text:
\section*{Introduction}

Let $X$ be a smooth projective variety.  A coherent sheaf $E\in \Coh
X$ is called \emph{quasi-exceptional} if $\ext ^i(E,E) =0$ for all
$i>0$. A coherent sheaf $E\in \Coh X$ is called \emph{tilting} if it
is quasi-exceptional, $E$ Karoubian generates the derived category
$D^b(X)$ and the algebra $\hom _X(E, E)$ has finite global dimension.

Let $Q_n$ be an $n$-dimensional quadric defined over an algebraically
closed field $k$ of characteristic $p>2$. Assume for simplicity that
$n\ge 3$.  Let $F: Q_n\to Q_n$ be the absolute Frobenius morphism and
let $F^s$ be the composition of $s$ absolute Frobenius morphisms.

\begin{Theorem}\label{main}
Let $s$ be a positive integer. Then $F^s_*\O_{Q_n}$ is a tilting
bundle if and only if one of the following holds:
\begin{enumerate}
\item $s=1$ and $p>n$,
\item $s=2$, $n=4$ and $p=3$,
\item $s\ge 2$, $n$ is odd and $p\ge n$.
\end{enumerate}
\end{Theorem}

The above theorem is a summary of Corollaries \ref{t1}, \ref{t2} and
\ref{t3}. In case of $s=1$ and $n=3,4$ the theorem (see also Corollary
\ref{quasi-exc}) gives the main result of \cite{Sa1}. In case of $s=1$
A. Samokhin in \cite{Sa2} proved a related result for $p\gg 0$ but
using a completely different method.

In fact, we prove a much stronger result than stated above: we
determine the decomposition of Frobenius push-forwards of line bundles
on quadrics.  In case of projective spaces one can easily compute the
corresponding decomposition using the Horrocks splitting criterion
(see, e.g., \cite[Lemma 2.1]{Ra}; one can also perform another direct
computation as in \cite[Proposition 4.1]{HKR}). We use a similar
strategy in the case of quadrics although it is much more difficult in
carrying out as we need to prove some non-trivial vanishing and
non-vanishing theorems for cohomology of Frobenius pull-backs of
spinor bundles.

\medskip

Let $X$ be a smooth projective variety defined over an algebraically
closed field $k$ of characteristic $p>0$. Let $X^{(s)}$ denotes the
$s$-th Frobenius twist of $X$, i.e., $X$ with the $k$-structure
defined by the fiber product of $X$ over the $s$-th Frobenius morphism
on $k$. Then the sheaf $\Diff_X$ of $k$-algebras of differential
operators admits the so called \emph{$p$-filtration}
$$\Db ^1_X\subset \Db ^2_X\subset \dots \subset \Db^s_X=\Hom
_{X^{(s)}}(\O_X, \O_X)\subset \dots \subset {\Diff}_X=\lim
_{\mathop{\leftarrow}_{s}}\Db ^s_X.$$ Let us also set $\db^s_X=\hom
_{X^{(s)}}(\O_X, \O_X)$. By definition $\db^s_X$ is an algebra, finite
dimensional over $k$.  By \cite[Proposition 3.4]{HKR} the above
theorem implies the following corollary:

\begin{Corollary}
If one of the conditions 1, 2, 3 of Theorem \ref{main} holds then
there is a canonical triangulated equivalence between the bounded
derived category $D^b(\Coh (\Db _{Q_n}^s))$ of coherent $\Db
_{Q_n}^s$-modules and the bounded derived category $D^b(\db
_{Q_n}^s-{\mathrm{mod}})$ of $\db _{Q_n} ^s$-modules.
\end{Corollary}

This is an analogue of the derived localization theorem of Beilinson
and Bernstein. Note that contrary to some expectations this equivalence
does not hold in any characteristic if $n$ is even and $s\ge 3$.

Let us recall that a variety $X$ is called \emph{D-quasi-affine} if
any $\O_X$-quasi-coherent $\Diff _X$-module is $\Diff_X$-generated by
its global sections. $X$ is \emph{D-affine} if it is D-quasi-affine and
$H^i (\Diff _{Q_n})=0$ for $i>0$.

In characteristic zero Beilinson and Bernstein (see \cite{BB}) proved
that flag varieties for semisimple algebraic groups are D-affine. In
\cite{Ha} Haastert proved that in positive characteristic the
projective space and the full flag variety for $\SL _3$ are D-affine but
no other examples of smooth projective D-affine varieties are
known. We prove that all quadrics are D-quasi-affine (see Proposition
\ref{quasi-aff}). Together with Theorem \ref{main} this implies the
following corollary:

\begin{Corollary}
If $n$ is odd and $p\ge n$ then $Q_n$ is D-affine.
\end{Corollary}

Let us note that contrary to \cite[Proposition]{AK}, non-vanishing of
$H^i (\Db ^s_{Q_n})$ for even $n$ and $s\ge 3$ does not imply that
$H^i (\Diff _{Q_n})\ne 0$ (there is an error in the last two lines of
proof of \cite[Proposition]{AK} as the natural inclusions $\Db ^s\to
\Db^r$ do not agree with the maps from \cite[Lemma]{AK}).  In
particular, our results do not immediately imply that other quadrics
are not D-affine.

\medskip

We do not consider the case of characteristic $2$ although our methods
apply also in this case. We skip it as this case needs a separate,
quite long treatment. Let us just note that on the $3$-dimensional
quadric already $F^4_*\O_{Q_3}$ is not quasi-exceptional (but
$F^i_*\O_{Q_3}$ is quasi-exceptional if $i\le 3$).

\medskip

The structure of the paper is as follows. In Section 1 we give a few
interpretations of spinor bundles, compute some related cohomology and
show some dualities. In Section 2 we study Hilbert functions of some
algebras that occur to describe decompositions of Frobenius
push-forwards of line bundles on a quadric. In Section 3 we use these
results to prove some vanishing and non-vanishing theorems for
Frobenius pull-backs of spinor bundles. In Section 4 we use them for
computation of decompositions of Frobenius push-forwards of line
bundles on a quadric and we prove Theorem \ref{main}.

\section{Spinor bundles}

\subsection{Spinor bundles via representation theory}

Let $k$ be an algebraically closed field of characteristic $p>2$.  Let
us recall that $\Spin (n+2)$, $n\ge 1$, is a connected, semi-simple,
simply connected $k$-group, isogenous to $\SO (n+2)$.

We fix a maximal torus $T\subset \Spin (n+2)$.  If $n=2m+1$ or $n=2m$
then $\Spin (n+2)$ has $m+1$ simple roots $\alpha _1, \dots,
\alpha_{m+1}$.  Let us recall that a smooth $n$-dimensional quadric
$Q_n\subset \PP^N$, $N=n+1$, is a homogeneous space $\Spin
(n+2)/P(\alpha_1)$.

If $n=2m+1$ then the Dynkin diagram of $\Spin (n+2)$ is of type
$B_{m+1}$. If $n=2m$ then the Dynkin diagram is of type $A_1\times
A_1$ for $m=1$, $A_3$ for $m=2$ and $D_{m+1}$ if $m\ge 3$.

Let $\lambda_i\in X^*(T)$ be the fundamental weights defined by
$2\langle\lambda_i,\alpha_j \rangle/\langle\alpha_i,\alpha_j
\rangle=\delta_{ij}$.

For $n=2m$ there are two spin representations of the Levi
quotient of $P(\alpha_1)$ (which is of the same type as $\Spin (n)$)
with highest weight $\lambda_m$ and $\lambda_{m+1}$. They are of
dimension $2^{m-1}$.  If $n=2m+1$ then there is one spin
representation of the Levi quotient of $P(\alpha_1)$ with highest
weight $\lambda_{m+1}$. It is of dimension $2^{m}$.

Duals of vector bundles on $Q_n$ associated to the principal
$P(\alpha_1)$-bundle $\Spin (n+2)\to \Spin (n+2)/P(\alpha_1)$ via
these representations are called \emph{spinor bundles} and denoted by
$\Sigma$ if $n$ is odd and $\Sigma_{-}$ and $\Sigma_{+}$ if $n$ is
even.  The determinant of any spinor bundle on $Q_n$, $n\ge 3$ is
isomorphic to $\O _{Q_n}(-2^{[\frac{n-3}{2}]})$. Let us note the
following useful isomorphisms:
$$
\begin{array}{cl}
\Sigma ^*\simeq \Sigma (1) \quad &\hbox{if $n=2m+1$, } \\
{\Sigma}_{-} ^*\simeq {\Sigma}_{-} (1) \quad \hbox{ and }\quad {\Sigma}_{+}^*\simeq
{\Sigma_{+}}(1) \quad &\hbox{if $n=4m$, } \\
{\Sigma}_{-} ^*\simeq {\Sigma_{+}} (1) \quad \hbox{ and }\quad {\Sigma}_{+}^*\simeq
\Sigma_{-} (1) \quad &\hbox{if $n=4m+2$. } \\
\end{array}
$$ We leave it to the reader to verify these isomorphisms using the
well known computation of the centre of the Spin group.

Let us recall that a vector bundle $E$ on a smooth $n$-dimensional
hypersurface $X=(f=0)\subset \PP^{n+1}=\Proj\, S$ is called
\emph{arithmetically Cohen-Macaulay} (ACM for short), if it has no
intermediate cohomology, i.e., $H^i(X, E(t))=0$ for $0<i<n$ and all
$t$. A vector bundle $E$ is ACM if and only if the corresponding
graded $S/(f)$-module is maximal Cohen-Macaulay.

Let us also recall that a vector bundle is called strongly slope (semi)stable
if all its Frobenius pull-backs are slope (semi)stable.

\begin{Theorem} \label{coh-spin}
A spinor bundle $\Sigma$ on $Q_n\subset \PP^N$ is a strongly slope
stable ACM bundle. Moreover, $h^0(Q_n, \Sigma (t))=0$ for $t\le 0$ and
$h^0(Q_n, \Sigma (t))=2^{[\frac{N}{2}]}{{t+n-1}\choose {n}}$ for $t\ge
1$.
\end{Theorem}

\begin{proof}
Let us first note that spin representations are irreducible (this
follows from \cite[Part II, Corollary 5.6]{Ja}). Therefore by
\cite[Theorem 2.1]{Bi} spinor bundles are slope stable. Strong slope
stability follows in a standard way from inequality $\mu _{\max}(\Omega _{Q_n})<0$.

The fact that spinor bundles are ACM can be proven in the same way as
\cite[Theorem 2.3]{Ot}. The last part of the theorem will be obvious
later (use sequences (\ref{(i')}), (\ref{(ii')}) and (\ref{(iii')})).
\end{proof}

\subsection{Spinor bundles via matrix factorization}

Theorem \ref{coh-spin} implies that spinor bundles correspond to
irreducible maximal Cohen--Macaulay modules on an affine cone over the
quadric (or equivalently to indecomposable matrix factorizations of
the equation of the quadric). Below we give an explicit construction
of spinor bundles using matrix factorization.

As a special case of Kn\"orrer's periodicity theorem (see
\cite[Theorem 3.1]{Kn}) we get the following theorem:

\begin{Theorem} \label{ACM-Q}
Any ACM bundle on a smooth projective quadric defined over an
algebraically closed field of characteristic $p\ne 2$ is a direct sum
of line bundles and twisted spinor bundles.
\end{Theorem}

If $n\le 2$ then the Picard group is not generated by $\O_{Q_n}(1)$
and any ACM bundle on $Q_n$ is isomorphic to a direct sum of line
bundles $\O_{Q_n}(i)$ and spinor bundles twisted by some $\O
_{Q_n}(i)$. If $n\ge 3$ then any direct sum of line bundles and
twisted spinor bundles is ACM.

\medskip

Let us set $\varphi_0=\psi_0=(x_0)$ and let us define inductively
pairs of matrices
$$\varphi_{m+1}=\left(
\begin{array}{cc}
\varphi_{m} & x_{2m+1}I_{2^{m}\times 2^{m}}\\
x_{2m+2}I_{2^{m}\times 2^{m}}& -\psi _m\\
\end{array}
\right)$$
and
$$\psi_{m+1}=\left(
\begin{array}{cc}
\psi_{m} & x_{2m+1}I_{2^{m}\times 2^{m}}\\
x_{2m+2}I_{2^{m}\times 2^{m}}& -\varphi _m\\
\end{array}
\right) .$$
Then the pair $(\varphi_m,\psi_m)$ is a matrix factorization of
$x_0^2+x_1x_2+\dots +x_{2m-1}x_{2m}$.

Let us set $\varphi'_0=(x_1)$, $ \psi'_0=(x_2)$ and as above define
inductively pairs of matrices
$$\varphi'_{m+1}=\left(
\begin{array}{cc}
\varphi'_{m} & x_{2m+1}I_{2^{m}\times 2^{m}}\\
x_{2m+2}I_{2^{m}\times 2^{m}}& -\psi' _m\\
\end{array}
\right)$$
and
$$\psi'_{m+1}=\left(
\begin{array}{cc}
\psi_{m}' & x_{2m+1}I_{2^{m}\times 2^{m}}\\
x_{2m+2}I_{2^{m}\times 2^{m}}& -\varphi'_m\\
\end{array}
\right) .$$
Then the pair $(\varphi_m',\psi_m')$ is a matrix factorization of
$x_1x_2+\dots +x_{2m-1}x_{2m}$.

Let $i:Q_n \hookrightarrow \PP^N$ be the above defined embedding of a
quadric ($n=2m$ or $n=2m+1$). Then we have the following short exact
sequences of sheaves on $\PP^N$:
\begin{equation}
0\to \O_{\PP^N}^{2^{m+1}}(-1)\mathop{\to}^{\varphi_{m+1}}
\O_{\PP^N}^{2^{m+1}}\to i_*\Sigma (1)\to 0,\label{(i')}
\end{equation}
if $n=2m+1$, and
\begin{equation}
0\to \O_{\PP^N}^{2^{m}}(-1)\mathop{\to}^{\varphi'_m}
\O_{\PP^N}^{2^{m}}\to i_*\Sigma _{-}(1)\to 0,\label{(ii')}
\end{equation}
\begin{equation}
0\to \O_{\PP^N}^{2^{m}}(-1)\mathop{\to}^{\psi'_m}
\O_{\PP^N}^{2^{m}}\to i_*\Sigma _{+}(1)\to 0.\label{(iii')}
\end{equation}
if $n=2m$.

Using the above description we get the following short exact sequences
of vector bundles:
\begin{equation}
0\to \Sigma \to \O_{Q_n}^{2^{m+1}}\to \Sigma (1)\to 0,\label{(i)}
\end{equation}
if $n=2m+1$, and
\begin{equation}
0\to \Sigma_{-}\to \O_{Q_n}^{2^{m}}\to \Sigma _{+}(1)\to 0,\label{(ii)}
\end{equation}
\begin{equation}
0\to \Sigma _{+}\to \O_{Q_n}^{2^{m}}\to \Sigma _{-}(1)\to 0,\label{(iii)}
\end{equation}
if $n=2m$. It should be noted that the above explicit presentations
allow for computer calculations of cohomology groups of Frobenius pull
backs of spinor bundles.

\begin{Lemma} \label{tensor-spin}
For any spinor bundles $\Sigma_1, \Sigma _2$ on $Q_n$
we have $H^1(Q_n, \Sigma_1 \otimes \Sigma _2(t))=0$ if $t\ne 0$.
Moreover,
$$
h^1(Q_n, \Sigma_1 \otimes \Sigma _2)=\left\{
\begin{array}{cl}
1& \quad\hbox{if $n=2m+1$,}\\
1& \quad\hbox{if $n=4m$ and $\Sigma_1\not\simeq \Sigma_2$,}\\
0& \quad\hbox{if $n=4m$ and $\Sigma_1\simeq \Sigma_2$,}\\
1& \quad\hbox{if $n=4m+2$ and $\Sigma_1\simeq \Sigma_2$,}\\
0& \quad\hbox{if $n=4m+2$ and $\Sigma_1\not \simeq \Sigma_2$.}\\
\end{array}\right.
$$ In particular, for any $0<i<n$ there exist spinor bundles
$\Sigma_1, \Sigma_2$ such that $\ext ^i(\Sigma_1(i), \Sigma_2)\ne 0$.
\end{Lemma}

\begin{proof}
As spinor bundles on a quadric are strongly stable we see that $\hom
(\Sigma_1, \Sigma_2(t))=0$ if $t<0$ or if $t=0$ and $\Sigma_1$ and
$\Sigma_2$ are not isomorphic. This remark, together with sequences
(\ref{(i)}), (\ref{(ii)}), (\ref{(iii)}), imply the second part of the
lemma.

To prove the first assertion note that by Lemma \ref{2-spin-duality}
for $s=0$ there exist spinor bundles $\ti \Sigma_1, \ti \Sigma _2$
such that
$$H^1(Q_n, \Sigma_1 \otimes \Sigma _2(t))\simeq H^1(Q_n, \ti \Sigma_1
\otimes \ti \Sigma _2(-t))^*.$$ So it is sufficient to note that by
Theorem \ref{coh-spin} sequences (\ref{(i)}), (\ref{(ii)}),
(\ref{(iii)}) imply that $H^1(\Sigma_1 \otimes \Sigma _2(t))$ is a
quotient of $H^0(\Sigma_1' \otimes \Sigma _2(t+1))$ for some spinor
bundle $\Sigma_1'$. This last cohomology group vanishes for $t<0$ as
we can write it as $\hom (\Sigma '', \Sigma_2(t))$ for some spinor
bundle $\Sigma''$.

The last part of the lemma follows from isomorphisms
$${\ext} ^i(\Sigma_1(i), \Sigma_2)\simeq H^i(\Sigma_1^*\otimes
\Sigma_2(-i))\simeq H^1(\Sigma_1'\otimes \Sigma_2)$$ for some spinor
bundle $\Sigma_1'$.
\end{proof}

\begin{Corollary} \label{dir-sum1}
Let $E$ be an ACM bundle on $Q_n$ and let $0<i<n$ be a fixed integer.
If for all spinor bundles $\Sigma$ on $Q_n$ we have
$H^i(Q_n, E\otimes \Sigma (t))=0$ for all $t\in \ZZ$ then $E$ is a direct sum
of line bundles.
\end{Corollary}

\begin{proof}
Using sequences (\ref{(i)}), (\ref{(ii)}), (\ref{(iii)}) and Theorem
\ref{coh-spin} we see that $H^i(E\otimes \Sigma (t))=H^1(E\otimes
\Sigma ' (t+i-1))$ for some spinor bundle $\Sigma'$.  Then the
required assertion follows from Theorem \ref{ACM-Q} and Lemma
\ref{tensor-spin}.
\end{proof}

\subsection{Some dualities}

For a non-negative integer $s$ let us set $q=p^s$ and
$d_{n,s}=(n-1)\frac{q-1}{2}$.

\begin{Lemma}\label{spin-duality}
Let $\Sigma$ be a spinor bundle on $Q_n$. Then for all $0<i<n$ there
exists some spinor bundle $\ti \Sigma$ such that for all integers
$j$ we have the following duality
$$H^i(Q_n, (F^s)^*\Sigma (d_{n,s}-(i-1)q+j)) \simeq H^i(Q_n,
(F^s)^*\ti\Sigma (d_{n,s}-(i-1)q-j-1))^*.$$
\end{Lemma}

\begin{proof}
Let us first prove the lemma for $i=1$.  Taking Frobenius pull backs of
sequences (\ref{(i)}), (\ref{(ii)}), (\ref{(iii)}) and twisting by
$\O_{Q_n}(j)$ we get the following isomorphisms
$$H^i(Q_n, (F^s)^*\Sigma (q+j))\simeq H^{i+1}(Q_n, (F^s)^*\Sigma _1 (j)),$$
for some spinor bundle $\Sigma _1$, $0<i<n-1$ and all integers $j$.
Hence we get
$$H^1(Q_n, (F^s)^*\Sigma (j))\simeq H^{2}(Q_n, (F^s)^*\Sigma_1(j-q))\simeq
\dots \simeq H^{n-1}(Q_n, (F^s)^*\Sigma_{n-2}(j-(n-2)q))$$ for some spinor
bundles $\Sigma _1, \dots , \Sigma _{n-2}$. Now  using the Serre duality we have
$$H^{n-1}(Q_n, (F^s)^*\Sigma_{n-2}(j-(n-2)q))\simeq H^1(Q_n, (F^s)^*\ti \Sigma
((n-1)(q-1)-j-1))^*$$ for some spinor bundle $\ti \Sigma$, which
proves the lemma for $i=1$.

In general, there exist some spinor bundles $\Sigma_1, \ti
\Sigma_1$ and $\ti \Sigma$ such that
$$\begin{array}{rl}
H^i(Q_n, (F^s)^*\Sigma (j))\simeq & H^1(Q_n, (F^s)^*\Sigma _1(j+(i-1)q))
\simeq H^1(Q_n, (F^s)^*\ti\Sigma_1 (2d_{n,s}-(i-1)q-j-1))^*\\
\simeq & H^i(Q_n, (F^s)^*\ti\Sigma (2d_{n,s}-2(i-1)q-j-1))^*.
\end{array}$$
\end{proof}

From the proof of the lemma it is clear that we can easily determine
dependence of $\ti \Sigma$ on $\Sigma$ but we need to consider some
cases depending on $n \pmod{4}$.  More precisely, $\ti \Sigma =\Sigma$
if $n$ is odd or $n$ is divisible by $4$ and $\ti \Sigma$ is the
opposite spinor bundle otherwise (at least for $i=1$).
\medskip

Similarly as above we have the following duality:

\begin{Lemma}\label{2-spin-duality}
Let $\Sigma_1$ and $\Sigma _2$ be spinor bundles on $Q_n$ (possibly
equal).  Then for all $0<i<n$ there exists some spinor bundles
$\ti \Sigma_1$ and $\ti \Sigma _2$ such that for all integers $j$ we
have the following duality
$$H^i(Q_n, \Sigma_1\otimes (F^s)^*\Sigma_2 (d_{n,s}-(i-1)q+j)) \simeq
H^i(Q_n,\ti \Sigma_1\otimes (F^s)^*\ti\Sigma_2
(d_{n,s}-(i-1)q-j))^*.$$
\end{Lemma}

\begin{proof}
The proof of the lemma is similar to that of Lemma \ref{spin-duality},
and so we just sketch it for $i=1$.  We can easily show the following
isomorphisms
$$H^1(Q_n, \Sigma_1\otimes (F^s)^*\Sigma_2 (j))\simeq \dots \simeq
H^{n-1}(Q_n, \Sigma_1'\otimes (F^s)^*\Sigma_2' (j-(n-2)q))$$ for some
spinor bundles. Finally, using the Serre duality we have
$$H^{n-1}(Q_n, \Sigma_1'\otimes (F^s)^*\Sigma_2' (j-(n-2)q))\simeq
H^1(Q_n, \ti \Sigma_1\otimes (F^s)^*\ti\Sigma_2 ((n-1)(q-1)-j))^*.$$
\end{proof}

Similarly as above one can easily find $\ti \Sigma_1$ and $\ti
\Sigma_2$ corresponding to $\Sigma_1$ and $\Sigma _2$.

\subsection{D-quasi-affinity of quadrics}

Let $L$ be a line bundle on a smooth projective variety $X$ and let
$\Diff _X(L)$ be the $\O_X$-bimodule of differential operators from
$L$ to $L$ (see \cite[1.1]{Ha}).  $X$ is called
\emph{D(L)-quasi-affine} if any $\O_X$-quasi-coherent $\Diff
_X(L)$-module is $\Diff_X(L)$-generated by its global sections.  We
say that $X$ is \emph{D(L)-affine} if it is D(L)-quasi-affine and for
any $\O_X$-quasi-coherent $\Diff _X(L)$-module $\M$ we have $H^i
(X,\M)=0$ for $i>0$.

\begin{Proposition} \label{quasi-aff}
Let $j$ be a non-negative integer. Then any smooth projective quadric
$Q_n$ is $D(\O_{Q_n}(j))$-quasi-affine.
\end{Proposition}

\begin{proof}
The proof is analogous to the proof of \cite[Satz 3.1]{Ha}.  By
\cite[Proposition 2.3.3]{Ha} it is sufficient to show that for any
integer $t$ the module $\Diff (\O_{Q_n}(j))\otimes \O_{Q_n}(-t)$ is
$\Diff_X(\O_{Q_n}(j))$-generated by its global sections.  Therefore by
\cite[Proposition 2.3.4]{Ha} it is sufficient to show that for any
positive integer $t$ there exists $s_0$ such that for every $s\ge s_0$
$(F^s_*(\O_{Q_n}(j+t)))^*$ is globally generated as an
$\O_{Q_n}$-module.  Let us take as $s_0$ any integer such that
$p^{s_0}>j+t$. By Theorem \ref{ACM-Q} we can write
$$F^s_*(\O_{Q_n}(j+t))\simeq \bigoplus \O _{Q_n}(a_i)\oplus \bigoplus
\Sigma_i(b_i)$$ for some spinor bundles $\Sigma_i$ and some integers
$a_i$ and $b_i$. Hence we need to show that all $a_i$ and $b_i$ are
non-positive.

Note that $\O _{Q_n}(a_i)\hookrightarrow F^s_*(\O_{Q_n}(j+t))$ gives
rise to a non-zero map $(F^s)^*(\O _{Q_n}(a_i))=\O _{Q_n}(p^s a_i)\to
\O_{Q_n}(j+t)\subset \O_{Q_n}(p^{s_0}-1)$, so $a_i\le 0$.  Similarly,
$\Sigma _i(b_i)\hookrightarrow F^s_*(\O_{Q_n}(j+t))$ gives rise to a
non-zero map $(F^s)^*\Sigma_i(p^s b_i)\to \O_{Q_n}(j+t)\subset
\O_{Q_n}(p^{s_0}-1)$. This gives rise to a section of
$(F^s)^*\Sigma_i^*(p^{s_0}-1-p^sb_i)$. Now it is sufficient to show
that $H^0((F^s)^*\Sigma_i^*(-1))=0$ as then $b_i\le 0$.  But this can
be easily shown by restricting to quadrics of lower dimension and
induction on the dimension $n$ (cf.~the proof of Theorem
\ref{vanishing}).
\end{proof}

\section{Hilbert functions of some algebras}

In this section we study Hilbert functions of some finite dimensional
algebras in positive characteristic. Their geometric meaning will
become clear in Sections 3 and 4 (see the proof of Corollary
\ref{dir-sum-lb} and Theorem \ref{decomposition}).

\begin{Proposition} \label{diff-new}
Let $k$ be a field of characteristic $p>2$. Let  $0\le e<p$  be an
integer. Then for any $d\le (N+1)\cdot \frac{p-1}{2}-e$ the $d$th
grading of the ideal $((x_0^p,\dots,
x_N^p):(\sum_{i=0}^Nx_i^2)^{e})$ of $k[x_0,\dots ,x_N]$ is
contained in $(x_0^p,\dots, x_N^p, \sum_{i=0}^Nx_i^2)_d$.
\end{Proposition}

\begin{proof}
We need to prove that if for some homogeneous polynomial $h\in
k[x_0,\dots ,x_N]_d$ with $d+e\le (N+1)\cdot \frac{p-1}{2}$ we
have
\begin{equation}
(\sum_{i=0}^Nx_i^2)^{e} h=\sum_{i=0}^N g_i x_i^p \label{eq1}
\end{equation}
for some homogeneous polynomials $g_i\in k[x_0,\dots ,x_N]$ then
$$h\in (x_0^p,\dots, x_N^p, \sum_{i=0}^Nx_i^2).$$
We can assume that $h=h_0+h_1x_0+...+h_{p-1}x_0^{p-1}$, where
$h_0, ... , h_{p-1}$ are polynomials in $x_1,..., x_N$. Moreover,
we can also assume that all $h_i$s  are of degree less than $p$ in
each variable. Let us set $y=\sum_{i=1}^Nx_i^2$. To simplify
exposition we divide the proof into two cases: first we deal with
the ``even'' part of $h$ and later we deal with the ``odd'' part
of $h$.

\medskip

Let us set
$$
W_{2j}={e\choose j}h_0+{e\choose j-1}y h_2+\dots + {e\choose 0}y^j
h_{2j}
$$
for $j=0,\dots ,\frac{p-1}{2}$. Note that in $\sum_{i=0}^N g_i
x_i^p$ treated as a polynomial in $x_0$, coefficients at $x_0^i$
for $i<p$ are in the ideal $(x_1^p,..., x_N^p)$. Comparing
coefficients in (\ref{eq1}) at even powers of $x_0$ we get
therefore the  following equalities:
\begin{equation}
\label{eq2} y^{e-j}W_{2j}  \equiv 0\pmod{(x_1^p,\dots ,x_N^p)}
\end{equation}
for $j=0,\dots, \frac{p-1}{2}$ (note that if $j>e$ then all the
terms of  $W_{2j}$ are divisible by $y^{j-e}$ and we still have a
polynomial on the left hand side of the above equality).

\begin{Lemma} \label{h_j}
We have
$$y^jh_{2j}=(-1)^{j}{e+j-1\choose e-1}W_0+ (-1)^{j-1}{e+j-2\choose e-1}W_2+\dots
+ (-1)^{0}{e-1\choose e-1}W_{2j}.
$$
\end{Lemma}

\begin{proof}
Let $a_0, ..., a_{p-1}$ be arbitrary elements in a fixed ring $R$.
Consider the following equality of formal power series in
$R[[x]]$:
$$a_0+a_1x +\dots +a_{p-1}x^{p-1}=(1+x^2)^{-e}\left( (1+x^2)^e(a_0+a_1x +\dots +a_{p-1}x^{p-1})\right).$$
Now use expansion
$$(1+x^2)^{-e}=\sum _{j\ge 0} {-e \choose j}x^{2j}=\sum _{j\ge 0} (-1)^j{e +j-1 \choose j}x^{2j}$$
and multiply this by expansion of the product
$(1+x^2)^e(a_0+a_1x_0 +\dots +a_{p-1}x_0^{p-1})$. Then comparing
appropriate coefficients of the product we compute $a_j$. The
assertion in the lemma is a special case of such equality when
comparing coefficients at even powers of $x$.
\end{proof}

\begin{Lemma} \label{combination}
If $j\ge e-\frac{p+1}{2}$ then $ W_{2j}$  is a linear combination
of  $W_{{2j+2}}$, \dots , $W_{p-1}$ and
$y^{\frac{p-1}{2}-j}h_{p-1-2j}$, \dots , $y^{\frac{p-1}{2}}
h_{p-1}$.
\end{Lemma}

\begin{proof}
Set $m=\frac{p-1}{2}-j$. By Lemma \ref{h_j} we have
$$\sum _{k\le j } (-1)^{k} {e+s-1-k \choose e-1}W_{2k}
=  y^{s} h_{s}-\sum _{k>j } (-1)^{k} {e +s-1-k\choose e-1}W_{2k}$$
for $s=m, ..., m+j$. The determinant of this linear system of
equations is up to sign equal to
$$\left|
\begin{array}{ccc}
{e-1+m\choose {e-1} } & ... & {e-1+m-j\choose {e-1}}\\
...&...& ...\\
{e-1+m+j\choose {e-1}}& ... & {e-1+m \choose {e-1}}\\
\end{array} \right|= \left|
\begin{array}{ccc}
{e-1+m-j\choose {e-1-j} } & ... & {e-1+m-j\choose {e-1}}\\
...&...& ...\\
{e-1+m\choose {e-1-j}}& ... & {e-1+m \choose {e-1}}\\
\end{array} \right|= \left|
\begin{array}{ccc}
{e-1+m-j\choose {e-1-j} } & ... & {e-1+m-j\choose {e-1}}\\
...&...& ...\\
{e-1+m-j\choose {e-1-2j}}& ... & {e-1+m-j \choose {e-1-j}}\\
\end{array} \right|$$
where the first equality follows by inductive subtracting columns,
and the second equality follows by inductive subtracting rows. The
last determinant can be computed and by \cite{HT} it is equal to
$$\prod _{i=0}^{j}
\frac{ (e-j+m-1+i)\,! \, i\,!}{ (m-1+i)\,!   \, (e-i)\,!}.$$ It is
non-zero when $e+m-1\le p-1$, which is equivalent to $j\ge
e-\frac{p+1}{2}$. In this case we can solve the above linear
system of equations obtaining the required assertion.
\end{proof}

\begin{Lemma} \label{bound2}
If  $e-\frac{p-1}{2}\le j \le \frac{p-1}{2}$ then there exists a
polynomial $W'_{2j}$ such that $W_{2j}-y^{p-e+j}W'_{2j}$ lies in
the ideal $(x_1^p,\dots ,x_N^p)$.
\end{Lemma}

\begin{proof}
Assume that the assertion of the lemma is false and choose the
largest $j \le \frac{p-1}{2}$ for which it fails. If $j\ge e$ then
we get a contradiction with (\ref{eq2}) as we know that
$W_{2j}\equiv 0\pmod{(x_1^p,\dots ,x_N^p)}$. Therefore $j<e$. By
assumption $j\ge e-\frac{p-1}{2}$ and for every $m>j$ there exists
a polynomial $W'_{2m}$ such that $W_{2m}\equiv
y^{p-e+m}W'_{2m}\pmod{(x_1^p,\dots ,x_N^p)}$. By Lemma
\ref{combination} there exist constants
$b_0,...,b_{\frac{p-1}{2}}$ such that
\begin{equation} \label{eq3}
W_{2j} = b_{0} y^{\frac{p-1}{2}-j}h_{p-1-2j}+ \dots+ b_{j}
y^{\frac{p-1}{2}} h_{p-1} + b_{j+1} W_{{2j+2}} +\dots +
b_{\frac{p-1}{2}}W_{p-1}.
\end{equation}
If $2j\le e- \frac{p-1}{2}$ then $\frac{p-1}{2}-j\ge p-e+j$ and
for the polynomial
$$W_{2j} '= b_0
y^{\frac{p-1}{2}-j-(p-e+j)}h_{p-1-2j}+ \dots+ b_j
y^{\frac{p-1}{2}-(p-e+j)} h_{p-1}+b_{j+1} yW'_{{2j+2}} +\dots +
b_{\frac{p-1}{2}}y^{\frac{p-1}{2}-j} W'_{p-1}.
$$
we have $W_{2j}\equiv y^{p-e+j}W'_{2j}\pmod{(x_1^p,\dots
,x_N^p)}$.  Therefore $2j> e- \frac{p-1}{2}$. We have
$$ y^{e-j} W_{2j} \equiv
0\pmod{(x_1^p,\dots ,x_N^p)}.
$$
and from equations (\ref{eq2}) we see that $ y^{e-j} W_{2m} \equiv
0\pmod{(x_1^p,\dots ,x_N^p)}$ for $m>j$. Hence by (\ref{eq3}) we
have
$$
y^{e+\frac{p-1}{2}-2j} (b_0 h_{p-1-2j}+ \dots+ b_j y^{j}
h_{p-1})\equiv 0\pmod{(x_1^p,\dots ,x_N^p)}.$$ But we know that
$0\le e+\frac{p-1}{2} -2j<p$ and
$$\left( e+\frac{p-1}{2} -2j\right)+\deg h_{p-1-2j}=d+e-\frac{p-1}{2}\le N\frac{p-1}{2},$$
so we can apply the induction assumption. Therefore there exists a
polynomial $P$ such that
$$b_0h_{p-1-2j}+ \dots+ b_j
y^{j} h_{p-1}\equiv y^{p-e+j-(\frac{p-1}{2} -j)} P
\pmod{(x_1^p,\dots ,x_N^p)} .
$$
If we multiply this equality by $y^{\frac{p-1}{2}-j}$ and use
equation (\ref{eq3}) and we see that
$$W_{2j}\equiv y^{p-e+j}\left(P+b_{j+1} yW'_{2j+2}+\dots + b_{\frac{p-1}{2}}y^{\frac{p-1}{2}-j}W_{p-1}'\right)
\pmod{(x_1^p,\dots ,x_N^p)}.
$$
This gives a contradiction with our choice of $j$.
\end{proof}

\begin{Lemma}
$$\sum _{j=0}^{\frac{p-1}{2}} (-1)^jy^jh_{2j}={{e+\frac{p-1}{2}}\choose e}W_0+
{{e+\frac{p-1}{2} -1}\choose e}W_2+\dots + {e\choose e}W_{p-1}.$$
\end{Lemma}

\begin{proof}
Use  Lemma \ref{h_j} and equality
$${n\choose n}+{n+1\choose n}+\dots +{n+m\choose n}={n+m+1\choose n+1}.$$
\end{proof}

Since ${{e+\frac{p-1}{2} -j}\choose e}$ vanishes if
$j<e-\frac{p-1}{2}$ we have
$$
h_0+h_2x_0^2+...+h_{p-1}x_0^{p-1}\equiv \sum
_{j=0}^{\frac{p-1}{2}} (-1)^jy^jh_{2j} = \sum _{j \ge
e-\frac{p-1}{2}} {{e+\frac{p-1}{2} -j}\choose e}W_{2j}
\pmod{(x_1^p,\dots x_N^p, \sum_{i=0}^N x_i^2)}.
$$
Now by Lemma \ref{bound2} for any $j \ge e-\frac{p-1}{2}$ there
exists $W'_{2j}$ such that $W_{2j}\equiv
y^{p-e+j}W'_{2j}\pmod{(x_1^p,\dots ,x_N^p)}$. But $p-e+j\ge
\frac{p+1}{2}$ and
$$y^{\frac{p+1}{2}}\equiv \pm \, x_0^{p+1}\equiv 0 \pmod{(x_0^p,\dots x_N^p, \sum_{i=0}^N x_i^2)}.$$
Therefore $ h^{\rm even}=h_0+h_2x_0^2+...+h_{p-1}x_0^{p-1} $
belongs to the ideal $(x_0^p,\dots x_N^p, \sum_{i=0}^N x_i^2)$.

\bigskip
In a similar way we deal with the remaining part of $h$. First,
let us set
$$
W_{2j+1}={e\choose j}h_1+{e\choose j-1}y h_3+\dots + {e\choose
0}y^j h_{2j+1}
$$
for $j=0,\dots ,\frac{p-3}{2}$. Comparing coefficients in
(\ref{eq1}) at odd powers of $x_0$ we get
\begin{equation}
\label{eq2'} y^{e-j}W_{2j+1}  \equiv 0\pmod{(x_1^p,\dots ,x_N^p)}.
\end{equation}
for $j=0,\dots ,\frac{p-3}{2}$.

\begin{Lemma} \label{h_j-odd}
We have
$$y^jh_{2j+1}=(-1)^{j}{e+j-1\choose e-1}W_1+ (-1)^{j-1}{e+j-2\choose e-1}W_3+\dots
+ (-1)^{0}{e-1\choose e-1}W_{2j+1}.
$$
\end{Lemma}

\begin{Lemma} \label{combination-odd}
If $j\ge e-\frac{p+1}{2}$ then $ W_{2j+1}$  is a linear
combination of  $W_{{2j+3}}$, \dots , $W_{p-2}$ and
$y^{\frac{p-3}{2}-j}h_{p-2-2j}$, \dots , $y^{\frac{p-3}{2}}
h_{p-2}$.
\end{Lemma}

\begin{Lemma} \label{bound2-odd}
If  $e-\frac{p+1}{2}\le j \le \frac{p-3}{2}$ then there exists a
polynomial $W'_{2j+1}$ such that $W_{2j+1}-y^{p-e+j}W'_{2j+1}$
lies in the ideal $(x_1^p,\dots ,x_N^p)$.
\end{Lemma}

\begin{Lemma}
$$\sum _{j=0}^{\frac{p-3}{2}} (-1)^jy^jh_{2j+1}={{e+\frac{p-3}{2}}\choose e}W_1+
{{e+\frac{p-3}{2} -1}\choose e}W_3+\dots + {e\choose e}W_{p-2}.$$
\end{Lemma}

Since ${{e+\frac{p-3}{2} -j}\choose e}$ vanishes if
$j<e-\frac{p+3}{2}$, we have
$$
h_1+h_3x_0^2+...+h_{p-2}x_0^{p-3}\equiv \sum
_{j=0}^{\frac{p-3}{2}} (-1)^jy^{j}h_{2j+1} = \sum _{j \ge
e-\frac{p+1}{2}} {{e+\frac{p-3}{2} -j}\choose e}W_{2j+1}
\pmod{(x_1^p,\dots x_N^p, \sum_{i=0}^N x_i^2)}.
$$
Now by Lemma \ref{bound2-odd} for any $j \ge e-\frac{p+1}{2}$
there exists $W'_{2j+1}$ such that $W_{2j+1}\equiv
y^{p-e+j}W'_{2j+1}\pmod{(x_1^p,\dots ,x_N^p)}$. But $p-e+j\ge
\frac{p-1}{2}$ and
$$y^{\frac{p-1}{2}}\equiv \pm \, x_0^{p-1} \pmod{(x_0^p,\dots x_N^p, \sum_{i=0}^N x_i^2)}.$$
Therefore $ h^{\rm odd}=h_1x_0+h_3x_0^3+...+h_{p-2}x_0^{p-2} $
belongs to the ideal $(x_0^p,\dots x_N^p, \sum_{i=0}^N x_i^2)$.

Since $h=h^{\rm even}+h^{\rm odd}$, this finishes proof of
the proposition.
\end{proof}

\medskip

The following corollary of Proposition \ref{diff-new} is the main
step in our proof of Theorem \ref{vanishing}:

\begin{Proposition} \label{diff}
Let $k$ be a field of characteristic $p>2$. Let  $0\le e<p$  be an
integer. Then for any $d\le (N+1)\cdot \frac{p-1}{2}-e$ we have
$$((x_0^p,\dots, x_N^p):(\sum_{i=0}^Nx_i^2)^{e})_d=(x_0^p,\dots,
x_N^p,(\sum_{i=0}^Nx_i^2)^{p-e})_d.
$$
in $k[x_0,\dots ,x_N]$.
\end{Proposition}

\begin{proof}
We need to prove that for every homogeneous degree $d$ (with
$d+e\le (N+1)\cdot \frac{p-1}{2}$) polynomial $h$ such that
$$(\sum_{i=0}^Nx_i^2)^{e} h\equiv 0 \pmod{(x_0^p,\dots x_N^p)}$$
we have $h\in(x_0^p,\dots, x_N^p,(\sum_{i=0}^Nx_i^2)^{p-e})_d. $

If $e=0$ then the assertion is trivial so we can assume that $e\ge
1$.

The proof is by induction on $N$. For $N=0$ we have $2(d+e)\le
p-1$. But if $h\ne 0$ then counting degrees we get $d+2e\ge p$, a
contradiction. Now assume that the assertion holds for polynomials
in $N$ variables (i.e., for $N-1$).

If $e=p-1$ then the required assertion follows from Proposition
\ref{diff-new}. Otherwise, by Proposition \ref{diff-new} there
exists a polynomial $h'$ such that
$$h\equiv (\sum_{i=0}^Nx_i^2)h'\pmod{(x_0^p,\dots x_N^p)}.$$
Therefore
$$(\sum_{i=0}^Nx_i^2)^{e+1} h'\equiv (\sum_{i=0}^Nx_i^2)^{e} h\equiv 0
\pmod{(x_0^p,\dots x_N^p)}$$ and now we can again apply
Proposition \ref{diff-new}, since $e+1+\deg h'=e+\deg h\le
(N+1)\cdot \frac{p-1}{2}$. Continuing in this way till new $e$
becomes $p-1$, we get the required assertion.
\end{proof}

\medskip

Let us set $S=k[x_0, \dots , x_{N}]$, where $\char k=p$ and $N=n+1$.
Let $q=p^s$ for some non-negative integer $s$. Set
$I=(x_0^2+\dots+x_{N}^2,x_1^q,\dots ,x_{N}^q)$ and $A=S/I$. Let us
recall that
$$
\left(\frac{1-t^q}{1-t}\right) ^N=\sum _{i=0}^{\infty}\alpha_{i,N} t^i,$$
where $$\alpha_{i,N}=\sum _{j=0}^N(-1)^j
{{N}\choose {j}} {{i-jq+N-1}\choose {N-1}} ,$$
where we set ${a\choose b}=0$ if $a<b$ or $b<0$.
The Hilbert series of $A$ can be computed as
$$h_A(t)=\sum \dim A_i\, t^i=h_S(t) (1-t^2)(1-t^q)^N=(1+t)\left(
\frac{1-t^q}{1-t}\right) ^N.$$
Therefore
$$\dim A_i=\alpha_{i,N}+\alpha_{i-1,N}.$$ Note that $\dim_k A=2q^N$
(e.g., by the Bezout's theorem). The following lemma will be used in
the proof of Lemma \ref{sum-B}.

\begin{Lemma} \label{p^n}
For every integer $i$ we have
$$\sum _{j\in \ZZ}\dim A_{i+jq}=2q^n.$$
\end{Lemma}

\begin{proof} By definition
$$(1+t+\dots +t^{q-1})\sum _{i=0}^{\infty}\alpha_{i,n}t^i=\sum _{i=0}^{\infty}\alpha _{i,N}t^i.$$
This gives $\alpha_{i,N}=\alpha_{i,n}+\dots +\alpha_{i-q+1,n}$.
Hence
$$\sum _{j\in \ZZ}\alpha_{i+jq,N}=\sum _{j\in \ZZ}\alpha _{j,n}=q^n.$$
Now the lemma follows from equality $\dim A_l=\alpha_{l,N}+\alpha_{l-1,N}.$
\end{proof}

\medskip

Let us set $d_{N,s}= \frac{1}{2}n(q-1)$ and
$$\gamma_N(i)=\frac{1}{2^N} \sum_{j\in \ZZ}(-1)^j\dim A_{d_{N,s}+i+jq}$$
for $i\in \ZZ$.  In principle, we could write the formulas for
$\gamma_N(i)$ using formulas for $\dim A_j$, but the obtained formulas
are rather useless and we need different formulas. Let us first define
inductively some sequences of numbers and functions.  Set $w_{0}=1$
and assume we have defined integers $w_0, ..., w_k$. Then we set
$$F_{k}(i)=\sum _{j=0}^{k}(-1)^j w_{k-j} {{i+j} \choose {2j+1}}$$
and
$$w_{k+1}=\sum _{j=0}^{k}(-1)^j w_{k-j} {{\frac{q+1}{2}+j} \choose
{2j+2}}.$$ Similarly, set $u_{0}=0$ and assume we have defined $u_0,
...,u_k$.  Then we set
$$G_{k}(i)=(-1)^k {{i+k} \choose {2k}}+ \sum _{j=0}^{k}(-1)^j u_{k-j}
{{i+j} \choose {2j+1}}$$
and
$$u_{k+1}=(-1)^k {{\frac{q+1}{2}+k}\choose {2k+1}}+ \sum
_{j=0}^{k}(-1)^j u_{k-j} {{\frac{q+1}{2}+j} \choose {2j+2}}.$$

\begin{Lemma} \label{Bl-Cl}
We have $\gamma_N(0)=0$, $\gamma_N(i)>0$ for $i=1,...,q-1$, and
$\gamma _N(q-i)=\gamma _N(i)$. Moreover, for $i=1,..., \frac{q-1}{2}$ we have
$$\gamma_{N}(i)=\left\{
\begin{array}{lc}
F_k(i)&\hbox{if $N=2k+2$,}\\
G_k(i)&\hbox{if $N=2k+1$.} \\
\end{array}
\right.$$
\end{Lemma}

\begin{proof}
Using the same method as in the proof of Lemma
\ref{p^n} we get the following recurssion:
$$2\gamma_N (i)=\sum _{|j|\le \frac{q-1}{2}}\gamma _{n}(i+j).$$
Then one can see that
$$\gamma _1(i)=\left\{ \begin{array}{cl}
0&\hbox{if $i=0$,}\\
1&\hbox{if $i=1,\dots , q-1$,}\\
\end{array}\right.$$
$\gamma _N(0)=0$, $\gamma _N(q-i)=\gamma _N(i)$ for $i=1,..., q-1$ and
$$\gamma _N(i)=\sum _{j=1}^i\gamma _n\left( \frac{q+1}{2} -j\right)$$
for $i=1,..., \frac{q-1}{2}.$ Using this recurrsion and the formula
$$\sum _{j=k}^n{j\choose k}={{n+1}\choose {k+1}},$$ one can easily
prove the lemma by induction.
\end{proof}

\medskip

Set $B=A/(x_0^q)$ and $C=A/(I:x_0^q)$. Note that $A$ and $C$ are
graded $0$-dimensional Gorenstein rings (in fact, $A$ is a complete
intersection ring).
From the definition of $A$ one can easily see that $\dim A_i= 0$ if
and only if $i<0$ or $i>N(q-1)+1$. As $C_i\subset A_{i+q}$, this implies that $C_i=0$ if $i<0$
or $i>N(q-1)+1-q=n(q-1).$

From now on we consider the case $s=1$, i.e., $q=p$.

\begin{Proposition} \label{C=B}
If $p>2$ then $C_d=B_{d}$ for $d\le d_N=\frac{1}{2}n(p-1).$
\end{Proposition}

\begin{proof}
Since $x_0^{2p}=(-\sum_{i\ge 1}x_i^2)^p$ in $S$, we see that
$I+(x_0^p)\subset I+(I:x_0^p)$. Our assertion is equivalent to the fact
that this inclusion is  equality in gradings $d \le d_N$.
Therefore it is sufficient to show that if $g_0\in (I:x_0^p)_d$, $d\le d_N$
then $g_0 \in I+x_0^p$.

Assume that $g_0\in (I:x_0^p)_d$ for some $d\le d_N$. Then there exist
some homogeneous polynomials $g_i$, $i=1,\dots, N$ and $h$, such that
$$g_0x_0^p=\sum _{i=1}^Ng_ix_i^p+(\sum _{i=0}^N x_i^2) h.$$
Then by Proposition \ref{diff} we can write $h$ as
$$h=\sum _{i=0}^N x_i^p h_i+(\sum _{i=0}^N x_i^2)^{p-1}h'$$
for some homogeneous polynomials $h_i$, $i=0,...,N$ and $h'$.
Then
$$g_0x_0^p=\sum _{i=1}^N(g_i+(\sum _{i=0}^N x_i^2)h_i)x_i^p+(\sum
_{i=0}^N x_i^2)h_0x_0^p + (\sum _{i=0}^N x_i^{2p})h'$$ and hence
$$(g_0-x_0^ph_0-(\sum _{i=0}^N x_i^2)h')x_0^p=\sum _{i=1}^N(g_i+(\sum
_{i=0}^N x_i^2)h_i+x_i^ph')x_i^p.$$ Now comparing coefficients of both
sides treated as polynomials in $x_0$ we see that $g_0-x_0^ph_0-(\sum
_{i=0}^N x_i^2)h'\in (x_1^p, \dots , x_N^p)$, which finishes the
proof.
\end{proof}

\medskip

The above proposition allows us to compute the Hilbert functions of
$B$ and $C$:
\begin{Lemma} \label{Hilb-B}
$$\dim {B_{i}}=\left\{
\begin{array}{lc}
\sum_{j\ge 0}(-1)^j\dim A_{i-jp} &\hbox{if $i\le d_N+p$,}\\
\dim B_{2d_N-i}&\hbox{if $i\ge d_N+p$.}\\
 \end{array}\right.$$
In particular, $B_i=C_i$ if $i\le d_N$ or $i\ge d_N+p$. We also have
$\dim C_{2d_N-i}=\dim C_i$ for all integers $i$.  Moreover, $B_i\ne 0$
($C_i\ne 0$) if and only if $0\le i\le 2d_N=n(p-1)$.
\end{Lemma}

\begin{proof}
Since we have short exact sequences
$$0\to C_{i}\mathop{\to}^{x_0^p} A_{i+p}\to B_{i+p}\to 0,$$
Proposition \ref{C=B} implies that
$$\dim {B_{i+p}}-\dim {A_{i+p}}= -\dim C_i= -\dim B_i $$ for $i\le
d_N$. This allows us to compute $\dim B_i$ for $i\le d_N+p$ and $\dim
C_i$ for $i\le d_N$.  An easy calculation implies that $\dim C_i$ is
increasing for $i\le d_N$. Let us recall that $C$ is Gorenstein and
hence the Hilbert function of $C$ is symmetric.  Hence the above
remark together with $C_{2d_N+1}=0$ imply that $\dim C_i=\dim
C_{2d_N-i}$.

Assume that $i\ge d_N$. Then using once more the above short exact
sequence we get
$$\dim {B_{i+p}}-\dim {A_{i+p}}= -\dim C_i= -\dim B_{2d_N-i}.$$
Finally, we can use the duality of the graded Gorenstein ring $A$ to
get
$$\dim B_{i+p}= \dim A_{i+p}-\dim B_{2d_N-i}=\dim A_{2d_N-i}-\dim
B_{2d_N-i}=\dim B_{2d_N-i-p}.$$
The remaining part of the lemma follows from Proposition \ref{C=B} and
equality of dimensions $\dim B_i=\dim C_i$ for $i\ge d_N+p$.
\end{proof}

\medskip

\begin{Lemma}\label{sum-B}
Let $l$ be an integer. Then
$$\sum _{i\in \ZZ} \dim B_{l+ip}=p^n+2^n\gamma_{N}(l_0),$$
where $l_0$ is the unique integer such that
$0\le l_0<p$ and $l\equiv d_N+l_0 \pmod{p}$,
and  $\gamma _N (\cdot )$ is as in Lemma  \ref{Bl-Cl}.
\end{Lemma}

\begin{proof}  Let us set $l_1=d_N+l_0$.
By Lemma \ref{Hilb-B}
we have
\begin{eqnarray*}
&\sum _{i\in \ZZ } \dim A_{l+i p}= \sum _{i\in \ZZ} (\dim B_{l+ip}+
\dim C_{l+ip-p})= 2 \sum _{i\in  \ZZ} \dim B_{l+ip}
-(\dim B_{l_1}-\dim C_{l_1}).\\
\end{eqnarray*}
But
\begin{eqnarray*}
&\dim B_{l_1}-\dim C_{l_1}=\sum_{j\ge 0}(-1)^j\dim
A_{l_1-jp}+\sum_{j\ge 0}(-1)^{j+1}\dim A_{2d_N-l_1-jp}\\ &=\sum_{j\ge
0}(-1)^j\dim A_{l_1-jp}+\sum_{j\le -1}(-1)^{j}\dim A_{l_1-jp}=
\sum_{j\in \ZZ}(-1)^j\dim A_{l_1-jp}.\\
\end{eqnarray*}
Therefore by Lemma \ref{p^n}
$$\sum _{i\in \ZZ} \dim B_{l+ip}=p^n+\frac{1}{2}\sum_{j\in \ZZ}(-1)^j\dim A_{l_1-jp},$$
which together with Lemma \ref{Bl-Cl} proves the required equality.
\end{proof}

\section{Vanishing and non-vanishing theorems}

In this section we prove some basic vanishing and non-vanishing
theorems for cohomology of twisted Frobenius pull-backs of spinor
bundles.

Let us set $\psi_1= \Omega_{\PP^N}(1)|_{Q_n}$.

\begin{Proposition} \label{psixspin}
For any spinor bundle $\Sigma$ on $Q_n$ we have
$$h^1(Q_n, \psi_1\otimes \Sigma (t))=\left\{
\begin{array}{cl}
0\quad&\hbox{if $t\ne 0$,}\\
2\rk \Sigma=2^{{[\frac{N}{2}]}}\quad&\hbox{if $t= 0$.}\\
\end{array}
\right.$$
\end{Proposition}

\begin{proof}
Every spinor bundle $\Sigma$ is ACM and so it fits into the following
short exact sequence of sheaves on $\PP^N$
\begin{equation}
0\to \O_{\PP^N}^{2^{[N/2]}}(-1)\to \O_{\PP^N}^{2^{[N/2]}}\to
i_*\Sigma(1) \to 0,\label{(*)}
\end{equation}
where $i: Q_n\hookrightarrow \PP^N$ is the embedding (see (\ref{(i')}), (\ref{(ii')}) and (\ref{(iii')})).
Tensoring this sequence with $\Omega^1_{\PP^N}(t)$ and using standard Bott formulas
for cohomology of twists of $\Omega^1_{\PP^N}$ on $\PP^N$ we get the
result.
\end{proof}

\medskip

\begin{Corollary}
If $E$ is arithmetically Cohen-Macaulay on $Q_n$ then it is a direct
sum of line bundles if and only if
$$\sum _{t\in \ZZ} h^1(E\otimes \psi_1(t))=\rk E.$$
\end{Corollary}

\begin{proof}
By Theorem \ref{ACM-Q} any ACM bundle on $Q_n$ is isomorphic to a
direct sum of line bundles $\O_{Q_n}(i)$ and spinor bundles twisted by
some $\O _{Q_n}(i)$. By Proposition \ref{psixspin} we see that
$$\sum _{t\in \ZZ} h^1(\Sigma (i) \otimes \psi_1(t))=2 \rk \Sigma.$$
On the other hand
$$h^1(Q_n, \psi_1 (t))=\left\{
\begin{array}{cl}
0\quad&\hbox{if $t\ne -1$,}\\
1\quad&\hbox{if $t=-1$,}\\
\end{array}
\right. $$ so if $\sum _{t\in \ZZ} h^1(E\otimes \psi_1(t))=\rk E$
then $E$ cannot contain any twists of spinor bundles as direct summands.
\end{proof}

\medskip

Let us recall that $d_N=(N-1)\frac{p-1}{2}=n\frac{p-1}{2}$.

\begin{Corollary} \label{dir-sum-lb}
$F_*(\O_{Q_n}(t))$ is a direct sum of line bundles if and only if
$t-d_{N}$ is divisible by $p$.
\end{Corollary}

\begin{proof}
By \cite[Theorem 4]{BM} the graded $S$-module $\bigoplus _{t\in \ZZ}
H^1(Q_n, F^*\psi_1(t-p))$ is isomorphic as a graded $S$-module to
$B$. But $H^1(Q_n, F^*\psi_1(t+ip))=H^1(Q_n, F_*(\O_{Q_n}(t))\otimes
\psi_1(i)),$ so by the above corollary the fact that
$F_*(\O_{Q_n}(t))$ is a direct sum of line bundles is equivalent to
equality
$$\sum _{i\in \ZZ} \dim B_{t+ip}=p^n.$$
Now the required assertion follows from Lemma \ref{sum-B}.
\end{proof}

The following vanishing theorem allows to compute the decomposition of
Frobenius push-forwards of line bundles:

\begin{Theorem} \label{vanishing}
Let $\Sigma$ be a spinor bundle on $Q_n$. Then for $0<i<n$ we have
$H^i(Q_n, F^*\Sigma (t)) =0$ if $t\le d_N-ip$ or $t\ge d_N-(i-1)p$.
\end{Theorem}

\begin{proof}
By Lemma \ref{spin-duality} it is sufficient to prove vanishing of
$H^i(Q_n, F^*\Sigma (t))$ for $t\ge d_N-(i-1)p$. By Corollary
\ref{dir-sum-lb} and the projection formula we have
$$H^i(Q_n, F^*\Sigma (d_N+tp))=H^i(Q_n, F_*(\O_{Q_n}(d_N))\otimes \Sigma
(t))=0$$ for any integer $t$. In particular, $H^i(Q_n, F^*\Sigma
(d_N-(i-1)p))=0$.  Now the proof is by induction on $n$.

For $n=2$, $\Sigma=\O_{\PP^1 \times \PP^1}(-1,0)$ or
$\Sigma=\O_{\PP^1\times \PP^1}(0,-1)$ and in both cases it is easy
to check the required assertion. Assume that the theorem holds for
quadrics of dimension less than $n$.
Let us recall that the restriction of the spin representation of
$\Spin (2m+1)$ to $\Spin (2m)$ is the sum of the two spin
representations of $\Spin (2m)$.  Similarly, the restriction of either
spin representation of $\Spin (2m)$ to $\Spin (2m-1)$ is the spin
representation. Therefore the restriction of a spinor bundle to a
hypersurface quadric is either a spinor bundle or a direct sum of two
spinor bundles.  Using the long cohomology sequence for the short
exact sequence
$$0\to F^*\Sigma (t)\to F^*\Sigma (t+1)\to F^*\Sigma (t+1)|_{Q_{n-1}}\to 0$$
and the induction assumption we see that for $t\ge d_n-(i-1)p-1$ we have
a surjection
$$H^i(Q_n, F^*\Sigma (t))\to H^i(Q_n, F^*\Sigma (t+1))\to 0.$$
In particular, vanishing of $H^i(Q_n, F^*\Sigma (d_N-(i-1)p))$ implies vanishing
of $H^i(Q_n, F^*\Sigma (t))$ for $t\ge d_N-(i-1)p$.
\end{proof}

The above vanishing theorem implies that the Frobenius pull-back of
the spinor bundle on $Q_3$ is very similar to an instanton
bundle. More precisely, we have the following proposition:

\begin{Proposition}
Let us set $E=F^*\Sigma(\frac{p-1}{2})$ on $Q_3$. Then $E$ is the
cohomology of the monad
$$ 0\to \O (-1)^b\to \Sigma ^{b+1}\to \O ^b\to 0,$$
where $b=h^1(E)=-\chi (E).$
\end{Proposition}

\begin{proof}
The proof is an application of Horrock's killing technique and it is
quite similar to the proof of \cite[Proposition 1.1]{OS}.
We leave the details to the reader.
\end{proof}

\begin{Corollary} \label{non-vanishing}
Let $\Sigma$ be a spinor bundle on $Q_n$. Then for $0<i<n$ we have
$H^i(Q_n, F^*\Sigma (t)) \ne 0$ if $d_N-ip <t< d_N-(i-1)p$.
\end{Corollary}

\begin{proof}
If $H^i(Q_n, F^*\Sigma (t)) = 0$ for some $d_N-ip <t< d_N-(i-1)p$, then
by  Theorem \ref{vanishing}
$$H^i(Q_n, F_*(\O_{Q_n} (t))\otimes \Sigma (j))=H^i(Q_n, F^*\Sigma
(t+jp)) = 0$$ for all integers $j$.

Let us note that if $n$ is even then there exists an automorphism $a:
Q_n\to Q_n$ such that $a^*\Sigma_{\pm}\simeq \Sigma_{\mp}$. Therefore
cohomology groups of $F^*\Sigma_{+}(t+jp)$ and $F^*\Sigma_{-}(t+jp)$ are the
same.  In particular, the above vanishing holds for all spinor bundles
on $Q_n$ and we can apply Corollary \ref{dir-sum1}.  But then we get
contradiction with Corollary \ref{dir-sum-lb}.
\end{proof}

\begin{Theorem} \label{2-vanishing}
Let $\Sigma_1, \Sigma_2$ be spinor bundles on $Q_{n}$, $n\ge 2$.  Then
for any $0<i<n$ we have $H^i(Q_n, \Sigma _1\otimes F^*\Sigma _2(t))=0$ if
$t\le d_N-ip$ or $ t\ge d_N-(i-1)p+1$.
\end{Theorem}

\begin{proof}
For simplicity of notation let us consider only odd dimensional
quadrics $Q_n$, $n=2m+1$ (the proof in the even dimensional case is
essentially the same). As before we can easily reduce to the case
$i=1$ (the proof of Lemma \ref{2-spin-duality} gives vanishing of
higher intermediate cohomology groups). Let us note the following
short exact sequences:
\begin{equation}\label{seq1}
0\to F^*\Sigma (t-p)\to \O _{Q_n} (t-p)^{2^{m+1}}\to F^*\Sigma (t)\to 0
\end{equation}
and
\begin{equation}\label{seq2}
0\to \Sigma (t-p)\to \O _{Q_n} (t-p)^{2^{m+1}}\to \Sigma (t-p+1)\to 0.
\end{equation}
 Using the long cohomology sequnces for appropriate twists we get
the following exact sequences:
$$0\to H^0(\Sigma \otimes F^*\Sigma (t-p))\to H^0(\Sigma (t-p))^{\oplus 2^{m+1}}\to
H^0(\Sigma \otimes F^*\Sigma (t))\to H^1(\Sigma \otimes F^*\Sigma (t-p)) \to 0
$$
for all $t$, and by Theorem \ref{vanishing}
\begin{equation}\label{seq3}
0\to H^0(\Sigma \otimes F^*\Sigma (t-p))\to H^0(F^*\Sigma (t-p))^{\oplus 2^{m+1}} \to
H^0(\Sigma \otimes F^*\Sigma (t-p+1))\to H^1(\Sigma \otimes F^*\Sigma (t-p)) \to 0
\end{equation}
for $t\le d_N$.  Using these sequences we get the following
recurrence equation
\begin{equation}\label{recur}
h^0(\Sigma \otimes F^*\Sigma (t))=2^{m+1}(h^0(\Sigma
(t-p))-h^0(F^*\Sigma (t-p))) +h^0(\Sigma \otimes F^*\Sigma (t-p+1))
\end{equation}
for $t\le d_N$. Let us first prove that
\begin{equation}\label{seq4}
h^0(\Sigma \otimes F^*\Sigma (t))+h^0(\Sigma \otimes F^*\Sigma
(t+1))=2^{m+1}h^0(F^*\Sigma (t)) \label{h+h=h}
\end{equation}
for $t\le d_N-1$.  We prove it by induction on $t$ starting with very
negative $t$ for which the equality is obvious. By (\ref{recur}) and
the induction assumption for $t\le d_N-1$ the left hand side of (\ref{h+h=h}) is equal
to
$$
\begin{array}{rl}
&2^{m+1}(h^0(\Sigma (t-p))-h^0(F^*\Sigma
(t-p))+h^0(\Sigma(t-p+1))-h^0(F^*\Sigma (t-p+1)))\\ &+h^0(\Sigma
\otimes F^*\Sigma (t-p+1))+h^0(\Sigma \otimes F^*\Sigma (t-p+2))=\\
=&2^{m+1}(h^0(\Sigma (t-p))-h^0(F^*\Sigma (t-p))+h^0(\Sigma
(t-p+1))).\\
\end{array}
$$
But sequences (\ref{seq1}) and (\ref{seq2}) imply that for $t\le d_N$
$$  \begin{array}{rl}
h^0(F^*\Sigma (t-p))+h^0(F^*\Sigma (t))&=2^{m+1} h^0(\O_{Q_n}(t-p))=
h^0(\Sigma (t-p))+h^0(\Sigma (t-p+1)),\\
\end{array}$$
which proves the required assertion.

Now let us note that (\ref{seq3}) and (\ref{seq4}) imply vanishing of
$H^1(\Sigma \otimes F^*\Sigma (t))$ for $t\le d_N-p$.  By Lemma
\ref{2-spin-duality} this implies vanishing $H^1(\Sigma \otimes
F^*\Sigma (t))$ for $t\ge d_N+1$.
\end{proof}

\begin{Corollary} \label{2-non-vanishing}
Let us set $S_n=\Sigma$ if $n$ is odd and $S_n=\Sigma_{+}\oplus \Sigma_{-}$ if
$n$ is even.  Then for any $0<i<n$ we have $H^i(Q_n, S_n \otimes F^*S_n
(t))\ne 0$ if $d_N-ip+1\le t\le d_N-(i-1)p$.
\end{Corollary}

\begin{proof}
As before it is sufficient to prove the corollary for $i=1$.  The
proof is by induction on $n$. For $n=2$ the group $H^1(S_2 \otimes
F^*S_2(t))$ contains $H^1(\O_{{\PP^1}\times {\PP^1}}(t, t-p-1))$ as a
direct summand and this cohomology group is non-zero if $0\le t\le
p-1$.

Assume we know the statement in dimensions less than $n$.  Then we
have an exact sequence
$$H^1(S_n\otimes F^*S_n(t-1))\to H^1(S_n\otimes F^*S_n(t))\to
H^1(S_n\otimes F^*S_n(t)|_{Q_{n-1}})\to H^2(S_n\otimes F^*S_n(t-1)).$$
This sequence and Theorem \ref{2-vanishing} imply that
$h^1(S_n\otimes F^*S_n(t-1)) \le h^1(S_n\otimes F^*S_n(t))$
if $t\ge d_{n}+1$, so by Lemma \ref{2-spin-duality} it is sufficient to prove
that $H^1(S_n\otimes F^*S_n(d_N))\ne 0$.
But Lemma \ref{2-spin-duality} implies also that
$$H^2(S_n\otimes F^*S_n(d_N-p))\simeq H^1(S_n\otimes F^*S_n(d_N))
\simeq H^1(S_n\otimes F^*S_n(d_N-p+1))^*,$$ so the above sequence
applied for $t=d_N-p+1$ gives by the induction assumption
non-vanishing of $H^1(S_n\otimes F^*S_n(d_N))$.
\end{proof}

\section{Decomposition of Frobenius push-forwards}

If $E$ is an ACM bundle then $F_*E$ is also an ACM bundle. In
particular, if $n\ge 3$ then Frobenius push-forwards of line bundles
and twisted spinor bundles split into a direct sum of line bundles and
twisted spinor bundles. In this section we study the corresponding
decompositions.

Let $S_n$ be as in Corollary \ref{2-non-vanishing}. Let $q=p^s$ for
some non-negative integer $s$.

\begin{Proposition} \label{splitting}
\begin{enumerate}
\item If $F^s_*(\O_{Q_n}(j))$ contains $\O_{Q_n}(-t)$ as a direct summand then
$0\le tq+j\le n(q-1)$.
\item If $F^s_*(S_n(j))$ contains $\O_{Q_n} (-t)$ as a direct summand
then  $1\le tq+j\le n(q-1)$.
\end{enumerate}
\end{Proposition}

\begin{proof}

(1) If $F^s_*(\O_{Q_n}(j))$ contains $\O_{Q_n}(-t)$ as a direct summand then
$$0\ne H^0(\O_{Q_n})\subset H^0(F^s_*(\O_{Q_n}(j))\otimes
\O_{Q_n}(t))=H^0(\O_{Q_n}(j+tq)),$$ which implies that $tq+j\ge
0$. Similarly,
$$0\ne H^n(\omega_{Q_n})\subset H^n(F^s_*(\O_{Q_n}(j))\otimes
\O_{Q_n}(t-n))=H^n(\O_{Q_n}(j+(t-n)q))$$ which implies that $tq+j\le
n(q-1)$.

(2) If $F^s_*(S_n(j))$ contains $\O_{Q_n}(-t)$ as a direct summand then
$$0\ne H^0(\O_{Q_n})\subset H^0(F^s_*(S_n(j))\otimes \O_{Q_n}(t))=H^0(S_n(j+tq)),$$
which implies that $tq+j\ge 1$. Similarly, by the Serre duality
$$0\ne H^n(\omega_{Q_n})\subset H^n(F^s_*(S_n(j))\otimes
\O_{Q_n}(t-n))=H^n(S_n(j+(t-n)q))=(H^0(S_n^*(n(q-1)-tq-j)))^*$$
which implies that $tq+j\le n(q-1)$.
\end{proof}

\begin{Proposition} \label{splitting2}
\begin{enumerate}
\item $F_*(\O_{Q_n}(j))$ contains $\Sigma(-t)$ as a direct summand if and
only if $d_N-p+1\le tp+j\le d_N-1$.  In this case, $F_*(\O_{Q_n}(j))$
contains $S_n(-t)$.  In particular,  $F_*(\O_{Q_n}(j))$ contains
at most one twist of a spinor bundle.
\item $F_*(S_n(j))$ contains $\Sigma (-t)$ as a direct summand if and
only if $d_N-p+1\le tp+j\le d_N$. In this case, $F_*(S_n(j))$
contains also $S_n(-t)$.  In particular,  $F_*(S_n(j))$ contains
exactly one twist of a spinor bundle.
\end{enumerate}
\end{Proposition}

\begin{proof}

(1) If $F_*(\O_{Q_n}(j))$ contains $\Sigma(-t)$ as a direct summand
then by symmetry (see the proof of Corollary \ref{non-vanishing}) it
also contains $S_n(-t)$. By Lemma \ref{tensor-spin} this happens if
and only if $H^1(Q_n, F^*\Sigma (j+tp))=H^1(F_*(\O_{Q_n}(j)) \otimes
\Sigma (t))\ne 0$ so the assertion follows from Theorem
\ref{vanishing} and Corollary \ref{non-vanishing}.

(2) If $F_*(S_n(j))$ contains $\Sigma(-t)$ as a direct summand
then by symmetry it also contains $S_n(-t)$. By Lemma
\ref{tensor-spin} this happens if and only if $H^1(Q_n, S_n\otimes F^*S_n
(tp+j))=H^1(F_*(S_n(j)) \otimes S_n (t))\ne 0$ so the assertion follows from
Theorem \ref{2-vanishing} and Corollary \ref{2-non-vanishing}.
\end{proof}

\begin{Corollary}\label{quasi-exc}
For any line bundle $\L$ on $Q_n$, $n\ge 3$, the bundle $F_*\L$ is
quasi-exceptional.
\end{Corollary}

\begin{proof}
The assertion follows from Lemma \ref{tensor-spin} and Propositions \ref{splitting}
and \ref{splitting2}.
\end{proof}

\medskip

Let us fix an integer $0\le j<p$. By Propositions \ref{splitting} and
\ref{splitting2} we can write $$F_*(\O_{Q_n}(d_N+j))=\bigoplus \O
_{Q_n}(-t)^{a_t}\oplus S_n (1)^b,$$ where $b=0$ if $j=0$ and $a_t=0$
if $|tp+j|>d_N$.

\begin{Theorem} \label{decomposition}
If $|tp+j|\le d_N$ then $0<a_t=\dim C_{d_N+tp+j}$.
Moreover,  $b=2^{[N/2]}\gamma _N(j)$, where $\gamma _N(\cdot )$ is as in
Lemma \ref{Bl-Cl}.
\end{Theorem}

\begin{proof} By Proposition \ref{psixspin} we have
$$h^1(F_*(\O_{Q_n}(d_N+j))\otimes \psi_1(t-1))=\left\{
\begin{array}{cl}
a_{0}+2^{[n/2]+1}b&\quad \hbox{ if $t=0$,}\\
a_t&\quad \hbox{ if $t\ne 0$.}\\
\end{array}
\right.$$
On the other hand, we have
$$h^1(F_*(\O_{Q_n}(d_N+j))\otimes
\psi_1(t-1))=h^1(F^*\psi_1(d_N+(t-1)p+j)) =\dim B_{d_N+tp+j},$$ so
$a_t=\dim B_{d_N+tp+j}=\dim C_{d_N+tp+j}$ for $t\ne 0$.  Comparing
ranks in the decomposition we get $\sum _{t\in \ZZ}
a_t+2^{[n/2]}b=p^n$. Therefore
$$\sum \dim B_{d_N+ip+j}=\sum _{t\in \ZZ}
a_t+2^{[n/2]+1}b=2^{[n/2]}b+p^n$$ and Lemma \ref{sum-B} implies that
$b=2^{[N/2]}\gamma _N(j)$. By the proof of Lemma \ref{sum-B}
$$a_0=\dim B_{d_N+j}-2^N\gamma _N(j)=\dim B_{d_N+j}- (\dim
B_{d_N+j}-\dim C_{d_N+j})=\dim C_{d_N+j},$$
which finishes the proof.
\end{proof}

\medskip

Since $F_*(\O_{Q_n}(j+tp))\simeq F_*(\O_{Q_n}(j))\otimes \O_{Q_n}(t)$,
the above theorem gives the decomposition for all Frobenius push
forwards of line bundles on $Q_n$ ($n\ge 3$).

We can also compute the decomposition of $F_*(S_n(j))$ along the
following lines.  By (\ref{seq2}) and (\ref{seq4}) we have an exact
sequence
$$
0\to H^1(S_n\otimes F^*S_n (t))\to H^1(F^*S_n (t))^{\oplus 2^{[N/2]}} \to
H^1(\Sigma \otimes F^*S_n (t+1))\to H^2(S_n \otimes S_n (t))
$$
for $t\le d_N-1$. By Theorem \ref{2-vanishing} the last cohomology
group vanishes for $t\ge d_N-p+1$. By the same theorem and the proof
of Corollary \ref{2-non-vanishing} we also have an exact sequence
$$0\to H^1(S_n\otimes F^*S_n(d_N-p+1))\to H^1(S_n\otimes
F^*S_n(d_N-p+1)|_{Q_{n-1}})\to H^1(S_n\otimes F^*S_n(d_N-p+1))^*\to
0.$$ Together with Lemmas \ref{tensor-spin} and \ref{2-spin-duality}
this is sufficient to determine the required decomposition.  By
induction, this also gives decomposition of $F^s_*(\O_{Q_n}(j))$.  We
skip the actual computation as it is long and it will not be used in
the following.

\begin{Corollary} \label{strong-splitting}
$F^s_*(\O_{Q_n}(j))$ contains $\O_{Q_n}(-t)$ as a direct summand if
and only if $0\le tq+j\le n(q-1)$.
\end{Corollary}

\begin{proof}
By Proposition \ref{splitting} we need only to show that some line
bundles appear in the decomposition. The proof is by induction on
$m$. For $s=1$ the required assertion follows from Theorem
\ref{decomposition}.  Assume that $0\le tp^{s+1}+j\le n(p^{s+1}-1)$
for some $t$.  Then $0\le tp^{s}+j/p\le n(p^{s}-1)+n-n/p$. Therefore
there exist an integer $l$ such that $0\le tp^{s}-l\le n(p^{s}-1)$ and
$-j/p\le l\le n-(n+j)/p$.  Then $\O_{Q_n}(-t)$ is a direct summand of
$(F^m)_*(\O_{Q_n}(-l))$ and $\O_{Q_n}(-l)$ is a direct summand of
$F_*(\O_{Q_n}(j))$.
\end{proof}

\begin{Corollary}\label{t1}
$F_*\O_{Q_n}$ is a tilting bundle if and only if $p> n$.
\end{Corollary}

\begin{proof}
By Theorem \ref{decomposition} $F_*\O_{Q_n}$ contains $\O _{Q_n}(-i)$
as a direct summand if and only if $0\le ip\le n(p-1)$. Moreover,
$F_*\O_{Q_n}$ contains at most one twist of $S_n$. If $p< n$ then
$i\le n-2$ as $n(p-1)<p(n-1)$. This implies that $F_*\O_{Q_n}$
Karoubian generates a proper subcategory of $D^b(Q_n)$ generated by at
most $(n-1)$ line bundles and one twist of spinor bundles.  If $p=n$
then $F_*\O_{Q_n}$ is a direct sum of line bundles so it does not
generate $D^b(Q_n)$. On the other hand, if $p> n$ then $F_*\O_{Q_n}$
contains as direct summands $\O_{Q_n},\dots , \O_{Q_n}(-n+1)$ and one
twist of $S_n$, so it is tilting.
\end{proof}

\begin{Corollary}\label{t2}
If $n=2m$, $m\ge 2$ and $s\ge 2$ then $F^s_*\O_{Q_n}$ is
quasi-exceptional only if $m=2$, $p=3$ and $s=2$. In this case
$F^s_*\O_{Q_n}$ is also tilting.
\end{Corollary}

\begin{proof}
First, let us assume that $m>2$ or $p>3$. Note that $F_*\O_{Q_n}$ or
$F_*\O_{Q_n}(-1)$ contain $S_n(-m+\lceil m/p \rceil)$ as a direct summand. Set
$l_0=[n-n/p]$. By assumption $l_0>m$ so $F_*\O_{Q_n}(-l_0)$ contains
$S_n(-m)$ as a direct summand.  Note that $F_*\O_{Q_n}$ contains
$\O_{Q_n}$, $\O_{Q_n}(-1)$ and $\O_{Q_n}(-l_0)$ as direct summands, so
$F^2_*\O_{Q_n}$ contains as direct summands both $S_n(-m+\lceil m/p\rceil)$ and
$S_n(-m)$. Then Lemma \ref{tensor-spin} implies that $F^2_*\O_{Q_n}$ is
not quasi-exceptional. Since $F^s_*\O_{Q_n}$ contains as a direct
summand $F^2_*\O_{Q_n}$ it is also not quasi-exceptional.

Now assume that $m=2$ and $p=3$. Then $F^2_*\O_{Q_4}$ contains as
direct summands only $\O_{Q_4},\dots , \O_{Q_4}(-3)$ and $S_4(-1)$.
But $F_*(\O_{Q_4}(-3))$ contains $S_4(-2)$ as a direct summand and hence
$F^s_*\O_{Q_4}$ is not quasi-exceptional for $s\ge 3$.
\end{proof}

\begin{Corollary} \label{t3}
Assume that $n=2m+1$ and $s\ge 2$. If $p\ge n$ then $F^s_*\O_{Q_n}$
is a tilting bundle. If $p<n$ then $F^s_*\O_{Q_n}$ is not
quasi-exceptional.
\end{Corollary}

\begin{proof}
It is easy to see that if $p\ge n$ then $F_*(\O_{Q_n}(-j))$ for
$0\le j\le 2m$ contain as direct summands only line bundles and
$\Sigma (-m)$. Moreover, $F_*(\Sigma (-m))$ contains as direct
summands only line bundles and $\Sigma (-m)$. This allows easily to
check that $F^s_*\O_{Q_n}$ is a tilting bundle.\

Now assume that $p<n$. Let us note that by Corollary
\ref{strong-splitting} $F_*\O_{Q_n}$ always contains $\O_{Q_n}(-t)$
for $0\le t\le m$, as $0\le mp\le n(p-1)=mp+(m+1)(p-2)+1$. But by
Proposition \ref{splitting2} $F_*(\O_{Q_n}(-m))$ contains
$\Sigma (-m)$ as a direct summand and $F_*(\O_{Q_n}(-m+\frac
{p+1}{2}))$ contains $\Sigma (-m+1)$ as a direct summand.  This
implies that $F^2_*\O_{Q_n}$, and hence $F^s_*\O_{Q_n}$, are not
quasi-exceptional.
\end{proof}

\bigskip

{\bf Acknowledgements.}

The author would like to thank A. Samokhin for providing the original
motivation for this paper by writing \cite{Sa1} and for stimulating
discussions.

The paper was written while the author's stay in the Max Planck
Institute f\"ur Mathematik in Bonn. The author would like to thank
this institution for the hospitality.

In the preparation of this paper the author used the computer algebra
system {\sc Singular} by G.-M. Greuel, G. Pfister, and H. Sch\"onemann
to check some results (e.g., Theorem \ref{vanishing} in dimensions
$3,4,5$ and in low characteristic) and in some cases to guess the
results (see Proposition \ref{C=B}).  All traces of actual computer
calculations were carefully removed from the paper.

Finally, the author would like to thank his student, Piotr
Achinger, who pointed out a gap in the previous proof of
Proposition \ref{diff-new}.

redpajama_set_name: RedPajamaArXiv


Jurassic Park trips are producing limits of Lingcod and big Pacific Snapper (rockfish) ! Salmon has slowed down.
Specializing in Group and Individual Therapy for FDS (Fishing Deficiency Syndrome).
MEGA-BITE is the premier 6-pack charter boat located in the Santa Cruz Yacht Harbor and we fish the Monterey Bay. We provide all types of charters aboard our spacious 32x12 foot 630 horsepower Sport Sedan. We can take up to six people fishing for salmon, halibut, lingcod, white seabass, albacore tuna and pacific snapper (rockcod), on the water picnics, marine and wildlife eco-tours, team building, private "Day on The Bay" parties. We are very family friendly (no smoking and cussing) and we can teach anyone how to catch fish. That little someone needs a little self-esteem or self-confidence, we are the champions! You want value. You want comfort. You want integrity and professionalism. You've come to the right place. We do it all and we do it good!
Copyright © 2006 - 2019 Monterey Bay Charters and Santa Cruz Charters. All rights reserved.


When: March 9-10. Various times.
Suitable for ages 4 months – under 4s and their grown-ups.
Big Imaginations and Turned On Its Head present … Sponge.
Sponge is a soft, bouncy adventure through playtime and dreamtime set to a funky 1970s soundtrack, including disco favourites The Hustle and Car Wash.
Follow a dad and daughter as they take you on a journey through springy environments, fascinating in shape, texture and touch.
Supported by The Spark Arts.


Big Cat Fatalities- Anyone see, any REAL danger ?
At least Something on the news against PETA.
Emo is the featured "Grown Up Puppy" on The Daily Puppy!
People are EATING foxes here in the UK!
anyone follow the casey anthony case?
Two Metropolitan Police dogs die in car.
Anyone catching the Westminster tonight?
Gorilla that walks like a human?!
Being Fired From Your Own Sanctuary????


Boy do I have a mess in the shack – several computers not very well coordinated due to a split vision and having been acquired over too much time.
My Flex-5000A with PowerSDR runs on a 4-5 year old Dell tower that really isn't adequate. Some of my peeves with PowerSDR can be lain at the feet of a marginal computer. Dual core Windows-XP box with not really that much going for it, the highlight is having dual screens and the requisite Firewire port.
Elsewhere in the house I do all writing, and all my "home work" from work, on an early Aluminum MacBook. This is the smallest & slowest MacBook but is running the latest Mountain Lion.
For portable operations I retired a now defunct IBM Thinkpad and replaced it with a decent Windows XP-Pro larger Thinkpad. Nothing fancy but when a decent battery was added it is close to a portable version of the Dell Tower.
Retired, but lingering around are a MacMini (early series) that my youngest bloated the hard disk gaming and won't run. Have been meaning to see if I can coax the machine back to life, at least enough to rescue pictures.
An older G5 iMac is around too – it was one that the power supply had the counterfeit capacitor issue, fixed under "quiet recall" and hasn't really been used since.
Then my newest computer is running on my project's desk – a Raspberry Pi. A neat little ARM based Linux computer, it has been running as a "burn-in" on the desk.
There are enough parts and licenses to put together a few wobbly generic PC's with either Linux installs or using one of the Window's Licenses if Microsoft will issue a key for replacement hardware.
Then after that I get into things I do occasionally, rather than regularly.
The new iMac really is a looker – and my Apple experience is much more positive than my Windows experience.
But then there are the Flex-5000A's PowerSDR and Firewire needs.
Presently I am leaning to upgrading the shack's Dell with a roll-you-own Windows-7 box AND replace the MacBook with an iMac running Parallels/Win-7.
The MacPro is pricy enough that it is not really justifiable for home use, but wow it would be sweet.
Back at Dayton I bought a Begali CW Machine for Code Practice.
Finally getting around to setting it up.
I've found an appropriate keyboard and will wire up a Kent Key and get started.


text:
F1 Challenge Game Review
Reviewed by Kidzworld on Dec 27, 2006
Summers here and the time is right for racing in the streets! If youre a demon for all things that go vroom, let F1 Challenge 99-02 satisfy that need for speed.
Summer's here and the time is right for racing in the streets! If you're a demon for all things that go vroom, let F1 Challenge '99-'02 satisfy that need for speed. EA Sports puts your PC in the game - and the pits - with the best darn F (formula) 1 (one) title on the market.
It's Pole Position For F1 Challenge '99-'02
There is so much coolness that it's hard to know where to begin. There's the funky changes in weather (that track sure does get slippery,) the cool texture of the sky and reflections on the cars which make it look totally realistic. Most of all, Challenge '99-'02 captures all the sounds of F1. If you've ever had the chance to hang near the pits, you are very familiar with what sounds like a billion bees nesting in your brain. That 130 plus decibels really takes a toll on the old eardrums after a few days, which is exactly how long it's gonna take you to really get behind the wheel of F1Challenge '99-'02. Practice is what it's all about. And now, for the question you have been waiting to be answered... how good are the crashes? Excellent! Lots of flying metal and car flips - but nobody dies. A good day at the track.
How Hard Is F1 Challenge '99-'02?
It's hard! Think of it as learning how to drive the real deal - that doesn't happen in an hour.
Age Rating: Everyone.
EA Game Review - Need For Speed: Porsche Unleashed
EA Game Review - NBA Street Vol. 2 Video Game
More Games and Gear!
Tags: game pics games sports one pc reviews ea racing race challenge arts images car electronic track speed driving screenshots formula
Hyrule Warriors: Age of Calamity Nint...
Kingdom Hearts Melody of Memory PlayS...
Pikmin 3 Deluxe Nintendo Switch Game ...
Games Videos
Games Poll
F1 Challenge '99-'02
Best Thing About Racing Games?
The real racing action.
Major crash action.
No injuries, no fatalities!

redpajama_set_name: RedPajamaCommonCrawl


It's easy to choose a thoughtful gift for loved ones who like to garden with our selection of gardener gifts. This assortment of gifts for the gardener has something for everyone. Beautifully-made hand tools, planters and garden beds are perfect for vegetable and flower gardeners. For novice gardeners, we have a range of gardening books on different topics. Skin care products such as lotion and soap also make wonderful gifts for gardeners as they can use them after working in the garden. For unique gifting options, take a look at our vintage wash sink with stand and vintage galvanized watering can. These charming garden accessories are the perfect combination of function and beauty.


text:
Wrightspeed debuts its first electric powertrain garbage truck in California
Josh Hittleman
Wrightspeed debuted its first electric powertrain system in a collection vehicle owned by the California-based Ratto Group and will retrofit at least 14 more vehicles over the next year, as reported by Trucks.com.
Terms weren't disclosed, but the contract's estimated worth is $3 million to $5 million. A new truck that can meet California's strict air emission standards could go for at least $500,000 and Ratto Group's COO Lou Ratto said he expected fuel savings of at least 50%.
The system was designed specifically for the waste industry and generates power as the vehicle brakes during frequent stops. A 66,000-pound truck will travel for 24 miles on battery power and can then run on standard fuel through a range extender.
This new technology — developed by Tesla co-founder Ian Wright — was a big hit when used in a Mack Trucks display at WasteExpo in June and has a lot of potential for the waste industry. The company has raised about $40 million in funding and has also been working with FedEx. California-based Motiv credits itself with releasing North America's first electric collection truck in 2014, and China's BYD recently debuted a full line of electric sanitation vehicles in Beijing, but the concept still hasn't fully taken off in the U.S.
New federal emissions standards for medium- and heavy-duty vehicles, which cover new models as of 2019, will make this type of technology even more appealing in coming years. Many waste companies have been switching to vehicles that run on natural gas, especially as reducing diesel emissions in environmental justice communities has become a higher priority. Los Angeles is mandating the use of these vehicles in the contract terms of its pending franchise system.
As other cities such as New York consider this model some haulers have said the cost of new trucks is hard to manage. Though as Ratto Group subsidiary North Bay Corp. is currently experiencing in Santa Rosa, the cost of fines for not meeting vehicle standards can sometimes be even more expensive.
San Francisco Chronicle Electric garbage trucks? Wrightspeed delivers
Trucks.com Wrightspeed Debuts First Fleet of Range-Extended Electric Refuse Trucks
Filed Under: Collections & Transfer
Latest in Collections & Transfer
How mattress recyclers adapt to evolving consumer trends
By Karine Vann • Jan. 15, 2021
Packaging, PPE and surgical supplies: How COVID-19 is pushing hospitals to reduce waste
By Deborah Abrams Kaplan • Jan. 12, 2021
Transportation Secretary Elaine Chao resigns following insurrection at Capitol
By Sarah Fuller • Jan. 07, 2021


The toefl test measures how well students use english, not just their knowledge of the language but also measures how well you read, listen, speak and write in english and use these skills together in the university classroom. Because it is a valid and reliable test with unbiased, objective scoring, the toefl test confirms that a student has the english language skills necessary to succeed in an academic setting.
WHY SHOULD STUDENTS TAKE THE TOEFL TEST?
The TOEFL test gives students the opportunity to prove they can communicate ideas effectively by simulating university classroom and student life communication. The language used in the test reflects real-life english-language usage in university lectures, classes, and laboratories. It is the same language professors use when they discuss coursework or concepts with students. It is the language students use in study groups and everyday university situations, such as buying books at the bookstore.
More than 9,000 colleges, agencies and other institutions in over 130 countries accept TOEFL scores. For the USA, singapore and australia, TOEFL ibt test scores can be used to satisfy the required language component for a student. As per the new regulations, toefl does not meet the visa requirements in the UK and Canada.
Hence it is not recommended.
FREE SCORE RECIPIENTS SELECTION Until 10 p.m. (local test center time) on the day prior to the test date.
RESCHEDULING FEES You can reschedule or cancel your registration up to 4 full days before your test date. The fee to reschedule is US $60 and must be paid before you can register for another date.
RETAKE THE TEST You can retake the test as many times as you wish, but you cannot take it more than once in a 12 day period.
SCORE RECEIPT BY UNIVERSITY Score reports are sent electronically or mailed to recipients approximately 13 days after the test date.
Each of the six tasks is rated from 0 to 4. The sum is converted to a scaled score of 0 to 30.


text:
Home Chevron down icon Rotation and Frequency CD
Rotation and Frequency CD
by Slick Shoes
Slick Shoes, the legendary Orange County pop punk band is back after nearly 20 years. The band dominated the late 90s and early 2000s, before ending in 2006, but their influence has carried on through two full decades. After releasing a live album last year, they're back with "Rotation & Frequency," their first full length album since 2003.


There is a serialized object injection vulnerability in the Akeeba Joomla update component functionality in versions <= 2.5.25, 3.0-3.2.5, and 3.3.0-3.3.4 which can lead to remote code execution (RCE). The vulnerability exists as a result of flawed sanitization of superglobal variables which store request data. The functionality intended to prevent tampering of the setup parameters can be bypassed enabling a modified PHP serialized object to be passed via a user-controlled parameter. The serialized object causes a zip file containing a malicious file to be downloaded by the update script and extracted to the server. This malicious file can then be requested by the attacker resulting in RCE. The vulnerability requires the 'restoration.php' file to be present in the 'com_joomlaupdate' directory to be successfully exploited.
An unauthenticated remote user sends a request containing a crafted serialized PHP object via the 'factory' parameter and a 'task' parameter value of 'stepRestore'.
The Joomla instance processes the request using the data contained within the serialized object to download and extract an external archive hosted by the attacker.
The server responds with a '200' response containing JSON-encoded status data indicating the successful status of the update request.
The unauthenticated remote attacker requests the extracted file.
The attacker must be able to send crafted packets to the victim host.
Joomla should be updated to the latest version at the earliest opportunity as later versions of the update component address and resolve the serialized object injection vulnerability. If updating Joomla is not immediately possible, virtual patching should be implemented to prevent serialized object injection.


Viewers will take a dive back into Blue Planet later this year.
The BBC has announced that a new series of the seminal nature documentary Blue Planet is being made.
The original series, which was narrated by Sir David Attenborough, aired in 2001 and featured never-before-seen footage of the world's marine life.
When it was first shown, it became a landmark in nature filmmaking and was watched by more than 12 million people.
Blue Planet II will air later this year, 16 years after the original series, with Attenborough once again taking the narrating helm.
The BBC claims it has spent 4 years filming footage for the new series, which uses new camera and filming innovations in order to get even closer to the marine life.
Highlights from the series will include footage of orcas and even the discovery of a new species of crab - dubbed the "Hoff crab" due to its hairy chest.
Executive producer of the series, James Honeyborne, said that the series would focus on scientific breakthroughs.
"The oceans are the most exciting place to be right now, because new scientific discoveries have given us a new perspective of life beneath the waves.
Blue Planet II will be broadcast on BBC One later this year.


text:
Leinster Senior Cup
Setanta Cup
UEFA Intertoto Cup
Friends of St Pat's
Thursday February 28th, 2019 @ 11:30
Saints And Scholars Programme
We hare happy to share an update on our football numeracy programme which we are running with a number of primary schools in the area.
Prior to our game against Cork City, our Community Officer David Morrissey was accompanied by Brendan Clarke in a visit to Loreto College in Crumlin who completed the 10-week programme.
The kids asked Brendan all about his career to date, playing football as a kid and supporting St Pat's.
We would like to thank Mr O'Conghaile and the 3rd Class at the school for all their hard work throughout the course, we have no doubt that the club has gained a lot of new fans in the school and we look forward to working with the school again in the future.
Additionally, we have begun working with Our Lady of the Wayside Primary School in Bluebell.
Bluebell is an area where Pat's have traditionally had a strong tradition and we look forward to ensuring this passes on to the next generation.
We would like to thank RS Ireland for their support in the production of our new Saints and Scholars booklets, which have been designed and produced by Joe O'Rourke and look fantastic.
Other News Headlines
2020 Saints In The Community Report
Saints Add Smith And Promote U19's To First Team
Benson And Desmond Sign-up For 2021
Mountney Joins Saints For 2021
Saints Crowned U19 Champions
WATCH OUR U19 FINAL LIVE: Pat's-Bohs (7.30)
Ronan Coughlan Joins For 2021
Tonight's U17 Shield Final Postponed
Jamie Lennon Talks 2021 With Pat's
Tweets by @stpatsfc
Shamrock Rovers 18 48
Bohemian FC 18 37
Dundalk 18 26
Sligo Rovers 18 25
Waterford FC 18 24
St Patrick's Athletic 18 21
Derry City 18 20
Finn Harps 18 20
Shelbourne 18 19
Cork City 18 11
St Patrick's Athletic FC
Founded in 1929 the 'Saints' play in the Premier Division of the SSE Airtricity League at Richmond Park, in Inchicore Dublin 8, Ireland.
125 Emmet Road, Inchicore, Dublin 8, D08 E0Y7, Ireland
info@stpatsfc.com
www.stpats.com
© Newtonheath Company Ltd (a private company limited by shares) trading as St Patrick's Athletic FC. Registered in Ireland. Company Registration no. 189439.


My Skill - UML, Rational Rose, Use Case Diagrams, Sequence Diagrams, SRS, FSD.
I as a company deals in - Telecom Product .
	Coordinating between Client, Analyst Team and Development Team for fast Closure of various project enhancements.
	Got SRS sign off from the client for the project.
	UAT sign off is the next milestone after phase 1 completion.
	Working on integration approach documents for phase 2 with Oracle Financials.
	The project is for billing of Leased circuits that BSNL provides to big Corporate.
	UAT Sign Off achieved for the BSNL WOMS project.
	Involved in FSD Design and implementing the design through coordination with the development team.
	Involved in presenting understanding sessions on FSD to the development, QA and Documentation teams.
	Client communication for various areas.
	Leading team towards the project success.
	Mentored a team of 4 people, 1 Business Analyst, 3 Developers.
	Giving functional inputs to the developers to achieve the desired feature as required by the client.
Gateway TechnoLabs Pvt. Ltd, Ahmedabad.
	Worked in the maintenance and customization of the ERP used by the company.
	Worked on upgradation of the SQL Forms version 2.3 to 3.0 to 4.5 to Forms version 10g.
	Working on database upgradation from Oracle 6.0 to 1Og.
	Worked on Pro*C reports.
	Designed and implemented Forms and Reports.
	Generating various MIS reports.
	Written Pro*c reports for Form 16, Profit and Loss A/C, Balance sheets and various other HRM and Financial Reports.
	Generated Dept wise Profit and Loss reports using Pro*C.
	Lead programmer in Government projects to design the system for implementation of crop protection in the Gujarat Agricultural University.
	Designed and coded Prototype systems for the said project.
	Technologies Used: ASP .net/VB .net + SQL Server 2000.
	Client: Department of Agriculture, Government of Gujarat.
	Created the complete SRS report for the Project and presented it before various renowned delegates and officials from the Gujarat Agriculture Department including the Chief Secretary (Agriculture).
	Worked on an ecatalogue Project for a Singapore based client named Epiphanyworks Pvt. Ltd., Singapore.
	Verified design revisions and component layouts.
	Technology Used: ASP 3.0, html, java script, Access 2000..


1437.02.26 Lijsbet Mandemakers is pardoned for something not specified.
1437.ca Testimony and a decision in a dispute between Tielman van Buschusen and the ambacht of Zoeterwoude concening whether or not Zoeterwoude had paid Tielman the amount to be collected but which was never turned over to the clerk of Rijnland. Because a credible witness sided with Zoeterwoude, so did the hoogheemraden.
1437.ca Testomony and a decision in a dispute between Tielman van Buschusen and the ambacht of Leiderdorp concening whether or not Leiderdorp had paid Tielman the amount to be collected but which was never turned over to the clerk of Rijnland. Because the witness sided with Tielman, do did the hoogheemraden, though Leiderdorp could appeal the decision with new evidence within eight days.


The hotel features a Honeymoon Villa which offers panoramic views over the sea, a private pool, a king-size bed with a walk-in closet, a fireplace and many other amenities which will make your honeymoon as romantic and memorable as possible.
For an utterly unique and luxurious honeymoon experience surrounded by lush greenery and beautiful nature, this hotel's Honeymoon Suite is nothing short of ideal, since it offers comfortable accommodation, quaint interiors, access to the spa and magnificent views.
The hotel offers a Honeymoon Suite which features a king-sized bed with electric blankets, a relaxing spa, garden, a private and quiet location and many more amenities that will make your honeymoon as romantic as possible.
Set amidst lush natural landscapes and boasting elegant suites in a rustic-luxe style with king-size beds and ample space, this hotel is a truly ideal destination for a romantic and unique honeymoon experience.
This adult-only Balinese style retreat offers the perfectly private and romantic backdrop for a honeymoon that will remain engraved in both you and your partner's memories.
The hotel features the Magnifique Romance Package which includes accommodation in a luxurious room with pool or river views, a cheese plate and champagne, daily breakfast, a gondola cruise, a gourmet breakfast for two, late checkout and many other amenities that will make your honeymoon one of a kind.
The hotel features various one bedroom suites, with lake or garden views and elegantly appointed with king-size beds and many amenities, making them ideal for a romantic honeymoon getaway.
Located right in front of the ocean, this hotel features the Romantic Hotel Package which includes luxurious beachfront accommodation, a bottle of sparkling wine, a cheese platter or chocolate covered strawberries and many other amenities that will create the most romantic atmosphere for the two of you.


text:
UVM Humanitarian Mapping Club Helping With Harvey Relief
As Hurricane Harvey neared the coast of Texas… The University of Vermont Humanitarian Mapping Club was hard at work throwing a party… But not the kind of party you would think.
Noah Ahles a UVM Research Specialist says that the first thing the group did when Hurricane Harvey hit is we looked at the available data in the area. The available data is on Open Street Maps… A website similarly to Google Earth but without certain restrictions.
According to Ahles "Google is a for profit company that doesn't share their data very much. So what we're trying to do is create a data set that anyone one can access without any red tape at all. So once we actually do the edits at our mapping parties its uploaded on to the open street maps which is kind of like the Wikipedia of maps anybody can edit it, anybody can download it for free to use for whatever purpose they would like."
This data can be used by relief workers and agencies such as the World Bank for damage assessment or disaster recovery.
Lindsay Barbieri a graduate student at UVM says the data can be used to "get a better sense of how many buildings were affected the size of the building that were affected the roads that were affected… these sorts of really critical information for aid."
In Houston the club discovered deficiencies in building data. So this group got together, took satellite images and mapped out specifics that aid workers can use for better accuracy during storm recovery.
"There weren't enough building datasets to be able to really quickly do certain types of analysis for damage assessment, figuring out how many buildings are in the flooding area… that kind of stuff." Says Ahles
And hurricanes aren't the only natural disaster that the UVM Humanitarian Mapping Club has assisted with. They've also helped with the Ebola outbreak as well as the Malaria outbreak.
As for the party they threw… It's called a mapping party. 25 students got together to map almost 4,000 buildings in Houston. And they did it in just two hours.
A Vermont community is leaning on each other after fire kills two boys
Local News / 1 year ago
Racial Justice Rally in Burlington calls for change


It was easy to lose track of how much time I spent wandering along the banks of the Seine. From Notre Dame all the way down to the Eiffel Tower is a UNESCO designated World Heritage Site. And for good reason! Never minding the long history in that part of Paris, that stretch of the river is just one beautiful piece of architecture after another. There's no question I'd like to go back some day and, with more time, explore Paris further afield, but the banks of the Seine was the part of Paris I haunted most in my short stop there.
I was pleasantly surprised one afternoon when I stumbled upon this section of the banks of the Seine just below the famous Eiffel Tower. Upon wandering down there, I found what appeared to be boats where people actually lived rather than the obviously commercial piers further upriver. In fact, the man strolling in the left hand of the frame in this picture was returning to the beautiful red boat in the foreground. Imagine living with that mammoth work of art towering above you and the Seine below your feet. Must be a beautiful life as it certainly makes for an enchanting print of Paris.
And his pretty red home made for a beautiful way to frame that landmark tower as it pierced the wintry Parisian skies. It's interesting how time changes our attitudes. When the tower was being erected on the banks of the Seine over a hundred years ago, it was meant to be temporary and it was still met with public outrage. Artists in particular said it spoiled the beautiful skyline of Paris. A century later, it still stands, a temporary work of architecture that it's now hard to imagine Paris without!


Evil corporate minions and government agents missed a fantastic opportunity to get a good look at a huge chunk of annoying infosec practitioners and researchers last weekend in Wellington.
Gathered at the Michael Fowler Centre auditorium were many of the super-skilled people who find holes and mistakes in the best laid IT plans and designs along with those who came to learn more at the Kiwicon 2038AD security conference.
Around 2,400 people attended Kiwicon, the biggest hackiest, cyber security conference in the southern hemisphere until next year at least when the even larger BSides in Canberra kicks off.
I for one am glad that Bad Actors didn't take security through obscurity to the extreme in Wellington.
Mainly because the infosec researchers are sorely needed to protect us against copious amounts of poorly secured systems that are now part of our lives thanks to the internet, personal computers, smartphones and electronics everywhere - but also because I was in the audience and would like to come back next year.
Kiwicon's a big, important regional security conference that most people won't have heard of. This is a shame in many ways, although looking around the motley crew of attendees it's clear that they'd be challenging people, with a Low Acceptance Factor in more conservative environments.
The LAF extends to infosec researchers' work. Things have got better and more insightful tech companies accept vulnerability reports and handsomely reward researchers who find bugs in their products and services - and disclose them responsibly so that they can be fixed before they're exploited.
Nevertheless, researchers are still likely to receive legal letters from companies who take their work the wrong way. Thanks to ambiguous and overly broad computer security laws everywhere in the world there's always the chance of early morning fun for researchers as cops kick in their doors and take them away to holding cell parties to await lengthy trials on matters the authorities often don't understand.
That's one reason why many use pseudonymous nicknames like Metlstorm, bogan, sput, moloch and mandatory, as attention around their work is a two-edged sword.
There's definitely shades of grey to many people in the cyber security industry.
One of the conference organisers joked about the infamous Rawshark being in attendance, knowing full well that the police might take it seriously and come and visit.
Rawshark is the undisputed Master of Operational Security, along with mysterious AI botherder and spammer turned good, Bismillah, who was rumoured to be at Kiwicon too.
On a serious note, infosec remains a male-dominated field. It has been tarnished by sexual misconduct against women, especially during conferences and events. That's unacceptable and the Kiwicon organisers sought to crack down on such rubbish behaviour and to ensure everyone felt safe and comfortable.
"Don't be an asterisk-hole," summarises Kiwicon's code of conduct. Everyone seemed to follow that, creating a respectful and relaxed atmosphere mixed with a cheesy 90s cyberpunk theme that the younger folks tapped into, much to the surprise of old folks arranging the conference.
Yes, the first generations of infosec practitioners have grown up. Mike Forbes and Adam Boileau said that between them and the other Kiwicon organisers, they have 15 kids which is way more than 2007 when the conference started.
Instead of leaving it at that, Kiwicon ran a day-long Kuracon programme for kids that was staffed volunteers. I brought two children who were taught lock-picking, coding basics, building little electronic robots, and taking apart old computer gear. Kuracon is fantastic initiative that other conference organisers should copy.
Unwrapping the hacker-proof tinfoil from my devices after the conference, the takeaway from Kiwicon is just how tenacious researchers are. They'll spend weeks cracking problems that other people give up on in a day or two. Black hat hackers do the same. Since IT is part of everyone's life now, that's a great reminder of how insecure it can be, and why we need security researchers.
My thoughts and thanks (don't do prayers) go to the hardworking volunteers who have been organising Kiwicon over the years. They took a breather last year, but here's hoping they'll recharge for 2019.


● Should have good experience working in Spring Boot.
● Good experience in RESTful API / Web Services.
● Experience with Design Patterns.
pressure in an independent environment.
● Education: Bachelors or Masters in Computer Science.


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You are here: Inicio » Solar Thermal News & Events » What is the Resale Value of My Solar Home?
FOR IMMEDIATE RELEASE – Does the solar system installed in your home increases the property value and helps you when selling your home?
Yes! The American Appraisal Institute estimated that any home features that save in operating costs would increase the home's value by twenty times the amount saved. Every homeowners who have installed solar systems in their property experience an increase in resale price of the home generated from the systems they have installed.
The question now is," How much will my property value increase from installing a solar system?" It really depends on your system. The more your system is able cut in operating cost, the higher the value increase will be. Considering the energy savings once the system is installed, the initial cost is always recaptured when the home sells. According to recent studies, solar thermal systems save more energy than PV Systems, which translates to solar thermal systems can increase your home value more than PV Systems.
Ed Gunn from SunMaxx Solar, one of the leading solar companies in the solar thermal industry, acknowledged that solar thermal systems can add up to $20,000 in property value to your home. Examples of such solar thermal systems are offered by SunMaxx Solar Inc., featuring some of the most efficient solar thermal systems in the entire industry. One of the many reasons to go solar, despite the fact that it is a home improvement that will pay for itself, is that the initial cost invested will recuperate with value added to the home.


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A bit late for that, I was all for trading him this past season. Take what you can get IMO.
Beane loves long shots. I'd prefer to see more sure-shots. Lots of potential though.
Definitely don't release, trade for what you can get, draft a young stud. He's getting up there and has lost a step imo.
A bit off topic here, but if you put Josh Allen in KC do you think they're in the game with everything else the same?
I'm not that excited. I'm anticipating one or two mid-tier signings on the offensive side of the ball.
Am I the only one that gets a little jealous when a QB has all day to throw? Either the Rams have an amazing oline or the Saints dline isn't very good. Perhaps both.
Top 4 Scoring teams are in the Conference Champ games..


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Heterobathmia pseuderiocrania is een vlinder uit de familie van de Heterobathmiidae. De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1979 door Kristensen & Nielsen.

Heterobathmiidae

redpajama_set_name: RedPajamaWikipedia


I have been fortunate enough to have been able to make my living "telling stories" for many years. As an actor, documentary producer, print journalist and business writer, my work has taken me to every continent on earth not covered in ice, where I've found myself continually gathering more questions than answers. Still, it has always been reading and writing fiction that lights my heart and mind. I am convinced that only in fiction can we truly explore the possibilities of our lives, to reveal the deepest truths in our souls.