text:
\section{Introduction}
Let $M$ be a factor of type $\mathrm{II}_1$ with a normalized trace $\tau$.
Murray and von Neumann introduced the fundamental group ${\mathcal F}(M)$ of $M$ in \cite{MN}.
The fundamental group of $M$ is defined as the set of the numbers $\tau \otimes Tr(p)$
for some projection $p \in M_n(M)$ such that $pM_n(M)p$ is isomorphic to $M$ where
$Tr$ is the usual unnormalized trace on $M_n(\mathbb{C})$.
They showed that if $M$ is  hyperfinite, then ${\mathcal F}(M) = {\mathbb R_+^{\times}}$.
Since then there has been many works on the computation of fundamental groups.
Voiculescu \cite{Vo} showed that ${\mathcal F}(L(\mathbb{F}_{\infty}))$ of the group factor
of the free group $\mathbb{F}_{\infty}$ contains the positive rationals and
Radulescu proved that
${\mathcal F}(L(\mathbb{F}_{\infty})) = {\mathbb R}_+^{\times}$ in
\cite{Ra}. Connes \cite{Co} showed that if $G$ is an ICC group with property
(T), then  ${\mathcal F}(L(G))$ is a countable group. Popa
showed that any countable subgroup of $\mathbb R_+^{\times}$
can be realized as the fundamental group of some factor of type $\mathrm{II}_1$ in \cite{Po1}.
Furthermore Popa and Vaes \cite{PV} exhibited a large family $\mathcal{S}$
of subgroups of $\mathbb{R}_{+}^\times$, containing $\mathbb{R}_{+}^\times$
itself, all of its countable subgroups, as well as uncountable subgroups with
any Hausdorff dimension in $(0,1)$, such that for each $H\in\mathcal{S}$
there exist many free ergodic measure preserving actions of $\mathbb{F}_{\infty}$
for which the associated $\mathrm{II}_1$ factor $M$ has the fundamental group equal to $H$.

Popa's results are based on the study of rigidity properties of II$_1$ factors
$L^{\infty} (X)\rtimes_{T} G$ and orbit equivalence relations $\mathcal{R}_{G,T}$ arising from
(free) ergodic measure preserving actions $T$ of countable groups $G$ on probability measure
spaces $(X, \mu)$ via the group measure space construction in \cite{MN1}.
(See also \cite{FM1} and \cite{FM2} for its generalization.)
The fundamental group of $\mathcal{R}_{G, T}$ can also be defined as the set of numbers
$\mu \times \delta (Y)$ for some measurable set $Y$ in $X\times \{1,..,n\}$ such that
$\mathcal{R}_{G, T}^{n}|_{Y}$ is orbit equivalent to $\mathcal{R}_{G, T}$ where
$\delta$ is the counting measure. It is easy to see that $\mathcal{F}(\mathcal{R}_{G, T})\subseteq
\mathcal{F}(L^{\infty}(X) \rtimes_{T} G)$. This inclusion may be strict. (See 6.1 in \cite{Pspectral}.)

The result in \cite{CFW} implies $\mathcal{F}(\mathcal{R}_{G, T})=\mathbb{R}_{+}^\times$
whenever $G$ is amenable.
Gefter and Golodets \cite{GG} showed that there exist ergodic measure preserving actions having
trivial fundamental group before Popa's result \cite{PoAnn}.
Moreover Gaboriau \cite{G} showed that if $\mathcal{R}_{\mathbb{F}_{n}, T}$ is an orbit equivalence
relation arising from a free ergodic measure preserving action $T$ of a non-amenable free group of
finite rank, then $\mathcal{F}(\mathcal{R}_{\mathbb{F}_{n}, T})=\{1 \}$.
On the contrary, Popa and Vaes showed that for each $H\in\mathcal{S}$
there exist many orbit equivalence relations $\mathcal{R}_{\mathbb{F}_{\infty}, T}$ arising from
free ergodic measure preserving actions $T$ of $\mathbb{F}_{\infty}$ such that
$\mathcal{F}(\mathcal{R}_{\mathbb{F}_{\infty}, T})=H$.
For a countable discrete group $G$, it is an interesting problem to consider how $G$
affect $\mathcal{F}(\mathcal{R}_{G,T})$.

Watatani and the author introduced the fundamental group $\mathcal{F}(A)$
of a simple unital $C^*$-algebra $A$ with a normalized trace $\tau$
based on the computation  of Picard groups by Kodaka \cite{kod1}, \cite{kod2} and \cite{kod3}.
The fundamental group ${\mathcal F}(A)$ is
defined as the set of
the numbers $\tau \otimes Tr(p)$ for some projection
$p \in M_n(A)$ such that $pM_n(A)p$ is isomorphic to $A$.
We computed the fundamental groups of several $C^*$-algebras and showed
that any countable subgroup of $\mathbb{R}_+^\times$
can be realized as the fundamental group of a separable simple unital $C^*$-algebra
with a unique trace in \cite{NW} and \cite{NW2}.
Note that the fundamental groups of separable simple unital $C^*$-algebras are countable.

We consider topological dynamical systems on the Cantor set in this paper.
Giordano, Putnam and Skau classified Cantor minimal $\mathbb{Z}$-systems up to (topological)
orbit equivalence \cite{GPS1}.
They also introduced the notion of strong orbit equivalence for Cantor minimal
$\mathbb{Z}$-system and showed that two Cantor minimal $\mathbb{Z}$-system are strong orbit
equivalent if and only if their associated $C^*$-algebras are isomorphic.
Giordano, Matui, Putnam and Skau extends the classification up to orbit equivalence of
Cantor minimal systems arising from actions of finitely generated abelian groups
\cite{GMPS1} and \cite{GMPS2}.

In this paper we introduce the fundamental group of a uniquely ergodic
Cantor minimal $G$-system $\mathcal{R}_{G, \varphi}$ where $G$ is a countable discrete group.
The fundamental group $\mathcal{F}(\mathcal{R}_{G, \varphi})$ of a uniquely ergodic Cantor
minimal $G$-system $\mathcal{R}_{G, \varphi}$ is defined as the set of numbers
$\mu \times \delta (U)$ for some clopen set $U$ in $X\times \{1,..,n\}$ such that
$\mathcal{R}_{G, \varphi}^{n}|_{U}$ is (topologically) orbit equivalent to
$\mathcal{R}_{G, \varphi}$ where $\mu$ is a unique invariant probability measure and $\delta$ is
the counting measure. We show that $\mathcal{F}(\mathcal{R}_{G, \varphi})$ is a countable
multiplicative subgroup of $\mathbb{R}_{+}^\times$.
Note that we cannot show this theorem in the same way as in the case of ergodic measure
preserving actions of countable groups on probability measure spaces because we do not know
whether the analogous result of Hopf equivalence theorem (see, for example, Proposition 3.3
in \cite{FM1}) is true for Cantor minimal $G$-systems. Moreover we do not know whether
$\mu\times \delta (U_1 )=\mu\times \delta (U_2)$ implies that
$\mathcal{R}_{G, \varphi}^{n}|_{U_1}$ is orbit equivalent to
$\mathcal{R}_{G, \varphi}^{n}|_{U_2}$ in general.

We show that for any unital countable subring $R$ of $\mathbb{R}$,
there exists a Cantor minimal $\mathbb{Z}$-system $\mathcal{R}_{\mathbb{Z}, \varphi}$ such that
$\mathcal{F}(\mathcal{R}_{\mathbb{Z}, \varphi})=R_{+}^\times$.
Conversely, we also show that if $\mathcal{R}_{G, \varphi}$ arises from a free action of a finitely
generated abelian group, then there exists a unital countable subring $R$ of $\mathbb{R}$ such
that $\mathcal{F}(\mathcal{R}_{G, \varphi})=R_{+}^\times$.
Therefore $\{9^n:n\in\mathbb{Z}\}$ cannot be realized as the
fundamental group of a Cantor minimal system in this class.

We do not know whether there exists a relation between orbit
equivalence of Cantor minimal $G$-systems and $C^*$-isomorphism of associated $C^*$-algebras
in general.
We show that if $\mathcal{R}_{\mathbb{Z}^n, \varphi}$ arises from a free minimal action $\varphi$
of $\mathbb{Z}^n$ on a Cantor set $X$ , then $\mathcal{F}(C(X)\rtimes_{\varphi} \mathbb{Z}^n)
\subseteq \mathcal{F}(\mathcal{R}_{\mathbb{Z}^n, \varphi})$.
Note that we do not know whether $C(X)\rtimes_{\varphi} \mathbb{Z}^2$ belongs to classifiable
classes by $K$-groups.
We also show that there exists a Cantor minimal $\mathbb{Z}$-system $R_{\mathbb{Z}, \varphi}$
such that $\mathcal{F}(C(X)\rtimes_{\varphi} \mathbb{Z})\neq \mathcal{F}(\mathcal{R}_{\mathbb{Z}, \varphi})$.

In section \ref{sec:examples}
we compute fundamental groups of several uniquely ergodic Cantor minimal systems.
Some computations suggest that the fundamental groups of Cantor minimal systems have
arithmetical flavor compared with the case of ergodic measure preserving actions on
probability measure spaces.

\section{Cantor minimal systems}\label{sec:minimal system}
In this section we review some definitions and results on Cantor minimal systems.
See, for example, \cite{GMPS2} and \cite{GPS2} for details.
Let $X$ be the Cantor set, that is, $X$ is a compact separable totally disconnected metric
space without isolated points. For a free action $\varphi$ of a countable discrete group $G$ on $X$
by homeomorphisms (we call this dynamical system the \textit{Cantor $G$-system}),
an orbit equivalence relation $\mathcal{R}_{G, \varphi}$ is defined by
$$
\mathcal{R}_{G, \varphi} =\{(x,\varphi_{g}(x))\in X\times X:x\in X, g\in G \} .
$$
For two orbit equivalence relations $\mathcal{R}_1$ on $X_1$ and
$\mathcal{R}_2$ on $X_2$, we say that
$\mathcal{R}_1$ is \textit{(topologically) orbit equivalent} to
$\mathcal{R}_2$ if there exists a homeomorphism $F$ of $X_1$ onto
$X_2$ such that $F\times F(\mathcal{R}_1)=\mathcal{R}_2$.
We denote by $\mathrm{Aut}(\mathcal{R})$ the set of
homeomorphism of $X$ such that $F\times F(\mathcal{R})=\mathcal{R}$.
An orbit equivalence relation $\mathcal{R}$ on $X$ is said to be \textit{minimal} if
$\mathcal{R}[x]=\{y\in X:(x,y)\in \mathcal{R}\}$ is dense in $X$
for any $x\in X$.
An \textit{$\mathcal{R}$-invariant measure} is a Borel measure $\mu$ on $X$ such
that $\mu$ is $h$-invariant for any $h\in \mathrm{Homeo}(X)$ satisfying
$(x,h(x))\in \mathcal{R}$ for all $x\in X$.
Let $M(\mathcal{R})$ denote the set of $\mathcal{R}$-invariant probability measures.
An orbit equivalence relation $\mathcal{R}$ is said to be \textit{uniquely ergodic} if
$M(\mathcal{R})$ has exactly one element.
These definitions coincide with usual definitions for topological dynamical systems.

Assume that $M(\mathcal{R})\neq \phi$.
Let $D_{m}(X, \mathcal{R})$ be a quotient group of $C(X,\mathbb{Z})$
by
$
\{f\in C(X,\mathbb{Z}):\int_{X}fd\mu =0\;\mathrm{for}\;\mathrm{all}\;\mu \in M
(\mathcal{R}) \}
$.
This countable abelian group $D_m(\mathcal{R})$ has an order structure
such that the positive cone $D_m(\mathcal{R})^{+}$ is
the set of all $[f]$, where $f\geq 0$. For a clopen set $U$ of $X$, we denote by
$1_{U}$ the characteristic function on $U$, that is, $1_{U} (x)=1$ if
$x\in U$ and
$1_{U} (x)=0$ if $x\notin U$.
We say that two triples $(D_m(\mathcal{R}_1),D_m(\mathcal{R}_1)^{+}, [1]_{X_1})$ and
$(D_m(\mathcal{R}_2),D_m(\mathcal{R}_2)^{+}, [1]_{X_2})$
are \textit{isomorphic} if there exists an order isomorphism $\psi$ of
$D_m(\mathcal{R}_1)$ onto $D_m(\mathcal{R}_2)$ such that
$\psi ([1_{X_1}]) =[1_{X_2}]$.
A triple $(D_m(\mathcal{R}),D_m(\mathcal{R})^{+}, [1]_{X})$ is an orbit equivalence invariant
for Cantor minimal systems.
Giordano, Putnam, Matui and Skau actually showed that this triple is a complete invariant
for AF relations and the case where $G$ is a finitely generated abelian group.
(See Theorem 2.5 in \cite{GMPS2}.)
\begin{rem}
If $R$ is uniquely ergodic with an $R$-invariant probability measure $\mu$, then the map
$T_{\mu}$ of $D_m(\mathcal{R})$ into $\mathbb{R}$ defined  by $T_{\mu}([f])=\int_{X}fd\mu$
is an order isomorphism of $D_m(\mathcal{R})$ onto $T_{\mu}(D_m(\mathcal{R}))\subseteq \mathbb{R}$
with $T_{\mu}([1]_{X})=1$. Moreover $T_{\mu}(D_m(\mathcal{R}))$ is an additive subgroup of
$\mathbb{R}$ generated by $\{\mu (U):U\text{ is a clopen set in }X \}$ because for any
$f\in C(X,\mathbb{Z})$, there exist clopen sets $U_1,..,U_n$ in $X$ and
integers $m_1,..,m_n$ such that $f=\sum_{k=1}^nm_k 1_{U_k}$.
\end{rem}
For a Cantor system $\mathcal{R}_{G, \varphi}$ and a natural number $n$,
define an orbit equivalence relation $\mathcal{R}_{G, \varphi}^n$
on $X\times \{1,..,n \}$ by $\{((x,i),(y,j)) :(x,y)\in\mathcal{R}_{G, \varphi}, i,j\in \{1,..,n \} \}$.
Note that $X\times \{1,..,n \}$ is a Cantor set and $\mathcal{R}_{G, \varphi}^n$ arises from
an action $\tilde{\varphi}$ of $G\times \mathbb{Z}/n\mathbb{Z}$ on $X\times \mathbb{Z}/n\mathbb{Z}$.
If $\mathcal{R}_{G, \varphi}$ is a Cantor minimal system, then
$\mathcal{R}_{G, \varphi}^n$ is minimal.
It can be easily checked that for any $\mathcal{R}_{G, \varphi}^n$-invariant
probability measure
$\nu$ on $X\times \{1,..,n\}$, there exists
an $\mathcal{R}_{G, \varphi}$-invariant probability measure $\mu$ on $X$ such that
$\nu =\mu\times \frac{1}{n}\delta$ where $\delta$ is the counting measure on $\{1,..,n\}$.
\begin{pro}
Let $\mathcal{R}_{G, \varphi}$ be a Cantor system and $n$ a natural number.
Then
$$(D_m(\mathcal{R}_{G, \varphi}^n), D_m(\mathcal{R}_{G, \varphi}^n)^+, [1_{X\times \{1,..,n \}}])
\cong
(D_m(\mathcal{R}_{G, \varphi}), D_m(\mathcal{R}_{G, \varphi})^{+}, n[1_{X}]).
$$
\end{pro}
\begin{proof}
Let $\psi$ be an order homomorphism of $D_m(\mathcal{R}_{G, \varphi})$ to
$D_m(\mathcal{R}_{G, \varphi}^n)$ such that $\psi ([1_{V}]) =[1_{V\times \{1\}}]$
for any clopen set $V$ in $X$.
It is easy to see that $\psi$ is well-defined and injective.
For any clopen set $U$ in $X\times\{1,..,n\}$, there exist
clopen sets $V_1$,..,$V_n$ in $X$ such that $U=\cup_{k=1}^nV_k\times \{k \}$.
Since $[1_{V_k\times\{k\}}]=[1_{V_k\times\{1\}}]$ in
$D_m(\mathcal{R}_{G, \varphi}^n)$ for any $1\leq k\leq n$, we see that
$\psi$ is surjective and $n\psi ([1_{X}]) =[1_{X\times\{1,..,n\}}]$.
Therefore we obtain the conclusion.
\end{proof}
Let $U$ be a clopen set in $X$.
Define an orbit equivalence relation $\mathcal{R}_{G, \varphi}|_{U}$
by $\mathcal{R}_{G, \varphi}\cap U\times U$.
If $\mathcal{R}_{\mathbb{Z}, \varphi}$ is a Cantor minimal $\mathbb{Z}$-system,
then $\mathcal{R}_{\mathbb{Z}, \varphi}|_{U}$ is equal to the induced system
$\mathcal{R}_{\mathbb{Z}, \varphi_{U}}$ in \cite{GPS1}.
It is easy to see that if $\mathcal{R}_{G, \varphi}$ is minimal, then
$\mathcal{R}_{G, \varphi}|_{U}$ is minimal.
\begin{pro}\label{pro:inclusion}
Let $\mathcal{R}_{G, \varphi}$ be a Cantor minimal system and $U$ a clopen set in $X$. Then
there exists a bijective map $\pi$ of $M(\mathcal{R}_{G, \varphi})$ onto
$M(\mathcal{R}_{G, \varphi}|_{U})$ such that $\pi (\mu)=\mu |_{U}$.  Moreover we have
$$(D_m(\mathcal{R}_{G, \varphi}|_{U}), D_m(\mathcal{R}_{G, \varphi}|_{U})^+,
[1_{U}])
\cong
(D_m(\mathcal{R}_{G, \varphi}), D_m(\mathcal{R}_{G, \varphi})^{+}, [1_{U}]).
$$
\end{pro}
\begin{proof}
It is easy to see that $\pi$ is well-defined.
Since $X$ is compact and $\mathcal{R}_{G, \varphi}$ is minimal,
there exist $g_1,..,g_n\in G$ such that $X=\cup_{k=1}^n\varphi_{g_k}(U)$.
Put $U_{k}:=\varphi_{g_k}(U)\backslash (\cup_{i=1}^{k-1}\varphi_{g_i}(U))$
for $1\leq k\leq n$.
Then we have
$\mu (V)=\sum_{k=1}^n\mu (\varphi_{g_k^{-1}}(V\cap U_k))$ for any
$\mu\in M(\mathcal{R}_{G, \varphi})$ and any clopen set $V$ in $X$.
Since $\varphi_{g_k^{-1}}(V\cap U_k)$ is a clopen set in $U$, $\mu$ is determined by
$\mu |_{U}$. Therefore $\pi$ is injective.
For any $\nu\in M(\mathcal{R}_{G, \varphi}|_U)$, define a measure $\mu$ on $X$ by
$\mu (V)=\sum_{k=1}^n\nu (\varphi_{g_k^{-1}}(V\cap U_k))$
for any clopen set $V$ in $X$.
We shall show that $\mu$ is $\mathcal{R}_{G, \varphi}$-invariant.
Note that for $1\leq k,j\leq n$, $g\in G$ and a clopen set $V$ in $X$,
there exist homeomorphisms $h_{k,j,g}$ of $U$ such that
$h_{k,j,g}(x)=x$ if $x\notin \varphi_{g_k^{-1}}(U_k\cap\varphi_{g}(U_j\cap V))$
and $h_{k,j,g}(x)=\varphi_{g_j}^{-1}\circ\varphi_{g}^{-1}\circ\varphi_{g_k}(x)$ if
$x\in \varphi_{g_k^{-1}}(U_k\cap\varphi_{g}(U_j\cap V))$.
It is clear that $(x, h_{k,j,g}(x))\in \mathcal{R}_{G, \varphi}|_U$ for any $x\in U$.
For any $g\in G$ and any clopen set $V$ in $X$, we have
\begin{align*}
\mu (\varphi_{g}(V))
& =\sum_{k=1}^n \nu (\varphi_{g_k^{-1}}(U_k \cap \varphi_{g} (V))) \\
& =\sum_{k=1}^n \nu (\varphi_{g_k^{-1}}(U_k \cap \varphi_{g} (\cup_{j=1}^n U_j\cap V))) \\
& =\sum_{k=1}^n \sum_{j=1}^n \nu (\varphi_{g_k^{-1}}(U_k\cap \varphi_{g} (U_j\cap V))) \\
& = \sum_{k=1}^n \sum_{j=1}^n \nu (h_{k,j,g} (\varphi_{g_k^{-1}}(U_k\cap \varphi_{g} (U_j\cap V)))) \\
& =\sum_{k=1}^n \sum_{j=1}^n \nu (\varphi_{g_j^{-1}}( \varphi_{g^{-1}}(U_k)\cap U_{j}\cap V)) \\
&=\sum_{j=1}^n\nu (\varphi_{g_j^{-1}}(\varphi_{g^{-1}} (\cup_{k=1}^n U_k)\cap U_j\cap V)) \\
& = \mu (V).
\end{align*}
Therefore we see that $\pi$ is surjective.

Define a homomorphism $\psi$ of $D_m(\mathcal{R}_{G, \varphi})|_{U}$ to
$D_m(\mathcal{R}_{G, \varphi})$ by $\psi ([1_{W}]) = [1_{W}]$
for any clopen set $W$ in $U$.
Then it can be easily checked that $\psi$ is an order isomorphism by a similar
argument.
\end{proof}
\section{Analogous result of Brown's Lemma}
In this section we shall show the analogous result of Lemma 2.5 in \cite{B}.
Let $\mathcal{R}_{G, \varphi}$ be a Cantor minimal system.
Define an orbit equivalence relation $\mathcal{R}_{G, \varphi}^\infty$
on $X\times \mathbb{N}$ by $\{((x,i),(y,j)) :(x,y)\in\mathcal{R}_{G, \varphi}, i,j\in \mathbb{N} \}$.
Note that $X\times \mathbb{N}$ is a locally compact separable totally disconnected metric
space without isolated points and $\mathcal{R}_{G, \varphi}^\infty$
arises from an action $\tilde{\varphi}$ of $G\times \mathbb{Z}$ on $X\times \mathbb{Z}$.
It can be easily checked that if $\mathcal{R}_{G, \varphi}$ is a uniquely ergodic
Cantor minimal system, then $\mathcal{R}_{G, \varphi}^\infty$ is minimal and
has unique (up to scalar multiple) $\mathcal{R}_{G, \varphi}^\infty$-invariant measure.
\begin{lem}\label{lem:lemma of key lemma}
Let $\mathcal{R}_{G, \varphi}$ be a Cantor minimal system on $X$ and
$U$ a clopen set in $X$. Then there exists
an injective continuous open map $F$ of $X\times \mathbb{N}$ to
$U\times \mathbb{N}$ such that $(y_1, y_2)\in\mathcal{R}_{G, \varphi}^{\infty}$ if and only if
$(F (y_1),F (y_2))\in\mathcal{R}_{G, \varphi}^{\infty}$.
\end{lem}
\begin{proof}
Since $X$ is compact and $\mathcal{R}_{G, \varphi}$ is minimal,
there exist $g_1,..,g_n\in G$ such that $X=\cup_{k=1}^n\varphi_{g_k}(U)$.
Put $U_{k}:=\varphi_{g_k}(U)\backslash (\cup_{i=1}^{k-1}\varphi_{g_i}(U))$
for $1\leq k\leq n$.
Let $\Psi$ be a bijective map of $\{1,..,n\}\times \mathbb{N}$ onto
$\mathbb{N}$.
Define a map $F$ of $X\times\mathbb{N}$ to $U\times\mathbb{N}$ by
$F ((x,i)) =(\varphi_{g_k^{-1}} (x),\Psi (k,i))$ if $x\in U_k$ for any $1\leq k\leq n$.
Then it can be easily checked that $F$ is an injective continuous open map such that
$(y_1, y_2)\in\mathcal{R}_{G, \varphi}^{\infty}$ if and only if
$(F (y_1),F (y_2))\in\mathcal{R}_{G, \varphi}^{\infty}$.
\end{proof}
The following lemma is the analogous result of Lemma 2.5 in \cite{B}.
\begin{lem}\label{lem:Brown}
Let $\mathcal{R}_{G, \varphi}$ be a Cantor minimal system on $X$ and
$U$ a clopen set in $X$. Then there exists a homeomorphism $\Phi$ of $U\times \mathbb{N}$
onto $X\times \mathbb{N}$ such that
$\Phi\times \Phi((\mathcal{R}_{G, \varphi}|_{U})^\infty) =\mathcal{R}_{G, \varphi}^\infty$ and
$\Phi(U\times \{1 \})= U\times \{1 \}\subseteq X\times \{1\}$.
\end{lem}
\begin{proof}
We shall construct a homeomorphism $\Phi$ of
$U\times \mathbb{N}\times\mathbb{N}$ onto $X\times \mathbb{N}\times\mathbb{N}$.
First, for $(x,i,1)\in U\times\mathbb{N}\times\mathbb{N}$,
define $\Phi ((x,i,1))=(x,i,1)\in X\times \mathbb{N}\times\mathbb{N}$.
Let $F$ be an injective continuous open map of $X\times \mathbb{N}$ to $U\times \mathbb{N}$
in Lemma \ref{lem:lemma of key lemma}.
Put $E_1=F((X\times \mathbb{N})\backslash (U\times \mathbb{N}))$.
Define $\Phi ((x,i,2))= (F^{-1}((x,i)), 1)$ if $(x,i)\in E_1$ and
$\Phi ((x,i,2))= (x,i,2)$ if $(x,i)\in (U\times\mathbb{N})\backslash E_1$.
Put $E_2:=F((X\times \mathbb{N})\backslash ((U\times\mathbb{N})\backslash E_1))$.
In a similar way, define $\Phi (((x,i),3)) =(F^{-1}((x,i)),2)$ if
$(x,i)\in E_2$ and $\Phi ((x,i),3)= (x,i,3)$ if $(x,i)\in
(U\times\mathbb{N})\backslash E_2$.
We construct inductively $\Phi$ as follows:
Let $E_n:=F((X\times \mathbb{N})\backslash((U\times\mathbb{N})\backslash E_{n-1}))$, and define
$\Phi ((x,i,n+1))= (F^{-1}((x,i)),n)$ if $(x,i)\in E_n$ and
$\Phi ((x,i,n+1))= (x,i,n+1)$ if $(x,i)\in (U\times \mathbb{N})\backslash E_n$.
By a Cantor-Bernstein type argument, we obtain the conclusion.
\end{proof}
\section{Fundamental group}
Let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
with an $\mathcal{R}_{G, \varphi}$-invariant probability measure $\mu$.
Put
$$
\mathcal{F}(\mathcal{R}_{G, \varphi}):=
\{\mu\times \delta (U):U\text{ is a clopen set in }X\times [n]\text{ such that }
\mathcal{R}_{G, \varphi}\underset{OE}{\sim} \mathcal{R}_{G, \varphi}^n|_{U}\}
$$
where $\delta$ is the counting measure on $[n]=\{1,..,n\}$.
\begin{thm}\label{thm:fundamental group}
Let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
with an $\mathcal{R}_{G, \varphi}$-invariant probability measure $\mu$. Then
$\mathcal{F}(\mathcal{R}_{G, \varphi})$ is a countable multiplicative subgroup of
$\mathbb{R}_{+}^\times$.
\end{thm}
\begin{proof}
Put
$$
\mathfrak{S}(\mathcal{R}_{G, \varphi})
=\{\lambda : \mu\times \delta\circ F=\lambda \mu\times \delta ,F\in
\mathrm{Aut}(\mathcal{R}_{G, \varphi}^\infty) \}.
$$
It is easy to see that $\mathfrak{S}(\mathcal{R}_{G, \varphi})$ is a multiplicative subgroup
of $\mathbb{R}_{+}^\times$. We shall show that
$\mathcal{F}(\mathcal{R}_{G, \varphi})=\mathfrak{S}(\mathcal{R}_{G, \varphi})$.

Let $\lambda\in \mathfrak{S}(\mathcal{R}_{G, \varphi})$, then there exists
a homeomorphism $F$ of $X\times\mathbb{N}$ such that
$\mu\times\delta \circ F = \lambda \mu\times\delta$ and
$F\times F (\mathcal{R}_{G, \varphi}^\infty)=\mathcal{R}_{G, \varphi}^\infty$.
We see that $\mathcal{R}_{G, \varphi}|_{X\times\{1\}}$ is orbit equivalent to
$\mathcal{R}_{G, \varphi}^\infty|_{F(X\times\{1\})}$.
Note that $\mathcal{R}_{G, \varphi}^\infty|_{X\times\{1\}}$ is orbit equivalent to
$\mathcal{R}_{G, \varphi}$.
Since $F(X\times\{1\})$ is compact and $F(X\times\{1\})\subseteq\cup_{i=1}^\infty
\{(x,k)\in X\times\mathbb{N}:k=i \}$, there exist a natural number $n$ and
a clopen set $U$ in $X\times \{1,..,n\}$ such that $F(X\times\{1\})=U$.
Therefore $\lambda\in \mathcal{F}(\mathcal{R}_{G, \varphi})$.

Conversely, let $\lambda\in \mathcal{F}(\mathcal{R}_{G, \varphi})$.
There exist a natural number $n$, a clopen set $U$ in $X\times \{1,..,n\}$
and a homeomorphism $h$ of $X$ onto $U\times\{1,..,n\}$ such that
$h\times h (\mathcal{R}_{G, \varphi})= \mathcal{R}_{G, \varphi}^n|_{U}$.
By Lemma \ref{lem:Brown}, there exist a homeomorphism $\Phi$ of $U\times\mathbb{N}$
onto $X\times\{1,..,n\}\times \mathbb{N}$.
Let $\Psi$ be a bijective map of $\{1,..,n\}\times \mathbb{N}$ onto
$\mathbb{N}$, and define a homeomorphism $F$ of $X\times\mathbb{N}$
by $F=(id\times \Psi )\circ\Phi \circ (h\times id)$.
It can be easily checked $F\in\mathrm{Aut}(\mathcal{R}_{G, \varphi}^\infty)$
and $\mu\times \delta (F(X\times\{1\}))= \mu\times \delta (U)=\lambda$.
Therefore $\lambda\in \mathfrak{S}(\mathcal{R}_{G, \varphi})$.

It is clear that $\mathcal{F}(\mathcal{R}_{G, \varphi})$ is countable because
$X$ is the Cantor set.
\end{proof}
\begin{Def}
Let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
with an $\mathcal{R}_{G, \varphi}$-invariant probability measure $\mu$.
We call
$\mathcal{F}(\mathcal{R}_{G, \varphi})$ the fundamental group of $\mathcal{R}_{G, \varphi}$,
which is a multiplicative (countable) subgroup of $\mathbb{R}^{\times}_{+}$.
\end{Def}
\begin{rem}\label{rem:definition}
(i) We need not assume that $\varphi$ is free.
Moreover we can define the fundamental group of \'etale equivalence relations on
Cantor sets by Proposition 2.3 in \cite{GPS2}.
But the associated $C^*$-algebra
$C(X)\rtimes_{\varphi} G$ ( $C_{r}^*(\mathcal{R})$ ) may not be simple.
In this paper, we assume that $\varphi$ is free.
\ \\
(ii) We do not know $\mu\times \delta (U_1 )=\mu\times \delta (U_2)$ implies that
$\mathcal{R}_{G, \varphi}^{n}|_{U_1}$ is orbit equivalent to $\mathcal{R}_{G, \varphi}^{n}|_
{U_2}$ in general. It is known that if $\mathcal{R}_{G, \varphi}$ is a Cantor minimal
$\mathbb{Z}^m$-system, then $\mathcal{R}_{G, \varphi}$ satisfies the
analogous result of Hopf equivalence theorem. See, for example,
\cite{LO} (Theorem 3.20) and \cite{M} (Theorem 6.11). \ \\
(iii) Since $M(\mathcal{R}_{G, \varphi})=\{\mu\}$, we may regard $\mathcal{R}_{G, \varphi}$ as
an orbit equivalence relation $\mathcal{R}_{G, T_{\varphi}}$ arising from an ergodic measure
preserving action $T_{\varphi}$ on a probability measure space $(X, \mu )$.
It is clear that
$\mathcal{F}(\mathcal{R}_{G, \varphi})\subseteq \mathcal{F}(\mathcal{R}_{G, T_{\varphi}})$.
\end{rem}
For an additive subgroup $E$ of $\mathbb{R}$ containing 1,
we define the positive inner multiplier group $IM_+(E)$ of $E$  by
$$
IM_+(E) = \{t \in {\mathbb R}_+^{\times} \ | t \in E, t^{-1}  \in E, \text{ and }
        tE = E \}.
$$
By Lemma 3.6 in \cite{NW}, there exists a unital subring $R$ of $\mathbb{R}$ such that
$IM_+(E)=R^\times_{+}$.
Hence not all countable subgroups of $\mathbb{R}_{+}^{\times}$
arise as  $IM_+(E)$. For example,
$\{9^n \in \mathbb{R}_{+}^{\times} \ |
n \in {\mathbb Z} \}$ does not arise as $IM_+(E)$
for any additive subgroup $E$ of $\mathbb{R}$ containing 1.

\begin{pro}\label{pro:subseteq}
Let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
with an $\mathcal{R}_{G, \varphi}$-invariant probability measure $\mu$.
Then
$$
\mathcal{F}(\mathcal{R}_{G, \varphi})\subseteq IM_{+}(T_{\mu}(D_m(\mathcal{R}_{G, \varphi})))
=IM_{+}(\{\int_{X} fd\mu :[f]\in D_m(\mathcal{R}_{G, \varphi})\}).
$$
\end{pro}
\begin{proof}
Let $\lambda\in\mathcal{F}(\mathcal{R}_{G, \varphi})$.
For any clopen set $U$ in $X\times\{1,..,n\}$, there exist
clopen sets $V_1,..,V_n$ in $X$ such that $U=\cup_{k=1}^nV_{k}\times \{k\}$.
Hence Theorem \ref{thm:fundamental group} implies that
$\lambda ,\lambda^{-1} \in T_{\mu}(D_m(\mathcal{R}_{G, \varphi}))$.
By the proof of Theorem \ref{thm:fundamental group}, there exist
homeomorphisms $F_1$ and $F_2$ of $X\times\mathbb{N}$ such that
$\mu\times\delta \circ F_1 =\lambda \mu\times\delta$ and
$\mu\times\delta \circ F_2 =\lambda^{-1}\mu\times\delta$.
Let $V$ be a clopen set in $X$.
Then $\mu\times\delta (F_1(V\times \{1\}))= \lambda\mu (V)$ and
$\mu\times\delta (F_2(V\times \{1\}))= \lambda^{-1}\mu (V)$.
Therefore we see that
$\lambda T_{\mu}(D_m(\mathcal{R}_{G, \varphi}))\subseteq T_{\mu}(D_m(\mathcal{R}_{G, \varphi}))$
and $\lambda^{-1} T_{\mu}(D_m(\mathcal{R}_{G, \varphi}))\subseteq T_{\mu}(D_m(\mathcal{R}_{G, \varphi}))$
because $F_1(V\times \{1\})$ and $F_2(V\times \{1\})$ are compact.
Consequently $\lambda\in IM_{+}(T_{\mu}(D_m(\mathcal{R}_{G, \varphi})))$.
\end{proof}
We shall show some examples.
\begin{ex}\label{ex:elementary}
Let $p$ be a prime number, and let $X=\{0,1,..,p-1\}^{\mathbb{N}}$.
Define $\varphi$ by the addition of $(1,0,0,...)$ with carry over the right
and $\varphi ((p-1,p-1,...))=(0,0,...)$.
Then $\mathcal{R}_{\mathbb{Z},\varphi}$ is a uniquely ergodic Cantor minimal system
with $T_{\mu}(D_m(\mathcal{R}))=\mathbb{Z}[\frac{1}{p}]$.
Put $U=\{(x_n)_n\in X:x_1=0\}$.
It is easy to see that $\mathcal{R}_{\mathbb{Z},\varphi}|_U$ is orbit equivalent to
$\mathcal{R}_{\mathbb{Z},\varphi}$ and $\mu (U)=\frac{1}{p}$.
(In particular, the induced homeomorphism $\varphi_{U}$ is topologically conjugate to $\varphi$.)
Hence
$$
\{p^n:n\in\mathbb{Z}\}\subseteq \mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi})
\subseteq IM_{+}(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi})))= \{p^n:n\in\mathbb{Z}\}.
$$
Therefore $\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi})=\{p^n:n\in\mathbb{Z}\}$.
\end{ex}
\begin{ex}
Let $p$ and $q$ be prime numbers such that $gcd(p,q)=1$, and let
$X=\{0,..,p-1\}^{\mathbb{N}}\times\{0,..,q-1\}^{\mathbb{N}}$.
Define $\varphi_1$ (resp. $\varphi_2$) by $\varphi_1 ((x,y))= (\tilde{\varphi_1}(x), y)$
(resp. $\varphi_2 ((x,y))= (x,\tilde{\varphi_2}(y))$) for $(x,y)\in \{0,..p-1\}^{\mathbb{N}}\times
\{0,..q-1\}^{\mathbb{N}}$
where $\tilde{\varphi_1}$ and $\tilde{\varphi_2}$ are the homeomorphisms in
Example \ref{ex:elementary}.
Then $\mathcal{R}_{\mathbb{Z}^2, (\varphi_1,\varphi_2)}$ is a uniquely ergodic Cantor minimal
system with $T(D_m(\mathcal{R}))=\mathbb{Z}[\frac{1}{p}, \frac{1}{q}]$.
Put $V=\{(x_n)_n\in \{0,..,p-1\}^\mathbb{N}:x_1=0\}$ and $U=V\times \{0,..,q-1\}^{\mathbb{N}}$.
Then $\mathcal{R}_{\mathbb{Z}^2, (\varphi_1,\varphi_2)}|U$
is equal to $\mathcal{R}_{\mathbb{Z}^2,(\varphi_{1U},\varphi_2)}$ where $\varphi_{1U}$ is the
induced homeomorphism of $\varphi_1$ on $U$.
Hence it is easy to see that $\mathcal{R}_{\mathbb{Z}^2, (\varphi_1,\varphi_2)}|U$ is
orbit equivalent to $\mathcal{R}_{\mathbb{Z}^2, (\varphi_1,\varphi_2)}$.
Therefore we see that
$\mathcal{F}(\mathcal{R}_{\mathbb{Z}^2, (\varphi_1,\varphi_2)})=\{p^nq^m:n,m\in\mathbb{Z}\}$.
by a similar argument as in Example \ref{ex:elementary}.
\end{ex}
\begin{ex}
If $\mathcal{R}_{\mathbb{F}^n, \varphi}$ is a uniquely ergodic Cantor minimal system
arising from a free action of a non-amenable free group of finite rank,
then $\mathcal{F}(\mathcal{R}_{\mathbb{F}^n, \varphi})=\{1\}$ by Remark \ref{rem:definition}
(iii) and \cite{G}.
Note that we do not know whether there exists such a example.
\end{ex}
We shall show more interesting examples in the next section.
Giordano-Matui-Putnam-Skau classification theorem enable us to compute
many examples.
\begin{thm}\label{thm:computation}
Let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
arising from a free action of a finitely generated abelian group.
Then
$$\mathcal{F}(\mathcal{R}_{G, \varphi})=IM_{+}(T_{\mu}(D_m(\mathcal{R})))=IM_{+}(\{\int_{X} fd\mu :[f]\in
D_m(\mathcal{R}_{G, \varphi})\}).
$$
\end{thm}
\begin{proof}
By the result in \cite{GMPS2}, $\mathcal{R}_{G, \varphi}$ is orbit equivalent to
a Cantor minimal $\mathbb{Z}$-system. Hence we may assume that $G=\mathbb{Z}$.

We have
$\mathcal{F}(\mathcal{R}_{G, \varphi})\subseteq
IM_{+}(T_{\mu}(D_m(\mathcal{R}_{G, \varphi})))$  by Proposition \ref{pro:subseteq}.
Conversely, let $\lambda$ be an element in
$IM_{+}(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi})))$ with $\lambda\leq 1$.
Then we see that there exists a clopen set $U$ in $X$ such that
$\mu (U)=\lambda$ by Lemma 2.5 in \cite{GW}.
Define an additive homomorphism $\psi$ of $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi}))$
by $\psi (t)= \lambda t$. Then $\psi$ is an order isomorphism and $\psi (1)=\lambda$.
Proposition \ref{pro:inclusion} implies that
$$(D_m(\mathcal{R}_{\mathbb{Z},\varphi}|_{U}),D_m(\mathcal{R}_{\mathbb{Z},\varphi}|_{U})_{+},
[1_{U}]) \cong
(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi})),T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi}))_{+},
\lambda).$$
Hence
$\mathcal{R}_{\mathbb{Z},\varphi}|_{U}$ is orbit equivalent to
$\mathcal{R}_{\mathbb{Z},\varphi}$ by Theorem 2.2 in \cite{GPS1}
because
$\mathcal{R}_{\mathbb{Z},\varphi}|_{U}$ is a Cantor minimal $\mathbb{Z}$-system
(see Section \ref{sec:minimal system} and \cite{GPS1}).
Therefore $\lambda\in \mathcal{F}(\mathcal{R}_{G, \varphi})$.
\end{proof}
\begin{cor}
If $R$ is a countable unital subring of $\mathbb{R}$,
then there exists a uniquely ergodic Cantor minimal $\mathbb{Z}$-system
$\mathcal{R}_{\mathbb{Z}, \varphi}$ such that $\mathcal{F}(\mathcal{R}_{\mathbb{Z}, \varphi})=R_{+}^\times$.
Conversely if $\mathcal{R}_{G, \varphi}$ arises from a uniquely ergodic free action of
a finitely generated abelian group, then there exists a countable unital subring $R$ of $\mathbb{R}$
such that $\mathcal{F}(\mathcal{R}_{G, \varphi})=R_{+}^\times$.
\end{cor}
\begin{proof}
Let $R$ be a countable unital subring of $\mathbb{R}_{+}^\times$.
By Corollary 6.3 in \cite{HPS}, there exists a uniquely ergodic Cantor minimal $\mathbb{Z}$-system
$\mathcal{R}_{\mathbb{Z}, \varphi}$ such that
$$(D_m(\mathcal{R}_{\mathbb{Z},\varphi}),D_m(\mathcal{R}_{\mathbb{Z},\varphi})_{+},
[1_{X}])\cong (R,R_{+},1).$$
Therefore we see that $\mathcal{F}(\mathcal{R}_{G, \varphi})=R_{+}^\times$
by Theorem \ref{thm:computation}.

Conversely, let $\mathcal{R}_{G, \varphi}$ be a uniquely ergodic Cantor minimal system
arising from a free action of a finitely generated abelian group.
Then we have $\mathcal{F}(\mathcal{R}_{G, \varphi})=IM_{+}(T_{\mu}(D_m(\mathcal{R})))$
by Theorem \ref{thm:computation}.
Therefore Lemma 3.6 in \cite{NW} implies that there exists a countable unital subring $R$ of $\mathbb{R}$
such that $\mathcal{F}(\mathcal{R}_{G, \varphi})=R_{+}^\times$.
\end{proof}
By the theorem above, $\{9^n:n\in\mathbb{Z}\}$ cannot be realized as the
fundamental group of a uniquely ergodic Cantor minimal system arising from a
free action of a finitely generated abelian group.

Let $A$ be a simple $C^*$-algebra with a unique trace $\tau$.
The fundamental group $\mathcal{F}(A)$ of $A$ is defined by the set
$$
\{ \tau\otimes Tr(p) \in \mathbb{R}^{\times}_{+}\ | \
 p \text{ is a projection in } M_n(A) \text{ such that } pM_n(A)p  \cong A \}
$$
where $Tr$ is the unnormalized trace on $M_n(\mathbb{C})$.
We refer the reader to \cite{Bla} for basic facts of operator algebras.
We denote by
$\tau_*$ the map $K_0(A)\rightarrow \mathbb{R}$ induced by a trace $\tau$ on $A$.
By Proposition 3.7 in \cite{NW}, we have
$\mathcal{F}(A)\subseteq IM_{+}(\tau_{*}(K_0(A)))$.
If $\varphi$ is a uniquely ergodic free minimal action of $G$ on $X$,
then the associated $C^*$-algebra $C(X)\rtimes_{\varphi}G$ is a simple $C^*$-algebra
with unique trace.
We do not know whether there exists a relation between orbit
equivalence of Cantor minimal $G$-systems and $C^*$-isomorphism of associated $C^*$-algebras
in general. But we have the following theorem.
\begin{thm}
Let $\mathcal{R}_{\mathbb{Z}^n, \varphi}$ be a uniquely ergodic Cantor minimal system
arising from a free action of a finitely generated free abelian group $\mathbb{Z}^n$.
Then $\mathcal{F}(C(X)\rtimes_{\varphi}\mathbb{Z}^n)\subseteq \mathcal{F}
(\mathcal{R}_{\mathbb{Z}^n, \varphi})$.
\end{thm}
\begin{proof}
The Gap Labeling Theorem (Theorem D.1 in \cite{MS}, see also \cite{Bell}, \cite{BO} and \cite{KP})
implies that
$$
T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^n, \varphi}))=\tau_{*}(K_0(C(X)\rtimes_{\varphi}\mathbb{Z}^n)).
$$
By Theorem \ref{thm:computation},
$\mathcal{F}(\mathcal{R}_{\mathbb{Z}^n, \varphi})=IM_{+}(T(D_m(\mathcal{R}_{\mathbb{Z}^n, \varphi})))$.
Hence we see that
$\mathcal{F}(C(X)\rtimes_{\varphi}\mathbb{Z})\subseteq \mathcal{F}(\mathcal{R}_{\mathbb{Z},
\varphi})$ because we have
$\mathcal{F}(C(X)\rtimes_{\varphi}\mathbb{Z}) \subseteq IM_{+}(\tau_{*}(K_0(C(X)\rtimes_{\varphi}
\mathbb{Z}^n)))$ by Proposition 3.7 in \cite{NW}.
\end{proof}
Note that we do not know whether $C(X)\rtimes_{\varphi} \mathbb{Z}^2$ belongs to classifiable
classes by $K$-groups.
We shall show that there exists a uniquely ergodic Cantor minimal $\mathbb{Z}$-system
$\mathcal{R}_{\mathbb{Z}, \varphi}$
such that $\mathcal{F}(C(X)\rtimes_{\varphi}\mathbb{Z})\neq \mathcal{F}
(\mathcal{R}_{\mathbb{Z}, \varphi})$.
By Corollary 6.3 in \cite{HPS} and \cite{EHS},
there exists  a uniquely ergodic Cantor minimal $\mathbb{Z}$-system
$\mathcal{R}_{\mathbb{Z}, \varphi}$
such that
$$
K_0(C(X)\rtimes_{\varphi}\mathbb{Z})=\{(\frac{b}{9^a},c) \in \mathbb{R} \times \mathbb{Z} \ |
\ a,b,c\in\mathbb{Z},b\equiv c\; \mathrm{mod}\; 8\},
$$
$$
K_0(C(X)\rtimes_{\varphi}\mathbb{Z})_{+}=\{(\frac{b}{9^a},c)\in K_0(A):\frac{b}{9^a}>0\}\cup \{(0,0)\}
\ \ \text{and} \ \   [1_A]_0=(1,1).
$$
Then we see that
$\mathcal{F}(\mathcal{R}_{\mathbb{Z}, \varphi})=\{3^n:n\in\mathbb{Z}\}$ and
$\mathcal{F}(C(X)\rtimes_{\varphi}\mathbb{Z})=\{9^n:n\in\mathbb{Z}\}$.
See Example 3.12 in \cite{NW2}.

\section{Examples}\label{sec:examples}
First we shall consider a generalization of Example \ref{ex:elementary}.
Let $\{n_i\}_{i\in\mathbb{N}}$ be a sequence of natural numbers, each greater than or equal to 2
and let $X=\Pi_{i\in\mathbb{N}}\{0,..,n_i-1\}$. Define
a homeomorphism $\varphi$ by the addition of $(1,0,0,...)$ with carry over the right
and $\varphi ((n_1-1,n_2-1,...))=(0,0,...)$.
Then $\mathcal{R}_{\mathbb{Z},\varphi}$ is a uniquely ergodic Cantor minimal system.
This system is called the \textit{odometer system} associated with $\{n_i\}_{i\in\mathbb{N}}$.
For each prime number $p$, define $\epsilon_{p}$ in
$\mathbb{Z}_{+}\cup\{\infty\}$ by
$\sup\{m:p^m\;\mathrm{divides}\;\Pi_{i=1}^kn_{i}\;\mathrm{for}\;\mathrm{some}\;k\in\mathbb{N}\}$.
Let $S$ be the set of prime numbers and
$N:=\prod_{p\in S}p^{\epsilon_{p}}$. We call $N$ the \textit{super natural number}.
It is known that two odometer systems $\varphi_1$ and $\varphi_2$ are topologically
conjugate if and only if their super natural numbers are equal.
\begin{pro}
Let $\mathcal{R}_{\mathbb{Z},\varphi}$ be an odometer system associated with
$\{n_i\}_{i\in\mathbb{N}}$.
Then
$$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi})=
\{p_1^{m_1}\cdot\cdot\cdot p_{n}^{m_n}:n\in\mathbb{N}, \epsilon_{p_i}=\infty ,m_i\in\mathbb{N} \}
$$
where
$\epsilon (p)=\sup\{m:p^m\;\mathrm{divides}\;\Pi_{i=1}^kn_{i}\;\mathrm{for}\;\mathrm{some}\;k\in\mathbb{N}\}$.
\end{pro}
\begin{proof}
Since $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi}))=\{\frac{n}{q}:n\in\mathbb{Z},\;q=p^m\;\mathrm{where}\;p\;\mathrm{is}\;\mathrm{a}\;\mathrm{prime}\;
\mathrm{number}\;\mathrm{and}\\ 0\leq m\leq \epsilon_{p}, m<\infty \}$,
Theorem \ref{thm:computation} implies the conclusion.
\end{proof}
We shall consider fundamental groups of certain Denjoy systems.
A \textit{Denjoy homeomorphism} $\varphi$ is an homeomorphism of $\mathbb{T}$ such that
the rotation number $\rho (\varphi )$ is irrational and $\varphi$ is not conjugate
to a pure rotation. Such a homeomorphism has a unique minimal set $\Sigma$ which is
homeomorphic to the Cantor set. The restriction of a Denjoy homeomorphism to its unique
minimal set $\Sigma$ is called the \textit{Denjoy system}.
It is known that a Denjoy system $\mathcal{R}_{\mathbb{Z},\varphi |_{\Sigma}}$ is a uniquely ergodic
Cantor minimal system.
We refer the reader to \cite{Mar}, \cite{MSY} and \cite{Put} for details of Denjoy systems.
It is known that for a Denjoy homeomorphism $\varphi$ there exists an orientation preserving
surjective continuous map $h$ of $\mathbb{T}$ to $\mathbb{T}$ such that $h\circ \varphi =R_{\rho (\varphi)}
\circ h$ where $R_{\rho (\varphi)}$ is the rotation such that
$R_{\rho (\varphi)}([t])=[t+\rho (\varphi)]$. (We regard $\mathbb{T}$ as
$\{e^{2\pi it}:t\in \mathbb{R} \}$.)
With notation as above, put $Q(\varphi)=h(\{z\in \mathbb{Z}:\sharp (h^{-1}(h(z)))\neq 1\})$.
The set $Q(\varphi )$ is is countable and $R_{\rho (\varphi)}$-invariant.
Conversely, any countable $R_{\rho (\phi)}$-invariant subset of $\mathbb{T}$ can be realized
as $Q(\varphi)$ of some Denjoy homeomorphism $\varphi$.
Markley showed that two Denjoy homeomorphisms $\varphi_1$ and $\varphi_2$ are conjugate
via an orientation preserving conjugating map
if and only if $\rho (\varphi_1)=\rho (\varphi_2)$ and
$Q(\varphi_1 )=R_{\theta} (Q(\varphi_2))$ for some $\theta\in (0,1]$.
\begin{pro}\label{pro:Denjoy}
Let $\theta$ be an irrational number in $[0,1]$ and $\varphi_{\theta}$ a Denjoy homeomorphism with
$\rho (\varphi_{\theta})=\theta$ and $Q(\varphi_{\theta})=\{e^{2\pi in\theta }:n\in\mathbb{Z}\}$.
Then $\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})$ is a singly
generated group.
More precisely, if $\theta$ is not a
quadratic number, then
$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})=\{1\}$.
If $\theta$ is a quadratic number,
then $\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})
= \{\epsilon _0^n  \in R^\times_+ \ | \ n \in \mathbb{Z} \}$
and the generator $\epsilon _0$ is given by
$\epsilon _0 =\frac{t+u\sqrt{D_\theta}}{2}$,
where
$D_\theta$ is the  discriminant of $\theta$ and $t,u$ are the positive
integers satisfying one of the Pell
equations $t^2-D_\theta u^2=\pm 4$ and $\frac{t+u\sqrt{D_\theta}}{2}$ exceeds $1$
and is least.
\end{pro}
\begin{proof}
By theorem \ref{thm:computation}, we see that
$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})=IM_{+}
(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})))$.
Since we have $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}}))=\mathbb{Z}+\mathbb{Z}\theta$
(see, for example, \cite{Put}), the same argument as in \cite{NW}(Corollary 3.18) implies
the conclusion.
\end{proof}
We shall show some examples.
\begin{ex}
Let $\theta =\frac{-1+\sqrt{5}}{2}$. Then
$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})=\{(\frac{1+\sqrt{5}}{2})^n:n
\in \mathbb{Z}\}$.
\end{ex}
\begin{ex}
Let $\theta =\sqrt{5}-2$. Then
$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})=\{(\sqrt{5}-2)^n:n
\in \mathbb{Z}\}$.
\end{ex}
\begin{ex}
Let $\theta =\frac{1}{\sqrt{5}}$. Then
$\mathcal{F}(\mathcal{R}_{\mathbb{Z},\varphi_{\theta} |_{\Sigma}})=\{(\sqrt{5}-2)^n:n
\in \mathbb{Z}\}$.
\end{ex}

We shall consider a natural generalization to $\mathbb{Z}^2$-actions of Denjoy systems.
Let $\theta_1$ and $\theta_2$ be irrational numbers and $C$ an countable invariant subset
of $\mathbb{T}$ under the rotations $R_{\theta_1}$ and $R_{\theta_2}$.
We can construct a Cantor set $X$ by disconnecting the circle $\mathbb{T}$ along
any point of $C$. Moreover there exist commutating homeomorphisms $\varphi_1$ and
$\varphi_2$ of $X$, which are extensions of $R_{\theta_1}$ and $R_{\theta_2}$ respectively.
See chapter 3 in \cite{CFS}, \cite{FH}, Example 4.4 in \cite{GPS3} and \cite{Put} for details
and precise definition.
It is known that this Cantor $\mathbb{Z}^2$-system is a uniquely ergodic
Cantor minimal system. Moreover the invariant probability measure of this Cantor minimal
system comes from the normalized Haar measure on $\mathbb{T}$.
We call this Cantor system the \textit{Denjoy $\mathbb{Z}^2$-system} associated with
$(\theta_1, \theta_2, C)$.
We need algebraic number theory of cubic fields to compute some examples of
fundamental groups of Denjoy $\mathbb{Z}^2$-systems.
We refer the reader to \cite{SW} for algebraic number theory of cubic fields.
\begin{ex}
Let $\mathcal{R}_{\mathbb{Z}^2,\varphi}$ be a Denjoy $\mathbb{Z}^2$-system associated with
$\theta_1=\sqrt[3]{2}$, $\theta_2=\sqrt[3]{4}$ and
$C=\{e^{2\pi i(n\sqrt[3]{2}+m\sqrt[3]{4})}:n,m\in\mathbb{Z}\}$.
Then we have $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^2,\varphi}))=
\mathbb{Z}+\mathbb{Z}\sqrt[3]{2}+\mathbb{Z}\sqrt[3]{4}$.
Since $\mathbb{Z}+\mathbb{Z}\sqrt[3]{2}+\mathbb{Z}\sqrt[3]{4}$ is the ring of
integers on a cubic field $\mathbb{Q}(\sqrt[3]{2})$,
$IM_{+}(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^2,\varphi} )))$ is the positive part of
the unit group of $\mathbb{Q}(\sqrt[3]{2})$, which is equal to
$\{(1+\sqrt[3]{2}+\sqrt[3]{4})^n:n\in\mathbb{Z}\}$. (See Theorem 13.6.2 in \cite{SW}.)
Therefore Theorem \ref{thm:computation} implies
$$\mathcal{F}(\mathcal{R}_{\mathbb{Z}^2, \varphi})=\{(1+\sqrt[3]{2}+\sqrt[3]{4})^n:n\in\mathbb{Z}\}.$$
\end{ex}
\begin{ex}
Let $\mathcal{R}_{\mathbb{Z}^2,\varphi}$ be a Denjoy $\mathbb{Z}^2$-system associated with
$\theta_1=\sqrt[3]{3}$, $\theta_2=\sqrt[3]{9}$ and
$C=\{e^{2\pi i(n\sqrt[3]{3}+m\sqrt[3]{9})}:n,m\in\mathbb{Z}\}$.
Then we have $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^2,\varphi}))=
\mathbb{Z}+\mathbb{Z}\sqrt[3]{3}+\mathbb{Z}\sqrt[3]{9}$.
Since $\mathbb{Z}+\mathbb{Z}\sqrt[3]{3}+\mathbb{Z}\sqrt[3]{9}$ is the ring of
integers on a cubic field $\mathbb{Q}(\sqrt[3]{3})$,
$IM_{+}(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^2,\varphi})))$ is the positive part of
the unit group of $\mathbb{Q}(\sqrt[3]{3})$, which is equal to
$\{(4+3\sqrt[3]{2}+2\sqrt[3]{4})^n:n\in\mathbb{Z}\}$.
(See \cite{SW} or Appendix B.3 in \cite{cohen}.)
Therefore Theorem \ref{thm:computation} implies
$$\mathcal{F}(\mathcal{R}_{\mathbb{Z}^2, \varphi})=\{(4+3\sqrt[3]{2}+2\sqrt[3]{4})^n:n\in\mathbb{Z}\}.$$
\end{ex}
\begin{ex}
Let $\mathcal{R}_{\mathbb{Z}^2,\varphi}$ be a Denjoy $\mathbb{Z}^2$-system associated with
$\theta_1=2\cos{\frac{2\pi}{7}}$, $\theta_2=4\cos^2{\frac{2\pi}{7}}$ and
$C=\{e^{2\pi i(n2\cos{\frac{2\pi}{7}}+m4\cos^2{\frac{2\pi}{7}})}:n,m\in\mathbb{Z}\}$.
Then we have $T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z}^2,\varphi}))
=\mathbb{Z}+\mathbb{Z}2\cos{\frac{2\pi}{7}}+\mathbb{Z}4\cos^2{\frac{2\pi}{7}}$.
Note that $2\cos{\frac{2\pi}{7}}$ is a root of the equation $x^3+x^2-2x-1=0$.
Since $\mathbb{Z}+\mathbb{Z}2\cos{\frac{2\pi}{7}}+\mathbb{Z}4\cos^2{\frac{2\pi}{7}}$ is the
ring of integers on a cubic field $\mathbb{Q}(2\cos{\frac{2\pi}{7}})$,
$IM_{+}(T_{\mu}(D_m(\mathcal{R}_{\mathbb{Z},\varphi |_{\Sigma}})))$ is the positive part of
the unit group of $\mathbb{Q}(2\cos{\frac{2\pi}{7}})$, which is equal to
$\{(-1+2\cos{\frac{2\pi}{7}}+4\cos^2{\frac{2\pi}{7}})^n(2-4\cos^2{\frac{2\pi}{7}})^m:n,m\in
\mathbb{Z}\}$.
(See Appendix B.4 in \cite{cohen}.)
Therefore Theorem \ref{thm:computation} implies
$$\mathcal{F}(\mathcal{R}_{\mathbb{Z}^2, \varphi})=\{(-1+2\cos{\frac{2\pi}{7}}+
4\cos^2{\frac{2\pi}{7}})^n(2-4\cos^2{\frac{2\pi}{7}})^m:n,m\in\mathbb{Z}\}.$$
\end{ex}

\section*{Acknowledgments}
The author would like to thank Professor Matui for informing him about the Gap Labeling
Theorem. He is also grateful to Professor Watatani for his constant encouragement.

redpajama_set_name: RedPajamaArXiv


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text:
Bailey's Harbor Light Seeing The Light
Bailey's Harbor, Wisconsin
The Door Peninsula is a long, thin finger of land which juts north into Lake Michigan. French voyageurs experiencing violent seas in the passage between the peninsula and Washington Island named the passage Porte des Mort" which translates as "Door of the Dead." When deciding on a name for a new county to include the entire peninsula in 1851, the Wisconsin Legislature decided to drop the latter part of the name, and settled on Door County.
In the fall of 1848, Captain Justice Bailey encountered a fierce October storm while traversing the Eastern Shore of the peninsula, and carrying a full complement of passengers, decided to seek shelter from the worsening storm. Pulling into a sheltered bay, and dropping anchor to ride out the storm, the Captain took the opportunity to explore some of the shore. Finding limestone, pine, maple and beech trees, Bailey reported his discovery to Mr. Alanson Sweet, the owner of the shipping company for whom Bailey was employed.
In 1849, Sweet sent a crew to the Harbor to construct a pier, sawmill, quarry and group of houses for his workers. That winter 2,500 cords of lumber were shipped from the harbor. Sweet managed to convince the State Legislature that Bailey's Harbor was destined to grow, and the village was named as the County Seat.
Sweet then petitioned the Federal Government for the construction of a lighthouse to assist vessels making their way into the harbor. Congress responded favorably and awarded Sweet the contract to build the lighthouse. Selecting a one thousand foot by two hundred and fifty foot island as the best location for the tower, Sweet's crew began construction in 1852. With lake levels extremely low that year, they were able to transport all the building materials directly to the island.
Built of native stone from Sweet's quarry, the fifty-two foot tower was capped with a birdcage-style lantern room, and equipped with a fixed white Sixth Order Fresnel lens. As was usual in lantern rooms of this style, the lens was supported in the center of the tower with no platform or catwalk surrounding the lantern room. The lens was later upgraded to a fixed white Fifth Order Fresnel in 1858.
As a result of significant deterioration, in 1866 the Lighthouse Board recommended that the light be rebuilt, and $15,000 was appropriated for the project. However, at the last minute, it was determined that the harbor would be better served through the construction of range lights on the west shore, and the coastline would be better served through the construction of a light on Cana Island. Thus the decision was made to abandon the light as soon as the range lights were constructed.
At the close of the shipping season on December 1, 1869 the light was displayed for the last time, and the station closed. At some time thereafter, the island was sold into private ownership.
Click Here to see a complete listing of all Baileys Harbor Light keepers compiled by Phyllis L. Tag of Great Lakes Lighthouse Research.
At the North end of Baileys' Harbor, turn at the Sandpiper Restaurant on the corner of Hwy 57. Follow Ridges Road past the yacht club, to the pint where the road takes a sharp left turn. Right at the turn you will see the lake ahead at the end of a gravel road. Drive down this gravel road approximately 300 feet, and you will see the lighthouse on the island to your right.
GPS Coordinates: 45° 3'21.03"N x 87° 5'49.81"W
Inventory of Historic Light Stations, National Parks Service, 1994 .
Wisconsin Handbook, 1997, Thomas Huhti.
Lighthouse Board annual reports, 1851 - 1870.
Numerous emails with Tom Williams, ex-President of Jacksonport Historical Society.
Old photograph of station courtesy of Annie Peil of Orphan Annie's Schoolhouse Inn.
Personal observation at Bailey's Harbor, 09/11/2000.

redpajama_set_name: RedPajamaCommonCrawl


text:
Search continues for Scottish businessman missing off Egyptian coast
Search continues for Scottish businessman missing off Egyptian coast /node/1610181/middle-east
Colin Finnie's yacht, Simba, was discovered only 350 miles from his destination on Dec. 1, 2019. (boatwatch.org)
https://arab.news/6hbjb
CAIRO: Egyptian authorities are searching for a missing businessman from Scotland whose yacht was found near the Egyptian coastal city of Marsa Alam.
Colin Finnie was on a solo voyage from Australia, traversing the Indian Ocean, to Port Ghalib, in the Southern Red Sea city of Marsa Alam.
His yacht, Simba, was discovered only 350 miles from his destination on Dec. 13 – six days after he last contacted his loved ones on Dec. 7.
The Egyptian Coastguard found the unmanned 45-foot-long yacht after it apparently had come into contact with a coral reef.
According to the Boat Watch website, his transponder – a device used to broadcast the location of vessels – had last "pinged" on 25 October, when it was at Christmas Island.
Finnie's wife appealed to everyone with information on her husband to help her, after she expressed her concern in a post on Facebook.
She said: "My husband, Colin Finnie was heading towards Port Ghalib in the sailing boat Simba, a Nauticat 44.
"I lost contact with him last Saturday 7th when he was 300 [nautical] miles away from his destination. He was in storms and low on fuel.
"I have been in contact with Foreign Office and Coastguard and the Egyptian authorities are actively looking for him. I know it's a long shot but I am trying everything. Sorry to post bad news."
A spokeswoman for the Foreign Office said: "We are supporting the family of a British man reported missing while sailing in the Red Sea and are in contact with Egyptian authorities."
Topics: Egypt Marsa Alam Scotland
Art exhibition marries Egyptian and Saudi 'Faces and Places'


text:
Show ALL Forums > Art/Music > Blues Artists Similar to Stevie Ray Vaughan Home login
mrcs84
Blues Artists Similar to Stevie Ray VaughanPage 1 of 2 (1, 2)
you might just want to plug him into Pandora.com and see what you find.
Ready4SomethingFun
Blues Artists Similar to Stevie Ray Vaughan
Try Corey Stevens (good start is his album Blue Drops of Rain) and a band called Indigenous. They are an American Indian group that are amazing. Their Live @ Pachyderm Studios album is great & the studio stuff, while not as great is still very good. Circle is my personal favorite of the studio LPs.
Another lesser known guy to check out is Eric Jerardi. He is a local Dayton Ohio artist but a few of his LPs are available nationally, and are almost always available on Amazon or Ebay. He injects a bit of the bluesy soul of SRV into almost everything he sings and plays.
iam4stringer
Check out Joe Bonamassa.Great player, he has been around for a long while, just not well known all over.
Oh yes, You may like The Derek Trucks band, or Trucks and Tedeschi, and one of my personal favourites Govt Mule.
newdejavu
Check out Omar and the Howlers.
2findU
Leslie West is another one to add to that list. Anyone like Govn't Mule?
number451
joe bonamassa, as already mentioned. the best blues player alive today
THEKidWicked
fabulous thunderbirds.
tinangel53
The ONE and the ONLY Elmore James!!!! After that HMMMMMM.......maybe Tinsley Ellis......all the rest are just that....the rest.
Tayfire
THEKidWicked mentioned The Fabulous Thunderbirds good band. It was Steve-rays elder brother Jimmy who was the original guitarist of the band and taught his younger brother how to play and went solo in 1990 who is still recording and playing today.
JvrodZ
http://www.youtube.com/watch?v=YkhNobd3HC8&feature=related
You can search a few vids of him and Stevie as well...
michele6259
Johnny Lee Hooker was a great "guitar' based artist.
although not "guitar" based like SRV I still love me some Gregg Allman.
Bachelor_Numero_Uno
OK, all the artists mentioned so far by other posters are great. But except for Kenny Wayne & Jeff Healey (listen to Healey's covers of "Confidence Man", & "Hoochie Coochie Man"!), they're not really channeling Stevie Ray. If you're specifically looking for artists that sound very similar to Stevie Ray, then check out these guys:
-Chris Duarte "Texas Sugar/Strat Magick" CD, especially the song "Big-Legged Woman". I think Chris's guitar playing comes the closest to Stevie's in terms of fat tone, attack, & passion. This album is like Stevie Ray meets the Chili Peppers!
-Doyle Bramhall (senior) (1949 - 2011). This guy is Doyle Bramhall II's father, & was 5 years older than Stevie Ray. He was the drummer for Stevie's brother Jimmie in the Chessmen, & Stevie looked up to him. Doyle senior was also a great singer, & wrote several songs for Stevie that Stevie recorded. Stevie got his singing style from Doyle senior, it's hard to tell them apart. Listen to Doyle's original version of "Life By the Drop", "Lost in the Congo", "Fourty Four", & "Tortured Soul".
-Cetan Clawson's song "I Wanna Try" with Chris Layton from Double Trouble on drums, & Scott Nelson of Kenny Wayne Shepherd on bass. Also the live version of "Slingin" on YouTube. This kid looks like a cross between a young Bob Dylan dressed like Jimi Hendrix, & plays like a cross between Hendrix & Stevie Ray.
-Fredrick Strand Halland, a young Norwegian Guitarist, strongly influenced by Stevie Ray. Lots of Stevie Ray covers by him on YouTube.
-Philip Sayce, a Canadian. He's the former guitarist for Melissa Etheridge, & former second Guitarist for Jeff Healey, & has a Stevie-influenced style, albeit drenched in a lot of reverb. Check out his performance at the NAMM guitar show, with Yngwee Malsteem in the audience (Yngwee thinking "I weel Keel you") – on the YouTube video titled "YouTube - Yngwie Checking Out Philip Sayce @ Fender_NAMM 2009.mp4".
Philip also got praised by Bon Jovi when Bon Jovi saw him perform with Melissa Etheridge. See YouTube video "JON BON JOVI REACTS TO PHILIP SAYCE AT GRAMMY EVENT"
-John Mayer also does a good live imitation of Stevie Ray on several Youtube videos.
That is the best advice. I have found artists similar to other artists I like using Pandora.
Page 1 of 2 (1, 2)
Show ALL Forums > Art/Music > Blues Artists Similar to Stevie Ray Vaughan


The proportion of women in high earning jobs has remained static for the past seven years, a study has shown.
Women account for a quarter of higher rate taxpayers - those declaring an income between £43,000 and £150,000 - in each of the last seven financial years, said law firm Clyde & Co.
Heidi Watson, employment partner at Clyde & Co, said: "In recent years there has been a raft of industry and government initiatives launched to help increase the number of women in senior positions, so it's disappointing and somewhat surprising not to see any progress at all in the proportion of high earners."
Ms Watson said the lack of women in senior positions was a main reason for the gender pay gap, although more companies were taking action as a result of having to report any pay differences between men and women.
A separate study found that the UK's gender pay gap was wider than in most European countries.
Consulting firm Korn Ferry said the difference was mainly caused by fewer women than men min high-paying jobs.
The gap falls to one of the lowest in Europe when measured against men and women doing the same job in a company, research showed.


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On October 7, 1998, Matthew Shepard, a 21-year-old student at the University of Wyoming, was brutally attacked and tied to a fence in a field outside of Laramie, Wyo. and left to die, finally succumbing to his wounds on October 12, 1998. In commemoration of the 20th anniversary of his death, please join us for the first ever Perspectives Series, bringing together community partners to open doors to conversation through education, storytelling and creativity.
Our series will conclude with a performance of Considering Matthew Shepard, Conspirare's GRAMMY-nominated choral work that is an intricate, beautiful and bold voice in response to one of America's most notorious anti-gay hate crimes.
Join Equality Michigan for training and conversation designed to give participants an introduction into LGBTQ communities.
Join us for a screening of Matt Shepard is a Friend of Mine, an award-winning, critically acclaimed documentary feature film that tells the story of Matt's life through his friends, family and those who were close to him throughout the years.
Join Here: Say Storytelling and invited storytellers from our local community to hear funny and painful stories alike about identity, acceptance, advocacy, dissent…the stories we don't tell each other.


text:
Publication date: 22 September 2022
Effects of 60Co γ ray radiation on the transmission characteristics of interconnection structures for 3D packaging
Youxin Zhang, Yang Liu, Rongxing Cao, Xianghua Zeng and Yuxiong Xue
Concerning the radiation effects on the three-dimensional (3D) packaging in space environment, this study aims to investigate the influence of the total dose effect on the…
Concerning the radiation effects on the three-dimensional (3D) packaging in space environment, this study aims to investigate the influence of the total dose effect on the transmission characteristics of high-frequency electrical signals using experimental and simulation methods.
This work carries out the irradiation test of the specimens and measures their S21 parameters before and after irradiation. A simulation model describing the total dose effect was built based on the experimental test results. And, the radiation hardening design is evaluated by the simulation method.
The experimental results demonstrate that the S21 curve of the interconnection decreases with the increase of the irradiation dose, indicating that the total dose effect leads to the decline of its signal transmission characteristics. According to the simulation results, decreasing the height of the through silicon via (TSV), increasing the radius of the TSV, reducing the length of Si and increasing the number of grounded through silicon via have positive effects on improving the radiation resistance of the interconnection structure.
This work investigates the effect of radiation on the transmission characteristics of interconnection structures for 3D packaging and proposes the hardening design methods. It is meaningful for improving the reliability of 3D packaging in space applications.
Microelectronics International, vol. ahead-of-print no. ahead-of-print
3D packaging
Effects of Sn-Ag-x leveling layers on the microstructure and mechanical properties of SnBi low-temperature solder joints
Yang Liu, Yuxiong Xue, Min Zhou, Rongxing Cao, Xianghua Zeng, Hongxia Li, Shu Zheng and Shuang Zhang
The purpose of this paper is to investigate the effects of Sn-Ag-x leveling layers on the mechanical properties of SnBi solder joints. Four Sn-Ag-x (Sn-3.0Ag-0.5Cu…
The purpose of this paper is to investigate the effects of Sn-Ag-x leveling layers on the mechanical properties of SnBi solder joints. Four Sn-Ag-x (Sn-3.0Ag-0.5Cu, Sn-0.3Ag-0.7Cu, Sn-0.3Ag-0.7Cu-0.5 Bi-0.05Ni and Sn-3.0Ag-3.0 Bi-3.0In) leveling layers were coated on Cu pads to prepare SnBi/Sn-Ag-x/Cu solder joints. The microstructure, hardness, shear strength and fracture morphology of solder joints before and after aging were studied.
The interfacial brittleness of the SnBi low-temperature solder joint is a key problem affecting its reliability. The purpose of this study is to improve the mechanical properties of the SnBi solder joint.
Owing to the addition of the leveling layers, the grain size of the ß-Sn phase in the SnBi/Sn-Ag-x/Cu solder joint is significantly larger than that in the SnBi/Cu eutectic solder joint. Meanwhile, the hardness of the solder bulk in the SnBi/Cu solder joint shows a decrease trend because of the addition of the leveling layers. The SnBi/Cu solder joint shows obvious strength drop and interfacial brittle fracture after aging. Through the addition of the Sn-Ag-x layers, the brittle failure caused by aging is effectively suppressed. In addition, the Sn-Ag-x leveling layers improve the shear strength of the SnBi/Cu solder joint after aging. Among them, the SnBi/SACBN/Cu solder joint shows the highest shear strength.
This work suppresses the interfacial brittleness of the SnBi/Cu solder joint after isothermal aging by adding Sn-Ag-x leveling layers on the Cu pads. It provides a way to improve the mechanical performances of the SnBi solder joint.
Soldering & Surface Mount Technology, vol. 34 no. 3
DOI: https://doi.org/10.1108/SSMT-08-2021-0052
SnBi
Sn-Ag-x
Leveling layer
Earlycite article (1)


text:
Basics of Android Rooting and Custom Roms
AndroidTricks
An Intro to Andriod Bootloader:
Every Android phone has a bootloader that instructs the kernel to boot normally. But you need to understand one thing here that as Android OS is an open source and is available on a variety of different hardware, every manufacturer has their own version of bootloader specific for the hardware present in it and even the developer-friendly Nexus devices shipped with a locked bootloader (Nexus devices and a couple tablets are easily unlocked with a single command).In fact, a lot bootloaders are locked and encrypted, meaning simple commands do the things.
Why are Bootloaders Locked?
A bootloader is usually locked on an Android device because although its an open source OS, still the manufacturers want you to stick to their Android OS version specifically designed for the device.With a locked bootloader on Android devices, it is virtually impossible to flash a Custom ROM and forced attempts void warranty as well as usually end up in bricks. Therefore, the first step is to always unlock the bootloader.
A look into Custom Roms
A ROM, for those who don't know, is an operating system build that runs on your device with basic applications such as an address book, calendar, camera, etc.A major advantage of Android smartphones is that they can be operated by third-parties and not only with the original ROM. Custom ROMs replace the pre-installed version of Android on devices. The substitution is made by a variation of the AOSP.
But what is AOSP?
AOSP means Android Open Source Program. It is a version of the open source code of Android, developed by Google to offer a pure version of the system, which is available to anyone. It can be modified by developers without the need to follow standards of Google applications.Thanks to this project, the birth of custom ROMs like CyanogenMod, Paranoid Android, MIUI and others have come about.
ALSO READ Zoomcar : Self Drive Car Rental Service
What is a kernel?
A kernel isn't something unique to Android — iOS and MacOS have one, Windows has one, BlackBerry's has one, in fact operating systems have one.
The kernel is the piece of software that bridges the gap between the operating system (and installed apps) and the hardware in the device. Everything you do on your device involves the kernel. Increasing the volume while playing a video? Android doesn't directly tell the speakers to increase output. Instead, it tells the kernel that it wants to increase the volume, and the kernel talks to the speaker to increase its output.
But why the middle man? In very simple terms, it makes Android a lot more flexible. Google can worry about creating all of the functionality in Android, but not have to worry about how that is exactly executed on each device. Since there are so many devices that have all sorts of hardware, Google can't worry about providing compatibility and drivers with each device. It's the manufacturer's job to create the kernel that has all the drivers necessary to make everything work on the device.
Do you want the best performance out of your Android device? Or maybe you want the best battery life? If you've already rooted your device or even installed a custom ROM onto it, it may be a good idea to check out the realm of custom kernels.
A look into Custom RomsAn Intro to Andriod BootloaderWhat is a kernel?what is AOSP?Why are Bootloaders Locked?
How to set up anti theft alarm in Android
Process of rooting Android Mobile easily
Process of rooting Android Mobile easily - TechieSwag 6 years ago
[…] Also Read-Basics of Android Rooting […]


text:
Q: MongoDB filtering and sorting an array I would like to find by a certain _id and then filter the author,subject,content and date and also do a sort({timestamp:-1}).
My data looks like this:
{
    "_id": "1",
    "message": [{
        "_id": {
            "$oid": "609cbe47a45b594af4bc6b5a"
        },
        "author": "Dr. Jameson",
        "subject": "Lab Results Uploaded",
        "content": "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nulla sceleri...",
        "timestamp": {
            "$date": "2021-05-13T05:51:03.793Z"
        },
        "lastDateRead": {
            "$date": "2021-05-13T05:51:03.793Z"
        }
    }, {
        "_id": {
            "$oid": "609cbe47a45b594af4bc6b5b"
        },
        "author": "Dr. Johnson",
        "subject": "Lab Results",
        "content": "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nulla sceleri...",
        "timestamp": {
            "$date": "2021-05-13T05:51:03.793Z"
        },
        "lastDateRead": {
            "$date": "2021-05-13T05:51:03.793Z"
        }
    }]
}

So I need to check if _id the top level is 1 and then go into messages and filter the rest based off a variable and pass the rest back.
Current code which finds by id correctly
async function findPatientById(client, id) {
    const result = await client.db("any").collection("any").findOne({ _id: id});
    if (result) {
        console.log(`Found in the collection with the id '${id}'`);
        console.log(result);
    } else {
        console.log(`None found with the id'${id}'`);
    }
}

So if I passed "Dr. Jameson" as a variable and id =1 it would give
{
    "_id": {
        "$oid": "609cbe47a45b594af4bc6b5a"
    },
    "author": "Dr. Jameson",
    "subject": "Lab Results Uploaded",
    "content": "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nulla sceleri...",
    "timestamp": {
        "$date": "2021-05-13T05:51:03.793Z"
    },
    "lastDateRead": {
        "$date": "2021-05-13T05:51:03.793Z"
    }
}

A: In MongoDB you can't do it with .find(), you need to use .aggregate() thru which you can perform more complex reads.
Try this:
db.collection.aggregate([
  { $match: { _id: "1" } }, //filter docs
  { $unwind: "$message" }, // unwind array
  { $match: { "message.author": "Dr. Jameson" }}, //Add more filters
  { $sort: { "message.timestamp": -1 } },
  { $group: { _id: "$_id", message: { $push: "$message" } } } // Re-group messages array
])

Alternative to $unwind you can use $filter stage to achieve same results. Don't forget to use proper indexes
Ref : MongoDB-Aggregation-Documentation
Test : MongoDB-Playground

redpajama_set_name: RedPajamaStackExchange


Thank you for visiting. My name is Sarah and I would like to welcome you to Hands of Healing.
I specialize in energy healing, yoga and wellness workshops.
I have created my business because I care deeply about helping people on their journeys, spiritual connection and healing on all levels.
I am a proud member of the 'Discover Healthy Living Group'.
We don't share an office- we share a common desire which is to help you get well, and find the best possible practitioners to get you there!


Such men are slaves, not of our Lord Christ but of their own appetites; and by their smooth and flattering speech they deceive the hearts of the unsuspecting.
In Romans 16, Paul closed his letter with a warning to watch out for false teachers. In verse 17, he talked about those who teach things "contrary to the teaching which you learned." That phrase literally means "alongside the teaching." Paul was saying, "Be careful of people who teach alongside the true gospel." You see, outright heresy is very easy to recognize. We can usually spot those who deny the deity of Christ, the inspiration of the Scripture, the blood atonement of Jesus for our sins, or the Second Coming. But Satan is also very subtle. Sometimes the way he deceives people is by changing Christian truth just a little. False teaching often talks about Jesus, the Bible, the crucifixion, and the resurrection, but it varies from the gospel in a slight way, just slight enough to cause people to miss Heaven.
These kinds of teachings say, "Yes, Jesus died on the cross, and we celebrate His death through the sacraments. And yes, to become a Christian you need to trust in Jesus–but you also need to take the sacraments, and you need to be baptized, and you need to keep the holy ordinances of the church." If you are trusting in anything other than Christ and Christ alone for your salvation, then you are lost, the Bible says. Paul said in Galatians 1:9, "If any man is preaching to you a gospel contrary to what you received, he is to be accursed!" He was talking about the Judaizers, who preached a false gospel. They accepted Jesus, but they said, "To be a Christian, you must trust in Jesus and also keep the Law and do these things." That is false teaching.
All cults can be defined this way: it is either the Bible plus something, or the Bible minus something. Deuteronomy 13:1-3 teaches us about false teachers. Through Moses, God said, "If a prophet or a dreamer of dreams arises among you and gives you a sign or a wonder, and the sign or the wonder comes true, concerning which he spoke to you, saying, 'Let us go after other gods (whom you have not known) and let us serve them,' you shall not listen to the words of that prophet or that dreamer of dreams; for the Lord your God is testing you to find out if you love the Lord your God with all your heart and with all your soul." If somebody is able to perform a sign or wonder but is teaching something contrary to God's Word, then do not listen to him.
You may say about a person, "If he worked a miracle, then he must be from God." No, Satan also has the power to perform miracles. The Bible says that during the Tribulation, the Antichrist and the False Prophet will work signs and wonders in order to deceive many. And remember, the magicians in Pharaoh's court had the ability to work miracles and signs. No, the way we judge whether a prophet is truly from God is not by the signs and wonders he performs; it is by his fidelity to the Word of God. That is what Paul was saying here. Beware of false teachers.


The Co-operative Group has joined an ethical living app called CoGo which links shoppers to local stores that are aligned with their values, on the back of the consumer trend towards more ethical living.
A total of 286 London Co-op stores now feature on the CoGo app which is also linked to consumers' payment cards, allowing it to show customers how their spending in selected stores is "contributing to good".
Co-op customer director Ali Jones said: "At Co-op we've been taking action on what matters most for decades.
"The products we sell impact millions of people, communities around the world, as well as the planet and that's why we work hard to make our products different, we always have done. To make them good for our customers, and the planet we share.
A Co-op spokeswoman confirmed that CoGo was being trialled in the London stores with no plans to extend it beyond Co-op owned stores for now.
The Co-op, which has pledged to make all packaging recyclable by 2023, launched its sustainable 'Future of Food' strategy to 2030 last year.
According to HIM's latest Future of Convenience report, plastic pollution and responsible retailing are among the key drivers of change that retailers are going to have to get to grips with to meet the needs of modern shoppers.
New data from Central England Co-op also suggests that shoppers are increasingly turning to Fairtrade products.
Data from the society shows that shoppers spent £7.4m on Fairtrade products at Central England Co-op Food Stores in 2018, 14% more than in the previous 12 months.
The figures also reveal an increase in customer spending on Fairtrade products during the 2018 Fairtrade Fortnight itself, up 12% from 2017, showing the positive effects the annual awareness campaign, currently running between 25 February to 10 March, has on shoppers.
The Co-operative Group is set to install SmartWater fog cannons in stores in an attempt to combat retail crime.


Over the past year or so I have become very fond of handmade and homemade skin care products. Love using products that were made by people that love what they do, so when Emily offered to send me a few of her products to review I was thrilled.
Today I'm going to be sharing some lovely products from Emily's Ardor.
I know I have mentioned how much I love lavender.I find the smell really soothing and it helps calm my headaches. So these butter bars are so nice to have sitting on my desk.
If you don't know what butter bars are, they are bars of solid lotion. These particular butter bars are made with essential oils, shea butter, beeswax, and coconut oil. They are super moisturizing, easy to use, and unlike moisturizer from a bottle, there is no waste because there is none left in the container that you can't get to.
Have I mentioned how much I love coffee? If you follow me on Instagram no doubt you already know that I'm a big fan of caffeine, but I'm not just a fan of coffee in liquid form, oh no, I love the smell of coffee, and I love coffee in skincare, so a vanilla latte scrub is right up my alley.
I have been using this every other morning, the smell wakes up my brain and the caffeine in the scrub wakes up my skin, so overall it is a perfect start to the day. This scrub is made with organic coffee, vanilla beans, and coconut oil.
I only have one little problem with this scrub, and that is that it is a little messy and there tends to be quite a bit of escaped scrub that doesn't make it to my face.
This scrub smells so yummy! I feel like apples tend to be a bit overshadowed by pumpkin this time of year, but I do have a soft spot for apples. This scrub leaves my skin feeling so soft and smooth, and the smell makes me want a real apple pie.
My final thoughts are that I absolutely love all three of these products, they are simple, refreshing, and made with love, so there really is no way you could go wrong with any of them.
Thank you so much to Emily for sending me these lovely products, it was such a pleasure working with her. Be sure to check out her shop (here).
Haha, it's adorable, isn't it?
You should try them sometime, they are a great new discovery for me.
aw it is soooo important to support small business – im such an advocate for always shopping in independant stores. and these look absolutely goooorgeous, i love how theyve been packaged up, sooo cute ♥♥ theres something you get from a product made 'with love' that you will never ever get from buying from a supermarket / large retailer!
I love the scents too, they are so nice and natural.
The vanilla latte scrub is the only one that's really messy, but it so worth it. The apple one smells incredible.
These look so nice and I am digging the all-natural ingredients! The lavendar would be my favorite.
Congrats on the new blog Hannah!
They smell good enough to eat too, although I doubt they would taste very good.
These sound like such lovely products made with such care.I could really do with the lavender one for on my desk for headaches too.


The year's best horror movie is now streaming for free!
'Hereditary' is 2018's horror movie with hype. That hype can be disastrous if you're seeing the movie late. Here's why you should see it anyway.
Netflix Movies Scorecard: What's Winning, From Rom-Coms to Sci-Fi?
You have never, ever, ever seen the ex-Naked Brother like this before.
The co-star of Jumanji: Welcome to the Jungle and the upcoming Dude has had a long road from being a naked brother.
Alex Wolff, Alex Fitzalan, Camila Perez, and Sebastian Chacon will star in the adaptation of the Michael Hassan novel.


You may have to pay a portion of the cost before the plumber works on the job, but never pay the total cost until you know the job has been completed correctly. You should know the plumber did what he promised before you pay him. If your toilet happens to be clogged, and the plunger is not working, and your water level inside of the toilet appears low, you may fix the problem by dumping warm water from a bucket into your toilet from about waist level. Do this every time the water level dips back down again. Check the floors in your bathroom for any give in order to be sure that there is no damage in the toilet to determine floor damage. You can save money by noticing the problem as early as possible. Pour baking soda followed by vinegar down your bathtub drain every month. Plug the drain and allow the chemical reaction to occur. Wait until the chemical reaction runs its course, then flush with boiling water. This procedure should open up the pipes by clearing out hair and soap scum. Ensure that the overflow holes in them. More information about Recommended Gas Hot Water Heater Service And Repair Company Sausalito CA.
Tankless water heaters only heat water constantly hot. This saves lots of money on your energy bill because you don't heat water that isn't used. The heat from the sun can be easy to ignore when you are focused on what you're doing, but that does not mean that it becomes any less dangerous. Copper pipes can expand and allow hot water is run through them. Save yourself big time and money by grouping your plumbing problems into the same call. Check the floor and/or wall to see if any loose tiles for water damage evidence.
If you are looking for Inexpensive Natural Gas Line Repair Company Olivehurst CA, sports tickets and theater seats at the best prices click here.


Lowest cost quality charger for NiMH batteries.
This a semi-custom -delta V detecting fast charger with a microprocessor controlled charge, sense LED display and preset time-out.
Some of the chargers time-out after 6 hours, limiting the size of the battery pack to 6AH or less. Others time-out after 16 hours, allowing battery packs up to 16AH to be charged. See the chart below.
The chargers in stock all have either the "plug" side of the "Tamiya" connector or the power connector kit. For the Tamiya connector the "receptacle" side is on the battery pack.
Plug of the Tamiya connector system (female or receptical side). For more information on this connector system click on the drawing.
ª Listed are the connector types which are on the shelf (see below). We can change these connectors for you in our shop if needed. See the connector discussion below for details.
These are known in the RC world as the "Tamiya Connector." Note the polarity. Mating connectors are available at Radio Shack or electronic distributors.
Several of our wall mount chargers are now available with a switchable power-plug style connector.
Inexpensive wall mount adapter with peak charge control circuit.


text:
"Chapter Seventy-Four: Wicked Little Town" is the seventeenth episode of the fourth season of the American television series Riverdale and the seventy-fourth episode of the series overall. The episode was written by Tessa Leigh Williams, directed by Antonio Negret and choreographed by Heather Laura Gray. It centered around the stage musical Hedwig and the Angry Inch by Stephen Trask.

It aired on The CW on April 15, 2020. Originally, the episode was scheduled for April 8, 2020, but was delayed due to the COVID-19 pandemic. According to Nielsen Media Research, it was watched by 0.54 million viewers.

The episode also introduced The Archies, a fictional band featuring Archie Comics's characters, for only the second time in a live action adaptation after the television film Archie: To Riverdale and Back Again in 1990.

It is the third musical episode in the series, after "Chapter Thirty-One: A Night to Remember" (season 2: Carrie: The Musical) and "Chapter Fifty-One: Big Fun" (season 3: Heathers: The Musical). It was later followed by "Chapter Ninety-One: The Return of the Pussycats" (season 5; on Josie and the Pussycats),  and "Chapter Ninety-Four: Next to Normal" (season 5; on Next to Normal) and "Chapter One Hundred and Twelve: American Psychos" (season 6; on American Psycho: The Musical).

Plot
Residents of Riverdale receive more videotapes of their houses. The school's Variety Show is coming up. Kevin wants to sing a song from the musical Hedwig and the Angry Inch, but Principal Honey believes the show to be too inappropriate. Archie signs up himself, Jughead, Betty, and Veronica to play as a band, "The Archies." Archie sees Hiram struggle lifting weights at the gym because of his disease, but he brushes off Archie's concerns.

Betty tutors Jughead to get him caught up in his classes so he can graduate at the same time, but he is not taking it seriously and instead investigates the videotapes while she thinks he is studying. Cheryl and Toni devise a plan to try to get Honey to accept the show by having everyone perform songs from Hedwig, but in response he threatens to ban students from Senior Prom. As a peaceful protest, all of the students show up to school dressed up as the character Hedwig. Honey cancels the Variety Show entirely instead of conceding.

Hiram collapses and Archie goes to meet Veronica. He reveals that he suspected Hiram wasn't going to the doctor enough, and Veronica gets angry with him for not telling her or making Hiram go himself. Meanwhile, Betty finds out that Jughead has been watching the videotapes rather than doing homework, and they have a fight. She shows up at rehearsal alone, and Archie and Betty end up kissing.

Later, Veronica and Jughead apologize for their blowups, as Betty and Archie look at each other nervously from their windows. Veronica hosts the Variety Show at Le Bonne Nuit and the Archies perform there.

Watching the tapes, Jughead sees a person tied up wearing a Jughead face mask who is brutally murdered by another person in a Betty face mask.

Cast and characters

Starring
 KJ Apa as Archie Andrews
 Lili Reinhart as Betty Cooper
 Camila Mendes as Veronica Lodge
 Cole Sprouse as Jughead Jones
 Marisol Nichols as Hermione Lodge
 Madelaine Petsch as Cheryl Blossom
 Mädchen Amick as Alice Cooper
 Mark Consuelos as Hiram Lodge
 Casey Cott as Kevin Keller / Hedwig
 Skeet Ulrich as F.P. Jones
 Charles Melton as Reggie Mantle
 Vanessa Morgan as Toni Topaz

Guest starring
 Kerr Smith as Principal Holden Honey
 Wyatt Nash as Charles Smith
 Drew Ray Tanner as Fangs Fogarty
 Jordan Connor as Sweet Pea
 Trinity Likins as Jellybean Jones
 Martin Cummins as Tom Keller
 Alvin Sanders as Pop Tate

Music

On April 15, 2020, WaterTower Music released the musical soundtrack from the "Wicked Little Town" episode performed by cast members.

All tracks from Hedwig and the Angry Inch were written by Stephen Trask.

Reception

Ratings
In the United States, the episode received a 0.2 percent share among adults between the ages of 18 and 49. It was watched by 0.54 million viewers.

References

External links

2020 American television episodes
Musical television episodes
Riverdale (2017 TV series) episodes
Fiction set in 2020
Hedwig and the Angry Inch

redpajama_set_name: RedPajamaWikipedia


How has current US foreign policy impacted globalization and trade in the world economy?
The United States has traditionally been a leader in the effort to increase globalization and free trade throughout the world. However, our current administration has taken a different path. This seminar will explore how changes in trade policies have affected different industries, the US economy at large and the role of the United States in foreign investment and development. Who benefits from our trade policies and who does not? Are tariffs the best way to protect American jobs? Do international trade agreements lead to an increase in trade and production or do they just limit future economic policy decisions? We will look at the economics behind the headlines.


Well i posted on here a while back that i was working on an online stat tacking app for BM well im almost done with the basic set up. Im looking for a test league to run the app its super simple to update one file upload and done... takes about 5 seconds. im looking for a mature league that has been around awhile.
i have league leaders sort able stats for pitching, hitting, and fielding also have individual categories with the top 5 or 10 players league wide, Im working on the player and team pages next they should be pumped rite out as most of the functions are already built.
If you are interested Email or send me a PM on here an Email will get my attention faster.


text:
Call or Text: (919) 263-2500
DWI Defense
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DWI Punishments
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Traffic Tickets & Violations
Speeding Charges
Common Traffic Violations
Revoked Driver's License
The vast majority of DWI charges in North Carolina involve alcohol, but an increasing number of these cases involve other impairing substances. People are being charged with DWI based on proof of recent use of marijuana before a traffic stop or for having been on prescription medication in their possession. The Government's burden is still the same, to prove beyond a reasonable doubt that someone was under the influence of an impairing substance at a time relevant to driving a vehicle. Because judges and juries are less informed about how certain substances can effect a person's mental and physical capabilities, this burden can sometimes be raised or lowered in drug DWI cases. Hiring a Raleigh or Wake County DWI lawyer familiar with these charges is the most important decision a person can make it their case.
Any substance a person ingests that can noticeably impair their normal mental or physical faculties can become the basis for a drug DWI charge if they operate a motor vehicle. This covers more than illegal substances; many people are charged even if they have taken prescription medication in the prescribed dosages. An officer is only concerned with whether the believe you are impaired and care very little about a person having legal permission to take any substance.
Drug DWI cases are generally more difficult for the State to prove. Unlike alcohol related charges where there is a legal limit as to how much can be in a person's system, there are no legal guidelines as to how much of each impairing substance is enough to impair a particular person. Nor are there roadside sobriety tests designed to measure impairment in non-alcohol cases. For the State to create a case against a person for drug DWI in North Carolina, they usually utilize two tools that are only used rarely in other cases: officers trained as drug recognition evaluators and blood tests.
Drug Recognition Evaluators
Many prosecutors and defense attorneys mislabel these officers as drug recognition experts. That could not be further from the truth. For an officer to qualify as a drug recognition evaluator (DRE), they will have completed roughly 90 hours of specialized training. For a person to be qualified as an expert in a specific field in North Carolina, they must at least be published, have had their work peer reviewed, and use methods that are generally used in the media or scientific community. Unfortunately, the government has created this title for the officers without ever having any of their studies or data published, peer reviewed, or vetted by any scientists or doctors. Never the less, this is the system used in North Carolina, and a person charged with a drug DWI should know what to expect during a possible drug recognition evaluation.
The drug recognition evaluation is a twelve-step test that is conducted after someone is arrested for suspicion of a drug related DWI. It is always conducted by a DRE and almost always occurs at the jail. The end result of this evaluation is the DRE forming an opinion as to whether the person's mental or physical faculties are noticeable impaired by a impairing substance.
The DRE Evaluation
A DRE will follow a twelve-step process to determine whether someone is impaired on a substance other than alcohol. All of these evaluations are supposed to happen in the same order and manner. Below is an overview of these steps:
1. An Initial Breath Test for Alcohol – Officers will want to be sure that alcohol is not the impairing substance. If a suspect registers a high enough number on a breath test, the DRE evaluation will not occur. Click here for more information on breath tests.
2. DRE Interview of the Arresting Officer – The evaluator will discuss the investigation with the arresting officer. In most of these cases, the suspect will admit to the officer what substances they have previously taken. The DRE will ask the arresting officer about anything relevant to a person's possible drug intake.
3. Preliminary Examination and First Pulse – The DRE conducts a preliminary examination of the suspect. The most important portion of this examination for the DRE is asking the suspect which drugs they've recently consumed. The answer to that question will basically dictate how the DRE will formulate his final opinion; it is incredibly rare for a DRE to form an opinion as to which substance someone has taken that is different from what they have admitted. The evaluator will also make notes about the suspect's demeanor, speech, coordination, and eyes. The suspect's pulse is also checked three times.
4. Eye Examination – The DRE examines the suspect for horizontal gaze nystagmus (HGN), vertical gaze nystagmus (VGN) and a lack of convergence. For a more detailed explanation of nystagmus, please click here. A lack of convergence occurs when the eyes are unable to meet toward the bridge of his nose when a stimulus is moved in. It is commonly believed by DRE's that marijuana causes this. Depressants, inhalants, and phencyclidine (PCP), the so-called DIP drugs, may cause HGN.
5. Divided Attention Sobriety Tests – The DRE then administers four psychophysical tests that are similar to roadside tests sometimes given in alcohol related DWIs: the Romberg Balance, the Walk and Turn, the One Leg Stand, and the Finger to Nose tests. The DRE will attempt to determine whether a subject's mental and physical faculties are impaired by administering these tests. Click here for more information on sobriety tests.
6. Vital Signs and Second Pulse – The DRE takes the subject's blood pressure, temperature and pulse a second time. Some drug categories may elevate the vital signs, and others may lower them. Vital signs can be a valuable indicator of which substances are possible in a person's system. Unfortunately, what the DRE training material has labeled as high or low for these categories in different from the standards in the medical community.
7. Dark Room Eye Examination – An estimation of the subject's pupil sizes are made under three different lighting conditions with a pupilometer (literally a paper card with different circular holes cut out of it) to determine whether the pupils are dilated, constricted, or normal. Some drugs categories increase or decrease pupil size. The DRE also checks for the pupil's reaction to light.
8. Muscle Tone Test – The DRE examines the suspect's muscle tone as certain categories of drugs may cause the muscles to become rigid or flaccid. Other categories may cause the muscles to become very loose and flaccid.
9. Check for Injection Sites and Third Pulse – The DRE will examine the suspect's body for injection sites, indicators recent use of certain types of drugs. This usually includes arms and hands but may include feet in certain scenarios. The DRE also takes the subject's pulse fora final time.
10. Subject's Statements and Other Observations – The DRE will asks the suspect a series of questions regarding their drug use after reading them their Miranda warnings. This will generally not warrant different answers from those given earlier about drug use.
11. Opinion of the Evaluator – The DRE will evaluate all the previous steps and, based on the totality of the circumstances, will form an opinion as to which substance or substances someone has consumed and whether the the person is noticeable impaired on those drugs. Almost all DRE's will utilize a chart called the DRE Matrix to help form their opinion. A copy of the Matrix is included below.
Attorney Saad was professional and compassionate. His office quickly responded to all of my requests and concerns. I am very grateful for his assistance in my legal matter and would highly recommend him to anyone.
Mr Saad got my case dismissed in one Court appearance he is the best and I'm glad I chose him to be my attorney he says true to his word.
Mr. Saad has gotten me out of seemingly impossible cases, would recommend to anybody wanting to get the best results on their case or ticket.
Mr. Saad is accommodating, reliable, upfront, thorough, and proactive. He made me aware of all options before we chose what was best for me considering the issues I had run into. He is professional, punctual, understanding, and a fantastic communicator. I highly recommend that you hire Steven Saad as your attorney- you will not be disappointed.
Steven Saad is the best young lawyer in town. He and his staff are very professional and he works magic. Sometimes friends of mine and myself don't understand how he does it but we are always very pleased with the results!
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