# A characterization of real matrix semigroups

Benedict Bauer
University of Vienna
<EMAIL_ADDRESS>

Stefan Gerhold
TU Wien
<EMAIL_ADDRESS>Financial support from the Austrian Science Fund
(FWF) under grants P 30750 and Y 1235 is gratefully acknowledged.

###### Abstract

We characterize all real matrix semigroups satisfying a mild boundedness
assumption, without assuming continuity. Besides the continuous solutions of
the semigroup functional equation, we give a description of solutions arising
from non-measurable solutions of Cauchy’s functional equation. To do so, we
discuss the primary decomposition and the Jordan–Chevalley decomposition of a
matrix semigroup. Our motivation stems from a characterization of all multi-
dimensional self-similar Gaussian Markov processes, which is given in a
companion paper.

MSC 2020: 39B22, 47D03, 15A16

Keywords: Matrix semigroup, matrix exponential, primary decomposition,
Jordan–Chevalley decomposition, Cauchy’s functional equation

##  1 Introduction

It is a classical fact that all continuous matrix-valued functions
$g:[0,\infty)\to\mathbb{R}^{n\times n}$ satisfying the semigroup property

$\displaystyle g(x+y)$ $\displaystyle=g(x)g(y),\quad x,y\geq 0,$ (1.1)
$\displaystyle g(0)$ $\displaystyle=1,$ (1.2)

are given by the maps $g(x)=\exp(Mx)$, where $M\in\mathbb{R}^{n\times n}$ is
arbitrary. See, e.g., Problem 2.1 in [4] and the subsequent discussion. To the
best of our knowledge, few authors have considered non-continuous solutions of
(1.1) in the multidimensional case $d>1$. Kuczma and Zajtz [8] determine all
matrix semigroups $(g(x))_{x\in\mathbb{R}}$ where $x\mapsto g(x)$ is
measurable. Zajtz [12] characterizes the matrix semigroups
$(g(x))_{x\in\mathbb{Q}}$ indexed by rational numbers. The one-dimensional
case, which is equivalent to Cauchy’s functional equation,

$f(x)+f(y)=f(x+y),\quad f:\mathbb{R}\to\mathbb{R},$ (1.3)

is well studied, on the other hand (see [1, 2]). Our motivation to investigate
non-continuous matrix semigroups stems from probability theory: In a companion
paper, we develop a characterization of all self-similar Gaussian Markov
processes. By self-similarity, the bivariate covariance function of such a
multi-dimensional stochastic process can be transformed to a matrix function
of a single argument, which must satisfy (1.1). See Section 2 for some more
details. Our main result (Theorem 2.6) determines all solutions of (1.1) which
satisfy a mild boundedness assumption. In Section 3, we provide a
decomposition of the vector space into invariant subspaces, and establish some
useful properties of the decomposition. The main proofs (in Section 5) are
preceded by a discussion of the Jordan–Chevalley decomposition of a semigroup,
in Section 4. In Appendix A, we give two auxiliary results on Cauchy’s
functional equation.

##  2 Preliminaries and main result

To motivate our investigation, we first recall some facts about Gaussian
Markov processes (see, e.g., [9, 10] for more information). Let
$X=(X_{t})_{t\geq 0}$ be a $d$-dimensional real centered Gaussian process. The
covariance of $X$ is a matrix-valued function $(s,t)\mapsto
R(s,t)\in\mathbb{R}^{d\times d}$ satisfying

$v^{\mathrm{T}}R(s,t)u=\mathbb{E}[X_{s}^{\mathrm{T}}v\,X_{t}^{\mathrm{T}}u],\quad
s,t\geq 0,\ u,v\in\mathbb{R}^{d},$

and uniquely characterizes the law of the process. Suppose that $X$ is self-
similar, and that $R(1,1)=\mathrm{id}$ is the identity matrix. In terms of the
covariance function, self-similarity means that $R(as,at)=a^{2H}R(s,t)$ for
$a,s,t>0$ and some self-similarity parameter $H>0$. By a classical criterion
[3, Theorem V.8.1], $X$ is a Markov process if and only if its covariance
function satisfies

$\displaystyle R(s,t)R(t,t)^{-1}R(t,u)=R(s,u),\quad 0\leq s\leq t\leq u.$
(2.1)

Upon introducing $g(x):=R(e^{-x},1)$, self-similarity allows to reduce (2.1)
to

$g(x)g(y)=g(x+y),\quad x,y,\geq 0.$

This observation has been used in dimension $d=1$ to prove that certain
Gaussian processes do not have the Markov property (see, e.g., [10, Theorem
2.3]). Unifying and generalizing these results in a companion paper, we obtain
a classification of all $d$-dimensional self-similar Gaussian Markov
processes. A full classification requires finding all solutions of (1.1),
without assuming continuity. The boundedness assumption 2.5 stated below
causes no problems, though. To state our main result, define the rotation
matrix

$Q(\theta)=\begin{pmatrix}\cos(\theta)&\sin(\theta)\\\
-\sin(\theta)&\cos(\theta)\end{pmatrix},\quad\theta\in\mathbb{R},$

and, for even $k$ and a function $\nu:\mathbb{R}\to\mathbb{R}$, the block-
diagonal matrix

$Q_{k}^{\nu}(x):=\begin{pmatrix}Q(\nu(x))&&0\\\ &\ddots&\\\
0&&Q(\nu(x))\end{pmatrix}\in\mathbb{R}^{k\times k}$ (2.2)

consisting of $k/2$ rotation matrices. In our statements, $\nu$ will denote
some non-measurable (equivalently, non-continuous) solution of (1.3). The
following assumption is in force throughout the paper.

######  Assumption 2.1.

In the following, $V$ denotes a real $d$-dimensional vector space equipped
with an inner product $\langle\cdot,\cdot\rangle$.

######  Definition 2.2.

We write $L(V)$ for the set of linear maps from $V$ to itself. The operator
norm induced by the inner product $\langle\cdot,\cdot\rangle$ is denoted by
$\|\cdot\|_{\mathrm{op}}$. If the basis is clear from the context, we will
identify elements of $L(V)$ and $d\times d$ matrices.

######  Definition 2.3.

We say that $g:\mathbb{R}_{\geq 0}\to L(V)$ is a semigroup if
$g(0)=\operatorname{id}$ and

$g(x+y)=g(x)g(y)$

for all $x,y\geq 0$. Semigroups acting on $V^{\mathbb{C}}$, the
complexification of $V$, are defined by the same property.

######  Definition 2.4.

We say that a semigroup $(g(x))_{x\geq 0}$ in $L(V)$ is elementary if there
exists an orthonormal basis such that

$g(x)=\exp(Mx),\ x\geq 0,\quad\text{or}\quad g(x)=Q_{d}^{\nu}(x)\exp(Mx),\
x\geq 0,$

for some non-continuous $\nu$ satisfying Cauchy’s equation (1.3) and some
matrix $M\in\mathbb{R}^{d\times d}$.

######  Assumption 2.5.

The semigroup $g(x)_{x\geq 0}$ satisfies $\|g(x)\|_{\mathrm{op}}\leq f(x)$,
$x\geq 0$, where $f:[0,\infty)\to\mathbb{R}$ is locally bounded, right-
continuous at $0$ and satisfies $f(0)=1$.

We can now state our main theorem, which will be proven at the end of Section
5.

######  Theorem 2.6.

Let $(g(x))_{x\geq 0}$ be a semigroup satisfying Assumption 2.5. Then there
exists an orthogonal decomposition $V=\bigoplus_{i=1}^{n}V_{i}$ such that each
$V_{i}$ is invariant under $g(x)$, and either $g(x)$ is elementary on $V_{i}$
or $g(x)|_{V_{i}}=0$ for $x>0$.

##  3 Primary decomposition of semigroups

######  Definition 3.1.

Let $S$ be a linear operator on the vector space $V$. A subspace $U\subseteq
V$ is called $S$-invariant if $S$ maps $U$ into $U$.

The primary decomposition theorem from linear algebra [11, Theorem 1.5.1]
decomposes a vector space into invariant subspaces for a given operator. On
each subspace, the operator has a single real eigenvalue or a pair of
conjugate complex eigenvalues. Instead of a single operator, we need the
following version for semigroups.

######  Theorem 3.2 (Primary decomposition).

For any semigroup $(g(x))_{x\geq 0}$ of linear maps acting on $V$ there exists
a decomposition $V=\bigoplus_{i=1}^{n}V_{i}$ with $\dim(V_{i})\geq 1$ such
that each $V_{i}$ is $g(x)$-invariant for all $x\geq 0$, and for all $i$ one
of the following holds:

  * (1)

For all $x\geq 0$, $g(x)$ has one eigenvalue $\lambda(x)\geq 0$ on $V_{i}$,

  * (2)

For all $x\geq 0$, $g(x)$ has eigenvalues
$\lambda(x),\overline{\lambda}(x)\in\mathbb{C}$ on $V_{i}$. They may coincide
(and thus be real) for some values of $x$, but not for all $x\geq 0$.

###### Proof.

It is known that the primary decomposition extends to commuting sets of
matrices. Indeed, the decomposition $V=\bigoplus_{i=1}^{n}V_{i}$ into
invariant subspaces follows from Theorem 5 on p. 40 in [7], applied to the
span of the semigroup $(g(x))_{x\geq 0}$. Thus, we only need to argue why (1)
or (2) follows from the semigroup property. Assume that each $g(x)$ has only
one eigenvalue $\lambda(x)\in\mathbb{R}$ on some $V_{i}$. Since the $g(x)$
commute, they share a common eigenvector $v_{i}$, and thus we have
$\lambda(x)\lambda(y)=\lambda(x+y)$ for $x,y\geq 0$. Suppose
$\lambda(x_{0})<0$ for some $x_{0}>0$. Then we have
$\lambda(x_{0})=\lambda(x_{0}/2)^{2}$, and hence
$\lambda(x_{0}/2)\not\in\mathbb{R}$, contradicting $\lambda(x)\in\mathbb{R}$
for all $x\geq 0$. Hence $\lambda(x)\in\mathbb{R}_{\geq 0}$ for all $x\geq 0$.
∎

######  Theorem 3.3.

Consider a semigroup $g=(g(x))_{\geq 0}$ acting on $V^{\mathbb{C}}$, the
complexification of $V$. Then there exists a decomposition
$V^{\mathbb{C}}=\bigoplus_{i=1}^{n}V_{i}$ with $\dim(V_{i})\geq 1$ such that,
for each $i$ and all $x\geq 0$, the space $V_{i}$ is $g(x)$-invariant and
$g(x)$ has only one eigenvalue $\lambda(x)\in\mathbb{C}$ on $V_{i}$.

###### Proof.

This is an immediate consequence of Theorem 5 on p. 40 in [7] (cf. the
preceding proof). ∎

######  Definition 3.4.

We call the decomposition from Theorem 3.2 simultaneous real primary
decomposition (SRPD), omitting “w.r.t. $g$” if the semigroup is clear from the
context. The component $V_{i}$ is of first type in case (1), and of second
type in case (2). Similarly, the simultaneous primary decomposition (SPD) is
the decomposition from Theorem 3.3.

######  Lemma 3.5.

Let $g$ be a semigroup acting on $V$, and let $V_{i}\subseteq V$ be a subspace
of first type from the SRPD of $V$. Then there exists a common eigenvector
$v\in V$ such that $g(x)v=\lambda(x)v$ for all $x>0$.

###### Proof.

We present an algorithm which yields the subspace of common eigenvectors. If
$V_{i}=:V_{i}^{(0)}$ itself is this subspace, then we are done. Otherwise
there exists $x_{0}>0$ such that the eigenspace $V_{i}^{(1)}$ of $g(x_{0})$ is
a strict subspace of $V_{i}$. For any $y>0$ and any $v\in V_{i}^{(1)}$ we have

$\displaystyle 0=g(y)0$ $\displaystyle=g(y)[\lambda(x_{0})v-g(x_{0})v]$
$\displaystyle=\lambda(x_{0})[g(y)v]-g(x_{0})[g(y)v].$

It follows that $V_{i}^{(1)}$ is $g(x)$-invariant for all $x>0$. Now either
$V_{i}^{(1)}$ consists of common eigenvectors, or we can again find $x_{1}>0$
such that the eigenspace $V_{i}^{(2)}$ of $g(x_{1})$ in $V_{i}^{(1)}$ is a
strict subspace of $V_{i}^{(1)}$. Repeating this argument yields a sequence of
nontrivial subspaces $V_{i}^{(n)}$ whose dimensions are strictly decreasing,
hence it has to terminate. Clearly, the final vector space in this sequence is
the space of common eigenvectors of the semigroup. ∎

######  Lemma 3.6.

Let $g$ be a semigroup acting on $V$, and let $V_{i}$ be a subspace from the
SRPD of $V$ such that there exists $x_{0}>0$ with $\lambda(x_{0})=0$. Then we
have $\lambda(x)=0$ for all $x>0$. Furthermore
$V_{i}\subseteq\operatorname{ker}(g(x))$ for all $x>0.$

###### Proof.

By the previous lemma there exists a common eigenvector $v\in V_{i}$. (Since
commuting matrices are simultaneously triangularizable it follows that they
share a common eigenvector $v\in V^{\mathbb{C}}$.) We have

$\lambda(x+y)v=g(x+y)v=g(x)g(y)v=\lambda(x)\lambda(y)v.$

Hence $\lambda$ satisfies $\lambda(x)\lambda(y)=\lambda(x+y)$. Since
$\lambda(x_{0})=0$ we have $\lambda(x_{0}/n)=0$ for all $n\in\mathbb{N}$ since
$\lambda(x_{0}/n)^{n}=0$. Let $y>0$, then there exists $n\in\mathbb{N}$ such
that $x_{0}/n<y$. We obtain

$\lambda(y)=\lambda(x_{0}/n)\lambda(y-x_{0}/n)=0.$

Since $\lambda(y)=0$ for all $y>0$ it follows that the characteristic
polynomial of $g(y)$ satisfies $\chi_{g(y)}(Z)=Z^{\operatorname{dim}(V_{i})}$
and hence, by the Cayley–Hamilton theorem,
$g(y)^{\operatorname{dim}(V_{i})}\equiv 0$. For $v\in V_{i}$ we obtain

$g(y)v=g(y/{\operatorname{dim}(V_{i})})^{\operatorname{dim}(V_{i})}v=0.\qed$

######  Corollary 3.7.

Let $g$ be a semigroup acting on $V$, with SRPD $V=\bigoplus_{i=1}^{n}V_{i}$.
For any $x>0$ we have $\operatorname{ker}(g(x))=\bigoplus_{i\in I}V_{i}$,
where

$I=\\{i\,|\,V_{i}\text{ of type 1 with }\lambda\equiv 0\\}.$

###### Proof.

If $V_{i}\subseteq\operatorname{ker}(g(x))$, then $0=\lambda(x)\in\mathbb{R}$
for all $x\geq 0$ and hence $V_{i}$ is of type 1. By the previous lemma we
have

$\operatorname{ker}(g(x))\supseteq\bigoplus_{i\in I}V_{i}.$

If the inclusion was strict, then there would exist $V_{j}$ with $j\not\in I$
such that $\operatorname{ker}(g(x))\cap V_{j}\not=\emptyset$. But since $g(x)$
is invertible on $V_{j}$ this gives a contradiction, hence we have equality. ∎

######  Corollary 3.8.

Let $g$ be a semigroup acting on $V$. Then there exists a decomposition
$V=V_{1}\oplus V_{2}$, such that $g(x)|_{V_{1}}$ is invertible for all $x\geq
0$ and $g(x)|_{V_{2}}\equiv 0$ for $x>0$.

##  4 Multiplicative Jordan–Chevalley decomposition of semigroups

Due to Corollary 3.8, from now on we assume in most of our statements that
$g(x)$ is invertible for all $x\geq 0$. A standard result from linear algebra,
the Jordan–Chevalley decomposition, asserts that any matrix $A$ can be
uniquely decomposed as $A=D+N$, where $D$ is diagonalizable, $N$ is nilpotent
and $D$ and $N$ commute. If $A$ is invertible, then we can express it as
$A=D(\operatorname{id}+D^{-1}N):=DT$ with $T$ unipotent and commuting with
$D$.

######  Definition 4.1.

For an invertible linear map $A$ on $\mathbb{K}^{d}$ with
$\mathbb{K}\in\\{\mathbb{R},\mathbb{C}\\}$, the multiplicative decomposition
$A=DT$ into commuting factors with $D$ diagonalizable and $T$ unipotent, is
called the multiplicative Jordan–Chevalley decomposition.

For background on the (multiplicative) Jordan–Chevalley decomposition, we
refer to Section 15.1 in [6]. We now analyze the structure of the
multiplicative Jordan–Chevalley decomposition of a semigroup.

######  Theorem 4.2.

Let $(g(x))_{x\geq 0}$ be a semigroup of invertible linear maps acting on
$\mathbb{K}^{d}$ with $\mathbb{K}\in\\{\mathbb{R},\mathbb{C}\\}$ and let
$g(x)=D(x)T(x)$ be the multiplicative Jordan–Chevalley decomposition of each
$g(x)$. Then $((D(x))_{x\geq 0})$ and $(T(x))_{x\geq 0}$ each form a
semigroup, and the two families commute with each other, i.e.
$T(x)D(y)=D(y)T(x)$ for all $x,y\geq 0$.

###### Proof.

We can w.l.o.g. assume that $\mathbb{K}=\mathbb{C}$, since by uniqueness of
the Jordan–Chevalley decomposition $D(x)\in\mathbb{R}^{d\times d}$ and
$T(x)\in\mathbb{R}^{d\times d}$ if $g(x)\in\mathbb{R}^{d\times d}$ for $x\geq
0$. Take the SPD from Theorem 3.3, $V=\bigoplus_{i=1}^{k}V_{i}$, so that each
$g(x)$ has only one eigenvalue on $V_{i}$. Denote by
$g(x)|_{V_{i}}=D_{i}(x)T_{i}(x)$ the multiplicative Jordan–Chevalley
decomposition $g(x)$ restricted to $V_{i}$. Denote by $\lambda_{i}(x)$ the
eigenvalue of $g(x)$ on $V_{i}$. Clearly,
$D_{i}(x)=\lambda_{i}(x)\operatorname{id}$. Since the $(g_{i}(x))_{x\geq 0}$
are a commuting family of matrices, they share a common eigenvector $v_{i}\in
V_{i}$. We have

$\lambda_{i}(x)\lambda_{i}(y)v_{i}=g_{i}(x)g_{i}(y)v_{i}=g_{i}(x+y)v_{i}=\lambda_{i}(x+y)v_{i},$

and hence $(D_{i}(x))_{x\geq 0}$ is a semigroup. Since each $D_{i}(x)$ is a
multiple of the identity, it commutes with every linear map and hence

$T_{i}(x)T_{i}(y)=\frac{1}{\lambda_{i}(x)\lambda_{i}(y)}g_{i}(x)g_{i}(y)=\frac{1}{\lambda_{i}(x+y)}g_{i}(x+y)=T_{i}(x+y),$

which shows that $(T_{i}(x))_{x\geq 0}$ is also a semigroup. The result then
follows, since by uniqueness $(\bigoplus D_{i}(x))\otimes(\bigoplus T_{i}(x))$
is the multiplicative Jordan–Chevalley decomposition of $g(x)$. ∎

######  Theorem 4.3.

Let $(g(x))_{x\geq 0}$ be a semigroup of invertible linear maps acting on $V$
and let $g(x)=D(x)T(x)$ be its multiplicative Jordan–Chevalley decomposition.
Then there exist commuting real diagonalizable linear maps $J(x)$ and
commuting real nilpotent linear maps $N(x)$ satisfying

  1. 1.

$J(x)+J(y)=J(x+y)$

  2. 2.

$N(x)+N(y)=N(x+y)$

  3. 3.

$J(x)N(y)=N(y)J(x)$ for all $x,y\geq 0$

such that

$D(x)=\exp(J(x)),\quad T(x)=\exp(N(x))$

and

$g(x)=\exp(J(x)+N(x)).$

###### Proof.

Let $V=\bigoplus_{i=1}^{n}V_{i}$ be the SRPD and let
$g_{i}(x):=g(x)|_{V_{i}}$. Assume first that $V_{i}$ is of first type and
$\lambda_{i}(x)$ is the single positive eigenvalue of $g_{i}(x)$. Then the
multiplicative Jordan–Chevalley decomposition on $V_{i}$ is
$g_{i}(x)=\lambda_{i}(x)\operatorname{id}T_{i}(x)$. Define
$J_{i}(x):=\log(\lambda_{i}(x))\operatorname{id}$. Since
$T_{i}(x)-\operatorname{id}$ is nilpotent, we have

$N_{i}(x):=\log(T_{i}(x))=\sum_{k=1}^{d-1}\frac{1}{k}(\operatorname{id}-T_{i}(x))^{k}.$

Notice that since the logarithm converges for all unipotent matrices the
exponential map between the Lie algebra of nilpotent matrices and the Lie
group of unipotent matrices is bijective (see p. 35 in [5]). Since $T_{i}(x)$
is a semigroup we have

$\exp(N_{i}(x))\exp(N_{i}(y))=\exp(N_{i}(y))\exp(N_{i}(x))=\exp(N_{i}(x+y)).$

Rewrite this as

$\exp(N_{i}(y))^{-1}\exp(N_{i}(x))\exp(N_{i}(y))=\exp\big{(}\exp(N_{i}(y))^{-1}N_{i}(x)\exp(N_{i}(y))\big{)}=\exp(N_{i}(x)).$

Since $\exp(N_{i}(y))^{-1}N_{i}(x)\exp(N_{i}(y))$ is nilpotent and $\exp$ is
bijective we have

$\exp(N_{i}(y))^{-1}N_{i}(x)\exp(N_{i}(y))=N_{i}(x).$

For any $t\in\mathbb{R}$ set $N_{i}^{t}(x):=t\operatorname{id}+N_{i}(x)$.
$N_{i}^{t}(x)$ is invertible for $t\not=0$. Clearly

$N_{i}^{t}(x)\exp(N_{i}(y))=\exp(N_{i}(y))N_{i}^{t}(x).$

Applying the same idea again yields

$\exp(N_{i}^{t}(x)^{-1}N_{i}(y)N_{i}^{t}(x))=\exp(N_{i}(y)).$

Again by uniqueness we obtain

$N_{i}^{t}(x)N_{i}(y)=N_{i}(y)N_{i}^{t}(x)$

or equivalently

$N_{i}(x)N_{i}(y)=N_{i}(y)N_{i}(x).$

By commutativity we obtain

$\exp(N_{i}(x)+N_{i}(y))=\exp(N_{i}(x))\exp(N_{i}(y))=\exp(N_{i}(x+y))$

and hence, again by uniqueness, we have $N_{i}(x)+N_{i}(y)=N_{i}(x+y)$. It
follows that $J_{i}(x)$ and $N_{i}(x)$ satisfy the required conditions.

In the second case $V_{i}$ is of second type. By Theorem 3.3 and since
$g_{i}(x)$ is real, $V_{i}$ decomposes over $\mathbb{C}$ as $V_{i}=U_{1}\oplus
U_{2}$ where $g_{i}(x)$ has only one eigenvalue on each $U_{j}$ and $U_{1}$ is
isomorphic to $U_{2}$ with isomorphism given by $u\in
U_{1}\mapsto\overline{u}\in U_{2}$. By taking the principal branch of the
logarithm, in the same manner as in the real case we obtain commuting $J,N$ on
$U_{1}$ and $\overline{J},\overline{N}$ on $U_{2}$ satisfying Cauchy’s
equation such that

$g_{i}(x)=\exp(J(x)+N(x))\oplus\exp(\overline{J}(x)+\overline{N}(x)).$

Notice that $J(x)=(\mu(x)+\mathbf{i}\nu(x))\operatorname{id}$, where
$\mathbf{i}=\sqrt{-1}$ and $\nu(x)$ satisfies Cauchy’s functional equation on
$\mathbb{R}/[-\pi,\pi)$. By Lemma A.1, we can lift any solution on
$\mathbb{R}/[-\pi,\pi)$ to a solution on $\mathbb{R}$ such that linearity is
preserved. From now on denote by $\nu(x)$ this lift. Hence if we choose any
basis of $U_{1}$ and its complex conjugate on $U_{2}$ we obtain that
$g_{i}(x)$ is similar to

$g_{i}(x)\sim\exp\left[\begin{pmatrix}J(x)+N(x)&0\\\
0&\overline{J}(x)+\overline{N}(x)\end{pmatrix}\right].$

Taking the similarity transform with the matrix

$A:=\frac{1}{\sqrt{2}}\begin{pmatrix}\operatorname{id}&\mathbf{i}\cdot\operatorname{id}\\\
\mathbf{i}\cdot\operatorname{id}&\operatorname{id}\end{pmatrix},$

where $\operatorname{id}=\operatorname{id}_{\operatorname{dim}(U_{1})}$, we
obtain

$A\begin{pmatrix}J(x)+N(x)&0\\\
0&\overline{J}(x)+\overline{N}(x)\end{pmatrix}A^{-1}\\\
=\begin{pmatrix}\operatorname{Re}(J(x)+N(x))&\operatorname{Im}(J(x)+N(x))\\\
-\operatorname{Im}(J(x)+N(x))&\operatorname{Re}(J(x)+N(x))\end{pmatrix}\in
L(V_{i}).$

Since matrix similarity over $\mathbb{C}$ is equivalent to matrix similarity
over $\mathbb{R}$ for two real matrices, there exists a real matrix $B_{i}$ on
$V_{i}$ such that

$\displaystyle
g_{i}(x)=\exp\left[B_{i}\begin{pmatrix}\operatorname{Re}(J(x))&\operatorname{Im}(J(x))\\\
-\operatorname{Im}(J(x))&\operatorname{Re}(J(x))\end{pmatrix}B_{i}^{-1}+B_{i}\begin{pmatrix}\operatorname{Re}(N(x))&\operatorname{Im}(N(x))\\\
-\operatorname{Im}(N(x))&\operatorname{Re}(N(x))\end{pmatrix}B_{i}^{-1}\right].$
(4.1)

Setting

$\displaystyle
J_{i}(x):=B_{i}\begin{pmatrix}\operatorname{Re}(J(x))&\operatorname{Im}(J(x))\\\
-\operatorname{Im}(J(x))&\operatorname{Re}(J(x))\end{pmatrix}B_{i}^{-1}=B_{i}\begin{pmatrix}\operatorname{id}\mu(x)&\operatorname{id}\nu(x)\\\
-\operatorname{id}\nu(x)&\operatorname{id}\mu(x)\end{pmatrix}B_{i}^{-1}$

and

$N_{i}(x):=B_{i}\begin{pmatrix}\operatorname{Re}(N(x))&\operatorname{Im}(N(x))\\\
-\operatorname{Im}(N(x))&\operatorname{Re}(N(x))\end{pmatrix}B_{i}^{-1},$

we see that $J_{i}$ and $N_{i}$ satisfy the desired conditions, with
$D_{i}(x)=\exp(J_{i}(x))$ and $T_{i}(x)=\exp(N_{i}(x))$ by uniqueness of the
Jordan–Chevalley decomposition. The direct sums $\bigoplus_{i=1}^{n}J_{i}(x)$
and $\bigoplus_{i=1}^{n}N_{i}(x)$ give the required matrices. Furthermore in
the case where there exists $x>0$ for which $g_{i}(x)$ has a complex
eigenvalue $\lambda(x)=\mu(x)+\mathbf{i}\nu(x)$ we have

$g_{i}(x)=B_{i}\underbrace{\begin{pmatrix}\cos(\nu(x))\operatorname{id}&\sin(\nu(x))\operatorname{id}\\\
-\sin(\nu(x))\operatorname{id}&\cos(\nu(x))\operatorname{id}\end{pmatrix}}_{U_{i}(x)}B_{i}^{-1}\exp(\mu(x)\operatorname{id}+B_{i}N(x)B_{i}^{-1}),$

where $U_{i}(x)\in SO(\operatorname{dim}(V_{i}))$. Recall the matrix defined
in (2.2). By changing the order of the basis, we have that $U_{i}$ is similar
to the block diagonal matrix (recall the notation (2.2))

$\displaystyle U_{i}(x)\sim\begin{pmatrix}Q(\nu(x))&&\\\ &\ddots&\\\
&&Q(\nu(x))\end{pmatrix}=Q_{\operatorname{dim}(V_{i})}^{\nu}(x).$ (4.2)

Hence

$g_{i}(x)=\tilde{B}_{i}Q_{\operatorname{dim}(V_{i})}^{\nu}(x)\tilde{B}_{i}^{-1}\exp(\mu(x)\operatorname{id}+B_{i}N(x)B_{i}^{-1}),$
(4.3)

where $\tilde{B}_{i}$ is the composition of $B_{i}$ with some permutation
matrix $P$. ∎

##  5 Proof of the main result

After providing some final preparatory results, this section ends with the
proof of Theorem 2.6. Consider again the SRPD from Theorem 3.2. Now on each
$V_{i}$, $g(x)$ has either one positive eigenvalue
$\lambda_{i}(x)=e^{\mu_{i}(x)}$ or two complex conjugate eigenvalues
$\lambda_{i}(x)=e^{\mu_{i}(x)+\mathbf{i}\nu_{i}(x)}$ and
$\overline{\lambda_{i}}(x)=e^{\mu_{i}(x)-\mathbf{i}\nu_{i}(x)}$, where it is
possible that $\nu_{i}(x)=\pi$ for some values of $x$. If $V_{i}$ is of first
type, set $\nu_{i}(x)=0$. Each $\nu_{i}$ is a solution to Cauchy’s functional
equation. Consider then the set $\\{f|f=\nu_{i}\text{ or }f=-\nu_{i}\text{ for
some }1\leq i\leq n\\}$ and partition it into equivalent solutions, according
to Definition A.2. This then gives a partition of the index set
$\\{1,\dots,n\\}=\bigcup_{l=1}^{k}I_{l}$ in the following manner: If
$\nu_{i}\sim\nu_{j}$ or $\nu_{i}\sim-\nu_{j}$, then $i,j$ are in the same
subset of the partition. This is well-defined, since if $f\sim g$ then
$-f\sim-g$. Set $W_{l}:=\bigoplus_{i\in I_{l}}V_{i}$.

######  Definition 5.1.

We call the decomposition $V=\bigoplus_{i=1}^{k}W_{i}$ the partitioned SRPD of
$V$.

Furthermore associate with each $W_{j}$ one solution $\eta_{j}$ such that
$\eta_{j}\sim\nu_{i}$ with $i\in I_{j}$. If $\eta_{j}$ is linear we always
take $\eta_{j}\equiv 0$. Notice that for $i\neq j$ $\eta_{i}\not\sim\eta_{j}$,
hence there can be at most one $W_{i}$ with $\eta_{i}=0$ and furthermore this
is the only $W_{i}$ which can have odd dimension since it contains all $V_{j}$
of type 1 (recall Definition 3.4).

######  Theorem 5.2.

Let $(g(x))_{x\geq 0}$ be a semigroup of invertible linear maps acting on $V$
and denote by $V=\bigoplus_{i=1}^{k}W_{i}$ its partitioned SRPD. Denote by
$\eta_{i}$ the solution associated with $W_{i}$. If $\eta_{i}$ is non-
continuous then there exists a change of basis $A_{i}$ on $W_{i}$ such that

$g(x)|_{W_{i}}=A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}\exp(G_{i}(x))$

where $d_{i}=\operatorname{dim}(W_{i})$ and

$A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}G_{i}(y)=G_{i}(y)A_{i}Q_{d_{i}}^{\nu_{i}}(x)A_{i}^{-1}$

for all $x,y\geq 0$. If $\eta_{i}=0$ then

$g(x)|_{W_{i}}=\exp(G_{i}(x)).$

In both cases $G_{i}(x):W_{i}\to W_{i}$ is a commuting family of matrices on
$W_{i}$ satisfying Cauchy’s functional equation. Furthermore if $v$ is a
common eigenvector of $(G_{i}(x))_{x\geq 0}$ such that
$G_{i}(x)v=[\mu(x)+\mathbf{i}\nu(x)]v$, then $\nu$ is linear in $x$.

###### Proof.

Assume first that $\eta_{i}$, the associated solution of Cauchy’s functional
equation, is non-continuous. For each $W_{i}$ we have the decomposition
$W_{i}=\bigoplus_{j\in I_{i}}V_{j}$ where $V_{j}$ are subspaces from the SRPD.
Since $\eta_{i}$ is non-continuous, each $V_{j}$ in the direct sum has to be
of second type, hence by (4.3) on each $V_{j}$ we have

$g_{j}(x)=\tilde{B}_{j}Q_{\operatorname{dim}(V_{j})}^{\nu_{j}}(x)\tilde{B}_{j}^{-1}\exp\big{(}\mu(x)\operatorname{id}+B_{j}N(x)B_{j}^{-1}\big{)}.$

By definition of $\eta_{i}$ we have $\eta_{i}\sim\nu_{j}$ and hence there
exists $c_{j}\in\mathbb{R}$ such that $\nu_{j}(x)=\eta_{i}(x)+c_{j}x$. Let

$C_{j}:=c_{j}\begin{pmatrix}L&&\\\ &\ddots&\\\ &&L\end{pmatrix}$

with

$L:=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},$

where $C_{j}$ is of dimension
$\operatorname{dim}(V_{j})\times\operatorname{dim}(V_{j})$. Then we have

$\displaystyle g_{j}(x)$
$\displaystyle=\tilde{B}_{j}Q_{\operatorname{dim}(V_{j})}^{\nu_{j}}(x)\tilde{B}_{j}^{-1}\exp\big{(}\mu_{j}(x)\operatorname{id}+B_{j}N_{j}(x)B_{j}^{-1}\big{)}$
$\displaystyle=\tilde{B}_{j}Q_{\operatorname{dim}(V_{j})}^{\eta_{i}}(x)\exp(C_{j}x)\tilde{B}_{j}^{-1}\exp\big{(}\mu_{j}(x)\operatorname{id}+B_{j}N_{j}(x)B_{j}^{-1}\big{)}$
$\displaystyle=\tilde{B}_{j}Q_{\operatorname{dim}(V_{j})}^{\eta_{i}}(x)\tilde{B}_{j}^{-1}\exp\big{(}\tilde{B}_{j}C_{j}x\tilde{B}_{j}^{-1}+\mu_{j}(x)\operatorname{id}+B_{j}N_{j}(x)B_{j}^{-1}\big{)}.$

Set
$\tilde{G}_{j}(x):=\tilde{B}_{j}C_{j}x\tilde{B}_{j}^{-1}+\mu_{j}(x)\operatorname{id}+B_{j}N_{j}(x)B_{j}^{-1}$.
Then

$A_{i}:=\bigoplus_{j\in I_{i}}\tilde{B}_{j}\quad\text{and}\quad
G_{i}(x):=\bigoplus_{j\in I_{i}}\tilde{G}_{j}(x)$ (5.1)

satisfy

$g(x)|_{W_{i}}=\bigoplus_{j\in
I_{i}}g_{j}(x)=A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}\exp(G_{i}(x)).$

By construction the common eigenvalues of $\exp(G_{i}(x))$ satisfy
$\lambda(x)=e^{\mu_{j}(x)\pm\mathbf{i}c_{j}x}$. If $\eta_{i}\equiv 0$, then by
Theorem 4.3 there exists $G_{i}(x)$ such that $g(x)|_{W_{i}}=\exp(G_{i}(x))$.
By the definition of $W_{i}$, it follows that the imaginary parts of the
eigenvalues of $G_{i}(x)$ have to be equivalent to $\eta_{i}\equiv 0$, hence
they have to be linear. ∎

######  Theorem 5.3.

Let $g$ be a non-degenerate semigroup acting on $V$ which satisfies Assumption
2.5. Then there exists a matrix $M$ and a semigroup $S(x)\in SO(d)$, where
$SO(d)$ is the set of special orthogonal matrices, such that

$g(x)=S(x)\exp(Mx),\quad x\geq 0,$

with $M$ commuting with $S(x)$.

###### Proof.

Consider the SRPD $V=\bigoplus_{i=1}^{n}V_{i}$ from Theorem 3.2, and define
$g_{i}(x):=g(x)|_{V_{i}}$. Clearly,
$\|g_{i}(x)\|_{\operatorname{op}}\leq\|g(x)\|_{\operatorname{op}}\leq f(x)$.
For each eigenvalue $\lambda_{i}(x)=e^{\mu_{i}(x)+\mathbf{i}\nu_{i}(x)}$ on
$V_{i}$ we have

$|\lambda_{i}(x)|\leq\|g_{i}(x)\|_{\operatorname{op}}\leq f(x).$

Since $\lambda_{i}(x)\lambda_{i}(y)=\lambda_{i}(x+y)$, we have
$|\lambda_{i}(x)||\lambda_{i}(y)|=|\lambda_{i}(x+y)|$. Moreover,
$\mu_{i}(x)=\log(|\lambda_{i}(x)|)$, and so $\mu_{i}(x)\leq\log(f(x))$ for
$x\geq 0$ and $\mu_{i}(x)+\mu_{i}(y)=\mu_{i}(x+y)$. It follows that $\mu_{i}$
is locally bounded and hence that $\mu_{i}(x)=a_{i}x$ for some $a_{i}\geq 0$.
As in the proof of Theorem 4.3, on $V_{i}$ we have
$g_{i}(x)=\exp(J_{i}(x))T_{i}(x)$ where $J_{i}(x)$ is diagonalizable with
eigenvalues $a_{i}x\pm\mathbf{i}\nu_{i}(x)$, and thus

$\|T_{i}(x)\|_{\operatorname{op}}\leq f(x)e^{-a_{i}x}.$

As in Theorem 4.3, set

$N_{i}(x):=\log(T_{i}(x))=\sum_{k=1}^{d-1}\frac{1}{k}(\operatorname{id}-T_{i}(x))^{k}.$

Then, since $\|\operatorname{id}-T_{i}(x)\|_{\operatorname{op}}\leq
1+f(x)e^{-a_{i}x}$, we obtain

$\displaystyle\|N_{i}(x)\|_{\operatorname{op}}\leq\sum_{k=1}^{d-1}\frac{1}{k}\|\operatorname{id}-T_{i}(x)\|_{\operatorname{op}}^{k}\leq\sum_{k=1}^{d-1}\frac{1}{k}\big{(}1+f(x)e^{-a_{i}x}\big{)}^{k}=:F(x),$
(5.2)

where $F(x)$ is again locally bounded. Since the operator norm in finite
dimensions is equivalent to the $L^{\infty}$-norm, it follows that each entry
of $N_{i}(x)$ is locally bounded and satisfies Cauchy’s functional equation.
Thus, there exists a nilpotent linear map $P_{i}$ such that $N_{i}(x)=P_{i}x$.
Let $V=\bigoplus_{i=1}^{k}W_{i}$ be the partitioned SRPD of $V$. Assume first
that the solution associated with $W_{i}$ is $\eta_{i}\equiv 0$. Since
$W_{i}=\bigoplus_{j\in I_{i}}V_{j}$ we have $g(x)|_{W_{i}}=\bigoplus_{j\in
I_{i}}\exp(J_{j}(x)+P_{j}x)$. The real part of the eigenvalues of $J_{j}(x)$
is linear in $x$ and by Theorem 5.2 also the complex parts have to be linear
since $\nu_{i}\equiv 0$. Since $J_{j}(x)$ is diagonalizable and all its
eigenvalues are continuous in $x$, $J_{j}(x)$ is continuous in $x$ and hence
$J_{j}(x)=\tilde{M}_{j}x$. Setting $M_{i}:=\bigoplus_{j\in
I_{i}}(\tilde{M}_{j}+P_{j})$ and $S_{i}(x)=\operatorname{id}$ yields
$g(x)|_{W_{i}}=S_{i}(x)\exp(M_{i}x)$. By Theorem 5.2 and (5.1), we have for
$W_{i}$ with $\eta_{i}$ non-continuous

$g(x)|_{W_{i}}=A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}\exp(G_{i}(x))$

with

$\displaystyle G_{i}(x)$ $\displaystyle=\bigoplus_{j\in
I_{i}}\big{(}\tilde{B}_{j}C_{j}x\tilde{B}_{j}^{-1}+\mu_{j}(x)\operatorname{id}+B_{j}N_{j}(x)B_{j}^{-1}\big{)}$
$\displaystyle=\bigoplus_{j\in
I_{i}}\big{(}\tilde{B}_{j}C_{j}\tilde{B}_{j}^{-1}+a_{j}\operatorname{id}+B_{j}P_{j}B_{j}^{-1}\big{)}x.$

Hence, setting $M_{i}:=\bigoplus_{j\in
I_{i}}(\tilde{B}_{j}C_{j}\tilde{B}_{j}^{-1}+a_{j}\operatorname{id}+B_{j}P_{j}B_{j}^{-1})$,
we have $G_{i}(x)=M_{i}x$. Thus

$g(x)|_{W_{i}}=A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}\exp(M_{i}x).$

Next we show that $S_{i}(x):=A_{i}Q_{d_{i}}^{\eta_{i}}(x)A_{i}^{-1}$ is an
isometry on $W_{i}$. Since $\|g(x)\|_{\operatorname{op}}\leq f(x)$, we have

$\displaystyle\langle v,v\rangle-\frac{1}{f(x)^{2}}\langle
g(x)v,g(x)v\rangle\geq 0$ (5.3)

for any $v\in V$. Hence, for $v\in W_{i}$ we have

$\langle
v,v\rangle-\frac{1}{f(x)^{2}}\big{\langle}\exp(M_{i}x)S_{i}(x)v,\exp(M_{i}x)S_{i}(x)v\big{\rangle}\geq
0.$

Fix an arbitrary $x_{0}\geq 0$, and set $\theta_{0}:=\eta_{i}(x_{0})$. The
graph of $\eta_{i}$ is dense in $\mathbb{R}\times\mathbb{R}$ (see Appendix A),
and so there is a sequence of positive reals $x_{n}$ with
$\lim_{n\to\infty}x_{n}=0$ such that $\lim\eta_{i}(x_{n})=\theta_{0}$. Then,
for $v\in W_{i}$ we obtain

$\displaystyle 0$ $\displaystyle\leq\lim\Big{(}\langle
v,v\rangle-\frac{1}{f(x_{n})^{2}}\big{\langle}\exp(M_{i}x_{n})S_{i}(x_{n})v,\exp(M_{i}x_{n})S_{i}(x_{n})v\big{\rangle}\Big{)}$
$\displaystyle=\langle v,v\rangle-\langle S_{i}(x_{0})v,S_{i}(x_{0})v\rangle,$

where $\lim\frac{1}{f(x_{n})}=1$ since $f$ is right-continuous at $0$. Choose
$(\tilde{x}_{n})$ such that $\lim\tilde{x}_{n}=0$ and
$\lim\eta_{i}(\tilde{x}_{n})=-\theta_{0}$. Then, since $S(x_{0})v\in W_{i}$,
we obtain

$\displaystyle 0$ $\displaystyle\leq\lim\Big{(}\langle
S(x_{0})v,S(x_{0})v\rangle-\frac{1}{f(\tilde{x}_{n})^{2}}\big{\langle}\exp(M_{i}\tilde{x}_{n})S_{i}(\tilde{x}_{n})S_{i}(x_{0})v,\exp(M_{i}\tilde{x}_{n})S_{i}(\tilde{x}_{n})S_{i}(x_{0})v\big{\rangle}\Big{)}$
$\displaystyle=\langle S_{i}(x_{0})v,S_{i}(x_{0})v\rangle-\langle v,v\rangle,$

since $Q_{d_{i}}(-\theta_{0})Q_{d_{i}}(\theta_{0})=\operatorname{id}$. Hence
we obtain

$\langle v,v\rangle=\langle S(x_{0})v,S(x_{0})v\rangle,$

which implies that $S(x_{0})$ is an isometry on $W_{i}$. Next we show that all
$W_{i}$ are pairwise orthogonal. Let $v\in W_{i}$ and $u\in W_{j}$ for $i\neq
j$. By Lemma A.3, there exists w.l.o.g. a sequence $\hat{x}_{n}$ such that
$\lim\hat{x}_{n}=0$, $\lim\eta_{i}(\hat{x}_{n})=\pi$ and
$\lim\eta_{j}(\hat{x}_{n})=\theta\in(-\pi,\pi)$. Hence $\lim
S(\hat{x}_{n})|_{W_{i}}=-\operatorname{id}$ and $\lim
S(\hat{x}_{n})|_{W_{j}}=A_{j}Q_{d_{j}}(\theta)A_{j}^{-1}$ with
$\theta\in(-\pi,\pi)$. We have the identity
$Q_{d_{j}}(\phi)+Q_{d_{j}}(-\phi)=2\cos(\phi)\operatorname{id}$. Applying this
to $\phi:=\theta/2$ we obtain

$\operatorname{id}+Q_{d_{j}}(\theta)=2\cos(\theta/2)Q_{d_{j}}(\theta/2).$

Hence we have
$(\operatorname{id}+Q_{d_{j}}(\theta))^{-1}=(2\cos(\theta/2))^{-1}Q_{d_{j}}(-\theta/2)$.
Set $\tilde{u}:=A_{j}^{-1}(\operatorname{id}+Q_{d_{j}}(\theta))^{-1}A_{j}u$.
Then $\tilde{u}\in W_{j}$, and applying (5.3) to $v+\tilde{u}$ and taking the
limit along $\hat{x}_{n}$ yields

$\displaystyle 0$ $\displaystyle\leq\langle
v+\tilde{u},v+\tilde{u}\rangle-\langle\lim S(\hat{x}_{n})(v+\tilde{u}),\lim
S(\hat{x}_{n})(v+\tilde{u})\rangle$
$\displaystyle=\|v\|^{2}+\|\tilde{u}\|^{2}-\|\lim S(\hat{x}_{n})v\|^{2}-\|\lim
S(\hat{x}_{n})\tilde{u}\|^{2}+2\langle v,\tilde{u}\rangle-2\langle\lim
S(\hat{x}_{n})v,\lim S(\hat{x}_{n})\tilde{u}\rangle$ $\displaystyle=2\langle
v,\tilde{u}\rangle-2\langle\lim S(\hat{x}_{n})v,\lim
S(\hat{x}_{n})\tilde{u}\rangle,$

where the last equality follows from the fact that each $S(\hat{x}_{n})$ is an
isometry on $W_{i}$ and $W_{j}$. Since the inequality above also holds for
$v-\tilde{u}$, we obtain the equality

$\displaystyle 0=$ $\displaystyle\langle v,\tilde{u}\rangle-\langle\lim
S(\hat{x}_{n})v,\lim S(\hat{x}_{n})\tilde{u}\rangle$ $\displaystyle=$
$\displaystyle\langle v,\tilde{u}\rangle+\langle
v,A_{j}Q_{d_{j}}(\theta)A_{j}^{-1}\tilde{u}\rangle=\langle
v,A_{j}(\operatorname{id}+Q_{d_{j}}(\theta))A_{j}^{-1}\tilde{u}\rangle=\langle
v,u\rangle.$

Since $v,u$ were arbitrary, this shows orthogonality. Hence the decomposition
$V=\bigoplus_{i=1}^{k}W_{i}$ is orthogonal, and since $S_{i}(x)$ is an
isometry on $W_{i}$, it follows that $S(x):=\bigoplus_{i=1}^{k}S_{i}(x)$ is in
$SO(d)$. Setting $M=\bigoplus_{i=1}^{k}M_{i}$, we obtain

$g(x)=S(x)\exp(Mx).\qed$

######  Corollary 5.4.

Assume that $S(x)=AQ_{d}^{\nu}(x)A^{-1}$ is a semigroup with $A$ being an
invertible matrix such that $S(x)$ is an isometry for each $x\geq 0$. Then
there exists an orthogonal matrix $U$ such that

$S(x)=UQ_{d}^{\nu}(x)U^{\mathrm{T}}.$

###### Proof.

One can easily verify the identities
$S(x)+S(x)^{\mathrm{T}}=2\cos(\nu(x))\operatorname{id}$ and
$\sin(\nu(y))S(x)-\sin(\nu(x))S(y)=\sin(\nu(x)-\nu(y))\operatorname{id}$.
Choose $x>0$ such that $\sin(\nu(x))\neq 0$. Such an $x$ clearly exists since
$\nu$ is linear on $\mathbb{Q}x$. Choose any $v\in V$ of unit length and set

$u:=\frac{S(x)v-\cos(\nu(x))v}{\sin(\nu(x))}.$

The definition of $u$ is invariant under the choice of $x$ as long as
$\sin(\nu(x))\neq 0$. This can be seen by noticing that

$\displaystyle\sin(\nu(y))S(x)-\sin(\nu(x))S(y)$
$\displaystyle=\sin(\nu(x)-\nu(y))\operatorname{id}$
$\displaystyle\Longleftrightarrow\quad\frac{S(x)v-\cos(\nu(x))v}{\sin(\nu(x))}$
$\displaystyle=\frac{S(y)v-\cos(\nu(y))v}{\sin(\nu(y))}.$

Hence we obtain $S(x)v=\cos(\nu(x))v+\sin(\nu(x))u$. We have

$\langle u,v\rangle=\frac{1}{\sin(\nu(x))}\langle
S(x)v-\cos(\nu(x))v,v\rangle=\frac{1}{\sin(\nu(x))}\big{(}\langle
S(x)v,v\rangle-\cos(\nu(x))\big{)}=0,$

where the last equality follows from

$2\langle S(x)v,v\rangle=\langle S(x)v,v\rangle+\langle
S(x)^{\mathrm{T}}v,v\rangle=\langle
S(x)v+S(x)^{\mathrm{T}}v,v\rangle=2\cos(\nu(x)).$

Similarly we can show that $u$ is also of unit length. Set
$H:=\operatorname{Span}(u,v)$. Clearly, $H$ is invariant under $(S(x))_{x}$
and hence so is $H^{\perp}$ since each $S(x)\in SO(d)$. In this manner we can
construct an orthonormal basis, and we denote by $U$ the matrix associated
with this change of basis. Then we have

$Ug(x)U^{\mathrm{T}}=UAS(x)A^{-1}U^{\mathrm{T}}\exp(UMU^{\mathrm{T}}x)=Q_{d}^{\nu}(x)\exp(UMU^{\mathrm{T}}x).\qed$

######  Lemma 5.5.

Suppose that the semigroup $g$, acting on $V$, satisfies Assumption 2.5. Let
$V=V_{1}\oplus V_{2}$ be the decomposition of Corollary 3.8 such that
$g|_{V_{1}}$ is invertible. Then, for any $v\in V_{1}$ we have

$\lim_{x\to 0}\frac{\|g(x)v\|}{\|v\|}=1.$

###### Proof.

By Theorem 5.3, we have $g(x)|_{V_{1}}=S(x)\exp(Mx)$. Since $S(x)\in
SO(V_{1})$, we obtain

$\langle g(x)v,g(x)v\rangle=\langle\exp(Mx)v,\exp(Mx)\rangle,\quad v\in
V_{1},$

which is continuous in $x$. Hence

$\lim_{x\to 0}\|g(x)v\|=\lim_{x\to 0}\|\exp(Mx)v\|=\|v\|.\qed$

######  Corollary 5.6.

Suppose that the semigroup $g$, acting on $V$, satisfies Assumption 2.5. Then
the decomposition from Corollary 3.8 is orthogonal.

###### Proof.

Let $V=V_{1}\oplus V_{2}$ with $g(x)|_{V_{1}}$ being non-degenerate and
$g(x)|_{V_{2}}\equiv 0$. Assume $V_{1}$ is not orthogonal to $V_{2}$. Then
there exists $v\in V_{1}$ such that $p_{V_{2}}(v)\neq 0$, where $p_{V_{2}}$
denotes the orthogonal projection onto $V_{2}$. By Lemma 5.5, we have

$\lim_{x\to 0}\frac{\|g(x)v\|}{\|v\|}=1.$

Calculating the same limit for $v-p_{V_{2}}(v)$ we obtain

$\displaystyle\lim_{x\to
0}\frac{\|g(x)(v-p_{V_{2}}(v))\|}{\|v-p_{V_{2}}(v)\|}=\lim_{x\to
0}\frac{\|g(x)v\|}{\|v\|}\frac{\|v\|}{\|v-p_{V_{2}}(v)\|}=\frac{\|v\|}{\|v-p_{V_{2}}(v)\|}.$

Since $\|v\|^{2}=\|v-p_{V_{2}}(v)\|^{2}+\|p_{V_{2}}(v)\|^{2}$ and
$\|p_{V_{2}}(v)\|\neq 0$, we have $\frac{\|v\|}{\|v-p_{V_{2}}(v)\|}>1$. Hence

$\lim_{x\to 0}\frac{\|g(x)(v-p_{V_{2}}(v))\|}{\|v-p_{V_{2}}(v)\|}>1,$

but this contradicts $\|g(x)\|_{\operatorname{op}}\leq f(x)$. Hence
$V_{1}\perp V_{2}$. ∎

###### Proof of Theorem 2.6.

By Corollary 5.6, we have the orthogonal decomposition
$V=\tilde{V}_{1}\oplus\tilde{V}_{2}$ with $g(x)|_{\tilde{V}_{2}}\equiv 0$ and
$g(x)|_{\tilde{V}_{1}}$ non-degenerate. Applying Theorem 5.3 to
$g(x)|_{\tilde{V}_{1}}$ yields the result. ∎

##  Appendix A Cauchy’s functional equation

It is classical that all continuous solutions of the equation (1.3),
$f(x)+f(y)=f(x+y)$, are linear, and that the non-linear solutions are not
continuous, even not Lebesgue measurable, and have dense graphs. For this, and
further references, we refer to [2, Section 1.1]. In this section, we provide
two auxiliary results on Cauchy’s equation. They concern lifting solutions
from an interval to the real line, resp. the joint behavior of two solutions
that differ by a non-linear function.

######  Lemma A.1.

Let $f:\mathbb{R}\to\mathbb{R}/[-a,a)$ be a solution to Cauchy’s functional
equation (1.3) on $\mathbb{R}/[-a,a)$ with $a>0$. Then there exists a solution
$\tilde{f}:\mathbb{R}\to\mathbb{R}$ of Cauchy’s functional equation such that
$f(x)\equiv\tilde{f}(x)\mod[-a,a)$. The solution $\tilde{f}$ is linear if and
only if $f$ is linear.

###### Proof.

Take a Hamel basis $(r_{i})_{i\in I}$ of $\mathbb{R}$ such that
$r_{i}\in[-a,a)$ for every $i\in I$. This is clearly possible by rescaling
every basis element if necessary. For any $x\in\mathbb{R}$ there exists a
finite subset $I_{x}\subset I$ and $c_{i}\in\mathbb{Q}$ such that
$x=\sum_{i\in I_{x}}c_{i}r_{i}$. Set $\tilde{f}(x)=\sum_{i\in
I_{x}}c_{i}f(r_{i})$, then clearly $f(x)\equiv\tilde{f}(x)\mod[-a,a)$. If $f$
is linear then clearly $\tilde{f}$ is linear as well. If $f$ is not linear
then there exist two basis elements $r_{1}$ and $r_{2}$ such that
$f(r_{1})-\frac{r_{1}}{r_{2}}f(r_{2})\neq 0$, and hence $\tilde{f}$ is also
not linear. ∎

######  Definition A.2.

We say that two solutions $\nu$ and $\eta$ of Cauchy’s functional equation are
equivalent if $\nu-\eta$ is linear.

######  Lemma A.3.

Let $f,g:\mathbb{R}\to\mathbb{R}$ be two non-equivalent solutions of Cauchy’s
functional equation. Then there exists a sequence $(x_{n})_{n\geq 0}$ in
$\mathbb{R}_{\geq 0}$, converging to $0$, such that either
$\lim_{n}f(x_{n})=\pi$ and $\lim_{n}g(x_{n})=\theta$ with $|\theta|<\pi$ or
vice versa.

###### Proof.

Choose $x,y\in\mathbb{R}_{>0}$ such that the two vectors $v_{1}=(x,f(x),g(x))$
and $v_{2}=(y,f(y),g(y))$ are linearly independent and
$\frac{f(x)}{x}-\frac{f(y)}{y}\neq\frac{g(x)}{x}-\frac{g(y)}{y}$. This is
possible, since

$\frac{f(x)}{x}-\frac{f(y)}{y}=\frac{g(x)}{x}-\frac{g(y)}{y}$

for all $x,y>0$ would imply that $f-g$ is linear. Assume w.l.o.g. that
$\left|\frac{f(x)}{x}-\frac{f(y)}{y}\right|>\left|\frac{g(x)}{x}-\frac{g(y)}{y}\right|$.
Since $f$ and $g$ are both linear on $\mathbb{Q}x$ and $\mathbb{Q}y$ and
$v_{1}$ and $v_{2}$ are linearly independent, there exist sequences
$q_{n}\in\mathbb{Q}$ and $r_{n}\in\mathbb{Q}$ such that $q_{n}x-r_{n}y\geq 0$
for every $n$, $\lim q_{n}x-r_{n}y=0$ and
$\lim_{n}f(q_{n}x-r_{n}y)=\lim_{n}q_{n}f(x)-r_{n}f(y)=\pi$. We show that
$x_{n}=q_{n}x-r_{n}y$ has the required property. Clearly
$\lim\frac{r_{n}}{q_{n}}=\frac{x}{y}$ and

$\pi=\lim_{n}q_{n}f(x)-r_{n}f(y)=\lim_{n}q_{n}\left(f(x)-\frac{r_{n}}{q_{n}}f(y)\right)=\left(f(x)-\frac{x}{y}f(y)\right)\lim
q_{n}.$

Hence

$\displaystyle|\lim q_{n}g(x)-r_{n}g(y)|=$
$\displaystyle\left|\left(g(x)-\frac{x}{y}g(y)\right)\lim q_{n}\right|$
$\displaystyle=$
$\displaystyle\left|\pi\left(\frac{g(x)}{x}-\frac{g(y)}{y}\right)/\left(\frac{f(x)}{x}-\frac{f(y)}{y}\right)\right|<\pi.\qed$

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