# Generating functions for irreversible Hamiltonian systems

Dan Goreac, Jonas Kirchhoff and Bernhard Maschke

#### Abstract

The definition of conservative-irreversible functions is extended to smooth
manifolds. The local representation of these functions is studied and reveals
that not each conservative-irreversible function is given by the weighted
product of almost Poisson brackets. The biquadratic functions given by
conservative-irreversible functions are studied and reveal a possibility for
an algebraic framework on arbitrary and in particular complex algebras.

#### Keywords

thermodynamic systems, geometric methods, metriplectic systems

Corresponding author: Jonas Kirchhoff
Dan Goreac
Shandong University Weihai, School of Mathematics and Statistics, 180 Wenhua W
Road, 264209 Weihai, Shandong, P.R. China
LAMA, Univ Gustave Eiffel, UPEM, Univ Paris Est Creteil, CNRS, 5 Boulevard
Descartes, 77454 Marne-la-Vallée, France
E-mail<EMAIL_ADDRESS>
Jonas Kirchhoff
Institut für Mathematik, Technische Universität Ilmenau, Weimarer Straße 25,
98693 Ilmenau, Germany
E-mail<EMAIL_ADDRESS>
Bernhard Maschke
Lab. d’Automatique et de Genie des Procedes, Universite Claude Bernard Lyon-1,
43 Boulevard du 11 Novembre 1918, 69622 Villeurbanne Cedex, Auvergne-Rhone-
Alpes, France
E-mail<EMAIL_ADDRESS>
Jonas Kirchhoff thanks the Technische Universität Ilmenau and the Freistaat
Thüringen for their financial support as part of the Thüringer
Graduiertenförderung. Bernhard Maschke acknowledges support by by IMPACTS
(ANR-21-CE48-0018).

##  1 Introduction

Hamiltonian systems defined on symplectic [14, 1] or, more general, Poisson
manifolds [16, 3] provide a well-established geometric theory of classical
mechanics. For dynamical systems arising from irreversible thermodynamic
systems, the classical Poisson geometry fails, as the effect of the entropy of
the system, which characterises reversible and irreversible processes [15],
has to be taken into account. To this end, several different generalisations
of Poisson geometry have been proposed.

One approach, which was pioneered by [11, 17, 8] is to assume that the
dissipating vector field depends, analogously to the Hamiltonian vector field,
linearly on the exterior derivative of the entropy function. This results in a
Leibniz bracket [21], i.e. a biderivation on $\mathcal{C}^{\infty}(M)$, which
is usually assumed to be symmetric and non-negative, whose Leibniz vector
fields with respect to the entropy model the irreversible entropy creation. To
ensure that the flow of the resulting vector field satisfies the laws of
thermodynamics, the non-interaction conditions that the entropy function is a
Casimir of the (almost) Poisson bracket and the Hamiltonian energy function is
a Casimir of the non-negative Leibniz bracket. The resulting class of systems
is known as metriplectic systems [18, 9], or the GENERIC (general equation for
non-equilibrium reversible-irreversible coupling) framework [24].

A second possibility is to assume that the dissipating vector field depends,
possibly non-linearly, on the Hamiltonian function [5, 2]. Then, in addition
to the noninteraction condition that the entropy is a Casimir of the (almost)
Poisson bracket, the requirement that the contraction of the dissipating
vector field with the Hamiltonian vanishes and with the entropy is non-
negative ensures that the flow satisfies the laws of thermodynamics. Taking
the free energy as the Hamiltonian function of the system, one sees that
metriplectic systems can due to the noninteraction condition be viewed as a
particular class of these systems.

Yet a third possibility of extending the Poisson bracket are irreversible
Hamiltonian systems, where the dissipating vector field is a Hamiltonian
vector field with respect to a multiplied with the contraction of itself with
the entropy and a non-negative scalar factor depending on the exterior
derivative of the Hamiltonian, or a convex combination of those. The flow of
the resulting vector field fulfils, provided that the entropy function is a
Casimir of the original Poisson bracket, automatically the laws of
thermodynamics.

In the present note, we consider a geometric structure which assigns a
dissipating vector field to an arbitrary Hamiltonian function and an arbitrary
entropy function so that the resulting flow automatically fulfils the laws of
thermodynamics. For simplicity, we assume that this assignment is linear in
the entropy and quadratic in the Hamiltonian function, and depends only on the
exterior derivatives, resulting in a generalised bracket operation. This
construction yields the conservative-irreversible functions as defined in
[12]. We study the properties and demonstrate how conservative-irreversible
functions may be used to combine the three approaches to thermodynamic systems
in one framework.

The paper is organised as follows. In section 2, we collect a list of
properties that we wish from a generalised bracket operation form
thermodynamics; the resulting definition being conservative-irreversible
functions. Then, following [12], we characterise the 4-contravariant tensor
fields which are uniquely associated to these functions. In particular, we may
show that a particular symmetrisation of the metriplectic 4-brackets
introduced in [20] yields conservative-irreversible functions.

In section 3, we give the local representation of the tensor fields associated
with the conservative-irreversible functions. Having identified simple
conservative-irreversible functions, which are, up to scalar multiplication
and arrangement of the arguments, tensor squares of almost Poisson brackets,
as an important subclass, we use local representations to study the
decomposability of an arbitrary conservative-irreversible function into simple
functions. In particular, this implies a one-to-one correspondence between
conservative-irreversible functions and metriplectic 4-brackets.

In the section 4, we characterize the biquadratic functions, which are induced
by conservative-irreversible functions.

In particular, these biquadratic function define the entropy dynamics of a new
class of irreversible Hamiltonian systems, which are defined using
conservative-irreversible functions in section 5. We demonstrate how these
systems unify, in some sense, metriplectic systems, the systems of Beris and
Edwards and the original irreversible Hamiltonian systems of Ramirez et al.

##  2 Conservative-irreversible functions on manifolds

###  2.1 Generalised brackets for thermodynamics

We study extensions Hamiltonian systems defined on a smooth almost Poisson
manifold [4] $(M,\\{\cdot,\cdot\\})$ where the Hamiltonian vector field is
perturbed by a vector field $\Xi(s,h)$, i.e.

$\displaystyle X_{s,h}:=\\{\cdot,h\\}+\Xi(s,h),$ (2.1)

for a functions $h,s\in\mathcal{C}^{\infty}(M)$, called the Hamiltonian and
the entropy function. Furthermore it is assumed that the either weak
noninteraction condition $\\{s,h\\}=0$ or the strong noninteraction condition
$\\{s,\cdot\\}\equiv 0$ holds [19]. The function
$\Xi:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathfrak{X}(M)$,
where $\mathfrak{X}(M):=\Gamma\mathcal{T}M$ denotes the space of global vector
fields, is to be chosen so that the flow of the vector field (2.1) satisfies
the laws of thermodynamics, i.e.

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}s\circ x\geq
0=\tfrac{\mathrm{d}}{\mathrm{d}t}h\circ x$

for each integral curve $x\in\mathcal{C}^{\infty}(I,M)$,
$I\subseteq\mathbb{R}$ an open interval, of (2.1). Due to the noninteraction
condition this is the case if, and only if,

$\displaystyle\iota_{s}\Xi(s,h)\geq 0=\iota_{h}\Xi(s,h),$ (2.2)

where $\iota$ denotes the contraction of a function with a vector field.

In this paper, we study the geometric structure on which the functions $\Xi$
with (2.2) independently of the choice of the almost Poisson bracket
$\\{\cdot,\cdot\\}$ are based on. To derive a meaningful geometric structure,
we restrict ourself to “simple” functions $\Xi$. In particular, $\Xi$ should
only depend on local data, i.e. for each
$s,s^{\prime},h,h^{\prime}\in\mathcal{C}^{\infty}(M)$ and $U\subseteq M$ open
with $s|_{U}-s^{\prime}|_{U}=h|_{U}-h^{\prime}|_{U}=0$, we have
$\Xi(s,h)|_{U}=\Xi(s^{\prime},h^{\prime})|_{U}$. It seems natural to assume
that $\Xi$ is bilinear, thus giving a ternary bracket operation on
$\mathcal{C}^{\infty}(M)$, and fulfil
$\iota_{f}\Xi(g,\cdot)=\iota_{g}\Xi(f,\cdot)$ for all
$f,g\in\mathcal{C}^{\infty}(M)$. Then, however, we have

$\displaystyle\forall s,h\in\mathcal{C}^{\infty}(M):\iota_{s}\Xi(s,h)\geq
0\geq-\iota_{s}\Xi(s,h)=\iota_{s}\Xi(s,-h)\geq 0$

and hence $\Xi$ vanishes identically. Therefore, we instead assume that $\Xi$
is linear in the first and quadratic in the third argument. Lastly, we impose
the condition that $\Xi$ is invariant under locally constant shifts of its
arguments, i.e. $\Xi(s,h)=\Xi(s+c,h+c^{\prime})$ for all
$s,h,c,c^{\prime}\in\mathcal{C}^{\infty}(M)$ with
$\mathrm{d}c=\mathrm{d}c^{\prime}=0$, which represents the fact that energy
and entropy of a given system may only be known up to some unknown (locally)
constant perturbation. This defines the geometric structure of thermodynamic
systems, analogously to Poisson geometry for classical mechanics. When we
recall that vector fields are precisely the derivations on the algebra of
smooth functions, then the functions $\Xi$ which we have described are
precisely covered by the following definition, which is the direct
generalisation of [12, Definition 3] to manifolds.

######  Definition 2.1.

Let $M$ be a smooth manifold. We call a function
$E:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M)$
conservative-irreversible if, and only if, there exists a function
$e:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M)$ so
that

$\displaystyle\forall f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=e(f,g,h,h)$
(2.3)

and, for all $f,g,h\in\mathcal{C}^{\infty}(M)$,

  1. (i)

$e$ is is a derivation, i.e. it is $\mathbb{R}$-linear and fulfils the Leibniz
rule, in each argument,

  2. (ii)

$e(f,g,\cdot,\cdot)$ is symmetric,

  3. (iii)

$e(\cdot,\cdot,h,h)$ is symmetric and pointwise positive semi-definite,

  4. (iv)

$e(h,\cdot,h,h)$ vanishes identically,

######  Remark 2.2.

Each conservative-irreversible function $E$ induces the function

$\displaystyle\Xi_{E}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathfrak{X}(M),\qquad(s,h)\mapsto
E(\cdot,s,h)$

which has exactly the properties described after (2.2). Conversely, each such
function $\Xi$ induces the conservative-irreversible function

$\displaystyle
E_{\Xi}:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M),\qquad(f,s,h)\mapsto\iota_{f}\Xi(s,h);$

and these constructions are inverse to each other, i.e. $E_{\Xi_{E}}=E$ and
$\Xi_{E_{\Xi}}=\Xi$.

###  2.2 Conservative-irreversible functions

In this section, we study the properties of conservative-irreversible
functions. Before we do that, we ensure that the class of conservative-
irreversible functions is non-trivial, when the dimension of the underlying
manifold exceeds one.

######  Example 2.3.

Let $M$ be a smooth manifold and let

$\displaystyle\\{\cdot,\cdot\\}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$

be a skew-symmetric biderivation, i.e. an almost Poisson bracket on $M$.
Define

$\displaystyle
e:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M),\qquad(f,s,h,g)\mapsto\frac{1}{2}\left(\\{s,h\\}\\{f,g\\}+\\{f,h\\}\\{s,g\\}\right).$

Since $\\{\cdot,\cdot\\}$ is a derivation in both arguments, $e$ is a
derivation in each argument. By construction, $e(f,s,\cdot,\cdot)$ and
$e(\cdot,\cdot,h,h)=\\{\cdot,h\\}\\{\cdot,h\\}$ are symmetric, and
$e(s,s,h,h)=\\{s,h\\}^{2}$ implies that the latter is non-negative for all
$f,s,h\in\mathcal{C}^{\infty}(M)$. Lastly, we observe

$\displaystyle e(h,\cdot,h,h)=\underbrace{\\{h,h\\}}_{=0}\\{\cdot,h\\}=0$

and hence the function

$\displaystyle
E:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M),\qquad(f,s,h)\mapsto\\{s,h\\}\\{f,h\\}$

is conservative-irreversible. Finally, when the almost Poisson bracket is non-
trivial, then $E$ is non-trivial; and the existence of non-trivial almost
Poisson brackets in dimension at least two implies that there are non-trivial
conservative-irreversible functions.

######  Remark 2.4.

The conservative-irreversible functions constructed in Example 2.3 are closely
related to the irreversible Hamiltonian systems of [23] defined on a smooth
manifold $M$, whose dissipating vector field has the form

$\displaystyle\Xi_{s,h}:=(\lambda\circ\mathrm{d}h)\\{s,h\\}\\{\cdot,h\\}$

for a Poisson bracket $\\{\cdot,\cdot\\}$ and a smooth function
$\lambda\in\mathcal{C}^{\infty}(\mathcal{T}^{*}M,\mathbb{R}_{>0})$.

We collect in the next lemmata some properties of the function $e$ given in
(2.3).

######  Lemma 2.5.

The functions $e$ of Definition 2.1 are uniquely determined by their
associated conservative-irreversible functions with (2.3).

###### Proof.

Let
$e,e^{\prime}:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M)$
be linear and symmetric in the third and fourth argument. Assume further that

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):e(f,g,h,h)=e^{\prime}(f,g,h,h).$

Then, we have, for all $f,g,h,k\in\mathcal{C}^{\infty}(M)$,

$\displaystyle 0$ $\displaystyle=e(f,g,h+k,h+k)-e^{\prime}(f,g,h+k,h+k)$
$\displaystyle=e(f,g,h,h)-e^{\prime}(f,g,h,h)+e(f,g,h,k)+e(f,g,k,h)+e(f,g,k,k)-e^{\prime}(f,g,h,k)$
$\displaystyle\quad-e^{\prime}(f,g,k,k)-e^{\prime}(f,g,k,h)$
$\displaystyle=2(e(f,g,h,k)-e^{\prime}(f,g,h,k))$

and hence $e(f,g,h,k)=e^{\prime}(f,g,h,k)$. ∎

######  Lemma 2.6.

Let $e:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M)$ be
linear in each argument. Then, $e(\cdot,\cdot,h,h)$ is symmetric for all
$h\in\mathcal{C}^{\infty}(M)$ if, and only if, $e(\cdot,\cdot,h,k)$ is
symmetric for all $h,k\in\mathcal{C}^{\infty}(M)$.

###### Proof.

The sufficiency is obvious; we show necessity. By linearity in the last two
arguments, we have, for all $f,g,h,k\in\mathcal{C}^{\infty}(M)$,

$\displaystyle e(f,g,h,k)$
$\displaystyle=\frac{1}{2}(e(f,g,h+k,h+k)-e(f,g,h,h)-e(f,g,k,k))$
$\displaystyle=\frac{1}{2}(e(g,f,h+k,h+k)-e(g,f,h,h)-e(g,f,k,k))=e(g,f,h,k).$

∎

In the following lemma, we characterise the tensor fields that generate
conservative-irreversible functions.

######  Lemma 2.7.

A function
$E:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M)$ is
conservative-irreversible if, and only if, there exists a uniquely defined
contravariant tensor field $\varepsilon\in\Gamma(T^{4}(M))$ so that

$\displaystyle\forall f,g,h\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall x\in
M:E(f,g,h)(x)=\varepsilon_{x}(\mathrm{d}f_{x},\mathrm{d}g_{x},\mathrm{d}h_{x},\mathrm{d}h_{x})$
(2.4)

and, for all $x\in M$, $\varepsilon_{x}$ has the properties

  1. (a)

$\varepsilon_{x}(\alpha,\beta,\gamma,\delta)=\varepsilon_{x}(\alpha,\beta,\delta,\gamma)$
for all $\alpha,\beta,\gamma,\delta\in\mathcal{T}_{x}^{*}M$,

  2. (b)

$\varepsilon_{x}(\alpha,\beta,\gamma,\delta)+\varepsilon_{x}(\delta,\beta,\alpha,\gamma)+\varepsilon_{x}(\gamma,\beta,\delta,\alpha)=0$
for all $\alpha,\beta,\gamma,\delta\in\mathcal{T}_{x}^{*}M$,

  3. (c)

$\varepsilon_{x}(\cdot,\cdot,\gamma,\gamma)$ is symmetric and positive semi-
definite for all $\alpha\in\mathcal{T}^{*}_{x}(M)$

###### Proof.

It is well-known that all $k$-derivations on $\mathcal{C}^{\infty}(M)$, i.e.
functions of $k$ argument that are linear and fulfil the Leibniz rule in each
argument, are given by $k$-contravariant tensor fields
$\xi\in\Gamma(T^{k}(M))$, where $T^{k}(M)$ denotes the vector bundle of
$k$-contravariant tensors, see e.g. [7, Section 3.18]. Furthermore, in view of
the lemma 2.5, there is a one-to-one correspondence between conservative-
irreversible functions and functions
$e:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M)$ with
the properties (i)–(iv) of Definition 2.1. Hence the relation

$\displaystyle\forall f,g,h,k\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall x\in M:e(f,g,h,k)(x)=\varepsilon_{x}(df_{x},dg_{x},dh_{x},dk_{x})$

establishes a one-to-one correspondence between 4-contravariant tensor fields
$\varepsilon\in\Gamma(T^{4}(M))$ and functions
$e:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{4}\to\mathcal{C}^{\infty}(M)$ with
the property (i) of Definition 2.1. The proof of the equivalence of the
properties (ii)–(iv) of Definition 2.1 for $e$ and the properties (a)–(c) of
the associated tensor field $\varepsilon$ is analogous to the proof of [12] to
which we refer the reader. ∎

We show now, that the functions $e$ associated to conservative-irreversible
functions by (2.3) are symmetric in the first and second pair of arguments.

######  Lemma 2.8.

Let $e:\left(\mathcal{C}^{\infty}(M)\right)^{4}\to\mathcal{C}^{\infty}(M)$
fulfil the properties (i)–(iv) of Definition 2.1. Then,
$e(f,g,h,k)=e(h,k,f,g)$ for all $f,g,h,k\in\mathcal{C}^{\infty}(M)$.

###### Proof.

Let $E:\left(\mathcal{C}^{\infty}(M)\right)^{3}\to\mathcal{C}^{\infty}(M)$ be
the conservative-irreversible function associated to $e$, and let
$\varepsilon\in\Gamma(T^{4}(M))$ the (unique) tensor field with (2.4).
Equivalently, we have

$\displaystyle\forall f,g,h,k\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall x\in
M:e(f,g,h,k)(x)=\varepsilon_{x}(\mathrm{d}f_{x},\mathrm{d}g_{x},\mathrm{d}h_{x},\mathrm{d}k_{x}).$

Thus, Lemma 2.7 (b) yields in particular

$\displaystyle\forall
f,g,h,k\in\mathcal{C}^{\infty}(M):e(f,g,h,k)+e(h,g,k,f)+e(k,g,f,k)=0.$

Let $f,g,h,k\in\mathcal{C}^{\infty}(M)$. Using the symmetry of $e$ in the
first two and in the last two arguments, we see

$\displaystyle e(f,h,k,g)$ $\displaystyle=e(h,f,k,g)$
$\displaystyle=-e(k,f,g,h)-e(g,f,h,k)$ $\displaystyle=-e(f,k,h,g)-e(g,f,h,k)$
$\displaystyle=e(h,k,g,f)+e(g,k,f,h)-e(g,f,h,k)$
$\displaystyle=e(h,k,f,g)+e(k,g,f,h)-e(f,g,h,k)$

and therefore

$\displaystyle 0$ $\displaystyle=e(f,h,k,g)-e(k,g,f,h)+e(f,g,h,k)-e(h,k,f,g)$
$\displaystyle=-e(k,h,g,f)-e(g,h,f,k)+e(f,g,h,k)+e(h,g,k,f)+e(f,g,h,k)-e(h,k,f,g)$
$\displaystyle=2(e(f,g,h,k)-e(h,k,f,g)).$

This completes the proof of the Lemma. ∎

###  2.3 Relation with metriplectic 4-brackets

Recently, Morrison and Updike [20] proposed a very similar bracket description
called metriplectic four-bracket as a geometric formulation for dissipative
systems, in particular for metriplectic systems. These brackets, whose
properties are motivated by Riemannian curvature tensors, are defined as
follows.

######  Definition 2.9 ([20, p. 6]).

A metriplectic four-bracket on the manifold $M$ is a function
$(\\!(\cdot,\cdot,\cdot,\cdot)\\!):\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$
with the properties

  1. (i)

$(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is a derivation in each argument

  2. (ii)

$(\\!(f,g,\cdot,\cdot)\\!)$ and $(\\!(\cdot,\cdot,f,g)\\!)$ are skew-symmetric
for all $f,g\in\mathcal{C}^{\infty}(M)$

  3. (iii)

$(\\!(f,g,h,k)\\!)=(\\!(h,k,f,g)\\!)$ for all
$f,g,h,k\in\mathcal{C}^{\infty}(M)$

  4. (iv)

$(\\!(f,g,h,k)\\!)+(\\!(f,h,k,g)\\!)+(\\!(f,k,g,h)\\!)=0$ for all
$f,g,h,k\in\mathcal{C}^{\infty}(M)$

  5. (v)

$(\\!(f,g,f,g)\\!)(x)\geq 0$ for all $f,g\in\mathcal{C}^{\infty}(M)$ and $x\in
M$

The dissipative bracket given by $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ and a
Hamiltonian function $h\in\mathcal{C}^{\infty}(M)$ is the bracket
$(\cdot,\cdot)_{h}:=(\\!(\cdot,h,\cdot,h)\\!)$.

Evidently, the symmetries of conservative-irreversible tensors and
metriplectic four-brackets are, at first glance, slightly different; and so
are their motivations: Conservative-irreversible functions are a general class
of tensor fields with easy to study symmetries so that the flow of the vector
field $e(\cdot,s,h,h)$ satisfies the laws of thermodynamics for all functions
$s,h\in\mathcal{C}^{\infty}(M)$ and each conservative-irreversible function
$e$. This raises the question of the relation of metriplectic four-brackets
and the four-tensors associated to conservative-irreversible functions. It is
easy to see the symmetrisation of metriplectic four tensors in the second and
fourth argument fulfil the conditions (i)–(iv) satisfied by the conservative-
irreversible functions in the Definition 2.1.

######  Lemma 2.10.

Let $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ be a metriplectic four-bracket on the
manifold $M$. The function

$\displaystyle
e:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$
$\displaystyle\to\mathcal{C}^{\infty}(M),$ $\displaystyle(f,g,h,k)$
$\displaystyle\mapsto\frac{1}{2}\big{(}(\\!(f,h,g,k)\\!)+(\\!(f,k,g,h)\\!)\big{)}$

has the properties (i)–(iv) in the definition 2.1.

###### Proof.

Since $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is a derivation in each argument, it
is readily seen that $e$ is a derivation in each argument; and $e$ is
symmetric in the last two arguments by construction, so that $e$ fulfils the
properties (i) and (ii) of Definition 2.1. Let $h\in\mathcal{C}^{\infty}(M)$.
Then, the symmetry of $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ in the first and the
last two arguments implies, for all $f,g\in\mathcal{C}^{\infty}(M)$,

$\displaystyle
e(f,g,h,h)=\frac{1}{2}\big{(}(\\!(f,h,g,h)\\!)+(\\!(f,h,g,h)\\!)\big{)}=\frac{1}{2}\big{(}(\\!(g,h,f,h)\\!)+(\\!(g,h,f,h)\\!)\big{)}=e(g,f,h,h),$

so that $e(\cdot,\cdot,h,h)$ is symmetric. Furthermore,
$e(f,f,h,h)=\frac{1}{2}\big{(}(\\!(f,h,f,h)\\!)+(\\!(f,h,f,h)\\!)\big{)}=(\\!(f,h,f,h)\\!)$
is pointwise non-negative, so that $e$ fulfils (iii) of Definition 2.1.
Finally, skew-symmetry of $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ implies

$\displaystyle
e(h,f,h,h)=\frac{1}{2}\big{(}(\\!(h,h,f,h)\\!)+(\\!(h,h,f,h)\\!)\big{)}=(\\!(h,f,h,h)\\!)=0$

so that $e$ fulfils the property (iv) of Definition 2.1. ∎

This observation implies, in particular, that the functions
$H(f,g,h):=(\\!(f,h,g,h)\\!)$ considered by Morrison and Updike are
conservative-irreversible functions. The converse relation, i.e. for each
conservative-irreversible function $H$ there exists a metriplectic four-
bracket $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ so that
$H(f,g,h)=(\\!(f,h,g,h)\\!)$ for all $f,g,h\in\mathcal{C}^{\infty}(M)$, is not
obvious. In the special case that there is some almost Poisson bracket
$\\{\cdot,\cdot\\}$ with $E(f,g,h)=\\{g,h\\}\\{f,h\\}$ for all
$f,g,h\in\mathcal{C}^{\infty}(M)$, it is straightforward to verify that

$\displaystyle\\{\\!\\{\cdot,\cdot,\cdot,\cdot\\}\\!\\}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$
$\displaystyle\to\mathcal{C}^{\infty}(M),$ $\displaystyle(f,g,h,k)$
$\displaystyle\mapsto\\{f,h\\}\\{g,k\\}$

is a metriplectic four-bracket, see [20]. This leads to the definition of
simple conservative-irreversible functions as follows.

######  Definition 2.11.

We call a conservative-irreversible function $E$ simple if, and only if, there
exist an almost Poisson bracket $\\{\cdot,\cdot\\}$ and a non-negative
function $\lambda:M\to\mathbb{R}_{\geq 0}$ so that

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=\lambda\\{f,h\\}\\{g,h\\}.$ (2.5)

The 4-contravariant tensor field associated to simple conservative-
irreversible functions is, up to scalar multiplication and rearranging of the
arguments, the tensor square of a Poisson bivector. This raises the question:
Given any two 2-contravariant tensor fields, when is their tensor product (up
to the same rearrangement of the arguments) the tensor field of a
conservative-irreversible function? The answer is the following.

######  Proposition 2.12.

Let $M$ be a smooth manifold and let
$[\cdot,\cdot]_{1},[\cdot,\cdot]_{2}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$
be biderivations. Then, the function

$\displaystyle
E:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),\qquad(f,g,h)\mapsto[f,h]_{1}[g,h]_{2}$
(2.6)

is conservative-irreversible if, and only if, it is simple.

###### Proof.

Sufficiency is evident; we show necessity. Let
$\Lambda^{1},\Lambda^{2}\in\Gamma(T^{2}(M))$ be the tensor fields associated
to $[\cdot,\cdot]_{1}$ and $[\cdot,\cdot]_{2}$, respectively. Using local
coordinates, we may use the methods of the proof of [12, Theorem 4.2] to see
that $[\cdot,\cdot]_{1}$ is skew-symmetric, and hence an almost Poisson
bracket. Further, we see that $\Lambda_{x}^{1}$ and $\Lambda_{x}^{2}$ are
linearly dependent for all $x\in M$. Therefore, there exists a function
$\lambda:M\to\mathbb{R}$ so that

$\displaystyle\forall
f,g,h,k\in\mathcal{C}^{\infty}(M):[f,h]_{1}[g,h]_{2}=\lambda[f,h]_{1}[g,h]_{1},$

which shows the assertion. ∎

We show that the closure of simple conservative-irreversible functions under
locally finite sums is in one-to-one correspondence with the symmetrisations
of metriplectic four-brackets.

######  Lemma 2.13.

Let $E$ be a conservative-irreversible function so that there is a family
$(E_{i})_{i\in I}$ of simple conservative-irreversible functions with
$E=\sum_{i\in I}E_{i}$ so that

$\displaystyle\forall f,g,h\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall K\subseteq M\leavevmode\nobreak\
\text{compact}:\left|\left\\{i\in\mathbb{N}\,\big{|}\,E_{i}(f,g,h)|_{K}\not\equiv
0\right\\}\right|<\infty,$

i.e. the sum is locally finite. Then, there exists a metriplectic four-bracket
$(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ so that

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=(\\!(f,h,g,h)\\!).$ (2.7)

###### Proof.

Consider, for each $i\in I$, functions $\lambda_{i}$ and Poisson brackets
$\\{\cdot,\cdot\\}_{i}$ so that
$E_{i}(f,g,h)=\lambda_{i}\\{f,h\\}_{i}\\{g,h\\}_{i}$ for all
$f,g,h\in\mathcal{C}^{\infty}(M)$. Define the functions

$\displaystyle(\\!(\cdot,\cdot,\cdot,\cdot)\\!)_{i}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),(\\!(f,g,h,k)\\!):=\lambda_{i}\\{f,g\\}_{i}\\{h,k\\}_{i}.$

so that $(\\!(f,h,g,h)\\!)_{i}=E_{i}(f,g,h)$ for all
$f,g,h\in\mathcal{C}^{\infty}(M)$. The observation

$\displaystyle\forall
f,g,h,k\in\mathcal{C}^{\infty}(M):(\\!(f,g,h,k)\\!)_{i}=\frac{1}{2}\left(E_{i}(f,h,g+k)-E_{i}(f,h,g)-E_{i}(f,h,k)\right)$
(2.8)

ensures that $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is well-defined. Since
$\\{\cdot,\cdot\\}_{i}$ is a Poisson bracket,
$(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is indeed a metriplectic four-bracket, see
[20]. Since the support of families $(E_{i}(f,g,h))_{i\in I}$ is locally
finite for all $f,g,h\in\mathcal{C}^{\infty}(M)$, we conclude in view of (2.8)
that $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)=\sum_{i\in
I}(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is well-defined; and since the vector
space of metriplectic four-brackets is closed under arbitrary locally finite
sums, $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ is a metriplectic four-bracket with
(2.7). ∎

In Definition 2.11, we have not required that $\lambda$ is smooth. In the
following example, we construct an almost Poisson bracket $\\{\cdot,\cdot\\}$,
which vanishes almost nowhere, and a function $\lambda$, which is not smooth,
so that the induced function $E(f,g,h)=\lambda\\{f,h\\}\\{g,h\\}$ is
conservative-irreversible.

######  Example 2.14.

It is well-known that the function

$\displaystyle f:\mathbb{R}\to\mathbb{R},\quad
x\mapsto\begin{cases}e^{-\frac{1}{x^{2}}},&x\neq 0,\\\ 0,&x=0.\end{cases}$

is smooth. Consider

$\displaystyle\lambda:\mathbb{R}\to\mathbb{R},\quad x\mapsto\left|x\right|.$

Of course, the restriction of $\lambda$ to
$\mathbb{R}\setminus\left\\{0\right\\}$ is smooth and hence $\lambda f$ is
smooth on $\mathbb{R}\setminus\left\\{0\right\\}$. Furthermore, we have

$\displaystyle\forall
x\in\mathbb{R}\setminus\left\\{0\right\\}:\frac{\lambda(x)f(x)-\lambda(0)f(0)}{x}=\mathrm{sgn}(x)f(x)$

and hence $\lambda f$ is differentiable in $0$ with vanishing derivative.
Since all derivatives of $f$ in zero vanish, we find that the derivative of
$\lambda f$,

$\displaystyle\mathrm{d}(\lambda
f)=\begin{cases}\mathrm{sgn}(x)f(x)+\lambda(x)\mathrm{d}f(x),&x\neq 0,\\\
0,&x=0,\end{cases}$

is indeed smooth and the $i$-th derivative of $\lambda f$,
$i\in\mathbb{N}^{*}$, may be determined as

$\displaystyle\mathrm{d}^{i}(\lambda
f)=\begin{cases}\lambda(x)d^{i}f(x)+i\,\mathrm{sgn}(x)\mathrm{d}^{i-1}f(x),&x\neq
0,\\\ 0,&x=0.\end{cases}$

Analogously, it may be shown that the product of $\lambda$ with the smooth
function $g:=f\circ x^{2}$ is smooth. Consider the smooth matrix field

$\displaystyle J:\mathbb{R}^{3}\to\mathbb{R}^{3\times 3},\quad
x\mapsto\begin{bmatrix}0&f(x_{1})&0\\\ -f(x_{1})&0&g(x_{1})\\\
0&-g(x_{1})&0\end{bmatrix}.$

Since $J$ is pointwise skew-symmetric, the associated bracket
$\\{\cdot,\cdot\\}_{J}$ is almost Poisson. Since $\lambda f$ and $\lambda g$
are smooth functions, $\\{\cdot,\cdot\\}_{\lambda J}$ is also an almost
Poisson bracket. Then, we see that

$\displaystyle
E:\mathcal{C}^{\infty}(\mathbb{R}^{2})\times\mathcal{C}^{\infty}(\mathbb{R}^{2})\times\mathcal{C}^{\infty}(\mathbb{R}^{2})\to\mathcal{C}^{\infty}(\mathbb{R}^{2}),\quad(f,g,h)\mapsto\lambda\\{f,h\\}_{J}\\{g,h\\}_{J}=\\{f,h\\}_{\lambda
J}\\{g,h\\}_{J},$

is a conservative-irreversible function. In particular, we see that the
function $\lambda$ in (2.5) does not need to be necessarily smooth.

The representation (2.5) is, in general, not unique. Of course, we may
multiply the Poisson bivector fields with $\sqrt{\frac{1}{\mu}}$, where $\mu$
is an everywhere positive smooth function, and multiply $\lambda$ with $\mu$.
A different non-uniqueness of the representation (2.5) may be observed, if the
tensor field of the conservative-irreversible function vanishes on an open
set, as the next example examines.

######  Example 2.15.

Recall the smooth Urysohn lemma: Let $M$ be a smooth manifold and let
$K,S\subseteq M$ be disjoint closed sets with $K$ compact. Then, there exists
a function $u\in\mathcal{C}^{\infty}(M)$ so that $u|_{K}\equiv 1$ and
$u|_{S}\equiv 0$. Any such function $u$ is called smooth Urysohn function. Let
$\dim M\geq 2$ and $K_{1},K_{2}\subseteq M$ be nonempty, compact sets which
possess disjoint open neighbourhoods $U_{1}$ and $U_{2}$, respectively. Such
sets exist in nonempty open subsets of $\mathbb{R}^{\dim M}$ and thus in $M$.
Let $\varphi$ and $\psi$ be smooth Urysohn functions separating $K_{1}$ and
$M\setminus U_{1}$ and $K_{2}$ and $M\setminus U_{2}$, respectively. Let
$\\{\cdot,\cdot\\}$ be a Poisson bracket whose Poisson bivector field does not
vanish on $K_{1}\cup K_{2}$ and put
$[\cdot,\cdot]_{1}:=\varphi\\{\cdot,\cdot\\}$ and
$[\cdot,\cdot]_{2}:=\psi\\{\cdot,\cdot\\}$. Since $\\{\cdot,\cdot\\}$ is a(n
almost) Poisson bracket, and since the almost Poisson brackets are a sub-
module of the $\mathcal{C}^{\infty}(M)$-module of $\mathbb{R}$-bilinear
functions on $\mathcal{C}^{\infty}(M)$, $[\cdot,\cdot]_{1},[\cdot,\cdot]_{2}$
are almost Poisson brackets with non-vanishing bivector field. By construction
of $\varphi$ and $\psi$, we have

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}:[f,h]_{1}[g,h]_{2}=\varphi[f,h]_{2}[g,h]_{2}=\psi[f,h]_{1}[g,h]_{1}\equiv
0.$

As a consequence, there are uncountably many representations (2.5) of the
trivial conservative-irreversible function.

##  3 Local representations

In this section, we study the representations of conservative-irreversible
functions in local coordinates. As a consequence of Lemma 2.7, the local
matrix fields associated to conservative-irreversible functions are readily
characterised.

######  Proposition 3.1.

A function
$E:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M)$ is
conservative-irreversible if, and only if, for each patch of local coordinates
$x\in\mathbb{R}^{n}$ on a nonempty open set $U\subseteq M$, there exists a
smooth tensor field $\varepsilon:U\to\mathbb{R}^{n\times n\times n\times n}$
with the properties

  1. (i)

$\varepsilon_{i,j,k,\ell}(x)+\varepsilon_{k,j,\ell,i}(x)+\varepsilon_{\ell,j,i,k}(x)=0$,

  2. (ii)

$\varepsilon_{i,j,k,\ell}(x)-\varepsilon_{i,j,\ell,k}(x)=0$,

  3. (iii)

$\left[\sum_{k,\ell=1}^{n}e_{i,j,k,\ell}(x)y_{k}y_{\ell}\right]_{i,j=1}^{n}$
symmetric and positive semi-definite for all $y\in\mathbb{R}^{n}$

for all $i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}:=\left\\{1,\ldots,n\right\\}$
and $x\in U$, so that, for all $f,g,h\in\mathcal{C}^{\infty}(M)$ and $x\in U$,

$\displaystyle
E(f,g,h)(x)=\sum_{i,j,k,\ell=1}^{n}\varepsilon_{i,j,k,\ell}(x)\partial_{x_{i}}f(x)\partial_{x_{j}}g(x)\partial_{x_{k}}h(x)\partial_{x_{\ell}}h(x).$
(3.1)

###### Proof.

Since $E$ is associated to a four-contravariant tensor field by (2.4), there
exist smooth local tensor fields $\varepsilon$ with (3.1). Sufficiency and
necessity of the properties (i)–(iii) for $E$ being conservative-irreversible
are a direct consequence of [12, Proposition 8, section 3]. ∎

Of particular interest are the simple conservative-irreversible functions,
i.e. those that are (up to multiplication with a scalar factor and order of
the arguments) the tensor square of an almost Poisson bracket. In the
following, we want to answer the question: Are there non-simple conservative-
irreversible functions? If the answer to this question is affirmative, then
this raises the question of the relation between simple and general
conservative-irreversible functions. In particular, contains the span of the
former with respect to $\mathcal{C}^{\infty}(M)$ the latter or are the general
conservative-irreversible functions truly different? To answer these
questions, we calculate local bases of the smallest subbundle of the tensor
bundle containing all conservative-irreversible functions. As a first step, we
calculate the fibre-wise dimension of the vector bundle given by the
properties (i) and (ii) and the symmetry of (iii) in Proposition 3.1.

######  Lemma 3.2.

Let $n\in\mathbb{N}^{*}$. The dimension of the subvector space

$\displaystyle V_{3}:=\left\\{e\in\mathbb{R}^{n\times n\times n\times
n}\,\big{|}\,\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq
n}:e_{i,j,k,\ell}+e_{k,j,\ell,i}+e_{\ell,j,i,k}=e_{i,j,k,\ell}-e_{i,j,\ell,k}=e_{i,j,k,\ell}-e_{j,i,k,\ell}=0\right\\}$

of $\mathbb{R}^{n\times n\times n\times n}$ is
$\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right)$.

###### Proof.

Consider the action of $S_{4}$, the group of permutations of
$\mathbb{N}^{*}_{\leq 4}$, on $(\mathbb{N}^{*}_{\leq n})^{4}$ given by

$\displaystyle\pi\cdot\alpha:=(\alpha_{\pi(1)},\ldots,\alpha_{\pi(4)})$

for all $\alpha\in(\mathbb{N}^{*}_{\leq n})^{4}$ and $\pi\in S_{4}$. Recall
that $S_{4}$ is generated by the permutations $\pi_{1}:=(34)$, $\pi_{2}:=(12)$
and $\pi_{3}:=(14)(34)$. Therefore, for all $i,j,k,\ell\in\mathbb{N}_{\leq
n}^{*}$,

$\displaystyle S_{i,j,k,\ell}$
$\displaystyle:=\left\\{x\in(\mathbb{N}^{*}_{\leq
n})^{4}\,\big{|}\,\left\\{i,j,k,\ell\right\\}=\left\\{x_{1},x_{2},x_{3},x_{4}\right\\}\right\\}$
$\displaystyle=\left\\{\pi\cdot(i,j,k,\ell)\,\big{|}\,\pi\in S_{4}\right\\}$
$\displaystyle=\left\\{\pi_{i_{1}}\cdots\pi_{i_{k}}\cdot(i,j,k,\ell)\,\big{|}\,i_{1},\ldots,i_{k}\in\mathbb{N}_{\leq
3}^{*},k\in\mathbb{N}_{0}\right\\}.$

Furthermore, the conditions

$\displaystyle\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}:\quad
e_{i,j,k,\ell}+e_{k,j,\ell,i}+e_{\ell,j,i,k}$ $\displaystyle=0,$
$\displaystyle e_{i,j,k,\ell}-e_{i,j,\ell,k}$ $\displaystyle=0,$
$\displaystyle e_{j,i,k,\ell}-e_{j,i,k,\ell}$ $\displaystyle=0$

are equivalent to

$\displaystyle\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}\leavevmode\nobreak\
\forall\alpha\in\left\\{i,j,k,\ell\right\\}^{4}:\quad
e_{\alpha}+e_{\pi_{3}\cdot\alpha}$
$\displaystyle=-e_{\pi_{3}^{2}\cdot\alpha},$ (3.2) $\displaystyle e_{\alpha}$
$\displaystyle=e_{\pi_{2}\cdot\alpha},$ $\displaystyle e_{\alpha}$
$\displaystyle=e_{\pi_{1}\cdot\alpha}.$

Let $i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}$ and $e\in V_{3}$. Then, (3.2)
yields that the values $e_{\alpha}$, $\alpha\in S_{i,j,k,\ell}$, are uniquely
determined by $e_{(i,j,k,\ell)},e_{\pi_{3}\cdot(i,j,k,\ell)}$ and
$e_{\pi_{3}\cdot\pi_{1}\cdot\pi_{2}\cdot(i,j,k,\ell)}$. If $i=j=k=\ell$, then
(3.2) yields that $e_{\alpha}=0$ for all $\alpha\in
S_{i,i,i,i}=\left\\{(i,i,i,i)\right\\}$. If
$\left|\left\\{i,j,k,\ell\right\\}\right|=2$, then we may assume that $j\neq
i=k=\ell$, since $S_{i,j,k,\ell}$ is invariant under permutation of $i,j,k$
and $\ell$. Then, we find

$\displaystyle-2e_{i,i,j,i}=-e_{\pi_{3}\cdot(j,i,i,i)}-e_{\pi_{3}^{2}\cdot(j,i,i,i)}=e_{j,i,i,i}=e_{i,j,i,i}=\frac{1}{3}(e_{i,j,i,i}+e_{\pi_{3}\cdot(i,j,i,i)}+e_{\pi_{3}^{2}\cdot(i,j,i,i)})=0.$

Therefore, we conclude that $e_{\alpha}=0$ for all $\alpha\in S_{i,j,k,\ell}$.
If $\left|\left\\{i,j,k,\ell\right\\}\right|=3$, then we may without loss of
generality assume that $k=\ell$. Thus, (3.2) implies

$\displaystyle
2e_{k,i,k,j}=e_{k,i,k,j}+e_{k,i,j,k}=-e_{j,i,k,k}=-e_{i,j,k,k}=e_{k,j,k,i}+e_{k,j,i,k}=2e_{k,j,k,i}$

and we conclude that the $e_{\alpha}$, $\alpha\in S_{i,j,k,\ell}$, are
uniquely determined by $e_{i,j,k,k}$. Let
$\left|\left\\{i,j,k,\ell\right\\}\right|=4$. Then, (3.2) implies

$\displaystyle 0$
$\displaystyle=-(-e_{i,j,k,\ell}-e_{\ell,i,k,j})-e_{k,j,\ell,i}+e_{\ell,i,k,j}+(-e_{i,j,k,\ell}-e_{k,j,\ell,i})$
$\displaystyle=2(e_{\ell,i,k,j}-e_{k,j,\ell,i})=2(e_{\pi_{3}\cdot(i,j,k,\ell)}-e_{\pi_{3}\cdot\pi_{1}\cdot\pi_{2}\cdot(i,j,k,\ell)}).$

Therefore, we conclude

$\displaystyle\dim V_{3}$
$\displaystyle=\binom{n}{2}+\binom{n}{3}+2\binom{n}{4}$
$\displaystyle=\frac{1}{2}n(n-1)\left(1+\frac{1}{3}(n-2)+\frac{1}{61}(n-2)(n-3)\right)$
$\displaystyle=\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right)$

∎

In the case of symmetric and positive semi-definite real matrices, their
linear span is the space of symmetric matrices. An analogous property can be
observed for the convex subset of $V_{3}$ whose elements are characterised by
the additional non-negativity condition of Proposition 3.1 (iii).

######  Lemma 3.3.

Let $n\in\mathbb{N}^{*}$. The convex cone

$\displaystyle S:=\left\\{e\in\mathbb{R}^{n\times n\times n\times
n}\,\left|\,\begin{array}[]{l}\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq
n}:e_{i,j,k,\ell}+e_{k,j,\ell,i}+e_{\ell,j,i,k}=e_{i,j,k,\ell}-e_{i,j,\ell,k}=e_{i,j,k,\ell}-e_{j,i,k,\ell}=0,\\\
\forall
x\in\mathbb{R}^{n}:\left[\sum_{k,\ell=1}^{n}e_{i,j,k,\ell}x_{k}x_{\ell}\right]_{i,j=1}^{n}\leavevmode\nobreak\
\text{positive\leavevmode\nobreak\ semi-definite}\end{array}\right.\right\\}$

spans the vector space $V_{3}$ of Lemma 3.2

###### Proof.

Denote with $\mathfrak{T}_{k}(S):=\left\\{x\in
S^{k}\,\big{|}\,x_{1}<\cdots<x_{k}\right\\}$ the set of increasing $k$-tuples
in a given totally ordered set $S$. Define the functions

$\displaystyle a:\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq n})\to
V_{3},\quad(a_{(\iota,\kappa)})_{i,j,k,\ell}:=\begin{cases}2,&(i,j,k,\ell)\in\left\\{(\iota,\iota,\kappa,\kappa),(\kappa,\kappa,\iota,\iota)\right\\},\\\
-1&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\kappa,\kappa,\iota),(\iota,\kappa,\iota,\kappa),\\\
(\kappa,\iota,\iota,\kappa),(\kappa,\iota,\kappa,\iota)\end{array}\right\\},\\\
0&\text{else},\end{cases}$ $\displaystyle
b:\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq n})\to
V_{3},\quad(b_{(\iota,\kappa,\lambda)})_{i,j,k,\ell}:=\begin{cases}2,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\kappa,\lambda,\lambda),(\kappa,\iota,\lambda,\lambda),\\\
(\lambda,\lambda,\kappa,\iota),(\lambda,\lambda,\iota,\kappa)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\lambda,\iota,\lambda,\kappa),(\lambda,\iota,\kappa,\lambda),\\\
(\lambda,\kappa,\lambda,\iota),(\lambda,\kappa,\iota,\lambda),\\\
(\kappa,\lambda,\lambda,\iota),(\kappa,\lambda,\iota,\lambda),\\\
(\iota,\lambda,\lambda,\kappa),(\iota,\lambda,\kappa,\lambda)\end{array}\right\\},\\\
0,&\text{else},\end{cases}$ $\displaystyle
c:\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\to
V_{3},\quad(c_{(\iota,\kappa,\lambda,\mu)})_{i,j,k,\ell}:=\begin{cases}1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\kappa,\lambda,\mu),(\iota,\kappa,\mu,\lambda),\\\
(\kappa,\iota,\lambda,\mu),(\kappa,\iota,\mu,\lambda),\\\
(\mu,\lambda,\kappa,\iota),(\mu,\lambda,\iota,\kappa),\\\
(\lambda,\mu,\kappa,\iota),(\lambda,\mu,\iota,\kappa)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\mu,\kappa,\iota,\lambda),(\mu,\kappa,\lambda,\iota),\\\
(\kappa,\mu,\iota,\lambda),(\kappa,\mu,\lambda,\iota),\\\
(\iota,\lambda,\mu,\kappa),(\iota,\lambda,\kappa,\mu),\\\
(\lambda,\iota,\mu,\kappa),(\lambda,\iota,\kappa,\mu)\end{array}\right\\},\\\
0,&\text{else},\end{cases}$

and

$\displaystyle d:\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\to
V_{3},\quad(d_{(\iota,\kappa,\lambda,\mu)})_{i,j,k,\ell}:=\begin{cases}1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\lambda,\kappa,\mu,\iota),(\lambda,\kappa,\iota,mu),\\\
(\kappa,\lambda,\mu,\iota),(\kappa,\lambda,\iota,\mu),\\\
(\mu,\iota,\lambda,\kappa),(\mu,\iota,\kappa,\lambda),\\\
(\iota,\mu,\lambda,\kappa),(\iota,\mu,\kappa,\lambda)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\mu,\kappa,\iota,\lambda),(\mu,\kappa,\lambda,\iota),\\\
(\kappa,\mu,\iota,\lambda),(\kappa,\mu,\lambda,\iota),\\\
(\lambda,\iota,\kappa,\mu),(\lambda,\iota,\mu,\kappa),\\\
(\iota,\lambda,\kappa,\mu),(\iota,\lambda,\mu,\kappa)\end{array}\right\\},\\\
0,&\text{else}.\end{cases}$

Evidently, the set
$B:=\left\\{a_{A},b_{B},c_{C},d_{D}\,\big{|}\,A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n}),T\in\mathfrak{P}_{3}(\mathbb{N}^{*}_{\leq
n}),C,D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\right\\}$ is linear
independent and has cardinality
$\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right)$.
Therefore, $B$ is a basis of $V_{3}$ and it remains to show that $B\subseteq
S$. Let $(\iota,\kappa)\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq n})$. Then, we
have, for all $x,y\in\mathbb{R}^{n}$,

$\displaystyle\sum_{i,j,k,\ell=1}^{n}(a_{(\iota,\kappa)})_{i,j,k,\ell}x_{i}x_{j}y_{k}y_{\ell}=2x_{\iota}^{2}y_{\kappa}^{2}+2x_{\kappa}^{2}y_{\iota}^{2}-4x_{\iota}x_{\kappa}y_{\iota}y_{\kappa}=2(x_{\iota}y_{\kappa}-x_{\kappa}y_{\iota})^{2}\geq
0$

and hence $a_{(\iota,\kappa)}\in S$. Let
$(\iota,\kappa,\lambda)\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq n})$ and
consider
$e^{(\iota,\kappa,\lambda)}:=a_{(\iota,\lambda)}+a_{(\kappa,\lambda)}+b_{(\iota,\kappa,\lambda)}$.
Then, we have, for all $x,y\in\mathbb{R}^{n}$,

$\displaystyle\sum_{i,j,k,\ell=1}^{n}e^{(\iota,\kappa,\lambda)}_{i,j,k,\ell}x_{i}x_{j}y_{k}y_{\ell}$
$\displaystyle=2(x_{\iota}y_{\lambda}-x_{\lambda}y_{\iota})^{2}+2(x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})^{2}+4(x_{\iota}y_{\lambda}-x_{\lambda}y_{\iota})(x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})$
$\displaystyle=2(x_{\iota}y_{\lambda}-x_{\lambda}y_{\iota}+x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})^{2}\geq
0$

and thus $e^{(\iota,\kappa,\lambda)}\in S$. Let
$(\iota,\kappa,\lambda,\mu)\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})$ and
consider
$e^{(\iota,\kappa,\lambda,\mu)}:=a_{(\iota,\mu)}+a_{(\kappa,\lambda)}+c_{(\iota,\kappa,\lambda,\mu)}$
and
$\widetilde{e}^{(\iota,\kappa,\lambda,\mu)}:=a_{(\iota,\kappa)}+a_{(\lambda,\mu)}+d_{(\iota,\kappa,\lambda,\mu)}$.
Then, for all $x,y\in\mathbb{R}^{n}$,

$\displaystyle\sum_{i,j,k,\ell=1}^{n}e^{(\iota,\kappa,\lambda,\mu)}_{i,j,k,\ell}x_{i}x_{j}y_{k}y_{\ell}$
$\displaystyle=2(x_{\iota}y_{\mu}-x_{\mu}y_{\iota})^{2}+2(x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})^{2}+4(x_{\iota}y_{\mu}-x_{\mu}y_{\iota})(x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})$
$\displaystyle=2(x_{\iota}y_{\mu}-x_{\mu}y_{\iota}+x_{\kappa}y_{\lambda}-x_{\lambda}y_{\kappa})^{2}\geq
0$

and

$\displaystyle\sum_{i,j,k,\ell=1}^{n}\widetilde{e}^{(\iota,\kappa,\lambda,\mu)}_{i,j,k,\ell}x_{i}x_{j}y_{k}y_{\ell}$
$\displaystyle=2(x_{\iota}y_{\kappa}-x_{\kappa}y_{\iota})^{2}+2(x_{\lambda}y_{\mu}-x_{\mu}y_{\lambda})^{2}+4(x_{\kappa}y_{\iota}-x_{\iota}y_{\kappa})(x_{\lambda}y_{\mu}-x_{\mu}y_{\lambda})$
$\displaystyle=2(x_{\kappa}y_{\iota}-x_{\iota}y_{\kappa}+x_{\lambda}y_{\mu}-x_{\mu}y_{\lambda})^{2}\geq
0.$

Therefore,
$e^{(\iota,\kappa,\lambda,\mu)},\widetilde{e}^{(\iota,\kappa,\lambda,\mu)}\in
S$. Since
$\left\\{a_{A},e^{B},e^{C},\widetilde{e}^{D}\,\big{|}\,A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n}),B\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq
n}),C,D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\right\\}$ is linear
independent and $S\subseteq V_{3}$, we conclude

$\displaystyle\dim V_{3}=\left|\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n})\right|+\left|\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq
n})\right|+2\left|\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\right|\leq\dim
S\leq\dim V_{3}$

and hence $\dim S=\dim
V_{3}=\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right).$
∎

As a consequence of this calculation, the following result is derived.

######  Proposition 3.4.

The restriction of the vector bundle structure of $T^{4}M$ to the set

$\displaystyle B:=\left\\{\varepsilon\in T^{4}_{x}M\,\big{|}\,x\in
M\leavevmode\nobreak\
\forall\alpha,\beta,\gamma,\delta\in\mathcal{T}_{x}^{*}M:\begin{array}[]{l}\varepsilon(\alpha,\beta,\gamma,\delta)=\varepsilon(\beta,\alpha,\gamma,\delta)=\varepsilon(\alpha,\beta,\delta,\gamma),\\\
\varepsilon(\alpha,\beta,\gamma,\delta)+\varepsilon(\gamma,\beta,\delta,\alpha)+\varepsilon(\delta,\beta,\alpha,\gamma)=0\end{array}\right\\}$

is the smallest subbundle of $T^{4}M$ so that each tensor field $\varepsilon$
with the properties (a)–(c) of Lemma 2.7 is a smooth section of $B$.

###### Proof.

It is evident that $B$ is a subvector bundle of $T^{4}M$ and that all tensor
fields associated to conservative-irreversible functions are sections of $B$.
Let $B^{\prime}$ denote the smallest subvector bundle of $T^{4}M$ with the
latter property. Using an atlas of local trivialisations and a subordinate
partition of unity, we see that for each $x\in M$ and each $\varepsilon\in
T^{4}_{x}M$ with the properties (a)–(c) of Lemma 2.7, there exists a tensor
field $\varepsilon^{\prime}\in\Gamma(T^{4}M)$ with
$\varepsilon^{\prime}_{x}=\varepsilon$ which has the properties (a)–(c) of
Lemma 2.7. Analogously to Lemma 3.3, we conclude that the rank of $B^{\prime}$
is $\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right)$.
With Lemma 3.2, we see that the rank of $B$ is
$\frac{1}{2}n(n-1)\left(\frac{1}{6}n^{2}-\frac{1}{2}n+\frac{4}{3}\right)$.
Since $B^{\prime}\subseteq B$, we conclude that $B^{\prime}=B$. This shows the
proposition. ∎

We may use the basis used in the proof of Lemma 3.3 to show that there are
non-simple conservative-irreversible functions.

######  Proposition 3.5.

There exists $e\in S$ so that there is no $J\in\mathbb{R}^{n\times n}$ so that

$\displaystyle\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq
n}:e_{i,j,k,\ell}=J_{i,k}J_{j,\ell}+J_{i,\ell}J_{j,k}$ (3.3)

if, and only if, $n\geq 3$.

###### Proof.

“$\implies$” If $n=1$, then $S=\left\\{0\right\\}$ and each $e$ has the
representation (3.3) with $J=0$. If $n=2$, then $V_{3}$ is one-dimensional and
each $e\in S$ has the representation (3.3) with

$\displaystyle J=\alpha\begin{bmatrix}0&1\\\ -1&0\end{bmatrix}$

for some $\alpha\in[0,\infty)$.

“$\impliedby$” Let $n\geq 3$ and consider $e\in V_{3}$ defined by

$\displaystyle\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq
n}:e_{i,j,k,\ell}=\begin{cases}2,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(1,1,2,2),(2,2,1,1),\\\
(1,1,3,3),(3,3,1,1)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(1,2,2,1),(1,2,1,2),\\\
(2,1,1,2),(2,1,2,1),\\\ (1,3,3,1),(1,3,1,3),\\\
(3,1,3,1),(3,1,1,3)\end{array}\right\\},\\\ 0,&\text{else}.\end{cases}$

Then, we have, for all $x,y\in\mathbb{R}$,

$\displaystyle\sum_{i,j,k,\ell=1}^{n}e_{i,j,k,\ell}x_{i}x_{j}y_{i}y_{k}=2(x_{1}y_{2}-x_{2}y_{1})^{2}+2(x_{1}x_{3}-x_{3}x_{1})^{2}\geq
0$

and hence $e\in S$. Seeking a contradiction, assume that $e$ has a
representation (3.3). Then, we have

$\displaystyle\frac{1}{2}e_{1,1,2,2}=\frac{1}{2}e_{1,1,3,3}=J_{1,2}^{2}=J_{1,3}^{2}=1$

and thus

$\displaystyle 0=e_{1,1,2,3}=2J_{1,2}J_{1,3}\neq 0.$

This shows that $e$ does not have a representation (3.3). ∎

######  Corollary 3.6.

There exists a conservative-irreversible function on $\mathbb{R}^{n}$ which is
not simple in standard coordinates if, and only if, $n\geq 2$.

###### Proof.

By [12, Definition 3.2], there exists for each conservative-irreversible
function
$E:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$
a unique smooth $e:\mathbb{R}^{n}\to S$ so that

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=\sum_{i,j,k,\ell=1}^{n}e_{i,j,k,\ell}\partial_{i}f\partial_{j}g\partial_{k}h\partial_{\ell}h.$
(3.4)

In [12, Theorem 4.2], it was shown that $E$ is simple if, and only if, there
is a pointwise skew-symmetric function $J:\mathbb{R}^{n}\to\mathbb{R}^{n\times
n}$ so that

$\displaystyle\forall i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}\leavevmode\nobreak\
\forall
x\in\mathbb{R}^{n}:e_{i,j,k,\ell}(x)=J_{i,k}(x)J_{j,\ell}(x)+J_{i,\ell}(x)J_{j,k}(x).$

By Proposition 3.5, we conclude that for $n\leq 2$ each conservative-
irreversible function is simple. If $n\geq 3$, then there exists $e_{0}\in S$
that has no representation (3.3). Then, the constant function $x\mapsto e_{0}$
is smooth and generates by (3.4) a conservative-irreversible function which is
not simple. ∎

In the proof of the following Lemma, we show that each element of the basis
constructed in Lemma 3.3 is the tensor representation of a simple
conservative-irreversible function. In particular, this implies the following
result.

######  Lemma 3.7.

Each conservative-irreversible function may be locally represented by the
weighted sum of simple conservative-irreversible functions.

###### Proof.

In the following, we use the notation introduced in the proof of Lemma 3.3.
Let $(\iota,\kappa)\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq n})$ and define
$J^{(\iota,\kappa)}\in\mathbb{R}^{n\times n}$ as

$\displaystyle\forall i,j\in\mathbb{N}^{*}_{\leq
n}:J_{i,j}^{(\iota,\kappa)}:\begin{cases}1,&(i,j)=(\iota,\kappa),\\\
-1,&(j,i)=(\iota,\kappa),\\\ 0,&\text{else}.\end{cases}$

Then, we have, for all $i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}$,

$\displaystyle
J_{i,k}^{(\iota,\kappa)}J_{j,\ell}^{(\iota,\kappa)}+J_{i,\ell}^{(\iota,\kappa)}J_{j,k}^{(\iota,\kappa)}=\left.\begin{cases}2,&i=j=\iota\leavevmode\nobreak\
\wedge\leavevmode\nobreak\ k=\ell=\kappa,\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\kappa,\iota,\kappa),(\iota,\kappa,\kappa,\iota),\\\
(\kappa,\iota,\iota,\kappa)\end{array}\right\\},\\\
0,&\text{else},\end{cases}\right\\}=(\alpha_{\left\\{(\iota,\kappa)\right\\}})_{i,j,k,\ell}.$

Let $(\iota,\kappa,\lambda)\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq n})$ and
define $K^{(\iota,\kappa,\lambda)}\in\mathbb{R}^{n\times n}$ by

$\displaystyle\forall i,j\in\mathbb{N}^{*}_{\leq
n}:K_{i,j}^{(\iota,\kappa,\lambda)}:=\begin{cases}1,&(i,j)\in\left\\{(\iota,\lambda),(\kappa,\lambda)\right\\},\\\
-1,&(i,j)\in\left\\{(\lambda,\iota),(\lambda,\kappa)\right\\},\\\
0,&\text{else}.\end{cases}$

Then, we have, for all $i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}$,

$\displaystyle
K_{i,k}^{(\iota,\kappa,\lambda)}K_{j,\ell}^{(\iota,\kappa,\lambda)}+K_{i,\ell}^{(\iota,\kappa,\lambda)}K_{j,k}^{(\iota,\kappa,\lambda)}=\left.\begin{cases}2,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\iota,\lambda,\lambda),(\lambda,\lambda,\iota,\iota),\\\
(\kappa,\kappa,\lambda,\lambda),(\lambda,\lambda,\kappa,\kappa),\\\
(\iota,\kappa,\lambda,\lambda),(\kappa,\iota,\lambda,\lambda),\\\
(\lambda,\lambda,\iota,\kappa),(\lambda,\lambda,\kappa,\iota)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\lambda,\lambda,\iota),(\iota,\lambda,\iota,\lambda),\\\
(\lambda,\iota,\lambda,\iota),(\lambda,\iota,\iota,\lambda),\\\
(\lambda,\iota,\lambda,\kappa),(\lambda,\iota,\kappa,\lambda),\\\
(\lambda,\kappa,\lambda,\iota),(\lambda,\kappa,\iota,\lambda),\\\
(\kappa,\lambda,\lambda,\iota),(\kappa,\lambda,\iota,\lambda),\\\
(\iota,\lambda,\lambda,\kappa),(\iota,\lambda,\kappa,\lambda)\end{array}\right\\},\\\
0,&\text{else},\end{cases}\right\\}=e^{(\iota,\kappa,\lambda)}_{i,j,k,\ell}.$

Let lastly $(\iota,\kappa,\lambda,\mu)\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq
n})$ and define the matrices
$M^{(\iota,\kappa,\lambda,\mu)},N^{(\iota,\kappa,\lambda,\mu)}\in\mathbb{R}^{n\times
n}$ with

$\displaystyle\forall i,j\in\mathbb{N}^{*}_{\leq
n}:M_{i,j}^{(\iota,\kappa,\lambda,\mu)}:=\begin{cases}1,&(i,j)\in\left\\{(\iota,\mu),(\kappa,\lambda)\right\\},\\\
-1,&(i,j)\in\left\\{(\mu,\iota),(\lambda,\kappa)\right\\},\\\
0,&\text{else},\end{cases}$

and

$\displaystyle\forall i,j\in\mathbb{N}^{*}_{\leq
n}:N_{i,j}^{(\iota,\kappa,\lambda,\mu)}:=\begin{cases}1,&(i,j)\in\left\\{(\kappa,\iota),(\lambda,\mu)\right\\},\\\
-1,&(i,j)\in\left\\{(\iota,\kappa),(\mu,\lambda)\right\\},\\\
0,&\text{else}.\end{cases}$

Then, we have, for all $i,j,k,\ell\in\mathbb{N}^{*}_{\leq n}$,

$\displaystyle
M_{i,k}^{(\iota,\kappa,\lambda,\mu)}M_{j,\ell}^{(\iota,\kappa,\lambda,\mu)}+M_{i,\ell}^{(\iota,\kappa,\lambda,\mu)}M_{j,k}^{(\iota,\kappa,\lambda,\mu)}=\left.\begin{cases}2,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\iota,\mu,\mu),(\mu,\mu,\iota,\iota),\\\
(\kappa,\kappa,\lambda,\lambda),(\lambda,\lambda,\kappa,\kappa)\end{array}\right\\},\\\
1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\kappa,\lambda,\mu),(\iota,\kappa,\mu,\lambda),\\\
(\kappa,\iota,\lambda,\mu),(\kappa,\iota,\mu,\lambda),\\\
(\mu,\lambda,\kappa,\iota),(\mu,\lambda,\iota,\kappa),\\\
(\lambda,\mu,\kappa,\iota),(\lambda,\mu,\iota,\kappa)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\mu,\kappa,\iota,\lambda),(\mu,\kappa,\lambda,\iota),\\\
(\kappa,\mu,\iota,\lambda),(\kappa,\mu,\lambda,\iota),\\\
(\iota,\lambda,\mu,\kappa),(\iota,\lambda,\kappa,\mu),\\\
(\lambda,\iota,\mu,\kappa),(\lambda,\iota,\kappa,\mu)\end{array}\right\\},\\\
0,&\text{else},\end{cases}\right\\}=e^{(\iota,\kappa,\lambda,\mu)}_{i,j,k,\ell}$

and

$\displaystyle
N_{i,k}^{(\iota,\kappa,\lambda,\mu)}N_{j,\ell}^{(\iota,\kappa,\lambda,\mu)}+N_{i,\ell}^{(\iota,\kappa,\lambda,\mu)}N_{j,k}^{(\iota,\kappa,\lambda,\mu)}=\left.\begin{cases}2,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\iota,\iota,\kappa,\kappa),(\kappa,\kappa,\iota,\iota),\\\
(\mu,\mu,\lambda,\lambda),(\lambda,\lambda,\mu,\mu)\end{array}\right\\},\\\
1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\lambda,\kappa,\mu,\iota),(\lambda,\kappa,\iota,mu),\\\
(\kappa,\lambda,\mu,\iota),(\kappa,\lambda,\iota,\mu),\\\
(\mu,\iota,\lambda,\kappa),(\mu,\iota,\kappa,\lambda),\\\
(\iota,\mu,\lambda,\kappa),(\iota,\mu,\kappa,\lambda)\end{array}\right\\},\\\
-1,&(i,j,k,\ell)\in\left\\{\begin{array}[]{c}(\mu,\kappa,\iota,\lambda),(\mu,\kappa,\lambda,\iota),\\\
(\kappa,\mu,\iota,\lambda),(\kappa,\mu,\lambda,\iota),\\\
(\lambda,\iota,\kappa,\mu),(\lambda,\iota,\mu,\kappa),\\\
(\iota,\lambda,\kappa,\mu),(\iota,\lambda,\mu,\kappa)\end{array}\right\\},\\\
0,&\text{else}.\end{cases}\right\\}=\widetilde{e}^{(\iota,\kappa,\lambda,\mu)}_{i,j,k,\ell}.$

In the proof of Lemma 3.3, we have seen that

$\displaystyle
B=\left\\{a_{A},e^{B},e^{C},\widetilde{e}^{D}\,\big{|}\,A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n}),B\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq
n}),C,D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})\right\\}$

is a basis of $V_{3}$. Let $E$ be a conservative-irreversible function with
associated representation (3.4) for some
$e\in\mathcal{C}^{\infty}(\mathbb{R}^{n},S)$. Since $B$ is a basis of $V_{3}$,
there exist
$\alpha_{A},\beta_{B},\gamma_{C},\delta_{D}\in\mathcal{C}^{\infty}(\mathbb{R}^{n},\mathbb{R})$,
$A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq n})$,
$B\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq n})$,
$C,D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq n})$, so that

$\displaystyle e=\sum_{A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n})}\alpha_{A}a_{A}+\sum_{B\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq
n})}\beta_{B}e^{B}+\sum_{C\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq
n})}\gamma_{C}e^{C}+\sum_{D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq
n})}\delta_{D}\widetilde{e}^{D}$ (3.5)

For each skew-symmetric matrix $J\in\mathbb{R}^{n\times n}$, we denote with

$\displaystyle\\{\cdot,\cdot\\}_{J}:=(f,g)\mapsto(\partial f)^{\top}J(\partial
g)$

the Poisson bracket induced by $J$. Then, the representation (3.5) and our
calculations above yield

$\displaystyle\forall f,g,h\in\mathbb{R}^{n}:E(f,g,h)$
$\displaystyle=\sum_{A\in\mathfrak{T}_{2}(\mathbb{N}^{*}_{\leq
n})}2\alpha_{A}\\{f,h\\}_{J^{A}}\\{g,h\\}_{J^{A}}+\sum_{B\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq
n})}2\beta_{B}\\{f,h\\}_{K^{B}}\\{g,h\\}_{K^{B}}$
$\displaystyle\quad+\sum_{C\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq
n})}2\gamma_{C}\\{f,h\\}_{M^{C}}\\{g,h\\}_{M^{C}}+\sum_{D\in\mathfrak{T}_{4}(\mathbb{N}^{*}_{\leq
n})}2\delta_{D}\\{f,h\\}_{N^{D}}\\{g,h\\}_{N^{D}}.$

This shows that $E$ is indeed the weighted sum of simple conservative-
irreversible functions. ∎

We have shown that any conservative-irreversible function is the weighted sum
of simple conservative-irreversible functions. One might expect that all
scalar factors are non-negative. This is, however, not necessarily the case,
as the following example shows.

######  Example 3.8.

We show that the weights constructed in the proof do not necessarily need to
be positive. Let $n\geq 3$ and
$(i,j,k)\in\mathfrak{T}_{3}(\mathbb{N}^{*}_{\leq n})$. Consider
$e:=a_{(i,k)}+a_{(j,k)}-\frac{1}{2}e^{(i,j,k)}$. Then, we have, for all
$x,y\in\mathbb{R}^{n}$,

$\displaystyle
f(x,y):=\frac{1}{2}\sum_{i,j,k,\ell=1}^{n}e_{i,j,k,\ell}x_{i}x_{j}y_{i}y_{j}$
$\displaystyle=(x_{i}y_{k}-x_{k}y_{i})^{2}+(x_{j}y_{k}-x_{k}y_{j})^{2}$
$\displaystyle\quad-\frac{1}{2}((x_{i}y_{k}-x_{k}y_{i})+(x_{j}y_{k}-x_{k}y_{j}))^{2}$
$\displaystyle=\frac{1}{2}(x_{i}y_{k}-x_{k}y_{i})^{2}+\frac{1}{2}(x_{j}y_{k}-x_{k}y_{j})^{2}-(x_{i}y_{k}-x_{k}y_{i})(x_{j}y_{k}-x_{k}y_{j})$
$\displaystyle=\frac{1}{2}((x_{i}y_{k}-x_{k}y_{i})-(x_{j}y_{k}-x_{k}y_{j}))^{2}\geq
0.$

Therefore, $e\in S$ and the associated function

$\displaystyle
E(f,g,h):=\sum_{i,j,k,\ell}e_{i,j,k,\ell}\partial_{i}f\partial_{j}g\partial_{k}h\partial_{\ell}h$

is conservative-irreversible.

As a consequence of Lemma 3.7, we shall now conclude that each conservative-
irreversible function is generated by a metriplectic four-bracket. In
particular, the tensor fields which are uniquely associated to conservative-
irreversible functions are precisely the symmetrisations of metriplectic four-
brackets in the second and third argument.

######  Proposition 3.9.

A function
$E:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$
is a conservative-irreversible function if, and only if, there exists a
metriplectic four-bracket $(\\!(\cdot,\cdot,\cdot,\cdot)\\!)$ on $M$ with

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=(\\!(f,h,g,h)\\!)$ (3.6)

###### Proof.

Sufficiency is a direct consequence of Lemma 2.10; we show the necessity. By
Lemma 3.7, there exists an open covering $(U_{i})_{i\in I}$ of $M$ and a
family $((E_{i,1},\ldots,E_{i,k_{i}}))_{i\in I}$ of simple conservative-
irreversible functions on $U_{i}$ with

$\displaystyle\forall f,g,h\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall i\in
I:E(f,g,h)|_{U_{i}}=E_{i,1}(f,g,h)|_{U_{i}}+\cdots+E_{i,k_{i}}(f,g,h)|_{U_{i}}.$

Consider a (locally finite) partition of unity $(\varphi_{j})_{j\in J}$
subordinate to the covering $(U_{i})_{i\in I}$ and define
$\overline{E}_{i,\ell}$ as the continuation of $E_{i,j}$ on $M$ by zero, $i\in
I$, $\ell\in\left\\{1,\ldots,k_{i}\right\\}$. Choose, for each $j\in J$,
$\iota(j)\in I$ with $\mathrm{supp}\,\varphi_{j}\subseteq U_{\iota(j)}$. Then,
we have $E=\sum_{j\in
J}\varphi_{j}\sum_{\ell=1}^{k_{\iota(j)}}\overline{E}_{\iota(j),\ell}$ and the
sum is a locally finite sum of simple conservative-irreversible functions. In
view of Lemma 2.13, $E$ has the representation (3.6) for a metriplectic four-
bracket. ∎

##  4 Biquadratic functions

Let $E:\big{(}\mathcal{C}^{\infty}(M)\big{)}^{3}\to\mathcal{C}^{\infty}(M)$ be
a conservative-irreversible function. Due to Lemma 2.6, $E$ is symmetric in
the first two entries and induces therefore the function

$\displaystyle\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}_{E}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),\qquad(f,g)\mapsto
E(f,f,g),$

which is biquadratic, i.e. a quadratic function (see Definition 4.1) in each
entry. Conversely, to each biquadratic function
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$, there is a four-linear
function (unique, if symmetry in the first two and second two entries is
imposed) $\varepsilon$ with
$\left\\{\\!\left\\{f,g\right\\}\\!\right\\}=\varepsilon(f,f,g,g)$. In this
section, we characterise the quadratic functions whose uniquely associated
symmetric bilinear functions are derivations in each argument. The proofs use
general properties of algebras so that this characterisation is not only valid
for the algebra of smooth real-valued functions, but for any unital
commutative algebra over a field $k$ with $1_{k}+1_{k}\neq 0_{k}$, i.e. the
characteristic is not two. As a corollary, we obtain a characterisation of the
biquadratic functions associated to conservative-irreversible functions.
Recall the elementary definition of quadratic functions.

######  Definition 4.1 (see [10]).

Let $V,W$ be vector spaces over the same field $k$. A function $q:V\to V$ is
quadratic if, and only if, $q$ is $k$-homogeneous of degree two, i.e.

$\displaystyle\forall\lambda\in k\leavevmode\nobreak\ \forall x\in V:q(\lambda
x)=\lambda^{2}q(x),$

and the function

$\displaystyle\beta:V\times V\to W,\qquad(x,y)\mapsto q(x+y)-q(x)-q(y)$ (4.1)

is bilinear.

When the field $k$ is fixed and there is no danger of confusion, we will refer
to homogeneity as degree-two-homogeneity.

######  Remark 4.2.

Unless $k$ is a field of characteristic two, it is easy to see that a function
$q:V\to W$ is quadratic if, and only if, there exists a bilinear function
$b:V\times V\to W$ so that $q(x)=b(x,x)$ for all $x\in V$: Indeed, if $q$ is
quadratic, then $\beta$ given in (4.1) fulfils

$\displaystyle\beta(x,x)=q(x+x)-q(x)-q(x)=((1_{k}+1_{k})^{2}-(1_{k}+1_{k}))q(x)=(1_{k}+1_{k})q(x);$

and the converse implication is straightforward. In characteristic two,
however, $1_{k}+1_{k}=0_{k}$ and hence $q$ cannot be reconstructed as the
concatenation of the associated bilinear function $\beta$ and the diagonal
embedding of $V$ into $V\times V$. For details about the theory of quadratic
function in characteristic two, see e.g. [6, 13].

Evidently, not each degree-two-homogeneous function is quadratic: all squares
of norms are homogeneous of degree two, but not all norms are induced by inner
products. A well-known characterisation of quadratic forms is the following.

######  Lemma 4.3 (see [10, Theorem 1.1]).

Let $V$, $W$ be vector spaces over the same field $k$. A function $q:V\to W$
is quadratic if, and only if, $q$ fulfils the inclusion-exclusion principle

$\displaystyle\delta(x+y+z)-\delta(x+y)-\delta(x+z)-\delta(y+z)+\delta(x)+\delta(x)+\delta(y)=0,$
(4.2)

and the generalized binomial equation

$\displaystyle\forall x,y\in V\leavevmode\nobreak\ \forall\lambda\in
k:q(\lambda x+y)=\lambda^{2}q(x)+\lambda(q(x+y)-q(x)-q(y))+q(y).$ (4.3)

A combination of the binomial formula with the Leibniz formula characterises
quadratic functions whose associated bilinear functions are biderivations in
characteristic unequal to two.

######  Proposition 4.4.

Let $A$ be a unital commutative algebra with identity $\mathbf{1}$ over the
field $k$ with characteristic not two. A function $\delta:A\to A$ is quadratic
and the associated bilinear function

$\displaystyle\Delta:A\times A\to
A,\qquad(x,y)\mapsto\delta(x+y)-\delta(x)-\delta(y)$

is a biderivation if, and only if, $\delta$ fulfils the inclusion-exclusion
principle (4.2) and, for all $x,y,z\in A$ and $\lambda\in k$,

$\displaystyle\delta(xy+\lambda z)$ $\displaystyle=xy\delta(x+y)+\lambda
x\delta(y+z)+\lambda
y\delta(x+z)+\lambda(\lambda\cdot\mathbf{1}-x-y)\delta(z)$ (4.4)
$\displaystyle\quad+y(y-x-\lambda\cdot\mathbf{1})\delta(x)+x(x-y-\lambda\cdot\mathbf{1})\delta(y).$

###### Proof.

“$\implies$” Due to [10, Theorem 1.1], $\delta$ fulfils the inclusion-
exclusion principle (4.2). Since $2_{k}:=1_{k}+1_{k}\neq 0_{k}$, the
associated bilinear function $\Delta$ fulfils
$q(x)=\frac{1}{2_{k}}\Delta(x,x)$ for all $x\in A$. Due to bilinearity,
Leibniz rule and symmetry, we have

$\displaystyle\delta(xy+z)$ $\displaystyle=\frac{1}{2_{k}}\Delta(xy+z,xy+z)$
$\displaystyle=\frac{1}{2_{k}}\big{(}\Delta(xy,xy)+2\Delta(xy,z)+\Delta(z,z)\big{)}$
$\displaystyle=\frac{1}{2_{k}}\big{(}x^{2}\Delta(y,y)+2xy\Delta(x,y)+y^{2}\Delta(x,x)+2x\Delta(y,z)+2y\Delta(x,z)+\Delta(z,z)\big{)}$
$\displaystyle=\frac{1}{2_{k}}\big{(}x^{2}\Delta(y,y)+y^{2}\Delta(x,x)+\Delta(z,z)+xy(\Delta(x+y,x+y)-\Delta(x,x)-\Delta(y,y))$
$\displaystyle\quad+y(\Delta(x+z,x+z)-\Delta(x,x)-\Delta(z,z))+x(\Delta(y+z,y+z)-\Delta(y,y)-\Delta(z,z))\big{)}$
$\displaystyle=xy\delta(x+y)+y\delta(x+z)+x\delta(y+z)+(1-y-x)\delta(z)+x(x-y-1)\delta(y)$
$\displaystyle\quad+y(y-x-1)\delta(x).$

and hence $\delta$ fulfils indeed (4.4).

“$\impliedby$” We proceed in steps.

Step 1: We show that $\delta(\mathbf{1})=\delta(\mathbf{1}+\mathbf{1})=0$.
Putting $x=y=z=\mathbf{1}$ and $\lambda=0_{k}$ in (4.4) yields

$\displaystyle\delta(\mathbf{1})=\delta(\mathbf{1}\cdot\mathbf{1}+0_{k}\cdot\mathbf{1})=\delta(\mathbf{1}+\mathbf{1}).$

Therefore, putting $x=y=z=\mathbf{1}$ and $\lambda=1_{k}$ in (4.4) yields

$\displaystyle\delta(\mathbf{1})=\delta(\mathbf{1}+\mathbf{1})$
$\displaystyle=\delta(\mathbf{1}+\mathbf{1})+\delta(\mathbf{1}+\mathbf{1})+\delta(\mathbf{1}+\mathbf{1})+1_{k}\cdot(\mathbf{1}-\mathbf{1}-\mathbf{1})\delta(\mathbf{1})+\mathbf{1}(\mathbf{1}-\mathbf{1}-\mathbf{1})\delta(1)$
$\displaystyle\quad+\mathbf{1}(\mathbf{1}-\mathbf{1}-\mathbf{1})\delta(1)$
$\displaystyle=\delta(\mathbf{1})+\delta(\mathbf{1})+\delta(\mathbf{1})-\delta(\mathbf{1})-\delta(\mathbf{1})-\delta(\mathbf{1})=0.$

Step 2: We show that $\delta(x+\mathbf{1})=\delta(x)$ for all $x\in A$. Let
$x\in A$. Then, we have

$\displaystyle\delta(x+\mathbf{1})$
$\displaystyle=\delta(\mathbf{1}\cdot\mathbf{1}+1_{k}\cdot x)$
$\displaystyle=\delta(\mathbf{1}+\mathbf{1})+\delta(x+\mathbf{1})+\delta(x+\mathbf{1})-\delta(x)-\delta(\mathbf{1})-\delta(\mathbf{1})$
$\displaystyle=\delta(x+\mathbf{1})+\delta(x+\mathbf{1})-\delta(x).$

This implies $\delta(x)=\delta(x+\mathbf{1})$.

Step 3: We show that

$\displaystyle\forall x,y\in A\leavevmode\nobreak\ \forall\lambda\in
k:\delta(\lambda
x+y)=\lambda^{2}\delta(x)+\lambda(\delta(x+y)-\delta(x)-\delta(y))+\delta(y).$

Let $x,y\in A$ and $\lambda\in k$. Then,

$\displaystyle\delta(\lambda x+y)$ $\displaystyle=\delta(\mathbf{1}\cdot
y+\lambda x)$ $\displaystyle=y\delta(y+\mathbf{1})+\lambda
y\delta(x+\mathbf{1})+\lambda\delta(x+y)+\lambda(\lambda\cdot\mathbf{1}-y-\mathbf{1})\delta(x)+(\mathbf{1}-y-\lambda\cdot\mathbf{1})\delta(y)$
$\displaystyle\quad+y(y-x-\lambda\cdot\mathbf{1})\delta(\mathbf{1})$
$\displaystyle=y\delta(y)+\lambda
y\delta(x)+\lambda\delta(x+y)+\lambda^{2}\delta(x)-\lambda
y\delta(x)-\lambda\delta(x)+\delta(y)-y\delta(y)-\lambda\delta(y)$
$\displaystyle=\lambda^{2}\delta(x)+\lambda(\delta(x+y)-\delta(x)-\delta(y))+\delta(y)$

Step 4: We have verified that $\delta$ fulfils the generalised binomial
formula (4.3). Therefore, [10, Theorem 1.1] implies that $\delta$ is a
quadratic function. In particular, the associated function $\Delta$ is
bilinear.

Step 5: We show that $\Delta$ is a derivation in each argument. Due to
linearity and symmetry, it suffices to show that $\Delta$ fulfils the Leibniz
rule in the first argument. Let $x,x^{\prime},y\in A$. Then, we have

$\displaystyle\Delta(xx^{\prime},y)$
$\displaystyle=\delta(xx^{\prime}+y)-\delta(xx^{\prime})-\delta(y)$
$\displaystyle=xx^{\prime}\delta(x+x^{\prime})+x\delta(x^{\prime}+y)+x^{\prime}\delta(x+y)+(\mathbf{1}-x-x^{\prime})\delta(y)+x^{\prime}(x^{\prime}-x-\mathbf{1})\delta(x)$
$\displaystyle\quad+x(x-x^{\prime}-\mathbf{1})\delta(x^{\prime})-xx^{\prime}\delta(x+x^{\prime})-x\delta(x^{\prime})-x^{\prime}\delta(x)-(\mathbf{1}-x-x^{\prime})\delta(0)$
$\displaystyle\quad-x^{\prime}(x^{\prime}-x-\mathbf{1})\delta(x)-x(x-x^{\prime}-\mathbf{1})\delta(x^{\prime})-\delta(y)$
$\displaystyle=x\delta(x^{\prime}+y)-x\delta(x^{\prime})-x\delta(y)+x^{\prime}\delta(x+y)-x^{\prime}\delta(x)-x^{\prime}\delta(y)$
$\displaystyle=x\Delta(x^{\prime},y)+x^{\prime}\Delta(x,y).$

This completes the proof of the proposition. ∎

In characteristic two, however, Proposition 4.4 remains no longer true, as the
following proposition shows.

######  Proposition 4.5.

Let $A$ be a unital commutative algebra with identity $\mathbf{1}$ over the
field $k$ with characteristic two. A function $q:A\to A$ fulfils (4.4), if,
and only if, $q\equiv 0$.

###### Proof.

Evidently, if $q\equiv 0$, then (4.4) is fulfilled. Conversely, let $q$ fulfil
(4.4). Putting $x=y=z=0$ and $\lambda=0_{k}$, we see that $q(0)=0$. Using the
calculation of Step 5 in the proof of Proposition 4.4, we see that
$(x,y)\mapsto q(x+y)-q(x)-q(y)$ is a derivation in the first argument.
Therefore, we find, for all $x,y\in A$,

$\displaystyle 0$
$\displaystyle=q(xy+xy)-q(xy)-q(xy)=x(q(xy+y)+q(y)+q(xy))+y(q(xy+x)+q(x)+q(xy)).$

Putting $x=1$ and rearranging the terms using $a=-a$ for all $a\in A$, we
conclude

$\displaystyle\forall y\in A:q(y)=y(q(y+\mathbf{1})-q(y)-q(\mathbf{1})).$

In particular, we find

$\displaystyle
q(\mathbf{1})=\mathbf{1}(q(\mathbf{1}+\mathbf{1})-q(\mathbf{1})-q(\mathbf{1}))=q(\mathbf{1}+\mathbf{1})=q(0)=0.$

Therefore, we have, for all $y\in A$,

$\displaystyle
q(y)=q(\mathbf{1}y)=y(q(y+\mathbf{1})-q(y)-q(\mathbf{1}))+q(y)+y^{2}q(\mathbf{1})=q(y)+q(y)=0.$

∎

In particular, Proposition (4.5) yields that the implication “$\impliedby$” of
Proposition 4.4 holds true in characteristic zero. This can also be seen from
the observation that we have made no argument that requires $1_{k}+1_{k}\neq
0_{k}$. In the following example, we construct a non-zero quadratic function
$\delta:\mathbb{F}_{2}[X]\to\mathbb{F}_{2}[X]$, which is the concatenation of
a biderivation with the diagonal embedding $\mathbb{F}_{2}[X]\hookrightarrow
F_{2}[X]\times\mathbb{F}_{2}[X]$. In particular, this shows that the
implication “$\implies$” of Proposition 4.4 fails in characteristic two.

######  Example 4.6.

Consider the ring of polynomials
$\mathbb{F}_{2}[X]=\left\\{p\in\mathbb{F}_{2}^{\mathbb{N}_{0}}\,\left|\,\sup\left\\{i\in\mathbb{N}_{0}\,\big{|}\,p_{i}\neq
0\right\\}<\infty\right.\right\\}$ as a commutative, unital algebra over
$\mathbb{F}_{2}$. Since the monomials $X^{n}$, $n\in\mathbb{N}_{0}$ are a
basis of $\mathbb{F}_{2}[X]$, there is a unique linear function

$\displaystyle d:\mathbb{F}_{2}[X]\to\mathbb{F}_{2}[X],\qquad
X^{n}\mapsto(n\\!\\!\mod 2)X^{n-1},\quad n\in\mathbb{N}_{0}.$

This function fulfils, for all $n,m\in\mathbb{N}_{0}$,

$\displaystyle X^{n}d(X^{m})+X^{m}d(X^{n})$ $\displaystyle=(m\\!\\!\mod
2)X^{n}X^{m-1}+(n\\!\\!\mod 2)X^{m}X^{n-1}$ $\displaystyle=((m+n)\\!\\!\mod
2)X^{n+m-1}=d(X^{n+m})=d(X^{n}X^{m})$

and hence $d$ is a derivation. In particular, the function

$\displaystyle\widetilde{d}:\mathbb{F}_{2}[X]\times\mathbb{F}_{2}[X]\to\mathbb{F}_{2}[X],\qquad(x,y)\mapsto
d(x)d(y)$

is a biderivation. The associated quadratic function $q=d(\cdot)^{2}$ does not
vanish, and the associated bilinear function $\Delta(x,y)=q(x+y)-q(x)-q(y)$
vanishes for all $x,y\in\mathbb{F}_{2}[X]$ and is therefore a biderivation.

We are now able to characterise the biquadratic functions induced by
conservative-irreversible functions. To this end, we introduce the following
objects.

######  Definition 4.7.

Let $M$ be a smooth manifold. A function
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}:\mathcal{C}^{\infty}\left(M\right)\times\mathcal{C}^{\infty}\left(M\right)\to\mathcal{C}^{\infty}\left(M\right)$
is a _square quasi-Poisson structure_ (SQPS) if, and only if, for each
$h\in\mathcal{C}^{\infty}\left(M\right)$, the functions
${}_{h}\delta:=\left\\{\\!\left\\{h,\cdot\right\\}\\!\right\\}$ and
$\delta_{h}:=\left\\{\\!\left\\{\cdot,h\right\\}\\!\right\\}$ fulfil

  1. (i)

$\delta_{h}(f),{}_{h}\delta(f)$ are pointwise non-negative for all
$f\in\mathcal{C}^{\infty}\left(M\right)$,

  2. (ii)

$\delta_{h}(\cdot)$ and ${}_{h}\delta(\cdot)$ are _$h$ -translation
invariant_, i.e. $\delta_{h}(\cdot+h)=\delta_{h}$ and
${}_{h}\delta(\cdot+h)={}_{h}\delta(\cdot)$, for all
$h\in\mathcal{C}^{\infty}(M)$,

  3. (iii)

$\delta_{h}$ and ${}_{h}\delta$ fulfil the inclusion-exclusion principle
(4.2),

  4. (iv)

$\delta_{h}$ and ${}_{h}\delta$ fulfil (4.4).

If (only) $\delta_{h}$ (respectively ${}_{h}\delta$) is required to be
$h$-translation invariant, then we call
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ left (respectively
right) SQPS.

Now, we see that conservative-irreversible functions are uniquely given by
left SQPS.

######  Proposition 4.8.

A function
$E:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$
is a conservative-irreversible function if, and only if, there is a left SQPS
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ so that

$\displaystyle\forall
f,g,h\in\mathcal{C}^{\infty}(M):E(f,g,h)=\frac{1}{2}(\\{\\!\\{f+g,h\\}\\!\\}-\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h\right\\}\\!\right\\});$
(4.5)

and this SQPS is unique.

###### Proof.

“$\implies$” Let $E$ be a conservative-irreversible function and define

$\displaystyle\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),\quad(f,g)\mapsto
E(f,f,g).$

Let $e$ be the unique representation (2.3) of $E$. Due to linearity and
symmetry in the first two entries of $e$ and hence in the first two entries of
$E$, we have, for all $f,g,h\in\mathcal{C}^{\infty}(M)$

$\displaystyle
E(f,g,h)=\frac{1}{2}(E(f+g,f+g,h)-E(f,f,h)-E(g,g,h))=\frac{1}{2}(\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h\right\\}\\!\right\\})$

and hence the function $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$
fulfils (4.5). It remains to show that
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a left SQPS. From
Definition 2.1 (iii), we conclude

$\displaystyle\forall f,g\in\mathcal{C}^{\infty}(M)\leavevmode\nobreak\
\forall x\in
M:\left\\{\\!\left\\{f,g\right\\}\\!\right\\}(x)=E(f,f,g)(x)=e(f,f,g,g)(x)\geq
0.$

and hence $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ has the
property (i) of Definition 4.7. Let $f,h\in\mathcal{C}^{\infty}(M)$. In view
of Definition 2.1 (iv) and the symmetry of $E$ in the first two entries, we
have

$\displaystyle\left\\{\\!\left\\{f+h,h\right\\}\\!\right\\}$
$\displaystyle=E(f+h,f+h,h)=E(f,f,h)+E(f,h,h)+E(h,f,h)+E(h,h,h)$
$\displaystyle=E(f,f,h)=\left\\{\\!\left\\{f,h\right\\}\\!\right\\}$

and hence $\left\\{\\!\left\\{\cdot,h\right\\}\\!\right\\}$ is $h$-translation
invariant. Further, Proposition 4.4 implies that
$\left\\{\\!\left\\{\cdot,h\right\\}\\!\right\\}$ fulfils Definition 4.7 (iii)
and (iv). This shows that
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is indeed a left SQPS.

“$\impliedby$” Let $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ be a
left SQPS with (4.5). Define

$\displaystyle e:\left(\mathcal{C}^{\infty}(M)\right)^{4}$
$\displaystyle\to\mathcal{C}^{\infty}(M),$ $\displaystyle(f,g,h,k)$
$\displaystyle\mapsto\frac{1}{4}\big{(}\left\\{\\!\left\\{f+g,h+k\right\\}\\!\right\\}+\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+\left\\{\\!\left\\{f,k\right\\}\\!\right\\}+\left\\{\\!\left\\{g,h\right\\}\\!\right\\}+\left\\{\\!\left\\{g,k\right\\}\\!\right\\}$
$\displaystyle\hskip
28.45274pt-\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,k\right\\}\\!\right\\}-\left\\{\\!\left\\{f,h+k\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h+k\right\\}\\!\right\\}\big{)}.$

Due to Proposition 4.4, we have, for all $f,g,h\in\mathcal{C}^{\infty}(M)$,

$\displaystyle E(f,g,h)$
$\displaystyle=\frac{1}{2}(\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h\right\\}\\!\right\\})$
$\displaystyle=\frac{1}{4}\big{(}4\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}+2\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+2\left\\{\\!\left\\{g,h\right\\}\\!\right\\}-2\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-4\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-4\left\\{\\!\left\\{g,h\right\\}\\!\right\\}\big{)}$
$\displaystyle=\frac{1}{4}\big{(}\left\\{\\!\left\\{f+g,2h\right\\}\\!\right\\}+2\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+2\left\\{\\!\left\\{g,h\right\\}\\!\right\\}-2\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f,2h\right\\}\\!\right\\}-\left\\{\\!\left\\{g,2h\right\\}\\!\right\\}\big{)}$
$\displaystyle=e(f,g,h,h)$

and hence $e$ generates $E$ with (2.3). It remains to verify that $e$ fulfils
Definition 2.1 (i)–(iv). The symmetry of $e$ in the first two arguments and in
the last two arguments is a direct consequence of the commutativity of the
addition in $\mathcal{C}^{\infty}(M)$. In view of Definition 4.7 (i) and
Proposition 4.4, we have, for all $f,h\in\mathcal{C}^{\infty}(M)$ and $x\in
M$,

$\displaystyle e(f,f,h,h)(x)$
$\displaystyle=\frac{1}{4}(\left\\{\\!\left\\{2f,2h\right\\}\\!\right\\}(x)+4\left\\{\\!\left\\{f,h\right\\}\\!\right\\}(x)-2\left\\{\\!\left\\{2f,h\right\\}\\!\right\\}(x)-2\left\\{\\!\left\\{f,2h\right\\}\\!\right\\}(x))$
$\displaystyle=\frac{1}{4}(16+4-8-8)\left\\{\\!\left\\{f,h\right\\}\\!\right\\}(x)=\left\\{\\!\left\\{f,h\right\\}\\!\right\\}(x)\geq
0$

This shows that $e$ fulfils Definition 2.1 (ii) and (iii). Since
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a left SQPS, we have,
for all $f,h\in\mathcal{C}^{\infty}(M)$,

$\displaystyle e(h,f,h,h)$
$\displaystyle=\frac{1}{4}(\left\\{\\!\left\\{f+h,2h\right\\}\\!\right\\}+2\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+2\left\\{\\!\left\\{h,h\right\\}\\!\right\\}-2\left\\{\\!\left\\{f+h,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f,2h\right\\}\\!\right\\}-\left\\{\\!\left\\{h,2h\right\\}\\!\right\\})$
$\displaystyle=\frac{1}{4}(\left\\{\\!\left\\{f,2h\right\\}\\!\right\\}+2\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+0-2\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-\left\\{\\!\left\\{2h\right\\}\\!\right\\}-0)=0$

and hence $e$ fulfils Definition 2.1 (iv). Finally, Proposition 4.4 implies
that $e$ is a derivation in each argument. This shows that $E$ is a
conservative-irreversible function. ∎

In particular, we conclude that the distinction between left and right SQPS is
artificial: left SQPS are right SQPS and vice versa.

######  Corollary 4.9.

Each left SQPS is symmetric, and therefore a SQPS.

###### Proof.

Let $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ be a SQPS. Then, the
function

$\displaystyle e:\left(\mathcal{C}^{\infty}(M)\right)^{4}$
$\displaystyle\to\mathcal{C}^{\infty}(M),$ $\displaystyle(f,g,h,k)$
$\displaystyle\mapsto\frac{1}{4}\big{(}\left\\{\\!\left\\{f+g,h+k\right\\}\\!\right\\}+\left\\{\\!\left\\{f,h\right\\}\\!\right\\}+\left\\{\\!\left\\{f,k\right\\}\\!\right\\}+\left\\{\\!\left\\{g,h\right\\}\\!\right\\}+\left\\{\\!\left\\{g,k\right\\}\\!\right\\}$
$\displaystyle\hskip
28.45274pt-\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,k\right\\}\\!\right\\}-\left\\{\\!\left\\{f,h+k\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h+k\right\\}\\!\right\\}\big{)}$

fulfils the properties (i)–(iv) of Definition 2.1. In view of Lemma 2.8, we
conclude, for all $f,g\in\mathcal{C}^{\infty}(M)$,

$\displaystyle\left\\{\\!\left\\{f,g\right\\}\\!\right\\}=e(f,f,g,g)=e(g,g,f,f)=\left\\{\\!\left\\{g,f\right\\}\\!\right\\}.$

Thus, $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is symmetric; and
hence
$\left\\{\\!\left\\{\cdot,h\right\\}\\!\right\\}=\left\\{\\!\left\\{h,\cdot\right\\}\\!\right\\}$
for all $h\in\mathcal{C}^{\infty}(M)$. This shows that
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a SQPS. ∎

In definition 4.7, SQPS are not assumed to be symmetric. Since each SQPS is a
left SQPS, Corollary 4.9 implies that each SQPS is symmetric. An alternative
proof, which does not use the one-to-one correspondence of SQPS and
conservative-irreversible systems is the following.

######  Lemma 4.10.

A function
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M)$
is a SQPS if, and only if,
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a symmetric left
SQPS.

###### Proof.

“$\impliedby$” By definition, a left SQPS
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a SQPS if, and only
if, the function

$\displaystyle[[\cdot,\cdot]]:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),\quad(f,g)\mapsto\left\\{\\!\left\\{g,f\right\\}\\!\right\\}$

is also a left SQPS. If $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$
is a symmetric left SQPS, then
$[[\cdot,\cdot]]=\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ and
hence the latter is a SQPS.

“$\implies$” Conversely, let
$\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ be a SQPS. In
particular, $\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}$ is a left
SQPS and it remains to show symmetry. Let $f,g\in\mathcal{C}^{\infty}(M)$.
Proposition 4.4 implies that
$\left\\{\\!\left\\{\cdot,0\right\\}\\!\right\\}\equiv 0$ and hence we
conclude with property (ii) of Definition 4.7

$\displaystyle\left\\{\\!\left\\{f,g\right\\}\\!\right\\}$
$\displaystyle=\left\\{\\!\left\\{f+g,g\right\\}\\!\right\\}=\left\\{\\!\left\\{f+g,f+g-f\right\\}\\!\right\\}$
$\displaystyle=\left\\{\\!\left\\{f+g,f+g\right\\}\\!\right\\}+\left\\{\\!\left\\{f+g,f-f\right\\}\\!\right\\}+\left\\{\\!\left\\{f+g,g-f\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,f\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,g\right\\}\\!\right\\}$
$\displaystyle\quad-\left\\{\\!\left\\{f+g,-f\right\\}\\!\right\\}$
$\displaystyle=0+0+\left\\{\\!\left\\{f+g,g-f\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,f\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,g\right\\}\\!\right\\}-\left\\{\\!\left\\{f+g,f\right\\}\\!\right\\}$
$\displaystyle=\left\\{\\!\left\\{2g,g-f\right\\}\\!\right\\}-2\left\\{\\!\left\\{g,f\right\\}\\!\right\\}-\left\\{\\!\left\\{f,g\right\\}\\!\right\\}$
$\displaystyle=4\left\\{\\!\left\\{g,g-f\right\\}\\!\right\\}-\left\\{\\!\left\\{f,g\right\\}\\!\right\\}-2\left\\{\\!\left\\{g,f\right\\}\\!\right\\}$
$\displaystyle=4\left\\{\\!\left\\{g,f\right\\}\\!\right\\}-\left\\{\\!\left\\{f,g\right\\}\\!\right\\}-2\left\\{\\!\left\\{g,f\right\\}\\!\right\\}=2\left\\{\\!\left\\{g,f\right\\}\\!\right\\}-\left\\{\\!\left\\{f,g\right\\}\\!\right\\},$

from which the symmetry follows. ∎

As a consequence, we may characterise simple conservative-irreversible
functions through their associated SQPS.

######  Lemma 4.11.

A conservative-irreversible function $E$ is simple if, and only if, the
associated SQPS with (4.5) is, up to multiplication with a non-negative
function, the square of an almost Poisson bracket.

###### Proof.

Necessity is evident; we show sufficiency. Let $\lambda:M\to\mathbb{R}$ and
let $\\{\cdot,\cdot\\}$ be an almost Poisson bracket so that

$\displaystyle\left\\{\\!\left\\{\cdot,\cdot\right\\}\\!\right\\}:\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)\to\mathcal{C}^{\infty}(M),\quad(f,g)\mapsto\lambda\\{f,g\\}^{2}$

is an SQPS. Then, the associated conservative-irreversible function $E$
fulfils, for all $f,g,h\in\mathcal{C}^{\infty}(M)$,

$\displaystyle E(f,g,h)$
$\displaystyle=\frac{1}{2}(\left\\{\\!\left\\{f+g,h\right\\}\\!\right\\}-\left\\{\\!\left\\{f,h\right\\}\\!\right\\}-\left\\{\\!\left\\{g,h\right\\}\\!\right\\})$
$\displaystyle=\lambda\frac{1}{2}(\\{f+g,h\\}^{2}-\\{f,h\\}^{2}-\\{g,h\\}^{2})=\lambda\\{f,h\\}\\{g,h\\},$

and hence $E$ is indeed simple. ∎

##  5 Dynamical systems of conservative-irreversibly functions

Let $M$ be a smooth manifold and $E$ a conservative-irreversible function on
$M$. Each pair of generating functions
$(s,h)\in\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$ induces the
dynamical system

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}f\circ x=E(f,s,h)\circ
x\qquad\text{for all}\leavevmode\nobreak\ f\in\mathcal{C}^{\infty}(M),$ (5.1)

i.e. a smooth curve $x\in\mathcal{C}^{\infty}(M)$ is a trajectory of the
dynamical system (5.1) if, and only if, for each observable
$f\in\mathcal{C}^{\infty}(M)$ the curve $f\circ x$ fulfils the equation (5.1).
Using the associated tensor field $\varepsilon\in\Gamma(T^{4}(M))$, we may
coin the following definition.

######  Definition 5.1.

Let $E$ be a conservative-irreversible function on a manifold $M$. A vector
field $X$ is called conservative-irreversible with generating functions
$(s,h)\in\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$ if, and only
if, $X=E(\cdot,s,h)$; we write $X_{(s,h)}:=E(\cdot,s,h)$.

Analogously to Hamiltonian systems defined on Poisson manifolds, the dynamical
system (5.1), written implicitly via the infinitesimal action of the solutions
on all observables, can then be recognised as the dynamical system given by
the vector field $X_{(s,h)}$.

######  Lemma 5.2.

Let $s,h\in\mathcal{C}^{\infty}$. A curve $x\in\mathcal{C}^{\infty}(I,M)$ is a
trajectory of (5.1) if, and only if, $x$ is an integral curve of $X_{(s,h)}$.

###### Proof.

“$\impliedby$” Let $x$ be an integral curve of $X_{(s,h)}$ and
$f\in\mathcal{C}^{\infty}(M)$. Then, we have

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}f\circ
x=\left\langle\mathrm{d}f\circ
x,\tfrac{\mathrm{d}}{\mathrm{d}t}x\right\rangle=\left\langle\mathrm{d}f\circ
x,X\circ
x\right\rangle=\varepsilon(\mathrm{d}f,\mathrm{d}s,\mathrm{d}h,\mathrm{d}h)\circ
x=E(f,s,h)\circ x$

and thus $x$ is a trajectory of the dynamical system (5.1).

“$\implies$” Conversely, let $x\in\mathcal{C}^{\infty}(I,M)$,
$I\subseteq\mathbb{R}$ a nonempty open interval, be a trajectory of the
dynamical system (5.1) and let $t_{0}\in I$. Let
$\varphi_{1},\ldots,\varphi_{n}\in\mathcal{C}^{\infty}(M)$ so that
$(\varphi_{1},\ldots,\varphi_{n})$ is a diffeomorphism in a neighbourhood of
$x(t_{0})$. Then, we have in a neighbourhood $J\subseteq I$ of $t_{0}$

$\displaystyle\forall i\in\mathbb{N}^{*}_{\leq
n}:\tfrac{\mathrm{d}}{\mathrm{d}t}\varphi_{i}\circ x=E(\varphi_{i},s,h)\circ
x=(X_{(s,h)})_{i}\circ x,$

where $(X_{(s,h)})_{i}$ is the $i$-th coordinate with respect to the local
coordinate system $(\partial_{\varphi_{1}},\ldots,\partial_{\varphi_{n}})$.
This shows that $x$ is indeed an integral curve of $X_{(s,h)}$. ∎

In the following lemma, we verify that the dynamical system (5.1) fulfils the
laws of thermodynamics. Additionally, we show that the entropy function $s$
is, in general, not preserved.

######  Lemma 5.3.

Let $E$ be a conservative-irreversible function on $M$,
$s,h\in\mathcal{C}^{\infty}(M)$ and let $x$ be a solution of (5.1). Then, we
have

  1. (i)

$\tfrac{\mathrm{d}}{\mathrm{d}t}h\circ x\equiv 0$, i.e. (generalised) energy
is preserved along trajectories.

  2. (ii)

$\tfrac{\mathrm{d}}{\mathrm{d}t}s\circ x\geq 0$, i.e. (generalised) entropy is
irreversibly produced along trajectories.

Furthermore, we see

  1. (iii)

If $E\neq 0$, then there exists generating function
$s_{0},h_{0}\in\mathcal{C}^{\infty}(M)$ so that
$\tfrac{\mathrm{d}}{\mathrm{d}t}s_{0}\circ x\neq 0$ for some solution $x$ of
(5.1).

###### Proof.

Let $E$ be a conservative-irreversible function and
$s,h\in\mathcal{C}^{\infty}(M)$. The properties (i) and (ii) are direct
consequences of Definition 2.1 (iii) and (iv):

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}h\circ x=E(h,s,h,h)\circ x\equiv
0$

and

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}s\circ x=E(s,s,h)\circ x\geq 0.$

Now, let $E\neq 0$. Due to symmetry of $E$ in the first two arguments, there
exists $s_{0},h_{0}\in\mathcal{C}^{\infty}(M)$ and $x_{0}\in M$ so that
$E(s_{0},s_{0},h_{0})(x_{0})\neq 0$. Consider an integral curve $x$ of the
vector field $X_{(s_{0},h_{0})}$ with $x(0)=x_{0}$. Then, we have

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}(s\circ
x)(0)=E(s_{0},s_{0},h_{0})(x_{0})\neq 0.$

This completes the proof. ∎

Using the conservative irreversible functions, an irreversible Hamiltonian
system may be defined in an alternative and slightly more general way to [23,
22] as follows.

######  Definition 5.4.

Let $M$ be a smooth manifold with Poisson bracket $\\{\cdot,\cdot\\}$ and
conservative-irreversible function $E$. An irreversible Hamiltonian system
with generating functions
$(s,h)\in\mathcal{C}^{\infty}(M)\times\mathcal{C}^{\infty}(M)$ is the
dynamical system

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}f\circ
x=\\{f,h\\}+E(f,s,h)\qquad\text{for all}\leavevmode\nobreak\
f\in\mathcal{C}^{\infty}(M),$ (5.2)

with either the weak noninteraction condition $\\{s,h\\}=0$ or the strong
noninteraction condition $\\{s,\cdot\\}\equiv 0$. The vector field
$\widehat{X}_{(s,h)}:=X_{h}+X_{(s,h)}$, where $X_{h}$ denotes the Hamiltonian
vector field $X_{h}=\\{\cdot,h\\}$ and $X_{(s,h)}$ is the conservative-
irreversible vector field generated by $(s,h)$, is called irreversible
Hamiltonian vector field.

Lastly, we demonstrate how irreversible Hamiltonian systems may be viewed as
particular classes of metriplectic systems as well as the general systems of
Beris and Edwards, thereby unifying, in some sense, these approaches.

Let $M$ be a smooth manifold with Poisson bracket $\\{\cdot,\cdot\\}$ and
conservative-irreversible function $E$. We show how the irreversible
Hamiltonian vector fields can be viewed simultaneously as metriplectic systems
and as systems of Beris and Edwards. Let $s,h\in\mathcal{C}^{\infty}(M)$ so
that the noninteraction condition $\\{s,h\\}=0$ is fulfilled. Define

$\displaystyle[\cdot,\cdot]_{BE}:=E(\cdot,s,\cdot)$

Then, $[\cdot,\cdot]_{BE}$ is a derivation in the first argument and fulfils,
by construction,

$\displaystyle[h,h]_{BE}=0\leq[s,h]_{BE}$

so that the system (5.2) written as

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}f\circ
x=\\{f,h\\}+[s,h]_{BE}\qquad\text{for all}\leavevmode\nobreak\
f\in\mathcal{C}^{\infty}(M),$

has the form which was considered by [2]. Secondly, consider the brackets

$\displaystyle[\cdot,\cdot]_{M}:=E(\cdot,\cdot,h).$

Definition 2.1 (iii) guarantees that this is a non-negative bracket and

$\displaystyle[h,\cdot]_{M}=E(h,\cdot,h)\equiv 0$

shows that $[\cdot,\cdot]_{M}$ fulfils the strong noninteraction condition of
metriplectic systems. Since the (weak or strong) noninteraction condition on
the almost Poisson bracket are the same as for irreversible Hamiltonian
systems, (5.2) written as

$\displaystyle\tfrac{\mathrm{d}}{\mathrm{d}t}f\circ
x=\\{f,h\\}+[f,s]_{M}\qquad\text{for all}\leavevmode\nobreak\
f\in\mathcal{C}^{\infty}(M),$

is a metriplectic system.

Insofar, irreversible Hamiltonian systems are indeed simultaneously particular
metriplectic systems and Beris-Edwards-systems. However, this view is a bit
flawed. The brackets $[\cdot,\cdot]_{BE}$ and $[\cdot,\cdot]_{M}$ depend
explicitly on $s$ and $h$, respectively. Therefore, a perturbation of the
generating functions must automatically be a perturbation of the underlying
brackets in this point of view, which is not the case for general systems.

##  6 Conclusion

We have studied conservative-irreversible functions on smooth, finite-
dimensional manifolds. These functions are uniquely associated to a class of
four-tensor-fields and a class of bi-quadratic functions, which were fully
characterised by intrinsic properties. As an interesting subclass, we
recognised simple conservative-irreversible systems, which are induced by
(almost) Poisson brackets and which were the original motivation of the study
of these systems. Using local representations, we were able to show that each
conservative-irreversible function is a locally finite, but in general not
trivial, sum of simple conservative-irreversible functions, which revealed the
relation to the recently suggested class of metriplectic four-tensors.

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