text:
\section{Introduction}

\footnotetext[1]{The term ``splayed'' means ``spread out'' or ``made oblique''. So  being ``splayed'' for a hypersurface $D=D_1 \cup D_2$ should indicate that the components $D_1$ and $D_2$  are ``spread out'' the ambient space.}

Let $M_1, M_2$ be two submanifolds of a complex manifold $S$. Then $M_1$ and $M_2$ intersect \emph{transversally} at a point $p \in S$ if their respective tangent spaces add up to the tangent space of
$S$ at $p$, that is,
\begin{equation} \label{Equ:transversalsmooth}
T_pM_1 + T_pM_2=T_pS.
\end{equation}

Transversality is a fundamental concept in algebraic geometry as well as in differential geometry.
However, in many applications (e.g., embedded resolution of singularities, Hodge structures) one needs a notion of transversality for more than two subspaces. This
leads to the notion of \emph{normal crossings}, which means that several smooth components cross in a transversal way. Another way to phrase this is that the union of several smooth subspaces is
isomorphic
to a union of coordinate subspaces. \\

In this article we study a natural generalization of the concept of transversal intersection of two hypersurfaces in complex manifolds allowing singular components. The geometric idea is that two
singular
hypersurfaces $D_1$ and $D_2$ in a complex manifold $S$ intersect transversally at a point $p$ if their ``tangent spaces'' fill out the whole space and the ideal of their intersection is reduced. The
notion of tangent space for singular hypersurfaces can be made precise by means of logarithmic derivations. In algebraic terms this means: one
can find local coordinates at $p$ such that the
defining equations of the $D_i$ can be chosen in separated variables. We call a union of such transversally
intersecting hypersurfaces  a \emph{splayed  divisor}.  This concept has already appeared in different contexts under different names, for example, J.~Damon \cite{Damon96} called germs of the form $(V_1 \times \ensuremath{\mathbb{C}}^m) \cup (\ensuremath{\mathbb{C}}^n \times V_2) \subseteq \ensuremath{\mathbb{C}}^{n+m}$, where $V_1 \subseteq \ensuremath{\mathbb{C}}^n$ and $V_2 \subseteq \ensuremath{\mathbb{C}}^m$, a \emph{product-union}, also see Remark \ref{Rem:Damon}. For an example of a splayed divisor in a threedimensional $S$, see Figure \ref{fig:splayedcusp}.  \\
\begin{figure}[!h]
\begin{tabular}{c@{\hspace{1.5cm}}c}
\includegraphics[width=0.39 \textwidth]{splayedcusp_new.jpg}&
\includegraphics[width=0.39 \textwidth]{notsplayedcusp_new.jpg}
\end{tabular}
\caption{ \label{fig:splayedcusp}  $D=\{x(y^2-z^3)=0\}$ (left) is splayed and  $D'=\{x(x+y^2-z^3)=0\}$ (right) is not splayed.}
\end{figure}

In the first part of this work three different characterizations of splayed divisors are shown. Suppose that a divisor $D \subseteq S$ is given by $(D,p)=(D_1,p) \cup (D_2,p)=\{gh=0\}$. First we
consider the Jacobian ideal $J_{gh}$ (the ideal generated by the partial derivatives of $gh$) of $D$ at $p$. It is clear that for a splayed $D$, the Jacobian ideal satisfies
$$ J_{gh}=gJ_h+hJ_g,$$
when the defining equations $g$ and $h$ are chosen in separated variables.
We show (Proposition \ref{Thm:Leibnizprod}) that this property already characterizes splayed divisors. \\
Further, for a splayed divisor $D$, the ideal $((gh)+J_{gh})$ defining its singular locus can
be written as the intersection of the ideals defining the singular loci of the $D_i$ and of the ideal defining the
intersection $D_1 \cap D_2$:
$$((gh)+J_{gh})=(g,h) \cap ((g)+J_g) \cap ((h)+J_h).$$
In Proposition \ref{Thm:splayedgeom} the sufficiency of this condition is proved. \\
Finally, equation (\ref{Equ:transversalsmooth}) can be directly translated to singular subspaces via
K.~Saito's logarithmic derivations \cite{Saito80},
namely, $D$ is splayed at $p$ if and only if
$$ \Der_{S,p}(\log D_1) + \Der_{S,p}(\log D_2)=\Der_{S,p}.$$
This is shown in Proposition \ref{Thm:splayedder}. \\

In the second part, two applications are considered. First
the relationship between splayed divisors and free divisors is studied. Free divisors are a generalization of normal crossing divisors and appear frequently in different areas of
mathematics, for example, in deformation theory as discriminants or in combinatorics as free hyperplane arrangements, see \cite{Aleksandrov86, OrlikTerao92, Buchweitz06, MondSchulze10, BuchweitzMond,
Saito81} for more examples.
Then we give a partial answer to a question of H.~Hauser about the characterization of normal crossing divisors by their Jacobian ideals: it is shown that if $D=\bigcup_{i=1}^n D_i$ is locally the
union of smooth
irreducible components then $D$ has normal crossings if and only if $D$ is locally free and its Jacobian ideal is radical. \\
The computation of singularity invariants of splayed divisors is very interesting, in particular one should be able to deduce them from singularity invariants of their splayed components. We start
with computing the Hilbert--Samuel polynomial $\chi_{D,p}$ of a splayed divisor $(D,p)=(D_1,p) \cup (D_2,p)$ and find that it satisfies an additivity relation: from the exact sequence
$$ 0 \rightarrow \mc{O}_{D,p} \rightarrow \mc{O}_{D_1,p} \oplus \mc{O}_{D_2,p}
\rightarrow \mc{O}_{D_1 \cap D_2,p} \ensuremath{\rightarrow} 0$$
we deduce
$$\chi_{D,p}(t)+\chi_{D_1 \cap D_2,p}(t)=\chi_{D_1,p}(t)+\chi_{D_2,p}(t).$$
This additivity relation does not characterize
splayed divisors, an example therefore is given at the end.

\section{Setting and triviality lemma} \label{Sec:preliminaries}

We work in the complex analytic category. The main objects of our study are divisors (=hypersurfaces) in complex manifolds. We write $(S,D)$ for a fixed divisor $D$ in an $n$-dimensional complex
manifold $S$.
Denote by $p$ a point in $S$ and by $x=(x_1, \ldots, x_n)$ the complex coordinates of $S$ around $p$. The divisor $(D,p)$ will then be defined locally by an equation $\{
h_p(x_1, \ldots, x_n)=0 \}$ where $h_p \in \mc{O}_{S,p} \cong \ensuremath{\mathbb{C}}\{x_1, \ldots, x_n\}$ (if the context is clear we leave the subscript $_p$). Note that we will always assume that $h$ is reduced! To
simplify the notation regarding splayed divisors, sometimes the coordinates are denoted by $(x_1, \ldots, x_k,y_{k+1}, \ldots, y_n)$ and $h(x,y) \in \ensuremath{\mathbb{C}}\{x,y\}$.
The divisor $D$ has \emph{normal crossings} at a point $p$ if one can find complex coordinates $(x_1, \ldots, x_n)$ at $p$ such that the defining equation $h$ of $D$ is $h=x_1 \cdots x_k$ for $k
\leq n$. We also say that $(D,p)$ is a \emph{normal crossing divisor}.
The \emph{Jacobian ideal} of $h$ is denoted by $J_{h}=(\ensuremath{\partial}_{x_1} h, \ldots, \ensuremath{\partial}_{x_n}h) \subseteq \mc{O}_{S,p}$.
The image of $J_h$ under the  canonical epimorphism that sends $\mc{O}_{S,p}$ to $\mc{O}_{D,p}=\mc{O}_{S,p}/(h)$ is denoted by $\widetilde{J_h}$.
The associated analytic coherent ideal sheaves are
denoted by $J \subseteq \mc{O}_{S}$ and $\widetilde J$ in $\mc{O}_{D}$.
The singular locus $\Sing D$ of $D$ is defined by the ideal sheaf $\widetilde J \subseteq \mc{O}_{D}$. The singular locus $(\Sing D,p)$ is defined by the local ring
$\mc{O}_{\Sing D, p}=\mc{O}_{S,p}/((h) + J_h)$. \\

The next lemma yields an ideal-theoretic characterization of Cartesian product structure. It is used frequently and can be found in various
different formulations in the literature (for example in \cite{dJP,Saito80,GH,CNM96}).

\begin{lem}[Triviality lemma] \label{Lem:saitotriv}
Let $(S,p)$ locally be isomorphic to $(\ensuremath{\mathbb{C}}^{n+m},0)$ and denote
$\mc{O}_{S,p}=\ensuremath{\mathbb{C}}\{ x_1, \ldots, x_n, y_1, \ldots, y_m\}$ and let $h(x_1, \ldots, x_n,$ $y_1, \ldots , y_m)$ be
an element of $\mc{O}_{S,p}$. Then the following are equivalent: \\
(a) The ideal $(\ensuremath{\partial}_{y_1}h, \ldots, \ensuremath{\partial}_{y_m}h)$ is contained in the ideal $(h,
\ensuremath{\partial}_{x_1}h, \ldots, \ensuremath{\partial}_{x_n}h)$. \\
(b) There exists a local biholomorphic map $\varphi: (\ensuremath{\mathbb{C}}^{n+m},0) \rightarrow
(\ensuremath{\mathbb{C}}^{n+m},0)$ and a holomorphic $v(x,y) \in \mc{O}_{S,p}^*$ such that
$$\varphi(x,y)=(\varphi_1(x,y), \ldots, \varphi_n(x,y),y_1, \ldots, y_m),$$
$\varphi(x,0)=(x,0)$, $v(x,0) \equiv 1$ and $h \circ \varphi (x,y)=v(x,y)h(x,
0)$.  \\
This means that $D=\{h(x,y)=0\}$ is locally at $p$ isomorphic to some $(D' \times \ensuremath{\mathbb{C}}^m, (0,0))$ where $D'$ is locally contained in $\ensuremath{\mathbb{C}}^n$. \\
\end{lem}

\begin{rem}
 This lemma is called ``triviality lemma'' because a strong\-er form of it characterizes local analytic triviality: if in $(a)$ all partial derivatives $\ensuremath{\partial}_{y_i}h$ are even contained in
$(x_1, \ldots, x_n, y_1, \ldots, y_m)(\ensuremath{\partial}_{x_1}h, \ldots, \ensuremath{\partial}_{x_n}h)$, then one can show that $(D,p) \cong (D_0 \times \ensuremath{\mathbb{C}}^m,(p',0))$ where $D_0=\{ h(x,0)=0\}$ is the ``fibre'' at the origin.
\end{rem}

\begin{proof}
See for example \cite{Saito80}.
\end{proof}

The objects of our studies are divisors that are unions of Cartesian products. We call them splayed because they fill out the whole space.

\begin{defin} \label{Def:produkt}
Let $D$ be a divisor in a complex manifold $S$, $\dim S=n$. The
divisor $D$ is called \emph{splayed} at a point $p \in S$ (or $(D,p)$ is \emph{splayed}) if one can find coordinates $(x_1, \ldots, x_n)$ at $p$ such that $(D,p)=(D_1,p) \cup (D_2,p)$ is defined by
$$h(x)=h_1(x_1, \ldots, x_k) h_2(x_{k+1}, \ldots, x_n),$$
$1\leq k \leq n-1$, where $h_i$ is the defining reduced equation of $D_i$. Note that the $h_i$ are not necessarily irreducible. The
\emph{splayed components} $D_1$ and $D_2$ are not unique. Splayed means that $D$ is the union of two products: since $h_1$ is independent of $x_{k+1}, \ldots, x_n$, the divisor $D_1$ is locally at $p$
isomorphic to a product $(D'_1,0) \times (\ensuremath{\mathbb{C}}^{n-k},0)$, where $(D'_1,0) \subseteq (\ensuremath{\mathbb{C}}^k,0)$ (and similar for $D_2$).
\end{defin}

\begin{exa} (1) Let $(D,0)$ be the divisor in $(\ensuremath{\mathbb{C}}^2,0)$ defined by $h_1 h_2=x (y-x^2)$. Since $D$ has normal crossings at the origin, $D$ is splayed. \\
(2) The divisor $D=\{ (x-y^2)zw=0\}$ is splayed in $(\ensuremath{\mathbb{C}}^4,0)$ but its splayed components are not unique, for example $h_1=x-y^2$ and $h_2=zw$ or $h_1=(x-y^2)w$ and $h_2=z$. \\
(3) The divisor $D=\{ (xz(x+z-y^2)=0\}$ is not splayed in $(\ensuremath{\mathbb{C}}^3,0)$.
\end{exa}

\begin{rem}  \label{Rem:Damon}
The concept of splayed divisors was also introduced by J. Damon in the context of free divisors  under the name product union, see \cite{Damon96}. In the theory of hyperplane arrangements one
studies the analogue concept, under various names: e.g., in \cite{OrlikTerao92} splayed arrangements are called reducible and in \cite{Budur10} they are referred to as decomposable.
\end{rem}

\section{Criteria for splayed divisors} \label{Sec:criteria}

Let $(D,p)$ be a divisor in $(S,p)$. If the decomposition into irreducible components of $D$ is known, then there are effective methods to test whether $D$ has normal crossings at $p$, see
\cite{Bodnar04}. The idea here is to find linearly independent tangent vectors, that is, to linearize the problem and thus reduce it to linear algebra. The algorithm of Bodn{\'a}r does
not work in order to test whether $(D,p)$ is splayed, since the irreducible components may be singular themselves at $p$.  \\
In Proposition \ref{Thm:Leibnizprod} it is shown that $D=D_1 \cup D_2=\{
gh=0\}$ is splayed at $p$ if and only if, up to multiplying $g, h$ with units, its
Jacobian ideal
$J_{gh}$ of $(D,p)$ is equal to $gJ_h+hJ_g$. This property can be tested in computer algebra systems like {\sc Singular}. The second characterization is more geometrical. In Proposition \ref{Thm:splayedgeom} we show that for a splayed
divisor $\Sing D$ is the scheme-theoretic union of the intersection of $D_1$ and $D_2$, and of the singular loci $\Sing D_1$ and $\Sing D_2$.
The third criterion (Proposition \ref{Thm:splayedder}) is nearer to Bodn{\'a}r's normal crossings test. Therefore we use the notion of
logarithmic
derivations and differential forms (in the sense of K.~Saito \cite{Saito80}), which will be a main ingredient in the section about freeness of splayed divisors.

\subsection{Jacobian ideal characterization 1 - algebra} \label{Sub:leibniz}

Here a characterization of splayedness by Jacobian Ideals is shown, which can be easily checked in concrete examples.

\begin{defin}
Let $D_1=\{g=0\}$, $D_2=\{h=0\}$ and $D=\{gh=0\}$ at $p$ be defined as above. We say that $J_{gh}$ satisfies the \emph{Leibniz property} if
$$J_{gh}=gJ_h + hJ_g.$$
\end{defin}

If $D$ is splayed then it is clear that its Jacobian ideal satisfies the Leibniz property. To establish the other implication, first an intermediate ideal-theoretic characterization of splayedness is
proven (Lemma
\ref{Lem:produktallg}).

\begin{lem}  \label{Lem:produktallg}
Let $\dim S=n$ and at a point $p=(x_1, \ldots, x_n)$ denote by $\calo=\mc{O}_{S,p}=\ensuremath{\mathbb{C}}\{x_1, \ldots, x_n\}$ the local ring at $p$. Let $D_1=\{g(x)=0\}$, $D_2=\{h(x)=0\}$ and
$D=\{ gh(x)=0\}$ be divisors, where we assume that $g,h \in \mc{O}$ are reduced and have no common factors. Then $(D,p)=(D_1,p) \cup D_2,p)$ is splayed if and only if
\begin{equation}   \label{Equ:splayeddurchschnitt}
(g) \cap ((gh)+J_{gh})= g((h) + J_h).
\end{equation}
\end{lem}

\begin{proof}
First note that the inclusion $(g) \cap ((gh)+J_{gh})\subseteq  g((h) + J_h)$ always holds since $g$ and $h$ have no common factors. \\
If $D$ is splayed, we can suppose without loss of generality that $D_1=\{g(x,0)=0\}$ and $D_2=\{h(0,y)=0\}$ where  $(x,y)=(x_1, \ldots, x_k,
y_{k+1},\ldots, y_n)$. Then it is easy to see, via the Leibniz property of $J_{gh}$, that $g((h)+J_h)$ is contained in $(g) \cap ((gh)+J_{gh})$.
For the other direction, suppose that  (\ref{Equ:splayeddurchschnitt}) holds.
In this case it is easy to see that (\ref{Equ:splayeddurchschnitt}) is preserved under local isomorphisms of $(S,p)$.
One may assume without loss of generality that $h(x_1, \ldots, y_n)=h(0, \ldots,0,$ $y_{k+1}, \ldots, y_n)$ and that
\begin{equation} \label{Equ:nichtinm}
 \ensuremath{\partial}_{y_i}h \not \in (h,\ensuremath{\partial}_{y_{k+1}}h, \ldots, \widehat{ \ensuremath{\partial}_{y_{i}}h}, \ldots,
\ensuremath{\partial}_{y_n}h)
\end{equation}
for all $i \in \{ k+1, \ldots, n\}$. (If not so, suppose that e.g. $\ensuremath{\partial}_{x_1}h$ is contained in $(h,\ensuremath{\partial}_{x_{2}}h, \ldots, \ensuremath{\partial}_{x_n}h)$. Then by the triviality lemma there exists a locally biholomorphic
map
$\varphi: (S,p) \rightarrow (S,p)$  such that
$h \circ \varphi(x)=v(x) h(0, x_{2}, \ldots, x_n)$, with $v \in \mc{O}^*$. Then
 set $\widetilde h:= h(0,x_{2}, \ldots, x_n)$  and $\widetilde g:= g \circ \varphi$. The divisor defined by $\widetilde g \cdot \widetilde h$ is clearly isomorphic to $D$, and similarly $D_1$ and
$D_2$ are defined by $\widetilde g$ and $\widetilde h$.) \\
Now let $h(y)=h(0, \ldots, 0, y_{k+1}, \ldots, y_n)$ such that (\ref{Equ:nichtinm}) holds. From (\ref{Equ:splayeddurchschnitt}) it follows that $g (\ensuremath{\partial}_{y_i}h)$ is contained in $((gh)+J_{gh})$ and by
definition $g \ensuremath{\partial}_{y_i}h + h \ensuremath{\partial}_{y_i}g$ is also contained in this ideal. Hence $h(\ensuremath{\partial}_{y_i}g) \in ((gh)+J_{gh})$ for all $k+1 \leq i \leq n$.
Thus we can write
$$h (\ensuremath{\partial}_{y_i}g)= a_i gh + \sum_j a_{ij}(g \ensuremath{\partial}_{y_j}h + h \ensuremath{\partial}_{y_j}g)+\sum_j b_{ij}h\ensuremath{\partial}_{x_j}g$$
where the summation indices run over the valid range. This equation can be rearranged as
\begin{equation} \label{Equ:splayed2}
 h(\ensuremath{\partial}_{y_i}g -  a_i g - \sum_j a_{ij} \ensuremath{\partial}_{y_j}g - \sum_j b_{ij}\ensuremath{\partial}_{x_j}g)  =g \sum_j a_{ij} (\ensuremath{\partial}_{y_j}h).
\end{equation}
Since $g$ and $h$ have by assumption no common factors, we conclude from (\ref{Equ:splayed2}) that $\ensuremath{\partial}_{y_i}g -  a_i g - \sum_j a_{ij} \ensuremath{\partial}_{y_j}g - \sum_j b_{ij}\ensuremath{\partial}_{x_j}g \in (g)$ and that $\sum_j
a_{ij} (\ensuremath{\partial}_{y_j}h)=c_i h$ for any $ k+1 \leq i \leq n$ and some $c_i \in \mathcal{O}$. Then (\ref{Equ:nichtinm}) implies that any $a_{ij} \in \mf{m}$: if this were not the case, that is, if
some $a_{im} \in \mc{O}^*$, then via the equation $\sum_j a_{ij} (\ensuremath{\partial}_{y_j}h)=c_i h$ one can express $\ensuremath{\partial}_{y_m}h$ in terms of $h$ and the other $\ensuremath{\partial}_{y_j}h$, which is a contradiction to
(\ref{Equ:nichtinm}). Hence it follows that
$$\ensuremath{\partial}_{y_i}g \in (g, \ensuremath{\partial}_{x_1}g, \ldots, \ensuremath{\partial}_{x_k}g)+\mf{m}(\ensuremath{\partial}_{y_{k+1}}g, \ldots, \ensuremath{\partial}_{y_n}g)$$
for any $k+1 \leq i \leq n$. By an application of Nakayama's lemma we obtain
$$\ensuremath{\partial}_{y_i}g \in (g, \ensuremath{\partial}_{x_1}g, \ldots, \ensuremath{\partial}_{x_k}g) \text{ for all }i=k+1, \ldots,
n.$$
By the triviality lemma  there exists a locally biholomorphic $\psi: (S,p) \ensuremath{\rightarrow} (S,p)$ with $\psi(x,y)=(\psi_1(x,y), \ldots, \psi_k(x,y),
y_{k+1}, \ldots, y_n)$
 such that $g \circ \psi$ is equal to $vg(x_1, \ldots, x_{k},0)$ for a unit $v \in \calo$. Set
$\tilde g:=v^{-1}(g \circ \psi)$ and $\tilde h:=h \circ
\psi=h=h(0, \ldots, 0, y_{k+1}, \ldots, y_n)$. By
construction $\tilde g \tilde h$ defines a splayed divisor that is isomorphic to
$D$.
\end{proof}

\begin{proposition}  \label{Thm:Leibnizprod}
Let $(S,D)$ be a complex manifold $S$, $\dim S=n$, together with a divisor $D \subseteq S$ that is locally at a point defined by $\{gh=0\}$, where $g$ and $h$ are reduced elements
of $\mc{O}_{S,p}$ that are not necessarily irreducible but have no common
factor. Then $D=\{ g=0\} \cup \{h=0\}$ is  splayed at $p$ if $J_{gh}$ satisfies the Leibniz property
$$ J_{gh}=gJ_h + h J_g.$$
Conversely, if $D=\{ gh = 0\}$ is splayed and $g$ and $h$ are chosen in different variables, then $J_{gh}$ satisfies the Leibniz property. This means that up to possible multiplication of $g$ and $h$ with units, $J_{gh}$ satisfies the Leibniz property.
\end{proposition}

\begin{proof}
As already remarked, if $D$ is splayed then we can choose the defining equations $g$ and $h$ in separated variables and it is clear that $J_{gh}=gJ_h+hJ_g$. Conversely, it is enough to show the equality $(g) \cap ((gh)+J_{gh})=g((h)+J_h)$ from lemma \ref{Lem:produktallg}. Since
$(g) \cap ((gh)+J_{gh})$ is always contained in $g((h)+J_h)$ it remains to check the other inclusion: take a $\beta \in g((h)+J_h)$. Then a  straightforward calculation shows that
$\beta$ is also contained in $(g) \cap ((gh)+J_{gh})$.
\end{proof}

As pointed out by P.~Aluffi, $D=\{ gh =0\}$ is splayed if and only if
\begin{equation} \label{Equ:leibnizmod}
(gh)+J_{gh}=(gh)+gJ_h+hJ_g,
\end{equation}
regardless of multiplying $g$ and $h$ by units. Equation (\ref{Equ:leibnizmod}) can also easily be shown with lemma \ref{Lem:produktallg}. For a different approach to this characterization of splayedness, see \cite{AluffiFaber12}.

\begin{exa} Let $D$ be the divisor in $\ensuremath{\mathbb{C}}^3$ given at a point $p$ by $x(x+y^2-z^3)$. Then $D$ is the union of two smooth components $D_1=\{h=x+y^2-z^3=0\}$ and $H=\{g=x=0\}$. The ideal
$(gh)+J_{gh}=J_{gh}=(2x+y^2-z^3, xy,xz^2)$ is strictly contained in $(gh)+gJ_h+hJ_g=gJ_h+hJ_g=(x,y^2-z^3)$.  Thus $D$ is not a splayed divisor (see Figure \ref{fig:splayedcusp}).
\end{exa}

\subsection{Jacobian ideal characterization 2 - geometry} \label{Sub:geometry}
Now we characterize a splayed divisor $D \subseteq S$, locally at a point $p$ given by a $gh \in \mc{O}_{S,p}$, by $\mc{O}_{\Sing D,p}=\mc{O}_{S,p}/((gh)+J_{gh})$. This characterization
 reflects the geometry of $(D,p)$, namely the two splayed components meeting transversally.

\begin{proposition}  \label{Thm:splayedgeom}
The divisor $D=D_1 \cup D_2$, defined at $p$ as above, is splayed if and only if
\begin{equation} \label{Equ:splayedgeom}
((gh)+J_{gh})=(g,h) \cap ((g)+J_g) \cap ((h)+J_h).
\end{equation}
\end{proposition}

\begin{proof}
Recall here from lemma \ref{Lem:produktallg} that a divisor $\{gh=0\}$ is splayed if and only if (\ref{Equ:splayeddurchschnitt}) holds.
First suppose that (\ref{Equ:splayedgeom}) holds. A straightforward calculation shows (\ref{Equ:splayeddurchschnitt}). \\
For the other implication we use Grauert's division theorem (for the notation and statement see \cite[Theorem 7.1.9]{dJP}): Suppose that $D=\{gh=0\}$ is splayed. Then without loss of generality $g(x,y)=g(x)$ and
$h(x,y)=h(y)$ in $\ensuremath{\mathbb{C}}\{x_1, \ldots, x_k, y_{k+1},$ $
\ldots, y_n\}$. Clearly $((gh)+J_{gh})
\subseteq (g,h) \cap ((g)+J_g) \cap ((h)+J_h)$. So let $\alpha$ be an element of the right-hand side, that is, $\alpha=ag+bh=cg+\sum_{i=1}^k a_i \ensuremath{\partial}_{x_i}g$ for some $a,b,c, a_i \in \mc{O}$.
Then also $\alpha
-ag=bh=(c-a)g+\sum_{i=1}^ka_i \ensuremath{\partial}_{x_i}g$ is contained in $((g)+J_g)$. By Grauert's division theorem there exist some $\tilde a, \tilde a_i, r, \tilde r_i$ such that for all
$i=1, \ldots, k$ one has $c-a=\tilde a h +r$ and $a_i=\tilde a_i h +r_i$  and the leading monomial $L(h)$ does not divide any monomial of the unique remainders $r, r_i$. Then write
$$(b-\tilde a g - \sum_{i=1}^k \tilde a_i (\ensuremath{\partial}_{x_i}g))h=rg +\sum_{i=1}^k r_i (\ensuremath{\partial}_{x_i}g).$$
Since $h$ only depends on $y$ and $g$ only on $x$, it follows that $L(h)$ does also not divide any of the monomials of the right-hand side of the equation. But this is only possible
if $(b-\tilde a g - \sum_{i=1}^k \tilde a_i (\ensuremath{\partial}_{x_i}g))h=0$.
It follows that $b$ is contained in $((g)+J_g)$. Interchanging the roles of $g$ and $h$
yields that $a \in ((h)+J_h)$ and thus $\alpha \in (gh, gJ_h +hJ_g)$. The Leibniz property of $J_{gh}$ implies that $\alpha \in ((gh)+ J_{gh})$.
\end{proof}

\begin{rem}
Here it can be seen that for splayed divisors the Jacobian ideal is the intersection of the two ideals defining the singular loci of the splayed components $D_1$ and $D_2$ plus the intersection of
$D_1$ and
$D_2$. For two smooth divisors $D_1$ and $D_2$ this means that $D=D_1 \cup D_2$ is a splayed divisor if and only if the scheme-theoretical intersection $D_1\cap D_2$ is smooth, which
is in turn equivalent to saying that $D_1$ and $D_2$ intersect transversally, (cf.~\cite{LiLi,FaberHauser10}). Note that for varieties of arbitrary codimension the condition that the
scheme-theoretical intersection is smooth, leads to the formulation of \emph{clean intersection}, see \cite{LiLi}.
\end{rem}

\begin{exa}   \label{Ex:splayedcusp}
Let $D \subseteq \ensuremath{\mathbb{C}}^3$ be given at the origin by $gh=x(x+y^2+z^3)=0$. Then $D$ is not splayed at the origin. Here the intersection of the two components is given by the ideal $(g,h)=(x,y^2+z^3)$. Also
consider the divisor $D' \subseteq \ensuremath{\mathbb{C}}^3$ that is given by $g'h'=x(y^2+z^3)$. Clearly $D'$ is splayed at the origin and the intersection of the two components is given by the ideal
$(g',h')=(x,y^2+z^3)$, see Figure \ref{fig:splayedcusp}.
\end{exa}

\subsection{Logarithmic derivation characterization - tangency} \label{Sub:logder}
As already mentioned in the introduction,  two submanifolds of a manifold intersect transversally at a point if their respective tangent spaces at that point together generate the tangent space of the
ambient manifold. In this section we will translate this definition for splayed divisors with the help of logarithmic derivations (Proposition \ref{Thm:splayedder}). Logarithmic derivations
were introduced by K.~Saito in \cite{Saito80} and are a useful tool in studying tangent behaviour for singular varieties. They lead to the definition of free divisors, which will be studied in more
detail in section \ref{Sub:freediv}.

\subsubsection{A very brief recap of Saito's theory of free divisors (\cite{Saito80})}
Let $D$ be a divisor in $S$ defined at $p$ by $D=\{h=0\}$. A \emph{logarithmic vector field} (or \emph{logarithmic derivation}) (along $D$) is a holomorphic vector field  on $S$, that is, an element
of $\Der_{S}$, satisfying
one of the two equivalent
conditions: \\
(i) For any smooth point $p$ of $D$, the vector $\delta(p)$ of $p$ is tangent to $D$, \\
(ii) For any point $p$, where $(D,p)$ is given by $h=0$, the germ $\delta(h)$ is contained in the ideal
$(h)$ of  $\mc{O}_{S,p}$. The module of germs of logarithmic derivations (along $D$) at $p$ is denoted by
\[  \Der_{S,p}(\log D)=\{  \delta: \delta \in \Der_{S,p} \text{ such that }\delta h \in (h)  \}, \]
The $\Der_{S,p}(\log D)$ are the stalks at points $p$ of the sheaf $\Der_S(\log D)$ of $\mc{O}_{S}$-modules.
Similarly we define logarithmic differential forms: a meromorphic $q$-form $\omega$ is logarithmic (along
$D$) at a point $p$ if $\omega h$ and $hd\omega$ are holomorphic in an
open neighbourhood around $p$. We write
\[\Omega^q_{S,p}(\log D)= \{ \omega: \omega \text{ germ of a logarithmic $q$-form at $p$} \},\]
One can show that $\Der_{S,p}(\log D)$ and $\Omega^1_{S,p}(\log D)$ are reflexive $\mc{O}_{S,p}$-modules dual to each other (see \cite{Saito80}). One says that $(D,p)$ is \emph{free} or that $D$ is
free at $p$ if
$\Der_{S,p}(\log D)$ resp. $\Omega_{S,p}^1((\log D)$ is a free $\calo_{S,p}$-module. In section \ref{Sub:freediv} free divisors are discussed in more detail and the following theorem is used, which
makes it possible to test whether $D$ is free (cf.~\cite[Thm.1.8]{Saito80}):

\begin{thm}[Saito's criterion] \label{Thm:Saito}
Let $(S,D)$, $p$ and $h$ be as defined in the introduction. The $\mc{O}_{S,p}$-module $\Der_{S,p}(\log D)$ is free if and only if there exist $n$ vector fields $\delta_i=\sum_{j=1}^n
a_{ij}(x) \ensuremath{\partial}_{x_j}$
in $\Der_{S,p}(\log D)$, $i=1, \ldots, n$, such that $\det(a_{ij}(x))$ is equal to $h$ up to an invertible factor. Moreover, then the vector fields $\delta_1, \ldots, \delta_n$ form a basis for
$\Der_{S,p}(\log
D)$. \\
\end{thm}

\begin{rem}
The module $\Der_{S,p}(\log D)$ naturally carries the structure of a Lie algebra. This was considered under the name \emph{tangent algebra} in \cite{HM93}, where also many properties of
algebraic and analytic varieties were characterized in terms of tangent algebras. The Lie algebra structure of logarithmic derivations is also considered in \cite{GrangerSchulze09,Sekiguchi08}.
\end{rem}

\subsubsection{Logarithmic derivations and splayed divisors}

\begin{proposition}  \label{Thm:splayedder}
Let $(D,p)=(D_1, p) \cup (D_2,p) \subseteq (S,p)\cong (\ensuremath{\mathbb{C}}^n,0)$ be a divisor such that $(D_1,p)=\{ g=0\}$ and $(D_2,p)=\{h=0\}$ for some $g,h$ not necessarily irreducible but with no common factor.
Then $D$ is
splayed at $p$ if and only if
\begin{equation}\Der_{S,p}(\log D_1)+ \Der_{S,p}(\log D_2)=\Der_{S,p}.
 \label{Equ:splayednew}
\end{equation}
\end{proposition}

\begin{proof}
As above, we denote by $(x,y):=(x_1, \ldots, x_k, y_{k+1}, \ldots, y_n)$ the coordinates at $p$ and by $\calo=\ensuremath{\mathbb{C}}\{x,y\}$ the corresponding local ring. If $D$ is splayed then without loss of generality one has $g=g(x,0)$
and $\Der_{S,p}(\log D_1)$ is generated by some $\delta_1, \ldots, \delta_m$, $m \geq k$, $\ensuremath{\partial}_{y_{k+1}},\ldots, \ensuremath{\partial}_{y_n}$. Similarly $h=h(0,y)$ and $\Der_{S,p}(\log D_2)$ is generated by some
$\ensuremath{\partial}_{x_1}, \ldots, \ensuremath{\partial}_{x_k}, \varepsilon_{1}, \ldots, \varepsilon_l$ for some $l \geq n-k$. Then clearly we have
$$ \Der_{S,p}(\log D_1)+\Der_{S,p}(\log D_2)  ={}_\calo\!\langle \ensuremath{\partial}_{x_1}, \ldots, \ensuremath{\partial}_{x_k}, \ensuremath{\partial}_{y_{k+1}}, \ldots, \ensuremath{\partial}_{y_n}\rangle = \Der_{S,p}.$$
For the other implication we may assume that $g(x,y)=g(x,0)$ and
$$\ensuremath{\partial}_{x_i}g \not \in (\ensuremath{\partial}_{x_1}g, \ldots, \widehat{\ensuremath{\partial}_{x_i}g}, \ldots, \ensuremath{\partial}_{x_k}g)$$
 for all $i \leq k$ by a similar argument as in lemma \ref{Lem:produktallg}.
Then it is clear that $\Der_{S,p}(\log D_1)$ can be generated by some $\delta_1, \ldots, \delta_m$ and $\ensuremath{\partial}_{y_{k+1}}, \ldots, \ensuremath{\partial}_{y_n}$, where $m \geq k$ and the coefficients of the $\delta_i$ only
depend on $x_1, \ldots, x_k$ and lie in the maximal ideal $\mf{m} \subseteq \calo$. \\
Using $\ensuremath{\partial}_{x_i} \in {}_\calo \langle \varepsilon_1, \ldots, \varepsilon_l, \delta_1, \ldots, \delta_m, \ensuremath{\partial}_{y_{k+1}}, \ldots, \ensuremath{\partial}_{y_n} \rangle$ and computing $\ensuremath{\partial}_{x_i}h$ we obtain (similar to the proof
of lemma \ref{Lem:produktallg}) that
$$(\ensuremath{\partial}_{x_1}h, \ldots, \ensuremath{\partial}_{x_k}h) \subseteq (h, \ensuremath{\partial}_{y_{k+1}}h, \ldots, \ensuremath{\partial}_{y_n}h).$$
An application of the triviality lemma shows that $D_2$ can be chosen independently of the variables $x_1, \ldots, x_k$, and thus that $D$ is splayed.
\end{proof}

\begin{question}
In view of Proposition \ref{Thm:splayedder} and the duality of $\Der_{S,p}(\log D)$ and $\Omega^1_{S,p}(\log D)$ we ask: can one express the splayedness of a divisor $(D_1 \cup D_2,p)$ in
terms of $\Omega^1_{S,p}(\log D_1)$, $\Omega^1_{S,p}(\log D_2)$ and $\Omega_{S,p}^1$?
\end{question}

\section{Applications}

\subsection{Free divisors and normal crossings} \label{Sub:freediv}

In some sense free divisors are a generalization of normal crossing divisors: their modules of tangent vector fields are free. However, irreducible (non-smooth) free divisors are highly singular,
that is, they are non-normal and their singular loci are Cohen--Macaulay of codimension 1 in the divisor, see \cite{Aleksandrov90,Simis06}. Here we consider the relationship between free and splayed
divisors and in particular
between splayed and normal crossing divisors. First we show that a splayed divisor is free if and only if its splayed components are free (Proposition \ref{Prop:produktfreiallg}). Then we turn to normal
crossing divisors, in particular to the problem of characterizing normal crossing divisors by their Jacobian ideal. This problem was stated by H.~Hauser and considered in \cite{Faber11}, where a
general answer was found. Here we use the ideal theoretic characterizations of splayed divisors in order to show that a divisor consisting of smooth irreducible components has normal crossings at a
point $p$ if and only if it is free at $p$ and its Jacobian ideal is radical (Corollary \ref{Cor:smoothcomp}). \\

First let us describe the structure of $\Der_{S,p}(\log D)$ for splayed divisors in a different way: let $S,T$ be complex manifolds of dimensions $n,m$ and suppose that $(S\times
T,0)\cong (\ensuremath{\mathbb{C}}^{n+m},0)$, with complex coordinates $(x,y)=(x_1, \ldots, x_n,$ $y_1,
\ldots, y_m)$ at the origin. Let $(D^x_1,0)$ be a divisor in $(S,0)$,
which is defined by a reduced $g'(x) \in \mc{O}_{S,0}\cong \ensuremath{\mathbb{C}}\{x_1, \ldots,
x_n\}$ and which has a logarithmic derivation module over $\ensuremath{\mathbb{C}}\{x\}$ denoted by
$\Der_{S,0}(\log D^x_1)$. Then we may consider the cylinder over $D^x_1$ in the
$T$-direction in $(S\times T,0)$, namely the hypersurface $D_1$ defined by $g(x,y)=g(x,0):=g'(x) \in \ensuremath{\mathbb{C}}\{x,y\}$. It is easy to see that
$$\Der_{S\times T,0}(\log D_1)=(\Der_{S,0}(\log
D^x_1)\otimes_{\ensuremath{\mathbb{C}}\{x\}}\ensuremath{\mathbb{C}}\{x,y\})\oplus (\Der_{T,0}
\otimes_{\ensuremath{\mathbb{C}}\{y\}}\ensuremath{\mathbb{C}}\{x,y\}).$$

Similarly define $D_2^y$ and $D_2$ with equations $h'(y)=h(x,y)$ and also $\Der_{S\times T,0}(\log D_2)$.
Thus we define
the (splayed) divisor $D=D_1 \cup D_2$ in $S \times T$ that is given at $0$ by
the equation $gh=0$. Since $g$ and $h$ have separated variables, there is a natural
splitting of $\Der_{S \times T,0}(\log D)$:
{\small \begin{equation}  \label{Equ:splayedvectorfields}
 \Der_{S\times T,0}(\log D)\!=(\Der_{S,0}(\log D^x_1)\otimes_{\ensuremath{\mathbb{C}}\{x\}}\ensuremath{\mathbb{C}}\{x,y\})
\oplus (\Der_{T,0}(\log D^y_2)\otimes_{\ensuremath{\mathbb{C}}\{y\}}\ensuremath{\mathbb{C}}\{x,y\}).
\end{equation}
} \normalsize

\begin{proposition} \label{Prop:produktfreiallg}
 Let $(D,p)=(D_1,p) \cup (D_2,p)$ be a splayed divisor in $S \cong \ensuremath{\mathbb{C}}^{n+m}$ defined as above.
The divisor $D=\{g(x)h(y)=0\}$ is
free if and only if $D_1=\{g(x)=0\}$ and $D_2=\{h(y)=0\}$ are both free.
\end{proposition}

\begin{proof}
If both $D_1$ and $D_2$ are free then there exist bases of
$\Der_{S\times T,p}(\log D_1)$ and $\Der_{S\times T,p}(\log D_2)$ of the form
$$\delta_1=\sum_{i=1}^n a_{1i} \ensuremath{\partial}_{x_i}, \ldots, \delta_n=\sum_{i=1}^n a_{ni}
\ensuremath{\partial}_{x_i}, \delta_{n+1}=\ensuremath{\partial}_{y_1}, \ldots, \delta_{n+m}=\ensuremath{\partial}_{y_m}$$
and
$$\varepsilon_1=\ensuremath{\partial}_{x_1}, \ldots, \varepsilon_n=\ensuremath{\partial}_{x_n}, \varepsilon_{n+1}=\sum_{i=1}^mb_{n+1,i}\ensuremath{\partial}_ {y_i}, \ldots,
\varepsilon_{n+m}=\sum_{i=1}^mb_{n+m,i}\ensuremath{\partial}_ {y_i}.$$
It is easy to see that any $\delta_i$ for $1 \leq i \leq n$ and any $\varepsilon_j$ for $n+1 \leq j \leq n+m$ is also an element of $\Der_{S\times T,p}(\log
D)$.
By Saito's criterion (Thm.~\ref{Thm:Saito}) it follows that $\delta_1, \ldots, \delta_n,
\varepsilon_{n+1},
\ldots, \varepsilon_{n+m}$ form a basis of $\Der_{S \times T,p}(\log D)$. Conversely, suppose that $\Der_{S\times T,p}(\log D)$ is free.
Since $\Der_{S\times T,p}(\log D)$ is free, it follows by (\ref{Equ:splayedvectorfields})
that $\Der_{S,0}(\log D^x_1)\otimes_{\ensuremath{\mathbb{C}}\{x\}}\ensuremath{\mathbb{C}}\{x,y\}$ and
$\Der_{T,0}(\log D^y_2)\otimes_{\ensuremath{\mathbb{C}}\{y\}}\ensuremath{\mathbb{C}}\{x,y\}$ are projective
$\mc{O}_{S\times T,p}$-mod\-ules. Since the notion of projective and free module over regular
local rings coincide,
these two
modules are even free.
\end{proof}

\subsubsection{Radical Jacobian ideal - normal crossings}

In \cite{Faber11} the following problem (proposed by H.~Hauser) was considered: suppose that $D \subseteq S$, $\dim S=n$, is a divisor that is locally at a point $p \in S$ given by $\{h=0\}$. Can we
determine if $D$ has normal crossings at $p$ by the knowledge of its Jacobian ideal $J_h$? \\
It was shown that $D$ has normal crossings in case $D$ is free, the Jacobian ideal is radical and the normalization of $D$ is smooth (see
\cite[Thm.~2.4]{Faber12}).   \\

Here we use splayed divisors to show this theorem for the case where $(D,p)=\bigcup_{i=1}^m(D_i,p)$ is a union of \emph{smooth} irreducible components $D_i$.  Note that in this special case the
hypothesis on the normalization is not needed. Moreover, the problem can be reduced to the case where $(D,p)$ is irreducible (Proposition \ref{Prop:produktfreireduktion}).\\

\begin{rem}  \label{Rmk:BrianconSkoda}
One can show, using the theorem of Brian{\c{c}}on--Skoda that $J_h$ radical implies that $h$ is already contained in $J_h$, see  \cite{Faber12}.
\end{rem}

\begin{lem} \label{Lem:produktfreiradikal}
 Let $D=\{gh=0\}$ be splayed at a point $p=(x_1, \ldots, x_k,$ $y_{k+1}, \ldots, y_n)$ in $S \cong \ensuremath{\mathbb{C}}^{n}$ and denote by $D_1=\{g=g(x,0)=0\}$ and $D_2=\{h=h(0,y)=0\}$ its splayed components.
The
Jacobian ideal of $D$, denoted by $J_{gh}$ is radical if and only if both $J_h$ and
$J_g$ are also radical.
\end{lem}

\begin{proof}
Suppose that $J_g$ and $J_h$ are radical. Similarly to the proof of Proposition \ref{Thm:splayedgeom} one concludes that
the ideals  $J_{gh}$ and $(g,h)\cap J_g \cap J_h$ are equal. We compute the radical
$$\sqrt{J_{gh}}=\sqrt{(g,h)\cap J_g \cap J_h}=\sqrt{(g,h)} \cap \sqrt{J_g} \cap
\sqrt{J_h}=(g,h)\cap J_g \cap J_h=J_{gh},$$
where the third equality holds because of our assumptions.
Conversely, suppose
that $J_{gh}=\sqrt{J_{gh}}$. Then $gh$ is an element of $J_{gh}$ (cf.~Remark  \ref{Rmk:BrianconSkoda}).
Localization of $\ensuremath{\mathbb{C}}\{x,y\}$ in $g$ yields $(J_{gh})_g=((h)+J_h)_g$, which is radical, since
$J_{gh}$ is radical. Note that for an ideal $I \subseteq \ensuremath{\mathbb{C}}\{x,y\}$, we denote by $I_g$ the localization of $I$ in $g$ (cf.~\cite{Matsumura86}). Using that $((h)+J_h)_g$ is an intersection of prime
ideals in the localization, a direct computation shows that any element of $\sqrt{((h)+J_h)}$ is already contained in $((h)+J_h)$.
Similarly one proves
$((g)+J_g)=\sqrt{((g)+J_g)}$.
\end{proof}

\begin{lem} \label{Lem:produktradikal}
Let $D\subseteq S$ be a divisor given at $p \in S$ by $\{ gh=0\}$ with $gh \in \mc{O}_{S,p}$ reduced and suppose that $J_{gh}$ is radical. Then
$$J_{gh}=gJ_h+hJ_g.$$
In particular, $D=D_1 \cup D_2$ defined by $D_1=\{g=0\}$ and $D_2=\{h=0\}$ is splayed at $p$.
\end{lem}

\begin{proof}
Consider the element $g\ensuremath{\partial}_{x_i}(gh)$, which is contained in $J_{gh}$ for any $i=1, \ldots, n$. Since $J_{gh}$ is radical, $gh$ is contained in $J_{gh}$, which forces $g^2(\ensuremath{\partial}_{x_i}h)$ to be in
$J_{gh}$. Using that $J_{gh}$ is an intersection of prime ideals $\mf{p}_j$, $j=1, \ldots, k$, one obtains that either $g$ or $\ensuremath{\partial}_{x_i}h$ is contained in any $\mf{p_j}$. Hence $g(\ensuremath{\partial}_{x_i}h) \in
J_{gh}$. This yields
$$J_{gh}=(g \ensuremath{\partial}_{x_1}(h), \ldots, g \ensuremath{\partial}_{x_n}(h), h \ensuremath{\partial}_{x_1}(g), \ldots, h \ensuremath{\partial}_{x_n}(g))=g
J_h + h J_g.$$
By Proposition \ref{Thm:Leibnizprod} $(D,p)$ is splayed.
\end{proof}

\begin{exa}
Splayed divisors need not have radical Jacobian ideals, as the following example shows. Let $D$ be the divisor in $(\ensuremath{\mathbb{C}}^3,0)$ with coordinates $(x,y,z)$ at
$0$, that is defined by $gh=x(y^2+z^3)$. Then clearly $(D,0)$ is splayed. The
Jacobian ideal is $J_{gh}=(y^2+z^3,xy,xz^2)=(y,z^2) \cap (x,y^2+z^3)$, which is
not radical. Note that $D$ is a free divisor.
\end{exa}

\begin{proposition} \label{Prop:produktfreireduktion}
Let $D=D_1 \cup D_2$ be a divisor in an $n$ dimensional complex manifold $S$ and let $D$, $D_1$ and $D_2$ at a point $p \in S$ be defined by the equations  $gh$,  $g$ and $h$, respectively. Suppose
that $J_{gh}$ is radical. Then $D$ is splayed and $J_h$ and
$J_g$ are also radical. If moreover $D$ is free at $p$ then also $D_1$ and $D_2$
are free at $p$.
\end{proposition}

\begin{proof}
This follows from proposition \ref{Thm:Leibnizprod}, proposition \ref{Prop:produktfreiallg} and lemma \ref{Lem:produktradikal}.
\end{proof}

\begin{cor}  \label{Cor:smoothcomp}
Let $(S,D)$ be a complex manifold, $\dim S=n$, together with a divisor $D \subseteq S$ and suppose that locally at a point $p \in S$ the divisor $(D,p)$ has a  decomposition into irreducible
components $\bigcup_{i=1}^m
(D_i,p)$ such that each $(D_i,p)$ is smooth. Let the corresponding equation of $D$ at $p$ be $h=h_1 \cdots h_m$.
If $D$ is free at $p$ and $J_h=\sqrt{J_h}$ then $D$ has normal crossings at $p$.
\end{cor}

\begin{proof}
We use induction on $n$. If $n=2$, then an explicit computation or an application of the theorem of Mather--Yau (\cite{MY}) shows the assertion.
Now suppose the assertion is true for divisors in manifolds of dimension $n-1$. For a smooth component $D_1$ of $D$, one can
find local coordinates $(x_1, \ldots, x_n)$ such that $D_1=\{x_1
=0\}$. Proposition \ref{Prop:produktfreireduktion} shows that the divisor $(D \setminus D_1):=h_2 \cdots
h_m$ is also free and has a radical Jacobian ideal. Moreover, $D $ is locally
splayed, that is, $D \setminus D_1$ is locally isomorphic to some divisor depending
only on the last $n-1$ coordinates. Thus by induction hypothesis $D \setminus
D_1$ is isomorphic to a normal crossings divisor $y_2 \cdots y_m=0$, where the $y_i$ are the result of a coordinate transformation of $(x_1, \ldots, x_n)$ such that $x_1=y_1$.
This implies that $m \leq n-1$. Hence $D$ is isomorphic to the normal
crossings divisor $x_1 y_2 \cdots y_m$.
\end{proof}

\subsection{Hilbert--Samuel polynomials} \label{Sub:HSP}

Splayed divisors are particularly interesting for
computational reasons. We start here a study of properties of splayed divisors by considering their Hilbert--Samuel polynomials. We find that multiplicities behave the same for splayed as for
non--splayed divisors but that the
Hilbert--Samuel polynomials for splayed divisors are additive, which means the
following: let $(D,p)=(D_1,p) \cup (D_2,p)$ be a splayed divisor at a point $p$ in a complex manifold. Then from the exact sequence
$$ 0 \rightarrow \mc{O}_{D,p} \rightarrow \mc{O}_{D_1,p} \oplus \mc{O}_{D_2,p}
\rightarrow \mc{O}_{D_1 \cap D_2,p} \ensuremath{\rightarrow} 0$$
follows
$$\chi_{D,p}+\chi_{D_1 \cap D_2,p}=\chi_{D_1,p} + \chi_{D_2,p},$$
where $\chi_{D,p}$ denotes the Hilbert--Samuel polynomial of $D$ at $p$. Since one can compute Hilbert--Samuel polynomials of divisors in an easy way (\cite[Lemma 4.2.20]{dJP}), this yields a method
to compute $\chi_{D_1 \cap D_2,p}$. \\

The Hilbert--Samuel function and the Hilbert--Samuel polynomial
do not only depend on the module one is considering but also on a chosen filtration.
However, the degree and leading coefficient of the Hilbert--Samuel polynomial are
independent of the filtration.
We use notation from \cite{Matsumura86} and
\cite{dJP}. \\

Let $R$ be a noetherian local ring with maximal ideal $\mf{m}$. Let $I \subseteq R$ be an ideal and let $M$ be a module over $R$. A set $\{ M_n\}_{n \geq 0}$ of submodules of $M$ is called an
$I$-\emph{filtration} of $M$ if $M=M_0 \supset M_1 \supset M_2 \supset \ldots$ and $IM_n \subseteq M_{n+1}$ for all $n \geq 0$. \\
In the following we will always suppose that $M$ is a finitely generated $R$ module. Let $\mf{q}$ be an $\mf{m}$-primary ideal of $R$ and let $\{M_i\}$ be a $\mf{q}$-filtration. Then the
\emph{Hilbert--Samuel function of the filtration $\{M_i\}$} is
$$HS_{\{M_i\}}: \ensuremath{\mathbb{N}} \rightarrow \ensuremath{\mathbb{N}}, d \mapsto \mathrm{length}_{R/ \mf{m}}M/ M_d.$$
If $M_i=\mf{m}^i M$ then we simply write $HS_M$. One can show that there exists a polynomial $\chi_{\{
M_i \}}$ with rational coefficients such that $HS_{\{M_i\}}(d)=\chi_{\{M_i\}}(d)$
for $d$ sufficiently large. We call $\chi_{M}^\mf{q}:=\chi_{\{\mf{q}^n M\}_{n
\geq 0}}$ the \emph{Hilbert--Samuel polynomial of $M$ with respect to $\mf{q}$}.
The degree $d$ of $\chi_{M}^\mf{q}(k)=\sum_{i=0}^d a_i k^i$ only depends on $M$
and not on $\mf{q}$. \\

When considering Hilbert--Samuel polynomials of modules over a local ring
$\mc{O}=\ensuremath{\mathbb{C}}\{x_1, \ldots, x_n\}$ with respect to the maximal ideal $\mf{m}$, one can use
 standard bases to simplify computations.
For definitions and notation about standard bases we refer to \cite{dJP}. Here we only need the following facts: \\

{\bf Fact (i)} Take the degree lexicographical ordering $<$. Then for an ideal $I \subseteq \calo$ one has
$$HS_{\calo/I}(k)=HS_{\calo/L(I)}(k),$$
where $L(I)$ denotes the leading ideal of $I$. In particular,
$\calo/I$ and $\calo/L(I)$ have the same Hilbert--Samuel polynomial with
respect to $\mf{m}$. \\
{\bf Fact (ii)} Let $f, g \in \calo$ and assume that $L(f)$ and $L(g)$ are coprime. Then $L((f,g))=(L(f),L(g))$, that is, $f,g$ are a standard basis of the ideal
$(f,g)$. \\

In contrast to the graded case, the Hilbert--Samuel polynomial is not additive on exact sequences, one has a certain error polynomial, whose degree can be determined with a theorem of Flenner and
Vogel, see \cite{FlennerVogel93}:

\begin{thm}[Flenner--Vogel] \label{Thm:FlennerVogel}
Let $(R, \mf{m})$ be a noetherian local ring, $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ an exact sequence of  finitely generated $R$-modules and $\mf{q}$ an $\mf{m}$-primary
ideal.  Denote further
$$\mathrm{Gr}_\mf{q}(N)=\bigoplus_{i=0}^\infty \mf{q}^iN/\mf{q}^{i+1}N$$
 the associated graded module of a finite $R$-module $N$. Then the following holds: \\
(a) $\mathrm{supp} \ker (\mathrm{Gr}_\mf{q}(M') \rightarrow \mathrm{Gr}_\mf{q}(M))=\mathrm{supp} \ker (\mathrm{Gr}_\mf{q}(M)/\mathrm{Gr}_\mf{q}(M') \rightarrow \mathrm{Gr}_\mf{q}(M''))$. \\
(b) Denote by $d$ the dimension of these supports. Then for all there is a polynomial $S$ of degree $d-1$ such that for $n \gg
0$ we have
$$ S(n):=\chi_{M'}^\mf{q}(n) + \chi_{M''}^\mf{q}(n)-\chi_{M}^\mf{q}(n).$$
In particular, if
$d=0$, then $S=0$.
\end{thm}

Before turning to splayed divisors, consider the additivity problem of the Hilbert--Samuel polynomial for arbitrary finitely generated modules over a local ring $(R, \mf{m})$.
If
$$0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0$$
is an exact sequence of finitely generated $R$-modules and $\mf{q}$ an $\mf{m}$-primary ideal, then
\begin{equation} \label{Equ:HSPerror} 0 \rightarrow N/(\mf{q}^n M \cap N) \rightarrow M/\mf{q}^nM \rightarrow (M/N)/\mf{q}^n(M/N) \rightarrow 0
\end{equation}

is an exact sequence, which implies
$$\chi_{M}^\mf{q}=\chi_{M/N}^\mf{q}+\chi_{\{\mf{q}^nM \cap N\}}.$$
Here $\chi_{\{\mf{q}^nM \cap N \}}$ denotes the Hilbert--Samuel polynomial with respect to the filtration $N/(\mf{q}^nM \cap N)$ and in general one has $\chi_{N}^\mf{q} \neq \chi_{\{\mf{q}^nM \cap N\}}$.
However, for split exact sequences the Hilbert--Samuel polynomial is always additive:

\begin{lem} \label{Lem:hilbertdirektesumme}
Let $(R, \mf{m})$ be a local ring and let $M, N$ be finitely generated $R$-modules. Consider the exact sequence
$$0 \rightarrow N \rightarrow N \oplus M \rightarrow M \rightarrow 0.$$
Then $\chi_{N \oplus M}^\mf{q}=\chi_{M}^\mf{q}+ \chi_{N}^\mf{q}$ for any $\mf{m}$-primary ideal $\mf{q}$.
\end{lem}

\begin{proof}
A direct computation shows that $N \cap \mf{q}^n (N\oplus M)=\mf{q}^n N$ for any
$n$. Using the exact sequence (\ref{Equ:HSPerror}) the assertion follows.
\end{proof}

Let now
$D=D_1 \cup D_2\subseteq \ensuremath{\mathbb{C}}^{n}$ be a not necessarily splayed divisor that is locally at a point $p=(x)$ defined by $h_1(x) h_2(x) \in \mc{O}=\ensuremath{\mathbb{C}}\{x\}$, with components $(D_1,p)=\{
h_1(x)=0\}$ and $D_2=\{h_2(x)=0\}$.
The multiplicities of $D_i$ at $p$ are denoted by $m_p(D_i):=\mf{e}(\mc{O}_{S,p}/(h_i), \mf{m})$.
The Hilbert--Samuel polynomial of $D_i$ at $p$ is denoted by
$\chi_{D_i,p}:=\chi_{\mc{O}/(h_i)}^\mf{m}$, and similarly the multiplicity and Hilbert--Samuel polynomial for $D$.

\begin{rem}
In order to compute the Hilbert--Samuel polynomial of $\mc{O}/I$ for any ideal
$I \subseteq \mc{O}$ we can consider $\mc{O}/I$ either as ring or as an
$\mc{O}$-module. This does not make a difference for the Hilbert--Samuel
functions, since they only depend on the graded structure of $\mc{O}/I$.
\end{rem}

It is well-known that multiplicities of divisors are additive, more precisely one can show that for $D$ and $D_i$ defined as above, $m_p(D)= m_p(D_1) + m_p(D_2)$ (see e.g. \cite{dJP}). For
splayed divisors the additivity also holds for Hilbert--Samuel polynomials:

\begin{proposition} \label{Prop:HSPadditiv}
Let $(D,p)=(D_1,p) \cup (D_2,p)$ be splayed at $p \in S$, where $(S,p) \cong (\ensuremath{\mathbb{C}}^n,0)$. Then the Hilbert--Samuel polynomials of the components $D_1$ and $D_2$ are additive, that is,
$$\chi_{D,p}(t)+\chi_{D_1 \cap D_2,p}(t)=\chi_{D_1,p}(t)+\chi_{D_2,p}(t).$$
\end{proposition}

\begin{proof}
Denote by $\mc{O}:=\ensuremath{\mathbb{C}}\{x,y\}=$ $\ensuremath{\mathbb{C}}\{x_1, \ldots, x_k,$ $y_{k+1},\ldots, y_n\}$ the coordinate ring of $(\ensuremath{\mathbb{C}}^n,0)$.
There is an exact sequence (see e.g. \cite{GreuelPfister08})
\begin{equation} \label{Diag:multiplicitiesexakt}
0 \rightarrow \mc{O} /( h_1 h_2) \rightarrow \mc{O}/ (h_1) \oplus \mc{O}/
(h_2) \rightarrow \mc{O}/(h_1,h_2) \rightarrow 0.
\end{equation}
With facts (i) and (ii) from above the question can be reduced to leading
ideals because (\ref{Diag:multiplicitiesexakt}) remains exact if we just consider the leading ideals.
The divisor $D$ is splayed, so we can assume that it is defined by $g(x)h(y)$. Choosing any valid monomial
ordering one finds
 $L(g)=x^\alpha$, $L(h)=y^\beta$, $L((gh))=x^\alpha y^\beta$, and by fact (ii) from above follows $L((g,h))=(x^\alpha, y^\beta)$. From
 Lemma \ref{Lem:hilbertdirektesumme} it follows that
$$\chi_{\mc{O}/(x^\alpha)}^\mf{m}+\chi_{\mc{O}/(y^\beta)}^\mf{m}=\chi_{\mc{O}/
(x^\alpha) \oplus \mc{O}/
(y^\beta)}^\mf{m}.$$
Thus it remains to prove that the Hilbert--Samuel polynomials with respect to $\mf{m}$ of
the exact sequence
$$0 \rightarrow \mc{O}/(x^\alpha \cdot y^\beta) \rightarrow \mc{O}/(x^\alpha)
\oplus \mc{O}/(y^\beta) \rightarrow \mc{O}/(x^\alpha, y^\beta)\rightarrow 0$$
are additive. In order to apply the theorem of Flenner--Vogel we
show that the map
$$ \xymatrix{ \mathrm{Gr}_\mf{m}(\mc{O}/(x^\alpha y^\beta)) \ar[r]^-\varphi &
\mathrm{Gr}_\mf{m}(\mc{O}/(x^\alpha) \oplus \mc{O}/(y^\beta))}$$
is injective. The map $\varphi$ clearly preserves the degree, so
it is
enough to show the assertion for a homogeneous element of degree $d$.
Therefore take
some $\overline{a} \in \mf{m}^d (\mc{O}/(x^\alpha y^\beta))/ \mf{m}^{d+1}
(\mc{O}/(x^\alpha y^\beta))$.
Consider $\overline{a}$ as an element in $\mc{O}$: from Grauert's division
theorem follows that $\overline{a}$ can be written as $r + \alpha_1 x^\alpha + \alpha_2
y^\beta$, where $r$ is the unique remainder from the division through
$x^\alpha$ and $y^\beta$ and $\alpha_1 \in \mc{O}$ is not divisible by $y^\beta$
and
$\alpha_2$ is not divisible by $x^\alpha$.
Suppose that $\varphi(\overline{a})=(0,0)$. Then
 write $\varphi(\overline{a})=(r + \alpha_2
y^\beta, r + \alpha_1 x^\alpha)$ in $\mc{O}/(x^\alpha) \oplus \mc{O}/(y^\beta)$. In $\mc{O}$ this reads as $r +
\alpha_2 y^\beta = c x^\alpha$ and $r + \alpha_1 x^\alpha = c' y^\beta$ for
some $c,c' \in \mc{O}$ with the right order. It follows that $r \in (x^\alpha , y^\beta)$. But $r$ is the unique
remainder from the division through the standard basis $(x^\alpha, y^\beta)$, so
$r=0$ in $\mc{O}$.
Since $x^\alpha, y^\beta$ are clearly a regular sequence in $\mc{O}$, their
syzygies are trivial and from the conditions on $\alpha_1, \alpha_2$ it follows
that $\alpha_1=\alpha_2=0$. This implies the injectivity of $\varphi$.
Hence by the theorem of Flenner--Vogel, the remainder polynomial
$S$ is zero and the assertion of the proposition follows.
\end{proof}

In general the Hilbert--Samuel polynomial of a divisor $(D,p)=(D_1,p)\cup
(D_2,p)$ is not additive, as is seen in the following example.

\begin{exa}
By Lemma 4.2.20 of \cite{dJP} one can explicitly compute the Hilbert--Samuel
polynomial of $\mc{O}/(f)$, where $\mc{O}=\ensuremath{\mathbb{C}} \{x_1, \ldots, x_n\}$ and
$\mathrm{ord}(f)=m$, namely
\begin{equation} \chi_{\mc{O}/(f)}^\mf{m}(d)=\sum_{j=1}^m {n+d-j-1 \choose n-1}. \label{Eq:HShyper}   \end{equation}
Consider now $(\ensuremath{\mathbb{C}}^2,0)$ with coordinate ring $\mc{O}=\ensuremath{\mathbb{C}}\{x,y\}$ and with
$h_1=x^2-y$ and $h_2=y$. Then the germ of the divisor $(D,0)=(D_1,0)\cup (D_2,0)$ that is locally
given by $\{y(x^2-y)=0\}$ with $(D_1,0)=\{x^2-y=0\}$
and $(D_2,0)=\{y=0\}$ is not splayed. The intersection $(D_1 \cap D_2,0)$ is locally given by
the ideal $(x^2,y)$ and coordinate ring $\mc{O}/(x^2,y)=\ensuremath{\mathbb{C}}\{x\}/(x^2)$. By
formula (\ref{Eq:HShyper}) we can compute the Hilbert--Samuel polynomials of $D,D_1,D_2$ and
$D_1 \cap D_2$ and obtain $\chi_{D,p}(t)=2t-1$, which is clearly not equal to
$\chi_{D_1,p}(t) +\chi_{D_2,p}(t)- \chi_{D_1 \cap D_2,p}(t)= t + t -2$.
\end{exa}

One might ask if the additivity of the Hilbert--Samuel polynomials
characterizes splayed divisors. However, here is a counterexample to this assertion:

\begin{exa}
Consider $D
\subseteq \ensuremath{\mathbb{C}}^3$ locally defined by $gh=(x^2-y^3)(y^2-x^2z)$. Then $(D,p)$ is
the union of the cylinder over a cusp $(D_1,p)$ and of the Whitney Umbrella
$(D_2,p)$. Clearly $(D,p)$ is not splayed (use for example the
Leibniz--property). However, $L(g)=x^2$ and $L(h)=y^2$, so the leading
monomials of $g$ and $h$ are coprime and one can repeat the argument in the
proof of the preceding proposition to find that
$$ \chi_{D,p}+ \chi_{D_1 \cap D_2,p}=\chi_{D_1,p} + \chi_{D_2,p}.$$
\end{exa}

\subsection*{Acknowledgments} I thank my advisor H. Hauser for getting me interested in this topic and helping me with comments and suggestions. I thank D.~Mond and L.~Narv{\'a}ez Macarro for several
discussions and comments on this work, and also P.~Aluffi, M.~Granger, H.~Kawanoue, S.~Perlega, D.~B.~Westra and the  referee for helpful comments on an earlier version of this paper. In particular I thank O.~Villamayor and his
research group at
the Universidad Aut{\'o}noma
de Madrid for their hospitality and for many discussions in which the basis for the results presented here was laid.

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


Blog — S'cool Moves, Inc.
Intervention within a tiered framework like RTI or MTSS is what S'cool Moves is all about. Often confused with other movement programs, S'cool Moves is much more than movement or exercise. It's specifically targeted intervention that has 5 key elements essential for implementation success.
Stepping outside of my education comfort zone, I have a conversation with an international superstar in the field of social work and therapy for trauma survivors. Because this engaging interview was packed full of great information, I edited the video and made two Cognitive Confetti videos. In fifteen minutes, get up to speed with the latest research in regulation.
On the run? You can watch this video in three minutes or less and learn how helping a friend paint her office led to this week's Tuesday Tip.
To the tune of Jingle Bells, here we go...Push those walls. Push those walls. Push them every day...I've put together a short little 3 minute video showing different ways schools are using Wall Push-ups. How are you using them? We want to know. Please post if you have an idea to share with us.
Blue light and the dangers to our vision is a real threat to our children and ourselves. This is an edited version from my original one hour interview with Deg'er Saner from Reticare. If you need research, I've included it here along with the video. Please share and let's save our children's eyes and our own!
Here's your Tuesday Tip to flip the learning switch for students in your classrooms or clinics. It's a three letter word that changes everything. Enjoy my first videoscribe, the program that animates concepts on a white board. I created a short three minute scribe. Shalea and I had fun choosing images and trying to figure out the ins and outs of the animation software.
We all know about Common Core State Standards but what I mean by "core" is a wee bit different than what policymakers mean by "core." This lesson includes a demonstration video to use with children in classrooms or clinics, a brief Debinar explaining the theory behind core movement, Minute Moves Core Out The Door Routine, Minute Moves Core On The Floor Routine, and Family Letters to send home in English and Spanish.
This Staff Chat includes a "Debinar" highlighting why using the Figure 8 is an important strategy for supporting student goals. How does this work? Deborah Sunbeck, Ph.D. provides the evidence base needed to validate using Figure 8s in our daily activities. Dr. Jill Bolte Taylor discusses serial and parallel processing information in her Ted talk. I cover this briefly in this video.
If you are new to S'cool Moves and would like more information, view our introductory videos, research, and more. This month, we are focusing on teaching students and staff the I Can Calm Myself poster moves.
In this post, we've included the link to the I Can Calm Myself* routine including "Why it Works" on the back side. For home collaboration, Family Letter #1 in English and Spanish are included to send home as you see fit.
Who is missing from your collaboration team when it comes to supporting struggling readers? Let's get those readers off the bench and into the game. Read more by clicking on the title of this blog post. You'll be directed to the International Literacy Association's Blog area.
Amy, we are so honored that you asked S'cool Moves to share in your journey of love to South Africa. When I received this email from you, my heart filled with so much hope and gratitude. Your mission to support the volunteers and children teach us about the difference that one person can make in the lives of many.
After doing an Online Blended Course with therapists from New Beginnings School in New Jersey, the therapists wanted to share what they had learned with their entire staff. After the therapists completed their staff training, I received an email from Jessica saying how excited she was about the training and that the staff was on board to begin implementing S'cool Moves. Her excitement led me to ask if I could interview her and share how she managed to get an entire staff on board with only a TWO HOUR presentation!
As collaboration becomes the holy grail of excellence in education, the calls and emails I receive highlight the challenges of sorting through education mandates, various intervention frameworks, and a host of bureaucratic processes. Teachers and support staff, wanting to hitch their wagons to best practice collaboration protocol find it difficult to travel through the rocky terrain, prickly brush, and many obstacles that make the journey frustrating and tiring.
I just finished doing three separate training days with a fabulous group of therapists from Cincinnati Public Schools. We had engaging conversations. Many challenges with collaboration surfaced through our discussions. As the final assignment, I asked the therapists to write out their challenges and brainstorm possible solutions. ...I'm sharing their great work with you today.
Do you ever feel the need to slip away into some quiet place where you can take a walk in nature or decompress with an activity that helps you regroup?
This week, I applaud a school in Oregon that is providing a quiet, yet effective space for children to diffuse and regroup when they are struggling to maintain their cool and focus control in the classroom.
I took my daughter and her friend to see Disney on Ice's Frozen. By now, I think just about everyone knows the chorus line, "Let it go. Let it go . . . the cold never bothered me anyway!" Yes, the song, "Let it Go" was voted the worst song ever by parents because they got so tired of hearing it played over and over and over and over again by their children. Back at the MODA Center in Portland, as the big moment arrived and the long awaited "Let It Go" song was sung to the glee of every child there, my mind (which often goes in directions that no mind should go) started thinking about the theme of the song . . . letting go. When do we let go when it comes to collaborating with others?
Now THAT got your attention, didn't it? In today's blog post I'll be sharing a strategy that is highly effective but also nets you more F's in school. No, I don't mean F's as in the lowest obtainable score one can achieve. I mean F's as in Fun, Fun, and more Fun. When I interviewed the teams of successful collaborators, the word "fun" was often mentioned, meaning collaboration wasn't some drawn out, snoozefest where nothing got accomplished and no one agreed on anything. Collaboration was an inspired work of art causing the wheels of instruction to turn smoothly and creating an environment that made the teams want to come to school and hang out with each other. Through collaboration, they put the "F" back into their day . . . Fun! One of the strategies they described, I ended up calling "The One-for-All Strategy." In my mind, I could see the Disney Three Musketeers (Mickey, Donald, and Goofy) with their swords held high squealing, "All for Fun and Fun for All!
Today a physical therapist called me from Ohio wanting to know more about our products, but what she really needed from me were some insights that I've gained from doing what I do for the last twenty years. She was a therapist making the huge shift from clinical practice to educational practice in schools.
I am very interested in using this program but want to get the best "bang for my buck." I am sure some of your resources are better suited for my needs than others. I am also looking for some help getting started with the program.
Okay, yes, oxygen. Maybe a winning lottery ticket, or the missing sock? Something else?
TIME. IT'S ONE THING WE ALL NEED MORE OF ON A DAILY BASIS.
Enter Amy Spence, occupational therapist from Dayton, Ohio. Amy's research study provides some insights to help us answer the TIME question.
Dr. Spence's research results provided evidence-based data showing that doing S'cool Moves activities for 15 minutes per day for 3-4 days a week leads to improvement in on-task classroom behavior in six weeks.
I've been connecting with the groups focusing on dyslexia and trying to understand why they have a problem with movement-based learning techniques for children diagnosed with dyslexia. One dyslexia "expert" has gone so far as to call Brain Gym and all movement-based learning programs snake oil.
The reason the dyslexia groups are so opposed to movement-based learning programs is because of the claims many of them make. It is really important that if we advocate for movement in the classroom, we are very clear what type of movement we are advising and not make claims about movement healing dyslexia and other learning challenges.
I get asked this question a lot. It makes me think about our name: S'cool Moves. While I love our name, I think sometimes that it may not adequately represent the depth of academic research and theory we cover in our workshops.
In a new Mayo Clinic study, researchers examined the physical act of reading to see if practicing eye movements in school could lead to better reading fluency. I was pleased to find the following video because in my position as a site literacy coordinator for twelve years, I used the King-Devick Saccadic Eye Movement Test to evaluate if the physical skills of reading were contributing to my students challenges with reading. I found the test to be highly reliable in separating the students with physical challenges from the students who had other issues, such as phonemic awareness or comprehension.


My new Gold Severum, this was a few days ago before that nasty parasite ich appeared.
awwwww wook at the wittle severum:p i remember when i first got mine he looked like that, hes only about 4.5 inches now but man is it a huge difference from that size to now.
Scottish fish- if I remember correctly - doesn't have a quarantine only because he isn't allowed to get another tank.


Our paperback collections feature the finest selection of books to captivate young readers. From classics to contemporary favorites, this set of 40 books was hand selected with care for your Grade 1 classroom.
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text:
Yayoi Kusama inside Kusama's Peep Show or Endless Love Show (1966). All images reproduced in our newly updated Yayoi Kusama monograph
When Yayoi Kusama created her first ever Infinity Room
Going to her show at the Broad? Then uncover the history behind these hugely popular artworks before you do
In 1968, in a press release issued to promote an Alice in Wonderland-themed nude happening in Central Park, the Japanese-born artist Yayoi Kusama wrote, "Like Alice, who went through the looking-glass, I, Kusama, (who have lived for years in my famous, specially-built room entirely covered by mirrors), have opened up a world of fantasy and freedom."
It's hard to verify this claim, as with many of the provocative public statements Kusama made around this time. However, the artist had certainly been building mirrored environments in her Manhattan studio since at least 1965.
"In her first installation incorporating mirrors, her 1965 work Infinity Mirror Room: Phalli's Field, Kusama arranged hundreds of soft, phallic forms in a roughly 25 square-metre mirrored room," writes author and curator Catherine Taft in our newly updated Kusama monograph. "Meant to be engaged with, the stuffed protuberances – which were multiplied through infinite reflection – enveloped the viewer, creating an almost psychosexual encounter with one's own body and image."
Yayoi Kusama, Infinity Mirror Room (Phalli's Field) (1965)
Indeed, this idea of an infinite, all-encompassing artwork was something that Kusama had explored before, in her large-scale Infinity Net paintings of the late 1950s and early 1960s. She had also kitted out rooms with similarly repetitive motifs, including polka dots and penises. However, it was with her following room work that Kusama streamlined her mirror and light works, in a 1966 work called Peep Show or Endless Love Show.
"A mirrored hexagonal room with coloured lights that flashed in time to piped-in rock and roll, Peep Show, like its bawdy namesake, was experienced by viewers through slots located at eye level," writes Taft. "In an onanistic twist, rather than ogling an anonymous 'star' on the Peep Show's stage, the only image one saw was one's own – reflected ad infinitum in the mirrored walls, surrounded by blinking lights, for all the world like a kinetic marquee."
Yayoi Kusama, Kusama's Peep Show or Endless Love Show (1966)
The work was in keeping with the psychedelic light shows of the period, as well as the seamier side of the sexual revolution; a similarly mirrored environment appears in Self Obliteration, the 1968 film Kusama made with Jud Yalkut. Look out for some deeply groovy body painting, which takes place inside Kusama's mirrored environment, towards the end of this short movie, below.
Infinity Rooms became a less–prominent aspect of Kusama's work following her return to Japan in 1977 and it wasn't until 1993, when she was chosen to represent Japan at the Venice Biennale, that the art world got a good look into one of her Infinity Rooms again, this time decorated with a polka-dot theme, and with Ms Kusama herself in attendance at points, handing out miniature pumpkins.
Yayoi Kusama, Infinity Mirrored Room - Love Forever (1996)
Since then, she has recreated different Infinity Rooms at various sites around the world, to critical and popular acclaim. A show of six Infinity Rooms opens at the Broad in LA this weekend, and advance tickets have sold out already; while big crowds are bound to greet the installation of two new Infinity Rooms at David Zwirner's Chelsea gallery next month. Once highly avant-garde, these newer infinity rooms have proven to be remarkably popular with pop stars such as Adele and Katy Perry, as well as ordinary gallerygoers, who will queue for hours to spend a few minutes inside these environments.
Why? Well, perhaps because, unlike the earlier versions, these later Infinity Rooms are less places of mind-blowing self-obliteration, and more sites for quiet contemplation which, in a push and pull world even Kusama could never have imagined is something everyone - art lover or simple aesthete - can do with now and again.
Yayoi Kusama, Obliteration of Eternity (2009)
"Mirrored rooms like Fireflies on the Water (2000) and Infinity Mirrored Room: The Souls of Millions of Light Years Away (2013) are comprised of tiny points of cool-hued LED lights floating in the dark like a city seen from space or a galaxy viewed from earth," Taft writes. "Love is Calling (2013), on the other hand, is a cavernous environment employing shifting coloured light that fills the interiors of soft, dotted, stalagmite and stalactite forms. Aftermath of Obliteration of Eternity (2009) is a mirrored chamber of golden-glowing lantern shapes that appear to float on their own like ceremonial luminarias.
These installations typically have a low platform on which the viewer stands while alone in the room with the mirrored door shut, surrounded by infinity. The floor is sometimes filled with a shallow pool of water that further reflects the lights and adds a natural element to the space, separating the viewer from the infinite walls and lights beyond. These are introspective spaces for perceiving, listening and feeling."
For greater insight into this extraordinarily talented artist order a copy of our newly updated Yayoi Kusama book here.
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"Yayoi Kusama transcended the art world to become a fixture of popular culture, in a league with Andy Warhol, David Hockney, and Keith Haring." —The New York Times
Kusama is internationally renowned for her groundbreaking work on themes such as infinity, self-image, sexuality, and compulsive repetition. A well-known name in the Manhattan scene of the 1960s, Kusama's subsequent work combined Psychedelia and Pop culture with patterning, often resulting in participatory installations and series of paintings. This revised and expanded edition of the 2000 monograph, which is arguably still one of the most comprehensive studies on her work to date, has been augmented by an essay by Catherine Taft and a collection of new poems by the artist.
Akira Tatehata is a poet and art critic who has been awarded the 2005 Takami Jun Award for Literature.
Laura Hoptman is a curator at the Museum of Modern Art in New York.
Udo Kultermann was a curator and art historian who taught at Washington University in St. Louis.
Catherine Taft is deputy director of LAXART in Los Angeles.
"When people look back at Kusama's work decades from now, they'll see that her idea of creation and infinity has an eternal endurance." —Marc Jacobs
"The effect of Kusama's work is both so complex and simple. It is produced by the interaction of the two close somewhat parallel planes, at points merging at the surface and at others diverging slightly but powerfully." —Donald Judd
"Yayoi Kusama comes with a pedigree of film and orchestrated happenings and performance in New York during the 1960s and the 1970s that was every bit as avant-garde as Warhol."—The Art Review Power 100
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"Still one of the most comprehensive studies of this intriguing artist [...] allows readers to traverse from 'Happenings' in 1960s New York through to Kusama's current work, which encompasses diverse media from paintings to sculpture. And, for the first time, a collection of poems by Kusama are published to accompany the colourful illustrations, compelling essays and interviews by curators and critics."—TheCultureTrip.com
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"Filled with visual treasures... Now includes poems by the artist."—Harper's Bazaar
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"Richly illustrated... Explores Kusama's varied career to the present, through informative essays written from diverse scholarly and critical perspectives, as well as interviews with the artist, and selections from her lively poetry and fiction."—The British Museum Magazine
"A fascinating artist."—Visi (South Africa)
"A dose of inspiration (that also looks good on the coffee table)."—Vogue Online
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redpajama_set_name: RedPajamaCommonCrawl


Risk Profiling, Risk Management and Investments Eben Maré, Head of Fixed Income, ABSA Asset Management Understanding your investments and investment process, including your attitude to risk management, plus carrying out risk profiling, is essential to attain longer-term goals such as retirement plans.
IBM recently estimated that we generate approximately 2.5 quintillion bytes of data daily (one followed by 18 zeros – roughly the equivalent of 57.5 billion 32GB iPads).
"Our favourite holding period is forever." – Warren Buffet.
According to the NYSE factbook, in the 1960s stock average holding periods amounted to around eight years – currently it is believed to be of the order of several days.
There are several interpretations one can draw from these two statements.
Buffet's quote refers to investment conviction and ignoring short-term 'noise'.
The NYSE factbook statement, however, is about flow of information, uncertainty, short-term goals, liquidity and changes in the investment landscape and investors' attitudes to bearing longer-term risks.
Modern investors undoubtedly face a proliferation of investment products and ideas. However, they also benefit from the diversification afforded by liquidity in international stock, bond, currency, and commodity markets as well as the access thereto.
The flow of information serves to create liquidity in the financial markets – one could argue, however, that investors are prone to misinterpretation of information and the likely effects of current events on long-term investment outcomes. Take as an example the amount of confusion and anxiety that has arisen as a result of the so-called Brexit decision.
Another example would be the effects of zero and negative interest rate policies (so-called ZIRP and NIRP, respectively) currently maintained by many central banks – how are investors to calculate the long-term effects of policies which have served to elevate asset price valuations without delivering meaningful extra economic growth?
Policies are one source of uncertainty but it is equally important to try to understand the effect of longer-term trends; interest rate curves are presently low in many parts of the world, this is partly reflective of low interest rate policies maintained by many central banks as referred to above, however, investor confidence and limited profitable opportunities have kept the so-called equilibrium real interest rate subdued.


text:
Prosecutor: Capitol Rioter Planned to 'Take Hostages'
Man Charged With DUI Has Only Caffeine in System
Joseph Schwab's lawyer says it's 'absurd'
By Michael Harthorne, Newser Staff
Posted Dec 24, 2016 12:55 PM CST
Too much soda or coffee perhaps? (AP Photo/Paul Sakuma, file)
(Newser) – "I've never seen this before," California attorney Stacey Barrett tells the Guardian. "I've never even heard of it." Barrett's client, 36-year-old Joseph Schwab, is facing a misdemeanor DUI charge—apparently over caffeine. Schwab was pulled over in August 2015 in Solano County while driving home from his job installing glass, the Fairfield Daily Republic reports. An agent from the state department of alcoholic beverage control said Schwab cut her off and was driving erratically. Barrett says a breathalyzer test at the scene showed a 0.00% alcohol level, but Schwab was booked into jail regardless. His blood was screened twice, testing negative for all sorts of drugs. The only positive result: caffeine.
A forensic toxicologist says he hasn't heard of anyone being charged with DUI for caffeine in his 41 years on the job, and Barrett calls the whole thing "absurd." Schwab's toxicology report didn't specify how much caffeine was in his system, and he says he's not sure he even consumed anything with caffeine that day. The county's chief deputy DA says the DUI charge isn't based on caffeine but has offered no other evidence, Barrett says. It took the county nearly 10 months to bring charges against Schwab, who says the delay makes it harder for him to defend himself, such as by finding the crew he was working with that day to testify on his behalf. With the trial finally about to start, Barrett is seeking to get the case dismissed. (This guy drank four or five energy drinks a day, then turned yellow.)


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HHS.gov President's Council on Sports, Fitness & Nutrition
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Kristina Harder
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PCSFN Executive Director
Kristina Harder is the Executive Director of the President's Council on Sports, Fitness & Nutrition. The President's Council advises the President, through the Secretary of Health and Human Services, on the progress of developing a national strategy to expand participation in youth sports and to promote good nutrition for all Americans. Harder is responsible for overseeing the administration of the programs and strategic plan of the organization.
Sports have always been an integral part of Kristina's life. She grew up with a dad who was an All-American basketball player at the University of Michigan, and was encouraged to play sports from a young age. She competed in basketball, volleyball, swimming, and soccer while dabbling in tennis and water polo, and culminated her athletic career playing varsity volleyball at the University of Pennsylvania. She also combined her love for sports with her faith by helping to lead the Fellowship of Christian Athletes (FCA) group at both her high school and college.
Kristina began working at the Department of Health and Human Services (HHS) in the Immediate Office of the Secretary before joining the Office of the Assistant Secretary for Health, first as Deputy Chief of Staff and most recently as Chief of Staff to ADM Brett Giroir.
Kristina is a native of Orange County, California and received her bachelor's degree from the University of Pennsylvania in Communications and Public Service, with a minor in Hispanic Studies. She went on to receive a Master's degree in Public Policy (MPP) from the University of Edinburgh in Scotland.
Kristina and her husband Brandon reside in Washington, D.C. Together they serve at their local church and lead mission trips to Kenya and Tanzania, serving orphans and victims of sex trafficking through health and leadership education, sports and fitness classes, relationship building, spiritual development, and construction projects.
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Content last reviewed on March 2, 2020
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Divisions of HHS commonly use websites, blog entries, and social media posts to issue communications with regulated parties. While these communications may provide the public with helpful information they cannot, by themselves, impose binding new obligations on regulated entities.


I ain't here to make no friends"
I'm right back at it again"
Said, "If you can't make it here you won't make it there"
Don't wanna hear about it, yeah!
I ain't here to make no..."
Composição: Jeremy McKinnon · Esse não é o compositor? Nos avise.


Located in the heart of Wigwam Creek South, this gorgeous 4 bedroom 2 bathroom home is great for families of all sizes. This home is insulated from main arterial streets reducing traffic and noise. Within walking distance from Wigwam Creek Middle School, this home is perfect for families who want the small town feel of a close knit community. The owners of this home, have truly loved it, and it shows throughout the home's maintenance and upgrades. Enjoy this home's split floor plan providing privacy in the master retreat, and a more unique flow to the home. The 4th bedroom is currently being used as a office/den, but has a full closet, and even though the doorway is not enclosed, doing so would be a breeze. This home has Tile in all the right places, and Carpet where you want it. The tile provides ease for cleaning, and minimal wear in the high traffic areas. The entry living room was just redone with high-end laminate floors! Enjoy entertaining and hanging out with the family in the open kitchen and living room concept. This kitchen is fully loaded with a built in microwave, dishwasher, and breakfast bar height counter tops. Enjoy cheaper energy bills with this home's North and South exposure. The end of the street is a cul-de-sac, bringing the traffic to a minimum giving you peace of mind letting the kids play in the neighborhood.


When developments occur that involves extension of City water and sewer mains, the owners are requested to petition for annexation into the city limits. A Petition for Annexation (PDF) must be signed by all owners of the property, and if a corporation, by the proper officers with the corporate seal.
A boundary survey prepared by a licensed surveyor is required along with a written metes and bounds description for annexation of property into the Sanford City Limits. The survey should tie to existing city limits and show existing and new city limit boundaries. The following items must be included on the map.
Two Mylar copies of the map are required. The written legal description should be provided on diskette in a file compatible with Windows and Microsoft Office so that City staff will not have to retype it.
The process normally takes 3 City Council meetings, including a public hearing on the petitioned annexation. After the property is annexed, an additional public hearing may be required to establish an appropriate city zoning classification if the property was previously in Lee County's zoning jurisdiction.
The City Council normally meets on the first and third Tuesday of each month. Public hearings by the Council and Planning Board for zoning matters are conducted on the third Tuesday. An annexation petition needs to be submitted ten days in advance of the Council meeting. This lead-time allows opportunity for review by the Law and Finance Committee, if needed prior to the Council meeting.
Please note: The applicant shall be responsible for all fees in regard to recordation and advertising of annexation requests.


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An Analysis of Post-Traumatic Stress Disorder in Reign Over Me, a Movie by Mike Binder
Topics: Post Traumatic Stress Disorder
Analysis, Pages 4 (931 words)
Post-Traumatic Disorder in Reign Over Me The movie Reign Over Me is about a man named Charlie Fineman. Charlie used to be a practicing dentist, but his whole life was turned upside down when he lost his wife, three daughters, and poodle during 9/11. One day while Charlie is on the street when his old roommate from college, Allen Johnson, sees him and tries yelling for him, but Charlie does not stop. Allen becomes very concerned about Charlie, and soon learns that something is very wrong with Charlie, and he sets out to help his old friend.
It is very clear that Charlie is suffering from Post Traumatic Stress Disorder. Charlie has experienced all four general types of persistent symptoms since his family died in 9/11. When Charlie confronts his deceased wife's parents after the court hearing, he admits to intrusive hallucinations and illusions where he still sees his family and wife in his day-to-day life. Charlie, also, constantly practices avoiding anything to do with his past life, including; avoiding his wife's parents, quitting his job as a dentist, refusing to acknowledge he ever had a family, and escaping into music when he sees Allen with his own family.
He also has periods of constant negative thoughts and mood, especially when he tried to commit suicide by cop.
Charlie also has trouble sleeping, often staying up all night to play video games, and he has random angry outbursts, especially when any aspect of his family or past life is brought to his attention.
writer-Charlotte
Proficient in: Post Traumatic Stress Disorder
" Amazing as always, gave her a week to finish a big assignment and came through way ahead of time. "
Charlie meets many of the DSM-5s diagnostic criteria, such as; learning that a traumatic even occurred to a close family member, recurrent and intrusive memories of the traumatic event, avoidance of external reminders of the events, persistent negative emotional state, a diminished interest in significant activates, irritable behavior and angry outbursts, sleep disturbance and reckless and self-destructive behavior. Charlie has been living with all of these symptoms for several years, and they have significantly impacted his social, occupational and overall functioning. Charlie's hippocampus is probably working harder than normal, and he likely has abnormal amounts of the neurotransmitters norepinephrine and serotonin. Charlie's experience convinced him that the world is a dangerous place, and he is unable to control anything happening to him. Charlie's past traumatic event of his parents dying could have made him more vulnerable to developing PTSD. Charlie coped with his family dying by dissociating himself from his past life.
Charlie's family dying became a unconditioned stimulus, and anything that remind him of them became the conditioned stimulus, which induced his powerful and conditioned emotional responses; avoiding any feelings associated with his family, and refusing to emotional connect to anybody until Allen. When Charlie experienced his traumatic event, he responded by detaching himself from his past life. This left him with no social support. In the psycho feedback loop Charlie's anxiety caused him to disassociate himself from the trauma, and in return caused a conditioned emotional response to stimuli associated with his family. This response coupled with the social feedback loop lead Charlie to have a complete lack of social support after he detached himself from his past life. And the neurological factors could have made Charlie more susceptible, in the first place, to develop PTSD. In treating Charlie I would aim to relieve his symptoms by helping him deal with the trauma he's experienced.
Rather than avoiding the trauma and any reminders of it, as he has for many years, treatment would encourage him to recall and process the emotions and sensations he felt when he first heard his family had died. Charlie needs an outlet for the emotions he's been bottling up, and treatment should help restore his sense of control and reduce the hold that the memory of his family passing away has on his life. I would focus on getting to understand his thoughts and feelings about his family, helping him work through feelings of guilt, self-blame, and isolation he feels, and help him learn how to cope with and control his intrusive memories. I would start with Charlie on a trauma focused cognitive-behavioral therapy to slowly expose him to his thoughts, feelings, and the situations that remind him of his family. Also, to start to identify his distorted thinking about the event, and about his regret over his last conversation with his wife. I think family therapy would help him reconnect with his in-laws and to help his in-laws better understand how Charlie feels and what he's going through. This should help them all communicate better and work through their relationship. Since Charlie has shown some signs of anxiety and depression, I may start him on an antidepressant, such as Prozac or Zoloft to help him feel less depressed, worried and on edge.
I feel as if all this treatments, combined with giving Charlie the time to fully cope with his trauma would help him return to the normal life he had before the loss of his family. The story that Reign Over Me tells is compelling and touching. It speaks for the many survivors of 9/11 who became indirect victims after the tragedy. Many of these people have a voice, but have lost it. Many lost something that day; friends innocence, a sense of security. Reign Over Me delicately handles all of this, along with Post Traumatic Stress Disorder. I feel, as of Reign Over Me does a good job of portraying Post Traumatic Stress Disorder to a point. However, no one can truly capture in a short movie all the feelings and emotions that somebody experiencing any mental illness will have.
An Analysis of Post-Traumatic Stress Disorder in Reign Over Me, a Movie by Mike Binder. (2021, Dec 27). Retrieved from https://paperap.com/an-analysis-of-post-traumatic-stress-disorder-in-reign-over-me-a-movie-by-mike-binder/
An Analysis of the Causes of Post-Traumatic Stress Disorder (PTSD) Pages: 3 (747 words)
Post traumatic Stress Disorder Essay Pages: 9 (2507 words)
The Suffering of Billy Pilgrim from Post-Traumatic Stress Disorder in Slaughterhouse Five, a Novel by Kurt Vonnegut Pages: 2 (585 words)
The Impact of Post Traumatic Stress Disorders on American Soldiers Pages: 3 (814 words)
Overview of Dissociative Identity Disorder and Narcissistic Personality Disorder Pages: 2 (419 words)
Mike Wells Photographer Pages: 2 (351 words)
Mike Tyson Daughter Exodus Funeral Pages: 7 (1901 words)
Why School by Mike Rose Pages: 5 (1419 words)
Reign of Terror Pages: 2 (596 words)
Better to reign in Hell than to serve in Heaven Apologies to Pages: 5 (1241 words)


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MA Top 30 Professional of the Year: Terell Johnson
December 31, 2021 | By Zachary Lewis
Chicago Philharmonic
One day after his appointment as executive director of the Chicago Philharmonic on July 1, Terell Johnson attended a recording session by his new band in its home venue, the Harris Theater. Coming on the heels of the double pandemic of the last year-and-a-half—COVID-19 and a heightened awareness, with cause, of social injustice—the relevance of the repertoire was stunning. Recorded for streaming from the Harris Theater website, Redemption, as it is titled, is a collection of spirituals and gospel songs by singer/ composer Adrian Dunn in memory of George Floyd and other Black men killed by police.
Also featuring an arrangement of "Lift Every Voice and Sing" by Dunn and the Adrian Dunn Singers, and the 2019 violin concerto Glory, by Marcus Norris, the performance epitomized for Johnson the projects he had in mind for the orchestra's future. "It was a remarkable experience and a testament to the Philharmonic's ambition to transform the classical music landscape into an everexpanding inclusive space," Johnson said, noting that the concert was the first he'd seen in which all participants "looked like me."
Not that Johnson is new to pushing for change in classical music. Before his appointment, he chaired the equity, diversity, inclusion, and belonging committee at Miami's New World Symphony (NWS), also serving as director of business development and community engagement. Last April, he moderated "Being Black in America," a panel discussion with pianist Awadagin Pratt and Human Rights Watch U.S. Executive Director Nicole Austin-Hillery, joined by NWS fellows.
Now Johnson leads a group that stands apart in a different way: The Chicago Philharmonic, with origins that date to 1979, is a self-governed orchestra that operates without a union contract and holds no formal auditions. Instead, Artistic Director and Principal Conductor Scott Speck maintains a group of some 200 professionals to contact as needed. Johnson called this unique governance model, along with the group's eclectic musical vision, "our super power…it's what inspired me to join the team."
Life during the pandemic got even more interesting for Johnson when he was accepted into the LEAD fellowship, a twoyear program grooming leaders of color offered by the Sphinx Organization. Said Donna Milanovich, who preceded Johnson as the Chicago Philharmonic's executive director, "I can think of no better leader," for the orchestra. "His business acumen and love of the transformative power of music is sure to take [it] to new heights."