# The Lie algebraic structure of colored networks

Fahimeh Mokhtari
Jan A. Sanders
Department of Mathematics, Faculty of Sciences
Vrije Universiteit, De Boelelaan 1111, 1081 HV Amsterdam, The Netherlands

###### Abstract

In the computation of the normal form of a colored network vector field,
following the semigroup(oid) approach in [19], one would like to be able to
say something about the structure of the Lie algebra of linear colored network
vector fields. Unlike the purely abstract approach in [10], we describe here a
concrete algorithm that gives us the Levi decomposition. If we apply this
algorithm to a given subalgebra, it does put the elements in the subalgebra in
the block form given by the Levi decomposition, but this need not be the Levi
decomposition of the given subalgebra.

We show that for $N$-dimensional vector fields with ${C}$ colors (different
functions describing different types of cells in the network) this Lie algebra
$\mathfrak{net}_{{C},N}$ is isomorphic to the semidirect sum of a semisimple
part, consisting of two simple components $\mathfrak{sl}_{C}$ and
$\mathfrak{sl}_{{B}}$, with ${B}=N-{C}$, which we write as a block-matrix and
a solvable part, consisting of two elements representing the identity
$\mathsf{{C}}$ in $\mathfrak{{c}}\simeq\mathfrak{gl}_{C}$ and $\mathsf{{B}}$
in $\mathfrak{{b}}\simeq\mathfrak{gl}_{{B}}$, and an abelian algebra
$\mathfrak{{a}}\simeq\mathfrak{Gr}({C},N)$, the Grassmannian, consisting of
the ${C}$-dimensional subspaces of $\mathbb{R}^{N}$. The methods in this paper
can be immediately applied to study the linear maps of colored networks.

##  1 Introduction

######  Remark 1.1.

General background and motivation for the study of linear network vector
fields with ${C}=1$, the homogeneous case, can be found in [9]. Although we
follow the semigroupoid approach (cf. [19]), we want to emphasize that the
linear transformation that we use to obtain the feedforward block form can be
applied without any knowledge of this approach. Readers who are only
interested in the application of the given algorithmic approach can get quite
far by going back and forth between the crucial Algorithm 4.6, Definition 4.9
and the Examples in §8.

######  Remark 1.2.

We remark here that in the literature (cf. [7]) the homogeneous case (${C}=1$)
is seen as opposite to the fully inhomogeneous case (${C}=N$). We show in this
paper that they can also be considered as neighbors, since the ${C}=1$ case
turns out to be isomorphic to the ${C}=N-1$-case, cf. Theorem 5.18.

In [19] an approach has been given to compute normal forms of network vector
fields. This was based on the observation that from the given vector field a
semigroup(oid) could be constructed. By representing the elements of the
semigroup(oid) as matrices in $\mathfrak{gl}_{N}(\mathbb{R})$, where $N$ is
the (finite) number of cells in the network, and each cell has an output value
$x_{i},i=1,\ldots,N$, one finds a Lie algebra $\mathfrak{g}$. The vector field
can then be considered as a vector field on the dual of $\mathfrak{g}$,
$\mathfrak{g}^{\star}$ and the usual operations like Jordan-Chevalley
decomposition (cf. [4, 3]) and the construction of an
$\mathfrak{sl}(2,\mathbb{R})$ containing $\mathsf{N}$ and commuting with
$\mathsf{S}$, can be performed according to the Jacobson-Morozov theorem [2,
13, 5]. The formulation of the Jacobson-Morozov theorem was given by Morozov
[18], the correct proof by Jacobson [12]. Since Jacobson-Morozov depends on
the Lie algebra being reductive, that is to say, it is isomorphic to the
direct sum of a semisimple Lie algebra and its center, we may have to extend
the Fundamental Network to the Total Network, where every possible
configuration of arrows is considered. In such a construction, the numbers of
elements that we need to consider can grow quickly, but if we construct a
basis in $\mathfrak{gl}_{N}(\mathbb{R})$, this number can be brought down
again.

We will show in this paper that the dimension of the network Lie algebra
$\mathfrak{net}_{{C},N}$, where ${C}$ is the number of colors, $1\leq{C}\leq
N$, equals ${B}^{2}+{B}{C}+{C}^{2}$, with ${B}=N-{C}$ (Some readers may want
to simplify this expression, others may want to conjecture the structure of
the Lie algebra from it). We notice that this result is as simple as it could
be: it does not depend on how many cells have a given color, only the totals
$N$ and ${C}$ play a role;

The Levi decomposition (cf. [16, p.101]) is given by solvable part

$\displaystyle\begin{bmatrix}\mathsf{{C}}&0\\\
\mathfrak{{a}}&\mathsf{{B}}\end{bmatrix},$

where $\mathfrak{{a}}$ is abelian and has dimension ${B}{C}$, $\mathsf{{C}}$
and $\mathsf{{B}}$ are elements such that
$\mathsf{{B}}+\mathsf{{C}}=\mathsf{I}\in\mathcal{Z}(\mathfrak{net}_{{C},N})$,
the center of $\mathfrak{net}_{{C},N}$, and the semisimple part

$\displaystyle\begin{bmatrix}\hat{\mathfrak{{c}}}&0\\\
0&\hat{\mathfrak{{b}}}\end{bmatrix}$ $\displaystyle\equiv$
$\displaystyle\begin{bmatrix}\mathfrak{sl}_{C}&0\\\
0&\mathfrak{sl}_{{B}}\end{bmatrix}.$

The proof of all this consists of several steps. First we choose a basis for
the matrices of maps from on color space to another. Writing out the basis in
terms of two-tensors suggests to us a new choice of basis, where we group
together the colors and the cocolors. Using this new basis we can now
explicitly construct the Lie subalgebras $\mathfrak{{a}},\mathfrak{{b}}$ and
$\mathfrak{{c}}$ and show that they are isomorphic to $\mathfrak{Gr}_{{C},N}$,
$\mathfrak{gl}_{{B}}$ and $\mathfrak{gl}_{{C}}$, respectively.

If $\mathfrak{g}$ is a subalgebra of $\mathfrak{net}_{{C},N}$, one can
generate the transformations on $\mathfrak{g}$ by the normalizer

$\displaystyle\mathsf{norm}_{\mathfrak{net}_{{C},N}}(\mathfrak{g})$
$\displaystyle=$
$\displaystyle\\{x\in\mathfrak{net}_{{C},N}|[x,y]\in\mathfrak{g},\forall
y\in\mathfrak{g}\\}.$

This is a matter of taste: one can be restrictive and generate the
transformations on $\mathfrak{g}$ by $\mathfrak{g}$, or one can be less
restrictive and allow more transformations (and usually a normal form with
less terms). This is comparable with the situation in Hamiltonian mechanics,
where one might require the transformations to be symplectic. But for some
questions it is allowed to use more general transformations. The important
thing is to always explicitly mention which choices have been made in any
particular theory or computation.

At the initial stage of our research, some of the choices (like the bases and
the linear map) were inspired by symbolic calculations in Form [15] and Maple
[1].

In §2 we give two examples with three cells and one, respectively two, colors.
In §3 we start with the semigroupoid formalism to determine a basis for the
Lie algebra $\mathfrak{net}_{{C},N}$ of linear network vector fields with
dimension $N$ and ${C}$ colors. We formulate an algorithm to fix the choice of
colors and cocolors, which tries to keep a possible feedforward structure
intact. In §4 we introduce a simple invertible transformation that puts the
Jacobi-matrix of any given linear network differential equation in the desired
block form. In §5 we compute the structure constants of
$\mathfrak{net}_{{C},N}$ and define an involution
$\theta:\mathfrak{net}_{{C},N}\rightarrow\mathfrak{net}_{{B},N}$. In §6 we
describe $\mathfrak{net}_{{C},N}$ in terms of its subalgebras and we determine
the Levi decomposition. In §7 we show the existence of a dual pair within
$\mathfrak{net}_{{C},N}$. Finally we give in §8 a number of examples, with $N$
ranging from two to eight, followed by the conclusions in §10.

######  Notation 1.3.

Two kind of colors play a role here: the colors of the colored network, which
we will call $c$-colors, and the colors associated to the position in each
$f^{i}$. These we call $\sigma$-colors. In our pictures of the networks, we
use the following conventions. The big balls are the nodes, with $f^{1}$ red-,
$f^{2}$ violet- and $f^{3}$ blue-colored, where the $c$-colors are of course
almost completely arbitrary and chosen for their suitability to convey the
following information. The small balls  denote the exit point of a arrow from
a red cell, and the $\sigma$-color of the arrow indicates the position in
$f^{i}$, while the thickness of the arrow is determined by the $c$-color $i$:
$>$, $>$, $>$, the  denote the arrival point of the arrow and the  denote an
blue arrow going out and back in again (selfinteraction). The differential
equations and the picture are completely equivalent, if the order of the
$\sigma$-colors is known.

######  Acknowledgement 1.4.

The authors want to thank Vincent Knibbeler for his careful reading of an
earlier version and pointing out errors in the theoretical setup. After
correcting these, we could finally give proof of what we previously hoped to
be obvious.

##  2 Some motivating examples

###  2.1 $N=3,{C}=1$

######  Example 2.1.

In [19, §11.2] the following $3$-cell example is treated:

$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}1}),$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}1}),$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}3}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}3},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}1}).$

Or, equivalently (cf. Notation 1.3),

123$>$$>$$>$$<$$x_{1}$$x_{2}$$x_{3}$

We let, according to the theory developed in [19], $\sigma_{1}=(1,2,3)$,
$\sigma_{2}=(1,1,2)$ and $\sigma_{3}=(1,1,1)$. One verifies that the
$\sigma_{i},i=1,2,3$ define a monoid $\Sigma$, with identity $\sigma_{1}$ and
zero $\sigma_{3}$. The Fundamental Network with monoid $\tilde{\Sigma}=\Sigma$
is described by $f(X_{1},X_{2},X_{3})$ and the linearized system is

$\displaystyle\dot{\mathbf{x}}$ $\displaystyle=$
$\displaystyle\mathsf{A}\mathbf{x},$

where $\mathsf{A}=\sum_{i=1}^{3}a_{i}\mu(\sigma_{i})$ with

$\displaystyle\mu(\sigma_{1})$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix},\quad\mu(\sigma_{2})=\begin{pmatrix}1&0&0\\\ 1&0&0\\\
0&1&0\end{pmatrix},\quad\mu(\sigma_{3})=\begin{pmatrix}1&0&0\\\ 1&0&0\\\
1&0&0\end{pmatrix}$

and $a_{i}=\frac{\partial f}{\partial X_{i}}(0,0,0)$. We call the Lie algebra
of all such matrices (with the usual commutator) $\mathfrak{net}_{1,3}$, or,
for $N$-cell networks with ${C}$ colors, $\mathfrak{net}_{{C},N}$. Using the
standard theory, one can now construct a polynomial (with coefficients in
$\mathbb{R}(a_{1},a_{2},a_{3})$ with total degree one) such that
$\mathsf{N}=p(\mathsf{A})$ is nilpotent and commutes with the semisimple
$\mathsf{S}=\mathsf{A}-\mathsf{N}$. One usually needs the underlying Lie
algebra to be reductive (that is, it can be split into a semisimple part and
its center), but in this case that is not necessary, since the construction
can be carried out on the $\sigma_{i}$s themselves. We do this by formally
defining an additional structure on $\Sigma$ by
$\mu(b_{i}\sigma_{i}+b_{j}\sigma_{j})=b_{i}\mu(\sigma_{i})+b_{j}\mu(\sigma_{j}),b_{i},b_{j}\in\mathbb{R}$.
We find that $\mathsf{N}=a_{2}(\mu(\sigma_{2})-\mu(\sigma_{3}))$ and
$\mathsf{S}=a_{1}\mu(\sigma_{1})+(a_{2}+a_{3})\mu(\sigma_{3})$ is the
$\mathsf{S}+\mathsf{N}$-decomposition. We return to this in Example 3.4.

The usual normal form theory proceeds as follows: first the general nonlinear
part of $f(X_{1},X_{2},X_{3})$ is projected on $\ker\mathsf{ad}_{\mathsf{S}}$.
Then, restricting our attention to $\ker\mathsf{ad}_{\mathsf{S}}$, we
construct $\mathsf{M}$ and $\mathsf{H}$ such that
$\langle\mathsf{N},\mathsf{H},\mathsf{M}\rangle$ is a Jacobson-Morozov triple
(or, as Jim Murdock called it, a triad; smart as this terminology is chosen in
this context, we stick to the original terminology, which was in use long
before $\mathfrak{sl}_{2}$ started to play a role in normal form theory.
Kostant refers to this as an S-triple in [14], but this terminology seems not
to have been taken up) containing a given nilpotent. This triple need not to
be contained in $\mathfrak{net}_{{C},N}$, but still allows us to define the
nilpotent normal form in the same fashion as described in [17]. To formulate
the versal normal form theory will need more work in this case and will not be
attempted here. The
$\ker\mathsf{ad}_{\mathsf{S}}\cap\ker\mathsf{ad}_{\mathsf{M}}$ will now be the
space of normal forms, being the complement to $\mathrm{im}\
\mathsf{ad}_{\mathsf{A}}$. The construction of the $\mathfrak{sl}_{2}$-triple
relies on the Jacobson-Morozov theorem and supposes the underlying Lie algebra
to be reductive.

###  2.2 $N=3$, ${C}=2$

######  Example 2.2.

We consider the following equation, taken from [7, §5.2], see also [10], but
interpreted differently, that is, not fully inhomogeneous, which leads to a
slightly restricted case of Example 8.3:

$\displaystyle\dot{X}$ $\displaystyle=$ $\displaystyle f^{1}(X),\quad
X\in\mathbb{R}^{2}$ $\displaystyle\dot{y}$ $\displaystyle=$ $\displaystyle
f^{2}(X,y),\quad y\in\mathbb{R}$

We write down the general form

$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}2},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2}),$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2}),$
$\displaystyle{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\dot{x}}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}3}$
$\displaystyle=$
$\displaystyle{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}f}^{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}2}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2};{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}3}).$

Or, equivalently (cf. Notation 1.3),

123$>$$>$$>$$<$$<$$<$$x_{1}$$x_{2}$$x_{3}$

We continue this example, slightly generalized, in Example 8.3.

######  Remark 2.3.

In [10, §5.2] the equation is seen as a $3$-color system

$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\dot{x}}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1}$
$\displaystyle=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f}^{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1};{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}2},{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2}),$
$\displaystyle{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}\dot{x}}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2}$
$\displaystyle=$
$\displaystyle{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}f}^{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}2}({\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}2};{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}1},{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}1};{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}2}),$
$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\dot{x}}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}3}$
$\displaystyle=$
$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}f}^{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x}_{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}1};{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}x}_{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}2};{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}x}_{\color[rgb]{.5,0,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,0,.5}3}).$

######  Remark 2.4.

If one has a concrete differential equation, how does one see whether it is a
colored network equation? This question is difficult to answer, since there
might be many possibilities. But is it a relevant question? Probably not,
since the network description should be part of the modeling. If there is no
colored network to begin with (except $N={C}$), there is also no need to stay
in the colored network framework. Despite this, if one can give a nontrivial
description of the equation as a colored network equation, the transformations
to be described in this paper can still be used to simplify the Jacobi-matrix
without knowledge of the eigenvalues. This is comparable to Gauß-elimination,
where division is trivial due to the relations between the coefficients
following from the coloring. The situation can be compared to that in
Hamiltonian mechanics: a given equation might be identifiable as a Hamiltonian
equation and one might be able to immediately draw conclusions from this fact,
like conservation of energy, even if there is no physical reason why the
system would be Hamiltonian.

##  3 Structure theory in the colored case

######  Remark 3.1.

The $\sigma$-notation only plays a role in the theoretical formulation. For
the application of the theory to any given example, it is enough to apply the
Algorithms 3.6 and 4.6 and Definition 4.9. The Algorithms 3.6 and 4.6 work for
general colored networks, they are not required to be closed under taking
brackets.

While in [19] the approach is to start with a limited given number of
$\sigma$s and extend this to a semigroup(oid), the approach here is to work
from the other side and consider all possible $\sigma$s. We then try to reduce
the number by going to a matrix representation of the $\sigma$s and construct
a basis for the representation. Once we have a basis, we construct explicitly
a new basis in which the structure of the Lie algebra can be easily
determined. The transformations that are used can also be used if we work with
a limited number of $\sigma$s; in that case one works in a subalgebra of the
general network Lie algebra.

We consider differential equations of the following type. To every color (as
in §2) $c\in\\{1,\ldots,{C}\\}$, we assign a dimension $n_{c}\in\mathbb{N}$,
the number of cells with color $c$, and we write $\bar{n}_{c}=n_{c}-1$ to
simplify the notation, and index sets $R_{c}=\\{0,\ldots,\bar{n}_{c}\\}$ and
$\bar{R}_{c}=\\{1,\ldots,\bar{n}_{c}\\}$. We have coordinates
$\mathbf{u}_{i}^{c}$ and differential equations

$\displaystyle\dot{\mathbf{u}}_{i}^{c}$ $\displaystyle=$ $\displaystyle
f^{c}(\mathsf{U}^{c,k_{1}}_{i,l_{1}};\mathsf{U}^{c,k_{2}}_{i,l_{2}};\cdots;\mathsf{U}^{c,k_{M_{c}}}_{i,l_{M_{c}}}),\quad
i=0,\ldots,\bar{n}_{c},\quad 1\leq c,k_{1},\ldots,k_{M_{c}}\leq{C},$

where each $\mathsf{U}^{c,k_{j}}_{i,l_{j}}$ denotes a list of coordinates with
color $k_{j}$ of length $l_{j}$. Let $\Delta_{c}=\sum_{j=1}^{M_{c}}l_{j}$ be
the degree of the color $c$, that is the number of arguments of $f^{c}$. In
the pictures this is the total number of incoming arrows;
$\Delta=\sum_{c=1}^{C}\Delta_{c}$ is the number of parameters in the Jacobi
matrix.

######  Remark 3.2.

The $\mathsf{U}$-notation is only intended to give a general definition of a
colored network differential equation.

Our $N$-dimensional space, with $N=\sum_{c=1}^{C}n_{c}$ has then coordinates
$\mathcal{N}={[}\mathbf{u}_{j}^{c}{]}^{c=1,\ldots,{C}}_{j\in R_{c}}$. We order
this list lexicographically,
$\mathbf{u}_{j}^{i}\prec\mathbf{u}_{j^{\prime}}^{i^{\prime}}$ if
$i<i^{\prime}$ or, if $i=i^{\prime},j<j^{\prime}$. We now split
$\mathcal{N}=\mathcal{{C}}\cup\mathcal{{B}}$ as follows.

$\displaystyle\mathcal{{C}}$ $\displaystyle=$
$\displaystyle{[}\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{2},\cdots,\mathbf{u}_{0}^{{C}}{]},$
$\displaystyle\mathcal{{B}}$ $\displaystyle=$
$\displaystyle{[}\mathbf{u}_{j}^{c}{]}^{c=1,\ldots,{C}}_{j\in\bar{R}_{c}}.$

We chose the subindex in the definition of $\mathcal{{C}}$ to be $0$ here,
because we know beforehand that it is always present. We identify
$\mathcal{{C}}$ with the list of colors, and call the elements in
$\mathcal{{B}}$ cocolors. Here the co in cocolors can be at first seen as the
co of complement, later we will think of this co in its usual dual meaning,
but this will require some work. At the moment we can only give an a
posteriori explanation for this splitting in colors and cocolors, which turns
out to be very natural.

In any given example with coordinates $x_{1},\cdots,x_{N}$ we have to identify
these with the $\mathbf{u}_{j}^{c}$. This may involve some rather arbitrary
choices, but if we use Algorithm 3.6 much of this arbitrariness will
disappear. While the choices are arbitrary in the sense that they do not
influence the promised $\mathfrak{{a}},\mathfrak{{b}},\mathfrak{{c}}$-block
form, the algorithm may improve the end result. For instance, in Examples 2.1
and 8.5 it preserves the feedforward structure, with nice consequences for the
application of both the Jordan-Chevalley decomposition and the Jacobson-
Morozov theorem.

Let ${B}=N-{C}$ be the cardinality of $\mathcal{{B}}$. The following
definition will be used extensively in §5, where it simplifies matters
considerably, but it turns out that the proofs given before that are very
complicated in this new notation, since the position in the list
$\mathcal{{B}}$ does not give any general information. Let the
$\mathbf{u}_{i}^{k}$ be a basis of $\mathbb{R}^{n_{k}}$ and
$\bar{\mathbf{u}}^{i}_{k}$ its standard dual basis, a basis of
$\mathbb{R}^{n_{k}\star}$, the dual of $\mathbb{R}^{n_{k}}$.

######  Definition 3.3.

We define
$\mathsf{u}_{k}=\mathbf{u}_{0}^{k},\mathsf{u}^{k}=\bar{\mathbf{u}}^{0}_{k}$
and
$\mathsf{v}_{l}=\mathbf{u}_{i}^{k},\mathsf{v}^{l}=\bar{\mathbf{u}}^{i}_{k},i\in\bar{R}_{k}$,
where $l$ is the position of $\mathbf{u}_{i}^{k}$ in the list $\mathcal{{B}}$.

######  Example 3.4.

In our motivating Example 2.1, this leads to:
$\mathcal{{C}}=\\{\mathbf{u}_{0}^{1}\\}$ and
$\mathcal{{B}}=\\{\mathbf{u}_{1}^{1},\mathbf{u}_{2}^{1}\\}$, with
$N=3,{C}=1,{B}=2$, $n_{1}=3$ and $R_{1}=\\{0,1,2\\}$, $\bar{R}_{1}=\\{1,2\\}$.
We now have to identify $x_{1},x_{2},x_{3}$ with
$\mathbf{u}_{0}^{1},\mathbf{u}_{1}^{1},\mathbf{u}_{2}^{1}$ and
$\mathsf{u}_{1},\mathsf{v}_{1},\mathsf{v}_{2}$.

######  Remark 3.5.

We remark here that we make two kinds of choices here. First the choice of the
$\mathbf{u}_{0}^{1}$ color-coordinate, second the ordering of the cocolor
coordinates for each color. In this particular case, we can just say
$x_{1}=\mathbf{u}_{0}^{1},x_{2}=\mathbf{u}_{1}^{1}$ and
$x_{3}=\mathbf{u}_{2}^{1}$ and this happens to be an excellent choice. But if
we make another choice, we might easily lose the upper triangular structure of
the Jacobi-matrix, which is determined by the feedforward structure.

To this end we suggest the following algorithm for $c=1,\ldots,{C}$. In this
algorithm we assign $\mathbf{u}^{c}$ coordinates to the $x_{i}$ coordinates
and in the meantime define new $\mathsf{u},\mathsf{v}$ coordinates as defined
in Definition 3.3. Why two sets of new coordinates? Both have their
advantages. It turned out that some statements were easy to formulate in one
set, other statements in the other set:

######  Algorithm 3.6.

Let $q=1,q_{1}=1$ and $c=1$, $\mathcal{S}=\emptyset$ and
$\mathcal{N}=\\{1,\dots,N\\}$.

While $\mathcal{N}\neq\emptyset$,

determine an equation $\dot{x}_{i}$ with color $c$ with the largest number of
selfinteractions. Let $x_{i}=\mathbf{u}_{0}^{c}=\mathsf{u}_{c}$ (This has the
effect that we create zeros in the solvable part $\mathfrak{{a}}$, to be
defined later).

Let $\mathcal{S}=\mathcal{S}\cup\langle i\rangle$ and
$\mathcal{N}=\mathcal{N}\setminus\langle i\rangle$.

While the cardinality of $\mathcal{S}<n_{c}$,

count for each $j\in\mathcal{N}$ such that $x_{j}$ has color $c$, the number
of arguments $\varpi_{j}$ that $\dot{x}_{j},j\in\mathcal{N}$, has not in
common with $\dot{x}_{i}$ on any given position. Now determine $k$, such that
$\varpi_{k}$ is minimal (or maximal, your choice may depend on which side of
the diagonal you want make zero as much as possible) among the remaining
$\varpi_{j},j\in\mathcal{N}$. If there is more than one $k$ such that
$\varpi_{k}$ is extremal, choose the $k$ with the highest position of the
first $x_{i}$ among the arguments of $\dot{x}_{k}$. If that does not decide
things, make an arbitrary choice of $k$. Let
$x_{k}=\mathbf{u}_{q_{c}}^{c}=\mathsf{v}_{q}$ and $q_{c}=q_{c}+1,q=q+1$. Let
$\mathcal{S}=\mathcal{S}\cup\langle k\rangle$ and
$\mathcal{N}=\mathcal{N}\setminus\langle k\rangle$. If the cardinality of
$\mathcal{S}$ equals $n_{c}$, put $c=c+1,q_{c}=1$ and $\mathcal{S}=\emptyset$.

######  Example 3.7.

Let us go through this algorithm for Example 2.1. We have $c=1$,
$\mathcal{S}=\emptyset$ and $\mathcal{N}=\\{1,\dots,3\\}$. We see that
$\dot{x}_{1}$ has the maximal number of selfinteractions. So
$x_{1}=\mathbf{u}_{0}^{1}=\mathsf{u}_{1}$ and $\mathcal{S}=\\{1\\}$ and
$\mathcal{N}=\\{2,3\\}$. Then we put $q=1$ and compute $\varpi_{2}=1$ and
$\varpi_{3}=2$. Thus $x_{2}=\mathbf{u}_{1}^{1}=\mathsf{v}_{1}$. and
$\mathcal{S}=\\{1,2\\}$ and $\mathcal{N}=\\{3\\}$. Then we put $q=2$ and
$x_{3}=\mathbf{u}_{2}^{1}=\mathsf{v}_{2}$. Then $\mathcal{S}=\\{1,2,3\\}$ and
$\mathcal{N}=\emptyset$ and we are done.

######  Remark 3.8.

In §8, we refer to Algorithm 3.6 all the time, but we remark here that in most
cases it does not do anything useful (nor any harm). This may have to do with
the particular choice of examples, and we would recommend to use it anyway.
But in the feedforward case it is doing very well, cf. Example 8.5. For other
types of structure, one may want to develop other algorithms to assign the
$x_{i}$ to $\mathsf{u},\mathsf{v}$-coordinates.

Since we have

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f(x_{1},x_{1},x_{1}),$ $\displaystyle\dot{x}_{2}$ $\displaystyle=$
$\displaystyle f(x_{2},x_{1},x_{1}),$ $\displaystyle\dot{x}_{3}$
$\displaystyle=$ $\displaystyle f(x_{3},x_{2},x_{1}),$

we have $\Delta_{1}=3$ and

$\displaystyle\mathsf{U}^{1,1}_{0,3}$ $\displaystyle=$
$\displaystyle[x_{1},x_{1},x_{1}]=[\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{1}]=[\mathsf{u}_{1},\mathsf{u}_{1},\mathsf{u}_{1}],$
$\displaystyle\mathsf{U}^{1,1}_{1,3}$ $\displaystyle=$
$\displaystyle[x_{2},x_{1},x_{1}]=[\mathbf{u}_{1}^{1},\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{1}]=[\mathsf{v}_{1},\mathsf{u}_{1},\mathsf{u}_{1}],$
$\displaystyle\mathsf{U}^{1,1}_{2,3}$ $\displaystyle=$
$\displaystyle[x_{3},x_{2},x_{2}]=[\mathbf{u}_{2}^{1},\mathbf{u}_{1}^{1},\mathbf{u}_{1}^{1}]=[\mathsf{v}_{2},\mathsf{v}_{1},\mathsf{v}_{1}].$

We return to this in Example 8.2.

######  Example 3.9.

Let $n_{1}=2$ and $n_{2}=1$, as in Example 2.2 and [19, Example 12.4], that
is, $N=3$ and ${C}=2,{B}=1$, with
$\mathcal{{C}}=\\{\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{2}\\}$ and
$\mathcal{{B}}=\\{\mathbf{u}_{1}^{1}\\}$. $R_{1}=\\{0,1\\}$,
$\bar{R}_{1}=\\{1\\}$, $R_{2}=\\{0\\}$. The coordinates are, following
Algorithm 3.6,

$\displaystyle
x_{1}=\mathbf{u}_{0}^{1}=\mathsf{u}_{1},x_{2}=\mathbf{u}_{1}^{1}=\mathsf{v}_{1},x_{3}=\mathbf{u}_{0}^{2}=\mathsf{u}_{2}.$

Since we have

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(x_{1},x_{1},x_{2},x_{2}),$ $\displaystyle\dot{x}_{2}$ $\displaystyle=$
$\displaystyle f^{1}(x_{2},x_{1},x_{1},x_{2}),$ $\displaystyle\dot{x}_{3}$
$\displaystyle=$ $\displaystyle f^{2}(x_{1},x_{2};x_{3}),$

we find that $\Delta_{1}=4$ and $\Delta_{2}=3$ and, since each
$\mathsf{U}^{c,k_{j}}_{i,l_{j}}$ denotes a list of coordinates with color
$k_{j}$ of length $l_{j}$,

$\displaystyle\mathsf{U}^{1,1}_{0,4}$ $\displaystyle=$
$\displaystyle[x_{1},x_{1},x_{2},x_{2}]=[\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{1},\mathbf{u}_{1}^{1},\mathbf{u}_{1}^{1}]=[\mathsf{u}_{1},\mathsf{u}_{1},\mathsf{v}_{1},\mathsf{v}_{1}],$
$\displaystyle\mathsf{U}^{1,1}_{1,4}$ $\displaystyle=$
$\displaystyle[x_{2},x_{1},x_{1},x_{2}]=[\mathbf{u}_{1}^{1},\mathbf{u}_{0}^{1},\mathbf{u}_{0}^{1},\mathbf{u}_{1}^{1}]=[\mathsf{v}_{1},\mathsf{u}_{1},\mathsf{u}_{1},\mathsf{v}_{1}],$
$\displaystyle\mathsf{U}^{2,1}_{2,2}$ $\displaystyle=$
$\displaystyle[x_{1},x_{2}]=[\mathbf{u}_{0}^{1},\mathbf{u}_{1}^{1}]=[\mathsf{u}_{1},\mathsf{v}_{1}],$
$\displaystyle\mathsf{U}^{2,2}_{3,1}$ $\displaystyle=$
$\displaystyle[x_{3}]=[\mathbf{u}_{0}^{2}]=[\mathsf{u}_{2}].$

We return to this in Example 3.11.

The remaining part of this section is devoted to theoretical matters. If one
is only interested in putting a particular example in block form, one can skip
it.

######  Definition 3.10.

Let $\Sigma_{k}^{l}$ be the set of maps with domain $R_{l}$ taking their
values from $R_{k}$. We can think of $\sigma\in\Sigma_{k}^{l}$ as an
$n_{l}$-tuple $(k^{\sigma}_{0},\ldots,k^{\sigma}_{\bar{n}_{l}})_{k}^{l}$, with
$k^{\sigma}_{j}\in R_{k}$ for $j\in R_{l}$. Then the cardinality of
$\Sigma_{k}^{l}$ is $n_{k}^{n_{l}}$. We define a map
$\nu:\Sigma_{k}^{l}\rightarrow\mathbb{R}^{n_{k}}\otimes\mathbb{R}^{n_{l}\star}=\mathrm{Hom}(\mathbb{R}^{n_{l}},\mathbb{R}^{n_{k}})$
as follows. To each $\sigma\in\Sigma_{k}^{l}$ we assign a $2$-tensor

$\displaystyle\nu(\sigma)=\sum_{j=0}^{\bar{n}_{l}}\mathbf{u}_{k^{\sigma}_{j}}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}\in\mathbb{R}^{n_{k}}\otimes\mathbb{R}^{n_{l}\star}\subset\mathfrak{gl}_{N},$

mapping $\mathbf{u}_{q}^{l}$ to $\mathbf{u}_{\sigma(q)}^{k}$.

######  Example 3.11.

Let $n_{1}=2$ and $n_{2}=1$. Then
$\Sigma^{1}_{1}=\\{(0,0)^{1}_{1},(0,1)^{1}_{1},(1,0)^{1}_{1},(1,1)^{1}_{1}\\}$,
$\Sigma_{1}^{2}=\\{(0)^{2}_{1},(1)^{2}_{1}\\}$,
$\Sigma_{2}^{1}=\\{(0,0)^{1}_{2}\\}$ and $\Sigma^{2}_{2}=\\{(0)^{2}_{2}\\}$.

$\displaystyle\nu((0,0)^{1}_{1})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},\quad\nu((0,1)^{1}_{1})=\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},$
$\displaystyle\nu((1,0)^{1}_{1})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},\quad\nu((1,1)^{1}_{1})=\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},\quad\\#\Sigma_{2}^{2}=n_{1}^{n_{1}}=2^{2}=4,$
$\displaystyle\nu((0)_{1}^{2})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{0}_{2},\quad\nu((1)_{1}^{2})=\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{0}_{2},\quad\\#\Sigma_{1}^{2}=n_{1}^{n_{2}}=2^{1}=2,$
$\displaystyle\nu((0,0)_{2}^{1})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{2}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{0}^{2}\otimes\bar{\mathbf{u}}^{1}_{1},\quad\\#\Sigma_{2}^{1}=n_{2}^{n_{1}}=1^{2}=1,$
$\displaystyle\nu((0)^{2}_{2})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{2}\otimes\bar{\mathbf{u}}^{0}_{2},\quad\\#\Sigma^{1}_{1}=1^{1}=1.$

We continue this example in Example 3.16, where we reduce the number of cases
from $8=4+2+1+1$ to $7$ by choosing a basis for each $\Sigma_{k}^{l}$.

######  Lemma 3.12.

The $\Sigma_{k}^{l}$ together form a semigroupoid $\Sigma$ and $\nu$ is a
faithful antirepresentation of $\Sigma$ in $\mathfrak{gl}_{N}$.

###### Proof.

Let $\sigma\in\Sigma_{k}^{l}$ and $\bar{\sigma}\in\Sigma_{l}^{m}$, so that
their composition exists. Then we compute
$\bar{\sigma}\cdot\sigma\in\Sigma_{k}^{m}$:

$\displaystyle\bar{\sigma}\cdot{\sigma}$ $\displaystyle=$
$\displaystyle(l^{\bar{\sigma}}_{0},\ldots,l^{\bar{\sigma}}_{\bar{n}_{m}})^{l}_{m}\cdot(k^{\sigma}_{0},\ldots,k^{\sigma}_{\bar{n}_{l}})^{k}_{l}$
$\displaystyle=$
$\displaystyle(k^{\sigma}_{l^{\bar{\sigma}}_{0}},\ldots,k^{\sigma}_{l^{\bar{\sigma}}_{\bar{n}_{m}}})^{k}_{m},$

that is,
$k^{\bar{\sigma}\cdot{\sigma}}_{i}=k^{\sigma}_{l^{\bar{\sigma}}_{i}}=k^{\sigma}_{\bar{\sigma}(i)},i=0,\cdots,\bar{n}_{m}$.
Then

$\displaystyle\nu(\sigma)\cdot\nu(\bar{\sigma})$ $\displaystyle=$
$\displaystyle\sum_{i_{1}=0}^{\bar{n}_{l}}\mathbf{u}_{k^{\sigma}_{i_{1}}}^{k}\otimes\bar{\mathbf{u}}^{i_{1}}_{l}\cdot\sum_{i_{2}=0}^{\bar{n}_{m}}\mathbf{u}_{l^{\bar{\sigma}}_{i_{2}}}^{l}\otimes\bar{\mathbf{u}}^{i_{2}}_{m}=\sum_{i_{1}=0}^{\bar{n}_{l}}\sum_{i_{2}=0}^{\bar{n}_{m}}\delta_{l^{\bar{\sigma}}_{i_{2}}}^{i_{1}}\mathbf{u}_{k^{{\sigma}}_{i_{1}}}^{k}\otimes\bar{\mathbf{u}}^{i_{2}}_{m}=\sum_{i=0}^{\bar{n}_{m}}\mathbf{u}_{k^{{\sigma}}_{l^{\bar{\sigma}}_{i}}}^{k}\otimes\bar{\mathbf{u}}^{i}_{m}=\sum_{i=0}^{\bar{n}_{m}}\mathbf{u}_{k^{\bar{\sigma}\cdot\sigma}_{i}}^{k}\otimes\bar{\mathbf{u}}^{i}_{m}$
$\displaystyle=$ $\displaystyle\nu(\bar{\sigma}\cdot\sigma).$

∎

######  Corollary 3.13.

We define the (Lie) algebra $\mathfrak{net}_{C,N}$ as follows. It is the
$\mathbb{R}$-span of the matrices
$\nu(\sigma)\in\mathfrak{gl}_{N},\sigma\in\Sigma_{k}^{l},k,l=1,\ldots,{C}$. We
have seen in Lemma 3.12 that the product of two of these is again in
$\mathfrak{net}_{C,N}$ and therefore $\mathfrak{net}_{C,N}$ is a (Lie)
subalgebra of $\mathfrak{gl}_{N}$. In the sequel we restrict our attention to
the Lie algebra $\mathfrak{net}_{C,N}$. We leave it to the the reader, if
necessary, to derive the structure constants for the algebra. This should pose
no problems if the methods of Section 5 are followed.

######  Definition 3.14.

We fix a basis (cf. Theorem 3.15) $\mathcal{{B}}_{k}^{l}$ for
$\nu(\Sigma_{k}^{l})$ as follows:

$\displaystyle
0<k,l\leq{C},\left\\{\begin{array}[]{lrclc}j=0,&\nu^{l,0}_{k,0}(\mathbf{u})&=&\nu(([0]_{n_{l}})^{l}_{k})=\mathbf{u}_{0}^{k}\otimes\sum_{q\in
R_{l}}\bar{\mathbf{u}}^{q}_{l},&\\\
j\in\bar{R}_{k},&\nu^{l,i}_{k,j}(\mathbf{u})&=&\nu(([0]_{i},j,[0]_{\bar{n}_{l}-i})^{l}_{k})=\nu^{l,0}_{k,0}(\mathbf{u})+(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})\otimes\bar{\mathbf{u}}^{i}_{l},&\quad
i\in R_{l},\end{array}\right.$

where $[0]_{p}$ is a row zeros of length $p$.

######  Theorem 3.15.

$\mathcal{{B}}_{k}^{l}$ is a basis, that is, every
$\nu(\sigma),\sigma\in\Sigma_{k}^{l}$, can be expressed in terms of
$\mathcal{{B}}_{k}^{l}$:

$\displaystyle\nu(\sigma)$ $\displaystyle=$ $\displaystyle\sum_{j\in
Q}\nu^{l,j}_{k,k^{\sigma}_{j}}(\mathbf{u})+(1-\\#Q)\nu^{l,0}_{k,0}(\mathbf{u}),$

where $Q$ is the set of all $j\in R_{l}$ such that $k^{\sigma}_{j}>0$, and the
elements in $\mathcal{{B}}_{k}^{l}$ are linearly independent.

###### Proof.

Let $\sigma=(k^{\sigma}_{0},\ldots,k^{\sigma}_{\bar{n}_{l}})^{k}_{l}$. If
$j>0$,

$\displaystyle\mathbf{u}_{k^{\sigma}_{j}}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}$
$\displaystyle=$
$\displaystyle\nu^{l,j}_{k,k^{\sigma}_{j}}(\mathbf{u})-\nu^{l,0}_{k,0}(\mathbf{u})+\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}.$

Let $Q$ be the set of all $j$ such that $k^{\sigma}_{j}>0$ and let $\bar{Q}$
be its complement in $R_{l}$.

$\displaystyle\nu(\sigma)$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{\bar{n}_{l}}\mathbf{u}_{k^{\sigma}_{j}}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}$
$\displaystyle=$ $\displaystyle\sum_{j\in
Q}\mathbf{u}_{k^{\sigma}_{j}}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}+\sum_{j\in\bar{Q}}\mathbf{u}_{k^{\sigma}_{j}}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}$
$\displaystyle=$ $\displaystyle\sum_{j\in
Q}\left(\nu^{l,j}_{k,k^{\sigma}_{j}}(\mathbf{u})-\nu^{l,0}_{k,0}(\mathbf{u})+\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}\right)+\sum_{j\in\bar{Q}}\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}$
$\displaystyle=$ $\displaystyle\sum_{j\in
Q}\left(\nu^{l,j}_{k,k^{\sigma}_{j}}(\mathbf{u})-\nu^{l,0}_{k,0}(\mathbf{u})\right)+\sum_{j\in
R_{l}}\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{j}_{l}$ $\displaystyle=$
$\displaystyle\sum_{j\in
Q}\nu^{l,j}_{k,k^{\sigma}_{j}}(\mathbf{u})+(1-\\#Q)\nu^{l,0}_{k,0}(\mathbf{u}).$

This shows that an arbitrary $\nu(\sigma)$ can be expressed as a linear
combination of the elements in $\mathcal{{B}}_{k}^{l}$. The linear indepence
of these elements is easy to see. ∎

######  Example 3.16.

The basis $\mathcal{{B}}^{2}_{2}$ in Example 3.11 is given by

$\displaystyle\nu((0,0)^{1}_{1})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},\quad\nu((0,1)^{1}_{1})=\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{1}_{1},$
$\displaystyle\nu((1,0)^{1}_{1})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{1}^{1}\otimes\bar{\mathbf{u}}^{0}_{1}+\mathbf{u}_{0}^{1}\otimes\bar{\mathbf{u}}^{1}_{1}.$

Together with $\Sigma_{1}^{2},\Sigma_{2}^{1}$ and $\Sigma_{2}^{2}$ this gives
us the basis of all linear network maps in this case.

######  Lemma 3.17.

The dimension of $\mathfrak{net}_{{C},N}$ is
${C}^{2}+{B}{C}+{B}^{2}=N^{2}-{B}{C}$.

###### Proof.

For this proof we look at Definition 3.14. The term $1$ corresponds to
$i=j=0$, the $\bar{n}_{k}$ to the number of $j\in\bar{R}_{k}$ and $n_{l}$ to
the number of $i\in R_{l}$:

$\displaystyle\mathrm{dim}\
\mathfrak{net}_{{C},N}=\sum_{k,l=1}^{C}\left(1+\bar{n}_{k}n_{l}\right)=\sum_{k=1}^{C}\left({C}+\bar{n}_{k}N\right)={C}^{2}+(N-{C})N={B}^{2}+{B}{C}+{C}^{2}.$

∎

######  Remark 3.18.

In the case that the number of colors ${C}$ equals the dimension $N$ (implying
$n_{i}=1$ for all $i=1,\ldots,{C}$), this reduces to

$\displaystyle\nu^{l,0}_{k,0}$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{l}\otimes\bar{\mathbf{u}}^{0}_{k}=\mathsf{u}_{l}\otimes\mathsf{u}^{k}.$

There are $N^{2}$ elements and $\mathfrak{net}_{N,N}=\mathfrak{gl}_{N}$.

######  Remark 3.19.

It will turn out that the case ${C}=N$ is a bit exceptional, even though it is
the case of we are most familiar with, that of the ordinary differential
equations (or maps). This will show up when we compute the Levi decomposition
of $\mathfrak{net}_{{C},N}$.

##  4 Subalgebra decomposition of $\mathfrak{net}_{{C},N}$

In this section we define an invertible linear transformation of
$\mathbb{R}^{N}$, mapping each colored subspace $\mathbb{R}^{n_{c}},1\leq
c\leq{C}$ onto itself. After this transformation it is not so difficult to
define subalgebras of $\mathfrak{net}_{{C},N}$ and to determine their Lie
algebraic type. We start our discussion with the definition of a Lie bracket
on the $2$-tensors and corresponding representations. In the linear case this
comes down to the usual Lie bracket of square matrices and their action on
(co)vectors, but the construction also applies to the nonlinear case, cf.
Remark 4.2.

######  Definition 4.1.

Let $\mathsf{e}_{i}$ be a basis of $\mathsf{V}$, and let the $\mathsf{e}^{j}$
be the standard dual basis, that is,
$\mathsf{e}^{j}\cdot\mathsf{e}_{i}=\delta^{j}_{i}$. Let
$\mathsf{x},\mathsf{y}\in\mathsf{V}\otimes\mathsf{V}^{\star}$. On the space of
two-tensors we can define a product $\trianglerighteq$ (matrix multiplication)
by:

$\displaystyle\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{i_{2}}\trianglerighteq\mathsf{e}_{j_{1}}\otimes\mathsf{e}^{j_{2}}$
$\displaystyle=$
$\displaystyle(\mathsf{e}^{i_{2}}\cdot\mathsf{e}_{j_{1}})\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{j_{2}}=\delta^{i_{2}}_{j_{1}}\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{j_{2}}$

and

$\displaystyle\mathsf{x}\trianglerighteq\mathsf{y}$ $\displaystyle=$
$\displaystyle\sum_{i_{1},i_{2}}x_{i_{1},i_{2}}\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{i_{2}}\trianglerighteq\sum_{j_{1},j_{2}}y_{j_{1},j_{2}}\mathsf{e}_{j_{1}}\otimes\mathsf{e}^{j_{2}}=\sum_{i_{1},i_{2},j_{1},j_{2}}x_{i_{1},i_{2}}y_{j_{1},j_{2}}(\mathsf{e}^{i_{2}}\cdot\mathsf{e}_{j_{1}})\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{j_{2}}$
$\displaystyle=$
$\displaystyle\sum_{i_{1},i_{2},j_{1},j_{2}}x_{i_{1},i_{2}}y_{j_{1},j_{2}}\delta^{i_{2}}_{j_{1}}\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{j_{2}}=\sum_{i,j}\sum_{k}x_{i,k}y_{k,j}\mathsf{e}_{i}\otimes\mathsf{e}^{j}.$

The associator  (cf. [8]) of $\trianglerighteq$ is a 3-tensor
$\alpha(\mathsf{x},\mathsf{y},\mathsf{z})=\mathsf{x}\trianglerighteq(\mathsf{y}\trianglerighteq\mathsf{z})-(\mathsf{x}\trianglerighteq\mathsf{y})\trianglerighteq\mathsf{z}$.
This is also written as $[\mathsf{x},\mathsf{y},\mathsf{z}]$ in the
literature. The proofs of the following two Lemmas 4.3 and 4.4 are left out,
since they are a matter of writing out the definitions and some trivial
mathematics.

######  Remark 4.2.

The use of the associator is a bit of overkill in the linear case, but will
prove useful when we consider nonlinear vector fields, replacing
$\mathsf{V}\otimes\mathsf{V}^{\star}$ by
$\mathsf{S(V)}\otimes\mathsf{V}^{\star}$, the space of polynomial vector
fields, extending the action of $\mathsf{V}^{\star}$ to $\mathsf{S(V)}$ by
derivation.

######  Lemma 4.3.

The associator of $\trianglerighteq$ is symmetric in its second and third
argument (in the nonlinear case this follows from the symmetry of the second
derivatives).

######  Lemma 4.4.

$[\mathsf{x},\mathsf{y}]=\mathsf{x}\trianglerighteq\mathsf{y}-\mathsf{y}\trianglerighteq\mathsf{x}$
defines an antisymmetric bracket that satisfies the Jacobi identity, in other
words: it defines a Lie algebra, isomorphic to $\mathfrak{gl}(\mathsf{V})$.

######  Definition 4.5.

We define a representation (cf. Definition 4.1) of
$\mathsf{V}\otimes\mathsf{V}^{\star}$ on $\mathsf{V}$ and $\mathsf{V}^{\star}$
as follows:

$\displaystyle\left\\{\begin{matrix}\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{i_{2}}\triangleright\mathsf{e}_{j}&=&(\mathsf{e}^{i_{2}}\cdot\mathsf{e}_{j})\
\mathsf{e}_{i_{1}}=\delta^{i_{2}}_{j}\ \mathsf{e}_{i_{1}},\\\
\mathsf{e}_{i_{1}}\otimes\mathsf{e}^{i_{2}}\triangleright\mathsf{e}^{j}&=&-(\mathsf{e}^{j}\cdot\mathsf{e}_{i_{1}})\
\mathsf{e}^{i_{2}}=-\delta^{j}_{i_{1}}\
\mathsf{e}^{i_{2}},\end{matrix}\right.$

in other words, a matrix multiplying a (co)vector.

The following Algorithm is a crucial step to obtain the subalgebra
decomposition. It is motivated by the choice of the basis in Definition 3.14.

######  Algorithm 4.6.

We now define for $1\leq k\leq{C}<N$ a color-preserving map from
$\mathbb{R}^{n_{k}}\oplus\mathbb{R}^{\star n_{k}}$ to itself:

$\displaystyle\begin{pmatrix}\mathbf{x}_{0}^{k}\\\ \mathbf{x}_{1}^{k}\\\
\vdots\\\ \mathbf{x}_{\bar{n}_{k}}^{k}\\\ \bar{\mathbf{x}}^{0}_{k}\\\
\bar{\mathbf{x}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{x}}^{\bar{n}_{k}}_{k}\end{pmatrix}$ $\displaystyle=$
$\displaystyle\exp(-\sum_{c=1}^{C}\mathbf{u}_{0}^{c}\otimes\sum_{q=1}^{\bar{n}_{c}}\bar{\mathbf{u}}^{q}_{c}\triangleright)\begin{pmatrix}\mathbf{u}_{0}^{k}\\\
\mathbf{u}_{1}^{k}\\\ \vdots\\\ \mathbf{u}_{\bar{n}_{k}}^{k}\\\
\bar{\mathbf{u}}^{0}_{k}\\\ \bar{\mathbf{u}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{u}}^{\bar{n}_{k}}_{k}\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{k}\\\
\mathbf{u}_{1}^{k}-\mathbf{u}_{0}^{k}\\\ \vdots\\\
\mathbf{u}_{\bar{n}_{k}}^{k}-\mathbf{u}_{0}^{k}\\\
\bar{\mathbf{u}}^{0}_{k}+\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{u}}^{q}_{k}\\\
\bar{\mathbf{u}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{u}}^{\bar{n}_{k}}_{k}\end{pmatrix},$

where the $\bar{\mathbf{u}}^{\cdot}_{k}$ are a dual basis to the
$\mathbf{u}_{\cdot}^{k}$.

######  Lemma 4.7.

The $\bar{\mathbf{x}}^{\cdot}_{k}$ are a standard dual basis to the
$\mathbf{x}_{\cdot}^{k}$.

###### Proof.

$\displaystyle\bar{\mathbf{x}}^{0}_{k}\cdot\mathbf{x}_{0}^{k}$
$\displaystyle=$
$\displaystyle(\bar{\mathbf{u}}^{0}_{k}+\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{u}}^{q}_{k})\cdot\mathbf{u}_{0}^{k}=\bar{\mathbf{u}}^{0}_{k}\cdot\mathbf{u}_{0}^{k}=1,$
$\displaystyle\bar{\mathbf{x}}^{0}_{k}\cdot\mathbf{x}_{j}^{k}$
$\displaystyle=$
$\displaystyle(\bar{\mathbf{u}}^{0}_{k}+\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{u}}^{q}_{k})\cdot(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})=\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{u}}^{q}_{k})\cdot\mathbf{u}_{j}^{k}-\bar{\mathbf{u}}^{0}_{k}\cdot\mathbf{u}_{0}^{k}=\sum_{q=1}^{\bar{n}_{k}}\delta^{q}_{j}-1=1-1=0,\quad
j=1,\cdots,\bar{n}_{k}$
$\displaystyle\bar{\mathbf{x}}^{i}_{k}\cdot\mathbf{x}_{0}^{k}$
$\displaystyle=$
$\displaystyle\bar{\mathbf{u}}^{i}_{k}\cdot\mathbf{u}_{0}^{k}=0,\quad
i=1,\cdots,\bar{n}_{k}$
$\displaystyle\bar{\mathbf{x}}^{i}_{k}\cdot\mathbf{x}_{j}^{k}$
$\displaystyle=$
$\displaystyle\bar{\mathbf{u}}^{i}_{k}\cdot(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})=\delta^{i}_{j},\quad
i,j=1,\cdots,\bar{n}_{k}$

∎

######  Corollary 4.8.

The $\bar{\mathbf{x}}^{\cdot}_{\cdot}$ are a standard dual basis to the
$\mathbf{x}_{\cdot}^{\cdot}$.

The inverse is given by

$\displaystyle\begin{pmatrix}\mathbf{u}_{0}^{k}\\\ \mathbf{u}_{1}^{k}\\\
\vdots\\\ \mathbf{u}_{\bar{n}_{k}}^{k}\\\ \bar{\mathbf{u}}^{0}_{k}\\\
\bar{\mathbf{u}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{u}}^{\bar{n}_{k}}_{k}\end{pmatrix}$ $\displaystyle=$
$\displaystyle\exp(\sum_{c=1}^{C}\mathbf{x}_{0}^{c}\otimes\sum_{q=1}^{\bar{n}_{c}}\bar{\mathbf{x}}^{q}_{c}\triangleright)\begin{pmatrix}\mathbf{x}_{0}^{k}\\\
\mathbf{x}_{1}^{k}\\\ \vdots\\\ \mathbf{x}_{\bar{n}_{k}}^{k}\\\
\bar{\mathbf{x}}^{0}_{k}\\\ \bar{\mathbf{x}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{x}}^{\bar{n}_{k}}_{k}\end{pmatrix}=\begin{pmatrix}\mathbf{x}_{0}^{k}\\\
\mathbf{x}_{1}^{k}+\mathbf{x}_{0}^{k}\\\ \vdots\\\
\mathbf{x}_{\bar{n}_{k}}^{k}+\mathbf{x}_{0}^{k}\\\
\bar{\mathbf{x}}^{0}_{k}-\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{x}}^{q}_{k}\\\
\bar{\mathbf{x}}^{1}_{k}\\\ \vdots\\\
\bar{\mathbf{x}}^{\bar{n}_{k}}_{k}\end{pmatrix}.$

This formula will be used in Section 8, since it allows us to compute the
Jacobi-matrix, given in terms of the $\mathbf{u}$-coordinates, to one given in
$\mathbf{x}$-coordinates, and then in $\mathsf{x},\mathsf{y}$-coordinates (as
defined in Algorithm 3.6), to finally arrive at the
$\mathfrak{{a}},\mathfrak{{b}},\mathfrak{{c}}$ notation to be introduced next.

######  Definition 4.9.

Let $k$ and $l$ be colors and $1\leq j\leq\bar{n}_{k}$ and $1\leq
i\leq\bar{n}_{l}$. Then we define the following subalgebras (cf. Lemma 4.10):

  * •

The space spanned by the
$\mathfrak{{a}}^{l,0}_{k,j}=\mathbf{x}_{j}^{k}\otimes\bar{\mathbf{x}}^{0}_{l}$
will be denoted by $\mathfrak{{a}}\subset\mathfrak{net}_{{C},N}$.

  * •

The space spanned by the
$\mathfrak{{b}}^{l,i}_{k,j}=\mathbf{x}_{j}^{k}\otimes\bar{\mathbf{x}}^{i}_{l}$
will be denoted by $\mathfrak{{b}}\subset\mathfrak{net}_{{C},N}$.

  * •

The space spanned by the
$\mathfrak{{c}}^{l,0}_{k,0}=\mathbf{x}_{0}^{k}\otimes\bar{\mathbf{x}}^{0}_{l}$
will be denoted by $\mathfrak{{c}}\subset\mathfrak{net}_{{C},N}$.

  * •

The space spanned by the
$\mathfrak{{o}}^{l,i}_{k,0}=\mathbf{x}_{0}^{k}\otimes\bar{\mathbf{x}}^{i}_{l}$
will be denoted by $\mathfrak{{o}}\subset\mathfrak{gl}_{N}$.

It will turn out that $\mathfrak{{a}}$ is contained in the solvable part of
the Lie algebra $\mathfrak{net}_{{C},N}$. The space $\mathfrak{{o}}$ is the
complement of $\mathfrak{net}_{{C},N}$ in $\mathfrak{gl}_{N}$.

######  Lemma 4.10.

The $\mathfrak{{a}},\mathfrak{{b}},\mathfrak{{c}}$ span the linear network
vector fields:

$\displaystyle\nu^{l,0}_{k,0}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\mathfrak{{c}}^{l,0}_{k,0},$
$\displaystyle\nu^{l,i}_{k,j}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\mathfrak{{c}}^{l,0}_{k,0}+\mathfrak{{b}}^{l,i}_{k,j},$
$\displaystyle\nu^{l,i}_{k,0}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\mathfrak{{c}}^{l,0}_{k,0}+\mathfrak{{a}}^{l,0}_{k,j}-\sum_{q=1}^{\bar{n}_{k}}\mathfrak{{b}}^{k,q}_{l,j}.$

One checks that this relation is invertible. Combined with Theorem 3.15 this
allows us to express $\nu(\sigma)$ in terms of $\mathfrak{{a}},\mathfrak{{b}}$
and $\mathfrak{{c}}$ explicitly.

###### Proof.

Recall that for $1\leq k,l\leq{C}$,

$\displaystyle\nu^{l,0}_{k,0}(\mathbf{u})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{k}\otimes\sum_{q=0}^{\bar{n}_{l}}\bar{\mathbf{u}}^{q}_{l},$
$\displaystyle\nu^{l,0}_{k,j}(\mathbf{u})$ $\displaystyle=$
$\displaystyle\nu^{l}_{k,0}(\mathbf{u})+(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})\otimes\bar{\mathbf{u}}^{0}_{l},\quad
j\in\bar{R}_{k},$ $\displaystyle\nu^{l,i}_{k,j}(\mathbf{u})$ $\displaystyle=$
$\displaystyle\nu^{l}_{k,0}(\mathbf{u})+(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})\otimes\bar{\mathbf{u}}^{i}_{l},\quad
i\in\bar{R}_{l},j\in\bar{R}_{k}.$

Then, using
$\exp(\mathsf{w}\trianglerighteq)(\mathsf{x}\otimes\mathsf{y})=\exp(\mathsf{w}\triangleright)\mathsf{x}\otimes\exp(\mathsf{w}\triangleright)\mathsf{y}$,

$\displaystyle\nu^{l,0}_{k,0}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{0}_{l}+\sum_{q=1}^{\bar{n}_{l}}\mathbf{u}_{0}^{k}\otimes\bar{\mathbf{u}}^{q}_{l}$
$\displaystyle=$
$\displaystyle\mathbf{x}_{0}^{k}\otimes(\bar{\mathbf{x}}^{0}_{l}-\sum_{q=1}^{\bar{n}_{l}}\bar{\mathbf{x}}^{q}_{l})+\sum_{q=1}^{\bar{n}_{l}}\mathbf{x}_{0}^{k}\otimes\bar{\mathbf{x}}^{q}_{l}=\mathbf{x}_{0}^{k}\otimes\bar{\mathbf{x}}^{0}_{l}=\mathfrak{{c}}^{l,0}_{k,0},$
$\displaystyle\nu^{l,0}_{k,j}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\nu^{l}_{k,0}(\mathbf{x})+(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})\otimes\bar{\mathbf{u}}^{0}_{l}$
$\displaystyle=$
$\displaystyle\mathfrak{{c}}^{l,0}_{k,0}+\mathbf{x}_{j}^{k}\otimes(\bar{\mathbf{x}}^{0}_{l}-\sum_{q=1}^{n_{l}}\bar{\mathbf{x}}^{q}_{l})=\mathfrak{{c}}^{l,0}_{k,0}+\mathfrak{{a}}^{l,0}_{k,j}-\sum_{q=1}^{n_{l}}\mathfrak{{b}}^{k,q}_{l,j},\quad
j\in\bar{R}_{l},$ $\displaystyle\nu^{l,i}_{k,j}(\mathbf{x})$ $\displaystyle=$
$\displaystyle\nu^{l,0}_{k,0}(\mathbf{u})+(\mathbf{u}_{j}^{k}-\mathbf{u}_{0}^{k})\otimes\bar{\mathbf{u}}^{i}_{l}$
$\displaystyle=$
$\displaystyle\mathfrak{{c}}^{l,0}_{k,0}+\mathbf{x}_{j}^{k}\otimes\bar{\mathbf{x}}^{i}_{l}=\mathfrak{{c}}^{l,0}_{k,0}+\mathfrak{{b}}^{l,i}_{k,j},\quad
i\in\bar{R}_{l},j\in\bar{R}_{k}.$

∎

######  Remark 4.11.

The semigroup(oid) approach was taken in [19] with the explicit goal to work
within the network and not allow transformations with generators outside the
network. Yet the generators
$\mathbf{u}_{0}^{k}\otimes\sum_{q=1}^{\bar{n}_{k}}\bar{\mathbf{u}}^{q}_{k}$
are not in the network. Are we cheating here, in the sense that we ignore our
categorical approach here? The subalgebras
$\mathfrak{{a}},\mathfrak{{b}},\mathfrak{{c}}$ that we construct can be
expressed in the original network basis by applying the inverse transformation
and are as such objects within the network. So we are not cheating, but follow
the analogy of solving real differential equations using complex numbers. We
(implicitly) use the fact that the exponential is a Lie algebra homomorphism.

######  Theorem 4.12.

$\mathfrak{net}_{{C},N}=\mathfrak{{a}}\oplus\mathfrak{{b}}\oplus\mathfrak{{c}}$
and $\mathfrak{gl}_{N}=\mathfrak{net}_{{C},N}\oplus\mathfrak{{o}}$ (as vector
spaces ).

###### Proof.

See Lemma 4.10 and check that the dimension is correct. ∎

######  Notation 4.13.

We let, for a given equilibrium $\mathbf{x}_{0}$,

$\displaystyle f^{i}_{j}=\frac{\partial f^{i}}{\partial
x_{j}}(\mathbf{x}_{0}),\quad i=1,\ldots,{C}.$

If there is only one function involved (${C}=1$), one may drop the superindex.

######  Theorem 4.14.

When ${C}=1$, $N$ arbitrary, and the number of selfinteractions equals
$\Delta_{1}$, that is, the color has the equation

$\displaystyle\dot{\mathsf{x}}_{1}=f(\mathsf{x}_{1},\cdots,\mathsf{x}_{1}),$

then the Jacobi-matrix of the full system has zeros in the
$\mathfrak{{a}}$-part of the matrix and the coefficient of
$\mathfrak{{c}}^{1}_{1}$ is $\sum_{j=1}^{\Delta_{1}}f_{j}$. In general, if the
number of selfinteraction terms is less than $\Delta_{1}$, let $\iota_{1}$ be
the number of $\mathsf{y}$-variables in the equation for $\mathsf{x}_{1}$.
Then there are at most $\iota_{1}$ nonzero terms in $\mathfrak{{a}}$.

###### Proof.

This follows from Algorithm 4.6. ∎

See Examples 2.1, 8.2 and 8.5 for an illustration of this theorem. In Example
9.1 one has $\iota_{1}=2$.

##  5 The Lie algebra $\mathfrak{net}_{{C},N}$

We are almost done now that the notation and definitions are fixed. The
problem is reduced to simple matrix calculations to compute the structure
constants, observe that they are familiar looking and then draw our
conclusions. We will use the indices as defined in Definition 3.3 to simplify
writing out the proofs. Although simple to obtain, the computation of the
structure constants will come in handy when we come to the dual pair in §7.

###  5.1 The Lie subalgebras $\mathfrak{{b}}$ and $\mathfrak{{c}}$

######  Lemma 5.1.

Let $\mathfrak{{c}}$ be the Lie algebra spanned by the
$\mathfrak{{c}}^{c_{2}}_{c_{1}},0<c_{i}\leq{C}$ as defined in Definition 4.9,
with Lie bracket as defined in Definition 4.1. Then the structure constants
are given by

$\displaystyle[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{c}}^{c_{4}}_{c_{3}}]=\delta^{c_{2}}_{c_{3}}\mathfrak{{c}}^{c_{4}}_{c_{1}}-\delta^{c_{4}}_{c_{1}}\mathfrak{{c}}^{c_{2}}_{c_{3}}$

and $\mathfrak{{c}}\simeq\mathfrak{gl}_{C}$.

###### Proof.

We define
$\mathsf{x}_{k}=\mathbf{x}_{0}^{k},\mathsf{x}^{k}=\bar{\mathbf{x}}^{0}_{k}$
and
$\mathsf{y}_{l}=\mathbf{x}_{i}^{k},\mathsf{y}^{l}=\bar{\mathbf{x}}^{i}_{k},i\in\bar{R}_{k}$,
where $l$ is the position of $\mathbf{x}_{i}^{k}$ in the list $\mathcal{{B}}$.
Then

$\displaystyle[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{c}}^{c_{4}}_{c_{3}}]$
$\displaystyle=$
$\displaystyle[\mathsf{x}_{c_{1}}\otimes\mathsf{x}^{c_{2}},\mathsf{x}_{c_{3}}\otimes\mathsf{x}^{c_{4}}]=\delta^{c_{2}}_{c_{3}}\mathsf{x}_{c_{1}}\otimes\mathsf{x}^{c_{4}}-\delta^{c_{4}}_{c_{1}}\mathsf{x}_{c_{3}}\otimes\mathsf{x}^{c_{2}}=\delta^{c_{2}}_{c_{3}}\mathfrak{{c}}^{c_{4}}_{c_{1}}-\delta^{c_{4}}_{c_{1}}\mathfrak{{c}}^{c_{2}}_{c_{3}}.$

∎

######  Lemma 5.2.

Let $\mathsf{{C}}=\sum_{c=1}^{{C}}\mathfrak{{c}}^{c}_{c}$. Then
$[\mathsf{{C}},\mathfrak{{c}}^{c_{2}}_{c_{1}}]=0$.

###### Proof.

$[\sum_{c=1}^{{C}}\mathfrak{{c}}^{c}_{c},\mathfrak{{c}}^{c_{2}}_{c_{1}}]=\sum_{c=1}^{{C}}(\delta^{c}_{c_{1}}\mathfrak{{c}}^{c_{2}}_{c}-\delta^{c_{2}}_{c}\mathfrak{{c}}^{c}_{c_{1}})=\mathfrak{{c}}^{c_{2}}_{c_{1}}-\mathfrak{{c}}^{c_{2}}_{c_{1}}=0$.
∎

We have now shown that:

######  Theorem 5.3.

The Lie subalgebra $\mathfrak{{c}}$, of $\mathfrak{net}_{{C},N}$ is isomorphic
as a Lie algebra to $\mathfrak{gl}_{C}(\mathbb{R})$, with $\mathsf{{C}}$ as
the identity matrix. The Lie subalgebra
$\hat{\mathfrak{{c}}}=\mathfrak{{c}}/\langle\mathsf{{C}}\rangle\simeq\mathfrak{sl}_{C}$
will turn out to be part of the semisimple component of
$\mathfrak{net}_{{C},N}$. In the case ${C}=N$, this gives a complete
description of the network Lie algebra.

######  Notation 5.4.

The elements in the space spanned by the
$\mathfrak{{b}}^{k,i}_{l,j}=\mathbf{x}_{j}^{l}\otimes\bar{\mathbf{x}}^{i}_{k}$
will be denoted by
$\mathsf{y}_{b_{1}}\otimes\mathsf{y}^{b_{2}}=\mathfrak{{b}}^{b_{2}}_{b_{1}},1\leq
b_{1},b_{2}\leq{B}$.

######  Lemma 5.5.

Let $\mathfrak{{b}}$ be the Lie subalgebra spanned by the
$\mathfrak{{b}}^{b_{2}}_{b_{1}},0<b_{i}\leq{B}$ as defined in Definition 4.9,
with Lie bracket as defined in Definition 4.1. Then the structure constants
are given by

$\displaystyle[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{b}}^{b_{4}}_{b_{3}}]=\delta^{b_{2}}_{b_{3}}\mathfrak{{b}}^{b_{4}}_{b_{1}}-\delta^{b_{4}}_{b_{1}}\mathfrak{{b}}^{b_{2}}_{b_{3}}$

and $\mathfrak{{b}}\simeq\mathfrak{gl}_{B}$. Furthermore,
$[\mathfrak{{b}},\mathfrak{{c}}]=0$.

###### Proof.

See the proof of Lemma 5.2, replacing $\mathsf{x}$ by $\mathsf{y}$. ∎

Let us now define

######  Definition 5.6.

Let $\mathsf{{B}}=\sum_{b=1}^{{B}}\mathfrak{{b}}^{b}_{b}$.

######  Lemma 5.7.

$[\mathsf{{B}},\mathfrak{{b}}]=0$.

###### Proof.

See the proof of Lemma 5.1, replacing $\mathsf{x}$ by $\mathsf{y}$. ∎

######  Theorem 5.8.

The Lie subalgebra $\mathfrak{{b}}$ of $\mathfrak{net}_{{B},N}$ is isomorphic
as a Lie algebra to $\mathfrak{gl}_{{B}}(\mathbb{R})$, with $\mathsf{{B}}$ as
the identity matrix. The Lie subalgebra
$\hat{\mathfrak{{b}}}=\mathfrak{{b}}/\langle\mathsf{{B}}\rangle\simeq\mathfrak{sl}_{{B}}$
will turn out to be part of the semisimple component of
$\mathfrak{net}_{{C},N}$.

######  Corollary 5.9.

It follows from  Corollary 4.8 that $[\mathfrak{{c}},\mathfrak{{b}}]=0$; we
show in §7 that ${\mathfrak{{c}}}\oplus\langle\mathsf{{B}}\rangle$ and
${\mathfrak{{b}}}\oplus\langle\mathsf{{C}}\rangle$ form a dual pair.

###  5.2 The Lie subalgebra $\mathfrak{{a}}$

######  Notation 5.10.

The elements in the space spanned by the
$\mathfrak{{a}}^{k}_{l,j}=\mathbf{x}_{j}^{l}\otimes\bar{\mathbf{x}}^{0}_{k}$
will be denoted by
$\mathsf{y}_{b_{1}}\otimes\mathsf{x}^{c_{1}}=\mathfrak{{a}}^{c_{1}}_{b_{1}},1\leq
b_{1},\leq{B}),1\leq c_{1}\leq{C}$.

######  Lemma 5.11.

The subalgebra $\mathfrak{{a}}$ is abelian, that is,
$[\mathfrak{{a}},\mathfrak{{a}}]=0$.

######  Lemma 5.12.

Structure constants are given by
$[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{a}}^{c_{1}}_{b_{3}}]=\delta^{b_{2}}_{b_{3}}\mathfrak{{a}}^{c_{1}}_{b_{1}}$.
We let $\mathfrak{{a}}^{c_{1}}$ be the space spanned by
$\mathfrak{{a}}^{c_{1}}_{b_{1}}$, then each $\mathfrak{{a}}^{c_{1}}$ is a
standard representation of $\mathfrak{{b}}$.

###### Proof.

One has
$[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{a}}^{c_{1}}_{b_{3}}]=[\mathsf{y}_{b_{1}}\otimes\mathsf{y}^{b_{2}},\mathsf{y}_{b_{3}}\otimes\mathsf{x}^{c_{1}}]=(\mathsf{y}^{b_{2}}\cdot\mathsf{y}_{b_{3}})\mathsf{y}_{b_{1}}\otimes\mathsf{x}^{c_{1}}=\delta^{b_{2}}_{b_{3}}\mathfrak{{a}}^{c_{1}}_{b_{1}}$.
∎

######  Lemma 5.13.

Structure constants are given by
$[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{a}}^{c_{3}}_{b_{1}}]=-\delta^{c_{3}}_{c_{1}}\mathfrak{{a}}^{c_{2}}_{b_{1}}$,
that is, if we let $\mathfrak{{a}}_{b_{1}}$ be the space spanned by
$\mathfrak{{a}}^{c_{1}}_{b_{1}}$, then $\mathfrak{{a}}_{b_{1}}$ is a standard
representation of $\mathfrak{{c}}$.

###### Proof.

$[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{a}}^{c_{3}}_{b_{1}}]=[\mathsf{x}_{c_{1}}\otimes\mathsf{x}^{c_{2}},\mathsf{y}_{b_{1}}\otimes\mathsf{x}^{c_{3}}]=-\delta^{c_{3}}_{c_{1}}\mathfrak{{a}}^{c_{2}}_{b_{1}}$.
∎

######  Lemma 5.14.

$[\mathsf{{C}},\mathfrak{{a}}^{c_{1}}_{b_{1}}]=-\mathfrak{{a}}^{c_{1}}_{b_{1}}$.

###### Proof.

$[\mathsf{{C}},\mathfrak{{a}}^{c_{1}}_{b_{1}}]=\sum_{c=1}^{{C}}[\mathfrak{{c}}^{c}_{c},\mathfrak{{a}}^{c_{1}}_{b1}]=-\sum_{c=1}^{{C}}\delta^{c_{1}}_{c}\mathfrak{{a}}^{c}_{b_{1}}=-\mathfrak{{a}}^{c_{1}}_{b_{1}}$.
∎

######  Corollary 5.15.

Its dimension is $\mathrm{dim}\
\mathfrak{{a}}=\sum_{i=1}^{{C}}\sum_{j=1}^{{B}}1={B}{C}$.

######  Lemma 5.16.

$[\mathsf{{B}},\mathfrak{{a}}^{c_{1}}_{b_{1}}]=\mathfrak{{a}}^{c_{1}}_{b_{1}}$.

###### Proof.

$[\mathsf{{B}},\mathfrak{{a}}^{c_{1}}_{b_{1}}]=\sum_{b=1}^{{B}}[\mathfrak{{b}}^{b}_{b},\mathfrak{{a}}^{c_{1}}_{b_{1}}]=\sum_{b=1}^{{B}}\delta^{b}_{b_{1}}\mathfrak{{a}}^{c_{1}}_{b}=\mathfrak{{a}}^{c_{1}}_{b_{1}}$.
∎

######  Corollary 5.17.

Let $\mathsf{I}=\mathsf{{C}}+\mathsf{{B}}$. Then
$[\mathsf{I},\mathfrak{net}_{{C},N}]=0$, that is,
$\mathsf{I}\in\mathcal{Z}(\mathfrak{net}_{{C},N})$.

###  5.3 The involution
$\theta:\mathfrak{net}_{{C},N}\simeq\mathfrak{net}_{{B},N}$

######  Theorem 5.18.

Let
$\mathfrak{net}_{{C},N}=\mathfrak{{a}}\oplus\mathfrak{{b}}\oplus\mathfrak{{c}}$
and
$\mathfrak{net}_{{B},N}=\mathfrak{\bar{a}}\oplus\mathfrak{\bar{b}}\oplus\mathfrak{\bar{c}}$,
with $\mathrm{dim}\ \mathfrak{\bar{c}}={B}^{2}$, etc. Define

$\displaystyle\theta:\mathfrak{net}_{{C},N}\rightarrow\mathfrak{net}_{{B},N}$

by

$\displaystyle\theta\mathfrak{{a}}^{c_{1}}_{b_{1}}$ $\displaystyle=$
$\displaystyle\mathfrak{\bar{a}}^{b_{1}}_{c_{1}},$
$\displaystyle\theta\mathfrak{{b}}^{b_{2}}_{b_{1}}$ $\displaystyle=$
$\displaystyle-\mathfrak{\bar{c}}^{b_{1}}_{b_{2}},$
$\displaystyle\theta\mathfrak{{c}}^{c_{2}}_{c_{1}}$ $\displaystyle=$
$\displaystyle-\mathfrak{\bar{b}}^{c_{1}}_{c_{2}},$

with $c_{i}=1,\ldots,{C}$ and $b_{i}=1,\ldots,{B}$ for $i=1,2$. Then $\theta$
is an involutative Lie algebra isomorphism.

###### Proof.

Observe that $\theta$ preserves the dimension $N={C}+{B}$. Furthermore,

$\displaystyle\theta[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{c}}^{c_{4}}_{c_{3}}]$
$\displaystyle=$
$\displaystyle\delta^{c_{2}}_{c_{3}}\theta\mathfrak{{c}}^{c_{4}}_{c_{1}}-\delta^{c_{4}}_{c_{1}}\theta\mathfrak{{c}}^{c_{2}}_{c_{3}}=-\delta^{c_{2}}_{c_{3}}\mathfrak{\bar{b}}^{c_{1}}_{c_{4}}+\delta^{c_{4}}_{c_{1}}\mathfrak{\bar{b}}^{c_{3}}_{c_{2}}=[\mathfrak{\bar{b}}^{c_{1}}_{c_{2}},\mathfrak{\bar{b}}^{c_{3}}_{c_{4}}]=[\theta\mathfrak{{c}}^{c_{2}}_{c_{1}},\theta\mathfrak{{c}}^{c_{4}}_{c_{3}}],$
$\displaystyle\theta[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{b}}^{b_{4}}_{b_{3}}]$
$\displaystyle=$
$\displaystyle\delta^{b_{2}}_{b_{3}}\theta\mathfrak{{b}}^{b_{4}}_{b_{1}}-\delta^{b_{4}}_{b_{1}}\theta\mathfrak{{b}}^{b_{2}}_{b_{3}}=-\delta^{b_{2}}_{b_{3}}\mathfrak{\bar{c}}^{b_{1}}_{b_{4}}+\delta^{b_{4}}_{b_{1}}\mathfrak{\bar{c}}^{b_{3}}_{b_{2}}=[\mathfrak{\bar{c}}^{b_{1}}_{b_{2}},\mathfrak{\bar{c}}^{b_{3}}_{b_{4}}]=[\theta\mathfrak{{b}}^{b_{2}}_{b_{1}},\theta\mathfrak{{b}}^{b_{4}}_{b_{3}}],$
$\displaystyle\theta[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{a}}^{c_{3}}_{b_{1}}]$
$\displaystyle=$
$\displaystyle-\delta^{c_{3}}_{c_{1}}\theta\mathfrak{{a}}^{c_{2}}_{b_{1}}=-\delta^{c_{3}}_{c_{1}}\mathfrak{\bar{a}}^{b_{1}}_{c_{2}}=-[\mathfrak{\bar{b}}^{c_{1}}_{c_{2}},\mathfrak{\bar{a}}^{b_{1}}_{c_{3}}]=[\theta\mathfrak{{c}}^{c_{2}}_{c_{1}},\theta\mathfrak{{a}}^{c_{3}}_{b_{1}}],$
$\displaystyle\theta[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{a}}^{c_{1}}_{b_{3}}]$
$\displaystyle=$
$\displaystyle\delta^{b_{2}}_{b_{3}}\theta\mathfrak{{a}}^{c_{1}}_{b_{1}}=\delta^{b_{2}}_{b_{3}}\mathfrak{\bar{a}}^{b_{1}}_{c_{1}}=-[\mathfrak{\bar{c}}^{b_{1}}_{b_{2}},\mathfrak{\bar{a}}^{b_{3}}_{c_{1}}]=[\theta\mathfrak{{b}}^{b_{2}}_{b_{1}},\theta\mathfrak{{a}}^{c_{1}}_{b_{3}}].$

This takes care of all the nonzero brackets. ∎

###  5.4 The Lie subalgebra $\mathfrak{{o}}$

The following lemma is the dual of Lemma 5.11.

######  Lemma 5.19.

The subalgebra $\mathfrak{{o}}$ is abelian, that is,
$[\mathfrak{{o}},\mathfrak{{o}}]=0$.

######  Lemma 5.20.

$[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{o}}^{b_{3}}_{c_{1}}]=-\delta^{b_{3}}_{b_{1}}\mathfrak{{o}}^{b_{2}}_{c_{1}}$.
We let $\mathfrak{{o}}_{c_{1}}$ be the space spanned by
$\mathfrak{{o}}^{b_{1}}_{c_{1}}$, then each $\mathfrak{{o}}_{c_{1}}$ is a
standard representation of $\mathfrak{{b}}$.

###### Proof.

One has
$[\mathfrak{{b}}^{b_{2}}_{b_{1}},\mathfrak{{o}}^{b_{3}}_{c_{1}}]=[\mathsf{y}_{b_{1}}\otimes\mathsf{y}^{b_{2}},\mathsf{x}_{c_{1}}\otimes\mathsf{y}^{b_{3}}]=-(\mathsf{y}^{b_{3}}\cdot\mathsf{y}_{b_{1}})\mathsf{x}_{c_{1}}\otimes\mathsf{y}^{b_{2}}=-\delta^{b_{3}}_{b_{1}}\mathsf{x}_{c_{1}}\otimes\mathsf{y}^{b_{2}}=-\delta^{b_{3}}_{b_{1}}\mathfrak{{o}}^{b_{2}}_{c_{1}}$.
∎

The following Lemma is the dual of Lemma 5.12.

######  Lemma 5.21.

$[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{o}}^{b_{1}}_{c_{3}}]=\delta^{c_{2}}_{c_{3}}\mathfrak{{o}}^{b_{1}}_{c_{1}}$,
that is, if we let $\mathfrak{{o}}^{b_{1}}$ be the space spanned by
$\mathfrak{{o}}^{b_{1}}_{c_{1}}$, then $\mathfrak{{o}}^{b_{1}}$ is a standard
representation of $\mathfrak{{c}}$.

###### Proof.

$[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{o}}^{b_{1}}_{c_{3}}]=[\mathsf{x}_{c_{1}}\otimes\mathsf{x}^{c_{2}},\mathsf{x}_{c_{3}}\otimes\mathsf{y}^{b_{1}}]=\delta^{c_{2}}_{c_{3}}\mathfrak{{o}}^{b_{1}}_{c_{1}}$.
∎

######  Lemma 5.22.

$[\mathfrak{{o}}^{b_{1}}_{c_{1}},\mathfrak{{a}}^{c_{2}}_{b_{2}}]=\delta^{b_{1}}_{b_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}-\delta^{c_{2}}_{c_{1}}\mathfrak{{b}}^{b_{1}}_{b_{2}}$.

###### Proof.

$[\mathfrak{{o}}^{b_{1}}_{c_{1}},\mathfrak{{a}}^{c_{2}}_{b_{2}}]=[\mathsf{x}_{c_{1}}\otimes\mathsf{y}^{b_{1}},\mathsf{y}_{b_{2}}\otimes\mathsf{x}^{c_{2}}]=\delta^{b_{1}}_{b_{2}}\mathsf{x}_{c_{1}}\otimes\mathsf{x}^{c_{2}}-\delta^{c_{2}}_{c_{1}}\mathsf{y}_{b_{2}}\otimes\mathsf{y}^{b_{1}}=\delta^{b_{1}}_{b_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}-\delta^{c_{2}}_{c_{1}}\mathfrak{{b}}^{b_{1}}_{b_{2}}$.
∎

##  6 Conclusion: structure of the Lie algebra

######  Theorem 6.1.

Assume $0<{C}<N$. Then the Levi decomposition of $\mathfrak{net}_{{C},N}$, is
given by a solvable part

$\displaystyle\begin{bmatrix}\langle\mathsf{{C}}\rangle&0\\\
\mathfrak{{a}}&\langle\mathsf{{B}}\rangle\end{bmatrix}$

and the semisimple part

$\displaystyle\begin{bmatrix}\mathfrak{{c}}/\langle\mathsf{{C}}\rangle&0\\\
0&\mathfrak{{b}}/\langle\mathsf{{B}}\rangle\end{bmatrix}\simeq\begin{bmatrix}\mathfrak{sl}_{C}&0\\\
0&\mathfrak{sl}_{{B}}\end{bmatrix}.$

The Lie algebra $\mathfrak{net}_{{C},N}$ is reductive if $\mathfrak{{a}}=0$.
If $\mathfrak{{a}}\neq 0$, it might still be removable by coordinate
transformations, especially when all eigenvalues are different. The dimension
of $\mathfrak{{a}}$ is (cf. Corollary 5.15 )

$\displaystyle{B}{C}$ $\displaystyle=$ $\displaystyle\mathrm{dim}\
\mathfrak{Gr}_{{C},N}.$

The dimension of the semisimple part is ${B}^{2}+{C}^{2}-2$ and of the
solvable part ${C}{B}+2$, where the $2$ is the dimension of the first homology
$H_{1}(\mathfrak{net}_{{C},N},\mathfrak{net}_{{C},N})=\mathfrak{net}_{{C},N}/[\mathfrak{net}_{{C},N},\mathfrak{net}_{{C},N}]\simeq\langle\mathsf{{B}},\mathsf{{C}}\rangle$,
cf. Corollary 6.2. We can view an element of $\mathfrak{net}_{{C},N}$ as a
block matrix:

$\displaystyle\begin{bmatrix}\begin{pmatrix}\mathsf{{c}}^{1}_{1}&\cdots&\mathsf{{c}}^{1}_{{C}}\\\
\vdots&\ddots&\vdots\\\
\mathsf{{c}}^{{C}}_{1}&\cdots&\mathsf{{c}}^{{C}}_{{C}}\end{pmatrix}&\begin{pmatrix}0&\cdots&0\\\
\vdots&\ddots&\vdots\\\ 0&\cdots&0\end{pmatrix}\\\
\begin{pmatrix}\mathsf{{a}}^{1}_{1}&\cdots&\mathsf{{a}}^{1}_{{C}}\\\
\vdots&\ddots&\vdots\\\
\mathsf{{a}}^{{B}}_{1}&\cdots&\mathsf{{a}}^{{B}}_{{C}}\end{pmatrix}&\begin{pmatrix}\mathsf{{b}}^{1}_{1}&\cdots&\mathsf{{b}}^{1}_{{B}}\\\
\vdots&\ddots&\vdots\\\
\mathsf{{b}}^{{B}}_{1}&\cdots&\mathsf{{b}}^{{B}}_{{B}}\end{pmatrix}\end{bmatrix}=\begin{bmatrix}\mathfrak{{c}}&0\\\
\mathfrak{{a}}&\mathfrak{{b}}\end{bmatrix}=\begin{bmatrix}\mathfrak{gl}_{C}&0\\\
\mathfrak{Gr}_{{C},N}&\mathfrak{gl}_{{B}}\end{bmatrix}.$

There are at most $\frac{N}{2}+1$ nonisomorphic $\mathfrak{net}_{{C},N}$ for
any given $N\in\mathbb{N}$, Moreover, $\min({C},{B})$ and $N$ fix the
isomorphy class. Every colored network has a Lie algebra that is isomorphic to
a subalgebra of one of these $\mathfrak{net}_{{C},N}$. The classification of
these subalgebras is in general nontrivial, cf. [20].

######  Corollary 6.2.

The first homology can now be computed:

$\displaystyle
H_{1}(\mathfrak{net}_{{C},N},\mathfrak{net}_{{C},N})=\langle\mathsf{{B}},\mathsf{{C}}\rangle.$

###### Proof.

The semisimple part is acyclic, and
$\mathfrak{{a}}=[\mathsf{{B}},\mathfrak{{a}}]$. We see that $\mathrm{dim}\
H_{1}(\mathfrak{net}_{{C},N},\mathfrak{net}_{{C},N})=2$, ‘explaining’ this
factor in the formula for the dimension of the semisimple and solvable part.
If ${C}=N$, this factor would have been $1$ since

$\displaystyle
H_{1}(\mathfrak{net}_{N,N},\mathfrak{net}_{N,N})=\langle\mathsf{{C}}\rangle=\langle\mathrm{Id}_{\mathbb{R}^{N}}\rangle.$

∎

##  7 Dual pair

######  Definition 7.1.

A dual pair in a Lie algebra $\mathfrak{g}$ is a pair of subalgebras
$\mathfrak{g}_{i},i=0,1$, such that
$\mathcal{Z}(\mathfrak{g}_{i+1\mod{2}})=\mathfrak{g}_{i}$, where the
irreducible representations are formed by tensorproducts of irreducible
representations of the respective elements in the pair.

######  Theorem 7.2.

The subalgebras $\langle\mathsf{{B}}\rangle\oplus\mathfrak{{c}}$ and
$\langle\mathsf{{C}}\rangle\oplus\mathfrak{{b}}$ form a dual pair, that is,
they are the center of one another:

$\displaystyle\mathcal{Z}(\langle\mathsf{{B}}\rangle\oplus{\mathfrak{{c}}})=\langle\mathsf{{C}}\rangle\oplus{\mathfrak{{b}}}$

and

$\displaystyle\mathcal{Z}(\langle\mathsf{{C}}\rangle\oplus{\mathfrak{{b}}})=\langle\mathsf{{B}}\rangle\oplus{\mathfrak{{c}}}.$

###### Proof.

Let

$\displaystyle\mathsf{A}=\sum_{c_{1},c_{2}=1}^{C}\mathsf{{c}}^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}+\sum_{c_{1}=1}^{C}\sum_{b_{1}=1}^{{B}}\mathsf{{a}}^{c_{1}}_{b_{1}}\mathfrak{{a}}^{b_{1}}_{c_{1}}+\sum_{b_{1},b_{2}=1}^{{B}}\mathsf{{b}}^{b_{1}}_{b_{2}}\mathfrak{{b}}^{b_{2}}_{b_{1}}.$

be a generic element in $\mathfrak{net}_{{C},N}$, and let
$\beta\mathsf{{B}}+\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}$
be a generic element of $\langle\mathsf{{B}}\rangle\oplus{\mathfrak{{c}}}$. We
compute
$[{\beta\mathsf{{B}}+\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}},\mathsf{A}]$:

$\displaystyle[{\beta\mathsf{{B}}+\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}},\mathsf{A}]$
$\displaystyle=$
$\displaystyle\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\sum_{{c_{3}},{c_{4}}=1}^{C}\mathsf{{c}}^{{c_{3}}}_{{c_{4}}}[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{c}}^{{c_{4}}}_{{c_{3}}}]+\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\sum_{{c_{3}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\mathsf{{a}}^{{c_{3}}}_{{b_{4}}}[\mathfrak{{c}}^{c_{2}}_{c_{1}},\mathfrak{{a}}^{{b_{4}}}_{{c_{3}}}]+\beta[\mathsf{{B}},\sum_{{c_{3}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\mathsf{{a}}^{{c_{3}}}_{{b_{4}}}\mathfrak{{a}}^{{b_{4}}}_{{c_{3}}}]$
$\displaystyle=$
$\displaystyle\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\sum_{{c_{3}},{c_{4}}=1}^{C}\mathsf{{c}}^{{c_{3}}}_{{c_{4}}}(\delta^{c_{2}}_{{c_{3}}}\mathfrak{{c}}^{{c_{4}}}_{c_{1}}-\delta^{{c_{4}}}_{c_{1}}\mathfrak{{c}}^{c_{2}}_{{c_{3}}})+\sum_{c_{1},c_{2}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{c_{2}}\sum_{{c_{3}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\mathsf{{a}}^{{c_{3}}}_{{b_{4}}}\delta^{c_{2}}\mathfrak{{a}}^{{b_{4}}}_{c_{1}}+\beta\sum_{{c_{3}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\mathsf{{a}}^{{c_{3}}}_{{b_{4}}}\mathfrak{{a}}^{{b_{4}}}_{{c_{3}}}$
$\displaystyle=$
$\displaystyle\sum_{{c_{4}},c_{1},{c_{2}}=1}^{C}\bar{\mathsf{{c}}}^{c_{1}}_{{c_{2}}}\mathsf{{c}}^{{c_{2}}}_{{c_{4}}}\mathfrak{{c}}^{{c_{4}}}_{c_{1}}-\mathsf{{c}}^{{c_{1}}}_{c_{2}}\bar{\mathsf{{c}}}^{c_{2}}_{{c_{4}}}\mathfrak{{c}}^{{c_{4}}}_{{c_{1}}}+\sum_{c_{1},{c_{2}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\bar{\mathsf{{c}}}^{c_{1}}_{{c_{2}}}\mathsf{{a}}^{{c_{2}}}_{{b_{4}}}\mathfrak{{a}}^{{b_{4}}}_{c_{1}}+\beta\sum_{{c_{3}}=1}^{C}\sum_{{b_{4}}=1}^{{B}}\mathsf{{a}}^{{c_{3}}}_{{b_{4}}}\mathfrak{{a}}^{{b_{4}}}_{{c_{3}}}.$
$\displaystyle=$
$\displaystyle\sum_{{c_{1}},{c_{2}}=1}^{C}[\bar{\mathsf{{c}}},\mathsf{{c}}]_{c_{2}}^{c_{1}}\mathfrak{{c}}^{{c_{2}}}_{{c_{1}}}+\sum_{c_{1}=1}^{C}\sum_{{b_{1}}=1}^{{B}}\left((\bar{\mathsf{{c}}}+\beta)\cdot\mathsf{{a}}\right)_{b_{1}}^{c_{1}}\mathfrak{{a}}^{{b_{1}}}_{c_{1}}.$

For $\mathsf{A}$ to be in
$\mathcal{Z}(\langle\mathsf{{B}}\rangle\oplus\hat{\mathfrak{{c}}})$, this
expression should be zero for all
$\beta\mathsf{{B}}+\sum_{c_{1},{c_{2}}=1}^{C}\bar{\mathsf{{c}}}^{b_{1}}_{{c_{2}}}\mathfrak{{c}}^{{c_{2}}}_{b_{1}}$.
This implies that the $\mathsf{{c}}$-matrix has to commute with all
$\bar{\mathsf{{c}}}$-matrices, that is, it equals a multiple of the identity
of $\mathfrak{{c}}$. In other words,
$\mathsf{{c}}^{c_{1}}_{c_{2}}=\alpha\delta^{c_{1}}_{c_{2}}$ and

$\displaystyle\sum_{c_{1},c_{2}=1}^{C}\mathsf{{c}}^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}$
$\displaystyle=$
$\displaystyle\sum_{c_{1},c_{2}=1}^{C}\alpha\delta^{c_{1}}_{c_{2}}\mathfrak{{c}}^{c_{2}}_{c_{1}}=\sum_{c_{1}=1}^{C}\alpha\mathfrak{{c}}^{c_{1}}_{c_{1}}=\alpha\mathsf{{C}}.$

The vectors spanning $\mathsf{{a}}$ have to be eigenvectors of all
$\bar{\mathsf{{c}}}$-matrices for all eigenvalues $-\beta$. This forces the
$\mathsf{{a}}^{c}_{{b}},c=1,\ldots,{C},b=1,\ldots,{B}$ to be zero. We conclude
that

$\displaystyle\mathcal{Z}(\langle\mathsf{{B}}\rangle\oplus{\mathfrak{{c}}})=\langle\mathsf{{C}}\rangle\oplus{\mathfrak{{b}}}.$

The dual statement, in the sense of Theorem 5.18,

$\displaystyle\mathcal{Z}(\langle\mathsf{{C}}\rangle\oplus{\mathfrak{{b}}})=\langle\mathsf{{B}}\rangle\oplus{\mathfrak{{c}}},$

has an analogous proof. ∎

##  8 Examples

In this section, we write out the relevant change of coordinates for given
$2\leq N\leq 8$ and ${C}$. Higher dimensional examples would be just as
mechanical, but take up lots of space, too much for our taste. Then we
introduce a vector field and compute its Jacobi-matrix in terms of
$\mathsf{x},\mathsf{y}$-coordinates, and express the result in terms of
$\mathfrak{{a}},\mathfrak{{b}}$ and $\mathfrak{{c}}$, illustrating how by
simple means a concrete equation can be fit in the theoretical description of
the network Lie algebra $\mathfrak{net}_{{C},N}$. This can then be used as a
starting point for the bifurcation analysis. Notice that the Algorithms 3.6
and 4.6 that define the Network Elimination, do not in any way depend on the
eigenvalues of the linearization, this dependence only comes in when we do the
Jordan-Chevalley decomposition and construct the Jacobson-Morozov triple.

###  8.1 Example with $N=2$

####  8.1.1 Example with ${C}=1$

The relation between the coordinates is given by

$\displaystyle\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\end{pmatrix}$
$\displaystyle=$
$\displaystyle\exp(\mathbf{x}_{0}^{1}\otimes\bar{\mathbf{x}}^{1}_{1}\triangleright)\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}\\\ \bar{\mathbf{x}}^{0}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\end{pmatrix}=\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}+\mathbf{x}_{0}^{1}\\\
\bar{\mathbf{x}}^{0}_{1}-\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\end{pmatrix}=\begin{pmatrix}\mathsf{x}_{1}\\\
\mathsf{y}_{1}+\mathsf{x}_{1}\\\ \mathsf{x}^{1}-\mathsf{y}^{1}\\\
\mathsf{y}^{1}\end{pmatrix}.$ (8.1)

######  Example 8.1.

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{1}},{x_{1}},{x_{2}},{x_{2}}),$ $\displaystyle\dot{x}_{2}$
$\displaystyle=$ $\displaystyle f^{1}({x_{2}},{x_{1}},{x_{1}},{x_{2}}),$

with picture

12$>$$>$$<$$<$$x_{1}$$x_{2}$

The Jacobi-matrix is

$\displaystyle\begin{pmatrix}{f^{1}_{2}}+{f^{1}_{1}}&{f^{1}_{4}}+{f^{1}_{3}}\\\
{f^{1}_{3}}+{f^{1}_{2}}&{f^{1}_{4}}+{f^{1}_{1}}\end{pmatrix}.$

The relation between the coordinates (using Algorithm 3.6) is given by

$\displaystyle\begin{pmatrix}x_{2}\\\ x_{1}\\\
\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\ \mathsf{v}_{1}\\\
\end{pmatrix}.$

Using Algorithm 4.6 we obtain:

$\displaystyle\mathsf{J}_{\mathfrak{{c}}}$ $\displaystyle=$
$\displaystyle({f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}){\mathfrak{{c}}^{1}_{1}},$
$\displaystyle\mathsf{J}_{\mathfrak{{a}}}$ $\displaystyle=$
$\displaystyle({f^{1}_{3}}+{f^{1}_{2}}){\mathfrak{{a}}^{1}_{1}},$
$\displaystyle\mathsf{J}_{\mathfrak{{b}}}$ $\displaystyle=$
$\displaystyle(-{f^{1}_{3}}+{f^{1}_{1}}){\mathfrak{{b}}^{1}_{1}}.$

with matrix

$\displaystyle\begin{bmatrix}{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}&0\\\
{f^{1}_{3}}+{f^{1}_{2}}&-{f^{1}_{3}}+{f^{1}_{1}}\end{bmatrix}.$

The Jacobian is

$\displaystyle({f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}})(-{f^{1}_{3}}+{f^{1}_{1}}).$

###  8.2 Examples with $N=3$

####  8.2.1 Example with ${C}=1$

The relation between the coordinates is given by

$\displaystyle\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\mathbf{u}_{2}^{1}\\\ \bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\\\
\bar{\mathbf{u}}^{2}_{1}\end{pmatrix}$ $\displaystyle=$
$\displaystyle\exp(\mathbf{x}_{0}^{1}\otimes\sum_{q=1}^{2}\bar{\mathbf{x}}^{q}_{1}\triangleright)\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}\\\ \mathbf{x}_{2}^{1}\\\ \bar{\mathbf{x}}^{0}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{2}_{1}\end{pmatrix}=\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}+\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{2}^{1}+\mathbf{x}_{0}^{1}\\\
\bar{\mathbf{x}}^{0}_{1}-(\bar{\mathbf{x}}^{1}_{1}+\bar{\mathbf{x}}^{2}_{1})\\\
\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{2}_{1}\end{pmatrix}=\begin{pmatrix}\mathsf{x}_{1}\\\
\mathsf{y}_{1}+\mathsf{x}_{1}\\\ \mathsf{y}_{2}+\mathsf{x}_{1}\\\
\mathsf{x}^{1}-(\mathsf{y}^{1}+\mathsf{y}^{2})\\\ \mathsf{y}^{1}\\\
\mathsf{y}^{2}\end{pmatrix}.$ (8.2)

######  Example 8.2.

In [19, §11.2] the following $3$-cell example with $\Delta_{1}=3$ is treated:

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{1}},{x_{1}},{x_{1}}),$ $\displaystyle\dot{x}_{2}$ $\displaystyle=$
$\displaystyle f^{1}({x_{2}},{x_{1}},{x_{1}}),$ $\displaystyle\dot{x}_{3}$
$\displaystyle=$ $\displaystyle f^{1}({x_{3}},{x_{2}},{x_{1}}),$

with picture

123$>$$>$$>$$<$$x_{1}$$x_{2}$$x_{3}$

The Jacobi-matrix is

$\displaystyle\begin{pmatrix}{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}&0&0\\\
{f^{1}_{3}}+{f^{1}_{2}}&{f^{1}_{1}}&0\\\
{f^{1}_{3}}&{f^{1}_{2}}&{f^{1}_{1}}\end{pmatrix}.$

The relation between the coordinates (using Algorithm 3.6) is given by

$\displaystyle\begin{pmatrix}x_{1}\\\ x_{3}\\\ x_{2}\\\
\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\mathbf{u}_{2}^{1}\\\ \end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\
\mathsf{v}_{1}\\\ \mathsf{v}_{2}\\\ \end{pmatrix}.$

Using Algorithm 4.6 we obtain:

$\displaystyle\mathsf{J}_{\mathfrak{{c}}}$ $\displaystyle=$
$\displaystyle({f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}){\mathfrak{{c}}^{1}_{1}},$
$\displaystyle\mathsf{J}_{\mathfrak{{b}}}$ $\displaystyle=$
$\displaystyle{f^{1}_{1}}{\mathfrak{{b}}^{1}_{1}}+{f^{1}_{2}}{\mathfrak{{b}}^{1}_{2}}+{f^{1}_{1}}{\mathfrak{{b}}^{2}_{2}},$

with matrix

$\displaystyle\begin{bmatrix}{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}&0&0\\\
0&{f^{1}_{1}}&0\\\ 0&{f^{1}_{2}}&{f^{1}_{1}}\end{bmatrix}.$

The Jacobian is

$\displaystyle({f^{1}_{1}})^{2}({f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}).$

####  8.2.2 Example with ${C}=2$

The relation between the coordinates is given by

$\displaystyle\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\mathbf{u}_{0}^{2}\\\ \bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\\\
\bar{\mathbf{u}}^{0}_{2}\end{pmatrix}$ $\displaystyle=$
$\displaystyle\exp(\sum_{c=1}^{2}\mathbf{x}_{0}^{c}\otimes\sum_{q=1}^{\bar{n}_{c}}\bar{\mathbf{x}}^{q}_{c}\triangleright)\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}\\\ \mathbf{x}_{0}^{2}\\\ \bar{\mathbf{x}}^{0}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{0}_{2}\end{pmatrix}=\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}+\mathbf{x}_{0}^{1}\\\ \mathbf{x}_{0}^{2}\\\
\bar{\mathbf{x}}^{0}_{1}-\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{0}_{2}\end{pmatrix}=\begin{pmatrix}\mathsf{x}_{1}\\\
\mathsf{y}_{1}+\mathsf{x}_{1}\\\ \mathsf{x}_{2}\\\
\mathsf{x}^{1}-\mathsf{y}^{1}\\\ \mathsf{y}^{1}\\\
\mathsf{x}^{2}\end{pmatrix}.$

######  Example 8.3.

We write down the basis form

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(x_{1},x_{1},x_{2},x_{2};x_{3}),$ $\displaystyle\dot{x}_{2}$
$\displaystyle=$ $\displaystyle f^{1}(x_{2},x_{1},x_{1},x_{2};x_{3}),$
$\displaystyle\dot{x}_{3}$ $\displaystyle=$ $\displaystyle
f^{2}(x_{1},x_{2};x_{3}).$

Or, equivalently (cf. Notation 1.3),

123$>$$>$$>$$<$$>$$<$$<$$<$$x_{1}$$x_{2}$$x_{3}$

Using Algorithm 3.6 we obtain the identification

$\displaystyle\begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\ x^{1}\\\ x^{2}\\\
x^{3}\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\mathbf{u}_{2}^{1}\\\ \bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\\\
\bar{\mathbf{u}}^{2}_{1}\end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\
\mathsf{v}_{1}\\\ \mathsf{u}_{2}\\\ \mathsf{u}^{1}\\\ \mathsf{v}^{1}\\\
\mathsf{u}^{2}\end{pmatrix}.$

Applying the transformation we obtain

$\displaystyle\dot{\mathsf{x}}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{2})-f^{1}(\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{2}),$
$\displaystyle\dot{\mathsf{y}}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{2}),$
$\displaystyle\dot{\mathsf{x}}_{1}$ $\displaystyle=$ $\displaystyle
f^{2}(\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{2}).$

The Jacobi-matrix is spanned by the basis of the $\sigma$s:

$\displaystyle\mathsf{J}$ $\displaystyle=$
$\displaystyle(f^{1}_{1}\mathbf{u}_{0}^{1}+f^{1}_{2}\mathbf{u}_{0}^{1}+(f^{1}_{3}+f^{1}_{4})\mathbf{u}_{1}^{1}+f^{1}_{5}\mathbf{u}_{0}^{2})\otimes\bar{\mathbf{u}}^{0}_{1}$
$\displaystyle+$
$\displaystyle(f^{1}_{1}\mathbf{u}_{1}^{1}+f^{1}_{2}\mathbf{u}_{0}^{1}+f^{1}_{3}\mathbf{u}_{0}^{1}+f^{1}_{4}\mathbf{u}_{1}^{1}+f^{1}_{5}\mathbf{u}_{0}^{2})\otimes\bar{\mathbf{u}}^{1}_{1}$
$\displaystyle+$
$\displaystyle(f^{2}_{1}\mathbf{u}_{0}^{1}+f^{2}_{2}\mathbf{u}_{1}^{1}+f^{1}_{3}\mathbf{u}_{0}^{2})\otimes\bar{\mathbf{u}}^{0}_{2}.$

Then

$\displaystyle\mathsf{J}$ $\displaystyle=$
$\displaystyle(f^{1}_{1}\mathsf{x}_{1}+f^{1}_{2}\mathsf{x}_{1}+(f^{1}_{3}+f^{1}_{4})(\mathsf{y}_{1}+\mathsf{x}_{1})+f^{1}_{5}\mathsf{x}_{2})\otimes(\mathsf{x}^{1}-\mathsf{y}^{1})$
$\displaystyle+(f^{1}_{1}(\mathsf{y}_{1}+\mathsf{x}_{1})+f^{1}_{2}\mathsf{x}_{1}+f^{1}_{3}\mathsf{x}_{1}+f^{1}_{4}(\mathsf{y}_{1}+\mathsf{x}_{1})+f^{1}_{5}\mathsf{x}_{2})\otimes\mathsf{y}^{1}$
$\displaystyle+(f^{2}_{1}\mathsf{x}_{1}+f^{2}_{2}(\mathsf{y}_{1}+\mathsf{x}_{1})+f^{2}_{3}\mathsf{x}_{2})\otimes\mathsf{x}^{2}$
$\displaystyle=$
$\displaystyle((f^{1}_{1}+f^{1}_{2}+f^{1}_{3}+f^{1}_{4})\mathsf{x}_{1}+(f^{1}_{3}+f^{1}_{4})\mathsf{y}_{1}+f^{1}_{5}\mathsf{x}_{2})\otimes\mathsf{x}^{1}$
$\displaystyle+(f^{1}_{1}-f^{1}_{3})\mathsf{y}_{1}\otimes\mathsf{y}^{1}$
$\displaystyle+((f^{2}_{1}+f^{2}_{2})\mathsf{x}_{1}+f^{2}_{2}\mathsf{y}_{1}+f^{2}_{3}\mathsf{x}_{2})\otimes\mathsf{x}^{2}$
$\displaystyle=$
$\displaystyle(f^{1}_{1}+f^{1}_{2}+f^{1}_{3}+f^{1}_{4})\mathfrak{{c}}^{1}_{1}+f^{1}_{5}\mathfrak{{c}}^{1}_{2}+(f^{2}_{1}+f^{2}_{2})\mathfrak{{c}}^{2}_{1}+f^{2}_{3}\mathfrak{{c}}^{2}_{2}+(f^{1}_{1}-f^{1}_{3})\mathfrak{{b}}^{1}_{1}+(f^{1}_{3}+f^{1}_{4})\mathfrak{{a}}^{1}_{1}+f^{2}_{2}\mathfrak{{a}}^{2}_{1}.$

The Jacobi-matrix is

$\displaystyle\mathsf{J}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}f^{1}_{1}+f^{1}_{2}+f^{1}_{3}+f^{1}_{4}&f^{2}_{1}+f^{2}_{2}&0\\\
f^{1}_{5}&f^{2}_{3}&0\\\
f^{1}_{3}+f^{1}_{4}&f^{2}_{2}&f^{1}_{1}-f^{1}_{3}\end{bmatrix}.$

Notice that in the feedforward case, with $f^{1}_{5}=0$, the eigenvalues are
on the diagonal.

###  8.3 Example with $N=4$

####  8.3.1 Example with ${C}=2$

The relation between the coordinates is given by

######  Example 8.4.

Let

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(x_{1},x_{2};x_{3},x_{4}),$ $\displaystyle\dot{x}_{2}$ $\displaystyle=$
$\displaystyle f^{1}(x_{2},x_{1};x_{4},x_{3}),$ $\displaystyle\dot{x}_{3}$
$\displaystyle=$ $\displaystyle f^{2}(x_{3},x_{4};x_{1},x_{2}),$
$\displaystyle\dot{x}_{4}$ $\displaystyle=$ $\displaystyle
f^{2}(x_{4},x_{3};x_{2},x_{1}),$

with $\Delta_{1}=\Delta_{2}=4$ and picture (cf. Notation 1.3)

1234$>$$>$$>$$<$$<$$>$$>$$>$$>$$<$$<$$<$$x_{1}$$x_{2}$$x_{3}$$x_{4}$

Using Algorithm 3.6 we obtain the identification

$\displaystyle\begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\ x_{4}\\\ x^{1}\\\
x^{2}\\\ x^{3}\\\ x^{4}\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\
\mathbf{u}_{1}^{1}\\\ \mathbf{u}_{0}^{2}\\\ \mathbf{u}_{1}^{2}\\\
\bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\\\
\bar{\mathbf{u}}^{0}_{2}\\\
\bar{\mathbf{u}}^{1}_{2}\end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\
\mathsf{v}_{1}\\\ \mathsf{u}_{2}\\\ \mathsf{v}_{2}\\\ \mathsf{u}^{1}\\\
\mathsf{v}^{1}\\\ \mathsf{u}^{2}\\\ \mathsf{v}^{2}\end{pmatrix}.$

Applying the transformation we obtain

$\displaystyle\dot{\mathsf{x}}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{2},\mathsf{y}_{2}+\mathsf{x}_{2})-f^{1}(\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{y}_{2}+\mathsf{x}_{2},\mathsf{x}_{2}),$
$\displaystyle\dot{\mathsf{y}}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}(\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1},\mathsf{y}_{2}+\mathsf{x}_{2},\mathsf{x}_{2}),$
$\displaystyle\dot{\mathsf{x}}_{2}$ $\displaystyle=$ $\displaystyle
f^{2}(\mathsf{x}_{2},\mathsf{y}_{2}+\mathsf{x}_{2},\mathsf{x}_{1},\mathsf{y}_{1}+\mathsf{x}_{1})-f^{2}(\mathsf{y}_{2}+\mathsf{x}_{2},\mathsf{x}_{2},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1}),$
$\displaystyle\dot{\mathsf{y}}_{2}$ $\displaystyle=$ $\displaystyle
f^{2}(\mathsf{y}_{2}+\mathsf{x}_{2},\mathsf{x}_{2},\mathsf{y}_{1}+\mathsf{x}_{1},\mathsf{x}_{1}).$

The Jacobi-matrix is

$\displaystyle\mathsf{J}$ $\displaystyle=$
$\displaystyle(f^{1}_{1}\mathbf{u}_{0}^{1}+f^{1}_{2}\mathbf{u}_{1}^{1}+f^{1}_{3}\mathbf{u}_{0}^{2}+f^{1}_{4}\mathbf{u}_{1}^{2})\otimes\bar{\mathbf{u}}^{0}_{1}+(f^{1}_{1}\mathbf{u}_{1}^{1}+f^{1}_{2}\mathbf{u}_{0}^{1}+f^{1}_{3}\mathbf{u}_{1}^{2}+f^{1}_{4}\mathbf{u}_{0}^{2})\otimes\bar{\mathbf{u}}^{1}_{1}$
$\displaystyle+$
$\displaystyle(f^{2}_{1}\mathbf{u}_{0}^{2}+f^{2}_{2}\mathbf{u}_{1}^{2}+f^{2}_{3}\mathbf{u}_{0}^{1}+f^{2}_{4}\mathbf{u}_{1}^{1})\otimes\bar{\mathbf{u}}^{0}_{2}+(f^{2}_{1}\mathbf{u}_{1}^{2}+f^{2}_{2}\mathbf{u}_{0}^{2}+f^{2}_{3}\mathbf{u}_{1}^{1}+f^{2}_{4}\mathbf{u}_{0}^{1})\otimes\bar{\mathbf{u}}^{1}_{2}$
$\displaystyle=$
$\displaystyle(f^{1}_{1}\mathsf{x}_{1}+f^{1}_{2}(\mathsf{y}_{1}+\mathsf{x}_{1})+f^{1}_{3}\mathsf{x}_{2}+f^{1}_{4}(\mathsf{y}_{2}+\mathsf{x}_{2}))\otimes\mathsf{x}^{1}+(f^{1}_{1}\mathsf{y}_{1}-f^{1}_{2}\mathsf{y}_{1}+f^{1}_{3}\mathsf{y}_{2}-f^{1}_{4}\mathsf{y}_{2})\otimes\mathsf{y}^{1}$
$\displaystyle+$
$\displaystyle(f^{2}_{1}\mathsf{x}_{2}+f^{2}_{2}(\mathsf{y}_{2}+\mathsf{x}_{2})+f^{2}_{3}\mathsf{x}_{1}+f^{2}_{4}(\mathsf{y}_{1}+\mathsf{x}_{1}))\otimes\mathsf{x}^{2}+(f^{2}_{1}\mathsf{y}_{2}-f^{2}_{2}\mathsf{y}_{2}+f^{2}_{3}\mathsf{y}_{1}-f^{2}_{4}\mathsf{y}_{1})\otimes\mathsf{y}^{2}$
$\displaystyle=$ $\displaystyle
f^{1}_{1}\mathfrak{{c}}^{1}_{1}+f^{1}_{2}(\mathfrak{{a}}^{1}_{1}+\mathfrak{{c}}^{1}_{1})+f^{1}_{3}\mathfrak{{c}}^{1}_{2}+f^{1}_{4}(\mathfrak{{a}}^{1}_{2}+\mathfrak{{c}}^{1}_{2})+f^{1}_{1}\mathfrak{{b}}^{1}_{1}-f^{1}_{2}\mathfrak{{b}}^{1}_{1}+f^{1}_{3}\mathfrak{{b}}^{1}_{2}-f^{1}_{4}\mathfrak{{b}}^{1}_{2}$
$\displaystyle+$
$\displaystyle(f^{2}_{1}\mathfrak{{c}}^{2}_{2}+f^{2}_{2}(\mathfrak{{a}}^{2}_{2}+\mathfrak{{c}}^{2}_{2})+f^{2}_{3}\mathfrak{{c}}^{2}_{1}+f^{2}_{4}(\mathfrak{{a}}^{2}_{1}+\mathfrak{{c}}^{2}_{1})+f^{2}_{1}\mathfrak{{b}}^{2}_{2}-f^{2}_{2}\mathfrak{{b}}^{2}_{2}+f^{2}_{3}\mathfrak{{b}}^{2}_{1}-f^{2}_{4}\mathfrak{{b}}^{2}_{1}$
$\displaystyle=$
$\displaystyle(f^{1}_{1}+f^{1}_{2})\mathfrak{{c}}^{1}_{1}+(f^{2}_{3}+f^{2}_{4})\mathfrak{{c}}^{2}_{1}+(f^{1}_{3}+f^{1}_{4})\mathfrak{{c}}^{1}_{2}+(f^{2}_{1}+f^{2}_{2})\mathfrak{{c}}^{2}_{2}$
$\displaystyle+$
$\displaystyle(f^{1}_{1}-f^{1}_{2})\mathfrak{{b}}^{1}_{1}+(f^{2}_{3}-f^{2}_{4})\mathfrak{{b}}^{2}_{1}+(f^{1}_{3}-f^{1}_{4})\mathfrak{{b}}^{1}_{2}+(f^{2}_{1}-f^{2}_{2})\mathfrak{{b}}^{2}_{2}$
$\displaystyle+$ $\displaystyle
f^{1}_{2}\mathfrak{{a}}^{1}_{1}+f^{2}_{4}\mathfrak{{a}}^{2}_{1}+f^{1}_{4}\mathfrak{{a}}^{1}_{2}+f^{2}_{2}\mathfrak{{a}}^{2}_{2}.$

Then the matrix is

$\displaystyle\begin{bmatrix}f^{1}_{1}+f^{1}_{2}&f^{2}_{3}+f^{2}_{4}&0&0\\\
f^{1}_{3}+f^{1}_{4}&f^{2}_{1}+f^{2}_{2}&0&0\\\
f^{1}_{2}&f^{2}_{4}&f^{1}_{1}-f^{1}_{2}&f^{2}_{3}-f^{2}_{4}\\\
f^{1}_{4}&f^{2}_{2}&f^{1}_{3}-f^{1}_{4}&f^{2}_{1}-f^{2}_{2}\end{bmatrix}.$

###  8.4 Example with $N=5$

####  8.4.1 Example with ${C}=1$

$\displaystyle\begin{pmatrix}\mathsf{u}_{1}\\\ \mathsf{v}_{1}\\\
\mathsf{v}_{2}\\\ \mathsf{v}_{3}\\\ \mathsf{v}_{4}\\\ \mathsf{u}^{1}\\\
\mathsf{v}^{1}\\\ \mathsf{v}^{2}\\\ \mathsf{v}^{3}\\\
\mathsf{v}^{4}\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\
\mathbf{u}_{1}^{1}\\\ \mathbf{u}_{2}^{1}\\\ \mathbf{u}_{3}^{1}\\\
\mathbf{u}_{4}^{1}\\\ \bar{\mathbf{u}}^{0}_{1}\\\ \bar{\mathbf{u}}^{1}_{1}\\\
\bar{\mathbf{u}}^{2}_{1}\\\ \bar{\mathbf{u}}^{3}_{1}\\\
\bar{\mathbf{u}}^{4}_{1}\end{pmatrix}$ $\displaystyle=$
$\displaystyle\exp(\mathbf{x}_{0}^{1}\otimes\sum_{q=1}^{4}\bar{\mathbf{x}}^{q}_{1}\triangleright)\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}\\\ \mathbf{x}_{2}^{1}\\\ \mathbf{x}_{3}^{1}\\\
\mathbf{x}_{4}^{1}\\\ \bar{\mathbf{x}}^{0}_{1}\\\ \bar{\mathbf{x}}^{1}_{1}\\\
\bar{\mathbf{x}}^{2}_{1}\\\ \bar{\mathbf{x}}^{3}_{1}\\\
\bar{\mathbf{x}}^{4}_{1}\end{pmatrix}=\begin{pmatrix}\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{1}^{1}+\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{2}^{1}+\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{3}^{1}+\mathbf{x}_{0}^{1}\\\
\mathbf{x}_{4}^{1}+\mathbf{x}_{0}^{1}\\\
\bar{\mathbf{x}}^{0}_{1}-\sum_{q=1}^{4}\bar{\mathbf{x}}^{q}_{1}\\\
\bar{\mathbf{x}}^{1}_{1}\\\ \bar{\mathbf{x}}^{2}_{1}\\\
\bar{\mathbf{x}}^{3}_{1}\\\
\bar{\mathbf{x}}^{4}_{1}\end{pmatrix}=\begin{pmatrix}\mathsf{x}_{1}\\\
\mathsf{y}_{1}+\mathsf{x}_{1}\\\ \mathsf{y}_{2}+\mathsf{x}_{1}\\\
\mathsf{y}_{3}+\mathsf{x}_{1}\\\ \mathsf{y}_{4}+\mathsf{x}_{1}\\\
\mathsf{x}^{1}-\sum_{q=1}^{4}\mathsf{y}^{q}\\\ \mathsf{y}^{1}\\\
\mathsf{y}^{2}\\\ \mathsf{y}^{3}\\\ \mathsf{y}^{4}\end{pmatrix}.$

######  Example 8.5.

We take our example from [21]:

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{1}},{x_{2}},{x_{3}},{x_{4}},{x_{5}}),$ $\displaystyle\dot{x}_{2}$
$\displaystyle=$ $\displaystyle
f^{1}({x_{2}},{x_{5}},{x_{4}},{x_{5}},{x_{5}}),$ $\displaystyle\dot{x}_{3}$
$\displaystyle=$ $\displaystyle
f^{1}({x_{3}},{x_{4}},{x_{5}},{x_{5}},{x_{5}}),$ $\displaystyle\dot{x}_{4}$
$\displaystyle=$ $\displaystyle
f^{1}({x_{4}},{x_{5}},{x_{5}},{x_{5}},{x_{5}}),$ $\displaystyle\dot{x}_{5}$
$\displaystyle=$ $\displaystyle
f^{1}({x_{5}},{x_{5}},{x_{5}},{x_{5}},{x_{5}}),$

with picture

12345$>$$>$$>$$>$$<$$<$$<$$>$$>$$>$$<$$<$$<$$<$$>$$<$$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$

The Jacobi-matrix is

$\displaystyle\begin{pmatrix}{f^{1}_{1}}&{f^{1}_{2}}&{f^{1}_{3}}&{f^{1}_{4}}&{f^{1}_{5}}\\\
0&{f^{1}_{1}}&0&{f^{1}_{3}}&{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{2}}\\\
0&0&{f^{1}_{1}}&{f^{1}_{2}}&{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}\\\
0&0&0&{f^{1}_{1}}&{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}\\\
0&0&0&0&{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}\end{pmatrix}.$

The relation between the coordinates (using Algorithm 3.6) is given by

$\displaystyle\begin{pmatrix}x_{5}\\\ x_{1}\\\ x_{2}\\\ x_{3}\\\
x_{4}\end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\ \mathbf{u}_{1}^{1}\\\
\mathbf{u}_{2}^{1}\\\ \mathbf{u}_{3}^{1}\\\
\mathbf{u}_{4}^{1}\end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\
\mathsf{v}_{1}\\\ \mathsf{v}_{2}\\\ \mathsf{v}_{3}\\\
\mathsf{v}_{4}\end{pmatrix}.$

Using Algorithm 4.6 we obtain:

$\displaystyle\mathsf{J}_{\mathfrak{{c}}}$ $\displaystyle=$
$\displaystyle({f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}){\mathfrak{{c}}^{1}_{1}},$
$\displaystyle\mathsf{J}_{\mathfrak{{b}}}$ $\displaystyle=$
$\displaystyle{f^{1}_{1}}{\mathfrak{{b}}^{1}_{1}}+{f^{1}_{2}}{\mathfrak{{b}}^{1}_{2}}+{f^{1}_{1}}{\mathfrak{{b}}^{2}_{2}}+{f^{1}_{3}}{\mathfrak{{b}}^{1}_{3}}+{f^{1}_{1}}{\mathfrak{{b}}^{3}_{3}}+{f^{1}_{4}}{\mathfrak{{b}}^{1}_{4}}+{f^{1}_{3}}{\mathfrak{{b}}^{2}_{4}}+{f^{1}_{2}}{\mathfrak{{b}}^{3}_{4}}+{f^{1}_{1}}{\mathfrak{{b}}^{4}_{4}},$

with matrix

$\displaystyle\mathsf{J}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}&0&0&0&0\\\
0&{f^{1}_{1}}&0&0&0\\\ 0&{f^{1}_{2}}&{f^{1}_{1}}&0&0\\\
0&{f^{1}_{3}}&0&{f^{1}_{1}}&0\\\
0&{f^{1}_{4}}&{f^{1}_{3}}&{f^{1}_{2}}&{f^{1}_{1}}\end{bmatrix}.$

The Jacobian is

$\displaystyle({f^{1}_{1}})^{4}({f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}).$

The Jordan-Chevalley decomposition is obvious, and Jacobson-Morozov can
immediately be applied.

##  9 Example with $N=8$

###  9.1 Example ${C}=1$

######  Example 9.1.

We consider [6, Example 4.9] with the following equations

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{1}},{x_{2}},{x_{3}},{x_{4}},{x_{5}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{2}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{2}},{x_{6}},{x_{4}},{x_{8}},{x_{2}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{3}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{3}},{x_{5}},{x_{7}},{x_{3}},{x_{8}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{4}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{4}},{x_{2}},{x_{7}},{x_{4}},{x_{8}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{5}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{5}},{x_{6}},{x_{3}},{x_{8}},{x_{5}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{6}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{6}},{x_{6}},{x_{8}},{x_{8}},{x_{6}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{7}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{7}},{x_{8}},{x_{7}},{x_{7}},{x_{8}},{x_{6}},{x_{7}},{x_{8}}),$
$\displaystyle\dot{x}_{8}$ $\displaystyle=$ $\displaystyle
f^{1}({x_{8}},{x_{6}},{x_{7}},{x_{8}},{x_{8}},{x_{6}},{x_{7}},{x_{8}}).$

The Jacobi-matrix is

$\displaystyle\begin{pmatrix}{f^{1}_{1}}&{f^{1}_{2}}&{f^{1}_{3}}&{f^{1}_{4}}&{f^{1}_{5}}&{f^{1}_{6}}&{f^{1}_{7}}&{f^{1}_{8}}\\\
0&{f^{1}_{5}}+{f^{1}_{1}}&0&{f^{1}_{3}}&0&{f^{1}_{6}}+{f^{1}_{2}}&{f^{1}_{7}}&{f^{1}_{8}}+{f^{1}_{4}}\\\
0&0&{f^{1}_{4}}+{f^{1}_{1}}&0&{f^{1}_{2}}&{f^{1}_{6}}&{f^{1}_{7}}+{f^{1}_{3}}&{f^{1}_{8}}+{f^{1}_{5}}\\\
0&{f^{1}_{2}}&0&{f^{1}_{4}}+{f^{1}_{1}}&0&{f^{1}_{6}}&{f^{1}_{7}}+{f^{1}_{3}}&{f^{1}_{8}}+{f^{1}_{5}}\\\
0&0&{f^{1}_{3}}&0&{f^{1}_{5}}+{f^{1}_{1}}&{f^{1}_{6}}+{f^{1}_{2}}&{f^{1}_{7}}&{f^{1}_{8}}+{f^{1}_{4}}\\\
0&0&0&0&0&{f^{1}_{6}}+{f^{1}_{5}}+{f^{1}_{2}}+{f^{1}_{1}}&{f^{1}_{7}}&{f^{1}_{8}}+{f^{1}_{4}}+{f^{1}_{3}}\\\
0&0&0&0&0&{f^{1}_{6}}&{f^{1}_{7}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{1}}&{f^{1}_{8}}+{f^{1}_{5}}+{f^{1}_{2}}\\\
0&0&0&0&0&{f^{1}_{6}}+{f^{1}_{2}}&{f^{1}_{7}}+{f^{1}_{3}}&{f^{1}_{8}}+{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{1}}\end{pmatrix}.$

The relation between the coordinates (using Algorithm 3.6) is given by

$\displaystyle\begin{pmatrix}x_{8}\\\ x_{1}\\\ x_{2}\\\ x_{3}\\\ x_{4}\\\
x_{5}\\\ x_{6}\\\ x_{7}\\\ \end{pmatrix}=\begin{pmatrix}\mathbf{u}_{0}^{1}\\\
\mathbf{u}_{1}^{1}\\\ \mathbf{u}_{2}^{1}\\\ \mathbf{u}_{3}^{1}\\\
\mathbf{u}_{4}^{1}\\\ \mathbf{u}_{5}^{1}\\\ \mathbf{u}_{6}^{1}\\\
\mathbf{u}_{7}^{1}\\\ \end{pmatrix}=\begin{pmatrix}\mathsf{u}_{1}\\\
\mathsf{v}_{1}\\\ \mathsf{v}_{2}\\\ \mathsf{v}_{3}\\\ \mathsf{v}_{4}\\\
\mathsf{v}_{5}\\\ \mathsf{v}_{6}\\\ \mathsf{v}_{7}\\\ \end{pmatrix}.$

Using Algorithm 4.6 we obtain:

$\displaystyle\mathsf{J}_{\mathfrak{{c}}}$ $\displaystyle=$
$\displaystyle+({f^{1}_{8}}+{f^{1}_{7}}+{f^{1}_{6}}+{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}){\mathfrak{{c}}^{1}_{1}},$
$\displaystyle\mathsf{J}_{\mathfrak{{a}}}$ $\displaystyle=$
$\displaystyle+({f^{1}_{6}}+{f^{1}_{2}}){\mathfrak{{a}}^{1}_{6}}+({f^{1}_{7}}+{f^{1}_{3}}){\mathfrak{{a}}^{1}_{7}},$
$\displaystyle\mathsf{J}_{\mathfrak{{b}}}$ $\displaystyle=$
$\displaystyle+{f^{1}_{1}}{\mathfrak{{b}}^{1}_{1}}+{f^{1}_{2}}{\mathfrak{{b}}^{1}_{2}}+({f^{1}_{5}}+{f^{1}_{1}}){\mathfrak{{b}}^{2}_{2}}+{f^{1}_{2}}{\mathfrak{{b}}^{4}_{2}}+{f^{1}_{3}}{\mathfrak{{b}}^{1}_{3}}+({f^{1}_{4}}+{f^{1}_{1}}){\mathfrak{{b}}^{3}_{3}}+{f^{1}_{3}}{\mathfrak{{b}}^{5}_{3}}$
$\displaystyle+{f^{1}_{4}}{\mathfrak{{b}}^{1}_{4}}+{f^{1}_{3}}{\mathfrak{{b}}^{2}_{4}}+({f^{1}_{4}}+{f^{1}_{1}}){\mathfrak{{b}}^{4}_{4}}+{f^{1}_{5}}{\mathfrak{{b}}^{1}_{5}}+{f^{1}_{2}}{\mathfrak{{b}}^{3}_{5}}+({f^{1}_{5}}+{f^{1}_{1}}){\mathfrak{{b}}^{5}_{5}}-{f^{1}_{2}}{\mathfrak{{b}}^{1}_{6}}$
$\displaystyle-{f^{1}_{2}}{\mathfrak{{b}}^{3}_{6}}-{f^{1}_{2}}{\mathfrak{{b}}^{4}_{6}}+({f^{1}_{5}}+{f^{1}_{1}}){\mathfrak{{b}}^{6}_{6}}-{f^{1}_{2}}{\mathfrak{{b}}^{7}_{6}}-{f^{1}_{3}}{\mathfrak{{b}}^{1}_{7}}-{f^{1}_{3}}{\mathfrak{{b}}^{2}_{7}}-{f^{1}_{3}}{\mathfrak{{b}}^{5}_{7}}-{f^{1}_{3}}{\mathfrak{{b}}^{6}_{7}}$
$\displaystyle+({f^{1}_{4}}+{f^{1}_{1}}){\mathfrak{{b}}^{7}_{7}},$

with matrix

$\displaystyle\begin{bmatrix}{f^{1}_{8}}+{f^{1}_{7}}+{f^{1}_{6}}+{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}}&0&0&0&0&0&0&0\\\
0&{f^{1}_{1}}&0&0&0&0&0&0\\\
0&{f^{1}_{2}}&{f^{1}_{5}}+{f^{1}_{1}}&0&{f^{1}_{2}}&0&0&0\\\
0&{f^{1}_{3}}&0&{f^{1}_{4}}+{f^{1}_{1}}&0&{f^{1}_{3}}&0&0\\\
0&{f^{1}_{4}}&{f^{1}_{3}}&0&{f^{1}_{4}}+{f^{1}_{1}}&0&0&0\\\
0&{f^{1}_{5}}&0&{f^{1}_{2}}&0&{f^{1}_{5}}+{f^{1}_{1}}&0&0\\\
{f^{1}_{6}}+{f^{1}_{2}}&-{f^{1}_{2}}&0&-{f^{1}_{2}}&-{f^{1}_{2}}&0&{f^{1}_{5}}+{f^{1}_{1}}&-{f^{1}_{2}}\\\
{f^{1}_{7}}+{f^{1}_{3}}&-{f^{1}_{3}}&-{f^{1}_{3}}&0&0&-{f^{1}_{3}}&-{f^{1}_{3}}&{f^{1}_{4}}+{f^{1}_{1}}\end{bmatrix}.$

The Jacobian is

$\displaystyle({f^{1}_{1}})({f^{1}_{8}}+{f^{1}_{7}}+{f^{1}_{6}}+{f^{1}_{5}}+{f^{1}_{4}}+{f^{1}_{3}}+{f^{1}_{2}}+{f^{1}_{1}})({f^{1}_{4}}{f^{1}_{5}}-{f^{1}_{2}}{f^{1}_{3}}+{f^{1}_{1}}{f^{1}_{5}}+{f^{1}_{1}}{f^{1}_{4}}+{f^{1}_{1}}^{2})^{3}.$

##  10 Concluding remarks

The algorithmic approach presented in this paper is but a first step towards
the classification of bifurcations of colored networks. Obvious areas of
further research are the computation of Jordan-Chevalley decompositions with
many variables, the generalization of the Jacobson-Morozov construction to the
situation of colored networks, the construction of versal deformations of
organizing centers and the nonlinear network normal form description.

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