\documentclass[reqno]{amsart}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 39, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{7mm}}

\begin{document}

\title[\hfilneg EJDE-2004/39\hfil Structure of group invariants]
{Structure of group invariants \\ of a quasiperiodic flow}

\author[Lennard F. Bakker\hfil EJDE-2004/39\hfilneg]
{Lennard F. Bakker}

\address{Department of Mathematics\\
Brigham Young University\\ 292 TMCB \\
Provo, UT  84602 USA}
\email{bakker@math.byu.edu}

\date{}
\thanks{Submitted July 02 2002. Published March 22, 2004.}
\subjclass[2000]{37C55, 37C80, 20E34, 11R04}
\keywords{Generalized symmetry, quasiperiodic flow, semidirect product}

\begin{abstract}
  It is shown that the multiplier representation of the generalized
  symmetry group of a quasiperiodic flow induces a semidirect product
  structure on certain group invariants (including the generalized
  symmetry group) of the flow's smooth conjugacy class.
\end{abstract}

\maketitle
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\numberwithin{equation}{section}

\section{Introduction}

The generalized symmetry group, $S_\phi$, of a smooth flow
$\phi:\mathbb{R}\times T^n\to T^n$ is the collection of all diffeomorphisms
of $T^n$ that map the generating vector field of $\phi$ to a uniformly
scaled copy of itself (see next section for definitions).
The multiplier representation of $S_\phi$ is the one-dimensional linear
representation
\[
\rho_\phi:S_\phi\to\mathbb{R}^*\equiv\mathop{\rm GL}(\mathbb{R})
\]
 that takes a generalized symmetry $R\in S_\phi$ to its unique multiplier
$\rho_\phi(R)$ (Theorem 2.8 in \cite{BC}), the multiplier being the scalar by
which the generating vector field of $\phi$ is uniformly scaled by $R$.
For each subgroup $\Lambda$ of the multiplier group $\rho_\phi(S_\phi)$,
 the multiplier representation induces the short exact sequence of groups,
 \[
\mathop{\rm id}{}_{T^n}\to \ker \rho_\phi\to  \rho_\phi^{-1}(\Lambda)
\stackrel{j_\Lambda}{\to } \Lambda\to  1,
\]
in which $\mathop{\rm id}_{T^n}$ is the identity diffeomorphism of $T^n$,
$\ker \rho_\phi\to\rho_\phi^{-1}(\Lambda)$ is the canonical monomorphism,
and $j_\Lambda:\rho_\phi^{-1}(\Lambda)\to\Lambda
\cong\rho_\phi^{-1}(\Lambda)/\ker \rho_\phi$ is
$\rho_\phi\vert\rho_\phi^{-1}(\Lambda)$. This short exact sequence indicates
that $\rho_\phi^{-1}(\Lambda)$ is a group extension of $\ker \rho_\phi$
by the Abelian group $\Lambda$. When $\phi$ is a quasiperiodic flow on $T^n$,
it will be shown that
 \begin{enumerate}
\item[(i)] every element of $\rho_\phi(S_\phi)$ is a real algebraic integer
of degree at most $n$ (Corollary \ref{solve4}),
\item[(ii)] $\ker \rho_\phi\cong T^n$ (Corollary \ref{solve7}),
\item[(iii)] every $R\in S_\phi$ with $\rho_\phi(R)=-1$ is an involution
(Corollary \ref{solve8}),
\item[(iv)] $\rho_\phi(S_\phi)$ is isomorphic to an Abelian subgroup of
 $\mathop{\rm GL}(n,\mathbb{Z})$ (Theorem \ref{rep2}), and
\item[(v)] for each subgroup $\Lambda<\rho_\phi(S_\phi)$ there is a splitting map $h_\Lambda:\Lambda\to\rho_\phi^{-1}(\Lambda)$ for the extension (Theorem \ref{rep4}).
\end{enumerate}
The main result (Theorem \ref{rep5}) is that
\[
\rho_\phi^{-1}(\Lambda)=\ker \rho_\phi \rtimes_\Gamma h_{\Lambda}(\Lambda)
\]
for every $\Lambda<\rho_\phi(S_\phi)$; that is, $\rho_\phi^{-1}(\Lambda)$
is the semidirect product of $\ker \rho_\phi$ by $h_{\Lambda}(\Lambda)$
corresponding to the conjugating homomorphism
$\Gamma:h_\Lambda(\Lambda)\to\mathop{\rm Aut}(\ker \rho_\phi)$.

\section{Multipliers and Quasiperiodic Flows}

A generalized symmetry of a (smooth, i.e.\ $C^\infty$) flow $\phi$ on the
$n$-torus $T^n$ ($n\geq 2$) is an $R\in\mathop{\rm Diff}(T^n)$ (the group of
smooth diffeomorphisms on $T^n$) for which there exists an
$\alpha\in\mathbb{R}^*$ such that
\[
R\phi(t,\theta)=\phi\big(\alpha t,R(\theta)\big)\quad\mbox{for all
$t\in \mathbb R$ and all }\theta\in T^n.
\]
This condition is $R\phi_t=\phi_{\alpha t}R$ for all $t\in\mathbb{R}$, where
$\phi_t$ is the diffeomorphism of $T^n$ defined by
 $\phi_t(\theta)=\phi(t,\theta)$. A generalized symmetry of $\phi$ is
 characterized by its action on the generating vector field $X$ of $\phi$,
which vector field is defined by
\[
X(\theta)=\frac{d}{dt} \phi_t (\theta)\Big|_{t=0},\quad\theta\in T^n.
\]
 (In what follows, $\mathbf{T}$ is the tangent functor, and
$R_*X=\mathbf{T}RXR^{-1}$ is the push-forward of $X$ by $R$.)

\begin{theorem}\label{char}
An $R\in\mathop{\rm Diff}(T^n)$ is a generalized symmetry of a flow
$\phi$ on $T^n$ if and only if there exists a unique $\alpha\in \mathbb{R}^*$
such that $R_*X=\alpha X$.
\end{theorem}

For the proof of this theorem, see Proposition 1.4  and Lemma 2.7 in
\cite{BC}.

The generalized symmetry group, $S_\phi$, of a flow $\phi$ on
$T^n$ is the collection of all the generalized symmetries of
$\phi$. The Abelian group
 $F_\phi=\{\phi_t:t\in\mathbb{R}\}\subset\mathop{\rm Diff}(T^n)$ generated
by $\phi$ is a subgroup of the normal subgroup $\ker \rho_\phi$ of $S_\phi$.
On the other hand, $S_\phi$ is the group theoretic normalizer of $F_\phi$ in
$\mathop{\rm Diff}(T^n)$ (Theorem 2.5 \cite{BC}).

The unique $\alpha$ attached to an $R\in S_\phi$ in Theorem \ref{char} is
$\rho_\phi(R)$, the multiplier of $R$. An $R\in S_\phi$ with $\rho_\phi(R)=1$
is known as a (classical) symmetry of $\phi$ (p.8 \cite{LR}); the symmetry
group of $\phi$ is $\ker \rho_\phi=\rho_\phi^{-1}(\{1\})$. An $R\in S_\phi$
with $\rho_\phi(R)=-1$ is called a reversing symmetry (p.4 \cite{LR}); if
$R^2=\mathop{\rm id}_{T^n}$, then $R$ is a reversing involution or a classical
time-reversing symmetry of $\phi$; the reversing symmetry group of $\phi$ is
$\rho_\phi^{-1}(\{1,-1\})$ (p.8 \cite{LR}). An $R\in S_\phi$ with
$\rho_\phi(R)\ne\pm1$, if it exists, is another type of symmetry of $\phi$.
Two flows $\phi$ and $\psi$ are smoothly conjugate if and only if there is a
$V\in\mathop{\rm Diff}(T^n)$ such that $V\phi_t=\psi_tV$ for all $t\in\mathbb{R}$. (This is equivalent to $V_*X=Y$ where $X$ is the generating vector field for $\phi$, and $Y$ is the generating vector field for $\psi$.) A flow $\phi$ on $T^n$ with generating vector field $X$ is quasiperiodic if and only if there exists a $V\in\mathop{\rm Diff}(T^n)$ such that $V_*X$ is a constant vector field whose coefficients are independent over $\mathbb{Q}$ (see pp.79-80 \cite{HB}). (Recall that real numbers $a_1,a_2,\dots ,a_n$ are independent over $\mathbb{Q}$ if for $m=(m_1,m_2,\dots ,m_n)\in\mathbb{Z}^n$, the equation $\sum^n_{j=1}m_ja_j=0$ implies that $m_j=0$ for all $j=1,2,\dots ,n$.) The frequencies of a quasiperiodic flow $\phi$ generated by a constant vector field $X$ are the components of $X$.

\begin{example}\label{example1}
\rm Identify $T^3$ with $S^1\times S^1\times S^1$ where
$S^1=\mathbb{R}/\mathbb{Z}$. Let $\theta=(\theta_1,\theta_2,\theta_3)$
be global coordinates on $T^3$. The quasiperiodic flow $\phi$ on $T^3$
generated by vector field
\[
X=\frac{\partial}{\partial\theta_1}+7^{1/3}\frac{\partial}{\partial\theta_2}+
7^{2/3}\frac{\partial}{\partial\theta_3}
\]
is
\[ \phi_t(\theta)=\phi(t,\theta_1,\theta_2,\theta_3)=\big(\theta_1+t,\theta_2+7^{1/3}t,\theta_3+7^{2/3}t\big),
\]
where the addition in the components of $\phi$ is mod $1$.
For each $c=(c_1,c_2,c_3)\in T^3$, the translation
\[
R_c(\theta_1,\theta_2,\theta_3)=\big(\theta_1+c_1,\theta_2+c_2,\theta_3+c_3\big)
\]
of $T^3$ is a symmetry of $\phi$ because
\[
R_c\phi(t,\theta_1,\theta_2,\theta_3)
 = \big(\theta_1+c_1+t,\theta_2+c_2+7^{1/3}t,\theta_3+c_3+7^{2/3}t\big)
= \theta(t, R_c(\theta_1,\theta_2,\theta_3).
\]
The involution $N(\theta_1,\theta_2,\theta_3)=(-\theta_1,\theta_2,\theta_3)$
of $T^3$ is a reversing symmetry of $\phi$ because
\[
N\phi(t,\theta_1,\theta_2,\theta_3)
 = (-\theta_1-t,-\theta_2-7^{1/3}t,-\theta_3-7^{2/3}t\big)
 = \phi\big(-t,N(\theta_1,\theta_2,\theta_3)\big).
\]
\end{example}

\begin{theorem}\label{mult}
If $\phi$ is a quasiperiodic, then $\{1,-1\}<\rho_\phi(S_\phi)$.
\end{theorem}

\begin{proof}
Suppose $\phi$ is quasiperiodic. Then there is a $V\in\mathop{\rm
Diff}(T^n)$ such that $Y=V_*X$ is a constant vector field. Let
$\psi$ be the flow generated by $Y$. For any $t\in\mathbb{R}$,
the diffeomorphism $\psi_t$ satisfies $(\psi_t)_*Y=Y$, so that
$1\in\rho_\psi(S_\psi)$. On the other hand, the map $N:T^n\to T^n$
defined by $N(\theta)=-\theta$ satisfies $N_*Y=-Y$, so that
$-1\in\rho_\psi(S_\psi)$. The flows $\phi$ and $\psi$ are smoothly
conjugate because $Y=V_*X$. This implies that
$\rho_\phi(S_\phi)=\rho_\psi(S_\psi)$ (Theorem 4.2 \cite{BC}), and
so
$\{1,-1\}<\rho_\phi(S_\phi)$.\end{proof}

\begin{theorem}\label{nonabelian}
If $\phi$ is quasiperiodic and $\Lambda$ is a nontrivial subgroup
of $\rho_\phi(S_\phi)$, then $\rho_\phi^{-1}(\Lambda)$ is
non-Abelian, and hence the generalized symmetry group of $\phi$
and the reversing symmetry group of $\phi$ are
non-Abelian.
\end{theorem}

\begin{proof} Suppose $\phi$ is quasiperiodic and $\Lambda$ is a nontrivial
subgroup of $\rho_\phi(S_\phi)$. Then there is an $R\in S_\phi$
such that $\alpha=\rho_\phi(R)\ne1$. Thus
$R\phi_1=\phi_{\alpha}R$. If $\phi_1=\phi_{\alpha}$, then $\phi$
would be periodic. Thus, $\rho_\phi^{-1}(\Lambda)$ is non-Abelian.
By Theorem \ref{mult}, both $\rho_\phi(S_\phi)$ and
$\rho_\phi\big(\rho_\phi^{-1}(\{1,-1\})\big)$ contain $-1$, so
that $S_\phi=\rho_\phi^{-1}(\rho_\phi(S_\phi))$ and
$\rho_\phi^{-1}(\{1,-1\})$ are both non-Abelian.
\end{proof}

For any $\Lambda<\rho_\phi(S_\phi)$, $\rho_\phi^{-1}(\Lambda)$ is an
invariant of the smooth conjugacy class of $\phi$ in the sense
that if $\phi$ and $\psi$ are smoothly conjugate, then
$\rho_\phi^{-1}(\Lambda)$ and $\rho_\psi^{-1}(\Lambda)$ are
conjugate subgroups of $\mathop{\rm Diff}(T^n)$ (Theorem 4.3
\cite{BC}). Because a quasiperiodic flow $\phi$ is smoothly
conjugate to a quasiperiodic flow $\psi$ generated by a constant
vector field, the group structure of $\mathop{\rm id}_{T_n}\to\ker
\rho_\phi\to\rho_\phi^{-1}(\Lambda)\to\Lambda\to 1$ is determined
by that of $\mathop{\rm id}_{T^n}\to\ker
\rho_\psi\to\rho_\psi^{-1}(\Lambda)\to\Lambda\to 1$. Attention is
therefore restricted to a quasiperiodic flow $\phi$ generated by a
constant vector field $X$.

\section{Lifting the Generalized Symmetry Equation}

The generalized symmetry equation of a flow $\phi$ on $T^n$ is the
equation $R_*X=\alpha X$ that appears in Theorem \ref{char}.
Lifting it from $\mathbf{T}T^n$  to $\mathbf{T}\mathbb{R}^n$, the
universal cover of $\mathbf{T}T^n$, requires lifting the
diffeomorphism $R$ of $T^n$ to a diffeomorphism of $\mathbb{R}^n$,
and lifting the vector field $X$ on $T^n$ to a vector field on
$\mathbb{R}^n$. The covering map $\pi:\mathbb{R}^n\to T^n$ is a
local diffeomorphism for which
\[
\pi(x+m)=\pi(x)
\]
for any $x\in\mathbb{R}^n$ and any $m\in\mathbb{Z}^n$. Let $R:T^n\to T^n$ be a
continuous map. A {\it lift} of $R\pi:\mathbb{R}^n\to T^n$ is a continuous map
$Q:\mathbb{R}^n\to\mathbb{R}^n$ for which $R\pi=\pi Q$. Since $\pi$ is a fixed map,
$Q$ is also said to be a lift of $R$. Any two lifts of $R$ differ by a deck
transformation of $\pi$, which is a translation of $\mathbb{R}^n$ by an
$m\in\mathbb{Z}^n$.

\begin{theorem}\label{lift3}
Let $R:T^n\to T^n$ and $Q:\mathbb{R}^n\to\mathbb{R}^n$. Then $Q$
is a lift of a diffeomorphism $R$ of $T^n$ if and only if $Q$ is a
diffeomorphism of $\mathbb{R}^n$ such that {\rm a)} for any
$m\in\mathbb{Z}^n$, $Q(x+m)-Q(x)$ is independent of
$x\in\mathbb{R}^n$, and {\rm b)} the map $l_Q(m)=Q(x+m)-Q(x)$ is
an isomorphism of $\mathbb{Z}^n$.
\end{theorem}

The proof of this theorem uses standard arguments in topology, we omit it.

The canonical projections $\tau_{\mathbb{R}^n}:\mathbf{T}\mathbb{R}^n\to\mathbb{R}^n$
and $\tau_{T^n}:\mathbf{T}T^n\to T^n$
are smooth. The former is a lift of the latter,
\[
\tau_{T^n}\mathbf{T}\pi=\pi\tau_{\mathbb{R}^n},
\]
which lift sends $w\in\mathbf{T}_x\mathbb{R}^n$ to $x\in\mathbb{R}^n$.
The covering map $\mathbf{T}\pi:\mathbf{T}\mathbb{R}^n \to \mathbf{T}T^n$ is a
local diffeomorphism.A vector field on $T^n$ is a smooth map
$Y:T^n\to\mathbf{T}T^n$ such that $\tau_{T^n}Y=\mathop{\rm id}_{T^n}$.
A vector field on $\mathbb{R}^n$ is a smooth map
$Z:\mathbb{R}^n\to\mathbf{T}\mathbb{R}^n$ such that
$\tau_{\mathbb{R}^n}Z=\mathop{\rm id}_{\mathbb{R}^n}$.

\begin{lemma}\label{lift4}
If $Y$ is a vector field on $T^n$, then there is only one lift of $Y$ that is a
vector field on $\mathbb{R}^n$.
\end{lemma}

\begin{proof}
Let $x_0\in\mathbb{R}^n$, $\theta_0\in T^n$ be such that
$Y\pi(x_0)=Y(\theta_0)$. Let
$w_{x_0}\in\mathbf{T}_{x_0}\mathbb{R}^n$ be the only vector such
that $\mathbf{T}\pi(w_{x_0})=Y(\theta_0)$. By the Lifting Theorem
(Theorem 4.1, p.143 \cite{BR}), there exists a unique lift
$Z:\mathbb{R}^n\to\mathbf{T}\mathbb{R}^n$ such that
$Y\pi=\mathbf{T}\pi Z$ and $Z(x_0)=w_{x_0}$. It needs only be
checked that this $Z$ is a vector field. Because $Y$ is a vector
field on $T^n$, $Z$ is a lift of $Y\pi$, and $\tau_{\mathbb{R}^n}$
is a lift of $\tau_{T^n}$, it follows that
\[
\pi(x)=\tau_{T_n}Y\pi(x)=\tau_{T^n}\mathbf{T}\pi
Z(x)=\pi\tau_{\mathbb{R}^n}Z(x).
\]
So the difference $x-\tau_{\mathbb{R}^n}Z(x)$ is a discrete valued map.
Because $\mathbb{R}^n$ is connected, this difference is a constant
(see Proposition 4.5, p.10 \cite{BR}). This constant is zero because
$\tau_{\mathbb{R}^n}Z(x_0)=x_0$, and so
$\tau_{\mathbb{R}^n}Z=\mathop{\rm id}_{\mathbb{R}^n}$.
The equation $Y\pi=\mathbf{T}\pi Z$ implies that $Z$ is smooth because $\pi$
and $\mathbf{T}\pi$ are local diffeomorphisms and because $Y$ is smooth.
The choice of the only vector $w\in\mathbf{T}_{x_0+m}\mathbb{R}^n$ for any
$0\ne m\in\mathbb{Z}^n$ such that $\mathbf{T}\pi(w)=Y(\theta_0)$ would lead to
a lift $Z_m$ of $Y$ that is not a vector field on $\mathbb{R}^n$ because
$\tau_{\mathbb{R}^n}Z_m(x)=x+m$. The collection $\{Z_m:m\in\mathbb{Z}\}$,
with $Z_0=Z$, accounts for all the lifts of $Y$ by the uniqueness of the lift
and the uniqueness of the vector $w$. Therefore $Z$ is the only lift of $Y$
that is a vector field on $\mathbb{R}^n$.
\end{proof}

For a vector field $X$ on $T^n$, let $\hat X$ denote the only lift
of $X$ that is a vector field on $\mathbb{R}^n$ as described in
Lemma \ref{lift4}; $\hat X$ satisfies $X\pi=\mathbf{T}\pi \hat X$.
For a diffeomorphism $R$ of $T^n$, let $\hat R$ be a lift of $R$;
the lift $\hat R$ is a diffeomorphism of $\mathbb{R}^n$ (by
Theorem \ref{lift3}) for which $R\pi=\pi\hat R$.

\begin{lemma}\label{lift5}
The only lift of the vector field $R_*X$ on $T^n$ that is a vector field on
$\mathbb{R}^n$ is $\hat R_*\hat X$.\end{lemma}

\begin{proof}A lift of $R_*X$ is $\hat R_*\hat X$ because
\begin{align*}
\mathbf{T}\pi\hat R_*\hat X
& =\mathbf{T}\pi\mathbf{T}\hat R\hat X\hat R^{-1}
= \mathbf{T}(\pi \hat R)\hat X\hat R^{-1}
= \mathbf{T}(R\pi)\hat X\hat R^{-1} \\
& = \mathbf{T}R\mathbf{T}\pi\hat X\hat R^{-1}
= \mathbf{T}RX\pi\hat R^{-1}
= \mathbf{T}RXR^{-1}\pi = R_*X\pi.
\end{align*}
By definition, $\hat R_*\hat X$ is a vector field on $\mathbb{R}^n$.
By Lemma \ref{lift4}, it is the only lift of $R_*X$ that is a vector field
on $\mathbb{R}^n$.\end{proof}

\begin{lemma}\label{lift6}
For any $\alpha\in\mathbb{R}^*$, the only lift of the vector field
$\alpha X$ on $T^n$ that is a vector field on $\mathbb{R}^n$ is
$\alpha\hat X$.\end{lemma}

\begin{proof}A lift of $\alpha X$ is
$\alpha\hat X$ because
$\mathbf{T}\pi(\alpha\hat X)=\alpha\mathbf{T}\pi\hat X=\alpha X\pi$.
Only one lift of $\alpha X$ is a vector field (Lemma \ref{lift4}), and
$\alpha\hat X$ is this lift.
\end{proof}

\begin{theorem}\label{lift7}
Let $X$ be a vector field on $T^n$, $\hat X$ the lift of $X$ that is a vector
field on $\mathbb{R}^n$, $R$ a diffeomorphism of $T^n$, $\hat R$ a
lift of $R$, and $\alpha$ a nonzero real number. Then $R_*X=\alpha
X$ if and only if $\hat R_*\hat X=\alpha\hat X$.
\end{theorem}

\begin{proof} Suppose that $R_*X=\alpha X$. By
Lemma \ref{lift5}, $\hat R_*\hat X$ is a lift of $R_*X$:
$\mathbf{T}\pi\hat R_*\hat X=R_*X\pi$. By Lemma \ref{lift6},
$\alpha \hat X$ is a lift of $\alpha X$: $\mathbf{T}\pi(\alpha
\hat X)=\alpha X\pi$. Then
\[
\mathbf{T}\pi\big(\hat R_*\hat
X-\alpha\hat X\big)=\big(R_*X-\alpha X\big)\pi=\mathbf{0}_{T^n}\pi,
\]
where $\mathbf{0}_{T^n}$ is the zero vector field on $T^n$. So
$\hat R_*\hat X-\alpha \hat X$ is a lift of $\mathbf{0}_{T^n}$. The only lift of
$\mathbf{0}_{T^n}$ that is a vector field on $\mathbb{R}^n$ is
$\mathbf{0}_{\mathbb{R}^n}$, the zero vector field on $\mathbb{R}^n$.
By Lemma \ref{lift5} and Lemma \ref{lift6}, the difference
$\hat R_*\hat X-\alpha\hat X$ is a vector field on $\mathbb{R}^n$. By
Lemma \ref{lift4}, $\hat R_*\hat X-\alpha\hat X=\mathbf{0}_{\mathbb{R}^n}$.
Thus, $\hat R_*\hat X=\alpha\hat X$.Suppose that $\hat R_*\hat X=\alpha\hat X$.
Then
\begin{align*}
R_*X\pi& = \mathbf{T}RXR^{-1}\pi=
\mathbf{T}RX\pi\hat R^{-1} =  \mathbf{T}R\mathbf{T}\pi\hat X\hat R^{-1} \\
& =  \mathbf{T}(R\pi)\hat X\hat R^{-1} =
\mathbf{T}(\pi\hat R)\hat X\hat R^{-1}=
\mathbf{T}\pi\mathbf{T}\hat R\hat X\hat R^{-1} \\
& =
\mathbf{T}\pi\hat R_*\hat X = \mathbf{T}\pi(\alpha\hat X)=
\alpha\mathbf{T}\pi\hat X=  \alpha X\pi.
\end{align*}
The surjectivity of $\pi$ implies that $R_*X=\alpha X$.
\end{proof}

\section{Solving the Lifted Generalized Symmetry Equation}

The lift of $R_*X=\alpha X$ is an equation on
$\mathbf{T}\mathbb{R}^n$ of the form $Q_*\hat X=\alpha\hat X$ for
$Q\in\mathop{\rm Diff}(\mathbb{R}^n)$. With global coordinates
$x=(x_1,x_2,\dots ,x_n)$ on $\mathbb{R}^n$, the diffeomorphism $Q$
has the form
\[
Q(x_1,x_2,\dots ,x_n)=(f_1(x_1,x_2,\dots ,x_n),\dots ,f_n(x_1,x_2,\dots ,x_n))
\]
for smooth functions $f_i:\mathbb{R}^n\to\mathbb{R}$, $i=1,\dots ,n$.
Let $\theta=(\theta_1,\theta_2,\dots ,\theta_n)$ be global coordinates on $T^n$
such that $\theta_i=x_i$ mod 1, $i=1,2,\dots ,n$. If
\[
X(\theta)=a_1\frac{\partial}{\partial \theta_1}+
a_2\frac{\partial}{\partial \theta_2}+\dots+a_n\frac{\partial}{\partial \theta_n}
\]
for constants $a_i\in\mathbb{R}$, $i=1,\dots ,n$, then
\[
\hat X(x)= a_1\frac{\partial}{\partial x_1}+ a_2\frac{\partial}{\partial x_2}
+\dots +a_n\frac{\partial}{\partial x_n},
\]
so that $Q_*\hat X=\alpha\hat X$ has the form
\[
\sum^n_{j=1}a_j\frac{\partial f_i}{\partial x_j}=\alpha a_i,\ i=1,\dots ,n.
\]
This is an uncoupled system of linear, first order equations which is readily
solved for its general solution.

\begin{lemma}\label{solve1}
For real numbers $a_1,a_2,\dots ,a_n$ and $\alpha$ with $a_n\ne0$, the general
solution of the system of $n$ linear partial differential equations
\[
\sum^n_{j=1}a_j\frac{\partial f_i}{\partial x_j}=\alpha a_i,\ i=1,\dots ,n
\]
is
\[f_i(x)=\alpha\frac{a_i}{a_n}x_n + h_i\Big(x_1-\frac{a_1}{a_n}x_n,x_2
-\frac{a_2}{a_n}x_n,\dots ,x_{n-1}-\frac{a_{n-1}}{a_n}x_n\Big),
\]
for arbitrary smooth functions $h_i:\mathbb{R}^{n-1}\to\mathbb{R}$, $i=1,\dots ,n$.
\end{lemma}

\begin{proof}
For each $i=1,\dots ,n$, consider the initial value problem
\begin{gather*}
\sum^n_{j=1}a_j\frac{\partial f_i}{\partial x_j}=\alpha a_i \\
x_j(0,s_1,s_2,\dots ,s_{n-1}) = s_j \hbox{\rm\ for\ }j=1,\dots ,n-1 \\
x_n(0,s_1,s_2,\dots ,s_{n-1}) = 0 \\
f_i(0,s_1,s_2,\dots ,s_{n-1}) = h_i(s_1,s_2,\dots ,s_{n-1})
\end{gather*}
for parameters $(s_1,s_2,\dots ,s_{n-1})\in\mathbb{R}^{n-1}$ and initial data
$h_i:\mathbb{R}^{n-1}\to\mathbb{R}$. Using the method of characteristics
(see \cite{JO} for example), the solution of the initial value problem in
parametric form is
\begin{gather*}
x_j(t,s_1,s_2,\dots ,s_{n-1})  =  a_jt+s_j\hbox{ for }j=1,\dots ,n-1 \\
x_n(t,s_1,s_2,\dots ,s_{n-1}) =  a_nt \\
f_i(t,s_1,s_2,\dots ,s_{n-1})  =  \alpha a_it +h_i(s_1,s_2,\dots ,s_{n-1}).
\end{gather*}
The coordinates $(x_1,x_2,\dots ,x_n)$ and the parameters
$(t,s_1,s_2,\dots ,s_{n-1})$ are related by
\[
 \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_{n-1} \\ x_n \end{bmatrix}
 =\begin{bmatrix}a_1 & 1 & 0 & 0 & \dots & 0 \\a_2 & 0 & 1 & 0 & \dots & 0 \\
 a_3 & 0 & 0 & 1 & \ldots & 0 \\\vdots & \vdots & \vdots & \vdots & \ddots &
 \vdots \\ a_{n-1} & 0 & 0 & 0 & \ldots & 1 \\
a_n & 0 & 0 & 0 & \ldots & 0\end{bmatrix}
\begin{bmatrix} t \\ s_1 \\ s_2 \\ \vdots \\ s_{n-2} \\ s_{n-1} \end{bmatrix}
\]
The determinant of the  $n\times n$ matrix is $(-1)^na_n$, which is nonzero by
hypothesis. Inverting the matrix equation gives
\[
\begin{bmatrix} t \\ s_1 \\ s_2 \\ \vdots \\ s_{n-2} \\ s_{n-1} \end{bmatrix}
=\begin{bmatrix}0 & 0 & \ldots & 0 & 0 & 1/a_n \\1 & 0 & \ldots & 0 & 0 & -a_1/a_n
\\0 & 1 & \ldots & 0 & 0 & -a_2/a_n \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & 0
& -a_{n-2}/a_n \\0 & 0 & \ldots & 0 & 1 & -a_{n-1}/a_n \end{bmatrix}
\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_{n-1} \\ x_n\end{bmatrix}
\]
Substitution of the expressions for $t$ and the $s_i$'s in terms of the $x_i$'s
into
\[
f_i(x_1,x_2,\dots ,x_n)=\alpha a_it+h_i(s_1,s_2,\dots ,s_{n-1})
\]
gives the desired form of the general solution. \end{proof}

\begin{lemma}\label{solve2}
If $a_1,a_2,\dots ,a_n$ are independent over $\mathbb{Q}$, then
\[
J=\Big\{\Big(m_1-\frac{a_1}{a_n}m_n,\dots ,m_{n-1}-\frac{a_{n-1}}{a_n}m_n\Big):
m_1,dots ,m_n\in\mathbb{Z}\Big\}
\]
is a dense subset of $\mathbb{R}^{n-1}$.\end{lemma}

\begin{proof}Suppose $a_1,a_2,\dots ,a_n$ are independent over $\mathbb{Q}$.
This implies that none of the $a_i$'s are zero. In particular, $a_n\ne0$.
Consider the flow
\[
\psi_t(\theta_1,\dots ,\theta_{n-1},\theta_n)=(\theta_1-(a_1/a_n)t,\dots ,
\theta_{n-1}-(a_{n-1}/a_n)t,\theta_n-t)
\]
on $T^n$ which is generated by the vector field
\[
Y=-\frac{a_1}{a_n}\frac{\partial}{\partial \theta_1}-\frac{a_2}{a_n}
\frac{\partial}{\partial \theta_2}-\dots -\frac{a_{n-1}}{a_n}
\frac{\partial}{\partial \theta_{n-1}}-\frac{\partial}{\partial \theta_n}.
\]
The coefficients of $Y$ are independent over $\mathbb{Q}$ because
$a_1,a_2,\dots ,a_n$ are independent over $\mathbb{Q}$ and
\[
m_1a_1+\dots +m_na_n=0\Leftrightarrow -m_1\frac{a_1}{a_n}-\dots -m_{n-1}
\frac{a_{n-1}}{a_n}-m_n=0.
\]
So the orbit of $\psi$ through any point $\theta_0\in T^n$,
\[
\gamma_\psi(\theta_0)=\{\psi_t(\theta_0):t\in\mathbb{R}\},
\]
is dense in $T^n$ (Corollary 1, p. 287 \cite{AR}).The submanifold
\[
P=\{(\theta_1,\dots ,\theta_{n-1},\theta_n):\theta_n=0\}
\]
of $T^n$, which is diffeomorphic to $T^{n-1}$, is a global Poincar\'e section
for $\psi$ because $X(\theta)\not\in\mathbf{T}_\theta P$ for every $\theta\in P$
and because $\gamma_\psi(\theta_0)\cap P\ne\emptyset$ for every $\theta_0\in T^n$.
Define the projection $\wp:T^n \to T^{n-1}$ by
\[
\wp(\theta_1,\theta_2,\dots ,\theta_{n-1},\theta_n)=(\theta_1,\theta_2,\dots ,
\theta_{n-1})
\]
and the injection $\imath:T^{n-1}\to T^n$ by\[\imath(\theta_1,\theta_2,\dots ,
\theta_{n-1})=(\theta_1,\theta_2,\dots ,\theta_{n-1},0).
\]
The Poincar\'e map induced on $\wp(P)$ by $\psi$ is given by
$\bar\psi=\wp\psi_1\imath$ because $\psi_1(\theta_0)\in P$ when $\theta_0\in P$.
For any $\kappa\in\mathbb{Z}$, $\bar\psi^\kappa=\wp\psi_{\kappa}\imath$.
So, for instance, with $0=(0,0,\dots ,0)\in T^n$ and $\bar 0=\wp(0)$,
\[
\wp\big(\gamma_\psi(0)\cap P\big)
 =  \{\bar\psi^\kappa(\bar 0):\kappa\in\mathbb{Z}\}
 =  \Big\{\Big(-\frac{a_1}{a_n}\kappa,-\frac{a_2}{a_n}\kappa,\dots ,
-\frac{a_{n-1}}{a_n}\kappa\Big):\kappa\in\mathbb{Z}\Big\},
\]
where for each $i=1,\dots ,n-1$, the quantity $-(a_i/a_n)\kappa$ is taken mod 1.
With $\bar\pi:\mathbb{R}^{n-1}\to T^{n-1}$ as the covering map,
\[
J=\bar\pi^{-1}\big(\wp(\gamma_\psi(0)\cap P)\big).
\]
If $\wp(\gamma_\psi(0)\cap P)$ were dense in $\wp(P)$, then $J$ would be dense
in $R^{n-1}$ because $\bar\pi$ is a covering map.
(That is, if $\wp(\gamma_\psi(0)\cap P)\cap[0,1)^{n-1}$ is dense in the fundamental
domain $[0,1)^{n-1}$ of the covering map $\bar\pi$, then by translation,
it is dense in $\mathbb{R}^{n-1}$.)Define $\chi:\mathbb{R}\times T^{n-1}\to T^n$ by
\[
\chi(t,\theta_1,\theta_2,\dots ,\theta_{n-1})=\psi\big(t,\imath(\theta_1,\theta_2,
\dots ,\theta_{n-1})\big).
\]
The map $\chi$ is a local diffeomorphism by the Inverse Function Theorem because
\[
\mathbf{T}\chi=\begin{bmatrix}-a_1/a_n & 1 & 0 & \ldots & 0 \\-a_2/a_n & 0 & 1 &
\ldots & 0 \\\vdots & \vdots & \vdots & \ddots & \vdots \\-a_{n-1}/a_n & 0 & 0
&\ldots & 1 \\-1 & 0 & 0 & \ldots & 0\end{bmatrix}
\]
has determinant of $(-1)^{n+1}$. Let $O$ be a small open subset of $\wp(P)$.
For $\epsilon>0$, the set $O_\epsilon=(-\epsilon,\epsilon)\times O$ is an open
subset in the domain of $\chi$. For $\epsilon$ small enough, the image
$\chi(O_\epsilon)$ is open in $T^n$ because $\chi$ is a local diffeomorphism.
By the denseness of $\gamma_\psi(0)$ in $T^n$, there is a point $\theta_0$ in
$\chi(O_\epsilon)\cap\gamma_\psi(0)$. By the definition of $\chi(O_\epsilon)$,
there is an $\bar\epsilon\in(-\epsilon,\epsilon)$ and a $\bar\theta_0\in O$ such
that $\chi(\bar\epsilon,\bar\theta_0)=\theta_0$.
Thus $\imath(\bar \theta_0)\in\gamma_\psi(0)$, and so $\wp(\gamma_\psi(0)\cap P)$
intersects $O$ at $\bar\theta_0$. Since $O$ is any small open subset of $\wp(P)$,
the set $\wp(\gamma_\psi(0)\cap P)$ is dense in $\wp(P)$.
\end{proof}

\begin{theorem}\label{solve3}
If $\alpha\in\mathbb{R}^*$ and the coefficients of
$X=\sum_{i=1}^n a_i\partial/\partial\theta_i$ are independent over $\mathbb{Q}$,
then for each $R\in\mathop{\rm Diff}(T^n)$ that satisfies $R_*X=\alpha X$ there
exist $B=(b_{ij})\in\mathop{\rm GL}(n,\mathbb{Z})$ and $c\in\mathbb{R}^n$ such that
\[\hat R(x)=Bx+c
\]
for $x=(x_1,x_2,\dots ,x_n)$, in which\[b_{in}= \alpha\frac{a_i}{a_n}-\sum_{j=1}^{n-1}b_{ij}\frac{a_j}{a_n},\ i=1,\dots ,n.
\]\end{theorem}

\begin{proof}
Suppose that the $a_1,a_2,\dots ,a_n$ are independent over $\mathbb{Q}$. For
$\alpha\in\mathbb{R}^*$, suppose that $R\in\mathop{\rm Diff}(T^n)$ is a solution
of $R_*X=\alpha X$. A lift $\hat R$ of $R$ is a diffeomorphism of $\mathbb{R}^n$
by Theorem \ref{lift3}. The lift of $X$ that is a vector field on $\mathbb{R}^n$
is $\hat X=\sum^n_{i=1}a_i(\partial/\partial x_i)$. By Theorem \ref{lift7},
 $\hat R$ is a solution of $\hat R_*\hat X=\alpha\hat X$. With global coordinates
 $(x_1,x_2,\dots ,x_n)$ on ${\mathbb R }^n$ write
\[
 \hat R(x)=(f_1(x_1,\dots ,x_n),\dots ,f_n(x_1,\dots ,x_n)).
\]
In terms of this coordinate description, the equation $\hat R_*\hat X=\alpha X$
written out is
\[
\sum_{j=1}^n a_j\frac{\partial f_i}{\partial x_j}=\alpha a_i,\ i=1,\dots ,n.
\]
The independence of the coefficients of $\hat X$ over $\mathbb{Q}$ implies that
$a_n\ne0$. By Lemma \ref{solve1}, there are smooth functions
$h_i:\mathbb{R}^{n-1}\to\mathbb{R}$, $i=1,\dots ,n$, such that
\[
f_i(x_1,\dots ,x_n)=\alpha\frac{a_i}{a_n}x_n+h_i(s_1,s_2,\dots ,s_{n-1})
\]
where
\[
s_i=x_i-\frac{a_i}{a_n}x_n,\ i=1,\dots ,n-1.
\]
By Theorem \ref{lift3}, $\hat R(x+m)-\hat R(x)$ is independent of $x$ for each
$m\in\mathbb{R}^n$. This implies for each $i=1,\dots ,n$ that
\begin{align*}
&  f_i(x+m)-f_i(x)\\
& =f_i(x_1+m_1,x_2+m_2,\dots ,x_n+m_n)-f_i(x_1,x_2,\dots ,x_n) \\
& =\alpha\frac{a_i}{a_n}m_n + h_i\big(s_1+m_1-\frac{a_1}{a_n}m_n,\dots ,s_{n-1}
+m_{n-1}-\frac{a_{n-1}}{a_n}m_n\big)
- h_i(s_1,\dots ,s_{n-1})
\end{align*}
is independent of $x$ for every $m=(m_1,m_2,\dots ,m_n)\in\mathbb{Z}^n$.
This independence means that $f_i(x+m)-f_i(x)$ is a function of $m$ only.
So for each $j=1,\dots ,n-1$,
\begin{align*}
0 & =  \frac{\partial}{\partial x_j}\big[f_i(x_1+m_1,x_2+m_2,\dots,x_n+m_n)
-f_i(x_1,x_2,\dots ,x_n)\big] \\
& = \frac{\partial h_i}{\partial s_j}\Big(s_1+m_1-\frac{a_1}{a_n}m_n,\dots ,
s_{n-1}+m_{n-1}-\frac{a_{n-1}}{a_n}m_n\Big)
-\frac{\partial h_i}{\partial s_j}\big(s_1,\dots ,s_{n-1}\big).
\end{align*}
So, in particular
\[
\frac{\partial h_i}{\partial s_j}\Big(m_1-\frac{a_1}{a_n}m_n,\dots ,m_{n-1}
-\frac{a_{n-1}}{a_n}m_n\Big)= \frac{\partial h_i}{\partial s_j}\big(0,\dots
,0\big)
\]
for all $(m_1,\dots ,m_n)\in\mathbb{Z}^n$. By Lemma \ref{solve2}, the set
\[
\Big\{\Big(m_1-\frac{a_1}{a_n}m_n,\dots ,m_{n-1}-\frac{a_{n-1}}{a_n}m_n\Big): m_1,\dots
,m_n\in\mathbb{Z}\Big\}
\]
is dense in $\mathbb{R}^{n-1}$, which together with the smoothness of $h_i$
implies that $\partial h_i/\partial s_j$ is a constant. Let this constant be
$b_{ij}$ for $i=1,\dots ,n$, $j=1,\dots ,n-1$. By Taylor's Theorem,
\[
h_i(s_1,\dots ,s_{n-1})=c_i + \sum_{j=1}^{n-1} b_{ij}s_j
\]
for constants $c_i\in\mathbb{R}$. Thus,
\begin{align*}
f_i(x_1,\dots ,x_n)
& =  c_i + \alpha\frac{a_i}{a_n}x_n+\sum_{j=1}^{n-1} b_{ij}
\Big(x_j-\frac{a_j}{a_n}x_n\Big) \\
& =  c_i + \sum_{j=1}^{n-1}b_{ij}x_j+\Big(\alpha\frac{a_i}{a_n}
-\sum_{j=1}^{n-1}b_{ij}\frac{a_j}{a_n}\Big)x_n.
\end{align*}
For each $i=1,2,\dots ,n$, set
\[
b_{in}=\alpha\frac{a_i}{a_n}-\sum_{j=1}^{n-1}b_{ij}\frac{a_j}{a_n}\Big.
\]
Then for each $i=1,2,\dots ,n$,
\[
f_i(x_1,x_2,\dots ,x_n)=c_i + \sum_{j=1}^n b_{ij}x_j.
\]
So $\hat R$ has the form $\hat R(x)=Bx+c$ where $B=(b_{ij})$ is an $n\times n$
matrix, and $c\in\mathbb{R}^n$.By Theorem \ref{lift3}, the map
$l_{\hat R}(m)=\hat R(x+m)-\hat R(x)$ is an isomorphism of $\mathbb{Z}^n$.
By the formula for $f_i$ derived above,
\[
f_i(x_1+m_1,\dots ,x_n+m_n)-f_i(x_1,x_2,\dots ,x_m)=\sum_{j=1}^n b_{ij}m_j
\]
for each $i=1,2,\dots ,n$. This implies that
$l_{\hat R}(m)=Bm$.
Since $l_{\hat R}$ is an isomorphism of $\mathbb{Z}^n$, it follows that
$B\in\mathop{\rm GL}(n,\mathbb{Z})$.
\end{proof}

Theorem \ref{solve3} restricts the search for lifts of generalized symmetries of
a quasiperiodic flow on $T^n$ to affine maps on $\mathbb{R}^n$ of the form
$Q(x)=Bx+c$ for $B\in{\rm GL}(n,\mathbb{Z})$ and $c\in\mathbb{R}^n$.
For an affine map of this form, the difference
\[
Q(x+m)-Q(x)=B(x+m)+c-(Bx+c)=Bm
\]
is independent of $x$, and the map $l_Q(m)=Q(x+m)-Q(x)$ is an isomorphism of
$\mathbb{Z}^n$, so that $Q$ is a lift of a diffeomorphism $R$ on $T^n$ by
Theorem \ref{lift3}. If $Q$ is a solution of $Q_*\hat X=\alpha\hat X$, then by
Theorem \ref{lift7}, $R$ is a solution of $R_*X=\alpha X$, so that by
Theorem \ref{char}, $R\in S_\phi$.The following two corollaries of
Theorem \ref{solve3} restrict the possibilities for the multipliers of the
generalized symmetries of a quasiperiodic flow on $T^n$. One restriction employs
the notion of an {\it algebraic integer}, which is a complex number that is a root
of a monic polynomial in the polynomial ring $\mathbb{Z}[z]$. If $m$ is the
smallest degree of a monic polynomial in $\mathbb{Z}[z]$ for which an algebraic
integer is a root, then $m$ is the {\it degree} of that algebraic integer
(Definition 1.1, p.1 \cite{RM}).

\begin{corollary}\label{solve4}
If $\phi$ is a quasiperiodic flow on $T^n$ with generating vector field
$X=\sum^n_{i=1}a_i\partial/\partial\theta_i$, then each
$\alpha\in\rho_\phi(S_\phi)$ is a real algebraic integer of degree at most $n$,
and $\rho_\phi(S_\phi)\cap\mathbb{Q}=\{1,-1\}$.
\end{corollary}

\begin{proof}
For each $\alpha\in\rho_\phi(S_\phi)$ (which is real) there is an $R\in S_\phi$
such that $\rho_\phi(R)=\alpha$. By Theorem \ref{solve3} there is a
$B\in\mathop{\rm GL}(n,\mathbb{Z})$ such that $\mathbf{T}\hat R=B$.
Then by Theorem \ref{char} and Theorem \ref{lift7},
\[
B\hat X=\hat R_*\hat X=\alpha\hat X.
\]
So, $\alpha$ is an eigenvalue of $B$ (and $\hat X$ is an eigenvector of $B$.)
The characteristic polynomial of $B$ is an n-degree monic polynomial in
$\mathbb{Z}[z]$:
\[
z^n+d_{n-1}z^{n-1}+\dots +d_1z+d_0.
\]
Thus $\alpha$ is a real algebraic integer of degree at most $n$. The value of $d_0$
is $\hbox{\rm det}(B)$, which is a unit in $\mathbb{Z}$
(Theorem 3.5, p.351 \cite{HU}). The only units in $\mathbb{Z}$ are $\pm 1$.
So the only possible rational roots of the characteristic polynomial of $B$ are
$\pm1$ (Proposition 6.8, p.160 \cite{HU}). This means that
$\rho_\phi(S_\phi)\cap\mathbb{Q}\subset\{1,-1\}$. But
$\rho_\phi(S_\phi)\cap\mathbb{Q}\supset\{1,-1\}$ by Theorem \ref{mult}.
Thus, $\rho_\phi(S_\phi)\cap\mathbb{Q}=\{1,-1\}$.
\end{proof}

The other restriction on the possibilities for the multipliers of any generalized
symmetries of $\phi$ employs linear combinations over $\mathbb{Z}$ of pair wise
ratios of the entries of the ``eigenvector'' $\hat X$ (which entries are the
frequencies of $\phi$).\begin{corollary}\label{solve5}If $\phi$ is a quasiperiodic
flow on $T^n$ with generating vector field
$X=\sum^n_{i=1}a_i\partial/\partial\theta_i$, then for any
$\alpha\in\rho_\phi(S_\phi)$ there exists a
$B=(b_{ij})\in\mathop{\rm GL}(n,\mathbb{Z})$ such that
\[
\alpha=\sum_{j=1}^n b_{ij}\frac{a_j}{a_i},\ \ i=1,\dots ,n.
\]
\end{corollary}

\begin{proof}
Suppose that $\alpha\in\rho_\phi(S_\phi)$. Then there is an $R\in S_\phi$ such that
$\alpha=\rho_\phi(R)$. By Theorem \ref{solve3}, there is a
$B=(b_{ij})\in{\rm GL}(n,\mathbb{Z})$ such that
$\mathbf{T}\hat R=B$ with
\[
b_{in}= \alpha\frac{a_i}{a_n}-\sum_{j=1}^{n-1}b_{ij}\frac{a_j}{a_n},\quad
  i=1,\dots ,n.
\]
Solving this equation for $\alpha$ gives
\[
\alpha=\sum_{j=1}^n b_{ij}\frac{a_j}{a_i},\ \  i=1,\dots ,n.
\]
\end{proof}

The multiplier group of any quasiperiodic flow $\phi$ always contains $\{1,-1\}$
as stated in Theorem \ref{mult}. For each $t\in\mathbb{R}$, the diffeomorphism
$\phi_t$ is in $S_\phi$ by definition. A lift of $\phi_t$ is
$\hat \phi_t(x)=Ix+t\hat X$, where $I=\delta_{ij}$ is the $n\times n$
identity matrix, so that by Corollary \ref{solve5},
\[
\alpha=\sum^n_{j=1}\delta_{ij}\frac{a_j}{a_i}=\frac{a_i}{a_i}=1
\]
for each $i=1,\dots ,n$. A lift of the reversing involution $N$ defined in the
proof of Theorem \ref{mult} is $\hat N(x)=-Ix$, so that by
 Corollary \ref{solve5},
 \[
\alpha=-\sum_{j=1}^n \delta_{ij}\frac{a_j}{a_i}=-\frac{a_i}{a_i}=-1
\]
for each $i=1,\dots ,n$. Corollary \ref{solve5} enables a complete description
of all symmetries and reversing symmetries of $\phi$.

\begin{theorem}\label{solve6}
Suppose that $\phi$ is a quasiperiodic flow on $T^n$ with generating vector field
$X=\sum^n_{i=1}a_i\partial/\partial\theta_i$. If $\rho_\phi(R)=\pm1$ for an
$R\in S_\phi$, then there is $c\in\mathbb{R}^n$ such that
$\hat R(x)=\rho_\phi(R)Ix+c$.
\end{theorem}

\begin{proof} Let $R\in S_\phi$. By Theorem \ref{solve3} there exists a
$B=(b_{ij})\in\mathop{\rm GL}(n,\mathbb{Z})$ and a $c\in\mathbb{R}^n$ such that
$\hat R(x)=Bx+c$. By Corollary \ref{solve5}, the entries of $B$ satisfy
\[
\rho_\phi(R)=\sum_{j=1}^n b_{ij}\frac{a_j}{a_i}
\]
for each $i=1,2,\dots ,n$. By hypothesis, $\rho_\phi(R)=\pm 1$.
Then for each $i=1,2,\dots ,n$,
\[
b_{i1}a_1+\dots +(b_{ii}\mp1)a_i+\dots +b_{in}a_n=0.
\]
By the independence of $a_1,a_2,\dots ,a_n$ over $\mathbb{Q}$, $b_{ij}=0$ when
$i\ne j$ and $b_{ii}=\rho_\phi(R)$ for all $i=1,2,\dots ,n$. Therefore,
$\hat R(x)=\rho_\phi(R)Ix+c$.\end{proof}

\begin{corollary}\label{solve7}
If $\phi$ is a quasiperiodic flow on $T^n$, then $\ker \rho_\phi\cong T^n$.
\end{corollary}

\begin{proof}
Let $R\in S_\phi$ such that $\rho_\phi(R)=1$. By Theorem \ref{solve6},
$\hat R(x)=Ix+c$ for some $c\in\mathbb{R}^n$. Now, for any $c\in\mathbb{R}^n$,
the $Q\in\mathop{\rm Diff}(T^n)$ induced by $\hat Q(x)=Ix+c$ satisfies
$Q_*X=X$ by Theorem \ref{lift7} because $\hat Q_*\hat X=\hat X$. So,
by Theorem \ref{char}, $Q\in\ker \rho_\phi$. Since $c$ is arbitrary,
$Q\pi=\pi\hat Q$, and $\pi(\mathbb{R}^n)=T^n$, it follows that
$\ker \rho_\phi\cong T^n$.
\end{proof}

\begin{corollary}\label{solve8}
If $\phi$ is a quasiperiodic flow on $T^n$,  then every reversing symmetry of
$\phi$ is an involution.\end{corollary}

\begin{proof}Suppose $R\in S_\phi$ is
a reversing symmetry. By Theorem \ref{solve6}, $\hat R(x)=-Ix+c$ for some
$c\in\mathbb{R}^n$, and so $\hat R^2(x)=Ix$.
This implies that $R^2=\mathop{\rm id}_{T^n}$.
\end{proof}

\begin{example}\label{example2} \rm
Recall the quasiperiodic flow $\phi$ on $T^3$ and its generating vector field
\[
X=\frac{\partial}{\partial\theta_1}+7^{1/3}\frac{\partial}{\partial\theta_2}+
7^{2/3}\frac{\partial}{\partial\theta_3}
\]
from Example \ref{example1}. By Corollary \ref{solve7}, the symmetry group of
$\phi$ is exactly the group of translations $\{R_c:c\in T^n\}$ on $T^n$,
where $R_c(\theta)=\theta+c$. By Corollary \ref{solve8}, every reversing
symmetry of $\phi$ is an involution. In particular, this implies that the
reversing symmetry group of $\phi$ is a semidirect product of the symmetry
group of $\phi$ by the $\mathbb{Z}_2$ subgroup generated by reversing
involution $N(\theta)=-\theta$ (see p.8 in \cite{LR}). Are there symmetries
of $\phi$ with multipliers other than $\pm 1$? The $\mathop{\rm GL}(3,\mathbb{Z})$
matrix
\[
B=(b_{ij})=\begin{bmatrix}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 7 & 0 & -2\end{bmatrix}
\]
induces a $Q\in\mathop{\rm Diff}(T^3)$ by Theorem \ref{lift3}. Since
\[
\hat Q_*\hat X=\mathbf{T}\hat Q\hat X=B\hat X=\big( -2+7^{1/3}\big)\hat X,
\]
Theorem \ref{lift7} implies that $Q_*X=(-2+7^{1/3})X$. Hence, by Theorem \ref{char},
$Q\in S_\phi$. The  number $-2+7^{1/3}$ is $\rho_\phi(Q)$, the multiplier of
$Q$, is an algebraic integer of degree at most $3$ by Corollary \ref{solve4},
and satisfies
\[
-2+7^{1/3}=\sum_{j=1}^3 b_{ij}\frac{a_j}{a_i}, \quad i=1,2,3,
\]
by Corollary \ref{solve5}. (The matrix $B$ was found by using Theorem 3.1 in
\cite{BA1}, a result which characterizes the matrices in
$\mathop{\rm GL}(3,\mathbb{Z})$ inducing generalized symmetries of a quasiperiodic
flow generated by a vector field of a certain type, of which $X$ above is.)
Since $S_\phi$ is a group and $\rho_\phi:S_\phi\to\mathbb{R}^*$ is a homomorphism,
it follows for each $k\in\mathbb{Z}$ that $Q^k\in S_\phi$ with
$\rho_\phi(Q^k)=\big(\rho_\phi(Q)\big)^k=(-2+7^{1/3})^k$, and that
$NQ^k\in S_\phi$ with $\rho_\phi(NQ^k)=-(-2+7^{1/3})^k$.
\end{example}

\section{A Splitting Map for the Extension}

For a quasiperiodic flow $\phi$ on $T^n$, Theorem \ref{solve3} implies that
$\mathbf{T}\hat R\in\mathop{\rm GL}(n,\mathbb{Z})$ for every $R\in S_\phi$. Set
\[
 \Pi_\phi=\{B\in\mathop{\rm GL}(n,\mathbb{Z}): \hbox{ there is }R\in
 S_\phi\hbox{ for which }B=\mathbf{T}\hat R\},
\]
and define a map $\nu_\phi:\Pi_\phi\to\rho_\phi(S_\phi)$ by
$\nu_\phi(B)=\rho_\phi(R)$ where $R\in S_\phi$ with $\mathbf{T}\hat R=B$.

\begin{lemma}\label{rep0} If $\phi$ is a quasiperiodic flow on $T^n$ with
generating vector field $X$, then $\nu_\phi$ is well-defined.
\end{lemma}

\begin{proof}
Let $B\in\Pi_\phi$, and suppose there are $R,Q\in S_\phi$ with
$\mathbf{T}\hat R=B=\mathbf{T}\hat Q$. Then $RQ^{-1}\in S_\phi$ and
$\hat R\hat Q^{-1}$ is a lift of $RQ^{-1}$ for which
$\mathbf{T}(\hat R\hat Q^{-1})=BB^{-1}=I$. Hence $\hat R\hat Q^{-1}(x)=Ix+c$
for some $c\in\mathbb{R}^n$. This implies that $(\hat R\hat Q^{-1})_*\hat X=\hat X$,
so that by Theorem \ref{lift7}, $(RQ^{-1})_*X=X$.
By Theorem \ref{char}, $\rho_\phi(RQ^{-1})=1$. Because $\rho_\phi$ is a homomorphism,
 $\rho_\phi(R)=\rho_\phi(Q)$.\end{proof}

 \begin{lemma}\label{rep1}
If $\phi$ is a quasiperiodic flow on $T^n$ with generating vector field $X$, then
$\Pi_\phi$ is a subgroup of $\mathop{\rm GL}(n,\mathbb{Z})$.\end{lemma}

\begin{proof}
Let $B,C\in\Pi_\phi$. Then there are $R,Q\in S_\phi$ such that
$\mathbf{T}\hat R=B$ and $\mathbf{T}\hat Q=C$. The latter implies that
$\mathbf{T}\hat Q^{-1}=(\mathbf{T}\hat Q)^{-1}=C^{-1}$. Then
$BC^{-1}=\mathbf{T}\hat R\mathbf{T}\hat Q^{-1}=\mathbf{T}(\hat R\hat Q^{-1})$.
The diffeomorphism $x\to \hat R\hat Q^{-1}x$ of $\mathbb{R}^n$ satisfies conditions
a) and b) of Theorem \ref{lift3}, and so is a lift of a diffeomorphism $V$ of $T^n$.
Let $\alpha=\rho_\phi(R)$ and $\beta=\rho_\phi(Q)$.
Then $\rho_\phi(Q^{-1})=\beta^{-1}$ because $\rho_\phi$ is a homomorphism, and so
$(\hat Q^{-1})_*\hat X=\beta^{-1}\hat X$. Thus,
$\mathbf{T}(\hat R\hat Q^{-1})\hat X=(\hat R\hat Q^{-1})_*\hat X
=\alpha\beta^{-1}\hat X$. By Theorem \ref{lift7}, $V_*X=\alpha\beta^{-1}X$, so that
 by Theorem \ref{char}, $V\in S_\phi$. The lifts $\hat R\hat Q^{-1}$ and $\hat V$
of $V$ differ by a deck transformation of $\pi$, so that
$BC^{-1}=\mathbf{T}(\hat R\hat Q^{-1})=\mathbf{T}\hat V$. Therefore,
$BC^{-1}\in \Pi_\phi$.\end{proof}

\begin{theorem}\label{rep2}
If $\phi$ is a quasiperiodic flow on $T^n$ with generating vector field $X$, then
$\nu_\phi$ is an isomorphism and $\Pi_\phi$ is an Abelian subgroup of
$\mathop{\rm GL}(n,\mathbb{Z})$.\end{theorem}

\begin{proof}
Let $B,C\in\Pi_\phi$. Then there are $R,Q\in S_\phi$ such that $\mathbf{T}\hat R=B$
and $\mathbf{T}\hat Q=C$. Let $\alpha=\rho_\phi(R)$ and $\beta=\rho_\phi(Q)$.
By Theorem \ref{char} and Theorem \ref{lift7},
$\mathbf{T}\hat R\hat X=\alpha\hat X$ and $\mathbf{T}\hat Q\hat X=\beta\hat X$.
By Lemma \ref{rep1}, $BC\in \Pi_\phi$, so that there is a $V\in S_\phi$ such that
$\mathbf{T}\hat V=BC$. Hence,
$\hat V_*\hat X=\mathbf{T}\hat V\hat X=BC\hat X=\alpha\beta\hat X$. By
Theorem \ref{lift7} and Theorem \ref{char}, $\rho_\phi(V)=\alpha\beta$.
Thus, $\nu_\phi(BC)=\alpha\beta=\nu_\phi(B)\nu_\phi(C)$. By definition, $\nu_\phi$
is surjective, and by Theorem \ref{solve6}, $\ker \nu_\phi=\{I\}$. Therefore,
$\nu_\phi$ is an isomorphism. The multiplier group $\rho_\phi(S_\phi)$ is Abelian
because it is a subgroup of the Abelian group $\mathbb{R}^*$. Thus $\Pi_\phi$ is
Abelian.\end{proof}

A splitting map for the short exact sequence,
\[
\mathop{\rm id}{}_{T^n}\to \ker \rho_\phi\to  \rho_\phi^{-1}(\Lambda)
\stackrel{j_\Lambda}{\to } \Lambda\to  1,
\]
is a homomorphism $h_\Lambda:\Lambda\to\rho_\phi^{-1}(\Lambda)$ such that
$j_\Lambda h_\Lambda$ is the identity isomorphism on $\Lambda$.
Take for $h_\Lambda$ the map where for each $\alpha\in\Lambda$, the image
 $h_\Lambda(\alpha)$ is the diffeomorphism in $\rho_\phi^{-1}(\Lambda)$
induced by the $\mathop{\rm GL}(n,\mathbb{Z})$ matrix $\nu_\phi^{-1}(\alpha)$.

\begin{theorem}\label{rep4}
If $\phi$ is a quasiperiodic flow on $T^n$, then $h_\Lambda$ is a splitting map for
the extension $\mathop{\rm id}_{T^n}\to\ker \rho_\phi\to\rho_\phi^{-1}(\Lambda)
\to\Lambda\to 1$ for each $\Lambda<\rho_\phi(S_\phi)$.\end{theorem}

\begin{proof}
For arbitrary $\alpha,\beta\in\Lambda$, set $R=h_\Lambda(\alpha)$,
$Q=h_\Lambda(\beta)$, and $V=h_\Lambda(\alpha\beta)$. Then
$\hat R(x)=\nu_\phi^{-1}(\alpha)x$, $\hat Q(x)=\nu_\phi^{-1}(\beta)x$, and
$\hat V(x)=\nu_\phi^{-1}(\alpha\beta)x$. By Theorem \ref{rep2}, $\nu_\phi^{-1}$
is an isomorphism, so that $\hat V(x)=\nu_\phi^{-1}(\alpha)\nu_\phi^{-1}(\beta)x$.
Because
\begin{align*}
h_\Lambda(\alpha)h_\Lambda(\beta)\pi(x)
& = RQ\pi(x)= \pi \hat R\hat Q(x)= \pi\nu_\phi^{-1}(\alpha)\nu_\phi^{-1}(\beta)x \\
& = \pi\nu_\phi^{-1}(\alpha\beta)x= \pi \hat V(x)= V\pi(x)
= h_\Lambda(\alpha\beta)\pi(x),
\end{align*}
and because $\pi$ is surjective,
$h_\Lambda(\alpha)h_\Lambda(\beta)=h_\Lambda(\alpha\beta)$. Let
$B=\mathbf{T}\hat R=\nu_\phi^{-1}(\alpha)$. Then $\nu_\phi(B)=\rho_\phi(R)$, so that
\[
 j_\Lambda h_\Lambda(\alpha)=j_\Lambda(R)=\rho_\phi(R)=\nu_\phi(B)
 =\nu_\phi(\nu_\phi^{-1}(\alpha))=\alpha.
\]
Therefore, $h_\Lambda$ is a splitting map for the extension.\end{proof}

\begin{theorem}\label{rep5}
If $\phi$ is a quasiperiodic flow on $T^n$, then
\[
\rho_\phi^{-1}(\Lambda)=\ker \rho_\phi \rtimes_\Gamma h_{\Lambda}(\Lambda)
\]
for each $\Lambda<\rho_\phi(S_\phi)$, where
$\Gamma:h_\Lambda(\Lambda)\to\mathop{\rm Aut}(\ker \rho_\phi)$ is the conjugating
homomorphism. Moreover, if $\Lambda$ is a nontrivial subgroup of
$\rho_\phi(S_\phi)$, then $\Gamma$ is nontrivial.
\end{theorem}

\begin{proof}
By Theorem \ref{rep4}, $h_\Lambda$ is a splitting map for the extension
\[
\mathop{\rm id}{}_{T^n}\to \ker \rho_\phi\to  \rho_\phi^{-1}(\Lambda)
\stackrel{j_\Lambda}{\to } \Lambda\to  1.
\]
Thus, $\rho_\phi^{-1}(\Lambda)=\big(\ker \rho_\phi\big)\big(h_\Lambda(\Lambda)\big)$
and $\ker \rho_\phi\cap h_\Lambda(\Lambda)={\rm id}_{T^n}$
(Theorem 9.5.1, p.240 \cite{SC}). Since $\ker \rho_\phi$ is a normal subgroup of
$\rho_\phi^{-1}(\Lambda)$, then
$\rho_\phi^{-1}(\Lambda)=\ker \rho_\phi\rtimes_\Gamma h_\Lambda(\Lambda)$ where
$\Gamma:h_\Lambda(\Lambda)\to\mathop{\rm Aut}(\ker \rho_\phi)$ is the
conjugating homomorphism (see p.21 in \cite{AB}). If $\Gamma$ is the trivial
homomorphism, then $\rho_\phi^{-1}(\Lambda)$ is Abelian since $\ker \rho_\phi$ is
Abelian by Corollary \ref{solve7} and $h_\Lambda(\Lambda)$ is Abelian by
Theorem \ref{rep2} (see p.21 in \cite{AB}). But $\rho_\phi^{-1}(\Lambda)$ is
non-Abelian by Theorem \ref{nonabelian} whenever $\Lambda$ is a nontrivial
subgroup of $\rho_\phi(S_\phi)$.
\end{proof}

\begin{example}\rm
For the quasiperiodic flow $\phi$ on $T^3$ with frequencies $1$, $7^{1/3}$, and
$7^{2/3}$, it was shown in Example \ref{example2} that
$\alpha=-2+7^{1/3}\in\rho_\phi(S_\phi)$. The set
\[
\Lambda=\{(-1)^j\alpha^k:j\in\{0,1\}, k\in\mathbb{Z}\}
\]
is a nontrivial subgroup of $\rho_\phi(S_\phi)$ that is isomorphic to
$\mathbb{Z}_2\times\mathbb{Z}$. By Theorem \ref{rep5} and Corollary \ref{solve7},
\[
\rho_\phi^{-1}(\Lambda)\cong T^3\rtimes_\Gamma \big(\mathbb{Z}_2\times\mathbb{Z}\big),
\]
where $\Gamma$ is the (nontrivial) conjugating homomorphism. In particular,
every element of $\rho_\phi^{-1}(\Lambda)$ can be written uniquely as $R_c N^jQ^k$
where $R_c\in\hbox{\rm  ker}\rho_\phi$ is a translation by $c$ on $T^n$
(as defined in Example \ref{example1}), $N$ is the reversing involution
(as defined Example \ref{example1}), and $Q$ is the generalized symmetry of $\phi$
whose multiplier is $\alpha$ (as defined in Example \ref{example2}). Thus
\[
\rho_\phi^{-1}(\Lambda) = \{R_cN^jQ^k:c\in T^n, j\in\{0,1\},k\in\mathbb{Z} \}.
\]
\end{example}

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