3.1 $\ell ^{2}$ formal inverses and growth

In this section we concentrate on conditions which guarantee that $f$ has a $\ell ^{2}$ formal inverse. If $f$ is lopsided, then by standard Banach algebra arguments it has an $\ell ^{1}$ formal inverse. If $f$ is semi-lopsided, but not lopsided, then it can be written as $f=m(1-x)$ with $m\in {\mathbb {Z}}$, $x\in {\mathbb {Q}}(G)$ and $|\hat {x}|\in \operatorname {Prob}(G).$ So we focus on conditions that guarantee that if $1-x\in {\mathbb {R}}(G)$ with $|\hat {x}|\in \operatorname {Prob}(G),$ then $1-x$ has a $\ell ^{2}$ formal inverse. Formally, one considers the geometric series $(1-x)^{-1}=\sum _{n}x^{n}$ and attempts to prove that this converges in $\ell ^{2}.$ The following two lemmas will be helpful in this regard. The proofs we give of them utilize the machinery of tracial von Neumann algebras, which require us to recall a few preliminaries. To ensure that the core ideas of this section remain in the foreground, we have relegated the proof of these two lemmas to Appendix A.

As an application of Lemma 3.1, in the well-balanced case we can completely characterize when $1-x$ has an $\ell ^{2}$ formal inverse as well as compute what this inverse has to be.

Proof. For a natural number $N\geq 1,$ set $\xi _{N}=\sum _{n=0}^{N}x^{n}.$

(i) Suppose $1-x$ has an $\ell ^{2}$ inverse $\xi .$ Then

\[ \|\xi_{N}\delta_{1}-\xi\|_{2}\!=\!\|\xi_{N}(1-x)\xi-\xi\|_{2}\!=\!\|x^{N+1}\xi\|_{2}\to_{N\to\infty}0, \]

the last step following from Lemma 3.1.

Conversely, suppose that $\sum _{n=0}^{\infty }x^{n}\delta _{1}$ converges conditionally, and set $\xi =\sum _{n=0}^{\infty }x^{n}\delta _{1}.$ Then

\[ (1-x)\xi=\lim_{N\to\infty}(1-x)\xi_{N}=\lim_{N\to\infty}\delta_{1}-x^{N+1}\delta_{1}=\delta_{1}, \]

by Lemma 3.1.

(ii) First, suppose that $\sum _{n,m\geq 0}\tau ((y^{*})^{m}y^{n})<\infty .$ For $N\geq M\geq 1,$ we have that

\[ \|\xi_{N}-\xi_{M}\|_{2}^{2}=\sum_{M\leq n,m\leq N}\tau((x^{*})^{m}x^{n})\leq \sum_{M\leq n,m\leq N}\tau((y^{*})^{m}y^{n}). \]

So

\[ \lim_{M\to\infty}\sup_{N\geq M}\|\xi_{N}-\xi_{M}\|_{2}=0, \]

by the dominated convergence theorem. Hence $\xi _{N}$ is a Cauchy sequence and thus converges. Hence, by part (i) we know that $1-x$ has an $\ell ^{2}$ formal inverse.

Conversely, suppose that $\xi$ is an $\ell ^{2}$ formal inverse to $1-x$ and that $\hat {x}\geq 0.$ Let $\xi _{N}$ be defined as in the first half of the proof. Then

\[ \|\xi_{N}\delta_{1}-\xi\|_{2}\to 0. \]

Hence,

\[ \|\xi\|_{2}^{2}=\lim_{N\to\infty}\|\xi_{N}\|_{2}^{2}=\lim_{N\to\infty}\sum_{0\leq n,m\leq N}\tau((x^{*})^{m}x^{n})=\sum_{n,m\geq 0}\tau((x^{*})^{m}x^{n}). \]

So

\[ \sum_{n,m\geq 0}\tau((x^{*})^{m}x^{n})<\infty. \]

The above result is much easier to say when $\hat {x}\geq 0$ and $x$ is self-adjoint.

Proof. By Lemma 3.3, we know that $f$ has an $\ell ^{2}$ formal inverse if and only if

\[ \sum_{n,m\geq 0}\tau(x^{n+m})<\infty. \]

Since the above sum is easily seen to be $\sum _{k=0}^{\infty }(k+1)\tau (x^{k}),$ the proof is complete.

The combination of Lemmas 3.2 and 3.4 reduces our problem to showing that if $x\in {\mathbb {R}}(G)$ with $\hat {x}\in \operatorname {Prob}(G),$ then $\tau ((({x^{*}+x})/{2})^{k})$ decays quickly. Recall that $\mu =\hat {x}$ is a probability measure on $G,$ so $\nu =({\mu ^{*}+\mu })/{2}$ is also a probability measure on $G,$ which is now symmetric. Given such a measure, one can form the random walk on $G$ which is a $G$-valued discrete time process $(X_{n})_{n=0}^{\infty }$ with $X_{0}=1$ and so that $X_{n+1}=X_{n}S_{n},$ where $(S_{n})_{n=1}^{\infty }$ are independent, random elements of $G$ each with distribution $\nu .$ It is easily seen that $\tau ((({x^{*}+x})/{2})^{k})$ is the probability that $X_{k}=1$, that is, that this random walk returns to the identity after $k$ steps. There are well-known results, due to Varopoulos, which give a precise relation between the decay rate of this probability and the growth of the group $G.$ Because of this, we easily obtain the following result.

Proof. Let $m=\tau (f),$ and write $f=m(1-x).$ In each case, we examine the invertibility of $(1-x).$

(i) This is well known, but we repeat the proof here. In this case, $\|\hat {x}\|_{1}<1,$ and so by standard Banach algebra theory we know that $1-\hat {x}$ has a $\ell ^{1}$-convolution inverse $\xi .$ By definition, this means that $(1-x)\xi =\xi (1-x)=\delta _{1}.$

(ii) By Lemmas 3.3 and 3.2, we may assume that $x=x^{*}$ and that $\hat {x}\geq 0.$ By Lemma 3.4, it suffices to show that $\sum _{k}k\tau (x^{k})<\infty .$ Set $S=\operatorname {supp}(\hat {x}).$ By assumption there is a constant $C>0$ with $|(S\cup \{1\})^{n}|\geq Cn^{5}$ for all $n\in {\mathbb {N}}.$ By [Reference VaropoulosVar94, Theorem 3] this implies that there is a constant $A>0$ so that $\tau (x^{k})\leq Ak^{-5/2}$ for all $k\in {\mathbb {N}}.$ Hence

\[ \sum_{k}k\tau(x^{k})\leq A\sum_{k}k^{{-}3/2}<\infty. \]

We remark that if $x=x^{*}\in {\mathbb {R}}(G)$ with $\hat {x}\in \operatorname {Prob}(G),$ and $\langle \operatorname {supp}(\hat {x}) \rangle$ has polynomial growth of degree $d,$ then there is a constant $B>0$ so that $\tau (x^{k})\geq Bk^{-d/2}$ for all $d\in {\mathbb {N}}$ (see [Reference AlexopoulosAle02, Corollary 1.9]). Thus, in the self-adjoint case, the assumption that $\langle \operatorname {supp}(\hat {x}) \rangle$ has either superpolynomial growth or polynomial growth of degree at least $5$ is optimal. Unfortunately, we do not know if the assumption that $\langle \operatorname {supp}(\hat {x}) \rangle$ has either superpolynomial growth or polynomial growth of degree at least $5$ is optimal in the case that $p$ is not self-adjoint. That is to say, it is possible that there is a group $G,$ and a $x\in {\mathbb {R}}(G)$ with $\hat {x}\in \operatorname {Prob}(G),$ so that $\langle \operatorname {supp}(\hat {x}) \rangle$ has polynomial growth of degree at most $4,$ and so that $1-x$ has an $\ell ^{2}$ formal inverse.

At this stage we have addressed why the growth assumption on $G$ is needed. In the next subsection we will address why the assumption of orderability of $G$ is important.

3.2 Orderability and Fourier transforms

In the previous section we saw that we could put mild assumptions on the growth of $G$ in order to guarantee that any semi-lopsided $f$ whose support generates $G$ has an $\ell ^{2}$ inverse. We thus turn to addressing the second part of exhibiting $m_{X_{f}}$ as a factor of a Bernoulli measure: finding conditions on $G,\nu$ so that the Fourier transform of $(\Theta _{\xi })_{*}(\nu ^{\otimes G})$ is $1_{{\mathbb {Z}}(G)f}.$

It may be helpful to sketch what the difficulty is here. Suppose $\nu \in \operatorname {Prob}({\mathbb {Z}})$ has mean zero and a finite second moment, and that $f$ has an $\ell ^{2}$ formal inverse $\xi .$ Set $\mu _{\xi }=(\Theta _{\xi })_{*}(\nu ^{\otimes G}).$ Let $\alpha \in {\mathbb {Z}}(G)$ with $\alpha \notin {\mathbb {Z}}(G)f$. Then

\[ \hat{\mu}_{\xi}(\alpha)=\prod_{g\in G}\hat{\nu}((\alpha \xi)(g)). \]

Note that this is an absolutely converging product. So this product will be zero if and only if $\hat {\nu }((\alpha \xi )(g_{0}))=0$ for some $g_{0}\in G.$ So we try to find such a $g_{0}$ (which will depend upon $\alpha$). As $\nu \in \operatorname {Prob}({\mathbb {Z}}),$ we know that $\hat {\nu }$ is identically $1$ on ${\mathbb {Z}},$ so we must necessarily find a $g_{0}$ so that $(\alpha \xi )(g_{0})\notin {\mathbb {Z}}.$ Fortunately, the fact that $\xi$ is an $\ell ^{2}$ formal inverse to $f$ and that $\alpha \notin {\mathbb {Z}}(G)f$ is enough to guarantee that there is a $g_{0}$ with $(\alpha \xi )(g_{0})\notin {\mathbb {Z}}.$ So now we have forced $\hat {\nu }((\alpha \xi )(g_{0}))$ to be less than $1$ in absolute value. Of course, this is not enough. We need to force $\hat {\nu }((\alpha \xi )(g_{0}))$ to be zero. So we need to find probability measures on ${\mathbb {Z}}$ whose Fourier transforms vanish reasonably often. Given $m\in {\mathbb {N}},$ it is not hard to exhibit a $\nu \in \operatorname {Prob}({\mathbb {Z}})$ with mean zero and finite second moment which has $\hat {\nu }=0$ on $(({1}/{m}){\mathbb {Z}})\cap {\mathbb {Z}}^{c}$, for example,

\[ \nu=\begin{cases} u_{\{{-}k,\ldots,k\}}, & \text{if }m=2k+1\text{ is odd},\\ u_{\{{-}m,\ldots,m\}{\setminus}\{0\}} & \text{if }m\text{ is even}. \end{cases} \]

So we need to ensure that there is some $g_{0}\in G$ so that the $(\alpha \xi )(g_{0})$ is not an integer, and that its denominator is ‘not too big’. The following lemma helps us in this regard by allowing us to assume that the coefficients of $\alpha$ are not very big.

Proof. Let $S=\operatorname {supp}(\hat {f}){\setminus} \{1\}.$ Let $\xi$ be a $c_{0}$ formal right inverse of $f$. Since $\alpha \xi \in c_{0}(G,{\mathbb {R}})$ we may write $\alpha \xi =x+\hat {c}$ where $x\in c_{0}(G)\cap [-1/2,1/2)^{G}$ and $c\in {\mathbb {Z}}(G).$ Right-multiplying by $f$, we have that

\[ \hat{\alpha}=xf+\widehat{cf}, \]

and this shows that $xf\in c_{0}(G,{\mathbb {R}})\cap {\mathbb {Z}}^{G}=c_{c}(G,{\mathbb {Z}}).$ So we may write $xf=\hat {\beta }$ for some $\beta \in {\mathbb {Z}}(G).$ We show that $\beta$ has the desired properties in each case.

(i) This case is divide into two subcases. First, suppose that $f$ is lopsided, so $\sum _{s\in S}|\hat {f}(s)|< m.$ Then for every $g\in G,$

\[ |\hat{\beta}(g)|=\biggl|mx(g)+m\sum_{s\in S}x(gs^{{-}1})\hat{f}(s)\biggr|\leq \frac{1}{2}\biggl(m+\sum_{s\in S}|\hat{f}(s)|\biggr)< m. \]

So $\hat {\beta }(g)\in (-m,m)\cap {\mathbb {Z}}=\{-(m-1),\ldots ,m-1\}.$ Thus $\|\hat {\beta }\|_{\infty }\leq m-1.$

If $f$ is not lopsided, then $\sum _{s\in S}|\hat {f}(s)|=m,$ and the fact that $\sum _{s\in S}\hat {f}(s)< m$ implies that we can choose an $s_{0}\in S$ so that $\hat {f}(s_{0})<0.$ Fix $g\in G.$ Then

\[ \hat{\beta}(g)=mx(g)+\sum_{s\in S}x(gs^{{-}1})\hat{f}(s)<\frac{m}{2}+\frac{1}{2}\sum_{s\in S}|\hat{f}(s)|=m. \]

Additionally,

\[ \hat{\beta}(g)=mx(g)+x(gs_{0}^{{-}1})\hat{f}(s_{0})+\sum_{s\in S,s\ne s_{0}}x(gs^{{-}1})\hat{f}(s). \]

The fact that $\hat {f}(s_{0})<0$ and $x(gs_{0}^{-1})\in [-1/2,1/2)$ implies that $x(gs^{-1})\hat {f}(s_{0})>-\frac {1}{2}|\hat {f}(s_{0})|.$ Thus

\[ \hat{\beta}(g)>{-}\frac{m}{2}-\frac{1}{2}|\hat{f}(s_{0})|-\frac{1}{2}\sum_{s\in S,s\ne s_{0}}|\hat{f}(s)|={-}m. \]

So $-m<\hat {\beta }(g)< m$ and as in the first subcase the fact that $\hat {\beta }$ is integer-valued implies that $\|\hat {\beta }\|_{\infty }\leq m-1.$

(ii) In this case, we must have that $\hat {f}(s)>0$ for every $s\in S.$ Fix $g\in G.$ Since $\hat {f}(s)>0$ for every $s\in S,$

\[ \hat{\beta}(g)=mx(g)+\sum_{s\in S}x(gs^{{-}1})\hat{f}(s)<\frac{1}{2}\biggl(m+\sum_{s\in S}\hat{f}(s)\biggr)=m. \]

Similarly,

\[ \hat{\beta}(g)\geq \frac{-m}{2}-\frac{1}{2}\sum_{s\in S}\hat{f}(s)={-}m. \]

So $\hat {\beta }(g)\in [-m,m)\cap {\mathbb {Z}}=\{-m,\ldots ,m-1\}.$ It simply remains to show that if $\hat {\beta }(g)=-m,$ then $\hat {\beta }(gs^{-1})<0$ for every $s\in S$.

So suppose that $\hat {\beta }(g)=-m.$ Then

\[ \hat{\beta}(g)=mx(g)+\sum_{s\in S}x(gs^{{-}1})\hat{f}(s). \]

Since $\hat {f}(s)> 0$ for all $s\in S,$ $\sum _{s\in S}\hat {f}(s)=m,$ and $x\in [-1/2,1/2)^{G},$ the fact that $\hat {\beta }(g)=-m$ forces $x(gs^{-1})=-1/2$ for every $s\in S.$ Now fix $s\in S.$ Using that $\hat {f}(t)>0$ for all $t\in S,$ and that $x\in [-1/2,1/2)^{G},$ we have

\[ \hat{\beta}(gs^{{-}1})={-}\frac{m}{2}+\sum_{t\in S}x(gs^{{-}1}t^{{-}1})\hat{f}(t)<{-}\frac{m}{2}+\frac{1}{2}\sum_{t\in S}\hat{f}(t)=0. \]

So $\hat {\beta }(gs^{-1})<0.$

We now explicitly discuss where orderability comes into play. We will work with something slightly more general than a left-invariant total order on $G.$

For example, it is easy to exhibit a left-invariant partial order on ${\mathbb {F}}_{2}=\langle a,b \rangle .$ Let $P$ be the set of elements of ${\mathbb {F}}_{2}$ which are products of $a,b$ (no $a^{-1},b^{-1}$ occur in its word decomposition), with the convention that $1\notin P.$ We can then define a partial order $\preceq$ by demanding that $h_{1}\preceq h_{2}$ if and only if $h_{1}^{-1}h_{2}\in P\cup \{1\}.$ This is a partial order on ${\mathbb {F}}_{2},$ and it makes the generators order positive (i.e. larger than $1$ in this partial order). It is a fact that there is a left-invariant total order on ${\mathbb {F}}_{2}$ which make the generators order positive, but it is harder to construct.

To prove the claim, we first note that if (a) holds then the claim is trivial, so we may assume that (b) holds. Note that $(gs^{-1})^{-1}g=s\in P$ for every $s\in S.$ So $gs^{-1}\prec g$ and $gs^{-1}\ne g$ for every $s\in S.$ So by minimality of $g$ we must have that $\hat {\alpha }(gs^{-1})=0$ for every $s\in S$. So (b) now implies that $|\hat {\alpha }(g)|\leq m-1.$

We now return to the proof of the lemma. Let $x=-({1}/{m})\sum _{s\in S}\hat {f}(s)s,$ so $f=m(1-x).$ Then by Lemma 3.3 we have that

\[ (\alpha\xi)(g)=\frac{\hat{\alpha}(g)}{m}+\frac{1}{m}\sum_{n=1}^{\infty}(\alpha x^{n})(g). \]

Fix $n\geq 1$. Then

\[ (\alpha x^{n})(g)=\sum_{s\in S^{n}}\alpha(gs^{{-}1})x^{n}(s). \]

Let $s\in S^{n}.$ Then $s\in P$ and $s\ne 1$ since $S\subseteq \{h\in H:h\succ 1\}.$ Thus $(gs^{-1})^{-1}g=s\in P.$ So $gs^{-1}\preceq g$ and $gs^{-1}\ne g,$ since $s\ne 1.$ By minimality we thus have that $\alpha (gs^{-1})=0$ for all $s\in S^{n},$ and thus $(\alpha x^{n})(g)=0$ for all $n\geq 1.$ So by the claim,

\[ (\alpha\xi)(g)=\frac{\hat{\alpha}(g)}{m}\in \biggl\{-\frac{(m-1)}{m},-\frac{(m-2)}{m},\ldots,\frac{m-1}{m}\biggr\}. \]

Since $g\in \operatorname {supp}(\hat {\alpha }),$ it follows that

\[ (\alpha\xi)(g)\in \biggl\{-\frac{(m-1)}{m},-\frac{(m-2)}{m},\ldots,\frac{m-1}{m}\biggr\}{\setminus} \{0\}\subseteq \frac{1}{m}{\mathbb{Z}}\cap {\mathbb{Z}}^{c}.\]

Proof. Let $\xi$ be the $\ell ^{2}$ formal inverse to $f,$ and $m=\tau (f).$ First assume that $m$ is an odd integer, and write $m=2k+1.$ We can let $\Theta _{\xi }\colon \{-k,\ldots ,k\}^{G}\to {\mathbb {T}}^{G}$ be defined as in Theorem 2.1 corresponding to $\nu =u_{\{-k,\ldots ,k\}}.$ Let $\mu =(\Theta _{\xi })_{*}(\nu ^{\otimes G}).$ So for every $\alpha \in {\mathbb {Z}}(G)$,

\[ \hat{\mu}(\alpha)=\prod_{g\in G}\hat{\nu}((\alpha\xi)(g))=\prod_{g\in G}\frac{\sin(m\pi(\alpha\xi)(g))}{m\sin(\pi(\alpha\xi)(g))}. \]

Suppose $\alpha \in {\mathbb {Z}}(G)f,$ and write $\alpha =\beta f.$ Then $\alpha \xi =\hat {\beta }\in c_{c}(G,{\mathbb {Z}}).$ Hence we have $(\alpha \xi )(g)\in {\mathbb {Z}}$ for all $g\in G.$ Since $\hat {\nu }(l)=1$ for every $l\in {\mathbb {Z}},$ we have that $\hat {\mu }(\alpha )=1.$ Suppose that $\alpha \in {\mathbb {Z}}(G),$ but $\alpha \notin {\mathbb {Z}}(G)f.$ Then by Lemma 3.8, there is some $g\in G$ so that $(\alpha \xi )(g)\in ({1}/{m}){\mathbb {Z}} \cap {\mathbb {Z}}^{c}.$ So $\hat {\nu }((\alpha \xi )(g))=0$ (since $\hat {\nu }(t)={\sin (m\pi t)}/{m\sin (\pi t)}$), and thus $\hat {\mu }(\alpha )=0.$ Hence we know that $\hat {\mu }=1_{{\mathbb {Z}}(G)f},$ and this is equivalent to saying that $\mu =m_{X_{f}}.$

Now assume that $m$ is even. Let $\nu =u_{\{-m,\ldots ,m\}{\setminus} \{0\}}$. Then

\[ \hat{\nu}(t)=\begin{cases} 1, & \text{if }t\in {\mathbb{Z}},\\ \dfrac{1}{2m}\biggl(\dfrac{\sin(\pi(2m+1)t)}{\sin(\pi t)}-1\biggr), & \text{if }t\notin {\mathbb{Z}}. \end{cases} \]

Moreover, it is direct to check that $\nu$ has mean zero. Let $\Theta _{\xi }\colon (\{-m,\ldots ,m\}{\setminus} \{0\})^{G}\to {\mathbb {T}}^{G}$ be defined as in Theorem 2.1 for this $\nu$, and set $\mu =(\Theta _{\xi })_{*}(\nu ^{\otimes G}).$ Then

\[ \hat{\mu}(\alpha)=\prod_{g\in G}\hat{\nu}((\alpha\xi)(g)) \]

for all $\alpha \in {\mathbb {Z}}(G).$ First suppose that $\alpha \in {\mathbb {Z}}(G)f.$ Then as in the case that $m$ is odd, we know that $\alpha \xi \in c_{c}(G,{\mathbb {Z}}),$ and thus $\hat {\mu }(\alpha )=1.$ Now assume that $\alpha \in {\mathbb {Z}}(G),$ but that $\alpha \notin {\mathbb {Z}}(G)f.$ As in the case that $m$ is odd, we may find a $g\in G$ so that $(\alpha \xi )(g)\in ({1}/{m}){\mathbb {Z}}\cap {\mathbb {Z}}^{c}.$ For such a $g$ we have that $\hat {\nu }((\alpha \xi )(g))=0,$ and thus $\hat {\mu }(\alpha )=0.$ So $\hat {\mu }=1_{{\mathbb {Z}}(G)f},$ and thus $\mu =m_{X_{f}}.$

As we will see in the next section, our assumptions imply that the kernel $N$ of $G\curvearrowright X_{f}$ is finite, and that $G/N\curvearrowright (X_{f},m_{X_{f}})$ is essentially free. Thus, by [Reference Ornstein and WeissOW87], if $G$ is amenable, then we can say that $G/N\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift.

4.1 The amenable case

Corollary 3.10 is most striking when $G$ is amenable, since in this case being a factor of a Bernoulli shift implies that the action is a Bernoulli shift (at least when one quotients by the kernel of the action). This is a consequence of Ornstein theory. However, as Ornstein theory only applies to free actions of groups we should first observe that the actions we are considering are free after modding out by the kernel.

So in the amenable case, we can always conclude that if $G\curvearrowright (X_{f},m_{X_{f}})$ is a factor of a Bernoulli shift. In most of our examples, $G$ is additionally torsion-free. In this case, we have that $N$ is trivial, that is, the algebraic action is faithful. If $G$ is assumed torsion-free, there is a more elementary proof that $G\curvearrowright (X_{f},m_{X_{f}})$ is essentially free by appealing to [Reference HayesHay19a, Appendix A]. This does not use the Feit–Thompson theorem as in [Reference Tucker-DrobTuc15]. By Proposition 4.1 and the extension of Ornstein theory to amenable groups (see [Reference Ornstein and WeissOW87]), if $G$ is amenable and $G\curvearrowright (X_{f},m_{X_{f}})$ is a factor of a Bernoulli shift, and if $N$ is the kernel of the action $G\curvearrowright X_{f},$ then $G/N\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to a Bernoulli shift (this follows from [Reference Ornstein and WeissOW87, III.4 Proposition 1, III.6 Theorem 2, III.5 Corollary 5]). We will give several examples of Bernoulli principal algebraic actions of amenable groups shortly, but let us first point out explicitly here that this annoyance with finite normal subgroups can occur.

Proposition 4.2 Let $G$ be a countable, discrete group and let $N\triangleleft G$ be finite with $N\ne \{1\}.$ Let $b$ be an integer, and let

\[ f=1+b\sum_{n\in N}n. \]

Then $f$ has an inverse in ${\mathbb {Q}}(G),$ and $N$ is the kernel of $G\curvearrowright X_{f}.$ Further, the induced action $G/N\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to the Bernoulli action $G/N\curvearrowright (\{1,\ldots ,1+b|N|\},u_{1+b|N|})^{G/N}.$

Proof. Let $x=\sum _{n\in N}n,$ $e=({1}/{|N|})x,$ so $e^{2}=e$ and we can write $f$ as

\[ f=(1-e)+(1+b|N|)e. \]

Simple calculations show that $f$ has an inverse in ${\mathbb {Q}}(G)$ given by

\[ \phi=(1-e)+(1+b|N|)^{{-}1}e. \]

Moreover, it is easy to check that the normality of $N$ implies that $f,$ and thus $\phi ,$ is central in ${\mathbb {Q}}(G).$

To check that $N$ acts trivially on $X_{f},$ let $n\in {\mathbb {N}}.$ It then suffices to show that $(n-1){\mathbb {Z}}(G)\subseteq {\mathbb {Z}}(G)f.$ So let $\alpha \in {\mathbb {Z}}(G).$ Then by direct calculation,

\[ (n-1)\alpha\phi=(n-1)\phi\alpha=(n-1)\alpha\in{\mathbb{Z}}(G). \]

So $(n-1)\alpha =(n-1)\alpha \phi f\in {\mathbb {Z}}(G)f.$ So $N$ acts trivially on $X_{f}.$

It simply remains to prove that $G/N\curvearrowright (X_{f},m_{X_{f}})$ is Bernoulli. Let $J$ be the left ideal in ${\mathbb {Z}}(G)$ generated by $\{n-1:n\in {\mathbb {N}}\}$; by normality of $N$ this is in fact a two-sided ideal. It is easy to see that we have an isomorphism of ${\mathbb {Z}}(G)$ modules $\Psi \colon {\mathbb {Z}}(G)/J\to {\mathbb {Z}}(G/N)$ given by $(\Psi (\alpha +J))^{\widehat {}}(gN)=(\alpha x)^{\widehat {}}(g)$. Since we already saw that ${\mathbb {Z}}(G)f\supseteq J,$ we get a natural ${\mathbb {Z}}(G)$-modular surjection $\pi \colon {\mathbb {Z}}(G)/J\to {\mathbb {Z}}(G)/{\mathbb {Z}}(G)f,$ and so we have a natural ${\mathbb {Z}}(G)$-modular surjection ${\mathbb {Z}}(G/N)\to {\mathbb {Z}}(G)/{\mathbb {Z}}(G)f$ given by $q=\pi \circ \Psi ^{-1}.$

It remains to compute the kernel of $q.$ Note that if $\alpha \in {\mathbb {Z}}(G),$ then $\alpha \in {\mathbb {Z}}(G)f$ if and only if $\alpha \phi \in {\mathbb {Z}}(G).$ By a direct computation, $\phi =1-({b}/({1+b|N|}))x.$ So

\[ \alpha\phi=\alpha-\frac{b}{1+b|N|}\alpha x \]

and thus $\alpha \phi \in {\mathbb {Z}}(G)$ if and only if $({b}/({1+b|N|}))\alpha x\in {\mathbb {Z}}(G).$ By definition, this means that $({b}/({1+b|N|}))\Psi (\alpha )\in {\mathbb {Z}}(G/N).$ Since $b,1+b|N|$ are easily seen to be relatively prime, this is the same as saying that $\Psi (\alpha )\in (1+b|N|){\mathbb {Z}}(G/N).$ So we have shown that $\ker (q)=(1+b|N|){\mathbb {Z}}(G/N).$

So we have exhibited an isomorphism ${\mathbb {Z}}(G)/{\mathbb {Z}}(G)f\cong {\mathbb {Z}}(G/N)/(1+b|N|){\mathbb {Z}}(G/N)$ of ${\mathbb {Z}}(G)$-modules, and applying Pontryagin duality shows that

\[ G\curvearrowright (X_{f},m_{X_{f}})\cong G\curvearrowright ({\mathbb{Z}}/(1+b|N|){\mathbb{Z}},u_{{\mathbb{Z}}/(1+b|N|){\mathbb{Z}}})^{G/N}. \]

So $G/N\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to the Bernoulli shift $G/N\curvearrowright (\{1,\ldots ,1+b|N|\}, u_{1+b|N|})^{G/N}.$

Let us proceed to give several examples of semi-lopsided elements whose corresponding actions are Bernoulli. In many examples our groups will be torsion-free, and so it is a consequence of Proposition 4.1 that the actions we are considering are essentially free. We will thus not explicitly reference that these actions are essentially free before applying Ornstein theory. In later examples, we will consider groups with torsion and will give an explicit argument that these actions are faithful (and thus free by Proposition 4.1).

Since we know in these examples that the actions are isomorphic to Bernoulli, it is nice to know what Bernoulli shift they are. Since Bernoulli shifts over amenable groups are completely classified by their entropy by [Reference OrnsteinOrn70a, Reference OrnsteinOrn70b, Reference Ornstein and WeissOW87] (see [Reference BowenBow10, Reference BowenBow12, Reference SewardSew18, Reference StepinSte75] for the more general fact that Bernoulli shifts over sofic groups are completely classified their entropy), once we compute the entropy of these actions we will know what Bernoulli shift they are isomorphic to. By [Reference HayesHay16, Proposition 2.2] we know that once $f$ has an $\ell ^{2}$ formal inverse, then $f$ is injective as a convolution operator $\ell ^{2}(G)\to \ell ^{2}(G).$ So by [Reference Li and ThomLT14] (see also [Reference HayesHay16] for the sofic case), the entropy of $G\curvearrowright (X_{f},m_{X_{f}})$ is $\log \operatorname {det}_{L(G)}(f).$ In Appendix A.3 (see Corollary A.15) it is shown that if $f\in {\mathbb {Z}}(G)$ is semi-lopsided, and if $\operatorname {supp}(\hat {f})=\{1\}\cup S$, where $S\subseteq P$ for some positive semigroup $P\subseteq G$, then $\log \operatorname {det}_{L(G)}(f)=\log \tau (f)$. So in all of the examples we are considering, we know that the entropy of $G\curvearrowright (X_{f},m_{X_{f}})$ is $\log \tau (f).$

Example 1 Let $G$ be the Heisenberg group of upper triangular matrices with integer entries and all diagonal entries $1.$ Then $G$ has polynomial growth of order 4 by the Bass–Guivarc'h formula. So our results only allow us to use lopsided elements. The group $G$ has a presentation $G=\langle a,b,c:[c,a]=1,[c,b]=1,[a,b]=c \rangle .$ Every element of $G$ can be uniquely represented as $a^{n}b^{m}c^{k}$ for integers $n,m,k\in {\mathbb {Z}}.$ Consider the following three conditions:

1. 1. $n_{1}<n_{2}$;

2. 2. both $n_{1}=n_{2}$ and $m_{1}<m_{2}$;

3. 3. all three of $n_{1}=n_{2}$ and $m_{1}=m_{2}$ and $k_{1}<k_{2}$.

If any one of 1–3 hold, then $x^{n_{1}}y^{m_{1}}z^{k_{1}}<x^{n_{2}}y^{n_{2}}z^{k_{2}}$.

It can be checked that this is a total order which is left- and right-invariant. By Corollary 3.10 and Ornstein theory, if

• – $f=k+na+mb$ with $n,m,k\in {\mathbb {Z}}$ and $|n|+|m|< k$, or

• – $f=k+na+mb+lc$ with $n,m,l,k\in {\mathbb {Z}}$ and $|n|+|m|+|l|< k$.

then $G\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to $G\curvearrowright (\{1,\ldots ,k\},u_{k})^{G}.$ Since we are only considering lopsided elements, if we set $P=\{g\in G:g>1\},$ then we can in fact take any pairwise distinct $x_{1},\ldots ,x_{k}\in P,$ any $m,n_{1},\ldots ,n_{k}\in {\mathbb {Z}}$ with $\sum _{j}|n_{j}|< m,$ and then

\[ f=m+\sum_{j=1}^{k}n_{j}x_{j} \]

will be such that $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (m),$ provided $k\geq 2$. The reason for the restriction $k\geq 2$ is that once $k\geq 2$ we have that $\langle \{x_{i}^{-1}x_{j}:1\leq i,j\leq k\} \rangle$ is a nontrivial subgroup of $G,$ and as such this subgroup is infinite because $G$ is torsion-free.

Example 2 We can generalize the previous example slightly. For $N\geq 1,$ let $H_{N}$ be the group of upper triangular $(N+2)\times (N+2)$ matrices with integer entries, $1$s on the diagonal, and so that all other nonzero entries are on the first row or the last column. For $1\leq j,k\leq n,$ let $E_{j,k}$ be the matrix defined by $(E_{j,k})_{rs}=\delta _{r=j}\delta _{k=s}.$ Define $c\in G$ by $c=\operatorname {id}+E_{1,N+2},$ and for $1\leq i\leq N,$ define $a_{i},b_{i}$ by $a_{i}=\operatorname {id}+E_{1,i+1},$ $b_{i}=\operatorname {id}+E_{i+1,N+2}.$ Then:

• – $\langle a_{1},\ldots ,a_{N},b_{1},\ldots ,b_{N},c \rangle =G;$

• – $\langle a_{1},\ldots ,a_{N} \rangle ,\langle b_{1},\ldots ,b_{N} \rangle$ are abelian;

• – $c$ is central in $G;$

• – $[a_{i},b_{j}]=1$ if $i\ne j;$

• – $[a_{i},b_{i}]=c.$

Every element of $g\in G$ can be uniquely represented as

\[ a_{1}^{n_{1}}b_{1}^{n_{2}}a_{2}^{n_{3}}b_{2}^{n_{4}}\cdots a_{N}^{n_{2N-1}}b_{n}^{n_{2N}}c^{n_{2N+1}} \]

for integers $n_{1},\ldots ,n_{N},$ $m_{1},\ldots ,m_{N},$ $l\in {\mathbb {Z}}.$ We can order $G$ lexicographically as before:

\[ a_{1}^{n_{1}}b_{1}^{n_{2}}a_{2}^{n_{3}}b_{2}^{n_{4}}\cdots a_{N}^{n_{2N-1}}b_{N}^{n_{2N}}c^{n_{2N+1}}< a_{1}^{m_{1}}b_{1}^{m_{2}}a_{2}^{m_{3}}b_{2}^{m_{4}}\cdots a_{N}^{m_{2N-1}}b_{N}^{m_{2N}}c^{m_{2N+1}} \]

if $n_{j}< m_{j}$ when $1\leq j\leq 2N+1$ is the minimal index such that $n_{j}\ne m_{j}.$ It can again be checked that this is both left- and right-invariant. If we set $P=\{g\in G:g>1\},$ then as before we can take $x_{1},\ldots ,x_{k}\in P$, $m,m_{1},\ldots ,m_{k}\in {\mathbb {Z}}$ with $\sum _{j}|m_{j}|< m$ and $f=m+\sum _{j}m_{j}x_{j}$ will have $G\curvearrowright (X_{f},m_{X_{f}})$ being a Bernoulli shift with entropy $\log (m).$

The growth of $H_{N}$ is $2(N+1)$ by the Bass–Guivarc’h formula, and so once $N\geq 2$ we even have semi-lopsided but not lopsided examples. For example, with

\[ f=(2N+1)-\biggl(\sum_{j=1}^{N}a_{j}+\sum_{j=1}^{N}b_{j}+c\biggr) \]

we have that $G\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to a Bernoulli shift with base entropy $\log (2N+1)$ (provided $N\geq 2$). As we show in Appendix B.1, since $f$ is well balanced, we know that $f$ does not have an $\ell ^{1}$ inverse. We show there as well, using that $G$ is amenable, that $\lambda (f)$ is not invertible.

We can of course add signs here and consider, for example,

\[ f=(2N+1)-\biggl(\sum_{j=1}^{N}\varepsilon_{j}a_{j}+\sum_{j=1}^{N}\sigma_{j}b_{j}+\alpha c\biggr) \]

for any $\varepsilon _{1},\ldots ,\varepsilon _{N},$ $\sigma _{1},\ldots ,\sigma _{N}$, $\alpha$ in $\{\pm 1\}.$ We then still have that $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (2N+1).$ Since $c\in \langle a_{1},\ldots ,a_{N},b_{1},\ldots ,b_{N} \rangle ,$ we can similarly consider

\[ f=m+\biggl(\sum_{j=1}^{N}m_{j}a_{j}+\sum_{j=1}^{N}l_{j}b_{j}\biggr) \]

where $\sum _{j=1}^{N}|m_{j}|+\sum _{j=1}^{N}|l_{j}|\leq m,$ and $m_{1},\ldots ,m_{N},\ l_{1},\ldots , l_{N}$ are not zero. Then $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (m).$

Similar examples can be given by taking products of the generators. For example, if $N=2,$ we may consider

\[ f=4-(a_{1}+a_{2}+a_{1}b_{1}+a_{2}b_{2}). \]

Suppose that

is an exact sequence of groups. Then if $H,K$ are equipped with left-invariant partial orders, we can equip $G$ with a left-invariant partial order as follows. We say that $g_{1}\preceq g_{2}$ if either

• – $\pi (g_{1}),\pi (g_{2})$ are comparable and $\pi (g_{1})\prec \pi (g_{2})$, or

• – $g_{1}^{-1}g_{2}\in H$ with $g_{1}^{-1}g_{2}\succeq 1.$

In particular, if $K,H$ are left-orderable, then $G$ is left-orderable.

Example 4 Fix an $n\in {\mathbb {N}},n\geq 2$, and let $G=BS(1,n)=\langle a,b:aba^{-1}=b^{n} \rangle .$ There is a natural map $BS(1,n)\to {\mathbb {Z}}$ given by $a\mapsto 1,b\mapsto 0,$ and it is direct to see that the kernel is isomorphic to ${\mathbb {Z}}(1/n).$ Thus $G$ is left-orderable and there is a left-invariant ordering $<$ on $BS(1,n)$ which makes $a,b>1.$ So if we set

\[ f=m+na+kb \]

with $m,n,k$ nonzero integers such that $|n|+|k|\leq m,$ then $G\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to a Bernoulli shift with entropy $\log (m).$

Example 5 Suppose that $G$ is polycyclic. This means that we have a chain of groups

\[ \{1\}=G_{0}\triangleleft G_{1}\triangleleft G_{2}\cdots G_{n-1}\triangleleft G_{n}=G \]

so that $G_{i}/G_{i-1}\cong {\mathbb {Z}}$ for all $1\leq i\leq n.$ Suppose we choose $x_{i}\in G_{i}$ so that $G_{i}=x_{i}G_{i-1}.$ Then it is possible to find a left-invariant order $<$ on $G$ so that $x_{i}>1$ for all $i.$ So if we choose $(n_{i})_{i=1}^{k}\in ({\mathbb {Z}}{\setminus} \{0\})^{k},$ $m\in {\mathbb {Z}}$ with $\sum _{i=1}^{k}|n_{i}|< m,$ then $f=m+\sum _{i}n_{i}x_{i}$ has $G\curvearrowright (X_{f},m_{X_{f}})$ isomorphic to a Bernoulli shift with entropy $\log (m).$ If $G$ is either superpolynomial growth, or polynomial growth of degree at least $5,$ then we can allow $\sum _{i=1}^{k}|n_{i}|\leq m$ to force $G\curvearrowright (X_{f},m_{X_{f}})$ to be isomorphic to a Bernoulli shift with entropy $\log (m).$

It often happens that $G$ has exponential growth. For example, suppose that $A\in SL_{2}({\mathbb {Z}})$ has no eigenvalues on the unit circle. Then ${\mathbb {Z}}^{2}\rtimes _{A}{\mathbb {Z}}$ is of exponential growth. Let $a=(e_{1},0),b=(e_{2},0),c=(0,1)$. Then for any $l,n,k,m\in {\mathbb {Z}}{\setminus} \{0\}$ with $|n|+|k|\leq m,$ setting $f=m-(la+nb+kc)$, we have that $G\curvearrowright (X_{f},m_{X_{f}})$ is isomorphic to a Bernoulli shift with entropy $\log (m).$

Suppose $X$ is a Polish space, and that $\preceq$ is a partial order on $X$ so that $\{y\in X:y\preceq x\}$ is closed for every $x\in X.$ Suppose that $G\curvearrowright X$ faithfully by order-preserving homeomorphisms. Then we can define a left-invariant partial order on $G$ as follows. Let $(x_{n})_{n}$ be a dense sequence in $X$. We then say that $g\prec h$ if

• – $\{n:gx_{n},hx_{n} \text { are comparable and not equal}\}$ is not empty;

• – if $m=\min \{n:g x_{n},hx_{n}\text { are comparable and not equal}\}$, then $g x_{m}\prec h x_{m}.$

If $\preceq$ is a total order on $X,$ then it is easy to check that this gives a total order on $G.$

It is a folklore result that this construction characterizes left-orderable groups: namely, a countable group is left-orderable if and only if it embeds into the group of order-preserving homeomorphisms of ${\mathbb {R}}.$ See, for example, [Reference GhysGhy01, Theorem 6.8].

The order described above seems fairly abstract. However, since we are allowed to prescribe the first few terms of our sequence $(x_{n})_{n},$ it makes it relatively straightforward to construct orders which make certain generators bigger than $1$.

Example 6 All the preceding examples were solvable. We can consider non-solvable examples, in fact groups of intermediate growth. Let $a,b,c,d$ be the homeomorphisms of ${\mathbb {Z}}^{{\mathbb {N}}}$ defined recursively by

\[ a(x_{1},x_{2},x_{3},\ldots)=(1+x_{1},x_{2},\ldots), \]

\[ b(x_{1},x_{2},\ldots)=\begin{cases} (x_{1},a(x_{2},x_{3},\ldots)), & \text{if }x_{1}\text{ is even},\\ (x_{1},c(x_{2},x_{3},\ldots)), & \text{if }x_{1}\text{ is odd}, \end{cases} \]

\[ c(x_{1},x_{2},\ldots)=\begin{cases} (x_{1},a(x_{2},x_{3},\ldots)), & \text{if }x_{1}\text{ is even},\\ (x_{1},d(x_{2},x_{3},\ldots)), & \text{if }x_{1}\text{ is odd}, \end{cases} \]

\[ d(x_{1},x_{2},\ldots)=\begin{cases} (x_{1},x_{2},x_{3},\ldots), & \text{if }x_{1}\text{ is even},\\ (x_{1},b(x_{2},x_{3},\ldots)), & \text{if }x_{1}\text{ is odd} . \end{cases} \]

Let $G$ be the group generated by $a,b,c,d.$ Notice that $G$ preserves the lexographic ordering $<$ on ${\mathbb {Z}}^{{\mathbb {N}}}.$ Fix a dense sequence $(a_{n})_{n=1}^{\infty }$ in ${\mathbb {Z}}^{{\mathbb {N}}}$ with $a_{1}=(0,0,\ldots ,0), a_{2}=(1,0,\ldots ,0).$ Define an order $<$ on $G$ by saying that $g< h$ if when we set $n=\min \{l\in {\mathbb {N}}:g(a_{l})\ne h(a_{l})\},$ then $g(a_{n})< h(a_{n}).$ Then $<$ is a left-invariant order on $G$ which makes $a,b,c,d$ all bigger than $1.$ Also by [Reference GrigorchukGri85] this group has intermediate growth. So for any $m,m_{1},m_{2},m_{3},m_{4}\in {\mathbb {Z}}{\setminus} \{0\}$ with $\sum _{j=1}^{4}|m_{j}|\leq m,$ and with $f=m+m_{1}a+m_{2}b+m_{3}c+m_{4}d,$ we have that $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (m).$

This group was defined in [Reference Grigorchuk and MachíGM93], and its orderability was first shown there. We have followed the exposition in [Reference NavasNav08, § 1.1].

Our ability to use partial orders is in fact nontrivial, and we can construct examples where the acting group is not torsion-free (and thus not left-orderable). For this, it will be helpful to switch to positive semigroups instead of partial orders.

Suppose $H$ is a group equipped with a left-invariant partial order $\preceq ,$ and let $P$ be the corresponding positive semigroup. Suppose that $K\curvearrowright H$ by automorphisms and that $K\cdot P\subseteq P.$ We can then define a positive semigroup $P_{G}$ in $G=H\rtimes K$ by $P_{G}=\{(h,k):h\in P,k\in K\}.$ From this positive semigroup we get a left-invariant partial order as described before. In many cases, $K$ is finite and so $G$ is not left-orderable.

A particular example is the case of generalized wreath products. For a group $H$ and a set $I,$ we will use $H^{\oplus I}=\{h\in H^{I}:h(i)=1\text { for all but finitely many }i\}.$ Let $K$ be a group acting on a set $I,$ and $H$ another group. We then let $K\curvearrowright H^{\oplus I}$ by permuting the coordinates of $H$ (using the action $H\curvearrowright I$). The generalized wreath product $H\wr _{I}K$ is then the semidirect product $H^{\oplus I}\rtimes K.$ Suppose that $H$ has a left-invariant partial order $\preceq .$ We may then define a partial order $\preceq$ on $H^{\oplus I}$ by saying that if $h=(h_{i})_{i\in I},h'=(h'_{i})_{i\in I}\in H^{\oplus I},$ then $h\preceq k$ if and only if $h_{i}\preceq h'_{i}$ for all $i\in I.$ This is clearly invariant under the action of $K$ on $H$ for all $i\in I,$ and thus induces an order on $H\wr _{I}K$ by the above construction. Since this construction sometimes produces groups which are not torsion-free, we need to take some care in applying Proposition 4.1 to have our actions be essentially free. The following lemma will do most of the work for us. This lemma is surely well known, but we include a proof for completeness.

We will use this to give one more example of a Bernoulli shift where the acting group is torsion-free, and then one more where it is not.

Let $S_{k}$ act on $\{1,\ldots ,k\}$ in the natural way. In $G_{0}={\mathbb {Z}}\wr _{\{1,\ldots ,k\}} S_{k}$ one has a left-invariant partial order $\preceq$ so that $\{g\in G_{0}:g\succ 1\}=\{(x,\sigma ):x\in ({\mathbb {N}}\cup \{0\})^{k},x\ne 0,\sigma \in S_{k}\}.$ So if $H\leq S_{k}$ is arbitrary, we may consider $G={\mathbb {Z}}^{k}\rtimes H$ as a group with a left-invariant partial order. So we can obtain similar modifications of Example 8. Let $S$ be a set of generators for $H,$ and let $\mathcal {O}\subseteq ({\mathbb {N}}\cup \{0\})^{k}{\setminus} \{0\}$ be such that the group generated by $H\mathcal {O}$ is all of ${\mathbb {Z}}^{k}.$ Let $(m_{x})_{x\in O},(l_{y,s})_{(y,s)\in \mathcal {O}\times S},m\in ({\mathbb {Z}}{\setminus} \{0\})^{k}$ be such that $\sum _{x}|m_{x}|+\sum _{y,s}|l_{y,s}|\leq m.$ Set

\[ f=m+\sum_{x}m_{x}(x,\operatorname{id})+\sum_{(y,s)\in \mathcal{O}\times S}l_{y,s}(y,s)\in {\mathbb{Z}}(G). \]

If $k\geq 5,$ then $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (m).$ If $k\leq 4,$ then as long as we assume that $\sum _{x}|m_{x}|+\sum _{y,s}|l_{y,s}|< m,$ we still have that $G\curvearrowright (X_{f},m_{X_{f}})$ is a Bernoulli shift with entropy $\log (m).$

Of course the possibilities here are endless, and one can consider other wreath products $H\wr _{I}K$ where $H$ is left-orderable. For example, one can take $H$ to be the Heisenberg group, or other polycyclic groups.

4.2 Nonamenable examples

More generally, any residually free group is left-orderable. This includes fundamental groups of compact surfaces without boundary whose genus is larger than $1$. Additionally, if $G$ is residually free, and we have an explicit family of homomorphisms $\pi _{n}\colon G\to {\mathbb {F}}_{r(n)}$ for integers $r(n)\in {\mathbb {N}}$ so that $\bigcap _{n}\ker (\pi _{n})=\{1\},$ then we get an explicit left-invariant partial order on $G.$ Thus in many cases, we can explicitly describe semi-lopsided elements $f\in {\mathbb {Z}}(G)$ so that $G\curvearrowright (X_{f},m_{X_{f}})$ is a factor of a Bernoulli shift. More generally by [Reference VinogradovVin49], we also have that free products of left-orderable groups are left-orderable.

Example 10 For an integer $n\geq 3,$ consider the braid group $B_{n}$ which has the following presentation:

\[ B_{n}= \langle \sigma_{1},\ldots,\sigma_{n-1}:\sigma_{i}\sigma_{i+1}\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1},\text{ for }1\leq i\leq n-2,\text{ and } \sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}\text{ if }|i-j|\geq 2 \rangle. \]

Dehornoy proved (see [Reference DehornoyDeh94]) that $B_{n}$ has a left-invariant order $<$ which is now called the Dehornoy order. This ordering is uniquely defined by saying that for all $1\leq i\leq n-2$ we have $\beta _{0}\sigma _{i}\beta _{1}>1$ for all $\beta _{0},\beta _{1}\in \langle \sigma _{i+1},\ldots ,\sigma _{n-1} \rangle .$ Let $m,m_{1},\ldots ,m_{n-1}\in {\mathbb {Z}}{\setminus} \{0\},$ and $f=m+\sum _{j=1}^{n-1}m_{j}\sigma _{j}.$ Since $n\geq 3,$ we know that $B_{n}$ contains a free group on two generators and thus has exponential growth. So if $\sum _{j}|m_{j}|\leq |m|,$ we have that $G\curvearrowright (X_{f},m_{X_{f}})$ is a factor of a Bernoulli shift.

More generally, let $S$ be a compact surface with a finite set of punctures (potentially empty) and nonempty boundary, and let $G$ be the mapping class group of $S.$ Then by [Reference Rourke and WiestRW00], we know that $G$ is left-orderable (see also [Reference Short and WiestSW00]). In many cases, we can explicitly describe a left-invariant order on $G$ and, as before, this allows us to explicitly produces semi-lopsided $f\in {\mathbb {Z}}(G)$ with $G\curvearrowright (X_{f},m_{X_{f}})$ a factor of a Bernoulli shift.