text:
\begin{document}

\begin{abstract}
  We compute generators and relations for the basic algebra of a
  non-semisimple singular block of the restricted enveloping algebra of
  $\ssl_3$ over a field of characteristic $p>3$. Working directly with
  the basic algebra we compute its centre and the internal degree zero
  part of its first Hochschild cohomology, and show its Verma modules are Koszul.
\end{abstract}

\maketitle
Let $A$ be a finite-dimensional algebra over an algebraically closed
field $k$.  The basic algebra of $A$ is a $k$-algebra all of whose
simple modules are one-dimensional which is Morita equivalent to $A$; it
can be thought of as the smallest algebra with the same representation
theory as $A$.
The small size of the basic algebra tends to make calculations
considerably easier.

This paper deals with restricted enveloping algebras of classical Lie
algebras.   While the basic algebras for non-semisimple blocks of
restricted $\ssl_2$ have long been known (and are reviewed in Section
\ref{hoch_section}), to the best of my knowledge no other nontrivial
examples have been computed, although some endomorphism algebras of projectives
are constructed in \cite[Chapter 19]{AJS}. Here we find generators and relations for the
basic algebras of non-semisimple singular blocks
of the restricted enveloping algebra $u$ of $\ssl_3(k)$ over an
algebraically closed field of
characteristic $p>3$, and some of their properties.
The simple modules of
$u$ are labelled by their highest weights; singular here means that the
block has a simple module whose highest weight $\lambda$ satisfies
$\langle \lambda + \rho, \alpha\rangle=0 \mod p$ for some root $\alpha$,
where $\rho$ is the half-sum of the positive roots for $\ssl_3(k)$.

The method is as follows.  Thanks to work of Simon Riche \cite{Riche} we
know that each block of $u$ admits a $\mathbb{Z}_{\geq 0}$-grading
with respect to which it is a Koszul algebra.  We show in Section
\ref{verma_section} that the Verma
modules in our block are graded modules with respect to this grading.
This allows us to
use a version of the Brauer-type reciprocity of \cite{Holmes} to
find the graded structure of the projectives in our block, and hence the
Hilbert series for the basic algebra and of its Koszul dual.

In Section \ref{sec:action} we show, using the Frobenius homomorphism for the hyperalgebra of
$\ssl_3(k)$, that the basic algebra admits a non-trivial action of
$\ssl_3(k)$ by
derivations.  We determine this action on the degree one part of the basic
algebra, which allows us to calculate its relations in Theorem
\ref{presentation}.  In Section
\ref{hoch_section} we use
this presentation of the basic algebra to calculate its centre and the
internal degree zero part
of its first
Hochschild cohomology.  Finally in Section \ref{kos} we identify the
modules for the basic algebra that correspond to the Verma modules in
our block, and prove that they are Koszul.

\section{Notation and definitions} \label{notation_section}
\subsection{Algebras} Let $\mathbb{F}$ be a field.  The Lie algebra $\ssl_3(\mathbb{F})$ is generated by elements $E_i, F_i, H_i$
for $i=1,2$ subject to the relations

\begin{center}
  \begin{tabular}{llllll}
    (r1) & $[H_1,H_2]=0$ & (r2) & $[H_i,E_i]=2E_i$ & (r3) &
    $[H_i,F_i]=-2F_i$ \\
  (r4) & $[H_i,E_j]=-E_j$ & (r5) & $[H_i,F_j]=F_j$ & (r6) &
  $[E_i,F_j]=0$ \\
  (r7) &
  $[E_i,[E_1,E_2]]=0$ &(r8) & $[F_i,[F_2,F_1]]=0$ &(r9) & $[E_i,F_i]=H_i$
\end{tabular}
\end{center}

for $i=1,2$ and $j \neq i$.   We write $E_{12}$ for $[E_1,E_2]$ and
$F_{12}$ for $[F_2,F_1]$.

Let $U_{\ZZ}$ be the Kostant $\ZZ$-form of the universal enveloping
algebra of $\ssl_3(\CC)$: this is the subring generated by all
elements of the form $E_i^{(n)}:= E_i^n/n!$ and $F_i^{(n)}:= F_i^n/n!$.
Let $k$ be an algebraically closed field of characteristic $p>3$ and
$D = U_\ZZ \otimes_\ZZ k$ be the hyperalgebra or algebra of
distributions \cite[I.7, II.1.12]{Jantzen_book}.  We will abuse notation by
writing $E_i^{(n)}$ for the element $E_i^{(n)} \otimes 1$ of $D$ and so
on.  The restricted enveloping algebra $u$ of $\ssl_3(k)$ is defined to
be the subalgebra of $D$ generated by the $E_i$ and $F_i$ for $i=1,2$,
it is a
Hopf algebra of dimension $p^8$
isomorphic to the quotient of the universal enveloping algebra
$U(\ssl_3(k))$ by the two-sided ideal generated by $E_i^p, F_i^p,
E_{12}^p, F_{12}^p, H_i^{p}-H_i$ for $i=1,2$.  We write $b^+$ for the
subalgebra of $u$ generated by the $E_i$ and $H_i$ for $i=1,2$, and
$b^-$ for the subalgebra generated by the $F_i$ and $H_i$ for $i=1,2$.

Let $D_2$ be the subalgebra of $D$ generated by the $E_i^{(n)}$ and
$F_i^{(n)}$ for $n<p^2$.  It has a basis consisting of all monomials
\[ F_1^{(a_1)} F_2^{(a_2)} F_{12}^{(a_3)} {H_1 \choose a_4}{H_2 \choose
a_5}
E_1^{(a_6)} E_2^{(a_7)} E_{12}^{(a_8)} \]
where $0 \leq a_i <p^2$ and ${H_i \choose n}$ is the image in $D_2$ of
the element
\[   \frac{ H_i (H_i-1) \cdots (H_i-n+1)}{n!} \]
of $U_\ZZ$.

There is a surjective algebra homomorphism $\Fr: D_2 \to u$ such that
\[ \Fr(E_i^{(r)}) = \begin{cases} E_i^{(r/p)} & \textrm{if } p\mid r \\
  0 & \textrm{otherwise} \end{cases} \]
and similarly for $F_i^{(r)}$, called the Frobenius map.  The
kernel of $\Fr$ is the two-sided ideal generated by the augmentation
ideal of $u$.

\subsection{Gradings} Let $\Phi= \{ \pm \alpha_1,\pm\alpha_2, \pm(\alpha_1+\alpha_2)\}$ be the $\ssl_3$ root system with simple roots $\alpha_1$ and
$\alpha_2$, and let $\varpi_1$ and $\varpi_2$ be the fundamental
weights. Let $X$ be the weight lattice, the free abelian group on the
$\varpi_i$, which we identify with $\ZZ^2$ by $\varpi_1 = (1,0)$ and
$\varpi_2=(0,1)$.  Then $U_\ZZ$,
$D$ and $u$ are $X$-graded with $E_i$ in degree $\alpha_i$ and
$F_i$ in degree $-\alpha_i$.  We refer to this as the weight
grading and
write $|x| = \alpha$ if $x$ is a homogeneous element of degree $\alpha
\in X$.  Let $W$ be the Weyl group of $\Phi$, generated by reflections
$s_i$ in the hyperplane perpendicular to $\alpha_i$, so $W$ acts by
automorphisms on the restricted Lie algebra $\ssl_3(k)$ and hence on $u$.  Thinking of $\ssl_3(k)$
as traceless $3\times 3$ matrices over $k$, this action comes from
conjugation by permutation matrices.

By \cite{Riche}, $u$ admits a $\ZZ_{\geq 0}$-grading with respect to
which it is a Koszul algebra in the sense of \cite{BGS}. We write $u_i$ for the $i$th
graded piece, and refer to this as the $K$-grading on $u$.  Koszulity implies that $u$ is generated as an algebra by
$u_0$ and $u_1$, and that $u_{>0} = \bigoplus_{i>0}u_i$ is the
Jacobson radical $J$ of $u$.

\subsection{Modules} All modules considered in this paper will be
left-modules unless otherwise stated.

If $M=\bigoplus _{\alpha \in X} M_\alpha$ is a
weight-graded $u$-module and $w \in W$ we write $^wM$ for the
weight-graded $u$-module with $(^wM)_\alpha= M_{w(\alpha)}$ for any
$\alpha \in X$ and action $x \cdot m = w(x)m$ for $x \in u$ and $m \in
M$.   If  $i,j \in \ZZ$ then $M[i,j]$ is the graded module which has the
same underlying vector space and action as $M$, but with grading
determined by $M[i,j]_\alpha = M_{\alpha-(i,j)}$ for any $\alpha \in X$.
We write
$\Hom_u^{\textrm{wt}}$ and $\Ext_u^{\textrm{wt}}$ for the Hom and Ext
functors in the category of
finite-dimensional weight-graded $u$-modules, so an element $f$ of
$\Hom_u^{\textrm{wt}}(M,N)$ sends $M_\alpha$ to $N_\alpha$ for all
$\alpha \in X$.  If $\Hom_u$ denotes the
hom-functor in the ungraded category then for finite-dimensional
weight-graded $u$-modules $M$ and $N$,\begin{equation}\label{decomp} \Hom_u(M,N) = \bigoplus_{\alpha \in X}
  \Hom_u^{\textrm{wt}}(M,N[\alpha]) \end{equation}
with a similar result for Ext.

If $M=\bigoplus _{i \in \ZZ} M_i$ is a $K$-graded module, we write $M[i]$ for the $K$-graded module
with the same underlying vector space and action as $M$, but with
grading determined by $M[i]_j = M_{i-j}$.  We write $\Hom_u^K$ and $\Ext_u^{K}$ for the Hom and Ext functors in the
category of finite-dimensional $K$-graded modules.  Again,
finite-dimensionality implies
\begin{equation}\label{Kdecomp}
\Hom_u(M,N) = \bigoplus_{i\in \ZZ} \Hom_u^{K}(M,N[i])
\end{equation}
for graded modules $M,N$, with a similar result for Ext.  We say that a
$u$-module $M$ is \textbf{$K$-gradable} if it admits a decomposition $M
= \bigoplus_{i \in \ZZ} M_i$ as vector spaces such that $u_i M_j
\subseteq M_{i+j}$ for all $i$ and $j$.  The simple $u$-modules are
$K$-gradable since $u_{>0}$ acts as zero on any simple if $i>0$, and the
projective $u$ modules are $K$-gradable \cite{Gordon}.

\section{Verma modules} \label{verma_section}
\subsection{Structure of Verma modules} Let $0 \leq t_1,t_2 <p$ be integers and let $\alpha = (t_1,t_2) \in X$.  Write
$k_\alpha$ for the one-dimensional $b^+$-module on which $E_i$ acts are
zero and $H_i$ as the image in $k$ of $t_i$.  The Verma module $M(\alpha)$ is defined to be
the induced module $u\otimes_{b^+} k_\alpha$, it has a unique simple
quotient module written $L(\alpha)$.  We regard $k_\alpha$ as a
weight-graded $b^+$-module concentrated in degree $\alpha$; this induces
a weight-grading on the Verma modules
since $b^+$ is a weight-graded subalgebra of $u$. The simple modules $L(\alpha)$ are also
weight-graded.  An element $m$ of a weight-graded module $M$ is called a
highest weight vector if it is

an eigenvector for the action of the $H_i$ and $E_1m=E_2m=0$; each
$M(\alpha)$ is generated by a highest weight vector $v_\alpha = 1\otimes
1$, and it has a universal property that implies it surjects onto any
module generated by a highest weight vector with weight $\alpha$. The
dual Verma module $M^-(\alpha)$
is $M(\alpha)^*=\Hom_k(M(\alpha),k)$ as a vector space, with weight grading
$M^-(\alpha)_\beta = (M(\alpha)_\beta)^*$ and action $(x \cdot f)(m) =
f(t(x) m)$ for $m \in M(\alpha),x \in u$ and where $t$ is the antigraded
antiautomorphism of $u$ such that $t(E_i)=F_i$ and $t(F_i)=E_i$ for
$i=1,2$ \cite[Remark after Theorem 5.1]{Holmes}.
In our case, $M^-(\alpha)$ is isomorphic to the $Z^-(\alpha)$ of
\cite{Holmes} as explained in the proof of their Theorem 5.1.  Since the
simple $u$-modules are self-dual, $M^-(\lambda)$ has simple socle
$L(\lambda)$.

For each $0 \leq a \leq p-2$ there is a non-semisimple singular block of
$u$ containing the simple modules $L(p-1,a), L(a,p-2-a)$ and $L(p-2-a,
p-1)$.  From now on we fix such an $a$ and write \[\mu=(p-1,a),\,\,\,
\lambda=(a,p-2-a),\,\,\, \gamma=(p-2-a,p-1).\]  The submodule lattices
for the Verma modules corresponding to these weights were determined in
\cite[Theorems 2.4, 2.5, 2.6]{Xi} and \cite[Theorems 3.3, 3.4]{Irving}.
The following diagrams display these submodule lattices; beneath each
composition factor $L$ is an element of the Verma module which is a
highest weight vector for $L$ modulo the sum of the submodules below it
in the lattice.

\[
  \xymatrixrowsep{0.2cm} \xymatrixcolsep{0.2cm}
  \xymatrix{
    & \txt{$L(\gamma)$\\$v_\gamma$}   \ar@{-}[dr] \ar@{-}[dl]& \\
     \txt{$L(\lambda)[-p,p]$\\$F_1^{(p-1-a)}v_\gamma$} \ar@{-}[dr] & & \ar@{-}[dl]
    \txt{$L(\lambda)[0,-p]$\\$\ad(F_2^{(p)})(F_1^{(p-1-a)})v_\gamma$}\\
    & \txt{$L(\mu)[-p,0]$ \\ $F_2^{(p-1-a)}F_1^{(p-1-a)}v_\gamma$} & } \]

\[
  \xymatrixrowsep{0.2cm} \xymatrixcolsep{0.2cm}
  \xymatrix{
    & \txt{$L(\mu)$\\$v_\mu$} \ar@{-}[dr] \ar@{-}[dl]& \\
    \txt{$L(\lambda)[-p,0]$\\$\ad(F_1^{(p)})(F_2^{(a+1)})v_\mu$} \ar@{-}[dr]
    & & \ar@{-}[dl] \txt{$L(\lambda)[p,-p]$\\$F_2^{(a+1)}v_\mu$} \\
    & \txt{$L(\gamma)[0,-p]$\\$F_1^{(a+1)}F_2^{(a+1)} v_\mu$} & } \]
\[
  \xymatrixrowsep{0.2cm} \xymatrixcolsep{0.2cm}
  \xymatrix{
    & \txt{$L(\lambda)$\\$v_\lambda$} \ar@{-}[dr] \ar@{-}[dl]& \\
    \txt{$L(\gamma)[-p,0]$\\$F_1^{(a+1)}v_\lambda$} \ar@{-}[dr] & & \ar@{-}[dl]
    \txt{$L(\mu)[0,-p]$\\$F_2^{(p-1-a)}v_\lambda$} \\
    &\txt{$
      L(\lambda)[-p,-p]$\\$F_2^{(a+1)}\ad(F_1^{(p)})(F_2^{(p-1-a)})v_\lambda$} & } \]

The weight-graded composition structure of the dual Verma modules
follows immediately from these.

\subsection{Verma modules are $K$-gradable}

 \begin{lem} \label{ext1}
  $\Ext^1_u(L(\gamma),L(\mu))=\Ext^1_u(L(\mu),L(\gamma))=\Ext^{1}_u(L(\lambda),L(\lambda))=0$.
\end{lem}
\begin{proof}

   Consider the first of these Ext groups.  Using the decomposition
   (\ref{decomp}) it is enough to prove that each of the weight-graded Ext
   groups $\Ext^{\textrm{wt},1}_u(L(\gamma),  L(\mu)[\beta])$ for
   $\beta \in
   X$ are zero.  Let $\supp L(\alpha)$ denote the support of $L(\alpha)$ in
   $X$, a convex set.  If
   \[ 0 \to L(\mu)[\beta] \to M \to L(\gamma) \to 0 \]
   is a non-split exact sequence of graded modules, there is a vertex of $\supp
   L(\gamma)$ lying outside or on the boundary of $\beta + \supp
   L(\mu)$.

 Since $^wL(\alpha)\cong L(\alpha)$ for any $w\in W$ and any $\alpha$, by twisting this
 extension with some Weyl group element if necessary we may assume that
 an element of $M$ whose weight corresponds to this vertex is a
 highest weight vector generating $M$.  Thus $M$ is a quotient of
 $M(\gamma)$, which is impossible by the structure of $M(\gamma)$
 given above. The other Ext groups follow similarly.
\end{proof}

\begin{prop}
  Let $\alpha \in \{\lambda,\gamma,\mu\}$.  Then $M(\alpha)$ and
  $M^-(\alpha)$ are
  $K$-gradable.
\end{prop}

\begin{proof}
We give the proof for $M(\alpha)$, the proof for $M^-(\alpha)$ being
analogous.  The projective $P(\alpha)$ is $K$-graded, generated in degree
zero.  Choose a surjection $F:P(\alpha) \to M(\alpha)$ and let
$M(\alpha)_0 =
F(P(\alpha)_0)$, a
$u_0$-submodule of $M(\alpha)$.  Let $M(\alpha)_2 = \soc_u(
M(\alpha))$.

The quotient map $\rad M(\alpha) \to \rad M(\alpha) / \soc M(\alpha)$ is
split as a map of $u_0$-modules because $u_0$ is semisimple; let
$M(\alpha)_1 \subset M(\alpha)$ be the image
of a splitting map.
As $u_0$-modules,
\[ M(\alpha)|_{u_0} = M(\alpha)_0 \oplus
M(\alpha)_1 \oplus M(\alpha)_2 \] and $M(\alpha)_1$ is isomorphic as a $u_0$-module to the
restriction of the two middle composition factors of $M(\alpha)$ to
$u_0$.  We claim that
this is a
$K$-grading on $M(\alpha)$.  Since we had a decomposition of
$u_0$-modules, the only thing we need check is that $u_1 M(\alpha)_0 \subseteq
M(\alpha)_1$.

Let $l_0 \in M(\alpha)_0, x \in u_1$ and suppose $xl_0=l_1+l_2$ for $l_i \in
M(\alpha)_i$.  We will show $l_2=0$.  Suppose not, let $\soc M(\alpha)
\cong
L(\pi)$, and
write $u_0 = \bigoplus _\beta M_\beta$ where $M_\beta $ is a matrix
algebra over $k$ corresponding to the simple module $L(\beta)$.
Pick $p \in
M_\pi$ such that $p l_2=l_2$.  By Lemma~\ref{ext1},
$JP(\alpha)/J^2P(\alpha)$ has no $L(\pi)$ summands, so $M_\pi$ acts as
zero on $P(\alpha)_1=u_1P(\alpha)_0$ and on the $u_0$-module $M(\alpha)_1$.

Let $\hat l_0 \in P(\alpha)_0$ be such that $F(\hat l_0) = l_0$.  Then
$F(px \hat l_0) =l_2\neq 0$ so $p\cdot x \hat l_0 \neq 0$, contradicting
that $M_\pi$ acts as zero on $u_1 P(\alpha)_0$.  \end{proof}

Henceforth $M(\alpha)$ will have the $K$-grading constructed above, so that
its top composition factor is in degree zero, and $M^-(\alpha)$ will be
given the $K$-grading so that its socle $L(\alpha)$ is in degree zero and
its head in degree $-2$.
Thus the multiplicity of $L(\alpha)[i]$
as a composition factor of $M(\beta)$ equals the multiplicity of
$L(\alpha)[-i]$ as a composition factor of $M^-(\beta)$.

\section{Projective modules and Hilbert series} \label{proj_sec}
In this section we study the structure of the indecomposable
projective modules in our block.

\subsection{$K$-graded Verma filtrations}
We will determine the composition
factors of these
projectives in the $K$-graded module category using a Brauer-type
reciprocity result.  The results of  \cite{Holmes}
cannot be used directly since $b$ is not a $K$-graded subalgebra of $u$, however the
proofs of the reciprocity results follow the arguments of \cite{Holmes}
closely.

\begin{defn} A $K$-graded module $M$ is said to have a $K$-graded Verma filtration if there
is a sequence $0 = M_0 \subset M_1 \subset \cdots \subset M_s =
M$ of $K$-graded submodules with each quotient $M_r/M_{r-1}$
isomorphic in the $K$-graded category to $M(\alpha_r)[n_r]$ for some $\alpha_r \in X$ and some $n_r
\in \ZZ$.\end{defn}
Write
$[M : M(\alpha_r)[n_r]]$ for the number of $M_r/M_{r-1}$ which are
isomorphic as $K$-graded modules to $M(\alpha_r)[n_r]$
(it is not clear yet that this is
independent of the filtration chosen), and $[M:L(\alpha)[n]]$ for the
multiplicity of $L(\alpha)[n]$ as a composition factor of $M$ in the
category of $K$-graded modules.
Let $P(\alpha)$ be a projective cover of $L(\alpha)$.

\begin{lem}
  Let $\alpha \in \{\lambda, \mu, \gamma\}$.  Then  $P(\alpha)$ has a
  $K$-graded Verma filtration.\end{lem}

\begin{proof}
 More generally, suppose that $M$ is any $u$-module in our block which is
 projective on restriction to $b^-$.  By \cite[Theorem 4.4]{Holmes}
 there is an ungraded injection $\iota: M(\beta) \to M$ for some $\beta
 \in \{\lambda, \mu, \gamma\}$.
 We may decompose $\iota$ as a sum of $K$-graded
 homomorphisms
 $\iota_j$.  A homomorphism $M(\beta) \to M$ is injective if and
 only if it is non-zero
 on the simple socle of $M(\beta)$, and not all the $\iota_j$ can be
 zero on the socle since $\iota$ isn't.  Thus one of the $\iota_j$ is a
 $K$-graded injective module homomorphism $M(\beta) \to M$.  The
 result follows by induction on the dimension of $M$.
\end{proof}

\begin{prop}
  Let $\alpha, \beta\in \{\lambda, \gamma, \mu\}$ and $i \in \ZZ$.   Then
  \[ [P(\alpha) : M(\beta)[i]] = [M(\beta)[-i] : L(\alpha)]. \]
\end{prop}

\begin{proof}
  We first show for any $K$-graded module $M$ with a $K$-graded Verma
  filtration that
  \[ [M : M(\beta)[i] ] = \dim \Hom_u^K (M, M^-(\beta)[i] ). \]
The proof is by induction on the length of the Verma filtration.
By \cite[Lemma 4.2]{Holmes}, $\Ext^{n}_u(M(\alpha),M^-(\beta))$ is
isomorphic to $k$ if $n=0$ and $\alpha=\beta$ and is zero otherwise.
Using the Ext-version of (\ref{Kdecomp}) it follows that if $n\neq 0$ or
$\alpha \neq \beta$
then
$\Ext^{K,n}_u(M(\alpha),M^-(\beta)[i])=0$, and that there is a unique
$i$ such that
$\Hom^K_u(M(\alpha), M^-(\alpha)[i])$ is nonzero.  Since $M(\alpha)$ has
simple top $L(\alpha)$ in $K$-degree zero and $M^-(\alpha)$ has simple
socle $L(\alpha)$ in $K$-degree zero, this is $i=0$
This proves the case where the Verma filtration has length one.

Now let
\[ 0 \to X \to M \to M(\alpha)[r] \to 0 \]
be an exact sequence of graded modules, where $X$ has a $K$-graded
Verma filtration.  For any $\beta, j$ we can apply $\Hom_u^K( -,
M^-(\beta)[j])$ and get a long exact sequence in which the Ext groups
vanish by \cite[Lemma 4.2]{Holmes} again.  So
\[ 0 \to \Hom^K_u(M(\alpha)[r], M^-(\beta)[j]) \to \Hom^K_u(M,
M^-(\beta)[j]) \to \Hom^K_u(X, M^-(\beta)[j]) \to 0 \]
is exact.
$[M: M(\beta)[i]]$ is the same as $[X: M(\beta)[i]]$ except if $\beta =
\alpha, i=r$ when it is one larger.  The result follows by induction.

Take $M = P(\alpha)$.  Then $\dim \Hom^K_u(P(\alpha), M^-(\beta)[i])$ is
the multiplicity of $L(\alpha)$ as a composition factor of
$M^-(\beta)[i]$, which is equal to the multiplicity of $L(\alpha)$ as a
composition factor of $M(\beta)[-i]$ by the remark at the end of the
previous section.
\end{proof}
Applying this result and the structure of the Vermas given previously
gives the factors in a $K$-graded Verma filtration of our projectives.
\begin{center} \begin{tabular}{l|l}
    Projective & $K$-graded Verma factors\\
    \hline
    $P(\lambda)$ &$M(\lambda)$, $M(\mu)[1]$,
    $M(\mu)[1]$,$M(\gamma)[1]$,
    $M(\gamma)[1]$, $M(\lambda)[2]$
    \\
    $P(\gamma)$ &$M(\gamma)$, $M(\lambda)[1]$, $M(\mu)[2]$ \\
    $P(\mu)$ &$M(\mu)$,$M(\lambda)[1]$,  $M(\gamma)[2]$
  \end{tabular} \end{center}

\subsection{Hilbert series and the basic algebra}

\begin{defn} The Hilbert series of our block is the $3\times 3$
matrix $P(t)$ with rows and columns labelled by $\gamma, \lambda, \mu$
whose $\alpha,\beta$ entry is $\sum_i t^i \dim \Hom_u^K (P(\alpha)[i],
P(\beta))$.
\end{defn}

Equivalently, the $\alpha,\beta$ entry of $P(t)$ is the
power series in $t$ whose coefficient of $t^i$ is the multiplicity of
$L(\alpha)[i]$ as a composition factor of $P(\beta)$.  The $K$-graded structure of
the Verma modules determined in Section \ref{verma_section} combined with the table
above immediately give:
\begin{thm}\label{hilb}  The Hilbert series of our block is \[ P(t) = \begin{pmatrix}
    1+t^2+t^4 & 3t+3t^3 & 3t^2 \\
    3t+3t^3 & 1+10t^2+t^4 & 3t+3t^3 \\
    3t^2 & 3t+3t^3 & 1+t^2+t^4 \end{pmatrix}. \]
\end{thm}
The determinant of $P(t)$ is $(t-1)^6(t+1)^6$.
We record the following corollary on the graded structure of the
projectives for future use:

\begin{cor}\label{first_second}
The first and second $K$-graded pieces of the projectives in our block are
as follows.
\begin{center}
  \begin{tabular}{l|lll}
    $\alpha$ & $\lambda$ & $\gamma$ & $\mu$ \\
    \hline
    $P(\alpha)_1$ & $L(\gamma)^{\oplus 3} \oplus L(\mu)^{\oplus 3}$ &
    $L(\lambda)^{\oplus 3}$ & $L(\lambda)^{\oplus 3}$ \\
    $P(\alpha)_2$ & $L(\lambda)^{\oplus 10}$  & $L(\gamma) \oplus
    L(\mu)^{\oplus 3}$ & $L(\mu)\oplus L(\gamma)^{\oplus 3}$
  \end{tabular}
\end{center}
\end{cor}

\begin{defn} The basic algebra of our block is $E = \End_u
  (P(\mu)\oplus P(\gamma)\oplus P(\lambda))^{\textrm{op}}$.\end{defn}
Since $E$ is Morita
equivalent \cite[\S 2.2]{BensonI} to a block of the Koszul algebra $u$, it is also Koszul
with respect to the grading arising from the $K$-grading on the
projectives
\cite[F.3]{AJS}.  If we write $e_\alpha$ for the element of $E$ which is
the identity map on $P(\alpha)$ and zero on the other two summands, then
the coefficient of $t^i$ in the $\alpha, \beta$ entry of the Hilbert series above equals the
dimension of $e_\beta E_i e_\alpha$.

\begin{thm}
The Hilbert series of the Koszul dual \[E^! \cong \Ext_u^*(L(\mu)\oplus L(\gamma)\oplus L(\lambda),
L(\mu)\oplus L(\gamma)\oplus L(\lambda) )\] is:
\[  (t-1)^{-4}(t+1)^{-4}\begin{pmatrix}
1+4t^2+t^4 & 3t(t^2+1) & 6t^2 \\
3t(t^2+1) & 1+4t^2 + t^4 & 3t(t^2+1) \\
6t^2 & 3t(t^2+1) & 1+4t^2 + t^4 \end{pmatrix}. \]
\end{thm}

\begin{proof}
By \cite[Lemma 2.11]{BGS} the Hilbert series of $E^!$ is $P(-t)^{-T}$,
so the result follows from Theorem \ref{hilb}. \end{proof}
As power series, the diagonal entries are $\sum_k (k+1)^3 t^{2k}$ and
the off-diagonal entries are $\frac{1}{2}\sum_k
(k+1)(k+2)(2k+3)t^{2k+1}$ and $\sum_k (k+2)(k+1)k t^{2k}$.

\subsection{Weight-graded Verma filtrations}
Each projective has a filtration by shifts of Verma
modules in the weight-graded category, and we write
$[P(\alpha):M(\beta)[\delta]]$ for the number of factors in a
weight-graded Verma filtration of $P(\alpha)$ isomorphic in the
weight-graded category to $M(\beta)[\delta]$.
In \cite[Theorem 5.1]{Holmes} the authors prove a result determining
these multiplicities
in a category of graded $u$-modules, but it is stated only
for a $\ZZ$-grading obtained by flattening the weight grading. Their
proofs go through in the weight-graded category however, or we can
instead apply the $n=1$ case of \cite[Satz 3.8]{Jantzen_uber}, to get:

\begin{prop}
  Let $\alpha, \beta \in \{\lambda, \gamma, \mu\}$ and let $\delta \in
  X$.  Then
  \[  [P(\alpha):M(\beta)[\delta]] =
  [M(\beta)[\delta]:L(\alpha)]. \]
\end{prop}
Again, using the structure of the Vermas given earlier gives:
\begin{center}\begin{tabular}{l|l}
  Projective & Weight-graded Verma factors \\ \hline
  $P(\gamma)$ & $M(\gamma), M(\lambda)[p,0],
M(\mu)[0,p]$ \\
  $P(\lambda)$ & $M(\lambda),
M(\mu)[p,0], M(\mu)[-p,p], M(\gamma)[p,-p],
M(\gamma)[0,p], M(\lambda)[p,p]$
  \\
  $P(\mu)$ & $M(\mu), M(\lambda)[0,p], M(\gamma)[p,0].$
\end{tabular}\end{center}
This determines the composition structure of the projectives in the
weight-graded category.  Figures~\ref{Pgamma}, \ref{Pmu} and
\ref{Plambda} display these structures, where a double circle represents
a composition factor of multiplicity two and so on, and we have written
$\alpha$ instead of $L(\alpha)$.  The coloured diamonds join the
composition factors of a term in the Verma filtration of $P(\alpha)$,
blue, orange and teal diamonds are $M(\lambda)$, $M(\gamma)$ and
$M(\mu)$ respectively.
\begin{figure}

\begin{tikzpicture}
\node[fill=black,circle,inner sep=1pt,label=above:$\gamma$] at (0,0) {};
\node[fill=black,circle,inner sep=1pt,label=above:{$\mu[0,p]$}] at (0,1) {};
\node[fill=black,circle,inner sep=1pt,label=below:{$\lambda[0,-p]$}] at (0,-1) {};
\node[fill=black,circle,inner sep=1pt,label=right:{$\lambda[p,0]$}] at ({0.5*sqrt(3)}, 0.5) {};
\node[fill=black,circle,inner sep=1pt,label=left:{$\mu[-p,0]$}] at ({-0.5*sqrt(3)}, -0.5) {};
\node[fill=black,circle,inner sep=1pt,label=right:{$\mu[p,-p]$}] at ({0.5*sqrt(3)}, -0.5) {};
\node[fill=black,circle,inner sep=1pt,label=left:{$\lambda[-p,p]$}] at ({-0.5*sqrt(3)}, 0.5) {};
\draw [gray] (0,0) circle (3pt);
\draw [gray] (0,0) circle (5pt);
\draw [gray] ({0.5*sqrt(3)}, 0.5) circle (3pt);
\draw [gray] (0, -1) circle (3pt);
\draw [gray] ({-0.5*sqrt(3)}, 0.5) circle (3pt);
\draw[color=blue]
(0.02,-0.02) -- (0.02,-1) -- ({0.5*sqrt(3)},-0.5) -- ({0.5*sqrt(3)},0.48) -- cycle;
\draw[color=orange]
(-0.02,-0.02) -- ({-0.5*sqrt(3)}, 0.48) -- ({-0.5*sqrt(3)},-0.5) --
(-0.02,-1) -- cycle;
\draw[color=teal]
(0,0.02) -- ({0.5*sqrt(3)},0.52) -- (0,1) -- ({-0.5*sqrt(3)},0.52) -- cycle;
\end{tikzpicture}

\caption{$P(\gamma)$} \label{Pgamma} \end{figure}
\begin{figure}

\begin{tikzpicture}
  \node[fill=black,circle,inner sep=1pt,label=below:$\mu$] at (0,0) {};
  \node[fill=black,circle,inner sep=1pt,label=above:{$\lambda\lbrack
  0, p\rbrack$}] at (0,1) {};
  \node[fill=black,circle,inner sep=1pt,label=below:{$\gamma[0,-p]$}] at (0,-1) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\gamma[p,0]$}] at ({0.5*sqrt(3)}, 0.5) {};
  \node[fill=black,circle,inner sep=1pt,label=left:{$\lambda[-p,0]$}] at ({-0.5*sqrt(3)}, -0.5) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\lambda[p,-p]$}] at ({0.5*sqrt(3)}, -0.5) {};
  \node[fill=black,circle,inner sep=1pt,label=left:{$\gamma[-p,p]$}] at ({-0.5*sqrt(3)}, 0.5) {};
\draw [gray] (0,0) circle (3pt);
\draw [gray] (0,0) circle (5pt);
\draw [gray] ({0.5*sqrt(3)}, -0.5) circle (3pt);
\draw [gray] (0, 1) circle (3pt);
\draw [gray] ({-0.5*sqrt(3)}, -0.5) circle (3pt);
\draw [color=blue]
(-0.02,0.02)--(-0.02,1)--({-0.5*sqrt(3)},0.5)--({-0.5*sqrt(3)},-0.48)--cycle;
\draw [color=orange]
(0.02,
0.02)--(0.02,1)--({0.5*sqrt(3)},0.5)--({0.5*sqrt(3)},-0.48)--cycle;
\draw[color=teal]
(0,-0.02)--({0.5*sqrt(3)},-0.52)--(0,-1)--({-0.5*sqrt(3)},-0.52)--cycle;
\end{tikzpicture}

\caption{$P(\mu)$} \label{Pmu} \end{figure}
\begin{figure}

\begin{tikzpicture}
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0.1cm]180:{$\lambda$}}]  at (0,0) {};
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0.15cm]90:{$\gamma[0,p]$}}] at (0,1) {};
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0.15cm]270:{$\mu[0,-p]$}}] at (0,-1) {};
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0cm]10:{$\mu[p,0]$}}] at ({0.5*sqrt(3)}, 0.5) {};
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0cm]250:{$\gamma[-p,0]\,\,$}}] at ({-0.5*sqrt(3)}, -0.5) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\gamma[p,-p]$}] at ({0.5*sqrt(3)}, -0.5) {};
  \node[fill=black,circle,inner sep=1pt,label={[label
  distance=0cm]140:{$\mu[-p,p]\,\,$}}] at ({-0.5*sqrt(3)}, 0.5) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\lambda[2p,-p]$}] at ({sqrt(3)},0) {};
  \node[fill=black,circle,inner sep=1pt,label=left:{$\lambda[-2p,p]$}] at ({-sqrt(3)},0) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\lambda[p,p]$}] at ({0.5*sqrt(3)}, 1.5) {};
  \node[fill=black,circle,inner sep=1pt,label=left:{$\lambda[-p,-p]$}] at ({-0.5*sqrt(3)}, -1.5) {};
  \node[fill=black,circle,inner sep=1pt,label=right:{$\lambda[p,-2p]$}] at ({0.5*sqrt(3)}, -1.5) {};
  \node[fill=black,circle,inner sep=1pt,label=left:{$\lambda[-p,2p]$}] at ({-0.5*sqrt(3)}, 1.5) {};
\draw [gray] (0,0) circle (2pt);
\draw [gray] (0,0) circle (3pt);
\draw [gray] (0,0) circle (4pt);
\draw [gray] (0,0) circle (5pt);
\draw [gray] (0,0) circle (6pt);
\draw [gray] ({0.5*sqrt(3)}, -0.5) circle (3pt);
\draw [gray] (0, 1) circle (3pt);
\draw [gray] ({-0.5*sqrt(3)}, -0.5) circle (3pt);
\draw [gray] ({0.5*sqrt(3)}, 0.5) circle (3pt);
\draw [gray] (0, -1) circle (3pt);
\draw [gray] ({-0.5*sqrt(3)}, 0.5) circle (3pt);

\pgfmathsetmacro{\EEE}{0.04}
\draw [color=blue]
($(0,0)+\EEE/3*({0.5*sqrt(3)},1.5)$)--($(0,1)+\EEE/3*({0.5*sqrt(3)},1.5)$)
-- ({0.5*sqrt(3)},1.5) --
($({0.5*sqrt(3)},0.5)+\EEE/3*({0.5*sqrt(3)},1.5)$) -- cycle;
\draw[color=teal]
($(0,0)+\EEE*(1,0)$)--($({0.5*sqrt(3)},0.5)+\EEE*(1,0)$)
-- ({sqrt(3)},0) --
($({0.5*sqrt(3)},-0.5)+\EEE*(1,0)$) -- cycle;
\draw[color=orange]
($(0,0)+\EEE/3*({0.5*sqrt(3)},-1.5)$)--($(0,-1)+\EEE/3*({0.5*sqrt(3)},-1.5)$)
-- ({0.5*sqrt(3)},-1.5) --
($({0.5*sqrt(3)},-0.5)+\EEE/3*({0.5*sqrt(3)},-1.5)$) -- cycle;
\draw[color=blue]
($(0,0)-\EEE/3*({0.5*sqrt(3)},1.5)$)--($(0,-1)-\EEE/3*({0.5*sqrt(3)},1.5)$)
-- ({-0.5*sqrt(3)},-1.5) --
($({-0.5*sqrt(3)},-0.5)-\EEE/3*({0.5*sqrt(3)},1.5)$) -- cycle;
\draw[color=teal]
($(0,0)-\EEE*(1,0)$)--($({-0.5*sqrt(3)},-0.5)-\EEE*(1,0)$)
-- ({-sqrt(3)},0) --
($({-0.5*sqrt(3)},0.5)-\EEE*(1,0)$) -- cycle;
\draw[color=orange]
($(0,0)-\EEE/3*({0.5*sqrt(3)},-1.5)$)--($(0,1)-\EEE/3*({0.5*sqrt(3)},-1.5)$)
-- ({-0.5*sqrt(3)},1.5) --
($({-0.5*sqrt(3)},0.5)-\EEE/3*({0.5*sqrt(3)},-1.5)$) -- cycle;
\end{tikzpicture}

\caption{$P(\lambda)$}\label{Plambda}\end{figure}

\section{Action by derivations}\label{sec:action}
The aim of this section is to show that there is a non-trivial action of
$\ssl_3(k)$ by derivations on the basic algebra of our block: that is, a
non-zero Lie homomorphism from $\ssl_3(k)$ into the Lie algebra of derivations of
$E$.  Rather than work with the basic algebra as defined
in the previous section, we use an equivalent definition.  For $\alpha
\in \{\lambda,\gamma,\mu\}$ choose idempotents $\eps_\alpha$ belonging
to a set of primitive orthogonal idempotents
of $u$ such that the image of $\eps_\alpha$ in $u/\rad(u)$ is a highest
weight vector for $L(\alpha)$.  Let $\eps = \eps_\lambda+\eps_\gamma +
\eps_\mu$.  Then $E$ is isomorphic to
$\eps u \eps$ \cite[Lemma 1.3.3 (ii)]{BensonI}.

For elements $x,y$ of some algebra we write $\ad(x)(y)$ for $xy-yx$.
The following formulas holding in $D_2$ follow readily from
\cite[\S26.2]{Humphreys}:
\[
\begin{array}{ll}
  \ad(E_i^{(p)}) ( F_i) = (H_i+1)E_i^{(p-1)} & \ad(F_i^{(p)})(E_i) =
  -F_i^{(p-1)}(H_i+1) \\
  \ad(E_1 ^{(p)}) (E_2) = E_1^{(p-1)} E_{12} & \ad(F_1 ^{(p)})(F_2) =
-F_1^{(p-1)}F_{12} \\
\ad(E_i^{(p)})(F_{12}) = -F_2 E_i^{(p-1)} & \ad(F_i ^{(p)})(
E_{12})=F_i^{(p-1)}E_2 \\
\ad(E_i^{(p)})(H_j) = 0 & \ad(F_i ^{(p)})(H_j) = 0
\end{array}
\]
for all $i,j$.  Thus the derivations $\ad(E_i^{(p)})$ and
$\ad(F_i^{(p)})$ of $D_2$, and the Lie subalgebra of $\der(u)$ they
generate, preserve $u$.
Unlike in the rank one case, this does not make $u$ into an
$\ssl_3(k)$-module,  the problem being that $\Fr: D_2 \to u$ is not obviously split as an
algebra homomorphism.
For example, although $E_i^{(p)}$ and
$F_i^{(p)}$ generate a subalgebra of $D$ isomorphic to $u(\ssl_2(k))$, the
elements $[E_1^{(p)}, F_1^{(p)}]$ and $[E_2^{(p)}, F_2^{(p)}]$ do not
commute in $D_2$: their commutator is a non-zero element of $u$.  In this
way we get a
Lie subalgebra of the first Hochschild cohomology $\der(u)/\ider(u)$
isomorphic to $\ssl_3(k)$, but because the Levi splitting theorem fails in
characteristic $p$ the surjection from the corresponding subalgebra of
$\der(u)$ onto $\ssl_3(k)$ is not guaranteed to split.  Some of the other
$\ssl_3(k)$ relations do hold in the hyperalgebra:

\begin{lem}\label{D2rels} In $D_2$,
  \begin{enumerate}
    \item $[E_i^{(p)},[E_1^{(p)},E_2^{(p)}]] =
  0=[F_i^{(p)},  [F_1^{(p)}, F_2^{(p)}]]$ for $i=1,2$.
\item $[ [E_i^{(p)}, F_i^{(p)}], E_i^{(p)}]=2E_i^{(p)}$ and $[
    [E_i^{(p)}, F_i^{(p)}],F_i^{(p)}=-2F_i^{(p)}$ for $i=1,2$.
\end{enumerate}
\end{lem}

\begin{proof}
  We prove the $E$ version of the first identity for $i=1$ only.
  In any universal enveloping algebra, if $X,Y$ are Lie algebra elements
  such that $[X,Y]=Z$ commutes with $X$ and $Y$ then
  \[ [X^{(n)},Y^{(m)}] = \sum_{i=1}^{\min (m,n)} (-1)^{i+1}
X^{(n-i)}Y^{(m-i)}Z^{(i)}. \]
Applying this twice gives
\begin{equation*} [E_1^{(p)}, [E_1^{(p)},E_2^{(p)}]] = \sum_{i=1}^{p-1}
  \sum_{j=1}^{p-i} (-1)^{i+j} E_1^{(2p-i-j)}E_2^{(p-i-j)}E_{12}^{(j)}
  {i+j \choose i} {2p-i-j \choose p-i}. \end{equation*}
The product of binomial coefficients vanishes mod $p$.
The second identity is proved using the same method employed for its
quantum analogue in \cite[Lemma 1]{Chin}.
\end{proof}

The Lie subalgebra of $\der(u)$ generated by the derivations
$\ad(E_i^{(p)})$ and $\ad(F_i^{(p)})$ preserves
the radical of $u$:
\begin{lem}
  Let $d \in D_2, x \in \rad(u)$ and suppose that $dx-xd \in u$.  Then
  $dx - xd \in \rad(u)$.
\end{lem}

\begin{proof}
  For each simple $u$-module $L$ there is a simple $D$-module $L'$
  such that $L'|_u \cong L$.  The result follows by the characterisation
  of $\rad(u)$ as the set of elements annihilating all simple
  $u$-modules.
\end{proof}

\begin{lem} \label{inJ}
  Let $x$ be a weight-homogeneous element of $u$ with weight degree $\pm
  p\alpha_i$.  Then $x\eps \in J\eps$ and $\eps x \in \eps J$.
\end{lem}

\begin{proof}
  For $\alpha \in \{\mu, \gamma,\lambda\}$ we have
  $u\eps_\alpha/J\eps_\alpha\cong L(\alpha)$ with  $\eps_\alpha+ J\eps$ being a
  highest weight vector.  This is killed by any element of the given
  weight degree.
\end{proof}

  Each derivation $\delta$ of $u$ gives rise to
a derivation $\delta_\eps$ of $\eps u \eps$ by $\delta_\eps(x):= \eps \delta(x)
\eps$ which satisfies $\delta_\eps(\eps)=0$.  Let $e_i$ and $f_i$ be the
derivations of
$\eps u \eps$ arising in this way from $\ad(E_i^{(p)})$ and $\ad(F_i^{(p)})$.
\begin{lem}\label{add} If $j \in J$ then $e_i (\eps j \eps) = \eps
  \ad(E_i^{(p)})(j) \eps$ modulo $\eps J^2 \eps$. \end{lem}

\begin{proof}
  $e_i (\eps j \eps) = \eps \ad(E_i^{(p)})(\eps) j \eps + \eps
  \ad(E_i^{(p)})(j) \eps + \eps j \ad(E_i^{(p)})(\eps) \eps$, and the
  first and last terms on the right lie in $\eps J^2 \eps$ by the
  previous lemma. \end{proof}

$\eps u \eps$, being Koszul, is isomorphic to the graded algebra arising
from its radical filtration $\eps u \eps \supset \eps J \eps \supset
\eps J^2 \eps \supset \cdots$.  We write $\gr(\eps u \eps)$ for this
graded algebra, so
\[ \gr(\eps u \eps) = \frac{\eps u \eps}{\eps J \eps} \oplus \frac{\eps
J \eps}{\eps J^2 \eps} \oplus \cdots \]

A radical-preserving derivation $\delta$  of $\eps u \eps$ induces a
grading-preserving derivation $\delta'$ of $\gr(\eps u \eps)$.
Define a linear map $\phi: \ssl_3(k) \to \der(\gr (\eps u \eps))$ by
putting
$\phi(E_i)=e_i'$, $\phi(F_i)=f_i'$, $\phi(H_i)=[e_i',f_i']$,

$\phi(E_{12})=[e_1',e_2']$,
$\phi(F_{12})=[f_2',f_1']$.

\begin{lem}
  $\phi$ is a Lie homomorphism.
\end{lem}

\begin{proof}
  We check the relations in $\ssl_3(k)$ hold for the $\phi(E_i)$ and
  $\phi(F_i)$.  Any relation which
  holds in $D_2$ when $E_i$ is replaced by $E_i^{(p)}$, $F_i$ by
  $F_i^{(p)}$ is satisfied: for example $[E_1,F_2]=0$ and
  \[ [\phi(E_1),\phi(F_2)](\eps j \eps) \equiv \eps [\ad(E_1^{(p)}),
  \ad(F_2^{(p)})](j) \eps \equiv \eps \ad([E_1^{(p)},F_2^{(p)}])(j) \eps
\equiv 0 \]
modulo $\eps J^2 \eps$, using Lemma~\ref{add}. Using Lemma 4.2, this
leaves only the relations (r1), (r2) and (r3) from Section
\ref{notation_section}.
The first of these follows from the other relations
using the Jacobi identity.  To get (r2) and (r3) note that $[ [E_1^{(p)}, F_1^{(p)}], E_2^{(p)}] + E_2^{(p)}$
lies in the kernel of $\Fr$, and in fact in $u$.  Thus
$-\ad(E_2^{(p)})$ and $\ad( [ [E_1^{(p)}, F_1^{(p)}], E_2^{(p)}])$
differ by an inner derivation $\ad(x)$ of $u$, where $|x|=p\alpha_2$.
The relation holds by Lemma \ref{inJ}, as do the other relations of this
form.
\end{proof}

If we multiply the weight degrees in $\ssl_3(k)$ by $p$ then $\phi$ respects the weight grading.

\begin{lem}
As representations of $\ssl_3(k)$ via $\phi$,
\[ \frac{\eps_\lambda J \eps_\mu}{\eps_\lambda J^2 \eps_\mu} \cong
  \frac{\eps_\gamma J \eps_\lambda}{\eps_\gamma J^2 \eps_\lambda} \cong
  L(0,1)\,\,\, \frac{\eps_\lambda J \eps_\gamma}{\eps_\lambda J^2
  \eps_\gamma}
  \cong
    \frac{\eps_\mu J \eps_\lambda}{\eps_\mu J^2 \eps_\lambda}
    \cong
  L(1,0). \]
\end{lem}
\begin{proof}
 We begin with the case $\eps_\lambda J \eps_\gamma / \eps_\lambda
 J^2 \eps_\gamma$.    Consider  $\eps_\lambda
 F_1^{(p-1-a)}\eps_\gamma$
This element lies in $\eps_\lambda J \eps_\gamma \setminus \eps_\lambda
 J^2 \eps_\gamma$
since under the surjection $u\eps_\gamma \to M(\gamma)$ induced
by $\eps_\gamma \mapsto v_\gamma$ it gets sent to an element with non-zero
$M(\gamma)_1$ component by the structure of $M(\gamma)$ given in Section
\ref{verma_section}.
Applying $\phi(F_2)$  to $\eps_\lambda F_1^{(p-1-a)}\eps_\gamma+
\eps_\lambda J^2 \eps_\gamma$ sends it to $\eps_\lambda
\ad(F_2^{(p)})(F_1^{(p-1-a)}) \eps_\gamma + \eps_\lambda J^2 \eps_\gamma$,
which is non-zero again by considering the map $u\eps_\gamma \to
M(\gamma)$.
This proves the claim about non-trivial action.

$(\eps_\lambda J \eps_\gamma) / (\eps_\lambda J^2
\eps_\gamma)$ is three-dimensional, and $P(\gamma)$ has only
$L(\lambda)$ composition factors shifted to weight
degrees $[p,0], [-p,p], [0,-p]$.  Since $\ssl_3(k)$ acts non-trivially
and respects this grading, the
only possibility is that $(\eps_\lambda J \eps_\gamma) / (\eps_\lambda
J^2
\eps_\gamma)$ is isomorphic to the simple module $L(1,0)$.

The $e_\lambda J e_\mu$ case can be dealt with similarly, using a
homomorphism to $M(\mu)$.  For $e_\gamma J e_\lambda$ we need to
construct a new $u$-module.  Let $M$ be the $b^+$-module which is
$M(\gamma)[-p,p]\oplus k_\lambda$ as an $X$-graded vector space, where
$k_\lambda = k\langle 1_\lambda\rangle$ is in weight degree $\lambda$,
with $b^+$ action on the $M(\gamma)[-p,p]$ summand defined by
restricting the $u$-action, $E_2\cdot 1_\lambda=0$ and $E_1 \cdot
1_\lambda = F_1^{(p-2-a)}v_\gamma$. We have a short exact sequence
\[ 0 \to M(\gamma)[p,-p]|_{b^+} \to M \to k_\lambda \to 0\]
which by Eckmann-Shapiro corresponds to a short exact sequence of
$u$-modules
\[ 0 \to M(\gamma)[p,-p] \to N \to M(\lambda) \to 0 \]
Let $\hat{v_\lambda}$ be a preimage in $N$ of $v_\lambda \in
M(\lambda)$, so that there is a surjection $P(\lambda) \to N$ sending
$\eps_\lambda$ to $\hat{v_\lambda}$.  Using the formula at the start of
Section \ref{sec:action}, we get that
$\ad(E_1^{(p)})(F_1^{(a+1)})\hat{v_\lambda}$ is a nonzero scalar
multiple of $v_\gamma$.  This can be used to show $E_1$ acts
non-trivially on $e_\gamma J e_\lambda/ e_\gamma J^2 e_\lambda$ as
before, and the $e_\mu J e_\lambda$ case is similar.
\end{proof}

\section{Presentation of the basic algebra} \label{sec:pres}

We now determine a presentation of
the basic algebra $E$.  Let $\mathbbm{k} = E_0 = k\langle e_\lambda, e_\gamma,
e_\mu\rangle$ where $e_\alpha$ is the identity map on $P(\alpha)$ and zero on
the other summands.  Let $V = E_1$.

By Koszulity $E$ is generated
over $\mathbbm{k}$ by $V$, and by the previous
section $E$ admits a grading-preserving $\ssl_3(k)$-action by derivations
such that as $\ssl_3(k)$-modules,
\[ e_\lambda V e_\gamma \cong e_\mu V e_\lambda \cong L(1,0) \,\,\,
e_\gamma V e_\lambda \cong e_\lambda V e_\mu \cong L(0,1). \]
Again by Koszulity, $E$ is quadratic in the sense that $E \cong
T_{\mathbbm{k}}(V)/(R)$ where $T_{\mathbbm{k}}$ denotes the tensor
algebra and $R\subseteq  V \otimes _\mathbbm{k} V$.  Because $\ssl_3(k)$ is
acting by grading-preserving derivations, $R$ is in fact a
$\ssl_3(k)$-submodule.

The $\alpha, \beta$ entry of the following table records the
$\ssl_3(k)$-module structure on $e_\alpha V\otimes_\mathbbm{k} V e_\beta$:

\begin{center}
\begin{tabular}{c|ccc}
  & $\gamma$ & $\lambda$ & $\mu$ \\
  \hline
 $\gamma$& $L(0,0)\oplus L(1,1)$ & &$L(0,1)\oplus L(2,0)$ \\
 $\lambda$ & & $L(0,0)^{\oplus 2} \oplus L(1,1)^{\oplus 2}$ & \\
 $\mu$ &$ L(1,0)\oplus L(0,2)$ & & $L(0,0) \oplus L(1,1)$ \end{tabular}
\end{center}

Write $v_{1,0}$, $w_{1,0}$ for highest weight vectors of $e_\lambda V
e_\gamma$ and $e_\mu V e_\lambda$, and $v_{0,1}, w_{0,1}$ for highest
weight vectors of $e_\gamma V e_\lambda$ and $e_\lambda V e_\mu$
respectively.

Corollary \ref{first_second} tells us the dimensions of each $e_\alpha
E_2 e_\beta$. For example, $e_\gamma E_2 e_\gamma$ is one-dimensional,
and so $e_\gamma Re_\gamma$ must be a $\ssl_3(k)$-submodule of
codimension one in $e_\gamma V\otimes_\mathbbm{k} V e_\gamma$, and therefore consists of the $L(1,1)$ summand from
the top row of the
previous table, which is generated as a $\ssl_3(k)$-module by
$v_{0,1}\otimes v_{1,0}$. Similarly, $R$ contains exactly the  $L(2,0)$
summand on the top row of the above table, and the $L(0,2)$ and
$L(1,1)$ summands on the bottom row.

We still need to identify $e_\lambda R e_\lambda$, which must be an
eight-dimensional submodule of $e_\lambda V\otimes_\mathbbm{k}
Ve_\lambda$ by Corollary \ref{first_second}.  It is therefore isomorphic
to $L(1,1)$ as an $\ssl_3(k)$-module, and so is generated as an
$\ssl_3(k)$-module by some element of
the form $r v_{1,0}\otimes v_{0,1} + s w_{0,1}\otimes w_{1,0}$ for $r,s \in
k$.

Suppose $r=0$.  Then in the quadratic dual algebra
\[ E^! \cong \Ext^*_u (L(\lambda)\oplus L(\gamma)\oplus
L(\mu),L(\lambda)\oplus L(\gamma)\oplus L(\mu)) \]
we have $v_{1,0}^* \cdot v_{0,1}^* = 0$.  Now $v_{1,0}^*$ corresponds to
a weight-graded extension of $L(\lambda)$ by $L(\gamma)[p,0]$ and
$v_{01}^*$ corresponds to a weight-graded extension of $L(\gamma)$ by
$L(\lambda)[0,p]$. If this product were zero in the Ext-ring, there
would be a weight-graded uniserial module with top $L(\lambda)$, middle
composition factor $L(\gamma)[-p,0]$ and socle $L(\lambda)[-p,-p]$ as in
\cite[Proposition 2.3(a)]{BC}.

Such a module is necessarily highest weight, hence a quotient
of $M(\lambda)$, but the structure of $M(\lambda)$ given in Section
\ref{verma_section} shows it has no such quotient. Therefore $r\neq 0$, and similarly $s \neq 0$. By
rescaling $v_{0,1}$ and the $\ssl_3(k)$-submodule it generates, we may
assume $r=s=1$.

We have proved:

\begin{thm} \label{presentation}
  $E\cong T_\mathbbm{k}(V)/(R)$, where $V$ and $\mathbbm{k}$ are as
  above and $R$ is the $\ssl_3(k)$-submodule of $V\otimes_\mathbbm{k} V$
  generated by $v_{0,1}\otimes v_{1,0}$,  $w_{1,0}\otimes v_{1,0}$, $w_{1,0}
  \otimes w_{0,1}$, $v_{01}\otimes w_{0,1}$, $v_{1,0}\otimes v_{0,1} +
  w_{0,1}\otimes w_{1,0}$.
\end{thm}
The following table displays $R$ as a submodule of
$V\otimes_\mathbbm{k}V$:
\begin{center}
  \begin{tabular}{l|lcl}
    & $\gamma$ & $\lambda$ & $\mu$ \\ \hline
    $\gamma$ & $L(1,1)$ & & $L(2,0)$ \\
    $\lambda$& & Diagonal $L(1,1)$ & \\
    $\mu$ & $L(0,2)$ & & $L(1,1)$
  \end{tabular}
\end{center}

\section{Hochschild cohomology} \label{hoch_section}
Any non-semisimple block of $u(\ssl_2(k))$ is Morita equivalent to a
smash product $\Gamma = kC_2 \ltimes k[x,y]/(x^2,y^2)$ where the cyclic
group $C_2=\langle g \rangle$ acts by $gxg=-x$ and $gyg=-y$.  This can
be used to show that the centre $Z(\Gamma)$ has dimension 3 and
$\HH^1(\Gamma) \cong \mathfrak{gl}_2(k)$ as a Lie algebra.

Since our block algebra is $K$-graded, its Hochschild cohomology is
bigraded, once by homological degree and once from this $K$-grading,
referred to as the internal grading.
In
this section we calculate the centre of our basic algebra and the
internal degree zero part of the first Hochschild cohomology, showing
that as in the $\ssl_2$ case, it is isomorphic to the general linear Lie
algebra.

Let $\Lambda = T_\mathbbm{k}(V)/(R)$.  Let $v_{-1,0}=F_1F_2v_{0,1}$, $v_{1,-1}=F_2v_{0,1}$,
$v_{-1,1}=-F_1v_{1,0}$, $v_{0,-1}=F_2F_1v_{1,0}$ in $\Lambda$, and make similar
definitions for the $w$s.
The relations for $\Lambda$ are then
\begin{center}
  \begin{tabular}{ll}
    $e_\gamma R e_\gamma$ & $v_\alpha v_{-\beta} = 0$ for  $\alpha \neq
    \beta$, \\
    &  $v_\alpha v_{-\alpha} = v_\beta v_{-\beta}$ for all $\alpha,\beta$. \\
    $e_\mu R e_\mu$ &  $w_{-\alpha} w_\beta=0$ for $\alpha \neq \beta$, \\
    &  $w_{-\alpha} w_{\alpha}=w_{-\beta} w_{\beta}$ for all
    $\alpha,\beta$. \\
    $e_\mu R e_\gamma$ & $w_{-\alpha} v_{-\beta} + w_{-\beta}
    v_{-\alpha} =0$ for all $\alpha,\beta$. \\
$ e_\gamma R e_\mu$ &
 $v_\alpha w_\beta + v_\beta w_\alpha =0$ for all $\alpha,\beta$. \\
$e_\lambda R e_\lambda$ &
$v_{-\alpha} v_\beta + w_\beta w_{-\alpha}$ for   $\beta \neq \alpha$, \\
&
$v_{-\alpha}v_{\alpha}+w_{\alpha}w_{-\alpha}=v_{-\beta}v_{\beta}+w_{\beta}w_{-\beta}$
for all $\alpha,\beta$
 \end{tabular}\end{center}
where $\alpha,\beta \in \{(1,-1), (-1,0), (0,1)\}$.
Let
\begin{center}
\begin{tabular}{llll} $\mathcal{V} = \sum v_\alpha v_{-\alpha}$ &
 $ \mathcal{W} =\sum
w_{-\alpha} w_\alpha $ \\
$z_{\gamma\lambda}  = 3v_{0,1}v_{0,-1}
+\mathcal{V}-2\mathcal{W}$ & $ z_{\mu\lambda} = 3w_{0,-1}w_{0,1} +
\mathcal{W}-2\mathcal{V}$\end{tabular}
\end{center}
where both sums are over $\alpha \in \{(-1,1), (1,0),
(0,-1)\}$.  Let  $z_\alpha$ span the socle of $\Lambda e_\alpha$ for $\alpha =
\lambda, \mu, \gamma$.

\begin{prop}The centre $Z(\Lambda)$ has basis $1, z_\lambda, z_\mu, z_\gamma,
  z_{\mu\lambda}, z_{\gamma\lambda}$.\end{prop}

\begin{proof}
We first show that the given elements are central.   Each of the elements
$z_\alpha$ lie in the centre because they are killed on both sides by
any element of $V$. Using the relations above:
\begin{center}
  \begin{tabular}{l|llll}
    $v_x$ & $v_x \mathcal{V}$ & $\mathcal{V}v_x$ & $v_x\mathcal{W}$ &
    $\mathcal{W}v_x$ \\
    \hline
    $e_\lambda V e_\gamma$ & 0 & $v_x v_{-x} v_x$ & 0 & $2v_x v_{-x}
    v_x$ \\
    $e_\gamma V e_\lambda$ & $v_x v_{-x}v_x$ & 0 & $2v_xv_{-x}v_x$ & 0
    \\
    \multicolumn{5}{c}{ }  \\
    $w_x$ & $w_x \mathcal{V}$ & $\mathcal{V}w_x$ & $w_x\mathcal{W}$ &
    $\mathcal{W}w_x$ \\
    \hline
    $e_\lambda V e_\mu$ & $2w_xw_{-x}w_x$ & 0& $w_xw_{-x}w_x$ & 0
     \\
     $e_\mu V e_\lambda$ & 0 & $2w_xw_{-x}w_x$ & 0 & $w_xw_{-x}w_x$.
  \end{tabular}
\end{center}
From these it follows easily that $z_{\gamma \lambda}$ and
$z_{\mu\lambda}$ are central.

Any element of the centre lies in $e_\lambda \Lambda e_\lambda +
 e_\gamma \Lambda e_\gamma + e_\mu \Lambda e_\mu$, since this is the condition to
 commute with the idempotents $e_\mu, e_\gamma$ and $e_\lambda$.
Since it is homogeneous with respect to the $K$-grading, it can be
non-zero only in degrees $0,2,4$ by considering the Hilbert series for
$\Lambda$.  Furthermore derivations preserve the centre of an algebra,
so $Z(\Lambda)$ is a
$\ssl_3(k)$-submodule.

From the Hilbert series, $e_\alpha \Lambda_2 e_\alpha$ is
one-dimensional for $\alpha=\mu, \gamma$ and $L(1,1)\oplus
L(0,0)^{\oplus 2}$ as an $\ssl_3(k)$-module for $\alpha = \gamma$.
The element $v_{10}v_{01}$ does not commute with $v_{10}$, thus the
degree two part of the centre is contained in the trivial summands. No
nonzero element of the two-dimensional subspace spanned by $v_{0,1}v_{0,-1}$ and
$w_{1,0}w_{-1,0}$ commutes with both $v_{01}$ and $w_{10}$, so the degree two part of the centre has dimension at most two.
It is therefore spanned by
$z_{\gamma\lambda}$ and $z_{\mu\lambda}$.  Any element of $\Lambda_4$ is
in the span of $z_\lambda$, $z_\mu$ and $z_\gamma$, so they certainly
form a basis of the degree four part of the centre.  Lastly a central
element of $\Lambda_0$ must be a scalar multiple of the identity, so we
are done.
\end{proof}

We already know some non-inner $K$-grading-preserving derivations of $\Lambda$, namely those
induced by the action of $\ssl_3(k)$ and the grading derivation $\Delta$
defined by $\Delta(l) = nl$ for $l \in \Lambda_n$.

\begin{prop} The internal degree zero part of $\HH^1(\Lambda)$ is
  isomorphic to $\mathfrak{gl}_3(k)$ and is spanned by the
  image of $\Delta$ together with the images of the derivations arising
  from the $\ssl_3(k)$-action.
\end{prop}
\begin{proof}
  Let $\delta$ be a non-inner derivation of $\Lambda$ which preserves
  the $K$-grading.

  Since $\delta$ preserves $\Lambda_0=\langle
  e_\lambda,e_\gamma,e_\mu\rangle$ and the $e_\alpha$ are idempotent, it
  must act as zero on $\Lambda_0$. Therefore $\delta$
  is determined by its
  action on $\Lambda_1$, and
  must send $e_\alpha\Lambda_1 e_\beta$ to
  $e_\alpha\Lambda e_\beta$ for each $\alpha,\beta$.

  So $\delta$ is
  determined by the four linear maps $\delta_{\alpha\beta} \in
\End_k(e_\alpha
  \Lambda_1 e_\beta)$, for $(\alpha, \beta) = (\gamma, \lambda), (\lambda,
  \gamma), (\mu,\lambda), (\lambda, \mu)$.  Any such maps extend
  uniquely to a derivation of $T_\mathbbm{k}(V)$.  The condition for
  $\delta$ to be a derivation is that these maps, when extended to a
  derivation on $T_{\mathbbm{k}}(V)$, must preserve $R \subseteq V
  \otimes _{\mathbbm{k}} V$.  Identify $e_\gamma V e_\lambda$ and
  $e_\lambda V e_\mu$ with $L(0,1)$, and $e_\lambda V e_\gamma$ and
  $e_\mu V e_\lambda$ with $L(0,1)^*$.  Then $e_\mu R e_\gamma$ is the
  symmetric square of $L(0,1)$, and if $a$ and $b$ are linear
  endomorphisms of $L(0,1)$ then $1\otimes a + b \otimes 1$ preserves
  the symmetric square of $L(0,1)$ if and only if $a$ and $b$ differ by
  a scalar multiple of the identity.

  Thus $\delta_{\mu \lambda}- \delta_{\lambda \gamma}$ and $\delta_{\gamma \lambda} -
  \delta_{\lambda \mu}$ are scalar multiples of the identity.

  Write $\delta_{\mu \lambda} = rI + x$ and $\delta_{\lambda \mu}=
  sI + y$ where $I$ is the identity, $r,s\in k$ and $x$ and $y$ are the
  linear endomorphisms of $L(0,1)$ and $L(0,1)^*$ induced by the action
  of elements
  $X,Y \in \ssl_3(k)$.  Note that under our
  identifications, $e_\mu R e_\mu$ is
  the kernel of the evaluation map $\nu: L(0,1) \otimes L(0,1)^* \to k$.
  We claim that the endomorphism \[(\delta_{\mu \lambda}-rI)\otimes 1 + 1 \otimes
  (\delta_{\lambda \mu}-sI)=x\otimes 1 + 1 \otimes y\] has image contained in $\ker \nu$.
  Note $L(0,1)\otimes L(0,1)^* = \ker \nu \oplus L(0,0)$ where
  the trivial summand is spanned by $v=\sum v_{ij}\otimes v_{ij}^*$, and
  that $\ker \nu$ belongs to $R$ so is preserved by this endomorphism
  since it differs from the action of $\delta$ by a scalar multiple of
  the identity.  Both $x\otimes 1$ and $1\otimes y$ act on $v$ by scalar
  multiplication by their trace, which is zero, proving the claim.

  This shows that for
  any $u \in L(0,1)$ and $f \in L(0,1)^*$,
  \[ f(x)(u) +
  (y(f))(u) = 0\]
  that is, $(Y\cdot f) (u) = f(-X\cdot u)$ for all $f,u$.  But by
  definition of the action on the dual space, $(Y\cdot f)(u) = f(-Y\cdot
  u)$.  It follows
  $X=Y$.  We now have
  \begin{align*}
  \delta_{\mu \lambda} &= rI + \rho(X) &    \delta_{\lambda \gamma} &=r'I +
  \rho(X) \\ \delta_{\lambda \mu} &=sI + \rho' (X)  &  \delta_{\gamma
  \lambda} & = s'I + \rho' (X) \end{align*} for some $X \in \ssl_3(k)$ and
$r,r',s,s' \in k$
  where $\rho$ and $\rho'$ are the representations corresponding to
  $L(0,1)$ and its dual.

  The only part of $R$ we
  are yet to consider is $e_\lambda R e_\lambda$, but
  $\delta$ preserves this if and only if $r+s=r'+s'$.  Adding
  \[((r+s)/2+r')\ad(e_\gamma) + ( ( s-r)/2)\ad(e_\mu)\] shows
  $\delta$ is $((r+s)/2)\Delta$ plus the action of an element of
  $\ssl_3(k)$.

  The inner derivations of degree zero are spanned by the
  $\ad(e_\alpha)$s which are linearly independent of
  $\Delta$ and the $\ssl_3(k)$-action. Any linear dependence could only
  involve the $H_i$ and $\Delta$ action because the $\ad(e_\alpha)$
  preserve the weight grading.  $\Delta$ and the $\ad(e_\alpha)$ act as
  a scalar on $e_\lambda V e_\gamma$, but no nonzero linear combination
  of the $H_i$ does, so the $H_i$ cannot be involved.  Finally $\Delta$
  is not a linear combination of $\ad(e_\gamma)$ and $\ad(e_\mu)$ since
  it acts by the same scalar on $e_\gamma V e_\lambda$ as it does on
  $e_\lambda V e_\mu$.

\end{proof}

\section{Verma modules in this block are Koszul} \label{kos}
Recall that a graded module $M$ over a Koszul algebra
$\Gamma$ is called Koszul if
it has a graded projective resolution
$P^* \to M$ such that $P^i = \Gamma P^i_i$. Such a projective resolution
is called linear. Equivalently $M$ is Koszul if $\Ext^{i,j}_\Gamma (M,
(\Gamma/\Gamma_{>0}))$ is zero unless $i=j$, where $i$ is the
homological degree and $j$ the internal degree arising from the
$\ZZ_{\geq 0}$-grading on $\Gamma$.
Koszulity implies $M
\cong \Gamma \otimes_\mathbbm{k} M_0 / H$ where $H \subseteq \Gamma_1\otimes M_0$.
In this section we show that the Verma modules $M(\lambda)$, $M(\gamma)$
and $M(\mu)$ are Koszul.

$\Lambda$ has an $X$-grading obtained by putting the $e_\alpha$ in
degree $(0,0)$ and the $v_{i,j}$ and $w_{i,j}$ in degree $(pi,pj)$.  We
refer to this as the weight grading of $\Lambda$.

\begin{lem}
  Under the Morita equivalence between our block of $u$ and $\Lambda$,
  the Verma modules $M(\lambda), M(\gamma), M(\mu)$ correspond to
  \[M_\lambda=\frac{\Lambda e_\lambda }{
\Lambda v_{0,1}+\Lambda v_{1,-1}+\Lambda w_{-1,1} + \Lambda
w_{1,0}},M_\gamma=\frac{\Lambda e_\gamma }{ \Lambda v_{1,0}},
M_\mu=\frac{\Lambda e_\mu}{ \Lambda w_{0,1}} \]
respectively.
\end{lem}

\begin{proof}
Let $S_\alpha$ be the weight-graded simple $\Lambda$-module $\Lambda
e_\alpha/J(\Lambda)e_\alpha$.  The Morita correspondent of $M(\gamma)$
is a quotient $\Lambda e_\gamma/I_\gamma$ with weight-graded composition
factors $S_\gamma, S_\lambda[-p,p], S_\lambda[0,-p], S_\mu[-p,0]$.
So $I_\gamma$ must contain $v_{1,0}$, but $\Lambda e_\gamma / \Lambda
v_{1,0}$ already has only four composition factors, so it must be the
Morita correspondent of $M(\gamma)$.  The other correspondences follow similarly.
\end{proof}

$M_\gamma$ has a basis consisting of the images of $e_\gamma, v_{-1,1},
v_{0,-1}$ and $w_{-1,1}v_{0,-1} = - w_{0,-1}v_{-1,1}$, $M_\mu$ has
a basis consisting of the images of $ e_\mu, w_{-1,0}, w_{1,-1}$ and $
v_{1,-1}w_{-1,0} = - v_{-1,0}w_{1,-1}$, and $M_\lambda$ has a basis
consisting of the images of $e_\lambda, w_{0,-1}, v_{-1,0},
v_{0,-1}v_{-1,0}=-w_{-1,0}w_{0,-1}$.

Put $I = \Lambda v_{0,1}+\Lambda w_{1,0} $ and $M = \Lambda e_\lambda /
I$, regarded as a $K$-graded module with $e_\lambda +I$ in degree zero.

\begin{lem}
  There is an exact sequence of $K$-graded modules
  \begin{equation}\label{M_seq} 0 \to M_\gamma[1] \oplus M_\mu [1]
    \stackrel i \to M \stackrel \pi \to M_\lambda \to 0\end{equation}
  where $i(e_\gamma + \Lambda v_{1,0})=v_{1,-1}+I$, $i(e_\mu+\Lambda
  w_{0,1})= w_{-1,1} + I$  and $\pi$ is the quotient
  map. \end{lem}

\begin{proof}
  $i$ is well-defined because
  $w_{0,1}w_{-1,1}=-v_{-1,1}v_{0,1} \in I$ and $v_{1,0} v_{1,-1} =
  -w_{1,-1}w_{1,0} \in I$, and $\im i$ is the kernel of the quotient
  map $M \to M_\lambda$.  We only need to show $i$ is injective,
  and it is enough to prove $i$ is injective on the socle $S_\mu
  \oplus S_\gamma$ of $M_\gamma \oplus M_\mu$.  Since the two simple
  summands are non-isomorphic (or by considering the weight degree), it is
  enough simply to show that $i$ is nonzero on $\soc M_\gamma$ and on
  $\soc M_\mu$.

  $i(\soc M_\mu)$ is spanned by the image of $v_{1,-1}w_{-1,0}w_{-1,1}$
  in $M$, which is homogeneous of $K$-degree 3 and weight degree $[-p,0]$.
  We want to show this element does not lie in $I=\Lambda
  v_{0,1}+\Lambda w_{1,0}$.  We claim $\Lambda v_{0,1}$ has no nonzero
  element in this $K$- and
  weight-degree.  To get such an element, we would have to multiply $v_{0,1}$ by two of the
  generators $v_{i,j},w_{i,j}$, and the first has to be $v_{0,-1}$ as
  all other products are either zero or too far from $[-p,0]$ (it helps
  to look at Figure~\ref{Plambda} to see this).  The second must then be
  $v_{-1,0}$ if we are to end up at weight degree $[-p,0]$, but
  $v_{-1,0}v_{0,-1}v_{0,1} =0$ because $v_{-1,0}v_{0,-1}=0$.  Similarly,
  $\Lambda w_{1,0}$ has no nonzero $K$-degree 3 weight degree $[-p,0]$
  element: the only possible product is $v_{-1,0} w_{-1,0} w_{1,0}$ as
  before, but $v_{-1,0}w_{-1,0}=0$.

  A similar argument shows that $i$ is injective on $\soc M_\gamma$,
  completing the proof.
\end{proof}

Let $\Omega (M) = (\Lambda e_\mu \oplus \Lambda
e_\gamma)/\Lambda(w_{0,1}+v_{1,0})$.

\begin{lem}
  There are exact sequences of $K$-graded modules
  \begin{gather} \label{OM_seq}0 \to M_\lambda [1]\stackrel \psi \to \Omega(M)
    \stackrel \phi \to M_\gamma \oplus
    M_\mu \to 0 \\
  \label{M_res} 0 \to M[1] \stackrel i \to \Lambda e_\mu[1] \oplus \Lambda
  e_\gamma[1]
    \stackrel j \to \Lambda
    e_\lambda \to M \to 0
  \end{gather}
\end{lem}

\begin{proof}
  $\phi( (a,b)+\Lambda(w_{0,1}+v_{1,0}) = (a + \Lambda w_{0,1},
  b+\Lambda v_{1,0}) $ is well-defined and surjective, with kernel
  \[ \ker \phi = \frac{\Lambda v_{1,0} + \Lambda
  w_{0,1}}{\Lambda(v_{1,0}+w_{0,1})} \]
  generated by the image of $v_{1,0}$.  Therefore $M_\lambda$ surjects
  onto $\ker \phi$ via the map $\psi$ induced by $e_\lambda \mapsto v_{1,0} +
  \Lambda(v_{1,0}+w_{0,1})$.  We only need $\psi$ to be injective.  If we
  show that $\dim \Omega(M) \geq 12$ then $\ker \phi$ has dimension at
  least four, but since it is a quotient of $M_\lambda$ it will be
  exactly four and hence $\psi$ will have been shown to be injective.

Letting $\Lambda e_\lambda \to M$ be the quotient map and defining $j$
by $j(e_\mu) =w_{1,0}$, $j(e_\gamma) =v_{0,1}$, we get an exact sequence
\[ \Lambda e_\gamma[1] \oplus \Lambda e_\mu[1] \stackrel j \to \Lambda
e_\lambda \to M \to 0.\]
$v_{1,0}+w_{0,1}\in \ker j$ so $j$
factors through $\Omega(M)[1]$, showing that $\dim \Omega(M) \geq \dim \im
j = \dim \Lambda e_\lambda - \dim M = 12$.  This completes the proof of
the exactness of (\ref{OM_seq}), so $\dim \Omega(M)=12$.

$\dim \im j = 12$, so $\dim \ker j = 12$, and so
$\Lambda(v_{1,0}+w_{0,1})$, which has dimension 12 as $\Omega(M)$ does,
must be all of  $\ker j$.  Therefore $M[1]$ surjects onto $\ker j$ by
$i(e_\lambda+ I) =  v_{1,0}+w_{0,1}$ (this is well-defined).  Since
$\dim M =12$, $i$ must be injective.  This gives exactness of
(\ref{M_res}).
\end{proof}

\begin{cor} $M$ and $\Omega(M)$ are Koszul. \end{cor}
\begin{proof}
  Splicing shifted copies
  of (\ref{M_res}) produces a linear projective
 resolution of $M$. Koszulity of $\Omega(M)$ follows by truncating and
 shifting the
 same resolution.
\end{proof}

\begin{thm}
  $M_\lambda, M_\gamma$ and $M_\mu$ are Koszul.
\end{thm}

\begin{proof}
  Let $S = S_\mu\oplus S_\lambda\oplus S_\gamma$. We will show that for every $n$, the graded vector spaces
  $\Ext^{n,m}_\Lambda(M_\lambda, S)$ and
  \[\Ext^{n,m}_\Lambda(M_\gamma \oplus M_\mu, S) \cong
    \Ext^{n,m}_\Lambda(M_\gamma , S)\oplus \Ext^{n,m}_\Lambda(M_\mu,S)
  \]
  are zero unless $m=n$.  This is certainly true for $n=0$.

   Consider the long
  exact sequence obtained by applying $\Hom_\Lambda(-, S)$ to
  (\ref{M_seq}). Let $\omega$
   be the connecting homomorphism: it corresponds to Yoneda multiplication by
   the short exact sequence (\ref{M_seq}) and is therefore homogeneous
   of $K$-degree one. For each $n$ we get a
  short exact sequence
  \[ 0 \to \frac{\Ext^n_\Lambda(M_\gamma \oplus M_\mu, S)}{\ker \omega} \to
    \Ext^{n+1}_\Lambda(M_\lambda,S) \stackrel{\pi^*}\to \im \pi^* \to
  0\]
in which the first map, induced by $\omega$, increases $K$-degree by one,
and the second preserves $K$-degree.  $M$ is Koszul, so $\im \pi^*
\subseteq \Ext^{n+1}_\Lambda(M, S)$ is zero outside $K$-degree $n+1$. It
follows that if $\Ext^n_\Lambda(M_\gamma \oplus M_\mu, S)$ is zero
outside $K$-degree $n$ then $\Ext^{n+1}_\Lambda(M_\lambda, S)$ is zero
outside $K$-degree $n+1$.

Similar short exact sequences arising from the long exact sequence
obtained by applying $\Hom_\Lambda(-,S)$ to (\ref{OM_seq}) show
that if $\Ext^{n}_\Lambda(M_\lambda, S)$ is zero outside $K$-degree
$n$ then $\Ext^{n+1}_\Lambda(M_\gamma \oplus M_\mu, S)$ is zero outside
$K$-degree $n+1$.
Since the result holds for $n=0$, this completes the proof.
\end{proof}

\bibliography{irregular_blocks}
\bibliographystyle{amsalpha}
\end{document}

config: arxiv


TITLE: necessary conditions for a quadric surface to be ruled (over a field of char 2)
QUESTION [2 upvotes]: Given a quadric surface $Q$ over a field $F$ of characteristic $2$, assume it is irreducible and reduced, we say it is ruled, if $Q$ is birational to $C \times \mathbb{P}^1$ for some $C$.
A sufficient condition for such a quadric to be ruled is that there is a rational point (use the projection from this point).
My question is, is this also necessary?
People have studied ruledness of a quadric hypersurface over fields of characteristic other than $2$. Maybe there are also results in characteristic $2$. In that case, I would also be very interested in knowing a reference.

REPLY [1 votes]: The condition you state is not a necessary condition.  You may find much more about these types of questions in Manin's "Cubic Forms".
First of all, over a finite field every quadric hypersurface has a rational point by Chevalley's theorem (or you can probably reduce this case to Wedderburn's earlier theorem).  Thus, assume that the field is infinite.  Then, by Bertini's Theorem, for a sufficiently general hyperplane section $C$ of your surface, $C$ is a smooth plane conic.
For the numerical polynomial $P(n) = n+1$, computing Hilbert polynomials with respect to the restriction to $Q$ of $\mathcal{O}_{\mathbb{P}^3}(1)$, the Hilbert scheme $\text{Hilb}^P_{Q/F}$ is a finite, flat scheme over $C$ of degree $2$.  Of course $C$ is geometrically simply connected.  Thus, either the Hilbert scheme is isomorphic to two disjoint copies of $C$ (the "split" case), or the Hilbert scheme is isomorphic to $C\times_{\text{Spec}\ F}\text{Spec}\ E$ for a degree $2$ separable extension of $F$ (the "non-split case").
In the split case, $Q$ is isomorphic to the product surface $C\times C$ embedded into $\mathbb{P}^3$ by the complete linear system of the Cartier divisor $\Delta(C)\subset C\times C$. If the surface $Q$ has an $F$-point, then that $F$-point projects to an $F$-point of $C$.  Thus, for every smooth, geometrically integral curve $C$ over $F$ of arithmetic genus $0$ that has no $F$-point, $Q=C\times C$ is an example of a smooth quadric surface that has no $F$-point.  On the other hand, identifying $C$ with either one of the connected components of $\text{Hilb}^P_{Q/F}$, also $Q$ is isomorphic to $C\times \mathbb{P}^1$.
Edit. The OP contacted me.  There is a mistake in what I wrote above.  It is still correct that for every conic $C$ with no $F$-rational point, the surface $Q=C\times C$ is birational to $C\times \mathbb{P}^1$ yet admits no $F$-rational point.  However, for a singular quadric surface $Q$ and a smooth hyperplane section $C$, the Hilbert scheme $\text{Hilb}^P_{Q/F}$ may fail to be an étale double cover of $C$.  The Hilbert scheme may be geometrically nonreduced (everywhere).  So the answer above is quite incomplete: it does not address the non-split case, and it does not address the nonreduced case.


text:
\begin{document}

\maketitle

\small

\begin{center}
  Bahçeşehir University, Faculty of Engineering\\ and Natural Sciences,
  Istanbul, Turkey\\
  {\it e-mail:} \href{mailto:selcuk.kayacan@eng.bau.edu.tr}{selcuk.kayacan@eng.bau.edu.tr}
\end{center}

\begin{abstract}
  A rack is a set with a binary operation such that left multiplications are automorphisms of the set and a quandle is a rack satisfying a certain condition. For a finite connected rack the cycle type of the permutation defined by left multiplication by an element is independent from the chosen element. This cycle type is called the profile of the rack. Hayashi conjectured, in the profile of a finite connected quandle, the length of a cycle must divide the length of the largest cycle. In this paper, we prove Hayashi's Conjecture in some particular cases.

  \smallskip
  \noindent 2010 {\it Mathematics Subject Classification.} Primary: 20N99;\\ Secondary: 08A99

  \smallskip
  \noindent Keywords: Finite connected rack; profile of a rack

\end{abstract}

\section{Introduction}

Racks and quandles are algebraic objects that are mostly studied in the context of knot theory. The defining properties of those objects are in a sense compatible with the Reidemeister moves which makes them useful to define knot invariants. For example, in \cite{Joy82} Joyce introduced \emph{knot quandle} and showed that it is a complete invariant for knots and in \cite{FR92} Fenn and Rourke proved that the \emph{fundamental rack} is a complete invariant for irreducible framed links in a $3$-manifold. There are various notational conventions used to define racks and quandles in literature. In this paper we define racks and quandles in the following way.

A \emph{rack} $X$ is a set together with a binary operation $\triangleright\colon X\times X\to X$ satisfying the following two axioms:
\begin{itemize}
\item[\textbf{(A1)}] for all $x,y,z\in X$ we have $x \triangleright (y \triangleright z) = (x \triangleright y) \triangleright (x \triangleright z)$
\item[\textbf{(A2)}] for all $x,z\in X$ there is a unique $y\in X$ such that $x \triangleright y = z$
\end{itemize}
A rack $X$ is a \emph{quandle} if, additionally, it satisfies the following axiom:
\begin{itemize}
\item[\textbf{(A3)}] for all $x\in X$ we have $x\triangleright x = x$
\end{itemize}

Let $G$ be a group and for any two elements $a,b\in G$ let $a\triangleright b := aba^{-1}$. Then, the group $G$ as well as any conjugacy class of $G$ satisfies those three axioms; hence equipped with the conjugation operation $\triangleright$ each one of them is an example of a quandle. More generally, we call a subset $X$ of $G$ \emph{conjugation rack} if it is a rack with the conjugation operation inherited by the group.

In this paper we only consider finite racks, although some of the statements are also valid for infinite racks. It is customary to identify elements of a finite rack $X$ with integers $1,2,\dots,n$ if $X$ has $n$ elements. By axiom (A2), the map $\phi_x\colon y\to x\triangleright y$ is a permutation hence is an element of the symmetric group $S_{n}$ of degree $n$. A permutation $\sigma\in S_{n}$ is an automorphism of $X$ if and only if $\sigma(x\triangleright y) = \sigma(x)\triangleright \sigma(y)$ for every $x,y\in X$. By axiom (A1), the map $\phi_x$ is an automorphism of $X$.

The set of all automorphisms of $X$, denoted $\mathsf{Aut}(X)$, form a subgroup of $S_{n}$ which is called the \emph{automorphism group} of $X$. The map $\Phi\colon x\mapsto \phi_x$ defines a rack morphism from $X$ to $S_{n}$, where $S_{n}$ is considered as a quandle with conjugation operation (see Proposition~\ref{prop:act}). The \emph{inner automorphism group} of $X$, denoted $\mathsf{Inn}(X)$, is the subgroup of $\mathsf{Aut}(X)$ generated by the elements of $\Phi(X)$. Let $\sigma$ be an automorphism of $X$. Since $\sigma\phi_x\sigma^{-1} = \phi_{\sigma(x)}$ for every $x\in X$, we see that $\mathsf{Inn}(X)$ is a normal subgroup of $\mathsf{Aut}(X)$. In general $\mathsf{Inn}(X)$ and $\mathsf{Aut}(X)$ may be different groups.

A rack $X$ is \emph{faithful} if the map $\Phi\colon X\to \mathsf{Inn}(X)$ is injective. Suppose $X$ is faithful. Obviously, the map $\Phi$ is an isomorphism between $X$ and $\Phi(X)$, hence $X$ is a quandle. Moreover, the center of the inner automorphism group is trivial and the action of $\mathsf{Inn}(X)$ on $\Phi(X)$ by conjugation is faithful (see Proposition~\ref{prop:act}).

A rack $X$ is \emph{connected} if the action of its inner automorphism group on $X$ itself is transitive. Suppose $X$ is a connected rack. In that case, any two elements of $\Phi(X)$ can be conjugated by an element of $\mathsf{Inn}(X)$. Since in symmetric group conjugate elements have the same cycle type, connectedness of $X$ implies each element of $\Phi(X)$ has the same cycle type. This cycle type $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ is called the \emph{profile} of $X$. Here $\lambda_s$ $(0\leq s\leq t)$ are ordered increasingly. Notice that $x\triangleright x = x$ for every $x\in X$ if $X$ is a quandle which means in the profile $\lambda$ of a quandle $X$ the value of the least element $\lambda_0$ must always be $1$. The purpose of this paper is to prove some special cases of the following conjecture which was stated by Hayashi originally for quandles (see \cite[Conjecture~1.1]{Hay13}).

\begin{conj}\label{conj:hay}
  For a finite connected rack $X$ with profile $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$, each $\lambda_s$\, $(0\leq s \leq t)$ divides $\lambda_t$.
\end{conj}

Let $G$ be a finite group, $\alpha$ be an automorphism of $G$, and $H$ be a subgroup of $G$ pointwise fixed by $\alpha$, i.e., $\alpha(h) = h$ for every $h\in H$. One can easily verify that the coset space $G/H$ with the operation $xH\triangleright yH := x\alpha(x^{-1}y)H$ satisfies quandle axioms. It is called the \emph{homogeneous quandle} and denoted by $(G,H,\alpha)$. Any finite connected quandle is isomorphic to a homogeneous quandle. Therefore, given the complete list of transitive groups of degree $n$, it is possible to determine the complete list of non-isomorphic connected quandles of $n$ elements. This method is used by Vendramin to compute all connected quandles having less than $48$ elements. The list is available in \textsf{Rig}~\cite{rig}, a \textsf{GAP} package designed for computations related to racks and quandles. In Table~\ref{table:sq} we present permutations of a quandle which is the fourth quandle with 12 elements in Vendramin's list. The reader may easily verify that the following partition
$$  \{1,5,9\},\; \{2,6,10\},\; \{3,7,11\},\; \{4,8,12\}  $$
formed by taking the first three numbers in each row of the table is a block system for the action of the inner automorphism group on the rack itself. We shall prove that Conjecture~\ref{conj:hay} holds when the inner automorphism group acts on the rack primitively (see Theorem~\ref{thm:prim}).

\begin{table}[h!]
\centering
\caption{Permutations of SmallQuandle(12,4) in \textsf{Rig}}
\label{table:sq}
\begin{tabular}{rll}
  $\phi_1$ & :     &\quad $(1)(5,9)(2,4,3)(6,12,7,10,8,11)$   \\
  $\phi_2$ & :     &\quad $(2)(6,10)(1,3,4)(5,11,8,9,7,12)$   \\
  $\phi_3$ & :     &\quad $(3)(7,11)(1,4,2)(5,12,6,9,8,10)$   \\
  $\phi_4$ & :     &\quad $(4)(8,12)(1,2,3)(5,10,7,9,6,11)$   \\
  $\phi_5$ & :     &\quad $(5)(1,9)(6,8,7)(2,12,3,10,4,11)$   \\
  $\phi_6$ & :     &\quad $(6)(2,10)(5,7,8)(1,11,4,9,3,12)$   \\
  $\phi_7$ & :     &\quad $(7)(3,11)(5,8,6)(1,12,2,9,4,10)$   \\
  $\phi_8$ & :     &\quad $(8)(4,12)(5,6,7)(1,10,3,9,2,11)$   \\
  $\phi_9$ & :     &\quad $(9)(1,5)(10,12,11)(2,8,3,6,4,7)$   \\
  $\phi_{10}$ & :  &\quad $(10)(2,6)(9,11,12)(1,7,4,5,3,8)$   \\
  $\phi_{11}$ & :  &\quad $(11)(3,7)(9,12,10)(1,8,2,5,4,6)$   \\
  $\phi_{12}$ & :  &\quad $(12)(4,8)(9,10,11)(1,6,3,5,2,7)$   \\
\end{tabular}
\end{table}

Let $A$ be a finite abelian group and $\alpha$ be an automorphism of $A$. The corresponding homogeneous quandle $(A,1,\alpha)$ is called an \emph{affine quandle}. Let $\beta(x) := x - \alpha(x)$ for every $x\in A$. Observe that in  $(A,1,\alpha)$ the quandle operation is defined by
$$x\triangleright y := \alpha(y-x) + x = \alpha(y) + \beta(x) = \beta(x-y) + y$$
and the orbit of $y$ under the action of inner automorphism group is $y + \beta(A)$. Suppose  $(A,1,\alpha)$ is connected. Then $\beta$ must be a bijection of $A$. Moreover, the inner automorphism group is isomorphic to a cyclic extension of $A$, hence is solvable. Now, consider the profile of $(A,1,\alpha)$. Kajiwara and Nakayama proved in \cite{KN16} that there must be an element of $A$ whose orbit size under the action of $\langle\phi_x\rangle$ is the order of $\phi_x$. Therefore, Hayashi's Conjecture holds for connected affine quandles.

Let $C$ be a conjugacy class of a symmetric group $S_d$. Hence, the inner automorphism group of $C$ is either isomorphic to $A_d$ or $S_d$. In particular $\mathsf{Inn}(C)$ is not solvable if $d\geq 5$. Let $x$ be an element of $C$ and let $\ell$ be the order of $x$ in $S_d$. To prove Hayashi's Conjecture holds for $C$, one possible way is to find an element $y$ of $C$ so that $x^kyx^{-k}\neq y$ for $1\leq k < \ell$. However, this method seems to be not applicable for other classes of groups as we don't have an explicit description of the elements of conjugacy class $C$. We shall prove that Conjecture~\ref{conj:hay} holds for the conjugacy classes in a symmetric group using a similar approach (see Theorem~\ref{thm:sym}).

\section{Preliminaries}\label{sec:pre}

The reader may refer to \cite{AG03} for the basic theory of racks and in particular to Lemma~1.7, Lemma~1.9 and Proposition~3.2 of the same paper for alternative statements and proofs of the following Proposition.

\begin{prop}\label{prop:act}
  Let $X$ be a rack. Then following statements hold.
  \begin{enumerate}[(i)]
  \item The map $\Phi\colon X\to \mathsf{Inn}(X)$ defines a rack morphism between $X$ and $\Phi(X)$ which is an isomorphism if $X$ is faithful.
  \item If $X$ is faithful, then the action of $\mathsf{Inn}(X)$ on $X$ by automorphisms and on $\Phi(X)$ by conjugation are isomorphic and the center of $\mathsf{Inn}(X)$ is trivial.
  \item If $X$ is a conjugation rack and $G := \langle X \rangle$, then the map $\Phi\colon X\to \Phi(X)$ extends to a map
  $$\Phi\colon G\to \mathsf{Inn}(X),$$
  which is a group homomorphism and $\mathsf{Inn}(X)\cong G/Z(G)$.
  \item The rack $X$ is connected and faithful if and only if it is isomorphic, as a quandle, to a conjugacy class $C$ of a group $G$ so that $G = \langle C\rangle$ and $Z(G) = 1$.
  \end{enumerate}

\end{prop}

\begin{proof}
  \emph{(i)}\; Observe that for every $x,y,z\in X$ the equality $\phi_x(y\triangleright z) = \phi_x(y)\triangleright \phi_x(z)$ holds by axiom (A1). In other words  $\phi_x\phi_y = \phi_{x\triangleright y}\phi_x$ holds for any two elements $x,y\in X$. However, this means the map $\Phi$ takes $x\triangleright y$ into $\phi_x\phi_y\phi_x^{-1} = \phi_x\triangleright\phi_y$ and so $\Phi$ is a rack morphism between $X$ and $\Phi(X)$. Further if $\Phi$ is injective then $X$ and $\Phi(X)$ would be isomorphic quandles.

  \emph{(ii)}\; Suppose $X$ is faithful. Pick an automorphism $\sigma\in \mathsf{Inn}(X)$. Since, for any $x\in X$,
  $$ \Phi(\sigma\cdot x) = \Phi(\sigma(x)) = \phi_{\sigma(x)} = \sigma\phi_x\sigma^{-1} = \sigma\cdot \Phi(x), $$
  we see that the map $\Phi$ defines an isomorphism between the actions of $\mathsf{Inn}(X)$ on $X$ and on $\Phi(X)$. Now, if $\sigma$ lies in the center of $\mathsf{Inn}(X)$, then $\phi_{\sigma(x)} = \sigma\phi_x\sigma^{-1} = \phi_x$ for every $x\in X$ implying $\sigma$ is the identity map on $X$ since the map $\Phi\colon x\mapsto \phi_x$ is a bijection.

  \emph{(iii)}\; Consider the map $\Phi\colon G\to \mathsf{Inn}(X)$ taking an element $g:=y_j\dots y_2y_1$ of $G$, written as the product of some elements $y_i\in X$ for $1\leq i\leq j$, to an element $\phi_g$ of the inner automorphism group $\mathsf{Inn}(X)$, where $\phi_g$ is defined as the composition
  $$\phi_g := \phi_{y_j}\dots\phi_{y_2}\phi_{y_1}.$$
  Since, for any element $x\in X$ we have
  $$ gxg^{-1} = (y_j\triangleright\dots(y_2\triangleright(y_1\triangleright x))\dots) = \phi_{y_j}\dots\phi_{y_2}\phi_{y_1}(x), $$
  the map $\Phi$ is well-defined and defines a surjective group homomorphism. Clearly, $g\in \mathsf{Ker}(\Phi)$ if and only if $\phi_g$ is the identity map which is the case if and only if $gxg^{-1} = x$ for every generator element $x \in X$.

  \emph{(iv)}\; Suppose $X$ is faithful and connected. Since $X$ is faithful, the rack $X$ is isomorphic to the conjugation rack $\Phi(X)$ and since $X$ is connected, conjugation rack $\Phi(X)$ is a conjugacy class of the inner automorphism group $\mathsf{Inn}(X)$. By the previous part the center of $\mathsf{Inn}(X)$ is trivial.

  Conversely, suppose $X$ is isomorphic to a conjugacy class $C$ of a group $G$ so that $G = \langle C\rangle$ and $Z(G) = 1$. By the previous part, the rack $C$ is connected and faithful. Therefore $X$ is faithful and connected as well.
\end{proof}

\begin{lem}[see {\cite[Lemma~1.21]{AG03}}]\label{lem:fiber}
    Let $X$ and $Y$ be two racks and let $f\colon X\to Y$ be a surjective rack homomorphism. Then, for any $y_1,y_2\in Y$, the cardinalities of the fibers $f^{-1}(y_2)$ and $f^{-1}(y_1\triangleright y_2)$ are equal. In particular, every fiber of $Y$ has the same cardinality if $Y$ is connected.
\end{lem}

\begin{proof}
  We show that there exist a bijective function $B\colon X\to X$ such that
  $$ B(f^{-1}(y_2)) \subseteq f^{-1}(y_1\triangleright y_2) \text{ and } B(f^{-1}(y_1\triangleright y_2)) \subseteq f^{-1}(y_2).$$
  Let $B(x) := a\triangleright x$, where $f(a) = y_1$. Observe that $B^{-1} = \phi_a^{-1}\in \mathsf{Inn}(X)$. Suppose $x\in f^{-1}(y_2)$. Then $f(a\triangleright x) = f(a) \triangleright f(x) = y_1\triangleright y_2$. That is, $B(x)\in f^{-1}(y_1\triangleright y_2)$. Next, suppose $x\in f^{-1}(y_1\triangleright y_2)$. Then $f(B^{-1}(x)) = f\phi_a^{-1}(x) = \phi_{f(a)}^{-1}f(x) = \phi_{y_1}^{-1}(y_1\triangleright y_2) = y_2$. That is, $B^{-1}(x)\in f^{-1}(y_2)$.
\end{proof}

\begin{rem}
Let $X$ be a rack and take two elements $x,y$ of $X$. Let $\lambda_y^x$ be the orbit size of $y$ and $\bar{\lambda}_y^x$ be the orbit size of $\phi_y$ under the actions of $\langle\phi_x\rangle$. By Lemma~\ref{lem:fiber}, $\Phi^{-1}$ yields a partition of $X$ into a disjoint union of fibers of the same size. Observe that for any integer $k$ the element $\phi_x^k(y)$ of $X$ does not lie in the fiber $\Phi^{-1}(\phi_y)$ unless $\phi_x^k\phi_y\phi_x^{-k} = \phi_y$. Therefore $\bar{\lambda}_y^x$ divides $\lambda_y^x$.
\end{rem}
Let $X$ be a finite connected rack whose profile is $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ and let $k$ be an integer. For a permutation $\sigma$ of $X$, the \emph{$k$-part} of $\sigma$ is the set of elements of $X$ which appear in a cycle of length $k$ in the permutation $\sigma$. And the \emph{$\lambda_s$-part} of $X$ is the multiset formed by the $\lambda_s$-parts of elements of $\Phi(X)$. For example, if $X$ is the fourth quandle with 12 elements in Vendramin's list, using Table~\ref{table:sq}, we see that the $2$-part of $X$ is
$$ \{ 5,9,6,10,7,11,8,12,1,9,2,10,3,11,4,12,1,5,2,6,3,7,4,8 \}.  $$

\begin{lem}\label{lem:multi}
  Let $X$ be a finite connected rack  with profile $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$. Then the number of occurrences of each element of $X$ in the $\lambda_s$-part of $X$ is same and equal to $a_s\lambda_s/f$ for each $\lambda_s$\, $(0\leq s \leq t)$, where $f$ is the cardinality of a fiber of $\Phi$.
\end{lem}

\begin{proof}
  Let $\sigma$ be an element of $\mathsf{Inn}(X)$. Conjugating each element of $\Phi(X)$ with $\sigma$ yields a permutation of $\Phi(X)$. Since $\sigma\phi_x\sigma^{-1} = \phi_{\sigma(x)}$, if $\sigma(u) = v$ for some distinct elements $u$ and $v$ of $X$ then the number of occurrences of $u$ and $v$ in the  $\lambda_s$-part of $X$ would be the same. The claim follows from the facts that $X$ is connected and the cardinality of $\Phi(X)$ is $n/f$, where $n$ is the number of elements in $X$.
\end{proof}

Next, we shall review Conjecture~\ref{conj:hay} in group theoretical terms. Let $X$ be a finite connected rack and fix an element $x\in X$. Let $F:=\langle\phi_x\rangle$ be the subgroup of $G:=\mathsf{Inn}(X)$ generated by $\phi_x$ and $H:=C_G(\phi_x)$ be the centralizer of $\phi_x$ in $G$. Clearly, $F\leq H$. The action of $G$ on $\Phi(X)$ by conjugation is equivalent to the action of $G$ on the coset space $G/H$ by left multiplication. Let $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ be the profile of $\Phi(X)$. Then, the permutation $\phi_{\phi_x}$ would correspond to the permutation
$$ (H)(\phi_x\phi_{y_1} H,\dots,\phi_x^{\lambda_1}\phi_{y_1} H)\dots (\phi_x\phi_{y_q} H,\dots,\phi_x^{\lambda_s}\phi_{y_q} H)\dots (\phi_x\phi_{y_r} H,\dots,\phi_x^{\lambda_t}\phi_{y_r} H), $$
where $\phi_{y_q}$ $(0\leq q \leq r = \sum_{s=0}^ta_s\lambda_s)$ are some representative elements from the cycles of $\phi_{\phi_x}$. Let $q := (\sum_{i=0}^{s-1}a_i\lambda_i) + j$ for some $1\leq j \leq a_s$. It should be clear that
$$ \lambda_s = \frac{| F \phi_{y_q} H |}{|H|}. $$
More importantly, the subgroup $\langle \phi_x^{\lambda_s}\rangle$ of $F$ is also a subgroup of $\phi_{y_q}H\phi_{y_q}^{-1}$.

\begin{prop}\label{prop:re}
  Let $X$ be a finite connected rack and $x$ be an element of $X$. Let $F$ be the subgroup of $\mathsf{Inn}(X)$ generated by $\phi_x$ and $H$ be the centralizer of $\phi_x$ in $\mathsf{Inn}(X)$. Let $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ be the profile of $X$. The following statements hold:
  \begin{enumerate}[(i)]
  \item If $F$ intersects $\phi_yH\phi_y^{-1}$ trivially for some $\phi_y\in \Phi(X)$, then $\lambda_s$ divides $\lambda_t$ for each  $0\leq s \leq t$.
  \item Conversely, if $\lambda_s$ divides $\lambda_t$ for each  $0\leq s \leq t$ and $X$ is faithful, then $F$ intersects one of the conjugates of $H$ trivially.
  \end{enumerate}
\end{prop}

\begin{proof}
  \emph{(i)}\; Let $k$ be an integer and $\bar{\lambda} = (\bar{\lambda}_0^{a_0},\bar{\lambda}_1^{a_1},\dots,\bar{\lambda}_{t'}^{a_{t'}})$ be the profile of $\Phi(X)$. By the assumption
$$ (\phi_x\phi_{y} H,\phi_x^2\phi_{y} H,\dots,\phi_x^{k}\phi_{y} H)  $$
is an orbit under the action of $F$ if and only if $k$ equals to the order of $\phi_x$. Then, clearly, $k$ equals to $\bar{\lambda}_{t'}$. From the remark following Lemma~\ref{lem:fiber} we see that $k$ divides $\lambda_s$ for some $0\leq s \leq t$ so that the order of $\phi_x$ is equal to $\lambda_t$.

  \emph{(ii)}\; Since $\lambda_s$ divides $\lambda_t$ for each  $0\leq s \leq t$, we see that the order of $\phi_x$ is $\lambda_t$. By Proposition~\ref{prop:act} we know that $X$ and $\Phi(X)$ have the same profiles. Therefore, the order of $\phi_{\phi_x}$ is $\lambda_t$ as well. From this we conclude that there exists an element $\phi_y$ of $\Phi(X)$ so that $\phi_yH$ lies in an orbit of size $\lambda_t$ under the action of $F$, which means the intersection of $F$ with the centralizer of $\phi_{\phi_y(x)}$ is trivial.
\end{proof}

\section{Main results}\label{sec:main}

Let $G$ be a group acting transitively on a set $\Omega$. A partition of $\Omega$ form a block system if the following property is satisfied: for any element $g$ of $G$ an element of the partition is either mapped to itself by $g$ or to another element of the partition. In that case we call the elements of the partition \emph{blocks}. A block system is trivial if either the partition consists of the whole set or it consists of the singletons. We say $G$ acts \emph{primitively} on $\Omega$ if the only block systems are the trivial block systems.

Let $X$ be a finite connected rack with profile $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ and $k$ be an integer. For a permutation $\sigma$ of $X$, the \emph{$\tilde{k}$-part} of $\sigma$ is the set of elements of $X$ which appear in a cycle whose length is a divisor of $k$ in the permutation $\sigma$. Let $x,y$ be two elements of $X$. If an automorphism $\sigma$ of $X$ centralizes $\phi_y$, then as a permutation $\sigma$ preserves $\lambda_s$-part of $\phi_y$ for each $\lambda_s$\, $(0\leq s \leq t)$. Let $K$ be the $\tilde{k}$-part of $\phi_x$ and $L:=X\setminus K$ be the complement of $K$ in $X$. If $\phi_x^k$ centralizes $\phi_y$, then $\phi_y$ centralizes $\phi_x^k$ as well. However, that means as a permutation $\phi_y$ preserves the set $K$ as well as the set $L$. Consider again the fourth quandle with 12 elements in Vendramin’s list (see Table~\ref{table:sq}). Its profile is $\lambda = (1^1,2^1,3^1,6^1)$ so that $\tilde{2}$-part of $\phi_i$ has $3$ elements whereas $\tilde{3}$-part has $4$ elements. Now, if $\phi_j$ centralizes $\phi_i^2$ then $\tilde{2}$-parts of $\phi_i$ and $\phi_j$ must be the same since the cycle type of $\phi_i^2$ is $(1^3,3^3)$. Moreover, for this example the set of $\tilde{2}$-parts of permutations $\phi_j$ form a block system for the action of the inner automorphism group. However, we don't know any general method for relating $\tilde{k}$-parts with a block system.

Let $H$ be the stabilizer of an element $x\in \Omega$. Then, the action of $G$ on  $\Omega$ is primitive if and only if the stabilizer $H$ is a maximal subgroup of $G$. To see this, observe that if there exists a proper subgroup $K$ of $G$ that contains $H$ strictly, then the cosets of $K$ would be unions of some of the cosets of $H$ in $G$. However, this determines a non-trivial block system for $G/H$ under the action of $G$ by left multiplication.

\begin{thm}\label{thm:prim}
  Let $X$ be a finite connected rack with profile $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$. If the inner automorphism group $\mathsf{Inn}(X)$ acts primitively on $X$, then $\lambda_s$ divides $\lambda_t$ for each $0\leq s \leq t$.
\end{thm}

\begin{proof}
  By assumption $\mathsf{Inn}(X)$ acts primitively on $X$. Since the fibers of $\Phi\colon X\to \Phi(X)$ form a partition of $X$ which is a block system for the action of $\mathsf{Inn}(X)$, the racks $X$ and $\Phi(X)$ are isomorphic and have the same profiles. Moreover, by Proposition~\ref{prop:act} the center of $\mathsf{Inn}(X)$ is trivial.

  To prove the Theorem suppose contrarily $\lambda_s$ does not divide $\lambda_t$ for some $0\leq s\leq t$. We want to derive a contradiction. Let us fix an element $x\in X$ and let $y$ be an element of $X$ different from $x$. Let $F:=\langle\phi_x\rangle$ and $F':=\langle\phi_y\rangle$. Also, let $H$ be the centralizer of $F$ in $G:=\mathsf{Inn}(X)$ and $H'$ be the centralizer of $F'$. Notice that $H'$ is a conjugate of $H$. Consider the intersection $I:= F\cap H'$. Using Proposition~\ref{prop:re}(i) contrapositively we see that $I$ is a non-trivial subgroup of $G$. Now $I$ centralizes both $F'$ and $H$, so it lies in the center of $\langle F', H\rangle$. Since $G$ acts on $X$ primitively by assumption, we see that $H$ is a maximal subgroup of $G$. Clearly, we can choose $y$ so that $F'$ is not contained in $H$. However that means $I$ lies in the center of $G$ which is a contradiction.
\end{proof}

Recall that in a symmetric group two elements are conjugate if and only if they share the same cycle type. Let $C$ be a conjugacy class in a symmetric group $S_d$. Recall that a conjugacy class generates a normal subgroup of the group. Therefore, if $d\geq 5$, as a quandle the inner automorphism group $\mathsf{Inn}(C)$ of $C$ is isomorphic to the symmetric group $S_d$ when $C$ is the conjugacy class of an odd permutation and it is isomorphic to the alternating group $A_d$ when $C$ is the conjugacy class of an even permutation different from the identity permutation. Observe that $C$ is faithful when $d\geq 5$ by Proposition~\ref{prop:act}, since the centers of $A_d$ and $S_d$ are trivial.

\begin{thm}\label{thm:sym}
  Let $C$ be a conjugacy class in a symmetric group $S_d$ which is connected as a quandle. If $\lambda = (\lambda_0^{a_0},\lambda_1^{a_1},\dots,\lambda_t^{a_t})$ is the profile of $C$, then $\lambda_s$ divides $\lambda_t$ for each $0\leq s \leq t$.
\end{thm}

\begin{proof}
  It is clear that the Theorem holds when $C$ is the conjugacy class of a central element. So we assume in the rest of the proof that $C$ is the conjugacy class of a non-central element.

  First, we prove the Theorem when the degree of $S_d$ is equal to three or four. If $d=3$, there are two possibilities. Either $C$ is the conjugacy class of $(1,2)$ or it is the conjugacy class of $(1,2,3)$. In the first case, the inner automorphism group is $S_3$ and since the center of $S_3$ is trivial, $C$ is faithful by Proposition~\ref{prop:act}. The Theorem holds in this case since the order of $(1,2)$ is a prime number. The latter possibility cannot occur as the conjugacy class of $(1,2,3)$ is not connected as a rack.

  If $d=4$, there are four possibilities. The conjugacy class of $(1,2)(3,4)$ generates an abelian subgroup so it is not connected. But in all other three cases the conjugacy class is connected as a quandle. Since the order of an element of the conjugacy class is a prime power in all those cases the Theorem holds when the degree of the symmetric group is four.

  Next, suppose $d\geq 5$.  From the previous explanations we know that $C$ is faithful and the inner automorphism group is isomorphic to either $A_d$ or $S_d$ when $d\geq 5$. Now, fix an element $x$ of $C$ and let $\mu = (\mu_0^{b_0},\mu_1^{b_1},\dots,\mu_r^{b_r})$ be the cycle type of $x$ in $S_d$. Moreover, since the map $\Phi\colon C\to \mathsf{Inn}(C)$ extends to a group isomorphism, using  Proposition~\ref{prop:re}(i), it is enough to show that $\langle x\rangle$ intersects the centralizer of some element $y$ of $C$ trivially. Suppose $r\geq 1$. Let $y$ be an element of $C$ having the following property: For any $0\leq s\leq r$ the $\mu_s$-parts of $x$ and $y$ are not coinciding. If the value of $r\geq 1$, the existence of $y$ in $S_d$ or $A_d$ is clear. Let $\ell$ be the order of $x$ and $k$ be an integer dividing $\ell$. Obviously, $x^k$ does not centralize $y$ unless $k=\ell$. Hence the Theorem holds when $C$ is the conjugacy class of an element containing at least two cycles of different lengths in its cycle decomposition. Next, suppose $r=0$. In that case $d = b_0\mu_0$ and the order of $x$ is $\mu_0$. In turn, the order of $\phi_x$ is $\mu_0$ as well and we may assume $\mu_0$ is neither $1$ nor a prime power since the Theorem holds trivially in those cases. Recall that the centralizer of $x$ in $S_d$ is isomorphic to $C_{\mu_0}\wr S_{b_0}$, where $C_{\mu_0}$ is the cyclic group of order $\mu_0$. Moreover, for some conjugate $y$ of $x$ the intersection of $\langle x\rangle$ with the centralizer of $y$ is trivial when $\mu_0\geq 6$. This completes the proof.
\end{proof}

The proof of Theorem~\ref{thm:sym} can be adapted to work when $C$ is the conjugacy class of an alternating group. Notice that a conjugacy class of a simple group is necessarily connected as a quandle. The only complication arises when a conjugacy class of $A_d$ is not a conjugacy class of $S_d$. Actually this may happen. Let $x$ be an even permutation of $S_n$ whose cycle type is $\mu = (\mu_0^{b_0},\mu_1^{b_1},\dots,\mu_r^{b_r})$. The conjugacy class of $x$ in $S_n$ splits into two different conjugacy classes in $A_n$ if $\mu_s$ is an odd number and $b_s=1$ for every $0\leq s\leq r$. Otherwise, the conjugacy class of $x$ in $S_n$ is same with the conjugacy class of $x$ in $A_n$. Even in the former case, fixing an element $x$ of $C$, we can show that there exist an element $y$ of $C$ so that the centralizer of $y$ intersects $\langle x\rangle$ trivially.


TITLE: Finding eigenfunctions and eigenvalues from a differential equation
QUESTION [1 upvotes]: Consider the differential equation
$$X''(x)+\lambda X=0$$
on $0 \leq x \leq 1$with boundary conditions
$$X'(0)+X(0)=0  \ \ \ \ \text{and}  \ \ \ \ X(1)=0.$$
I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix.  Or, maybe I am totally wrong!
$\textbf{My first goal:}$

Find an eigenfunction associated with eigenvalue $\lambda=0.$

An eigenvalue $\lambda =0$ would mean that $X''(x)=0$.  This means that the solution takes the form
$$X(x)=Ax+B.$$
Since $X'(0)=A$ and $X(0)=B$,
$$X'(0)+X(0)=0 \iff A=-B.$$
Therefore an eigenfunction that works would be $\boxed{X_{0}(x)=-2x+2}.$
$\textbf{My second goal:}$

Find an expression for all eigenvalues $\lambda = \beta ^2>0.$

This one requires a little more work.  The solution to $X'(x)+\beta ^2 X=0$ takes the form
$$X(x)=A\cos\beta x +B\sin \beta x.$$
Taking derivatives, one easily finds that $X'(0)=B\beta$ and $X(0)=A.$ Thus we obtain
$$B\beta + A=0 \implies \beta= \frac{-A}{B}.$$
Finally, this gives $\boxed{\beta=\frac{A^2}{B^2}}$
What are your thoughts?  Thanks in advance!
Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$

REPLY [0 votes]: Since you want $X(1)=0$, you cat set $X'(1)$ to any non-zero constant ($0$ leads to $X\equiv 0$.) So I'd pick a value of $1$. Then
$$
   X_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}}
$$
Choosing a constant value for $X'(1)$ forces the limiting case as $\lambda\rightarrow 0$ to be the correct solution for $\lambda=0$ as well,  which is
$$
    X_{0} =  x-1.
$$
You can check that this is an eigenfunction. So $\lambda=0$ is an eigenvalue, with corresponding eigenfucntion $x-1$.
The general eigenvalue equation becomes
$$
        X_{\lambda}(0)+X_{\lambda}'(0)=0 \\
      -\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda})=0
$$
The limit at $\lambda\rightarrow 0$ is $0$. So $\lambda_0=0$ is an eigenvalue with $X_{0}=x-1$. For $\lambda\ne 0$, the solutions are zeros of
$$
      \tan(\sqrt{\lambda})=\sqrt{\lambda}.
$$
This is a transcendental equation. You can plot the graphs of $y=\tan(x)$ and $y=x$, and check the intersections of the graphs for $x \ge 0$. The negative values of $\sqrt{\lambda}$ can be ignored because they lead to duplicate values of $\lambda$.
The non-negative solutions are ordered as follows:
$$
       \sqrt{\lambda_0} = 0 < \sqrt{\lambda_1} < \frac{\pi}{2} < \sqrt{\lambda_2} < \frac{3\pi}{2} < \sqrt{\lambda_3} < \frac{5\pi}{2} < \cdots
$$
Asymptotically, $\sqrt{\lambda_n} \approx \frac{(2n-1)\pi}{2}$ or $\lambda_n \approx \frac{(2n-1)^2\pi^2}{4}$ as $n\rightarrow\infty$.


TITLE: Are elements in $A_n$ the squares of elements in $S_n$?
QUESTION [0 upvotes]: Tt is true that elements in $A_4$ are squares of elements in $S_4$.
Is it true for $A_n$?

REPLY [1 votes]: $A_n$ is the set of squares for $n\le 5$. See oeis.org/A003483.
In general, $A_n$  is the set of products of squares, that is, the subgroup generated by the squares.


TITLE: Algebra help, logarithms, calculus
QUESTION [3 upvotes]: I'm seriously out of my depth. I have very basic understanding of logarithms and calculus. Please could someone walk me through how to get from:
$$\frac{{d}p}{p} = -\frac{g}{R}\cdot \frac{dh}{T_{0}-\lambda h}$$
To:
$$\log p = \frac{g}{\lambda R}\log(T_{0}-\lambda h) + constant$$
I'd appreciate terminology of any manipulations done so I can Google them for more information. Thank you!

REPLY [2 votes]: Using Clayton's hint (integrating both sides)...
$$\int\frac{dp}{p}= \int\frac{-g}{R}\frac{dh}{T_0-\lambda h}$$ this gives
$$\log(p)+C= -\frac{g}{R(-\lambda)}\int\frac{1}{u}du = \frac{g}{\lambda R}$$  where $u= T_0-\lambda h$, which gives $du = -\lambda dh$ and $C$ is our constant term.
Finally when substituting $u$ back in and combining our constant terms we get
$$\log(p)= \frac{g}{\lambda R}\log(T_0-\lambda h) + C$$


TITLE: Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$
QUESTION [8 upvotes]: Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2}  \right) ^2 dx =  \lim_{ A \to \infty }  \int_{0}^{A} \left( \frac{x}{1+x^2}  \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2}  \right) ^2 dx $ but I don't have idea.

REPLY [0 votes]: We can use Parseval's theorem!
If $F$ is the Fourier transform of $f$, and $G$ is the Fourier transform of $g$, then
$$
\int_{-\infty}^\infty\overline{f(t)}g(t)\,dt=
\frac{1}{2\pi}\int_{-\infty}^\infty\overline{F(x)}G(x)\,dx.
$$
We have the Fourier transform pairs

$\displaystyle e^{-|t|}\longmapsto\frac{2}{1+\omega^2}$
$\displaystyle \frac{d}{dt}f(t) \longmapsto i\omega F(\omega)$

It is clear from those that
$$
f(t)=\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\longmapsto
\frac{ix}{1+x^2}=F(x),
$$
and conveniently $\overline{F(x)}F(x)=\frac{x^2}{(1+x^2)^2}$ is the integrand that we want.
Using this and Parseval's theorem, we see that
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
2\pi\int_{-\infty}^\infty\left(\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\right)^2dt,
$$
and since the integrand is even, one can write
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
\pi\int_0^\infty\left(\frac{d}{dt}\left(e^{-t}\right)\right)^2dt=
\pi\int_0^\infty e^{-2t}dt=\frac{\pi}{2}.
$$


TITLE: $A289B$ is divisible by $90$, then what is $A+B$?
QUESTION [3 upvotes]: It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?

My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
since it is divisible by $90$, the sum of the remainders of $A \times 1000$ and $2890$ being divided by 90, must be divisible by 90.
We have  $mod(2890, 90) = 10$.
Now we may find $A$ such that $mod(A \times 10000, 90) = 80$.
$$mod(10000, 90) = 10 \implies 10000 = (90 \times 111 + 10) $$
so $A$ must be $8$, since
$$ 8 \times 10000 + 2890 = 8 \times (90 \times 111 + 10) + (90 \times 32 + 10) = 90 \times (888+32) + 90 $$
is divided by 90.
$$A+B = 8$$

Is this the best approach..? Thanks.

REPLY [0 votes]: You are going to a lot of effort to avoid noting that if $A2890=A289*10$ is divisible by $90$ and $\gcd(9,10) = 1$ then $A289$ is divisible by $9$.
And $A289 = A28*10 + 9$ being divisible by $9$ means $A28*10 + 9 \equiv A28 \equiv A27 + 1 \equiv 100A + 27 + 1 \equiv A + 1 \mod 9$.  And $A289\equiv 0 \mod 9$ so $A+1 \equiv 0 \mod 9$ so $A = 8$.
So $A+B = 8+0= 8$.
That's assuming you are not aware that if a number is divisible by $9$ then the sum of it's digits is divisible by $9$.  Which can be proven the exact same way.
$n = \sum_{k=0}n a_k 10^k = \sum_{k=0} a_k (10^k - 1) + \sum_{k=0}a_k$ and as $(10^k - 1) = 9*(\underbrace{1111....111}_{k}) \equiv 0 \mod 9$, we may conclude that  $n \mod 9 \equiv \sum_{k=0}a_k \mod 9$.


TITLE: If $x = a + b$, and only $x$ is known, how to solve what is $a-b$?
QUESTION [0 upvotes]: If $x$ equals to $a+b$, how can I solve what is $a-b$, knowing only $x$? (approximation will do as well, if it cannot be solved exactly)

REPLY [2 votes]: Let $a = x + k$, $b = x - k$, where $k$ is anything; then $a-b = 2k$.  In other words, you can't say anything about $a-b$, knowing only the value of $a+b$.

REPLY [1 votes]: You cannot solve for $a-b$, but you can restrict the solutions to a line.
If $x = a+b$, then $b = x-a$, so $a-b = a-(x-a) = 2a-x.$
Therefore, all possible solutions will lie along the line $c = 2a-x$, where $x$ is your fixed known quantity.