#  $R_{\infty}$–property for groups commensurable to nilpotent quotients of
RAAGs

Maarten Lathouwers, Thomas Witdouck KU Leuven Kulak Kortrijk Campus,
Department of Mathematics. Email<EMAIL_ADDRESS>The author
was supported by a PhD fellowship of the Research Fund - Flanders (FWO), Grant
Number 1102424N.KU Leuven Kulak Kortrijk Campus, Department of Mathematics.
Email<EMAIL_ADDRESS>The author was supported by a PhD
fellowship of the Research Fund - Flanders (FWO), Grant Number 1153122N.

###### Abstract

Let $G$ be a group and $\varphi$ an automorphism of $G$. Two elements $x,y\in
G$ are said to be $\varphi$-conjugate if there exists a third element $z\in G$
such that $zx\varphi(z)^{-1}=y$. Being $\varphi$-conjugate defines an
equivalence relation on $G$. The group $G$ is said to have the
$R_{\infty}$–property if all its automorphisms $\varphi$ have infinitely many
$\varphi$-conjugacy classes. For finitely generated torsion-free nilpotent
groups, the so-called Mal’cev completion of the group is a useful tool in
studying this property. Two groups have isomorphic Mal’cev completions if and
only if they are abstractly commensurable. This raises the question whether
the $R_{\infty}$–property is invariant under abstract commensurability within
the class of finitely generated torsion-free nilpotent groups. We show that
the answer to this question is negative and provide counterexamples within a
class of 2-step nilpotent groups associated to edge-weighted graphs. These
groups are commensurable to 2-step nilpotent quotients of right-angled Artin
groups.

##  1 Introduction

Let $G$ and $H$ be groups. We say $G$ is abstractly commensurable with $H$ if
there exist finite index subgroups $G^{\prime}\subset G$ and
$H^{\prime}\subset H$ such that $G^{\prime}$ is isomorphic to $H^{\prime}$.
The starting point of this paper is the following question:

######  Question 1.

Is the $R_{\infty}$–property for groups an abstract commensurability
invariant?

It is known however that the answer to this question is negative. The easiest
counterexample is given by the fundamental group of the Klein bottle. This
group has the $R_{\infty}$–property, while it has a subgroup of index 2
isomorphic to ${\wiskunde Z}^{2}$ which is known not to have the
$R_{\infty}$–property, see [GW09, Section 2]. Note that the fundamental group
of the Klein bottle is torsion-free virtually abelian, but not nilpotent. A
new interesting question arises when we ask our groups to be finitely
generated torsion-free nilpotent.

######  Question 2.

Is the $R_{\infty}$–property for groups an abstract commensurability invariant
within the class of finitely generated torsion-free nilpotent groups?

The reason for considering finitely generated torsion-free nilpotent groups is
that there is a well-known abstract commensurability invariant within this
class of groups, namely the rational Mal’cev completion (see section 2.1). Two
finitely generated torsion-free nilpotent groups are abstractly commensurable
if and only if they have isomorphic rational Mal’cev completions. In an
attempt to answer Question 2, we arrived at the following observation. We
recall that a linear self-map on a finite-dimensional vector space is called
integer-like if its characteristic polynomial has integral coefficients and
constant term equal to $\pm 1$.

######  Proposition 1.1.

Let $G$ be a finitely generated torsion-free nilpotent group. All groups that
are abstractly commensurable to $G$ have the $R_{\infty}$–property if and only
if every integer-like automorphism on the associated Mal’cev rational Lie
algebra has an eigenvalue 1.

The actual answer to Question 2 turns out to be negative and we will present a
counterexample within the class of 2-step nilpotent quotients of right-angled
Artin groups. Using weight functions on the edges of the defining graphs, one
can define groups which are commensurable to these 2-step nilpotent quotients
of right-angled Artin groups, but which have considerably less induced
automorphisms on the abelianization. These groups are defined in section 4 and
the restrictions on their automorphisms are proven in section 5. The (to our
knowledge) smallest concrete counterexample to Question 2 is then presented in
6.4, which proves the following.

######  Theorem 1.2.

There exist finitely generated torsion-free 2-step nilpotent groups which do
not have the $R_{\infty}$–property but have for any $n\in{\wiskunde
N}\setminus\\{0,1\\}$ a subgroup of index $n$ which does have the
$R_{\infty}$–property.

##  2 Preliminaries

First, let us fix some notation. For any group $G$ we define the lower central
series of $G$ inductively by setting $\gamma_{1}(G)=G$ and
$\gamma_{i+1}(G)=[\gamma_{i}(G),G]$ (for all $i\in{\wiskunde N}_{0}$). For a
Lie algebra $L$, we use the same notation for the lower central series, i.e.
$\gamma_{1}(L)=L$ and $\gamma_{i+1}(L)=[L,\gamma_{i}(L)]$ (for all
$i\in{\wiskunde N}_{0}$).

###  2.1 Lie algebras associated to groups

In this section, we recall two well-known constructions to associate a Lie
algebra to a group and how they relate to one another.

#### The Lie ring construction

Let $G$ be a group. Denote the factors of the lower central series of $G$ by
$L_{i}(G):=\gamma_{i}(G)/\gamma_{i+1}(G)$. The quotients are abelian (and thus
${\wiskunde Z}$-modules). We define the Lie ring $L(G)$ as the direct sum of
these ${\wiskunde Z}$-modules, i.e.

$L(G):=\bigoplus_{i=1}^{\infty}L_{i}(G)$

and the Lie bracket on $L(G)$ is induced by setting (for $i,j\in{\wiskunde
N}_{0}$, $g\in\gamma_{i}(G)$ and $h\in\gamma_{j}(G)$)

$[g\gamma_{i+1}(G),h\gamma_{j+1}(G)]:=[g,h]\gamma_{i+j+1}(G).$

In order to turn $L(G)$ into a Lie algebra over a field $K$, it suffices to
tensor $L(G)$ with $K$. We define $L^{K}(G):=L(G)\otimes_{{\wiskunde
Z}}K=\oplus_{i=1}^{\infty}L_{i}^{K}(G)$ where
$L_{i}^{K}(G):=L_{i}(G)\otimes_{{\wiskunde Z}}K$ (for all $i\in{\wiskunde
N}_{0}$). Moreover, $L^{K}(G)$ is clearly ${\wiskunde N}_{0}$-graded (by using
the $L_{i}^{K}(G)$) and if $G$ is finitely generated, the $L_{i}^{K}(G)$ are
finite dimensional. The lower central series of $L^{K}(G)$ is given by

$\gamma_{i}(L^{K}(G))=\bigoplus_{j=i}^{\infty}L_{j}^{K}(G)$

for all $i\in{\wiskunde N}_{0}$.
Note that this way of associating a Lie algebra to a group is functorial. In
particular, any automorphism of the group $G$ induces a graded Lie algebra
automorphism of $L^{K}(G)$ (i.e. an automorphism which leaves the subspaces
$L_{i}^{K}(G)$ invariant). Indeed, fix an automorphism $\varphi\in{\rm
Aut}(G)$. Since the lower central series of $G$ is characteristic, we obtain
induced automorphisms $\varphi_{i}\in{\rm Aut}(L_{i}(G))$ (for all
$i\in{\wiskunde N}_{0}$). These ${\wiskunde Z}$-module automorphisms extend to
automorphisms $\overline{\varphi_{i}}:=\varphi_{i}\otimes{\rm Id}$ of
$L_{i}^{K}(G)$. Combining them yields a graded Lie algebra automorphism
$\overline{\varphi}\in{\rm Aut}_{g}(L^{K}(G))$, i.e. we set
$\overline{\varphi}(x)=\overline{\varphi_{i}}(x)$ for each $x\in
L_{i}^{K}(G)$.

#### The Mal’cev completion

A group $G$ is said to be radicable if the assignment $x\mapsto x^{n}$ defines
a bijection on $G$ for any non-zero integer $n$. If $G$ is a finitely
generated torsion-free nilpotent group, there exists a radicable group
$G^{\wiskunde Q}$ of which $G$ is a subgroup, such that for any $x\in
G^{\wiskunde Q}$ there is a positive integer $n>0$ such that $x^{n}\in G$. The
group $G^{\wiskunde Q}$ is nilpotent and unique, in the sense that, if there
exists another group with the same properties, it is isomorphic to
$G^{\wiskunde Q}$ with the isomorphism restricting to the identity on $G$. The
group $G^{\wiskunde Q}$ is called the rational Mal’cev completion of $G$.
The group $G^{\wiskunde Q}$ can also be given the structure of a nilpotent Lie
algebra over ${\wiskunde Q}$ by use of the Baker-Campbell-Hausdorff formula.
We will refer to this Lie algebra as the Mal’cev Lie algebra of $G$ and write
it as $M(G)$. Note that $G^{\wiskunde Q}$ and $M(G)$ have the same underlying
sets. It is well-known that two finitely generated torsion-free nilpotent
groups have isomorphic Mal’cev completions (or, equivalently, isomorphic
Mal’cev Lie algebras) if and only if they are abstractly commensurable.
Moreover, if $G$ and $H$ are finitely generated torsion-free nilpotent groups
with isomorphic finite index subgroups $G^{\prime}\subset G$ and
$H^{\prime}\subset H$, then there is a unique isomorphism between $M(G)$ and
$M(H)$ which restricts to the given isomorphism between $G^{\prime}$ and
$H^{\prime}$.
As with the graded Lie ring construction, the association of the rational
Mal’cev completion to a finitely generated torsion-free nilpotent group is
functorial. Any automorphism of $G$ extends uniquely to an automorphism of
$G^{\wiskunde Q}$ which in turn defines an automorphism of the Lie algebra
$M(G)$.

Next, we recall a known result that relates $L^{\wiskunde Q}(G)$ and $M(G)$
for a finitely generated torsion-free nilpotent group $G$. Let $L$ be any Lie
algebra. Similar to the Lie ring construction from above, one can define from
$L$ a new ${\wiskunde N}_{0}$-graded Lie algebra $\operatorname{gr}(L)$ by

$\operatorname{gr}(L)=\bigoplus_{i=1}^{\infty}\gamma_{i}(L)/\gamma_{i+1}(L)$

with Lie bracket defined by
$[X+\gamma_{i+1}(L),\,Y+\gamma_{j+1}(L)]=[X,Y]+\gamma_{i+j+1}(L)$ for
$i,j\in{\wiskunde N}_{0}$ and any $X\in\gamma_{i}(L)$ and $Y\in\gamma_{j}(L)$.
In [Qui68, Lemma 3.3] it is proven that for any finitely generated torsion-
free nilpotent group $G$, the injection of $G$ into its rational Mal’cev
completion $G^{\wiskunde Q}$ induces an isomorphism of graded Lie algebras

$L^{\wiskunde Q}(G)\cong\operatorname{gr}(M(G)).$ (2.1)

From this result, it is not hard to see that, given an automorphism of $G$,
the induced automorphisms on $L^{\wiskunde Q}(G)$ and $M(G)$ have the same
eigenvalues. Moreover, there is a very nice connection between the eigenvalues
of these induced automorphisms and the Reidemeister number of the original
automorphism.

######  Lemma 2.1 ([Rom11],[DG14]).

Let $G$ be a finitely generated nilpotent group and $\varphi\in{\rm Aut}(G)$.
Let $\overline{\varphi}\in{\rm Aut}_{g}(L^{K}(G))$ be the induced graded
automorphism on $L^{K}(G)$ and $\varphi^{\prime}\in{\rm Aut}(M(G))$ the
induced automorphism on $M(G)$. Then

$R(\varphi)=\infty\quad\Longleftrightarrow\quad\overline{\varphi}\text{ has
eigenvalue }1\quad\Longleftrightarrow\quad\varphi^{\prime}\text{ has
eigenvalue }1.$

###  2.2 (Nilpotent) partially commutative groups and Lie algebras

In this section, we introduce groups and Lie algebras that will be used
throughout this paper.

Let $\Gamma=(V,E)$ be a finite undirected simple graph with (finite) vertex
set $V$ and edge set $E\subset\\{\\{v,w\\}\mid v,w\in V\text{ and }v\neq
w\\}$. From now on, we always assume a graph $\Gamma$ to be finite, undirected
and simple. The free partially commutative group $A(\Gamma)$ associated to the
graph $\Gamma$ is defined by

$A(\Gamma):=\langle V\mid[v,w]=1\text{ for }v,w\in V\text{ with
}\\{v,w\\}\not\in E\rangle.$

These groups are also known as right-angled Artin groups or graph groups.
Sometimes the complement graph is used to define $A(\Gamma)$ (i.e. edges
become non-edges in our definition). Nevertheless, our convention will come in
handy later when generalising to groups associated to graphs with weights on
their edges. By taking the quotient of $A(\Gamma)$ with
$\gamma_{c+1}(A(\Gamma))$ (with $c>1$) we get a finitely generated torsion-
free c-step nilpotent group

$A(\Gamma,c):=A(\Gamma)/\gamma_{c+1}(A(\Gamma)).$

We will mostly use the 2-step nilpotent quotient and thus for simplicity, we
write

$G_{\Gamma}:=A(\Gamma,2).$

Next to a graph $\Gamma$, now also fix a field $K$. We can associate,
similarly as with $A(\Gamma)$, a Lie algebra $L^{K}(\Gamma)$ (over $K$) to the
graph $\Gamma$. Denote with $\mathfrak{f}^{K}(V)$ the free Lie algebra on $V$
over $K$ and with $I(\Gamma)$ the ideal of $\mathfrak{f}^{K}(V)$ generated by
the set of Lie brackets $\\{[v,w]\mid v,w\in V,\>\\{v,w\\}\not\in E\\}$. The
Lie algebra $L^{K}(\Gamma)$, which is also known as the free partially
commutative Lie algebra over $K$, is defined as the quotient of
$\mathfrak{f}^{K}(V)$ by $I(V)$. Similarly, we get the free $c$-step nilpotent
partially commutative Lie algebra $L^{K}(\Gamma,c)$ by taking the quotient of
$L^{K}(\Gamma)$ with $\gamma_{c+1}(L^{K}(\Gamma))$. Summarising, we defined

$L^{K}(\Gamma):=\frac{\mathfrak{f}^{K}(V)}{I(\Gamma)}\quad\text{and}\quad
L^{K}(\Gamma,c):=\frac{L^{K}(\Gamma)}{\gamma_{c+1}(L^{K}(\Gamma))}.$

The free Lie algebra $\mathfrak{f}^{K}(V)$ is canonically graded by defining
inductively $\mathfrak{f}_{1}^{K}(V):={\rm Sp}_{K}(V)$ (i.e. the vector space
span of $V$ over $K$) and
$\mathfrak{f}_{i+1}^{K}(V):=[\mathfrak{f}_{1}^{K}(V),\mathfrak{f}_{i}^{K}(V)]$
(for $i\in{\wiskunde N}_{0}$). Since the ideal $I(\Gamma)$ is a graded ideal
$\mathfrak{f}^{K}(V)$ it follows that $L(\Gamma,c)$ inherits an ${\wiskunde
N}_{0}$-grading from $\mathfrak{f}^{K}(V)$. We denote the subspaces that make
up this grading by $L_{i}^{K}(\Gamma,c)$ ($i\in{\wiskunde N}_{0}$). Note that
$L_{i}^{K}(\Gamma,c)=\\{0\\}$ for $i>c$.

A classical result from [DK92, Theorem 2.1] establishes a connection between
$A(\Gamma)$ and $L^{K}(\Gamma)$ using the Lie ring construction of the
previous section. Concretely, there is an isomorphism of graded Lie algebras

$L^{K}(\Gamma)\cong L^{K}(A(\Gamma))$ (2.2)

which is determined by taking the identity on the vertices $V$.

######  Remark 2.2.

Note that the subspace ${\rm Sp}_{K}(V)$ generates $L^{K}(\Gamma,c)$. In
particular, any graded automorphism $\varphi\in{\rm Aut}_{g}(L^{K}(\Gamma,c))$
is completely determined by its restriction to ${\rm Sp}_{K}(V)$. So if we
denote with $\pi:{\rm Aut}_{g}(L^{K}(\Gamma,2))\to{\rm GL}({\rm
Sp}_{K}(V)):\varphi\mapsto\varphi|_{{\rm Sp}_{K}(V)}$, then $\pi$ is an
isomorphism onto its image.

##  3 The $R_{\infty}$-property and commensurability

In this section, we prove Proposition 1.1 from the introduction and apply it
to the class of free nilpotent partially commutative groups.

###### Proof of Proposition 1.1.

Let $G$ be a finitely generated torsion-free nilpotent group, $G^{\wiskunde
Q}$ its Mal’cev completion and $M(G)$ the Mal’cev Lie algebra.

First, assume that every integer-like automorphism of $M(G)$ has an eigenvalue
1. Let $H$ be a group commensurable with $G$ and $\varphi$ any automorphism of
$H$. Then $\varphi$ induces an automorphism on the Mal’cev Lie algebra of $H$,
which is isomorphic to $M(G)$. Moreover, this automorphism must be integer-
like, thus by the assumption it has an eigenvalue 1. Using Lemma 2.1, we find
that $R(\varphi)=\infty$ and thus that $H$ has the $R_{\infty}$–property.

Conversely, assume that every group commensurable to $G$ has the
$R_{\infty}$–property. Let $\varphi$ be an integer-like automorphism of
$M(G)$. A property of such an automorphism is that there exists a basis
$\\{x_{1},\ldots,x_{n}\\}$ of $M(G)$ with respect to which $\varphi$ has a
matrix representation in ${\rm GL}_{n}({\wiskunde Z})$. Moreover, it follows
from the BCH formula that there exists a positive integer $k$ big enough, such
that

$H:={\rm Sp}_{\wiskunde Z}(\\{kx_{1},\ldots,kx_{n}\\})$

is a subgroup of $G^{\wiskunde Q}$. Note as well that $H$ is preserved by
$\varphi$. Clearly, $H$ is torsion-free and nilpotent since it is a subgroup
of $G^{\wiskunde Q}$. To check that $H$ is finitely generated, it suffices to
check that its abelianization is so, since $H$ is nilpotent. This follows
straightforwardly from the fact that the abelianizations of $M(G)$ (seen as
abelian group for ’+’) and $G^{\wiskunde Q}$ are isomorphic. At last, using
the BCH formula, one can easily verify that for any $x\in G^{\wiskunde Q}$,
there exists a non-zero integer $m$ such that $x^{m}$ lies in $H$. We thus
have that $G^{\wiskunde Q}$ is the Mal’cev completion of $H$. As a
consequence, $H$ is commensurable to $G$. Thus, by assumption, $H$ has the
$R_{\infty}$–property and $R(\varphi|_{H})=\infty$. Lemma 2.1 then implies
that the induced morphism on $M(H)=M(G)$, which must be equal to $\varphi$,
has an eigenvalue 1. This completes the proof. ∎

It is not clear to us whether in the statement of Proposition 1.1 one can
replace ‘the associated Mal’cev Lie algebra’ with ‘the associated graded Lie
algebra’. Equivalently, we do not have an answer to the following question.

######  Question 3.

Does there exist a finite-dimensional nilpotent rational Lie algebra $L$ for
which every integer-like automorphism has an eigenvalue 1, but for which
$\operatorname{gr}(L)$ does not have this property?

It is not hard to check that the other direction is true, namely if
$\operatorname{gr}(L)$ has the property then so does $L$.

###  3.1 Free abelian factors and $R_{\infty}$-property for commensurable
groups

In light of Question 2, we argue in this section that we can forget about the
(maximal) free abelian factors when stating the question if we are considering
2-step nilpotent groups, i.e. we prove that if $H$ is finitely generated
torsion-free 2-step nilpotent and $H^{\wiskunde Q}$ does not have an abelian
direct factor, then the following are equivalent for any $m\in{\wiskunde
N}_{0}$

None of the (fg. tf. 2-step nilpotent) groups commensurable with $H$ has the
$R_{\infty}$–property

$\Updownarrow$

None of the (fg. tf. 2-step nilpotent) groups commensurable with ${\wiskunde
Z}^{m}\times H$ has the $R_{\infty}$–property.

Recall that the adapted lower central series of a group $G$ is defined as the
series of isolators of the terms of the lower central series, i.e. the series
of terms $\sqrt[G]{\gamma_{i}(G)}=\\{g\in G\mid\exists\,l\in{\wiskunde
N}_{0}:\>g^{l}\in\gamma_{i}(G)\\}$ (for $i\in{\wiskunde N}_{0}$). If $G$ is
nilpotent, these are in fact characteristic subgroups in $G$. We will
sometimes omit the superscript $G$ if the group is clear. Note that
$H^{\wiskunde Q}$ does not have an abelian direct factor is equivalent with
$Z(H)=\sqrt[H]{\gamma_{2}(H)}$ if we are considering nilpotency class 2.

######  Lemma 3.1.

Let $G$ be a finitely generated torsion-free 2-step nilpotent group, then
$Z(G)=\sqrt{\gamma_{2}(G)}$ if and only if $G^{{\wiskunde Q}}$ does not have
an abelian direct factor.

###### Proof.

If $G^{{\wiskunde Q}}$ does not have an abelian factor, then Proposition 2.18
in [Sen22] tells us that $Z(G)=\sqrt{\gamma_{2}(G)}$. Hence, it suffices to
prove the converse.
Suppose that $Z(G)=\sqrt{\gamma_{2}(G)}$ and $G^{{\wiskunde Q}}\cong{\wiskunde
Q}^{l}\times G^{\prime}$ with $l\in{\wiskunde N}$. Since
$\gamma_{3}(G^{\prime})=\gamma_{3}(G^{{\wiskunde Q}})=1$, it follows that
$G^{\prime}$ is finitely generated and nilpotent of class at most 2. Thus we
obtain that

$\displaystyle\dim_{{\wiskunde Q}}(Z(G^{\prime}))$ $\displaystyle\leq
l+\dim_{{\wiskunde Q}}(Z(G^{\prime}))=\dim_{{\wiskunde Q}}(Z(G^{{\wiskunde
Q}}))=\dim_{{\wiskunde Q}}((Z(G))^{{\wiskunde Q}})$
$\displaystyle=\dim_{{\wiskunde Q}}((\sqrt{\gamma_{2}(G)})^{{\wiskunde
Q}})=\dim_{{\wiskunde Q}}(\gamma_{2}(G^{{\wiskunde Q}}))=\dim_{{\wiskunde
Q}}(\gamma_{2}(G^{\prime}))$ $\displaystyle\leq\dim_{{\wiskunde
Q}}(Z(G^{\prime})).$

In particular, this implies that $l=0$ and thus $G^{{\wiskunde Q}}$ does not
have an abelian direct factor if $Z(G)=\sqrt{\gamma_{2}(G)}$. ∎

Next, we argue that for a finitely generated torsion-free 2-step nilpotent
group its finite index subgroups are the only groups commensurable with it (up
to isomorphism). This follows directly by applying the next lemma.

######  Lemma 3.2.

Let $G$ be a finitely generated torsion-free 2-step nilpotent group and
$H\subset_{\text{fin}}G$. Then there exists some
$G^{\prime}\subset_{\text{fin}}H$ such that $G^{\prime}\cong G$.

###### Proof.

Fix an adapted set of generators of $G$
$\\{x_{1},\dots,x_{n},y_{1},\dots,y_{m}\\}$, i.e. such that
$\\{x_{1}\sqrt{\gamma_{2}(G)},\dots,x_{n}\sqrt{\gamma_{2}(G)}\\}$
(respectively $\\{y_{1},\dots,y_{m}\\}$) is a ${\wiskunde Z}$-basis for
$G/\sqrt{\gamma_{2}(G)}\cong{\wiskunde Z}^{n}$ (respectively
$\sqrt{\gamma_{2}(G)}\cong{\wiskunde Z}^{m}$). Hence, there are
$c_{ij,l}\in{\wiskunde Z}$ (for $l=1,\dots,m$) such that

$G=\Bigg{\langle}x_{1},\dots,x_{n},y_{1},\dots,y_{m}\>\bBigg@{3}|\>\begin{array}[]{l}[x_{i},x_{j}]=\prod_{l=1}^{m}y_{l}^{c_{ij,l}};\\\
[x_{i},y_{l_{1}}]=[y_{l_{1}},y_{l_{2}}]=1\forall
i,l_{1},l_{2}\end{array}\Bigg{\rangle}.$

Since $[G:H]<\infty$ there are $a_{i},b_{l}\in{\wiskunde N}_{0}$ (for all
$i,l$) such that $x_{i}^{a_{i}},y_{l}^{b_{l}}\in H$. Define
$k:=\text{lcm}_{i,l}\\{a_{i},b_{l}\\}$ and consider $G^{\prime}:=\langle
x_{1}^{k},\dots,x_{n}^{k},y_{1}^{k^{2}},\dots,y_{m}^{k^{2}}\rangle\subset G$.
Since $G$ is 2-step nilpotent, it follows that $G^{\prime}\cong G$. Note that
$x_{i}^{k},y_{l}^{k^{2}}\in H$ (for all $i,l$) and thus by Lemma 2.8 in
[Bau71] we obtain that $G^{\prime}\subset_{\text{fin}}G$. Hence, we can
conclude that $G^{\prime}\subset_{\text{fin}}H$ and $G^{\prime}\cong G$. ∎

######  Corollary 3.3.

If $G$ and $H$ are two finitely generated torsion-free 2-step nilpotent groups
that are commensurable, then there is some finite index subgroup
$N\subseteq_{\text{fin}}G$ such that $H\cong N$.

Next we argue that for a finitely generated torsion-free 2-step nilpotent
group, we can always factor out the (maximal) free abelian factor.

######  Lemma 3.4.

Let $H$ be a finitely generated torsion-free 2-step nilpotent group, then
$H\cong{\wiskunde Z}^{m}\times H^{\prime}$ (with $m\in{\wiskunde N}$) such
that

  1. (i)

$H^{\prime}$ is a finitely generated torsion-free 2-step nilpotent group

  2. (ii)

$Z(H^{\prime})=\sqrt[H^{\prime}]{\gamma_{2}(H^{\prime})}$

###### Proof.

Fix a ${\wiskunde Z}$-basis $\\{x_{1}Z(H),\dots,x_{n}Z(H)\\}$ (respectively
$\\{y_{1}\sqrt{\gamma_{2}(H)},\dots,y_{m}\sqrt{\gamma_{2}(H)}\\}$) of
$H/Z(H)\cong{\wiskunde Z}^{n}$ (respectively
$Z(H)/\sqrt{\gamma_{2}(H)}\cong{\wiskunde Z}^{m}$). Define
$H^{\prime}:=\langle x_{1},\dots,x_{n}\rangle\sqrt{\gamma_{2}(H)}$. One can
easily check that any element of $H^{\prime}$ can be uniquely written as
$\prod_{i=1}^{n}x_{i}^{t_{i}}\gamma$ (with $t_{i}\in{\wiskunde Z}$ and
$\gamma\in\sqrt{\gamma_{2}(H)}$) and thus $H\cong{\wiskunde Z}^{m}\times
H^{\prime}$.
Now we prove that
$\sqrt[H^{\prime}]{\gamma_{2}(H^{\prime})}=\sqrt[H]{\gamma_{2}(H)}=Z(H^{\prime})$.
Note that $\gamma_{2}(H)=\gamma_{2}(H^{\prime})$ and thus by using that
$\sqrt[H]{\gamma_{2}(H)}\subset H^{\prime}\subset H$ it readily follows that
$\sqrt[H^{\prime}]{\gamma_{2}(H^{\prime})}=\sqrt[H]{\gamma_{2}(H)}$. Since
$\sqrt[H]{\gamma_{2}(H)}\subset Z(H)\cap H^{\prime}\subset Z(H^{\prime})$, it
suffices to prove the other two inclusions. Since $H/Z(H)=\langle
x_{1}Z(H),\dots,x_{n}Z(H)\rangle$ and since $\langle
x_{1},\dots,x_{n}\rangle\subset H^{\prime}$, it follows that
$Z(H^{\prime})=Z(H)\cap H^{\prime}$. Now fix any
$h^{\prime}=\prod_{i=1}^{n}x_{i}^{t_{i}}\gamma\in Z(H)\cap H^{\prime}$ (where
$\gamma\in\sqrt[H]{\gamma_{2}(H)}\subset Z(H)$ and $t_{i}\in{\wiskunde Z}$).
Since $\gamma\in\sqrt[H]{\gamma_{2}(H)}\subset Z(H)$, we get that
$\prod_{i=1}^{n}x_{i}^{t_{i}^{\prime}}\in Z(H)$ and thus $t_{i}=0$ for all
$i$. Hence, $h^{\prime}\in\sqrt[H]{\gamma_{2}(H)}$ and thus we conclude that
$\sqrt[H^{\prime}]{\gamma_{2}(H^{\prime})}=\sqrt[H]{\gamma_{2}(H)}=Z(H^{\prime})=Z(H)\cap
H^{\prime}$. ∎

Any finitely generated group $G$ is polycyclic, i.e. it has a series
$1=G_{1}\lhd G_{2}\lhd\dots\lhd G_{n}=G$ with all $G_{i+1}/G_{i}$ cyclic. The
number of infinite cyclic factors in such a series is called the Hirsch number
$h(G)$ of $G$. The next Lemma gives some important properties.

######  Lemma 3.5.

Let $G$ be a finitely generated nilpotent group. Then the Hirsch number $h(G)$
of $G$ is well-defined, i.e. it does not depend on the chosen series.
Moreover, if $H\subset G$ and $N\lhd G$, then it holds that

  1. (i)

$h(G)\geq h(H)$

  2. (ii)

$h(G)=h(H)\>\Leftrightarrow\>[G:H]<\infty$

  3. (iii)

$h(G)=h(N)+h(G/N)$

  4. (iv)

$h(G)=0\>\Leftrightarrow\>|G|<\infty$

  5. (v)

if $G$ and $H$ are commensurable, then $h(G)=h(H)$.

######  Remark 3.6.

Remark that for a finitely generated torsion-free 2-step nilpotent group $H$,
it holds that

$h(Z(H))=h(\sqrt[H]{\gamma_{2}(H)})\quad\Longleftrightarrow\quad
Z(H)=\sqrt[H]{\gamma_{2}(H)}.$

Indeed, since $H$ is torsion-free 2-step nilpotent, it holds that
$\sqrt[H]{\gamma_{2}(H)}\subset Z(H)$. If moreover
$h(Z(H))=h(\sqrt[H]{\gamma_{2}(H)})$, then by Lemma 3.5 it follows that
$\sqrt[H]{\gamma_{2}(H)}\subset_{\text{fin}}Z(H)$. By Lemma 2.8 in [Bau71] it
holds that $\gamma_{2}(H)\subset_{\text{fin}}\sqrt[H]{\gamma_{2}(H)}$. Thus we
obtain that $\gamma_{2}(H)\subset_{\text{fin}}Z(H)$ and thus for any $g\in
Z(H)$ there is some $k\in{\wiskunde N}_{0}$ such that $g^{k}\in\gamma_{2}(H)$.
In particular, it holds that $Z(H)\subset\sqrt[H]{\gamma_{2}(H)}$ and thus
equality follows.

Now we have gathered all the tools to prove the result mentioned earlier.

######  Theorem 3.7.

Let $H$ be a finitely generated torsion-free 2-step nilpotent group with
$h(Z(H))=h(\sqrt[H]{\gamma_{2}(H)})$ (or equivalently
$Z(H)=\sqrt[H]{\gamma_{2}(H)}$) and $m\in{\wiskunde N}_{0}$. Then

None of the (fg. tv. 2-step nilpotent) groups commensurable with $H$ has the
$R_{\infty}$–property

$\Updownarrow$

None of the (fg. tv. 2-step nilpotent) groups commensurable with ${\wiskunde
Z}^{m}\times H$ has the $R_{\infty}$–property.

###### Proof.

We start by proving the implication from top to bottom. Fix any finitely
generated torsion-free 2-step nilpotent group $N$ which is commensurable with
$G:={\wiskunde Z}^{m}\times H$. By Corollary 3.3 we can assume that
$N\subset_{\text{fin}}G$. Using Lemma 3.4, we can write $N\cong{\wiskunde
Z}^{n}\times N^{\prime}$ (with $n\in{\wiskunde N}$ such that
$Z(N)\cong{\wiskunde Z}^{n}\times\sqrt[N]{\gamma_{2}(N)}$). We argue that
$n=m$. One can check that $\sqrt[G]{\gamma_{2}(G)}=\sqrt[H]{\gamma_{2}(H)}$.
Since two finitely generated torsion-free nilpotent groups are commensurable
if and only if their Mal’cev completions are isomorphic, we know that
$\sqrt[N]{\gamma_{2}(N)}$ and $\sqrt[G]{\gamma_{2}(G)}$ as well as $Z(N)$ and
$Z(G)$ are commensurable. Thus, using Lemma 3.5, we obtain that

$\displaystyle n$
$\displaystyle=h\Bigg{(}\frac{Z(N)}{\sqrt[N]{\gamma_{2}(N)}}\Bigg{)}=h(Z(N))-h(\sqrt[N]{\gamma_{2}(N)})$
$\displaystyle=h(Z(G))-h(\sqrt[G]{\gamma_{2}(G)})=m+h(Z(H))-h(\sqrt[H]{\gamma_{2}(H)})$
$\displaystyle=m.$

It follows that $N\cong{\wiskunde Z}^{m}\times N^{\prime}$ and thus (by
[BMO16, Proposition 5]) we have that

${\wiskunde Q}^{m}\times(N^{\prime})^{{\wiskunde Q}}\cong N^{{\wiskunde
Q}}\cong G^{{\wiskunde Q}}\cong{\wiskunde Q}^{m}\times H^{{\wiskunde Q}}.$

Hence, the associated Lie algebras ${\wiskunde Q}^{m}\times M(N^{\prime})$ and
${\wiskunde Q}^{m}\times M(H)$ are isomorphic. By Proposition 2.5 in [Der19]
it follows that $M(N^{\prime})\cong M(H)$, thus $(N^{\prime})^{{\wiskunde
Q}}\cong H^{{\wiskunde Q}}$, concluding that $N^{\prime}$ and $H$ are
commensurable. In particular, $N^{\prime}$ doesn’t have the
$R_{\infty}$–property. Hence, we can fix automorphisms
$\varphi_{N^{\prime}}\in{\rm Aut}(N^{\prime})$ and $\varphi_{{\wiskunde
Z}^{m}}\in{\rm Aut}({\wiskunde Z}^{m})$ with $R(\varphi_{N^{\prime}})<\infty$
and $R(\varphi_{{\wiskunde Z}^{m}})<\infty$. By for example [Sen21,
Proposition 2.4] we can now conclude that $R(\varphi_{{\wiskunde
Z}^{m}}\times\varphi_{N^{\prime}})=R(\varphi_{{\wiskunde
Z}^{m}})R(\varphi_{N^{\prime}})<\infty$. Hence, $N$ doesn’t have the
$R_{\infty}$–property.

Now we prove the other implication. Assume that $N^{\prime}$ is a finitely
generated torsion-free 2-step nilpotent group which is commensurable with $H$
and which has the $R_{\infty}$–property. Define $N:={\wiskunde Z}^{m}\times
N^{\prime}$. Since $H^{{\wiskunde Q}}\cong(N^{\prime})^{{\wiskunde Q}}$ and
$Z(H)=\sqrt[H]{\gamma_{2}(H)}$ it follows by Lemma 3.1 that
$Z(N^{\prime})=\sqrt[N^{\prime}]{\gamma_{2}(N^{\prime})}$ and
$(N^{\prime})^{{\wiskunde Q}}$ doesn’t have an abelian factor. Theorem 3.8 in
[Sen22] now tells us that ${\rm Spec}_{R}(N)={\rm Spec}_{R}({\wiskunde
Z}^{m})\cdot{\rm Spec}_{R}(N^{\prime})$. In particular, we can conclude that
$N$ has the $R_{\infty}$–property which concludes the proof. ∎

If $\Gamma(V,E)$ is a graph, then we use $V_{\text{iso}}\subset V$ to denote
the set of isolated vertices, i.e. vertices that are not adjacent to any other
vertex. Now we can apply Theorem 3.7 to our set-up of groups associated to
graphs.

######  Corollary 3.8.

Let $\Gamma(V,E)$ be a finite undirected simple graph. Denote with
$\tilde{\Gamma}$ the subgraph of $\Gamma$ on the vertex set $V\setminus
V_{\text{iso}}$. Then $G_{\Gamma}={\wiskunde Z}^{|V_{\text{iso}}|}\times
G_{\tilde{\Gamma}}$ and
$Z(G_{\tilde{\Gamma}})=\sqrt{\gamma_{2}(G_{\tilde{\Gamma}})}$. In particular,

None of the (fg. tv. 2-step nilpotent) groups commensurable with
$G_{\tilde{\Gamma}}$ has the $R_{\infty}$–property

$\Updownarrow$

None of the (fg. tv. 2-step nilpotent) groups commensurable with $G_{\Gamma}$
has the $R_{\infty}$–property.

######  Remark 3.9.

In an upcoming paper (see [DL]), we argue that all finitely generated torsion-
free 2-step nilpotent groups with Hirsch number at most six do not have the
$R_{\infty}$–property. Note that for all graphs $\Gamma(V,E)$ it holds that
$h(G_{\Gamma})=|V|+|E|$. Hence, if $|V|+|E|\leq 6$ then none of the groups
commensurable with $G_{\Gamma}$ has the $R_{\infty}$–property. Thus applying
Corollary 3.8 yields the stronger result that

if $\Gamma(V,E)$ is a finite undirected simple graph such that $|V\setminus
V_{\text{iso}}|+|E|\leq 6$, then none of the groups commensurable with
$G_{\Gamma}$ has the $R_{\infty}$–property.

For example, this holds for the graph depicted below. Note that for this graph
$G_{\Gamma}\cong{\wiskunde Z}\times H_{1}\times H_{1}$, where $H_{1}$ is the
Heisenberg group.

###  3.2 The case of free nilpotent partially commutative groups

In the final section of this paper, a result from [Wit23] is recalled (as
Theorem 6.1) which tells us for certain $c>1$ whether $A(\Gamma,c)$ has the
$R_{\infty}$–property or not. In this section, we explore how $R_{\infty}$ and
abstract commensurability behave within the class of free nilpotent partially
commutative groups. As it turns out, the free nilpotent partially commutative
group $A(\Gamma,c)$ (with $c>1$) contains a lot of information in the
following sense: if $A(\Gamma,c)$ has the $R_{\infty}$–property, then all
finitely generated torsion-free nilpotent groups commensurable with it must
have the $R_{\infty}$-property too. In fact, even a wider family of groups are
implied to have $R_{\infty}$, which is stated in the following proposition.

######  Proposition 3.10.

Let $\Gamma$ be a graph and $c>1$ an integer. The following are equivalent:

  1. (i)

$A(\Gamma,c)$ has the $R_{\infty}$–property.

  2. (ii)

Every integer-like automorphism of $L^{\wiskunde Q}(\Gamma,c)$ has an
eigenvalue 1.

  3. (iii)

Every group $H$ with $L^{\wiskunde Q}(H)\cong L^{\wiskunde Q}(\Gamma,c)$ has
the $R_{\infty}$–property.

###### Proof.

$(i)\Rightarrow(ii)$. Take any integer-like automorphism $\varphi$ of
$L(\Gamma,c)$. By Corollary 3.10 from [Wit23], there exists an automorphism
$\phi$ on $A(\Gamma,c)$ such that the induced automorphism $\overline{\phi}$
on $L^{\wiskunde Q}(\Gamma,c)$ has the same characteristic polynomial as
$\varphi$. Using Lemma 2.1 and the assumption, the automorphism
$\overline{\phi}$ has an eigenvalue 1 and hence $\varphi$ has an eigenvalue 1.

$(ii)\Rightarrow(iii)$. Let us write $\tilde{H}$ for the group
$H/\gamma_{c+1}(H)$. Clearly, we have the isomorphisms

$L^{\wiskunde Q}(\tilde{H})\cong\frac{L^{\wiskunde
Q}(H)}{\gamma_{c+1}(L^{\wiskunde Q}(H))}\cong\frac{L^{\wiskunde
Q}(\Gamma,c)}{\gamma_{c+1}(L^{\wiskunde Q}(\Gamma,c))}\cong L^{\wiskunde
Q}(\Gamma,c).$

Let $\varphi$ be any automorphism of $\tilde{H}$ and write
$\overline{\varphi}$ for the induced graded automorphism on $L^{\wiskunde
Q}(\Gamma,c)$ where we use the identification above. Note that
$\overline{\varphi}$ is integer-like. Thus, by assumption, it has an
eigenvalue 1. Lemma 2.1 then implies that $R(\varphi)=\infty$. Since $\varphi$
was an arbitrarily chosen automorphism of $\tilde{H}$, this proves that
$\tilde{H}$ has the $R_{\infty}$–property and by Lemma 1.1. from [GW09], so
does $H$.

$(iii)\Rightarrow(i)$. This follows trivially if we substitute $A(\Gamma,c)$
for $H$ in the assumption. ∎

######  Remark 3.11.

Note that any finitely generated torsion-free nilpotent group $H$, which is
abstractly commensurable to $A(\Gamma,c)$, satisfies (iii) of the above
proposition. Indeed, we get that $M(H)\cong M(A(\Gamma,c))$ and thus also
$L^{\wiskunde
Q}(H)\cong\operatorname{gr}(M(H))\cong\operatorname{gr}(M(A(\Gamma,c)))\cong
L^{\wiskunde Q}(A(\Gamma,c))\cong L^{\wiskunde Q}(\Gamma,c)$.

######  Remark 3.12.

Note that in (iii) of the above proposition one can, in general, not replace
the field ${\wiskunde Q}$ with a non-trivial field extension of it (which
would be a weaker assumption). Let us give a concrete example. Consider the
2-step nilpotent group $H$ with presentation

$H=\left\langle\begin{array}[]{l}x_{1},x_{2},x_{3},x_{4},\\\
y_{1},y_{2},y_{3}\end{array}\middle|\begin{array}[]{llll}\,[x_{1},x_{3}]=y_{1},&[x_{1},x_{4}]=y_{2},&[x_{3},x_{4}]=y_{3},&[y_{i},y_{j}]=1,\\\
\,[x_{2},x_{4}]=y_{1}^{2},&[x_{2},x_{3}]=y_{2},&[x_{1},x_{2}]=1\end{array}\right\rangle.$
(3.1)

This group does not have the $R_{\infty}$–property since the automorphism of
$H$ that is defined by

$\begin{array}[]{ll}x_{1}\mapsto x_{1}^{-1}x_{2}&y_{1}\mapsto
y_{1}^{3}y_{2}^{-2}\\\ x_{2}\mapsto x_{1}^{2}x_{2}^{-1}&y_{2}\mapsto
y_{1}^{-4}y_{2}^{3}\\\ x_{3}\mapsto x_{3}^{-1}x_{4}&y_{3}\mapsto y_{3}^{-1}\\\
x_{4}\mapsto x_{3}^{2}x_{4}^{-1}&\end{array}$

induces an automorphism on $L^{\wiskunde C}(H)$ which does not have 1 as an
eigenvalue (see Lemma 2.1). On the other hand, consider the graph $\Gamma$ as
drawn below.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$

By Theorem 1.4 in [Wit23], $G_{\Gamma}=A(\Gamma,2)$ has the
$R_{\infty}$–property. However, over the field $K={\wiskunde Q}(\sqrt{2})$, we
have a graded Lie algebra isomorphism $L^{K}(H)\cong L^{K}(\Gamma,2)$ which is
determined by

$\begin{array}[]{ll}x_{1}\mapsto v_{1}+v_{4}&x_{3}\mapsto v_{2}+v_{3}\\\
x_{2}\mapsto\sqrt{2}(v_{1}-v_{4})&x_{4}\mapsto\sqrt{2}(v_{2}-v_{3}).\end{array}$

##  4 Groups associated to edge-weighted graphs

###  4.1 Relations on the vertices and edges

In this section, we introduce notations and definitions used throughout the
rest of the paper. For a finite undirected simple graph $\Gamma=(V,E)$ we will
first define preorders on $V$ and $E$ (denoted with $\prec$) and induced
equivalence relations on $V$ and $E$ (denoted with $\sim$), for which the sets
of equivalence classes will be denoted by $\Lambda$ and $M$, respectively.
Next, we define the quotient graph
$\overline{\Gamma}=(\Lambda,\overline{E},\Psi)$ associated to $\Gamma$. At
last, we define partial orders on $\Lambda$ and $M$ (denoted with
$\mathbin{\overline{\prec}}$) and refine these orders to get total orders on
$V$ and $E$ (denoted with $<$). All definitions are illustrated by an example
at the end of the section.

Let $\Gamma=(V,E)$ be a finite undirected simple graph. Let $k:E\to{\wiskunde
N}_{0}$ be a map associating to each edge $e\in E$ a weight $k(e)\in{\wiskunde
N}_{0}$. We use $\Gamma(k)$ to denote the graph $\Gamma$ with weight function
$k$ on its edges. From now on, we abbreviate $\Gamma(1)$ to $\Gamma$ (where
$1:E\to{\wiskunde N}_{0}:e\mapsto 1$).
Let $v\in V$ be a vertex of the graph $\Gamma$. We define its open and closed
neighbourhood ($\Omega^{\prime}(v)$ and $\Omega(v)$ respectively) by setting

$\Omega^{\prime}(v):=\\{w\in V\mid\\{v,w\\}\in
E\\}\quad\text{and}\quad\Omega(v):=\Omega^{\prime}(v)\cup\\{v\\}.$

Next, we define a preorder $\prec$ (i.e. a reflexive and transitive relation)
on the vertex set $V$ by setting for any $v,w\in V$:

$v\prec w\quad\Leftrightarrow\quad\Omega^{\prime}(v)\subseteq\Omega(w).$

This preorder on $V$ induces a preorder $\prec$ on $E$ by defining for all
$\\{v,w\\},\\{v^{\prime},w^{\prime}\\}\in E$ that

$\\{v,w\\}\prec\\{v^{\prime},w^{\prime}\\}\quad\Longleftrightarrow\quad(v\prec
v^{\prime}\text{ and }w\prec w^{\prime})\text{ or }(v\prec w^{\prime}\text{
and }w\prec v^{\prime}).$

Using these preorders, we define an equivalence relation $\sim$ on $V$ and
$\sim$ on $E$ by setting for all $v,w\in V$ and $e,e^{\prime}\in E$ that

$\displaystyle v\sim w$ $\displaystyle\quad\Longleftrightarrow\quad v\prec
w\text{ and }w\prec v$ $\displaystyle e\sim e^{\prime}$
$\displaystyle\quad\Longleftrightarrow\quad e\prec e^{\prime}\text{ and
}e^{\prime}\prec e.$

Note that for two vertices $v,w\in V$ the expression $v\sim w$ is equivalent
to stating that the transposition $(v\>w)$ is an automorphism of the graph
$\Gamma$. If none of the transpositions is a graph automorphism of $\Gamma$
(or equivalently, no two vertices are equivalent), we say that $\Gamma$ is
transposition-free. We let $[v]$ and $[e]$ denote the equivalence classes of
$v\in V$ and $e\in E$, respectively. The equivalence classes on $V$ are called
the coherent components of $\Gamma$ and we denote the quotient space with
$\Lambda:=V/\sim$. The correspondent quotient space $E/\sim$ is written as
$M$. The map $\Psi:\Lambda\to{\wiskunde N}_{0}$ sends any coherent component
to the number of vertices it contains. Sometimes we use the abbreviation
$|\lambda|$ instead of $\Psi(\lambda)$. Remark that if two vertices are
connected via an edge, then all the edges between their corresponding coherent
components are contained in $E$, i.e. for all (not necessarily distinct)
$[v],[w]\in\Lambda$ it holds that

$\\{v,w\\}\in E\quad\Longrightarrow\quad\\{v^{\prime},w^{\prime}\\}\in E\text{
for {all} }v^{\prime}\in[v]\text{ and }w^{\prime}\in[w].$

Hence, the graph $\Gamma$ induces unambiguously a graph $\overline{\Gamma}$
with vertex set $\Lambda$ and edge set $\overline{E}$ defined by

$\overline{E}:=\\{\\{[v],[w]\\}\mid\\{v,w\\}\in E\\}.$

The triple $\overline{\Gamma}=(\Lambda,\overline{E},\Psi)$ is called the
quotient graph of $\Gamma$. Each edge of the quotient graph corresponds with a
collection of edges of $E$ (by considering the fibers of the map
$E\to\overline{E}:\\{v,w\\}\mapsto\\{[v],[w]\\}$). The next lemma shows us,
among other, that the map

$M\to\overline{E}:[\\{v,w\\}]\mapsto\\{[v],[w]\\}$ (4.1)

is a bijection. By consequence, we can identify $M$ with $\overline{E}$.

######  Lemma 4.1.

For any $e=\\{v,w\\}\in E$ it holds that

$[e]=\Big{\\{}\\{v^{\prime},w^{\prime}\\}\>\Big{|}\>v^{\prime}\in[v],w^{\prime}\in[w]\Big{\\}}.$

###### Proof.

The $\supseteq$-inclusion follows directly by using the definition of $\prec$
and $\sim$ on $E$. For the other inclusion let us fix some
$e^{\prime}=\\{v^{\prime},w^{\prime}\\}\in[e]$. Hence, it holds that

$\displaystyle(v^{\prime}\prec v\text{ and }w^{\prime}\prec w)$
$\displaystyle\text{ or }(v^{\prime}\prec w\text{ and }w^{\prime}\prec
v)\text{ and }$ $\displaystyle(v\prec v^{\prime}\text{ and }w\prec
w^{\prime})$ $\displaystyle\text{ or }(v\prec w^{\prime}\text{ and }w\prec
v^{\prime}).$

Note that if for example ($v^{\prime}\prec v\text{ and }w^{\prime}\prec w$)
and ($v\prec w^{\prime}\text{ and }w\prec v^{\prime}$) then $v\prec
w^{\prime}\prec w\prec v^{\prime}\prec v$ and thus it follows that $v\sim
w\sim v^{\prime}\sim w^{\prime}$. Hence, in this particular case $e^{\prime}$
is definitely contained in the right-hand side. Considering all the possible
cases, one can conclude the rest of the proof. ∎

Moreover, the preorders $\prec$ on the vertices and the edges induce partial
orders $\mathbin{\overline{\prec}}$ on $\Lambda$ and $M$ by setting for all
$v,w\in V$ and $e,e^{\prime}\in E$ that

$\displaystyle[v]\mathbin{\overline{\prec}}[w]$
$\displaystyle\quad\Longleftrightarrow\quad v\prec w$
$\displaystyle[e]\mathbin{\overline{\prec}}[e^{\prime}]$
$\displaystyle\quad\Longleftrightarrow\quad e\prec e^{\prime}.$

Using these partial orders, we can fix total orders

$V/\sim\,=\Lambda=\\{\lambda_{1},\dots,\lambda_{r}\\}\quad\text{and}\quad
E/\sim\,=M=\\{\mu_{1},\dots,\mu_{s}\\}$

with the property that $i\leq j$ if
$\lambda_{i}\mathbin{\overline{\prec}}\lambda_{j}$ and $i\leq j$ if
$\mu_{i}\mathbin{\overline{\prec}}\mu_{j}$. We can refine these orders to
total orders on $V$ and $E$ which we will both denote with $<$.

######  Example 4.2.

We explain the introduced notation using a concrete example. We consider the
graph $\Gamma$ in Figure 1(a).

1 (a) Graph $\Gamma$

2121 (b) Graph $\overline{\Gamma}$

2

$\lambda_{1}$

1

$\lambda_{3}$

2

$\lambda_{2}$

1

$\lambda_{4}$

$\mu_{3}$

$\mu_{1}$

$\mu_{4}$

$\mu_{2}$

(c) Graph $\overline{\Gamma}$ with notation

$v_{1}$

$v_{5}$

$v_{2}$

$v_{3}$

$v_{6}$

$v_{4}$

1

$e_{1}$

$e_{2}$

$e_{5}$

$e_{6}$

$e_{3}$

$e_{4}$

(d) Graph $\Gamma(k)$ with notation

Figure 1: Graphs $\Gamma(k)$ and $\overline{\Gamma}$ with introduced notation

We construct the quotient graph $\overline{\Gamma}$ and we display the size of
each coherent component next to the corresponding vertex of
$\overline{\Gamma}$ (see Figure 1(b)). Next, we look at the induced relation
$\mathbin{\overline{\prec}}$ on $\Lambda$. We could for example fix the total
order on $\Lambda$ as depicted in Figure 1(c). Indeed, the relations on
$\Lambda$ are

$\lambda_{1}\mathbin{\overline{\prec}}\lambda_{3},\>\lambda_{2}\mathbin{\overline{\prec}}\lambda_{3}\text{
and }\lambda_{2}\mathbin{\overline{\prec}}\lambda_{4}$

and thus this total order satisfies that $i\leq j$ if
$\lambda_{i}\mathbin{\overline{\prec}}\lambda_{j}$. We could have for example
also interchanged $\lambda_{1}$ and $\lambda_{2}$. Now we fix an order inside
each coherent component and get the total order on $V$ depicted in Figure
1(d).

The only relations on the edges are

$\\{v_{1},v_{5}\\}\sim\\{v_{2},v_{5}\\}\text{ and
}\\{v_{3},v_{6}\\}\sim\\{v_{4},v_{6}\\}$

and thus the induced partial order $\mathbin{\overline{\prec}}$ on
$\overline{E}$ is the identity relation. Hence, we could for example fix the
total order

$[\\{v_{1},v_{5}\\}]<[\\{v_{3},v_{6}\\}]<[\\{v_{1},v_{2}\\}]<[\\{v_{5},v_{6}\\}]$

on $\overline{E}$ and refine the order to the following total order on $E$

$\\{v_{1},v_{5}\\}<\\{v_{2},v_{5}\\}<\\{v_{3},v_{6}\\}<\\{v_{4},v_{6}\\}<\\{v_{1},v_{2}\\}<\\{v_{5},v_{6}\\}.$

###  4.2 Definition and structure of the groups $G_{\Gamma(k)}$

The total order on the vertices and edges allows us to construct and study
multiple groups associated to a graph $\Gamma$. For a graph $\Gamma(k)$ with
weight function $k:E\to{\wiskunde N}_{0}$, we define $G_{\Gamma(k)}$ by

$\displaystyle G_{\Gamma(k)}$ $\displaystyle:=\Bigg{\langle}V\cup
E\;\;\Bigg{|}\;\;[V,E]=[E,E]=1\text{ and }\forall v,w\in V\text{ with }v<w:$
$\displaystyle\hskip 150.00023pt[v,w]=\begin{cases}e^{k(e)}&\text{if
}e:=\\{v,w\\}\in E\\\ 1&\text{else}\end{cases}\Bigg{\rangle}.$

Note that $G_{\Gamma(k)}\cong G_{\Gamma}$ if $k:E\to{\wiskunde N}_{0}:e\mapsto
1$.

######  Remark 4.3.

The definition of $G_{\Gamma(k)}$ is independent of the choice of total order
on the vertices in the sense that one gets isomorphic groups for different
total orders. Indeed, let $\tilde{<}$ be another total order on the vertices
$V$. Then it is not hard to check that the map $\theta$, which is the identity
on $V$ and is defined for all $e=\\{v,w\\}\in E$ by

$\theta(e)=\begin{cases}e&\text{if }v<w\Leftrightarrow v\tilde{<}w\\\
e^{-1}&\text{otherwise},\end{cases}$

induces an isomorphism from the group defined using $<$ to the group defined
using $\tilde{<}$.

######  Example 4.4.

Consider the graph $\Gamma=(V,E)$ where $V=\\{v_{1},v_{2}\\}$ and
$E=\\{\\{v_{1},v_{2}\\}\\}$ and for any $n\in{\wiskunde N}_{0}$, let
$k:E\to{\wiskunde N}_{0}$ be the weight function on the singleton set $E$ with
value $n$.

$v_{1}$$v_{2}$$n$

Then for each $n\in{\wiskunde N}_{0}$, the group $G_{\Gamma(k)}$ is isomorphic
to the matrix group

$H_{n}=\left\\{\begin{pmatrix}1&nx&z\\\ 0&1&y\\\
0&0&1\end{pmatrix}\>\middle|\>x,y,z\in{\wiskunde Z}\right\\}$

where the isomorphism is given by

$v_{1}\mapsto\begin{pmatrix}1&n&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix},\quad
v_{2}\mapsto\begin{pmatrix}1&0&0\\\ 0&1&1\\\
0&0&1\end{pmatrix},\quad\\{v_{1},v_{2}\\}\mapsto\begin{pmatrix}1&0&1\\\
0&1&0\\\ 0&0&1\end{pmatrix}$

In particular, when $n=1$, $G_{\Gamma(k)}\cong G_{\Gamma}$ is isomorphic to
the Heisenberg group. It is well-known that the groups $H_{n}$ for
$n\in{\wiskunde N}_{0}$ are, up to isomorphism, all the possible groups
abstractly commensurable to $H_{1}$. Moreover, ${\rm
Spec}_{R}(H_{n})=2{\wiskunde N}_{0}\cup\\{\infty\\}$ for all $n\in{\wiskunde
N}_{0}$ (see for example [Rom11]). Hence, $H_{1}$ is an example of a finitely
generated torsion-free 2-step nilpotent group such that all groups
commensurable with $H_{1}$ have the same (infinite) Reidemeister spectrum.

It will be helpful to address the elements $e^{k(e)}\in G_{\Gamma(k)}$, so we
define

$E^{k}:=\\{e^{k(e)}\mid e\in E\\}.$

Recall that $V_{\text{iso}}\subset V$ denotes the set of isolated vertices of
$\Gamma$.

The next lemma provides insight into the structure of the group
$G_{\Gamma(k)}$.

######  Lemma 4.5.

Denote $V=\\{v_{1},\dots,v_{n}\\}$ and $E=\\{e_{1},\dots,e_{m}\\}$. Then any
element of $G_{\Gamma(k)}$ can be uniquely expressed as a product

$\prod_{i=1}^{n}v_{i}^{z_{i}}\prod_{j=1}e^{t_{j}}$

with $z_{i},t_{j}\in{\wiskunde Z}$. Moreover, the centre and the factors of
the (adapted) lower central series of $G_{\Gamma(k)}$ are given by

$\displaystyle Z(G_{\Gamma(k)})$ $\displaystyle=\left\langle
V_{\text{iso}}\cup E\right\rangle\cong\mathbb{Z}^{|V_{\text{iso}}\cup E|},$
$\displaystyle\gamma_{2}(G_{\Gamma(k)})$ $\displaystyle=\left\langle
E^{k}\right\rangle\cong{\wiskunde Z}^{|E|},\quad\quad$
$\displaystyle\frac{G_{\Gamma(k)}}{\gamma_{2}(G_{\Gamma(k)})}$
$\displaystyle\cong{\wiskunde Z}^{|V|}\times\bigtimes_{e\in E}\frac{{\wiskunde
Z}}{k(e){\wiskunde Z}},$ $\displaystyle\sqrt{\gamma_{2}(G_{\Gamma(k)})}$
$\displaystyle=\langle E\rangle\cong{\wiskunde Z}^{|E|},\quad\quad$
$\displaystyle\frac{G_{\Gamma(k)}}{\sqrt{\gamma_{2}(G_{\Gamma(k)})}}$
$\displaystyle=\bigtimes_{v\in V}\left\langle
v\sqrt{\gamma_{2}(G_{\Gamma(k)})}\right\rangle\cong{\wiskunde Z}^{|V|}.$

Moreover, the identity on $V$ induces an injective morphism
$\iota:G_{\Gamma}\xhookrightarrow{}G_{\Gamma(k)}$ with $G_{\Gamma}\cong{\rm
Im}(\iota)=\langle V\cup E^{k}\rangle$. In particular, $G_{\Gamma}$ is
isomorphic to the finite index subgroup $\langle V\cup E^{k}\rangle\subset
G_{\Gamma(k)}$ with index $\prod_{e\in E}k(e)$.

Note that $\gamma_{2}(G_{\Gamma(k)})\subset Z(G_{\Gamma(k)})$ and thus
$G_{\Gamma(k)}$ is a (finitely generated torsion-free) 2-step nilpotent group.
In section 2.1 we associated two Lie algebras to these kinds of groups. As it
turns out, in this case, they are isomorphic and do not depend on the chosen
weight function.

######  Lemma 4.6.

For any finite undirected simple graph $\Gamma=(V,E)$ and a weight function
$k:E\to{\wiskunde N}_{0}$ on its edges, the Lie algebras $L^{{\wiskunde
Q}}(\Gamma,2)$, $L^{{\wiskunde Q}}(G_{\Gamma})$, $L^{{\wiskunde
Q}}(G_{\Gamma(k)})$, $M(G_{\Gamma})$ and $M(G_{\Gamma(k)})$ are all isomorphic
and the isomorphisms are induced by the identity on the vertices.

###### Proof.

Note that any 2-step nilpotent rational Lie algebra $L$ is isomorphic to
$\operatorname{gr}(L)$ and that the isomorphism is canonical up to a choice of
vector space complement to $[L,L]$. Since for $M(G_{\Gamma})$ and
$M(G_{\Gamma(k)})$, such a complement is naturally given by the span of the
vertices, one gets (after combining with Equation (2.1)) the isomorphisms
$M(G_{\Gamma})\cong\operatorname{gr}(M(G_{\Gamma}))\cong L^{\wiskunde
Q}(G_{\Gamma})$ and
$M(G_{\Gamma(k)})\cong\operatorname{gr}(M(G_{\Gamma(k)}))\cong L^{\wiskunde
Q}(G_{\Gamma(k)})$. As one can check, all these isomorphisms restrict to the
identity on the vertices.

Next, since the group $G_{\Gamma}$ is a finite index subgroup of
$G_{\Gamma(k)}$, it follows (see section 3.10) that the associated Mal’cev Lie
algebras are isomorphic and that the isomorphism restricts to the inclusion of
$G_{\Gamma}$ in $G_{\Gamma(k)}$ and thus to the identity on the vertices.

At last, using Equation (2.2), we find that $L^{\wiskunde Q}(\Gamma,2)\cong
L^{\wiskunde Q}(G_{\Gamma})$ where the isomorphism restricts to the identity
on the vertices. This completes the proof. ∎

##  5 Automorphisms of $G_{\Gamma(k)}$

In this section, we describe (to some extent) the automorphisms of
$G_{\Gamma(k)}$. In the first part (section 5.1), we recall a result from
[DM21], where the automorphisms of $L^{K}(\Gamma,2)$ are described for any
field $K\subset{\wiskunde C}$. Since $L^{\wiskunde Q}(\Gamma,2)\cong
L^{\wiskunde Q}(G_{\Gamma(k)})$ for any weight function $k$ (see Lemma 4.6),
this result gives us information about the automorphisms of $G_{\Gamma(k)}$
(recall that any automorphism of $G_{\Gamma(k)}$ induces an automorphism on
$L^{\wiskunde Q}(G_{\Gamma(k)})$). In the second part (section 5.2), we argue
on the group level to get more restrictions on the possible automorphisms.
These restrictions are imposed by the weight function $k:E\to{\wiskunde
N}_{0}$.

###  5.1 Automorphisms of $L^{K}(\Gamma,2)$

In order to describe the automorphisms of $L^{K}(\Gamma,2)$, one needs some
extra maps and notation:

Recall the definition from the quotient graph
$\overline{\Gamma}=(\Lambda,\overline{E},\Psi)$ of $\Gamma=(V,E)$. We say that
$\sigma\in{\rm Aut}(\overline{\Gamma})$ is an automorphism of the quotient
graph if it is a bijection $\sigma:\Lambda\to\Lambda$ such that edges are
mapped to edges (i.e. $\sigma(\overline{E})=\overline{E}$) and the sizes of
the coherent components are invariant (i.e. $\Psi\circ\sigma=\Psi$). Note that
any graph automorphism $\sigma\in{\rm Aut}(\Gamma)$ induces a unique
automorphism $p(\sigma)\in{\rm Aut}(\overline{\Gamma})$ by setting
$p(\sigma)([v]):=[\sigma(v)]$ for any vertex $v\in V$. This gives a surjective
group homomorphism $p:{\rm Aut}(\Gamma)\to{\rm Aut}(\overline{\Gamma})$.

For any basis $X$ of a vector space $W$ over $K$ and any permutation $\sigma$
on the basis $X$, we write $P(\sigma)$ for the linear map on $W$ (or for the
corresponding matrix) defined by $P(\sigma)(x):=\sigma(x)$ (for all $x\in X$).

We define the linear endomorphisms $E_{vw}$ (for all vertices $v,w\in V$) on
${\rm Sp}_{K}(V)$ (for any $v^{\prime}\in V$) by

$E_{vw}(v^{\prime})=\begin{cases}v&\text{if }v^{\prime}=w\\\ 0&\text{if
}v^{\prime}\neq w\end{cases}.$

Denote with $I\in{\rm GL}({\rm Sp}_{K}(V))$ the identity map and with
$\mathcal{U}\subseteq{\rm GL}({\rm Sp}_{K}(V))$ the subgroup defined by

$\mathcal{U}:=\langle I+tE_{vw}\>\mid\>v\prec w,\>w\not\prec v,\>t\in
K\rangle.$

We use ${\rm GL}({\rm Sp}_{K}(\lambda))$ (for any $\lambda\in\Lambda$) to
denote the subgroup of ${\rm GL}({\rm Sp}_{K}(V))$ where $f\in{\rm GL}({\rm
Sp}_{K}(\lambda))$ induces a linear map (of ${\rm GL}({\rm Sp}_{K}(V))$) by
setting (for any $v\in V$)

$v\mapsto\begin{cases}f(v)&\text{if }v\in\lambda\\\ v&\text{if
}v\not\in\lambda\end{cases}.$

We have now introduced enough notation to describe the graded automorphisms of
$L^{K}(\Gamma,2)$ (with $K\subset{\wiskunde C}$). Recall Remark 2.2.

######  Theorem 5.1 ([DM21]).

Let $K\subset{\wiskunde C}$ be a field. Then the image of $\pi:{\rm
Aut}_{g}(L^{K}(\Gamma,2))\to{\rm GL}({\rm Sp}_{K}(V))$ is a linear algebraic
group given by

$\mathcal{G}_{\Gamma}:=P({\rm Aut}(\Gamma))\cdot\bBigg@\char
3\relax(\prod_{\lambda\in\Lambda}{\rm GL}({\rm Sp}_{K}(\lambda))\bBigg@\char
3\relax)\cdot\mathcal{U}$

where $\mathcal{U}$ is its unipotent radical. In particular, the matrix of any
element $f\in\mathcal{G}_{\Gamma}$ with respect to the basis $(V,<)$ has the
form (where the blocks correspond with the coherent components
$\lambda_{1}<\dots<\lambda_{r}$)

$P(\sigma)\cdot\begin{pmatrix}A_{1}&A_{12}&\ldots&A_{1r}\\\
0&A_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&A_{r-1r}\\\
0&\ldots&0&A_{r}\end{pmatrix}$

where $\sigma\in{\rm Aut}(\Gamma)$, $A_{i}\in{\rm GL}_{|\lambda_{i}|}(K)$,
$A_{ij}\in K^{|\lambda_{i}|\times|\lambda_{j}|}$ and $A_{ij}=0$ if
$\lambda_{i}\not\mathbin{\overline{\prec}}\lambda_{j}$. The permutation
$\sigma$ is not necessarily unique, but the induced permutation
$p(\sigma)\in{\rm Aut}(\overline{\Gamma})$ is unique. Moreover, the assignment
$f\mapsto p(\sigma)$ is a group homomorphism from $\mathcal{G}_{\Gamma}$ to
${\rm Aut}(\overline{\Gamma})$.

Recall that any automorphism $\varphi\in{\rm Aut}(G_{\Gamma(k)})$ induces an
automorphism $\overline{\varphi}\in{\rm Aut}_{g}(L^{\wiskunde
Q}(G_{\Gamma(k)}))\cong{\rm Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))$. Hence, by
Theorem 5.1 it induces a (unique) automorphism $p(\sigma)\in{\rm
Aut}(\overline{\Gamma})$. In the next section, we study which automorphisms of
$\overline{\Gamma}$ can arise in this way. Later, we use this (together with
Lemma 2.1) to derive some restrictions on an edge-weighted graph (with
weighted edges) such that the associated group has the $R_{\infty}$–property.

###  5.2 Automorphisms of $G_{\Gamma(k)}$

######  Notation 5.2.

Let $(B=\\{b_{1},\dots,b_{n}\\},<)$ be a finite totally ordered set (with
$b_{1}<\dots<b_{n}$) and $\theta:B\to{\wiskunde Z}$ a map. Then we define
$D(\theta)\in{\wiskunde Z}^{n\times n}$ to be the diagonal matrix with the
image of $\theta$ (ordered by $<$ on $B$) on its diagonal, i.e.

$D(\theta):=\text{diag}(\theta(b_{1}),\dots,\theta(b_{n}))=\begin{pmatrix}\theta(b_{1})&0&\ldots&0\\\
0&\theta(b_{2})&\ddots&\vdots\\\ \vdots&\ddots&\ddots&0\\\
0&\ldots&0&\theta(b_{n})\end{pmatrix}.$

If $B$ is a basis of a ${\wiskunde Z}$-module $M$, then we will sometimes
abuse notation and denote with $D(\theta)$ also the linear map on $M$ induced
by $D(\theta)(b_{i}):=\theta(b_{i})b_{i}$ (for all $i=1,\dots,n$).

######  Lemma 5.3.

Fix a finite undirected simple graph $\Gamma=(V,E)$ with weights
$k:E\to{\wiskunde N}_{0}$ on its edges. Let $\varphi\in{\rm
Aut}(G_{\Gamma(k)})$ and denote with $A\in{\rm GL}_{|E|}({\wiskunde Z})$ the
matrix of $\varphi|_{\gamma_{2}(G_{\Gamma(k)})}$ with respect to the
${\wiskunde Z}$-basis $(E^{k},<)$.
Then the matrix of $\varphi|_{\sqrt{\gamma_{2}(G_{\Gamma(k)})}}$ with respect
to $(E,<)$ is given by $D(k)AD(k)^{-1}$. In particular, $D(k)AD(k)^{-1}\in{\rm
GL}_{|E|}({\wiskunde Z})$.

###### Proof.

Since the induced automorphisms on $\gamma_{2}(G_{\Gamma(k)})$ and
$\sqrt{\gamma_{2}(G_{\Gamma(k)})}$ are restrictions of $\varphi$, we obtain
the following commuting diagram.

${\gamma_{2}(G_{\Gamma(k)})}$${\gamma_{2}(G_{\Gamma(k)})}$${\sqrt{\gamma_{2}(G_{\Gamma(k)})}}$${\sqrt{\gamma_{2}(G_{\Gamma(k)})}}$$\scriptstyle{\varphi|_{\gamma_{2}}}$$\scriptstyle{\iota}$$\scriptstyle{\huge{\circlearrowleft}}$$\scriptstyle{\iota}$$\scriptstyle{\varphi|_{\sqrt{\gamma_{2}}}}$

Recall that $\gamma_{2}(G_{\Gamma(k)})$ and $\sqrt{\gamma_{2}(G_{\Gamma(k)})}$
are both free abelian of rank $|E|$. By Lemma 4.5, the sets $E$ and
$E^{k}=\\{e^{k(e)}\mid e\in E\\}$ ordered with the total order $<$ give a
basis for the ${\wiskunde Z}$-modules $\sqrt{\gamma_{2}(G_{\Gamma(k)})}$ and
$\gamma_{2}(G_{\Gamma(k)})$, respectively. Hence, by using these bases we get
the commuting diagram of matrices

${{\wiskunde Z}^{|E|}}$${{\wiskunde Z}^{|E|}}$${{\wiskunde
Z}^{|E|}}$${{\wiskunde
Z}^{|E|}}$$\scriptstyle{A}$$\scriptstyle{D}$$\scriptstyle{\huge{\circlearrowleft}}$$\scriptstyle{D}$$\scriptstyle{B}$

where $A,B\in{\rm GL}_{|E|}({\wiskunde Z})$ are the matrix representations of
$\varphi|_{\gamma_{2}(G_{\Gamma(k)})}$ and
$\varphi|_{\sqrt{\gamma_{2}(G_{\Gamma(k)})}}$, respectively, and
$D\in{\wiskunde Z}^{|E|\times|E|}$ is the matrix representation of the
inclusion $\iota$. Note that $D=D(k)$ since a basis vector $e^{k(e)}$ of
$\gamma_{2}(G_{\Gamma(k)})$ is sent to the basis vector $e$ of
$\sqrt{\gamma_{2}(G_{\Gamma(k)})}$ but raised to the power $k(e)$. Thus the
commutative diagram implies that $BD(k)=D(k)A$. Since $k(e)\neq 0$ (for all
$e\in E$), these matrices are invertible over ${\wiskunde Q}$. Over
${\wiskunde Q}$, we indeed obtain that $B=D(k)AD(k)^{-1}$. ∎

######  Remark 5.4.

Recall that any automorphism of $G_{\Gamma(k)}$ is completely determined by
the image of the vertices. Indeed, by Lemma 5.3 the image of $E$ is fixed by
what happens on $\gamma_{2}(G_{\Gamma(k)})$ which on its turn is determined by
the image of the vertices. Since $V$ and $E$ generate $G_{\Gamma(k)}$ the
claim follows. Which matrices $A\in{\rm GL}_{|V|}({\wiskunde Z})$ on the
quotient of $G_{\Gamma(k)}$ with $\sqrt{\gamma_{2}(G_{\Gamma(k)})}$ induce an
automorphism on $G_{\Gamma(k)}$? One can prove that it suffices to require
that it induces an automorphism of $G_{\Gamma}$ and that the induced matrix on
$\sqrt{\gamma_{2}(G_{\Gamma(k)})}$ (which we described in Lemma 5.3) has
integer entries.
Note however that any such matrix $A\in{\rm GL}_{|V|}({\wiskunde Z})$ does not
induce a unique automorphism of $G_{\Gamma(k)}$. Indeed, one can always
multiply the image of a vertex with any element of
$\sqrt{\gamma_{2}(G_{\Gamma(k)})}$. Nevertheless, the induced automorphisms on
the Lie ring $L(G_{\Gamma(k)})$ are the same and they have the same
Reidemeister number. Hence, since we are interested in the Reidemeister
numbers of automorphisms of $G_{\Gamma(k)}$, we can assume without loss of
generality that $\varphi(\langle V\rangle)=\langle V\rangle$ for the
automorphisms we work with.

Using the description of the automorphisms of the Lie algebra $L^{\wiskunde
Q}(\Gamma,2)$ of above, we can describe to some extent the automorphisms of
$G_{\Gamma(k)}$ and in particular their induced maps on
$\gamma_{2}(G_{\Gamma(k)})$. In order to do so, we need the following
notation.

Let $\Gamma=(V,E)$ be a graph with an automorphism $\sigma\in{\rm
Aut}(\Gamma)$. Then $\sigma$ induces a permutation $\sigma_{E}$ on the edge
set by

$\sigma_{E}:E\to E:\\{v,w\\}\mapsto\\{\sigma(v),\sigma(w)\\}.$

If there is a total order $<$ on the vertices $V$, we say that an edge
$\\{v,w\\}\in E$ is an inversion of $\sigma$ if $v<w$, but
$\sigma(v)>\sigma(w)$. We then define the map
$\varepsilon_{\sigma}:E\to\\{-1,1\\}$ by setting for all $e\in E$

$\varepsilon_{\sigma}(e):=\begin{cases}-1&\text{if }e\text{ is an inversion of
}\sigma\\\ 1&\text{otherwise}.\end{cases}$

With this notation at hand, we can say the following about the automorphisms
of $G_{\Gamma(k)}$. Recall from section 4.1 that we defined a total order $<$
on both the vertices and the edges, which were a refinement of the orders on
$\Lambda=\\{\lambda_{1},\dots,\lambda_{r}\\}$ and
$M=\\{\mu_{1},\dots,\mu_{s}\\}$, respectively.

######  Lemma 5.5.

Let $\Gamma=(V,E)$ be a finite undirected simple graph with weight function
$k:E\to{\wiskunde N}_{0}$ on its edges and write $G=G_{\Gamma(k)}$. For any
automorphism $\varphi\in{\rm Aut}(G)$ it holds that

  1. (i)

with respect to the ordered ${\wiskunde Z}$-basis $(V,<)$, the induced
automorphism on $G/\sqrt{\gamma_{2}(G)}$ is represented by a matrix of the
form

$P(\sigma)\cdot\begin{pmatrix}A_{1}&A_{12}&\ldots&A_{1r}\\\
0&A_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&A_{r-1r}\\\
0&\ldots&0&A_{r}\end{pmatrix}$

where $\sigma\in{\rm Aut}(\Gamma)$, $A_{i}\in{\rm
GL}_{|\lambda_{i}|}({\wiskunde Z})$, $A_{ij}\in{\wiskunde
Z}^{|\lambda_{i}|\times|\lambda_{j}|}$ and $A_{ij}=0$ if
$\lambda_{i}\not\mathbin{\overline{\prec}}\lambda_{j}$.

  2. (ii)

with respect to the ordered ${\wiskunde Z}$-basis $(E^{k},<)$, the induced
automorphism on $\gamma_{2}(G)$ is represented by a matrix of the form

$P(\sigma_{E})\cdot
D(\varepsilon_{\sigma})\cdot\begin{pmatrix}B_{1}&B_{12}&\ldots&B_{1s}\\\
0&B_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&B_{s-1s}\\\
0&\ldots&0&B_{s}\end{pmatrix}$

where $B_{i}\in{\rm GL}_{|\mu_{i}|}({\wiskunde Z})$, $B_{ij}\in{\wiskunde
Z}^{|\mu_{i}|\times|\mu_{j}|}$ and $B_{ij}=0$ if
$\mu_{i}\not\mathbin{\overline{\prec}}\mu_{j}$.

The permutation $\sigma\in{\rm Aut}(\Gamma)$ is not necessarily unique, but
the induced permutation $p(\sigma)\in{\rm Aut}(\overline{\Gamma})$ is unique.
Moreover, the assignment $\varphi\mapsto p(\sigma)$ defines a group
homomorphism from ${\rm Aut}(G)$ to ${\rm Aut}(\overline{\Gamma})$.

###### Proof.

Fix $\varphi\in{\rm Aut}(G)$ and denote with $\overline{\varphi}\in{\rm
Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))$ the induced graded automorphism. By
Theorem 5.1 there exist $A_{i}\in{\rm GL}_{|\lambda_{i}|}({\wiskunde Q})$,
$A_{ij}\in{\wiskunde Q}^{|\lambda_{i}|\times|\lambda_{j}|}$ and $\sigma\in{\rm
Aut}(\Gamma)$ (with $p(\sigma)\in{\rm Aut}(\overline{\Gamma})$ unique) such
that the matrix of $\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V)}$ with
respect to $(V,<)$ equals

$P(\sigma)\cdot\underbrace{\begin{pmatrix}A_{1}&A_{12}&\ldots&A_{1r}\\\
0&A_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&A_{r-1r}\\\
0&\ldots&0&A_{r}\end{pmatrix}}_{=:U}$

where $A_{ij}=0$ if $\lambda_{i}\not\mathbin{\overline{\prec}}\lambda_{j}$.
Since $\overline{\varphi}$ is induced by the automorphism $\varphi\in{\rm
Aut}(G)$, the matrix $P(\sigma)U$ is by construction equal to the matrix of
the induced automorphism on $G/\sqrt{\gamma_{2}(G)}$ with respect to the
${\wiskunde Z}$-basis $(V/\sqrt{\gamma_{2}(G_{\Gamma(k)})},<)$ (where the
order is induced by $<$). In particular, $P(\sigma)U\in{\rm
GL}_{|V|}({\wiskunde Z})$. Since $P(\sigma)\in{\rm GL}_{|V|}({\wiskunde Z})$,
this implies that $U\in{\rm GL}_{|V|}({\wiskunde Z})$. Hence, we obtain that
$A_{i}\in{\rm GL}_{|\lambda_{i}|}({\wiskunde Z})$ and $A_{ij}\in{\wiskunde
Z}^{|\lambda_{i}|\times|\lambda_{j}|}$.
Now we derive the matrix of the induced automorphism on $\gamma_{2}(G)$ with
respect to the ordered ${\wiskunde Z}$-basis $(E^{k},<)$. Fix any edge
$e=\\{v,w\\}\in E$ (with $v<w$) and note that the Lie bracket $[v,w]=e^{k(e)}$
is sent by the Lie algebra morphism $P(\sigma)$ to

$[\sigma(v),\sigma(w)]=\sigma_{E}(e)^{\varepsilon_{\sigma}(e)k(\sigma_{E}(e))}.$

Thus the matrix (with respect to $(E^{k},<)$) of the morphism on
$\gamma_{2}(L^{\wiskunde Q}(\Gamma,2))$ that is induced by $P(\sigma)$ equals

$P(\sigma_{E})D(\varepsilon_{\sigma}).$

So it suffices to look at the morphism on $\gamma_{2}(L^{\wiskunde
Q}(\Gamma,2))$ induced by $U$. Fix any edge $e=\\{v,w\\}$ with $v<w$. Note
that the only non-zero matrices in the column of $U$ corresponding with $[v]$
are those matrices in the row corresponding with $[v^{\prime}]$ where
$[v^{\prime}]\mathbin{\overline{\prec}}[v]$ (or equivalently $v^{\prime}\prec
v$). Hence, there are $a_{v^{\prime}},b_{w^{\prime}}\in{\wiskunde Z}$ (for all
$v^{\prime}\prec v$ and $w^{\prime}\prec w$) such that

$Uv=\sum_{\begin{subarray}{c}v^{\prime}\in V\\\ v^{\prime}\prec
v\end{subarray}}a_{v^{\prime}}v^{\prime}\quad\text{and}\quad
Uw=\sum_{\begin{subarray}{c}w^{\prime}\in V\\\ w^{\prime}\prec
w\end{subarray}}b_{w^{\prime}}w^{\prime}.$

Thus we obtain that

$\displaystyle Ue^{k(e)}$
$\displaystyle=U[v,w]=\left[\sum_{\begin{subarray}{c}v^{\prime}\in V\\\
v^{\prime}\prec
v\end{subarray}}a_{v^{\prime}}v^{\prime},\sum_{\begin{subarray}{c}w^{\prime}\in
V\\\ w^{\prime}\prec
w\end{subarray}}b_{w^{\prime}}w^{\prime}\right]=\sum_{\begin{subarray}{c}v^{\prime}\in
V\\\ v^{\prime}\prec v\end{subarray}}\sum_{\begin{subarray}{c}w^{\prime}\in
V\\\ w^{\prime}\prec
w\end{subarray}}a_{v^{\prime}}b_{w^{\prime}}[v^{\prime},w^{\prime}]=\sum_{\begin{subarray}{c}\\{v^{\prime},w^{\prime}\\}\in
E\\\ \\{v^{\prime},w^{\prime}\\}\prec
e\end{subarray}}a_{v^{\prime}}b_{w^{\prime}}[v^{\prime},w^{\prime}]$
$\displaystyle\in\left\\{\sum_{\begin{subarray}{c}e^{\prime}\in E\\\
e^{\prime}\prec
e\end{subarray}}a_{e^{\prime}}(e^{\prime})^{k(e^{\prime})}\>\middle|\>a_{e^{\prime}}\in{\wiskunde
Z}\right\\}.$

Since $\mu_{i}\not\mathbin{\overline{\prec}}\mu_{j}$ if $i>j$, it follows that
the matrix (with respect to $(E,<)$) of the morphism on
$\gamma_{2}(L^{\wiskunde Q}(\Gamma,2))$ that is induced by $U$ is a block
upper triangular matrix

$\begin{pmatrix}B_{1}&B_{12}&\ldots&B_{1s}\\\ 0&B_{2}&\ddots&\vdots\\\
\vdots&\ddots&\ddots&B_{s-1s}\\\ 0&\ldots&0&B_{s}\end{pmatrix}$

where $B_{ij}\in{\wiskunde Z}^{|\mu_{i}|\times|\mu_{j}|}$ and $B_{ij}=0$ if
$\mu_{i}\not\mathbin{\overline{\prec}}\mu_{j}$. Hence, we obtain that the
matrix of $\overline{\varphi}|_{\gamma_{2}(L^{\wiskunde Q}(\Gamma,2))}$ has
the form

$P(\sigma_{E})\cdot
D(\varepsilon_{\sigma})\cdot\begin{pmatrix}B_{1}&B_{12}&\ldots&B_{1s}\\\
0&B_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&B_{s-1s}\\\
0&\ldots&0&B_{s}\end{pmatrix}.$

Since this matrix coincides, by construction, with the matrix of
$\varphi|_{\gamma_{2}(G)}\in{\rm Aut}(\gamma_{2}(G))$ it follows similarly as
before that $B_{i}\in{\rm GL}_{|\mu_{i}|}({\wiskunde Z})$.

By Theorem 5.1, it holds that $p(\sigma)\in{\rm Aut}(\overline{\Gamma})$ is
unique. The assignment $\varphi\to\overline{\varphi}$ from ${\rm Aut}(G)$ to
${\rm Aut}(L(\Gamma,2))$ is a group morphism since the association of a Lie
algebra to a group from section 2.1 is functorial. Using the last sentence of
Theorem 5.1, it follows that $\varphi\mapsto p(\sigma)$ defines a group
homomorphism from $G$ to ${\rm Aut}(\overline{\Gamma})$. ∎

As stated at the end of the lemma above, any automorphism $\varphi\in{\rm
Aut}(G_{\Gamma(k)})$ induces a unique permutation $p(\sigma)\in{\rm
Aut}(\overline{\Gamma})$. In general, one does not obtain every permutation of
${\rm Aut}(\overline{\Gamma})$ in this way, as the weight function
$k:E\to{\wiskunde N}_{0}$ imposes conditions on $p(\sigma)$. Before describing
these conditions, we introduce some necessary invariants of integer matrices
and recall their relation with the Smith normal form.

######  Definition 5.6.

Let $A\in{\wiskunde Z}^{n\times n}$ be a matrix and $l=1,\dots,n$. Then the
$l$-th determinant divisor $d_{l}(A)$ of $A$ equals the greatest common
divisor of the determinants of the $l\times l$ minors of $A$.
In particular, if $A=\text{diag}(a_{1},\dots,a_{n})$ is a diagonal matrix,
then $d_{l}(A)$ equals the greatest common divisor of all $l$-fold products of
diagonal elements, i.e. for any $l=1,\dots,n$ it holds that

$d_{l}(\text{diag}(a_{1},\dots,a_{n}))=\gcd\left\\{\prod_{i\in
I}a_{i}\>\middle|I\subset\\{1,\ldots,n\\}\text{ with }|I|=l\right\\}.$

The determinant divisors completely determine the Smith normal form of a
matrix in ${\wiskunde Z}$. This is summarized by the following lemma. Proofs
of these facts and an introduction to the Smith normal form of a matrix can be
found in most handbooks about basic algebra, for example in [Nor12, Part 1,
section 1].

######  Lemma 5.7.

Let $A,B\in{\wiskunde Z}^{n\times n}$ be two matrices, then the following are
equivalent:

  1. (i)

$A$ and $B$ have the same Smith normal form.

  2. (ii)

there exist $P,Q\in{\rm GL}_{n}({\wiskunde Z})$ such that $A=PBQ$.

  3. (iii)

$d_{i}(A)=d_{i}(B)$ for all $i=1,\dots,n$.

######  Notation 5.8.

Let $\Gamma=(V,E)$ be a graph with weight function $k:E\to{\wiskunde N}_{0}$
on its edges and quotient graph
$\overline{\Gamma}=(\Lambda,\overline{E},\Psi)$. Recall from section 4.1 that
any edge $e=\\{v,w\\}\in E$ gives both an equivalence class $\mu=[e]$ and an
edge in the quotient graph $\\{[v],[w]\\}$ which can be identified under the
map from Equation (4.1). The edges in the equivalence class $\mu$ have weights
$k|_{\mu}$ and thus the diagonal matrix with these weights on its diagonal
equals $D(k|_{\mu})$. For any such $\mu\in M$ and $l=1,\dots,|\mu|$ we denote
with $d_{l}(\mu)$ the $l$-th determinant divisor of the matrix $D(k|_{\mu})$,
i.e.

$d_{l}(\mu):=\gcd\left\\{\prod_{e\in I}k(e)\>\middle|\>I\subseteq\mu\text{
with }|I|=l\right\\}.$

Now we are ready to formulate a connection between the possible graph
automorphisms $\sigma\in{\rm Aut}(\Gamma)$ that can occur in the matrix
representation from Lemma 5.5 and the weight function $k$ on the edges of
$\Gamma$.

######  Lemma 5.9.

Let $\Gamma=(V,E)$ be a finite undirected simple graph with weight function
$k:E\to{\wiskunde N}_{0}$ on its edges. Suppose that $\varphi\in{\rm
Aut}(G_{\Gamma(k)})$ and let $\sigma\in{\rm Aut}(\Gamma)$ be a graph
automorphism which can occur in the matrix representation of $\varphi$ given
by Lemma 5.5. Let $e\in E$ be any edge, then for any $l=1,\dots,|[e]|$ it
holds that

$d_{l}([e])=d_{l}([\sigma_{E}(e)]).$

In particular, it holds that:

$\begin{cases}\underset{e^{\prime}\in[e]}{\gcd}(k(e^{\prime}))&=\underset{e^{\prime}\in[\sigma_{E}(e)]}{\gcd}(k(e^{\prime}))\\\
\vskip
9.0pt\cr\prod\limits_{e^{\prime}\in[e]}k(e^{\prime})&=\prod\limits_{e^{\prime}\in[\sigma_{E}(e)]}k(e^{\prime}).\end{cases}$

###### Proof.

By Lemma 5.5, we know that, with respect to the ordered basis $(E^{k},<)$, the
matrix representation of $\varphi$ restricted to $\gamma_{2}(G_{\Gamma(k)})$
is of the form

$P(\sigma_{E})\cdot
D(\varepsilon_{\sigma})\cdot\underbrace{\begin{pmatrix}B_{1}&B_{12}&\ldots&B_{1s}\\\
0&B_{2}&\ddots&\vdots\\\ \vdots&\ddots&\ddots&B_{s-1s}\\\
0&\ldots&0&B_{s}\end{pmatrix}}_{=:B}$

where $B_{i}\in{\rm GL}_{|\mu_{i}|}({\wiskunde Z})$, $B_{ij}\in{\wiskunde
Z}^{|\mu_{i}|\times|\mu_{j}|}$ and $B_{ij}=0$ if
$\mu_{i}\not\mathbin{\overline{\prec}}\mu_{j}$. Now Lemma 5.3 implies that

$D(k)\cdot P(\sigma_{E})\cdot D(\varepsilon_{\sigma})\cdot B\cdot
D(k)^{-1}\in{\rm GL}_{|E|}({\wiskunde Z}).$ (5.1)

Remark that $D(k)\cdot P(\sigma_{E})=P(\sigma_{E})\cdot D(k\circ\sigma_{E})$.
Indeed, viewing these matrices as linear maps on ${\rm Sp}(E)$, it holds for
any edge $e\in E$ that

$\displaystyle D(k)P(\sigma_{E})\,e$ $\displaystyle=D(k)\,\sigma_{E}(e)$
$\displaystyle=k(\sigma_{E}(e))\,\sigma_{E}(e)$
$\displaystyle=k(\sigma_{E}(e))P(\sigma_{E})\,e$
$\displaystyle=P(\sigma_{E})k(\sigma_{E}(e))\,e$
$\displaystyle=P(\sigma_{E})D(k\circ\sigma_{E})\,e.$

Using this and the fact that $D(k\circ\sigma_{E})$ and
$D(\varepsilon_{\sigma})$ commute, we find that

$D(k)\cdot P(\sigma_{E})\cdot D(\varepsilon_{\sigma})\cdot B\cdot
D(k)^{-1}=\Big{(}P(\sigma_{E})\cdot
D(\varepsilon_{\sigma})\Big{)}\cdot\Big{(}D(k\circ\sigma_{E})\cdot B\cdot
D(k)^{-1}\Big{)},$

which, by Equation (5.1) must lie in ${\rm GL}_{|E|}({\wiskunde Z})$. Since
$P(\sigma_{E})\cdot D(\varepsilon_{\sigma})$ is clearly a matrix in ${\rm
GL}_{|E|}({\wiskunde Z})$, we find that

$C:=D(k\circ\sigma_{E})\cdot B\cdot D(k)^{-1}\in{\rm GL}_{|E|}({\wiskunde
Z}).$

Recall that $B$ is a block upper triangular matrix and thus $C$ is also block
upper triangular. Note that if a block upper triangular matrix lies in ${\rm
GL}_{n}({\wiskunde Z})$, then so does every one of its blocks on the diagonal.
Thus by considering the block on the diagonal corresponding to
$[e]=\mu=\mu_{i}$, we find that

$D((k\circ\sigma_{E})|_{\mu})\cdot B_{i}\cdot D(k|_{\mu})\in{\rm
GL}_{|\mu|}({\wiskunde Z}).$

Since $B_{i}$ also lies in ${\rm GL}_{|\mu|}({\wiskunde Z})$, Lemma 5.7
implies that $D((k\circ\sigma_{E})|_{\mu})$ and $D(k|_{\mu})$ must have the
same Smith normal form or, equivalently, that for any $l=1,\ldots,|\mu|$ we
have

$d_{l}(\sigma(\mu))=d_{l}(D((k\circ\sigma_{E})|_{\mu}))=d_{l}(D(k|_{\mu}))=d_{l}(\mu).$

This proves the main claim of the lemma. By noticing for all $\mu\in M$ that

$\begin{cases}d_{1}(\mu)&=\underset{e^{\prime}\in\mu}{\gcd}(k(e^{\prime}))\\\
\vskip 9.0pt\cr
d_{|\mu|}(\mu)&=\prod\limits_{e^{\prime}\in\mu}k(e^{\prime}),\end{cases}$

we obtain the last claim of the statement. ∎

Recall that $M=E/\sim$ is the set of equivalence classes of edges and that for
any $\sigma\in{\rm Aut}(\Gamma)$ we write $\sigma_{E}$ for the induced
permutation on the edges. We define the following subgroup of ${\rm
Aut}(\Gamma)$

${\rm Aut}(\Gamma(k)):=\Big{\\{}\sigma\in{\rm
Aut}(\Gamma)\>\Big{|}\>\forall\mu\in M,1\leq
l\leq|\mu|:d_{l}(\mu)=d_{l}(\sigma_{E}(\mu))\Big{\\}}.$ (5.2)

Then, for any field $K\subset{\wiskunde C}$, define the following subgroup of
${\rm GL}({\rm Sp}_{K}(V))$

$\mathcal{G}_{\Gamma(k)}:=P({\rm Aut}(\Gamma(k)))\cdot\bBigg@\char
3\relax(\prod_{\lambda\in\Lambda}{\rm GL}({\rm Sp}_{K}(\lambda))\bBigg@\char
3\relax)\cdot\mathcal{U}.$

Clearly $\mathcal{G}_{\Gamma(k)}$ is a subgroup of $\mathcal{G}_{\Gamma}$ as
defined in Theorem 5.1. Moreover, we can prove the following lemma about
$\mathcal{G}_{\Gamma(k)}$, which will be useful in the following section.

######  Lemma 5.10.

Let $K$ be a subfield of ${\wiskunde C}$. The group $\mathcal{G}_{\Gamma(k)}$
is a linear algebraic group that decomposes as a semi-direct product
$\mathcal{L}\rtimes\mathcal{U}$ where

$\mathcal{L}:=P({\rm Aut}(\Gamma(k)))\cdot\bBigg@\char
3\relax(\prod_{\lambda\in\Lambda}{\rm GL}({\rm Sp}_{K}(\lambda))\bBigg@\char
3\relax)$

is a linearly reductive group and $\mathcal{U}$ is the unipotent radical of
$\mathcal{G}_{\Gamma(k)}$.

###### Proof.

To prove that $\mathcal{G}_{\Gamma(k)}$ is a linear algebraic group, it
suffices to show that it is Zariski closed as a subset of ${\rm GL}({\rm
Sp}(V))$ (with coordinates with respect to the basis of vertices $V$). This
can be easily seen to be the case since it can be written as a finite union of
Zariski closed subsets:

$\bigcup_{\sigma\in{\rm Aut}(\Gamma(k))}P(\sigma)\cdot\bBigg@\char
3\relax(\prod_{\lambda\in\Lambda}{\rm GL}({\rm Sp}_{K}(\lambda))\bBigg@\char
3\relax)\cdot\mathcal{U}.$

To see that $\mathcal{L}$ is linearly reductive, it suffices to show that its
Zariski-connected component at the identity is linearly reductive. This
Zariski connected component is equal to $\prod_{\lambda\in\Lambda}{\rm
GL}({\rm Sp}_{K}(\lambda))$. It must be linearly reductive as it is a direct
product of general linear groups, which are known to be linearly reductive.
That $\mathcal{U}$ is the unipotent radical of $\mathcal{G}_{\Gamma(k)}$
follows from the fact the $\mathcal{U}$ is the unipotent radical of
$\mathcal{G}$ (see Theorem 5.1). Indeed, $\mathcal{G}_{\Gamma(k)}$ and
$\mathcal{G}$ have the same Zariski-connected component at the identity and
thus the same unipotent radical. ∎

At last, from Lemma 5.9, we can now conclude the following corollary. The
field in this corollary is $K={\wiskunde Q}$. Recall that $\pi:{\rm
Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))\to{\rm GL}({\rm Sp}_{\wiskunde Q}(V))$ is
the restriction map.

######  Corollary 5.11.

Let $\Gamma=(V,E)$ be a finite undirected simple graph with weight function
$k:E\to{\wiskunde N}_{0}$ on its edges. For any automorphism $\varphi\in{\rm
Aut}(G_{\Gamma(k)})$ it holds that $\pi(\overline{\varphi})$ lies in
$\mathcal{G}_{\Gamma(k)}$, where $\overline{\varphi}$ denotes the induced
automorphism on $L^{\wiskunde Q}(\Gamma,2)$.

##  6 The $R_{\infty}$–property for the groups $G_{\Gamma(k)}$

For any group $G$ we define the $R_{\infty}$–nilpotency index as the least
integer $c$ such that $G/\gamma_{c+1}(G)$ has the $R_{\infty}$–property. If
such an integer does not exist, we define it to be infinite. Recall that by
construction $G_{\Gamma}=A(\Gamma)/\gamma_{3}(A(\Gamma))$. In [Wit23] the
author gives a lower and upper bound for the $R_{\infty}$–nilpotency index of
$A(\Gamma)$.

######  Theorem 6.1.

[Wit23, Theorem 1.6] If $\Gamma$ is a non-empty finite undirected simple
graph, then $A(\Gamma)$ has the $R_{\infty}$–property and its
$R_{\infty}$–nilpotency index $c$ satisfies

$\xi(\Gamma)\leq c\leq\Xi(c)$

where

$\xi(\Gamma):=\min_{\\{\lambda_{1},\lambda_{2}\\}\in\overline{E}}(|\lambda_{1}|+|\lambda_{2}|)\quad\text{and}$
$\Xi(\Gamma):=\min_{\\{\lambda_{1},\lambda_{2}\\}\in\overline{E}}c(\lambda_{1},\lambda_{2})\quad\text{with}\quad
c(\lambda_{1},\lambda_{2}):=\begin{cases}\max\left\\{2|\lambda_{1}|+|\lambda_{2}|,|\lambda_{1}|+2|\lambda_{2}|\right\\}&\text{if
}\lambda_{1}\neq\lambda_{2}\\\ 2|\lambda_{1}|&\text{if
}\lambda_{1}=\lambda_{2}\end{cases}.$

######  Theorem 6.2.

Let $\Gamma=(V,E)$ be a finite undirected simple graph and denote with
$\Gamma_{0}=(V_{0},E_{0})$ the subgraph of $\Gamma$ containing the vertices
belonging to a coherent component of size one. Then the following holds.

  1. (i)

If $E_{0}=\emptyset$, then $G_{\Gamma}$ does not have the
$R_{\infty}$–property.

  2. (ii)

If $\Gamma_{0}$ has edges and $m\in{\wiskunde N}\setminus\\{0,1\\}$, then
there exists a weight function $k:E\to{\wiskunde N}_{0}$ such that
$G_{\Gamma(k)}$ has the $R_{\infty}$–property and
$|G_{\Gamma(k)}:G_{\Gamma}|=m$.

  3. (iii)

If $\Gamma_{0}$ has edges and $\Gamma=\Gamma_{0}$, then $G_{\Gamma}$ has the
$R_{\infty}$–property. In particular, all finitely generated torsion-free
2-step nilpotent groups commensurable with $G_{\Gamma}$ have the
$R_{\infty}$–property.

###### Proof.

  1. (i)

If $V_{0}=\emptyset$, then all coherent components contain at least two
vertices. Hence, it follows that $\xi(\Gamma)\geq 4$. By Theorem 6.1 we get
that the $R_{\infty}$–nilpotency index of $A(\Gamma)$ is at least $4$. Thus
$G_{\Gamma}=A(\Gamma)/\gamma_{2}(A(\Gamma))$ does not have the
$R_{\infty}$–property.

Let us assume that $V_{0}\neq\emptyset$. By Lemma 6.2 in [Wit23] it follows
that there exist monic polynomials $p_{\lambda}(X)\in{\wiskunde Z}[X]$ (for
$\lambda\in\Lambda\setminus(V_{0}/\sim)$) of degree $|\lambda|\geq 2$ such
that, if $\alpha_{\lambda 1},\dots,\alpha_{\lambda|\lambda|}\in{\wiskunde C}$
are the zeros of $p_{\lambda}(X)$, it holds for all
$\lambda,\lambda^{\prime}\in\Lambda\setminus(V_{0}/\sim)$ (possibly equal),
$i=1,\dots,|\lambda|$ and $j=1,\dots,|\lambda^{\prime}|$ that

$\alpha_{\lambda i}\neq\pm 1,\>\alpha_{\lambda
i}\alpha_{\lambda^{\prime}j}\neq 1\quad\text{and}\quad\alpha_{\lambda
1}\dots\alpha_{\lambda|\lambda|}=-1.$ (6.1)

Denote for any $\lambda\in\Lambda$ with $A_{\lambda}\in{\rm
GL}_{|\lambda|}({\wiskunde Z})\subset{\rm GL}({\rm Sp}_{\wiskunde
Q}(\lambda))$ the linear map for which the matrix with respect to the
${\wiskunde Z}$-basis $(\lambda,<)$ is the companion matrix of
$p_{\lambda}(X)$ if $\lambda\not\in V_{0}/\sim$ or is
$\begin{pmatrix}-1\end{pmatrix}$ if $\lambda\in V_{0}/\sim$. Combining these
linear maps, leads to the linear map

$A:=\prod_{\lambda\in\Lambda}A_{\lambda}\in{\rm GL}_{|V|}({\wiskunde
Z})\subset\prod_{\lambda\in\Lambda}{\rm GL}({\rm Sp}_{\wiskunde Q}(\lambda)).$

Recall that we defined a projection map

$\pi:{\rm Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))\to\mathcal{G}_{\Gamma}:f\mapsto
f|_{{\rm Sp}_{\wiskunde Q}(V)}.$ (6.2)

Lemma 3.8 in [Wit23] argues that there exists an automorphism $\varphi\in{\rm
Aut}(G_{\Gamma})$ such that the induced automorphism
$\overline{\varphi}\in{\rm Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))$ projects onto
$\pi(\overline{\varphi})=A$. By construction, we know that
$\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V)}=\overline{\varphi}|_{{\rm
Sp}_{\wiskunde Q}(V_{0})}\times\overline{\varphi}|_{{\rm Sp}_{\wiskunde
Q}(V\setminus V_{0})}$ is diagonalizable. Hence, we can fix a basis of
eigenvectors $\mathcal{V}_{1}\subset{\rm Sp}_{\wiskunde Q}(V_{0})$ of
$\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V_{0})}$ and
$\mathcal{V}_{2}\subset{\rm Sp}_{\wiskunde Q}(V\setminus V_{0})$ of
$\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V\setminus V_{0})}$. Thus
$\mathcal{V}:=\mathcal{V}_{1}\cup\mathcal{V}_{2}$ is a basis of eigenvectors
of $\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V)}$ for ${\rm Sp}_{\wiskunde
Q}(V)$. Note that $[\mathcal{V},\mathcal{V}]$ is a generating set of
$\gamma_{2}(L^{\wiskunde Q}(\Gamma,2))$. Since $[{\rm Sp}_{\wiskunde
Q}(V_{0}),{\rm Sp}_{\wiskunde Q}(V_{0})]={\rm Sp}_{\wiskunde Q}(E_{0})$ is
trivial, it follows that

$[\mathcal{V},\mathcal{V}]=[\mathcal{V}_{1},\mathcal{V}_{2}]\cup[\mathcal{V}_{2},\mathcal{V}_{2}].$

Note that the non-zero brackets in $[\mathcal{V}_{1},\mathcal{V}_{2}]$ and
$[\mathcal{V}_{2},\mathcal{V}_{2}]$ are eigenvectors of $\overline{\varphi}$.
Moreover, the corresponding eigenvalues of the latter are products of two
(possibly equal) $\alpha_{\lambda i}$ in Equation (6.1) and thus are not equal
to 1. Since $\overline{\varphi}|_{{\rm Sp}_{\wiskunde Q}(V_{0})}=-{\rm Id}$,
the eigenvalues corresponding with the eigenvectors in
$[\mathcal{V}_{1},\mathcal{V}_{2}]$ are of the form $-\alpha_{\lambda i}$ and
are thus by Equation (6.1) also not equal to 1. The non-zero brackets in
$[\mathcal{V},\mathcal{V}]$ form a basis of eigenvectors of
$\overline{\varphi}$ for ${\rm Sp}_{\wiskunde Q}(E)=\gamma_{2}(L^{\wiskunde
Q}(\Gamma,2))$. Hence, $\mathcal{V}\cup[\mathcal{V},\mathcal{V}]$ forms a
basis of $L^{\wiskunde Q}(\Gamma,2)$ of eigenvectors of $\overline{\varphi}$
with all eigenvalues not equal to 1. By Lemma 2.1 it now follows that
$R(\varphi)<\infty$ which concludes the proof.

  2. (ii)

Fix any $m\in{\wiskunde N}\setminus\\{0,1\\}$ and any edge
$e_{0}=\\{v_{0},w_{0}\\}\in E_{0}$ (with $v_{0},w_{0}\in V_{0}$). Note that
the equivalence class $[e_{0}]=\\{\\{v_{0},w_{0}\\}\\}\subset E$ is a
singleton, by definition. Define $k:E\to{\wiskunde N}_{0}$ by setting

$k:E\to{\wiskunde N}_{0}:e\mapsto\begin{cases}m&\text{if
}e=\\{v_{0},w_{0}\\}\\\ 1&\text{if }e\neq\\{v_{0},w_{0}\\}\end{cases}.$

By Lemma 4.5 it follows that $|G_{\Gamma(k)}:G_{\Gamma}|=\prod_{e\in
E}k(e)=m$. We prove that $G_{\Gamma(k)}$ has the $R_{\infty}$–property.

First, recall the definition of ${\rm Aut}(\Gamma(k))$ in Equation (5.2) and
note that with the above choice of $k$, for any $\sigma\in{\rm
Aut}(\Gamma(k))$, it holds that $\sigma_{E}(e_{0})=e_{0}$. Next, take any
automorphism $\varphi\in{\rm Aut}(G_{\Gamma(k)})$. Recall the definition of
$\pi$ from Equation (6.2). Since $\pi$ is in fact bijective, it makes sense to
write $\pi^{-1}(g)$ for an element $g\in\mathcal{G}_{\Gamma}$.

By Corollary 5.11, we know that $\pi(\overline{\varphi})$ lies in the linear
algebraic group $\mathcal{G}_{\Gamma(k)}\subset\mathcal{G}_{\Gamma}$ (over
${\wiskunde Q}$). Thus, the same must hold for its semi-simple part
$\pi(\overline{\varphi})_{s}$. By [Hoc81, Theorem 4.3, chapter VIII] and Lemma
5.10, there must exist an $h\in\mathcal{G}_{\Gamma(k)}$ such that
$g:=h\pi(\overline{\varphi})_{s}h^{-1}$ lies in
$\mathcal{L}\subset\mathcal{G}_{\Gamma(k)}$. Note that $g$ is thus represented
by a matrix of the form

$P(\sigma)\begin{pmatrix}A_{1}&0&\ldots&0\\\ 0&A_{2}&\ddots&\vdots\\\
\vdots&\ddots&\ddots&0\\\ 0&\ldots&0&A_{r}\end{pmatrix},$

where $\sigma\in{\rm Aut}(\Gamma(k))$ (and thus $\sigma_{E}(e_{0})=e_{0}$) and
$A_{i}\in{\rm GL}_{|\lambda_{i}|}({\wiskunde Q})$ for all $i=1,\ldots,r$. Note
that from the fact that $\sigma_{E}(e_{0})=e_{0}$ and $[v_{0}],[w_{0}]$ are
singletons, the subspace ${\rm Sp}_{\wiskunde Q}\\{v_{0},w_{0}\\}$ is
preserved by $g$. Now, since $\pi(\overline{\varphi})$ is integer-like, so is
$g$ and since $g$ is given by a matrix over ${\wiskunde Q}$ of the above form,
it follows that each $A_{i}$ is integer-like. In particular, if $A_{i}$ has
dimension $1\times 1$, it must be equal to $(\pm 1)$. This implies that if we
restrict $g$ to ${\rm Sp}_{\wiskunde Q}\\{v_{0},w_{0}\\}$, we get with respect
to $v_{0},w_{0}$, the matrix representation

$\begin{pmatrix}a&0\\\
0&b\end{pmatrix}\quad\text{or}\quad\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\begin{pmatrix}a&0\\\ 0&b\end{pmatrix},$

with $a,b\in\\{-1,1\\}$. Let us prove that in either case, the automorphism
$\pi^{-1}(g)\in{\rm Aut}_{g}(L^{\wiskunde Q}(\Gamma,2))$ has an eigenvalue 1.

In the first case, if either $a$ or $b$ is equal to $1$, clearly $g$ has an
eigenvalue 1 and thus also $\pi^{-1}(g)$ has. If both $a$ and $b$ are equal to
$-1$, then the Lie bracket $[v_{0},w_{0}]$ will be an eigenvector of
$\pi^{-1}(g)$ with eigenvalue 1. Note that $[v_{0},w_{0}]$ is non-zero since
$\\{v_{0},w_{0}\\}\in E$.

In the second case, if $a=b=1$, then $v_{0}+w_{0}$ will be an eigenvector of
$g$ with eigenvalue 1. If $a=b=-1$, then $v_{0}-w_{0}$ will be an eigenvector
of $g$ with eigenvalue 1. In both cases we thus also get that $\pi^{-1}(g)$
has an eigenvalue 1. If $a$ and $b$ do not have the same sign, then the (non-
zero) Lie bracket $[v_{0},w_{0}]$ will be an eigenvector of $\pi^{-1}(g)$ with
eigenvalue 1.

We can thus conclude that $\pi^{-1}(g)$ always has an eigenvalue 1 and thus
since they have the same eigenvalues, the same holds for
$\overline{\varphi}\in L^{\wiskunde Q}(\Gamma,2)$. We can conclude by Lemma
2.1 that $R(\varphi)=\infty$ and thus, since $\varphi$ was an arbitrary
automorphism of $G_{\Gamma(k)}$, that $G_{\Gamma(k)}$ has the
$R_{\infty}$–property.

  3. (iii)

If $\Gamma_{0}=\Gamma$, then all coherent components are singletons (or
equivalently $\Gamma$ is transposition-free). In [Wit23, Theorem 1.4] it is
proven that $A(\Gamma)$ has $R_{\infty}$–nilpotency index 2. Thus $G_{\Gamma}$
has the $R_{\infty}$–property. By Proposition 3.10 the result follows.

∎

We give some examples where we can say more than the statement in Theorem 6.2.

######  Example 6.3 ($\Gamma_{0}\subsetneq\Gamma$ has edges and $G_{\Gamma}$
has the $R_{\infty}$–property).

Let us consider the next graph $\Gamma$. We also show its quotient graph
$\overline{\Gamma}$.
$\Gamma$000$\overline{\Gamma}$ In [DL23, Theorem 4.4 and Example 4.5], we
proved that $G_{\Gamma}$ has the $R_{\infty}$–property (where we used the
complement graph to represent $G_{\Gamma}$). Proposition 3.10 now tells us
that all groups commensurable with $G_{\Gamma}$ have the
$R_{\infty}$–property. In particular, for any weight function
$k:E\to{\wiskunde N}_{0}$ on the edges of $\Gamma$ it holds that
$G_{\Gamma(k)}$ has the $R_{\infty}$–property.

######  Example 6.4 ($G_{\Gamma}$ does not have the $R_{\infty}$–property, but
$G_{\Gamma(k_{n})}$ does).

We could for example consider the next graph with weight function
$k_{n}:E\to{\wiskunde N}_{0}$ (for any $n\in{\wiskunde N}\setminus\\{0,1\\}$)
on its edges. We fixed a total order of $V$, $E$ and $\Lambda=V/\sim$ as
introduced in section 4.1.

$v_{1}$

$v_{7}$

$v_{2}$

$v_{3}$

$v_{8}$

$v_{4}$

$v_{5}$

$v_{9}$

$v_{6}$

11$n$11111

$e_{1}$

$e_{2}$

$e_{7}$

$e_{3}$

$e_{4}$

$e_{8}$

$e_{5}$

$e_{6}$

(a) Graph $\Gamma=(V,E)$ with weights on its edges and with total orders for
$V$ and $E$

2

$\lambda_{1}$

1

$\lambda_{4}$

2

$\lambda_{2}$

1

$\lambda_{5}$

2

$\lambda_{3}$

1

$\lambda_{6}$

(b) Quotient graph $\overline{\Gamma}=(\Lambda,\overline{E},\Psi)$
with total order on $\Lambda$

For any $n\in{\wiskunde N}\setminus\\{0,1\\}$ we consider the group
$G_{\Gamma(k_{n})}$ as defined in section 4.2. By the proof of Theorem 6.2
(ii) we already know that $G_{\Gamma(k_{n})}$ has the $R_{\infty}$–property
for any $n\in{\wiskunde N}\setminus\\{0,1\\}$.
We now prove that $G_{\Gamma}$ does not have the $R_{\infty}$–property by
giving an explicit automorphism $\varphi\in{\rm Aut}(G_{\Gamma})$ with
$R(\varphi)<\infty$. Denote with
$\sigma:=(v_{1}\>v_{5})(v_{7}\>v_{9})(v_{2}\>v_{6})\in{\rm Aut}(\Gamma)$.
Define the matrix $A\in{\rm GL}_{2}({\wiskunde Z})$ and $B\in{\rm
GL}_{9}({\wiskunde Z})$ by

$A:=\begin{pmatrix}2&1\\\ 1&1\end{pmatrix}\quad\text{and}\quad
B:=\begin{pmatrix}0&0&A&0&0&0\\\ 0&A&0&0&0&0\\\ \mathds{1}_{2}&0&0&0&0&0\\\
0&0&0&0&0&-1\\\ 0&0&0&0&-1&0\\\
0&0&0&1&0&0\end{pmatrix}=P(\sigma)\cdot\begin{pmatrix}\mathds{1}_{2}&0&0&0&0&0\\\
0&A&0&0&0&0\\\ 0&0&A&0&0&0\\\ 0&0&0&1&0&0\\\ 0&0&0&0&-1&0\\\
0&0&0&0&0&-1\end{pmatrix}$

where the second equality for $B$ will illustrate Lemma 5.5. Define the
automorphism $\varphi\in{\rm Aut}(G_{\Gamma})$ by defining it on the
${\wiskunde Z}$-basis $(V/\gamma_{2}(G_{\Gamma}),<)$ of
$G_{\Gamma}/\gamma_{2}(G_{\Gamma})$ by the matrix $B\in{\rm GL}_{9}({\wiskunde
Z})$. One can derive that the matrix of $\varphi|_{\gamma_{2}(G_{\Gamma})}$
with respect to the ${\wiskunde Z}$-basis $(E,<)$ equals

$C:=\begin{pmatrix}0&0&-A&0&0\\\ 0&-A&0&0&0\\\ \mathds{1}_{2}&0&0&0&0\\\
0&0&0&0&-1\\\
0&0&0&1&0\end{pmatrix}=P(\sigma_{E})D(\varepsilon_{\sigma})\begin{pmatrix}\mathds{1}_{2}&0&0&0&0\\\
0&-A&0&0&0\\\ 0&0&-A&0&0\\\ 0&0&0&-1&0\\\ 0&0&0&0&1\end{pmatrix}$

where $\sigma_{E}:=(e_{1}\>e_{5})(e_{7}\>e_{8})(e_{2}\>e_{6})$ and
$\varepsilon_{\sigma}(e_{i})=\begin{cases}-1&\text{if }i=7,8\\\
1&\text{else}\end{cases}$. In particular, the eigenvalues of the induced
automorphism $\overline{\varphi}\in{\rm Aut}(L^{\wiskunde Q}(G_{\Gamma}))$ are
precisely the eigenvalues of the matrices $B$ and $C$. Denote with
$\alpha_{1}:=\frac{3-\sqrt{5}}{2}$ and $\alpha_{2}:=\frac{3+\sqrt{5}}{2}$ the
eigenvalues of $A$. One can calculate that $B$ has eigenvalues

$\alpha_{1},\alpha_{2},\pm\sqrt{\alpha_{1}},\pm\sqrt{\alpha_{2}},-1\text{ and
}\pm i$

and the eigenvalues of $C$ are

$-\alpha_{1},-\alpha_{2},\pm\sqrt{\alpha_{1}}i,\pm\sqrt{\alpha_{2}}i\text{ and
}\pm i.$

Since 1 is not an eigenvalue of $\overline{\varphi}$, Lemma 2.1 implies that
$R(\varphi)<\infty$ and thus $G_{\Gamma}$ does not have the
$R_{\infty}$–property.

######  Corollary 6.5.

There exist finitely generated torsion-free 2-step nilpotent groups which do
not have the $R_{\infty}$–property but admit for any $n\in{\wiskunde
N}\setminus\\{0,1\\}$ a subgroup of index $n$ that does have the
$R_{\infty}$–property.

## References

  * [Bau71] G. Baumslag.  Lecture Notes on Nilpotent Groups.  Amer. Math. Soc. Regional Conference Series No.2, 1971.
  * [BMO16] Gilbert Baumslag, Charles F. Miller, III, and Gretchen Ostheimer.  Decomposability of finitely generated torsion-free nilpotent groups.  Internat. J. Algebra Comput., 26(8):1529–1546, 2016.
  * [Der19] Jonas Deré.  The structure of underlying Lie algebras.  Linear Algebra Appl., 581:471–495, 2019.
  * [DG14] Karel Dekimpe and Daciberg Goncalves.  The $R_{\infty}$-property for free groups, free nilpotent groups and free solvable groups.  To appear in Bull. London Math. Soc., 2014.
  * [DK92] G. Duchamp and D. Krob.  The lower central series of the free partially commutative group.  Semigroup Forum, 45(3):385–394, 1992.
  * [DL] Karel Dekimpe and Maarten Lathouwers.  The $R_{\infty}$-property for finitely generated torsion-free 2-step nilpotent groups of small hirsch length.  In preparation 2023.
  * [DL23] Karel Dekimpe and Maarten Lathouwers.  The Reidemeister spectrum of 2-step nilpotent groups determined by graphs.  Comm. Algebra, 51(6):2384–2407, 2023.
  * [DM21] Jonas Deré and Meera Mainkar.  Anosov diffeomorphisms on infra-nilmanifolds associated to graphs.  Math. Nachr., 2021.
  * [GW09] Daciberg Gonçalves and Peter Wong.  Twisted conjugacy classes in nilpotent groups.  J. Reine Angew. Math., 633:11–27, 2009.
  * [Hoc81] Gerhard P. Hochschild.  Basic theory of algebraic groups and Lie algebras, volume 75 of Graduate Texts in Mathematics.  Springer-Verlag, New York-Berlin, 1981.
  * [Nor12] Christopher. Norman.  Finitely Generated Abelian Groups and Similarity of Matrices over a Field.  Springer Undergraduate Mathematics Series. Springer London, London, 1st ed. 2012. edition, 2012.
  * [Qui68] Daniel G. Quillen.  On the associated graded ring of a group ring.  J. Algebra, 10:411–418, 1968.
  * [Rom11] V. Roman’kov.  Twisted conjugacy classes in nilpotent groups.  J. Pure Appl. Algebra, 215(4):664–671, 2011.
  * [Sen21] Pieter Senden.  Twisted conjugacy in direct products of groups.  Comm. Algebra, 49(12):5402–5422, 2021.
  * [Sen22] Pieter Senden.  The reidemeister spectrum of direct products of nilpotent groups.  Preprint, https://arxiv.org/abs/2206.13853, 2022.
  * [Wit23] Thomas Witdouck.  The $R_{\infty}$-property for right-angled artin groups and their nilpotent quotients, 2023.