text:
\section{Introduction}

Elliott's classification of AF $C^*$-algebras ([2]) via the scaled,
ordered $\mathrm{K}_0$-group began what is now a widespread effort to
classify nuclear $C^*$-algebras via the Elliott
invariant.  In the case of a stably finite,
unital, simple $C^*$-algebra $A$ this invariant consists of the group $\mathrm{K}_{*} A = \mathrm{K}_0 A
\oplus \mathrm{K}_1 A$, the class of the unit of $A$ in $\mathrm{K}_{*} A$,
an order structure on $\mathrm{K}_{*} A$ (an element $[p] \oplus x$ is
positive if $[p]$ is positive in $\mathrm{K}_0 A$ and $x$ can be
represented as the $\mathrm{K}_1$-class of a unitary $u \in \mathrm{M}_l (A)$ such that $u u^*$ is
a sub-projection of $p$), the Choquet simplex of normalised traces $\mathrm{T} A$,
and the pairing between $\mathrm{K}_0 A$ and $\mathrm{T} A$ via evaluation.  In this
paper the invariant above will be denoted $\mathrm{Ell}(A)$.  Let $\mathrm{sr}(A)$
be the stable rank of $A$, as defined by Rieffel in [9].
$\mathrm{Ell}(-)$ has been particularly successful in classifying simple
$C^*$-algebras of stable rank one.  Until now, it was not
known whether this invariant would suffice for the classification of
stably finite $C^*$-algebras of stable rank greater than one.

An ordered group $(G,G^+)$ is said to be weakly unperforated if
$x \notin G^+$ and $nx \in G^+$ for some natural number $n$ implies that $nx = 0$.
We recall that the Elliott invariant of a simple nuclear unital
$C^*$-algebra $A$ is isomorphic to that of $A \otimes \mathcal{Z}$ whenever
$\mathrm{K}_{*} A$ is weakly unperforated ([4]).  If $A \cong A \otimes
\mathcal{Z}$, then we say that $A$ is $\mathcal{Z}$-stable.  Our main result is
the following:
\begin{thms}
For each natural number $n > 2$ there exists a simple, unital,
nuclear, separable,
infinite-dimensional, stably finite, non-$\mathcal{Z}$-stable
$C^*$-algebra $B_n$ such that $\mathrm{K}_* B_n$ is weakly
unperforated and $\mathrm{sr}(B_n) \in \{n+1,n+2\}$.  In particular,
\begin{displaymath}
\mathrm{Ell}(B_n) \simeq \mathrm{Ell}(B_n \otimes \mathcal{Z}).
\end{displaymath}
\end{thms}
Thus, $B_n$ and $B_n \otimes \mathcal{Z}$ constitute a counterexample
to the Elliott conjecture for the class of simple, nuclear,
infinite-dimensional, stably finite $C^*$-algebras.
We note that the existence of $B_n$ answers Question 1.5 of
[4] negatively;  the weak unperforation of the $\mathrm{K}_*$-group
does not imply that a simple, unital, nuclear, separable, infinite-dimensional
$C^*$-algebra absorbs $\mathcal{Z}$.

The title of this paper derives from the fact that the algebra $B_n$ of Theorem 1.1
has $\mathrm{sr}(B_n) \in \{n+1,n+2\}$ while, as we shall see,
$\mathrm{sr}(B_n \otimes \mathcal{Z}) \leq 2$.
It is possible (but purely speculative) that finer
invariants such as $\mathrm{K}$-theory with coefficients, the
semigroup of Murray-von Neumann equivalence classes
of projections, or higher algebraic K-theory will recover
stable rank, and so the independence of the title is only with respect to
the notion of $\mathrm{K}$-theory captured by $\mathrm{Ell}( - )$.

We conclude this section with an outline of the sequel.
Section 2 lists several theorems from [3], which are
applied in section 3 to construct the algebra $B_n$ of Theorem 1.1.
The general ideas of this latter section are also found in [3].
In section 4 $B_n$ is shown to have the properties claimed
in Theorem 1.1.

\emph{Acknowledgements.} The author would like to thank George Elliott
and Ping Wong Ng for their many helpful comments on earlier drafts of
this paper.  This research was funded by both the Israel Halperin
Graduate Award at the University of Toronto and by a Natural Sciences and
Engineering Research Council of Canada Postdoctoral Fellowship.

\section{Background and essential results}

We begin by reviewing the definition of the generalised mapping torus, due to
Elliott.  Let $C$, $D$ be $C^{*}$-algebras and let $\phi_{0}$, $\phi_{1}$ be
$*$-homomorphisms from $C$ to $D$.  Then the generalised mapping torus of
$C$ and $D$ with respect to $\phi_{0}$ and $\phi_{1}$ is
\begin{displaymath}
A:=\{(c,d)|d \in C([0,1];D), \ c \in C, \ d(0)=\phi_{0}(c), \ d(1)=\phi_{1}(c)\}
\end{displaymath}
We will denote $A$ by $A(C,D,\phi_{0}, \phi_{1})$ where appropriate for clarity.
We now list (without proof) some theorems of [3] which will be used in the sequel.

\begin{thms}[{Elliott and Villadsen ([3]), Theorem 2}]
The index map $b_{*} : K_{*}C \longrightarrow K_{1-*}SD = K_{*}D$ in the six term periodic
sequence for the extension
\begin{displaymath}
0 \longrightarrow SD \longrightarrow A \longrightarrow C \longrightarrow 0
\end{displaymath}
is the difference
\begin{displaymath}
\mathrm{K}_{*} \phi_{1} - \mathrm{K}_{*} \phi_{0} : \mathrm{K}_{*} C \longrightarrow \mathrm{K}_{*} D.
\end{displaymath}
Thus, the six-term exact sequence may be written as the short exact sequence
\begin{displaymath}
0 \longrightarrow \mathrm{Coker} b_{1-*} \longrightarrow \mathrm{K}_{*} A \longrightarrow \mathrm{Ker}
b_{*} \longrightarrow 0.
\end{displaymath}
In particular, if $b_{1-*}$ is surjective, then $\mathrm{K}_{*} A$ is isomorphic to its image,
$\mathrm{Ker} b_{*}$, in $\mathrm{K}_{*} C$.

Suppose that cancellation holds for $D$ --- i.e., that cancellation holds in the semigroup of
Murray-von Neumann equivalence classes of projections of projections in $D$ and
in matrix algebras over $D$ (equivalently, in $D \otimes \mathcal{K}$).  It follows that if $b_{1}$ is
surjective, so that $\mathrm{K}_{0} A \subseteq \mathrm{K}_{0} C$, then
\begin{displaymath}
{(\mathrm{K}_{0} A)}^{+} = {(\mathrm{K}_{0} C)}^{+} \cap \mathrm{K}_{0} A.
\end{displaymath}

The preceding conclusion also holds if cancellation is only known to hold for each pair of
projections in $D \otimes \mathcal{K}$ obtained as the images under the maps $\phi_{0}$ and
$\phi_{1}$ of a single projection in $C \otimes \mathcal{K}$.  (In other words, if two such such
projections in $D \otimes \mathcal{K}$ have the same $\mathrm{K}_{0}$ class then they should be
equivalent, assuming as before that $b_{1}$ is surjective.)
\end{thms}

\begin{thms}[{Elliott and Villadsen ([3]), Theorem 3}]

Let $A_{1}$ and $A_{2}$ be building block algebras as described above,

\begin{displaymath} A_{i}=A(C,D,\phi_{0}^{i},\phi_{1}^{i}),\ \ i=1,2.
\end{displaymath}

Let there be given four maps between the fibres,
\begin{displaymath}
\begin{array}{rrr}
\gamma : & C_{1} \rightarrow C_{2}, &  \\
\delta, \delta ' : & D_{1} \rightarrow D_{2}, & \ \ and, \\
\epsilon : & C_{1} \rightarrow D_{2}, &
\end{array}
\end{displaymath}
such that $\delta$, $\delta '$ and $\epsilon$ have mutually orthogonal images, and
\begin{displaymath} \delta \phi_{0}^{1} + \delta ' \phi_{1}^{1} + \epsilon = \phi_{0}^{2} \gamma,
\end{displaymath}
\begin{displaymath} \delta \phi_{1}^{1} + \delta ' \phi_{0}^{1} + \epsilon = \phi_{1}^{2} \gamma.
\end{displaymath}

Then there exists a unique map
\begin{displaymath} \theta : A_{1} \rightarrow A_{2},
\end{displaymath}
respecting the canonical ideals, giving rise to the map $\gamma : C_{1} \rightarrow
C_{2}$ between the quotients (or fibres at infinity), and such that for any $0 < s< 1$,
if $e_{s}$ denotes evaluation at $s$, and $e_{\infty}$ the evaluation at infinity,
\begin{displaymath} e_{s} \theta = \delta e_{s} + \delta ' e_{1-s} + \epsilon e_{\infty}.
\end{displaymath}

\end{thms}

\begin{thms}[{Elliott and Villadsen ([3]), Theorem 4}]

Let $A_{1}$ and $A_{2}$ be building block algebras as in Theorem 2.1.  Let $\theta : A_{1} \rightarrow
A_{2}$ be a homomorphism constructed as in Theorem 2.2, from maps $\gamma : C_{1} \rightarrow C_{2}$,
$\delta,\delta ' : D_{1} \rightarrow D_{2}$, and $\epsilon : C_{1} \rightarrow D_{2}$.

Let there be given a map $\beta : D_{1} \rightarrow C_{2}$ such that the composed map
$\beta \phi_{1}^{1}$ is a direct summand of the map $\gamma$, and such that the
composed maps $\phi_{0}^{2} \beta$ and $\phi_{1}^{2} \beta$ are direct summands of
the maps $\delta '$ and $\delta$, respectively.  Suppose that the decomposition of
$\gamma$ as the orthogonal sum of $\beta \phi_{1}^{1}$ and another map is such that
the image of the second map is orthogonal to the image of $\beta$.  (Note that this
requirement is automatically satisfied if $C_{1}$, $D_{1}$, and the map $\beta \phi_{1}^{1}$
are unital.)

It follows that, for any $0 < t < \frac{1}{2}$, the map $\theta : A_{1} \rightarrow A_{2}$
is homotopic to a map $\theta_{t} : A_{1} \rightarrow A_{2}$ differing from it only as follows:
the map $e_{\infty} \theta_{t}$ has the direct summand $\beta e_{t}$ instead of one of the
direct summands $\beta \phi_{0}^{1} e_{\infty}$ and $\beta \phi_{1}^{1} e_{\infty}$ of
$e_{\infty} \theta$, and for each $0 < s < 1$ the map $e_{s} \theta_{t}$ has either the
direct summand $\phi_{0}^{2} \beta e_{t}$ instead of the direct summand $\phi_{0}^{2} \beta e_{s}$
of $e_{s} \theta$, or the direct summand $\phi_{1}^{2} \beta e_{t}$ instead of the direct summand
$\phi_{1}^{2} \beta e_{s}$ of $e_{s} \theta$, or both.

Furthermore, let $\alpha : D_{1} \rightarrow C_{2}$ be any map homotopic to $\beta$ within the
hereditary sub-$C^{*}$-algebra of $C_{2}$ generated by the image of $\beta$.  Then the map
$\theta_{t}$ is homotopic to a map $\theta_{t}^{'} : A_{1} \rightarrow A_{2}$ differing from $\theta_{t}$
only in the direct summands mentioned, and such that $e_{\infty} \theta_{t}^{'}$ has the direct summand
$\alpha e_{t}$ instead of $\beta e_{t}$, and for each $0 < s < 1$, $e_{s} \theta_{t}^{'}$ has
either $\phi_{0}^{2} \alpha e_{t}$ instead of $\phi_{0}^{2} \beta e_{t}$, or $\phi_{1}^{2} \alpha e_{t}$
instead of $\phi_{1}^{2} \beta e_{t}$.

\end{thms}

\begin{thms}[{Elliott and Villadsen ([3]), Theorem 5}]

Let
\begin{displaymath} A_{1} \stackrel{\theta_{1}}{\rightarrow} A_{2} \stackrel{\theta_{2}}{\rightarrow} \cdots
\end{displaymath}
be a sequence of separable building block $C^{*}$-algebras,
\begin{displaymath} A_{i} = A(C_{i},D_{i},\phi_{0}^{i}, \phi_{1}^{i}), \ i=1,2, \ldots
\end{displaymath}
with each map $\theta_{i} : A_{i} \rightarrow A_{i+1}$ obtained by the construction of Theorem 2.2 (and thus
respecting the canonical ideals).  For each $i=1,2, \ldots$ let $\beta_{i} : D_{i} \rightarrow C_{i+1}$ be a
map verifying the hypotheses of Theorem 2.3.

Suppose that for every $i=1,2, \ldots$, the intersection of the kernels of the boundary maps $\phi_{0}^{i}$ and
$\phi_{1}^{i}$ from $C_{i}$ to $D_{i}$ is zero.

Suppose that, for each i, the image of each of $\phi_{0}^{i+1}$ and $\phi_{1}^{i+1}$ generates $D_{i+1}$ as a
closed two-sided ideal, and that this is in fact true for the restriction of $\phi_{0}^{i+1}$ and
$\phi_{1}^{i+1}$ to the smallest direct summand of $C_{i+1}$ containing the image of $\beta_{i}$.  Suppose
that the closed two-sided ideal of $C_{i+1}$ generated by the image of $\beta_{i}$ is a direct summand.

Suppose that, for each i, the maps $\delta_{i}^{'} - \phi_{0}^{i} \beta_{i}$ and
$\delta_{i} - \phi_{1}^{i} \beta_{i}$ from $D_{i}$ to $D_{i+1}$ are injective.

Suppose that, for each i, the map $\gamma_{i} - \beta_{i} \phi_{1}^{i}$ takes each non-zero direct summand
of $C_{i}$ into a subalgebra of $C_{i+1}$ not contained in any proper closed two-sided ideal.

Suppose that, for each i, the map $\beta_{i} : D_{i} \rightarrow C_{i+1}$ can be deformed---inside the
hereditary sub-$C^{*}$-algebra generated by its image---to a map $\alpha_{i} : D_{i} \rightarrow C_{i+1}$
with the following property:  There is a direct summand of $\alpha_{i}$, say ${\bar{\alpha}}_{i}$, such
that ${\bar{\alpha}}_{i}$ is non-zero on an arbitrary given element $x_{i}$ of $D_{i}$, and has image a
simple sub-$C^{*}$-algebra of $C_{i+1}$, the closed two-sided ideal generated by which contains the image
of $\beta_{i}$.

Choose a dense sequence $(t_{n})$ in the open interval $(0,\frac{1}{2})$, such that $t_{2n} = t_{2n-1},
n = 1,2,\ldots$.

Choose a sequence of elements $x_{3} \in D_{3}, \ x_{5} \in D_{5}, \ x_{7} \in D_{7}, \ldots$
(necessarily non-zero) with the following property:  For some countable basis for the topology of
the spectrum of each of $D_{1}, D_{2},\ldots$, and for some choice of non-zero element of the closed
two-sided ideal associated to each of these (non-empty) open sets, under successive application of the
maps $\delta_{i} - \phi_{1}^{i+1} \beta_{i}$ each one of these elements is taken into $x_{j}$ for all j
in some set $S \subseteq \{3,5,7,\ldots\}$ such that $\{t_{j}, j \in S \}$ is dense in $(0,\frac{1}{2})$.
Choose $\alpha_{j}$ as above such that ${\bar{\alpha}}_{j}(x_{j}) \neq 0$ for some direct summand
${\bar{\alpha}}_{j}$ of $\alpha_{j}$ for each $j \in \{3,5,7,\ldots\}$.  For each $j \in \{4,6,8,\ldots\}$
choose $\alpha_{j}$ with respect to the non-zero element $(\delta_{j-1}^{'} - \phi_{0}^{j} \beta_{j-1})
(x_{j-1})$ of $D_{j}$.  (If $j=1$ or $2$, choose $\alpha_{j} = \beta_{j}$.)

It follows that, if $\theta_{i}^{'}$ denotes the deformation of $\theta_{i}$ constructed in Theorem 4,
with respect to the point $t_{i} \in (0,\frac{1}{2})$ and the maps $\alpha_{i}$ and $\beta_{i}$ (and
a fixed homotopy of $\beta_{i}$ to $\alpha_{i}$), then the inductive limit of the sequence
\begin{displaymath} A_{1} \stackrel{\theta_{1}^{'}}{\rightarrow} A_{2} \stackrel{\theta_{2}^{'}}{\rightarrow}
\cdots
\end{displaymath}
is simple.

\end{thms}

\section{The construction of $B_n$}

We now specify $C^{*}$-algebras $A_{i} = A_{i}(C_{i}, D_{i}, \phi_{i}^{0}, \phi_{i}^{1})$ as in
Theorem 2.1, and maps $\delta_{i}$, $\delta_{i}^{'}$, $\gamma_{i}$, and $\beta_{i}$ satisfying
the hypotheses of Theorems 2.2, 2.3, and 2.4 in order to construct an inductive sequence
\begin{displaymath}
A_{1} \stackrel{\theta_{1}^{'}}{\longrightarrow} A_{2} \stackrel{\theta_{1}^{'}}{\longrightarrow} \cdots
\end{displaymath}
whose limit will be the $C^*$-algebra $B_n$ of Theorem 1.1.

Let $\mathrm{D}$ denote the closed unit disc in the complex numbers. Put
\begin{displaymath}
X_{i} = \mathrm{D}^{n} \times \mathrm{CP}^{n\sigma(1)} \times
\mathrm{CP}^{n\sigma(2)} \times \cdots \times \mathrm{CP}^{n\sigma(i)}
\end{displaymath}
--- the $\sigma(i)$ are natural numbers to be specified --- so that
\begin{displaymath}
X_{i+1} = X_{i} \times \mathrm{CP}^{n\sigma(i+1)},
\end{displaymath}
and let
\begin{displaymath}
\begin{array}{cc}
\pi_{i+1}^{1}: X_{i+1} \longrightarrow X_{i}, \  & \ \pi_{i+1}^{2}:
X_{i+1} \longrightarrow \mathrm{CP}^{n\sigma(i+1)}
\end{array}
\end{displaymath}
be the co-ordinate projections.

We will take $C_{i} = p_{i}(\mathrm{C}(X_{i}) \otimes \mathcal{K})p_{i}$, where $p_{i}$ is a projection in
$\mathrm{C}(X_{i}) \otimes \mathcal{K}$ to be specified.  Let $D_{i} =
C_{i} \otimes  \mathrm{M}_{k_{i} \mathrm{dim}(p_{i})}$,
where $k_{i}$ is a positive integer to be specified.  Define maps
\begin{displaymath}
\mu_{i}, \nu_{i}:C_{i} \longrightarrow C_{i} \otimes \mathrm{M}_{\mathrm{dim}(p_{i})}
\end{displaymath}
as follows:
\begin{displaymath}
\mu_{i}(a) = p_{i} \otimes a(x_{i}),
\end{displaymath}
\begin{displaymath}
\nu_{i}(a) = a \otimes 1_{\mathrm{dim}(p_{i})}.
\end{displaymath}
For $t \in \{0,1\}$, we will take $\phi_{i}^{t}$ to be the direct sum of $l_{i}^{t}$ copies of
$\mu_{i}$ and $k_{i} - l_{i}^{t}$ copies of $\nu_{i}$, where the $l_{i}^{t}$ are non-negative integers
to be specified.  All that we mention now is that we should have $l_{i}^{1} - l_{i}^{0} \neq 0$.
We need only specify the $\phi_{i}^{t}$ up to unitary equivalence, a fact we shall
exploit below.

By Theorem 2.1 we have that for any $e \in \mathrm{K}_{0}(C_{i})$,
\begin{displaymath}
\begin{array}{rl}
b_{0}(e) & = (l_{i}^{1} - l_{i}^{0})(\mathrm{K}_{0}(\mu_{i}) - \mathrm{K}_{0}(\nu_{i})) \\
         & = (l_{i}^{1} - l_{i}^{0})(\mathrm{dim}(e) \cdot \mathrm{K}_{0}(p_{i}) - \mathrm{dim}(p_{i}) \cdot e).
\end{array}
\end{displaymath}
Since $l_{i}^{1} - l_{i}^{0} \neq 0$ and since $\mathrm{K}_{0}C_{i}$ is a finitely generated free abelian group,
we have that $\mathrm{Ker}b_{0}$ is the largest subgroup
of $\mathrm{K}_{0}C_{i}$ containing $\mathrm{K}_{0}(p_{i})$ and
isomorphic to the integers.  In the sequel we will choose $p_{i}$
so that $\mathrm{K}_{0}(p_{i})$ in fact generates said subgroup.
Since $\mathrm{K}_{1}C_{i} = 0$ we have, by Theorem
2.1, that $\mathrm{K}_{0}A_{i}$ is isomorphic as an ordered group to
its image, $\mathrm{Ker}b_{0}$, in $\mathrm{K}_{0}C_{i}$,
considered as a sub ordered group.  The latter (with the choice of $p_i$
below) is isomorphic to the integers with the unique unperforated
order structure, and the image of $[1_{A_i}]$ is $[p_i]$.

Let $p_{1}$ be a projection corresponding to the vector bundle
\begin{displaymath}
\theta_{1} \times \xi_{n\sigma(1)},
\end{displaymath}
over $X_{1}$, where $\theta_{1}$ denotes the trivial line bundle of
dimension one over $\mathrm{D}$, $\xi_k$
denotes the universal line bundle over $\mathrm{CP}^{k}$ for every
natural number $k$, and $\sigma(1) = 1$.  We now
specify, inductively, the maps $\gamma_{i}:C_{i} \longrightarrow C_{i+1}$.  Consider first the map
\begin{displaymath}
\psi_{i} := \mathrm{id} \otimes 1
\end{displaymath}
from $\mathrm{C}(X_{i})$ to $\mathrm{C}(X_{i+1}) = \mathrm{C}(X_{i}
\times \mathrm{CP}^{n\sigma(i+1)}) = \mathrm{C}(X_{i}) \otimes \mathrm{C}(\mathrm{{}CP}^{n\sigma(i+1)})$,
where $1$ denotes the unit of $\mathrm{C}(\mathrm{CP}^{n\sigma(i+1)})$ and $\mathrm{id}$ denotes the identity map from
$\mathrm{C}(X_{i})$ to itself.

Consider also the map
\begin{displaymath}
\beta_{i}^{'} := {\pi_{i+1}^{2 *}}(\xi_{n\sigma(i+1)}) \cdot e_{x_{i}}
\end{displaymath}
from $\mathrm{C}(X_{i})$ to $\mathrm{C}(X_{i+1}) \otimes \mathcal{K}$, where
$e_{x_{i}}$ denotes evaluation at $x_{i}$.  All that we shall require of the
$x_{i}$ at this stage is that $\pi_{i+1}^{1}(x_{i+1}) = x_{i}$.

Now, inductively, let us take $\gamma_{i}$ to be the map from $C_{i}$
to $\mathrm{C}(X_{i+1}) \otimes \mathrm{M}_{2}(\mathcal{K})$ consisting
of the direct sum of the following two maps:  first, the restriction
to $C_{i} \subseteq \mathrm{C}(X_{i}) \otimes \mathcal{K}$ of the
tensor product of $\psi_{i}$ with the identity map from $\mathcal{K}$
to $\mathcal{K}$, and second, the map from $C_{i}$ to
$\mathrm{C}(X_{i+1}) \otimes \mathrm{M}_{q_{i}}(\mathcal{K})$
consisting of the composition of the map $\phi_{i}^{1}$ from $C_{i}$ to
$D_{i}$ with the direct sum of $q_{i}$ copies of the tensor product of the map $\beta_{i}^{'}$ with the
identity map from $\mathcal{K}$ to $\mathcal{K}$ (restricted to $D_{i}
\subseteq \mathrm{C}(X_{i}) \otimes \mathcal{K}$), where $q_{i}$ is to
be specified.  The induction consists in first considering the case
$i=1$ (as $p_{1}$ has already been chosen),
then setting $p_{2} = \gamma_{1}(p_{1})$, so that $C_{2}$ is
specified as the cut-down of $\mathrm{C}(X_{2}) \otimes \mathcal{K}$
by $p_{2}$, and continuing in this way.

With the maps $\gamma_{i}$ defined as above, we have that $p_{i}$
is a projection in $\mathrm{C}(X_{i}) \otimes \mathcal{K}$
corresponding to the vector bundle
\begin{displaymath}
\theta_{1} \times \xi_{n} \times \sigma(2) \xi_{n\sigma(2)} \times \cdots \times \sigma(i) \xi_{n\sigma(i)},
\end{displaymath}
where
\begin{displaymath}
\sigma(i) = \prod_{l=1}^{i-1} (\mathrm{mult}(\gamma_{l}) - 1).
\end{displaymath}
Notice that by the K{\"{u}}nneth formula (in Chapter 5 of [10], for instance) the classes
$[\theta_1]$, $[\xi_{n \sigma(1)}]$, \ldots, $[\xi_{n \sigma(i)}]$ are independent in
$\mathrm{K}^{0}(X_i)$ (we are abusing notation slightly here, using $[\xi_k]$
to represent the class of the induced bundle $\pi^{*}(\xi_k)$, where $\pi$ is
projection from $X_i$ onto $\mathrm{CP}^k$).  Suppose that $[p_{i}] = ky$ for some $k \in
\mathrm{Z}$, $y \in \mathrm{K}^0(X_i)$.  It follows from independence
that we have $[\xi_n] = ky^{'}$, $y^{'} \in
\mathrm{K}^0(\mathrm{CP}^n)$, whence $k = \pm 1$.  We conclude that
$[p_i]$ itself generates the subgroup of rational multiples of
$[p_i]$ in $\mathrm{K}^{0}X_{i}$, as desired.  Thus $\gamma_{i}$
induces an isomorphism of ordered groups from
$\mathrm{Ker}b_{0}$ at the $i^{\mathrm{th}}$ stage
to $\mathrm{Ker}b_{0}$ at the ${(i+1)}^{\mathrm{th}}$ stage.

Note that $\gamma_{i} - \beta_{i} \phi_{i}^{1}$ is non-zero, and so takes $C_{i}$ into a subalgebra of $C_{i+1}$
not contained in any proper closed two-sided ideal.

Next, we construct the maps $\delta_{i}, \delta_{i}^{'}:D_{i} \rightarrow D_{i+1}$, with orthogonal images,
such that
\begin{displaymath}
\begin{array}{c}
\delta_{i} \phi_{i}^{0} + \delta_{i}^{'} \phi_{i}^{1} = \phi_{i+1}^{0} \gamma_{i}, \\
\delta_{i} \phi_{i}^{1} + \delta_{i}^{'} \phi_{i}^{0} = \phi_{i+1}^{1} \gamma_{i},
\end{array}
\end{displaymath}
and $\phi_{i+1}^{0} \beta_{i}$ and $\phi_{i+1}^{1} \beta_{i}$ are direct summands of $\delta_{i}^{'}$ and
$\delta_{i}$, respectively.  To do this we shall have to modify $\phi_{i+1}^{0}$ and $\phi_{i+1}^{1}$ by
inner automorphisms.  This will not affect the $\mathrm{K}$-theory of $A$.

Note that, up to unitary equivalence, we have
\begin{displaymath}
e_{x_{i+1}} \gamma_{i} = \mathrm{mult}(\gamma_{i}) e_{x_{i}},
\end{displaymath}
where $\mathrm{mult}(\gamma_{i})$ denotes the factor by which
$\gamma_{i}$ multiplies dimension.  It follows that up to unitary equivalence
\begin{displaymath}
\begin{array}{rl}
\mu_{i+1} \gamma_{i} &  = p_{i+1} \otimes e_{x_{i+1}} \gamma_{i} \\
                     &  = \gamma_{i}(p_{i}) \otimes \mathrm{mult}(\gamma_{i}) e_{x_{i}} \\
                     &  = \mathrm{mult}(\gamma_{i}) \gamma_{i}(p_{i} \otimes e_{x_{i}}) \\
                     &  = \mathrm{mult}(\gamma_{i}) \gamma_{i} \mu_{i},
\end{array}
\end{displaymath}
and
\begin{displaymath}
\begin{array}{rl}
\nu_{i+1} \gamma_{i} &  = \gamma_{i} \otimes 1_{\mathrm{dim}(p_{i+1})} \\
                     &  = \mathrm{mult}(\gamma_{i}) \gamma_{i} \otimes 1_{\mathrm{dim}(p_{i})} \\
                     &  = \mathrm{mult}(\gamma_{i}) \gamma_{i} \nu_{i}.
\end{array}
\end{displaymath}

Take $\delta_{i}$ and $\delta_{i}^{'}$ to be $r_{i}$ and $s_{i}$ copies of $\gamma_{i}$, where $r_{i}$ and
$s_{i}$ are integers to be specified.  The conditions
\begin{displaymath}
\delta_{i} \phi_{i}^{0} + \delta_{i}^{'} \phi_{i}^{1} = \phi_{i+1}^{0} \gamma_{i}
\end{displaymath}
and
\begin{displaymath}
\delta_{i} \phi_{i}^{1} + \delta_{i}^{'} \phi_{i}^{0} = \phi_{i+1}^{1} \gamma_{i},
\end{displaymath}
understood up to unitary equivalence imply that
\begin{displaymath}
r_{i} \gamma_{i}(l_{i}^{t}\mu_{i} - (k_{i} - l_{i}^{t})\nu_{i}) + s_{i} \gamma_{i}(l_{i}^{1-t}\mu_{i} +
(k_{i} - l_{i}^{1-t})\nu_{i}) = (l_{i+1}^{t}\mu_{i+1} + (k_{i+1} - l_{i+1}^{t})\nu_{i+1}) \gamma_{i},
\end{displaymath}
again, up to unitary equivalence.  As $\mathrm{K}_{0}(\mu_{i})$
and $\mathrm{K}_{0}(\nu_{i})$ are independent, the above equation is
equivalent to the two equations
\begin{displaymath}
\begin{array}{rl}
r_{i} l_{i}^{t} + s_{i} l_{i}^{1-t} & = \mathrm{mult}(\gamma_{i}) l_{i+1}^{t}, \\
(r_{i} + s_{i}) k_{i} & = \mathrm{mult}(\gamma_{i}) k_{i+1}.
\end{array}
\end{displaymath}
Choose $r_{i} = 2 \mathrm{mult}(\gamma_{i})$ and $s_{i} = \mathrm{mult}(\gamma_{i})$, so that
\begin{displaymath}
k_{i+1} = 3 k_{i},
\end{displaymath}
and
\begin{displaymath}
l_{i+1}^{t} = 2 l_{i}^{t} + l_{i}^{1-t}.
\end{displaymath}

Take $k_{1} = 1$, $l_{1}^{1} = 1$, and $l_{1}^{0} = 0$.  Then $l_{i}^{1} - l_{i}^{0} \neq 0$ for
all $i$, as required.

Next, let us show that, up to unitary equivalence preserving the equations $\delta_{i} \phi_{i}^{t} +
\delta_{i}^{'} \phi_{i}^{1-t} = \phi_{i+1}^{t} \gamma_{i}$, $\phi_{i+1}^{1} \beta_{i}$ is a direct summand
of $\delta_{i} = 2 \mathrm{mult}(\gamma_{i}) \gamma_{i}$ and $\phi_{i+1}^{0} \beta_{i}$ is a direct summand of
$\delta_{i}^{'} = \mathrm{mult}(\gamma_{i}) \gamma_{i}$.

Note that $\phi_{i+1}^{t} \beta_{i}$ is a direct sum of $l_{i+1}^{t}$ copies of $p_{i+1} \otimes \beta_{i}$
and \mbox{${(k_{i+1} - l_{i+1}^{t})}$} copies of $\beta_{i}$, whereas $\delta_{i}$ and $\delta_{i}^{'}$ contain,
respectively, $q_{i} \mathrm{mult}(\gamma_{i})$ and $2 q_{i} \mathrm{mult}(\gamma_{i})$ copies of $\beta_{i}$.
Note also, that by Theorem 8.1.2 of [5], a trivial projection of dimension $\mathrm{dim}(p_{i+1}) + \frac{1}{2}
\mathrm{dim}X_{i+1}$ in $\mathrm{C}(X_{i+1}) \otimes \mathcal{K}$ contains a copy of
$p_{i+1}$.  Therefore, $2 \mathrm{dim}(p_{i+1}) + 2
\mathrm{dim}X_{i+1}$ copies of $\beta_{i}$ contain a copy of $p_{i+1} \otimes \beta_{i}$
(since $2 \mathrm{dim}(p_{i+1}) + 2\mathrm{dim}X_{i+1}$ copies of $\xi_{i+1}$ contain
a trivial projection of dimension $\mathrm{dim}(p_{i+1}) + \frac{1}{2}
\mathrm{dim}X_{i+1}$).  It follows that
$k_{i+1} (2 \mathrm{dim}(p_{i+1}) + 2 \mathrm{dim} X_{i+1})$ copies of $\beta_{i}$ contain a copy of
$\phi_{i+1}^{t} \beta_{i}$ when $t$ is equal to either $0$ or $1$.  By a copy of a given map from $D_{i}$ to
$D_{i+1}$ we mean another map obtained from it by conjugating by a partial isometry in $D_{i+1}$ with initial
projection the image of the unit.

Note that
\begin{displaymath}
k_{i+1}(2 \mathrm{dim} (p_{i+1}) + 2 \mathrm{dim} X_{i+1}) = 6 k_{i} \mathrm{mult}(\gamma_{i})(\mathrm{dim}(p_{i}) +
                                                                \mathrm{dim} X_{i}),
\end{displaymath}
and that $k_{i}$, $\mathrm{dim}(p_{i})$, and $\mathrm{dim}X_{i}$ have already been specified, and do not depend on $q_{i}$.
It follows that, with
\begin{displaymath}
q_{i} \geq 6 k_{i} (\mathrm{dim}(p_{i}) + \mathrm{dim} X_{i}),
\end{displaymath}
$q_i \mathrm{mult}(\gamma_{i})$ copies of $\beta_{i}$ contain a copy of $\phi_{i+1}^{t} \beta_{i}$, ($t \in \{0,1\}$).
In particular $\delta_{i}^{'}$ and $\delta_{i}$ contain copies of
$\phi_{i+1}^{0} \beta_{i}$ and $\phi_{i+1}^{1} \beta_{i}$, respectively.

With $q_{i}$ as above, let us show that for each $t=0,1$ there exists a unitary $u_{t} \in D_{i+1}$ such that
\begin{displaymath}
(\mathrm{Ad} u_{t}) \phi_{i+1}^{t} \gamma_{i} = \phi_{i+1}^{t} \gamma_{i},
\end{displaymath}
with $(\mathrm{Ad} u_{0}) \phi_{i+1}^{0} \beta_{i}$ a direct summand of $\delta_{i}^{'}$ and $(\mathrm{Ad} u_{1})
\phi_{i+1}^{1} \beta_{i}$ a direct summand of $\delta_{i}$.  In
other words, for each $t=0,1$, we must show that the
partial isometry constructed in the preceding paragraph, producing
a copy of $\phi_{i+1}^{t} \beta_{i}$ inside $\delta_{i}$
or $\delta_{i}^{'}$ may be chosen in such a way that it extends
to a unitary element of $D_{i+1}$ --- which in addition
commutes with the image of $\phi_{i+1}^{t} \gamma_{i}$.

        Let us consider the case $t=0$; the case $t=1$ is
similar.  First note that the unit of the image of
$\phi_{i+1}^{0} \beta_{i}$ --- the initial projection of
the partial isometry transforming $\phi_{i+1}^{0} \beta_{i}$
into a direct summand of $\delta_{i}^{'}$ --- lies in the
commutant of the image of $\phi_{i+1}^{0} \gamma_{i}$.
Indeed, this projection is the image by $\phi_{i+1}^{0} \beta_{i}$
of the unit of $D_{i}$, which, by construction,
is the image of the unit of $C_{i}$ by $\phi_{i}^{1}$.  The
property that $\beta_{i} \phi_{i}^{1}$ is a direct summand
of $\gamma_{i}$ implies that the image by $\beta_{i} \phi_{i}^{1}$
of the unit of $C_{i}$ commutes with the image of
$\gamma_{i}$.  The unit of the image of $\phi_{i+1}^{0} \beta_{i}$
therefore commutes with the image of
$\phi_{i+1}^{0} \gamma_{i}$, as desired.

        The final projection of the above partial isometry also
commutes with the image of $\phi_{i+1}^{0} \gamma_{i}$.
Indeed, it is the unit of the image of a direct summand of
$\delta_{i}^{'}$, and since $D_{i}$ is unital it is the image
of the unit of $D_{i}$ by this direct summand; since $C_{i}$ is
unital and $\phi_{i}^{1}: C_{i} \rightarrow D_{i}$ is
unital, the projection in question is the image of the unit of $C_{i}$
by a direct summand of $\delta_{i}^{'} \phi_{i}^{1}$.
But $\delta_{i}^{'} \phi_{i}^{1}$ is itself a direct summand
of $\phi_{i+1}^{0} \gamma_{i}$, and so the projection in
question is the image of the unit of $C_{i}$ by a direct
summand of $\phi_{i+1}^{0} \gamma_{i}$, and in particular
commutes with the image of $\phi_{i+1}^{0} \gamma_{i}$.

Note that both direct summands of $\phi_{i+1}^{0} \gamma_{i}$
under consideration ($\phi_{i+1}^{0} \beta_{i}
\phi_{i}^{1}$ and a copy of it) factor through the evaluation
of $C_{i}$ at the point $x_{i}$, and so are contained in the
largest such direct summand of $\phi_{i+1}^{0} \gamma_{i}$;
this largest direct summand, say $\pi_{i}$, is seen to exist
by inspection of the construction of $\phi_{i+1}^{0} \gamma_{i}$.
Since both projections under consideration (the images
of $1 \in C_{i}$ by the two copies of $\phi_{i+1}^{0} \beta_{i}
\phi_{i}^{1}$) are less than $\pi_{i}(1)$, to show that they
are unitarily equivalent in the commutant of the image of
$\phi_{i+1}^{0} \gamma_{i}$ (in $D_{i+1}$) it is sufficient to
show that they are unitarily equivalent in the commutant of the
image of $\pi_{i}$ in $\pi_{i}(1) D_{i+1} \pi_{i}(1)$.
Note that this image is isomorphic to $\mathrm{M}_{\mathrm{dim}
(p_{i})}(\mathrm{C})$.  By construction, the two projections in question are
Murray-von Neumann equivalent in $D_{i+1}$ and hence in $\pi_{i}(1)
D_{i+1} \pi_{i}(1)$, but all we shall use from this is
that they have the same class in $\mathrm{K}^{0} X_{i+1}$.
Note that the dimension of these projections is $(k_{i+1}
\mathrm{dim}(p_{i+1}))(k_{i} \mathrm{dim}(p_{i}))$, and that
the dimension of $\pi_{i}(1)$ is at least $l_{i+1}^{0}
{(\mathrm{dim}(p_{i+1}))}^{2}$.  Since the two projections
under consideration commute with $\pi_{i}(C_{i})$, and this is
isomorphic to $\mathrm{M}_{\mathrm{dim}(p_{i})}(\mathrm{C})$,
to prove unitary equivalence in the commutant of $\pi_{i}(C_{i})$ in
$\pi_{i}(1) D_{i+1} \pi_{i}(1)$ it is sufficient to prove
unitary equivalence of the product of these projections with
a fixed minimal projection of $\pi_{i}(C_{i})$, say $e$.  Since
$\mathrm{K}^{0} X_{i+1}$ is torsion free, the products of the two
projections under consideration with $e$ still have the same class
in $\mathrm{K}^{0} X_{i+1}$.  To prove that they are unitarily
equivalent in $e D_{i+1} e$, it is sufficient (and necessary)
to prove that both they and their complements inside $e$ are
Murray-von Neumann equivalent.  Since both the cut-down projections
and their complements inside $e$ have the same class in
$\mathrm{K}^{0} X_{i+1}$, to prove that they (i.e., the two pairs)
are equivalent it is sufficient, by Theorem 8.1.5 of [5], to show that
all four projections have dimension at least $\frac{1}{2}
\mathrm{dim}X_{i+1}$ (note that $\mathrm{dim} X_{i}$ is even).
Dividing the numbers above by $\mathrm{dim}(p_{i})$ (the order
of the matrix algebra), we see that the dimension of the
first pair of projections is $k_{i+1} k_{i} \mathrm{mult}(\gamma_{i})
\mathrm{dim}(p_{i})$, so that the dimension of the
second pair of projections is at least $\mathrm{mult}(\gamma_{i})
(l_{i+1}^{0} \mathrm{dim}(p_{i+1}) - k_{i+1} k_{i}
\mathrm{dim}(p_{i}))$.  By construction, $\mathrm{dim}(p_{i}) =
\frac{1}{2} \mathrm{dim} X_{i}$.  Since $k_{i+1} k_{i}$ is
non-zero for all $i$, the first inequality holds.  Since
$l_{i+1}^{0}$, the second inequality holds if $\mathrm{mult}(\gamma_{i})$
is strictly greater than $k_{i+1} k_{i}$.  Since $k_{i+1} k_{i} = 3
{k_{i}}^{2}$, and $k_{i}$ was chosen before $q_{i}$, we
may modify our choice of $q_{i}$ to ensure that $\mathrm{mult}(\gamma_{i})$ is sufficiently large.

This shows that the two projections in $D_{i+1}$ under consideration
are unitarily equivalent by a unitary in the commutant of
the image of $\phi_{i+1}^{0} \gamma_{i}$.  Replacing $\phi_{i+1}^{0}$
by its composition with the corresponding inner
automorphism, we may suppose that the two projections in question
are equal.  In other words $\phi_{i+1}^{0} \beta_{i}$ is
unitarily equivalent to the cut-down of $\delta_{i}^{'}$ by
the projection $\phi_{i+1}^{0} \beta_{i}(1)$.

Now consider the compositions of these two maps with $\phi_{i}^{1}$,
i.e., $\phi_{i+1}^{0} \beta_{i} \phi_{i}^{1}$
and the cut-down of $\delta_{i}^{'} \phi_{i}^{1}$ by the projection
$\phi_{i+1}^{0} \beta_{i}(1)$. Since both of these maps can be
viewed as the cut-down of $\phi_{i+1}^{0} \gamma_{i}$ by the
same projection, they are in fact the same map.

Therefore, any unitary inside the cut-down of $D_{i+1}$ by
$\phi_{i+1}^{0} \beta_{i}(1)$ taking $\phi_{i+1}^{0} \beta_{i}$
into the cut-down of $\delta_{i}^{'}$ by this projection ---
such a unitary is known to exist --- must commute with the
image of $\phi_{i+1}^{0} \beta_{i} \phi_{i}^{1}$, and hence
with the image of $\phi_{i+1}^{0} \gamma_{i}$ --- since this
commutes with the projection $\phi_{i+1}^{0} \beta_{i}(1)$.
The extension of such a partial unitary to a unitary $u_{0}$
in $D_{i+1}$ equal to one inside the complement of this projection
then belongs to the commutant of the image of
$\phi_{i+1}^{0} \gamma_{i}$, and transforms $\phi_{i+1}^{0}
\beta_{i}$ into the cut-down of $\delta_{i}^{'}$ by this
projection, as desired.

Inspection of the construction will show that the maps $\delta_{i}^{'}
- \phi_{i}^{0} \beta_{i}$ and $\delta_{i} - \phi_{i}^{1}
\beta_{i}$ are injective, as required in the hypotheses of Theorem 2.4.

Replacing $\phi_{i+1}^{t}$ with $(\mathrm{Ad} u_{t}) \phi_{i+1}^{t}$
and deforming the $\beta_i$ to other point evaluations $\alpha_i$
which are non-zero on a given element (as we may, since $X_i$ is
connected), we have completed the construction of the desired
inductive system $(A_{i}, \theta_{i})$ satisfying
the hypotheses of Theorem 2.4.  Thus, the limit $B_n$ of the
inductive system with deformed finite stage maps, $(A_{i}, \theta_{i}^{'})$,
is simple.  Notice
that $(\mathrm{K}_{0}B_n, [1_{B_n}]) = (\ensuremath{\mathsf{\mathbf{Z}}},1)$ --- the
$\theta_i^{'}$ are unital and  $(\mathrm{K}_{0}A_i, [1_{A_i}])
= (\ensuremath{\mathsf{\mathbf{Z}}},1)$ for every $i$ --- and that $B_n$ is separable, nuclear and
stably finite since each of the $A_i$ is ([1]).

\section{The main result}

In this section we prove Theorem 1.1 through a series of lemmas.  We
establish that $\mathrm{sr}(B_n) \in \{n+1,n+2\}$ (Lemma 4.1), that
$\mathrm{K}_*$ is weakly unperforated (Lemma 4.3), and that
$B_n$ does not absorb $\mathcal{Z}$ (Lemma 4.4).  Taken together,
these results show that $B_n$ is as claimed in Theorem 1.1.

\begin{lms}
\begin{displaymath}
\mathrm{sr}(B_n) \in \{n+1,n+2\}.
\end{displaymath}
\end{lms}

The proof will depend on some definitions and results which we review below.

For a unital $C^*$-algebra $A$ we let
\begin{displaymath}
\mathrm{Lg}_s(A) = \{(a_1,\ldots,a_s) \in A^s| a_1 A + \ldots + a_s A = A \}
\end{displaymath}
for every natural number $s$, and recall that the stable rank of $A$,
$\mathrm{sr}(A)$, is the
least natural number $s$ such that $\mathrm{Lg}_s(A)$ is dense in $A^s$.  If no
such natural number exists, we set $\mathrm{sr}(A) = \infty$ ([9]).
Note that if $(c_k,d_k)$ are elements of a generalised
mapping torus $A(C,D,\phi_0,\phi_1)$ for $k \in \{1,2,\ldots,n\}$ such
that
\begin{displaymath}
\mathrm{dist} ((c_1,c_2,\ldots,c_n),\mathrm{Lg}_n (C)) \geq \delta,
\end{displaymath}
then
\begin{displaymath}
\mathrm{dist} (((c_1,d_1),(c_2,d_2),\ldots,(c_n,d_n)),\mathrm{Lg}_n
(A)) \geq \delta.
\end{displaymath}
Indeed, one can check that
\begin{displaymath}
\|{(c,d)}\| := \mathrm{max} \{ \|{c}\|, \mathrm{sup}_{t \in [0,1]} \|{d(t)}\| \}
\end{displaymath}
defines the unique $C^*$-norm on $A(C,D,\phi_0,\phi_1)$.  Thus, if
$\mathrm{dist}(c,c^{'}) \geq \delta$ for $c,c^{'} \in C$, then
$\mathrm{dist}((c,d),(c^{'},d^{'})) \geq \delta$ for any $(c,d),(c^{'},d^{'})
\in A(C,D,\phi_0,\phi_1)$.

For the remainder of this proof, any notation with subscript $i$
refers, where applicable, to the corresponding object in section 3.
In order to show that $B_n$ has stable rank greater than $n$, we must
exhibit $n$ sequences of elements $A_i \ni a_{i,j} = \theta_{i1}^{'}(a_{1,j})
= (c_{i,j},d_{i,j})$, $1 \leq j \leq n$, $i \in \ensuremath{\mathsf{\mathbf{N}}}$, such that
\begin{displaymath}
\mathrm{dist} (((c_{i,1},d_{i,1}),(c_{i,2},d_{i,2}),\ldots,(c_{i,n},d_{i,n})), \mathrm{Lg_{n}}
(A_i)) \geq \delta > 0
\end{displaymath}
for all $i$.  From this it follows that
\begin{displaymath}
\mathrm{dist} ((\theta_{\infty 1}^{'}((c_{1,1},d_{1,1})),\ldots,
\theta_{\infty 1}^{'}((c_{1,n},d_{1,n}))),\mathrm{Lg}_n(B_n)) \geq
\delta,
\end{displaymath}
so that $\mathrm{sr}(B_n) > n$ by definition.  (Here $\theta_{\infty
  1}^{'}$ denotes the inclusion of $A_i$ into $B_n$.)
By the definition of the norm on the $A_i$, it will be enough to show that
\begin{displaymath}
\mathrm{dist}
((c_{i,1},c_{i,2},\ldots,c_{i,n}), \mathrm{Lg_{n}} (C_i))
\geq \delta > 0
\end{displaymath}
for all $i$.

We now review Theorem 7 of [11].  Let $e(\cdot)$ denote the Euler class
of a vector bundle.  Suppose that $C$ is a
$C^*$-algebra of the form
\begin{displaymath}
(r+q)(\mathrm{C}(M \times \mathrm{D}^n)
\otimes \mathcal{K})(r+q),
\end{displaymath}
where $M$ is a smooth oriented manifold, and $r$
and $q$ are orthogonal projections in $\mathrm{C}(M \times \mathrm{D}^n)
\otimes \mathcal{K}$ such that $r$ corresponds to the trivial line bundle
and $q$ corresponds to a vector bundle $\alpha$ for which ${e(\alpha)}^n \neq 0$.
Let $\pi: M \times \mathrm{D}^n \rightarrow \mathrm{D}^n$ be
projection onto $\mathrm{D}^n$, and let $f_j: \mathrm{D}^n \rightarrow \mathrm{D}$ be the
$j^{\mathrm{th}}$ co-ordinate projection.

\begin{thms}[{Villadsen ([11]), Theorem 7}]
Let $C$, $\pi$ and $f_j$ be as above, and let $\tilde{c} = (c_1,\ldots,c_n) \in C^n$ be
such that $r c_j r = (f_j \circ \pi)r$ for all $1 \leq j \leq n$.  Then,
$\mathrm{dist}(\tilde{c}, \mathrm{Lg}_n (C)) \geq 1$.
\end{thms}

\noindent
\textbf{Proof of Lemma 4.1:}
We wish to apply Theorem 4.2 above to the algebras $C_i$,
$i \geq 1$.  The sequel is similar to the proof of Theorem 8
in [11].  For all $i$, let $r_i$ denote
the sub-projection of the unit of $C_i$ corresponding to the
one-dimensional trivial sub-bundle of $\theta_1 \times \xi_n \times
\cdots \times \sigma(i) \xi_{n \sigma(i)}$.  Note that $p_i$ considered
as a vector bundle over $X_i$ is the Whitney sum of $r_i$ and a second
vector bundle, say $q_i$, and this second vector bundle has
${e(q_i)}^n \neq 0$.  Indeed,
\begin{displaymath}
q_i = \xi_{n} \times \sigma(2) \xi_{n\sigma(2)} \times \cdots \times \sigma(i) \xi_{n\sigma(i)},
\end{displaymath}
and $e(\omega \oplus \gamma) = e(\omega) e(\gamma)$ for any two vector bundles
$\omega$ and $\gamma$ over a fixed base space so that
\begin{displaymath}
{e(q_i)}^n = {e(\xi_n)}^n {e(\xi_{n \sigma(2)})}^{n \sigma(2)} \cdots
{e(\xi_{n \sigma(i)})}^{n \sigma(i)}.
\end{displaymath}
(We are, as before, abusing notation slightly, using $\xi_k$
to represent the bundle induced on $X_i$ by $\xi_k$ via projection
from $X_i$ onto $\mathrm{CP}^k$.)
Since the integral cohomology ring $\mathrm{H}^* (\mathrm{CP}^k)$ is generated by
$e(\xi_k)$ with the relation ${e(\xi_k)}^{k+1}=0$, we may conclude by
the K{\"{u}}nneth Theorem that ${e(q_i)}^n \neq 0$, as claimed.  Each $X_i$ is
of the form $M_i \times \mathrm{D}^n$ for some smooth oriented manifold
$M_i$, so the $C_i$ have the same form as the algebra $C$ of
Theorem 4.2.

Note that for any element $c \in C_i$ there exists an element $(c,d)
\in A_i$ for some suitable $d \in \mathrm{C}([0,1];D)$.
Let $\pi_{i}:X_{i} \rightarrow D^n$ be the co-ordinate projection,
and let $f_j:D^n \rightarrow D$ be projection onto the $j^{th}$
co-ordinate.  Let $a_{1,j} = (c_{1,j},d_{1,j})$ be elements of $A_1$
such that $c_{1,j} = (f_j \circ \pi_1) r_1$, $1 \leq j \leq n$.  For each $i \geq 2$, put
$a_{i,j} = \theta_{i-1}^{'} \circ \theta_{i-2}^{'} \circ \cdots \circ
\theta_{1}^{'} (a_{1,j})$.  Write $a_{i,j} = (c_{i,j},d_{i,j})$.

In section 3, the map $\gamma_i$ was constructed as the direct sum of
$\psi_i$ and a second map.  Let
$\psi_{i1}$ denote the composition $\psi_i \circ \psi_{i-1} \circ
\cdots \circ \psi_1$.  Note that $\psi_{i1}(r_1) = r_{i+1}$.
By Theorem 2.3
\begin{displaymath}
c_{i+1,j} = \psi_i(c_{i,j}) \oplus c_{i+1,j}^{'},
\end{displaymath}
where $c_{i+1,j}^{'}$ is an element
of the cut down of $C_i$ by $q_i$;  the deformation
of $\theta_i$ to $\theta_i^{'}$ is visible in the fibre at infinity
only in the perturbation of the image of the second direct summand of
$\gamma_i$ --- the image of $\psi_i$ remains unchanged.  Thus, by construction
\begin{displaymath}
r_{i+1} c_{i+1,j} r_{i+1} = \psi_{i1}(r_1) c_{i+1,j} \psi_{i1}(r_1) =
\psi_{i1}(c_{1,j}) = (f_j \circ \pi_i) \psi_{i1}(r_1) = (f_j \circ \pi_i) r_{i+1}.
\end{displaymath}
By Theorem 4.2 we conclude that
\begin{displaymath}
\mathrm{dist} ((c_{i+1,1},c_{i+1,2},\ldots,c_{i+1,n}),
\mathrm{Lg}_n(C_{i+1})) \geq 1.
\end{displaymath}
As noted above, this implies that the simple limit $B_n$ has
stable rank strictly greater than $n$.

We now show that $\mathrm{sr}(B_n) \leq n+2$.
Given an exact sequence $B \rightarrow A \rightarrow C$ of
$C^*$-algebras, Corollary 4.12 of [9] states that
\begin{displaymath}
\mathrm{sr}(A) \leq \mathrm{max} \{ \mathrm{sr}(B), \mathrm{sr}(C)+1 \}.
\end{displaymath}
Applying this formula to the exact sequence $\mathrm{S}D_i
\rightarrow A_i \rightarrow C_i$ we have
\begin{displaymath}
\mathrm{sr}(A_i) \leq \mathrm{max}\{
\mathrm{sr}(\mathrm{S}D_i),\mathrm{sr}(C_i)+1 \}.
\end{displaymath}
It is known that
\begin{displaymath}
\mathrm{sr}(p(\mathrm{C}(X) \otimes \mathcal{K})p) = \lceil \lceil \mathrm{dim}
  X / 2 \rceil / \mathrm{dim}p \rceil + 1 \nonumber
\end{displaymath}
whenever $X$ a compact Hausdorff space and $p$ is a projection in
$\mathrm{C}(X) \otimes \mathcal{K}$ ([7]).  Thus, $\mathrm{sr} (C_i) =
\mathrm{sr} (p_i (\mathrm{C}(X_i) \otimes \mathcal{K}) p_i) = n+1$ by
inspection of the dimensions of the $p_i$ and $X_i$.  Since
$\mathrm{S}D_i$ is an ideal in $D_i \otimes \mathrm{C}([0,1])$,
we have
\begin{displaymath}
\mathrm{sr}(\mathrm{S}D_i) \leq \mathrm{sr}(D_i \otimes
\mathrm{C}([0,1])) \leq \mathrm{sr}(D_i) + 1
\end{displaymath}
by Corollary 7.2 of [9].
Theorem 6.1 of [9] states that
\begin{displaymath}
\mathrm{sr}(\mathrm{M}_n (A))\leq \lceil (\mathrm{sr}(A)-1) / n \rceil
+ 1,
\end{displaymath}
so that
$\mathrm{sr}(D_i) = \mathrm{sr}(\mathrm{M}_{k_i \mathrm{dim}p_i}
\otimes C_i) \leq n+1$ for all $i$.
We conclude that $\mathrm{sr}(A_i) \leq n+2$, so that
$\mathrm{sr}(B_n) \leq n+2$ by Theorem 5.1 of [9].
Combining this with the fact that $\mathrm{sr}(B_n) \geq n+1$ yields
Lemma 4.1. \hfill $\Box$

\begin{lms}
The ordered group $\mathrm{K}_* B_n =\mathrm{K}_0 B_n \oplus \mathrm{K}_1
B_n$ is weakly unperforated.  Its order structure is the strict one
coming from the first direct summand $(\mathrm{K}_0 B_n, {\mathrm{K}_0
  B_n}^+) = (\ensuremath{\mathsf{\mathbf{Z}}},{\ensuremath{\mathsf{\mathbf{Z}}}}^+)$.
\end{lms}

\noindent
\textbf{Proof:}
Since $(\mathrm{K}_0 B_n, {\mathrm{K}_0 B_n}^+)$ is weakly
unperforated it will be enough to show that every element in
$\mathrm{K}_1 B_n$ is the $\mathrm{K}_1$-class of a
unitary element in $B_n$.  Since $\mathrm{K}_* B_n$ is the inductive
limit of the $\mathrm{K}_* A_i$, it will suffice to prove this assertion
for all $A_i$ with $i$ sufficiently large.  By the formulas and discussion in
the proof of Lemma 4.1, we know that $\mathrm{sr}(\mathrm{M}_{\mathrm{dim} p_i}(\mathrm{S}C_i))=2$ for
all $i$ sufficiently large.  Assume that $i$ is so large for the remainder of the proof.

From [1] and [9] we know that there is a bijection between elements of
$\mathrm{K}_1 \mathrm{S}D_i$ and the $\mathrm{K}_1$-classes of unitaries in $\mathrm{M}_3 \otimes \widetilde{
\mathrm{M}_{\mathrm{dim} p_i}(\mathrm{S}C_i)}$.  Furthermore, any unitary in this latter
algebra is homotopic to a unitary in $\widetilde{\mathrm{M}_{3 \mathrm{dim} p_i}(\mathrm{S}C_i)}$.
Unitaries in $\widetilde{\mathrm{M}_{3 \mathrm{dim} p_i}(\mathrm{S}C_i)}$ give rise to
unitaries in $\mathrm{S} D_i$, since $3 \leq k_i$ for all $i$.  Thus,
every element of $\mathrm{K}_1 \mathrm{S}D_i$ can be represented as the $\mathrm{K}_1$-class
of a unitary.  The map $\mathrm{K}_1 \iota$ induced by the inclusion
$\iota: \mathrm{S} D_i \rightarrow A_i$ is surjective (as
$\mathrm{K}_1 C_i = 0$) and
the desired conclusion for $A_i$ follows from functoriality. \hfill $\Box$

\begin{lms}
For $n > 2$, $B_n$ and $B_n \otimes  \mathcal{Z}$ are not isomorphic.
\end{lms}

\noindent
\textbf{Proof:}
We proceed by showing that $\mathrm{sr}(B_n \otimes \mathcal{Z}) \leq
2$, so that $\mathrm{sr}(B_n) \neq \mathrm{sr}(B_n \otimes \mathcal{Z})$.

The algebra $\mathcal{Z}$ is an inductive limit of prime dimension drop algebras
 $\mathrm{I}[p_i,p_i q_i, q_i]$, $i = 1,2,\ldots$, where $p_i \rightarrow
\infty$ and $q_i \rightarrow \infty$ as $i \rightarrow \infty$ (cf. [6]).  For any $C^*$-algebra
$A$ the algebra $\mathrm{I}[p_i,p_i q_i, q_i] \otimes A$ is a full
algebra of operator fields, so by Theorem 1.1 of [8] we have
\begin{displaymath}
\mathrm{sr}(\mathrm{I}[p_i,p_i q_i, q_i] \otimes A) \leq \mathrm{sup}_{t \in [0,1]}
\{ \mathrm{sr}(A_t \otimes \mathrm{C}([0,1])) \},
\end{displaymath}
where $A_t$ is the fibre of $\mathrm{I}[p_i,p_i q_i, q_i] \otimes A$ at $t \in [0,1]$.
Since each such fibre is one of $\mathrm{M}_{p_i}(A)$, $\mathrm{M}_{q_i}(A)$, or
$\mathrm{M}_{p_i q_i}(A)$ we may rewrite our estimate above as
\begin{eqnarray}
\mathrm{sr}(\mathrm{I}[p_i,p_i q_i, q_i] \otimes A) & \leq & \mathrm{max}
\{ \mathrm{sr}(\mathrm{M}_{p_i q_i}(A \otimes \mathrm{C}([0,1]))),
\nonumber \\
& & \ \ \mathrm{sr}(\mathrm{M}_{q_i}(A \otimes \mathrm{C}([0,1]))),
\mathrm{sr}(\mathrm{M}_{p_i}(A \otimes \mathrm{C}([0,1]))) \}. \nonumber
\end{eqnarray}
By Corollary 7.2 of [9] we have $\mathrm{sr}
(A \otimes C[0,1]) \leq \mathrm{sr}(A) + 1$.  By Theorem 6.1 of [9] we have
that $\mathrm{sr}(\mathrm{M}_n (A))
\leq \lceil (\mathrm{sr}(A)-1) / n \rceil + 1$.  Thus, there exists $i_0 \in
\mathrm{N}$ such that $\mathrm{sr}(\mathrm{M}_{p_i q_i}(A \otimes \mathrm{C}([0,1])))$,
$\mathrm{sr}(\mathrm{M}_{q_i}(A \otimes \mathrm{C}([0,1])))$ and
\mbox{$\mathrm{sr}(\mathrm{M}_{p_i}(A \otimes \mathrm{C}([0,1])))$} are all less than
or equal to two for $i \geq i_0$.  We conclude that
\begin{displaymath}
\mathrm{sr}(\mathrm{I}[p_i,p_i q_i, q_i] \otimes A) \leq 2
\end{displaymath}
for all $i \geq i_0$.  Finally, $B_n \otimes \mathcal{Z}$ is an
inductive limit of algebras of the form $\mathrm{I}[p_i,p_i q_i, q_i] \otimes B_n$,
all but finitely many of which have stable rank less than or equal to
two.  By Theorem 5.1 of [9], the limit $B_n \otimes  \mathcal{Z}$ must
have stable rank less than or equal to two, as claimed. \hfill $\Box$

Thus, we have established Theorem 1.1.  In closing, we note that given
two natural numbers $n$ and $m$ one may carry out the construction of
section 3 to produce algebras $B_n$ and $B_m$ which, if the parameters
$q_i$ are chosen to be the same for both constructions, will have
isomorphic Elliott invariants.  This shows that one can produce
simple, nuclear, infinite-dimensional, stably finite counterexamples
to the Elliott conjecture which lie entirely outside the class of
$\mathcal{Z}$ absorbing $C^*$-algebras.  The explicit calculation of
$\mathrm{Ell}(B_n)$ and $\mathrm{Ell}(B_m)$ is long and not
particularly illuminating.  We leave it to the reader.

\pagebreak

redpajama_set_name: RedPajamaArXiv


Back to work… with Filofax!?
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text:
Archives Home →
Absolutely Fabulists: Liberal Mythmaking
By Daniel H. Fernald
Following 9/11, there was a debate over the need for "inclusiveness" in a proposed 9/11 Memorial based on the iconic photograph of three white firemen raising an American flag at Ground Zero. At the far end of a broad spectrum of sentiments was the hope that "the artistic expression of diversity would supersede any concern over factual correctness"[i].
On this view, a memorial depicting a factual account of the event would have been racist, sexist, and divisive. The logic of diversity takes otherwise straightforward matters and twists them into politicized knots. Facts be damned.
Although there were only three firemen in the original "factually correct" picture, the more "diverse" version would have required a veritable battalion to cover all of the various interest groups. After all, it takes a village to raise a flag.
The evident absurdity of this disparagement as mere "factual correctness" of what normal people simply call "reality" should not encourage anyone to dismiss its advocate as a solitary voice, however. This is not an aberration -- quite the contrary. It is indeed emblematic of the liberal approach to reality, and a prime example of liberal mythmaking.
Karl Marx, the original voice of "hope and change," said it best in the eleventh of his "Theses on Feuerbach": "Philosophers have hitherto only interpreted the world in various ways; the point is to change it."
In this, he was merely extending the logic (such as it was) of others who had come before him, including René Descartes and Immanuel Kant, who made the human mind -- rather than the world -- the starting point for scientific inquiry.
The goal of these earlier thinkers was, broadly speaking, scientific rather than political. Even so, Marx would later use a Hegelian variation on Kant's position to make the political argument that external reality is an obstacle to be overcome, rather than a fact to be objectively, dispassionately, and scientifically understood. The good Marxist knows that the world is but raw matter to be reworked into the shape predetermined by the ideological mind.
Of course, putting ideas first and remaking the world in one's own image can be a good thing. Such is the story of the American Founding. Much later, the seemingly unreachable dream of true racial equality was "midwifed" from the realm of ideas into flesh-and-blood reality by the heroes and journeymen of the civil rights movement. The nation responded to Dr. King's words because he challenged the American people to live up to their ideals. He also drew from the deep well of our shared cultural and religious heritage.
What, then, distinguishes the Founders and Dr. King from the Rev. Jeremiah Wright and his Alinskyite White House protégé?
In a word: Memory.
What has historically distinguished conservative (or what used to be called "classical liberal") political thought from that of leftists is that conservatives remember their mistakes and consequently can learn from them. Conservatism adapts its ideology to contrary evidence; liberalism reinterprets contrary evidence to fit its ideology.
For example, our Founders crafted our Constitution using the failed Articles of Confederation as both inspiration and object lesson.
In contrast, no matter how many times statism fails, the left mystically assures itself -- and us -- that "this time" it will work, somehow.
Historical ignorance and misology -- i.e., the hatred of logic and reason -- are part and parcel of the liberal mindset. Remembering the past spotlights the left's repeated failures. Logic demonstrates the inevitability of those same failures.
The left is stuck in an endless cycle of error because it rejects negative feedback on its ideas. Like Kant and Marx, they know that their ideas are correct (even though they aren't). Thus, if the available data contradicts the ideas that they know to be true a priori, then reality must be wrong. If it weren't for liberals' insistence on inflicting their ideas on the rest of us, their devotion to repeatedly debunked notions would almost be funny. Almost.
We are living in the midst of the topsy-turvy spectacle of these absurd ideas come to life.
How many times have you heard someone utter the ludicrous, self-contradictory but commonplace "Well, Communism may not work, but it's a great idea"?
How often have crime statistics been either suppressed or attacked as "racist" because they revealed patterns of criminality that were deemed insufficiently "diverse"?
The War on Poverty, the Great Society, AFDC, and related programs have broken up the very communities they were supposed to strengthen. Yet liberals resist any substantive reform tooth and nail.
Now, in the ultimate suspension of reality, our government has passed a health care bill that does not provide health care. This same law will cost at least a trillion dollars, yet somehow will magically lower the deficit.
This leads to an inescapable conclusion: Although we may excuse some liberals as merely ignorant and irrational, contemporary liberalism itself is explicit in its duplicity and misology.
What can one say of a person who elevates considerations of "diversity" over facts, who blindly praises a failed ideology that has killed over 100 million people, who believes that accurate scientific data can be "racist," and who insists on spending trillions in order to weaken the bonds that hold together American families, particularly among the poor?
At a minimum, he is a fool. More likely, he is a liar with an agenda.
Liberals can no more arrive in their imagined Utopia by the light of their mythic worldview than I could get from Albuquerque to Houston with a map of Narnia.
In response to the repeated implosion of their misbegotten notions, liberals have to deny past failures by either forgetting them or falsely casting them as successes. Either approach requires the denial of memory and logic.
So we get a bowdlerized FDR, who supposedly saved us from the Great Depression, when he actually prolonged it.
We get a sanitized Bill Clinton, who was the victim of a vast right-wing conspiracy, rather than his own emotional and sexual incontinence.
And we even get a misunderstood Castro, who would have done such great things for Cuba if it hadn't been for those wicked Yanquis and their nasty embargo.
Because the liberal worldview cannot survive in the clear, open air of free and unfettered rational discourse, it moves in a subterranean world of dark conspiracies, in which every failure is first blamed on nebulous enemies (Halliburton, the Vast Right-Wing Conspiracy, etc.) and later either recast as a success or forgotten entirely. This requires the denial of both objective reality and reason itself.
For the liberal fabulists in their Ministry of Truth, the past that cannot be rewritten soon finds itself shoved down the nearest Memory Hole, to be consumed by the waiting fires below.
Daniel H. Fernald holds a Ph.D. in philosophy and rhetoric from Emory University and is currently an Associate Professor at Hankuk University of Foreign Studies in South Korea. He may be reached at professordhf@hotmail.com.
[i] Jonah Goldberg, "Facts & Firemen," National Review Online, January 18, 2002: www.nationalreview.com/goldberg/goldbergprint011802.html

redpajama_set_name: RedPajamaCommonCrawl


text:
Cell Signal Booster Guide
Open Box: Cel-Fi PRO for AT&T
by Cel-Fi
Powerful 3G/4G/LTE signal booster approved and authorized by AT&T.
Open Box: Cel-Fi PRO for AT&T has a rating of 5.0 stars based on 1 reviews.
Retail price: $695.00 You save: $69.50
External Antenna Port No external antenna port Includes external antenna port
Cel-Fi products come with a manufacturer's warranty against defects in workmanship and materials.
This is an open box item. The product will contain all original parts and accessories and is in like new condition. Open Box items may contain...
This is an open box item. The product will contain all original parts and accessories and is in like new condition. Open Box items may contain superficial scratches that do not affect performance. All open box units are sold AS IS and can be returned within 30 days from receipt of the product.
The Cel-Fi PRO is a powerful signal booster system that boosts 3G/4G/LTE voice and data for AT&T. The Cel-Fi PRO offers up to 100 dB of gain and coverage up to 13,000 sq ft.
The Cel-Fi PRO system is incredibly easy to set up. The simple plug and play installation means that anyone can install the device. There are no external cables, antennas, wires, or drills needed. The units simply plug-in to power.
The Cel-Fi PRO has an LCD color interface and is fully self-contained and self-configuring. It supports the Cel-Fi WAVE mobile application.
Cel-Fi WAVE Mobile App
Cel-Fi WAVE is a smartphone app that will help you get the best performance from your Cel-Fi PRO. The WAVE app is available for desktop, iOS, and Android. The app's dashboard allows you to optimize setup and download software updates.
Attractive, unobtrusive design
End-to-end cellular communication encryption without additional risk of vulnerability
Support for the Nextivity WAVE Portal
Unlocked: Cell phones do not need to be registered with Cel-Fi PRO to benefit.
Does not interfere with nearby Cel-Fi systems
LCD User Interface
Remote software update capability
Mounting brackets included with every unit
The Network Unit (NU) should be placed where the existing cellular signal is best (generally by a window). The Coverage Unit (CU) should be placed in the location you need improved signal. You can optimize coverage by moving the CU. The NU and CU communicate with each other wirelessly to provide the best signal improvement possible. As the distance between the units is increased, the CU adjusts its coverage bubble to provide the ideal amount of coverage.
Mounting brackets and hardware are provided to mount both (or either) the Coverage Unit and the Network Unit. The brackets come equipped with screws and inserts for mounting in standard drywall. The Network Unit has mounting holes on the back. The Coverage Unit has mounting holes on the bottom and on the back. The holes on the bottom of the Coverage Unit can be used for mounting the Coverage Unit to a ceiling.
The coverage area you can expect from any signal booster kit depends on the signal strength outside the building.
Outdoor Signal
Estimated coverage area
Strong outdoor signal (5 bars) up to 13,000 sq. ft. coverage
Medium outdoor signal (3-4 bars) up to 8,000 sq. ft. coverage
Weak outdoor signal (1-2 bars) up to 5,000 sq. ft. coverage
Like almost all of our products, the Open Box: Cel-Fi PRO for AT&T can be installed by anyone who feels comfortable doing simple home repairs and running cabling. If you're not comfortable doing that kind of work, you might consider finding a local electronics installer or asking a friend.
Unlimited free technical support:
Anytime you purchase from Waveform, you get unlimited free technical support from our signal experts. Whether it's installing a kit or troubleshooting a problem, we're here. In over 11 years of being in business we've helped over 25,000 customers install kits and improve their signal.
Have a question now? Call us from 7am to 5pm PST: 1-800-761-3041.
SKU OB-590NP34GWUS1ATUS1B14
OB-590NP34GWUS2ATUS1B14
Coverage 5000 sq ft (1 coverage with weak signal)
8000 sq ft (2 coverage with medium signal)
13000 sq ft (3 coverage with strong signal)
FCC notice Read full notice
This is a CONSUMER device. BEFORE USE, you MUST REGISTER THIS DEVICE with your wireless provider and have your provider's consent. Most wireless providers consent to the use of signal boosters. Some providers may not consent to the use of this device on their network. If you are unsure, contact your provider. You MUST operate this device with approved antennas and cables as specified by the manufacturer. Antennas MUST be installed at least 20 cm (8 inches) from any person. You MUST cease operating this device immediately if requested by the FCC or a licensed wireless service provider. WARNING. E911 location information may not be provided or may be inaccurate for calls served by using this device. Please note, the four largest carriers - AT&T, T-Mobile, Verizon and Sprint - and more than 90 regional carriers have given blanket consent for use of all boosters certified to the new FCC standards.
User manual Specifications
The Cell Signal Booster Guide
11 Best Cell Signal Boosters
Best AT&T Signal Boosters
Best Verizon Signal Boosters


Looking for prediction for today's match up of Phoenix at Portland today? The game time will be 10:00 PM ET, with the current odds set at O/U: 216.5 and spread of POR -13.5.
PHO currently comes into this game with a 4 – 20 record, and 8 – 16 record against the spread.
While POR has a 13 – 11 record coming into this matchup, with a record of 11 – 13 against the spread.


text:
December 2, 2009 / 5:04 PM / 10 years ago
Tax haven clamp down moving to enforcement phase-OECD
* Tax haven clamp-down moving to new enforcement phase
* Progress in implementing tax standards a "revolution".
* OECD is not seeking tax rate harmonisation
By Chris Vellacott
LONDON, Dec 2 (Reuters) - A global crack down on tax evasion in financial centres is moving into a more difficult enforcement phase, an Organisation for Economic Co-operation and Development official said on Wednesday.
Jeffrey Owens, director of the OECD's centre for tax policy and administration told delegates at a conference that progress made in implementing tougher disclosure requirements since the G20 summit in London in April amounted to a "revolution."
World leaders agreed at the G20 summit in London to crack down on tax evasion and banking secrecy and asked the OECD to publish lists of tax havens.
There are now 59 countries on the OECD's "white list" of jurisdictions that have implemented internationally agreed tax standards while 28 remain on the "grey list" having committed to the standards without having fully implemented them.
The grey list includes Malaysia, Uruguay and Chile though most of the countries are tiny Pacific and Caribbean island nations.
There are now no financial centres surveyed by the OECD which have not committed to the standards.
"Listing is never a pleasant process but it worked. The combination of the list and the threat of sanctions are the reason why we have seen so much progress over the last 10 months. Anybody who tells you different, they're wrong," Owens said.
The OECD will now focus on assessing the extent to which the standards are met in practice, and seek to close loopholes in implementation.
"If we see a small jurisdiction which has 12 agreements (on information disclosure) with countries with which they have no economic ties at all, to me that is not playing the game," he said.
Owens said the greater political will in G20 countries to curb banking secrecy in offshore financial centres for the purpose of tax evasion has led to more progress made in the last 10 months than all the achievements of the previous decade.
He stressed however, that while the OECD is clamping down on jurisdictions seeking a competitive advantage through tax secrecy, it is not seeking harmonisation of tax rates.
"There is no reason why Sweden should have the same tax system as the Cayman Islands. Setting minimum tax rates is not on our agenda, neither is impinging on national sovereignty. Basically what it's about is a level playing field," he said. (Reporting by Chris Vellacott; Editing by Ron Askew) (For the Hedge Hub blog: blogs.reuters.com/hedgehub) (For Global Investing: here) ((chris.vellacott@thomsonreuters.com; +44 (0) 20 7542 3987; Reuters Messaging chris.vellacott.thomsonreuters.com@reuters.net)


text:
Diese Liste stellt die Abgeordneten des 35. (und letzten) Landtags des Großherzogtums Hessen, gewählt im Jahr 1911 dar. Da die Wahl des 36. Landtags kriegsbedingt nicht erfolgte, wurden die gewählten Abgeordneten auch die Abgeordneten des 36. Landtags bis sein Mandat mit der Novemberrevolution 1918 endete.

Der Landtag bestand aus 2 Kammern.

Erste Kammer
Präsidium:
 Erster Präsident: Graf Emil von Schlitz, genannt von Görtz in Schlitz
 Zweiter Präsident: Fürst zu Solms-Hohensolms-Lich
 Dritter Präsident: Freiherr Cornelius Wilhelm von Heyl zu Herrnsheim
 Schriftführer: Fürst Isenburg-Birstein
 Schriftführer: Oberstallmeister Moritz Riedesel Freiherr zu Eisenbach

Mitglieder:
 Standesherren
 Friedrich Wilhelm zu Ysenburg und Büdingen
 Christian-Ernst zu Stolberg-Wernigerode
 Emich zu Leiningen
 Karl zu Solms-Hohensolms-Lich
 Franz-Josef zu Ysenburg-Büdingen
 Aloys zu Löwenstein-Wertheim-Rosenberg
 Alexander zu Erbach-Schönberg, vertreten durch Victor zu Erbach-Schönberg
 Wolfgang zu Ysenburg und Büdingen
 Jost-Christian zu Stolberg-Roßla-Ortenberg, vertreten durch Kuno zu Stolberg-Roßla-Ortenberg bzw. als Nacherbe Christoph Martin zu Stolberg-Roßla
 Georg Friedrich zu Solms-Braunfels, vertreten durch Friedrich zu Solms-Braunfels
 Georg Albrecht (IV.) zu Erbach-Erbach
 Graf Emil von Schlitz, genannt von Görtz in Schlitz
 Adalbert zu Erbach-Fürstenau
 Gustav Graf zu Ysenburg und Büdingen in Meerholz
 Karl Franz zu Solms-Rödelheim
 Georg Friedrich (Solms-Laubach)
 Der Senior der Familie Riedesel, Erbmarschall Ludwig Riedesel zu Eisenbach
 Der katholische Landesbischof Georg Heinrich Kirstein vertreten durch Domkapitular Ludwig Bendix
 Der evangelische Prälat Friedrich Flöring
 Für die Landesuniversität Gießen: Geheimrat Arthur Benno Schmidt
 Für die Technische Hochschule Darmstadt: Geheimer Baurat Alexander Koch
 Grundbesitzender Adel:
 Moritz Riedesel zu Eisenbach
 Philipp Wambold von Umstadt in Birkenau
 Vom Großherzog auf Lebenszeit berufen:
 Cornelius Wilhelm von Heyl zu Herrnsheim
 Erasmus Kittler
 Karl Kleinschmidt, Justizrat
 Louis Merck, Geheimer Kommerzienrat in Darmstadt
 Otto Gastell, Fabrikant in Mainz
 Carl Parcus, Präsident der Handelskammer Darmstadt
 Friedrich Stroh, Geheimer Kommerzienrat in Offenbach
 Wilhelm Gail, Geheimer Kommerzienrat in Gießen
 Ludwig Strecker, Geheimer Kommerzienrat in Mainz
 Wilhelm Joseph Valckenberg, Geheimer Kommerzienrat in Worms
 Philipp Hangen, Landgerichtspräsident in Mainz
 Karl Göttelmann, Oberbürgermeister in Mainz
 Als Vertreter des Handels, der Industrie, der Landwirtschaft und des Handwerks
 Franz Bamberger, Präsident der Handelskammer zu Mainz
 Wilhelm Haas, Vorsitzender der Landwirtschaftskammer für Hessen
 Jean Falk, Vorsitzender der Handwerkskammer

Zweite Kammer

Wahl
Die Wahl der zweiten Kammer erfolgte am 3. November 1911.

Präsidium

 Erster Präsident: Heinrich Köhler (NLP), Oberbürgermeister in Worms
 Zweiter Präsident: Gustav Korell
 Dritter Präsident: Adam Joseph Johann Schmitt (Zentrum)
 Schriftführer: Ernst Christian Finger
 Schriftführer: Carl Adam Damm (FVP)

Mitglieder

Quellen
 Hans Georg Ruppel, Birgit Groß: Hessische Abgeordnete 1820–1933. Biographische Nachweise für die Landstände des Großherzogtums Hessen (2. Kammer) und den Landtag des Volksstaates Hessen (= Darmstädter Archivschriften. Bd. 5). Verlag des Historischen Vereins für Hessen, Darmstadt 1980, ISBN 3-922316-14-X.
 Inhaltsverzeichnis über die Verhandlungen der 2. Kammer der Landstände, Liste der Mitglieder und Präsidium der zweiten Kammer Seite 3-7; Präsidium der ersten Kammer Seite 158
 Sitzverteilung

 35

redpajama_set_name: RedPajamaWikipedia


Beijing this week officially ended UnionPay's longstanding monopoly and opened up the CNY 55 trn ($8.4bn) Chinese credit card clearing market to domestic and foreign players. The move has been four years in the making since a 2012 WTO ruling judged that it was discriminatory of China to bar foreign electronic payment processors from handling yuan-denominated transactions. There are 5.4 billion cards in circulation in China, and some 48% of all the country's gross consumption and retail activities were conducted by card last year, according to the People's Bank of China that issued a joint statement with the China Banking Regulatory Commission explaining that the move was intended to encourage competition and open up the market.
"The principle… is to create a level playing field and attract more institutions to participate in market competition," a spokesman said. "In future, China's bank card clearing market will be pluralistic. Multiple brands would compete and offer different and differentiated services. This will allow the card market to sustainably renew itself and optimise its structures and developments.


When I was a kid, I had it all figured out. I'd be married by 25 and have all my kids by the time I was 30. Sure, it'd be difficult those first few years, but my parents would be around to help. By the time I was 40, I'd be a famous author, my kids would adore me, and life would be peachy-keen. Only my schedule didn't go according to plan.
In January of 1978 when I was almost 20, my dear stepfather died. The unexpectedness of his death jolted me and sent me into a tailspin that took years to recover from. Writing lost its pleasure for me. My plans to go study Chinese in Hong Kong became my only escape. However, as most of us know, you can't escape emotional tragedy for long. In Hong Kong, my loneliness was amplified with each hour. Everything seemed worthless in the face of mortality. I dropped out of school, came home, and found no solace there. Finally, I went back to Michigan and completed my college education. I was still frightened to set goals. After all, I could die at any moment. I worked odd jobs, fell into and out of relationships, and then I decided to go to grad school.
So by 25, I was entering school yet again, and marriage seemed far, far away. I earned a Masters in Social Work, moved to Washington, D.C., and got better odd jobs with bigger and more impressive names. But I still felt lost, as if in an adult adolescence. After a few years, I returned home again.
In Albuquerque, I realized all of my work had led me to public relations. It was my profession that helped focus other parts of my life. By now I was 30, still no marriage, kids, or book written or sold. But things were becoming more solid once again. I made tentative plans, set tentative goals, and inched my way back into believing in life and myself. I started writing seriously again.
At 33, I reconnected with someone I'd known since childhood. A romance ensued, and at 35, I was married. One doctor told me I was too old for children. I got pregnant and miscarried. I swirled into another tailspin with the fear that I'd never be able to have children because I'd waited too long. A new doctor said though my age might make me seem old for having babies, my body didn't act old to him. He urged me to try for a child once more, and I did. I gave birth to my first daughter when I was 37.
As I sat marveling at her beauty, my own creativity rekindled, and the idea for a mystery series was born. In the long midnight hours with my newborn, I worked on the outline for my first novel. The same year I completed the manuscript, my father died. Then at 40, I had another daughter. My mother died eight weeks later. My godfather died two months later. The realization that I'm at the top of my family's generation has had a different effect than the horror of the first death I experienced. Now I'm even more determined to live my life well, and to provide my children with the resilience to meet life's joys and tragedies and not be knocked down for as long and as hard as I was.
Oh, and at 44, I've written three novels. My third was the first one to sell. It will published around the time of my 46th birthday.
No, things haven't gone according to plan in my life. I missed many of those important chronological benchmarks. But I did become a wife, a parent, and an author. What has my roller coaster life taught me so far?
Well, making plans and having goals is like designing a roadmap. The destination is important, but the journey to getting there can go any number of ways. By having destinations on your life's roadmap, you can gauge your progress. For goal-oriented people like me, that's important. But I've also learned through my up-down life that it's difficult to force the details of the journey.
When I went to Hong Kong, I planned to stay for years. Though I left after eleven months, living there and learning what it was like to be a visible minority instilled a sensibility in my writing that would be difficult to gain any other way. I had my children much later than I anticipated. The delay allowed me to travel and have a profession before pulling back into the singularly intense experience of being a stay-at-home mother. I'm a more patient parent as a result, and I'm convinced it has made me a more grateful mother than I would have been had I had children in my late 20s.
In our generation, we were told to, "go with the flow." I never quite bought that concept because it was too vague. I didn't know if I was supposed to float along aimlessly, or grab every opportunity that came my way. A roadmap is priceless for developing your life's goals. But just as important is the realization that there are myriad routes on that map. Many yield more interesting and fruitful results than the ones you originally planned.


Most people experience the truth that, 'Happiness seems as minute as a grain of barley, while unhappiness seems as huge as a mountain'. Since most people do not know how to attain Anand (Bliss), each tries to acquire at least temporary happiness through the five senses, mind and intellect. However, happiness is temporary and of inferior quality, whereas Anand is eternal and of supreme quality. The science which teaches us how to acquire Anand is termed as the science of Spirituality. Anand is infinitely superior to object pleasure. If this is realised then one will certainly make efforts to acquire Anand instead of object pleasure.
Why is purashcharan performed before commencing chanting of mantra?
Why is (Omkar) Om mantra called as pranav mantra?
How does pranayam facilitate activation of the kundalini?
How to gain control over mind by practicing pranayam?
How does tratak help in increasing ones sattvikta?


Something that happens to many people, and especially when they're self-aware to a certain degree, is that they realize that they don't know it all. Now, I'm saying that anyone goes around claiming to know it all, although some do, but there comes a time when someone will do some soul searching and figure out, "Geez, I've got issues!" We all have certain things that we need to work on, and sometimes having people telling you won't be the key to what will make you search within and come to terms with whatever it is. And just because people will tell you that you have certain issues, it doesn't mean that you do. They might even be projecting their own issues onto you. However, many people are simply in denial of their issues and think of themselves as someone who frankly, is perfect in certain areas, only to later on find out the hard way, that they're not.
One of the hardest things in life to endure in oneself is coming to terms with the fact that maybe we were wrong. Being wrong about something that you might've said or done to someone can be a lot easier to fix at times, than when you find out that you were wrong about something that you've been feeling or thinking is right, and especially when you've been feeling or thinking this way for a long time. Working on our own issues can be one of the hardest things to do, but sometimes, coming to terms with the fact that we have certain issues can be even harder than the work ahead of us.
Once a person realizes that they have certain issues, you might think, "Hey, at least I'm not in denial, right?" But that's only the first step. However, it can be such a difficult step when you first come to terms with the fact that you have certain issues. We all see ourselves in a certain light, and despite how others may see us, it's great to think highly of ourselves and to feel good, confident, and have a good self-esteem. As well, we should all feel worthy of love and happiness. But one of the most important things that it takes in order to be a truly well rounded and evolved individual, is to know the difference between being confident, secure, humble, and to recognize your imperfections and things that you need to work on, from someone who's arrogant, you know, "overly confident," and who thinks that they're flawless, and who never looks within, when there's an issue in their life.
Arrogance is ugly, but so is when people aren't humble or never own up to their sh**. As well, when people aren't self-aware, it can be quite unappealing too. I mean think about it, who would you rather spend your time and hang out with the type of person who acts perfect, flawless, is arrogant, and has their sh** together, but who never owns up to their mistakes, etc., or the type of person who tries their best, is confident, humble, self-aware, owns up to their mistakes, and pushes forward in their life on every level; spiritually, emotionally, intellectually, and tries hard to improve themselves? I would definitely rather spend my time with the type who's knowingly imperfect, and who won't be too proud or cocky, and who sees the obvious, that they're imperfect.
Our personal issues were created because of many different reasons. Perhaps they were created as a result from what we'd seen growing up, how we were raised, seeing friends or loved one's who acted in certain ways, or even from people who have hurt us in life. But regardless, we end up having to deal with the issues that were created, despite how they were made, and if we don't, we'll never heal, and never become the best version of ourselves. Working on our issues isn't an easy task, but no one else is going to do it for us. And if we don't work on our issues, we won't grow and evolve into better human beings. And we might even pass our issues onto others or even our children if we're not careful. We need to acknowledge our issues, but also be willing to put the hard work into healing, improving, and hopefully permanently fixing our issues.
We can't live our lives blaming the world around us or even blaming certain specific people for they'd done to us. We need to own up to our issues, just as much as we need to own up to our mistakes. It doesn't matter how badly one might've been abused or hurt when it comes to where you are in life NOW. Right now is what matters, because the past is dead and gone, and reliving it or blaming those who might've hurt you for you having issues is only going to bring headache, heartache, turbulence, and a lack of spiritual growth to your life. You can't live your life in the best way and be happy or at peace, or even find a stable, loving, and happy relationship or even friendships for that matter, if you're not going to work on yourself. So be willing to look within, and acknowledge your needed areas of improvement.
Don't hide in a shell, blaming others for how you behave or think, because you make your own decisions in life, and therefore, you will have to face the consequences. The people who hurt you won't have to face them. They'll have to face consequences of their own I'm sure, but you are to blame for your own decisions and mistakes. You will be to blame if you mess up. And so what if you have a lot of issues! And Hell yeah, you should still be confident and love yourself. Feel good on your journey in life.
Remember, you won't always know what other people's "real issues" are, and many people are likely fighting their own battles, which in comparison to yours and whether you see their issues upfront or not, might be a Hell of a lot worse! But either way, don't compare yourself to someone else, and don't fret for the road ahead of you and how long it might take to heal or work on your issues. Start working on yourself. Start today, start now, and do so with the goal of being the best version of yourself, the new you, the new and improved you.
I wanted to add this one important point. When you mess up and make mistakes and it affects other people, you should own up to what you did, even if it wasn't intentionally. However, when you realize that something you're doing isn't right, or if you start to realize just how imperfect you really are, you shouldn't be ashamed, but you also don't have to tell the world. You don't owe anyone an explanation for how you feel inside or what issues that you discover are needing improvement.
The only person who you truly need to answer to is G-d, and then yourself. So pray to G-d, ask for guidance, but don't ever feel obligated to confess your issues to anyone else, especially if you're working on them. Don't let others have the opportunity to bring you down. Sharing personal issues with others can be helpful at times, but only if you're confiding in certain people who will truly have your best interest at heart. Remember, you don't owe anyone an explanation of how to want to improve yourself, only an explanation or even an apology for why you might've hurt someone.
Know that you're not alone, and everyone else has their own needed areas of improvement, so they're probably struggling in their own way. But you are never alone, and if you need help to sort things out or to help push you in the right direction, never be afraid to reach out to someone who you trust. Self help articles and books can be helpful, but confiding in someone who you trust can also be beneficial. Just pick and choose who you trust to confide in. Just because someone is a therapist, a family member, or a close friend, doesn't mean that they will be a good person for you to share such personal things with, so be cautious.
Be smart with who you share your innermost secrets and private matters with. People can be very judgmental or even claim to want to help you but then either hurt you, take advantage of you, or Heaven's forbid, give you bad advice. Last, but not least, try working on yourself on your own first and foremost, and praying, and then acknowledging your faults, apologizing to those who you might've hurt in life, and working hard on improving your issues. It may be a long road ahead, but the the battle will be worth it in the end.
i believe that the world should live by the principal " As above so below": you just can not live the happy life that is waiting for you with whole new opportunity. if you keep the issues buried inside you . If you are unhappy within. you'll be unhappy outside. Great post and keep doing good work.


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Stop Botox Testing on Mice Now
by: Alicia Mottur
recipient: Allergan, Inc. Corporate Headquarters, Irvine, California
Each year worldwide close to a million mice die after enduring the unbearable suffering of Botox testing, in which lethal doses of the drug are injected into groups of laboratory mice to test its potency.
Animal rights campaigners in the United States and Europe have long been urging the leading American pharmaceutical company, Allergan, which owns and markets Botox, to stop testing the product on mice, and to adopt instead one of the reliable, effective alternative tests not using animals.
The test currently used, known as the LD50 test, involves injecting the botulinum toxin type A (the active ingredient in Botox) into a group of mice in sufficient quantities to kill 50% of the animals. The toxin is injected into the mouse's abdomen, then researchers wait to see how many of the animals die over a period of 3 to 4 days. About 100 mice are normally used for each test, and each batch is often tested more than once. The mice are first assigned to one of various groups, and each group receives a different strength of the product in order to estimate the strength that kills half of the targeted group.
The antiquated LD50 test, first developed in the 1920's, is cruel and brutal. No pain relief is administered to the test animals, who either die immediately from suffocation after their diaphragms have been paralysed, or may languish over several days with varying degrees of paralysis before being euthanised.
U. S. and European authorities have identified a number of replacement tests NOT using animals that, once validated and adopted by drugs manufacturers, such as Allergan, would replace the LD50 test. In contrast to the LD50 test, these alternatives apply modern technology and insights from cutting-edge research in molecular biology and immunology into the toxic effects of botulinum toxin on the organism. The two most promising alternatives are the endopeptidase assay and the neural cell-based assay, also known as "SNAP-25" tests.
The U. S. Food and Drug Administration does not specifically require manufacturers of drugs using the botulinum toxin to employ the LD50 test; rather it requires the "performance of a sensitive potency assay."
Please urge Allergan to show courage and true leadership by validating and adopting one of these humane alternatives.
For further information on Botox and animal testing, please go to these links:
http://www.prijatelji-zivotinja.hr/index.en.php?id=1192
http://www.dailymail.co.uk/sciencetech/article-1320889/How-fashion-Botox-seen-number-mice-used-medical-experiments-double.html
http://www.mnn.com/earth-matters/animals/stories/botox-testing-blamed-for-rise-in-animal-experimentation
We the undersigned urge you to please encourage Allergan to abandon the testing of Botox on mice, and to instead validate and subsequently adopt the alternative SNAP-25 test not involving laboratory animals.
As we are sure you are aware, the currently used LD50 test causes unspeakable suffering to the animals, during which at least 50% of test mice die from paralysis and suffocation. The SNAP-25 test has been proven effective and accurate, and its application in the pharmaceutical industry is only awaiting the validation by market leaders, such as your company. It is estimated that about 600,000 laboratory mice die each year as a result of testing of the botulinum toxin for pharmaceutical purposes. A decision by Allergan to adopt an alternative, humane test method would represent a tremendous step forward in ending the suffering of millions of animals.
According to the National Institute of Health, both the Consumer Product Safety Commission and the Environmental Protection Agency discourage the use of the LD50 test, while the Food and Drug Administration does not require it.
Furthermore, as public opinion is increasingly on the side of protecting animals from suffering in product testing, a decision by Allergan to abolish animal testing in favor of humane alternatives could only reflect favorably on Allergan's business and reputation.
We hope you will consider this problem, and help your company to institute new methods addressing the tragic plight of countless fragile and sensitive animals who involuntarily bear the heaviest burden of our medical research.
Thank you in advance for your time and attention.


Gra o Tron - Puzzle Westeros 3D 540 el.
Unknown World. Each jigsaw puzzle piece is made out of plastic material that snaps together during assembly ! No glue required and when completed it forms a complete spherical puzzle map.


text:
You're Married With Kids & Have a Crush? Here's What to Do
Sure, it feels good when that cute parent at drop-off smiles at you. But if you're fixing your hair before you get out of the car, it's time to come clean.
By Daphne de Marneffe, Ph.D
LDprod/Shutterstock
A spark of attraction between two people is one of the great pleasures of life. But when you have a long-term partner and that spark is between you and someone else, things can get tricky.
Being able to deal in a healthy way with the attractive people who cross your path is an important skill for couples. People usually manage through a mental balancing act—letting themselves feel a little, yet not too much. But the years of early parenthood can be challenging. Postbaby, many of us don't feel at our most attractive. Marriages can feel more functional than romantic, and we don't always have time to nourish our own friendships or hobbies. Even when we try to open up with our partner, missed cues or rejected advances can shut us down.
RELATED: Staying Lovers While Raising Kids
When you least expect it, someone else might catch your eye, and suddenly you wake up and remember how you used to feel. Before you know it, daydreams start wafting through your mind, and you wonder whether you're in the grips of a crush. Perhaps you're spending time imagining the witty conversation that you might strike up with your son's soccer coach, or you're giving a lot more thought to what you'll wear on a day when you know you'll be meeting up with a certain colleague.
Most crushes are harmless and relatively fleeting. And there can even be some benefits. Here are three:
A crush lets you see yourself in a new way.
Life can sometimes feel like endless logistics and just putting one foot in front of the other. We're focused on our kids' moods, sleep schedules, screen time, schoolwork, and the rest. Having a crush helps us see ourselves in a new, more attractive, and more interesting light. That added sense of vitality can pervade every part of life, even trips to the grocery store.
A crush can add spice to your primary relationship.
One woman said to me in her therapy session, "You'd think that having a crush would make me feel guilty. But I actually think it makes me like all men better, including my husband!" Her openness to the charms of men was expanded, as was her own feeling of desirability. Both helped her become more amorous at home.
RELATED: 3 Get-Closer Tips for Otherwise Happy Couples
Feeling energized can jumpstart new possibilities.
Okay, so maybe you began going to yoga because of the cute teacher or got involved with a political cause because your charismatic neighbor was organizing it. But being open to people and new experiences is what ultimately brings us a sense of purpose and joy. Romantic feelings can expand your vision of what's possible and can get you out of a rut you didn't even know you were in.
Consider if it's gone too far
Of course, a crush isn't always harmless, and you may be playing with fire. Here are some ways to check yourself:
Know the difference between a harmless and a harmful crush.
The crucial distinction is whether your thoughts and feelings are getting more intense. Are you taking a momentary feeling and weaving a story about it? Are you starting to compare your real relationship with an idealized romance? One problem for people is that they think they're in control and then discover too late that their alternative fantasy relationship is sucking up vast amounts of attention and energy. Sometimes people realize they're feeling needy or empty, and don't want to give up their crush. The desire to feel appreciated is too strong, the impulse to resist the daydreams too weak. They begin to hide from themselves that they are making subtle choices that intensify their feelings, and they may even begin telling white lies or shading the truth.
RELATED: An Inside Peek Into Other Parents' Sex Lives
Try to be brutally honest with yourself.
Is asking that handsome stay-at-home dad out for a coffee really necessary to help make him feel a part of the community? Is complimenting your fitness instructor's appearance a bit over the line? Is a thank-you dinner for the consultant who helped advocate for you at work truly called for? Sometimes it's best to err on the side of caution and to find a friendly but less "out there" way to compliment or thank someone.
RELATED: 5 Ways to Rev Up Your Relationship Without a Date Night
Talk to your partner about it.
This may sound awkward, but it's one of the most effective ways to regain your balance. Of course, no one wants to hear that his partner has a crush on someone else. Telling your partner may seem like overkill ("If it means nothing, why bother him about it?"), and secretly, you may be reluctant to pour cold water on your infatuation. That's all the more reason to talk honestly. It prioritizes protecting your relationship, and it can even be bonding. The sooner the conversation happens, the easier it is.
"I'm really embarrassed to admit this," said Tina in a therapy session with her husband, James, "and I wouldn't want to hear something like this from you. But I realize I've developed a crush on Dan." Dan was their contractor, and he was in the middle of a small building project at their house. James felt hurt, angry, and threatened, and asked some pointed questions: "Have you told him? Have you touched him at all?" Tina said no, but she felt intensely ashamed and worried about her husband's reaction.
In the course of the difficult conversations that followed, I said to both of them, "I know this is really hard, but I'm proud of you for talking about it. Tina, confiding in James is a sign of your trust in him and respect for the relationship. James, it's very loving of you to listen without blowing up or shutting down." Tina's most important message was that she was sorry she hadn't protected their relationship better. It reassured James about how much she valued him.
If you are the one being told about your partner's crush, you should feel free to voice your hurt, fear, or anger. But try to steer clear of shaming your partner with blanket criticism or judgments about his character. Long-term relationships have many challenges, and it's how you face them together that determines how well you survive. After the painful initial conversations, Tina and James turned their attention to their mutual longing for more romance between them, and they committed to spending more quality time together.
For couples busy with kids, staying romantically and sexually engaged with each other can be a challenge. And it doesn't have to preclude the occasional crush. What's key is that when you feel the pull of someone other than your partner, you notice and appreciate, rather than intensify and pursue. If you can mentally enjoy the electricity of everyday encounters while still maintaining firm boundaries, it can help keep excitement alive at home.
Daphne de Marneffe, Ph.D., is a psychologist in the San Francisco Bay Area and the author of 'Maternal Desire: On Children, Love, and the Inner Life' and 'The Rough Patch: Marriage and the Art of Living Together.' Follow her on Twitter @DaphnedeMarneff.
This article originally appeared in Parents Magazine as 'How to Handle a Crush (Honestly).'
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Wouldn't you like to have a well-mannered dog as a member of your family? Most people would, but whether we like to admit it or not, most of our canine friends are in need of help when it comes to learning what we consider to be appropriate behavior. They do not always understand "our" rules and expectations, and they do not always understand what we are asking them to do. This lack of communication can cause frustration for both you and your pet, and this can unfortunately lead to bigger problems in the future. Providing training in order to give your dog the gift of understanding is the mission at Colonial Dog School. Only positive training methods are used because your best friend deserves it!
With group classes, in-home private lessons, behavior modification programs for both dogs and cats, plus evening, daytime and weekend hours, it couldn't be more convenient. All dogs can learn, and we take pride in making it an enjoyable experience for all involved.


Peter did all of the cabinetry in our house. He did the kitchen and bathrooms. His work was ok but he never came back to fix anything. He said he would but never showed up. Our cabinets have scratches on the fronts and have never been repaired. We were very dissatisfied.
peter ha millworking has not added any photos.