text:
\section{Introduction}

S.~Schanuel proposed the following conjecture while attending a course given by
S.~Lang at Columbia University in the 1960's. Most of the known results in the
transcendental number theory about the values of the exponential function are
encompassed by Schanuel's conjecture, and they can be derived as its consequence.

\begin{conj}[Schanuel]\label{Schanuel}
Let $x_1,\ldots,x_n \in {\mathbb C}$ be such that they are linearly independent
over ${\mathbb Q}$. Then the transcendence degree of the field
$$
{\mathbb Q}(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})
$$
over ${\mathbb Q}$ is at least $n$.
\end{conj}

For example, C.~Cheng et al. have shown in \cite{CDHHKMMMW} how to derive from
\conjref{Schanuel}, the linear disjointness of the two fields constructed over
${\mathbb Q}$ by adjoining repeatedly the algebraic closure of the field generated by the values
of the exponential and logarithm functions respectively.

The only known cases of \conjref{Schanuel} are $n=1$ and
$x_1,\ldots,x_n \in \overline{\mathbb Q}$ for general $n$. The $n=1$ case is a consequence
of the Hermite-Lindemann theorem, whereas the latter case is known as
the Lindemann-Weierstrass theorem. But these two special cases were
known much before the inception of this conjecture.

Schanuel's conjecture has been generalised to various other
contexts. The elliptic analogue of Schanuel's conjecture is well-studied.
Let $\Lambda$ be a lattice in ${\mathbb C}$ and $\wp$ denote the associated
Weierstrass $\wp$-function,
$$
\wp(z)=\wp(\Lambda;z):= \frac{1}{z^2}+ \sum_{\omega \in \Lambda \setminus \{0\}}
\left( \frac{1}{(z-\omega)^2} - \frac{1}{\omega^2} \right)
$$
for $z \in {\mathbb C} \setminus \Lambda$. The Weierstrass $\wp$-function is an
elliptic function with double poles at the points of $\Lambda$ and
holomorphic in ${\mathbb C} \setminus \Lambda$. Moreover, for all
$z \in {\mathbb C} \setminus \Lambda$, we have the relation
$$
\wp'(z)^2=4 \wp(z)^3 - 60 G_4(\Lambda) \wp(z) - 140 G_6(\Lambda).
$$
Here for $k \ge 2$,
$G_{2k}(\Lambda):= \sum_{\omega \in \Lambda \setminus \{0\}} \omega^{-2k}$
is the associated Eisenstein series of weight $2k$.
Let $g_2=60 G_4(\Lambda)$ and $g_3=140 G_6(\Lambda)$.
Then the modular invariant $j(\Lambda)$ is defined by
$$
j(\Lambda):=1728 \frac{g_2^3}{g_2^3-27g_3^2}.
$$
and the associated Weierstrass $\zeta$-function is defined by
$$
\zeta(z)=\zeta(\Lambda;z):= \frac{1}{z}+ \sum_{\omega \in \Lambda \setminus \{0\}}
\left( \frac{1}{z-\omega} + \frac{1}{\omega}+ \frac{z}{\omega^2} \right),
$$
where the series above converges absolutely and uniformly in any compact subset of
${\mathbb C} \setminus \Lambda$. Thus it is holomorphic  in ${\mathbb C} \setminus \Lambda$.
If $\omega_1,\omega_2$ denote the fundamental periods of $\Lambda$, then
the quasi-periods $\eta_1,\eta_2$ are defined by $\eta_i:=\zeta(z+\omega_i)-\zeta(z)$
for $i=1,2$. With these notations, the elliptic Schanuel conjecture reads as follows (see \cite{CB}) :

\begin{conj}[elliptic Schanuel]\label{eS}
Let $\Lambda$ be a lattice and $\wp,\zeta$ denote the associated
Weierstrass functions. Let $K$ be the field of endomorphisms of $\Lambda$
and $x_1,\ldots,x_n \in {\mathbb C} \setminus \Lambda$ such that they are
linearly independent over $K$. Then
$$
\text{\rm trdeg}_{\mathbb Q} {\mathbb Q}(g_2,g_3,\omega_1,\omega_2,\eta_1,\eta_2, x_1,\ldots,x_n,
\wp(x_1),\ldots,\wp(x_n),\zeta(x_1),\ldots,\zeta(x_n)) \ge 2n+ \frac{4}{[K:{\mathbb Q}]}.
$$
\end{conj}

Often the weaker statement
\begin{equation}\label{weak-elliptic}
\text{\rm trdeg}_{\mathbb Q} {\mathbb Q}(g_2,g_3, x_1,\ldots,x_n, \wp(x_1),\ldots,\wp(x_n)) \ge n
\end{equation}
is also considered for application of the elliptic Schanuel conjecture.
Here also the $n=1$ case is known and it can be deduced as a consequence of a
more general theorem of T. Schneider and S. Lang about transcendental values
of meromorphic functions. The analogue of Lindemann-Weierstrass theorem when
the Weierstrass $\wp$-function with algebraic invariants $g_2,g_3$
has complex multiplication, was proved independently by the first author
\cite{PP} and G. W\"ustholtz \cite{GW}.

We also have A. Grothendieck's period conjecture for an abelian variety $A$, defined
over $\overline{\mathbb Q}$. It states that the transcendence degree of the period matrix of $A$ is
the same as the dimension of the associated Mumford-Tate group. From the work
of P. Deligne \cite[Cor. 6.4, p.76]{PD} one gets that this dimension is at least an upper bound
for the transcendence degree of the period matrix. Y. Andr\'e \cite[Chap. 23]{YA} suggested
a more general conjecture which is now known as the generalised period conjecture.
This concerns periods of $1$-motives, which are defined over a subfield of ${\mathbb C}$, not
necessarily algebraic. In fact, Grothendieck's period conjecture can be seen as
a special case of Andr\'e's conjecture, using Deligne's work. Further,
C. Bertolin \cite{CB} has shown that this generalised period conjecture includes
Schanuel's conjecture as a special case.

We now consider the following weaker version of Andr\'e's generalised period
conjecture. Let $A({\mathbb C})$ be an abelian variety of dimension $g$ defined over $\overline{\mathbb Q}$
and $\exp_A: {\mathbb C}^g \to A({\mathbb C})$ denote the exponential map,
which is periodic with respect to a lattice $\Lambda_A$.
Let $\omega_1,\dots,\omega_g$ be a basis of the holomorphic
differential 1-forms and $\eta_1,\dots,\eta_g$ be a basis of the meromorphic
differential 1-forms with residue $0$ on $A$. Next let
$\gamma_1,\dots,\gamma_{2g}$ be a basis of the homology of $A$.
So the matrix of period $\tilde\Lambda_A$ is the $2g\times 2g$ matrix with entries
$\int_{\gamma_j}\omega_i$ and $\int_{\gamma_j}\eta_i$, $i=1,\dots,g$, $j=1,\dots,2g$,
while the matrix of the lattice $\Lambda_A$ is the $g\times 2g$ matrix with entries $\int_{\gamma_j}\omega_i$.

Let $u\in{{\mathbb C}}^g$ and $y = \exp_A(u)$. The relevant 1-motive here is $M = [{{\mathbb Z}} \to A]$, $1 \mapsto y$,
which is defined over ${\overline{\mathbb Q}}(\exp_A(u))$. Let $MT(M)$ denote its Mumford-Tate group.
The periods of the 1-motive $M$ include the periods of $A$ and the components of $u$
i.e. the numbers $\int_0^u\omega_i$, and also the integrals $\int_0^u\eta_i$. Let
$\zeta_A(u)$ denote the vector with components $\int_0^u\eta_i$, $i=1,\dots,g$.

\begin{conj}[weak abelian Schanuel]\label{aS}
With the notations as above,
let $\overline{\mathbb Q}(\tilde\Lambda_A)$ denote the field generated by the periods
and quasi-periods over $\overline{\mathbb Q}$. Let $u\in{{\mathbb C}}^g$ and $H$ be the smallest
algebraic subgroup of $A$ containing the point $\exp_A(u)$. Then
\begin{equation}\label{qp-2}
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u,\zeta_A(u)) \ge 2\dim(H).
\end{equation}
\end{conj}

In a discussion, Daniel Bertrand told us that it is possible to deduce
\conjref{aS} from Andr\'e's conjecture based on \cite[Proposition 1]{YA1}
and extending the methods of \cite{CB}. With due consent, we
reproduce an indication of this argument here.

The generalised period conjecture of Andr\'e for $M$ implies
$$
\text{\rm trdeg}_{\overline{\mathbb Q}}\overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u,\zeta_A(u)) \ge \dim MT(M).
$$
Since $A$ is defined over $\overline{\mathbb Q}$, Grothendieck's conjecture (which is a particular case
of Andr\'e's conjecture) gives $\text{\rm trdeg}_{\overline{\mathbb Q}}\overline{\mathbb Q}(\tilde\Lambda_A) = \dim(MT(A))$. Hence
\begin{equation}\label{tr-deg}
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u,\zeta_A(u))
\ge \dim MT(M) - \dim MT(A).
\end{equation}

If $U(M)$ denotes the unipotent radical of $MT(M)$, then $MT(M)/U(M)$ is the
(reductive) group $MT(A)$. Hence the right hand side of \eqref{tr-deg}
equals $\dim(U(M))$. Furthermore, by \cite[Proposition 1]{YA1}, $U(M)$ is known
to be equal to $H^1_{Betti}(H^\circ)$, where $H^\circ$ is the connected
component of $H$ containing the trivial element. Thus
$\dim(U(M))=2\dim(H)$ and we therefore get \eqref{qp-2}, i.e.
\begin{equation*}
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u,\zeta_A(u)) \ge 2\dim(H).
\end{equation*}
Thus \conjref{aS} follows from the generalised period conjecture of Andr\'e.

\begin{rmk}\label{rmk-aS-qp}\rm
Note that
$$
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u,\zeta_A(u))
\le \text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,\exp_{A}(u),u) + \dim(H).
$$
Hence from \eqref{qp-2} we can deduce,
\begin{equation}\label{qp-1}
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} \overline{\mathbb Q}(\tilde\Lambda_A,u, \exp_A(u)) \ge \dim(H),
\end{equation}
and therefore
\begin{equation}\label{qp-0}
\text{\rm trdeg}_{\overline{\mathbb Q}(\Lambda_A)} \overline{\mathbb Q}(\Lambda_A, u, \exp_A(u)) \ge \dim(H).
\end{equation}
So we get \eqref{qp-1} and \eqref{qp-0} as consequence of \eqref{qp-2}.
\end{rmk}

\begin{rmk}\rm
\conjref{aS} can further be considered for abelian varieties defined over a
subfield of ${\mathbb C}$, not necessarily algebraic. G. Vall\'ee \cite{GV} has formulated the
relevant statement from Andr\'e's generalised period conjecture, and his statement
includes \conjref{aS} as a special case.
\end{rmk}

We therefore have supporting evidence for considering \conjref{aS}.
With this conjecture in place we want to extend the results in C.~Cheng
et al.~\cite{CDHHKMMMW} to this setting, that is prove the linear disjointness of
the two fields defined below. For the sake of completeness we recall the definition
of linear disjointness (see \cite[Chap. VIII, \S3]{SL}).

\begin{defn}
Let $F$ be a field and $F_1,F_2$ two of its field extensions contained
in a larger field $G$. Then $F_1$ is said to be linearly disjoint ({\it resp.} free) from
$F_2$ over $F$ if any finite $F$-linearly ({\it resp.} algebraically) independent subset
of $F_1$ is also $F_2$-linearly ({\it resp.} algebraically) independent (as a subset of $G$).
\end{defn}

Though the above definition is asymmetric, it can be shown that the property
of being linearly disjoint ({\it resp.} free) is actually symmetric for $F_1$ and $F_2$.
It is easy to see that if $F_1$ and $F_2$ are linearly disjoint over $F$ then
$F_1 \cap F_2 = F$. Also if $F_1$ and $F_2$ are linearly disjoint over $F$ then
one can deduce that $F_1$ and $F_2$ are free over $F$
(see \cite[Chap. VIII, Prop. 3.2]{SL}). The converse is true in special
cases (see \cite[Chap. VIII, Theorem 4.12]{SL} and \lemref{l3} below).
The property of being free is also called as $F_1$ and $F_2$
being algebraically independent over $F$.

We now setup the relevant notations for our theorem. Recall, $A$ is an abelian
variety over $\overline{\mathbb Q}$ and $\tilde\Lambda_A$ denote the matrix of periods.
We consider two recursively defined sets ${\mathcal E},{\mathcal L}$. Let us define
\footnote{As previously for the field ${\mathbb Q}$, we denote with a bar $\overline{K}$
the algebraic closure in ${\mathbb C}$ of a subfield $K\subset{\mathbb C}$.}
$$
{\mathcal E} =\bigcup_{n \ge 0} {\mathcal E}_n \ \ \text{and} \ \ {\mathcal L} =\bigcup_{n \ge 0} {\mathcal L}_n
$$
where ${\mathcal E}_0=\overline{\mathbb Q}$, ${\mathcal L}_0 = \overline{{\mathbb Q}(\tilde\Lambda_A)}$ and for $n \ge 1$,
$$
{\mathcal E}_n = \overline{{\mathcal E}_{n-1}(\{\text{components of }\exp_A(u) : u \in {\mathcal E}_{n-1}^g\})}
$$
and
$$
{\mathcal L}_n = \overline{{\mathcal L}_{n-1}(\{\text{components of }u, \zeta_A(u) : \exp_A(u) \in A({\mathcal L}_{n-1})\})}.
$$
Often, as in \cite{GV}, for a point $P$ of $A({\mathbb C})$, one uses $\log_A P$
({\it resp.}  $\tilde\log_A P$) to denote the point
$u=\exp_A^{-1}(P)$ ({\it resp.} $(u,\zeta_A(u))$). By convention we take
$\exp_A(\tilde\log_A P)=\exp_A(\log_A P)=P$. Given a suitable set $S$,
we denote $\exp_A(S)$ ({\it resp.} $\log_A(S)$, $\tilde\log_A(S)$) the set of
elements $\exp_A(u)$ for $u\in S$ ({\it resp.} $\log_A(P)$, $\tilde\log_A(P)$
for $P\in S$). Then, by induction, one can see that for $n \ge 1$,
$$
{\mathcal E}_n = \overline{{\mathbb Q}(\text{components of } \exp_A({\mathcal E}_{n-1}^g))}
\ \ \text{and} \ \
{\mathcal L}_n = \overline{{\mathbb Q}(\Lambda_A,\text{components of } \tilde\log_A(A({\mathcal L}_{n-1})))}.
$$
Below we state our main theorem about ${\mathcal E}$ and ${\mathcal L}$,
where we take the field $G$ in Definition 1 to be ${\mathbb C}$.

\begin{thm}\label{lin-dis}
If \conjref{aS} is true, then ${\mathcal E}$ and ${\mathcal L}$ are linearly disjoint over $\overline{\mathbb Q}$.
Since ${\mathcal E},{\mathcal L}$ are algebraically closed, it is equivalent to say that they are
algebraically independent over $\overline{\mathbb Q}$.
\end{thm}

Proof of this theorem follows the structure of the proof of the main theorem of
\cite{CDHHKMMMW}, but a good part of it differs towards the end of our proof.
It will be interesting to consider similar problem for semi-abelian varieties
to encompass the cases treated here and in \cite{CDHHKMMMW}.

\begin{rmk}\rm
In \thmref{lin-dis}, one can consider abelian varieties defined over a
subfield of ${\mathbb C}$, not necessarily algebraic, as considered by G. Vall\'ee \cite{GV}.
However, the statement is not true as it stands. In \S4.1, we exhibit an elliptic curve
such that ${\mathcal E}_1 \cap {\mathcal L}_1 \supsetneq {\mathcal E}_0 \cap {\mathcal L}_0$ for the natural
candidate of ${\mathcal E}_0$ and ${\mathcal L}_0$. The difficulty in this case is
coming from the fact that we no more have the equality in Grothendieck's period
conjecture (also see \cite[Chap. 23.4]{YA}).
\end{rmk}

\begin{rmk}
\rm We get an immediate application
of \thmref{lin-dis} for elliptic curves over $\overline{\mathbb Q}$. See \S4 for more details.
\end{rmk}

\section{Intermediate lemmas}

In this section we deduce some intermediate results  to prove \thmref{lin-dis}.

\begin{lem}\label{l3}
Let $K_1,K_2$ be two sub-fields of ${\mathbb C}$, which are algebraically closed over $K_1 \cap K_2$.
Then they are algebraically independent over $K_1 \cap K_2$ if and only if they are linearly
disjoint over $K_1 \cap K_2$.
\end{lem}

\begin{proof}
It follows immediately from Theorem 4.12 and Proposition 3.2 of \cite[Chap. VIII]{SL}.
\end{proof}

\begin{lem}\label{l1}
Let $A_n$ be a finite subset of ${\mathcal E}_n$. Then there exists
a finite subset ${\mathcal A}$ of ${\mathcal E}_{n-1}$ such that
${\mathcal A} \cup A_n$ is algebraic over $\overline{\mathbb Q}(\text{components of }\exp_A({\mathcal A}^g))$.
\end{lem}

\begin{proof}
Since $A_n \subset {\mathcal E}_n$, for each $x \in A_n$, there exists a finite subset $C_x$ of
${\mathcal E}_{n-1}$ such that $x$ is algebraic over $\overline{\mathbb Q}(\text{components of }\exp_A(C_x^g))$.
Let $A_{n-1}:= \cup_{x \in A_n} C_x \subset {\mathcal E}_{n-1}$. Then $A_{n-1}$ is a finite
subset of ${\mathcal E}_{n-1}$.

We repeat the process to get sets $\{A_i\}_{0 \le i \le n-2}$
such that for each $i$, $A_i$ is a finite subset of ${\mathcal E}_i$ and $A_{i+1}$ is algebraic
over $\overline{\mathbb Q}(\text{components of }\exp_A(A_i^g))$. We take
${\mathcal A}:= \cup_{0 \le i \le n-1} A_i \subset {\mathcal E}_{n-1}$. Then
${\mathcal A} \cup A_n$ is algebraic over $\overline{\mathbb Q}(\text{components of }\exp_A({\mathcal A}^g))$.
\end{proof}

\begin{lem}\label{l2}
Let $C$ be a finite subset of ${\mathcal L}_n$. Then there exists
a finite set ${\mathcal C} \subset {\mathcal L}_{n}^{2g}$ with $\exp_A({\mathcal C}) \subset A({\mathcal L}_{n-1})$
such that the set $\{\text{components of }\exp_A({\mathcal C})\} \ \cup \ C$ is algebraic over
the field $\overline{\mathbb Q}(\tilde\Lambda_A,\text{components of }{\mathcal C})$.
\end{lem}

\begin{proof}
Since $C \subset {\mathcal L}_n$, for each $y \in C$, there exists a finite subset $D_y$ of
$A({\mathcal L}_{n-1})$ such that $y$ is algebraic over
$\overline{\mathbb Q}(\tilde\Lambda_A,\text{components of }\tilde\log_A(D_y))$.
Define $B_{n-1}:=\tilde\log_A(\cup_{y \in C}D_y)$, so that
$C$ is algebraic over $\overline{\mathbb Q}(\tilde\Lambda_A,\text{components of }B_{n-1})$.
Then $B_{n-1}\subset \tilde\log_A(A({\mathcal L}_{n-1})) \subset {\mathcal L}_{n}^{2g}$.
Hence $\exp_A(B_{n-1}) \subset A({\mathcal L}_{n-1})$,
i.e. the components of $\exp_A(B_{n-1})$ is a finite subset of ${\mathcal L}_{n-1}$.

We repeat this process for the components of $\exp_A(B_{n-1})$
in place of $C$ and so on, to get sets $\{B_i\}_{0 \le i \le n-2}$
such that for each $i$, $\exp_A(B_{i}) \subset A({\mathcal L}_{i})$ and components of
$\exp_A(B_{i+1})$ is algebraic over $\overline{\mathbb Q}(\tilde\Lambda_A,\text{components of }B_i)$.
We set ${\mathcal C}:= \cup_{0 \le i \le n-1} B_i$ to complete the proof.
\end{proof}

\section{Proof of \thmref{lin-dis}}

In view of \lemref{l3}, we show that ${\mathcal E}$ and ${\mathcal L}$ are algebraically independent over $\overline{\mathbb Q}$.
It is enough to prove that ${\mathcal E}_m$ and ${\mathcal L}_n$ are algebraically independent
over $\overline{\mathbb Q}$ for all $m,n$.

Now suppose that there exists a pair
$(m,n) \in {\mathbb N}^2$ such that ${\mathcal E}_m$ and ${\mathcal L}_n$ are not algebraically
independent over $\overline{\mathbb Q}$. We choose such a pair $(m,n)$ with the property that
if $(a,b) < (m,n)$, then ${\mathcal E}_a$ and ${\mathcal L}_b$ are algebraically independent over $\overline{\mathbb Q}$.
Here the ordering `$<$' is the ordering on ${\mathbb N}^2$ where $(a,b) < (m,n)$ if and only if either
$a \le m$ and $b < n$, or $a < m$ and $b \le n$. Clearly $m,n \ge 1$.

As ${\mathcal E}_m$ and ${\mathcal L}_n$ are not algebraically independent over $\overline{\mathbb Q}$,
there exists an element $\ell$ of ${\mathcal L}_n \setminus \overline{\mathbb Q}$
which is algebraic over ${\mathcal E}_m$ i.e. there exists
a finite subset $\{e_1,\ldots,e_k\}$ of non-zero elements of ${\mathcal E}_m$
such that $\ell$ is algebraic over $\overline{\mathbb Q}(e_1, \ldots, e_k)$.

Now from \lemref{l1}, we know that there exists
a finite subset ${\mathcal A}$ of ${\mathcal E}_{m-1}$ such that
${\mathcal A} \cup \{e_1, \ldots, e_k\}$ is algebraic over
$\overline{\mathbb Q}(\text{components of }\exp_A({\mathcal A}^g))$.
Similarly by \lemref{l2}, we have a finite set
${\mathcal C} \subset {\mathcal L}_{n}^{2g}$ with $\exp_A({\mathcal C}) \subset A({\mathcal L}_{n-1})$
such that $\{\text{components of }\exp_A({\mathcal C})\} \ \cup \ \{ \ell \}$
is algebraic over $\overline{\mathbb Q}(\tilde\Lambda_A,\text{components of }{\mathcal C})$.
Let $|{\mathcal A}^g|=n_1$ and $|{\mathcal C}|=n_2$.

Define ${\bf u_1}:=(u : u \in {\mathcal A}^g) \in TA^{n_1}$
by concatenating elements of ${\mathcal A}^g$ one after another. Here $TA^{n_1}$
denotes the tangent space of the abelian variety $A^{n_1}$. Let $A_1$
be the smallest algebraic subgroup of $A^{n_1}$
containing the point $\exp_{A^{n_1}}({\bf  u_1})$.
Similarly define ${\bf  u_2}:=( u :  u \in {\mathcal C}) \in TA^{n_2}$
Let $A_2$ be the smallest algebraic subgroup of $A^{n_2}$
containing the point $\exp_{A^{n_2}}({\bf  u_2})$.

Let
$$
K_1:=\overline{{\mathbb Q}({\bf  u_1},\exp_{A^{n_1}}({\bf  u_1}),\zeta_{A^{n_1}}({\bf  u_1}))}
\ \text{and} \
K_2:=\overline{{\mathbb Q}(\tilde\Lambda_A,{\bf  u_2},\exp_{A^{n_2}}({\bf  u_2}),\zeta_{A^{n_2}}({\bf  u_2}))}
$$
Then $\overline{\mathbb Q}(e_1, \ldots, e_k) \subset K_1 =
\overline{{\mathbb Q}(\exp_{A^{n_1}}({\bf  u_1}),\zeta_{A^{n_1}}({\bf  u_1}))}$ by \lemref{l1}
and $\overline{\mathbb Q}(\ell) \subset K_2 =
\overline{{\mathbb Q}(\tilde\Lambda_A,{\bf  u_2},\zeta_{A^{n_2}}({\bf  u_2}))}$
by \lemref{l2}.
By \conjref{aS}, we get
$$
\text{\rm trdeg}_{\overline{\mathbb Q}} K_1 \ge 2\dim(A_1) \ \text{and} \ \text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_2 \ge 2\dim(A_2)
$$
for $i=1,2$. Since $\exp_{A^{n_1}}({\bf  u_1}) \in A_1$
and ${\bf  u_2} \in TA_2$, we get that
$$
\text{\rm trdeg}_{\overline{\mathbb Q}} K_1 \le 2\dim(A_1)
$$
and
$$
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_2 \le 2\dim(TA_2) = 2\dim(A_2).
$$
Hence
$$
\text{\rm trdeg}_{\overline{\mathbb Q}} K_1 = 2\dim(A_1) \ \text{and} \ \text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_2 = 2\dim(A_2).
$$
We want to show that
\begin{equation}\label{to-prove}
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_1 K_2 = 2\dim(A_1 \times A_2)
=\text{\rm trdeg}_{\overline{\mathbb Q}} K_1+\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_2.
\end{equation}
Adding $\text{\rm trdeg}_{\overline{\mathbb Q}} \overline{\mathbb Q}(\tilde\Lambda_A)$ to both sides of \eqref{to-prove}, we would get
$$
\text{\rm trdeg}_{\overline{\mathbb Q}} K_1 K_2 =\text{\rm trdeg}_{\overline{\mathbb Q}} K_1+\text{\rm trdeg}_{\overline{\mathbb Q}} K_2.
$$
This would prove that the fields $K_1$ and $K_2$ are algebraically
independent over $\overline{\mathbb Q}$. We will thus get a contradiction to our assumption.

Define ${\bf  u_3}:=({\bf  u_1},{\bf  u_2})$.
Let $B$ be the smallest algebraic subgroup of $A^{n_1+n_2}$
containing the point $\exp_{A^{n_1+n_2}}({\bf  u_3})$.
Then by \conjref{aS}, we get
$$
\text{\rm trdeg}_{\overline{\mathbb Q}(\tilde\Lambda_A)} K_1 K_2 \ge 2\dim(B).
$$
Thus we are reduced to prove that $\dim(B)=\dim(A_1) + \dim(A_2)$.
If they are torsion subgroups then we have nothing to prove.
So we assume that at least one of $A_1$ and $A_2$ is not a torsion subgroup.

We first assume $A$ to be simple. For ${\bf  u_i} \in TA_i \hookrightarrow TA^{n_i}$,
we choose a basis and write ${\bf  u_i}=(u_{i1},\ldots,u_{in_i})$ with $u_{ij} \in TA$
for $i=1,2$ and $j=1,\ldots,n_i$. We consider any of the defining relation
for $TB$,
\begin{equation}\label{def-TB}
\sum_{1\le j \le n_1} \delta_{1j} x_{1j} - \sum_{1 \le j \le n_2} \delta_{2j} x_{2j}=0,
\end{equation}
where $\delta_{ij} \in End(A)$ for $i=1,2$ and $j=1,\ldots,n_i$. Let $u:=
\sum_{1\le j \le n_1} \delta_{1j} u_{1j} = \sum_{1 \le j \le n_2} \delta_{2j} u_{2j}$.
Then $u \in {\mathcal E}_{m-1}^g \cap {\mathcal L}_{n}^g$ as each $u_{1j} \in {\mathcal A}^g$
and $u_{2j} \in {\mathcal C}$. Thus $u \in \overline{\mathbb Q}^g$, by the choice of $m,n$.

On the other hand $\exp_A(u)=\sum_{1\le j \le n_1} \delta_{1j} \exp_A(u_{1j})
= \sum_{1 \le j \le n_2} \delta_{2j} \exp_A(u_{2j})$. For similar reason
$\exp_A(u) \in A({\mathcal E}_m) \cap A({\mathcal L}_{n-1})$. Again ${\mathcal E}_m \cap {\mathcal L}_{n-1} =\overline{\mathbb Q}$,
and hence $\exp_A(u) \in A(\overline{\mathbb Q})$. Thus,
$$
\text{\rm trdeg}_{\overline{\mathbb Q}} \overline{\mathbb Q}(u, \exp_A(u))=0
$$
Now if $H$ is the smallest algebraic subgroup of $A$ containing $\exp_A(u)$,
then $\dim(H)=0$, by \conjref{aS}, i.e. $H$ is torsion subgroup.
In particular, $n u \in \overline{\mathbb Q}^g$ and $\exp_A(nu)=0$, for a suitable integer $n$.

Thus, $\exp_A(nu)=\sum_{1\le j \le n_1} \delta_{1j} \exp_A(nu_{1j})
= \sum_{1 \le j \le n_2} \delta_{2j} \exp_A(nu_{2j})=0$. Now
$A_i$ is the smallest algebraic subgroup of $A^{n_i}$
containing the point $\exp_{A^{n_i}}({\bf  u_i})$ for $i=1,2$.
Thus, $\sum_{1\le j \le n_i} \delta_{ij} \exp_A(nx_{ij})=0$
for any point $\exp_{A^{n_i}}({\bf  x_i}) \in A_i$ for $i=1,2$.
Since at least one of $A_1$ or $A_2$ is not a torsion subgroup,
there exists $i\in\{1,2\}$, such that $\sum_{1\le j \le n_i} \delta_{ij} n x_{ij}=0$ on $TA_i$.
Hence $\sum_{1\le j \le n_i} \delta_{ij} x_{ij}=0$ on $TA_i$, and therefore
defining relations for $TB$ separate into disjoint relations defining $TA_1$
and $TA_2$. Thus we have $\dim(B)=\dim(A_1) + \dim(A_2)$.

Now we treat the case when $A$ is not a simple abelian variety. In this case,
we would like to write down the generic form of a defining relation for $TB$
and we show that it is a collection of relations of the form \eqref{def-TB}.
Then the proof will follow as above.

We suppose that $A^{n_1+n_2}$ is isogenous to $V_1^{r_1} \times \cdots \times V_l^{r_l}$,
where for $1 \le i \neq j \le l$, $V_i$ is an abelian variety not isogenous to $V_j$. Thus,
the tangent space $TA^{n_1+n_2}$ has the form
$$
\underbrace{TV_1\oplus \cdots \oplus TV_1}_{r_1 \text{ times}}
\oplus \cdots \oplus \underbrace{TV_l\oplus \cdots \oplus TV_l}_{r_l \text{ times}}.
$$
Now $B \subset A_1 \times A_2 \subset A^{n_1+n_2}$. Hence,
$A^{n_1+n_2}/B$ can be written in the form $V_1^{s_1} \times \cdots \times V_l^{s_l}$,
where for each $1 \le i \le l$, $s_i \le r_i$. Now $B$ is the kernel
of the natural map from $A^{n_1+n_2} \to A^{n_1+n_2}/B$.
So for this we find the corresponding map
$V_1^{r_1} \times \cdots \times V_l^{r_l} \to V_1^{s_1} \times \cdots \times V_l^{s_l}$,
for which $B$ is isogenous to the kernel.

Such a map is expressed as a block diagonal matrix of order
$(s_1+\cdots+s_l,r_1+\cdots+r_l)$. This matrix has diagonal blocks of order
$(s_i,r_i)$ with entries from $End(V_i)$, for each $1 \le i \le l$.
Now such a matrix acts on an element $(x_{11}, \ldots, x_{1r_1}, \ldots , x_{l1}, \ldots, x_{l r_l})$
of $TA^{n_1+n_2}$, written as a column. Under this action an element of $TB$ is mapped
to the zero vector i.e.
\begin{equation}\label{def-TB-2}
\left( \begin{array}{c c c c}
(\delta_{1jk})_{s_1 \times r_1} & (0)_{s_1 \times r_2} & \cdots & (0)_{s_1 \times r_l}\\
(0)_{s_2 \times r_1} & (\delta_{2jk})_{s_2 \times r_2} & \cdots & (0)_{s_2 \times r_l}\\
\vdots & \vdots & \ddots & \vdots\\
(0)_{s_l \times r_1} & (0)_{s_l \times r_2} & \cdots & (\delta_{ljk})_{s_l \times r_l}
\end{array} \right)
\left( \begin{array}{c} {{\bf x}_{1}} \\ {{\bf x}_{2}}\\ \vdots \\ {{\bf x}_{l}}
\end{array} \right)
 = {\bf 0},
\end{equation}
where for $1 \le i \le l$, ${{\bf x}_{i}}=\left( \begin{array}{c} {x_{i1}} \\ \vdots \\ x_{i r_i}
\end{array} \right)$.
Since the matrix is block diagonal, we get that
a defining relation for $TB$ is given as the relations of the form
$$
\sum_{k=1}^{r_i} \delta_{ijk} x_{ik} = 0
$$
for some $\delta_{ijk} \in End(V_i)$ with $1 \le i \le l, 1\le j \le s_i$ and $1 \le k \le r_i$.
This completes the proof.

\subsection{A special case}

Let us consider the following two sub-fields of ${\mathcal E}$ and ${\mathcal L}$ :
$$
{\mathcal E}' =\bigcup_{n \ge 0} {\mathcal E}_n' \ \ \text{and} \ \ {\mathcal L}' =\bigcup_{n \ge 0} {\mathcal L}_n'
$$
where ${\mathcal E}_0'=\overline{\mathbb Q}$, ${\mathcal L}_0' = \overline{\mathbb Q}(\Lambda_A)$ and for $n \ge 1$,
$$
{\mathcal E}_n' = \overline{{\mathcal E}_{n-1}'(\{\text{components of }\exp_A(u) : u \in {\mathcal E}_{n-1}^{'g}\})}
$$
and
$$
{\mathcal L}_n' = \overline{{\mathcal L}_{n-1}'(\{\text{components of }u : \exp_A(u) \in A({\mathcal L}_{n-1}')\})}.
$$
In fact, by induction, one can see that for $n \ge 1$,
$$
{\mathcal E}_n' = \overline{{\mathbb Q}(\text{components of } \exp_A({\mathcal E}_{n-1}^{'g}))}
\ \ \text{and} \ \
{\mathcal L}_n' = \overline{{\mathbb Q}(\Lambda_A,\text{components of } \exp_A^{-1}(A({\mathcal L}_{n-1}')))}.
$$
Arguments as in our proof  of \thmref{lin-dis} immediately yields the following :

\begin{thm}\label{lin-dis-1}
If \eqref{qp-0} is true, then ${\mathcal E}'$ and ${\mathcal L}'$ are linearly disjoint over $\overline{\mathbb Q}$.
\end{thm}

\section{An application}

Now we consider the two recursively defined sets ${\mathcal E},{\mathcal L}$
related to the Weierstrass $\wp$-function associated to a lattice
$\Lambda$ with algebraic invariants $g_2,g_3$, defined as follows
$$
{\mathcal E} =\bigcup_{n \ge 0} {\mathcal E}_n \ \ \text{and} \ \ {\mathcal L} =\bigcup_{n \ge 0} {\mathcal L}_n
$$
where ${\mathcal E}_0=\overline{\mathbb Q}, {\mathcal L}_0 = \overline{{\mathbb Q}(\Lambda)}$  and for $n \ge 1$,
$$
{\mathcal E}_n = \overline{{\mathcal E}_{n-1}(\{\wp(x) : x \in {\mathcal P}_{n-1} \setminus \Lambda\})}
\ \ \text{and} \ \
{\mathcal L}_n = \overline{{\mathcal L}_{n-1}(\{x : \wp(x) \in {\mathcal L}_{n-1} \cup \{\infty\}\})}.
$$
In fact, by induction, one can see that for $n \ge 1$,
$$
{\mathcal E}_n = \overline{{\mathbb Q}(\wp({\mathcal E}_{n-1} \setminus \Lambda))}
\ \ \text{and} \ \
{\mathcal L}_n = \overline{{\mathbb Q}(\wp^{-1}({\mathcal L}_{n-1} \cup \{\infty\}))}.
$$
As a corollary to \thmref{lin-dis}, we obtain the following result.

\begin{cor}\label{lin-dis-2}
If \conjref{eS} is true, then ${\mathcal E}$ and ${\mathcal L}$
are linearly disjoint over $\overline{\mathbb Q}$.
\end{cor}

In fact, in view of \thmref{lin-dis-1}, the conclusion of the above corollary also holds
if \eqref{weak-elliptic} is true. We know that when $g_2,g_3$ are algebraic,
$\omega \in \Lambda\setminus\{0\}$ is transcendental. Hence as a corollary
to \corref{lin-dis-2} we obtain that $\omega \notin {\mathcal E}$.
In particular, $\omega$ can not be a value of the $\wp$ function iterated
at an algebraic point.

\subsection{An example}

We now exhibit an example in the elliptic case which shows that
if the curve is not defined over $\overline{\mathbb Q}$, then the conclusion of our
\thmref{lin-dis} (or, \corref{lin-dis-2}) does not follow from the same
hypothesis.

For a fixed real algebraic irrational number $\alpha$, our aim is to find out a
lattice ${\mathbb Z} + {\mathbb Z} \tau$ such that $\wp(\tau;\alpha)=\wp({\mathbb Z}+{\mathbb Z}\tau;\alpha)$
can be written as a polynomial in $\tau$ with algebraic coefficients.

We first claim that we can find $\tau_0$ such that
$\frac{d}{d \tau} \wp(\tau;\alpha) |_{\tau=\tau_0} \neq \frac{\wp(\tau_0;\alpha)}{\tau_0}$.
If not, then we have
$\frac{d \wp(\tau;\alpha)}{\wp(\tau;\alpha)}|_{\tau=\tau_0} = \frac{d \tau}{\tau}|_{\tau=\tau_0}$
for all $\tau_0$ in the complex upper half plane. Thus $d (\log(\wp(\tau;\alpha)))|_{\tau=\tau_0}
=d (\log \tau)|_{\tau=\tau_0}$ i.e. $\wp(\tau_0;\alpha)=c_\alpha \tau_0$ for all $\tau_0$,
where $c_\alpha$ is a constant depending on $\alpha$. But this is not possible, as can be
checked from the $q$-expansion of $\wp$.

So we choose $\tau_0$ such that
$\frac{d}{d \tau} \wp(\tau;\alpha) |_{\tau=\tau_0} \neq \frac{\wp(\tau_0;\alpha)}{\tau_0}$
and denote the ratio $\frac{\wp(\tau_0;\alpha)}{\tau_0}$ by $\lambda_0$. We now choose
$\lambda$ close to $\lambda_0$ such that $\lambda \in \overline{\mathbb Q}$ and
$\frac{d}{d \tau} \wp(\tau;\alpha) |_{\tau=\tau_0} \neq \lambda$.
Consider the function $f(\tau)=\wp(\tau;\alpha)- \lambda \tau$. Then we get that
$f'(\tau)|_{\tau=\tau_0} \neq 0$. Hence $f$ has a local inverse at $\tau_0$,
say $g$, which is defined in a neibourghood of $f(\tau_0)$.
Choose $\beta \in \overline{\mathbb Q}$ sufficiently close to $f(\tau_0)$ and set $\tau_1=g(\beta)$.
Then $\beta=f(\tau_1)=\wp(\tau_1;\alpha)- \lambda \tau_1$. Thus $\wp(\tau_1;\alpha)$
can be written as a polynomial in $\tau_1$ with algebraic coefficients.

The elliptic Schanuel conjecture implies that $\tau_1$ is transcendental.
Indeed, if $\tau_1$ is a quadratic irrational, then the associated $j$ invariant
is algebraic. Hence $g_2,g_3$ are algebraically related and
$\eta_1,\eta_2$, satisfying Masser's relation, are algebraically dependent
over $\overline{{\mathbb Q}(g_2,g_3)}$. Now from \conjref{eS} we get a contradiction by
taking $n=1$ and $x_1=\alpha$. If $\tau_1$ is algebraic of degree
larger than 2, then \conjref{eS} gives a contradiction again for $n=1$ and
$x_1=\alpha$.

Now for this choice of $\tau_1$, we see that $\tau_1$ belongs to both ${\mathcal E}_1$ and ${\mathcal L}_1$,
where the tower of fields ${\mathcal E}_i$'s and ${\mathcal L}_i$'s are constructed as in the beginning of
this section, but with $\overline{\mathbb Q}$ replaced by the corresponding field of definition $\overline{{\mathbb Q}(g_2,g_3)}$.
However, we show below that $\tau_1$ is transcendental over ${\mathbb Q}(g_2,g_3)$.
This gives $\text{\rm trdeg}_{{\mathcal E}_0} {\mathcal E}_1 \cap {\mathcal L}_1 \ge 1$, which implies
${\mathcal E}_1 \cap {\mathcal L}_1 \neq {\mathcal E}_0$ and therefore ${\mathcal E}_1$ and ${\mathcal L}_1$
are not linearly disjoint over ${\mathcal E}_0$.

To prove that $\tau_1$ is transcendental over ${\mathbb Q}(g_2,g_3)$,
note that $1,\tau_1$ and $\alpha$ are ${\mathbb Q}$ linearly independent. Then the
elliptic Schanuel conjecture yields
$\text{\rm trdeg}_{\mathbb Q} {\mathbb Q}(g_2,g_3,1,\tau_1,\alpha,\wp(\tau_1;\alpha)) \ge 3$.
Now from our construction we see that
$\text{\rm trdeg}_{\mathbb Q} {\mathbb Q}(g_2,g_3,1,\tau_1,\alpha,\wp(\tau_1;\alpha))
=\text{\rm trdeg}_{\mathbb Q} {\mathbb Q}(g_2,g_3,\tau_1)$. Thus $\tau_1$ is transcendental
over ${\mathbb Q}(g_2,g_3)$.

\medskip

\noindent {\bf Acknowledgement:}
The authors would like to thank D. Bertrand for helpful
discussions and useful comments on an earlier version of this article.
The second and the third author would like to thank the Institut de
Math\'ematiques de Jussieu for hospitality during academic visits in the
frame of the IRSES Moduli and LIA. Research of the second author is also
supported by SERB-DST-NPDF grant vide PDF/2016/002938.

redpajama_set_name: RedPajamaArXiv


text:
Q: Deadlock prevention in a complicated system Suppose I have a system with 7 processes and 3 resources A, B, C. Suppose every process needs to read & write a different non-empty subset of {A, B, C}, how do I efficiently prevent deadlocks? i.e. cant put a mutex on the 3 resources together as a whole.

A: If the mutex support non-blocking acquisition (often referred as trylock), a deadlock-avoiding solution can be implemented without lock ordering/hierarchy. The algorithm is as follows:

*

*Acquire the lock for the first resource; here waiting is allowed.

*Once the first lock is acquired, try acquiring another one with try-lock.

*If try-lock fails (indicating that a resource is busy), release all previously acquired locks and go to the step 1.

*If try-lock succeeds, repeat from step 2 for all remaining resources.

Deadlock avoidance is achieved through non-blocking acquisition (never wait for a lock if you already hold one) and cooperation (be a good neighbor and release whatever you already acquired if you cannot continue without blocking).
Such a strategy is used in e.g. Boost for acquiring multiple locks at once. Actually, in Boost it's even more advanced: after releasing locks at step 3, they start waiting from the lock that was busy at the previous attempt.
Also, look at these SO topics: Acquire a lock on two mutexes and avoid deadlock, Multiple mutex locking strategies and why libraries don't use address comparison

A: These kind of problems are usually solved when we organize resources into some kind of hierarchy. You have three resources, {A, B, C}, and we can decide that A is on the top of hierarchy, and C is at the bottom. Now there are two approaches:
Pessimistic: if I need resource X then I must also acquire all resources that are above X, even if I don't need them at this time. The reason is that I might need them at some other time, while still holding the resource X.
Optimistic: if I need resource X then acquire it. If later I need resource X-below, then acquire it as well, no problem there. But, if after X I need resource X-above then tough luck, I can't have it while holding resource X, but I could release resource X, and then first acquire X-above, and after that X again.
If all processes respect the hierarchy of resources then deadlock cannot occur.

A: You need to decide on an order the resources will be acquired and then always acquire them in the same order.
Using your example, the resources should always be acquired in the order A, B, C. So, for example:
P1 => A
P2 => A C
P3 => C
P4 => B C
P5 => B
P6 => A B
P7 => A B C

This way there will never be a scenario where all the processes are holding one resource and waiting on a resource that another process has.
In a more simple scenario:
P1 => A B
P2 => B C

Here both process will get their first resource, but process P1 will block because P2 has B, However, P2 will be able to acquire C and will make progress. At some point it will release C which will allow P1 to proceed.
If you don't acquire them in a fixed order then you can get something like the following:
P1 => A B
P2 => B A

Now each process has locked on one resource and is waiting for the other process to release the resource it needs, causing a deadlock.

redpajama_set_name: RedPajamaStackExchange


Rebecca Alexander is a psychologist and writer. Rebecca wrote her first book aged nineteen, and since then has been runner up in the Mslexia novel writing competition and the Yeovil Literary Prize 2012. She is the author of the Jackdaw Hammond series of supernatural crime novels published by Del Rey, The Secrets of Life and Death (2013), The Secrets of Blood and Bone (2014) and The Secrets of Time and Fate (2016). She lives in Devon.


Police Quest casts the player as Sonny Bonds, a 15 year veteran police officer in the fictional town of Lytton, California. Assigned to traffic duty, Sonny investigates what appears to be a simple car crash but turns out to be a homicide. Sonny continues to give a traffic violation citation, single-handedly face a tough gang of drunken bikers, and make a DUI arrest. As the game progresses, he goes from patrol to temporary narcotics detective and finally takes on an undercover operation, tracking "The Death Angel", a murderous drug dealer named Jessie Bains. Sonny is assisted by "Sweet Cheeks" Marie, his former high school sweetheart who is now working as a prostitute. Jessie Bains' drug business has found its way into Lytton's school and one of the victims of an overdose is the daughter of Jack, a colleague of Sonny at the police department.
Written and designed by retired police officer Jim Walls, Police Quest was one of Sierra's first games which aimed to achieve authentic realism, making the game quite different from other classic Sierra adventure games where the task was to rescue damsels in distress, defeat evil wizards or save the universe. Police Quest was in fact so realistic that it was used as a real police training tool. The game comes with an indoctrination guide with police codes and procedures.
This game was made with Sierra's Adventure Game Interpreter engine (AGI), using text parser and was released for the IBM PC, Apple II, Amiga, Atari ST and Apple IIGS. It was remade in 1992 with 256-color VGA graphics and the SCI1.1 engine.


Midland TX – Fireworks fill the sky after a Midland RockHounds baseball game at Citibank Ballpark.
Like the drought-like conditions now, it s a decision Harris made three years ago to help limit the extra days and opportunity for young people, who may be unsupervised, to come out and purchase fireworks and shoot them off before the New year's celebration. It's been 86 days without rain; and the risk of wild fires is great.
Midland fire officials still want to remind residents that it is illegal to set off fireworks inside the city limits and can result in a maximum fine of $2,000. An additional citation of starting a fire, with another maximum fine of $2,000, could also be charged to an individual who is caught using fireworks in the city limits and a fire is started because of it.


text:
Why We Need Trust in Negotiations
The Trust Project at Northwestern University
Contributor / Jeanne Brett
DeWitt W. Buchanan, Jr., Professor of Dispute Resolution and Organizations
Professor of Management & Organizations
Director of Dispute Resolution Research Center
Kellogg School of Management / Negotiations
Part 1 / Signaling Trustworthiness (0:00)
Establishing trust in negotiations depends on each side showing an element of vulnerability. In cultures defined by a high level of trust, becoming vulnerable by revealing what is important happens quickly and creates reciprocity allowing an agreement to emerge based on each partner's priorities.
Part 2 / Positive Outcomes with More Trust & Vulnerability (1:30)
Two people arguing over an orange, for example, might want something very different from it—the rind or the juice. In negotiations, both parties can often have what they want as long as there is sufficient trust to reveal their true interests and underlying motivations.
BUMPER: Signaling Trustworthiness
I'd like to distinguish between swift trust and slow trust. Swift trust means I don't know you — I may not even know you by reputation — but I assume you're a professional. I assume that you are benevolent; I assume that you're trustworthy.
And I signal that information to you. And then you come back to me, usually, by reinforcing me and indicating that, yeah, I made the right call about you. It's hard to not fulfill someone's trusting expectations of you.
In slow trust, there's no assumption that the other party is trustworthy. In slow trust, it's slow; it takes time to develop trust. What that means is, people have to have experience with each other, gain familiarity with each other and have that experience where I make myself a little vulnerable and you don't take advantage of me.
Slow trust builds slowly over time. Swift trust happens quickly. One is just more efficient than the other in negotiations. So, it's understandable that many people are reluctant to take the swift-trust risk.
BUMPER: Positive Outcomes with More Trust & Vulnerability
Two sisters are both in the kitchen; they're both cooking; and they have need for an orange. And they only have one orange.
So, the sisters get into a fight: "I want the orange." "No, I want the orange; you can't have the orange." And there's no solution because half an orange is not going to satisfy either sister.
They get nowhere until they take that single issue of who gets the orange and say, "Why do you want the orange?" And then they learn that one sister wants the orange for the rind, and the other sister wants the orange for the juice.
Now, if they had just taken half an orange, neither one of their recipes would have come out, but by finding out why they wanted the orange — those are those interests in negotiations — they were able to both win, if you will.
What happens is, in negotiations — even if it looks like it's a single issue, or many negotiations are multi-issue — what you have to understand is where the other party is coming from, what's motivating the other party and what's most important to the other party.
You're not going to get everything you want in negotiation; you're too interdependent for that. But if you find out what's more important to the other party that's somewhat less important to you, then you can begin to make a trade-off.
So, why do negotiators need to trust? Because as soon as I start revealing what's important to me, you have the opportunity to take advantage of me.
So, I have to trust you that you won't take advantage of me, which means we coach negotiators to share a little information about interests and priorities and ask for some information — comparable information — in return.
And then you can get this reciprocity going — sharing information — understand where the other party is coming from, make those trade-offs and build high-quality agreements, like the two sisters and the orange.
Keep up with the latest insights on trust
How Culture Can Influence Whether Negotiators Cooperate: Key Findings
Jeanne Brett Negotiations Swift Trust,Reciprocity,Institutions and Context BUMPER: Trust and Social Sanctions
There are really three key papers to understand culture and negotiation strategy in the role of trust in that relationship. The first one is one that identifies swift trust, and this is Meyerson, Weick, and Kramer's work.
They were looking at film crews. And their observation was, these film crews coalesced very quickly and used a lot of cooperative behavior. They were assuming that the other parties in the team were professional, competent, and trustworthy.
BUMPER: Swift Trust in Negotiations
When we take a look at negotiation research and we think about Meyerson, Weick, and Kramer's work, we see a lot of similarities between the film crews that they studied and negotiations.
In the first place, negotiators come together and they typically don't know each other. When they're putting together a new business deal, they don't have a history. They may have some reputation that goes in front of them, but they don't have a personal history.
Meyerson and her colleagues propose that to get trust going, people have to wade in; they can't just sit back and wait for the other party to act in a trustworthy manner. But the way to get trust going is to initiate trust, and then that global norm of reciprocity kicks in and people trust back.
BUMPER: Trust and Social Sanctions
I want to talk about Toshio Yamagishi's research. He was studying Japanese and American participants and putting them in social dilemmas. Now, a social dilemma is a multiparty prisoner's dilemma.
If you know about prisoner's dilemmas, you know that they put the two prisoners in separate interrogation rooms, and they offer each of them a deal to squeal on the other. And if nobody squeals, everybody gets a light sentence. And if both people squeal on each other, they both get a really heavy sentence.
But the best outcome would be, is if I squeal on you and you don't squeal on me.
A social dilemma is the same phenomenon, only there are multiple parties.
And what Yamagishi did is he put his experiment and he varied in two dimensions. He measured his participants' level of trust, and he had high-trusting participants and low-trusting participants.
And then he put them in a situation of no sanctions or a situation of sanctions. By sanctions, what he was doing is he was giving them money to do the experiment, and if they failed to cooperate, the money would be taken away from them.
And he found that among the Japanese, in the no-sanction condition, cooperation was extremely low; in the high-sanction condition, cooperation was high—particularly there were no differences between high and low trusters in the sanctioning condition, but in the no-sanction condition, the high trusters were willing to cooperate more than the low trusters.
Yamagishi then went and did the same study using American participants. And the first thing he found was that American participants overall were more trusting on a trust scale than his Japanese participants.
But he still sorted them into high- and low-trust groups, and he still had the no-sanction condition and the sanction condition.
But here, he found that the Americans in the no-sanction condition, regardless of whether they were high trusters or low trusters, they cooperated. They also cooperated in the sanctioning condition.
And so, Yamagishi stepped back. He was, I think, very surprised at what his research showed at first. And then he had to figure out how to interpret it.
And his interpretation was that the sanctioning condition cued the Japanese participants to the kind of environment that they lived in, in their everyday social interactions because, in Japan, there are strong norms for cooperative behavior, there is social monitoring and social sanctioning for failure to cooperate.
What Yamagishi argued is that when you take away that monitoring and sanctioning, the Japanese participants had no basis for predicting or expecting the counterparties in this social dilemma to behave cooperatively.
And so, in a defensive posture, they competed in order to defend themselves.
When he looked at the American data again and started to interpret the American data, he says, "The US culture is very different." He said, "The US culture is much more open; there's much less monitoring and sanctioning."
And he said, "These American participants, they have an internal moral compass. And so, they don't have to be cued by the context to know whether to cooperate or not to cooperate; they cooperate on the basis of their internal moral compass."
Now, Yamagishi explained his results with a cultural concept called "collectivism versus individualism."
In individualist cultures, people's self-identity is a function of their accomplishments. In collectivist cultures, people's social identity is a function of whether or not they are adequately, in the eyes of others, fulfilling their social roles.
BUMPER: Cultural Tightness and Looseness
Michele Gelfand is another cultural psychologist, and she came along and said, "Hmm, I don't think so. I don't think it's collectivism and individualism that's explaining Yamagishi's results." She says, "I think it's something called cultural tightness–looseness."
She defines cultural tightness–looseness very similarly to the way Yamagishi described the differences between Japanese and US culture. She says, "In a tight culture—strong social norms, lots of monitoring, lots of sanctions for failure to fulfill those social norms. And in a loose culture," she says, "norms are looser. There's more flexibility; there's more improvisation in social interaction in everyday life."
So, Michele Gelfand and her colleagues collected data from 33 countries, and they correlated it with a whole lot of other psychological variables, but they also correlated it with data about history, geography in these countries.
And what they found is that tight cultures have a profile—they have a cultural profile—that is very different from the cultural profile of loose cultures.
And so, the way to understand culture and negotiation strategy is to understand both trust in that culture and to understand whether that culture is a tight culture or a loose culture.
Negotiating Across Cultures Depends on Trust: A Psychologist's Perspective
Jeanne Brett Negotiations Definitions,Swift Trust,Communication,Reciprocity, Institutions and Context I study culture and negotiation strategy, and trust is a very important concept for researchers like me. We define trust pretty much the way people in other disciplines define trust—with an emphasis on benevolence and integrity.
What we find is that negotiators who trust use strategy in a way that allows them to negotiate high-quality agreements. What they do is they share a little information about their own interests and priorities and they ask for information from their counterpart about their counterpart's interest and priorities.
Now, an interest in negotiation—that concept—refers to the reason underlying why negotiators are asking for what they want. So, it's the reason why you're asking for what you're asking: why do you want that? It answers the "why" question.
Priorities refer to, what issues are more important, and what issues are less important? Negotiators who trust and share information about interests and priorities, learn about the interests and priorities of their counterpart are able to trade things off and create high-quality agreements.
But of course to do that, you have to make yourself a little vulnerable because you're sharing information that the other party could take advantage of. And so, the key question in our research is, who trusts and why?
BUMPER: Trust across Cultures
And one answer to that question is culture. I'd like you to take a look at the chart that plots World Values Survey trust data by regions of the world.
The World Values Survey is run by a group of scholars that collect data from a representative sample of about a thousand people within a nation, ages 18 to 80.
And they come back every two or three years.
Their trust question is, "Most people could be trusted—yes or no?" And so, what you see in the chart is the percentage, averaged by nation, of people who say, "We can trust other people."
Before we look too carefully at the data in the chart, there are two things we need to address. One is, this is a single-issue question. And behavioral scholars like myself don't like single-issue questions; they worry about reliability.
But because we have data from these nations, collected with this exact same question over many, many different years, you can look at the reliability across time.
And when you look at the reliability of the trust question, it's over .90 on average across time. So, the question is reliable.
The other potential threat to validity of this question is that people in different cultures interpret trust differently when they ask a question, "Can people in your culture be trusted?"
But again, scholars have looked at this question, they ask people, "What do you believe in? What do you believe trust means?" and they get consistently the same answer over and over again—integrity and benevolence.
So, if we take a look at the data, you might infer—from seeing that trust is relatively high in the West and in East Asia, and relatively low in South Asia and in the Middle East and in Latin America—that East Asians and Westerners are going to be using a lot of strategy that shares information, and people in Latin America and in the Middle East and South Asia aren't.
And you'd only be half right.
It turns out that the research on negotiation in the West shows a lot of information sharing, a lot of cooperative behavior, but we don't see the same thing in East Asia.
And the question again is, why?
BUMPER: Swift Trust
It turns out, in the West, negotiators seem to use something that we call "swift trust." And the idea of swift trust was coined by some researchers, Meyerson, Weick, and Kramer.
And they were studying film crews. They looked at film crews and they said, "These are people who come together to do a highly interdependent job. They frequently don't know each other—maybe by reputation, but they haven't worked together in the past.
But they come together and they coalesce very rapidly and they are very cooperative. And why does that happen?" these researchers asked.
And they said, "It happens because they assume the other is a professional. They assume that the other is trustworthy."
And as soon as you signal that you think someone else is trustworthy, that has two or three implications. One is, they would have to be a cad not to follow up and step up to your expectations of trust.
But the other is, you're signaling something about yourself—that you're willing to make yourself vulnerable to the other party and believe that they are trustworthy.
In negotiation research, we see very much the same thing in our North American and Northern European data. Negotiators come to the table and make the assumption that the counterpart is trustworthy.
And they then go in and ask questions, get a little information about those very important interests and priorities, offer a little information in return. And in that counterpoint–point–counterpoint, they begin to justify the assumption of trust that they made in the first place, and they begin to solidify that trust because they have behavior that they see of the other party.
So, let's take a look at the data from East Asia. Consistently, in study after study, we find East Asians are extremely competitive in negotiation. And this, of course, is not what we would expect, looking at the trust data.
A really good answer to the question of why the East Asians do not seem to act in a trustworthy fashion in negotiation was given by Toshio Yamagishi, a Japanese psychologist.
He was studying social dilemmas. Now, a social dilemma is a multiparty prisoner's dilemma. It's a commons problem in economics.
The tension in a social dilemma is that if all the parties cooperate, everyone shares a larger piece of the pie, a larger set of resources. If, however, everyone competes, they share a smaller set of resources.
What it means is, for you, if you compete and the rest of us cooperate, you do really well, but we do very poorly. And what Yamagishi found is that his Japanese students in a social dilemma situation would cooperate as long as there were sanctions for failure to cooperate, or competition.
When he removed the sanctions, the Japanese students competed like crazy. Yamagishi then repeated his research in the United States with American students.
And he found, first of all, that the American students were much higher trusting than the Japanese were and that they would cooperate in both the situation where there were sanctions, but they would also cooperate pretty much at the same level in situations in which there were no sanctions.
So, Yamagishi explained his results by looking at Japanese society. And what he described Japanese society is a society that's very tightly controlled by social norms, by social monitoring, and by social sanctioning.
And he said what happened then in his research is that when monitoring and sanctioning was missing, in the situation of a social dilemma, the Japanese students competed.
But he described the United States culture as much looser in terms of norms, monitoring, and sanctioning.
And he said, "Look, the Americans have an internal moral compass that directs them to be cooperative. They don't need and don't expect the external controls that are characteristic of the Japanese society."
If we take a look a look again at the trust chart and look at the Middle East and South Asia and look at Latin America, we see pretty uniformly low trust. And that would predict that we would have a lot of difficulty in negotiations—not a lot of information sharing, not a lot of high-quality agreements.
There's much more limited negotiation data in the Middle East, South Asia, and Latin America. But it shows not a lot of cooperation, not a lot of trust, at least in the Middle East and in South Asia.
But the Latin American data are more interesting. The Latin American data show much more complexity.
For example, in some data that I have from Brazil—in fact, I've collected data from Brazilian managers several different times over the past 10 years—I get a lot of cooperation, a lot of information sharing, and really high-quality outcomes.
In some Spanish data, we saw that when the Spanish were negotiating with each other, they weren't sharing information and they got low-quality outcomes; compared to the US negotiators, when they're negotiating with each other, using more information sharing, getting higher-quality outcomes. You see that difference in trust.
But what was interesting, in the next part of the data, is when the Spanish were negotiating with the Americans. And there, the Americans continued to share information, ask for information, and the Spanish started to do that too.
And they got a lot more insight about what was important, more and less important, to their counterparts, and they negotiated much better deals in the intercultural negotiations with the Americans than they did in the intracultural negotiations with other Spanish counterparts.
So, looking at all this data, my research team concluded that trust isn't enough to explain the relationship between culture and negotiation strategy. Taking our cue from Yamagishi's research, we turn to some more recent work in cultural psychology by Michele Gelfand.
Michele Gelfand studies a concept that she calls "cultural tightness–cultural looseness." And it's very similar to the way Yamagishi described Japanese culture versus American culture.
A tight culture, according to Michele Gelfand, is one where there are strong social norms, there is monitoring of social norms, there is sanctioning of failure to conform to social norms.
And maybe it would help to have a little story to explain what that might look like if you've never been in Japan or East Asia.
One of my executive education students said, "I've got a story, professor." He said,
"My wife and I were living in Japan. I was an expat working for an American company over there. And we decided we wanted that full cultural experience, so we rented a house in a Japanese community, not an expat community."
And he said, "Our first major challenge was figuring out how to properly sort our recyclables because it's very strict in Japan."
And he said, "We tried really hard to get everything in the right bag. And then the recycle truck was coming the next night, and we would put everything out where it was supposed to be. And then we would see our neighbors coming out of their homes, going through our recycle bags, resorting it so that it was correct."
The neighbors thought it was important for the neighborhood that the recycling be done appropriately. So, here's a very strong set of norms, monitoring, and sanctioning.
They didn't really sanction directly this expat couple, but the expat couple was sufficiently embarrassed that they kept working on getting their recycling right.
So, Michele Gelfand and her colleagues, all around the world, collected cultural tightness–looseness data from participants—these were mostly teachers and university students.
They asked a series of questions about how culture operated, and how tight those norms were, how much sanctioning and monitoring was going on.
So, you see in this chart how those same countries fall out on cultural tightness–looseness. These are all the same nations that we show in the trust scale. And what do you see?
We see that Western culture is a loose culture (that's no surprise), East Asian culture is a tight culture (that's no surprise), the Middle East and South Asia are also a tight culture, but those interesting Latin Americans are a loose culture.
So, what we have in Latin America, and maybe the way to explain what we see in the negotiation data in Latin American, is that Latin Americans live in a loose culture.
That means they can't assume on a day-in, day-out basis, they can't predict other people's behavior. They don't have expectations the way the tight-culture East Asians or Middle Easterners do.
What they have to do is navigate a loose culture and navigate a low-trust, loose culture. What do they do? They certainly don't engage in swift trust—that assumption that everyone else is trustworthy.
What they do is they build slow trust; their focus is on building relationships. And they do all the things—getting to know you as a person, learning about your family, finding commonalities in interests in sports, in interests in food—to get a relationship going with you that has nothing to do with a negotiation.
But if they can build a common path with you personally, they build a relationship to be able to negotiate cooperatively.
And so, my colleagues and I think that there's more to culture and negotiation strategy than just trust. It's that interface between trust and cultural tightness–looseness that accounts for how negotiators in different parts of the world use strategy.
Differentiating Trust and Trustworthiness: A Sociologist's Perspective
Contributor / Bruce Carruthers
Bruce Carruthers Sociology Government,Institutions and Context,Legal Guarantees,Reputation Management,Social Psychology,Vulnerability As a social scientist, I'm very interested in trust as it concerns people because really society, human cooperation, human coordination, all of the activities that we do together, really depend on trust.
And you might kind of wonder, well, trust sounds a bit like faith? You know, we're just going to take people on faith.
And don't we have a bunch of ways of coordinating our activity and making sure that the left hand knows what the right hand is doing, and I can figure out what's going on vis-a-vis my employer or all the people that I interact with — and don't we have a bunch of formal coordination devices like contracts or instructions or standard operating procedures?
Why aren't those good enough? Why do we have to go beyond that and trust people? And there's a couple of reasons for this. And one of them is that, as wonderful as these formal devices are, they really do have limits.
And one of the reasons is that contracts and other instructions, lists, standard operating procedures, all of these devices — they're always incomplete; that is, the world is more complicated and unpredictable than we can anticipate.
And so, stuff will happen that will effect whatever it is you're doing with these other people. It will have an impact on your ability to execute whatever it is you're trying to do, and it's not going to be in the contract what's up.
You might think to yourself, "Well, maybe trust isn't such an issue if we go to the marketplace." Let's think about markets and capitalism and self-interest and competition — maybe that's a world in which contracts and other formal devices will be sufficient.
And once you've got an airtight contract, you don't have to worry about the personal character or the trustworthiness of the people you're dealing with, because you've got a good contract and you hired a good lawyer.
So, the most famous person who thought about this sphere, of course, was Adam Smith in his famous book The Wealth of Nations, which really did talk about the virtues of capitalist production and competitive markets and so forth.
And I think it's very telling that before he wrote The Wealth of Nations, Adam Smith wrote a book on The Theory of Moral Sentiments.
And in this previous book, he posited that people are linked through strong bonds of sympathy and empathy and trust, and that on top of this, we're able to have markets and capitalism and all that kind of fun stuff.
It was clear to me (and I think clear to Smith) that some measure of some baseline, some foundation of trust is very important even in social settings where we might think that the issue of trust can be solved or avoided.
You might want to ask, when does trust arise? I've talked about it as kind of ever-present — it's all over the place. But really, there's two elements that drive trust situations.
One of them is uncertainty, and the other is vulnerability — that is, people are uncertain about what others are going to do (they don't know what's going to happen in the future), and they're vulnerable to what those other people do to the extent that their interests and those other people's actions are intertwined.
So, the trick for dealing with a trust situation is really addressing these two key elements: either trying to deal with uncertainty by acquiring more information and learning (or trying to figure out) what is likely to happen in the future.
…Or by managing your vulnerabilities and thinking about ways to mitigate or reduce the impact (or the potential harm) that others' actions, future actions, could have on your interests.
So, this kind of sets up a generic recipe book for how to deal with trust situations. How do people trust?
People rely on a lot of heuristics, rules of thumb, to decide who is trustworthy and who is not, and that distinction is really important because you can't go through the world trusting everyone, and you can't function in the world if you trust no one.
And so, what you have to do at the simplest level, is kind of put everyone into two bins: there's people that are trustworthy; there is people who are not. And you want to be able to trust the trustworthy and avoid those who are not trustworthy
I'm going to offer a couple of distinctions that help clarify the discussion of trust. And one of them is the difference between trust and trustworthiness. And this really speaks to who is doing the trusting and who is being trusted.
One party trusts the other, and the other party may or may not be trustworthy — that is, they deserve the trust. But someone who is trustworthy may not be trusted, and someone who is trusting may end up trusting someone who is not trustworthy.
So, these two things have to be kept separate. Another distinction is the distinction between generalized and relational trust.
Generalized trust really speaks to the question of how you deal with strangers. Do you trust abstract institutions? Do you trust the average citizen that you might run into on the street?
That kind of a thing — where you're really dealing with someone with whom you have no relationship and about whom you have no prior information. What kind of ambient or generic level of trust do you have?
Relational trust is, what happens after you start to get to know someone? What happens after you start to develop a social relationship?
You have a history together; you have contracts; you have prior transactions. That is a very particular and non-anonymous form of trust, and it is really driven by the nature of the interaction that you have with that individual.
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redpajama_set_name: RedPajamaCommonCrawl


I was surprised at how well Wim Wenders summarized the first five years of Pope Francis' pontificate in his latest documentary, Pope Francis: A Man of His Word. In the end, it truly deepened my admiration for Pope Francis.
Throughout the film, Pope Francis is frequently compared to St. Francis of Assisi, his papal namesake, whose example of a simple lifestyle is reflected in the Pope's own life.
Pope Francis' simplicity has captured me since his election to the papacy in 2013. The film makes reference to this, pointing out that during his papal visit to the United States in 2015, the Pope was driven to his many events in a small black Fiat rather than a fancy limousine.
The documentary also discusses Pope Francis' views on the role of women in society. He said the integration of women in the Church's life is important and that advances in society cannot be made without them. This point has been highlighted by the Pope's appointments of women in roles in the Roman Curia and various pontifical congregations.
I have seen many papal events live-streamed through the Vatican News. I see people screaming in joy and from time to time, the Pope would kiss a child. However, watching the documentary, I realized the importance of Pope Francis' presence, especially in areas affected by pain and terror.
The documentary showed moving moments of Pope Francis off the island of Lampedusa in the Mediterranean Sea, where many refugees drowned while striving to find freedom. In another scene, Pope Francis was at a prayer service at the Jerusalem Holocaust Memorial.
The documentary also reminded viewers of his 2015 visit to the site of the World Trade Center 9/11 memorial in New York. His words and presence bring comfort to the suffering and unites people together in prayer and solidarity.
Scattered throughout the documentary are clips of Pope Francis talking to the audience. I really liked this, given that most people viewing the film may never get to meet Pope Francis in person. It is as if he were speaking directly to those in the theatre.
Pope Francis, who advocates that the Church be a "Church for the poor," emphasized three "T's" in one of the interactive clips — "Trabajo (work), tierra (earth) and techo (roof)." He called these basic rights which all people should be able to access.
The images of Pope Francis presented in Pope Francis: A Man of His Word captures the essence of his pontificate: a person of justice, simplicity and advocate for a Church of mercy, a Church for the poor. More than just saying these things, Pope Francis acts upon his words.
Watching this documentary only convinced me more that Pope Francis' actions are the strongest messages he can give to the world, and those actions make me admire this pontiff even more.


text:
Q: Terraform issue setting up VM logging 'Microsoft.EnterpriseCloud.Monitoring' and type 'MicrosoftMonitoringAgent' I'm trying to provision a windows VM with logging with Terraform, But I had several different errors trying to add the required components. The current error is
No version found in the artifact repository that satisfies the requested version '1.0' for VM extension with publisher 'Microsoft.EnterpriseCloud.Monitoring' and type 'MicrosoftMonitoringAgent'
resource "azurerm_log_analytics_workspace" "law" {
  name                      = "${local.vm_name}-law"
  location                  = azurerm_resource_group.rg.location
  resource_group_name       = azurerm_resource_group.rg.name
  sku                       = "PerGB2018"
  retention_in_days         = "30"
  internet_ingestion_enabled= true
  internet_query_enabled    = false
  tags                      = local.common_tags
}

resource "azurerm_log_analytics_solution" "vminsights" {
  solution_name         = "VMInsights"
  location              = azurerm_resource_group.rg.location
  resource_group_name   = azurerm_resource_group.rg.name
  workspace_resource_id = azurerm_log_analytics_workspace.law.id
  workspace_name        = azurerm_log_analytics_workspace.law.name
  tags                  = local.common_tags

  plan {
    publisher = "Microsoft"
    product   = "OMSGallery/VMInsights"
  }
}

resource "azurerm_virtual_machine_extension" "omsext" {
  name                  = "OMSExtension"
  virtual_machine_id    = azurerm_virtual_machine.iis-vm.id
  publisher             = "Microsoft.EnterpriseCloud.Monitoring"
  type                  = "MicrosoftMonitoringAgent"
  type_handler_version  = "1.0"
  auto_upgrade_minor_version = true

  settings = <<SETTINGS
    {
      "workspaceId": "${azurerm_log_analytics_workspace.law.workspace_id}"
    }
SETTINGS
  protected_settings = <<PROTECTED_SETTINGS
    {
      "workspaceKey": "${azurerm_log_analytics_workspace.law.primary_shared_key}"
    }
PROTECTED_SETTINGS

  tags                       = local.common_tags
}

resource "azurerm_virtual_machine_extension" "DAAgent" {
  name                       = "${local.vm_name}-daa"
  virtual_machine_id         = azurerm_virtual_machine.iis-vm.id
  publisher                  = "Microsoft.Azure.Monitoring.DependencyAgent"
  type                       = "DependencyAgentWindows"
  type_handler_version       = "9.10"
  auto_upgrade_minor_version = true
  tags                       = local.common_tags
}

    variable "iis_vm_image" {
      type        = map(string)
      description = "Virtual machine source image information"
      default     = {
        publisher = "MicrosoftWindowsServer"
        offer     = "WindowsServer"
        sku       = "2022-datacenter-azure-edition"
        version   = "latest"
      }
    }

I have tried the 1.0.x.x version, but then get an invalid version message

A: I see that you put type_handler_version  = "1.0" however the extension type  MicrosoftMonitoringAgent has version format 1.0.X.X
The Publisher, Type of Virtual Machine Extensions and version of the extension to use can be found using the Azure CLI, via:
az vm extension image list --location mylocation -o table

Here is an output of the above command:


The Department of Communication hosted an annual event that included workshops, panel discussions, and lectures. That year, the Virtual MLK Project was featured as a public exhibition, presentation, and panel discussion at COM Week 2015.
An opportunity for NC State students and faculty to experience the vMLK project commemorative exhibition. Members of the community were also in attendance.


text:
Q: Android Facebook integration user revoke permissions handling I am working on integrating our company's android application with facebook login via facebook sdk. I am able now to communicate with facebook to get an access token but I have a problem that raises two questions:
1- the specific one is: what to do if the user revokes the facebook application from his account in 'apps' tab?
2- the more general: how to handle permissions change in my application if the user change them in his account?


Other photos: Photo 1, Photo 2, Photo 3, Photo 4, Manufacturer Provided Image, Photo 6, Manufacturer Provided Image, Photo 8, Manufacturer Provided Image, Manufacturer Provided Image, Photo 11, Photo 12, Photo 13, Photo 14, Photo 15, Photo 16, Photo 17, Photo 18, Photo 19, Photo 20.
One of the nicer cruise/racers available today!
Anyone who has ever owned a Saber Yacht is familiar with the quality built into this boat.
But this yacht is not just about quality but also all about performance.
If you are looking for a quality vessel that sails great and is capable of taking you anywhere, then this is certainly the one to consider!!
She needs a new lover!!! Owner says "Lets sell"!!!
SYS Yacht Sales is pleased to assist you in the purchase of this vessel. This boat is centrally listed by Grove Yacht Sales LLC.


Cost: starting at $79 for the Guru. Sign up with my link & get 20% off!!!
It is always so exciting when my YogaClub box arrives! I love having yoga clothes sent to your door. Yoga clothes are just one of those things I hate trying on. So I either go shopping just for yoga stuff or I just skipped it. Problem Solved!!! I now just have them curated, shipped to my door so that I can try on in my own home and a HUGE discount. *High Five* Seriously, YogaClub is the greatest.
I review and have interacted with alot of subscription boxes over the last 1.5 years and it is nearly impossible to find one that does what Yoga Club does. They offer fabulous brands, fully curated and shipped to you at a big discount! Curation + shipping usually cost extra over MSRP and they actually give it to you cheaper than you can get on your own. LOVE!!!!
In this box I received a MPG Sport tank top in bright white in size small, MSRP $34. Also a Columbia Sports Bra in white in size small, MSRP $26. Then lastly, the Lukka Lux 3/4 Leggings in Mountain Print in size small, MSRP $120. So total MSRP for the 3 items is $180 and I only paid $79….that is a $101 savings AND it was sourced for me and shipped to my door all free to me!!!!!!! The leggings are slightly big on me so I am going to request and exchange for an extra small. If they don't have it I'm going to keep these and try to shrink them. They are just too cute to give up! Yoga Clothes Sent to Your Door for the win!


text:
// Copyright 2009 the Sputnik authors.  All rights reserved.
// This code is governed by the BSD license found in the LICENSE file.

/*---
info: Expression in "while" IterationStatement is bracketed with braces
es5id: 12.6.2_A6_T6
description: Checking if execution of "while 'hood' break" fails
negative: SyntaxError
---*/

//////////////////////////////////////////////////////////////////////////////
//CHECK#1
while 'hood' break;
//
//////////////////////////////////////////////////////////////////////////////

redpajama_set_name: RedPajamaGithub


Over the holidays while many people were concerned about over-eating, the French national assembly passed legislation that is being called the Photoshop law. The law requires all Photoshop retouching that is performed on models be clearly labeled. The Photoshop law is included with other legislation designed to safeguard the health of models used in fashion shows, commercials, advertisements, and photography. Photoshop is being regulated by requiring a disclosure when publishing an altered image that has been modified in such a way that it "makes the silhouette narrower or wider" of a model.
This is the first known legislation in Europe in which using Photoshop is illegal if it isn't fully disclosed. Violations of the Photoshop law are punishable by 6 months in prison along with a fine of 75,000 Euros. Israel also recently enacted a Photoshop law which requires disclosure of image manipulation of models, and also specifies a body mass index (BMI) for those working in the modeling profession.
The push to regulate Photoshop comes after many widely publicize incidents of models and celebrities being made to appear thinner through retouching. While excessive Photoshop retouching originally was criticized by advocates for maintaining a healthy body image, especially among young women, the cause has expanded to a much wider group. Some celebrities have criticized advertisers and publishers for the excessive use of Photoshop on their own images. Despite a movement towards more natural looking images, there remain almost weekly occurrences of celebrities sharing images that involve excessive use of Photoshop, generally used in an attempt to thin their appearance. Ironically at the same time one government is passing laws regulating the use of Photoshop, another government across the world was posting official images that were clearly manipulated.
This new Photoshop law is part of a more comprehensive set of legislation that seeks to also make certain that models are of good health. The original legislation also included a specific body mass index 9BMI) requirement, but this was replaced with a requirement that a physician certify that models are in good health.
One aspect of learning Photoshop is understanding when retouching is useful and acceptable. A Photoshop professional will work to improve images tone, contrast, lighting, an emphasis. Yet care should be placed into using Photoshop in a way that limits changes to the intent and appearance of an individual in any way that causes them to appear unnaturally thin. This Photoshop law attempts to address this by forcing retouching artists to label any work that changes the silhouette of a model.


A Hard Superhighway to Travel. The Superhighway project has been controversial from the day it was first announced publicly for many reasons. First, it was routed without regard to the negative impacts it would have on the Cross River National Park (CRNP) and a number of community forests in its path. The path chosen initially for the 260 kilometres Superhighway was carved out in a manner reminiscent of how Africa was partitioned at the Berlin Conference of 1884 – probably over tea and coffee, or as men hunted for game, and for territories. The path showed a disregard for the unique biodiversity of the region and was equally mindless of the climate impact that would ensue from the massive deforestation that the project was bound to cause. There was also no clarity about how the CRSG would ensure that this is not a white elephant project that would only promote the harvesting of timber from the forest and leave a scarred environment and impoverished communities in its wake.
This entry was posted in Analysis, Climate Change, Climate/Ecological Crimes, Culture, Environment, Food, Rights of Nature/Mother Earth. Bookmark the permalink.
You really make it seem so easy with your presentation however I to find this matter to be actually something that I feel I might never understand. It kind of feels too complex and extremely extensive for me. I am having a look forward for your subsequent post, I will attempt to get the dangle of it!


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Home » Nations » Soviet Union » Heads of State » Svernik (Švernik), Nikolaj
Николай Михайлович Шверник (Nikolaj Mihajlovič Švernik)
b. 7/19 May 1888, St. Petersburg, Russian Empire
d. 24 Dec 1970, Moscow, USSR
Title: Председатель Президиума Верховного Совета СССР (Chairman of the Presidium of the Supreme Soviet of the USSR)
Term: 19 Mar 1946 - 15 Mar 1953
Chronology: 19 Mar 1946, elected, 1st session of the 2nd USSR Supreme Soviet [1]
15 Mar 1953, discharged, resolution of the 4th session of the 3rd USSR Supreme Soviet [2]
Names/titles: Председатель Президиума Верховного Совета РСФСР (Chairman of the Presidium of the Supreme Soviet of the RSFSR) [4 Mar 1944 - 25 Jun 1946] (see details)
Biography:
Nikolay Shvernik joined the Russian Social Democratic Workers' Party in 1905 and was an activist of revolutionary movement in Samara. In 1924 Shvernik was appointed people's commissar for the workers' and peasants' inspection of the Russian SFSR and member of the Presidium of the party Central Control Commission. The 14th party congress elected (31 Dec 1925) Shvernik full member of the Central Committee. In 1925-26 he served as a secretary of the Leningrad province party organization. His party career was successful as he earned promotion to the post of secretary of the Central Committee (9 Apr 1926 - 16 Apr 1927) and became full member of the Orgburo (9 Apr 1926 - 16 Apr 1927). However, in 1927 Shvernik was dismissed from his position in the Secretariat and Orgburo and was transferred to the Urals, where he was made a secratery of the province party organization (1927-1928). He proved to be loyal supporter of the industrialization policy and returned to Moscow in 1929 as chairman of the Central Committee of the Metallurgist Trade Union. He was again made a candidate member of the Orgburo (17 Nov 1929 - 26 Jun 1930). After the 16th party congress, Shvernik was elected full member of the Orgburo (13 Jul 1930 - 18 Mar 1946) and a candidate member of the Secretariat (13 Jul 1930 - 26 Jan 1934). From that time his career was closely connected with Soviet trade unions. In 1930 he assumed the office of first secretary of the All-Union Central Council of Trade Unions (Jul 1930 - Mar 1944). Elected at parliamentary elections a member of the Supreme Soviet, Shvernik was involved in creation of the new Soviet parliament. He was elected chairman of the Soviet of Nationalities (12 Jan 1938 - 10 Feb 1946), one of the two chambers in bicameral Soviet parliament. Shvernik became candidate member of the Politburo (22 Mar 1939 - 5 Oct 1952) after the 18th party congress. During World War II, he was responsible for evacuation of Soviet industry to the eastern part of the Soviet Union. In 1944 he became the first deputy chairman of the USSR Presidium of the Supreme Soviet (1 Feb 1944 - 19 Mar 1946) and chairman of the Presidium of the Russian Supreme Soviet (4 Mar 1944 - 25 Jun 1946). During the second world war Shvernik also chaired the extraordinary commission on investigation of war crimes committed by the Nazis on the occupied territories of Soviet Union. On 19 Mar 1946, Shvernik replaced Mikhail Kalinin as chairman of the Presidium of the Supreme Soviet of the USSR (19 Mar 1946 - 15 Mar 1953). As a result of reformation of Politburo in October 1952, Shvernik was promoted to full membership (16 Oct 1952 - 5 Mar 1953) in the party Presidium, but Stalin's death interrupted Shvernik accession to party top leadership. The joint session of the Central Committee, the Council of Ministers and the Presidium of the Supreme Soviet (5 Mar 1953), made a recommendation on replacement of Shvernik as titular head of Soviet state and demoted him to candidate membership of the Presidium (5 Mar 1953 - 29 Jun 1957). The session of Supreme Soviet followed the recommendation and elected Kliment Voroshilov as new head of state on 15 Mar 1953. Shvernik again returned to the All-Union Central Council of Trade Unions as its chairman (Mar 1953 - Feb 1956). With the rise of Khrushchev, he regained in part his influential position as chairman of the Party Control Commission (Feb 1956 - Mar 1966) put in charge of political rehabilitation of victims of Stalinist purges. In 1957, Shvernik was made full member of the Presidium (29 Jun 1957 - 29 Mar 1966). Advanced age made him to retire from politics after the party's 23th congress in 1966. Biography source: [3, pp. 596-597]
[1] Ведомости Верховного Совета СССР, 1946, № 10.
[2] Ведомости Верховного Совета СССР, 1953, № 3.
[3] Государственная власть СССР. Высшие органы власти и управления и их руководители. 1923-1991 гг. Историко-биографический справочник / Сост. В.И.Ивкин. М.: РОССПЭН, 1999,


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Tooth decay is closely related to other health issues, some which are considered life altering. If a child is suffering the pain that a cavity can bring he won't be able to eat the nutritional foods that he needs to maintain his overall health. He may also find it hard to communicate in the classroom or be comfortable in social situations.
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A Cinecle View – The Walking Dad
Tony Marion
SPOILER ALERT. If you have yet to see the midseason finale of THE WALKING DEAD and wish to remain unsullied, go watch it first and come back here when you're caught up.
This is your final warning. Here be spoilers …
So, yeah, that happened. The cycloptic kid with the floppy hair and Hazzard County deputy hat revealed that he's been bitten by one of the undead, and, unlike Hershel before him, there's no hope of saving him through an amputation because EVERYONE NEEDS A TORSO.
Fandom is largely divided on this development, with some fans of the show and the comics irritated by what is a MAJOR SWERVE from the plot of the book as Carl has large storylines that will now never be realized (at least by him) in the television adaptation. While others, like me, think that variation will provide last remaining original character, Rick Grimes, with the motivation needed to evolve past the warrior/general he's become.
(The Walking Dead showrunner) Scott (Gimple) was trying to figure out why there was a hole between Rick slitting Negan's throat at the end of the 'All-Out War' arc and then there's the time jump and Negan is alive and in prison (in the comic) and Rick didn't kill Negan. Scott was trying to figure out how to bridge the gap between Rick not wanting to kill Negan and Rick also really wanting to kill Negan, which he does right now [in the show's story]. Scott's way to get around that was to make Carl this really humanitarian figure and person who could see the good in people and see that people can change and not everyone out there is bad. That's what Carl's talk to Rick was in this episode: There's no way that they can kill every one of the Saviors and not everyone is a bad person and there has to be some (other) way forward than just killing people.
– Chandler Riggs to The Hollywood Reporter
Surpisingly, the person who should be the most bummed, teenaged actor Chandler Riggs, who has spent nearly half of his life in the role, is actually the most centered …
It was devastating for me and my family because the show has been such a huge part of my life for so long. For a few days, we didn't know what to do; I just bought a house in Senoia [near where the show films in Georgia]. That was a big deal that I wouldn't be on anymore. I decided that I wanted to not go to college for at least a year and move to L.A. and focus on acting and music. It ended up being a great thing because now I get to do all kinds of other stuff that I haven't been able to do in the last eight years.
It's awesome that a 17 year old kid, who grew up on this show, with this cast and crew, and is now having his life (college and home ownership) turned upside down by a creative choice for the longevity of the program,could be so mature and centered in the face of such life changing
upheaval.
If only his 54 year old father, William Riggs, could have been so chill.
WHOA, TRIGGER! Easy, fella!
Sure, parents can be overprotective sometimes …
… but the Riggs social media tantrum transforms from Richard Williams-esque obsession to drive his daughter to G.O.A.T. status, to LaVar Ball-style child-parasitism when you see this …
By using his son's fame and hard work to charge fans of a television show with which he's only tangentially involved $29.99 a pop to show them where hundreds of other people actually work on said show, he's reinvented himself as some sort of gross Kris Jenner/Kenny Kramer hybrid and added his name to the list of the worst Hollywood parents ever.
Sure, the tour Facebook page is filled with great reviews of the service, but whether these people had a good time or not is irrelevant when viewed in the context that William Riggs' business model is based entirely off of being able to exploit his son's accomplishments, as I imagine the lack thereof being the real impetus for lashing out The Walking Dead showrunner Scott Gimple and AMC.
Cause let's face it; Chandler Riggs is going to be fine. He's had other acting credits before and since dad's gravy train pulled into the station … I mean, before he joined the cast of The Walking Dead, plus his music … the kid's star is on the rise.
As for dad? Well, he'll just have to wait for Chandler's next full time TV gig so he can resume his career by pulling a bus up alongside it and pointing out to hardcore fans where his son achieved things.
That is, unless he can find a way to Kit Culkin his younger son, Grayson, onto a hit genre show first.
Tony Marion is a writer and filmmaker who splits time between Lancaster, PA and Baltimore, MD. He lives for the work of Descendents (the band), Chuck Palahniuk and Rian Johnson. Check out the digital embodiment of procrastination he calls his website here.


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Holders of Ormat Technologies (ORA: NYSE) shares may feel somewhat blindsided by the cessation of operations at the Puna geothermal power plant on the Hawaii Island. Lava flowing from neighboring Kilauea volcano has overrun two of the geothermal wells as well as a warehouse and an electricity substation.
Ormat is not the only geothermal power company with publicly traded common stock. Innergex Renewable Energy (INE: TO) is began its first foray in geothermal energy with the acquisition of Alterra Power in early 2018, and Calpine Corporation was taken private in 2017 by the private equity fund Energy capital Partners. The actions left two fewer public stocks through which minority shareholders can take a stake geothermal power. Of the five remaining, geothermal power sources represent a significant portion of total power production for all except one. The Italian utility company Enel has a very strong track record to renewable power production, but is just getting involved in geothermal.
Notably, initial news reports related to the volcano and the Puna Geothermal facility appeared print and online media. The company remained silent on the topic until well after the initial volcanic eruptions before issuing updates through a news release service. Ormat has yet to submit an 8K filing with the SEC and waited until the week of this post to display updates about the Puna Geothermal facility on the corporate website. Sluggish communications with shareholders appear to be the norm for Ormat, a habit that has attracted a scrum of law firms investigating the company for breaches in fiduciary duty by management.
The more diversified Enel has been given something of a pass by investors. The rest of the geothermal power group has been punished by traders over the last few months. Ormat, with its high profile difficulties has seen its stock price trimmed by 18.7% over the last six months and reduced by 28.2% from the 52-week high sent in late January 2018. A good share of the sell-off was in initial reaction to the announcement of year-end financial 2017 results. The stock has been attempting a recovery until news of the Kilauea volcanic activity become more visible in the news.
Innergex Renewable Energy has also been given a thorough thrashing by traders, perhaps because of new exposure risk in geothermal power generation through Alterra Power. The stock has lost over 20%% value over the last three months. Polaris Infrastructure (PIF.TO) is focused on developing geothermal power in both North and South America. The company has been stripped of 28.7% of its market value over the last three months. Yet it seems Polaris price trends appear to be more in reaction to sales and earnings performance rather than growing awareness of the risks in geothermal power production.
Average forward price/earnings ratio for the green and renewable energy sector is 38.96. The forward price/earnings ratios for the S&P 600 Index is near 24.0 times. It seems plausible that renewable energy earnings higher multiples because of expectations for faster growth than most of the companies in S&P's index of small companies. The S&P 500 Index of large companies, against which Enel is best compared, merits an average price/earnings ratio of 16.5 times forward earnings.
Ormat's forward price/earnings ratio is higher than the rest of the group. However, this is to be expected given that the company's sales and earnings have been trimmed beginning in the current quarter with the loss Puna Geothermal plant closure. It might even be too simplistic to assume ORA shares are trading only in reaction to difficulties in Hawaii. Ormat has a restatement of financial results underway and the threat of delisting from the NYSE looming over its corporate head. There are many reasons for investors to worry about the company.
The valuation dynamic for Ormat is probably the same for each of the other geothermal power companies. There will be a mix of factors that influence valuation for stock. Yet, with the lava spouting out of Hawaii's Kilauea volcano it is certain that forewarnings of the risk of volcanic activity will be taken more seriously than ever.


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Cosmo Tech's CEO to Participate in Panel at Produrable
by Sarah Paronto | Mar 29, 2018 | Events
Comso Tech CEO and co-founder Hugues de Bantel was invited to participate in a round table panel next week at Produrable Paris on Thursday, 5 April. De Bantel is one out of five speakers who will be participating in the discussion of Water, Energy, Waste: The...
Three Questions for Michelle Alonda
by Sarah Paronto | Mar 28, 2018 | Three Questions
Michelle Alonda is a Technology Marketer at Cosmo Tech. She answered our questions from the Cosmo Tech headquarters in Lyon, France. What is your role at Cosmo Tech and what does a typical day look like for you? I prefer to talk about a typical month. For me, a...
How Energy Storage is Starting to Rewire the Electricity Industry
by The Conversation | Mar 23, 2018 | The Conversation
Eric Hittinger, Rochester Institute of Technology and Eric Williams, Rochester Institute of Technology The market for energy storage on the power grid is growing at a rapid clip, driven by declining prices and supportive government policies. Based on our research on...
What's Critical About Critical Infrastructure?
Wendy Steele, RMIT University; Karen Hussey, The University of Queensland, and Stephen Dovers, Australian National University Our cities and regions depend on the critical nodes and arteries that together comprise urban infrastructure systems. This includes energy,...
How to Make Smart Cities Human Again
Nick Malleson, University of Leeds and Alison Heppenstall, University of Leeds Huge quantities of networked sensors have appeared in cities across the world in recent years. These include cameras and sensors that count the number of passers by, devices to sense air...