text:
\section{Introduction}
\hspace{1.5 em}
We recall related results which are previously known.
T.Fujita conjectured that,
for a smooth projective variety $X$ and  an ample divisor $L$ on
$X$, the linear system $|K_X+mL|$ is free if $m \geq \dim X+1$.
Reider [Rdr] proved that this is the case if $\dim X=2$.
Ein and Lazarsfeld  [EL] proved that this is the case if $\dim X=3$.
Kawamata [K3] proved that this is the case if $\dim X=4$.
For a projective variety $X$ of dimension $2$ with some singularities,
Ein and Lazarsfeld [EL] and Matsushita [M] extended the result of Reider [Rdr] to singular cases.
Kawachi [KM] obtained the effective estimates for a normal surface $X$.
For a projective variety $X$ of dimension $3$ with some singularities,
Oguiso and Peternell [OP] proved that $|K_X +5L|$ is free
if $X$ is a projective threefold with only ${\mathbb Q}$-factorial terminal
Gorenstein singularities.
Ein, Lazarsfeld and Ma\c{s}ek [ELM] and Matsushita [M]
extended some of the results of Ein and Lazarsfeld [EL]
to projective threefolds with terminal singularities.

A strong version of Fujita's freeness conjecture: \\
Let $X$ be a normal projective variety of dimension $n$,
$x_0 \in X$ a smooth point, and $L$ an ample Cartier divisor.
Assume that there exist positive numbers $\sigma_p$ for $p=1,2, \cdots, n$
which satisfy the following conditions:
 (1) $\sqrt[p]{(L)^p \cdot W} \geq \sigma_p$ for
any subvariety $W$ of dimension $p$ which contains $x_0$,
(2)$\sigma_p \geq n$ for all $p$, and $\sigma_n> n$.
Then $|K_X+L|$ is free at $x_0$.

Fujita [F] proved that, if $\sigma_1 \geq 3$, $\sigma_2 \geq \sqrt{7}$,
and $\sigma_3 > \sqrt[3]{51}$, then $| K_X + L |$ is free at $x_0$.
Our results are generalizations of the following result of Kawamata [K3]:
Let $X$ be a normal projective variety of dimension $3$,
$L$ an ample Cartier divisor, and $x_0 \in  X$ a smooth point.
Assume that there are positive numbers $\sigma_p$
for $p = 1,2,3$ which satisfy the following conditions:
(1) $\sqrt[p]{(L)^p \cdot W} \geq \sigma_p$ for
any subvariety $W$ of dimension $p$ which contains $x_0$,
(2) $\sigma_1 \geq 3$, $\sigma_2 \geq 3$, and $\sigma_3 > 3$.
Then $| K_X + L |$ is free at $x_0$.

We shall prove the following results in this paper:
Let $X$ be a normal projective variety of dimension $3$,
$x_0 \in X$ a point, and $L$ an ample ${\mathbb Q}$-Cartier divisor
such that $K_X+L$ is a Cartier divisor at $x_0$.
Assume that $\sqrt[p]{(L)^p \cdot W} \geq \sigma_p$ for
any subvariety $W$ of dimension $p$ which contains $x_0$.\\
$(1)$(Theorem 3.1)
Let $x_0 \in X$ be a Gorenstein terminal singular point.
Assume that $\sigma_1 > (\sqrt[3]{2}+\sqrt{3})/\sqrt[3]{2}$,
$\sigma_2>\sqrt[3]{2}+\sqrt{3}$, and $\sigma_3>\sqrt[3]{2}+\sqrt{3}$.
Then $| K_X + L | $ is free at $x_0$.
$(\mathrm{Note\ that\ }3>\sqrt[3]{2}+\sqrt{3}>2.99)$\\
$(2)$(Theorem 3.3)
Let $x_0 \in X$ be a Gorenstein terminal singular
${\mathbb Q}$-factorial point.
Assume that $\sigma_1 \geq 2$, $\sigma_2 \geq 2 \sqrt{2}$, and
$\sigma_3 >  2 \sqrt[3]{2}$.
Then $| K_X + L | $ is free at $x_0$.\\
$(3)$(Theorem 3.4)
Let $x_0 \in X$ be a quotient singular point of type
$({1}/{r},{1}/{r},{1}/{r})$ for an integer $r$.
Assume that $\sigma_1 \geq {3}/{r}$, $\sigma_2 \geq {3}/{\sqrt{r}}$,
and $\sigma_3 > {3}/{\sqrt[3]{r}}$.
Then $| K_X + L |$ is free at $x_0$.\\
$(4)$(Theorem 3.6)
Let $x_0 \in X$ be a quotient singular point of type
$(1/r,1/r,-1/r)$ for an integer $r \geq 3$.
Assume that $\sigma_1 \geq 1+(1/r)$, $\sigma_2 \geq (1+(1/r))\sqrt{r+3}$,
and $\sigma_3 > (1+(1/r))\sqrt[3]{r+2}$.
Then $| K_X + L |$ is free at $x_0$.\\
$(5)$(Theorem 3.8)
Let $X$ be a projective threefold with only canonical singularities and
$x_0 \in X$ be a canonical and not terminal singular point.
Assume that $\sigma_1 \geq {3}$, $\sigma_2 \geq {3}$, and $\sigma_3 > {3}$.
Then $| K_X + L |$ is free at $x_0$.

The result (1), the result (5), and the result
for the smooth case [K3] imply the following:
Corollary (Corollary 3.9).
Let X be a projective variety of dimension $3$
and $H$ an ample Cartier divisor on $X$. Assume that $X$ has at most
canonical Gorenstein singularities. Then $|K_X+mH |$ is free if $m \geq 4$.
Moreover, if $(H^3) \geq 2$, then $|K_X +3H|$ is also free.
Note that Lee[L1, L2] also obtained a result similar to this corollary
independently.

We also give a series of examples (Example 3.5) which shows that
our estimate in $(3)$ is optimal for each $r$.
Note that the estimates for $\sigma_p$ in cases $(1), (2)$ and
$(3)$ are better than the one in the smooth case. On the other hand,
our estimate in $(4)$ is worse than in the smooth case, especially
when the indices $r$ being large, indeed,
$\sigma_3 \to +\infty$ if $r \to +\infty$.
In Example 3.7, we construct an infinite sequence $(X_r,L_r)
(r=1,2,3,\cdots)$ consisting of a $3$-fold $X_r$ having a singular point
of type $1/r(1,1,-1)$ and an ample ${\mathbb Q}$-Cartier divisor $L_r$
on $X_r$ which satisfies that $(L_r^3) \to +\infty$ if $r \to +\infty$,
$K_{X_r}+L_r$ is Cartier at the point,
but $|K_{X_r}+L_r|$ is not free at the point for each $r$.
This sequence shows that there exists no uniform estimate for $\sigma_3$
independent of the indices $r$ and also explains the reason
why $\sigma_3 \to +\infty$ under $r \to +\infty$ in our estimate $(4)$.

Our proof is very similar to the one for the smooth case given in [K3].
However this involves more  careful and  detailed analysis of multiplicities
and discrepancies. For example, important differences for proof in between
the case $(1)$ and the smooth case are: the discrepancy coefficient for $K_X$
under the blow up $x_0 \in X$ is $1$ in $(1)$,
while it is $2$ in the smooth case;
$\mathrm{mult}_{x_0}X=2$ in $(1)$, while it is $1$ in the smooth case
and the multiplicity $\mathrm{mult}_{x_0}S$ of
the minimal center $S$ (Definition 1.2) at $x_0$ is $1$, $2$ or $3$ in $(1)$
while $\mathrm{mult}_{x_0}S= 1$ or $2$ in the smooth case if $S$ is a surface.
We also notice that in cases $(3)$ and $(4)$ the discrepancy coefficients
for $K_X$ under the blow up $x_0 \in X$ are no more integers;
it is $3/r-1$ in $(3)$ and $1/r$ in $(4)$.
By this reason, we need to treat ${\mathbb Q}$-Cartier divisors
whose orders $\mathrm{ord}_{x_0}D$ at $x_0$ is of the form $d/r$ and
especially for Theorem 3.4, the case $(3)$, we need Lemma 2.3,
a generalization of [K3, Theorem 2.2] to ${\mathbb Q}$-Cartier divisors of
fractional orders at $x_0$.

Acknowledgment:
The author would like to express his thanks to
Professor Yujiro Kawamata
for his advice and warm encouragement.
Also the author would like to express his thanks to
Referee for his useful and kind comments.

 \section{Preliminaries}
\hspace{1.5 em}
Most of the results of this paper are the applications of
the following vanishing theorem:
\begin{thm}[{[K1, V]}]
Let $X$ be a smooth projective variety and
$D$ a ${\mathbb Q}$-divisor.
Assume that $D$ is nef and big, and that the support
of the difference $ {}^{\lceil} {D} { }^{\rceil} - D $ is a normal crossing
divisor. Then $ H^p (X, K_X + {}^{\lceil} {D} { }^{\rceil}) = 0 $
for $p > 0$.
\end{thm}

We recall notation of [K3](cf [KMM]).
 \begin{defn} \normalfont
 Let $X$ be a normal variety and $D =\Sigma_i d_i D_i$
 an effective ${\mathbb Q}$-divisor such that $K_X + D$ is ${\mathbb Q}$-Cartier.
 If $\mu:Y \rightarrow X$ is an embedded resolution of the pair $(X,D)$,
 then we can write
 \[ K_Y + F = {\mu}^* (K_X + D) \]
 with $F = {\mu}_{*}^{-1}D +\Sigma_j e_j E_j$
 for the exceptional divisors $E_j$.

 The pair $( X,D )$ is said to have only
 \textit{log canonical singularities} $(LC)\\
 (\mathrm{resp. } \mathit{kawamata\ log\ terminal\ singularities} (KLT))$
 if $d_i \leq 1 (\mathrm{resp.} < 1)$
 for all $i$ and  $e_j \leq 1(\mathrm{resp.} <1)$ for all $j$.

 A subvariety $W$ of $X$ is said to be a \textit{center of
 log canonical singularities} for the pair $( X,D )$,
 if there is a birational morphism from a normal variety
 $\mu:Y \rightarrow X$ and a prime divisor $E$ on $Y$
 with the coefficient $e \geq 1$
 such that ${\mu}(E) = W$.
 The set of all the centers of log canonical singularities is denoted
 by $CLC( X,D )$.
 The union of all the subvarieties in $CLC(X,D)$ is denoted by
 $LLC(X,D)$ and called the \textit{locus of log canonical
 singularities} for $(X,D)$.
 For a point $ x_0 \in X$, we define
 $CLC ( X,x_0,D ) = \{ W \in CLC( X,D ) ; x_0 \in W \}$.

 \end{defn}

We shall use the following propositions
proved by Kawamata [K3].
Then we shall control the singularities of
the minimal center of log canonical singularities
and replace the minimal center of log canonical
singularities by a smaller subvariety.
\begin{prop} [{[K3, 1.5,1.6]}]
Let $X$ be a normal variety and  $D$ an effective
${\mathbb Q}$-Cartier divisor such that $K_X + D $ is ${\mathbb Q}$-Cartier.
Assume that  $X$ is $KLT$ and $(X,D)$ is $LC$.
If $W_1,W_2 \in CLC(X,D)$ and $W$ is
an irreducible component of $W_1 \cap W_2$,
then $W \in CLC(X,D) $.
If $(X,D)$ is not  $KLT$ at a point $x_0 \in X $,
then there exists the unique minimal element $W_0$ of $CLC(X,x_0,D)$.
Moreover, $W_0$ is normal at $x_0$.
\end{prop}
\begin{prop}[{[K3, 1.9]}]
Let $x_0 \in X$, $D$ and $W_0$ be as in Proposition $1.3$.
Assume that $\dim W_0 = 2$.
Then $W_0$ has at most a rational singularity at $x_0$.
Moreover, if $W_0$ is singular at $x_0$, and if $D'$ is
an effective ${\mathbb Q}$-Cartier divisor on $X$ such that
${\mathrm{ord}}_{x_0} D'|_{W_0} \geq 1$, then $\{ x_0 \} \in CLC(X,x_0,D+D')$.
\end{prop}
\begin{Rema} \normalfont
The result in Proposition $1.4$ is extended to higher dimensions ([K4]).
\end{Rema}

\begin{prop}[{[K3, 1.10]}]
Let $x_0 \in X$, $D$ and $W_0$ be as in Proposition $1.3$.
Let $D_1$ and $D_2$ be effective ${\mathbb Q}$-Cartier divisors on $X$
whose supports do not contain $W_0$ and which  induce the same
${\mathbb Q}$-Cartier divisor on $W_0$.
Assume that $(X,D+D_1)$ is $LC$ at $x_0$ and
there exists an element of $CLC(X,x_0,D+D_1)$ which is properly
contained in $W_0$.
Then the similar statement holds for the pair $(X,D+D_2)$.
\end{prop}

\section{General method}
\hspace{1.5 em}
We can construct divisors which have high multiplicity at a given point
from the following lemma.
\begin{Lemm}
Let $X$ be a normal and complete variety of dimension $n$,
$L$ a nef and big ${\mathbb Q}$-Cartier divisor,
$x_0 \in X$ a point, and  $t$,$t_0$ rational numbers
such that $t>t_0>0$.
Then there exists an effective  ${\mathbb Q}$-Cartier divisor $D$ such that
$ D \sim_{\mathbb Q} t L $ and
\[ {\mathrm{ord}}_{x_0}D \geq (t_0+\epsilon )
 \sqrt[n]{ \frac{(L^n)}{ \mathrm{mult}_{x_0}X } } \]
which is a rational number for $0 \leq \epsilon
\ll \sqrt[n]{ \mathrm{mult}_{x_0}X/ (L^n)}$.
\end{Lemm}

\begin{proof}
We take $r \in {\mathbb Q}$ such that $rL$ is Cartier and
$t' \in {\mathbb Q}$ such that $t'=t/t_0 >1$.
By [K3, 2.1], there exists an effective ${\mathbb Q}$-Cartier
divisor $rD'$ such that
\[ {\mathrm{ord}}_{x_0}rD' \geq r \sqrt[n]{ \frac{(L^n)}{ \mathrm{mult}_{x_0}X } } .\]
Therefore we may take $D=t_0 D'$.
\end{proof}

We generalize [K3, 2.3] in which a given point is assumed to be
a Gorenstein $KLT$ point.
For our purpose, we need to consider the case
where a given point is a $KLT$ point.
The following proposition is the key of the proofs of our main results.
\begin{prop}
Let $X$ be a normal projective
variety of dimension $n$,
$x_0 \in X$ a $KLT$ point,
and $L$ an ample ${\mathbb Q}$-Cartier divisor such that
$K_X + L$ is Cartier at $x_0$.
Assume that there exists an effective ${\mathbb Q}$-Cartier divisor $D$
which satisfies the following conditions: \\
$(1)$ $D \sim_{\mathbb Q} t L $ for a rational number $ t<1$,\\
$(2)$ $(X,D)$ is $LC$ at $x_0$,\\
$(3)$ $\{x_0\} \in CLC(X,D)$.\\
Then  $| K_X + L | $ is free at $x_0$.
\end{prop}

\begin{proof}
(cf [K3 2.3])
Let $r=\min\{s \in {\mathbb N}$; $sL$ is Cartier\}
and $D'$ be a general member of $| mrL |$ for
$m \gg 0$ which passes through $x_0$.
Replacing $D$ by $(1- \epsilon_1)(D + \epsilon_2 D')$
for some $0 < \epsilon_i \ll 1/(rm)$,
we may assume that $x_0$ is an isolated point of
$LLC (X,D)$.
Let $\mu:Y \rightarrow X$ be an embedded resolution of the pair
$(X,D)$.
Then
\[ K_Y + E + F_1+F_2  = {\mu}^{*} ( K_X + D ) \]
where $E$ is a reduced divisor such that
$\mu( E ) = \{x_0\}$, $F_1$ is a divisor of the form
$\sum_{j} f_{1j} F_{1j}$
such that $f_{1j} < 1$ and $x_0 \in \mu(F_{1j})$,
and $F_2$ is a divisor of the form
$\sum_{i} f_{2i} F_{2i}$ such that $x_0 \notin \mu(F_{2i})$.
Then
\[ K_Y + ( 1- t ){\mu}^{*} L \sim_{\mathbb Q}
{\mu}^{*}( K_X + L ) - E - F_1-F_2 .\]
Thus
\[ H^1( Y,{\mu}^{*}( K_X + L )- E +
{}^{\lceil}{- F_1}^{\rceil} -F_{2}')=0\]
where $F_{2}' = {\mu}^{*}( K_X + L )
-{}^{\lceil}{{\mu}^{*}( K_X + L )-F_2}^{\rceil}$
and we obtain a surjection
\[ H^0(Y,{\mu}^{*}(K_X + L) + {}^{\lceil} {-{F_1}}^{\rceil} -F_{2}')
\rightarrow
H^0(E,{\mu}^{*} (K_X + L)) \cong {\mathbb C}. \]
Since ${}^{\lceil} {-F_1}^{\rceil}-F_{2}'$ is effective and exceptional
over a neighborhood of $x_0$,
\[ H^0( X, K_X + L ) \rightarrow H^0(E,{\mu}^{*}( K_X + L )) \]
is also surjective.
\end{proof}

We generalize [K3, 2.2]
in which ${\mathbb Q}$-divisors have integral order at $x_0$.
For our purpose, we need to treat ${\mathbb Q}$-Cartier divisors
of fractional orders at $x_0$.

\begin{Lemm}
Let $X$ be a normal projective variety of dimension $3$,
${x_0} \in X$ a quotient singular point of type
$({1}/{r},{1}/{r},{1}/{r})$ for an integer $r$,
$L$ an ample ${\mathbb Q}$-Cartier divisor such that $K_X + L$ is Cartier
at $x_0$,
$W$ a prime divisor with
${\mathrm{ord}}_{x_0} W = {d}/{r} \geq {1}/{r}$ for an integer $d$,
and $e$,$k$ positive rational numbers such that $de \leq 1$ and
$({k}/{r})^3 < {(L)^3}/{r^2}$.
Assume that there exists an effective ${\mathbb Q}$-divisor $D$
such that  $D \sim_{\mathbb Q} L$ and ${\mathrm{ord}}_{x_0} D \geq {k}/{r}$,
and moreover that $D \geq ekW$ for any such $D$.
Then there exists a real number $\lambda$ with $0 \leq \lambda < 1$
and $\lambda \leq max \{ 1-de ,(3de)^{-{1}/{2}} \}$
which satisfies the following condition:
if $k'$ is a positive rational number such that $k' > k$ and
\[ (\lambda \frac{k}{r} )^3 + ( \frac{1-de-\lambda }{ 1-\lambda } )^2
\{ (\frac{k'}{r} + \frac{\lambda de }{1-de-\lambda}\frac{k}{r} )^3
-(\frac{ \lambda k}{r} +
\frac{\lambda de}{1-de-\lambda}\frac{k}{r} )^3 \} < \frac{(L)^3}{r^2},\]
then there exists an effective ${\mathbb Q}$-divisor $D$ such that
$D \sim_{\mathbb Q} L$ and ${\mathrm{ord}}_{x_0} D \geq {k'}/{r}$.
(If $\lambda=1-de$,
then the left hand side
of the above inequality should be taken as a limit.)
\end{Lemm}
\begin{proof}
(cf [K3, 2.2])
We have $\mathrm{mult}_{x_0} X=r^2$.
Let $\bar{k}$ = sup \{$q$; there exists an effective ${\mathbb
Q}$-divisor $D$ such that $D \sim_{\mathbb Q} L$
and ${\mathrm{ord}}_{x_0} D={q}/{r}$ \}.
Let us define a function  $\phi(q)$ for $q \in {\mathbb Q}$
with  $0 \leq q < \bar{k}$ to be the largest real number such that
$D \geq \phi(q) W$ whenever $D \geq 0$, $D \sim_{\mathbb Q} L$ and
${\mathrm{ord}}_{x_0} D = {q}/{r}$.
Then $\phi$ is a convex function.
In fact, if ${\mathrm{ord}}_{x_0} D_i = {q_i}/{r}$ and $D_i
=( \phi(q_i) + \epsilon_i) W +$ other components for
$0 \leq \epsilon_i \ll 1$ and $i = 1$, $2$, then
${\mathrm{ord}}_{x_0} (t D_1 + (1-t) D_2 )= t ({q_1}/{r})+(1-t) ({q_2}/{r})$ and
$tD_1 + (1-t)D_2
= ( t(\phi(q_1) +\epsilon_1)+(1-t)(\phi(q_2)+\epsilon_2))W +$
other components,
hence $\phi(t q_1 +(1-t) q_2) \leq t \phi(q_1) +(1-t) \phi(q_2)$.
Since $\phi(k) \geq ek$, there exists a real number $\lambda$ such
that $0 \leq \lambda < 1$
and $\phi(q) \geq {e(q- \lambda k)}/({1-\lambda})$ for any $q$.

Let $m$ be a large and sufficiently divisible integer and
$\nu : H^0 (X,mL)\rightarrow O_{X,{x_0}}(mL)
\cong O_{X,{x_0}}$ the evaluation homomorphism.
We consider subspaces $V_i ={\nu}^{-1}({m_{x_0}}^{i} )$
of $H^0 (X,mL)$ for integers $i$
such that ${\lambda k m}/{r} \leq i \leq {k'm}/{r}$.
First, we have
\[ \mathrm{dim} V_{\lceil {\lambda km}/{r}\rceil}
\geq  \mathrm{dim} H^0 (X,mL) - r^2 \frac{ ( \lambda k m /r)^3}{3!}
+ \mathrm{lower\ terms\ in\ } m. \]
Let $D \in | mL |$ be a member corresponding to
$h \in V_i$ for some $i$.
Since we have $D \geq \phi(ir/m) m W$,
the number of conditions in order for $h \in V_{i+1}$ is at most
( the number of homogeneous polynomials of order
$i-\phi(ir/m)m ({d}/{r})$ in $3$ variables )
$\times$ $\mathrm{mult}_{x_0} X$, i.e.,
\[ r^2 \frac{ ( i- \phi(ir/m) m d/r )^2}{2!}+
\mathrm{lower\ terms\ in\ } m. \]
Therefore, we have $k' < \bar{k}$ because
\[ r^2 \frac{ (\lambda k m/r )^3}{3!} +
\sum_{i=\lceil {\lambda km/r}\rceil}^
{k'm/r-1}
r^2 \frac{ ( i - \frac{ e(ir/m -\lambda
k)}{(1-\lambda)} m {d}/{r})^2}{2!} +
\mathrm{lower\ terms\ in\ } m \]
\[= r^2 \frac{ (\lambda k m/r )^3 }{3!} +
r^2 \frac{ ( \frac{1-de-\lambda}{1-\lambda} )^2
\{ ( k'/r + \frac{ \lambda d e}{1-de-\lambda}{k}/{r})^3 -
( \lambda k/r +
\frac{ \lambda d e}{1-de-\lambda } {k}/{r})^3 \} m^3 }{3!}\]
\[+ \mathrm{lower\ terms\ in\ } m \]
\[< \frac{m^3 (L^3)}{3!} + \mathrm{lower\ terms\ in\ } m.\]
By [K3 2.2 last part], we have that $\lambda \leq {\max}
\{ 1-de,(3de)^{-{1}/{2}} \}$.
\end{proof}

The following theorem plays the important role in
the proofs of our main results.

\begin{thm}[{[A, Corollary 6]}]
Let $S$ be a normal surface, $x_0$ be a point of $S$.
Suppose $S$ has a rational singularity at $x_0$.
Let $Z$ be the fundamental cycle.

Then
\[ {\mathrm{mult}}_{x_0}S= -Z^2 \]
for all integers $k$,
\[ \dim m_{S,x_0}^k/ m_{S,x_0}^{k+1}
= k {\mathrm{mult}}_{x_0} S + 1  \]
where $m_{S,x_0}$ is the maximal ideal of $x_0$ in $S$ and
\[ \mathrm{emb} {\dim}_{x_0} S = {\mathrm{mult}}_{x_0} S + 1 .\]
\end{thm}

The following lemma plays the supporting role in
the proof of Theorem 3.6.

\begin{Lemm}
Let $(X,x_0)$ be a quotient singularity
${\mathbb C}^3/{\mathbb Z}_r(a/r,-a/r,1/r)$
with $0 < a < r$.
Let $S_i={\mathbb C}^2/{\mathbb Z}_r(i/r,1/r)$
for $i = a, -a$.
Then
\[{\mathrm{emb}}{\dim}_{x_0} X = {\mathrm{mult}}_{x_0} X + 2  .\]
\[ {\mathrm{mult}}_{x_0} X
={\mathrm{mult}}_{x_0} S_a +{\mathrm{mult}}_{x_0} S_{-a} .\]
\[ \mathrm{emb} {\dim}_{x_0} X
=\mathrm{emb} {\dim}_{x_0} S_a +\mathrm{emb} {\dim}_{x_0} S_{-a} .\]
\end{Lemm}
\begin{proof}
Let
$(xy)^{w}{x^s}{z^u}, (xy)^{w}{y^t}{z^u}  \in m_{X,x_0}^{n}/
m_{X,x_0}^{n+1}$
where $m_{X,x_0}$ is the maximal ideal of $x_0$ in $X$.
Then
${x^s}{z^u}, {y^t}{z^u}  \in m_{X,x_0}^{n-w}/
m_{X,x_0}^{n-w+1}$.
We consider that
${x^s}{z^u} \in m_{S_a,x_0}^{n-w}/
m_{S_a,x_0}^{n-w+1}$
where $m_{S_a,x_0}$ is the maximal ideal of $x_0$ in $S_a$
and that
${y^t}{z^u} \in m_{S_{-a},x_0}^{n-w}/
m_{S_{-a},x_0}^{n-w+1}$
where $m_{S_{-a},x_0}$ is the maximal ideal of $x_0$ in $S_{-a}$.
By Theorem 2.4., for $i= a, r-a$
\[ \dim m_{S_{i},x_0}^{n-w}/
m_{S_{i},x_0}^{n-w+1}
= (n-w) {\mathrm{mult}}_{x_0} S_{i} + 1 . \]
$m_{{S_a},x_0}^{n-w}/
m_{S_{a},x_0}^{n-w+1} \cap
m_{S_{-a},x_0}^{n-w}/
m_{S_{-a},x_0}^{n-w+1} = (z^r)^{n-w}$ for $0 \leq w < n$.
Then
\[ {\dim} m_{X,x_0}^{n}/m_{X,x_0}^{n+1}
\]
\[= \sum_{w=0}^{n-1}
 \{ \dim m_{S_{a},x_0}^{n-w}/ m_{S_{a},x_0}^{n-w+1}
 +\dim m_{S_{-a},x_0}^{n-w}/ m_{S_{-a},x_0}^{n-w+1}-1 \}+1
\]
\[= \sum_{w=0}^{n-1}
 \{ (n-w)({\mathrm{mult}}_{x_0} S_{a} + {\mathrm{mult}}_{x_0} S_{-a}) + 1 \}+1 \]
\[ = ({\mathrm{mult}}_{x_0} S_{a} + {\mathrm{mult}}_{x_0} S_{-a}) \frac{n^2}{2}
+({\mathrm{mult}}_{x_0} S_{a} + {\mathrm{mult}}_{x_0} S_{-a}+2) \frac{n}{2} +1\]
Hence  \[ {\mathrm{mult}}_{x_0} X ={\mathrm{mult}}_{x_0} S_{a}+{\mathrm{mult}}_{x_0} S_{-a} .\]
\[ {\mathrm{emb}}{\dim}_{x_0} X ={\mathrm{mult}}_{x_0} S_{a} +{\mathrm{mult}}_{x_0} S_{-a} +2 .\]
\end{proof}

\section{Main Theorem}
\hspace{1.5 em}
First we consider Gorenstein terminal singular points.
\begin{thm} Let $X$ be a normal projective variety of dimension $3$,
$x_0 \in X$ a Gorenstein terminal singular point, and
$L$ an ample ${\mathbb Q}$-Cartier divisor such that
$L$ is Cartier at $x_0$.
Assume that there are positive numbers $\sigma_p$
for $p = 1,2,3$ which satisfy the following conditions: \\
$(1)$ $\sqrt[p]{(L)^p \cdot W} \geq \sigma_p$ for
any subvariety $W$ of dimension $p$ which contains $x_0$,\\
$(2)$ $\sigma_1 > (\sqrt[3]{2}+\sqrt{3})/ \sqrt[3]{2}$,
$\sigma_2 > \sqrt[3]{2}+\sqrt{3}$,
and $\sigma_3 > \sqrt[3]{2}+\sqrt{3}$.\\
Then $| K_X + L |$ is free at $x_0$.
\end{thm}
\begin{proof}
Since $x_0$ is  a Gorenstein terminal singularity of dimension $3$,
$x_0$ is an isolated hypersurface
singularity of multiplicity 2 ([R, (1.1) Theorem]).
Let $U$ be a neighborhood at $x_0$.
Let $U \subset V$ be an embedding of $U$ as
a hypersurface in a smooth fourfold $V$, and
let $g:\widetilde{V} \longrightarrow V$
be the blowing-up of $V$ at $x_0$.
Let $\widetilde{U}$ be the proper transform of $U$
in $\widetilde{V}$, $f: \widetilde{U} \longrightarrow U$
the restriction of $g$ to $\widetilde{U}$,
and $F \subset \widetilde{V}$
the exceptional divisor of $g$.
Then
\[  K_{\widetilde{U}}= ( K_{\widetilde{V}} + \widetilde{U} ) |_{\widetilde{U}}
=(g^{*} K_V +3F + g^{*}U -2F )|_{\widetilde{U}}
=f^{*}K_U+ (F|_{\widetilde{U}}). \]
(cf.[ELM, Lemma 2.2 's proof, l.8--l.13])

Step 0.
Let $t$ be a rational number such that
$t>{2\sqrt[3]{2}}/{\sqrt[3]{(L^3)}}$.
Since $\sigma_3 > 2\sqrt[3]{2}$, we can take $t < 1$.
Let $t_0$ be a rational number such that
$t_0=2\sqrt[3]{2}/ \sqrt[3]{{(L^3)}}-\epsilon$
for $0 \leq \epsilon \ll \sqrt[3]{ 2/(L^3)}$.
By Lemma 2.1, there exists an effective ${\mathbb Q}$-Cartier
divisor $ D$
such that $ D \sim_{\mathbb Q} t L $ and
${\mathrm{ord}}_{x_0} D \geq (t_0 +\epsilon) \sqrt[3]{(L^3)/2}$.
Hence ${\mathrm{ord}}_{x_0} D =2$.

Let $c$ be the
\textit{log canonical threshold} of $ ( X,D )$ at $x_0$:
\[ c=\sup{ \{t \in {\mathbb Q}; \mbox{ $K_X +tD$ is $LC$ at $x_0$ } \} }. \]
Then $c \leq 1$.
Let $W$ be the minimal element of $CLC(X,x_0,cD)$.
If $W=\{x_0\}$, then $| K_X+L |$ is free at $x_0$ by Proposition 2.2,
since $c t <1$.\\

Step 1. We consider the case in which $W =C$ is a curve.
By Proposition 1.3, $C$ is normal at $x_0$, i.e., smooth at $x_0$.
Since $ t < 1$, we have
$c t +(1-c)<1$. Since $\sigma_1 \geq 2$, there exists a rational number $t'$
with $c t +(1-c)<t'<1 $ and an effective ${\mathbb Q}$-Cartier divisor $D_C' $
on $C$ such that  $D_C' \sim_{\mathbb Q}(t'-c t )L|_C$ and
${\mathrm{ord}}_{x_0}D_C' =2(1-c)$.
As in [K3, 3.1 Step1], there exists an effective ${\mathbb Q}$-Cartier
divisor $D'$ on $X$ such that $D'\sim_{\mathbb Q}(t'-c t )L$
and $ D'|_C=D_C'$.
Let $D_1'$ be a general effective ${\mathbb Q}$-Cartier divisor
on an affine neighborhood $U$ of $ x_0$ in $ X$ such that
$D_1'|_{C \cap U} = D_C'|_{C \cap U}$ and ${\mathrm{ord}}_{x_0}D_1'=2(1-c)$.
Then we have  ${\mathrm{ord}}_{x_0}(cD+D_1')=2$, hence $\{x_0\} \in CLC(U,cD+D_1')$.
Let
\[ c' =\sup{\{ t \in {\mathbb Q};
\mbox{ $K_X +(cD+tD_1')$ is $LC$ at $x_0$ } \} }. \]
Since $D_1'$ is chosen to be general, we have $c' > 0$.
We have an element $W'$ such that $W'\in CLC(X,x_0,cD+c'D_1')$ and
$W' \not\supseteq C$. By Proposition 1.3, $CLC(X,x_0,cD+c'D_1')$ has
an element which is properly contained in $C$.
By Proposition 1.6, we conclude that $(X,cD+c'D')$ is $LC$ at $ x_0$, and
$CLC(X,x_0,cD+c'D')$ has
an element which is properly contained in $ C$, i.e., $\{x_0\}$.\\

Step 2. We consider the case in which $W =S$ is a surface.
By Proposition 1.4, $S$ has at most a rational singularity at $x_0$.\\
Step2-1. We assume first that $S$ is smooth at $x_0$.
As in Step 1, we take a rational number $t'$,
an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$
and a positive number $c'$
such that $c t +(1-c)<t'<1$, $D' \sim_{\mathbb Q}(t'-c t )L$,
${\mathrm{ord}}_{x_0}D'|_S, =2(1-c)$, $(X,cD+c'D')$ is $LC$ at $x_0$, and
that the minimal element $W'$ of $CLC(X,x_0,cD+c'D')$
is properly contained in $S$.
Thus we have the theorem when $W'=\{x_0\}$.

We consider the case in which $W'=C$ is a curve.
Since $t, t' < 1 $, we have $c t + c'(t'-c t)+(1-c)(1-c') <1$.
As in Step 1, we take a rational number $t''$,
an effective ${\mathbb Q}$-Cartier divisor $D''$ on $X$
and a positive number $c''$ such that
$c t + c'(t'-c t)+(1-c)(1-c') <t''<1$,
$D''\sim_{\mathbb Q}(t''-c t-c'(t'-c t))L$,
${\mathrm{ord}}_{x_0}D''|_C=2(1-c)(1-c')$,
$(X,cD+c'D'+c''D'')$ is $LC$ at $x_0$ and
that the minimal element $W''$ of $CLC(X,x_0,cD+c'D'+c''D'')$ is
properly contained in $C$, i.e., $ \{ x_0 \} \in
CLC(X,x_0,cD+c'D'+c''D'')$.

Step 2-2. We assume that $S$ has a rational singularity at $x_0$.
Since the embedding dimension of $X$ at $x_0$ is $4$ and
$x_0$ is also a singular point of $S$,
the embedding dimension of $S$ at $x_0$ is $3$ or $4$.
Therefore $d: =\mathrm{mult}_{x_0}S= 2$ or $3$, because $x_0 \in S$ is
a rational singular point([A, Corollary 6]).
Since $t>{2\sqrt[3]{2}}/{\sqrt[3]{(L^3)}}$ and
$\sqrt[3]{(L^3)}> \sqrt[3]{2}+\sqrt{3}$
, we can take $t <{ 2 \sqrt[3]{2}/ (\sqrt[3]{2}+\sqrt{3})}$
, so $t/2 +\sqrt{3}/ \sigma_2< 1$.
Therefore, if $ c \geq {1}/{2}$, then
\[ c t + \frac{ 2 \sqrt{d} (1-c)}{\sigma_2} < 1 .\]
In this case, we can take a rational number $t'$ and
an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$
such that $ c t + 2\sqrt{d}(1-c)/\sigma_2 < t' < 1 $,
$D' \sim_{\mathbb Q}(t'-ct )L$ and
${\mathrm{ord}}_{x_0}D'|_S = 2 ( 1 - c )$,
and proceed as in Step 2-1.

On the other hand, if  $ c \leq {1}/{2}$, then
\[ c t + \frac{\sqrt{d}}{\sigma_2} < 1 .\]
We take $t'$ and $D'$ with $D' \sim_{\mathbb Q}(t'-ct ) L $ and
${\mathrm{ord}}_{x_0}D'|_S = 1$, by Proposition 1.4,
$ \{ x_0 \} \in CLC( X,x_0,cD+D' )$.
As in Step 1, there exists $ c' $ such that $ 1 \geq c' > 0 $,
$ ( X,x_0,cD+c'D' )$ is $LC$ at $x_0$,
and that the minimal element of $W'$ of $CLC( X,x_0,cD+c'D' )$
is properly contained in $S$.
If $W'=\{x_0\}$ , we have the theorem.

We consider the case in which  $W'= C$ is a curve.
We have
\[{\mathrm{ord}}_{x_0}( c D + c' D')|_S \geq 2 c + c'
= 2 - 2 ( 1 - c - \frac{c'}{2}). \]
Since $ \sqrt{d}/ \sigma_2  < t' - c t $,
we can take $t'-ct<{\sqrt{d}/(\sqrt[3]{2}+\sqrt{3})}$. Then,
\begin{eqnarray*}
& & c t + c'( t'-c t ) + \frac{2}{\sigma_1}
( 1 - c - \frac{c'}{2}) \\
&<& c ({ \frac{2 \sqrt[3]{2}}{\sqrt[3]{2}+\sqrt{3}}})
+ \frac{2 \sqrt[3]{2}}{\sqrt[3]{2}+\sqrt{3}} ( 1 - c )
+ c' ({ \frac{ \sqrt{d}}{\sqrt[3]{2}+\sqrt{3}}
-\frac{\sqrt[3]{2}}{\sqrt[3]{2}+\sqrt{3}}}) \leq 1
\end{eqnarray*}
The rest is the same as before.
\end{proof}

The following example shows
that the condition $\sigma_3 > 2 \sqrt[3]{2}$ in Theorem 3.1 is optimal.
\begin{exam}\normalfont
Let $X=\{ xy+z^2+t^2=0 \} \subset  {\mathbb P}^4$ and
$x_0=(0:0:0:0:1)$.
Then $x_0$ is a Gorenstein terminal singular point and
$K_X=\mathcal{O}(-3)$.
If $L=\mathcal{O}(3)$, then $|K_X+L|$ is free at $x_0$.
If $L=\mathcal{O}(2)$, then $|K_X+L|$ is not free at $x_0$ and
$L^3 X=16$.
\end{exam}

If we assume furthermore that $X$ is ${\mathbb Q}$-factorial at $x_0$,
we obtain better estimates for $\sigma_p$.
\begin{thm}
Let $X$ be a normal projective variety of dimension $3$, $L$ an
ample ${\mathbb Q}$-Cartier divisor, $x_0 \in X$ a Gorenstein terminal
${\mathbb Q}$-factorial singular point.
Assume that there are positive numbers ${\sigma}_p$ for $p = 1,2,3$
which satisfy the following conditions: \\
$(1)$ $\sqrt[p]{ (L)^p \cdot W } \geq {\sigma}_p $
for any subvariety $W$ of dimension $p$ which contains $x_0$,\\
$(2)$ ${\sigma}_1 \geq 2$, ${\sigma}_2 \geq 2\sqrt{2}$,
and ${\sigma}_3 > 2\sqrt[3]{2}$.\\
Then $| K_X + L |$ is free at $x_0$.
\end{thm}
\begin{proof}
Since ${\sigma}_1 \geq 2$
and ${\sigma}_3 > 2\sqrt[3]{2}$,
Steps 0,1 are the same as Steps 0,1 of the proof of Theorem 3.1.\\
Step 2. We consider the case in which $W = S$ is a surface.
Since $X$ is ${\mathbb Q}$-factorial, terminal Gorenstein at $x_0$,
$X$ is factorial at $x_0$([K2, Lemma 5.1]).
In particular, $S$ is a Cartier divisor at $x_0$.
Then we have $2 > \mathrm{ord}_{x_0} cD \geq \mathrm{ord}_{x_0} S$.
Hence we have $\mathrm{ord}_{x_0}S =1$ and $\mathrm{mult}_{x_0} S = 2$.
There exists a rational number $t'$
with $ct + (1 - c) < t' < 1$.
By Lemma 2.1 and ${\sigma}_2 \geq 2 \sqrt{2}$,
there exists an effective ${\mathbb Q}$-Cartier divisor $D_S$
on $S$ such that
$D_S \sim_{\mathbb Q} (t'-ct ) L|_S$
and ${\mathrm{ord}}_{x_0}D_S =2 (1 - c)$. \\
The rest is the same as Step 2-1 of the proof of Theorem 3.1.
\end{proof}

We consider non Gorenstein singular points in the following.

The following theorem is generalization to dimension of $3$
of the following result of Kawachi [KM]:
Let $S$ be a normal projective surface,
$x_0 \in S$ a quotient singular point of type $({1}/{r},{1}/{r})$
for an integer $r$ and $L$ a nef and big ${\mathbb Q}$-Cartier divisor
such that $K_X + L$ is Cartier at $x_0$.
If $LC \geq {2}/{r}$ for any curve $C$ through $x_0$
and $\sqrt{L^2} > 2/\sqrt{r}$, then $| K_X + L |$ is free at $x_0$.\\

\begin{thm}
Let $X$ be a normal projective variety of dimension $3$,
$x_0 \in X$ a quotient singular point
of type $({1}/{r},{1}/{r},{1}/{r})$
for an integer $r$,
and $L$ an ample ${\mathbb Q}$-Cartier divisor such that
$K_X + L$ is Cartier at $x_0$.
Assume that there are positive numbers ${\sigma}_p$ for $p = 1,2,3$
which satisfy the following conditions: \\
$(1)$ $\sqrt[p]{ (L)^p \cdot W} \geq {\sigma}_p$
for any subvariety $W$ of dimension $p$ which contains $x_0$,\\
$(2)$ ${\sigma}_1 \geq {3}/{r}$,
${\sigma}_2 \geq {3}/{\sqrt{r}}$, and
${\sigma}_3 >  {3}/{\sqrt[3]{r}}$.\\
Then $| K_X + L |$ is free at $x_0$.
\end{thm}

\begin{proof}
Since $X={\mathbb C}^3/{\mathbb Z}_r (1/r,1/r,1/r)$ around $x_0$,
setting $\mathcal{O}_{{\mathbb C}^3,0}={\mathbb C } \{ x, y, z \}$,
we have $\mathcal{O}_{X,x_0}
={ \mathcal{O}_{{\mathbb C}^3,0} }^{ {\mathbb Z}_r (1/r,1/r,1/r)}
={\mathbb C} \{ x^{a} y^{b} z^{c} | a +b + c =r, a \geq 0, b \geq 0, c \geq 0 \}$
and ${m}_{X,x_0}^n=(x^{a} y^{b} z^{c} | a + b +c =n r)$
where ${m}_{X,x_0}$ is the maximal ideal of $x_0$ in $X$. Therefore
\[ \dim_{\mathbb C} \mathcal{O}_{X,x_0}/{m}_{X,x_0}^n
= \sum_{k=0}^{n-1} {(kr+2)(kr+1)}/{2}=(r^2/3!)n^3
+ \mathrm{ lower \ terms \ in \ } n.\]
Hence ${\mathrm{mult}}_{x_0} X = r^2$.\\
Step 0. Let $t$ be a rational number such that
$t >  ({3}/{\sqrt[3]{r}}) / \sqrt[3]{({L}^3)} $.
Since ${\sigma}_3 > {3}/{\sqrt[3]{r}}$, we can take $t < 1$.
Let $t_0$ be a rational number such that
$ t_0 = ({3}/{\sqrt[3]{r}}) / \sqrt[3]{({L}^3)}-\epsilon$
for $0 \leq \epsilon \ll \sqrt[3]{r^2/(L^3)}$.
By Lemma 2.1, there exists an effective ${\mathbb Q}$-Cartier divisor
$D$ such that  $D \sim_{\mathbb Q} t L$ and
${\mathrm{ord}}_{x_0} D \geq (t_0+\epsilon)\sqrt[3]{(L^3)/r^2}$.
Hence ${\mathrm{ord}}_{x_0}D =3/r$.\\
Let $f:{\bar{X}} \rightarrow X$ be the blowing-up of $X$
at $x_0$, $E \subset {\bar{X}}$ the exceptional divisor of $f$
, and ${\bar{D}}$ the proper transform of $D$ in ${\bar{X}}$.
Then
\[ K_{\bar{X}} = {f}^{*}K_X + (3/r-1) E .\]
\[{f}^{*} D = \bar{D}+ (3/r) E . \]
Let $c$ be the log canonical threshold of $ ( X,D )$ at $x_0$:
\[ c =\sup{ \{t \in {\mathbb Q}; \mbox{ $K_X +t D$ is $LC$ at $x_0$ } \} }.\]
Then $c \leq 1$.
Let $W$ be the minimal element of $CLC(X,x_0,cD)$.
If $W=\{x_0\}$, then $| K_X+L |$ is free at $x_0$ by Proposition 2.2,
since $c t <1$.\\

Step 1. We consider the case in which $W =C$ is a curve.
By Proposition 1.3, $C$ is normal at $x_0$, i.e., smooth at $x_0$.
Since $\sigma_1 \geq 3/r$,
as in Step $1$ of the proof of Theorem $3.1$,
we take a rational number $t'$,
an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$
and a positive number $c'$
such that
$c t+(1-c) < t'<1$,
$D' \sim_{\mathbb Q}(t'-c t )L$,
${\mathrm{ord}}_{x_0}D'|_C =(3/r)(1-c)$,
$(X, c D +c' D')$ is $LC$ at $ x_0$, and
that the minimal element $W'$ of $CLC(X,x_0,cD+c'D')$
is properly contained in $ C, i.e., \{x_0\}$.\\

Step 2. We consider the case in which $W=S$ is a surface.
Let $U$ be a neighborhood at $x_0$,
and $h: \widetilde{U} \longrightarrow U$
its closure of local universal cover, \\
i.e., $U={\mathbb C}^3/{\mathbb Z}_r(1/r,1/r,1/r)$
and $\widetilde{U} ={\mathbb C}^3$.
Let $\widetilde{S}$ be $h^{*} (S |_{U} )$,
and $x_1$ be $h^{-1} (x_0)$.
Since $x_0 \in S$ is a $KLT$ point [K3, 1.7] and
$h$ is ramified only over $x_0$, $\widetilde{S} \subset {\mathbb C}^3$ is
also irreducible and $x_1 \in \widetilde{S}$ is at most $KLT$.
Moreover $\widetilde{S} \subset {\mathbb C}^3$ is a hypersurface,
$\widetilde{S}$ is also Gorenstein. Therefore,
$x_1 \in \widetilde{S}$ is a rational Gorenstein point, that is,
$x_1 \in \widetilde{S}$ is a smooth point or a rational double point.
If $x_1$ is a smooth point of $\widetilde{S}$,
then $S |_{U} \cong {\mathbb C}^2/{\mathbb Z}_r(1/r,1/r)$.
Hence we have $\mathrm{mult}_{x_0} S = r$.
If $x_1$ is a rational double point of $\widetilde{S}$,
that $\widetilde{S}$ is $A_{rn+1}$ type $(n \in {\mathbb Z}_{\geq 0})$,
i.e.,
$\widetilde{S} \cong \mathrm{Spec}{\mathbb C}[ x^{rn+2},y^{rn+2}, xy ]$,
since $\widetilde{S}$ is invariant by
the action of ${\mathbb Z}_r (1/r,1/r,1/r)$.
Since the action of ${\mathbb Z}_r$ on ${\mathbb C}^3$ is a scalar
multiplication by $\xi$, we may assume that
the equation of $x_1 \in \widetilde{S}$ is of the standard form
under the same coordinates of ${\mathbb C}^3$.
Then by looking the action of ${\mathbb Z}_r$ on the equation and
using the fact that the equation must be semi-invariant,
we conclude the result. Since
\[ S|_{U}=\mathrm{Spec}{\mathbb C}[
(x^{rn+2})^{r},(x^{rn+2})^{r-1}(x y),\ldots ,(x^{rn+2})(x y)^{r-1}, \]
\[(xy)^{r},(y^{rn+2})^{r}, \ldots, (y^{rn+2})(x y)^{r-1} ],\]
the embedding dimension of $S|_U$ is $2r +1$.
Hence we have ${\mathrm{mult}}_{x_0} S = 2r$ ([A, Corollary 6]).
Hence ${\mathrm{mult}}_{x_0} S = 2r$ and ${\mathrm{ord}}_{x_0} S = 2/r$
 or ${\mathrm{mult}}_{x_0} S =r$ and ${\mathrm{ord}}_{x_0} S = 1/r$.
Let $d:={\mathrm{mult}}_{x_0} S /r = 1$ or $2$.\\

Step 2-1.
We assume first that $d = 1$.
As in Step 1,
there exists a rational number $t'$
with $c t + (1 - c) < t' < 1$.
By Lemma 2.1 and ${\sigma}_2 \geq 3/ \sqrt{r}$,
there exists an effective ${\mathbb Q}$-Cartier divisor $D_S$
on $S$ such that
$D_S \sim_{\mathbb Q} (t'-c t ) L|_S$ and
${\mathrm{ord}}_{x_0}D_S =(3/r)(1 - c)$.\\
The rest of Step 2-1 is the same as Step 2-1 of the proof of
Theorem 3.1.\\

Step 2-2.
We assume that $d = 2$.
As in Step 2-1,
we take a rational number $t'$ with $c t +\sqrt{2}(1-c) < t'$
and  an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$ with
$D' \sim_{\mathbb Q}(t'-c t)L$ and
${\mathrm{ord}}_{x_0}D'|_S =(3/r)(1-c)$.
Here we need the factor $\sqrt{2}$
because $S$ has multiplicity $2r$ at $x_0$.
Then we take $0 < c' \leq 1$ such that
$(X, c D +c' D')$ is $LC$ and $CLC(X,x_0,cD+c'D')$ has
an element which is properly contained in $S$.

We shall prove that we may assume $c t +\sqrt{2}(1-c) < 1$.
Then we can take $t' < 1$ as in Step 2-1,
and the rest of the proof is the same.
For this purpose, we apply Lemma 2.3.
In argument of Steps 0 through 2-1, the number $t$ was chosen
under the only condition that $t<1$.
So we can take $t=1-\epsilon_1$, where the $\epsilon_n$ for
$n = 1,2,...$ will stand for very small positive rational numbers.
Then $k/r= 3/(r(1-\epsilon_1))
=(3/r) +\epsilon_2$ and $e=1/(3c)$.
This means the following:
for any effective $D \sim_{\mathbb Q} tL$,
if ${\mathrm{ord}}_{x_0} D \geq 3/r$,
then $cD \geq S$.
We look for $k'> 6/(3-\sqrt{2})$
so that there exists an effective ${\mathbb Q}$-Cartier divisor
$D \sim_{\mathbb Q} tL$ with $t < (3-\sqrt{2})/2$
and ${\mathrm{ord}}_{x_0} D \geq 3/r$.
The equation for $k'$ becomes
\[ {\lambda}^{3} +( \frac{1-2e-\lambda}{1-\lambda})^2
\{(\frac{2}{3-\sqrt{2}}+\frac{2 \lambda e}{1-2e-\lambda})^{3}
-(\lambda + \frac{2 \lambda e}{1-2e-\lambda})^{3} \} < 1.  \]
We have $\lambda \leq 1/ \sqrt{6e}$,
$1/3 \leq e \leq 1/2$, and
in particular, $0 \leq \lambda \leq 1/ \sqrt{2}$.
By [K3, 3.1 Step2-2],
we obtain a desired $D$, and can choose a new $t$
such that $t < (3-\sqrt{2})/2$.
Then we repeat the preceding argument from Step 0.
If we arrive at Step 2-2 again, then we have
$2/3 \leq c \leq 1$ and $ct +\sqrt{2}(1-c) < 1$.
\end{proof}

The following example shows that the conditions in Theorem 3.4
is best possible.

\begin{exam}\normalfont
Let $X={\mathbb P}(1,1,1,r)$ and $x_0=(0:0:0:1)$.
Then $x_0$ is a quotient singular point of type $(1/r,1/r,1/r)$ and
$K_X = \mathcal{O}(-3-r)$.
If $K_X + L$ is Cartier and $L$ is effective,
we have $L = \mathcal{O}(rk+3)$ $(k \in {\mathbb Z},rk+3 \geq 0)$.\\
If $L = \mathcal{O}(3)$,
then $| K_X + L |$ is not free at $x_0$.
Hence the condition ${\sigma}_3 > 3 /\sqrt[3]{r}$
is necessary in Theorem $3.4$.
\end{exam}

We consider non Gorenstein terminal singular points in the following.
\begin{thm}
Let $X$ be a normal projective variety of dimension $3$,
$x_0 \in X$ a quotient singular point
of type $({1}/{r},{1}/{r},{-1}/{r})$
for an integer $r \geq 3$,
and $L$ an ample ${\mathbb Q}$-Cartier divisor such that
$K_X + L$ is Cartier at $x_0$.
Assume that there are positive numbers ${\sigma}_p$ for $p = 1,2,3$
which satisfy the following conditions: \\
$(1)$ $\sqrt[p]{ (L)^p \cdot W} \geq {\sigma}_p$
for any subvariety $W$ of dimension $p$ which contains $x_0$,\\
$(2)$ ${\sigma}_1 \geq 1+(1/r)$,
${\sigma}_2 \geq (1+(1/r)){\sqrt{r+3}}$, and
${\sigma}_3 >  (1+(1/r)){\sqrt[3]{r+2}}$.\\
Then $| K_X + L |$ is free at $x_0$.
\end{thm}

\begin{proof}
Since
\[{\mathbb C}^3/{\mathbb Z}_r(1/r,1/r,-1/r) =\mathrm{Spec}{\mathbb C}[
x^{r}, x^{r-1}y ,\ldots, x y^{r-1},y^{r}, x z, y z, z^{r}],\]
by Lemma 2.5, ${\mathrm{mult}}_{x_0} X = {\mathrm{emb}}{\dim}_{x_0} X - 2 = r + 2$.
Hence ${\mathrm{mult}}_{x_0} X = r+2$.
Step 0. Let $t$ be a rational number such that
$t >  (1+(1/r))\sqrt[3]{r+2} / \sqrt[3]{({L}^3)}$.
Since ${\sigma}_3 > (1+(1/r))\sqrt[3]{r+2}$, we can take $t < 1$.
Let $t_0$ be a rational number such that
$t_0 = (1+(1/r))\sqrt[3]{r+2} / \sqrt[3]{({L}^3)}-\epsilon$
for $0 \leq \epsilon \ll  \sqrt[3]{(r+2)/(L^3)}$.
By Lemma 2.1, there exists an effective ${\mathbb Q}$-Cartier divisor
$D$ such that  $D \sim_{\mathbb Q} t L$ and
${\mathrm{ord}}_{x_0} D  \geq (t_0+\epsilon)\sqrt[3]{(L^3)/(r+2)}$.
Hence ${\mathrm{ord}}_{x_0}D=1+(1/r)$.\\
Let $f:{\bar{X}} \rightarrow X$ be the weighted blowing-up of $X$
at $x_0$, $E \subset {\bar{X}}$ the exceptional divisor of $f$
, and ${\bar{D}}$ be the proper transform of $D$ in ${\bar{X}}$.
Then
\[ K_{\bar{X}} = {f}^{*}K_X + (1/r) E .\]
\[{f}^{*} D = \bar{D}+ e E , e \geq  1+(1/r). \]
Let $c$ be the log canonical threshold of $ ( X,D )$ at $x_0$:
\[ c =\sup{ \{t \in {\mathbb Q}; \mbox{ $K_X +t D$ is $LC$ at $x_0$ } \} }.\]
Then $c \leq 1$.
Let $W$ be the minimal element of $CLC(X,x_0,cD)$.
If $W=\{x_0\}$, then $| K_X+L |$ is free at $x_0$ by Proposition 2.2,
since $c t <1$.\\

Step 1. We consider the case in which $W =C$ is a curve.
By Proposition 1.3, $C$ is normal at $x_0$, i.e., smooth at $x_0$.
Since $\sigma_1 \geq 1+(1/r)$,
as in Step $1$ of the proof of Theorem $3.1$,
we take a rational number $t'$,
an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$
and a positive number $c'$
such that
$c t+(1-c)<t'<1$,
$D' \sim_{\mathbb Q}(t'-ct )L$,
${\mathrm{ord}}_{x_0}D'|_C =( 1+(1/r) )(1-c)$,
$(X,cD+c'D')$ is $LC$ at $ x_0$, and
that the minimal element $W'$ of $CLC(X,x_0,cD+c'D')$ is
properly contained in $ C, i.e., \{x_0\}$.\\

Step 2. We consider the case in which $W=S$ is a surface.
Since the embedding dimension of $X$ at ${x_0}=r+4$,
the embedding dimension of $S$ at ${x_0} \leq r+4$.
Hence we have ${\mathrm{mult}}_{x_0} S \leq r+3$ ([A, Corollary 6]).
There exists a rational number $t'$
with $c t + (1 - c) < t' < 1$.
By Lemma 2.1 and ${\sigma}_2 \geq (1+(1/r))\sqrt{r+3}$,
there exists an effective ${\mathbb Q}$-Cartier
$D_S$ on $S$ such that
$D_S \sim_{\mathbb Q}(t'-ct ) L|_S$
and ${\mathrm{ord}}_{x_0}D_S =(1+(1/r))(1 - c)$. \\
The rest of Step 2 is the same as Step 2-1 of the proof of Theorem 3.1.
\end{proof}

The following example shows that the estimates in some terminal
singular cases is necessarily worse than
the one for the smooth case.
\begin{exam}\normalfont
Let $X={\mathbb P}(1,1,r-1,r)$ and $x_0=(0:0:0:1)$.
Then $x_0$ is a quotient singular point of type $(1/r,1/r,-1/r)$ and
$K_X = \mathcal{O}(-2r-1)$.
If $K_X+L$ is Cartier at $x_0$ and $L$ is effective,
we have $L =\mathcal{O}(rk+1)$ $(k \in {\mathbb Z},rk+1 \geq 0)$.
If $L =\mathcal{O}(r+1)$,
then $|K_X + L|$ is not free at $x_0$ and
$L^3 = (r+1)^3/r(r-1)$.
Hence if $r \geq 23$, then the condition ${\sigma}_3 >3$.
\end{exam}

We obtain the following theorem based on [K3 Theorem 3.1]
and [R, Main Theorem].
\begin{thm} Let $X$ be a projective variety of dimension $3$
with only canonical singularities,
$x_0 \in X$ be a canonical and not terminal singularity point,
$L$ an ample ${\mathbb Q}$-Cartier divisor on $X$
such that $K_X+L$ is Cartier at $x_0$.
Assume that there are positive numbers $\sigma_p$
for $p = 1,2,3$ which satisfy the following conditions: \\
$(1)$ $\sqrt[p]{(L)^p \cdot W} \geq \sigma_p$ for
any subvariety $W$ of dimension $p$ which contains $x_0$,\\
$(2)$ $\sigma_1 \geq 3$, $\sigma_2 \geq 3$, and $\sigma_3 > 3$.\\
Then $| K_X + L |$ is free at $x_0$.
\end{thm}

\begin{proof}
By a theorem of Reid [R, Main Theorem],
a partial resolution $f: Y \rightarrow X$
such that $K_Y =f^{*}K_X$, and
$Y$ has only terminal singularity points.
There exists a smooth point $y_0 \in f^{-1}(x_0)$.\\
Step 0.
Let $t$ be a rational number such that
$t >  3 / \sqrt[3]{({f^{*}L}^3)} $.
Since ${\sigma}_3 > {3}$, we can take $t < 1$.
Let $t_0$ be a rational number such that
$ t_0 = {3}/ \sqrt[3]{({f^{*}L}^3)}-\epsilon$
for $0 \leq \epsilon \ll \sqrt[3]{1/(L^3)}$.
By Lemma 2.1, there exists an effective ${\mathbb Q}$-Cartier divisor
$D$ such that  $D \sim_{\mathbb Q} t L$ and
${\mathrm{ord} }_{y_0} f^{*}D \geq (t_0+\epsilon)\sqrt[3]{(L^3)}$.
Hence ${\mathrm{ord}}_{y_0}f^{*}D =3$.\\
Let $g:{\bar{Y}} \rightarrow Y$ be the blowing-up of $Y$
at $y_0$, $E \subset {\bar{Y}}$ the exceptional divisor of $g$
, and ${\bar{f^{*}D}}$ the proper transform of $f^{*}D$ in ${\bar{Y}}$.
Then
\[ K_{\bar{Y}} = {g}^{*}K_Y + 2 E .\]
\[g^{*}f^{*} D = \bar{f^{*}D}+ 3 E . \]
\[ K_{\bar{Y}} +\bar{f^{*}D}+ E = {g}^{*}f^{*}(K_X +D) .\]
Let $c$ be the log canonical threshold of $ ( X,D )$ at $x_0$:
\[ c=\sup{ \{t \in {\mathbb Q}; \mbox{ $K_X +t D$ is $LC$ at $x_0$ } \} }.\]
Then $c \leq 1$.
Let $W$ be the minimal element of $CLC(X,x_0,cD)$.
If $W=\{x_0\}$, then $| K_X+L |$ is free at $x_0$ by Proposition 2.2,
since $c t <1$.\\

Step 1. We consider the case in which $W =C$ is a curve.
By Proposition 1.3, $C$ is normal at $x_0$, i.e., smooth at $x_0$.
Since $\sigma_1 \geq 3$,
as in Step $1$ of the proof of Theorem $3.1$,
we take a rational number $t'$,
an effective ${\Bbb Q}$-Cartier divisor $D'$ on $X$
and a positive number $c'$
such that
$c t+(1-c) < t'<1$,
$D' \sim_{\mathbb Q}(t'-c t )L$,
${\mathrm{ord}}_{x_0}D'|_C =3(1-c)$,
$(X, c D + c' D')$ is $LC$ at $ x_0$, and
that the minimal element $W'$ of $CLC(X,x_0,cD+c'D')$
is properly contained in $ C, i.e., \{x_0\}$.\\

Step 2. We consider the case in which $W=S$ is a surface.
We have $S' =f^{-1} S \in CLC(Y,y_0,f^{*}c D)$.
By Proposition 1.4,
$S'$ has at most a rational singularity at $y_0$.
Step 2-1.
We assume first that $S'$ is smooth at $y_0$.
As in Step 1,
there exists a rational number $t'$
with $c t + (1 - c) < t' < 1$.
By Proposition 2.1 and ${\sigma}_2 \geq 3$,
there exists an effective ${\mathbb Q}$-Cartier divisor $f^{*}(D_S)$
on $S'$ such that
$D_S \sim_{\mathbb Q} (t'-c t ) L|_S$ and
${\mathrm{ord}}_{y_0}f^{*}(D_S) =3(1 - c)$.\\
The rest of Step 2-1 is the same as Step 2-1 of the proof of
Theorem 3.1.\\
Step 2-2.
We assume that $d = 2$.
As in Step 2-1,
we take a rational number $t'$ with $c t +\sqrt{2}(1-c) < t'$
and  an effective ${\mathbb Q}$-Cartier divisor $D'$ on $X$ with
$D' \sim_{\mathbb Q}(t'-c t)L$ and
${\mathrm{ord}}_{y_0}f^{*}(D'|_S) =3(1-c)$.
Here we need the factor $\sqrt{2}$
because $S'$ has multiplicity $2$ at $y_0$.
Then we take $0 < c' \leq 1$ such that
$(X, c D +c' D')$ is $LC$ and $CLC(X,x_0,cD+c'D')$ has
an element which is properly contained in $S$.

We shall prove that we may assume $c t +\sqrt{2}(1-c) < 1$.
Then we can take $t' < 1$ as in Step 2-1,
and the rest of the proof is the same.
For this purpose, we apply Lemma 2.3.
In argument of Steps 0 through 2-1, the number $t$ was chosen
under the only condition that $t<1$.
So we can take $t=1-\epsilon_1$, where the $\epsilon_n$ for
$n = 1,2,...$ will stand for very small positive rational numbers.
Then $k= 3/(1-\epsilon_1)
=3 +\epsilon_2$ and $e=1/(3c)$.
This means the following:
for any effective $D \sim_{\mathbb Q} t L$,
if ${\mathrm{ord}}_{y_0} f^{*}D \geq 3$,
then $c f^{*}D \geq S'$.
We look for $k'> 6/(3-\sqrt{2})$
so that there exists an effective ${\mathbb Q}$-Cartier divisor
$D \sim_{\mathbb Q} t L$ with $t < (3-\sqrt{2})/2$
and ${\mathrm{ord}}_{y_0} f^{*}D \geq 3$.
The equation for $k'$ becomes
\[ {\lambda}^{3} +( \frac{1-2e-\lambda}{1-\lambda})^2
\{(\frac{2}{3-\sqrt{2}}+\frac{2 \lambda e}{1-2e-\lambda})^{3}
-(\lambda + \frac{2 \lambda e}{1-2e-\lambda})^{3} \} < 1.  \]
We have $\lambda \leq 1/ \sqrt{6e}$,
$1/3 \leq e \leq 1/2$, and
in particular, $0 \leq \lambda \leq 1/ \sqrt{2}$.
By [K2, 3.1 Step2-2],
we obtain a desired $f^{*}D$, and can choose a new $t$
such that $t < (3-\sqrt{2})/2$.
Then we repeat the preceding argument from Step 0.
If we arrive at Step 2-2 again, then we have
$2/3 \leq c \leq 1$ and $c t +\sqrt{2}(1-c) < 1$.
\end{proof}

Theorem 3.1, Theorem 3.8, and
the result for the smooth case [K3 Theorem 3.1]
imply the following corollary in which the estimate is better
than the one in [OP, Theorem 2]

\begin{Coro}
Let $X$ be a projective variety of dimension $3$
with only \\Gorenstein canonical singularities, and
$H$ an ample Cartier divisor. \\
Then $| K_X + m H |$ is free if $m \geq 4$.
Moreover, if $(H^3) \geq 2$, then
$| K_X + 3H |$ is also free.
\end{Coro}

We obtain the following corollary from Corollary 3.9 and [OP Theorem 1,3]
\begin{Coro}[{ cf [OP Theorem I,II]}]
 Let $(X,L)$ be a polarized canonical Calabi-Yau threefolds.
$(1)$ $| m L |$ gives a birational map when  $m \geq 5$.
$(2)$ $| m L |$ is free if $m \geq 4$.
$(3)$ $| m L |$ is very ample when $m \geq 10$.
\end{Coro}


text:
Curriculum vitae | Zuzana Beňušková
Short biography:
Zuzana Beňušková (1960) was born and lives in Bratislava, Slovakia. In 1983 she graduated in ethnology from the Comenius University in Bratislava and started to work at the Slovak Academy of Sciences, Institute of Ethnology, devoting to rituals, religiosity, ethnic minorities and the town ethnology. Since 1996 she also teaches at the Faculty of Philosophy of the University of Constantine Philosopher (UCP) in Nitra. Since 2004 she became the professor assistant and 2013 university professor. 2005 – 2011 she was the Head of the Department of Ethnology and Ethnomusicology at the UCP University. She is the author of over 80 papers, a successful publication project "The Traditional Culture of the Regions in Slovakia" (VEDA1998, 2005, DAJAMA 2010 under the title The Folk Culture), Religiozita a medzikonfesionálne vzťahy v lokálnom spoločenstve/The Religiosity and Inter–Confession Relations in a Local Community (ÚEt SAV 2004) and Tekovské Lužany : kultúrna a sociálna diverzita vidieckeho spoločenstva / Tekovské Lužany: cultural and social diversity of the rural community (UKF, Nitra 2011), Prečo sú Vianoce tradičné a nové/Why Christmas is traditional and new? (ÚEt SAV 2014), Občianske obrady na Slovensku/The civil rituals on Slovakia (ÚEt SAV 2017). The result ot the research of the history of Slovak ethnology is a series of audiovisual documents.
She was the editor of the journal Etnologické rozpravy (The Discussions on Ethnology) in 2002 –2003; she was 2002-2008 the chairperson of the Slovak Society of Ethnology (Národopisná spoločnosť Slovenska). Since 2006 – 2016 she has been a member of the Organizing Committee of Etnofilm Čadca international film festival.

redpajama_set_name: RedPajamaCommonCrawl


text:
/**
 * Angular Services tests
 */

var search_term = "search string";
var search_filter =  {isOpen : true,
                      sampleType : {name : 'Paint Cross Section'},
                      earliestDate : '1850',
                      latestDate : '1900'};
var search_items = 10; /*page size*/
var search_page = 1;   /*page number*/

var excel_url = "sample?fulltext=search string&sampletype=Paint Cross Section&startdate=1850&enddate=1900";
var search_url = "sample?pageSize=10&pageNum=1&fulltext=search string&sampletype=Paint Cross Section&startdate=1850&enddate=1900";
var searchsize_url = "sample?count=true&fulltext=search string&sampletype=Paint Cross Section&startdate=1850&enddate=1900";

describe('myApp.services', function() {

    /*check external interface exists and has a valid definition*/
    beforeEach(module('myApp.services'));

    describe("catsAPIservice", function() {

        var catsAPIservice;
        var httpBackend;

        /*get the service you want to test*/
        beforeEach(inject(function(_catsAPIservice_, $httpBackend) {
            catsAPIservice = _catsAPIservice_;
            httpBackend = $httpBackend;
        }));

        it("catsAPIservice.search should build the correct url from the arguments", function () {

            /*this will only get hit if catsAPIservice.search builds the url correctly
             * it will mock the http service and send a response */
            httpBackend.whenGET(search_url).respond("ok");

            catsAPIservice.search(search_term, search_filter, search_items, search_page).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.searchSize should build the correct url from the arguments", function () {

            httpBackend.whenGET(searchsize_url).respond("ok");

            catsAPIservice.searchSize(search_term, search_filter).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.createSample should build the correct request from the arguments", function () {

            var create_sample_url = "sample";
            var create_sample_data = "sample data";
            var create_sample_headers = {'Content-Type' : 'application/json'};

            httpBackend.whenPOST(create_sample_url, create_sample_data).respond("ok");

            catsAPIservice.createSample(create_sample_data).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.delete should build the correct request from the argument", function () {

            var delete_sample_id = '12345';
            var delete_sample_url = 'sample/' + delete_sample_id;

            httpBackend.whenDELETE(delete_sample_url).respond("ok");

            catsAPIservice.delete(delete_sample_id).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.createArtwork should build the correct request from the arguments", function () {

            var create_artwork_url = "artwork";
            var create_artwork_data = "artwork data";
            var create_artwork_headers = {'Content-Type' : 'application/json'};

            httpBackend.whenPOST(create_artwork_url, create_artwork_data).respond("ok");

            catsAPIservice.createArtwork(create_artwork_data).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.readArtwork should build the correct url from the argument", function () {

            var artwork_id = "55555";
            var artwork_url = "artwork/" + artwork_id;

            httpBackend.whenGET(artwork_url).respond("ok");

            catsAPIservice.readArtwork(artwork_id).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.getVocab should build the correct url from the argument", function () {

            var vocab_type = "pigments";
            var vocab_url = "vocab/" + vocab_type;

            httpBackend.whenGET(vocab_url).respond("ok");

            catsAPIservice.getVocab(vocab_type).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.Excel should build the correct url from the arguments", function () {

            var excel_headers = {'Accept' : 'application/vnd.openxmlformats'};

            httpBackend.whenGET(excel_url, excel_headers).respond("ok");

            catsAPIservice.Excel(search_term, search_filter).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.login should build the correct request from the arguments", function () {

            var email = "user@smk.dk";
            var password = "secret";
            var login_url = "/login?username=" + email + "&password=" + password;

            httpBackend.whenPOST(login_url).respond("ok");

            catsAPIservice.login(email, password).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.logout should build the correct request from the arguments", function () {

            var logout_url = "/logout";

            httpBackend.whenPOST(logout_url).respond("ok");

            catsAPIservice.logout().then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.loggedin should build the correct url from the arguments", function () {

            var loggedin_url = "/loggedin";

            httpBackend.whenGET(loggedin_url).respond("ok");

            catsAPIservice.loggedin().then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
        it("catsAPIservice.updateUser should build the correct request from the argument", function () {

            var update_url = "user";
            var data = "data";

            httpBackend.whenPOST(update_url, data).respond("ok");

            catsAPIservice.updateUser(data).then(function(response) {
                response.data.should.equal("ok");
            });
            httpBackend.flush();
        });
    });
    describe("state service", function() {

        var state;
        var data = {part1: "something", part2 : "something else"};

        beforeEach(inject(function(_state_) {
            state = _state_;
        }));

        it("there must be a more meaningful test than this", function () {
            state.temp = data;
            state.temp.should.equal(data);
        });
    });
});

redpajama_set_name: RedPajamaGithub


Images courtesy of Kunsthalle Mainz. Photos by Norbert Miguletz.
"For a long time my aunt on my mother's side have recorded TV snow, because she thought they contained subliminal messages from the enemies of Lebanon. She tried very hard to decode thesis messages but she has always failed. With time, she has become addicted to TV snow and she forgot that they were messages from the enemy. She silently record and archive TV snow. Maybe because she loves snow and in Beirut it never snows. Or maybe because she wanted to become a dancer and she found in synthesis reports her own choreographic scores.
TV snow provided the inspiration for Rabih Mroué's work Between Two Battles, which is also the title of the exhibition. Between Two Battles links an everyday experience in the lives of Lebanese inhabitants – grey static – with a specific figure: the artist's aunt. We too are familiar with the blackand- white snow, or 'noise'. Several decades ago it would appear on our screens at night after the end of transmission. There is a frequent but erroneous supposition that the pattern was formed from background radiation in space, where as in reality it is an abundance of overlapping vibrations in the device itself. Rabih Mroué takes up this technical phenomenon to spin a tale about a family member, Lebanon and its people, weather phenomena and the political situation.
The title, Between Two Battles, targets the pause, the (brief) intermission between two battles. It underlines the end of one clash while simultaneously announcing the next engagement, which will undubitably follow. There is no end in sight; the defining principle is strike and counterstrike. But for now we are 'between', when peace reigns supreme in a moment of respite, inviting us to reflect.
Rabih Mroué is all too familiar with both conflict and ceasefire. Born in Lebanon in 1967, he lived through—and was highly aware of—the country's civil war from 1975 to 1990. He experienced his own family members being threatened and even injured. These existential experiences have shaped his thinking and art up to the present day, with his works exploring political developments in Lebanon and the Middle East. Images of war and terror, personal experiences and their effect on the individual all remain recurrent themes. The resulting works are anything but documentary-like in style; he takes the facts he has gathered and develops complex tales, interweaving fictitious elements with real-life events and personal experiences. Shocking news reports, violent acts, grim pictures are paired with people's descriptions, which in some cases are the stuff of legend, although related in a sober tone. In this manner he contrasts the solemnity of factual reports and images from the press, radio, television or internet with his own stories and interpretations of events, and spoken in his own language. Inordinately distrustful of the way incidents are represented in the media, he questions the truth content of such acounts, tracing how a country's history is rewritten through scattered images and reports. At the same time he actively weaves together stories and the present day, using his works to spread his own personal viewpoint on various occurrences and developments.
Rabih Mroué has now become one of the most important contemporary Lebanese artists. With wide-ranging works that blend literature, theatre, performance and fine arts, he develops new and up-to-date ways of combining the various genres and their vocabularies. He questions the demarcation between theatre and art, the relationship between an artwork's space and form, and how this work addresses the viewer. How and with which media (video, installation, performance) can the viewer be reached and drawn in to the subject matter? How can individuality unfold when life itself is in constant danger? How are the different realities (that which is lived through, described, and experienced) interwoven? Rabih Mroué focuses on the political and cultural contexts of the ongoing conflicts, directing our attention both to the construction of identity and history, and the relevance of remembrance and forgetting. Thus he supports the complex—yet absolutely necessary—process of public perception, a process which directly influences the formation and character of the individual, society, politics, the economy, and art.


Inc. VAT: £88.74 Inc. VAT: Info UK VAT charged at 20%. Ex VAT £73.95 Ex VAT Info Outside of EU and EU VAT registered companies only (not UK).
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text:
Watch series Teen Mom 2 full online free » Best site to watch Movies free and TV series online is CipFlix » 123Movies.
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text:
Home VIRAL Ex-Vice President Alex Ekwueme Is Dead
Ex-Vice President Alex Ekwueme Is Dead
Former Vice President of Nigeria, Dr Alex Ifeanyichukwu Ekwueme is dead. A statement from his family signed by his brother and the traditional ruler of Oko in Anambra State, Igwe Laz Ekwueme said that he died by 10:00 pm in a London clinic, Sunday. The statement read: "Ekwueme family regrets to announce the peaceful passing away of their patriarch, the former Vice-President of the Federal Republic of Nigeria Dr. Alex Ifeanyichukwu Ekwueme GCON. "The sad event occurred at the London Clinic at 10:00 pm on Sunday 19th November
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text:
Madonna Receives Backlash After Posting Video of Her Son as George Floyd Tribute
22 Words Staff
On Monday, the tragic news came in that yet another black man has died after a police officer knelt on his neck for several minutes, causing him to suffocate to death.
As tributes flood in, Madonna has been receiving backlash for hers. Keep scrolling to see for yourself...
Things have taken a turn for the worst in the past few days...
People are hurt, angry, and devastated at the outcome of a police officer's fatal actions on a black man in Minneapolis.
Police brutality is one of society's most lethal issues.
About 1 in 1,000 black men and boys in America can expect to die at the hands of police, according to a new analysis of deaths involving law enforcement officers.
It's a harsh reality...
"That 1-in-1,000 number struck us as quite high," said study leader Frank Edwards, a sociologist at Rutgers University. "That's better odds of being killed by police than you have of winning a lot of scratch-off lottery games."
This makes black people twice as likely to die at the hands of a police officer than white people...
And nothing much has been done in the last few years, despite these harrowing statistics.
It has been predicted that around 1 in 65 deaths will be a black person at the hands of the police.
"This epidemic is disproportionately affecting black people," said Brittany Packnett, an activist and member of the White House task force on policing. "We are wasting so many promising young lives by continuing to allow this to happen."
The police are more likely to use racial profiling from criminal records...
And this is why black people are more likely to be randomly handcuffed, searched, and detained than white people.
This is not okay.
Yet nothing seems to be getting done about this imminent problem in our country.
Growing up in America as a black person isnot easy...
And sadly, things just seem to be getting worse, even during lockdown.
The tragic death of Ahmaud Arbery has made headlines in the last few weeks.
Arbery was shot dead by police in Brunswick back in February while he was jogging through the Satilla Shores neighborhood.
Cellphone footage showing the moment he was shot has prompted national outrage since it surfaced online...
The twenty-eight-second video shows Arbery jogging around a neighborhood on a sunny afternoon in late February. The footage ends with 2 loud gunshots.
After the video circulated on social media...
This man was hunted down like an animal and his life was thrown away like he was worthless this could have been me… https://t.co/X2pegmyiIJ
— Jay (@Jay)1588806057.0
A large crowd of protesters marched through the area where Arbery was killed and people have been demanding justice.
The pressure has been mounting on the police force to take action against the officers responsible for Arbery's death...
We're literally hunted EVERYDAY/EVERYTIME we step foot outside the comfort of our homes! Can't even go for a damn j… https://t.co/r2yPGUnb8c
— LeBron James (@LeBron James)1588802793.0
And even though the case is currently being investigated, attitudes clearly aren't changing.
Yetanother tragic incident occurred in our country this week involving a black man and 2 white police officers.
It happened in Minneapolis on Monday, and passersby filmed 2 police officers "arresting" a black man for an assumed drug-related incident.
Police found the man, who is believed to be in his forties, matching a suspect's description in his car.
When he supposedly "resisted arrest", one of the officers got him on the ground and pinned him there by pressing their knee into his neck.
This position is clearly very painful for the man who was identified as George Floyd.
"Please, please, I can't breathe," he can be heard begging as the officers remain standing.
In the disturbing video, the man's nose begins to bleed...
"My stomach hurts. My neck hurts. Everything hurts," he moans as passers-by begin to gather and repeatedly tell the officer to stop leaning on him.
Multiple witnesses are arguing with the officers when the man starts to lose consciousness...
"He's not f****** moving!" one witness shouts as another says, "Get the f*** off him, he isn't moving!" One of the officers responds with, "This is why you don't do drugs, kids."
After around 4 minutes of the officer kneeling on him, the man is completely unresponsive.
An ambulance arrives and the paramedics check his pulse and move his body onto a stretcher - he was later pronounced dead in the hospital.
Meanwhile, the 2 police officers have shown absolutely no emotion, regret, or urgency.
Which has caused global outrage.
The officers involved have been put on paid administrative leave, per department protocol.
Two other members of the Minneapolis Police Department who were involved in Monday's incident have also been fired, and the FBI has launched an investigation into the George Floyd's death.
Minneapolis Mayor, Jacob Frey addressed the incident in a press conference on Tuesday morning, referring to events in the video as "wrong at every level."
"Being black in America should not be a death sentence. For 5 minutes we watched as a white officer pressed his knee to the neck of a black man. For 5 minutes," he said. "When you hear someone calling for help, you are supposed to help. This officer failed in the most basic human sense. What happened on Chicago and 38th, this last night, is simply awful."
It has been yet another restless night for those in the state.
With the permission of the mayor, who allowed them to go out, thousands upon thousands of protesters took to the streets to express their hurt and their anger at the injustice that had taken place less than forty-eight hours ago.
As they took to the streets, police in riot gear pelted them with rubber bullets and tear gas.
In their hands, they had signs that read "I can't breathe," and "justice for George Floyd." When the police started attacking them, they ran for cover or doused their faces in milk to prevent the horrible after-effects of coming into contact with tear gas.
The protest took an ugly turn when both sides started hurling things at each other.
As the protest continued into the night, the peaceful protest started getting more and more intense with gas canisters and more bullets being used frivolously.
Footage taken during the protest revealed some pretty disturbing sights.
At one point, some protesters sit on the ground, while officers throw smoke bombs, tear gas, and flash grenades into the crowds, injuring many.
The photographs have been just as harrowing.
It seems as though the police in the area still don't understand that people are hurt. They're hurting, just like George Floyd's family, who demand that the officers involved get put on trial.
But as people get more and more restless, it seems as if a small minority of the protestors have turned to other methods.
In order to grab the attention of authorities, some of the people protesting George Floyd's murder have resorted to looting a Target store.
The store was closest to where George Floyd was murdered.
In footage released by 6ABC Philadelphia, there is evidence that some protestors can be seen stealing goods from the store including groceries and clothes.
This is not the answer.
Unfortunately, actions such as these will only distract from the respect George Floyd deserves. It is not a good idea to tarnish his legacy with a few minutes of criminal activity.
Here's the footage:
This does not take away from what the police have done to an innocent black man. It's 2020, yet lukewarm acceptance of black folks means more deaths at the hands of the state. Black lives do matter.
Tributes for George have been flooding in.
💔 #JusticeForGeorgeFloyd https://t.co/cFazgClwNe
— Kim Kardashian West (@Kim Kardashian West)1590592604.0
And Madonna was among the famous faces paying tribute.
But her tribute has faced immense backlash...
@Madonna Shut—and let me make this as clear as possible—the FUCK UP.
— Preston Mitchum (@Preston Mitchum)1590706960.0
The singer shared a video of her son David Banda dancing to Michael Jackson's "They Don't Care About Us," but it wasn't well received.
It was clear that the moment meant something to David.
@Madonna Why are you using your black child like this for internet clicks when someone else's black child was actua… https://t.co/nrv7hSvcEi
— Victoria Brownworth #LockThemUp (@Victoria Brownworth #LockThemUp)1590710114.0
But it was the pair's choice of dance that fell flat with many people online, with some even going as far as to suggest Madonna forced her son into making the video.
Many also raised the point that Madonna's son is a child of privilege.
And that they don't feel that Madonna's son, even though he is a young black man, can truly understand the experience of being black in America.
Others simply raised the point that what she's done isn't enough.
@Madonna I really appreciate you for allowing your son to dance away the racism for us.
— openly black executive producer of vibes (@openly black executive producer of vibes)1590705754.0
There was no holding back.
Brutal murder travels around the world my son David Dances to honor and pay tribute to George and His Family and al… https://t.co/Rwz5xaqXbJ
— Madonna (@Madonna)1590702336.0
Regardless of whether you think Madonna's video was in poor taste, we must all try and focus on getting justice for George Floyd.


Let's take a look back at 2014! Shall we?
I started off 2014 being a bit of a crabby crab. But really- on January 1st I began several weeks of sickness by finding myself with a stomach flu while in the car at the mall with our family. My poor brother in law is scarred for life! Haha! This was really spiritually hard time for me. I entered into the year with a very clear vision (read that, here) and being sick in bed for several weeks kicked my tail.
I still owe y'all a full renovation recap of this room.. it's on my blogging to do list for 2015! :) It's funny that I now can't remember life before a garbage disposal.. scrubbing that grout with a toothbrush.. and that yellow! Yikes! I'm so grateful that we got to complete this renovation!
We continued our Valentines Day tradition of a picture within a picture, I turned 23, and completed my real estate salesperson certification this month.
In March I created the illusion that I attend classes all the time- Sushi 101 at Central Market and a fullosophie class with The Southern Table! I also had the honor of being a bridesmaid in my friend Michelle's wedding.. her bridal shower was this month too! We also had our home photographed for The Everygirl!
Spring is my least favorite season, and it's reflected by the photos and events (or lack of!) I have for these months of the year. I'm already brainstorming ways to add a little bit of pep into the season! I worked with Winifred Paper on baby shower invitations, picked out fabrics for my tshirt quilt, and we had the blessing of opening up our home a few times, including one that was particularly special for me.
This month we prepared for and took our trip to Europe. By far my favorite two and a half weeks of the whole year. I love traveling with Ben SO much. We have decided that we want to take our next trip to Europe during the first year of our firstborn child's life (whenever that is!) Call us crazy, but the idea of taking our babe to a bunch of fun cities is exactly what we have in mind.. I grew up traveling a lot as a baby, and really want to make that something special in our little family, too!
Aaaaaand that's the first half of 2014! Stay tuned! Ps these blog posts are probably just for me so it's okay if you didn't get this far, haha!
Wow, what a great first half of 2014!! Your Europe photos were so amazing and made me long for London and Paris (also inspired me to have a photoshoot with my future hubby with the Eiffel Tower in the background!). Your home remodel and Evergirl contribution are also AWESOME! What a beautiful home! And your monthly goals were really great too! Inspired me to get my booty back into shape by being more productive. I've really enjoyed following your blog in 2014, Victoria! Have a blessed and joyful 2015!


Ultrasound offers a non-invasive, non-painful method for diagnosing many diseases. An ultrasound is different than an x-ray in that sound, rather than radiation, is used to create the image. This enables us to not only see the structure we are looking at, but to also see inside. Unlike conventional x-rays which can cause harm if used excessively, evidence shows that ultrasound is completely safe from these side-effects, and can be performed on the pet while they are fully awake. Ultrasound is an invaluable tool in assisting in the diagnosis of various diseases including heart disease, kidney and bladder disease, liver problems, and diseases of other major organs. We can also look inside the bladder for stones and growths.
Green Valley Animal Hospital is proud and excited to offer the latest in x-ray technology – digital radiology. Until recently, x-ray technology had remained basically unchanged for almost a hundred years. New digital x-ray systems have vastly improved our diagnostic abilities by using computers to aid in the capture and manipulation of the image (including magnification) to see certain conditions more clearly. It also means you can view your pet's x-rays on the computer screen in the exam room and we can e-mail x-rays to specialists for review. The results are excellent! In addition, we no longer have to use the chemicals associated with processing x-ray film – a plus for the environment and our staff! Now x-rays are safer, quicker, and much more useful. They can be stored in a digital format and sent over the web for consultation with a specialist if necessary. Green Valley Animal Hospital has invested in this revolutionary technology to better serve you and your four legged companions.


Very nice nearly 1 acre lot in Madison Twp. Rolling Meadows subdivision. County Water is available. Soil samples and septic system design have been completed. If you've been considering building your dream home, please take a look at this lot.
Directions Middletown Eaton Road, (St Rt 122), To North On Elk Creek Rd, To Left On Eunice Dr To The Corner.