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\title{The Restricted Weyl Group of the Cuntz Algebra and Shift Endomorphisms}

\author{Roberto Conti, Jeong Hee Hong\footnote{This work was supported by
National Research Foundation of Korea
Grant funded by the Korean Government (KRF--2008--313-C00039).} and
Wojciech Szyma{\'n}ski\footnote{This work was partially supported by the FNU Rammebevilling
grant `Operator algebras and applications' (2009--2011),  the NordForsk Research
Network `Operator algebra and dynamics' (2009--2011), and the Marie Curie Research
Training Network MRTN-CT-2006-031962 EU-NCG.}}

\date{{\small 22 June 2010}}
\maketitle

\renewcommand{\sectionmark}[1]{}

\vspace{7mm}
\begin{abstract}
It is shown that, modulo the automorphisms which fix the canonical diagonal MASA point-wise,
the group of those automorphisms of $\O_n$ which globally preserve both the diagonal and
the core UHF-subalgebra is isomorphic, via restriction, with the group of those homeomorphisms
of the full one-sided $n$-shift space which eventually commute along with their inverses with
the shift transformation. The image of this group in the outer automorphism group of $\O_n$
can be embedded into the quotient of the automorphism group of the full two-sided
$n$-shift by its center, generated by the shift. If $n$ is prime then this embedding is an
isomorphism.
\end{abstract}

\vfill\noindent {\bf MSC 2010}: 46L40, 37B10

\vspace{3mm}
\noindent {\bf Keywords}: Cuntz algebra, automorphism, MASA, UHF-subalgebra, shift, shift automorphism,
shift endomorphism

\newpage

\section{Introduction}

Investigations of endomorphisms of the simple $C^*$-algebras $\O_n$, \cite{Cun1}, were initiated
by Cuntz in his seminal paper \cite{Cun2}. A key tool for those investigations was provided by a fairly simple
observation that unital endomorphisms of $\O_n$ are in an explicit bijective correspondence with
unitary elements of this $C^*$-algebra. In particular, Cuntz studied the subgroup of automorphisms
of $\O_n$ which globally preserve the canonical diagonal MASA $\D_n$. He showed that the quotient
of this group by its normal subgroup consisting of those automorphisms which fix $\D_n$ point-wise (and
this is a maximal abelian subgroup of $\aut(\O_n)$) is discrete. It is natural to think of this quotient as an
analogue of the Weyl group. The Weyl group contains a natural interesting subgroup corresponding to
those automorphisms which also globally preserve the core UHF-subalgebra $\F_n$ of $\O_n$. Cuntz
proposed in \cite{Cun2} a problem of determining the structure of this restricted Weyl group.

Ever since, endomorphisms of $\O_n$ have been an active area of investigations both intrinsically and
in connection to many other areas. Not even attempting any exhausting overview of the relevant
literature, let us only mention two pieces of research which influenced our current study the most:
Conti-Pinzari work on Jones index for endomorphisms of $\O_n$, \cite{CP}, and Bratteli-J{\o}rgensen
work on iterated function systems and representations of Cuntz algebras, initiated in \cite{BJ1} and
\cite{BJ2}. More recently, quite significant progress has been achieved in the study of those endomorphisms
which preserve either the core UHF-subalgebra, \cite{CRS}, or the diagonal MASA, \cite{HSS}. A powerful,
novel combinatorial approach to the study of endomorphisms which globally preserve both $\F_n$ and
$\D_n$ has been developed in \cite{S}, \cite{CS} and \cite{CKS}.

The main result of this paper is an explicit and intrinsic description
of the restricted Weyl group of the Cuntz algebra $\O_n$ and its image in
the outer automorphism group $\out(\O_n)$ (which we call the restricted outer Weyl group of $\O_n$).
The way we achieve this is by
analyzing the action of the restricted Weyl group on the diagonal MASA, and by showing
that certain class of automorphisms of $\D_n$ admits extensions to permutative automorphisms
of $\O_n$ (and thus also of $\F_n$). In general, existence of such extensions is
not guaranteed, as demonstrated by \cite{Co}.
It turns out that the restricted outer Weyl group admits a natural embedding into
the group of shift automorphisms of the two-sided full shift (with the mod out center),
and for prime $n$ this embedding is actually an isomorphism. These facts
have profound implications. For example, they immediately imply that the
restricted outer Weyl group is residually finite.
For $n =2$, this result also provides a not unexpected answer to a
question left open in \cite{CS}. Namely, the restricted outer Weyl
group is not amenable (for $n \geq 3$ this was already shown earlier in \cite{S} and \cite{CS}).
It also follows from our results that the restricted Weyl group of $\O_n$ is big
enough to contain a copy of the group of shift automorphisms of the
one-sided full $n$-shift. The groups of shift automorphisms (both for the one-sided and
the two-sided shift) have been extensively studied in the literature as they reveal an intriguing and
highly nontrivial structure, see \cite{K,LM} and the literature cited therein.

One important outcome of our investigations is a powerful and neat link between two areas:
the study of automorphisms of the Cuntz algebras and symbolic dynamics. This kind of interaction
was pioneered by Cuntz and Krieger in \cite{CK}, but we believe its limits have not been
reached yet. On one hand, we expect that the nice results available in the literature
on symbolic dynamics could shed new light on some aspects of automorphisms of Cuntz
algebras, and possibly even more general classes of $C^*$-algebras.
But perhaps it is even more intriguing to speculate the other way round, as one could hope that
the algebraic environment that we unveil could provide new tools
and insight for attacking some of the problems in the dynamical systems setting.

\vspace{3mm}\noindent
{\bf Acknowledgements.}
We are utmost grateful to Mike Boyle for a very helpful and illuminating discussion of shift endomorphisms.

\section{Notation and preliminaries}

If $n$ is an integer greater than 1, then the Cuntz algebra $\O_n$ is a unital, simple,
purely infinite $C^*$-algebra generated by $n$ isometries $S_1, \ldots, S_n$, satisfying
$\sum_{i=1}^n S_i S_i^* = I$, \cite{Cun1}.
We denote by $W_n^k$ the set of $k$-tuples $\mu = (\mu_1,\ldots,\mu_k)$
with $\mu_m \in \{1,\ldots,n\}$, and by $W_n$ the union $\cup_{k=0}^\infty W_n^k$,
where $W_n^0 = \{0\}$. We call elements of $W_n$ multi-indices.
If $\mu \in W_n^k$ then $|\mu| = k$ is the length of $\mu$.
If $\mu = (\mu_1,\ldots,\mu_k) \in W_n$, then $S_\mu = S_{\mu_1} \ldots S_{\mu_k}$
($S_0 = 1$ by convention) is an isometry with range projection $P_\mu=S_\mu S_\mu^*$.
Every word in $\{S_i, S_i^* \ | \ i = 1,\ldots,n\}$ can be uniquely expressed as
$S_\mu S_\nu^*$, for $\mu, \nu \in W_n$ \cite[Lemma 1.3]{Cun1}.

We denote by $\F_n^k$ the $C^*$-subalgebra of $\O_n$ spanned by all words of the form
$S_\mu S_\nu^*$, $\mu, \nu \in W_n^k$, which is isomorphic to the
matrix algebra $M_{n^k}({\mathbb C})$. The norm closure $\F_n$ of
$\cup_{k=0}^\infty \F_n^k$, is the UHF-algebra of type $n^\infty$,
called the core UHF-subalgebra of $\O_n$, \cite{Cun1}. It is the fixed point algebra
for the gauge action of the circle group $\gamma:U(1)\rightarrow{\aut}(\O_n)$ defined
on generators as $\gamma_t(S_i)=tS_i$. For $k\in\Zb$, we denote by $\O_n^{(k)}
:=\{x\in\O_n:\gamma_t(x)=t^k x\}$,
the spectral subspace for this action. In particular, $\F_n=\O_n^{(0)}$.
The $C^*$-subalgebra of $\F_n$ generated by projections $P_\mu$, $\mu\in W_n$, is a
MASA (maximal abelian subalgebra) both in $\F_n$ and in $\O_n$. We call it
the diagonal and denote $\D_n$. The spectrum of $\D_n$ is naturally identified with
$X_n$ --- the full one-sided $n$-shift space. We also set $\D_n^k:=\D_n\cap\F_n^k$.
Throughout this paper we are interested in the inclusions
$$ \D_n \subseteq \F_n\subseteq \O_n.$$
The UHF-subalgebra $\F_n$ posseses a unique normalized trace, denoted $\tau$.
We will refer to the restriction of $\tau$ to $\D_n$ as to the canonical trace on $\D_n$.

We denote by $\SS_n$ the group of those unitaries in $\O_n$ which can be written
as finite sums of words, i.e., in the form $u = \sum_{j=1}^m S_{\mu_j}S_{\nu_j}^*$
for some $\mu_j, \nu_j \in W_n$. We also denote $\P_n=\SS_n\cap\U(\F_n)$. Then
$\P_n=\cup_k\P_n^k$, where $\P_n^k$ are permutation unitaries in $\U(\F_n^k)$.
That is, for each $u\in\P_n^k$ there is a unique permutation $\sigma$ of multi-indices
$W_n^k$ such that
\begin{equation}\label{permutunitary}
u = \sum_{\mu \in W_n^k} S_{\sigma(\mu)} S_\mu^*.
\end{equation}

As shown by Cuntz in \cite{Cun2}, there exists the following bijective correspondence
between unitaries in $\O_n$ and unital $*$-endomorphisms of $\O_n$ (whose collection we denote
by $\en(\O_n)$). A unitary $u$  in $\O_n$ determines an endomorphism $\lambda_u$ by
$$ \lambda_u(S_i) = u S_i, \;\;\; i=1,\ldots, n. $$
Conversely, if $\rho :\O_n\rightarrow \O_n$ is an endomorphism, then
$\sum_{i=1}^n\rho(S_i)S_i^*=u$ gives a unitary $u\in\O_n$
such that $\rho=\lambda_u$. If the unitary $u$ arises from a permutation $\sigma$ via the formula
\eqref{permutunitary}, the corresponding endomorphism will be
sometimes denoted by $\lambda_{\sigma}$.
Composition of endomorphisms corresponds to a `convolution'
multiplication of unitaries as follows:
\begin{equation}\label{convolution}
\lambda_u \circ \lambda_w = \lambda_{\lambda_u(w)u}
\end{equation}
We denote by $\varphi$ the canonical shift:
$$ \varphi(x)=\sum_iS_ixS_i^*, \;\;\; x\in\O_n. $$
If we take $u=\sum_{i, j}S_iS_jS_i^*S_j^*$ then $\varphi=\lambda_u$.
It is well-known that $\varphi$ leaves invariant both $\F_n$ and $\D_n$, and that $\varphi$
commutes with the gauge action $\gamma$. We denote by $\phi$ the standard left inverse
of $\varphi$, defined as $\phi(a)=\frac{1}{n}\sum_{i=1}^n S_i^* a S_i$.

If $u\in\U(\O_n)$ then for each positive integer $k$ we denote
\begin{equation}\label{uk}
u_k = u \varphi(u) \cdots \varphi^{k-1}(u).
\end{equation}
We agree that $u_k^*$ stands for $(u_k)^*$. If
$\alpha$ and $\beta$ are multi-indices of length $k$ and $m$, respectively, then
$\lambda_u(S_\alpha S_\beta^*)=u_kS_\alpha S_\beta^*u_m^*$. This is established through
a repeated application of the identity $S_i a = \varphi(a)S_i$, valid for all
$i=1,\ldots,n$ and $a \in \O_n$. If $u\in\F_n^k$ for some $k$ then,
following \cite{CP},  we call endomorphism $\lambda_u$ {\em localized}.

For algebras $A\subseteq B$ we denote by $\N_B(A)=\{u\in\U(B):uAu^*=A\}$ the normalizer
of $A$ in $B$ and by $A' \cap B=\{a \in A: (\forall b \in B) \; ab=ba\}$ the
relative commutant of $A$ in $B$. We also denote by $\aut(B,A)$ the collection of all those
automorphisms $\alpha$ of $B$ that $\alpha(A)=A$, and by $\aut_A(B)$ those
automorphisms of $B$ which fix $A$ point-wise.

\section{The restricted Weyl group of the Cuntz algebra}

Let $\alpha$ be an automorphism of $\D_n$. We denote by $\alpha_*$ the corresponding
homeomorphism of $X_n$. We say that $\alpha$ has {\em property (P)} if
there exists $m$ such that for all $k\geq m$ we have
\begin{equation}\label{propertyP}
\alpha\varphi^k(x)=\varphi^{k-m}\alpha\varphi^m(x)
\end{equation}
for all $x\in\D_n^1$. That is, $\alpha$ eventually commutes with the shift. Equivalently,
$\alpha\in\aut(\D_n)$ satisfies (P) with $m$ if the endomorphism $\alpha\varphi^m$ commutes
with the shift $\varphi$. We define
\begin{equation}\label{gn}
\Gf_n:=\{\alpha\in\aut(\D_n):\text{ both $\alpha$ and $\alpha^{-1}$ have property (P)}\}.
\end{equation}

\begin{lemma}\label{groupgn}
$\Gf_n$ is a subgroup of $\aut(\D_n)$.
\end{lemma}
{\em Proof.}
Let $\alpha$ and $\beta$ belong to $\Gf_n$. Take $m$ so large that both $\alpha$ and $\beta$
satisfy (P) with $m$. Also, let $r$ be so large that $\beta(\varphi^m(\D_n^1))$ is contained in
$\D_n^r$. Since $\D_n^r=\D_n^1\varphi(\D_n^1)\cdots\varphi^{r-1}(\D_n^1)$, there exist
linear functionals $f_\mu:\D_n^1\to\Cb$, $\mu\in W_n^r$, such that
\begin{equation}\label{betam}
\beta(\varphi^m(x))=\sum_{\mu\in W_n^r}f_\mu(x)P_{\mu_0}\varphi(P_{\mu_1})
\cdots \varphi^{r-1}(P_{\mu_{r-1}}).
\end{equation}
Thus for $k\geq 2m$ and $x\in\D_n^1$ we have
\begin{eqnarray*}
\alpha\beta\varphi^k(x) & = & \alpha\varphi^{k-m}\beta\varphi^m(x) \\
 & = & \alpha\varphi^{k-m}\sum_{\mu\in W_n^r} f_\mu(x)P_{\mu_0}\varphi(P_{\mu_1})
\cdots \varphi^{r-1}(P_{\mu_{r-1}}) \\
 & = & \sum_{\mu\in W_n^r} f_\mu(x)(\alpha\varphi^{k-m}(P_{\mu_0}))
 \cdots (\alpha\varphi^{k-m+r-1}(P_{\mu_{r-1}})) \\
 & = & \sum_{\mu\in W_n^r} f_\mu(x)(\varphi^{k-2m}\alpha\varphi^{2m}(P_{\mu_0}))
 \cdots (\varphi^{k-2m}\alpha\varphi^{2m+r-1}(P_{\mu_{r-1}})) \\
 & = & \varphi^{k-2m}\alpha\sum_{\mu\in W_n^r} f_\mu(x)\varphi^{2m}(P_{\mu_0})
 \cdots \varphi^{2m+r-1}(P_{\mu_{r-1}}) \\
 & = & \varphi^{k-2m}\alpha\beta\varphi^{2m}(x),
\end{eqnarray*}
since
$$ \sum_{\mu\in W_n^r} f_\mu(x)\varphi^{2m}(P_{\mu_0})
 \cdots \varphi^{2m+r-1}(P_{\mu_{r-1}}) =  \beta\varphi^{2m}(x) $$
by virtue of (\ref{betam}) and property (P) for $\beta$. Consequently, the product $\alpha\beta$
satisfies (P) with $2m$, and whence $\Gf_n$ is a group.
\hfill $\Box$

\medskip
We denote $\iGf_n=\{\ad(u)|_{\D_n}:u\in\P_n\}$. This is a normal subgroup of $\Gf_n$,
since for $u\in\P_n^k$ we have $\ad(u)\varphi^k=\varphi^k$. In what follows we agree that $\varphi^0=\id$.

\begin{lemma}\label{amk}
If $\alpha\in\aut(\D_n)$ then $\alpha\in\iGf_n$ if and only if there exist $m,k$ such that $\alpha\varphi^m=\varphi^k$.
\end{lemma}
{\em Proof.} Let $\alpha\in\aut(\D_n)$ and let $m,k$ be such that $\alpha\varphi^m=\varphi^k$.
Suppose first that $m\geq k$. Then $\alpha\varphi^{m-k}(x)=x$ for all $x\in\varphi^k(\D_n)$. Let
$r\geq k$ be so large that $\alpha\varphi^{m-k}(\D_n^k)\subseteq\D_n^r$. Since $\D_n^r=
\D_n^k\varphi^k(\D_n^{r-k})$ and $\alpha\varphi^{m-k}$ acts identically on $\varphi^k(\D_n^{r-k})$,
it follows that $\alpha\varphi^{m-k}(\D_n^r)\subseteq\D_n^r$. Since the map is injective
and the space finite dimensional, we have $\alpha\varphi^{m-k}(\D_n^r)=\D_n^r$. As $\alpha
\varphi^{m-k}$ acts identically on $\varphi^r(\D_n)$, it follows that there is a permutation unitary
$u\in\P_n^r$ such that $\alpha\varphi^{m-k}=\ad(u)|_{\D_n}$. In particular, $\alpha\varphi^{m-k}$ is
an automorphism of $\D_n$. Thus $m-k=0$ and hence $\alpha=\ad(u)|_{\D_n}$. If $m\leq k$ then
$\alpha^{-1}\varphi^k=\varphi^m$ and we argue in the same way. This proves one implication.
The other one is obvious.
\hfill$\Box$

\begin{lemma}\label{braiding}
If $\alpha\in\aut(\D_n)$ then there exists $\beta\in\aut(\D_n)$ such that $\alpha\varphi=\beta\varphi\alpha$
and $\beta(x)=\alpha(x)$ for all $x\in\D_n^1$.
\end{lemma}
{\em Proof.} Given $x\in\D_n$, there exist unique elements $x_1,\ldots,x_n$ in $\D_n$ such that
$x=\sum_{j=1}^n P_j\varphi(x_j)$. We define
\begin{equation}\label{beta}
\beta(x) = \alpha\left(\sum_{j=1}^n P_j\varphi(\alpha^{-1}(x_j))\right).
\end{equation}
It follows that $\beta$ is an automorphism of $\D_n$ satisfying the required braiding property.
\hfill$\Box$

\medskip
If $\alpha\in\aut(\D_n)$ and $\beta\in\aut(\D_n)$ is defined by formula (\ref{beta}) then we
call $\beta$ the {\em braiding automorphism} for $\alpha$.

\medskip
As shown in  \cite[Lemma 5]{S}, if $u\in\P_n$ and $\lambda_u$
is an automorphism of $\O_n$ then the restriction of $\lambda_u$ to $\D_n$ belongs
to $\Gf_n$. This can be further generalized, as follows.

\begin{theorem}\label{asym}
Let $\alpha \in \aut(\D_n)$. Then the following conditions are equivalent:
\begin{itemize}
\item[(1)] $\alpha^{-1}$ has property (P);
\item[(2)] there exists a permutation $u \in \P_n$ such that $\alpha = \lambda_u|_{\D_n}$.
\end{itemize}
In this case,
$\alpha$ extends to an endomorphism of $\F_n$ and
both $\alpha$ and $\alpha^{-1}$ are $\tau$-preserving.

\end{theorem}

{\em Proof.}
(2) $\Rightarrow$ (1):

By assumption, $\alpha$ can be written as
$\lim_{h\to\infty}{\rm Ad}(u_h)$ (pointwise norm limit)
and therefore $\alpha^{-1} = \lim_{h\to \infty}{\rm Ad}(u_h)^*$. Now,
if $u \in \P_n^r$, one has,
for all $k-1 \geq r$ and all $1 \leq i \leq n$,
$$\begin{aligned}
\alpha^{-1}(\varphi^{k-1}(S_i S_i^*))
& =  \lim_{h \to \infty} {\rm Ad} (u_h)^*(\varphi^{k-1}(S_i S_i^*)) \\
& =  \lim_{h \to \infty} \varphi^{h-1}(u^*) \ldots \varphi(u^*)u^*
   \varphi^{k-1}(S_i S_i^*)u \varphi(u) \ldots \varphi^{h-1}(u) \\
& =  \lim_{h \to \infty} \varphi^{h-1}(u^*) \ldots \varphi^{k-r}(u^*)
   \varphi^{k-r}(\varphi^{r-1}(S_i S_i^*))\varphi^{k-r}(u) \ldots \varphi^{h-1}(u) \\
& =  \varphi^{k-r}\Big(\lim_{h \to \infty}\varphi^{h-k+r-1}(u^*) \ldots u^*
   \varphi^{r-1}(S_i S_i^*)u \ldots \varphi^{h-k+r-1}(u)\Big) \\
& =  \varphi^{k-r}(\alpha^{-1}(\varphi^{r-1}(S_i S_i^*))) \ .
\end{aligned}$$
\\
(1) $\Rightarrow$ (2):
At first we observe that there exists a permutation unitary $u\in\P_n$
such that the braiding automorphism for $\alpha$ is of the form $\beta=\Ad(u)|_{\D_n}$.
Indeed, let $\alpha^{-1}$ satisfy (P) with $m$. Since $\alpha^{-1}\varphi^{m+1}=
\varphi \alpha^{-1}\varphi^m$, we have $\varphi^{m+1} = \alpha \varphi \alpha^{-1} \varphi^m
= \beta \varphi^{m+1}$. Therefore, there exists a permutation unitary $u\in\P_n$ such that
$\beta=\Ad(u)|_{\D_n}$, by Lemma \ref{amk}.

Now we verify that the unitary $u$ as above satisfies $\alpha=\lambda_u|_{\D_n}$. Note that,
by induction, $\alpha \varphi^k = (\beta\varphi)^k \alpha$ for all $k$. Therefore,
$\alpha\varphi^k = (\Ad(u)\varphi)^k\alpha = \Ad(u_k)\varphi^k\alpha$ for all $k$.
In particular, $\alpha\varphi^k(P_i) = \Ad(u_k)\varphi^k\alpha(P_i) = \Ad(u_{k+1})\varphi^k(P_i)$
for all $k$ and $i$.
Now we compute
$$\begin{aligned}
\alpha(P_{i_1}\varphi(P_{i_2}) \cdots \varphi^{k-1}(P_{i_k}))
& = \alpha(P_{i_1}) \alpha\varphi(P_{i_2}) \cdots \alpha \varphi^{k-1}(P_{i_k}) \\
& = \Ad(u)(P_{i_1}) \Ad(u_2)\varphi(P_{i_2}) \cdots \Ad(u_k)\varphi^{k-1}(P_{i_k}) \\
& = u_k P_{i_1} u_k^* u_k \varphi(P_{i_2})u_k^* \cdots u_k\varphi^{k-1}(P_{i_k})u_k^* \\
& = \Ad(u_k) (P_{i_1}\varphi(P_{i_2}) \cdots \varphi^{k-1}(P_{i_k})).
\end{aligned}$$
That is, $\alpha = \lambda_u|_{\D_n}$, as required.

If the above conditions are satisfied, then in particular $\alpha$ extends to an endomorphism
$\lambda_u|_{\F_n}$ of $\F_n$, Thus $\tau\lambda_u|_{\F_n}=\tau$ by uniqueness of trace
on $\F_n$ (or the fact that $\lambda_u|_{\F_n}$ is a point-wise limit of inner automorphisms).
Hence $\tau \alpha = \tau$ as well. This clearly implies also that $\tau \alpha^{-1}=\tau$.

\hfill $\Box$

\medskip
The preceding theorem provides a dynamical explanation
of the puzzling phenomenon observed in \cite{CS} through combinatorial arguments,
namely the coexistence of permutative automorphisms and proper endomorphisms of $\O_n$
restricting to automorphisms of the diagonal,
in the form of a different dynamics they induce on $X_n$.

\medskip
The following proposition is yet another slight generalization of \cite[Lemma 5]{S}. We omit the proof.

\begin{proposition}\label{Ploc}
Let $\lambda_u$ be a localized automorphism of $\O_n$.
Then, there exists some nonnegative integer $m$ such that, for all $k \geq m$ and
all $x\in\F_n^1$,
\begin{equation}
\lambda_u^{-1} \circ \varphi^k(x) = \varphi^{k-m} \circ \lambda_u^{-1} \circ \varphi^m(x) \ .
\end{equation}
\end{proposition}

\begin{theorem}\label{extendingautos}
If $\alpha\in\Gf_n$ then there exists a permutation unitary $u\in\P_n$ such that
$\alpha=\lambda_u|_{\D_n}$. If $w$ is any unitary in $\O_n$ such that $\lambda_w|_{\D_n}=\alpha$,
then $\lambda_w\in\aut(\O_n)$.
\end{theorem}
{\em Proof.} The first statement follows immediately from Theorem \ref{asym}.
Let $w\in\U(\O_n)$ be such that $\lambda_w|_{\D_n}=\alpha$. Then, by the same argument,
there exists a permutation unitary $v\in\P_n$ such that $\lambda_v|_{\D_n}=\alpha^{-1}$. Thus
$\lambda_w\lambda_v$ is an endomorphism of $\O_n$ acting identically on $\D_n$, and
consequently $\lambda_w\in\aut(\O_n)$ by \cite[Proposition 3.2]{Co}.
\hfill$\Box$

\begin{corollary}\label{uhf}
Each $\alpha\in\Gf_n$ can be extended to an automorphism of $\F_n$.

\end{corollary}

\medskip
By Theorem \ref{asym}, the restriction $r:\aut(\O_n,\D_n)\to\aut(\D_n)$
yields a group embedding $\lambda(\P_n)^{-1}\hookrightarrow\Gf_n$, see \cite{Cun2,CS}.
Since the restriction map $r:\aut(\O_n,\D_n)\to\aut(\D_n)$ is injective on $\lambda(\P_n)^{-1}$,
\cite{Cun2,CS}, Theorem \ref{extendingautos} yields the following.

\begin{corollary}\label{permautos}
The restriction $r:\lambda(\P_n)^{-1}\to\Gf_n$ is a group isomorphism.
\end{corollary}

We recall from \cite{Cun2} that $\aut(\O_n,\D_n)$ is the normalizer of $\aut_{\D_n}(\O_n)$ in $\aut(\O_n)$ and
it can be also described as the group $\lambda(\N_{\O_n}(\D_n))^{-1}$ of automorphisms of $\O_n$ induced by elements in the (unitary) normalizer $\N_{\O_n}(\D_n)$.
Furthermore, using \cite{Pow},
one can show that $\aut(\O_n,\D_n)$ has the structure of a semidirect product $\aut_{\D_n}(\O_n) \rtimes \lambda(\SS_n)^{-1}$ \cite{CS}.
In particular, the group $\lambda(\P_n)^{-1}$ is isomorphic with the quotient of
the group $\aut(\O_n,\D_n)\cap\aut(\O_n,\F_n)$ by its normal subgroup $\aut_{\D_n}(\O_n)$.
We call it the {\em restricted Weyl group} of $\O_n$, cf. \cite{Cun2,CS}. Thus, the preceding
corollary provides in a sense an answer to the question raised by Cuntz in \cite{Cun2}.

\begin{corollary}\label{sngn}
Let $u\in\SS_n$ be such that $\lambda_u(\D_n)=\D_n$ and $\lambda_u|_{\D_n}\in\Gf_n$. Then $u\in\P_n$.
\end{corollary}
{\em Proof.} By Theorem \ref{extendingautos}, there is $w\in\P_n$ such that $\lambda_w(x)=
\lambda_u(x)$ for all $x\in\D_n$. However, the restriction map $r:\lambda(S_n)^{-1}\to\aut(\D_n)$ is
injective, \cite{Cun2,CS}. Thus $u=w$ belongs to $\P_n$.
\hfill$\Box$

\section{The restricted outer Weyl group of the Cuntz algebra and shift endomorphisms}

Denote by $\operatorname{Inn}\lambda(\P_n)^{-1}$ the normal subgroup of $\lambda(\P_n)^{-1}$
consisting of all inner permutative automorphisms $\{\Ad(u):u\in\P_n\}$. We call the quotient
$\lambda(\P_n)^{-1}/\operatorname{Inn}\lambda(\P_n)^{-1}$ the {\em restricted outer Weyl group}
of $\O_n$. From Theorem \ref{extendingautos}, we get the following.

\begin{corollary}\label{weyl}
The restricted outer Weyl group of $\O_n$ is naturally isomorphic to the quotient $\Gf_n/\iGf_n$.
\end{corollary}
In what follows, if $\alpha\in\Gf_n$ then we denote its class in $\Gf_n/\iGf_n$ by $\overline{\alpha}$.
We denote by $\en(\D_n,\varphi)$ the semigroup of unital, injective $*$-homomorphisms from $\D_n$
into itself which commute with the shift. We define $E_n$ as the collection of all those
$\alpha\in\en(\D_n,\varphi)$ for which there exists an $m$ and a $\beta\in\en(\D_n,\varphi)$  such that
\begin{equation}\label{phiinverse}
\alpha\beta=\varphi^m.
\end{equation}
In such a case, we have $\alpha\beta\alpha=\varphi^m\alpha=\alpha\varphi^m$, and thus injectivity
of $\alpha$ implies that $\beta\alpha=\varphi^m$ as well. In particular, $\beta$ itself belongs to $E_n$.
$E_n$ is a subsemigroup of $\en(\D_n,\varphi)$ containing all powers of $\varphi$, $\Gf^0$,
as well as all endomorphisms $\alpha\varphi^m$ with $\alpha\in\Gf_n$ and suitably large $m$.

We note that if $\alpha\in E_n$ then $\alpha_*:X_n\to X_n$ is an open mapping. Indeed, if $U\subseteq
X_n$ is open then so is $V=\beta_*^{-1}(U)$, and $U=\beta_*(V)$ since $\beta_*$ is surjective.
Then $\alpha_*(U)=\alpha_*\beta_*(V)=\varphi_*^m(V)$ is open, since $\varphi_*^m$ is an
open mapping.

\begin{lemma}\label{shifttrace}
If $\alpha\in E_n$ then $\alpha_*$ acts bijectively on periodic words. Consequently,
$\tau\alpha(d)=\tau(d)$ for all $d\in\D_n$.
\end{lemma}
{\em Proof.} Say $x\in X_n$ has period $r$ if $\varphi_*^r(x)=x$. For each $r$, the set $X_n(r)$ of
all points with period $r$ is finite. Furthermore, $\varphi_*$ restricts to a {\em bijection} on each $X_n(r)$.
If $\mu\in W_n$ and $r\geq|\mu|$ then there are exactly $n^{r-|\mu|}$ words $x\in X_n(r)$
such that $P_\mu(x)=1$.

Now let $\alpha$ be in $E_n$, and let $\beta\in E_n$ be such that (\ref{phiinverse}) holds.
For each $r$ we have  $\alpha_*(X_n(r))\subseteq X_n(r)$, since $\alpha_*$ commutes with
$\varphi_*$. If $x,y\in X_n(r)$ and $\alpha_*(x)=\alpha_*(y)$ then also
$\varphi_*^m(x)=\beta_*\alpha_*(x)=\beta_*\alpha_*(y)=\varphi_*^m(y)$. Thus
$x=y$, since $\varphi_*$ acts bijectively on $X_n(r)$. Therefore, $\alpha_*$ yields a one-to-one
mapping from $X_n(r)$ to itself. By finiteness of $X_n(r)$, this map is bijective.

Now let $\mu\in W_n$ and let $\alpha(P_\mu)=\sum_{j=1}^k P_{\mu_j}$. By subdividing, if necessary,
we may assume that each $\mu_j$ is of the same length $r$ and that $r\geq|\mu|$. Then the number
of words $x\in X_n(r)$ such that $P_\mu(x)=1$ is the same as the number of words $y\in X_n(r)$
such that $P_{\mu_j}(y)=1$ for some $j$. Thus $k=n^{r-|\mu|}$, and consequently
$$ \tau\alpha(P_\mu)=\frac{k}{n^r}=\frac{1}{n^{|\mu|}}=\tau(P_\mu). $$
Since $\mu$ was arbitrary, the proposition follows.
\hfill$\Box$

\medskip
The following lemma is due to Mike Boyle, \cite{Boy}, although the proof given below is our own.

\begin{lemma}[M. Boyle]\label{boyle}
If $\alpha\in E_n$ then there exists a $k$ such that the mapping $\alpha_*$ is  $k$-to-one. This
$k$ divides a power of $n$. Thus if $n$ is prime then there exists an $r$ such that
$\alpha_*$ is $n^r$-to-one.
\end{lemma}
{\em Proof.} Let $\alpha,\beta\in E_n$ satisfy (\ref{phiinverse}). Since $\beta_*\alpha_*=
\varphi_*^m$ is $n^m$-to-one, each $y\in X_n$ has at most $n^m$
inverse images under $\alpha_*$. Let $k$ be the minimal cardinality of $\alpha_*^{-1}(y)$, and
let $\omega\in X_n$ be such that $\alpha_*^{-1}$ has exactly $k$ elements. Then the set
$\alpha_*^{-1}(\varphi_*^{-1}(\omega))=\varphi_*^{-1}(\alpha_*^{-1}(\omega)$ has $nk$
elements. By the minimality of $k$, this can only happen if each element of $\varphi_*^{-1}(\omega)$
has $k$ inverse images under $\alpha_*$. Let $\Omega$ be the smallest subset of $X_n$
containing $\omega$ and closed under taking inverse images of $\varphi_*$. It follows from the above
that for each $y\in\Omega$ the set $\alpha_*^{-1}(y)$ has $k$ elements. Clearly, $\Omega$ is
dense in $X_n$. Now let $y\in X_n$ be arbitrary, and let $\alpha_*^{-1}(y)$ have $k'$ elements.
Since $\alpha_*$ is an open mapping, there exists an open subset $V$ of $X_n$ containing $y$ whose
each element has at least $k'$ inverse images under $\alpha_*$. Since $V\cap\Omega\neq\emptyset$,
we have $k'=k$.

Applying the same reasoning to $\beta$ instead of $\alpha$ we get an $l$ such that $\beta_*$ is
$l$-to-one. Thus $kl=n^m$.
\hfill$\Box$

\medskip
We define an equivalence relation $\sim$ in $E_n$ as follows: $\alpha\sim\beta$ if there exists a $k$
such that either $\alpha=\beta\varphi^k$ or $\alpha\varphi^k=\beta$. Then we set
$$ \Ef_n = E_n/\sim. $$
By construction, $\Ef_n$ is a group. If $\alpha\in E_n$ then we denote its class in $\Ef_n$ by
$\overline{\alpha}$. Let $\aut(\Sigma_n)$ denote the group of automorphisms of the full two-sided
$n$-shift, and let $\la \sigma \ra$ be its subgroup generated by the two-sided shift $\sigma$.
The following proposition is well-known, but for completeness we include a proof.

\begin{proposition}\label{quotientiso}
The groups $\Ef_n$ and $\aut(\Sigma_n)/\la \sigma \ra$ are isomorphic.
\end{proposition}
{\em Proof.}
We can realize $\D_n$ as $\otimes_{i \in {\mathbb N}} \D_n^1$.
Likewise, we consider $\widetilde\D_n := \otimes_{i \in {\mathbb Z}} \D_n^1$,
equipped with the two-sided shift automorphism $\widetilde\varphi$.
The Gelfand spectrum of $\widetilde{\D}_n$ can be identified with $\Sigma_n$ and
the canonical embedding $\D_n \to \widetilde{\D}_n$ corresponds to the canonical surjection
$\Sigma_n \to X_n$.

Now, every shift invariant endomorphism $\rho$ of $\D_n$ canonically extends
to an endomorphism $\tilde\rho$ of $\widetilde\D_n$, uniquely determined
by the properties of being $\widetilde\varphi$-invariant
and restricting to $\rho$ on $\D_n \subset \widetilde\D_n$.
Of course, the shift endomorphism $\varphi$ extends to $\widetilde\varphi$.
If $\rho$ is injective then $\tilde\rho$ is injective too. Moreover, the map
$\rho \mapsto \tilde\rho$ gives a semigroup homomorphism from
${\rm End}(\D_n,\varphi)$ to ${\rm End}(\widetilde\D_n,\widetilde\varphi)$, which is clearly injective.

If $\rho \in E_n$ then it is easy to see that $\tilde\rho$ is surjective and thus it is an
automorphism of $\widetilde\D_n$. Therefore, we get an injective semigroup homomorphism
$E_n \to {\rm Aut}(\widetilde\D_n,\widetilde\varphi)$. Passing to quotients, the previous
map provides a well-defined and injective group homomorphism from $\Ef_n$ into
${\rm Aut}(\widetilde\D_n,\widetilde\varphi)/\langle \widetilde\varphi \rangle
\cong \aut(\Sigma_n)/\la \sigma \ra$.

In order to show surjectivity of this map, we observe that
given any $\widetilde\varphi$-commuting automorphism $\eta$ of $\widetilde\D_n$
there exists a nonnegative integer $k$ such that $\eta':=\widetilde\varphi^k \eta |_{\D_n}$
belongs to $E_n$ (use the fact that $\Sigma_n$ is the Cantor set). Furthermore, thanks to
uniqueness of the extension, one has $\widetilde\varphi^{-k}\tilde{\eta'}=\eta$ and the
proof is complete.
\hfill$\Box$

\begin{theorem}\label{endogroup}
There exists an embedding of $\Gf_n/\iGf_n$ into $\aut(\Sigma_n)/\la \sigma \ra$. If $n$ is
prime then these two groups are isomorphic.
\end{theorem}
{\em Proof.} By virtue of Proposition \ref{quotientiso}, we may replace $\aut(\Sigma_n)/\la \sigma \ra$
with $\Ef_n$.

 Let $\alpha\in\Gf_n$ satisfy (P) with $m$. Then $\alpha\varphi^m\in E_n$, and we
map $\alpha$ to $\overline{\alpha\varphi^m}$ in $\Ef_n$. Clearly, this definition does not depend on
the choice of $m$ and thus the map is well defined. One easily checks that this map is a group homomorphism.
If $\alpha\in\iGf_n$ then $\alpha\varphi^k=\varphi^k$ for sufficiently large $k$, and thus the image
of such $\alpha$ in $\Ef_n$ is the trivial element. Thus, the homomorphism $\Gf_n\to\Ef_n$ factors
though $\Gf_n/\iGf_n$, and we get a homomorphism $\Gf_n/\iGf_n\to\Ef_n$. The latter map is
injective. Indeed, let $\alpha\in\Gf_n$ satisfy (P) with $m$ and $\overline{\alpha\varphi^m}=
\overline{\id}$. Then there is a $k$ such that either $\alpha\varphi^{m+k}=\id$ or
$\alpha\varphi^m=\varphi^k$. In either case, $\alpha\in\iGf_n$ by Lemma \ref{amk}.

Now assuming $n$ prime we show that the map $\Gf_n/\iGf_n\to\Ef_n$ is surjective. Let $\alpha\in E_n$
and let $\beta\in E_n$ be such that identity (\ref{phiinverse}) holds. There is an $r$ such
that $\alpha_*$ is $n^r$-to-one, by Lemma \ref{boyle}. Since $\alpha_*$ is an open map
onto the Cantor set it admits a continuous section $f_1$,
\cite[Corollary 1.4]{M}. Space $X_n\setminus f_1(X_n)$ being open is completely metrizable and thus
the restriction of $\alpha_*$ to $X_n\setminus f_1(X_n)$ admits a section $f_2$. Continuing in this manner,
we arrive at a maximal set of independent, continuous sections.
For convinience, we label these sections with words from $W_n^r$.  That is, there exist continuous
functions $f_\mu:X_n\to X_n$, $\mu\in W_n^r$, such that $\alpha_* f_\mu=\id$ and
$\alpha_*^{-1}(x)=\{f_\mu(x) : \mu\in W_n^r \}$ for each $x\in X_n$. Define a mapping
$g:X_n \to X_n$ by $g(\mu x)=f_\mu(x)$. Then $g$ is a homeomorphism of $X_n$. Furthermore,
 we have $\alpha_* g(\mu x) = \alpha_* f_\mu(x) = x = \varphi_*^r(\mu x)$. Thus
$\alpha_* g = \varphi_*^r$ and hence $\alpha_* = \varphi_*^r g^{-1}$. Let $\psi\in\aut(\D_n)$
be such that $\psi_*=g^{-1}$. Then $\alpha = \psi \varphi^r$ and whence $\psi$ satisfies
condition (P).  Arguing in the same way, we obtain an $\eta\in\aut(\D_n)$ such that $\beta=
\eta\varphi^t$ for some $t$. Then $\varphi^m=\alpha\beta=\psi\eta\varphi^{r+t}$. Thus
$\psi\eta\in\iGf_n$ by Lemma \ref{amk}. Now $\psi^{-1}$ being a composition of $\eta$ and
an element from $\iGf_n$, itself satisfies condition (P). Consequently, $\psi$ belongs to $\Gf_n$.
Clearly, the homomorphism $\Gf_n/\iGf_n\to\Ef_n$ maps $\overline{\psi}$ to $\overline{\alpha}$,
and the proof is complete.
\hfill$\Box$

\begin{remarks}
\rm In view of the preceding theorem, the restricted outer Weyl group of $\O_n$
has a number of striking properties known to hold for $\aut(\Sigma_n)/\la \sigma \ra$, see \cite{K,LM}
and the references therein. This is immediate at least for $n$ prime.  For example, we now know
that in the case of $n=2$ the group $\Gf_n/\iGf_n$ is non-amenable (for $n\geq3$ this has been
already observed in \cite{S,CS}).  We wonder if the extensive theory of shift automorphisms could not
bring new insight into the structural properties of not just Cuntz algebras but also graph algebras and
possibly even a larger class of $C^*$-algebras. In particular, it appears to be an intriguing possibility of
translating some features of the beautiful analysis of the action of automorphisms on periodic points,
\cite{BK}, into a more algebraic setting.

In general, when $n$ is not prime, the embedding from Theorem \ref{endogroup}
is not surjective. This is due to existence of factorizations $\varphi=\alpha\beta$ with $\alpha,\beta\in
E_n$ and neither $\alpha$ nor $\beta$ being an automorphism, \cite{BM}. Then it is easy
to verify that $\overline{\alpha}$, $\overline{\beta}$ do not belong to the range of the embedding.
Nevertheless, even for $n$ not a prime, Theorem \ref{endogroup} sheds a lot of light on the structure of the
restricted outer Weyl group of $\O_n$.  In particular, it implies that $\Gf_n/\iGf_n$ is residually finite.
As an example, we give an elementary, self-contained proof of the fact that $\Ef_n$ is residually finite in
Proposition \ref{rf} below.
\end{remarks}

\begin{proposition}\label{rf}
The group $\Ef_n$ is residually finite.
\end{proposition}
{\em Proof.} Let $\per$ denote the set of periodic points in $X_n$.
If $\alpha\in E_n$ then $\alpha_*$ acts bijectively on $\per$,
Lemma \ref{shifttrace}. Thus the group $\Ef_n$ acts on the orbits of $\per$
under the action of $\varphi_*$. The restriction of this action to the orbits contained
in $X_n(r)$ yields a homomorphism from $\Ef_n$ into a finite permutation group.
Thus it suffices to show that if $\alpha\in E_n$ is not a power of the shift then
$\alpha_*$ moves at least one orbit.

Let $\alpha\in E_n$. Suppose that $\alpha_*$ fixes every orbit of periodic
points under the action of the shift. This means that for each periodic point $x$
there exists $k$ such that $\alpha_*(x)=\varphi_*^k(x)$. We claim that
$\alpha$ is a power of the shift. The proof involves the following five steps.

1. There is a map $a:W_n^r \to W_n^1$ such that if $y=\alpha_*(x)$ then
$y_1=a(x_1,\ldots,x_r)$, \cite{H}.

2. If $x_1=\ldots = x_r = j$ then $a(j,\ldots,j)=j$, for otherwise
$(j,j,\ldots)$ would be moved by $\alpha_*$, being a fixed point for
$\varphi_*$.

3. Let $Z$ be the set of all periodic points of the following form:
$$ (x_1,\ldots,x_t,w,\ldots,w)(x_1,\ldots,x_t,w,\ldots,w)\ldots, $$
where $t>2r$, the number of
$w$'s (in one block) is $d>t$, and $w\neq a(x_1,\ldots,x_r)$. Note that $Z$
is dense in $X_n$. By hypothesis on $\alpha_*$, the image under $\alpha_*$ of
such a periodic point from $Z$ is of the form:
$$ \begin{aligned}
\text{either}\;\;\; & (w,\ldots,w,x_1,\ldots,x_t,w,\ldots,w) (w,\ldots,w,x_1,\ldots,x_t,w,\ldots,w)\ldots \\
\text{or}\;\;\; & (x_p,\ldots,x_t,w,\ldots,x_1,\ldots,x_{p-1}) (x_p,\ldots,x_t,w,\ldots,x_1,\ldots,x_{p-1})\ldots
\end{aligned} $$
However, the former is impossible since $w\neq a(x_1,\ldots,x_r)$ and $t>r$.
Furthermore, there is a $k$ in $\{1,\ldots,r\}$ such that $x_k\neq w$, for
otherwise $a(x_1,\ldots,x_r)=w$, contrary to the assumption. Consequently,
$1 \leq p \leq 2r-1$.

4. By 3. above, for each $x\in Z$ there is a $0 \leq p \leq 2r-2$ such that
$\alpha_*(x)
= \varphi_*^p(x)$. Now let $x$ be an arbitrary point in $X_n$. Take a sequence
$x_k\in Z$ converging to $x$. Passing to a subsequence we may assume that
there is a fixed $p \leq 2r-2$ such that for each $k$ we have $\alpha_*(x_k)=
\varphi_*^p(x_k)$. By continuity, $\alpha_*(x)=\varphi_*^p(x)$.

5. For $j=0,1,\ldots,2r-2$ let $Y_j=\{ x\in X_n : \alpha_*(x)=\varphi_*^j(x) \}$.
By 4. above, $X_n = Y_0 \cup \ldots \cup Y_{2r-2}$, and each $Y_j$ is a closed set.
Thus there is $j$ such that $Y_j$ contains a non-empty open subset.
Then there exists a finite word $\mu$ such that
for all infinite words $x$ we have $\mu x \in Y_j$ and  $\alpha_*(\mu x)=
\varphi_*^j(\mu x)$. Since $\alpha_*$ commutes with $\varphi_*$, we have
$$ \alpha_*(x) = \alpha_*\varphi_*^{|\mu|}(\mu x) = \varphi_*^{|\mu|}\alpha_*(\mu x) =
\varphi_*^{j+|\mu|}(\mu x) =\varphi_*^j(x). $$
Since $x$ was arbitrary, $\alpha_* = \varphi_*^j$.
\hfill$\Box$

\section{Shift automorphisms}

Note that if $\alpha\in\aut(\D_n)$ satisfies (P) with $m=0$ then $\alpha_*$ is just an automorphism
of the full one-sided $n$-shift (see \cite{K}). The collection of all such automorphisms constitutes
a subgroup of $\Gf_n$, which we denote $\Gf_n^0$. In the case of $n=2$ we have $\Gf_2^0\cong\Zb_2$
(generated by the restriction of Archbold's flip-flop, \cite{Ar}) by \cite[Theorem 3.1.1]{K}. But
for $n\geq3$ the group $\Gf_n^0$ is infinite (see \cite[Chapter 3]{K} and \cite{BFK,H}).

\begin{example}\label{Kitchensexample}
\rm Consider an order two automorphism $\alpha$ of $\D_3$ such that $\alpha_*$ changes
subwords $13$ and $23$ (of any one-sided infinite word) into $23$ and $13$, respectively, as
in \cite[Example 3.3.10]{K}. Then $\alpha$ belongs to $\Gf_3^0$. Define
$$ u=P_{11}+P_{12}+P_{21}+P_{22}+P_3+S_{23}S_{13}^*+S_{13}S_{23}^*, $$
an order two permutation unitary in $\P_3^2$. We have
\begin{eqnarray*}
uP_1u^* & = & P_{11}+P_{12}+P_{23}, \\
uP_2u^* & = & P_{21}+P_{22}+P_{13}, \\
uP_3u^* & = & P_3.
\end{eqnarray*}
One checks that the unitary $u$ commutes with the three minimal projections in $\varphi(u\D_3^1u^*)$.
This implies (via an easy inductive argument) that the restriction of $\lambda_u$ to $\D_3$
commutes with the shift $\varphi$. It follows that $\alpha=\lambda_u|_{\D_3}$.
\hfill$\Box$\end{example}

We note, in passing, that among the three rooted trees associated with automorphism $\lambda_u$
from Example \ref{Kitchensexample} as in \cite[Section 4.1]{CS}, two are of height 2 and one is
of height 1 (cf. \cite[Section 2.1]{CKS}).

\medskip
By Theorem \ref{extendingautos}, we already know that each $\alpha\in\Gf_n^0$ may be extended
to an automorphism of $\O_n$. Below, we provide an alternative proof of this fact,
involving a thourough description of the underlying structure in this specific case and thus leading to
a more explicit construction of the required permutation unitary. We will need the following two lemmas.
A straightforward proof of the former is omitted.

\begin{lemma}\label{shiftext}
Let $\alpha \in \Gf_n^0$.
If $u$ is a unitary in $\O_n$ such that $u x u^* = \alpha(x)$ for all $x \in \D_n^1$
and $u \varphi^j(\alpha(x)) u^* = \varphi^j(\alpha(x))$ for all $x \in \D_n^1$ and all $j=1,\ldots,k-1$
then $u_k x u_k^* = \alpha(x)$ for all $x \in \D_n^k$.
\end{lemma}

\begin{lemma}\label{phicommute}
Each $\alpha\in\Gf_n^0$ commutes with the left-inverse $\phi$ of $\varphi$.
\end{lemma}
{\em Proof.} Choose $r$ so large that $\alpha(\D_n^1) \subseteq \D_n^r$.
We write $\alpha(P_j) = \sum_{\mu \in W(j)} P_\mu$, where $W(j) \subseteq W_n^r$ is
a subset of cardinality $n^{r-1}$, since $\tau\alpha=\tau$ by Lemma \ref{shifttrace}.
Clearly, the sets $\{W(j):j=1,\ldots,n\}$ form a partition of $W_n^r$.

Since $\alpha$ commutes with the shift $\varphi$, in order to show that $\alpha$ commutes with
its left inverse $\phi$ it suffices to prove that $1=\sum_{i=1}^n S_i^* \alpha(P_j) S_i$ for
all $j=1,\ldots,n$. This is equivalent to $\{(\mu_2,\ldots,\mu_r) :  \mu \in W(j)\} = W_n^{r-1}$.
Now the last claim will follow if we can show that
$\alpha(P_j) \varphi(P_{i_1} \cdots \varphi^{r-2}(P_{i_{r-1}})) \neq 0$
for all $i_1,\ldots,i_{r-1}$ and all $j$. But the last expression is different from $0$
if and only if
$P_j \varphi(\alpha^{-1}(P_{i_1} \cdots \varphi^{r-2}(P_{i_{r-1}})))$ is, and
this is clearly the case.
\hfill$\Box$

\begin{theorem}\label{shiftautos}
Let $\alpha$ be an automorphism of the full one-sided $n$-shift, i.e. $\alpha\in\Gf_n^0$.
Then there exists a permutation unitary $u\in\P_n$ such that $\alpha=\lambda_u|_{\D_n}$.
\end{theorem}
{\em Proof.} Let $\alpha(P_j) = \sum_{\mu \in W(j)} P_\mu$, as in Lemma \ref{phicommute}.
Then $\{(\mu_2,\ldots,\mu_r) \ | \ \mu \in W(j)\}=W_n^{r-1}$, for every $j=1,\ldots,n$.
Therefore, there exists a unique $u \in \P_n^r$ such that
$u P_{\mu} u^* = P_{\mu(j)}$ for all $\mu \in W_n^r$,
where $\mu_1 = j$ and
$\mu(j)$ is the unique multi-index in $W(j)$ such that
$$(\mu_2,\ldots,\mu_r)=(\mu(j)_2,\ldots,\mu(j)_r) \ . $$
It is straightforward to check that the conditions of Lemma \ref{shiftext} are satisfied
(for arbitrarily large $k$), and hence $\lambda_u|_{\D_n}=\alpha$.
\hfill$\Box$

\begin{remark}
\rm All the permutation unitaries $u \in \P_n^r$ as in the preceeding theorem have the
following general structure: there are $n$ functions $f_1, \ldots, f_n$ from
$W_n^{r-1}$ into $W_n^1$ such that for each $w \in W_n^{r-1}$
the map $j \mapsto f_j(w)$ is a permutation of $W_n^1$, and moreover
for all $\mu \in W_n^r$, $u P_\mu u^*=P_{\sigma_f(\mu)}$, where
$\sigma_f((\mu_1,\ldots,\mu_r))=(f_{\mu_1}(\mu_2,\ldots,\mu_r),\mu_2,\ldots,\mu_r)$.
Therefore,
$$u = \sum_{\mu\in W_n^r} S_{\sigma_f(\mu)}S_\mu^* \ . $$
\end{remark}

\begin{proposition}\label{outershiftauto}
If $\alpha\in\aut(X_n)$ is a non-trivial automorphism of the full one-sided $n$-shift and $\tilde{\alpha}$ is an
automorphism of $\O_n$ extending $\alpha$, then $\tilde{\alpha}$ is outer.
\end{proposition}
{\em Proof.} Suppose $\tilde{\alpha}=\Ad(u)$ for some $u\in\U(\O_n)$. Then $u\in\N_{\O_n}(\D_n)$ and
thus $u=vw$, where $v\in\SS_n$ and $w\in\U(\D_n)$, \cite{Pow}. But $\Ad(w)$ acts identically
on $\D_n$. Thus $\tilde{\alpha}=\Ad(v)$. Since $\tau\alpha=\tau$, we must have $v\in\P_n^k$ for
some $k$. But then $\varphi^k(x)=\Ad(v)\varphi^k(x)=\varphi^k\Ad(v)(x)$ for all $x\in\D_n$. Thus
$v=1$ and consequently $\tilde{\alpha}=\id$.
\hfill$\Box$

\medskip
We have shown, above, that every automorphism of $\D_n$ commuting with the shift extends to an
automorphism of $\O_n$. In this subsection we observe that in most cases such an extension cannot
commute with the shift $\varphi$.

\begin{proposition}
Let $\lambda_v$ be an endomorphism of $\O_n$. Then $\lambda_v$ commutes with $\varphi$
on $\O_n$ if and only if
\begin{equation}
v \varphi(v) \theta \varphi(v^*) = \varphi(v) \theta \ ,
\end{equation}
where $\theta$ is the flip unitary. If moreover $\lambda_v \in \aut(\O_n)$ then
$\lambda_v \varphi = \varphi \lambda_v$ if and only if $v\in\U(\F_n^1)$, i.e. $\lambda_v$ is
a Bogolubov automorphism of $\O_n$.
\end{proposition}
{\em Proof.}
The first statement follows easily from the composition rule of endomorphisms and the fact that
$\varphi = \lambda_\theta$, where $\theta \in \F_n^2$.

Since $\lambda_v \varphi = {\rm Ad}(v) \varphi \lambda_v$ for all unitaries $v \in \U(\O_n)$,
the assumption implies that ${\rm Ad}(v) \varphi \lambda_v(x) = \varphi \lambda_v(x)$
for all $x \in \O_n$ and thus, $\lambda_v$ being an automorphism,
$v \in \varphi(\O_n)' \cap \O_n = \F_n^1$.
\hfill $\Box$

\begin{proposition}
Let $u \in \U(\O_n)$ and suppose that $\lambda_u(\F_n)=\F_n$.
If $\lambda_u$ and $\varphi$ commute on $\F_n$
then $\lambda_u$ is a Bogolubov automorphism.
\end{proposition}
{\em Proof.}
Observe that indeed $u \in \F_n$ \cite{CRS}.
By a similar argument as in the previous proposition,
we get ${\rm Ad}(u) \varphi \lambda_u = \varphi \lambda_u$ on $\F_n$
and therefore $u \in \varphi(\F_n)' \cap \F_n = \F_n^1$.
\hfill $\Box$

\medskip
In the case $n=2$, if $u$ is a permutation for which

$\lambda_u \in \aut(\O_2)$ commutes with $\varphi$ on $\D_2$
then $u$ must be the flip automorphism, i.e. Bogolubov permutation.
(And then, in turn, $\lambda_u$ commutes with $\varphi$ on the whole of $\O_2$.)

However Example \ref{Kitchensexample} illustrates that this is not true anymore for $\D_n$ with $n>2$,
i.e. there are permutation automorphisms $\lambda_u$
(which therefore satisfy automatically $\lambda_u(\D_n)=\D_n$)
commuting with $\varphi$ on $\D_n$ which are not Bogolubov
(and thus, by the above proposition, they do not commute with $\varphi$ on $\F_n$,
let alone on $\O_n$).


TITLE: Oloid and sphericon: rolling develops entire surface
QUESTION [9 upvotes]: Wikipedia says that,

"The oloid is one of the only known objects, along with some members of the sphericon family, that while rolling, develops its entire surface."

Below are illustrations of these two objects:

   Link to video of oloid rolling.

   Link to video of sphericon rolling.

Two questions:
Q1. The phrasing of the Wikipedia sentence quoted above seems to
allow the interpretation that
it is unknown if there are other objects with the same rolling property.
Is this the case? Or is there a proof that these are the only such objects?
Q2. Does anyone know where I can find a proof of this rolling property for
either of these objects?

REPLY [8 votes]: Here is a more general construction, which contains both the sphericon and the oloid as particular cases.

Take two intersecting planes, $\alpha$ and $\beta$.
Take two compact convex regions $A\subset\alpha$ and $B\subset\beta$.
Make sure that $A$ is split in two non-empty regions by the plane $\beta$, and $B$ by the plane $\alpha$.
Let $a=A\cap\beta$, and $b=B\cap\alpha$. Both sets $a-b$ and $b-a$ are nonempty.
The desired object is the convex hull of the union of the two planar convex regions, $A\cup B$. Let $S$ be its boundary.

Of course, because we the convex hull, in condition 2 we can drop the condition that the two planar regions are convex, but this doesn't increase generality.
Proof that this works.
Let $\partial A$ be the boundary of $A$  in the plane $\alpha$, and $\partial B$ the boundary of $B$ in the plane $\beta$.
A. Condition 3 implies that for any point $P$ on $\partial A$, there are two points $Q_1$ and $Q_2$ on $\partial B$, which are separated by the plane $\alpha$, hence the segments $PQ_1$ and $PQ_2$ are also separated by $\alpha$. This also holds for $B$.
B. From A, follows that if $P\in\partial A\cap S$, there are two points $Q_1,Q_2\in\partial B\cap S$, so that the segments $PQ_1,PQ_2\subset S$.
C. From condition 3, and remarks A and B, and the definition of convex hull, it follows that $S$ is the union of segments $PQ$, where $P\in\partial A\cap S$, and $Q\in\partial B\cap S$. If both $P$ and $Q$ would belong to $\partial A\cap S$, then by condition 3, the segment $PQ$ would be in the interior of the convex hull, so we can rule this possibility out.
D. From condition 4, $\partial A\cap S$ and $\partial B\cap S$ are open curves.
E. From D, and since the curves $\partial A\cap S$ and $\partial B\cap S$ don't intersect, the surface $S-(\partial A\cup\partial B)$ is connected.
F. By using my answer to another question,
When is the hull of a space curve composed of developable patches?, the surface $S$ is made of developable patches. To apply that result, join the two open curves $\partial A\cap S$ and $\partial B\cap S$ by two curves which lie on $S$, so that the union is not self-intersected.
G. From C, E and F, the surface $S-PQ$ is developable, where $PQ$ is as in C.
H. During complete rolling cycle, the curve $\partial A\cap S$ is covered twice, once from one end to the other on one side, and once back, but on the other side. Similarly for $\partial B\cap S$.
Therefore, $S$ has the property that, when rolling, it develops its entire surface.
Examples.
For both the sphericon and the oloid, we take $\alpha\perp\beta$.
For oloid, take $A$ and $B$ two disks, so that each disk has its center on the boundary of the other disk.
For sphericon, take $A$ and $B$ two identical hemidisks, each being symmetric w.r.t. $\alpha\cap\beta$, and so that their centers coincide with their intersection.
The generalization made by Manfred Weis in his comment is also included here.
It is easy to morph the oloid and the sphericon. Take $A',B'$ two identical disks, each being symmetric w.r.t. $\alpha\cap\beta$. Take a plane $\gamma$ through their mass center, perpendicular on $\alpha$ and $\beta$. Cut the circles by $\gamma$, and take $A$ as the part of $A'$ which contains the center of $A'$, and similarly for $B$. It is easy to see that the oloid and sphericon are particular cases of this construction, and by varying the distance between the centers of the disks, we can morph one into the other.
Answers.
Q1. No, the sphericon and the oloid are not unique with these properties, there is an infinity of such objects. Perhaps not many were known.
Q2. In this answer, hopefully.

Update 1.
If the curves $\partial A\cap S$ and $\partial B\cap S$ are line segments, we obtain a tetrahedron. If they are each made of two segments, they can be chosen so that we get an octahedron.
In general, a convex polyhedron whose faces admit a Hamiltonian path, can be rolled so that it develops its entire surface, but is not necessarily obtained by the method I proposed.
One may want to rule out such cases, in which, after the rolling starts, the rolling is not unique, hence for which the contact between the object and the plane surface on which it is rolled is a (sur)face. One may want that at any point, the contact is along a segment. In this case, we have to add the condition that the curves $\partial A\cap S$ and $\partial B\cap S$ don't contain line segments.


TITLE: Weak limits with bounded nets II
QUESTION [1 upvotes]: This is a follow-up to a previous question that I forgot to add: Weak limits with bounded nets.
Let $H$ be a Hilbert space and suppose $\{\psi_{U}\}_{U\in \mathcal{U}}$ is a bounded net in $H$ such that $<v,\psi_{U}>$ converges for every $v\in H$. Why does this imply that $\{\psi_{U}\}$ converges weakly to some $\epsilon\in H$, i.e., that $<(\psi_{U}-\epsilon),v>\rightarrow 0$ for every $v\in H$?  Does it have something to the fact that the unit ball in $H$ is weakly compact?

REPLY [1 votes]: The collection $\{f_U\}_U$ of linear functionals defined by $f_U(v)=\langle v,\psi_U\rangle$ is pointwise convergent. Define $f:=\lim f_U.$ It is easy to see that $f$ is linear. Now, since $\{\psi_U\}_U$ is bounded, $\sup_U\|f_U(v)\|\le \|v\|\|\psi_U\|<\infty$ so the Banach-Steinhaus theorem implies that $\sup_U\|f_U\|=M<\infty.$ Then, $\sup_{x\le 1}\|f(x)\|=\sup_{x\le 1}\|\lim f_U(x)\|<M$ so $f$ is continuous. That is, $f\in L(H).$ To finish, we may apply the Riesz theorem to find an $h\in H$ such that $f(v)=\langle v,h \rangle$ so that $\langle v,\psi_U\rangle\to \langle v,h \rangle.$


TITLE: Find normal form of a subgroup of $(\mathbb{Q}^*, \cdot, 1)$
QUESTION [2 upvotes]: I have troubles solving this exercise for college:
Let $G$ = {$2^h3^k$ | $h,k \in \mathbb{Z}$} be a subgroup of $(\mathbb{Q}^*, \cdot, 1)$. Determine the normal form of $G$.
I tried finding an isomorphism between $G$ and $\mathbb{Z}_p$ $\oplus$ $\mathbb{Z}_n$ to connect it to its primary decomposition, but I had no success. What am I missing? Is this the correct way to proceed?
Sorry if I used some terms that make no sense: I am from Italy and I'm still learning specific terms for college level mathematics in English. Just ask me if you don't understand, I'll try to rewrite it the correct way.

REPLY [3 votes]: $\left(G,\cdot\right)$ is isomorphic to $\left(\mathbb{Z},+\right)\oplus\left(\mathbb{Z},+\right)$ (torsion-free decomposition), through $2^h3^k \mapsto (h,k)$.


TITLE: Can I think of a toy contour as any closed piecewise-smooth curve which is simple?
QUESTION [1 upvotes]: I feel Stein doesn't give the specific definition of toy contour, can I think of a toy contour as any closed piecewise-smooth curve which is simple?

REPLY [3 votes]: Yes, that's a valid way to think about it.
What's going on here is a theorem of topology:

The Jordan Curve Theorem. If $C$ is any continuous simple closed curve then $\mathbb R^2 - C$ has two connected components, one of which (called the "interior") is bounded and the other (called the "exterior") is unbounded.

It's pretty clear from what the author is saying (in the first paragraph of the passages you have shown us) that he/she does not want to spend the time to either state or prove the Jordan Curve Theorem. This is certainly a reasonable decision for a textbook author, because it is a very hard theorem to prove.
Instead, the words "toy contour" are used to mean any continuous simple closed curve for which the conclusion of the Jordan Curve Theorem is obvious. Examples of this would include any convex simple closed curve (an exercise). But he also wants some nonconvex examples such as the "keyhole" example, and the exercise is still pretty easy for those.
And, by the way, there is another theorem with the same hypotheses but still stronger conclusions than the Jordan Curve Theorem, it is known as the Schönflies Theorem. The author's comments in that first paragraph lead me to think that the concept of "toy contour" might also be referring to those stronger conclusions in some form or another.


TITLE: How to solve $xy'+y+x^4y^4e^x=0$?
QUESTION [3 upvotes]: I've tried a lot of tricks but I'm still not able to solve the first order differential equation
$$
xy'+y+x^4y^4e^x=0
$$
It's not an homogeneous or isobaric equation, it's not a simple linear equation and it's also not an exact equation.
For an exact differential equation I would need that:
$$\frac{\partial x}{\partial x}=\frac{\partial(y+x^4y^4e^x)}{\partial y},$$
which is not the case.
Now I've also tried to find some sort of integrating factor $\lambda(x,y)$ to make the above equation exact. But this yielded another difficult equation:
$$
x\frac{\partial \lambda}{\partial x}+(y+x^4y^4e^x)\frac{\partial \lambda}{\partial y}+4\lambda x^4y^3e^x=0
$$
I don't know how to proceed to solve the above differential equation ? I'm thinking of substituting a series, but I'm afraid this would also complexify my problem :(. All tips (substitutions, ...) are appreciated !

REPLY [1 votes]: You write: $xy'+y= - x^4 e^xy^4$. Consider two cases:
Case 1: $y=0$. It is easy to see that $y=0$ is a solution of the equation.
Case 2: $y\ne 0$. Dividing two both sides of the equation by $y^4$, we get
$$x.y^{-4}y+y^{-3}=-x^4.e^x.$$
Put $z=y^{-3}$. Then $z'=-3y^{-4}y'$. So, $y^{-4}.y'=-\frac{1}{3}z$. Then we get
$$-\frac{1}{3}xz'+z=-x^4e^x,$$
or
$$xz'-3z=3x^4e^x,$$
or
$$z'-\frac{3}{x}z=3x^3 e^x$$
Multiplying $\frac{1}{x^3}$ two sides of the latter equation, we get
$$\frac{1}{{{x^3}}}z' - \frac{3}{{{x^4}}}z = 3{e^x},$$
or you can write in the form
$$\frac{d}{{dx}}\left( {\frac{1}{{{x^3}}}.z} \right) = 3{e^x}.$$
This leads to
$$\frac{1}{{{x^3}}}z = \int {3{e^x}dx}  = 3{e^x} + C.$$
So, we get $z=3x^3e^x+Cx^3$. But $z=y^{-3}=\frac{1}{y^3}$, we deduce
$$y = \frac{1}{{\sqrt[3]{z}}} = \frac{1}{{\sqrt[3]{{3{x^3}{e^x} + C{x^3}}}}}.$$


TITLE: Is my interpretation as to why we use trigonometric functions so much in mechanics correct?
QUESTION [0 upvotes]: I tried to have my question answered reading a question related to this here but it didn't really do it for me. Our questions, however are very similar but I think mine asks a slightly different question, being that it is less general and I'm asking through its ubiquity in physics.
Trigonometric functions are everywhere in physics (especially mechanics) and it's gotten me scratching my head as to whether I could explain well why that is.
I'm going to whimsically attempt to explain why it appears in mechanics and, more broadly, when dealing with vectors. I do this because I feel like it's a concise way for me to explain what I think I know about something and lays any of my logical fallacies noticeably naked to be corrected. Please let me know if I'm wrong and about what:

Any line can be composed of two smaller line segments orthogonal to each other. This is basically axiomatic.
We arbitrarily put an arrow on our line to imply direction. It is in now a vector space. We choose to do vectors in such a vector space because nature appears to agree with the results of experiments that subsequently prove that using the rules of such a space produces accurate results.
Any vector can now be composed of two smaller vectors orthogonal to each other.
When thinking about mechanics, it is useful to isolate motion in the $x$ and $y$ dimensions, because they do not interfere with each other (this is relatively axiomatic..?)
With angle $\theta$ next to the origin, the $x$ component vector can be proven to be the cosine of $\theta$ and the $y$ component the sine of $\theta$.
Thus, when modeling the motion of objects that does not move in one direction solely, $x$ and $y$ vector components are a useful way to solve problems in mechanics because all vectors effecting an object can be broken into vectors either orthogonal or in the same dimension to each other. This means they can be considered individually, as things vectors with magnitudes in the same direction undergo arithmetic just like normal numbers do, as vectors of Rank 0 are act in the same way normal real number arithmetic is conducted.

REPLY [1 votes]: Nothing you said was wrong, but there were some things missing, and some extraneous things. Here are the most important parts of what you said, plus a little more information that makes the explanation more fundamental.
There are actually two reasons for the prevalence of trigonometric functions in physics. The first is
because we do a lot of trigonometry.
Specifically, if we choose some set of coordinates, whenever a vector* is not parallel to any coordinate vector, a right triangle can be drawn with the vector as the hypotenuse, with one leg parallel to one coordinate vector.
Assuming the vector has length $L$, and the right triangle that we just drew has an opening angle of $\theta$, we can say that the leg opposite the angle $\theta$ has length $L\sin{\theta}$, and the leg adjacent to the angle $\theta$ has length $L\cos{\theta}$. These two legs have physical meaning - the opposite leg is the component of the vector parallel to the chosen coordinate vector, and the adjacent leg is the projection of the vector onto the plane defined by the other two vectors. This decomposition of vectors can be continued until the result is parallel to one of the coordinate vectors, at which point we have a complete decomposition of the vector into components parallel to each of the coordinate vectors. Each time, the trigonometric functions are necessary to determine the lengths of each component.
This decomposition is enormously useful in classical mechanics mainly because many common systems are vulnerable to a clever choice of coordinates. If you choose your coordinates correctly, then, for example, the dynamics of the system might always be parallel to one of the coordinate vectors, greatly simplifying the analysis of the system.
The second reason is
because harmonic oscillators are everywhere.
The trigonometric functions are solutions to one of the simplest nontrivial second-order differential equations:
$$\frac{d^2 x}{dt^2} = -\omega x$$
This differential equation models an abstract physical system called a simple harmonic oscillator. Many complicated systems reduce to a simple harmonic oscillator when they are perturbed from their natural equilibrium by a small amount. Thus, the trigonometric functions describe the dynamics of a wide swath of problems, from a mass on a spring to electromagnetic radiation.
*The fact that the space we live in can be represented as a vector space is a fundamental assumption of classical mechanics.


TITLE: Linear algebra book, emphasis on linear transformations, first concrete then abstract
QUESTION [2 upvotes]: I'm looking for suggestions for linear algebra books, suitable for first year students, that emphasize linear transformations. Rigorous, with proofs and every detail, ideally starting out with concrete vector spaces, immediarely introducing linear transformations with lots of examples, a bit about linear subspaces, and then developing the subject under the motto "linear algebra = linear transformation theory", leading to abstract ideas and the usual material all presented as means to study linear transformations. Books along similar lines would be of interest.

REPLY [4 votes]: You might want to take a look at Linear Algebra Done Right. The Table of Contents and other information about the book are available at www.linear.axler.net. Linear Algebra Done Right very much emphasizes the linear transformation approach, so it might be suitable for you.


TITLE: Linear combination of inconsistent b and consistent b in Ax = b
QUESTION [1 upvotes]: It's a trivial problem but I'm not sure if my proof is valid it's number 13d in the following link: http://www.math.upenn.edu/ugrad/calc/m240/240la.pdf
For (a,b,c) = (0,0,1) you can trivially show the system is inconsistent by using row reduction, but I'm trying to utilize the hint. I wrote a proof using the hint, but I'm having lingering doubts if what I wrote is entirely correct.
Edited out what I had. I had very confusing notation and was treating the b's as x's.

REPLY [1 votes]: Your solution is fine but I think this is what you meant.
Let $(1,2,4) = u$, $(1,2,3) = v$ and $(0,0,1) = w$. Then $u - v = w$. Now assume there exists $r$ such that $Ar = w$. We have already shown that there is $p$ such that $Ap = u$. Hence $Ap - v = Ar$ i.e. $v = A(p-r)$ which is a contradiction.