text:
\section{Introduction}

The theory of $n$-ary generalizations of Lie algebras has been studied since
the late 1960's, beginning with fundamental work by Russian mathematicians; see
the survey articles by Kurosh \cite{Kurosh} and Baranovich and Burgin
\cite{BaranovichBurgin}. This theory also appeared naturally and independently
in various domains of theoretical physics. Indeed, the discovery of Nambu
mechanics \cite{Nambu} in 1973, as well as the work of Fillippov
\cite{Filippov1} in 1985, gave impulse to a significant development of this
theory. From a physical point of view, we mention the work of Takhtajan
\cite{Takhtajan}, Michor and Vinogradov \cite{MichorVinogradov}, de Azc\'arraga
and P\'erez-Bueno \cite{deAzcarragaPerezBueno}, Gautheron \cite{Gautheron},
Vaisman \cite{Vaisman}, Curtright and Zachos \cite{CurtrightZachos}, Curtright
et al.~\cite{Curtright+} and Ataguema et al.~\cite{Ataguema+}. From an
algebraic point of view, we mention the work of Kasymov \cite{Kasymov}, Ling
\cite{Ling}, Hanlon and Wachs \cite{HanlonWachs}, Gnedbaye \cite{Gnedbaye},
Bremner \cite{Bremner2}, Filippov \cite{Filippov2}, Bremner and Hentzel
\cite{BremnerHentzel1} and Pozhidaev \cite{Pozhidaev}. The latest development
in mathematical physics related to $n$-ary algebras is the influential work of
Bagger and Lambert \cite{BaggerLambert} and Gustavsson \cite{Gustavsson}, which
aims at a world-volume theory of multiple M2-branes. For a very recent
comprehensive survey of this entire area, referring to both the physical and
mathematical literature, see de Azc\'arraga and Izquierdo
\cite{deAzcarragaIzquierdo}.

The main difficulty in this theory is to find a useful generalization of the
Jacobi identity. There are two principal candidates: (1) the derivation (or
Filippov) identity, which states that the multiplication operators in the
algebra are derivations of the $n$-ary structure, and (2) the alternating sum
identity, which states that the alternating sum over all possible nested pairs
of operations is identically zero. For $n = 2$ both of these identities reduce
to the familiar Jacobi identity for Lie algebras.

A serious disadvantage of the theory of Filippov algebras ($n$-Lie algebras) is
that for $n \ge 3$ it has been shown by Ling \cite{Ling} that there is only one
simple finite-dimensional object over an algebraically closed field of
characteristic 0. That is, Filippov algebras generalize the 3-dimensional
simple Lie algebra to the $n$-ary case for $n \ge 3$, but not any of the other
simple Lie algebras. It seems that the definition of Filippov algebra is too
restrictive: the derivation identity is too strong. From this point of view,
the alternating sum identity is a good candidate for a weaker identity which
can still be naturally regarded as a generalization of the Jacobi identity. It
is an open problem to classify the alternating $n$-ary algebras which satisfy
the alternating sum identity, but it is clear from the results in Bremner and
Hentzel \cite{BremnerHentzel2} and the present paper that there is more than
one simple object, at least for $n = 3$ and $n = 4$.

Our approach is to use the representation theory of Lie algebras to construct
new alternating algebra structures and to discover natural $n$-ary
generalizations of Lie algebras. In particular, we study the exterior powers of
an irreducible representation of a simple Lie algebra; these are important
objects in invariant theory, algebraic geometry, and the theory of Lie groups
(see for example Fulton and Harris \cite{FultonHarris}). If $\Lambda^n V$, the
$n$-th exterior power of an irreducible representation $V$ of a simple Lie
algebra $L$, contains $V$ itself as a direct summand, then the projection
$\Lambda^n V \to V$ defines an alternating $n$-ary algebra structure on $V$
which is $L$-invariant in the sense that the derivation algebra of this $n$-ary
structure contains a subalgebra isomorphic to $L$. This approach was initiated
by Bremner and Hentzel \cite{BremnerHentzel2} in the case of the third exterior
powers of irreducible representations of the 3-dimensional simple Lie algebra
$\mathfrak{sl}_2(\mathbb{C})$. In this paper we extend this work to the fourth
exterior power.

We recall some basic information about representation theory of Lie algebras.
The simple Lie algebra $\sltwo$ has basis $\{ H, E, F \}$ and structure
constants
  \[
  [H,E] = 2E, \qquad [H,F] = -2F, \qquad [E,F] = H.
  \]
All other brackets follow from bilinearity and anticommutativity. For $n \in
\mathbb{Z}$, $n \ge 0$, the irreducible representation $V(n)$ of $\sltwo$ with
highest weight $n$ has dimension $n+1$; the action of $\sltwo$ with respect to
the basis $\{ \, v_{n-2i} \, | \, i=0,\hdots,n \, \}$ is
  \allowdisplaybreaks
  \begin{align}
  &
  H.v_{n-2i} = (n{-}2i) v_{n-2i},
  \label{Haction}
  \\
  &
  E.v_n = 0,
  \qquad
  E.v_{n-2i} = (n{-}i{+}1) v_{n-2i+2}
  \;
  (i=1,\hdots,n),
  \label{Eaction}
  \\
  &
  F.v_{n-2i} = (i{+}1) v_{n-2i-2}
  \;
  (i=0,\hdots,n-1) \, ,
  \qquad
  F.v_{-n} = 0.
  \label{Faction}
  \end{align}
Any finite dimensional irreducible representation of $\sltwo$ is isomorphic to
$V(n)$ for some $n$. Any finite dimensional representation of $\sltwo$ is
isomorphic to a direct sum of irreducible representations. The multiplicity of
$V(n)$ in its $k$-th exterior power $\Lambda^k V(n)$ is the dimension of the
vector space
  \[
  \mathrm{Hom}_{\sltwo}\big( \Lambda^k V(n), V(n) \big)
  \]
of $\sltwo$-invariant linear maps $P \colon \Lambda^k V(n) \to V(n)$. If this
multiplicity is positive then $P$ defines an alternating $k$-ary algebra
structure on $V(n)$,
  \[
  [x_1,\hdots,x_k] = P( x_1 \wedge \cdots \wedge x_k ),
   \]
which is $\sltwo$-invariant in the sense that the action of any $L \in \sltwo$
is a derivation of the $k$-ary multiplication: for any $x_1,\hdots,x_k \in
V(n)$ we have
  \[
  L.[x_1,\hdots,x_i,\hdots,x_k]
  =
  \sum_{i=1}^k
  [ x_1, \hdots, L.x_i, \hdots, x_k ],
  \]
For the representation theory of $\sltwo$ we refer to Humphreys
\cite{Humphreys}.

\section{Multiplicity formulas}

Bremner and Hentzel \cite{BremnerHentzel1} studied the case $k = 2$
corresponding to alternating binary algebra structures on $V(n)$. We have
  \[
  \dim \mathrm{Hom}_{\sltwo}\big( \Lambda^2 V(n), V(n) \big)
  =
  \begin{cases}
  1 &\text{if $n \equiv 2$ (mod 4)}, \\
  0 &\text{otherwise}.
  \end{cases}
  \]
In this case $n = 2$ gives the 3-dimensional adjoint representation of
$\sltwo$, $n = 6$ gives the 7-dimensional simple non-Lie Malcev algebra, and $n
= 10$ gives a new 11-dimensional anticommutative algebra satisfying a
polynomial identity of degree 7. Bremner and Hentzel \cite{BremnerHentzel2}
studied the case $k = 3$ corresponding to alternating ternary algebra
structures on $V(n)$. For $n = 6q + r$ ($0 \le r\le 5$) we have
  \[
  \dim \mathrm{Hom}_{\sltwo}\big( \Lambda^3 V(n), V(n) \big)
  =
  \begin{cases}
  q &\text{if $r = 0, 1, 2, 4$}, \\
  q+1 &\text{if $r = 3, 5$}.
  \end{cases}
  \]
The multiplicity is 1 for $n = 3, 5, 6, 7, 8, 10$; the corresponding $V(n)$
provide new examples of alternating ternary algebras.

In this paper we consider the case $k = 4$: we study alternating quaternary
algebra structures on $V(n)$ obtained from $\sltwo$-invariant linear maps
$\Lambda^4 V(n) \to V(n)$.  We first obtain a closed formula for the
multiplicity $\dim \mathrm{Hom}_{\sltwo} \big( \Lambda^4 V(n), V(n) \big)$
using a general approach which applies to arbitrary exterior powers.

\begin{theorem} \label{multiplicitytheorem}
If $n$ is odd then $\dim \mathrm{Hom}_{\sltwo}\big( \Lambda^4 V(n), V(n) \big)
= 0$. If $n$ is even then $n = 24 q + r$ with $0 \le r < 24$ ($r$ even) and we
have
  \begin{align*}
  &
  \dim \mathrm{Hom}_{\sltwo}\big( \Lambda^4 V(n), V(n) \big) =
  \\
  &
  \frac{1}{1152}
  \left\{\three
  \begin{array}{ll|ll}
  30n^2+96n &\text{if} \;\; r = 0, 16 &
  30n^2-120n+120 &\text{if} \;\; r = 2 \\
  30n^2+96n+288 &\text{if} \;\; r = 4, 12 &
  30n^2-120n+792 &\text{if} \;\; r = 6, 22 \\
  30n^2+96n-384 &\text{if} \;\; r = 8 &
  30n^2-120n+504 &\text{if} \;\; r = 10, 18 \\
  30n^2-120n+408 &\text{if} \;\; r = 14 &
  30n^2+96n-96 &\text{if} \;\; r = 20
  \end{array}
  \right.
  \end{align*}
\end{theorem}

\begin{proof}
A proof using P\'olya enumeration is given in Section \ref{proofsection}.
\end{proof}

\begin{example}
The multiplicities for $n = 24q + r$ ($0 \le q \le 9$) are given in Table
\ref{multiplicityexample}.
\end{example}

\begin{corollary}
The representation $V(n)$ of $\sltwo$ occurs in $\Lambda^4 V(n)$ with
multiplicity 1 (resp.~2) if and only if $n = 4$ or $n = 6$ (resp.~$n = 8$ or $n
= 10$).
\end{corollary}

\begin{proof}
The vertices of the parabolas in Theorem \ref{multiplicitytheorem} occur at
either $n = -8/5$ or $n = 2$, so for each $r$ the multiplicity is an increasing
function of $q$.
\end{proof}

For $n = 4, 6, 8, 10$ we use computational linear algebra to find all the
multilinear polynomial identities of degree $\le 7$ satisfied by the resulting
quaternary algebras.

\section{Quaternary algebra structures}

In this section we explain how to compute explicitly the decomposition of
$\Lambda^4 V(n)$ as a direct sum of irreducible representations of $\sltwo$,
together with an explicit multiplication table for the alternating quaternary
algebra structure on $V(n)$ obtained from a projection $\Lambda^4 V(n) \to
V(n)$. Recall that $V(n)$ has the vector space basis $\{ \, v_{n-2i} \mid i =
0, 1, \hdots, n \, \}$ and that the subscript on $v_{n-2i}$ is its weight: its
eigenvalue for the action of $H \in \sltwo$.

\begin{lemma} \label{multiplicityformula} \cite[Lemma 5.1]{BremnerHentzel2}
Let $M$  be an $\sltwo$-module with $\dim M < \infty$. For $n \in \mathbb{Z}$
let $M_n = \{ v \in M \, | \, H.v = nv \}$ be the subspace of all vectors of
weight $n$ together with $0$. For $n \ge 0$ the multiplicity of $V(n)$ in the
decomposition of $M$ as a direct sum of simple $\sltwo$-modules is $\dim M_n -
\dim M_{n+2}$.
\end{lemma}

\begin{definition} \label{definitiontensorbasis}
The \textbf{tensor basis} of $\Lambda^4 V(n)$ consists of $\binom{n{+}1}{4}$
\textbf{quadruples}:
  \[
  v_p \wedge v_q \wedge v_r \wedge v_s
  =
  \sum_{\sigma \in S_4}
  \epsilon(\sigma)
  \big(
  v_{\sigma(p)} \otimes v_{\sigma(q)} \otimes v_{\sigma(r)} \otimes v_{\sigma(s)}
  \big),
  \]
where $n \ge p > q > r > s \ge -n$ with $p,q,r,s \equiv n \, (\mathrm{mod}\,2)$
and $\epsilon\colon S_4 \to \{\pm1\}$ is the sign homomorphism. We usually
abbreviate $v_p \wedge v_q \wedge v_r \wedge v_s$ by $[p,q,r,s]$.  The action
of $L \in \sltwo$ satisfies the derivation property,
  \begin{equation} \label{commutingformula}
  \begin{array}{r}
  L . ( v_p \wedge v_q \wedge v_r \wedge v_s )
  =
  L . v_p \wedge v_q \wedge v_r \wedge v_s
  +
  v_p \wedge L . v_q \wedge v_r \wedge v_s
  \qquad
  \\
  {}
  +
  v_p \wedge v_q \wedge L . v_r \wedge v_s
  +
  v_p \wedge v_q \wedge v_r \wedge L . v_s,
  \end{array}
  \end{equation}
and hence the \textbf{weight} of the quadruple $T = [p,q,r,s]$ is $w(T) =
p+q+r+s$. The \textbf{standard order} of the quadruples is given by decreasing
weight, and within each weight by reverse lex order: $T = [p,q,r,s]$ precedes
$T' = [p',q',r',s']$ if and only if either $w(T) > w(T')$ or $w(T) = w(T')$ and
$t > t'$ where $t, t'$ are the components of $T, T'$ in the leftmost position
where the components are not equal.
\end{definition}

\begin{example} \label{tensorbasis}
The quadruples in the tensor basis of $\Lambda^4 V(n)$ for $n = 4, 6, 8$ are
displayed in Tables \ref{tensorweightvectorbasistable4},
\ref{tensorbasistable6}, \ref{tensorbasistable8} in standard order.
\end{example}

\begin{remark} \label{completedecomposition}
If we apply Lemma \ref{multiplicityformula} to Tables
\ref{tensorweightvectorbasistable4}, \ref{tensorbasistable6},
\ref{tensorbasistable8} then we obtain the decomposition of $\Lambda^4 V(n)$
for $n = 4, 6, 8$ as a direct sum of irreducible representations:
  \begin{align*}
  \Lambda^4 V(4)
  &\cong
  V(4),
  \qquad
  \Lambda^4 V(6)
  \cong
  V(12) \oplus V(8) \oplus V(6) \oplus V(4) \oplus V(0).
  \\
  \Lambda^4 V(8)
  &\cong
  V(20) {\oplus}
  V(16) {\oplus}
  V(14) {\oplus}
  2 V(12) {\oplus}
  V(10) {\oplus}
  2 V(8) {\oplus}
  V(6) {\oplus}
  2 V(4)

  {\oplus}
  V(0).
  \end{align*}
Similar computations establish the decomposition of $\Lambda^4 V(10)$:
  \begin{align*}
  \Lambda^4 V(10)
  &\cong
  V(28) \oplus
  V(24) \oplus
  V(22) \oplus
  2 V(20) \oplus
  V(18) \oplus
  3 V(16) \oplus
  2 V(14)
  \\
  &\quad
  \oplus
  3 V(12) \oplus
  2 V(10) \oplus
  3 V(8) \oplus
  V(6) \oplus
  3 V(4) \oplus
  V(0).
  \end{align*}
\end{remark}

The next step is to determine the highest weight vectors for the irreducible
summands of $\Lambda^4 V(n)$ as linear combinations of the quadruples in the
tensor basis.

\begin{lemma} \label{topvector}
The quadruple $[n,n{-}2,n{-}4,n{-}6]$ is the quadruple with highest weight in
$\Lambda^4 V(n)$ and is a highest weight vector for the summand $V(4n{-}12)$.
\end{lemma}

\begin{proof}
This follows directly from equations \eqref{Eaction} and
\eqref{commutingformula}.
\end{proof}

\begin{example}
For $n = 4$ we have $\Lambda^4 V(4) \cong V(4)$, and so the quadruple
$[4,2,0,-2]$ is the only highest weight vector in $\Lambda^4 V(4)$. If we
identify $[4,2,0,-2]$ with the highest weight vector $v_4$ of $V(4)$, and
repeatedly apply $F$ using equations \eqref{Faction} and
\eqref{commutingformula}, then we obtain the weight vectors of $\Lambda^4 V(4)$
corresponding to the basis vectors $v_2$, $v_0$, $v_{-2}$, $v_{-4}$ of $V(4)$:
  \begin{align*}
  &
  v_4 = [4,2,0,-2], \quad
  v_2 = F . v_4 = 4 [4,2,0,-4 ], \quad
  v_0 = \frac{1}{2!} F^2 . v_4 = 6 [4,2,-2,-4 ],
  \\
  &
  v_{-2} = \frac{1}{3!} F^3 . v_4 = 4 [4,0,-2,-4 ], \quad
  v_{-4} = \frac{1}{4!} F^4 . v_4 = [2,0,-2,-4 ].
  \end{align*}
The matrix expressing the weight vectors in $V(4)$ in terms of the quadruples
in $\Lambda^4 V(4)$ is $C = \mathrm{diag}(1,4,6,4,1)$. The matrix expressing
the quadruples in $\Lambda^4 V(4)$ in terms of the weight vectors in $V(4)$ is
$C^{-1} = \mathrm{diag}(1,\frac14,\frac16,\frac14,1)$. We now have the
structure constants for the $\sltwo$-invariant alternating quaternary algebra
structure on $V(4)$, which we denote by $[v_p,v_q,v_r,v_s]$:
  \begin{alignat*}{3}
  [v_4,v_2,v_0,v_{-2}] &= v_4,
  &\quad
  [v_4,v_2,v_0,v_{-4}] &= \tfrac14 v_4,
  &\quad
  [v_4,v_2,v_{-2},v_{-4}] &= \tfrac16 v_4,
  \\
  [v_4,v_0,v_{-2},v_{-4}] &= \tfrac14 v_4,
  &\quad
  [v_2,v_0,v_{-2},v_{-4}] &= v_4.
  \end{alignat*}
The LCM of the denominators of the coefficients is 12. Taking $a = \root 3 \of
{12}$ and setting $v'_t = v_t/a$, we obtain integral structure constants (see
Table \ref{structureconstants4}):
  \begin{alignat*}{3}
  [v'_4,v'_2,v'_0,v'_{-2}] &= 12 v'_4,
  &\quad
  [v'_4,v'_2,v'_0,v'_{-4}] &= 3 v'_4,
  &\quad
  [v'_4,v'_2,v'_{-2},v'_{-4}] &= 2 v'_4,
  \\
  [v'_4,v'_0,v'_{-2},v'_{-4}] &= 3 v'_4,
  &\quad
  [v'_2,v'_0,v'_{-2},v'_{-4}] &= 12 v'_4.
  \end{alignat*}
\end{example}

In general, for all other weights $w < 4n{-}12$ we need to find a basis for the
subspace of highest weight vectors of weight $w$ in $\Lambda^4 V(n)$. The
dimension of this subspace is the multiplicity of $V(w)$ as a summand of
$\Lambda^4 V(n)$.

\begin{definition} \label{definitionEmatrix}
Suppose that $4n{-}14 \ge w \ge 0$ ($w$ even). Let $d(w)$ be the dimension of
the weight space of weight $w$ in $\Lambda^4 V(n)$: the number of quadruples of
weight $w$. We define the matrix $E^{(n)}_w$ of size $d(w{+}2) \times d(w)$ by
setting the $(i,j)$ entry equal to the coefficient of the $i$-th quadruple of
weight $w{+}2$ in the expression for the action of $E \in \sltwo$ on the $j$-th
quadruple of weight $w$.  We call this the \textbf{E-action matrix} for weight
$w$ of $\Lambda^4 V(n)$; the nonzero vectors in its nullspace are the highest
weight vectors of weight $w$ in $\Lambda^4 V(n)$. We compute the row canonical
form (RCF) and extract the canonical integral basis (CIB) by setting the free
variables equal to the standard unit vectors, clearing denominators, and
canceling common factors.
\end{definition}

\begin{example} \label{highestweightvectors6}
For $\Lambda^4 V(6)$ we use the weight space basis of Table
\ref{tensorbasistable6} and obtain
  \allowdisplaybreaks
  \begin{align*}
  E^{(6)}_8
  &=
  \left[ \begin{array}{rr}
  2 & 4
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rr}
  1 & 2
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rr}
  -2 & 1
  \end{array} \right]
  \\[2pt]
  E^{(6)}_6
  &=
  \left[ \begin{array}{rrr}
  1 &\!\! 4 &\!\! . \\
  . &\!\! 2 &\!\! 5
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrr}
  1 &\!\! . &\!\! -10 \\
  . &\!\! 1 &\!\! 5/2
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrr}
  20 &\!\! -5 &\!\! 2
  \end{array} \right]
  \\[2pt]
  E^{(6)}_4
  &=
  \left[ \begin{array}{rrrr}
  4 &\!\! . &\!\! . &\!\! . \\
  1 &\!\! 3 &\!\! 5 &\!\! . \\
  . &\!\! . &\!\! 2 &\!\! 6
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrr}
  1 &\!\! . &\!\! . &\!\! . \\
  . &\!\! 1 &\!\! . &\!\! -5 \\
  . &\!\! . &\!\! 1 &\!\! 3
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrr}
  . &\!\! 5 &\!\! -3 &\!\! 1
  \end{array} \right]
  \\[2pt]
  E^{(6)}_0
  &=
  \left[ \begin{array}{rrrrr}
  2 &\!\! 5 &\!\! . &\!\! . &\!\! . \\
  . &\!\! 3 &\!\! . &\!\! 6 &\!\! . \\
  . &\!\! 1 &\!\! 4 &\!\! . &\!\! 6 \\
  . &\!\! . &\!\! . &\!\! 1 &\!\! 3
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrr}
  1 &\!\! . &\!\! . &\!\! . &\!\! 15 \\
  . &\!\! 1 &\!\! . &\!\! . &\!\! -6 \\
  . &\!\! . &\!\! 1 &\!\! . &\!\! 3 \\
  . &\!\! . &\!\! . &\!\! 1 &\!\! 3
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrr}
  -15 &\!\! 6 &\!\! -3 &\!\! -3 &\!\! 1
  \end{array} \right]
  \end{align*}
\end{example}

\begin{example} \label{highestweightvectors8}
For $\Lambda^4 V(8)$ we use the weight space basis of Table
\ref{tensorbasistable8} and obtain
  \allowdisplaybreaks
  \begin{align*}
  E^{(8)}_{16}
  &=
  \left[ \begin{array}{rr}
  4 &\three 6
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rr}
  1 &\three 3/2
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rr}
  -3 &\three 2
  \end{array} \right]
  \\[4pt]
  E^{(8)}_{14}
  &=
  \left[ \begin{array}{rrr}
  3 &\three 6 &\three . \\
  . &\three 4 &\three 7
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrr}
  1 &\three . &\three -7/2 \\
  . &\three 1 &\three 7/4
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrr}
  14 &\three -7 &\three 4
  \end{array} \right]
  \\[4pt]
  E^{(8)}_{12}
  &=
  \left[ \begin{array}{rrrrr}
  2 &\three 6 &\three . &\three . &\three . \\
  . &\three 3 &\three 5 &\three 7 &\three . \\
  . &\three . &\three . &\three 4 &\three 8
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrr}
  1 &\three . &\three -5 &\three  . &\three 14 \\
  . &\three 1 &\three 5/3 &\three . &\three -14/3 \\
  . &\three . &\three . &\three 1 &\three 2
  \end{array} \right]
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrr}
   15 &\three  -5 &\three 3 &\three  . &\three . \\
  -42 &\three 14 &\three . &\three -6 &\three 3
  \end{array} \right]
  \\[4pt]
  E^{(8)}_{10}
  &=
  \left[ \begin{array}{rrrrrr}
  1 &\three 6 &\three . &\three . &\three . &\three . \\
  . &\three 2 &\three 5 &\three 7 &\three . &\three . \\
  . &\three . &\three 3 &\three . &\three 7 &\three . \\
  . &\three . &\three . &\three 3 &\three 5 &\three 8 \\
  . &\three . &\three . &\three . &\three . &\three 4
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrrr}
  1 &\three . &\three . &\three . &\three 70 &\three . \\
  . &\three 1 &\three . &\three . &\three -35/3 &\three . \\
  . &\three . &\three 1 &\three . &\three 7/3 &\three . \\
  . &\three . &\three . &\three 1 &\three 5/3 &\three . \\
  . &\three . &\three . &\three . &\three . &\three 1
  \end{array} \right]
  \\
  &\qquad
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrrr}
  -210 &\three 35 &\three -7 &\three -5 &\three 3 &\three .   \end{array} \right]
  \\[4pt]
  E^{(8)}_8
  &=
  \left[ \begin{array}{rrrrrrrr}
  6 &\three . &\three . &\three . &\three . &\three . &\three . &\three . \\
  1 &\three 5 &\three . &\three 7 &\three . &\three . &\three . &\three . \\
  . &\three 2 &\three 4 &\three . &\three 7 &\three . &\three . &\three . \\
  . &\three . &\three . &\three 2 &\three 5 &\three . &\three 8 &\three . \\
  . &\three . &\three . &\three . &\three 3 &\three 6 &\three . &\three 8 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 3 &\three 5
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrrrrr}
  1 &\three . &\three . &\three . &\three . &\three . &\three . &\three . \\
  . &\three 1 &\three . &\three . &\three . &\three 7 &\three . &\three 56/3 \\
  . &\three . &\three 1 &\three . &\three . &\three -7 &\three . &\three -14 \\
  . &\three . &\three . &\three 1 &\three . &\three -5 &\three . &\three -40/3 \\
  . &\three . &\three . &\three . &\three 1 &\three 2 &\three . &\three 8/3 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three 5/3
  \end{array} \right]
  \\
  &\qquad
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrrrrr}
  . &\three -7 &\three 7 &\three 5 &\three -2 &\three 1 &\three . &\three . \\
  . &\three -56 &\three 42 &\three 40 &\three -8 &\three . &\three -5 &\three 3
  \end{array} \right]
  \\[4pt]
  E^{(8)}_6
  &=
  \left[ \begin{array}{rrrrrrrrr}
  5 &\three . &\three 7 &\three . &\three . &\three . &\three . &\three . &\three . \\
  1 &\three 4 &\three . &\three 7 &\three . &\three . &\three . &\three . &\three . \\
  . &\three 2 &\three . &\three . &\three 7 &\three . &\three . &\three . &\three . \\
  . &\three . &\three 1 &\three 5 &\three . &\three . &\three 8 &\three . &\three . \\
  . &\three . &\three . &\three 2 &\three 4 &\three 6 &\three . &\three 8 &\three . \\
  . &\three . &\three . &\three . &\three . &\three 3 &\three . &\three . &\three 8 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 2 &\three 5 &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 3 &\three 6
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrrrrrr}
  1 &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three -224/3 \\
  . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three . &\three 70/3 \\
  . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three 160/3 \\
  . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three -8/3 \\
  . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three -20/3 \\
  . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three 8/3 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three -5 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three 2
  \end{array} \right]
  \\
  &\qquad
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrrrrrr}
  224 &\three -70 &\three -160 &\three 8 &\three 20 &\three -8 &\three 15 &\three -6
  &\three 3
  \end{array} \right]
  \\[4pt]
  E^{(8)}_4
  &=
  \left[ \begin{array}{rrrrrrrrrrr}
  4 &\three . &\three 7 &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . \\
  1 &\three 3 &\three . &\three 7 &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . \\
  . &\three . &\three 5 &\three . &\three . &\three . &\three 8 &\three . &\three . &\three .
  &\three . \\
  . &\three . &\three 1 &\three 4 &\three 6 &\three . &\three . &\three 8 &\three . &\three .
  &\three . \\
  . &\three . &\three . &\three 2 &\three . &\three 6 &\three . &\three . &\three 8 &\three .
  &\three . \\
  . &\three . &\three . &\three . &\three 2 &\three 4 &\three . &\three . &\three . &\three 8
  &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three 5 &\three . &\three .
  &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 2 &\three 4 &\three 6
  &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three 3
  &\three 7
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrrrrrrrr}
  1 &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three -28 &\three .
  &\three 98 \\
  . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three . &\three 14 &\three .
  &\three -245/3 \\
  . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three 16 &\three .
  &\three -56 \\
  . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three -2 &\three .
  &\three 21 \\
  . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three -4 &\three .
  &\three 14/3 \\
  . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three 2 &\three .
  &\three -7 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three -10 &\three .
  &\three 35 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three 2 &\three .
  &\three -7 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1
  &\three 7/3 \\
  \end{array} \right]
  \\
  &\qquad
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrrrrrrrr}
  28 &\three -14 &\three -16 &\three 2 &\three 4 &\three -2 &\three 10 &\three -2
  &\three 1 &\three . &\three . \\
  -294 &\three 245 &\three 168 &\three -63 &\three -14 &\three 21 &\three -105
  &\three 21 &\three . &\three -7 &\three 3
  \end{array} \right]
  \\[4pt]
  E^{(8)}_0
  &=
  \left[ \begin{array}{rrrrrrrrrrrr}
  2 &\three 7 &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . &\three . \\
  . &\three 3 &\three 6 &\three . &\three . &\three 8 &\three . &\three . &\three . &\three .
  &\three . &\three . \\
  . &\three 1 &\three . &\three 6 &\three . &\three . &\three 8 &\three . &\three . &\three .
  &\three . &\three . \\
  . &\three . &\three 4 &\three . &\three . &\three . &\three . &\three 8 &\three . &\three .
  &\three . &\three . \\
  . &\three . &\three 1 &\three 3 &\three 5 &\three . &\three . &\three . &\three 8 &\three .
  &\three . &\three . \\
  . &\three . &\three . &\three . &\three 2 &\three . &\three . &\three . &\three . &\three 8
  &\three . &\three . \\
  . &\three . &\three . &\three . &\three . &\three 4 &\three . &\three 6 &\three . &\three .
  &\three . &\three . \\
  . &\three . &\three . &\three . &\three . &\three 1 &\three 3 &\three . &\three 6 &\three .
  &\three . &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three 4 &\three .
  &\three 7 &\three . \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three 2 &\three 5
  &\three . &\three 7 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three 2 &\three 4
  \end{array} \right]
  \xrightarrow{\mathrm{RCF}}
  \left[ \begin{array}{rrrrrrrrrrrr}
  1 &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . &\three   224 \\
  . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . &\three   -64 \\
  . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . &\three    16 \\
  . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three . &\three .
  &\three . &\three    12 \\
  . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three . &\three .
  &\three . &\three    -8 \\
  . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three . &\three .
  &\three . &\three    12 \\
  . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three . &\three .
  &\three . &\three    -1 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three . &\three .
  &\three . &\three   -8 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1 &\three .
  &\three . &\three  -\frac32 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three 1
  &\three . &\three 2 \\
  . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three . &\three .
  &\three 1 &\three 2
  \end{array} \right]
  \\
  &\qquad
  \xrightarrow{\mathrm{CIB}}
  \left[ \begin{array}{rrrrrrrrrrrr}
  -448 &\three 128 &\three -32 &\three -24 &\three 16 &\three -24 &\three 2 &\three 16
  &\three 3 &\three -4 &\three -4 &\three 2
  \end{array} \right]
  \end{align*}
\end{example}

\begin{definition} \label{weightvectorbasis}
We construct the \textbf{weight vector basis} of $\Lambda^4 V(n)$ as follows.
We first determine the highest weight vectors for each summand of $\Lambda^4
V(n)$ as described in Definition \ref{definitionEmatrix} and Examples
\ref{highestweightvectors6} and \ref{highestweightvectors8}. For each highest
weight vector $X$ (of weight $w$, say) we apply $F \in \sltwo$ repeatedly $w$
times to obtain weight vectors of weights $w{-}2, w{-}4, \hdots, -w$ forming a
basis of the summand isomorphic to $V(w)$:
  \[
  X, \quad F.X, \quad \frac{1}{2!} F^2.X, \quad \hdots, \quad \frac{1}{w!} F^w.X.
  \]
The set of all these weight vectors is the weight vector basis of $\Lambda^4
V(n)$. The \textbf{standard order} on this basis is as follows: We order the
weight vectors first by decreasing weight of the corresponding highest weight
vector and then by increasing power of $F$ within each summand. (When there is
more than one highest weight vector with the same weight, we order them as in
the canonical integral basis.)
\end{definition}

\begin{example}
The weight vector basis of $\Lambda^4 V(6)$ is displayed in Table
\ref{weightvectorbasistable6}. The weight vector basis of $\Lambda^4 V(8)$ is
displayed in Tables
\ref{weightvectorbasistable8part1}--\ref{weightvectorbasistable8part3}.
\end{example}

\begin{definition} \label{weightvectormatrix}
The \textbf{weight vector matrix} $C$ is the $\binom{n{+}1}{4} \times
\binom{n{+}1}{4}$ matrix which expresses the weight vector basis in terms of
the tensor basis: the $(i,j)$ entry is the coefficient of the $i$-th quadruple
in the $j$-th element of the weight vector basis.
\end{definition}

\begin{definition}
The \textbf{alternating quaternary algebra structure} on $V(n)$ is defined in
terms of structure constants as follows. The inverse $C^{-1}$ of the weight
vector matrix expresses the tensor basis in terms of the weight vector basis.
Let $[p,q,r,s]$ be the $j$-th quadruple in the tensor basis. Column $j$ of
$C^{-1}$ expresses $[p,q,r,s]$ as a linear combination of the elements of the
weight vector basis. Suppose that the highest weight vector for the summand of
$\Lambda^4 V(n)$ isomorphic to $V(n)$ is the $k$-th element of the weight
vector basis.  The entries of $C^{-1}$ in column $j$ and rows $i = k, \hdots,
k{+}n$ are the coefficients of the projection of $[p,q,r,s]$ onto the summand
isomorphic to $V(n)$.  Let $P \colon \Lambda^4 V(n) \to V(n)$ be this
surjective homomorphism of $\sltwo$-modules. The quadruple $[p,q,r,s]$ has
weight $p{+}q{+}r{+}s$, and the summand isomorphic to $V(n)$ has (at most) one
basis vector of this weight. Hence there is at most one nonzero entry in
$C^{-1}$ in column $j$ and rows $i = k, \hdots, k{+}n$. If all these entries
are zero then $P( v_p \wedge v_q \wedge v_r \wedge v_s ) = 0$. If there is a
nonzero entry, say in row $\ell$, then $P( v_p \wedge v_q \wedge v_r \wedge v_s
) = (C^{-1})_{i\ell} \, v_{p+q+r+s}$. The resulting alternating quaternary
algebra structure on $V(n)$ is denoted by $[v_p,v_q,v_r,v_s]$ and defined by
$[v_p,v_q,v_r,v_s] = P( v_p \wedge v_q \wedge v_r \wedge v_s )$.
\end{definition}

\begin{example}
For $n = 6$, from rows 23 to 29 of the matrix inverse we obtain the structure
constants for the alternating quaternary algebra structure on $V(6)$. We ignore
the equations for which $|p{+}q{+}r{+}s| > 6$ since in these cases the result
is obviously zero: there is no vector of the given weight in $V(6)$. The LCM of
the denominators of the coefficients is 120, so we can scale the basis vectors
of $V(n)$ by setting $v'_t = v_t / \root 3 \of {120}$ to obtain integral
structure constants. We use the shorthand notation $[p,q,r,s] = c$ for the
equation $[v'_p,v'_q,v'_r,v'_s] = c v'_{p+q+r+s}$. After making these changes
we obtain the structure constants in Table \ref{structureconstants6}.
\end{example}

\begin{remark}
In this paper we also study the alternating quaternary algebra structures on
$V(10)$ but in this case the tables of the tensor basis, the weight vector
basis, and the structure constants are too large to include.
\end{remark}

\section{Polynomial identities and computational methods}

\begin{definition}
The nonassociative polynomial $I(x_1,\dots, x_n)$ is a \textbf{polynomial
identity} for the algebra $A$ if $I( x_1, \hdots, x_n ) = 0$ for all
$x_1,\dots, x_n\in A$.
\end{definition}

We are concerned with multilinear polynomial identities of degree $n$ for an
alternating quaternary algebra.  This means that each term consists of a
coefficient and a monomial, where the monomial is some permutation of $n$
distinct variables $x_1$, $x_2$, $\hdots$, $x_n$ together with some association
type, by which we mean some placement of brackets representing the quaternary
operation. For any $k$-ary operation, the degree of a monomial has the form $n
= 1 + \ell(k{-}1)$ where $\ell$ is the number of occurrences of the operation
in the monomial. Thus for a quaternary operation the degree of a monomial is
congruent to 1 modulo 3.

In degree 4, we have only the single association type $[-,-,-,-]$; the
alternating property implies that we have only the single monomial
$[x_1,x_2,x_3,x_4]$. In degree 7, the alternating property implies that we have
only one association type $[[-,-,-,-],-,-,-]$ and only $\binom74 = 35$ distinct
multilinear monomials,
  \[
  [
  [x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)} ],
  x_{\sigma(5)}, x_{\sigma(6)}, x_{\sigma(7)}
  ],
  \]
where $\sigma \in S_7$ is a $(4,3)$-shuffle; that is, $1 \le \sigma(1) <
\sigma(2) < \sigma(3) < \sigma(4) \le 7$ and $1 \le \sigma(5) < \sigma(6) <
\sigma(7) \le 7$. In degree 10, the alternating property implies that we have
two association types,
  \[
  [[[-,-,-,-],-,-,-],-,-,-],
  \qquad
  [[-,-,-,-],[-,-,-,-],-,-],
  \]
and that the corresponding numbers of distinct multilinear monomials are
  \[
  \binom{10}{4,3,3} + \frac12 \binom{10}{4,4,2}
  =
  4200 + 1575
  =
  5775.
  \]
The number $T(\ell)$ of association types which involve $\ell$ occurrences of
an alternating quaternary product is equal to the number of rooted trees with
$\ell$ internal vertices in which each internal vertex has four children; see
Sloane \cite{Sloane}, sequence A036718. The first terms in this sequence are
  \[
  \begin{array}{lrrrrrrrrrrrrrrr}
  \ell & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
  T(\ell) & 1 & 1 & 1 & 2 & 4 & 9 & 19 & 45 & 106 & 260 & 643 & 1624 & 4138 & 10683
  \end{array}
  \]
The monomials $[[[a,b,c,d],e,f,g],h,i,j]$ and $[[a,b,c,d],[e,f,g,h],i,j]$
correspond respectively to the following trees:
  \begin{center}
  \Tree [ . [ . [ . $a$ $b$ $c$ $d$ ] $e$ $f$ $g$ ] $h$ $i$ $j$ ]
  \\
  \Tree [ . [ . $a$ $b$ $c$ $d$ ] [ . $e$ $f$ $g$ $h$ ] $i$ $j$ ]
  \end{center}

\subsection{Fill-and-reduce algorithm}

Suppose we wish to find all the multilinear polynomial identities of degree $n$
satisfied by an algebra $A$ of dimension $d$.  We assume that we have chosen a
basis of $A$ and that we know the structure constants with respect to this
basis. We write $m$ for the number of distinct multilinear monomials of degree
$n$, and we assume that these monomials are ordered in some way. We create a
matrix $X$ of size $(m+d) \times m$ and initialize it to zero; the columns of
$M$ correspond bijectively to the monomials. We choose two parameters $p$ and
$s$: we generate pseudorandom integers in the range 0 to $p{-}1$.

We perform the following ``fill-and-reduce'' algorithm until the rank of the
matrix $X$ has remained stable for $s$ iterations:
  \begin{enumerate}
  \item Generate $n$ pseudorandom elements $a_1, \dots, a_n$ of $A$:
      vectors of length $d$ in which the components are integers in the
      range 0 to $p{-}1$.
  \item For $j$ from 1 to $m$ do:
    \begin{enumerate}
    \item Evaluate monomial $j$ by setting the variable $x_k$ equal to
        the vector $a_k$ for $k = 1, \dots, n$ and using the structure
        constants for $A$, obtaining another vector $b$ of length $d$.
    \item Store $b$ as a column vector in column $j$ of $X$ in rows
        $m{+}1$ to $m{+}d$.
    \end{enumerate}
  \item Compute the row canonical form of $X$; the last $d$ rows are now
      zero.
  \end{enumerate}
After this process has terminated, if the nullspace of $X$ is not zero then it
contains candidates for polynomial identities satisfied by $A$. We usually find
that $s = 10$ is a sufficient number of iterations after the rank has
stabilized, but we use $s = 100$ to increase our confidence in the results. We
now compute the canonical integral basis of the nullspace.

\subsection{Module generators algorithm} We assume that we have the canonical
integral basis of the nullspace of the matrix $X$ used in the fill-and-reduce
algorithm. Let $r$ be the number of these basis vectors; these are the
coefficient vectors of polynomial identities satisfied by the algebra $A$.
These identities are linearly independent over $\mathbb{Q}$ but they are not
necessarily independent as generators of the $S_n$-module of identities. We
want to find a minimal set of module generators. We start by sorting the basis
vectors by increasing Euclidean norm. We create a new matrix $M$ of size
$(m{+}n!) \times m$ and initialize it to zero.

We perform the following algorithm for $k$ from 1 to $r$:
  \begin{enumerate}
  \item For $i$ from 1 to $n!$ apply permutation $i$ of the variables $\{
      x_1, \dots, x_n \}$ to basis identity $k$ and store the result in row
      $m{+}i$ of $M$. More precisely, for each nonzero coefficient $c$ of
      the identity, apply permutation $i$ to the corresponding monomial,
      use the alternating property to straighten the monomial, obtain a
      standard basis monomial (with index $j$ say) and store $\pm c$ in
      position $(m+i,j)$ of $M$ (straightening may introduce a sign
      change).
  \item Compute the row canonical form of $M$. If the rank has increased
      from the previous iteration, then we record basis identity $k$ as a
      new generator.
  \end{enumerate}

\section{Multiplicity 1: representation $V(4)$}

In this section and the next we describe computer searches for polynomial
identities satisfied by the two irreducible representations of $\sltwo$ which
admit an alternating quaternary structure which is unique up to a scalar
multiple; we determine all their identities of degree $7$, and show that there
are no new identities in degree 10. For all our calculations we use the
computer algebra system \texttt{Maple}, especially the packages
\texttt{LinearAlgebra} and \texttt{LinearAlgebra[Modular]}.

\begin{theorem} \label{TheoremV(4)}
The vector space of multilinear polynomial identities in degree $7$ for the
alternating quaternary structure on $V(4)$ has dimension 21.
\end{theorem}

\begin{proof}
We use the fill-and-reduce algorithm with $n = 7$, $d = 5$, $m = 35$, $p = 10$
and $s = 100$. We create a matrix $X$ of size $40\times 35$ consisting of an
upper block of size $35 \times 35$ and a lower block of size $5 \times 35$; the
columns are labeled by the ordered basis of multilinear monomials in degree $7$
for an alternating quaternary operation. We generate seven random elements of
$V(4)$ and evaluate the $35$ monomials on these seven elements. We put the $35$
resulting elements of $V(4)$ as column vectors into the lower block of the
matrix. Each of the last five rows of the matrix now contains a linear relation
that must be satisfied by the coefficients of any identity for the alternating
quaternary structure on $V(4)$. We repeat the fill-and-reduce process until the
rank of the matrix stabilizes. The rank reached 14 and did not increase
further. Therefore the nullspace of the matrix has dimension 21.
\end{proof}

\begin{theorem}
Every multilinear polynomial identity in degree 7 for the alternating
quaternary structure on $V(4)$ is a consequence of the alternating property in
degree 4 together with the quaternary derivation identity in degree 7:
  \begin{align*}
  &
  [a,b,c,[d,e,f,g]]
  =
  \\
  &
  [[a,b,c,d],e,f,g] + [d,[a,b,c,e],f,g] + [d,e,[a,b,c,f],g] + [d,e,f,[a,b,c,g]]
  \end{align*}
\end{theorem}

\begin{proof}
We use the module generators algorithm, slightly modified to use less memory.
We create a matrix of size $59 \times 35$ with an upper block of size $35
\times 35$ and a lower block of size $24 \times 35$. We generate all 5040
permutations of seven letters and divide them into 210 groups of 24
permutations. For each of the 21 basis identities, we perform the following
computation. For each group of permutations, we apply the corresponding 24
permutations to the identity, obtain 24 new identities which we store in the
lower block of the matrix, and compute the row canonical form of the matrix.
After all 210 groups of permutations have been processed, the rank of the
matrix is equal to the dimension of the module generated by all the identities
up to and including the current identity. After the first identity has been
processed, the rank of the matrix is 21, which is the same as the entire
nullspace; the rank does not increase as we process the remaining identities.
Therefore every identity is a consequence of the first identity, which has the
form
  \begin{align*}
  &
  [[a,b,c,d],e,f,g] - [[a,b,c,e],d,f,g] + [[a,b,c,f],d,e,g] - [[a,b,c,g],d,e,f]
  \\
  &
  + [[d,e,f,g],a,b,c]
  \end{align*}
Applying the alternating property of the quaternary product, we see that this
identity can be written in the stated form.
\end{proof}

\begin{remark}
The alternating property in degree 4 and the quaternary derivation identity
together define the case $n = 4$ of the variety of $n$-Lie algebras introduced
by Filippov \cite{Filippov1}. Thus the isomorphism $\Lambda^4 V(4) \cong V(4)$
makes $V(4)$ into an alternating quaternary algebra isomorphic to one of the
5-dimensional 4-Lie algebras in Filippov's classification of
$(n{+}1)$-dimensional $n$-Lie algebras.
\end{remark}

For $n = 7$ we can use rational arithmetic for these calculations since the
matrix $X$ is not large. We can extend these calculations to $n = 10$, but we
need to use modular arithmetic to save memory, since the matrix $X$ is very
large: it has 5775 columns (the number of alternating quaternary monomials in
degree 10). The fill-and-reduce algorithm stabilizes at rank 660, and so the
nullspace has dimension 5115. We need to determine which of these identities in
degree 10 are consequences of the quaternary derivation identity in degree 7,
which we denote by $D(a,b,c,d,e,f,g)$. Since this polynomial alternates in
$a,b,c$ we only need to consider six consequences in degree 10, using the
variables $\{a,b,c,d,e,f,g,h,i,j\}$:
  \begin{alignat*}{3}
  &
  D([a,h,i,j],b,c,d,e,f,g), &\;
  &
  D(a,b,c,[d,h,i,j],e,f,g), &\;
  &
  D(a,b,c,d,[e,h,i,j],f,g),
  \\
  &
  D(a,b,c,d,e,[f,h,i,j],g), &\;
  &
  D(a,b,c,d,e,f,[g,h,i,j]), &\;
  &
  [D(a,b,c,d,e,f,g),h,i,j].
  \end{alignat*}
We use a modification of the module generators algorithm to determine that
these identities generate a module of dimension 5115. Since this equals the
dimension of the nullspace from the fill-and-reduce algorithm, it follows that
the alternating quaternary structure on $V(4)$ satisfies no new identities in
degree 10; that is, every identity in degree 10 is a consequence of the known
identities in lower degrees. We used $p = 101$ for these calculations; since
the group algebra of $S_n$ is semisimple over any field of characteristic $p
> n$, and we are studying identities of degree $n = 10$, it follows that any
prime $p > 10$ would give the same dimensions.

\section{Multiplicity 1: representation $V(6)$}

\begin{theorem} \label{TheoremV(6)1}
The vector space of multilinear polynomial identities in degree 7 for the
alternating quaternary structure on $V(6)$ has dimension 1.
\end{theorem}

\begin{proof}
We use the fill-and-reduce algorithm with $n = 7$, $d = 7$, $m = 35$, $p = 10$
and $s = 100$. The details of the computations are similar to those described
in the proof of Theorem \ref{TheoremV(4)}. The rank reached 34 and did not
increase further. Therefore the nullspace of the matrix has dimension 1.
\end{proof}

\begin{theorem} \label{TheoremV(6)2}
Every multilinear polynomial identity in degree $7$ for the alternating
quaternary structure on $V(6)$ is a consequence of the alternating property in
degree 4 together with the quaternary alternating sum identity in degree 7:
  \[
  \sum_{\sigma \in S_7}
  \epsilon(\sigma) \,
  [[a^\sigma,b^\sigma,c^\sigma,d^\sigma],e^\sigma,f^\sigma,g^\sigma]
  \]
\end{theorem}

\begin{proof}
Since the nullspace has dimension 1, this is an immediate corollary of Theorem
\ref{TheoremV(6)1}; we do not need to apply the module generators algorithm.
\end{proof}

\begin{remark}
The referee provided the following alternative proof. The quaternary
alternating sum identity is an alternating multilinear function of 7 variables.
Evaluating this function on the 7-dimensional space $V(6)$ gives a map
$\alpha\colon \Lambda^7 V(6) \to V(6)$. But $\Lambda^7 V(6)$ is 1-dimensional
(it is isomorphic to $V(0)$ as an $\sltwo$-module), and $\alpha$ is invariant
under the action of $\sltwo$. Hence the image of $\alpha$ is an
$\sltwo$-submodule which has dimension 0 or 1. Since $V(6)$ is irreducible, it
has no submodule of dimension 1, and so $\alpha$ must be the zero map.
\end{remark}

\begin{remark}
It is shown in Bremner \cite{Bremner1} (Theorems 3 and 4) that the quaternary
alternating sum identity in degree 7 is satisfied by the following multilinear
operation (the alternating quaternary sum) in every totally associative
quadruple system,
  \[
  [x_1,x_2,x_3,x_4]
  =
  \sum_{\pi \in S_4}
  \epsilon(\pi) \,
  x_{\pi(1)} x_{\pi(2)} x_{\pi(3)} x_{\pi(4)},
  \]
and that the quaternary alternating sum identity of Theorem \ref{TheoremV(6)2}
is a consequence of the quaternary derivation identity of Theorem
\ref{TheoremV(4)}.
\end{remark}

We can extend these calculations to $n = 10$ using modular arithmetic. The
fill-and-reduce algorithm stabilizes at rank 1903, and so the nullspace has
dimension 3872. We need to determine which of these identities in degree 10 are
consequences of the quaternary alternating sum identity in degree 7, which we
denote by $S(a,b,c,d,e,f,g)$. Since this polynomial alternates in all 7
variables, we only need to consider two consequences in degree 10:
  \[
  S([a,h,i,j],b,c,d,e,f,g),
  \qquad
  [S(a,b,c,d,e,f,g),h,i,j].
  \]
We can use an elementary argument to find an upper bound on the dimension of
the submodule generated by these two identities. The first identity alternates
in $a,h,i,j$ and also in $b,c,d,e,f,g$; hence any permutation of the first
identity is equal, up to a sign, to one of $\binom{10}{4} = 210$ possibilities.
Similarly, the second identity alternates in $a,b,c,d,e,f,g$ and in $h,i,j$;
hence any permutation of the second identity is equal, up to a sign, to one of
$\binom{10}{7} = 120$ possibilities.  Altogether we see that the submodule
generated by these two identities has dimension at most 330. (In fact our
computations show that this submodule has dimension 329.) Since this is less
than the dimension of the nullspace from the fill-and-reduce algorithm, the
alternating quaternary structure on $V(6)$ satisfies new identities in degree
10 that are not consequences of the known identities in lower degrees. It is an
open problem to determine generators for the $S_{10}$-module of new identities
in degree 10.

\section{Multiplicity 2: representation $V(8)$}

In this section and the next we describe computer searches for polynomial
identities satisfied by the two irreducible representations of $\sltwo$ which
admit a two-dimensional space of alternating quaternary structures; we
determine all their identities of degree $7$.

Any $\sltwo$-module homomorphism $\Lambda^4 V(8) \to V(8)$ is a linear
combination of the structures $f$ and $g$ from Tables
\ref{structureconstants8part1} and \ref{structureconstants8part2}. Up to a
scalar multiple, we need to consider only the single structure $g$ and the
one-parameter family of structures $f + xg$ for $x \in \mathbb{C}$. For $g$ our
methods are similar to those used for $V(4)$ and $V(6)$. For $f + xg$ we need
to use the Smith normal form to determine the values of the parameter $x$ which
produce a nonzero nullspace. For this we use the Maple command
\texttt{linalg[smith]} instead of \texttt{LinearAlgebra[SmithForm]} since the
former is much more efficient than the latter.

\begin{theorem} \label{TheoremV(8)g}
The vector space of multilinear polynomial identities in degree 7 for the
alternating quaternary structure $g$ on $V(8)$ has dimension 1 and is spanned
by the quaternary alternating sum identity.
\end{theorem}

\begin{proof}
Similar to the proofs of Theorems \ref{TheoremV(6)1} and \ref{TheoremV(6)2}.
\end{proof}

\begin{theorem} \label{TheoremV(8)f+xg}
For any $x \in \mathbb{C}$, the vector space of multilinear polynomial
identities in degree 7 for the alternating quaternary structure $f + xg$ on
$V(8)$ has dimension 1 and is spanned by the quaternary alternating sum
identity.
\end{theorem}

\begin{proof}
In order to determine how the space of identities depends on the parameter $x$,
we use the Smith normal form of a matrix over the polynomial algebra
$\mathbb{C}[x]$. Since the computation of the Smith form performs not only row
operations but also column operations, we must fill the matrix using a suitable
number of trials, and then compute the Smith form once. In the general case, we
create a matrix of size $t(n{+}1) \times m$ where $n$ is the highest weight
(recall that $V(n)$ has dimension $n{+}1$) and $m$ is the number of multilinear
monomials in degree $d$; the matrix consists of $t$ blocks of size $(n{+}1)
\times m$. We choose $t$ so that $t(n{+}1) > m$ in order to guarantee that we
have enough nonzero rows in the matrix to eliminate false nullspace vectors. We
perform the following algorithm:
  \begin{enumerate}
  \item For $b$ from 1 to $t$ do:
    \begin{enumerate}
    \item Generate $d$ pseudorandom vectors of length $n{+}1$
        representing elements of $V(n)$.
    \item For $j$ from 1 to $m$ do:
      \begin{enumerate}
      \item Evaluate the $j$-th alternating quaternary monomial on
          the $d$ pseudorandom vectors to obtain another vector of
          length $n{+}1$ with components which are polynomials in
          the parameter $x$.
      \item Put the resulting vector into column $j$ of block $t$.
      \end{enumerate}
    \end{enumerate}
  \item Compute the Smith normal form of the matrix.
  \end{enumerate}
For $n = 8$ and $d = 7$ we have $m = 35$ and we choose $t = 4$. The entries of
the resulting $36 \times 35$ matrix are quadratic polynomials in the parameter
$x$ since each monomial involves two occurrences of the quaternary operation.
In the Smith normal form of the matrix, the diagonal entries are 1 (34 times)
and 0 (once). It follows that the matrix has a one-dimensional nullspace for
every value of $x$. In Bremner \cite{Bremner1} (Proposition 3, page 85) it is
shown that there a unique 1-dimensional $S_7$-submodule of the 35-dimensional
module with basis consisting of the alternating quaternary monomials in degree
7, and this submodule is spanned by the quaternary alternating sum identity.
Hence the nullspace basis does not depend on the value of the parameter $x$,
and this completes the proof. We checked this result independently by
evaluating the quaternary alternating sum identity on pseudorandom vectors for
the product $f + xg$ with indeterminate $x$ and verifying that the result was
zero.
\end{proof}

\begin{remark}
It is an open problem to determine whether the alternating quaternary
structures on $V(8)$ are isomorphic for all values of the parameter $x$.
\end{remark}

\section{Multiplicity 2: representation $V(10)$}

As in the previous section, any $\sltwo$-module homomorphism $\Lambda^4 V(10)
\to V(10)$ is a linear combination of two structures $f$ and $g$, and we
consider separately the single structure $g$ and the one-parameter family of
structures $f + xg$ for $x \in \mathbb{C}$.

\begin{theorem} \label{TheoremV(10)g}
The vector space of multilinear polynomial identities in degree 7 for the
alternating quaternary structure $g$ on $V(10)$ has dimension 0: every identity
is a consequence of the alternating properties in degree 4.
\end{theorem}

\begin{proof}
Similar to the proofs of Theorem \ref{TheoremV(8)g} except that the matrix
achieves the full rank of 35.
\end{proof}

\begin{theorem} \label{TheoremV(10)f+xg}
For $x = \frac54$, the vector space of multilinear polynomial identities in
degree 7 for the alternating quaternary structure $f + xg$ on $V(10)$ has
dimension 1 and is spanned by the quaternary alternating sum identity. For all
other $x \in \mathbb{C}$, the vector space of multilinear polynomial identities
in degree 7 for the alternating quaternary structure $f + xg$ on $V(10)$ has
dimension 0.
\end{theorem}

\begin{proof}
Similar to the proof of Theorem \ref{TheoremV(8)f+xg} except that now $n = 10$.
As before, the entries of the resulting $44 \times 35$ matrix are quadratic
polynomials in the parameter $x$. In the Smith normal form of this matrix, the
diagonal entries are 1 (28 times) and $x-\frac54$ (7 times). It follows that
the matrix has zero nullspace except in the case $x = \frac54$. We now
specialize to this value of $x$ and consider the structure $f + \frac54 g$; the
rest of the proof is similar to that of Theorems \ref{TheoremV(6)1} and
\ref{TheoremV(6)2}.
\end{proof}

\section{Proof of multiplicity formula} \label{proofsection}

In this section we prove the multiplicity formula of Theorem
\ref{multiplicitytheorem}.  We reduce the problem to a combinatorial question
and apply the theory of P\'olya enumeration.

\begin{lemma} \label{weightspacedimension} \cite[Lemma 5.2]{BremnerHentzel2}
Let $M = \Lambda^k V(n)$ be the $k$-th exterior power of $V(n)$. If $w \in
\mathbb{Z}$ with $kn \ge w \ge -kn$ and $w \equiv kn \, (\mathrm{mod} \, 2)$
then the dimension of the weight space $M_w$ is the number of sequences
$(w_1,w_2,\ldots,w_k) \in \mathbb{Z}^k$ satisfying
  \[
  n \ge w_1 > w_2 > \ldots > w_k \ge -n;
  \;\;
  w_1 + w_2 + \ldots + w_k = w;
  \;\;
  w_1, \hdots, w_k \equiv n \, (\mathrm{mod} \, 2).
  \]
\end{lemma}

We now specialize to $k = 4$ since we are interested in the fourth exterior
power. To compute the multiplicity of $V(n)$ as a direct summand of $\Lambda^4
V(n)$ using Lemmas \ref{multiplicityformula} and \ref{weightspacedimension}, we
must determine the number of quadruples $(p,q,r,s)$ satisfying
  \begin{equation}\label{conditions}
  n \ge p > q > r > s \ge -n;
  \quad
  p + q + r + s = w;
  \quad
  p, q, r, s \equiv n \, (\mathrm{mod} \, 2)
  \end{equation}
for $w = n$ and $w = n + 2$. Let $n$ be a non-negative integer and let $w$ be a
weight of $\Lambda^4 V(n)$: thus $w$ is an integer satisfying
  \[
  4n \ge w \ge -4n,
  \qquad
  w \equiv 0 \, (\mathrm{mod} \, 2 ).
  \]
For integers $p,q,r,s$ satisfying (\ref{conditions}) we define
  \[
  P' = p + n, \quad
  Q' = q + n, \quad
  R' = r + n, \quad
  S' = s + n.
  \]
Then $(P',Q',R',S')$ is a quadruple of even integers satisfying
  \[
  2n \ge P' > Q' > R' > S' \ge 0,
  \quad
  P' + Q' + R' + S' = W',
  \quad
  W' = w + 4n.
  \]
We need to count the number of partitions of $W'$ into four distinct
nonnegative even parts less than or equal to $2n$. We only need $W' = 5n$ and
$W' = 5n+2$ corresponding to $w = n$ and $w = n+2$. It is clear that if $n$ is
odd then there are no solutions in both cases, so $V(n)$ does not occur as a
summand of $\Lambda^4 V(n)$: the multiplicity is zero. Therefore we may assume
that $n$ is even and define
  \[
  P = \frac{p+n}{2}, \quad
  Q = \frac{q+n}{2}, \quad
  R = \frac{r+n}{2}, \quad
  S = \frac{s+n}{2}, \quad
  W = \frac{w+4n}{2}.
  \]
Then $(P,Q,R,S)$ is a quadruple of integers satisfying
  \[
  n \ge P > Q > R > S \ge 0,
  \qquad
  P + Q + R + S = W.
  \]

\begin{definition} \cite[page 612]{WuChao}
If $G$ is a subgroup of the symmetric group $S_n$ then the \textbf{cycle index}
of $G$ is the following polynomial in the indeterminates $x_1, x_2, \dots,
x_n$:
  \[
  Z_G{\left(x_1,x_2,\hdots,x_n\right)}
  =
  \frac{1}{|G|}\sum_{\sigma \in G}{x^{b_1}_1 x^{b_2}_2 \cdots x^{b_n}_n};
  \]
here $b_i$ is the number of cycles of length $i$ in the disjoint cycle
factorization of $\sigma$.
\end{definition}

\begin{lemma} \cite[page 36]{HararyPalmer} \label{book1}
The cycle index of the alternating group $A_n$ is
  \[
  Z_{A_n} \left( x_1, x_2, \hdots, x_n \right)
  =
  Z_{S_n} \left( x_1, x_2, \hdots, x_n \right)
  +
  Z_{S_n} \left( x_1, -x_2, \hdots, (-1)^{n-1} x_n \right).
  \]
\end{lemma}

\begin{proof}
The definition of cycle index gives
  \[
  Z_{A_n}\left(x_1,x_2,\hdots,x_n\right)
  =
  \frac{2}{n!}
  \bigg[
  \sum_{\sigma \in S_n}
  x^{b_1}_1 x^{b_2}_2 \cdots x^{b_n}_n
  -
  \!\!\!\!\!\!
  \sum_{\sigma\in S_n \backslash A_n}
  x^{b_1}_1 x^{b_2}_2 \cdots x^{b_n}_n
  \bigg].
  \]
Since $\sigma\in S_{n}\backslash A_{n}$ if and only if $\sigma$ has an odd
number of even length cycles, we get
  \allowdisplaybreaks
  \begin{align*}
  &
  Z_{A_n}{\left(x_1,x_2,\hdots,x_n\right)}
  =
  \\
  &
  \frac{2}{n!} \cdot \frac12
  \bigg[
  \sum_{\sigma \in S_n}
  x^{b_1}_1 x^{b_2}_2 \cdots x^{b_n}_n
  +
  \sum_{\sigma \in S_n}
  x_1^{b_1} (-x_2)^{b_2} \cdots \big( (-1)^{n-1} x_n \big)^{b_n}
  \bigg].
  \end{align*}
This completes the proof.
\end{proof}

The next result is the special case $k = 4$ of Theorem 2 in Wu and Chao
\cite{WuChao}; but note that we allow $0 \in S$.

\begin{proposition} \label{cycleindextheorem}
If $S$ is a set of non-negative integers then the number of partitions of an
integer $n$ into four distinct parts in $S$ is the coefficient of $x^n$ in
  \allowdisplaybreaks
  \begin{align*}
  &Z_{A_4}
  \left(
  \sum_{i\in S} x^i,\sum_{i\in S} x^{2i},\sum_{i\in S} x^{3i},\sum_{i\in S} x^{4i}
  \right)
  -
  Z_{S_4}
  \left(
  \sum_{i\in S} x^i,\sum_{i\in S} x^{2i},\sum_{i\in S} x^{3i},\sum_{i\in S} x^{4i}
  \right).
  \end{align*}
\end{proposition}

\begin{corollary} \label{polyacorollary}
If $S$ is a set of non-negative integers then the  number of partitions of a
positive integer $n$ into four distinct parts in $S$ is the coefficient of
$x^n$ in
  \[
  Z_{S_4}
  \left(
  \sum_{i \in S} x^i,
  - \sum_{i \in S} x^{2i},
  \sum_{i \in S} x^{3i},
  - \sum_{i \in S} x^{4i}
  \right).
  \]
\end{corollary}

\begin{proof}
Take $n = 4$ in Lemma \ref{book1}, set $x_j = \sum_{i \in S} x^{ji}$, and apply
Proposition \ref{cycleindextheorem}.
\end{proof}

\begin{definition} \label{Pdefinition}
For us $S = \{0,1,\hdots,n\}$ so we define the following polynomials:
  \[
  P_n(x)
  =
  Z_{S_4}
  \left(
  \sum_{i=0}^n x^i, - \sum_{i=0}^n x^{2i}, \sum_{i=0}^n x^{3i}, - \sum_{i=0}^n x^{4i}
  \right).
  \]
\end{definition}

\begin{lemma} \label{coefficientformula}
We have
  \[
  \left( \sum_{i=0}^n x^i \right)^t
  =
  \sum_{\ell=0}^{nt}
  \left[
  \sum_{k=0}^{\min \left( t, \left\lfloor \frac{\ell}{n+1} \right\rfloor \right)}
  (-1)^k
  \binom{t}{k}
  \binom{\ell{-}(n{+}1)k{+}t{-}1}{t{-}1}
  \right]
  x^\ell.
  \]
\end{lemma}

\begin{proof}
We use these three familiar identities:
  \[
  \frac{1{-}x^{n+1}}{1{-}x}
  =
  \sum_{i=0}^n x^i,
  \;
  \left( 1{-}x^{n+1} \right)^t
  =
  \sum_{k=0}^{t}
  (-1)^k
  \binom{t}{k}
  x^{(n+1)k},
  \;
  \frac{1}{(1{-}x)^t}
  =
  \sum_{j=0}^\infty
  \binom{j{+}t{-}1}{t{-}1}
  x^j.
  \]
We obtain
  \begin{equation} \label{combinedformula}
  \left(
  \sum_{i=0}^n x^i
  \right)^t
  =
  \frac{(1-x^{n+1})^t}{(1-x)^t}
  =
  \sum_{k=0}^t
  \sum_{j=0}^\infty
  (-1)^k
  \binom{t}{k}
  \binom{j{+}t{-}1}{t{-}1}
  x^{(n+1)k+j}.
  \end{equation}
We set $\ell = (n{+}1)k+j$ so that $\ell{-}j = (n{+}1)k$ and $\ell{-}j \equiv 0
\; (\mathrm{mod} \; n{+}1)$. We also have $k = (\ell{-}j)/(n{+}1)$ and so $k
\le \lfloor \ell/(n{+}1) \rfloor$. Substituting $j = \ell - (n{+}1)k$ in
\eqref{combinedformula}, and noting that $nt$ is the largest power of $x$, we
obtain the stated formula.
\end{proof}

\begin{definition}
We use the following notation:
  \[
  \Delta_m^n
   =
  \begin{cases}
  1 &\text{if $n \equiv 0 \, (\mathrm{mod} \, m)$}  \\
  0 &\text{otherwise}
  \end{cases},
  \qquad
  \Delta_{s,m}^n
  =
  \begin{cases}
  1 &\text{if $n \equiv s \, (\mathrm{mod} \, m)$} \\
  0 &\text{otherwise}
  \end{cases}.
\]
\end{definition}

\begin{definition} \label{alpha}
We consider the following integer-valued functions of $n$:
  \[
  \alpha(n) = \left\lceil \frac{n}{4} \right\rceil,
  \quad
  \beta(n) = \left\lceil \frac{3n}{4} \right\rceil,
  \quad
  \gamma(n) = \left\lfloor \frac{3n{-}2}{4} \right\rfloor,
  \quad
  \delta(n) = \left\lfloor \frac{5n}{6} \right\rfloor.
  \]
\end{definition}

\begin{proposition} \label{dimension5n}
For even $n \in \mathbb{Z}$, the number of solutions $P,Q,R,S \in \mathbb{Z}$
to
  \[
  n \ge P > Q > R > S \ge 0,
  \qquad
  P + Q + R + S = \frac{5n}{2},
  \]
equals
  \allowdisplaybreaks
  \begin{align*}
  &
  \frac{23}{1152} \, n^3
  -
  \frac{29}{96} \, n^2
  +
  \frac{1}{288}
  \Big( -36 \alpha(n) + 180 \beta(n) + 36 \gamma(n) + 27 \Delta_4^n - 167 \Big) n
  \\
  &
  +
  \frac{1}{24}
  \Big(
  6 \alpha(n)^2 - 6 \beta(n)^2 - 6 \gamma(n)^2
  + 12 \beta(n) - 12 \gamma(n) + 8 \delta(n)
  + 3 \Delta_4^n - 6 \Delta_8^n - 3
  \Big).
  \end{align*}
\end{proposition}

\begin{proof}
By Corollary \ref{polyacorollary}, we need to find the coefficient of
$x^{5n/2}$ in the polynomial $P_n(x)$ of Definition \ref{Pdefinition}. The
cycle index of $S_4$ is
  \[
  Z_{S_4}(x_1,x_2,x_3,x_4)
  =
  \frac{1}{24}
  \big(
  x_1^4 + 6 x_1^2 x_2 + 8 x_1 x_3 + 3x^2_2 + 6 x_4
  \big).
  \]
For the first four terms, we need to evaluate the following products:
  \[
  A = \bigg( \sum_{i=0}^n x^i \bigg)^4,
  \;
  B = \bigg( \sum_{i=0}^n x^i \bigg)^2 \sum_{i=0}^n x^{2i},
  \;
  C = \sum_{i=0}^n x^i \sum_{i=0}^n x^{3i},
  \;
  D = \bigg( \sum_{i=0}^n x^{2i} \bigg)^2.
  \]
Lemma \ref{coefficientformula} gives
  \[
  A
  =
  \sum_{\ell=0}^{4n}
  \sum_{k=0}^{\min\left( 4, \left\lfloor \frac{\ell}{n+1} \right\rfloor \right)}
  (-1)^k
  \binom{4}{k}
  \binom{\ell{-}(n{+}1)k{+}3}{3}
  x^\ell.
  \]
Similarly,
  \[
  B
  =
  \Bigg(
  \sum_{\ell=0}^{2n}
  \sum_{k=0}^{\min\left( 2, \left\lfloor \frac{\ell}{n+1} \right\rfloor \right)}
  (-1)^k
  \binom{2}{k}
  \big( \ell{-}(n{+}1)k{+}1 \big)
  x^\ell
  \Bigg)
  \sum_{i=0}^n x^{2i}.
  \]
The upper limit of $k$ is 0 for $0 \le \ell \le n$, and 1 for $n{+}1 \le \ell
\le 2n$. Hence
  \allowdisplaybreaks
  \begin{align*}
  B
  &=
  \Bigg(
  \sum_{\ell=0}^n
  \big( \ell{+}1 \big)
  x^\ell
  +
  \sum_{\ell=n+1}^{2n}
  \big[ (\ell{+}1){-}2(\ell{-}n) \big]
  x^\ell
  \Bigg)
  \sum_{i=0}^n x^{2i}
  \\
  &=
  \Bigg(
  \sum_{\ell=0}^n
  \big( \ell{+}1 \big)
  x^\ell
  +
  \sum_{\ell=n+1}^{2n}
  \big( 2n{-}\ell{+}1 \big)
  x^\ell
  \Bigg)
  \sum_{i=0}^n x^{2i}
  \\
  &=
  \sum_{\ell=0}^n
  \sum_{m=0}^n
  (\ell{+}1)
  x^{\ell+2m}
  +
  \sum_{\ell=n+1}^{2n}
  \sum_{m=0}^n
  (2n{-}\ell{+}1)
  x^{\ell+2m}.
  \end{align*}
We now set $p = \ell {+} 2m$, so that $\ell = p{-}2m$.  For $0 \le \ell \le n$
we have $0 \le p{-}2m \le n$ and so $\frac12 (p{-}n) \le m \le \frac12 p$, but
$m \in \mathbb{Z}$ so $\lceil \frac12 (p{-}n) \rceil \le m \le \lfloor \frac12
p \rfloor$; since also $0 \le m \le n$ we get $\max( 0, \lceil \frac12 (p{-}n)
\rceil ) \le m \le \min( n, \lfloor \frac12 p \rfloor )$. Similarly, for $n{+}1
\le \ell \le 2n$ we obtain $\max \left( 0, \left\lceil \frac12(p{-}2n)
\right\rceil \right) \le m \le \min \left( n, \left\lfloor \frac12
(p{-}(n{+}1)) \right\rfloor \right)$. Therefore
  \[
  B
  =
  \sum_{p=0}^{3n}
  \sum_{m=\max\left(0,\left\lceil\frac{p-n}{2}\right\rceil\right)}
  ^{\min\left(n,\left\lfloor\frac{p}{2}\right\rfloor\right)}
  \!\!\!\!\!\!\!\!
  \left(
  p{-}2m {+} 1
  \right)
  x^p
  +
  \sum_{p=n+1}^{4n}
  \sum_{m=\max\left(0,\left\lceil\frac{p-2n}{2}\right\rceil\right)}
  ^{\min\left(n,\left\lfloor\frac{p-(n+1)}{2}\right\rfloor\right)}
  \!\!\!\!\!\!\!\!
  \left(
  2n {-} (p{-}2m) {+} 1
  \right)
  x^p.
  \]
Using a similar change of index we obtain
  \allowdisplaybreaks
  \begin{align*}
  C
  &=
  \sum_{i=0}^n
  \sum_{j=0}^n
  x^{i+3j}
  =
  \sum_{p=0}^{4n}
  \sum_{m=\max(0,\lceil\frac{p-n}{3}\rceil)}^{\min(n,\lfloor\frac{p}{3}\rfloor)}
  \!\!\!\!\!\!
  x^p
  \\
  &=
  \sum_{p=0}^{4n}
  \left[
  \min\left(n,\left\lfloor\frac{p}{3}\right\rfloor\right){-}
  \max\left(0,\left\lceil\frac{p{-}n}{3}\right\rceil\right){+}1
  \right]
  x^p.
  \end{align*}
Replacing $x$ by $x^2$ in Lemma \ref{coefficientformula} gives
  \[
  D
  =
  \sum_{\ell=0}^n
  (\ell{+}1)
  x^{2\ell}
  +
  \sum_{\ell=n+1}^{2n}
  (2n{-}\ell{+}1)
  x^{2\ell}.
  \]
We now write
  \[
  E = \sum_{\ell=0}^n x^{4\ell},
  \]
and obtain
  \allowdisplaybreaks
  \begin{align*}
  &
  A - 6 B + 8 C + 3 D - 6 E
  \\
  &=
  \sum_{p=0}^{4n}
  \sum_{k=0}^{\min\left( 4, \left\lfloor \frac{p}{n+1} \right\rfloor \right)}
  (-1)^k
  \binom{4}{k}
  \binom{p{-}(n{+}1)k{+}3}{3}
  x^p
  \\
  &\quad
  - 6
  \Bigg[
  \sum_{p=0}^{3n}
  \sum_{m=\max\left(0,\left\lceil\frac{p-n}{2}\right\rceil\right)}
  ^{\min\left(n,\left\lfloor\frac{p}{2}\right\rfloor\right)}
  \!\!\!\!\!\!\!\!
  \left(
  p{-}2m {+} 1
  \right)
  x^p
  +
  \sum_{p=n+1}^{4n}
  \sum_{m=\max\left(0,\left\lceil\frac{p-2n}{2}\right\rceil\right)}
  ^{\min\left(n,\left\lfloor\frac{p-(n+1)}{2}\right\rfloor\right)}
  \!\!\!\!\!\!\!\!
  \left(
  2n {-} (p{-}2m) {+} 1
  \right)
  x^p
  \Bigg]
  \\
  &\quad
  +
  8
  \Bigg[
  \sum_{p=0}^{4n}
  \left[
  \min\left(n,\left\lfloor\frac{p}{3}\right\rfloor\right){-}
  \max\left(0,\left\lceil\frac{p{-}n}{3}\right\rceil\right){+}1
  \right]
  x^p
  \Bigg]
  \\
  &\quad
  +
  3
  \Bigg[
  \sum_{\ell=0}^n
  (\ell{+}1)
  x^{2\ell}
  +
  \sum_{\ell=n+1}^{2n}
  (2n{-}\ell{+}1)
  x^{2\ell}
  \Bigg]
  -
  6
  \sum_{\ell=0}^n x^{4\ell}.
  \end{align*}
We need the coefficient $T$ of $x^{5n/2}$ in the last equation:
  \allowdisplaybreaks
  \begin{align*}
  T
  &=
  \sum_{k=0}^{\left\lfloor\frac{5n}{2(n+1)}\right\rfloor}
  (-1)^k
  \binom{4}{k}
  \binom{\frac{5n}{2}{-}(n{+}1)k{+}3}{3}
  \\
  &\quad
  -
  6
  \Bigg[
  \sum_{m=\lceil\frac{3n}{4}\rceil}^n
  \left(\frac{5n}{2}{-}2m{+}1\right)
  +
  \sum_{m=\left\lceil\frac{n}{4}\right\rceil}^{\left\lfloor\frac{3n-2}{4}\right\rfloor}
  \left(2m{-}\frac{n}{2}{+}1\right)
  \Bigg]
  \\
  &\quad
  +
  8
  \left( \left\lfloor \frac{5n}{6} \right\rfloor {-} \frac{n}{2} {+} 1 \right)
  +
  3 \,
  \delta_4^n \left( 0 {+} \frac{3n}{4} {+} 1 \right)
  -
  6 \,
  \delta_8^n.
  \end{align*}
For $n = 0$ and $n = 2$ we get $T = 0$; this is expected since the
$\sltwo$-modules $V(0)$ and $V(2)$ have dimensions 1 and 3 respectively, so in
both cases $\Lambda^4 V(n)$ is $\{0\}$.  For $n \ge 4$ the upper limit of $k$
is 2, and we use the formula
  \[
  \sum_{m=a}^b m = \frac12 (b-a+1)(b+a).
  \]
We obtain
  \allowdisplaybreaks
  \begin{align*}
  T
  &=
  \binom{\frac{5n}{2}{+}3}{3}
  -
  4
  \binom{\frac{3n}{2}{+}2}{3}
  +
  6
  \binom{\frac{n}{2}{+}1}{3}
  -
  6
  \left(n{-}\left\lceil\frac{3n}{4}\right\rceil{+}1\right)
  \left(\frac{5n}{2}{+}1\right)
  \\
  &\quad
  +
  6
  \left(n{-}\left\lceil\frac{3n}{4}\right\rceil{+}1\right)
  \left(n{+}\left\lceil\frac{3n}{4}\right\rceil\right)
  -
  6
  \left(\left\lfloor\frac{3n{-}2}{4}\right\rfloor{-}\left\lceil\frac{n}{4}\right\rceil{+}1\right)
  \left(-\frac{n}{2}{+}1\right)
  \\
  &\quad
  -
  6
  \left(\left\lfloor\frac{3n{-}2}{4}\right\rfloor{-}\left\lceil\frac{n}{4}\right\rceil{+}1\right)
  \left(\left\lfloor\frac{3n{-}2}{4}\right\rfloor{+}\left\lceil\frac{n}{4}\right\rceil\right)
  +
  8 \left\lfloor\frac{5n}{6}\right\rfloor - 4 n + 8
  \\
  &\quad
  +
  3 \, \delta_4^n \left( \frac{3n}{4} {+} 1 \right)
  -
  6 \, \delta_8^n.
  \end{align*}
Expanding this and collecting terms with the same power of $n$ gives
  \allowdisplaybreaks
  \begin{align*}
  &
  \frac{23}{48} n^3
  -
  \frac{29}{4} n^2
  +
  \frac{1}{12}
  \Big(
  -36 \alpha(n) + 180 \beta(n) + 36 \gamma(n) + 27 \Delta_4^n - 167
  \Big)
  n
  \\
  &
  +
  \Big(
  6 \alpha(n)^2 - 6 \beta(n)^2 - 6 \gamma(n)^2
  + 12 \beta(n) + 8 \delta(n) - 12 \gamma(n)
  + 3 \Delta_4^n - 6 \Delta_8^n - 3
  \Big).
  \end{align*}
We check that this gives $T = 0$ for $n = 0$ and $n = 2$. Finally, we divide by
24.
\end{proof}

\begin{corollary} \label{dimension5ncorollary}
For even $n \in \mathbb{Z}$, write $n = 24 q + r$ with $q, r \in \mathbb{Z}$
and $0 \le r < 24$.  The dimension $\dim [ \Lambda^4 V(n) ]_n$, of the weight
space of weight $n$ in the $\sltwo$-module $\Lambda^4 V(n)$, is given in Table
\ref{multiplicitytable}. (Note the denominator 1152.)
\end{corollary}

\begin{proof}
The dimension is given by the formula of Proposition \ref{dimension5n}. The LCM
of the denominators of the functions $\alpha(n)$, $\beta(n)$, $\gamma(n)$,
$\delta(n)$ and the periods of the functions $\Delta_4^n$ and $\Delta_8^n$
equals 24. Hence the dimension is given by a cubic polynomial in $n$ which
depends on the remainder of $n$ modulo 24.  Considering each remainder $r$
separately, we obtain the polynomials in the first column of Table
\ref{multiplicitytable}.
\end{proof}

\begin{definition} \label{epsilon}
We consider the following integer-valued functions of $n$:
  \[
  \epsilon(n) = \left\lfloor\frac{3n}{4}\right\rfloor, \quad
  \zeta(n) = \left\lceil\frac{n{+}2}{4}\right\rceil, \quad
  \eta(n) = \left\lceil\frac{3n{+}2}{4}\right\rceil, \quad
  \theta(n) = \left\lfloor\frac{5n{+}2}{6}\right\rfloor.
  \]
\end{definition}

\begin{proposition} \label{dimension5n+2}
For even $n \in \mathbb{Z}$, the number of solutions $P,Q,R,S \in \mathbb{Z}$
to
  \[
  n \ge P > Q > R > S \ge 0,
  \qquad
  P + Q + R + S = \frac{5n{+}2}{2},
  \]
equals
  \allowdisplaybreaks
  \begin{align*}
  &
  \frac{23}{1152} \, n^3
  -
  \frac{21}{64} \, n^2
  +
  \frac{1}{288}
  \Big( 36 \epsilon(n) - 36 \zeta(n) + 180 \eta(n) + 27 \Delta_{4,2}^n - 254 \Big) n
  \\
  &
  +
  \frac{1}{48}
  \Big(
  - 12 \epsilon(n)^2 + 12 \zeta(n)^2 - 12 \eta(n)^2
  - 12 \epsilon(n) - 12 \zeta(n) + 36 \eta(n) + 16 \theta(n)
  \\
  &\qquad\qquad
  + 3 \Delta_{4,2}^n - 12 \Delta_{8,6}^n - 24
  \Big).
  \end{align*}
\end{proposition}

\begin{proof}
Similar to the proof of Proposition \ref{dimension5n}.
\end{proof}

\begin{corollary}
For even $n \in \mathbb{Z}$, write $n = 24 q + r$ with $q, r \in \mathbb{Z}$
and $0 \le r < 24$.  The dimension $\dim [ \Lambda^4 V(n) ]_{n+2}$, of the
weight space of weight $n{+}2$ in the $\sltwo$-module $\Lambda^4 V(n)$, is
given in Table \ref{multiplicitytable}. (Note the denominator 1152.)
\end{corollary}

\begin{proof}
Similar to the proof of Corollary \ref{dimension5ncorollary}.
\end{proof}

\begin{theorem}
The multiplicity $\mathrm{mult}(n) = \dim \mathrm{Hom}_{\sltwo}\big( \Lambda^4
V(n), V(n) \big)$ of the irreducible representation $V(n)$ of $\sltwo$ as a
summand of its fourth exterior power $\Lambda^4 V(n)$ is given in Table
\ref{multiplicitytable}. (Note the denominator 1152.)
\end{theorem}

\begin{proof}
Apply Lemma \ref{multiplicityformula} to Propositions \ref{dimension5n} and
\ref{dimension5n+2}.
\end{proof}

\begin{remark}
We can use Corollary \ref{polyacorollary} to obtain another proof of the
decompositions in Remark \ref{completedecomposition}. For $n = 4, 6, 8, 10$ we
compute the polynomial $P_n(x)$ from Definition \ref{Pdefinition}. In each
case, the coefficient of $x^{(w+4n)/2}$ is $\dim [ \Lambda^4 V(n) ]_w$, and we
then apply Lemma \ref{multiplicityformula} to find the multiplicity of $V(w)$
in $\Lambda^4 V(n)$:
  \begin{align*}
  P_4(x)
  &=
  x^{10} + x^9 + x^8 + x^7 + x^6,
  \\
  P_6(x)
  &=
  x^{18} + x^{17} + 2 x^{16} + 3 x^{15} + 4 x^{14} + 4 x^{13} + 5 x^{12} + 4 x^{11}
  + 4 x^{10} + 3 x^9 + 2 x^8
  \\
  &\quad
  + x^7 + x^6,
  \\
  P_8(x)
  &=
  x^{26} + x^{25} + 2 x^{24} + 3 x^{23} + 5 x^{22} + 6 x^{21} + 8 x^{20} + 9 x^{19}
  + 11 x^{18} + 11 x^{17}
  \\
  &\quad
  + 12 x^{16} + 11 x^{15} + 11 x^{14} + 9 x^{13} + 8 x^{12} + 6 x^{11} + 5 x^{10} + 3 x^9
  + 2 x^8 + x^7
  \\
  &\quad
  + x^6,
  \\
  P_{10}(x)
  &=
  x^{34} + x^{33} + 2 x^{32} + 3 x^{31} + 5 x^{30} + 6 x^{29} + 9 x^{28} + 11 x^{27}
  + 14 x^{26} + 16 x^{25}
  \\
  &\quad
  + 19 x^{24} + 20 x^{23} + 23 x^{22} + 23 x^{21} + 24 x^{20} + 23 x^{19} + 23 x^{18}
  + 20 x^{17}
  \\
  &\quad
  + 19 x^{16} + 16 x^{15} + 14 x^{14} + 11 x^{13} +  9 x^{12} + 6 x^{11} + 5 x^{10}
  + 3 x^9 + 2 x^8
  \\
  &\quad
  + x^7 + x^6.
  \end{align*}
\end{remark}

\section{Conclusion}

We recovered a 5-dimensional 4-Lie algebra from the isomorphism $\Lambda^4 V(4)
\cong V(4)$. This algebra satisfies the quaternary derivation identity $D$, and
hence also the quaternary alternating sum identity $S$. We found that the
identity $S$ is also satisfied by the unique structure on $V(6)$, every
structure $f + xg$ on $V(8)$, and the structure $f + \frac54 g$ on $V(10)$.  By
Bremner \cite{Bremner1} it is known that the quaternary alternating sum
operation in a totally associative quadruple system also satisfies $S$. This
raises the question whether the structures which satisfy $S$ can be embedded
into totally associative quadruple systems if the original associative
operation is replaced by the quaternary alternating sum. An affirmative answer
to this question would provide a partial generalization of the
Poincar\'e-Birkhoff-Witt theorem for Lie algebras; see Pozhidaev
\cite{Pozhidaev} for related work.  Simple associative $n$-tuple systems were
classified by Carlsson \cite{Carlsson}. In particular, for $n = 4$, any simple
associative quadruple system is isomorphic to a subspace of matrices of the
form
  \[
  \left[
  \begin{array}{ccc}
  0 & 0 & Z \\
  X & 0 & 0 \\
  0 & Y & 0
  \end{array}
  \right]
  \]
where $X$, $Y$, $Z$ have sizes $q \times p$, $r \times q$, $p \times r$
(respectively) and $p, q, r$ are positive integers. It would be very
interesting to determine whether any of the alternating quaternary algebras
presented in this paper are isomorphic to such a subspace of matrices under the
quaternary alternating sum.

\section*{Acknowledgements}

We thank the referee for a careful reading of the original version and for
bringing to our attention a number of errors and inconsistencies.  This work
forms part of the doctoral thesis of the second author, written under the
supervision of the first author. The first author was partially supported by
NSERC, the Natural Sciences and Engineering Research Council of Canada. The
second author was supported by a University Graduate Scholarship from the
University of Saskatchewan.

redpajama_set_name: RedPajamaArXiv


text:
Fire Probe
Posted on June 21, 2014 | Legal
COLOMBO, June 21, 2014 (AFP) – Sri Lankan police were investigating a fire which Saturday destroyed a Muslim-owned clothing store in a district where religious tensions have been heightened after anti-Muslim riots killed four people and wounded 80. Authorities have not ruled out arson at the upmarket No-Limit store in Panadura, 25 kilometres (16 miles) south of Colombo, police spokesman Ajith Rohana said.
"We are looking at the possibility of arson as well as an accidental fire," Rohana told AFP. "The fire started in the upper floor and we are still investigating."
The blaze came a day after Sri Lanka's Muslim minority shortened their Friday prayer services because of heightened religious tensions after clashes with hardline Buddhists earlier in the week in Kalutara district.
The district includes Panadura, and the popular and mainly Muslim coastal resorts of Alutgama and Beruwala, where religious riots earlier in the week killed four and wounded 80. Hundreds of homes and businesses were set on fire in the violence.
President Mahinda Rajapakse, who toured the riot-hit resort of Beruwala, said he was ordering a probe into "recent disturbances".
"I will be appointing a high-level panel to inquire into recent disturbances," the Pre

redpajama_set_name: RedPajamaCommonCrawl


text:
EditorialCo-Digestion Is a Key Component in Montpelier's Quest for Net Zero Energy Status
Co-Digestion Is a Key Component in Montpelier's Quest for Net Zero Energy Status
Septage and high-strength food processing waste provide a substantial source of renewable energy for the clean-water plant in Vermont's capital.
Appeared in print as "An Energy Boost"
By Steve Lund
Sustainable Operations
Renewable Energy Sustainability Dewatering Biogas Ovivo USA, LLC Vaughan Company, Inc. American Public Works Association (APWA) BDP Industries, Inc. PFAS
The biogas goes directly to dual-fuel boilers. The Montpelier plant produces more biogas than it can use in summer. The excess is flared (lower right of the photo), but the staff is developing a project to use more of the gas.
Interested in Energy?
Get Energy articles, news and videos right in your inbox! Sign up now.
Energy + Get Alerts
Taking in food waste has enabled a Vermont clean-water plant to upgrade its biosolids dewatering equipment and produce enough biogas to nearly eliminate fuel oil for heating buildings and digesters. It also helps the host city toward its goal of net-zero energy.
The project at the Montpelier Water Resource Recovery Facility also included the addition of a receiving station for high-strength organics.
The three digesters were also upgraded. One was converted to a blend tank to feed internally generated sludge and high-strength organic waste to the two other digesters. The blend tank and one of the digesters were fitted with new steel covers and linear motion mixers (Ovivo) and the third digester received a dual-membrane cover (also Ovivo) and a Vaughan Rotamix mixing system.
In 2021 the American Public Works Association honored the project as the Rural Communities Project of the Year/Environment.
LOTS OF SOLIDS
Even before it started accepting food waste, the Montpelier facility was handling a lot of solids. The plant takes in about one third of all septage in Vermont, which has a high percentage of its population on septic tanks. It also takes in sludge from other municipal treatment plants as well as landfill leachate.
"Our average flow is about 2 mgd, but because of the solids we take in, we actually operate like a 10 mgd facility if you look at how much solids we produce each day," says Christopher Cox, chief operator.
The septage and imported sludges go directly to two new incline screw presses (BDP Industries) for dewatering. The old belt presses produced material 16-18% solids; the new ones achieved 26% solids.
BOOSTING BIOGAS
"We really pushed to have the presses be fully automatic," says Cox, whose plant operates with a staff of four. "It's like a push of a button. We can run them into the night. Whoever is on call checks them once a night. We were running both belt presses eight hours. Now we're running two screw presses 12 to 16 hours a day."
The dewatered material goes to landfill, and the filtrate returns to the headworks. The digesters are reserved for the high-strength organic waste and the sludge produced on site.
"There's not a lot of organics in septage compared to FOG, dairy waste or brewery waste," says Cox. "Those have much higher organic concentrations. We're trying to make gas to offset fuel oil, so we don't want to waste any of that space digesting septage. It dewaters very well. It's a very fibrous material."
Tipping fees for receiving septage, sludges and leachate bring significant revenue, which increased from about $1 million a year to $1.25 million in the first full year of taking in waste from food processors, dairies, breweries and ice cream plants. The material boosted biogas production significantly. Previously the plant produced only enough to heat the digesters. Now it heats all the buildings as well.
The biogas goes directly to two dual-fuel boilers without refinement. "There's no treatment in between," says Cox. "All it needs is pressure and a good concentration of methane, which comes from a well-run anaerobic digester. We're not trying to strip out siloxanes or other things you normally remove when trying to make power.
"The simplicity of burning gas with no conditioning is pretty nice. Once you start cleaning the gas, you have a whole other process. The break-even point for a generator didn't make sense for the city." In winter all the gas is used to heat the digesters and buildings. In warmer weather the digesters produce excess gas, which at present is flared.
NEXT PHASE IN SIGHT
Montpelier has committed to becoming a net-zero energy city by 2030, so flaring excess methane is not in the long-range plan. The aim is to add a new process to use the methane but also to be flexible, in case methane production should drop.
"Not flaring is a huge priority," Cox says. "But if we can't figure out how to not flare, do it economically and have the correct payback, it's not worth doing. We're not going to do it and lose tons of money."
The city considered using the excess methane to fuel additional biosolids drying and make a fertilizer product, but land application is difficult now because of regulation of emerging contaminants, such as PFAS.
"At first we were all excited," Cox says. "We thought farmers could use it and maybe we could sell it. But that was a few years ago. The climate of managing biosolids in New England is just crazy now because of PFAS. Land application sites are going away."
Now the city is considering a process to gasify the sludge. The excess methane could preheat the dryer for that process, which would significantly reduce the volume of solids landfilled. "Everyone's competing to get material into the landfills," Cox says. "Anything we can do to reduce the amount of solids we produce will be a huge benefit."
Cox, who has a degree in environmental science from Colby-Sawyer College in New London, New Hampshire, has worked in the wastewater treatment industry for 10 years: "I got into the field thinking my job was to protect the environment by keeping the water clean, which is the main goal. But a lot of my job is figuring out how to get all the energy out of the waste we take in. It's exciting."


text:
Canada Basketball Mens National Team Canada falls to USA in final tune-up before the FIBA Basketball World...
Canada falls to USA in final tune-up before the FIBA Basketball World Cup
opbadmin
SYDNEY, AUSTRALIA (August 27, 2019) – The Canadian Senior Men's Basketball team dropped an 84-68 decision to the United States in their final exhibition game in Australia before heading to China for the FIBA Basketball World Cup 2019.
Click here for the box score.
Kyle Wiltjer scored a game-high 21 points, including 11 in the fourth quarter for Canada, while Khem Birch added 13 in the loss. The United States were led by Jaylen Brown's 19 points, as well as a 10-point, 15-rebound double-double from Myles Turner.
"You don't play great every night," head coach Nick Nurse said. "They'll be fine. We just flush this one away and get back to what we're doing."
Canada was without Oshae Brissett who missed the game due to injury, but saw the return of Andrew Nembhard and Brady Heslip. Nembhard left the game after a fall early in a win against New Zealand last week, while Heslip joined the team in Australia for the first time. Despite having the two back in the lineup, Canada was unable to overcome a slow start in the first quarter on Monday night.
"I don't think we played very well tonight," Nurse said. "I think we've been playing pretty well, with a lot of juice and spirit and energy and I think you could almost sum that game up in the first four or five possessions [of the game]."
Donovan Mitchell scored the first two baskets of the game to give USA an early 5-0 lead. A fadeaway from Kevin Pangos put Canada on the board at the 7:09 mark, but Team Canada struggled to see their shots fall in the early going and connected on just three more field goals the rest of the quarter. After starting just 3-for-17 from the floor, Canada finished 4-for-20 in the quarter, giving up 41 percent shooting to the United States to trail 20-9 after one.
Canada put together a 12-5 run that was capped by a Wiltjer jumper to get within five with two minutes remaining in the half, but Team USA closed the quarter on an 11-1 run to take a 46-31 lead at the break. Canada shot 27 percent from the floor in the first half, as the United States shot 48 percent. Neither team was particularly successful from deep, with Canada making 3-of-14 attempts from three, while the United States connected on just 2-of-8 attempts. Jaylen Brown led all scorers with 13 points in the half.
Khem Birch opened the scoring in the second half with a layup at the 8:26 mark, then followed that with a tip in on Canada's next possession. After a pair of free throws from Melvin Ejim trimmed Canada's deficit to 52-41 midway through the third, the United States scored eight straight to stretch the lead to 60-41 with three minutes remaining in the quarter as Canada called timeout. Out of the timeout, Wiltjer scored Canada's final six points of the quarter, but the Canadians went into the fourth trailing 64-47.
Wiltjer continued to light it up in the fourth, scoring 11 of his game-high 21 in the quarter. Despite Wiltjer's scoring boost off the bench, Canada was ultimately unable to overcome the early 11-point lead at the end of the first quarter, as Canada was outscored by just five points over the final three quarters.
"We're just learning a lot about each other," Wiltjer said. "Games are the best way to prepare and we've got a great coaching staff. The film doesn't lie. We're going to watch it tomorrow, clean up some things.
"We're really grateful we're able to come out here and prepare for the World Cup."
For first-time Team Canada head coach Nurse, the stay in Australia provided necessary time for players and coaching staff to get familiar.
"I've learned a lot about them," Nurse said. "I'm a new coach [for this team], and they're still trying to learn me, I think and we've made a lot of progress.
"I think our guys have made a lot of strides, play hard, have chemistry," he said. "We've been really competitive, fun to watch, we've defended, moved the ball, all that stuff. We've played great. We don't expect to play perfectly every night, or even to play well every night."
Canada finished 3-2 in its five-game exhibition slate in Australia, after splitting a pair of exhibition games played in Canada against Nigeria. The team will now head to China in advance of the FIBA Basketball World Cup 2019 which begins Aug. 31.
Aaron Best
Andrew Nembhard
Canada Senior Men's National Team
Duane Notice
FIBA Basketball World Cup 2019
FIBAWC
Gordie Herbert
International Basketball Series
Kaza Kajami-Keane
Kevin Pangos
Khem Birch
Melvin Ejim
O'Shae Brissett
Owen Klassen
Phil Scrubb
Thomas Scrubb
Toronto Basketball
USA Basketball
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The role of the deacon has always been to serve. In Act 6, when the disciples chose the first deacons, their primary duty was to serve. You remember the situation: There was a complaint because some of the widows felt they were being neglected by the disciples as far as the care they needed. The disciples on the other hand felt they could not leave their primary responsibilities of what God had called them to do – declaring and teaching the Word of God – to serve tables. It was then the decision was made to look for 7 men from the congregation who would be assigned to serve the widows so the disciples could continue their responsibilities. And as the rest of the saying goes, the rest is history.
Here at the church I pastor in New Orleans, our deacons have embraced their role as servants. Many of them serve by assisting elderly members of our congregation with various things like providing transportation to the grocery store or to a doctor's appointment, cutting grass at home, or providing other services as needed. Our deacons also visit a local nursing home once a month and hold a worship service for the residents there. During the Christmas season, they give each resident a gift bag that includes socks, lotion, and toiletries. Man, you would think they had gold and silver in those bags when you see the expressions on their faces as they open their gift bags!
Deacons as called to serve. Throughout the winter edition of the Deacon Magazine you will see examples of how deacons across the country are serving others in their congregations and communities. I encourage each reader to find ways this quarter to serve others in your church and community. I assure you, you will be blessed by serving others.


A prominent computer security expert was barred from boarding one United flight, removed from another, and detained by the FBI—all apparently over a tweet pointing out how easy United's security system would be to hack.
Chris Roberts—the founder of security analysis firm One World Labs—tweeted about vulnerabilities on his United flight's engine-indicating and crew-alert systems (EICAS) last Wednesday.
That same day he was reportedly pulled off a United flight and questioned by the FBI for more than four hours. His lawyers also claim the FBI seized his laptop, iPad, hard drives and other equipment without a warrant.
The FBI reportedly determined Roberts—who was warned not to access flight controls after publicly identifying vulnerabilities in AirBus and Boeing systems—meant no harm.
But either way, United isn't having it: Roberts was reportedly removed from a second flight while en route to a security conference in California. An airline spokesperson tells the AP that the airline feels "confident" about its system but still considers Roberts a threat.
Johnson did not respond to a follow-up question Sunday why Roberts would still be a threat if he couldn't, in fact, compromise United's control systems.
United tells reporters Roberts should receive a letter in the next two weeks explaining why he's been barred from the airline.


Check it out freaks! We're coming at you with another spot on another Horror soundtrack that you won't want to miss. You trusted us with DEATHGASM, and most of you found a new favorite movie. Trust us again, and keep an eye out for RUN, directed by Daniel Mark Young.
RUN will premiere 10/31/2016. Be on the lookout for it at Horror fests everywhere.
RUN was created by Viral Films UK.


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Matilda McCrear, African Slave born
Matilda McCrear
*The birth of Matilda McCrear in c1857 is celebrated on this date. She was a Black African (Yoruba) woman who was enslaved and transported to America.
Matilda McCrear was captured by slave traders in West Africa when she was just two years old and taken to the USA on the Clotilda, the last ship to transport enslaved Africans to America when it docked in Mobile, Alabama, in July 1860. McCrear was transported with her mother Gracie, her three elder sisters, and the man who would go on to be her stepfather. Her family was split up immediately when two of her brothers were left in West Africa. She would never know what happened to them.
On arrival in Alabama, she was bought by Memorable Walker Creagh along with her ten-year-old sister Sallie and her mother Gracie. Gracie was forcibly paired with Guy, another Clotilda survivor, while her two oldest daughters were bought by another slave owner and never seen again. She never married and was probably not a citizen of Africatown, Ala. Instead had a decades-long common-law marriage with a white-American (German), with whom she had 14 children. Even though she left West Africa when she was a toddler, throughout her life she wore her hair in a traditional Yoruba style, a style presumably taught to her by her mother. She also changed her surname from Creagh her former enslaver's spelling to McCrear.
When she was in her 70s and this is what led to her being interviewed by the Selma Times-Journal." Matilda was interviewed by journalist Octavia Wynn of the Selma Times-Journal newspaper because she had made the 15-mile journey from her home to Dallas County Courthouse in Selma to make a claim for compensation for herself and Redoshi Smith as survivors of the Clotilda. Matilda McCrear the last survivor of the middle passage to America died in Selma, Alabama, in January 1940, aged 83.
University of Newcastle.uk


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Home news Moderna amends its COVID-19 vaccine supply agreement with the UK Gov
Moderna amends its COVID-19 vaccine supply agreement with the UK Gov
Shreshtha Dhatrak
Moderna, Inc., a renowned biotechnology firm pioneering the field of messenger RNA therapeutics as well as vaccines to develop the next generation of medicines, has reportedly announced a new supply deal with the Government of United Kingdom to ship an extra two million mRNA-1273 doses, the company's COVID-19 vaccine candidate, to the country beginning March 2021.
The UK has now secured over seven million doses of Moderna mRNA-1273. This new confirmation arrives as the country continues its ongoing efforts to secure an effective and safe access to COVID-19 vaccines by setting up a wide portfolio of the world's most promising coronavirus vaccines.
Chief Executive Officer, Moderna, Stéphane Bancel stated that the company appreciates the partnership it has with the government of UK as with several other governments as well as other important partners across the globe. For almost 10 years, Moderna has made investments in developing a new platform for designing as well as manufacturing a novel class of vaccines that are mRNA-based. Moderna is proud of the advancements it has made on mRNA-1273 to date, which includes the positive interim assessment from the company's Phase 3 COVE clinical study.
On the 16th of November, the company announced that the NIH-appointed, independent Data Safety Monitoring Board for Phase 3 trial of mRNA-1273, Moderna's vaccine candidate for COVID-19, informed that the trial has fulfilled all the statistical criteria that was pre-specified within the clinical study protocol for efficacy, achieving a vaccine efficacy of approximately 94.5%.
The study, more commonly known as COVE study, has signed up over 30,000 individuals within the United States and is being carried out in collaboration with Biomedical Advanced Research & Development Authority (BARDA), which is a part of the U.S. Department of Health and Human Services, and the National Institute of Allergy and Infectious Diseases (NIAID), which is a part of National Institutes of Health (NIH).
On October 27th, 2020, the company received confirmation that Medicines and Healthcare products Regulatory Agency (MHRA), United Kingdom, has started the review process of the company's mRNA-1273.
Source credit: https://www.businesswire.com/news/home/20201129005403/en/Moderna-Announces-Amendment-to-Current-Supply-Agreement-with-United-Kingdom-Government-for-an-Additional-2-Million-Doses-of-mRNA-Vaccine-Against-COVID-19-mRNA-1273
Having completed her Post Graduate degree in Digital Marketing, Shreshtha always nurtured an innate passion for writing. She works as a content writer at JustPositivity.com and pens down news articles spanning numerous verticals. Her other interests include reading and travelling.


Safety is always our first concern when commuting by bicycle. I enjoy commuting by bicycle as often as my schedule permits. Remember that snowy days can create additional hazards for us on the roadways we share with motor vehicles. Metro Transit provides a safe alternative when the bike lanes have not been fully cleared of snow.
Some of the best advice is to position yourself as a motor vehicle would when waiting at an intersection. Hugging the curb near other vehicles can lead to an accident when inattentive drivers turn into you. This is the cause of 30% of our local bicycle vs. motor vehicle crashes.
Enjoy your winter commute and drive safe.