---
abstract: 'We show that for acylindrically hyperbolic groups $\Gamma$ (with no nontrivial finite normal subgroups) and arbitrary unitary representation $\rho$ of $\Gamma$ in a (nonzero) uniformly convex Banach space the vector space $H^2_b(\Gamma;\rho)$ is infinite dimensional. The result was known for the regular representations on $\ell^p(\Gamma)$ with $1<p<\infty$ by a different argument. But our result is new even for a non-abelian free group in this great generality for representations, and also the case for acylindrically hyperbolic groups follows as an application.'
author:
- 'Mladen Bestvina, Ken Bromberg and Koji Fujiwara[^1]'
bibliography:
- './../ref2.bib'
title: Bounded cohomology with coefficients in uniformly convex Banach spaces
---

Introduction
============

Quasi-cocycle and quasi-action
------------------------------

Let $G$ be a group and $E$ a normed vector space (usually complete, either over $\R$ or over $\mathbb C$). The linear or rotational part of an isometric $G$-action on $E$ determines a representation $\rho:G\to O(E)$ where $O(E)$ is the group of norm-preserving linear isomorphisms $E\to E$. We will refer to $\rho$ as a [*unitary representation*]{}. We will usually write $\rho(g)x$ as $g(x)$ or $gx$.

The translational part of the $G$-action is a [*cocycle*]{} (with respect to $\rho$). Namely the translational part is a function $F:G\to E$ that satisfies $$\label{cocycle}
F(gg')=F(g)+gF(g')$$ for all $g,g'\in G$. Going in the other direction, if $\rho$ is a unitary representation and $F$ a cocycle then the map $g\mapsto
(x\mapsto \rho(g)x+F(g))$ determines an isometric $G$-action on $E$. Note that $F(g^{-1})=-g^{-1}F(g)$.

For an isometric quasi-action of $G$ on $E$ the linear part will still be a unitary representation. However, the translational part $F$ will become a [*quasi-cocycle*]{} and will only satisfy up to a uniformly bounded error so that $$\label{quasi-cocycle}
\Delta(F):=\sup_{g,g'\in G}|F(gg')-F(g)-gF(g')|<\infty.$$ The quantity $\Delta(F)$ is the [*defect*]{} of the quasi-cocycle.

A basic question is if there are quasi-actions that are not boundedly close to an actual action. Such a quasi-action is [ *essential*]{}. Since quasi-actions determine unitary representations a more refined question is if there are essential quasi-actions for a given unitary representation.

The above discussion is perhaps more familiar in its algebraic form where it can be rephrased in terms of bounded cohomology. A quasi-cocycle $F$ can be viewed as 1-cochain in the group cohomology twisted by the representation $\rho$. Condition , is equivalent to the coboundary $\delta F$ being a bounded 2-cocycle and will therefore determine a cohomology class in $H^2_b(G;\rho)$, the second bounded cohomology group. Now this cocycle will clearly be trivial in the regular second cohomology group $H^2(G;\rho)$ as it is the coboundary of a 1-cochain. If the cochain $F$ is a bounded distance from a cocycle then $\delta F$ will also be trivial in $H^2_b(G;\rho)$ so we are interested in the kernel of the map $$H^2_b(G;\rho)\to H^2(G;\rho)$$ from bounded cohomology to regular cohomology. In particular this kernel is the vector space $QC(G;\rho)$ of all quasi-cocycles modulo the subspace generated by bounded functions and cocycles. We denote this quotient space $\widetilde{QC}(G;\rho)$. This is the vector space of [*essential*]{} quasi-cocycles and it is the main object of study of this paper.

For the trivial representation on $\R$ a cocycle is just a homomorphism to $\R$ and a quasi-cocycle is usually called a quasi-morphism. When $G= F_2$, the free group on two generators, Brooks [@brooks] gave a combinatorial construction of an infinite dimensional family of essential quasi-morphisms.

Uniformly convex Banach space and main result
---------------------------------------------

Following the work of Brooks, there is a long history of generalizations of this construction to other groups. Initially, the work focused on the trivial representation. See [@epstein-fujiwara; @bestvina-fujiwara; @rank1]. This was followed by generalizations to the same groups $G$ but with coefficients in the regular representation $\ell^p(G)$, $1\leq p<\infty$. See [@ursula.rank1; @ursula2].

In this paper we will extend this work to unitary representations in [*uniformly convex*]{} Banach spaces. Note that this essentially includes the previous cases since $\ell^p(G)$ is uniformly convex when $1<p<
\infty$.

If one is a bit more careful about how the counting is done then Brooks construction of quasi-morphisms can also be used to produce quasi-cocycles. In Brooks’ original work (i.e., for trivial representations) it is easy to see that the quasi-morphisms are essential. Here we will have to work harder to get the following result.

\[main:free2\] Let $\rho$ be a unitary representation of $F_2$ on a uniformly convex Banach space $E\neq 0$. Then $\dim\widetilde{QC}(F_2;\rho)=\infty$.

To show $\widetilde{QC}(F_2;\rho)$ is non-trivial is already hard. We will argue for a certain Brooks’ quasi-cocycle $H$ into a Banach space $E$, there exists a sequence of elements in $F_2$ on which $H$ is unbounded. For that we use that $E$ is uniformly convex in an essential way (Lemma \[unbounded\]). We also show those quasi-cocycle are not at bounded distance from any cocycle using that $E$ is reflexive (using Lemma \[bounded cocycle 2\]). Those two steps are the novel part of the paper. It seems that the uniform convexity is nearly a necessary assumption for the conclusion. See the examples at the end of this section.

Recently Osin [@osin2] (see also [@DGO]) has identified the class of [*acylindrically hyperbolic groups*]{} and this seems to be the most general context where the Brooks’ construction can be applied. Osin has shown that acylindrically hyperbolic groups contain [*hyperbolically embedded*]{} copies of $F_2$ and then applying work of Hull-Osin ([@hull.osin]) we have the the following corollary to Theorem \[main:free\]. See Section \[s:hyp-embed\] for the proof.

\[main:acylind\] Let $\rho$ be a unitary representation of an acylindrically hyperbolic group $G$ on a uniformly convex Banach space $E \neq 0$ and assume that the maximal finite normal subgroup has a non-zero fixed vector. Then $\dim \widetilde{QC}(G;\rho) = \infty$.

A wide variety of groups are acylindrically hyperbolic. In particular our results apply to the following examples. To apply our result, in all examples assume $G$ has no nontrivial finite normal subgroups, or more generally that for the maximal finite normal subgroup $N$ (see [@DGO]) we have that $\rho(N)$ fixes a nonzero vector in $E$.

-   $G$ is non-elementary word hyperbolic,

-   $G$ admits a non-elementary isometric action on a connected $\delta$-hyperbolic space such that at least one element is hyperbolic and WPD,

-   $G=Mod(S)$, the mapping class group of a compact surface which is not virtually abelian,

-   $G=Out(F_n)$ for $n\geq 2$,

-   $G$ admits a non-elementary isometric action on a $CAT(0)$ space and at least one element is WPD and acts as a rank 1 isometry. ([@rank1] for $\rho=\R$ and [@ursula.rank1] for $\rho=\ell^p(G)$, both under the assumption that the CAT(0) space $X$ and the action of $G$ are proper and $G$ contains a rank-1 element).

Recall that a Banach space is [*superreflexive*]{} if it admits an equivalent uniformly convex norm. It is observed in [@bfgm Proposition 2.3] that if $\rho:G\to E$ is a unitary representation with $E$ superreflexive, then there is an equivalent uniformly convex norm with respect to which $\rho$ is still unitary. Thus in Corollary \[main:acylind\] we may replace “uniformly convex” with “superreflexive”.

There is also a more direct approach to going from Theorem \[main:free\] to our the main theorem. The key point is that any group $G$ covered in the the main theorem acts on a quasi-tree such that there is a free group $F \subset G$ that acts properly and co-compactly on a tree isometrically embedded in the quasi-tree. This is done using the [*projection complex*]{} of [@bbf]. Using this one can apply the Brooks’ construction to produce quasi-cocycles that when restricted to the free group are exactly the quasi-cocycles of Theorem \[main:free\]. We carry this out in a separate paper ([@bc]).

Known examples with certain Banach spaces
-----------------------------------------

Here are some known vanishing/non-vanishing examples in the literature.

-   $E = \R$ and $\rho$ is trivial. In this case $H^2_b(G;\rho)$ is the usual bounded cohomology and quasi-cocycles are quasi-morphisms. As we said this case was known for various kinds of groups.

-   $E = \ell^p(G)$ and $\rho$ is the regular representation, see [@ursula.cat; @ursula]. When $1<p< \infty$, $\ell^p(G)$ is uniformly convex and our theorem applies. When $p = 1$ or $p=\infty$ then $\ell^p(G)$ is not uniformly, or even strictly, convex. However, for $p=1$ summation determines a $\rho$-invariant functional and one can produce a family of quasi-cocycles that when composed with the invariant functional are an infinite dimensional family of non-trivial quasi-morphisms in $\widetilde{QH}(G)$ implying that $\dim \widetilde{QC}(G;\ell^1(G)) = \infty$.

On the other hand,

-   When $p=\infty$ given any quasi-cocycle one can explicitly find a cococyle a bounded distance away so $\widetilde{QC}(G;\ell^\infty(G))=0$ for any group $G$.

-   If $G$ is countable and exact (e.g., $F_2$), then $H^2_b(G;\ell_0^\infty(G))=0$. In particular, $\widetilde{QC}(G;\ell^\infty_0(G))=0$ (Example \[ex.ell.0\]). Here $\ell^\infty_0(G)$ is the subspace of $\ell^\infty(G)$ consisting of sequences which are asymptotically 0.

There are also examples where $G$ is not acylindrically hyperbolic but where $\widetilde{QC}(G; \rho)$ is known to be non-zero for certain actions of $G$ on $\ell^p$ spaces.

-   If $G$ has a non-elementary action on a $CAT(0)$ cube complex then $\widetilde{QC}(G; \rho) \neq 0$ where $\rho$ is the representation of $G$ on the space of $\ell^p$-functions ($1 \leq p
      < \infty$) on a certain space where $G$ naturally acts ([@CFI]). Note that this class of groups is closed under products so it contains groups that aren’t acylindrically hyperbolic.

There are other examples where essentially nothing is known.

-   $E = \ell^1_0(G) \subset \ell^1(G)$ is the space of $\ell^1$-functions on $G$ that sum to zero and $\rho$ is the regular representation. Unlike with $\ell^1(G)$, $\ell^1_0(G)$ has no $\rho$-invariant functionals.

-   $E = {\mathcal B}(\ell^2(G))$ the space of bounded linear maps of $\ell^2(G)$ to itself. This example was suggested to us by N. Monod as the non-commutative analogue to $\ell^\infty(G)$.

Quasi-cocycles from trees {#s:free}
=========================

Fix $F_2= \langle a,b\rangle$ and choose a word $w\in F_2$. For simplicity we will assume that $w$ is cyclically reduced. Let $E$ be a normed vector space and $\rho:G\to O(E)$ a linear representation. Also choose a nonzero $e\in E$. We now set up some notation that will be convenient for what we will do later.

Let $[g,h]$ be an oriented segment in the Cayley graph for $F_2$ with generators $a$ and $b$. Then we write $[g,h] {\overset{\circ}{\subset}}[g', h']$ if $[g,h]$ is a subsegment of $[g',h']$ and the orientations of the two segments agree. We then define $$w_+(g) = \{h \in G | [h, hw] {\overset{\circ}{\subset}}[1,g]\}$$ and $$w_-(g) = \{h \in H | [h,hw] {\overset{\circ}{\subset}}[g,1]\}.$$

Now define a function $H=H_{w,e}:F_2\to E$ by $$H(g)=\sum_{h \in w_+(g)}h(e)-\sum_{h\in w_-(g)}h(e)$$

In other words, to a translate $h\cdot w$ we assign $h(e)$ when traversed in the positive direction, and $-h(e)$ when traversed in negative direction. Note that it follows that $H(g^{-1})=-g^{-1}H(g)$.

\[free-Brooks\] The function $H$ constructed above is a quasi-cocycle.

This is the standard Brooks argument. Consider the tripod spanned by $1,g,gf$. Call the central point $p$. We will see that contributions of copies of $w$ in the tripod that do not cross $p$ cancel out leaving only a bounded number of terms.

If $h\cdot w{\overset{\circ}{\subset}}[1,p]$ then $h(e)$ enters with positive sign in $H(g)$ and in $H(gf)$, so it cancels in the expression $H(gf)-H(g)$. Likewise, if $h\cdot w{\overset{\circ}{\subset}}[p,1]$ then $-h(e)$ enters both $H(g)$ and $H(gf)$, so it again cancels.

If $h\cdot w{\overset{\circ}{\subset}}[p,g]$ then $h(e)$ is a summand in $H(g)$. Since $h\cdot w{\overset{\circ}{\subset}}[gf,g]$ we also have $g^{-1}h\cdot w{\overset{\circ}{\subset}}[f,1]$, so $-g^{-1}h(e)$ is a summand in $H(f)$, and thus we have cancellation in $-H(g)-gH(f)$. There is similar cancellation if $h\cdot w{\overset{\circ}{\subset}}[g,p]$.

If $h\cdot w{\overset{\circ}{\subset}}[p,gf]$ or $[gf,p]$ then similarly to the previous paragraph there is cancellation in $H(gf)-gH(f)$.

After the above cancellations in the expression $H(gf)-H(g)-gH(f)$ the only terms left are of the form $\pm h(e)$ where $h(w)$ is contained in the tripod and contains $p$ in its interior. The number of such terms is clearly (generously) bounded by $6|w|$ so we deduce that $\Delta(H)\leq 6|w|\|e\|$.

Note that if $h\cdot w$ does not overlap $w$ for any $1\not=h \in F_2$, then $\Delta(H) \le 6\|e\|$. More generally, for a given $w$, write $w=a^nb$ as a word such that $|b|<|a|$ and $n>0$ is maximal. Then, $\Delta(H) \le 6(n+1)\|e\|$.

\[ab\] Suppose $w=ab$. Then $H(a^n)=H(b^n)=0$, while $H((ab)^n)=(1+ab+(ab)^2+\cdots+(ab)^{n-1})e\in E$. If the operator $1-ab:E\to E$ has a continuous inverse (i.e. if $1\in\mathbb C$ is not in the spectrum of $ab$) then $H$ is uniformly bounded on the powers of $ab$ since $(1-ab)H((ab)^n)=e-(ab)^n(e)$ has bounded norm. For example, this happens even for $E=\R^2$ when $\rho(ab)$ is a (proper) rotation.

On the other hand, for the representation $\ell^p(F_2)$ with $1\leq p<\infty$ and with $e\in\ell^p(F_2)$ defined by $e(1)=1$, $e(g)=0$ for $g\neq 1$, the quasi-cocycle $H$ is unbounded on the powers of $ab$.

Nontriviality of quasi-cocycles {#nontrivial}
===============================

In Brooks’ original construction of quasi-morphisms $F_2={\langle}a,b{\rangle}\to\R$ it is easy to see that the quasi-morphisms are nontrivial. Choosing $w$ to be a reduced word not of the form $a^m$ or $b^m$ it is clear that $H(w^n)$ will be unbounded while $H(a^n)$ and $H(b^n)$ will be zero. By this last fact if $G$ is a homomorphism that is boundedly close to $H$ then $G$ must be bounded on powers of $a$ and $b$ and therefore $G(a) = G(b) =0$. Since any homomorphism is determined by its behavior on the generators we have $G \equiv 0$ and the nontriviality of $H$ follows.

When the Brooks construction is extended to quasi-cocycles it is no longer clear that the quasi-cocycle is nontrivial. In particular if $H
= H_{w,e}$ it may be that $H(w^n)$ is bounded. See Examples \[ab\] and \[U2\]. In fact if $1$ is not in the spectrum of $\rho(w)$ then $H(w^n)$ will be bounded for all choices of vectors $e$. Even if $1$ is in the spectrum, when $e$ is chosen arbitrarily $H(w^n)$ may be bounded. To show that the Brooks quasi-cocycles are unbounded we will need to restrict to the class of [*uniformly convex*]{} Banach spaces and to look at a wider class of words than powers of $w$.

We will also have to work harder to show that a cocycle $G$ that is bounded on powers of the generators is bounded everywhere. In fact we cannot do this in general but instead will show that in a reflexive Banach space (which includes uniformly convex Banach spaces) either the cocycle is bounded or the original representation, when restricted to a non-abelian subgroup, has an eigenvector. In this latter case it is easy to construct many nontrivial quasi-cocycles.

Uniformly convex and reflexive Banach spaces
--------------------------------------------

We will use basic facts about Banach spaces. General references are [@bourbaki; @megginson]. The following concept was introduced by Clarkson [@clarkson].

A Banach space $E$ is [*uniformly convex*]{} if for every $\epsilon>0$ there is $\delta>0$ such that $x,y\in E$, $|x|\leq 1$, $|y|\leq 1$, $|x-y|\geq\epsilon$ implies $|\frac {x+y}2|\leq 1-\delta$.

The original definition in [@clarkson] replaces $|x|,|y|\leq 1$ above with equalities, but it is not hard to see that the two are equivalent.

\[UC basic\]

(i) $\ell^p$ spaces are uniformly convex for $1<p<\infty$ ([@clarkson]). $\ell^1$ and $\ell^{\infty}$ spaces are not uniformly convex and not reflexive.

(ii) A uniformly convex Banach space is reflexive (the Milman-Pettis theorem).

(iii) If $E$ is uniformly convex, then for any $R>0$ there are $\epsilon>0$ and $\mu>0$ so that the following holds. If $|v|\le R$ and $f:E\to\R$ is a functional of norm 1 with $f(v)=|v|$ and if $e$ is a vector of norm $\geq 1/2$ with $f(e)\geq -\mu$ then $|v+e|\geq
      |v|+\epsilon$.

We only prove (iii). Choose $\delta \in (0,1)$ so that $|x|,|y|\leq 1$, $|x-y|\geq
  \frac 1{2(R+1)}$ implies $|\frac{x+y}2|\leq 1-\delta$. Then choose $\epsilon,\mu>0$ so that $\epsilon<\frac 18$ and $\frac{\frac
    18-\frac {\mu}2}{\frac 18+\epsilon}>1-\delta$. Suppose $f,v,e$ satisfy the assumptions but $|v+e|<R+\epsilon$. If $|v| \le 1/8$ then $|v + e| \ge |e| - |v| \ge
  1/4 \ge |v| + 1/8$ and we are done. So assume that $|v| >1/8$. Then for $x=\frac v{|v|+\epsilon}$, $y=\frac {v+e}{|v|+\epsilon}$ we have $|x|,|y|\leq 1$ and $|x-y|\ge \frac1{2(|v|+1)} \geq\frac 1{2(R+1)}$, so we must have $|\frac {x+y}2|\leq 1-\delta$. Thus $$1-\delta\geq
  \bigg|\frac{x+y}2\bigg|=\bigg|\frac{v+e/2}{|v|+\epsilon}\bigg|\geq
  \frac {|v|-\frac{\mu}2}{|v|+\epsilon}\geq\frac{\frac
    18-\frac{\mu}2}{\frac 18+\epsilon}$$ since $f(v+e/2)=|v|+f(e)/2\geq |v|-\frac{\mu}2$ and $|f|=1$. This contradicts the choice of $\mu,\epsilon$.

\[fixed points\] Let $\rho$ be a unitary representation of a group $F$ on a reflexive Banach space $E$. If there is a linear functional $f$ and a vector $e
\in E$ such that the $F$-orbit of $e$ lies in the half space $\{f \ge
\mu\}$ with $\mu >0$ then there is an $F$-invariant vector $e'
\neq 0 \in E$ and a $F$-invariant functional $\phi$ with $\phi(e') \ge
\mu$. If $e$ is $F$-invariant, then we can take $e'=e$.

Let $\Lambda$ be the convex hull of the $F$-orbit of $e$ in the weak topology on $E$. Since $E$ is reflexive, $\Lambda$ is weakly compact. The convex hull $\Lambda$ is also $F$-invariant so by the Ryll-Nardzewski fixed point theorem it will contain an $F$-invariant vector $e'$. Since $e' \in \Lambda$, $f(e') \ge \mu$ and therefore $e' \neq 0$.

Since $e'$ is a functional on the reflexive Banach space $E^*$ and the $F$-orbit of $f$ will be contained in the half space $\{e' \ge \mu\}$ we similarly get a $F$-invariant vector $\phi \in E^*$ with $e'(\phi) = \phi(e') \ge \mu$.

Note that if $E$ contains a nonzero vector that is $F$-invariant, then the Hahn-Banach theorem supplies a functional that satisfies the conditions of the lemma and so there is also a nonzero $F$-invariant functional.

Detecting unboundedness
-----------------------

\[unbounded\] Let $\rho$ be any unitary representation of $F_2 = \langle a,b\rangle$ into a uniformly convex Banach space $E$. Then one of the following holds:

(i) for every $e\neq 0 \in E$ and any $1\neq w\in F_2$ not of the form $a^mb^n$ nor $b^ma^n$ the quasi-cocycle $H=H_{w,e}$ is unbounded on $F_2$, or

(ii) there is a free subgroup $F\subset F_2$ with $F\cong F_2$, a linear functional $g$, a vector $e$ and a $\mu>0$ such that the $F$-orbit of $e$ is contained in the half-space $\{g \le -\mu\}$. In particular, there is an $F$-invariant vector $e' \not=0$ in the half space.

We first make some observations about words in $F_2$. Given a word $w$ as in (i) we can find buffer words $B$ and $B'$ of the form $a^\ell b^\ell$ or $b^\ell a^\ell$ and a subgroup $F = \langle a^m, b^m \rangle$ with $m
>>\ell, |w|$ such that if $w' = BwB'$ and $y_1, y_2, \dots, y_n \in F$ then in the reduced word for the element $x=y_1 w' y_2 w' \cdots y_n w'$ there is exactly one copy of $w$ for each $w'$ and no other copies of either $w$ or $w^{-1}$. Note that the word $y_1 w' y_2 w' \cdots y_n w'$ may not be reduced and in its reduced version there may be cancellations in the $w'$. However, the buffer words will prevent these cancellations from reaching $w$. The restrictions on $w$ ensure that $w$ does not appear as a subword of some $y_i$. In particular, $|H(w')|=|e|$ and $H(xyw')=H(x)+xH(yw')=H(x)+xyH(w')$ for any $y\in F$.

For simplicity, normalize so that $|e|=1$, so $|H(w')|=1$. Assume that (ii) doesn’t hold, and that $H$ is bounded on $F_2$. Let $F_w$ be the set of words of the form $$y_1 w' y_2 w' \cdots y_n w' , (y_i  \in F)$$ and let $R= \underset{x\in
  F_w}{\sup} |H(x)|<\infty$. Let $\epsilon,\mu>0$ be as in Proposition \[UC basic\](iii). Choose an $x\in F_w$ such that $|H(x)| > R -
\epsilon$. We will find a $y \in F$ such that $|H(xyw')| > R$ to obtain a contradiction since $xyw' \in F_w$.

Let $\phi$ be a linear functional of norm 1 such that $\phi(H(x)) =
|H(x)|$. Let $\psi = \phi\circ x$. Since (ii) doesn’t hold, there exists a $y \in F$ with $\psi(yH(w')) > -\mu$. (We are applying the negation of (ii) not to $e$ but to $H(w')$, which is in the $F_2$-orbit of $e$, but it is easy to see that this follows from the corresponding fact for $e$ by replacing $F$ with a conjugate.) So, $\phi(xyH(w')) > -\mu$. Then by Proposition \[UC basic\](iii), $|H(xyw')| = |H(x) + xyH(w')| \ge
|H(x)| + \epsilon > R$, contradiction.

For an $F$-invariant vector in (ii), see the proof of Lemma \[fixed points\].

We give an application of Lemma \[unbounded\].

\[U2\] Choose an embedding $\rho:F_2\subset U(2)$ so that every nontrivial element is conjugate to a matrix of the form $$\left(\begin{matrix}
e^{2\pi it}&0\\
0&e^{2\pi is}\end{matrix}\right)$$ with $t,s,\frac ts$ all irrational. (Such representations can be constructed by noting that they form the complement of countably many proper subvarieties in $Hom(F_2,U(2))$.) Put $E={\Bbb C}^2$.

Then any $H=H_{w,e} $ with $0\not=e\in E, 1\not=w\in F_2$ is bounded on any cyclic subgroup, but many are globally unbounded. The second statement follows by noting that the orbit of any unit vector under a nontrivial cyclic subgroup is dense in a torus $S^1\times S^1\subset \mathbb
C^2$, so (ii) of Lemma \[unbounded\] fails, and (i) must hold. For the first statement, observe that for a fixed $g\in F$ the values $H(g^n)$ can be computed, up to a bounded error, by adding sums of the form $$U_n=u(e)+gu(e)+\cdots+g^{n-1}u(e)$$ one for every $g$-orbit of occurrences of $w$ or $w^{-1}$ along the axis of $g$. Applying $g$ we have $$g(U_n)=gu(e)+\cdots +g^nu(e)$$ and so $|g(U_n)-U_n|\leq 2|e|$, which implies that $|U_n|$ is bounded, since $g:\mathbb C^2\to\mathbb C^2$ moves every unit vector a definite amount. It follows $H(g^n)$ is bounded on $n$. This gives an isometric quasi-action of $F_2$ on $\mathbb C^2$ or $\R^4$ with unbounded orbits, but with every cyclic subgroup having bounded orbits.

In fact, since $H^1(F_2;\rho)\neq 0$, it follows that there are [ *isometric*]{} actions of $F_2$ on $\R^4$ with unbounded orbits and with every element fixing a point.

The following is our basic method of detecting bounded cocycles. In the presence of reflexivity of the Banach space, bounded isometric actions have fixed points. Thus a cocyle $G:F_2\to E$ is bounded if and only if for some $v\in E$ (a fixed point of the action) we have $G(g)=v-\rho(g)v$ for every $g\in F_2$.

\[bounded cocycle 2\] Let $\rho$ be a unitary representation of $F_2$ on a reflexive Banach space $E$ and $G$ a cocycle that is bounded on $\langle a^2,b \rangle$ and $\langle a^3, b \rangle$. Then one of the following holds.

(i) $G$ is bounded on $F_2$, or

(ii) There is a free subgroup $F\subset F_2$ with $F\cong F_2$ such that $\rho|_F$ fixes a nonzero vector in $E$.

The cocycle $G$ induces an action of $F_2$ on $E$ by affine isometries and the image of $G$ is the orbit of $0$ under this action. If the restriction of this action to $\langle a^2, b \rangle$ is bounded (with respect to the norm topology) then the convex hull of the orbit (in the weak topology) will be $\langle a^2, b \rangle$-invariant and compact since $E$ is reflexive so by the Ryll-Nardzewski fixed point theorem $\langle a^2,b \rangle$ will have a fixed point. Thus $Fix(a^2)\cap Fix(b)\neq\emptyset$. If this intersection is not a single point then (ii) holds since $\rho$ restricted to $F=\langle
a^2,b\rangle$ fixes the difference of any two vectors in the intersection. Similarly, (ii) holds if $Fix(a^3)\cap
Fix(b)\neq\emptyset$ is not a single point. If the two intersections coincide then the intersection point is fixed by both $a=a^3(a^2)^{-1}$ and $b$, so $G$ is bounded. If the intersections are distinct then $F=\langle a^6,b\rangle$ fixes two distinct points, so (ii) holds.

Detecting essentiality and proof of Theorem \[main:free2\]
----------------------------------------------------------

We now show that under suitable conditions our quasi-cocycles are essential. We consider two cases. If there is a free subgroup that fixes a nonzero vector $e\in E$, the argument essentially goes back to Brooks, since in this case we restrict to the trivial representation. This case is presented first.

\[fixed vector:free\] Let $\rho$ be a unitary representation of $F_2$ in a reflexive Banach space $E$ and let $F$ be a rank two free subgroup such that $\rho|_F$ has an invariant vector $e\neq 0$. Let $w'$ be a cyclically reduced word that is conjugate into $F$. Then quasi-cocycles of the form $H_{w, e}$ where $w$ is a reduced word that contains $w'$ as a subword span an infinite dimensional subspace of $\widetilde{QC}(F_2;\rho)$.

The word $w'$ can be empty and that is the case we use later.

After possibly conjugating $F$ we can assume that $w'$ is contained in $F$. Since $w'$ is cyclically reduced its axis contains the identity in the Cayley graph for $F_2$. This implies that the minimal $F$-tree also contains the identity and allows us to find cyclically reduced words $\alpha$ and $\beta$ in $F$ such that the concatenation $$w_k = w'\alpha^k \beta^k\alpha^k\beta^k$$ is cyclically reduced. Furthermore we can assume that $\alpha$ and $\beta$ generate $F$. Let $H_k=H_{w_k,e}$. By Lemma \[fixed points\] there exists an $F$-invariant (continuous) linear functional $\phi$ with $\phi(e) \ge
\mu>0$.

Then the restriction to $F$ of the composition $\phi \circ G$ with any co-cycle $G$ is a homomorphism, and similarly the restriction of the composition $\phi \circ H$ to $F$ with any quasi-co-cycle $H$ is a quasi-morphism.

We will show that the sequence $H_{k_0},H_{k_0+1},\cdots$ represents linearly independent elements in $\widetilde {QC}(F_2;\rho)$ for $k_0$ so large that all powers of $\alpha$ and $\beta$ appearing in $w'$ are $<<k_0$ in absolute value. Indeed, if $H=H_k-c_{k_0}H_{k_0}-\cdots -c_{k-1}H_{k-1}$, with $k_0 <k$, for any constants $c_i$ then the quasi-morphism $\phi \circ H$ on $F$ is 0 on the powers of $\alpha$ and $\beta$, so if a co-cycle $G$ is boundedly close $H$, then the homomorphism $\phi
\circ G$ on $F$ must be bounded, and therefore zero, on powers of $\alpha$ and $\beta$. Therefore $\phi \circ G$ is trivial when restricted to $F$. On the other hand a staightforward calculation shows that $\phi
\circ H(w^n_k) \ge n \mu$ so $\phi \circ H$ is unbounded on $F$ and $H$ and $G$ cannot be boundedly close. We showed that $H$ is non-trivial in $\widetilde {QC}(F_2;\rho)$, so $H_{k_0},H_{k_0+1},\cdots, H_k$ are linearly independent.

We now consider the opposite case when no reduction to the trivial representation is possible.

\[no fixed vector:free\] Let $\rho$ be a unitary representation of $F_2= \langle a, b \rangle$ on a uniformly convex Banach space and assume that no nonabelian subgroup of $F_2$ fixes a nonzero vector. Then for any fixed $e\neq 0$ and a cyclically reduced word $w'$ the quasi-cocycles of the form $H_{w, e}$ span an infinite dimensional subspace of $\widetilde{QC}(F_2;\rho)$, where $w$ range over cyclically reduced words that contain $w'$ as a subword.

The word $w'$ can be empty, and that is the case we use later.

We assume that $w'$ does not start with $b^{-1}$ or end with $a^{-1}$ (otherwise, swap $a$ and $b$ in the definition of $w_m$). Let $w_m=w'a^{5m}b^{5m}a^{7m}b^{7m}$, $m\geq 1$, and $gcd(m,6)=1$. By Lemma \[unbounded\], $H_m
=H_{w_m,e}$ is unbounded. Furthermore $H_m$ is 0 on the subgroups $\langle a^2,b\rangle$ and $\langle a^3,b\rangle$ listed in Lemma \[bounded cocycle 2\].

We claim that those $H_m$’s are linearly independent in $\widetilde{QC}(F_2;\rho)$. Fix $m$ and let $H=H_m-\sum_{i<m}c_iH_i$ for constants $c_i$. Then $H$ is also unbounded, since the $H_i$ for $i<m$ are visibly 0 on all words in $F_{w_m}$, the set given in the proof of Lemma \[unbounded\], but $H_m$ is unbounded on $F_{w_m}$. $H$ is bounded on $\langle a^2,b\rangle$ and $\langle a^3,b\rangle$.

Suppose $H$ differs from a cocycle $G$ by a bounded function. Then $G$ is also bounded on the subgroups $\langle a^2,b\rangle$ and $\langle a^3,b\rangle$, therefore $G$ is bounded on $F_2$ since (i) must holds in Lemma \[bounded cocycle 2\]. So, $H$ is bounded on $F_2$, contradiction. We showed that $H_i, i\le m$ are linearly independent in $\widetilde{QC}(F_2;\rho)$.

Theorem \[main:free2\] now follows immediately. If there is a rank two free subgroup $F$ in $F_2$ with an $F$-invariant vector $e\not=0$, then use Proposition \[fixed vector:free\] with $w'$ empty to produce an infinite dimensional subspace. Otherwise, use Proposition \[no fixed vector:free\] with $w'$ empty.

\[main:free\] Let $\rho$ be a unitary representation of $F_2$ on a uniformly convex Banach space $E\neq 0$. Then $\dim\widetilde{QC}(F_2;\rho)=\infty$.

We remark that Pascal Rolli has a new construction, different from the Brooks construction, that he showed in [@rolli] produces nontrivial quasi-cocycles on $F_2$ (and some other groups) when the Banach space $E$ is an $\ell^p$-space (or finite dimensional).

\[ex.ell.0\] To see the importance of uniform convexity we will look more closely at the examples $\ell^\infty(F_2)$ of bounded functions and $\ell^\infty_0(F_2)$ of bounded functions that vanish at infinity.

\(1) For the regular representation on $\ell^\infty(F_2)$ (or any group $G$) the constant functions determine an one-dimensional invariant subspace. In particular, any quasi-morphism canonically determines a quasi-cocycle with image in this invariant subspace. If the original quasi-morphism is essential one may expect that the associated quasi-cocycle is also essential. However, for any quasi-cocycle $H$ we can we define the function $H_0:F_2 \to \ell^\infty(F_2)$ by $H_0(g)(f) = H(f)(f) -
\rho(g)H(g^{-1}f)(f)=H(f)(f)-H(g^{-1}f)(g^{-1}f)$ and then we can check that $H_0$ is a cocycle (essentially it is the coboundary of the 0-cochain defined by the function $f \mapsto H(f)(f)$) and that $\|H
- H_0\|_\infty \le \Delta(H)$. In particular, $\widetilde{QC}(F_2;\ell^\infty(F_2))=0$ and $H^2_b(F_2;\ell^\infty(F_2)) = 0$.

\(2) For the regular representation of $F_2$ on $\ell^\infty_0(F_2)$ neither $F_2$ nor any non-trivial subgroup fixes a non-trivial subspace so we cannot, as in the $\ell^\infty(F_2)$ case, use quasi-morphisms to construct unbounded quasi-cocycles. Furthermore for some choices of the vector $e$, the quasi-cocyle $H_{w,e}$ will be bounded. For example if $e \in \ell^\infty_0(F_2)$ is defined by $$e(x) = \left\{ \begin{array}{ll} 1 & x = 1 \\ 0 & x \neq
  1 \end{array} \right.$$ then $\|H_{w,e}(x)\|_\infty = 0 \mbox{ or }
1$ depending on whether $x$ does or doesn’t contain a copy of $w$. More generally if $e \in \ell^1(F_2) \subset \ell^\infty_0(F_2)$ we have that $\|H_{w,e}(x)\|_\infty \le \|e\|_1$. On the other hand if we define $f \in \ell^\infty_0(F_2)$ by $$f(x) = \left\{ \begin{array}{ll} 1/n & x = w^{-n}, n>0\\ 0 &
  \operatorname{otherwise}\end{array} \right.$$ then $|H_{w,
  f}(w^n)(id)| = \sum_{i=1}^n 1/i$ so $\|H_{w,f}(w^n)\|_\infty$ is unbounded. We can still construct the cocycle $H_0$ as in the previous paragraph where $H=H_{w,f}$ but this cocycle will not lie in $\ell^\infty_0(F_2)$. This example emphasizes an inherent difficulty in extending our results to a wider class of Banach spaces.

Note that $H^n_b(G;\ell^{\infty}(G))=0 (n \ge 1)$ for any group $G$ [@monod Proposition 7.4.1] since $\ell^{\infty}(G)$ is a “relatively injective” Banach $G$-module [@monod Chapter II], so some assumption on the Banach space is necessary.

\(3) We also note that $H^2_b(G;\ell_0^{\infty}(G))=0$ for any countable, [*exact*]{} group (e.g. $G=F_2$, see [@ozawaicm]). This can be seen as follows. First, since $\ell^\infty(G)$ is a relatively injective Banach $G$-module, $H^n_b(G;\ell^{\infty}(G))=0$ for all $n>0$. From the long exact sequence in bounded cohomology ([@monod Proposition 8.2.1]) induced by the short exact sequence $0\to\ell_0^\infty(G)\to \ell^\infty(G)\to
\ell^{\infty}(G)/\ell_0^{\infty}(G) \to 0$, it suffices to show $H^1_b(G, \ell^\infty(G)/\ell_0^{\infty}(G))=0$. But this holds if $G$ is countable and exact [@ozawa3 Theorem 3]. We thank Narutaka Ozawa for pointing out his work to us.

To show $H^n_b(G;\ell_0^\infty(G))=0$ for all $n>1$ it suffices to know $$H^n_b(G, \ell^\infty(G)/\ell_0^{\infty}(G))=0$$ for all $n>0$. Ozawa informs us that this is also true.

Hyperbolically embedded subgroups {#s:hyp-embed}
=================================

Before proving our main theorem we need a couple of straightforward lemmas.

\[finite normal\] Let $\rho$ be a unitary representation of a group $G$ on $E$ and $K$ a finite normal subgroup. Let $E' \subset E$ be the closed subspace of $K$-invariant vectors and $\rho'$ the unitary representation of $G$ on $E'$ obtained by restriction. Then every (quasi)-cocyle in $QC(G;
\rho)$ is a bounded distance from a (quasi)-cocyle in $QC(G;\rho')$

We first define a linear projection $\pi: E \to E'$ by $$\pi(x) = \frac{1}{|K|} \sum_{k \in K} \rho(k) x.$$ If $H$ is a (quasi)-cocycle in $QC(G; \rho)$ then $\tilde H = \pi \circ H$ is a (quasi)-cocycle in $QC(G; \rho')$. We need to show that $\tilde H$ is at bounded distance from $H$.

Recall that $H$ is the translational part of an isometric $G$(-quasi)-action on $E$. By the normality of $K$ if two points in $E$ are in the same $G$-orbit then their $K$-orbits are (quasi)-isometric. Since $H(G)$ is the $G$-orbit of $0$ under this (quasi)-action and $H(Kg)$ is at bounded distance from the $K$-orbit of $H(g)$ we have that the $K$-orbits of points in the image of $H$ are uniformly bounded, and so $\pi$ moves points in $Im(H)$ a uniformly bounded amount.

\[invariant-extension\] The natural map $\widetilde{QC}(G; \rho') \to \widetilde{QC}(G;
\rho)$ is an isomorphism.

\[finite-extension\] Let $\rho$ be a unitary representation of $G \times K$ on $E$ such that $K$ is finite and $\rho$ restricted to the $K$-factor is trivial. Then there is a natural isomorphism from $\widetilde{QC}(G
\times K; \rho) \to \widetilde{QC}(G; \rho)$.

Given $H \in QC(G \times K; \rho)$ define $\tilde H \in QC(G; \rho)$ by $\tilde H(g) = H(g, id)$. The linear map defined by $H \mapsto \tilde H$ descends to a linear map $\widetilde{QC}(G \times K; \rho) \to \widetilde{QC}(G;
\rho)$. Any quasi-cocycle in $QC(G; \rho)$ determines a quasi-cocylce in $QC(G\times K; \rho)$ by extending it to be constant on the $K$-factor. This also descends to a map $\widetilde{QC}(G; \rho) \to
\widetilde{QC}(G \times K; \rho)$, which is an inverse of our first map since $\|H(g, k) - H(g, id)\| \le
\Delta(H) + C$ where $C = \max\{\|H(id,k)\| | k \in K\}$. Hence we have the desired isomorphism.

In [@DGO], Dahmani, Guiradel and Osin defined the notion of a [ *hyperbolically embedded subgroup*]{}. For convenience we recount the definition here. Let $G$ be a group, $H$ a subgroup and $X \subset G$ such that $X \cup H$ generates $G$. Let $\Gamma(G, X \cup H)$ be the Cayley graph with generating set $X \cup H$. Then $H$ is hyperbolically embedded in $G$ if

-   $\Gamma(G, X \cup H)$ is hyperbolic;

-   For all $n>0$ and $h \in H$ there are at most finitely many $h'
      \in H$ that can be connected to $h$ in $\Gamma(G,X \cup H)$ by a path of length $\le n$ with no edges in $H$.

A quasi-cocycle is [*anti-symmetric*]{} if $$H(g^{-1}) = -\rho(g^{-1})H(g).$$ A cocycle automatically satisfies this condition. Furthermore every quasi-cocycle is a bounded distance from an anti-symmetric quasi-cocycle. (Simply replace $H(g)$ with $\frac12(H(g) - \rho(g) H(g^{-1})$.) We have the following important theorem of Hull and Osin.

\[Hull-Osin\] Let $G$ be a group and $F$ a hyperbolically embedded subgroup. Then there exists a linear map $$\iota : QC_{as}(F; \rho) \to QC_{as}(G;\rho)$$ such that if $H \in
QC_{as}(F;\rho)$ then $H = \iota(H)|_F$. In particular, $\dim
\widetilde{QC}(F;\rho) \le \dim \widetilde{QC}(G;\rho)$.

The action of a group $G$ on a metric space $X$ is [*acylindrical*]{} if for all $B>0$ there exist $D,N$ such that if $x,y \in X$ and with $d(x,y)> D$ then there are at most $N$ elements $g \in G$ with $d(x,
gx)<B$ and $d(y,gy)<B$. A group $G$ is [*acylindrically hyperbolic*]{} if it has an acylindrical,, non-elementary, action on a $\delta$-hyperbolic space. To apply the previous theorem we need the following result of Dahmani-Guiradel-Osin and Osin:

\[hyp-embed\] Let $G$ be an acylindrically hyperbolic group and $K$ the maximal finite normal subgroup. Then $G$ contains a hyperbolically embedded copy of $F_2 \times K$.

Theorem \[hyp-embed\] is a combination of two theorems. In [@osin2 Theorem 1.2], Osin proves that an acylindrically hyperbolical group contains a non-degenerate hyperbolically embedded subgroup. In [@DGO Theorem 2.24], Dahmani-Guirardel-Osin show that if $G$ contains a non-degenerate hyperbolically embedded subgroup then it contains a hyperbolically embedded copy of $F_2 \times K$. We note that this latter theorem relies on the projection complex defined in [@bbf].

Let $E'\subset E$ be the subspace fixed by $K$ and $\rho'$ the restriction of $\rho$ to $E'$. By assumption $\dim E' > 0$. By Theorem \[hyp-embed\] there is a copy of $F_2 \times K$ hyperbolically embedded in $G$. By Lemma \[finite-extension\] and Theorem \[main:free\] we have that $\dim \widetilde{QC}(F_2 \times K; \rho')
= \dim \widetilde{QC}(F_2, \rho') = \infty$. Corollary \[invariant-extension\] implies that $\dim \widetilde{QC}(F_2 \times
K; \rho) = \infty$. The corollary then follows from Theorem \[Hull-Osin\].

[^1]: The first two authors gratefully acknowledge the support by the National Science Foundation. The third author is supported in part by Grant-in-Aid for Scientific Research (No. 23244005)
