# AUT-INVARIANT QUASIMORPHISMS ON FREE PRODUCTS

BASTIEN KARLHOFER

###### Abstract

Let $G=A\ast B$ be a free product of freely indecomposable groups. We
explicitly construct quasimorphisms on $G$ which are invariant with respect to
all automorphisms of $G$. We also prove that the space of such quasimorphisms
is infinite-dimensional whenever $G$ is not the infinite dihedral group. As an
application we prove that an invariant analogue of stable commutator length
recently introduced by Kawasaki and Kimura is non-trivial for these groups.

##  1 Introduction

The study of quasimorphisms on a given group $G$ is an important branch of
geometric group theory [Cal09] with quasimorphisms sharing deep relationships
with the underlying structure of the group $G$. For free groups $F_{n}$ the so
called counting quasimorphisms originating from the work of Brooks in [Bro81]
yield a wide variety of examples. His ideas have been developed further by
Calegari and Fujiwara who constructed unbounded quasimorphisms on non-
elementary hyperbolic groups [CaFu10]. For diffeomorphism groups of surfaces
many important constructions are given in [GaGh04]. There are also numerous
applications in symplectic geometry originating from work of Entov and
Polterovich [EnPo03]. Another fundamental paper on the geometry of
quasimorphisms and central extensions is [BaGh92].

In this paper we construct unbounded Aut-invariant quasimorphisms on free
products of groups. To achieve this we associate tuples of natural numbers we
call codes to each element in a free product $G=A\ast B$. Inspired by Brooks
counting quasimorphisms on free groups we then count these codes rather than
actual elements of the free product and verify that this indeed yields
quasimorphisms on $G$. We call them code quasimorphisms. We make use of an
explicit description of the automorphism group of a free product found in
[Gil87] to see in Proposition 4.11 that our code quasimorphisms are unbounded
and invariant with respect to all automorphisms of $G$ if $A$ and $B$ are not
infinite cyclic.

If one of the factors of $G=A\ast B$ is infinite cyclic, our code quasi-
morphisms are not necessarily invariant under a specific class of
automorphisms of $G$ which is called the class of transvections. So, we
slightly adjust the way we count codes for infinite cyclic factors and call
the resulting maps weighted code quasimorphisms. We show in Proposition 5.7
that these are unbounded and invariant with respect to all automorphisms of
$G$.

These two propositions together with an independent result for the free group
on two generators from [BrMa19, Theorem 2] comprise the following result in
Section 6 which is the main result of the paper.

######  Theorem 1.

Let $G=A\ast B$ be the free product of two non-trivial freely indecomposable
groups $A$ and $B$. Assume $G$ is not the infinite dihedral group. Then $G$
admits infinitely many linearly independent homogeneous Aut-invariant
quasimorphisms, all of which vanish on single letters.

The infinite dihedral group does not admit any unbounded quasimorphism since
all its elements are conjugate to their inverses. As a corollary of our
construction we immediately deduce the existence of stably unbounded Aut-
invariant norms on free products of two factors.

######  Corollary 1.1.

Let $G=A\ast B$ be the free product of two non-trivial freely indecomposable
groups and assume $G$ is not the infinite dihedral group. Then there exists a
stably unbounded Aut-invariant norm on $G$.

We denote the Aut-invariant stable commutator length, which was recently
introduced by Kawasaki and Kimura in [KaKi20], by
$\operatorname{scl}_{\operatorname{Aut}}$. As an application of our
construction we prove the following result in Section 7.

######  Theorem 2.

Let $G=A\ast B$ be a free product of freely indecomposable groups and assume
that $G$ is not the infinite dihedral group. Then there always exist elements
$g\in G$ with positive Aut-invariant stable commutator length
$\operatorname{scl}_{\operatorname{Aut}}(g)>0$.

##  2 Preliminaries

######  Definition 2.1.

For any group $G$ we denote by $\operatorname{Aut}(G)$ the group of all
automorphisms of $G$. Moreover, we denote the normal subgroup of inner
automorphisms by $\operatorname{Inn}(G)$ and define the group of outer
automorphisms of $G$ to be the quotient
$\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$.

######  Definition 2.2.

Let $G$ be a group. A map $\psi\colon G\rightarrow\mathbb{R}$ is called a
quasimorphism if there exists a constant $D\geq 0$ such that

$\displaystyle|\psi(g)+\psi(h)-\psi(gh)|\leq D\hskip 2.84526pt\text{for
all}\hskip 2.84526ptg,h\in G.$

The smallest number $D(\psi)$ with the above property is called the defect of
$\psi$. We call a quasimorphism homogeneous if it satisfies
$\psi(g^{n})=n\psi(g)$ for all $g\in G$ and all $n\in\mathbb{Z}$. If
$\psi(\varphi(g))=\psi(g)$ for all $g\in G$,
$\varphi\in\operatorname{Aut}(G)$, then $\psi$ is called Aut-invariant.

######  Definition 2.3.

Let $\psi\colon G\to\mathbb{R}$ be a quasimorphism. Then the homogenisation
$\bar{\psi}\colon G\to\mathbb{R}$ of $\psi$ is defined by
$\bar{\psi}(g)=\lim_{n\in\mathbb{N}}\frac{\psi(g^{n})}{n}$ for all $g\in G$.

######  Lemma 2.4 ([Cal09, p.18]).

The homogenisation $\bar{\psi}$ of a quasimorphism $\psi\colon G\to\mathbb{R}$
is a homogeneous quasimorphism. Moreover, it satisfies
$|\bar{\psi}(g)-\psi(g)|\leq D(\psi)$ for any $g\in G$.

######  Definition 2.5.

A function $\nu\colon G\to\mathbb{R}$ satisfying for all $g,h\in G$:

  * •

$\nu(g)\geq 0$,

  * •

$\nu(g)=0$ if and only if $g=1$,

  * •

$\nu(gh)\leq\nu(g)+\nu(h)$,

is called a norm on $G$. If in addition for all $g\in G$ and
$\varphi\in\operatorname{Aut}(G)$ it satisfies

  * •

$\nu(\varphi(g))=\nu(g)$,

then it is called Aut-invariant. The supremum $\nu(G)=\sup\\{\nu(g)\mid g\in
G\\}$ is called the diameter of the norm $\nu$. If $\nu(G)=\infty$, then $\nu$
is called unbounded. If there exists $g\in G$ such that
$\lim_{n\to\infty}\nu(g^{n})=\infty$, then $\nu$ is called stably unbounded.

######  Example 2.6.

Let $G$ be a group together with a generating set $S$. The word norm generated
by $S$ is the norm on $G$ defined by

$\displaystyle\nu_{S}(g)=\min\\{n\mid g=s_{1}\cdots s_{n}\hskip
2.84526pt\text{where}\hskip 2.84526ptn\in\mathbb{N}\hskip
2.84526pt\text{and}\hskip 2.84526pts_{i}\in S\hskip 2.84526pt\text{for
all}\hskip 2.84526pti\\}.$

If we assume additionally that the set $S$ is invariant under
$\operatorname{Aut}(G)$ then $\nu_{S}$ is Aut-invariant.

######  Lemma 2.7.

Let $\psi\colon G\to R$ be an Aut-invariant quasimorphism with unbounded
image, but bounded on a generating set $S$ of $G$. Then there exists a stably
unbounded Aut-invariant norm on $G$.

###### Proof.

By Lemma 2.4 we can assume that $\psi$ is homogeneous. The word norm
$\|.\|_{\bar{S}}$ on $G$ associated to the generating set
$\bar{S}=\\{\varphi(s)\mid s\in S,\varphi\in\operatorname{Aut}(G)\\}$ is
clearly Aut-invariant. Let $K$ be a positive bound for the absolute value of
$\psi$ on $S$. Write $g\in G$ as a product
$g=\varphi_{1}(s_{1})\cdots\varphi_{n}(s_{n})$ for some $n\in\mathbb{N}$ where
$s_{i}\in S$ and $\varphi_{i}\in\operatorname{Aut}(G)$ for all $i$. The
calculation

$\displaystyle|\psi(g)|=|\psi(\varphi_{1}(s_{1})\cdots\varphi_{n}(s_{n}))|\leq|\psi(\varphi_{1}(s_{1}))|+\dots+|\psi(\varphi_{n}(s_{n}))|+(n-1)D(\psi)\leq
n(K+D(\psi))$

shows that $\|g\|_{\bar{S}}\geq\frac{|\psi(g)|}{K+D(\psi)}$ for all $g\in G$.
It follows that $\|g^{k}\|_{\bar{S}}\geq k\cdot\frac{|\psi(g)|}{K+D(\psi)}$
for all $k\in\mathbb{N}$, $g\in G$. Since $\psi$ does not vanish everywhere,
$\|.\|_{\bar{S}}$ is a stably unbounded Aut-invariant norm on $G$. ∎

The following examples illustrate that the converse of Lemma 2.7 above is not
true and finding unbounded Aut-invariant quasi-morphisms is much more
difficult than finding unbounded Aut-invariant norms.

######  Example 2.8.

Let $\Sigma_{\infty}$ be the infinite symmetric group of finitely supported
bijections of the natural numbers. The cardinality of the support defines an
Aut-invariant norm of infinite diameter on $\Sigma_{\infty}$. However, any
element $g\in\Sigma_{\infty}$ has finite order. Therefore, no Aut-invariant
norm on $\Sigma_{\infty}$ is stably unbounded and any homogeneous quasi-
morphism vanishes on all of $\Sigma_{\infty}$. Consequently, by Lemma 2.4 any
quasimorphism on $\Sigma_{\infty}$ is bounded.

######  Example 2.9.

Let $G=\mathbb{Z}^{k}$ for $k\geq 1$. Since every $g\in G$ lies in the same
$\operatorname{Aut}(G)$-orbit as $g^{-1}$, it follows that any homogeneous
Aut-invariant quasimorphism vanishes on all of $G$. So any Aut-invariant
quasimorphism is bounded on $G$. For $k=1$ the standard absolute value defines
a stably unbounded Aut-invariant norm on $G$, whereas for $k\geq 2$ any Aut-
invariant norm on $G$ has finite diameter.

######  Example 2.10.

Let $G$ be the fundamental group of the Klein bottle
$G=\mathbb{Z}\ast_{2\mathbb{Z}}\mathbb{Z}$. Let $a$ and $b$ be generators of
the two infinite cyclic factors of $G$ in its above presentation. Consider the
Aut-invariant word norm $\nu_{S}$ generated by $S=\\{\varphi(a^{\pm
1}),\varphi(b^{\pm 1})\mid\varphi\in\operatorname{Aut}(G)\\}$. To see that
this norm is unbounded on $G$ we first note that commutator subgroup of $G$ is
a characteristic subgroup and $G/[G,G]=\mathbb{Z}/2\times\mathbb{Z}$, where
$\mathbb{Z}/2$ is a characteristic subgroup again. Consequently, the
projection map $p\colon G\to\mathbb{Z}$ sending $p(a)=p(b)=1$ maps the set $S$
to the Aut-invariant set $\\{\pm 1\\}$ in $\mathbb{Z}$, which generates a
stably unbounded Aut-invariant norm on $\mathbb{Z}$. Therefore, the word norm
$\nu_{S}$ generated by $S$ on $G$ is stably unbounded as well.

However, there is no unbounded Aut-invariant quasimorphism on $G$. If $\psi$
was such a quasimorphism, it could be chosen to be homogeneous by Lemma 2.4.
Let $\varphi$ be the automorphism inverting the generators $a$ and $b$. Then
$\varphi$ inverts the center $Z(G)=2\mathbb{Z}$ as well. Hence, $\psi$
vanishes on $Z(G)$. Similarly, every element of
$S=\\{(ab)^{n},(ba)^{n},(ab)^{n}a,(ba)^{n}b\hskip 2.84526pt|\hskip
2.84526ptn\in\mathbb{N}\\}$ belongs to the same Aut-orbit that its inverse
belongs to. So $\psi$ vanishes on $S$ as well. However, every element $g\in G$
can be written as a product $=zs$ where $z\in Z(G)$ and $s\in S$. Therefore,
$\psi$ is bounded on all of $G$.

######  Definition 2.11.

Let $G=\ast_{i\in I}G_{i}$ be a free product of a family of groups
$\\{G_{i}\\}_{i\in I}$ for some indexing set $I$. For each $i$ the factor
$G_{i}$ is a subgroup of $G$ via the canonical inclusion. An element of $G$
that belongs to one of the factors is called a letter of $G$. Any product of
letters is called a word in $G$. The product of any two letters belonging to
the same factor in $G$ can be replaced by the letter that represents their
product in that factor. Moreover, any identity letters appearing in a word can
be omitted without changing the element the word represents in $G$. Recall
that any element $g\in G$ has a unique presentation as a word, where no two
consecutive letters lie in the same factor and no identity letters appear.
Such a word is called reduced.

######  Lemma 2.12.

Let $I$ be a set of cardinality at least two. Let $G_{i}$ be a non-trivial
group for all $i$ and $G=\ast_{i\in I}G_{i}$ be their free product. Let
$\theta\colon G\to\mathbb{R}$ be a map whose absolute value is bounded on all
letters of $G$ by a constant $B\geq 0$. Assume that there exists a constant
$D\geq 0$ such that

$\displaystyle|\theta(w_{1}w_{2})-\theta(w_{1})-\theta(w_{2})|\leq D$

holds for all reduced words $w_{1},w_{2}$ for which their product $w_{1}w_{2}$
is a reduced word. Then the map $f\colon G\to\mathbb{R}$ defined by
$f(w)=\theta(w)-\theta(w^{-1})$ defines a quasimorphism of defect at most
$12D+6B$, which is bounded on all letters by $2B$.

###### Proof.

Any element in $G$ can be represented by a reduced word. So let $w_{1}$,
$w_{2}$ be reduced words. The word given by their product $w_{1}w_{2}$ is
reduced if and only if the last letter from $w_{1}$ belongs to a factor
different from the one that the first letter of $w_{2}$ belongs to. Indeed,
otherwise those two letters could be multiplied in their common factor and
replaced by their product to shorten the number of letters appearing in the
expression.

In order to bring $w_{1}\cdot w_{2}$ to its reduced form we first perform all
cancellations which form a word we call $c$. After all cancellations have
taken place the final potential reduction is to possibly replace a non-trivial
product of two letters $b$ and $d$ belonging to the same factor by a non-
trivial letter $x$ representing their product in that factor. Therefore, we
have two cases.

  * •

The reduced presentations of $w_{1}$ and $w_{2}$ are given by $w_{1}=ac$,
$w_{2}=c^{-1}e$ and $ae$ is the reduced presentation for $w_{1}\cdot w_{2}$.

  * •

The reduced presentations of $w_{1}$ and $w_{2}$ are given by $w_{1}=abc$,
$w_{2}=c^{-1}de$, where $b$ and $d$ are letters belonging to the same factor.
The reduced presentation of $w_{1}\cdot w_{2}$ is given by $axe$, where $x=bd$
is the letter representing the non-trivial product of $b$ and $d$.

We calculate for the second case that

$\displaystyle|f(w_{1}w_{2})-f(w_{1})$
$\displaystyle-f(w_{2})|=|f(axe)-f(abc)-f(c^{-1}de)|$ $\displaystyle=$
$\displaystyle\left|\theta(axe)-\theta(e^{-1}x^{-1}a^{-1})-\theta(abc)+\theta(c^{-1}b^{-1}a^{-1})-\theta(c^{-1}de)+\theta(e^{-1}d^{-1}c)\right|$
$\displaystyle\leq$
$\displaystyle|\theta(a)+\theta(x)+\theta(e)-\theta(e^{-1})-\theta(x^{-1})-\theta(a^{-1})-\theta(a)-\theta(b)-\theta(c)+\theta(c^{-1})$
$\displaystyle+\theta(b^{-1})+\theta(a^{-1})-\theta(c^{-1})-\theta(d)-\theta(e)+\theta(e^{-1})+\theta(d^{-1})+\theta(c)|+12D$
$\displaystyle=$
$\displaystyle|\theta(x)-\theta(x^{-1})-\theta(b)+\theta(b^{-1})-\theta(d)+\theta(d^{-1})|+12D$
$\displaystyle\leq$ $\displaystyle 6B+12D.$

The first case follows analogously. Since $w_{1}$, $w_{2}$ were arbitrary
reduced words and every element of $G$ can be written in its reduced form, $f$
is a quasimorphism of defect at most $6B+12D$. Since $\theta$ is bounded on
all letters by $B$, so is $f$ by $2B$. ∎

##  3 Aut-invariant quasimorphisms

######  Definition 3.1.

We call a group $G$ freely indecomposable if $G$ is non-trivial and not
isomorphic to any free product of the form $G_{1}\ast G_{2}$ where $G_{1}$,
$G_{2}$ are non-trivial groups.

Any free product of non-trivial groups has trivial center and contains
elements of infinite order. So every abelian group and every finite group is
freely indecomposable.

######  Lemma 3.2.

Let $\psi\colon G\to\mathbb{R}$ be a quasimorphism. Let
$\\{\varphi_{i}\\}_{i\in I}$ be a set of representatives for the elements of
$\operatorname{Out}(G)$. If $\psi$ is invariant under $\varphi_{i}$ for all
$i$, then its homogenisation $\bar{\psi}\colon G\to\mathbb{R}$ is invariant
under all automorphisms of $G$.

###### Proof.

The homogenisation $\bar{\psi}$ is constant on conjugacy classes [Cal09,
p.19]. By definition $\bar{\psi}$ is also invariant under the collection
$\\{\varphi_{i}\\}_{i\in I}$, since $\psi$ is. The result follows since any
element $\varphi\in\operatorname{Aut}(G)$ can be written as the composition of
some $\varphi_{j}$ with a conjugation. ∎

Consider the free product $G=G_{1}\ast G_{2}$ where $G_{i}$ is freely
indecomposable for $i=1,2$. Following the exposition in [Gil87, p.116] based
on results in [FoRa40] and [FoRa41] the automorphism group
$\operatorname{Aut}(G_{1}\ast G_{2})$ is generated by the following types of
automorphisms (1.-3.) if neither $G_{1}$ nor $G_{2}$ is infinite cyclic:

  1. 1.

Elements from $\operatorname{Aut}(G_{1})$ and $\operatorname{Aut}(G_{2})$ give
rise to automorphisms of $G_{1}\ast G_{2}$. These are called factor
automorphisms.

  2. 2.

Let $g\in G_{i}$ for some $i\in\\{1,2\\}$. Define the map $p_{g}\colon G\to G$
to be conjugation by $g$ on the letters of $G_{j}$ for $j\neq i$ and to be the
identity on all letters from the group $G_{i}$. This definition gives rise to
an automorphism of $G$ which is called a partial conjugation.

  3. 3.

If $G_{1}\cong G_{2}$ are isomorphic, interchanging the two factors is an
automorphism of $G$. Such an automorphism is called a swap automorphism.

If $G_{1}\cong\mathbb{Z}$ is infinite cyclic and the freely indecomposable
group $G_{2}$ is not, then $\operatorname{Aut}(G_{1}\ast G_{2})$ is generated
by the above automorphisms together with the following additional type of
automorphisms:

  1. 4.

Let $s$ be a generator of $G_{1}$ and let $a\in G_{2}$ be any element. Then a
transvection is the unique automorphism of $G_{1}\ast G_{2}$ defined to be the
identity on all letters from $G_{2}$ and maps $s\to as$ or $s\to sa$.

Following the above description of the group of automorphisms of a free
product of two factors we obtain:

######  Lemma 3.3.

Let $G_{1}$, $G_{2}$ be freely indecomposable groups such that $G_{2}$ is not
infinite cyclic. Then the outer automorphism group of their free product
$\operatorname{Out}(G_{1}\ast G_{2})$ is generated by the images of
$\operatorname{Aut}(G_{1})$, $\operatorname{Aut}(G_{2})$ in
$\operatorname{Out}(G_{1}\ast G_{2})$ together with a swap automorphism if
$G_{1}\cong G_{2}$ and the transvections if $G_{1}\cong\mathbb{Z}$.

###### Proof.

By the universal property of the free product of two groups any automorphism
is uniquely determined by its image on single letters. Let $h\in G_{1}$ and
denote conjugation by $h^{-1}$ on all of $G$ by $c_{h}$. Then

$\displaystyle(c_{h}\circ p_{h})(g)=\begin{cases}h^{-1}gh&\text{if }\hskip
2.84526ptg\in G_{1},\\\ g&\text{if }\hskip 2.84526ptg\in G_{2}.\end{cases}$

Thus, $p_{h}$ and the factor automorphism given by conjugation by $h^{-1}$ on
$G_{1}$ represent the same element in $\operatorname{Out}(G)$. Similarly, in
$\operatorname{Out}(G)$ partial conjugations on $G_{1}$ by elements from
$G_{2}$ represent the same elements that factor automorphisms from $G_{2}$ do.
Finally, any two choices of swap automorphism differ by a product of factor
automorphisms. ∎

######  Lemma 3.4.

Let $G$ be a group and $H\leq G$ be a characteristic subgroup with quotient
projection $p\colon G\rightarrow G/H$. Then for any unbounded Aut-invariant
quasimorphism $\psi\colon G/H\to\mathbb{R}$ the composition $\psi\circ p\colon
G\to\mathbb{R}$ is an unbounded Aut-invariant quasimorphism on $G$. Moreover,
linearly independent quasimorphisms on $G/H$ give rise to linearly independent
quasimorphisms on $G$.

###### Proof.

Clearly, $\psi\circ p$ is a quasimorphism. The Aut-invariance of $\psi\circ p$
on $G$ follows from the Aut-invariance of $\psi$ on $G/H$ together with the
fact that $H$ is characteristic. Finally, the statement about linear
independence follows from the surjectivity of the projection to the quotient.
∎

##  4 Code quasimorphisms

Recall that a tuple always refers to a finite sequence and so all tuples are
naturally ordered.

######  Definition 4.1.

Let $A$ and $B$ be groups. Write a given element $g\in A\ast B$ in its reduced
form. We assign two tuples of non-zero natural number that we will call codes
as follows. Let $(a_{1},\dots,a_{k})$ be the tuple of letters from $A$
appearing in the reduced form of $g$. We call $(a_{1},\dots,a_{k})$ the
A-tuple of $g$. Then we count how often any one letter of
$(a_{1},\dots,a_{k})$ appears consecutively. This yields a tuple of positive
numbers $\operatorname{\mbox{$A$}-code}(g)=(n_{1},n_{2},\dots,n_{r})$ which we
call the A-code of $g$. Similarly, we obtain the B-tuple, which is the tuple
of letters from $B$ appearing in the reduced form of $g$, and the B-code of
$g$, denoted $\operatorname{\mbox{$B$}-code}(g)$, by counting consecutive
appearances of letters in the B-tuple.

Note that $\operatorname{\mbox{$A$}-code}(g)$ and
$\operatorname{\mbox{$B$}-code}(g)$ might have very different length for
elements $g\in A\ast B$ in general.

######  Example 4.2.

Let $G=A\ast B$ where $A=\mathbb{Z}/5$ and $B$ is any group. Let $a\in A,b\in
B$ be non-trivial elements. Consider $g=a^{2}bababa^{4}baba$. The $A$-tuple of
$g$ is $(a^{2},a,a,a^{4},a,a)$ and therefore
$\operatorname{\mbox{$A$}-code}(g)=(1,2,1,2)$. However, the $B$-tuple of $g$
is $(b,b,b,b,b)$ and so $\operatorname{\mbox{$B$}-code}(g)=(5)$.

######  Remark 4.3.

The code of any element $g\in A\ast B$ is clearly invariant under all factor
automorphisms.

The following lemma is immediate.

######  Lemma 4.4.

The $\operatorname{\mbox{$A$}-code}$ and $\operatorname{\mbox{$B$}-code}$ of
$g^{-1}$ are the reversed A- and $\operatorname{\mbox{$B$}-code}$ of $g$ for
any $g\in A\ast B$. That is, let
$\operatorname{\mbox{$A$}-code}(g)=(n_{1},\dots,n_{k})$ and
$\operatorname{\mbox{$B$}-code}(g)=(m_{1},\dots,m_{\ell})$, then
$\operatorname{\mbox{$A$}-code}(g^{-1})=(n_{k},\dots,n_{1})$ and
$\operatorname{\mbox{$B$}-code}(g^{-1})=(m_{\ell},\dots,m_{1})$. ∎

In the spirit of Brooks counting quasimorphisms we will now define code
quasimorphisms, which are counting the occurrences of a string of natural
numbers in the $\operatorname{\mbox{$A$}-code}$ and
$\operatorname{\mbox{$B$}-code}$ associated to an element in the free product
$A\ast B$.

######  Definition 4.5 (Code quasimorphisms).

Let $k\geq 1$ and let $z=(n_{1},\dots,n_{k})$ be a tuple of non-zero natural
numbers $n_{1},\dots,n_{k}$ for some $k\in\mathbb{N}$. Let $C\in\\{A,B\\}$.
Define $\theta^{C}_{z}\colon A\ast B\to\mathbb{Z}_{\geq 0}$ to count the
maximal number of disjoint appearances of $z$ as a tuple of consecutive
numbers in the $\operatorname{\mbox{$C$}-code}$ for all $g\in A\ast B$.
Further, define the code quasimorphism

$\displaystyle f^{C}_{z}\colon A\ast B\to\mathbb{Z}\hskip
14.22636pt\text{by}\hskip
14.22636ptf^{C}_{z}(g)=\theta^{C}_{z}(g)-\theta^{C}_{z}(g^{-1})$

for all $g\in A\ast B$. Note that
$\theta^{C}_{z}(g^{-1})=\theta^{C}_{\bar{z}}(g)$ due to Lemma 4.4, where
$\bar{z}$ denotes the reversed tuple $(n_{k},\dots,n_{1})$. Consequently,
$f^{C}_{z}(g)$ can also be written as
$f^{C}_{z}(g)=\theta^{C}_{z}(g)-\theta^{C}_{\bar{z}}(g)$ for all $g\in G$.

######  Example 4.6.

Let $G=\mathbb{Z}/5\ast B$ and $g=a^{2}bababa^{4}baba$ for non-trivial $a\in
A,b\in B$ as in Example 4.2. For $z=(1,2)$ we calculate $\theta^{A}_{z}(g)=2$
and $\theta^{A}_{z}(g^{-1})=\theta^{A}_{\bar{z}}(g)=1$ and so
$f^{A}_{z}(g)=2-1=1$.

######  Example 4.7.

Let $G=\mathbb{Z}/5\ast B$ and $g=a^{4}bababa^{3}bababa^{3}$ for non-trivial
$a\in A,b\in B$. In this case $\operatorname{\mbox{$A$}-code}(g)=(1,2,1,2,1)$.
Then $\theta^{A}_{z}(g)=1$ for $z=(1,2,1)$ since we only count disjoint
occurrences. Similarly, $\theta^{A}_{z}(g^{-1})=\theta^{A}_{\bar{z}}(g)=1$ and
so $f^{A}_{z}(g)=0$.

######  Lemma 4.8.

Let $A$, $B$ be non-trivial groups and let $C\in\\{A,B\\}$. For a non-empty
tuple of non-zero natural numbers $z$ the map $f^{C}_{z}\colon A\ast
B\to\mathbb{Z}$ defines a quasimorphism that is bounded on letters and
invariant with respect to all factor automorphisms. Moreover,
$D(f^{C}_{z})\leq 30$.

###### Proof.

We want to apply Lemma 2.12 to deduce that $f^{C}_{z}$ is a quasimorphism.
Clearly, $|\theta^{C}_{z}(x)|\leq 1$ for all letters $x\in A\ast B$ and all
$z$. Let $w_{1},w_{2}$ be reduced words representing elements in $A\ast B$
such that their product $w_{1}w_{2}$ is reduced. That is, the last letter of
$w_{1}$ and the first letter of $w_{2}$ belong to different factors. Without
loss of generality we can assume $C=A$. Let
$\operatorname{\mbox{$A$}-code}(w_{1})=(n_{1},\dots,n_{k})$ and
$\operatorname{\mbox{$A$}-code}(w_{2})=(m_{1},\dots,m_{\ell})$. Let $x$ be the
last letter from $A$ in $w_{1}$ and let $y$ be the first letter from $A$ in
$w_{2}$. Then

$\displaystyle\operatorname{\mbox{$A$}-code}(w_{1}w_{2})=\begin{cases}(n_{1},\dots,n_{k},m_{1},\dots,m_{\ell})&\text{if}\hskip
2.84526ptx\neq y,\\\ (n_{1},\dots
n_{k-1},n_{k}+m_{1},m_{2},\dots,m_{\ell})&\text{if}\hskip
2.84526ptx=y.\end{cases}$

If $x\neq y$, then
$\theta^{C}_{z}(w_{1}w_{2})\in\\{\theta^{C}_{z}(w_{1})+\theta^{C}_{z}(w_{2}),\theta^{C}_{z}(w_{1})+\theta^{C}_{z}(w_{2})+1\\}$
since at most one of the disjoint occurrences of $z$ can involve numbers that
do not lie completely in the $\operatorname{\mbox{$A$}-code}$ of either
$w_{1}$ or $w_{2}$.

If $x=y$, then
$\theta^{C}_{z}(w_{1}w_{2})\geq\theta^{C}_{z}(w_{1})+\theta^{C}_{z}(w_{2})-2$
since $n_{k}$ and $m_{1}$ can each be contained in at most one occurrences of
$z$ in the $\operatorname{\mbox{$A$}-code}$ of $w_{1}$ and $w_{2}$. Moreover,
if an occurrence of $z$ in the $\operatorname{\mbox{$A$}-code}$ of
$w_{1}w_{2}$ involves $n_{k}+m_{1}$, then all other occurrences are fully
contained in either the $\operatorname{\mbox{$A$}-code}$ of $w_{1}$ or
$w_{2}$. Thus,
$\theta^{C}_{z}(w_{1}w_{2})\leq\theta^{C}_{z}(w_{1})+\theta^{C}_{z}(w_{2})+1$.

In both cases we conclude

$\displaystyle|\theta^{C}_{z}(w_{1}w_{2})-\theta^{C}_{z}(w_{1})-\theta^{C}_{z}(w_{2})|\leq
2,$

and it follows from Lemma 2.12 that $f^{C}_{z}$ is a quasimorphism of defect
$D(f^{C}_{z})\leq 30$.

Moreover, by Remark 4.3 the maps $\theta^{C}_{z}$ are invariant under all
factor automorphisms of $A\ast B$. Consequently,
$f^{C}_{z}=\theta^{C}_{z}-\theta^{C}_{\bar{z}}$ is invariant under factor
automorphisms as well. ∎

######  Definition 4.9.

A tuple of non-zero natural numbers $z=(n_{1},\dots,n_{k})$ is called generic
if $\bar{z}$ does not appear as a tuple of $k$ adjacent numbers in
$z^{2}=(n_{1},\dots,n_{k},n_{1},\dots n_{k})$.

######  Example 4.10.

Let $z=(n_{1},\dots,n_{k})$. If $k\leq 2$, $z$ is not generic. If $k\geq 3$
and the $n_{i}$ are pairwise distinct, then $z$ is generic. E.g. for
$z=(1,2,3)$ we have $\bar{z}=(3,2,1)$ does not appear in
$z^{2}=(1,2,3,1,2,3)$.

######  Proposition 4.11.

Let $A\ast B$ be a free product of two freely indecomposable groups $A$ and
$B$, neither of which is infinite cyclic. Then for any generic tuple of
natural numbers $z$ the following holds:

  1. 1.

if $A\ncong B$ and $C\in\\{A,B\\}$ is such that $C\ncong\mathbb{Z}/2$, then
the homogenisation $\bar{f}^{C}_{z}$ of the quasimorphism $f^{C}_{z}$ is an
unbounded Aut-invariant quasimorphism on $A\ast B$;

  2. 2.

if $A\cong B\ncong\mathbb{Z}/2$, then the sum
$\bar{f}^{A}_{z}+\bar{f}^{B}_{z}$ is an unbounded Aut-invariant quasimorphism
on $A\ast B$.

In both cases the space of homogeneous Aut-invariant quasimorphisms on $A\ast
B$ that vanish on letters has infinite dimension.

###### Proof.

First, consider the case $A\ncong B$. Since $A\ast B$ is not the infinite
dihedral group, at least one of the factors is not isomorphic to
$\mathbb{Z}/2$. Without loss of generality we assume $A\ncong\mathbb{Z}/2$.
Let $z$ be generic. By Lemma 4.8 the map $f^{A}_{z}$ defines a quasimorphism
invariant under all factor automorphisms. According to Lemma 3.3 this means
that $f^{A}_{z}$ is invariant under a full set of representatives for
$\operatorname{Out}(A\ast B)$. Therefore, the homogenisation $\bar{f}^{A}_{z}$
is invariant under all automorphisms of $A\ast B$ by Lemma 3.2. It remains to
check that $\bar{f}^{A}_{z}$ is unbounded, which is equivalent to checking
that $f^{A}_{z}$ itself is unbounded by Lemma 2.4.

Since $A\ncong\mathbb{Z}/2$, it satisfies $|A|\geq 3$ and we can choose two
distinct non-trivial elements $a_{1},a_{2}\in A$. Furthermore, choose a non-
trivial element $b\in B$. Let $z=(n_{1},\dots,n_{k})$ and choose
$m\in\mathbb{N}$ to be non-zero and distinct from all $n_{i}\in\mathbb{N}$. We
set

$\displaystyle
w_{0}=(a_{1}b)^{n_{1}}(a_{2}b)^{n_{2}}(a_{1}b)^{n_{3}}(a_{2}b)^{n_{4}}\dots(a_{s}b)^{n_{k}},$

where $s=1$ if $k$ is odd and $s=2$ if $k$ is even. Set

$\displaystyle w=\begin{cases}w_{0}&\text{if }k\text{ is even},\\\
w_{0}(a_{2}b)^{m}&\text{if }k\text{ is odd}.\end{cases}$

The $\operatorname{\mbox{$A$}-code}$ of $w$ is given by

$\displaystyle\operatorname{\mbox{$A$}-code}(w)=\begin{cases}(n_{1},\dots,n_{k})=z&\text{if
}k\text{ is even},\\\ (n_{1},\dots,n_{k},m)=(z,m)&\text{if }k\text{ is
odd}.\end{cases}$

Since $w$ starts and ends with letters from different groups, the reduced
expression of $w^{\ell}$ is the $\ell$-fold product of the word $w$ for all
$\ell\in\mathbb{N}$. Moreover, because the first letter from $A$ in $w$ is
$a_{1}$ and the last letter from $A$ is $a_{2}$, the
$\operatorname{\mbox{$A$}-code}$ of $w^{\ell}$ is

$\displaystyle\operatorname{\mbox{$A$}-code}(w^{\ell})=\begin{cases}(z,z,\dots,z)&\text{if
}k\text{ is even},\\\ (z,m,z,m,\dots,z,m)&\text{if }k\text{ is
odd}.\end{cases}$

Since $m$ is distinct from all $n_{i}$, $m$ can never appear in any occurrence
of $z$ or $\bar{z}$ in the $\operatorname{\mbox{$A$}-code}$ of $w^{\ell}$. So
$\theta^{A}_{z}(w^{\ell})=\ell$, whereas $\theta^{A}_{\bar{z}}(w^{\ell})=0$
since $z$ is generic. Consequently,

$\displaystyle
f^{A}_{z}(w^{\ell})=\theta^{A}_{z}(w^{\ell})-\theta^{A}_{\bar{z}}(w^{\ell})=\ell,$

which shows that $f^{A}_{z}$ is unbounded.

Second, consider the case $A\cong B$ and fix a choice of isomorphism. Let $z$
be generic. It holds that $|A|=|B|\geq 3$ since $A\ast B$ is not the infinite
dihedral group. Consider the swap isomorphism $s$ interchanging the factors
$A$ and $B$, where we use the fixed isomorphism from before to identify $A$
and $B$ with each other. Then the application of $s$ to any element $g$
interchanges the $\operatorname{\mbox{$A$}-code}$ and
$\operatorname{\mbox{$B$}-code}$ of $g$ with each other. This implies that the
sum $\theta^{A}_{z}+\theta^{B}_{z}$ is invariant under $s$ and consequently
the sum $f^{A}_{z}+f^{B}_{z}$ is invariant under $s$ as well. Again, by Lemma
4.8 $f^{A}_{z}$ and $f^{B}_{z}$ define quasimorphisms invariant under all
factor automorphisms and so does their sum $f^{A}_{z}+f^{B}_{z}$. According to
Lemma 3.3 this means that $f^{A}_{z}+f^{B}_{z}$ is invariant under a full set
of representatives for $\operatorname{Out}(A\ast B)$. Again, by Lemma 3.2 we
see that the homogenisation $\bar{f}^{A}_{z}+\bar{f}^{B}_{z}$ is invariant
under all automorphisms of $A\ast B$. It remains to verify unboundedness.

For this let $a_{1},a_{2}\in A$ and $b_{1},b_{2}\in B$ be non-trivial such
that $a_{1}\neq a_{2}$ and $b_{1}\neq b_{2}$. Pick a non-zero number
$m\in\mathbb{N}$ distinct from all $n_{i}\in\mathbb{N}$, where
$z=(n_{1},\dots,n_{k})$. As before, we set

$\displaystyle
w_{0}=(a_{1}b_{1})^{n_{1}}(a_{2}b_{2})^{n_{2}}(a_{1}b_{1})^{n_{3}}(a_{2}b_{2})^{n_{4}}\dots(a_{s}b_{s})^{n_{k}},$

where $s$ is 1 or 2 depending on whether $k$ is odd or even. We set

$\displaystyle w=\begin{cases}w_{0}&\text{if }k\text{ is even},\\\
w_{0}(a_{2}b_{2})^{m}&\text{if }k\text{ is odd}.\end{cases}$

Then the $\operatorname{\mbox{$A$}-code}$ and $\operatorname{\mbox{$B$}-code}$
of $w$ agree and are given by

$\displaystyle\operatorname{\mbox{$A$}-code}(w)=\operatorname{\mbox{$B$}-code}(w)=\begin{cases}(n_{1},\dots,n_{k})=z&\text{if
}k\text{ is even},\\\ (n_{1},\dots,n_{k},m)=(z,m)&\text{if }k\text{ is
odd}.\end{cases}$

Since $m$ is distinct from all $n_{i}$, $m$ can never appear in any occurrence
of $z$ or $\bar{z}$ in the $\operatorname{\mbox{$A$}-code}$ and
$\operatorname{\mbox{$B$}-code}$ of $w^{\ell}$. As in the first case,
$\theta^{A}_{z}(w^{\ell})=\theta^{B}_{z}(w^{\ell})=\ell$, whereas
$\theta^{A}_{\bar{z}}(w^{\ell})=\theta^{B}_{\bar{z}}(w^{\ell})=0$ since $z$ is
generic. Consequently,

$\displaystyle
f^{A}_{z}(w^{\ell})+f^{B}_{z}(w^{\ell})=\theta^{A}_{z}(w^{\ell})+\theta^{B}_{z}(w^{\ell})-\theta^{A}_{\bar{z}}(w^{\ell})-\theta^{B}_{\bar{z}}(w^{\ell})=2\ell,$

which shows that $f^{A}_{z}+f^{B}_{z}$ is unbounded and therefore its
homogenisation is the desired unbounded Aut-invariant quasimorphism on $A\ast
B$.

Finally, let us verify that the space of homogeneous Aut-invariant
quasimorphisms on $A\ast B$ that vanish on letters is infinite-dimensional.
Let $r\in\mathbb{N}$ and let $z_{1},\dots,z_{r}$ be generic tuples. Choose
$z_{r+1}$ be a 3-tuple whose entries are distinct non-zero natural numbers and
do not appear in any of the $z_{i}$; then $z_{r+1}$ is generic. It follows
from the above construction of the word $w$ for $z_{r+1}$ in both cases that
any linear combination of the associated quasimorphisms
$f^{A}_{z_{1}}+f^{B}_{z_{1}},\dots,f^{A}_{z_{r}}+f^{B}_{z_{r}}$ vanishes on
all powers of $w$. It follows that the same holds for any linear combination
of their homogenisations
$\bar{f}^{A}_{z_{1}}+\bar{f}^{B}_{z_{1}},\dots,\bar{f}^{A}_{z_{r}}+\bar{f}^{B}_{z_{r}}$.
Thus, $\bar{f}^{A}_{z_{r+1}}+\bar{f}^{B}_{z_{r1}}$ is not contained in the
subspace spanned by the first $r$ quasimorphisms. Clearly, the homogenisation
of any code quasi-morphism vanishes on all letters of $A\ast B$. Since
$r\in\mathbb{N}$ was arbitrary, it follows that the space of homogeneous Aut-
invariant quasimorphisms on $A\ast B$ that vanish on letters cannot have
finite dimension. ∎

##  5 Weighted code quasimorphisms

If one of the factors of a free product $A\ast B$ of freely indecomposable
groups happens to be infinite cyclic, the code quasimorphisms above are in
general not Aut-invariant since they are not necessarily invariant with
respect to transvections. Thus, we need to modify our original construction to
deal with infinite cyclic factors. Afterwards we will follow steps similar to
the previous section in order to establish their Aut-invariance.

######  Lemma 5.1.

Let $B$ be a non-trivial group and let $w$ be any word in $\mathbb{Z}\ast B$
such that $w$ only contains letters of the same sign from $\mathbb{Z}$ and
starts and ends with a non-zero letter from $\mathbb{Z}$. Then its unique
reduced form $w^{\prime}$ starts and ends with a letter from $\mathbb{Z}$ with
that given sign. Moreover, the sum over all letters in $w$ belonging to the
factor $\mathbb{Z}$ remains the same in its reduced form $w^{\prime}$.

###### Proof.

Any word in the free product is brought to its reduced form by successively
eliminating trivial letters and replacing two adjacent letters from the same
factor by their product in that factor. The sum of all letters from
$\mathbb{Z}$ stays the same because any two adjacent letters of $\mathbb{Z}$
are always replaced by their sum throughout the reduction process. The only
way to encounter an elimination of the first letter $a_{1}\in\mathbb{Z}$ or
the last letter $a_{n}\in\mathbb{Z}$ during the reduction process would be by
the occurrence of $-a_{1}$ or $-a_{n}$. This is not possible since $a_{1}$ and
$a_{n}$ are non-zero and all letters have the same sign by assumption. ∎

######  Definition 5.2 (Weighted $\mathbb{Z}$-code).

Let $B$ be freely indecomposable and $B\ncong\mathbb{Z}$. Write
$g\in\mathbb{Z}\ast B$ in reduced form. Let $(a_{1},\dots,a_{k})$ be the
$\mathbb{Z}$-tuple of $g$. We define a tuple $(x_{1},\dots,x_{\ell})$ of non-
zero natural numbers as follows. Consider the successive subsequences of
maximal length in $(a_{1},\dots,a_{k})$ consisting of integers all of the same
sign. For the $i$-th such sequence, we define $x_{i}$ to be the absolute value
of the sum of integers in that sequence. We call the tuple
$(x_{1},\dots,x_{\ell})$ the weighted $\mathbb{Z}$-code of $g$.

######  Example 5.3.

Let $B$ be a non-trivial group and let $b_{i}\in B$ be non-trivial elements.
Then the reduced word

$\displaystyle w=7b_{1}(-2)b_{2}(-4)b_{3}(-1)b_{4}9b_{5}2b_{6}(-3)$

has the $\mathbb{Z}$-tuple $(7,-2.-4,-1,9,2,-3)$ which yields the weighted
$\operatorname{\mathbb{Z}-code}$ $(7,7,11,3)$.

######  Definition 5.4 (Weighted code quasimorphisms).

Let $z=(n_{1},\dots,n_{k})$ be a tuple of non-zero natural numbers. We set
$\theta^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{Z}_{\geq 0}$ to count
the number of disjoint appearances of $z$ as a tuple of consecutive numbers
inside the weighted $\operatorname{\mathbb{Z}-code}$ of $g\in\mathbb{Z}\ast
B$. Define the weighted code quasimorphism

$\displaystyle f^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{Z}\hskip
14.22636pt\text{by}\hskip
14.22636ptf^{\mathbb{Z}}_{z}(g)=\theta^{\mathbb{Z}}_{z}(g)-\theta^{\mathbb{Z}}_{z}(g^{-1})$

for all $g\in\mathbb{Z}\ast B$. Note that we again have
$\theta^{\mathbb{Z}}_{z}(g^{-1})=\theta^{\mathbb{Z}}_{\bar{z}}(g)$ for all
$g$.

######  Lemma 5.5.

Let $z$ be a non-empty tuple of non-zero natural numbers. Then the counting
function $\theta^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{Z}_{\geq 0}$
satisfies

  1. 1.

$\theta^{\mathbb{Z}}_{z}(g^{-1})=\theta^{\mathbb{Z}}_{\bar{z}}(g)$ for all
$g\in\mathbb{Z}\ast B$,

  2. 2.

$|\theta^{\mathbb{Z}}_{z}(w_{1}w_{2})-\theta^{\mathbb{Z}}_{z}(w_{1})-\theta^{\mathbb{Z}}_{z}(w_{2})|\leq
2$ for all reduced words $w_{1},w_{2}$ in $\mathbb{Z}\ast B$ for which
$w_{1}w_{2}$ is a reduced word.

Moreover, $f^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{Z}$ defined for
all $g\in\mathbb{Z}\ast B$ by
$f^{\mathbb{Z}}_{z}(g)=\theta^{\mathbb{Z}}_{z}(g)-\theta^{\mathbb{Z}}_{z}(g^{-1})$
is a quasimorphism of defect $D(f^{\mathbb{Z}}_{z})\leq 30$.

###### Proof.

First, recall that the reduced form of $g^{-1}$ is obtained by inverting the
reduced form of $g$, which amounts to reversing the order and inverting all
letters. This means to obtain the weighted $\operatorname{\mathbb{Z}-code}$ of
$g^{-1}$ one needs to reverse the one of $g$. Consequently, counting the
number of disjoint occurrences of $z$ in the weighted
$\operatorname{\mathbb{Z}-code}$ of $g^{-1}$ amounts to counting the disjoint
occurrences of the reversed tuple $\bar{z}$ in the weighted
$\operatorname{\mathbb{Z}-code}$ of $g$ itself. This proves the first part.

Second, let $w_{1},w_{2}$ be written as reduced words with $\mathbb{Z}$-tuples
given by $(n_{1},\dots,n_{k})$ for $w_{1}$ and $(m_{1},\dots,m_{\ell})$ for
$w_{2}$ for integers $n_{i},m_{j}$. Let $(x_{1},\dots,x_{k^{\prime}})$ and
$(y_{1},\dots,y_{\ell^{\prime}})$ be the weighted
$\operatorname{\mathbb{Z}-code}$s of $w_{1}$ and $w_{2}$. By assumption there
is no cancellation or reduction in the product of their reduced expressions
representing $w_{1}w_{2}$. That means the last letter of $w_{1}$ and the first
letter of $w_{2}$ belong to different factors. Then

$\displaystyle\text{weighted
}\operatorname{\mathbb{Z}-code}(w_{1}w_{2})=\begin{cases}(x_{1},\dots,x_{k^{\prime}},y_{1},\dots,y_{\ell^{\prime}})&\text{if
}\operatorname{sgn}(n_{k})\neq\operatorname{sgn}(m_{1}),\\\
(x_{1},\dots,x_{k^{\prime}-1},x_{k^{\prime}}+y_{1},y_{2},\dots,y_{\ell^{\prime}})&\text{if
}\operatorname{sgn}(n_{k})=\operatorname{sgn}(m_{1}).\end{cases}$

If $\operatorname{sgn}(n_{k})\neq\operatorname{sgn}(m_{1})$, then
$\theta^{\mathbb{Z}}_{z}(w_{1}w_{2})\in\\{\theta^{\mathbb{Z}}_{z}(w_{1})+\theta^{\mathbb{Z}}_{z}(w_{2}),\theta^{\mathbb{Z}}_{z}(w_{1})+\theta^{\mathbb{Z}}_{z}(w_{2})+1\\}$
since at most one of the disjoint occurrences of $z$ can involve numbers that
do not lie completely in the weighted $\operatorname{\mathbb{Z}-code}$ of
either $w_{1}$ or $w_{2}$.

If $\operatorname{sgn}(n_{k})=\operatorname{sgn}(m_{1})$, then
$\theta^{\mathbb{Z}}_{z}(w_{1}w_{2})\geq\theta^{\mathbb{Z}}_{z}(w_{1})+\theta^{\mathbb{Z}}_{z}(w_{2})-2$
since only one occurrence of $z$ in the weighted
$\operatorname{\mathbb{Z}-code}$ of $w_{1}$ and $w_{2}$ can involve the first
or last number respectively. Moreover, if an occurrence of $z$ in the weighted
$\operatorname{\mathbb{Z}-code}$ of $w_{1}w_{2}$ involves
$x_{k^{\prime}}+y_{1}$, then all other occurrences are fully contained in the
weighted $\operatorname{\mathbb{Z}-code}$ of either $w_{1}$ or $w_{2}$. Thus,
$\theta^{\mathbb{Z}}_{z}(w_{1}w_{2})\leq\theta^{\mathbb{Z}}_{z}(w_{1})+\theta^{\mathbb{Z}}_{z}(w_{2})+1$.

In both cases we conclude that

$\displaystyle|\theta^{\mathbb{Z}}_{z}(w_{1}w_{2})-\theta^{\mathbb{Z}}_{z}(w_{1})-\theta^{C}_{z}(w_{2})|\leq
2.$

It follows from Lemma 2.12 that $f^{\mathbb{Z}}_{z}$ is a quasimorphism of
defect at most $30$. ∎

######  Lemma 5.6.

For all non-empty tuples $z$ the weighted code quasimorphism
$f^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{Z}$ is invariant under
factor automorphisms and transvections.

###### Proof.

It is immediate from the definition that the weighted
$\operatorname{\mathbb{Z}-code}$ of any element in the free product is
invariant under factor automorphisms. Let $x$ be a generator of the infinite
cyclic factor in $\mathbb{Z}\ast B$. Any transvection is defined to be the
identity on letters from $B$ and maps $x\to xy$ or $x\to yx$ for some non-
trivial element $y\in B$. Let us consider the transvection $\varphi$ uniquely
specified by $x\to xy$ and show that the weighted
$\operatorname{\mathbb{Z}-code}$ of any element in $\mathbb{Z}\ast B$ is
invariant under $\varphi$. Then it immediately follows that
$\theta^{\mathbb{Z}}_{z}$ and $f^{\mathbb{Z}}_{z}$ are invariant under
$\varphi$. The argument for transvections of the second kind will follow
analogously to the one we present now.

Let $w\in\mathbb{Z}\ast B$ be a reduced word such that its weighted
$\operatorname{\mathbb{Z}-code}$ has length one. This means that all letters
from $\mathbb{Z}$ in the reduced expression of $w$ have the same sign and the
weighted $\operatorname{\mathbb{Z}-code}$ is given by the image of $w$ under
the factor projection $\mathbb{Z}\ast B\to\mathbb{Z}$. Note that this factor
projection is invariant with respect to $\varphi$ and so the weighted
$\operatorname{\mathbb{Z}-code}$ of $\varphi(w)$ agrees with the one of $w$.
There cannot be any cancellations of letters from $\mathbb{Z}$ occurring.

Let us do a preliminary calculation to visualise the general case more easily.
Let $k,\ell$ be non-zero natural numbers and $b\in B$ non-trivial. Then

$\displaystyle\varphi(x^{k}bx^{-\ell})$
$\displaystyle=\varphi(x)^{k}b\varphi(x)^{-\ell}=(xy)^{k}b(y^{-1}x^{-1})^{\ell}=xy\dots
xyxyby^{-1}x^{-1}y^{-1}x^{-1}\dots y^{-1}x^{-1},$
$\displaystyle\varphi(x^{-k}bx^{\ell})$
$\displaystyle=\varphi(x)^{-k}b\varphi(x)^{\ell}=(y^{-1}x^{-1})^{k}b(xy)^{\ell}=y^{-1}x^{-1}\dots
y^{-1}x^{-1}bxy\dots xy.$

This shows that the letter from $B$ separating the positive and negative
powers of $x$ either remains $b$ or is a conjugate of $b$ in $B$ after
applying $\varphi$.

Let $w\in\mathbb{Z}\ast B$ be a reduced word with weighted
$\operatorname{\mathbb{Z}-code}$ of length $k\geq 2$. In $w$ we formally
gather all consecutive occurrences of powers of $x$ of the same sign and call
these sub-words $w_{i}$ for $i=\\{1,\dots,k\\}$. That is, we write the reduced
word $w$ uniquely as a product of reduced words as

$\displaystyle w=w_{1}b_{1}w_{2}b_{2}\dots w_{k-1}b_{k-1}w_{k},$

where the $b_{i}\in B$ are non-trivial and the $w_{i}$ are of of maximal
length such that all letters from $\mathbb{Z}$ inside any $w_{i}$ have the
same sign. Moreover, in this decomposition $w_{1}$ ends with a letter from
$\mathbb{Z}$, $w_{n}$ starts with a letter from $\mathbb{Z}$ and all other
$w_{i}$ start and end with letters from $\mathbb{Z}$. By the maximality of
$w_{i}$ all letters from $\mathbb{Z}$ occurring in $w_{i}$ have different
signs from the ones occurring in $w_{i+1}$ for all $i$.

We apply $\varphi$ to $w$ and obtain an a priori not necessarily reduced word,
which we rewrite in the previous block form as

$\displaystyle\varphi(w)=\varphi(w_{1})b_{1}\varphi(w_{2})b_{2}\dots\varphi(w_{k-1})b_{k-1}\varphi(w_{k})=w_{1}^{\prime}b_{1}^{\prime}w_{2}^{\prime}b_{2}^{\prime}\dots
w_{k-1}^{\prime}b_{k-1}^{\prime}w_{k}^{\prime},$

where $b_{i}^{\prime}=yby^{-1}$ if the letters from $\mathbb{Z}$ change sign
from positive to negative at $b_{i}$ and $b_{i}^{\prime}=b_{i}$ if they change
from negative to positive. Moreover, all letters from $\mathbb{Z}$ inside any
$w_{i}^{\prime}$ have the same sign again, $w_{1}^{\prime}$ ends with a letter
from $\mathbb{Z}$, $w_{n}^{\prime}$ starts with a letter from $\mathbb{Z}$ and
all other $w_{i}^{\prime}$ start and end with letters from $\mathbb{Z}$.

We observe that when bringing $\varphi(w)$ to its reduced form there cannot be
any cancellations of the letters $b_{i}^{\prime}$. This is because by Lemma
5.1 the letters that are adjacent to $b_{i}$ will always remain letters from
$\mathbb{Z}$ after the reduction procedure of all $w_{i}^{\prime}$. Indeed,
replacing all $w_{i}^{\prime}$ by their reduced forms $w_{i}^{\prime\prime}$
we see that the product

$\displaystyle
w^{\prime\prime}=w_{1}^{\prime\prime}b_{1}^{\prime}w_{2}^{\prime\prime}b_{2}\dots
w_{k-1}^{\prime\prime}b_{k-1}^{\prime}w_{k}^{\prime\prime}$

is the reduced representative of $\varphi(w)$ since the letters adjacent to
the $b_{i}^{\prime}$ are always letters from $\mathbb{Z}$. Consequently, no
cancellations in between letters of different signs from $\mathbb{Z}$ can
occur when bringing $\varphi(w)$ to its reduced form. The reduced words
$w_{i}^{\prime\prime}$ have the same weighted $\operatorname{\mathbb{Z}-code}$
as the original $w_{i}$ for all $i$. Therefore, the weighted
$\operatorname{\mathbb{Z}-code}$ of $\varphi(w)$ agrees with the weighted
$\operatorname{\mathbb{Z}-code}$ of $w$. ∎

######  Proposition 5.7.

Let $B$ be a freely indecomposable group which is not infinite cyclic. Then
for any generic tuple of natural numbers $z$ the homogenisation
$\bar{f}^{\mathbb{Z}}_{z}\colon\mathbb{Z}\ast B\to\mathbb{R}$ of the
quasimorphism $f^{\mathbb{Z}}_{z}$ is an unbounded Aut-invariant quasimorphism
on $\mathbb{Z}\ast B$. Moreover, the space of homogeneous Aut-invariant
quasimorphisms on $\mathbb{Z}\ast B$ that vanish on letters has infinite
dimension.

###### Proof.

By Lemma 5.5 $f^{\mathbb{Z}}_{z}$ is a quasimorphism, which is invariant under
factor automorphisms and transvections according to the previous Lemma 5.6.
Images of these automorphisms generate the outer automorphism group
$\operatorname{Out}(\mathbb{Z}\ast B)$ by Lemma 3.3 . Thus,
$f^{\mathbb{Z}}_{z}$ is invariant under a full set of representatives of all
outer automorphisms and so by Lemma 3.2 the homogenisation
$\bar{f}^{\mathbb{Z}}_{z}$ is invariant under
$\operatorname{Aut}(\mathbb{Z}\ast B)$. It remains to check that it is
unbounded, which is equivalent to $f^{\mathbb{Z}}_{z}$ itself being unbounded.

Since $z$ is generic, $z=(n_{1},\dots,n_{k})$ for some $k\geq 3$ where all
$n_{i}\in\mathbb{N}$ are non-zero. Let $b\in B$ be non-trivial and $m$ a
strictly positive integer number distinct from all $n_{i}$. Set
$w\in\mathbb{Z}\ast B$ to be

$\displaystyle w=\begin{cases}n_{1}b(-n_{2})bn_{3}b(-n_{4})\dots
b(-n_{k})b&\text{if }k\text{ is even},\\\ n_{1}b(-n_{2})bn_{3}b(-n_{4})\dots
b(-n_{k})bmb&\text{if }k\text{ is odd}.\end{cases}$

The weighted $\operatorname{\mathbb{Z}-code}$ of $w$ is given by

$\displaystyle\text{weighted
}\operatorname{\mathbb{Z}-code}(w)=\begin{cases}(n_{1},\dots,n_{k})=z&\text{if
}k\text{ is even},\\\ (n_{1},\dots,n_{k},m)=(z,m)&\text{if }k\text{ is
odd}.\end{cases}$

Since $w$ starts and ends with letters belonging to different factors, the
reduced expression of $w^{\ell}$ is the $\ell$-fold product of the word $w$
for all $\ell\in\mathbb{N}$. Moreover, since the first and last letter from
$\mathbb{Z}$ in $w$ have different signs the weighted
$\operatorname{\mathbb{Z}-code}$ of $w^{\ell}$ is

$\displaystyle\text{weighted
}\operatorname{\mathbb{Z}-code}(w^{\ell})=\begin{cases}(z,z,\dots,z)&\text{if
}k\text{ is even},\\\ (z,m,z,m,\dots,z,m)&\text{if }k\text{ is
odd}.\end{cases}$

Since $m$ is distinct from all $n_{i}$, $m$ cannot appear in any occurrence of
$z$ or $\bar{z}$ inside the weighted $\operatorname{\mathbb{Z}-code}$ of
$w^{\ell}$. So $\theta^{\mathbb{Z}}_{z}(w^{\ell})=\ell$, whereas
$\theta^{A}_{\bar{z}}(w^{\ell})=0$ since $\bar{z}$ is $z$ is generic.
Consequently,

$\displaystyle
f^{\mathbb{Z}}_{z}(w^{\ell})=\theta^{\mathbb{Z}}_{z}(w^{\ell})-\theta^{\mathbb{Z}}_{\bar{z}}(w^{\ell})=\ell,$

which shows that $f^{\mathbb{Z}}_{z}$ is unbounded.

Finally, let us verify that the space of homogeneous Aut-invariant
quasimorphisms on $\mathbb{Z}\ast B$ that vanish on letters is infinite-
dimensional. Let $r\in\mathbb{N}$ and let $z_{1},\dots,z_{r}$ be generic
tuples. Choose $z_{r+1}$ be a 3-tuple whose entries are distinct non-zero
natural numbers and do not appear in any of the $z_{i}$; then $z_{r+1}$ is
generic. It follows from the above construction of the word $w$ for $z_{r+1}$
that any linear combination of
$f^{\mathbb{Z}}_{z_{1}},\dots,f^{\mathbb{Z}}_{z_{r}}$ vanishes on all powers
of this $w$. It follows that the same holds for any linear combination of
their homogenisations
$\bar{f}^{\mathbb{Z}}_{z_{1}},\dots,\bar{f}^{\mathbb{Z}}_{z_{r}}$. Thus,
$\bar{f}^{\mathbb{Z}}_{z_{r+1}}$ is not contained in the subspace spanned by
the first $r$ quasimorphisms. Clearly, the homogenisation of any weighted code
quasi-morphism vanishes on all letters of $\mathbb{Z}\ast B$. Since,
$r\in\mathbb{N}$ was arbitrary, it follows that the space of homogeneous Aut-
invariant quasimorphisms on $\mathbb{Z}\ast B$ that vanish on letters cannot
have finite dimension. ∎

##  6 Applications of code quasimorphisms

###### Proof of Theorem 1.

By [BrMa19, Theorem 2] the space of homogeneous Aut-invariant quasimorphisms
on $\mathbb{Z}\ast\mathbb{Z}$ is infinite-dimensional. Inverting both
generators of the factors defines an automorphism which inverts all letters in
$\mathbb{Z}\ast\mathbb{Z}$. So any homogeneous Aut-invariant quasimorphism on
$\mathbb{Z}\ast\mathbb{Z}$ vanishes on all letters. For all other free
products of two factors Proposition 4.11 and Proposition 5.7 imply the
existence of infinitely many linearly independent homogeneous Aut-invariant
quasimorphisms, all of which vanish on letters. ∎

###### Proof of Corollary 1.1.

Let $A\ast B$ be a free product of two freely indecomposable groups which is
not the infinite dihedral group. By Theorem 1 there exist unbounded Aut-
invariant quasimorphisms on $A\ast B$ that are bounded on all letters. Since
$A\ast B$ is generated by letters, the result follows from Lemma 2.7. ∎

######  Remark 6.1.

If neither $A$ nor $B$ is infinite cyclic, then Corollary 1.1 can also be
deduced from the result given in [Mar20, Lemma 4.4] together with the explicit
description of the automorphism group given in Section 3.

######  Corollary 6.2.

Let $G=\ast_{i\in I}G_{i}$ be a free product of finitely many freely
indecomposable groups $G_{i}$. Assume there exist free factors $G_{j}$ and
$G_{k}$ with $j\neq k$ such that no free factors $G_{i}$ for
$i\notin\\{j,k\\}$ is isomorphic to $G_{j}$ or $G_{k}$ or is infinite cyclic.
Moreover, assume that $G_{j}$, $G_{k}$ are not both equal to $\mathbb{Z}/2$.
Then any unbounded Aut-invariant quasimorphism on $G_{j}\ast G_{k}$ gives rise
to an unbounded Aut-invariant quasimorphism on $G$. In particular, the space
of homogeneous Aut-invariant quasimorphisms on $G$ is infinite-dimensional.

###### Proof.

We claim that the projection $p\colon G\rightarrow G_{j}\ast G_{k}$ is Aut-
equivariant, i.e. any automorphism of $G$ descends via $p$ to an automorphism
of $G_{j}\ast G_{k}$. This is equivalent to $\ker(p)$ being a characteristic
subgroup of $G$. Once this is established, we apply Theorem 1 to $G_{j}\ast
G_{k}$ and conclude the proof by applying Lemma 3.4.

Let us now show that any automorphism of $G$ indeed descends to $G_{j}\ast
G_{k}$. By [Gil87] $\operatorname{Aut}(G)$ is generated by factor
automorphisms, swap automorphisms, partial conjugations and transvections
since inner automorphisms can be written as products of factor automorphisms
and partial conjugations. It is clear that all factor automorphisms and all
partial conjugations of $G$ descend to automorphisms of $G_{j}\ast G_{k}$ via
$p$. By our assumption there are no swap automorphisms permuting any other
free factors in $G$ with $G_{j}$ and $G_{k}$, so these descend to the quotient
as well. It only remains to check the transvections if $G_{j}$ or $G_{k}$
happen to be infinite cyclic. So let $G_{j}$ be infinite cyclic generated by
$x$ and let $a$ be a letter from a different factor $G_{\ell}$. If $\ell=k$,
then any transvection $\varphi_{a}$ defined by $\varphi_{a}(x)=ax$ or
$\varphi_{a}(x)=xa$ descends via $p$ to the same transvection on $G_{j}\ast
G_{k}$. If $\ell\neq k$, any such transvection descends to the identity on
$G_{j}\ast G_{k}$. In particular, it always descends via $p$. Since a
generating set of $\operatorname{Aut}(G)$ descends to automorphisms of the
quotient $G_{j}\ast G_{k}$, any element of $\operatorname{Aut}(G)$ does so.
Consequently, the map $p$ is Aut-equivariant. ∎

######  Corollary 6.3.

Let $H\to G\to A\ast B$ be an extension of a free product of freely
indecomposable groups $A$ and $B$ by a group $H$. Assume that $H$ is a
characteristic subgroup of $G$ and $A\ast B$ is not the infinite dihedral
group. Then the space of homogeneous Aut-invariant quasimorphisms on $G$ is
infinite-dimensional.

###### Proof.

The space of homogeneous Aut-invariant quasimorphisms on $A\ast B$ is
infinite-dimensional by Theorem 1. Therefore, the result follows from Lemma
3.4. ∎

######  Corollary 6.4.

Let $G_{1}\ast_{H}G_{2}$ be a free product of groups $G_{1},G_{2}$ amalgamated
over a common subgroup $H$ which is proper and central in both $G_{1}$ and
$G_{2}$. If $G_{1}/H$ and $G_{2}/H$ are freely indecomposable and not both
equal to $\mathbb{Z}/2$, the space of homogeneous Aut-invariant quasimorphisms
on $G_{1}\ast_{H}G_{2}$ is infinite-dimensional.

###### Proof.

By assumption $H\neq G_{1}$ and $H\neq G_{2}$ and so $H$ equals the center of
$G_{1}\ast_{H}G_{2}$. As such it is a characteristic subgroup of
$G_{1}\ast_{H}G_{2}$. Furthermore,
$\frac{G_{1}\ast_{H}G_{2}}{H}\cong\frac{G_{1}}{H}\ast\frac{G_{2}}{H}$. By
assumption $\frac{G_{1}}{H}\ast\frac{G_{2}}{H}$ is not isomorphic to the
infinite dihedral group and the result follows from Corollary 6.3 above. ∎

######  Example 6.5.

For $q\geq 3$, the Hecke groups $H_{q}\cong\mathbb{Z}/2\ast\mathbb{Z}/q$ admit
infinitely many linearly independent homogeneous Aut-invariant quasimorphisms
by Theorem 1.

######  Example 6.6.

By Corollary 6.4 the space of homogeneous Aut-invariant quasimorphisms on
$\operatorname{SL}(2,\mathbb{Z})$ is infinite-dimensional, since
$\operatorname{SL}(2,\mathbb{Z})$ is the amalgamated product
$\mathbb{Z}/4\ast_{\mathbb{Z}/2}\mathbb{Z}/6$.

######  Example 6.7.

The braid group $B_{3}$ admits infinitely many linearly independent
homogeneous Aut-invariant quasimorphism as well by Corollary 6.3. Indeed,
$B_{3}$ is the universal central extension of
$\operatorname{PSL}(2,\mathbb{Z})=\mathbb{Z}/2\ast\mathbb{Z}/3$ by
$\mathbb{Z}$.

######  Example 6.8.

Let $G_{p,q}=\mathbb{Z}\ast_{\mathbb{Z}}\mathbb{Z}$ be the free product of two
copies of the integers amalgamated over inclusions
$\iota_{1},\iota_{2}\colon\mathbb{Z}\to\mathbb{Z}$ which are multiplication by
$p$ and $q$. For coprime choices of $p$ and $q$ these are the so called knot
groups $K_{p,q}$ arising as the fundamental group of the complement of torus
knots. Then $G_{p,q}$ admits infinitely many linearly independent homogeneous
Aut-invariant quasimorphisms if $\min\\{|p|,|q|\\}\geq 2$ and
$\max\\{|p|,|q|\\}\geq 3$. We have seen in Example 2.10 that this is no longer
true for $p=q=2$.

######  Example 6.9.

$B_{3}\ast_{\mathbb{Z}}B_{3}$, the free product of $B_{3}$ with itself
amalgamated over their common center generated by the Garside element, admits
infinitely many linearly independent unbounded Aut-invariant quasimorphisms.
To prove this we cannot apply Corollary 6.4 directly since $B_{3}/\mathbb{Z}$
is not freely indecomposable. The center of $B_{3}\ast_{\mathbb{Z}}B_{3}$ is
again generated by the Garside element of each of the factors. This fits into
the short exact sequence

$\displaystyle\mathbb{Z}\to
B_{3}\ast_{\mathbb{Z}}B_{3}\to(B_{3}/\mathbb{Z})\ast(B_{3}/\mathbb{Z}).$

Finally,
$(B_{3}/\mathbb{Z})\ast(B_{3}/\mathbb{Z})=\operatorname{PSL}(2,\mathbb{Z})\ast\operatorname{PSL}(2,\mathbb{Z})=\mathbb{Z}/2\ast\mathbb{Z}/3\ast\mathbb{Z}/2\ast\mathbb{Z}/3$.
So Corollary 6.2 applies with $G_{j}=G_{k}=\mathbb{Z}/3$. Then the statement
for $B_{3}\ast_{\mathbb{Z}}B_{3}$ follows from Lemma 3.4.

##  7 Aut-invariant stable commutator length

For any group $G$ let $\operatorname{cl}_{G}$ denote the commutator length on
$[G,G]$, which is defined to be the minimal number of commutators required to
write a given element of the commutator subgroup. Let
$\operatorname{scl}_{G}(x)=\lim_{n}\frac{\operatorname{cl}(x^{n})}{n}$ denote
the stable commutator length of $x\in[G,G]$. It shares a deep relationship
with quasimorphisms on $G$ through the so called Bavard duality [Cal09]. We
now define the Aut-invariant (stable) commutator length; this is a special
case of the $\hat{G}$-invariant (stable) commutator length defined in
[KaKi20].

######  Definition 7.1.

Let $G\leq\hat{G}$ be a normal subgroup. Consider the subgroup
$[\hat{G},G]\leq G$ generated by commutators of the form $[F,g]$ and their
inverses where $F\in\hat{G}$ and $g\in G$. Then for $x\in[\hat{G},G]$ the
$\hat{G}$-invariant commutator length $\operatorname{cl}_{\hat{G},G}(x)$ is
defined to be the minimal length of an expression of $x$ as a product of
commutators $[F,g]$ and their inverses where $F\in\hat{G}$ and $g\in G$. The
$\hat{G}$-invariant stable commutator length $\operatorname{scl}_{\hat{G},G}$
for $x\in[\hat{G},G]$ is defined by
$\operatorname{scl}_{\hat{G},G}(x)=\lim_{n}\frac{\operatorname{cl}_{\hat{G}G}(x^{n})}{n}$.

Given any group $G$, its inner automorphism group $\operatorname{Inn}(G)$ is a
normal subgroup of $\operatorname{Aut}(G)$ and so the above definition applies
to $\operatorname{Inn}(G)$. If $G$ has trivial center, $G$ can be identified
with $\operatorname{Inn}(G)$. In this case we simplify the notation by
denoting the Aut(G)-invariant commutator length simply as
$\operatorname{cl}_{\operatorname{Aut}}$ and the Aut(G)-invariant stable
commutator length simply as $\operatorname{scl}_{\operatorname{Aut}}$.

Setting $\hat{G}=\operatorname{Aut}(G)$ the following lemma is proven in
[KaKi20, Lemma 2.1].

######  Lemma 7.2.

Let $G$ be a group with trivial center so that $G=\operatorname{Inn}(G)$. Let
$\phi$ be an homogeneous Aut-invariant quasimorphism on $G$. Then any
$x\in[\operatorname{Aut}(G),G]\leq G$ satisfies

$\displaystyle\operatorname{scl}_{\operatorname{Aut}}(x)\geq\frac{1}{2}\frac{|\phi(x)|}{D(\phi)}.$

∎

In fact, according to [KaKi20, Theorem 1.3] $\hat{G}$-invariant quasimorphisms
satisfy an analogue of the Bavard duality theorem if $[\hat{G},G]=G$. All free
products $A\ast B$ of freely indecomposable groups $A$ and $B$ have trivial
center and so the notions $\operatorname{cl}_{\operatorname{Aut}}$ and
$\operatorname{scl}_{\operatorname{Aut}}$ apply. However, free products often
fail to satisfy $[\operatorname{Aut}(G),G]=G$. We will use a constructive
approach rather than relying on an invariant analogue of Bavard’s duality in
the following.

######  Example 7.3.

For $D_{\infty}=\mathbb{Z}/2\ast\mathbb{Z}/2$ it holds that
$\operatorname{scl}_{\operatorname{Aut}}\equiv 0$. To see this denote the
generators of the $\mathbb{Z}/2$ factors by $a$ and $b$ and let $s$ be the
automorphism of $D_{\infty}$ interchanging $a$ and $b$. Then we calculate
$[s,a]=sas^{-1}a^{-1}=s(a)a^{-1}=ba$ and $[s,b]=ab=(ba)^{-1}$. Since the total
number of letters appearing in any expression of the form $[f,x]$ is always
even where $f\in\operatorname{Aut}(D_{\infty})$ and $x\in D_{\infty}$, it
holds that $[\operatorname{Aut}(D_{\infty}),D_{\infty})]\cong\mathbb{Z}$
generated by $ba$. In fact, any power $(ba)^{k}$ for $k\in\mathbb{Z}$ can be
written as a single commutator $[s,w]$, where $w$ is one of the two words of
length $|k|$. Thus, $\operatorname{cl}_{\operatorname{Aut}}$ is equal to one
for any non-trivial element in $[\operatorname{Aut}(D_{\infty}),D_{\infty})]$
and $\operatorname{scl}_{\operatorname{Aut}}$ vanishes.

######  Example 7.4.

Consider $G=\operatorname{PSL}(2,\mathbb{Z})=\mathbb{Z}/3\ast\mathbb{Z}/2$.
Then $\operatorname{Aut}(G)$ is generated by the set $C$ consisting of the
non-trivial factor automorphism of $\mathbb{Z}/3$ and conjugations by letters
of $\mathbb{Z}/3$ and $\mathbb{Z}/2$, since both free factors are abelian
groups. Consequently, $[\operatorname{Aut}(G),G]$ is normally generated by
commutators of the form $[c,g]=cgc^{-1}g^{-1}=c(g)g^{-1}$ for $c\in C$ and
$g\in G$. In all expressions $[c,g]$ the letter $b$ representing the non-
trivial element of the factor $\mathbb{Z}/2$ arises an even number of times.
Therefore, $b\notin[\operatorname{Aut}(G),G]$ and the latter is not the full
group $G$.

######  Lemma 7.5.

Let $G=A\ast B$ be a free product of freely indecomposable groups where at
least one of the factors is infinite cyclic. Then $[\operatorname{Aut}(G),G]$
has index at most 2 in $G$. Therefore, any unbounded quasimorphism on $G$ is
unbounded when restricted to $[\operatorname{Aut}(G),G]$.

###### Proof.

If an unbounded quasimorphism $q$ is bounded on a finite index subgroup $H\leq
G$, its homogenisation $\bar{q}$ vanishes on $H$. Then $\bar{q}$ descends to a
map of sets on the finite set $G/H$ implying that the image of $\bar{q}$ is
bounded and so was the image of $q$ to begin with. Therefore, any
quasimorphism $q$ with unbounded image cannot be bounded on $H$. So, it
remains to show that the index of $[\operatorname{Aut}(G),G]$ in $G$ is finite
to prove the lemma.

First, consider the case where $A$ and $B$ are both infinite cyclic and so $G$
can be identified with $F_{2}$, the free group of rank 2. Let $x$ and $y$ be
standard generators. Consider the automorphism $\varphi$ of $F_{2}$ defined by
$\varphi(x)=yx$ and $\varphi(y)=y$. Then $[\varphi,x]=\varphi(x)\cdot
x^{-1}=y$. So $\langle y\rangle\leq[\operatorname{Aut}(F_{2}),F_{2}]$. By
symmetry of the generating set it holds that $\langle
x\rangle\leq[\operatorname{Aut}(F_{2}),F_{2}]$ as well and it follows that
$[\operatorname{Aut}(F_{2}),F_{2}]=F_{2}$.

Second, consider the case where only one of the factors is infinite cyclic.
Without loss of generality assume $A=\mathbb{Z}$. Let $x$ denote a generator
of $A$. For any $b\in B$ we can define the transvection $\varphi_{b}$ on $x$
by $\varphi_{b}(x)=bx$ and by $\varphi_{b}(b^{\prime})=b^{\prime}$ for all
$b^{\prime}\in B$. Then $\varphi_{b}$ is an automorphism and satisfies
$[\varphi_{b},x]=\varphi_{b}(x)\cdot x^{-1}=b$ for all $b\in B$. Thus,
$B\leq[\operatorname{Aut}(G),G]$. Moreover, denoting the non-trivial factor
automorphism of $A=\mathbb{Z}$ by $f$ we compute $[f,x]=f(x)\cdot
x^{-1}=x^{-2}$ and deduce that $2\mathbb{Z}\leq[\operatorname{Aut}(G),G]$.
Therefore, $2\mathbb{Z}\ast B\leq[\operatorname{Aut}(G),G]$. Since
$[\operatorname{Aut}(G),G]$ is a normal subgroup, it holds that
$N\leq[\operatorname{Aut}(G),G]$ where $N$ is the normal closure of
$2\mathbb{Z}\ast B$. However, $G/N\leq\mathbb{Z}/2$ and so
$[\operatorname{Aut}(G),G]$ has at most index 2 in $G$. ∎

###### Proof of Theorem 2.

If one of the factors is infinite cyclic, Theorem 1 implies the existence of
an unbounded Aut(G)-invariant homogeneous quasimorphism, which is unbounded on
$[\operatorname{Aut}(G),G]$ according to Lemma 7.5. The statement then follows
from Lemma 7.2.

Assume from now on that neither $A$ nor $B$ is infinite cyclic. We will prove
the theorem by explicitly constructing an element in
$[\operatorname{Aut}(G),G]$ together with homogeneous Aut-invariant
quasimorphism which is non-trivial on that element.

The only non-trivial group with no non-trivial automorphisms is
$\mathbb{Z}/2$. Thus, one of the factors has to have a non-trivial
automorphism since $A\ast B$ is not the infinite dihedral group. Without loss
of generality we assume that $|\operatorname{Aut}(A)|\geq 2$. Let $a_{1}\in A$
such that $f(a_{1})=a_{2}\neq a_{1}$ for some $f\in\operatorname{Aut}(A)$.
Then we compute that $[f,a_{1}]=a_{2}a_{1}^{-1}$ is a non-trivial element in
$[\operatorname{Aut}(G),G]$. Since $[\operatorname{Aut}(G),G]$ is normal, it
holds for any $h\in B$ that
$ha_{2}a_{1}^{-1}h^{-1}\in[\operatorname{Aut}(G),G]$. Thus,
$(a_{2}a_{1}^{-1}ha_{2}a_{1}^{-1}h^{-1})^{k}\in[\operatorname{Aut}(G),G]$ for
any $k\in\mathbb{N}$ and $h\in B$.

First, assume that $A$ and $B$ are not isomorphic. Since
$|\operatorname{Aut}(A)|\geq 2$, it holds that $|A|\geq 3$ and we can choose a
non-trivial $a\neq a_{2}a_{1}^{-1}$ and fix some non-trivial $h\in B$. We
define for $n_{1},\dots,n_{\ell}\in\mathbb{N}$ the word

$\displaystyle
w=\prod_{i=1}^{\ell}\big{(}[a,h](a_{2}a_{1}^{-1}ha_{2}a_{1}^{-1}h^{-1})^{n_{i}}\big{)}\in[\operatorname{Aut}(G),G].$

We calculate

$\displaystyle\operatorname{\mbox{$A$}-code}(w)=\begin{cases}(1,1,2n_{1},1,1,2n_{2},\dots,1,1,2n_{\ell})&\text{if}\hskip
2.84526pta^{-1}\neq a_{2}a_{1}^{-1},\\\
(1,2n_{1}+1,1,2n_{2}+1,\dots,1,2n_{\ell}+1)&\text{if}\hskip
2.84526pta^{-1}=a_{2}a_{1}^{-1}.\end{cases}$

Set $z=\operatorname{\mbox{$A$}-code}(w)$. For all $n\in\mathbb{N}$ it holds
that $\operatorname{\mbox{$A$}-code}(w^{n})=(z,\dots,z)$. Choose $\ell\geq 3$
together with large and distinct $n_{1},\dots,n_{\ell}\in\mathbb{N}$, which
implies that $z$ is generic. It follows from Proposition 4.11 that
$\bar{f}^{A}_{z}$ is an Aut-invariant quasimorphism. By construction
$f^{A}_{z}$ satisfies $f^{A}_{z}(w^{n})=n$ for all $n\in\mathbb{N}$ and so
$\bar{f}^{A}_{z}(w)>0$ for our choice of $w\in[\operatorname{Aut}(G),G]$. The
statement then follows from Lemma 7.2.

Second, assume $A\cong B$. Let $s$ denote a swap automorphism. Set
$b_{i}=s(a_{i})$ for $i\in\\{1,2\\}$. Then the element
$[sfs^{-1},b_{1}]=b_{2}b_{1}^{-1}\in[\operatorname{Aut}(G),G]$ is non-trivial
and belongs to the factor $B$. Set $b=s(a)$. Then $a^{-1}=a_{2}a_{1}^{-1}$ is
equivalent to $b^{-1}=b_{2}b_{1}^{-1}$. For
$n_{1},\dots,n_{\ell}\in\mathbb{N}$ we define the word

$\displaystyle
w=\prod_{i=1}^{\ell}\big{(}[a,b](a_{2}a_{1}^{-1}b_{2}b_{1}^{-1})^{n_{i}}\big{)}\in[\operatorname{Aut}(G),G].$

Observe, that
$\operatorname{\mbox{$A$}-code}(w)=\operatorname{\mbox{$B$}-code}(w)$. As in
the previous case,

$\displaystyle\operatorname{\mbox{$A$}-code}(w)=\begin{cases}(1,1,n_{1},1,1,n_{2},\dots,1,1,n_{\ell})&\text{if}\hskip
2.84526pta^{-1}\neq a_{2}a_{1}^{-1},\\\
(1,n_{1}+1,1,n_{2}+1,\dots,1,n_{\ell}+1)&\text{if}\hskip
2.84526pta^{-1}=a_{2}a_{1}^{-1}.\end{cases}$

Set $z=\operatorname{\mbox{$A$}-code}(w)$ as before and choose $\ell\geq 3$
and $n_{1},\dots,n_{\ell}\in\mathbb{N}$ large enough and distinct, so that $z$
is generic. Again, we calculate $f^{A}_{z}(w^{n})=n=f^{B}_{z}(w^{n})$, which
implies that $(\bar{f}^{A}_{z}+\bar{f}^{B}_{z})(w)>0$. Proposition 4.11
implies that $\bar{f}^{A}_{z}+\bar{f}^{B}_{z}$ is an homogeneous Aut-invariant
quasimorphism and so applying Lemma 7.2 concludes the proof. ∎

We will now give a few more examples of free products $G$ where
$[\operatorname{Aut}(G),G]=G$ and therefore the notions of Aut-invariant
(stable) commutator length are defined on all of $G$.

######  Example 7.6.

For a product $G=A\ast B$ of two freely indecomposable perfect groups $A$ and
$B$ it holds that $[\operatorname{Aut}(G),G]=G$. Indeed, since $A$ perfect, it
holds that $A=[A,A]\leq[\operatorname{Aut}(G),G]$. Similarly,
$B\leq[\operatorname{Aut}(G),G]$ and $A$ and $B$ generate $G$ it follows that
$[\operatorname{Aut}(G),G]=G$.

######  Example 7.7.

$G=\mathbb{Z}/p\ast\mathbb{Z}/q$ satisfies $[\operatorname{Aut}(G),G]=G$ for
$p,q\geq 3$ prime. Let $m\colon\mathbb{Z}/p\to\mathbb{Z}/p$ be multiplication
by 2, which is a factor automorphism. Consider the standard generator
$1_{p}\in\mathbb{Z}/p$. It holds that $[m,1_{p}]=m(1_{p})-1_{p}=1_{p}$. Thus,
$\mathbb{Z}/p\leq[\operatorname{Aut}(G),G]$. Similarly,
$\mathbb{Z}/q\leq[\operatorname{Aut}(G),G]$ and so
$[\operatorname{Aut}(G),G]=G$.

######  Example 7.8.

Let $k,\ell\geq 2$. Then $G=A^{k}\ast B^{\ell}$ satisfies
$[\operatorname{Aut}(G),G]=G$ for all non-trivial abelian groups $A$ and $B$.
For simplicity of notation consider the case $k=2$. Let $\phi$ be the factor
automorphism defined by $\phi(a,0)=(a,a)$ and $\phi(0,a)=a$ for all $a\in A$.
Then $[\phi,(a,0)]=\phi(a,0)-(a,0)=(0,a)$ for all $a\in A$. Thus, $A\times
0\leq[\operatorname{Aut}(G),G]$. Analogously, $0\times
A\leq[\operatorname{Aut}(G),G]$. Since these factors generate $A^{2}$ it
follows that $A^{2}\leq[\operatorname{Aut}(G),G]$. Similarly,
$B^{\ell}\leq[\operatorname{Aut}(G),G]$ and so $[\operatorname{Aut}(G),G]=G$.

######  Example 7.9.

In all of the above examples for $G=A\ast B$ the equality
$[\operatorname{Aut}(G),G]=G$ is always derived by showing that both factors
$A$ and $B$ form subgroups of $[\operatorname{Aut}(G),G]$. Thus, any
combination of free factors appearing in the three examples above still
satisfies this equality, for example $A=\mathbb{Z}/p$ for $p\geq 3$ prime and
$B$ any freely indecomposable perfect group.

## Acknowledgements

I like to thank Jarek Kędra and Benjamin Martin for their continued support
and all their helpful comments. This work was partly funded by the Leverhulme
Trust Research Project Grant RPG-2017-159.

## References

  * [BaGh92] J. Barge, E. Ghys, Cocycles d’Euler et de Maslov, Math. Ann. 294, 1992, 235–265.
  * [Bav91] C. Bavard. Longueur stable des commutateurs, Enseign. Math. (2) 37(1-2), 1991, 109–150.
  * [BrMa19] M. Brandenbursky, M. Marcinkowski, Aut-invariant norms and Aut-invariant quasimorphisms on free and surface group, Commentarii Mathematici Helvetici, vol.94, No. 4, 2019, 661–687.
  * [Bro81] Robert Brooks, Some remarks on bounded cohomology, Riemann surfaces and related topics: Proceedings of the 1978 Stony Brook Conference (State Univ. New York, Stony Brook, N.Y., 1978) (Princeton, N.J.), Ann. of Math. Stud., vol. 97, Princeton Univ. Press, 1981, pp.53–63.
  * [CaFu10] D. Calegari, K. Fujiwara, Stable commutator length in word-hyperbolic groups, Groups, Geometry and Dynamics 4 (no. 1), 2010, 59–90.
  * [Cal09] D. Calegari, scl. Mathematical Society of Japan Memoirs: Volume 20, 2009.
  * [EnPo03] M. Entov, L. Polterovich, Calabi quasimorphism and quantum homology, International Mathematics Research Notices, 2003.
  * [Gil87] N.D. Gilbert, Presentations of the Automorphism Group of a Free Product, Proceedings of the London Mathematical Society, vol. 3-54, Issue 1, 1987, 115–140.
  * [GaGh04] J.M. Gambaudo, E. Ghys Commutators and diffeomorphisms of surfaces, Ergodic Theory Dynam. Systems 24, no. 5, 2004, 1591–1617.
  * [FoRa40] D.I. Fouxe-Rabinovitch, Über die Automorphismengruppen der freien Produkte I, Rec. Math. [Mat. Sbornik] vol. 8, 1940, 265–276.
  * [FoRa41] D.I. Fouxe-Rabinovitch, Über die Automorphismengruppen der freien Produkte II, Rec. Math. [Mat. Sbornik] vol. 9, 1940, 183–220.
  * [KaKi20] M. Kawasaki, M. Kimura $\hat{G}$-invariant quasimorphisms and symplectic geometry of surfaces, https://arxiv.org/abs/1911.10855
  * [Mar20] M. Marcinkowski, Aut-Invariant Word Norm on Right-Angled Artin and Right-Angled Coxeter Groups Michigan Math. Journal, Volume 69, Issue 2 (2020), 285–295.

BASTIEN KARLHOFER, UNIVERSITY OF ABERDEEN, UK.
EMAIL<EMAIL_ADDRESS>