖, then h = 0 .

Proof

If k = 1 , then h S means there exist α 1 , α 2 , , α r Z ( U × T ) such that

h = α 1 ( u 1 - f 1 ) e 1 + α 2 ( u 1 - f 1 ) e 2 + + α r ( u 1 - f 1 ) e r .

Since

h = h 1 e 1 + h 2 e 2 + + h r e r ,

we have h i = α i ( u 1 - f 1 ) . Note that h i Z T . It follows that α i ( u 1 - f 1 ) does not have any term involving u 1 . Suppose α i 0 for some 𝑖. Because f 1 Z T , we have deg u 1 ( α i u 1 ) > deg u 1 ( α i f 1 ) . Thus α i ( u 1 - f 1 ) has at least one term containing u 1 , which leads to a contradiction.

If the statement for k = n has been proved, then, for k = n + 1 , we have

h = i = 1 r j = 1 n + 1 α i , j ( u j - f j ) e i .

We choose an integer 𝑁 large enough such that u 1 N α i , j does not have any negative power of u 1 for all i , j . Then

u 1 N h = i = 1 r j = 1 n + 1 f 1 N α i , j ( u j - f j ) e i = : i = 1 r j = 1 n + 1 β i , j ( u j - f j ) e i ,

where β i , j = u 1 N α i , j . We regard β i , j ( u 1 ) as a polynomial of u 1 . Replacing u 1 by f 1 , we have

f 1 N h = i = 1 r j = 2 n + 1 β i , j ( f i ) ( u j - f j ) e i .

Note that f 1 N h i Z T for i = 1 , , r , and then, by the inductive assumption, f 1 N h i = 0 for all 𝑖. Since f 1 = 1 + c 1 , 1 t 1 + c 1 , 2 t 1 2 + + c i , d 1 - 1 t 1 d 1 - 1 + t 1 d 1 and t 1 has infinite order, it follows that f 1 is not a zero divisor in Z ( U × T ) . Thus h i = 0 for all 𝑖.

Therefore, h = 0 . This induction completes the proof. ∎

It follows that if a 1 h 1 a 2 h 2 a r h r = W F 1 such that h i Z T for all 𝑖, then h i = 0 as an element in Z ( U × T ) for every 𝑖. To convert it to 1, we only need those metabelian relations to commute all the conjugates of the a i . By Theorem 6.1, it will cost at most exponentially many relations with respect to the length of the word to kill the word.

Next, let w = W F 1 , and consider the minimal van Kampen diagram Δ over W F . There are two types of relations contain u i : (1) commutative relations [ u i , u j ] = 1 , [ u i , t s ] = 1 , j i , 1 s l ; (2) action relations a j u i = a j f i , 1 j r . Those cells, in the van Kampen diagram, form a u i -band.

Figure 3 
                  An example of a 
                        
                           
                              
                                 u
                                 1
                              
                           
                           
                           u_{1}
                        
                     -bands
Figure 3

An example of a u 1 -bands

We have some properties for u i -bands in a van Kampen diagram over W F .

Lemma 7.4

The following statements hold.

  1. The top (or bottom) path of a u i -band is a word 𝑤 such that all t s , u j for s , j i are in the same orientation, i.e. the exponents of each letter t s , u j are either all 1 or all −1. In particular,

    w = W F a 1 h 1 a 2 h 2 a r h r t 1 α 1 t l α l u 1 β 1 u k β k ,

    where h i Z ( U × T ) , sgn ( α i ) = sgn ( β j ) for all i , j , and α s (or β j ) is equal to the number of times that t s (resp. u j ) appears in 𝑤 for s , j i .

  2. u i -bands do not intersect each other. In particular, a u i -band does not self-intersect.

  3. If i j , then a u i -band intersects a u j -band at most one time.

Proof

(i) By the definition of a u i , all letters t s , u j , s , j i , of 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 Section 5.2).

(ii) Because there is no u i on the top or the bottom path of a u i -band, two u i -bands cannot intersect each other.

(iii) If i j and a u i -band intersects a u j -band, since the van Kampen diagram is a planer graph, by comparing the orientation, it is impossible for a u i -band to intersect a u j -band twice (or more). ∎

Lemma 7.5

Let f ( t ) = t d + c d - 1 t d - 1 + + c 1 t + 1 Z [ t ] , c i Z , d > 0 . Then there exists some α > 1 such that | ( f ( t ) ) n | > α n for all 𝑛.

Proof

We write ( f ( t ) ) n = i = 0 n d c n , i t i .

Consider the corresponding holomorphic function

g ( z ) = z d + c 1 , d - 1 z 1 , d - 1 + + c 1 , 1 z + 1 .

If there exists z 0 , | z 0 | = 1 such that | g ( z 0 ) | > 1 , we have

| g ( z 0 ) | n = | ( g ( z 0 ) ) n | = | i = 0 n d c n , i z 0 i | i = 0 n d | c n , i | = | f n | .

Then we are done.

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

1 = g ( 0 ) = 1 2 π i | z | = 1 g ( z ) z d z .

Take the modulus of both sides,

1 = | 1 2 π i | z | = 1 g ( z ) z d z | 1 2 π θ = 0 2 π | g ( e i θ ) | d θ 1 .

Therefore, | g ( z ) | = 1 for | z | = 1 almost everywhere. Let z = e i θ . We have

g ( e i θ ) = ( j = 0 d c 1 , j cos ( i θ ) ) + i ( j = 0 d c 1 , j sin ( j θ ) ) .

Then

( j = 0 d c 1 , j cos ( i θ ) ) 2 + ( j = 0 d c 1 , j sin ( j θ ) ) 2 = h = 0 d c 1 , j 2 + 2 j < k c 1 , j c 1 , k cos ( ( k - j ) θ ) = 1

holds for all 𝜃. But cos ( ( k - j ) θ ) is a polynomial with respect to cos θ , i.e. cos ( ( k - j ) θ ) = T k - j ( cos θ ) , where T m ( x ) is the 𝑚-th Chebyshev polynomial. The leading term of T m ( x ) is 2 m - 1 x m . Thus

h = 0 d c 1 , j 2 + 2 j < k c 1 , j c 1 , k T k - j ( cos θ ) = 1 for all θ .

Note the leading term of the left-hand side is 2 d - 1 cos d θ . That leads to a contradiction since the equation above has at most 𝑑 solutions for cos θ . ∎

Proof of Proposition 7.2

First, we show that the lower bound is exponential. Consider the word w = [ a 1 u 1 n , a 1 ] . By definition, the word 𝑤 is of length 2 n + 4 and w = W F 1 . Let Δ be a minimal van Kampen diagram with boundary label 𝑤. By comparing the orientation, u 1 -bands starting at the top left of Δ will end at either bottom left or top right. By Lemma 7.4, u 1 -bands do not intersect each other, so we can suppose at least half of the u 1 -bands starting at the top left end at the top right. See Figure 4, where the shaded areas are u 1 -bands.

Figure 4 
                        
                           
                              
                                 
                                    
                                       u
                                       1
                                    
                                 
                                 
                                 u_{1}
                              
                           -bands in Δ
Figure 4

u 1 -bands in Δ

We first claim that there are no cells containing t s , u j , s , j > 1 , on each u 1 -band. We denote the top and bottom path of the 𝑖-th u 1 -band from the top by γ i top and γ i bot , where i = 1 , 2 , , m , m > n 2 . Assuming a u i -band intersects one of the u 1 -band, again by Lemma 7.4 (ii), (iii), it can neither intersect a u 1 -band twice nor intersect itself. Thus it has to end all the way to the boundary of Δ, a contradiction.

If there exists a cell containing t s for s > 1 in the top most u 1 -band, then, by Lemma 7.4 (i), γ 1 top is a word a 1 h 1 a 2 h 2 a r h r t 1 α 1 t l α l . Thus γ 1 top and a 1 form a cycle 𝛾. We have

a 1 h 1 + 1 a 2 h 2 a r h r t 1 α 1 t l α l = 1 , α i 0 .

This leads to a contradiction since the image of the left-hand side in U × T is not trivial. And by the definition of a u 1 -band, if γ i top does not have any t s , s > 1 , neither does γ i bot . Next, consider two consecutive u 1 -bands. If γ i bot does not have t s , then, by the same argument, neither does γ i + 1 top . Therefore, the claim is true.

Denote the words of γ i top and γ i bot by w i top and w i bot respectively. Such words only consist of the a i and t 1 . Note that w i bot = w i + 1 top for i = 1 , , m - 1 . Since w 1 bot = a 1 - f 1 , by the same discussion as above, w i top = a 1 - f 1 i - 1 and w i bot = a 1 - f 1 i (Figure 4). Next, we focus on the number of a 1 in each w i bot , which is at least | f 1 i | . By Lemma 7.5, there exists some α > 1 such that | f 1 i | > α i . Therefore, the number of a 1 in w m top is at least α m - 1 . Since m > n 2 , the number of cells in the 𝑚-th u 1 -band is at least α [ n 2 ] . Thus the area of [ a 1 u 1 n , a 1 ] is at least α [ n 2 ] . It follows that the lower bound is exponential.

For the upper bound, as Theorem 6.1 suggests, all we need is to consider how to solve the membership problem of the submodule 𝑆, where 𝑆 is generated by { ( u i - f i ) e j 1 i k ,  1 j r } . Suppose w = 1 with | w | n ; the 𝑤 has a ordered form as

w = W F a 1 g 1 a 2 g 2 a r g r , g i Z ( U × T ) .

And the cost of converting 𝑤 to its ordered form is exponential with respect to 𝑛 as we showed in Section 5. Also note that deg ( g i ) , | g i | n for all 𝑖. Without loss of generality, we assume that all exponents of the u i are positive. The corresponding module element of 𝑤 is

g 1 e 1 + g 2 e 2 + + g r e r .

For each term t 1 α 1 t 2 α 2 t l α l u 1 β 1 u 2 β 2 u k β k , where α i Z , β i 0 , we replace u i by u i - f i + f i . Then we convert t 1 α 1 t 2 α 2 t l α l u 1 β 1 u 2 β 2 u k β k to a form

i = 1 k η i ( u i - f i ) + τ , η i Z ( U × T ) , τ Z T .

If | α 1 | + + | α l | + | β 1 | + + | β k | < n , then we have deg ( η i ) , deg ( τ ) < D n and | η i | , | τ | < D n , where D = max { d 1 , , d k , | f 1 | , , | f k | } . Therefore, replacing u i by u i - f i + f i in every term of 𝑤, we have

g 1 e 1 + g 2 e 2 + + g r e r = i = 1 r i = 1 k μ i , j ( u j - f j ) e i + ρ , μ i , j Z ( U × T ) , ρ Z T .

Since w = 1 , 𝜌 lies in the submodule 𝑆. By Lemma 7.3, ρ = 0 . Also note that deg ( μ i , j ) , deg ( ρ ) < D n , | μ i , j | , | ρ | < n D n . It follows from Lemma 5.7 that all module computations in the process cost exponentially many relations with respect to 𝑛. And it also costs at most exponentially many relations to convert 𝜌 to 0. Therefore,

w = W F i = 1 r j = 1 k a i μ i , j ( u j - f j ) ,

and the cost of converting is exponential with respect to 𝑛. And the area of the right-hand side is bounded by i , j | μ i , j | r k n D n . The upper bound is exponential. ∎

Theorem 1.5 follows immediately from Proposition 7.2.

Proof

Let 𝐴 be a free abelian group of finite rank, and let 𝑇 be a finitely generated abelian group. If the torsion-free rank of 𝑇 is greater than 0, by Proposition 7.2, A T embeds into W F , which has exponential Dehn function. If the torsion-free rank is 0 and A T can be first embedded into A ( T × Z ) , then the problem is reduced to the first case. ∎

7.2 Further comments

Theorem 6.1 shows that the metabelianness cost is at most exponential. By Theorem 6.4 and Lemma 6.8, if we write 𝐺 as an extension of two abelian groups 𝐴 by 𝑇, the complexity of the membership problem of the 𝑇-module 𝐴 gives the lower bound of δ ( n ) . It follows that, to construct a finitely presented metabelian group with Dehn function bigger than exponential function, the only hope is to find a complicated membership problem of a submodule in a free module over a group ring of a free abelian group because it is simply impossible to get anything harder than exponential anywhere else.

The first obstruction for us is the existence of such a membership problem. There is already a lot of work on the polynomial ideal membership problem, which is a special case of the membership problem over modules. For example, Mayr and Meyer showed that the lower space bound of a general polynomial ideal membership problem is exponential [19], though, in their work, the input of the problem includes the generating set of the ideal, which is different from the problem we are concerned with. Other results can be found in several surveys, such as [18, 20]. But it remains unknown whether or not there exists an integral coefficient polynomial ideal such that the time complexity of its membership problem is harder than exponential.

The second obstruction comes from the finitely-presentedness. Recall that a finitely generated metabelian group is finitely presented if and only if the module structure is tame [6]. Thus, even if we manage to find a complicated enough membership problem in some submodule, it may not give us a finitely presented metabelian group unless the module is tame.

Acknowledgements

I would like to thank my advisor Mark Sapir who encouraged me to study this question and instructed me. I also would like to thank Nikolay Romanovskiy, who kindly answered my question about an algorithm solving the membership problem in modules, and Alexander Olshanskiy, who pointed out a case I did not consider.

  1. Communicated by: Benjamin Klopsch

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Received: 2020-11-17
Published Online: 2021-05-19
Published in Print: 2021-09-01

© 2021 Walter de Gruyter GmbH, Berlin/Boston

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