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\begin{document}
\textwidth = 12cm \textheight = 19.2cm

\title[On the dimension and other
numerical invariants]{On the dimension and other
numerical invariants of algebras and vector products}


\author{Larissa Cadorin}
\thanks{}
\address{Larissa Cadorin\\
Mathematik Departement\\
ETH Zentrum\\
CH-8092 Z\"urich\\
Switzerland }
\email{cadorin@math.ethz.ch }
%%\urladdr{http://www.math.ethz.ch/$\sim$cadorin}
\urladdr{http://www.math.ethz.ch/\string~cadorin}

\author{Max-Albert Knus}
\thanks{}
\address{Max-Albert Knus\\
Mathematik Departement\\
ETH Zentrum\\
CH-8092 Z\"urich\\
Switzerland }
\email{knus@math.ethz.ch }
%%\urladdr{http://www.math.ethz.ch/$\sim$knus}
\urladdr{http://www.math.ethz.ch/\string~knus}

\author{Markus Rost}
\thanks{ }
\address{Markus Rost\\
Fakult\"at f\"ur Mathematik
Universit\"at Bielefeld\\
Postfach 100131\\
33501 Bielefeld\\
Germany  }
\email{rost@mathematik.uni-bielefeld.de }
%\urladdr{http://www.mathematik.uni-bielefeld.de/$\sim$rost}
\urladdr{http://www.mathematik.uni-bielefeld.de/\string~rost}





\keywords{tensor categories, composition algebras, vector
products} \subjclass[2000]{Primary 17A75; Secondary 81T99}

\begin{abstract}
Tensor categorical and diagrammatic techniques used in the theory
of knot invariants can be applied to compute the dimension
and other numerical invariants for certain algebraic structures
defined by tensor identities. These techniques are described, 
and applied to symmetric composition algebras and $3$-vector
products.
\end{abstract}

\maketitle

%%%%%%%%%%%%%%%%%%%  The main body  of the paper %%%%%%%%%%%%%%%%%%

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%\input intro.tex
%%%\input proc1.tex

\section*{Introduction }
Many algebraic structures defined on vector spaces occur
only in  specific dimensions, independent of the base field.
One classical example is given by Hurwitz's theorem for 
quadratic composition algebras, which
 implies that such algebras
 exist only in dimension $1$, $2$ (quadratic extensions),
$4$ (quaternions) and $8$ (octonions). Recall
 that  quadratic composition algebras are
 (not necessarily associative)
 algebras $V$ with unit $1$
over a field $F$ with a nonsingular
quadratic form $q: V \to F$, which is multiplicative,
i.e., $q(xy)=q(x)q(y)$ for all $x$, $y \in V$. 
The usual proofs of Hurwitz's theorem
are based on an explicit construction of the composition law
and its classification. 
(See for example Jacobson \cite{Jaccomp}, or the book \cite{KMRT}).
Moreover over algebraically closed fields the dimension is a 
numerical invariant which classifies quadratic composition algebras.

Vector products are also examples of algebraic structures which
 occur only in specific dimensions.
$2$-vector products are generalizations
of the classical vector product in the $3$-dimensional euclidian space.
It is known that they can only occur
 in dimension $3$ and $7$ 
(see \cite{Eckmann} or \cite{BrownGray}). 
Let $b\col V \times V \to F$  be a
nondegenerate, symmetric bilinear
 form on a vector space $V$. A~{$2$-vector product} 
is a skew-symmetric bilinear map
$V \times V \to V$, $(x,\, y) \mapsto x \times y$,  such that
$b( x \times y,x) = b( x \times y,y)=0$ and
\[
b(x \times y, x \times y)= \det \begin{vmatrix} b(x,x) & b(x,y) \\
b(y,x) & b(y,y) \end{vmatrix}
\]
for all $x,\,y \in V$.

Recently a new, tensor theoretical approach was given in
\cite{Ro:96a} and \cite{Bo98} for the calculation of the dimension
of spaces admitting $2$-vector products (see also \cite{Rost:04}).
A variation on the argument of \cite{Ro:96a} is given in
 \cite{Meyberg02}.
 The linearization  of the  relations satisfied by $2$-vector products gives
tensor identities and a universal tensor category
can be constructed for these relations. A canonical ring of
numerical invariants is attached to this category.
 One proves that the dimension $d$
of  the $2$-vector product algebra generates this ring of numerical
invariants  and satisfies
the identity $d(d-1)(d-3)(d-7)=0$.

 Inspired by knot theory
and quantum field theory (see for example \cite{barnatan95},
 \cite{Baez2002} or
 \cite{Turaev94}),
 a graphical incarnation
of the universal tensor category associated with $2$-vector products
is used in \cite{Bo98}. This graphical representation allows 
to do linear algebra and tensor calculus
in a handy way. The bilinear product
$V \times V \to V$, $(x,y) \mapsto x \times y$ is represented
as a graph 
connecting two horizontal lines:
%
\[
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
\put(15,30){\makebox(0,0){\tiny $x$}}
\put(35,30){\makebox(0,0){\tiny $y$}}
\put(25,0){\makebox(0,0){\tiny $x\times y$}}
%\put(-15,15){\makebox(0,0){$\mu= \ $}}
\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
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\put(35,25){\circle*{4}}
%\put(25,15){\circle*{4}}
\put(25,5){\circle*{4}}
\end{picture}}
\end{picture}
%
\qquad
\begin{array}{ccc} V &\tens& V \\
& \downarrow& \\
& V &
\end{array} \quad .\]
%
A single point on a horizontal line corresponds to one copy of $V$ 
and
a finite set of  $n$ points to the tensor product  $V^{\tens n}$
of the same number of copies of $V$. Since $V^{\tens 0}$ is 
identified
with the base field $F$,
 the symmetric bilinear form $b\, : V \times V \to F$
is represented by the graph
\[
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
\put(15,30){\makebox(0,0){\tiny $x$}}
\put(35,30){\makebox(0,0){\tiny $y$}}
%\put(-15,15){\makebox(0,0){$\beta=  $}}
\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
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\put(15,25){\circle*{4}}
\put(35,25){\circle*{4}}
\end{picture}}
\end{picture}
%
%
\qquad
\begin{array}{ccc} V &\tens& V \\
& \downarrow& \\
&F  &
\end{array} \quad . \]
Any multilinear map occurring in the tensor identities
satisfied by the $2$-vector product can
accordingly be represented by a graph connecting points on the
two horizontal lines and composition of maps corresponds to gluing
of diagrams. Closed graphs (i.e., without points
on the two horizontal lines) represent numerical invariants.
The tensor identities defining the $2$-vector product
translate into relations between formal sums of graphs and 
the above identity  for
the dimensions of vector products can be obtained by pure graph
computations.

 In this report we describe these diagrammatic
techniques for a class of algebras
and apply them for two specific algebraic structures: symmetric
compositions and $3$-vector products.

Symmetric composition algebras are algebras (without 1)
with a multiplicative, nonsingular,  quadratic form $q$ such that
the associated symmetric  bilinear form
$b(x,y) = \tfrac{1}{2}\big(q(x+y) -q(x) -q(y)\big)$ is associative:
\[
b(xy,z) =b(x,yz) \quad \text{for all $x,\,y,\, z \in V$.}
\]
They arise in a natural way from degree $3$ alternating algebras
and their structure is known (see \cite{EM} or\cite{KMRT}).
%
In Section \ref{symmetriccompositions} we prove that
 the ring of numerical invariants of the corresponding universal
 tensor category is generated by
two numerical invariants: the dimension $d$ and the norm
$e =q(c)$ of a Casimir element $c \in V$ 
(see Section \ref{tensors} for the 
definition of the Casimir element). 
These two invariants satisfy the
identities
\[
 (e -(d-2)^2)e =0 \ \ \text{and}\ \ (d-2)(d-8)(d-e) =0
\]
with possible values
 \[
(d,e)= (0,0),\, (1,1),\, (2,0),\, (4,4),\, (8,0) \quad \text{and}
\quad (8,36).
\]
For each pair $(d,e)$ there exists a symmetric composition algebra
with invariants $(d,e)$; further, over an algebraically closed field
the pair $(d,e)$  determines the symmetric composition algebra
up to isomorphism.
In particular the appearance of the second invariant 
$e$ permits to distinguish the
two known types of symmetric compositions in dimension~$8$. 
The computation of the ring of numerical invariants of
the universal tensor category of symmetric compositions  given here
is due to the third named author.

 A  $3$-vector product
is an alternating trilinear product
 \[
V \times V \times V \to V, \, (x,y,z) \mapsto xyz,
\]
such that
\[
b(xyz,xyz)=\det \begin{vmatrix}
b(x,x) & b(x,y) & b(x,z) \\
b(y,x) & b(y,y) & b(y,z) \\
b(z,x) & b(z,y) & b(z,z)
\end{vmatrix}
\]
%
for all $x,\,y,\, z \in V$. The structure of $3$-vector products
is known (or more generally  $r$-vector products, 
see \cite{Eckmann} or \cite{BrownGray}). In particular they
only occur in dimension  $4$ and $8$.
 The diagrammatic interpretation of $3$-vector products
given in Section \ref{3vectors} is new and will be part of the PhD thesis \cite{cadorin04}
 of the first author.
More applications of tensor and diagrammatic techniques to algebraic
structures, including a diagrammatic proof of Hurwitz's theorem,
  will be given in the thesis.
%
\section{Algebraic structures and tensors}\label{tensors}
%
Let $F$ be a field of characteristic different from $2$ and let $V$ be
 a finite dimensional vector space over $F$. Let $q:V \to F$ be a
quadratic form
on $V$ with polar form
$b(x,y)= \tfrac{1}{2}\bigl(q(x+y) -q(x) -q(y)\bigr)$.
We assume that $b$ is nonsingular, i.e., $b$ induces an isomorphism
$h: V \isom V^*=\Hom_F(V,F)$, $x \mapsto b(x,-)$.

 Let $V^{\tens m}= V \tens \cdots \tens  V$ ($m$ copies),
 $V^{\tens 0}=F$. We identify
$\bigl(V^{\tens m}\bigr)^* $ with $(V^*)^{\tens m}$ through the
canonical map $(V^*)^{\tens m}\to \bigl(V^{\tens m}\bigr)^*$given by
\[
(f_1 \tens \cdots \tens f_m)(v_1 \tens \cdots\tens  v_m)=
 f_1(v_1) \cdots f_m(v_m)
\]
for $f_i \in V^*$ and $v_i\in V$.
The dual  $\varphi^*: (V^{\tens n })^* \to
(V^{\tens m })^*$ of any linear map $\varphi:V^{\tens m}
\to V^{\tens n}$ then can be viewed as a linear map
$\varphi^*: (V^*)^{\tens n} \to (V^*)^{\tens m}$. We
denote by
$\varphi^t: V^{\tens n} \to V^{\tens m}$
the linear map $(h^{\tens m})^{-1}\circ \varphi^*\circ h^{\tens n}$
and call it the {\it transpose} of $\varphi$.
%
The bilinear form $b$ can be viewed as a {\it contraction}
 $b: V\tens V \to F$. The fact that $b$ is symmetric can be
expressed as $b =b \circ \tau$, where $\tau: V\tens V\to V\tens V$ is
the {\it switch} $\tau (x \tens y)= y\tens x$, $x,\,y \in V$.
We extend $b$ to
$b_k \col V^{\tens k} \tens V^{\tens k} \to F$, $k\geq 1$:
\[
b_k\bigl((u_1\otimes\cdots\otimes u_{k})
\tens(v_1\otimes\cdots\otimes v_{k})\bigr)
=\prod_{i=1}^k b(u_{i},v_{i}).
\]
 For $m\geq k$, we have  isomorphisms
\begin{equation}\label{functisov}
\Hom_F(V^{\tens n}, V^{\tens m})
\isom \Hom_F(V^{\tens (n+ k)}, V^{\tens (m-k)}), \ \theta\mapsto\theta^{[-k]}
\end{equation}
where
$\theta^{[-k]}$ is the composite
\[
\underbrace{V\otimes\cdots\otimes V}_{n+k\>{\rm factors}}
\mskip10mu{\buildrel\theta\otimes 1_{V^{\otimes k}}
\over{\mathrel-\joinrel\mathrel-\joinrel\mathrel-\joinrel\mathrel-\joinrel\longrightarrow}}
\mskip10mu\underbrace{V\otimes\cdots\otimes V}_{m+k\>{\rm factors}}
\mskip10mu{\buildrel 1_{V^{\otimes (m-k)}}\otimes b_k
\over{\mathrel-\joinrel\mathrel-\joinrel\mathrel-\joinrel\mathrel-\joinrel\mathrel-\joinrel
\mathrel-\joinrel\mathrel-\joinrel\longrightarrow}}
\mskip10mu\underbrace{V\otimes\cdots\otimes V}_{m-k\>{\rm factors}} \ .
\]
%

We assume that $b$ admits
an orthonormal basis \base, $b(e_i,e_j)= \delta _{ij}$,
$i,\,j =1, \ldots,\, d$, extending the base field if necessary.
We fix $f_i= b(e_i, -)$, $i=1, \ldots,\, d$, as a dual basis of
$V^*$, so that $h:V \to V^*$ is given by $e_i \mapsto f_i$. The map
$b^t\col F \to V\tens V$ is given by
\begin{equation}
b^t(1_F) = \sum_i e_i \tens e_i.
\end{equation}
It follows that $(b \circ b^t)(1_F)=\sum_ib(e_i,e_i) =d$,
 the dimension of $V$.
%\end{example}
%

A bilinear {\it multiplication} on $V$ is a linear map
$m: V \tens V \to V, \ x\tens y \mapsto xy$.
We call the pair $(V,m)$ an
{\it algebra} (over $F$).
 Let $(e_1, \ldots, e_d)$ be an orthonormal
 basis of $V$ and
let $m(e _i \tens e_j) = \sum_k m_{ijk}e_k$. It follows that
$m^t(e_k) = \sum_{i,j} m_{ijk} e_i \tens e_j$.
If the algebra $(V,m)$ admits  an
{\it associative (or invariant)} nonsingular
bilinear form, i.e.,
\begin{equation}
b(xy,z)= b(x,yz) \quad \text{for all $x,\,y,\, z \in V$},
\end{equation}
then
\begin{equation}\label{example2}
m^t(v) = \sum_i e_iv \tens e_i = \sum_ie_i \tens v e_i.
\end{equation}
%
For any $x\in V$, let $\ell_x\col V \to V$ be the linear map $y \mapsto xy$
and let $\Tr(\ell_x)$ be the trace of $\ell_x$.

\begin{lemma}\label{casimir}
Let $(V,b,m)$ be an algebra with a nonsingular associative bilinear
form and let $(e_1,\ldots,e_d)$ be an orthonormal basis.
The element $c= \sum_ie_ie_i \in~V$ is uniquely determined by
the property $\Tr(\ell_x)= b(c,x)$.
In particular it is independent of the choice of the orthonormal
 basis.
\end{lemma}
%
\begin{proof}
Uniqueness follows from the fact that $b$ is nonsingular.
Let $xe_i =\sum_jx_{ij}e_j$, so that $\Tr(\ell_x)=\sum_ix_{ii}$.
We have
\[
\sum_i b(x,e_ie_i) = \sum_i b(xe_i,e_i) =
 \sum_{i,j}x_{ij}b(e_j,e_i) =
 \sum_ix_{ii}.
\]
\end{proof}
%
The element $c$ is usually called a {\it Casimir element}.
The dimension and the  value $e=b(c,c) \in F$ are numerical invariants
attached to the algebra $(V,b,m)$
(where $b$ is associative with respect to $m$).

%For later use we denote the {\it opposite multiplication}
%$x\tens y\mapsto yx$, by $m^{op}$. We have $m^{op} =m\circ\tau$, whe%re

For simplicity we call elements of
$\Hom_F(V^{\tens n},V^{\tens m})$, for arbitrary $n$ and $m$,
{\it tensors}. For any triple $(V,b,m)$, $b$ a nonsingular
 symmetric bilinear form and $m$ a multiplication, we have an
{\it alphabet} of basic tensors consisting of
$1_V$, $b$, $m$,  the switch $\tau$ and the transposed $b^t$,
$m^t$.
Let $\mathfrak C=\big\{ c_i \in
\Hom_F(V^{\tens n_i},V^{\tens m_i})\big\}$
 be a set of tensors, usually  generated (through compositions and tensor
products) by elements from the basic alphabet.
We say that an algebra $(V,b,m)$ is {\it  of tensor type}
$\mathfrak C$\/ if the $c_i$ are identities for $(V,b,m)$, i.e., if
\[
c_i(x_1 \tens \ldots \tens x_{n_i}) =0
\]
 for all $i$ and all $x_k\in V$.
%
\begin{example}
 {\rm The vector product in $\mathbb R^3$,
$m\col x\tens y \mapsto x \times y$,
satisfies the identities
\begin{equation}\label{vector3}
\begin{array}{lcl}
 x \times y +y \times x &= &0 \\
x \times(y \times z) -\langle x,z\rangle y + \langle x,y\rangle z &=& 0
\end{array}
\end{equation}
for  $x,\,y,\, z \in V$, where $\langle x,y\rangle =b(x,y)$ is
the scalar product. The  tensors
are $m+m\circ \tau$  and $ m\circ (1_V \tens m)  -
(b\tens 1_V) \circ (1_V \tens  \tau ) + b\tens 1_V$ in
$\Hom_F(V^{\tens k},V)$, $k=2,\,3$.}
\end{example}
%
\section{Graphs and tangles}\label{tangles}
A {\it graph} $G$ is a finite $1$-dimensional  cell space. It consists
of a finite set $V$ of {\it vertices} ($0$-cells)
 and  a finite set $E$ of {\it edges} ($1$-cells) and is denoted by
$G=(V,E)$.  Circles are admitted.
 Every edge $e$  connects two vertices $\{v_1,v_2\}$, which
can possibly coincide. The two endpoints determine the edge
and we write $e = v_1v_2$.
Graphs are drawn as $2$-dimensional projections with the convention
that braidings are symmetric:\\[-15pt]
%\input text.sty
\[
\tau \quad= \begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
%\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
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\linethickness{0.3mm}  \qbezier(15,25)(20,20)(22,18)
\linethickness{0.3mm}  \qbezier(27,13)(30,10)(35,5)
\end{picture}}
\end{picture} =
%
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad } %\makebox(3,0){$Y$}}
%\put(5,25){\line(1,0){40}\quad }%\makebox(3,0){$X$}}
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\linethickness{0.3mm}  \qbezier(15,25)(30,10)(35,5)
\end{picture}}
\end{picture} \quad .
\]
\\ %[5pt]
%
A {\it path} $P$ is a graph with vertices
 $\{v_0,\ldots,v_k\}$ and edges
\[
\{v_0v_1,\,v_1v_2,\ldots, v_{k-1}v_k\}\]
 where the $v_i$ are pairwise
different, with the possible exception $v_0 =v_k$. We say that
$P$ is a path from $v_0$ to $v_k$, denote it
$P= v_0v_1\ldots v_k$ and call $k$ the {\it length} of the path.
A {\it refinement} of a graph $G$ is a  graph  obtained from $G$
by dividing some of the edges, i.e., by
replacing  these edges with paths having the same endpoints
and so that the intermediate points are not vertices of $G$.
Given a graph $G=(V,E)$, let $G'=(V',E')$ be the refinement obtained by adding
a vertex in the ``middle'' of each edge. The {\it star} ${\rm St}(v)$ of a vertex
$v$ of $G$ is the set of edges $e' \in E'$ which are incident with $v$ and the
{\it valency}  $\val(v)$ of $v$ is the cardinality of ${\rm St}(v)$.


The set of vertices  of a graph $G$ having  valency $1$ is the
{\it boundary }
$\partial G$ of $G$.
We consider graphs whose vertices have valency $\leq 3$. Let $v$ be a trivalent
vertex.
Any bijective map $\Phi_v$ of $\{1,2,3\}$ to the star   ${\rm St}(v)$ of $v$
 is an {\it ordering}
of the vertex $v$.  We use the ordering to equip $v$
with an {\it orientation}. The orientation is {\it positive} if
the image of $\Phi_v$ is (clockwise) an even permutation and is {\it negative}
otherwise. Orientations are indicated  by  black or white colorings:\\[-20pt]
%
\begin{equation} \label{positive}
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
%\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
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\linethickness{0.3mm}  \qbezier(25,15)(35,15)(35,25)
\linethickness{0.3mm}  \qbezier(25,4.5)(25,9.5)(25,14.5)
\put(25,15){\circle*{4}}
\end{picture}}
\end{picture}
%
\quad \text{(positive)}\, , \quad
%
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
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%\put(15,25){\circle*{4}}
%\put(35,25){\circle*{4}}
\put(25,15){\circle{4}}
%\put(25,5){\circle*{4}}
\end{picture}}
\end{picture}
 \quad \text{(negative)}\,.
%\input text.sty
%$$
%\pic2\quad\hbox{(positive),}\qquad\pic3\quad\hbox{(negative).}
%$$
\end{equation}\\
%
  If  no orientation is indicated we understand some orientation.

Let $X=\{x_1,\ldots,x_k\}$ and $Y= \{y_1,\ldots, y_\ell\}$ be
finite sets
(the empty set is not excluded).
 A {\it m-tangle} \footnote{Classical
tangles do not have trivalent edges. The ``m'' in the definition of a
m-tangle recalls that trivalent vertices represent  multiplications.}
$G: X \to Y$ is a graph such that $\partial G$ is the disjoint union
$ X \sqcup Y$.
The sets $X$ and $Y$ are represented by points on two superposed horizontal lines
and $G$ connects the two sets. The set $X$ on the upper line is
 the {\it source} of $G$ and
$Y$ on the lower line is the {\it target} of $G$. For example
%
%
\[
\tanglexample
\]\\
is a m-tangle from $X =\{x_1,\ldots,x_4\}$ to $Y=\{y_1,\ldots,y_5\}$.
%
 For any $X$ there is an obvious identity  m-tangle $I_X\col X \to X$.

 Exchanging source  and target of $G$  through a reflection along the
target line we get
the {\it transposed m-tangle} $G^t\col  Y \to X$ of $G$. The disjoint union
$G \sqcup G'\col X\sqcup X' \to Y\sqcup Y'$ of two m-tangles $G\col X \to Y$,
 $G'\col X' \to Y'$ is a m-tangle.  There is an obvious composition of
m-tangles
\[
(G \col Y \to Z) \times (H\col X \to Y) \mapsto G \circ H  \col  X \to Z
 \]
 defined by gluing the source of $H$ with the target of $G$.
  Two m-tangles with  identical refinements are  called
{\it isomorphic}  and the isomorphism class of a m-tangle
$G$ is denoted by $[G]$. Open paths are isomorphic to edges and closed ones
to circles.
Composition   passes over to isomorphism  classes and we may define
the {\it m-tangle category} $\mathcal T _1$: the objects of $\T _1 $
are finite  sets
$X$ and  a morphism $\gamma \col X \to Y \in \Mor_{\T_1}(X,Y)$ is
the  isomorphism class $[G]$ of a m-tangle
$G \col X \to Y$. The  composition of two morphisms is induced by the corresponding
composition of  m-tangles and $[I_X]$ is the identity element of $\Mor_{\T_1}(X,X)$.
The disjoint   union $\sqcup$ induces a {\it tensor product}\/
 $\square\col \T_1 \times \T_1 \to \T_1$. The tensor product
is associative and has the unit object $[\emptyset]$.
For pairs of objects $(X,\, Y)$, the transpose induces a map
$[G] \mapsto [G^t],\ \Mor_{\T_1}(X,Y) \to \Mor_{\T_1}(Y,X)$
such that $([G]\circ [H])^t =[H]^t \circ [G]^t$. Note that transposing
changes the orientation of $3$-valent vertices.

The  following  morphisms \\[-10pt]
%
\[
\begin{array}{ccccccccccccccccccccc}
I_{\{x\}} & =& \hspace{-11pt} \IdX\ , &\beta &=& \betta\ , \quad&  \mu
& =& \mult \\
\tau &=& \hspace{-16pt}\twistl, &\beta^t & =&  \bettat \ ,\quad
&  \mu^t
& =& \multtop \\
\end{array}
\]\\[10pt]
are  called {\it basic}. They constitute the {\it alphabet}
of $\T_1$.

\begin{remark}
Composing $\mu$ with $\tau$ we can change the orientation of the trivalent vertex of $\mu$. We have\\ \vspace{-25pt}
\begin{center}
\begin{picture}(45,30)
\put(-20,-12){
\begin{picture}(45,30)
\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
\linethickness{0.3mm}  \qbezier(15,5)(25,15)(35,25)
\linethickness{0.3mm}  \qbezier(15,25)(20,20)(22,18)
\linethickness{0.3mm}  \qbezier(27,13)(30,10)(35,5)
\put(15,25){\circle*{4}}
\put(35,25){\circle*{4}}
\put(15,5){\circle*{4}}
\put(35,5){\circle*{4}}
\put(-2,-8){\mult}
\put(54,4){\makebox(3,0){$=$}}
\put(60,2){\mulop}
\end{picture}}
\end{picture}
\end{center}
\end{remark}

\vspace{20pt}

\hspace*{-4ex}up to isomorphisms.
%
\begin{prop}\label{prop1}
The morphisms of the category $\T_1$ are generated up to isomorphisms
through composition
and disjoint union by  basic morphisms.
\end{prop}
%
Remark that refinements are needed in \eqref{prop1}, for example to separate
adjacent trivalent vertices.
Moreover, the decomposition of a morphism in basic ones
is not unique. In fact the alphabet is redundant. The morphism
$\mu^t$ can be deduced from the other basic morphisms.

The next step is to enrich the category
$\T_1$ to  a category $\T _2$ in such a way  that, for every pair of objects $X$ and $Y$,
 the set of morphisms $\Mor(X,Y)$ has a linear structure.
  Let $R=\mathbb Z[\frac{1}{2}]$. The objects of $\T_2$ are
finite  sets, as for $\T_1$. For
two finite sets $X$ and $Y$,
$ \Mor_{\T_2}(X,Y)$ is the free $R$-module generated by the set
$ \Mor_{\T_1}(X,Y)$ and composition is extended by linearity:
for $\gamma = \sum_i a_i[ G_i],\,  a_i \in R, [G_i] \in
\Mor_{\T_1}(X,Y)$ and \,
 $\rho = \sum_j b_j[ H_j],\,  b_i \in R, [H_j] \in
\Mor_{\T_1}(Y,Z)$, we have $\rho \circ\gamma= \sum_{i,j} a_ib_j[H_j\circ G_i]
\in \Mor_{\T_2}(X,Z)$. The identity of $X$ is by definition the
element $1\cdot [I_X]$. The tensor product~$\square$ also extends by
linearity:
\[
H\,\square\,  G= \sum_{i,j} a_ib_j[H_j\,\square\, G_i]
\]
and the extension is again associative with unit element
$1\cdot [\emptyset]$. Composition and tensor product coincide
on $\Mor_{\T_2}(\emptyset,\emptyset)$ and endow
$\Mor_{\T_2}(\emptyset,\emptyset)$ with the
structure of a free commutative $R$-algebra with $1$.

In addition to
generators we also can impose relations:
Let
\[
\Gamma =\{\gamma_i \in \Mor_{\T_2}(X_i,Y_i),\ i\in I\}
\]
be a  set of morphisms of $\T_2$. For each pair of objects
$(X,\, Y)$  of $ \T_2$, the set  $\Gamma$ generates
through compositions and tensor
products a submodule $M(X,Y)$ of $\Mor_{\T_2}(X,Y)$. We define a new
category $\T_\Gamma$ by setting
\[
\Obj(\T_\Gamma) = \Obj(\T_2) \quad \text{and}\quad
\Mor_{\T_\Gamma}(X,Y)=
\Mor_{\T_2}(X,Y)/M(X,Y).
\]
%
For any $\rho  \in \Mor_{\T_2}(X,Y)$, let $[\rho]$ be its class in
$\Mor_{\T_\Gamma}(X,Y)$. We get a projection functor
$\mathcal P\col \T_2 \to \T_\Gamma$ such that $\mathcal P(X)=X$ for
 $X \in \Obj(\T_2)$ and $\mathcal P(\rho) =[\rho]$ for
$\rho$ any morphism. In particular we have
$\mathcal P (\gamma_i)=0$ for each $\gamma_i \in \Gamma$.
We  say that the $\gamma_i$ are  {\it relations defining}
$\T_\Gamma$
and we call $\T_\Gamma$  the {\it m-tangle category  associated with the set of relations}  $\Gamma$. The algebra
$\Mor_{\T_2}(\emptyset,\emptyset)$ projects on
$E_\Gamma=\Mor_{\T_\Gamma}(\emptyset,\emptyset)$. We call $E_\Gamma$
the   {\it algebra of numerical invariants} of the category
$\T_\Gamma$. To compute $E_\Gamma$ is the main feature of the theory.
 We call a m-tangle category  associated with
 some set of tensor relations  a
 {\it m-tangle category of type }$\T_3$.
%
\begin{example}\label{example1}
 The simplest nontrivial element  of $E_\Gamma$ is the  circle\\[-10pt]
%
\[
d= \beta \circ\beta^t = \dinv \ .
\]\\[-5pt]
Another interesting
element of $E_\Gamma$ is the element\\[-15pt]
\[
e = \beta \circ (\mu \square \mu)\circ (\beta^t \square \beta^t)
 =  \einv\ \quad \quad  .
\]\\
(As we shall see the orientations do not play a r\^ole.)

A simple example is given by $\Gamma =\{\mu\}$, so that
$\mu =0$ in $\T_\Gamma$ is the only relation. Then  $d$
generates  $E_\Gamma$
 and $E_\Gamma$ is isomorphic
 to the polynomial ring $R[d]$.
\end{example}
%
For any of the categories $\T$  and any
object $X=\{x_1,\ldots x_n\}$, we have a contraction
 $\beta_X\col X\, \square\, X \to \emptyset$:
%
\[
\beta_X\bigl(\{x_1,\ldots, x_n\},\,\{ y_1,\ldots, y_n\}\bigr)=
\beta\bigl(\{x_1,y_1\} \bigr)\square \ldots \square
 \beta\bigl(\{x_n,y_n\}\bigr) \in \emptyset,
\]
where $\{y_1,\ldots, y_n\}$ is another copy of $X$.
As for vector spaces, the contraction induces
functorial isomorphisms
\begin{equation}\label{functiso}
\Mor_\tau(X, Y_1\square Y_2) \isom  \Mor_\tau(X\square Y_2, Y_1)
\end{equation}
given by $\theta \mapsto \theta'=
(I_{Y_1}\, \square\, \beta_{Y_2})
\circ (\theta \,\square\,I_{Y_2})$. For example the image of
the isomorphism class of
 $\mu^t \in  \Mor_\tau\big(\{x\}, \{y_1, y_2\}\big)$
in
 $\Mor_\tau\big(\{x,y_2\}, \{y_1\}\big)$ is the isomorphism class
of $\mu\circ \tau$.
%
\section{Representations of m-tangle categories}\label{representations}
%
Let $F$ be a field of characteristic different from $2$ and let
$\mathfrak V= (V,b,m)$ be an $F$-algebra.
The bilinear form is supposed to be associative with respect to $m$.
Let $\VV$ be the category of finite dimensional vector
 spaces over $F$.
Enlarging $F$ if necessary (which will not change the tensor
type of $\mathfrak V$),
 we may assume that $V$ has  an orthonormal basis \base.
%
Let  $I_{\{x\}},\, \beta,\, \beta^t, \,\mu,\, \mu^t$, and $\tau$ be
 the basic
generators (the alphabet) of the m-tangle categories $\T_1$
and $\T_2$  introduced in
 Section~\ref{tangles}.
%
\begin{thm}{\rm (Boos, \cite{Bo98})}\label{boos}
There exists a unique functor 
\[\FF \col \T_2 \to \VV\]
 such that
\begin{enumerate}
\item $\FF(\emptyset)= F$,  $\FF(X) =
 V_1 \tens V_2 \tens \ldots  \tens V_n$ for
$X =(x_1,\,x_2,\ldots,x_n)$ and $V_i$ denotes a copy of $V$ to which
$x_i$ is mapped,
\item $\FF(I_{\{x\}}) = 1_V$, $\FF(\tau) = \tau$,
\item $\FF(\beta) = b$, $\FF(\mu) = m$,
$\FF(\gamma^t) = \FF(\gamma)^t$ \/ and \/
$\FF(\gamma\, \square\, \rho)= \FF(\gamma) \tens \FF(\rho)$
for morphisms $\gamma$ and $\rho$ in $\T_2$.
\end{enumerate}
\end{thm}
%
\begin{proof}
It suffices to check that $\FF$ is uniquely defined
on the category~$\T_1$.
The extension to $\T_2$ follows by linearity.  The functor is clearly
uniquely defined on basic morphisms (the alphabet).   By Proposition \ref{prop1}
an extension to $\T_1$ will be unique. However, since there is no unique
decomposition of morphisms in basic  morphisms, the existence of an
extension is not obvious.  In particular there is only one type of open paths
up to isomorphisms, but open paths may have a priori different realizations.
We follow the clever  argument given in
\cite{Bo98}.
We first consider the case where $\T_1$ has no
$3$-valented vertices.
Connected components are paths and it suffices to
check that $\FF$ is well
 defined on paths. Paths are composed of elementary morphisms
of type $I_{\{x\}}$, $\beta$, $\beta^t$  and $\tau$. We have\\[5pt]
\begin{equation} \label{rigidity}
\ndgr \quad  \quad \ndgc \quad  \quad \quad  \quad\ndgl \\[10pt]
\end{equation}
up to isomorphism, since open paths are isomorphic.
 The image of the left diagram under
the functor $\FF$ is the linear map
$v \mapsto \sum_i e_ib(e_i,v)$  and the image of the right diagram is
$v \mapsto \sum_i b(v,e_i)e_i $. Both expressions are equal
 to $v$
since \base\ is an orthonormal basis\footnote{In a more abstract setting,
the two parts of \eqref{rigidity} corresponds to the two
so-called ``rigidity axioms''.}.
By induction on the length of a path,
this implies that $\FF$ is well defined on isomorphism classes of
 paths which do not contain braidings. In presence of braidings we
 have\\[-25pt]
\[
\begin{array}{ccccccc}
\btwist & \quad \text{or} & \bttwist
\end{array}
\]\\[15pt]
in $\T_1$. This corresponds in $\VV$ to the fact that the bilinear form $b$
is symmetric. Again by induction $\FF$ is well defined.
For the general situation it is convenient to reduce to the case where
the target of any morphism $\rho$ is empty. By \eqref{functiso}
 we have a canonical isomorphism
\begin{equation}\label{XYempty}
\Mor_{\T_2}(X,Y) \isom \Mor_{\T_2}(X \square Y, \emptyset)
\end{equation}
which  maps to
\[
\Hom_F(V^{\tens n}, V^{\tens m}) \isom
\Hom_F(V^{\tens n} \tens V^{\tens m}, F)
\]
in $\VV$  through the functor $\FF$ (see \eqref{functiso}).
 Thus we are reduced to  the case $Y = \emptyset$ in $\rho$.
%
Assume  that the morphism $\rho\col X \to \emptyset$ contains  $k$
trivalent vertices $P_1,\, \ldots P_k$.  Refining $\rho$ if
 necessary,
we may assume that all $P_i$ are centers of morphisms  $\mu_i$
of type $\mu$. Replacing
the negative oriented ones by $(\mu_i \circ \tau)\circ \tau$
we get graphs $\mu_i \circ \tau$ which are positively oriented
and
 we can assume that all orientations are positive.
Let $\overline{\rho}$ be the graph obtained from $\rho$ by deleting
all $\mu_i$.  Let
$(z_i^1,\,z_i^2,\,z_i^3)$ be the boundary of $\mu_i$
 (in a positive orientation) and let $Z$ be the finite ordered set
$Z= (z_1^1,\,z_1^2,\,z_1^3, \ldots, z_k^1,\,z_k^2,\,z_k^3)$. The
graph $\overline{\rho}$ defines uniquely a morphism
$\overline{\rho}\col X \to Z$ which does  not contain any $\mu_i$
(up to isomorphism). Thus $\FF$ can be uniquely extended to
$\overline{\rho}\col X \to Z $. Let
%
\begin{equation}\label{3bil}
 \overline{\mu} =  \multbar \qquad \qquad .
\end{equation}
\\[10pt]
%
Composing $\overline{\rho}$ with  $k$ copies of  the morphism
$\overline{\mu}$
we get back $\rho$. Thus, to conclude, we only have to show
that the  images under $\FF$ of the  two left morphisms in \eqref{3bil}
define the same multilinear maps in $\VV$. This follows
from the associativity of $b$, $b(x,yz)=b(xy,z)$.
\end{proof}
%

%
\begin{cor} Let $ \mathfrak V=(V,b,m)$ be an algebra of tensor type
 $\mathfrak C=\{c_i\}$ for tensors $c_i$ expressible through the alphabet
$1_V,\, m,\, m^t, b$, $b^t$ and $\tau$.
 Let $\Gamma= \{\gamma_i\}$ be a chosen set
of graphs in $\T_2$ such that  $\FF(\gamma_i)=c_i$
 for all $c_i \in \mathfrak C$
and let $\T_\Gamma$ be the corresponding category of type $\Gamma$.
There  is a unique functor
$\mathcal R_\Gamma \col \T_\Gamma \to \V$ such that
$\FF= \mathcal R_\Gamma \circ \mathcal P$.
\end{cor}

\begin{cor}
A necessary condition for the existence of an algebra of tensor
 type,  for a set  of tensors $\mathfrak C = \{c_i\}$,
is the existence  of an $R$-algebra  homomorphism
 $\Phi\col E_\Gamma \to F$, where $\Gamma= \{\gamma_i\}$ is a set
of relations in $\T_2$ such that  $\FF(\gamma_i)=c_i$
 for all $c_i \in \mathfrak C$.
\end{cor}
%
%
\begin{proof}
If such an algebra exists, the functor $\mathcal R_\Gamma$
induces a  homomorphism
$E_\Gamma =\Mor_{\T_\Gamma}(\emptyset,\emptyset) \to \Hom_F(F,F)=F$.
\end{proof}
%

Since
\[
\FF(\beta \circ\beta^t)(1_F) =
(b\circ b^t)(1_F) = \sum_ib(e_i,e_i)=\dim_F V\cdot 1_F
\]
the image  of the circle $d$ under the functor $\FF$
is the dimension of $V$. (It is convenient
to use the same symbol $d$ for the circle and the dimension.)
Rotation around the vertical symmetry axis gives the following
identities
%
\begin{equation} \label{casvet}
\mu \circ \beta^t= \pic28 =\pic29=\pic30.
\end{equation}
The image under $\mathcal R_{\Gamma}$ of $\mu \circ \beta^t$ is the Casimir element
$c =\sum_ie_ie_i$. Thus the element \\[-20pt]
\[
e =\pic84
\]\\[-10pt]
of $E_\Gamma$ is mapped under
$\mathcal R_{\Gamma}$ to $b(c,c)\cdot 1_F$,
 which we denoted also $e$ in Section~\ref{tensors}.

\begin{example}
Let $(V,\,b,\,m)$ be an algebra with $m=0$, as in
Example~\ref{example1}. The algebra
$E_\Gamma$ is the polynomial ring $R[d]$, where $d$ is the circle,
and $\mathcal R_\Gamma(d) = \dim_F\cdot 1_F$. Thus there are no obstruction to
the existence of such an algebra in any dimension $d$
(which is of course obvious!).
\end{example}
%
\begin{example}\label{r3}
Let $(\mathbb R^3, \times)$ be the classical vector product with
the identities \eqref{vector3}. With the techniques described in
the next section, it can be readily shown that for the corresponding
set $\Gamma$ of relations,
\[
E_\Gamma \isom R[d]/d(d-1)(d-3).
\]
 Thus a non trivial vector product satisfying the identities
\eqref{vector3} can  only exist in dimension $3$.
\end{example}

\section{Associative bilinear forms}

 It is clear from the last step of the proof of Theorem \ref{boos}
 that the associativity of the bilinear form plays an
essential r\^ole. The associativity has a nice diagrammatic
interpretation. A $120^{\rm o}$ degree clockwise
 rotation around the center
of the diagram $\mu$
describing a multiplication does not change the m-tangle.
 The rotation can also be
 viewed as  deformation:\\[-20pt]
%
\def\IdXsl{
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(-15,15){ \makebox(0,0){$1_X$ =\ }}
\linethickness{0.3mm} \qbezier(25,5)(25,15)(25,25)
\end{picture}}
\end{picture}}
%
\def\bettasl{
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
%\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
\linethickness{0.3mm}  \qbezier(15,25)(15,15)(25,15)
\linethickness{0.3mm}  \qbezier(25,15)(35,15)(35,25)
\put(15,25){\circle*{4}}
\put(35,25){\circle*{4}}
\end{picture}}
\end{picture}}
%
\def\bettatsl{
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
%\put(5,25){\line(1,0){40} \quad }% \makebox(3,0){$Y$}}
\linethickness{0.3mm}  \qbezier(15,5)(15,15)(25,15)
\linethickness{0.3mm}  \qbezier(25,15)(35,15)(35,5)
\put(15,5){\circle*{4}}
\put(35,5){\circle*{4}}
\end{picture}}
\end{picture}}
%
%
\def\multsl{
\begin{picture}(45,30)
\put(-5,-12){
\begin{picture}(45,30)
%\put(-15,15){\makebox(0,0){$\mu= \ $}}
\linethickness{0.3mm}  \qbezier(15,25)(15,15)(25,15)
\linethickness{0.3mm}  \qbezier(25,15)(35,15)(35,25)
\linethickness{0.3mm}  \qbezier(25,4.5)(25,9.5)(25,14.5)
%\put(15,25){\circle*{4}}
\put(35,25){\circle*{4}}
\put(25,15){\circle*{4}}
%\put(25,5){\circle*{4}}
\end{picture}}
\end{picture}}
%
%
\begin{equation}\label{rotation}
\begin{picture}(45,30)
\put(-25,-10){
\begin{picture}(45,30)
%\put(-15,15){\makebox(0,0){$\mu= \ $}}
%\put(5,5){\line(1,0){40}\quad  }%\makebox(3,0){$Y$}}
%\put(5,25){\line(1,0){40}\quad  }%\makebox(3,0){$X$}}
\linethickness{0.3mm}  \qbezier(15,25)(15,15)(25,15)
\linethickness{0.3mm}  \qbezier(25,15)(35,15)(35,25)
\linethickness{0.3mm}  \qbezier(25,4.5)(25,9.5)(25,14.5)
%\put(15,25){\circle*{4}}
%\put(35,25){\circle*{4}}
\put(25,15){\circle*{4}}
%\put(25,5){\circle*{4}}
\put(60,12){$\mapsto$}
\end{picture}}
\end{picture}
\begin{picture}(60,30)
\put(0,-20){
\begin{picture}(130,30)
\put(0,20){\IdXsl}
\put(10,0){\bettasl}
\put(20,20){\multsl}
\put(40,40){\bettatsl}
\put(50,20){\IdXsl}
%\put(25.25,5){\circle*{4}}

%\put(77.5,13){\circle*{4}}
%\put(88,-7){\line(1,0){60} }
%\put(183,-7){\line(1,0){60} }
\end{picture}}
\end{picture}
\end{equation}\\
and the two graphs in \eqref{rotation} are isomorphic. The image under
the functor $\FF$ of the graph on the right corresponds to the linear
map
\[ u \tens v \mapsto \sum_i b(u,v e_i)e_i, \quad u,\, v \in V
\]
which is the multiplication map $u \tens v \mapsto uv$
if the bilinear form $b$ is associative.
Thus the associativity of the bilinear form corresponds to the invariance
of a trivalent vertex under a cyclic permutation of the edges.

If the bilinear form is not associative with respect to the
 multiplication,
we need to consider  $6$ types of trivalent vertices. Taking the two
orientations in account, this can be done by distinguishing one edge
(which gives the output of the multiplication under the realization
functor $\FF$):\\[-20pt]
\[
\multnpp\quad .
\]\\
The $6$  different trivalent vertices are:\\[0pt]
\begin{equation}\label{sixpictures}
\begin{picture}(0,0)
\put(-140,-10){
\begin{picture}(45,30)
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 \end{picture}}
\end{picture}
\end{equation}\\
 and we get a much more complicated alphabet of basic
 m-tangles.
 This is the case  for example for quadratic
compositions with unity (\cite{cadorin04}).
Observe that the second m-tangle in \eqref{sixpictures}
is obtained from the first one through a deformation of type \eqref{rotation}.
A similar situation occurs for $3$-vector products,
see Section~\ref{3vectors}.


%%% Local Variables: 
%%% mode: latex
%%% TeX-master: t
%%% End: 

%%%\input proc2.tex
  %%%%%%%%%%%%%%%%%%
\section{Symmetric compositions}\label{symmetriccompositions}

%\input text.sty
A {\it symmetric composition algebra}
 is an algebra $(V,m)$ with a nonsingular associative
symmetric bilinear form $b$ such that
%
\begin{equation}\label{symcomp1}
 q(xy)=q(x)q(y), \quad q(x)= b(x,x) \quad
 \text{for all $x,\,y\in V$.}
\end{equation}
Symmetric composition algebras were considered independently
by \mbox{Petersson} \cite{P69}, Okubo \cite{O78}
and Faulkner \cite{F89}.

The linearization of \eqref{symcomp1} is
the identity
%
\begin{equation}\label{idsym}
b(xu,yv) +b(xv,yu) = 2b(x,y)b(u,v)
\end{equation}
%
for $x,\,y,\,u,\,v \in V$. Thus symmetric compositions are
algebras of tensorial type and there is a corresponding m-tangle
category $\T_{\symcomp}$ (Recall that the bilinear form is assumed
to be associative). As in Section~\ref{tangles}, trivalent
vertices are cyclically oriented.
%
The relation in the m-tangle category $\T_{\symcomp}$
corresponding to~\eqref{idsym}
is
\begin{equation}\label{symtangle}
\multabc \quad \  \ +  \multdef \qquad  = \quad 2\quad \multgh
\end{equation}\\[0pt]
%
Applying the isomorphism \eqref{XYempty} with $X=\{u,y\}$ and
 $Y=\{x,v\}$
to \eqref{symtangle} we get the relation
%
\begin{equation}\label{symrel}
\begin{array}{cccccccc}%\label{symrel}
{\tiny u \hspace{30pt} y}& & {\tiny u \hspace{30pt} y} &&
 \hspace{12pt} {\tiny u \hspace{28pt} y}\\[-15pt]
\pic12& + & \pic18& =& 2\cdot\pic4.\\[-15pt]
{\tiny x \hspace{30pt} v}& & {\tiny x \hspace{30pt} v} &&
 \hspace{12pt} {\tiny x \hspace{28pt} v}
\end{array}
\end{equation} %
%
in $\T_{\symcomp}$. 
%
\begin{remark}
It is curious to compare the relation \eqref{symrel} with 
the so-called $IHX$ identity in the theory of knot invariants
(see \cite{barnatan95}, p.~428). The $IHX$ identity reflects
the Jacobi identity of Lie algebras.
\end{remark}
%
Expressed in terms of the basic alphabet,
the relation \eqref{symrel} is
\[
\big(1 \square\beta \square1\big)\circ
 \big(( \tau \circ \mu^t) \square(\tau \circ \mu^t )\big) +
\mu^t \circ \mu \circ \tau  = 2\tau\ .
\]

Let $E_{\symcomp} =\Mor_{\T_{\symcomp}}(\emptyset,\emptyset)$
be the ring of numerical invariants of the category $\T_{\symcomp}$.
As in Section \ref{representations} we denote\\[-20pt]
$$d=\pic26,\qquad e=\pic84$$\\[-20pt]
in $E_{\symcomp}$.
%
\begin{thm}\label{thmsym}
Let $R=\mathbb Z[\frac{1}{2}]$ and
let J be the ideal of the polynomial ring $R[\bar{d},\bar{e}]$,
generated by the elements\\[-15pt]
%
\begin{align}
p_1 & = \bar{e}\bigl(\bar{e}-(2-\bar{d})^2\bigr)\label{p1} \\
p_2 & = (\bar{d}-2)(\bar{d}-8)(\bar{d}-\bar{e}). \label{p2}
\end{align}\\[-15pt]
%
The homomorphism $\Phi\col R[\bar{d},\bar{e}] \to
E_{\symcomp} =\Mor_{\T_{\symcomp}}(\emptyset,\emptyset)$
given by
$
\phi(\bar{d}) = d$, $\phi(\bar{e})=e $,
 induces  a surjective homomorphism of rings
\[
\overline{\Phi}\col R[\bar{d},\bar{e}]/J \to E_{\symcomp}.
\]
In particular the only possible values of the pair $(d,e)$ of
invariants are
\begin{equation}\label{solsym}
(d,e)= (0,0),\,(1,1),\,(2,0),\,(4,4),\,(8,0)\, \ \text{and}\ \
 (8,36).
\end{equation}
\end{thm}
%
\begin{remark}
We do not know whether the homomorphism $\overline{\Phi}$ in
Theorem~\ref{thmsym} is injective.  There is the isomorphism
$$
R[\bar{d},\bar{e}]/J \otimes {\mathbb Q} \simeq {\mathbb
Q}\times {\mathbb Q}\times {\mathbb Q}[\epsilon]/(\epsilon^2)\times
{\mathbb Q}\times {\mathbb Q}\times {\mathbb Q}
$$
given by
$$\bar{d}\mapsto (0,1,2,4,8,8), \qquad \bar{e}\mapsto
(0,1,\epsilon,4,0,36).
$$
On the other hand, since the $6$ pairs in \eqref{solsym} can be realized
by a symmetric composition algebra over~${\mathbb Q}$
(see Proposition \ref{6real}), there is a homomorphism
$$
E_{\symcomp} \to {\mathbb Q}^6
$$
with
$$
d\mapsto (0,1,2,4,8,8), \qquad e\mapsto
(0,1,0,4,0,36).
$$
It follows that the kernel of~$\overline {\Phi}_{\mathbb Q}$ is
contained in the $1$-dimensional radical of~$R[\bar{d},\bar{e}]/J
\otimes {\mathbb Q}$.
\end{remark}
%
\begin{proof}
We keep the  convention that if, in a plane graph, no
orientation is indicated we understand that the graph does not depend on
the choice of the orientation. For oriented trivalent
vertices we have the identities

 %\label{noorient}
$$\pic37 =\pic40  \label{(0.1)} \\
\pic38 =\pic39 \label{(0.2)}$$  induced by rotation along the
vertical symmetry axis.

To start with a proof, first consider the following consequence of
\eqref{symrel}:
\begin{equation}\label{(2.1)}
\pic50+\pic40=2\cdot\pic47=\ 2\cdot\pic41.
\end{equation}

This and \eqref{(0.1)} show that $\pic32$
is invariant under a reflection.
Together with \eqref{casvet} this means that in
such a graph the orientation at the
two vertices is inessential. So we may forget about them and we have

\begin{equation}\label{(2.2)}
\pic31=\pic32=\pic33=\pic34=\pic35.
\end{equation}

%\def\overset{#1 {\mathop{=\joinrel=\joinrel=}^{\hbox{\smalltextfont(#1)}}}

We have by \eqref{(0.1)} and \eqref{(2.1)}
\begin{equation}\label{(2.3)}
\pic37\overset{\eqref{(0.1)}}{=} \pic40\overset{\eqref{(2.1)}}{=}
 2\cdot\pic41-\pic52.
\end{equation}
%\bigskip


Identities \eqref{(0.2)} and \eqref{symrel}  show
$$
2\cdot \pic39 \overset{\eqref{(0.2)}}{=} \pic39 +\pic38 =
\pic54 +\pic55 \overset{\eqref{symrel}}{=}
2\cdot\pic42=2d\cdot\pic41.
$$
Hence we have
\begin{equation}\label{(2.4)}
\pic39=\pic38=\ d\cdot\pic41
\end{equation}
%\bigskip
%\goodbreak
which implies
\begin{equation}\label{(2.5)}
\pic87=d\cdot\pic88.
\end{equation}
By \eqref{(2.2)}  and \eqref{(2.4)}
 this equation is valid for any choice of orientation.
Relation \eqref{(2.1)} implies similarly
$$
\pic90+\pic92=\ 2\cdot\pic91.
$$
By \eqref{(2.5)} this gives
\begin{equation}\label{(2.5a)}
\pic96=(2-d)\cdot\pic86.
\end{equation}
Composing both sides with themselves yields
\[
\pic93=(2-d)^2\cdot\pic89=(2-d)^2e.
\]
In the next computation the second equation
 follows from \eqref{symrel}
$$ {\scriptsize
2\cdot\pic93=\pic94+ \pic95\overset{\eqref{symrel}}{=} 2\cdot\pic99.}
$$
We conclude  that $(2-d)^2e=e^2$  and get the first
relation $e(e-(2-d)^2)=0$.

%\vfill\eject
Going on,
%\bigskip
an application of \eqref{symrel} and of \eqref{(2.4)} (three times) gives
%$$
%\eqalign{\pic81&\overset{ R3 -\pic83+2\cdot\pic82\cr
%&\phantom{=\joinrel}=\phantom{\joinrel=}-d^2\cdot\pic66+2d\cdot\pic66.\cr}
%$$
$$
\pic81\overset{\eqref{symrel}}{=}
 -\pic83+2\cdot\pic82\overset{\eqref{(2.4)}}{=} -d^2\cdot\pic66+2d\cdot\pic66.
$$
Hence
\begin{equation}\label{(2.6)}
\pic81=(2-d)d^2.
\end{equation}
%\bigskip
A further application of \eqref{symrel} gives
$$
\pic=78+\pic76=2\cdot\pic77.
$$
Hence
$$\def\PiC{\vcenter{\lineskip7ptplus3pt
\null\vbox to-4pt{\hbox{\a\char\charno}\vss}\vphantom{\a\char106}\null}}
\pic106=2\cdot\pic107-\pic105.
$$
Composing this relation with itself shows
$$
\vcenter{\offinterlineskip
\null\vskip10ptplus3pt\hbox{\a\char103}\hbox{\a\char106}\vskip10ptplus3pt\null}
=4\cdot\vcenter{\offinterlineskip\hbox{\a\char104}\hbox{\a\char107}}
-4\cdot\vcenter{\offinterlineskip\hbox{\a\char104}\hbox{\a\char105}}
+\vcenter{\offinterlineskip\hbox{\a\char102}\hbox{\a\char105}}.
$$
%\bigskip
We compute the three terms on the right.
%\bigskip
\eqref{(2.3)} gives
%
\begin{equation}\label{(2.7)}
\pic27=\pic43=2\cdot\pic26-\pic84=2d-e.
\end{equation}
Then \eqref{(2.4)} and the last computation gives
$$
\vcenter{\offinterlineskip
\null\vskip7ptplus3pt\hbox{\a\char104}\hbox{\a\char105}
\vskip7ptplus3pt\null}
=d\cdot\pic70=d(2d-e).
$$
%
Applying \eqref{(2.3)} twice for the first equation, then
\eqref{(2.7)},\eqref{(2.3)},\eqref{symrel} for the
second equation and finally \eqref{(2.5)} for the third
equation gives
\[
\begin{array}{lll}
\vcenter{\null\vskip7ptplus3pt
\offinterlineskip\hbox{\a\char102}\hbox{\a\char105}\vskip7ptplus3pt\null}
&=\ 4\cdot\pic70-4\cdot\pic70\pic101+\pic100\pic70\pic101\\[-40pt]
&=\ 4(2d-e)-4\cdot\Big(2\pic100\pic97-\pic100\pic26\pic101\Big)\\[-20pt]
&\ \quad+\bigg(2\cdot\pic100\pic26\pic101-\pic100\pic98\pic101\bigg)\\[-10pt]
&=\ 4(2d-e)-4(2e-de)+(2de-d^2e).\\
&=\ 4(2d-e)+(2-d)(-4e+de).
\end{array}
\]
%
%Applying (2.3) twice for the first equation, then (2.7),(2.3),(R3) for the
%second equation and finally (2.5) for the third equation gives
%$$
%\eqalign{\vcenter{\null\vskip7ptplus3pt
%\offinterlineskip\hbox{\a\char102}\hbox{\a\char105}\vskip7ptplus3pt\null}
%&=4\cdot\pic70-4\cdot\pic70\pic101+\pic100\pic70\pic101\cr
%&=4(2d-e)-4\cdot\Big(2\pic100\pic97-\pic100\pic26\pic101\Big)\cr
%&\quad+\bigg(2\cdot\pic100\pic26\pic101-\pic100\pic98\pic101\bigg)\cr
%&=4(2d-e)-4(2e-de)+(2de-d^2e).\cr
%&=4(2d-e)+(2-d)(-4e+de).\cr
%
%$$
%
%
So far we have computed
\[
\begin{array}{ccc}
\vcenter{\offinterlineskip
\null\vskip10ptplus3pt\hbox{\a\char111}\hbox{\a\char112}\vskip10ptplus3pt\null}
\hspace{-20pt} &\makebox{$ \begin{array}{ccc}
&=4(2d-e)-4d(2d-e)+4(2d-e)+(2-d)(-4e+de)\\[3pt]
&=(2-d)(8d-8e+de).
\end{array} $}
\end{array}
\]
%
The right hand side is invariant under orientation reversing. Hence
\begin{equation}\label{(2.8)}
\vcenter{\offinterlineskip
\null\vskip10ptplus3pt\hbox{\a\char111}\hbox{\a\char112}\vskip10ptplus3pt\null}
=
\pic108\pic109
=
(2-d)(8d-8e+de).
\end{equation}
%
Next note
\begin{equation}\label{(2.9)}
\vcenter{\offinterlineskip
\null\vskip10ptplus3pt\hbox{\a\char111}\hbox{\a\char112}\vskip10ptplus3pt\null}
+\pic108\pic109\overset{\eqref{symrel}}{=}2\cdot\pic110=2\cdot\pic81.
\end{equation}
%
For the second equality note that the two graphs are the same,
they are just drawn
in a different way. Relations \eqref{(2.6)}, \eqref{(2.8)}
and \eqref{(2.9)} give after dividing by $2$
$$
\vcenter{\offinterlineskip
\null\vskip10ptplus3pt\hbox{\a\char111}\hbox{\a\char112}\vskip10ptplus3pt\null}
=\pic81=(2-d)d^2.
$$
Putting things together:
$$
0=(2-d)(8d-8e+de)-(2-d)d^2=(2-d)(8-d)(d-e).
$$
This is the second desired equality.
%\end{proof}


%\ifx\wholejob\!\endinput\fi























%%%%%%%%%%%%%%%%%%%%%%%%%%












%%% Local Variables:
%%% mode: latex
%%% TeX-master: t
%%% End:

%%%\input proc3.tex
%\input text.sty
To complete the proof of Theorem \ref{thmsym}
we next check that the two graphs $d$ and $e$ generate the ring
 of numerical invariants $E_{\symcomp} =
\Mor_{\T_{\symcomp}}(\emptyset,\emptyset)$.
We call the elements of $E_{\symcomp}$ {\it closed graphs}.
Since they are defined up to isomorphism, we  assume
that closed graphs have only  trivalent vertices. Observe that
a closed graph is a linear combination of ``real'' closed graphs.

\begin{definition}
Two graphs are called {\it equivalent}, if one can 
 be obtained from the 
other by a sequence of moves\\[-30pt]
\end{definition}
%
$$
\pic12\longleftrightarrow\pic18.
$$\\[-20pt]
%\defmopa{l}{length}
%
\begin{prop}\label{(3.2)}
For any graph\/ with at least one (trivalent) vertex
there is an equivalent one which contains one of the portions
$$
\pic39\qquad\hbox{or}\qquad \pic31.
$$
\end{prop}
%
\begin{proof}
By a {\it cycle} $S$ in a graph $\Gamma$ we mean a simple closed
path in $\Gamma$, where, in contrast with Section \ref{tangles},
all vertices have valence $3$.
 By $\Ll(S)$ we mean the number of vertices on $S$.

Fix some direction on $S$.
Then at each vertex $P$ we have an incoming edge, an outgoing
edge and the third edge not belonging to $S$. This triple determines
an orientation at the vertex $P$ which may be different from
the given orientation in $P$.
We now divide the  vertices on $S$  into two
classes: the set of vertices
for which this new orientation coincides with the given orientation
of the trivalent vertex and the others.
If two vertices belong to the same class, we say that 
they have the same 
orientation with respect to $S$.
In the equivalence class of a given graph we choose a graph $\Gamma$
which contains a cycle $S$ of minimal length.  
If $\Ll(S)=1$, the proposition is clear.
Suppose $\Ll(S)\geq2$.
For an edge $\sigma$ of $S$ the two endpoints have 
either the same orientation
or not: 
$$
\Sarcr3\qquad
\raise-3pt\hbox{or}
\qquad\Sarcl4
$$
In the second case we say that $\sigma$ is an alternating edge.
In the first case we could make a move
\begin{equation}\label{*}
\Sarcr3 
\raise-3pt\hbox{$\longrightarrow$}
\Sarcl5 
\end{equation}
and find a cycle of smaller length in a graph equivalent to $\Gamma$.
Since $\Ll(S)$ is minimal, any edge of $S$ is alternating.
Suppose that the endpoints $P$, $Q$ of an edge $\sigma$  of 
$\Gamma\setminus S$ 
lie both on $S$. Then the union of $\sigma$ with one of 
the arcs on $S$
between $P$ and $Q$ would have smaller length that $S$, except when 
each of these arcs have length $1$. But then $\Ll(S)=2$. 
The last two conclusions show that in case $\Ll(S)\geq3$
there is a portion as indicated by the first
of the following pictures:
$$
\Sarcr6 
\raise-8pt\hbox{$\longrightarrow$}\Sarc7
\raise-8pt\hbox{$\longrightarrow$}
\Sarcl8
$$
Now the two indicated moves would again give rise to a cycle of 
smaller length.
So we know $\Ll(S)\leq2$ and in case $\Ll(S)=2$ the edges of $S$ 
are alternating.
\end{proof} 

Now we consider graphs modulo the relation
\begin{equation}\label{(R0)}
\pic12+\pic18 =0.\end{equation}
%
\begin{prop}\label{(3.3)}
The ring of closed graphs modulo the relation \eqref{(R0)} 
is generated by 
the circle $d$ and  the closed graph $e$.
\end{prop}
\begin{proof}
Relation \eqref{(R0)} has the following consequence
 (see the proof of
 \eqref{(2.4)}):
\begin{equation}\label{(3.4)}
\pic39=0.
\end{equation}
Let $\Gamma$ be a closed graph. We have to show
that $\Gamma$ is a polynomial in $d$ and $e$. We may 
assume that $\Gamma$ is connected and argue by induction on
the number of vertices in $\Gamma$.

By Proposition \ref{(3.2)} we have $\Gamma=\pm\Gamma'$,
 where $\Gamma'$ contains one of the portions of
 Proposition~\ref{(3.2)}.
In the first case we may use \eqref{(3.4)} and 
reduce the number of vertices.
Let us consider the second case, i.e., $\Gamma$ contains a portion
$$
\eta=\pic31.
$$ 
We may assume that $\Gamma$ has more than two vertices, because
otherwise $\Gamma=e$.
 Relation \eqref{(R0)} gives
\begin{equation}\label{(3.5)}
\pic133=\pic129=-\pic130=-\mskip5mu\pic131.
\end{equation}
%
Iterating this relation three times yields
\begin{equation}\label{(3.6)}
\pic133=-\pic132.
\end{equation}
%\medskip
It follows that if we change the orientation of a vertex, we merely
change the sign of $\Gamma$. This is clear if the vertex is near $\eta$
as in \eqref{(3.6)}. But in general we may use \eqref{(3.5)} to move $\eta$ to the vertex,
then change the orientation and move back again using
\eqref{(3.5)}.
%\medskip
But if we don't have to care about orientations,
 then it is easy to 
use \eqref{(R0)} to produce a further cycle of length $1$. 
So $\Gamma=\pm\Gamma'$, where $\Gamma'$ contains a portion 
$$
\pic96.
$$
But this equals $(2-d)\cdot\pic86$, cf. \eqref{(2.5a)}.
\end{proof}
%
\begin{cor}\label{generatorssym}
The ring $E_{\symcomp}$ of closed graphs is generated by
$d$ and~$e$.
\end{cor}
\begin{proof}
The claim follows from a filtration argument, where filtration 
is by the number of vertices. More precisely, consider
for example the case of a cycle  $S$ in the closed graph $\Gamma$
containing two black vertices:
\begin{equation}\label{blackblack}
\Sarcr3 \quad .
\end{equation}
Instead of relation
\eqref{(R0)} which leads to steps
\begin{equation}\label{**}
\Sarcr3  
\raise-3pt\hbox{$\longrightarrow$ \quad --} 
 \Sarcl5 
\end{equation}
to reduce the length of $S$, 
we use the relation
\begin{equation}\label{generators}
\pic12 +  \pic18 = 2\cdot\pic4.
\end{equation}
We claim that for $S$ with minimal length, $\Ll(S) \leq 2$.
Assume that $\Ll(S) \geq 3$. Let $T$ be any cycle in $\Gamma$,
different from $S$, but connected with $S$ through
the two marked black vertices in \eqref{blackblack}.
 We have $\Ll(T) \geq \Ll(S)$. 
For each move induced by
\eqref{generators}, $T$ is replaced by a linear combination of
a cycle of length\/
$\Ll(T)-1$ (corresponding to the term $\pic18$) and  
 a cycle of length \/$\Ll(T) +\Ll(S)-4$
(corresponding to the term $\pic4$). If
$\Ll(T)=\Ll(S)$ we get in any case shorter cycles. If
$\Ll(T)>\Ll(S)$, $S$ is replaced by a linear combination of 
cycles which contains a shorter cycle.
\end{proof}
%
Theorem~\ref{thmsym} follows now from 
Corollary~\ref{generatorssym}.
\end{proof}
%
All values in Theorem~\ref{thmsym} can be realized:
%
\begin{prop}\label{6real}
 Let $F$ be an  algebraically closed field of
characteristic different from $2$ and $3$. For each 
value
$ (0,0)$, $(1,1)$, $(2,0)$, $(4,4)$, $(8,0)$ and $(8,36)$
of the pair $(d,e)$, 
there exists up to isomorphisms exactly one symmetric
composition algebra with the given value.
\end{prop}
%
\begin{proof}
 Symmetric
composition algebras are of two types
 (see for example \cite{OO81}, \cite{EM} or \cite{KMRT}, Chapter  VIII).
 The first one is related
to quadratic composition algebras with identity 
(''Hurwitz algebras''), which,
by Hurwitz' theorem, occur in dimension $1,\,2,\, 4$ and $8$.
Over an algebraically closed field, Hurwitz algebras are split
and there is only one type up to isomorphism for each  possible
dimension.
The bilinear form associated with the norm is not associative,
but satisfies identities of the form
\begin{equation}\label{bar}
b(xy,z) = b(x,z\bar y),
\end{equation}
where $x \mapsto \bar x$ it the conjugation map. It readily
follows from \eqref{bar} that, for the new multiplication
$x \star y = \bar x \cdot\bar y$,  the bilinear form is associative.
Thus we can associate to every Hurwitz algebra, a symmetric
composition algebra, which is usually called a 
{\it para-Hurwitz algebra}. Since $F$ is an algebraically 
closed field, the standard basis
$(1,i,j,\ldots )$  of quadratic extensions, resp. quaternions or
octonions, can be normalized so
 that  $i^2=j^2 = \ldots = -1$ and is orthonormal. We get for 
the Casimir element
\[
c = 1 +(d-1)(-1)=  (2-d)\cdot 1_F.
\]
Thus $e=b(c,c)=(d-2)^2$ and
the numerical invariants for para-Hurwitz algebras are
$(d,e)= (1,1),\, (2,0),\, (4,4)$ and $(8,36)$.

 The other type of symmetric compositions is related to 
separable associative cubic algebras and were first considered by 
Okubo (see \cite{O78} or \cite{KMRT}). Over an algebraically closed field
 we have three classes of separable associative cubic  algebras; 
$A=F$, $F^3$ and $M_3(F)$. We denote the multiplication of 
$A$ by $(x,y) \mapsto xy$. Let 
\[
p(X)= X^3 -T(x)X^2 + S(x)X -N(x)\cdot 1
\]
be the generic polynomial. The form $T$ is linear, $S$ is quadratic
and $N$ cubic. 
 Let 
$A^0= \{x \in A\ | \ T(x)=0\}$
  be the vector space of trace  zero elements in $A$. Let 
$q  = -\tfrac{1}{3}S$ and let $b$ be the associated symmetric
 bilinear form. Let $\omega$ be a primitive cubic root of $1$.
The bilinear form $b$ is associative with respect to the
multiplication defined  on $A^0$ by
$(x,y) \mapsto x\star y$
with
\[
x\star y = \tfrac{1-\omega}{3}xy + \tfrac{2+\omega}{3} yx
 -\tfrac{1}{3}T(xy)
\]
and $q(x \star y)=q(x)q(y)$ for all $x \in A^0$ (see for example
 \cite{KMRT}). 
Thus $(A^0, q,\star)$ is a symmetric composition algebra.
One checks that in all three cases  the Casimir element
is trivial, so that symmetric 
composition algebras
associated with cubic algebras have invariants 
$(d,e)= (0,0),\, (2,0)$ and $(8,0)$. The two algebras in
dimension $2$ (and  for which $e=0$) are isomorphic. 
In contrast, we have
two types of  symmetric compositions in dimension $8$ which are
not isomorphic,  since they have different $e$-invariant. 
\end{proof} 
%
 \begin{remark}
In dimension $8$, one class of symmetric composition algebras
(``type'' $G_2$) is associated
with octonions, and  the second (``type'' $A_2$)  is associated
with $3\times 3$  matrices. The existence of these two types
is related to the existence of two non conjugate trialitarian
actions of $S_3$ on ${\rm Spin}_8$ (see \cite{KMRT} or 
\cite{helgason78}, 
Chap. X, Exercise E.2).
 Observe that  relation \eqref{p1}
has two branches, one ($e=0$) corresponding to  type $A_2$ 
and the other one ($e=(d-2)^2$) to  type $G_2$. The fact that
the type of a symmetric algebra of dimension $8$
can be easily detected through the Casimir element is new.
\end{remark}

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\section{$3$-Vector products}\label{3vectors}

 Let $V$ be a $d$-dimensional vector space over a field $F$ of 
characteristic
zero.
 Let $b$  be a nondegenerate, symmetric bilinear
 form on $V$.

 An {\it $r$-vector product} in $V$ is an $r$-linear map
$$P_r: V^r=V\times V \times \cdots \times V \longrightarrow V, \qquad 1\le r\le n,$$
satisfying the following properties:
\begin{align}\label{rvector1}
 & b\big(P_r(v_1,\dots,v_r),v_i\big)=0 \quad \text{for} \quad
  v_i\in V,\, i=1,\,\ldots, r  \\
 & b\big(P_r(v_1,\dots,v_r),P_r(v_1,\dots,v_r)\big)=
\det\big(b(v_i,v_j)\big)_{i,j} \quad  \label{rvector2}
\end{align}
%
It follows from \eqref{rvector1} that the multilinear form
\begin{equation}\label{multilinf}
f(v_1,\dots,v_r,\, v_{r+1})= b\big(P_r(v_1,\dots,v_r),v_{r+1}\big)
\end{equation}
%
and that the $r$-vector product $P_r$ are alternating.
Let $r=3$ and denote $P_3(v_1,v_2,v_3)=v_1v_2v_3$.
 Relation \eqref{rvector2}  becomes
\begin{equation}\label{0}
b(xyz,xyz)=\det \left| \left(
\begin{array}{ccc}
b(x,x) & b(x,y) & b(x,z) \\
b(y,x) & b(y,y) & b(y,z) \\
b(z,x) & b(z,y) & b(z,z)
\end{array} \right) \right|.
\end{equation}
%
We have
%
\begin{equation}\label{1}
v_{\pi(1)}v_{\pi(2)}v_{\pi(3)}=
{\textrm sgn}( \pi)  v_1v_2v_3, \ v_i \in V
\end{equation}
and
\begin{equation}\label{2}
b(xyz,t)=-b(x,yzt) \quad \text{for $x,\,y,\,z$ and $t \in V$.}
\end{equation}
%
Linearizing the above identity \eqref{0} we get:
{\setlength\arraycolsep{2pt}
\begin{eqnarray}%\label{3}
\lefteqn{% \label{3}
\quad \quad b(uxt,vys)+ b(uxs,vyt)+b(uys,vxt)+b(uyt,vxs) = }\label{3}  \\
& =& 4b(x,y)b(u,v)b(s,t)-2b(x,y)b(u,s)b(v,t)-2b(x,y)b(u,t)b(v,s)   \nonumber\\
&&   - 2b(x,u)b(y,v)b(s,t)-2b(x,v)b(y,u)b(s,t)-2b(x,s)b(y,t)b(u,v) \nonumber \\
&&   - 2b(x,t)b(y,s)b(u,v)+b(x,u)b(y,t)b(v,s)+b(x,v)b(u,t)b(y,s)  \nonumber\\
&&   + b(x,u)b(y,s)b(v,t)+b(x,v)b(u,s)b(y,t)+b(x,s)b(y,u)b(t,v)  \nonumber\\
&&   + b(x,s)b(y,v)b(t,u)+b(x,t)b(y,u)b(s,v)+b(x,t)b(y,v)b(s,u)\nonumber
\end{eqnarray}}
\noindent
\hspace*{-1ex}for all  $u,\,v,\,x,\,y,\,s$ and $t \in V$.
In the terminology of Section~\ref{tangles},
the data $(V,P_3,b)$ is an algebra of tensor type, where the product
$m$ is replaced by the $3$-vector product $P_3$. The tensor 
 relations  are \eqref{1}, \eqref{2} and \eqref{3},
and we shall associate to  $(V,P_3,b)$ a series of
 m-tangle  categories  $\mathcal T_i$,
 $i=1,2,3$,  as in Section~\ref{tangles}.
In particular we have again basic m-tangles $I_{\{x\}}$,
 $\tau$, $\beta$
and $\beta^t$. The main difference
is that $3$-valented  vertices corresponding to products
are replaced by $4$-valented vertices corresponding to
$3$-vector products. 
%
Since
there is no associativity for the bilinear form
(a cyclic permutation of the variables in the
multilinear form \eqref{multilinf}
changes the sign), we 
distinguish one edge out of  every $4$-valented vertex,
 which corresponds to the product $xyz$:\\[-25pt]
%
\begin{equation}\label{3prod}
 \mu = \quad \threevecn\ .
\end{equation}\\
In a $4$-valented vertex, the (clockwise) orientation of
the three black edges will always  be assumed to be positive,
so that we do not  mark  $4$-valented vertices.
 Composition of m-tangles is as in Section~\ref{tangles},
with the rule that
dotted edges can only be composed with dotted ones
(because of \eqref{7.5} this can always be arranged).

We next introduce the relations corresponding to a
$3$-vector product in the m-tangle category $\T_2$:
the fact that the $3$-product is alternating leads to the relation
\\[-20pt]
\begin{equation}\label{7}
\threevecpi =\ \sgn (\pi) \quad \threevecnum
\end{equation}\\
 %
for any permutation $\pi \in S_3$ of three objects.
 Relation \eqref{2}  leads to\\
%
\begin{equation}\label{7.5}
\threeveconuma  = -\quad \threeveconumb = \quad\threeveconumd =  -
\quad \threeveconumc \ ,
\end{equation}\\
%
where the second m-tangle is obtained from the first through
a deformation similar to the one defined in \eqref{rotation}.
Relation \eqref{7}  implies that
\begin{equation}\label{zero}
\ii{3}=0 \quad \text{and} \quad \ii{3a}\ =\ \ii{3b}=0.
\end{equation}

The morphisms of $\T_1$ consists of m-tangles generated by
the alphabet
$I_{\{x\}}$, $\tau$, $\beta$, $\beta^t$ and $\mu$, those  of $\T_2$
are linear combinations with coefficients in $R=\mathbb Q$.
(We choose here the ring of coefficients to be $\mathbb Q$
to avoid computation difficulties.)
The linearized formula \eqref{3} has the following diagrammatic
translation:

{\setlength\arraycolsep{2pt}
\begin{eqnarray}\label{8}
 \lefteqn{\hspace{4ex}
\iiii{g1}\ + \ \iiii{g2}\ +\ \iiii{g3}\ +\ \iiii{g4}=
 \hspace{2cm} \ }   \\[7pt]
&=& 4\ \iiii{gg1}-\ 2\ \iiii{gg2}-\ 2\ \iiii{gg3}-\ 2\ \iiii{gg4}\ -\ 
2\ \iiii{gg5} \nonumber \\[7pt] 
& &-\ 2\ \iiii{gg6}\ -\ 2\ \iiii{gg7}\ +\ \iiii{gg8} \ +\ \iiii{gg9}\ 
+\ \iiii{gg10}\nonumber \\[7pt] 
& &+\ \iiii{gg11}\ + \ \iiii{gg12}\ +\ \iiii{gg13}\ +\ \iiii{gg14}\ 
+\ \iiii{gg15}.\nonumber
\end{eqnarray}}

%
% \begin{multline}\label{8}
%
%{\scriptsize
%\iiii{g1}\ + \ \iiii{g2}\ +\ \iiii{g3}\ +\ \iiii{g4}= \\[7pt]
%4\ \iiii{gg1}-\ 2\ \iiii{gg2}-\ 2\ \iiii{gg3}-\ 2\ \iiii{gg4}\\[7pt]
%\ -\ 2\ \iiii{gg5}\ -\ 2\ \iiii{gg6}\ -\ 2\ \iiii{gg7}\ +\ \iiii{gg8}\\[7pt]
%\ +\ \iiii{gg9}\ +\ \iiii{gg10}\ +\ \iiii{gg11}\ + \ \iiii{gg12}\\[7pt]
%\ +\ \iiii{gg13}\ +\ \iiii{gg14}\ +\ \iiii{gg15}.
%\end{multline}

The step $\T_2 \to \T_3$ is as  in Section~\ref{tangles}:
we define the
category $\T_{\3vec}$ of type $\T_3$ by going modulo the relations
\eqref{7}, \eqref{7.5}  and \eqref{8}.

 Let $\mathcal{V}$ be the category of  finite dimensional
vector spaces over $F$ with morphisms linear maps. As  for
Theorem~\ref{boos},
we have:
%
\begin{thm}
Let $V$ be a finite dimensional vector space over a field of
 characteristic zero.
If there exists a structure  $(V,P_3,b)$ of $3$-vector product
on $V$,
there is a unique
functor $\mathcal R \col \T_{\3vec} \to \mathcal{V}$
sending a point to $V$, tensor products to tensor products,
 $\tau$ to $\tau$, $\beta$ to the symmetric bilinear form $b$ and
 $\mu$ to the
$3$-multiplication $P_3$.
\end{thm}
%
The functor $\mathcal R$ induces a ring homomorphism 
\[
E_{\3vec}=
 \Mor_{A_{\3vec}}(\emptyset,\emptyset) \to~F\]
 and the
main aim is to compute
$E_{\3vec}$ (or at least part of it).
Here again  a simple  element of $E_{\3vec}$ is the circle\\[-23pt]
\[
\beta^t \circ\beta = \dinv \ .
\]
Since $\mathcal R(\beta^t \circ\beta) = \dim_F V$, we denote 
the circle  also by $d$.
%
\begin{thm}\label{larissa}
 Let \[
p(\bar{d})= \bar{d}(\bar{d}-1)(\bar{d}-2)(\bar{d}+2)(\bar{d}-4)(\bar{d}+4)(\bar{d}-8)\in \mathbb Q[\bar{d}]. 
\] 
The map $\mathbb Q[\bar{d}] \to E_{3vec}$ induced by $\bar{d} \mapsto d$
induces a
$\mathbb Q$-algebra homomorphism
\[
\mathbb Q[\bar{d}]/p(\bar{d}) \to E_{3vec}.
\]
%
\end{thm}
%

\begin{proof}
%
%
We verify that $p(d)=0$ in $E_{3vec}$.
 Closing  \eqref{8}  on the left (and using \eqref{zero})
gives
{\setlength\arraycolsep{2pt}
\begin{equation}\label{9}
\ix{-2.7}{.4}{15}+ \ix{-2.7}{.4}{7}\ = (2-d) \cdot \Bigg( 2\ \ix{-2.7}{.4}{8}\ -\
\ix{-2.7}{.4}{9} - \ix{-2.7}{.4}{4}\ \Bigg)
\end{equation}}
%
 and closing on the right:
\begin{equation}\label{ert}
\ix{-2.7}{.4}{6}=-(d-1)(d-2)\ \ \ix{-2.7}{.4}{16}=\ix{-2.7}{.4}{6a} \ .
\end{equation}\\
Equation \eqref{ert} yields immediately
\begin{equation}\label{palla}
\ix{-2.7}{.4}{5}=\ix{-2.7}{.4}{5a}=\ix{-2.7}{.4}{5b}=\ix{-2.7}{.4}{5c}
=-d(d-1)(d-2).
\end{equation}
Moreover one finds
\begin{equation}\label{q}
\hh{q}=d(d-1)^2(d-2)^2.
\end{equation}

We next close \eqref{9} with the graph \iiii{15a} on the right as follows:
\begin{equation}\label{9a}
\ix{-2.7}{.3}{100} + \ \ix{-2.7}{.3}{101}\ = (2-d) \cdot \Bigg( 2\ 
\ix{-2.7}{.3}{102}\ - \ix{-2.7}{.3}{103} - \ix{-2.7}{.3}{104} \Bigg).
\end{equation}
%\begin{equation}\label{9a}
%\ix{-2}{.2}{100} + \ \ix{-2}{.2}{101}\ = (2-d) \cdot \Bigg( 2\ \ix{-2}{.2}{102}\ - \ix{-2}{.2}{103} - \ix{-2}{.2}{104} \Bigg).
%\end{equation}
The fourth graph is zero by \eqref{zero}. The first and the second 
one can be redrawn and so we get the useful relation:
\begin{equation}\label{7a}
\ix{-4.2}{.4}{p}+\ix{-4.2}{.4}{r}=-3(d-2)\ 
\ix{-2.7}{.4}{5}=3d(d-1)(d-2)^2\,.
\end{equation}

We now look for an equation which 
permits to compute one of the two graphs on the left
 side of \eqref{7a}. The strategy is to compose all
 graphs in \eqref{8} on the top left and on the bottom right with  
copies of $\mu$. For example we get the graph\\
\begin{center} \iii{200} \end{center} 
from the first graph of  \eqref{8}.
This yields to
$$\ix{-6.5}{.4}{6double} + 3 \ \ix{-6.5}{.4}{201} = -12 \ \ \iii{6a}\ ,$$
hence (with using  \eqref{ert})
$$ 3 \ \ix{-6.5}{.4}{201} = -(d-1)(d-2)(d+2)(d-5) \ \iii{16a}\ .$$
Connecting the dotted lines we get the first graph of 
\eqref{7a}:
\begin{equation}\label{p}
\ix{-3.5}{.4}{202} = \ix{-4.2}{.4}{p} =
 -\frac{1}{3}d(d-1)(d-2)(d+2)(d-5)\, ,
\end{equation}
so that:
\begin{equation}\label{r}
\ix{-4.2}{.4}{r} = \frac{1}{3}d(d-1)(d-2)(d^2+6d-28).
\end{equation}
The graphs in \eqref{q}, \eqref{p} and
 \eqref{r}  play a central r\^ole in 
the next computations, since all other computed 
graphs will be linear combinations of these three.\\

A further application of \eqref{8} (compose \eqref{8} with $\mu$)  gives
\begin{equation}\label{babbinatali}
-\ii{12}+ \ii{13} + \ii{14} = (d-5)(d+2) \threevecol
\end{equation}
%By isomorphism \eqref{functiso}
Moving down the three points of the source, we obtain
$$ -\ \ii{12a}+ \ii{13a} + \ii{14a} = (d-5)(d+2)\  \ix{-5.5}{.5}{5meta} \ .$$
Then, composing this relation with itself:
\begin{eqnarray*}\label{10}
3\ \ix{-4.5}{.4}{A} + 6\ \ix{-4.5}{.4}{B}& =&  (d-5)^2(d+2)^2 \ \  \ix{-2.7}{.4}{5a}\\[-2ex]
 &=&  -d(d-1)(d-2)(d+2)^2(d-5)^2.
\end{eqnarray*}

The graph relation \eqref{8} can be collated in other ways
 with  graphs of
type \\[-30pt]
\[
\threevecol \quad \text{and} \quad \bettas\ ,
\] \\
to get on the left side of \eqref{8}  linear combinations of the diagrams
\[
\hhh{A}\ ,\  \hhh{B}\ ,\  \hhh{C} \quad \textrm{and}\quad \hhh{D}\ , 
\]
and on the right side linear combinations of the diagrams
\[
\hhh{p}\ ,\  \hhh{q} \quad \textrm{and}\quad \hhh{r}
\]
which are already computed. 
We get a system of $4$ equations
\begin{eqnarray*}
\hhh{A} + 2\ \hhh{B} + \hhh{C} &=& -d(d-1)(d-2)(d+2)(7d-26)\\
\hhh{A} - 2\ \hhh{C} + \hhh{D} &=& -6d(d-1)(d-2)^2(d-4)\\ 
2\ \hhh{B} + \hhh{C} + \hhh{D} &=& d(d-1)(d-2)(d^2+24d-100)\\ 
3\ \hhh{A} + 6\  \hhh{B}         &=& -d(d-1)(d-2)(d+2)^2(d-5)^2 
\end{eqnarray*}
which has the solution              
\begin{eqnarray*}
\hhh{A} &=& \frac{1}{3}d(d-1)(d-2)(d+2)(d^3-8d^2-37d+206)\\ 
\hhh{B} &=& -\frac{1}{3}d(d-1)(d-2)(d-4)(d+4)(d-8)(d+2)\\ 
\hhh{C} &=& \frac{1}{3}d(d-1)(d-2)(d-4)(d+4)(d-8)(d+2)\\ 
\hhh{D} &=& \frac{1}{3}d(d-1)(d-2)(d^4-6d^3-29d^2+168d-44) 
\end{eqnarray*}
Two other ways of collating \eqref{8} yield to the two equations:
$$\hhh{A}-\hhh{C}=\hhh{A} + 2\ \hhh{B} + \hhh{C}\ ,$$ 
$$\hhh{A} + 2\ \hhh{B} - \hhh{C}= \hhh{A}-\hhh{C}\ ,$$
which implies \quad
$\hhh{B}= - \  \hhh{C} = 0$,\, hence
\[
 -\frac{1}{3}d(d-1)(d-2)(d+2)(d-4)(d+4)(d-8)=0
\]
in $E$. 
\end{proof}
%
\begin{cor}
The possible values for the dimension 
 of a vector space admitting a $3$-vector product
are $0,\,1,\,2,\, 4$ and $8$. In dimension $0,\,1$ and $2$ the
product is degenerate.
\end{cor}
%
\begin{proof}
The existence of
a $3$-vector product in dimension $d^* > 0$ implies the existence of an 
evaluation map  $E_{\3vec} \to F$, sending $d$ to $d^*$. 
Such a map  induces in turn a
homomorphism $ \mathbb Q[\bar{d}]/p(\bar{d}) \to F$ 
(where $ Q[\bar{d}]/p(\bar{d}$)
is as in Theorem \ref{larissa}), sending again $\bar{d}$ to $d^*$.
Thus $d^*$ has to be equal to \/$0$, $1$, $2$, $4$ or $8$.
In dimensions \/$0$, $1$ and $2$ the product is trivial.
\end{proof}
%
\begin{remark}
The polynomial $p(\bar{d})$ also admits the negative 
zeroes $-2$ and~$-4$. 
We do not know if  there is another relation in $E_{\3vec}$
which would exclude these values.
\end{remark}









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\section*{Acknowledgements}
The first author thanks Professor B. Eckmann for suggesting
to analyse $3$-vector products with diagrammatic techniques.
Her research is partially supported by the Swiss National Science
Foundation, Grant 21-64712.01.



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\end{thebibliography}

\end{document}
