text:
\begin{document}

\begin{abstract}
	Let $\rho$ be an algebraic action of the additive group $\CC^+$
	on the three-dimensional affine space $\CC^3$.
	We describe the group $\Cent(\rho)$ of polynomial automorphisms of $\CC^3$
	that commute with $\rho$. A particular emphasis lies in the
	description of the automorphisms in $\Cent(\rho)$ coming from
	algebraic $\CC^+$-actions. As an application we prove
	that the automorphisms in $\Cent(\rho)$ that are
	the identity on the algebraic quotient of $\rho$ form a characteristic
	subgroup of $\Cent(\rho)$.
\end{abstract}

\maketitle

\section{\texorpdfstring{Introduction}{Introduction}}

Let $X$ be an affine algebraic variety.
A classification of the algebraic $\CC^+$-actions on
$X$ up to conjugacy in the automorphism group $\Aut(X)$ is only known for a few varieties $X$. For example when $X = \CC^2$ we have a classification:
every $\CC^+$-action is a modified translation up to conjugacy,
i.e. an action of the form $t \cdot (x, y) = (x + td(y), y)$ for a suitable polynomial $d$
(see \cite{Re1968Operations-du-grou}).
In contrast to the two-dimensional case, there is no classification known for the
$\CC^+$-actions on $\CC^3$. As a first step towards a classification, we
study in this article the centralizer of a $\CC^+$-action in $\Aut(\CC^3)$, i.e. the
group of automorphisms that commute with the $\CC^+$-action.

It is known that there is a bijective correspondence of algebraic
$\CC^+$-actions on $\CC^n$ and locally nilpotent derivations of the
polynomial ring in $n$-variables $\OO(\CC^n) = \CC[x_1, \ldots, x_n]$,
i.e. $\CC$-derivations of $\OO(\CC^n)$ such that
for every $f \in \OO(\CC^n)$ there exists an integer $n = n(f)$ such that $D^n(f) = 0$
(see \cite[sec. 1.5]{Fr2006Algebraic-theory-o}). The correspondence is given
by the exponential, i.e. the $\CC^+$-action corresponding to a locally
nilpotent derivation $D$ is given by
\[
	t \cdot (x_1, \ldots, x_n) =
	\Exp(t D)(x_1, \ldots, x_n) =
	\left( \sum_{i=0}^{\infty} \frac{t^i D^i(x_1)}{i!} , \ldots,
		\sum_{i=0}^{\infty} \frac{t^i D^i(x_n)}{i!} \right) \, .
\]

\begin{defn}
	An automorphism of the form $\Exp(D) \in \Aut(\CC^n)$ is called
	\emph{unipotent} where $D$ is a locally nilpotent derivation of
	$\OO(\CC^n)$. For a subset $S \subseteq \Aut(\CC^n)$ we denote
	by $S_u$ the unipotent automorphisms in $S$.
\end{defn}

\begin{rem}
An algebraic $\CC^+$-action $\rho$ on $\CC^n$ is uniquely determined
by the unipotent automorphism
$\rho_1 = ( (x_1, \ldots, x_n) \mapsto \rho(1, x_1, \ldots, x_n) )$
of $\CC^n$
and every unipotent automorphism of $\CC^n$ can be constructed in this way.
Thus $\rho \mapsto \rho_1$ is a bijection between algebraic $\CC^+$-actions
on $\CC^n$ and unipotent automorphisms of $\CC^n$. Moreover,
the centralizer of $\rho$ is the same as the centralizer of $\rho_1$.
\end{rem}

\begin{defn}
	Let $\bu = \Exp(D)$ be a unipotent automorphism of $\AA^n$ and let
	$f \in \ker D$. Then $f D$ is a locally
	nilpotent derivation and we call $\Exp(fD)$ a \emph{modification} of $\bu$.
	We then denote
	\[
		f \cdot \bu = \Exp(f D) \, .
	\]
	We call $\bu$ irreducible, if $\bu \neq \bid$ and
	the following holds: if $\bu = f \cdot \bu'$
	for some unipotent $\bu' \in \Aut(\AA^n)$
	and some $f \in \ker D$, then $f \in \CC^\ast$.
	If $\bu \neq \bid$, then there exists an irreducible
	$\bu' \in \Aut(\AA^n)$ such that $\bu = d \cdot \bu'$ for some
	$d \in \ker D$ and $\bu'$ is unique up to a modification
	by some element in $\CC^\ast$. We call then $\bu = d \cdot \bu'$
	a \emph{standard decomposition}.
\end{defn}

\begin{rem}
	If $\bu = \Exp(D) \in \Aut(\CC^n)$ is unipotent, then the ring of
	$\bu$-invariant polynomials $\OO(\CC^n)^\bu$ is equal to $\ker D$. The group
	of modifications of $\bu$ we denote then by $\OO(\CC^n)^\bu \cdot \bu$.
\end{rem}

In dimension $n=2$, \name{Shmuel Friedland} and \name{John Milnor} proved
that every automorphism of $\AA^2$ is conjugate
to a composition of generalized \name{H\'enon} maps

or to a triangular automorphism
(cf. \cite[Theorem~2.6]{FrMi1989Dynamical-properti}). In the first case,
\name{St\'ephane Lamy} showed that the centralizer of such an automorphism
is isomorphic to a semi-direct product of $\ZZ$ with a finite cyclic group $\ZZ_q$
(cf. \cite[Proposition~4.8]{La2001Lalternative-de-Ti}). In the second case,
assuming in addition that the automorphism is unipotent, it has
the form $\bu(x, y) = (x + d(y), y)$.
Thus, $\bu = d \cdot \bu'$ is a standard decomposition where
$\bu'(x, y) = (x+1, y)$. One can check
that the centralizer $\Cent(\bu)$ fits in the following split short exact
sequence
\begin{equation}
	\label{2dim.eq}
	1 \to \OO(\AA^2)^{\bu'} \cdot \bu'
	\hookrightarrow \Cent(\bu)
	\twoheadrightarrow \Aut(\AAone, \Gamma) \to 1
\end{equation}
where $\Aut(\AAone, \Gamma)$ denotes the automorphisms of $\AAone$
preserving the principal divisor $\Gamma = \div(d)$ in $\AAone$.

In dimension $n=3$, \name{Cinzia Bisi} proved that any automorphism $\g$ that commutes with a
so-called regular automorphism $\f$ satisfies $\g^m = \f^k$ for certain integers
$k, m$
(cf. \cite[Main~Theorem~1.1]{Bi2008On-commuting-polyn}).
As a counterpart to the regular automorphisms, one can regard
the unipotent automorphisms (a regular automorphism is always
algebraically stable and thus can not be unipotent). The work of
\name{David Finston} and \name{Sebastian Walcher}
\cite{FiWa1997Centralizers-of-lo} can be seen
as a first step in the study of the centralizer of a unipotent automorphism.
They explore the centralizer of a triangulable (locally nilpotent) derivation inside
the algebra of all derivations of the polynomial ring $\OO(\CC^3)$.

\section{\texorpdfstring{Statement of the main results}
	    {Statement of the main results}}

Let us recall briefly some notion and facts
of the theory of ind-groups
(see \cite[ch. IV]{Ku2002Kac-Moody-groups-t}
for an introduction).

A group $G$ is called an \emph{ind-group} if it is endowed with a filtration by
affine varieties $G_1 \subseteq G_2 \subseteq \ldots \,$, each one closed in the
next, such that $G = \bigcup_{i=1}^{\infty} G_i$ and such that the map
$G \times G \to G$, $(x, y) \mapsto x \cdot y^{-1}$
is a morphism of ind-varieties.
We then write $G = \varinjlim G_i$. For example, $\Aut(\AA^n)$
is an ind-group with the filtration
$\Aut(\AA^n)_1 \subseteq \Aut(\AA^n)_2 \subseteq \ldots \,$
where $\Aut(\AA^n)_i$ is the set of all automorphisms of degree $\leq i$
(see \cite{BaCoWr1982The-Jacobian-conje}).

We endow an ind-group $G = \varinjlim G_i$ with the following topology:
a subset $X \subseteq G$ is closed if and only if $X \cap G_i$ is closed in
$G_i$ for each $i$.

A subgroup $H$ of an ind-group $G = \varinjlim G_i$
is called \emph{algebraic}, if it is a closed subset of
some $G_i$. We call an element $g \in G$ \emph{algebraic}, if the closure
of the cyclic group $\langle g \rangle$ is an algebraic subgroup of $G$.

In order to state our main results we introduce some notation
and recall some facts about $\GGa$-actions on $\AA^3$.
Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent.
We denote by $\pi \colon \AA^3 \to \AA^3 \aquot \GGa$ the algebraic quotient
of the $\GGa$-action $(t, v) \mapsto (t \cdot \bu)(v)$, i.e. $\pi$ corresponds
to the inclusion $\OO(\CC^3)^\bu \subseteq \OO(\CC^3)$.

\begin{defn}
	Let $\bu = \Exp(D) \in \Aut(\AA^3)$ be unipotent.
	We call the ideal $\im D \cap \ker D$ of $\ker D$
	the \emph{plinth ideal}.
	By \cite[Theorem~1]{DaKa2009A-note-on-locally-} the plinth ideal
	is principal. We fix some generator of it and denote it by
	$a$. We denote further by
	\[
		\Gamma = \div(a)
	\]
	the principal divisor in $\AA^3 \aquot \GGa$
	corresponding to $a$ and call it the \emph{plinth divisor} of
	$D$ (respectively of $\bu$). As $\Gamma$ is completely
	determined by the closed subscheme $V(a) \subseteq \AA^3 \aquot \GGa$,
	we can and will identify this scheme with $\Gamma$.
\end{defn}

\begin{rem}
By \name{Miyanishi}'s Theorem (cf. \cite[Theorem~5.1]{Fr2006Algebraic-theory-o}),
the algebraic quotient $\AA^3 \aquot \GGa$ is isomorphic to $\AA^2$.
The restriction $\pi |_{\AA^3 \setminus \pi^{-1}(\Gamma)} \colon
\AA^3 \setminus \pi^{-1}(\Gamma) \to
\AA^2 \setminus \Gamma$ is a trivial principal $\GGa$-bundle
(cf. \cite[Principle~11]{Fr2006Algebraic-theory-o}).
\end{rem}

\begin{defn}
	\label{def:RelativeAuto}
	Let $\Gamma \subseteq \CC^2$ be an effective divisor.
	We denote by $\Aut(\AA^2, \Gamma)$ the subgroup of all
	$\g \in \Aut(\AA^2)$ such that the scheme-theoretic image
	$\g(\Gamma)$ is again $\Gamma$ and we denote by
	$\Iner(\AA^2, \Gamma)$ the subgroup of all
	$\g \in \Aut(\AA^2, \Gamma)$ such that the pullback to $\Gamma$ is
	the identity. If $\Gamma' \subseteq \CC^2$ is another effective divisor,
	then we denote the intersection
	$\Aut(\AA^2, \Gamma) \cap \Aut(\AA^2, \Gamma')$ by
	$\Aut(\AA^2, \Gamma, \Gamma')$.
\end{defn}

\begin{defn}
	Let $\Gamma = \sum_i n_i \Gamma_i$ be an effective divisor in $\AA^2$.
	We call $\Gamma$ a \emph{fence}, if
	$\Gamma_i \simeq \AAone$ for all $i$ and the $\Gamma_i$ are
	pairwise disjoint.
\end{defn}

\medskip

Let $\bu = d \cdot \bu'$ be a standard decomposition of a unipotent
$\bu \in \Aut(\AA^3)$ and let $\Gamma$, $\Gamma'$ be the
plinth divisors of $\bu$, $\bu'$ respectively.
We have an induced action of the centralizer
$\Cent(\bu)$ on the algebraic quotient
$\AA^3 \aquot \GGa$ that preserves $\Gamma$ and also $\Gamma'$.
This implies that there is
a sequence of ind-groups
\begin{equation}
	\label{3dim.eq}
	1 \to \OO(\AA^3)^{\bu'} \cdot \bu'
	\hookrightarrow \Cent(\bu) \stackrel{p}{\to}
	\Aut(\AA^3 \aquot \GGa, \Gamma, \Gamma') \, ,
\end{equation}
which is exact by Proposition~\ref{firstfam.prop}.
In contrast to the two-dimensional case (see \eqref{2dim.eq}), the homomorphism
$p$ is in general not surjective
(see \cite[Proposition~1]{St2012A-note-on-Automorp}).
Thus, in order to study $\Cent(\bu)$ we have to understand the image of $p$.

The description of $\Cent(\bu)$ is special in the case when $\bu$ is a
\emph{translation}, i.e. $\bu(x, y, z) = (x+1, y, z)$ for suitable coordinates $(x, y, z)$.
Note that $\bu$ is a translation if and only if its plinth divisor $\Gamma$ is empty.

\subsection{\texorpdfstring{Structure theorems for $\Cent(\bu)$}
	{Structure theorems for Cent(u)}}

\begin{mainprop}[cf. Proposition~\ref{modtranslation.prop}]
	Let $\buÊ\in \Aut(\AA^3)$ be a translation. Then
	\[
		1 \to \OO(\AA^3)^{\bu} \cdot \bu
		\hookrightarrow \Cent(\bu) \to
		\Aut(\AA^3 \aquot \GGa) \to 1
	\]
	is a split short exact sequence of ind-groups.
\end{mainprop}

\begin{mainthm}[cf. Theorem~\ref{main.thm1},
\ref{main.thm2} and Corollary~\ref{mainthm2.cor}]
	\label{unipotgeneral.mtm}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent and not a translation. Then
	\begin{enumerate}[i)]
		\item All elements in $\Cent(\bu)$ are algebraic.
		\item The set of unipotent elements $\Cent(\bu)_u \subseteq \Cent(\bu)$
		is a closed normal subgroup.
		\item There exists an algebraic
		subgroup $R \subseteq \Cent(\bu)$ consisting only
		of semi-simple elements such that
		$\Cent(\bu) \simeq \Cent(\bu)_u \rtimes R$
		as ind-groups.
	\end{enumerate}
\end{mainthm}

\begin{mainthm}[cf. Proposition~\ref{properties.prop} and
Theorem~\ref{main.thm2}]
	\lab{unipotfence.mthm}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent, not a translation and
	let $\bu = d \cdot \bu'$ be a standard decomposition.
	Let $\Gamma$ and $\Gamma'$ be the plinth divisors
	of $\bu$ and $\bu'$ respectively. Then,
	the sequence induced by \eqref{3dim.eq}
	\[
		1 \to \OO(\AA^3)^{\bu'} \cdot \bu' \hookrightarrow
		\Cent(\bu)_u \stackrel{p}{\to}
		\Aut(\AA^3 \aquot \GGa, \Gamma) \cap
		\Iner(\AA^3 \aquot \GGa, \Gamma')_u
		\to 1 \,
	\]
	is a split short exact sequence of ind-groups.
	Moreover, there exists an irreducible unipotent $\e \in \Aut(\AA^3)$ such
	that the restriction of
	$p$ to ${\OO(\AA^3)^{\langle \bu, \e \rangle} \cdot \e}$
	is an isomorphism of ind-groups
	($\OO(\CC^3)^{\langle \bu, \e \rangle}$
	is the subring of $\e$-invariant polynomials inside $\OO(\CC^3)^\bu$).
\end{mainthm}

Let us give some explanation of the last result.
If $\Gamma$ is not a fence, then the underlying variety
cannot be a union of orbits of a non-trivial $\GGa$-action on $\AA^2$.
Hence all unipotent automorphisms
of $\Cent(\bu)$ induce the identity on the algebraic quotient
and $\Aut(\AA^3 \aquot \GGa, \Gamma) \cap
\Iner(\AA^3 \aquot \GGa, \Gamma')_u$ is trivial.
Thus the result follows from \eqref{3dim.eq}.
So let us assume that $\Gamma$ is a non-empty fence.
There exists a proper non-empty open subset $U \subseteq \AAone$ such that
we have the following commutative diagram
\[
	\xymatrix@=1.5pc{
		\AA^3 \ar[d]_-{\pi} \ar@{}[r]|-{\supseteq} &
		\AA^3 \setminus \pi^{-1}(\Gamma) \ar@{->>}[d] \ar@{}[r]|-{\simeq} &
		(U \times \AAone) \times \AAone \ar@{->>}[d]^-{\pr}  \\
		\AA^2 \ar@{}[r]|-{\supseteq} &
		\AA^2 \setminus \Gamma \ar@{}[r]|-{\simeq} &
		U \times \AAone
	}
\]
where $\pr$ denotes the projection onto the first two factors (see Proposition~\ref{genAMS.prop}).
There exist coordinates $(u, v, w)$ of $(U \times \AAone) \times \AAone$ such that the automorphism induced
by $\bu$ on $(U \times \AAone) \times \AAone$ is given by
$(u, v, w) \mapsto (u, v, w+1)$. The unipotent automorphism
$(u, v, w) \mapsto (u, v+1, w)$ of $(U \times \AAone) \times \AAone$ extends to an
irreducible unipotent automorphism $\e$ on $\AA^3$
that commutes with $\bu$
(see subsec.~\ref{The_second_subgroup.subsec}). Thus, $\Cent(\bu)_u$ contains
$\OO(\AA^3)^{\langle \bu, \e \rangle} \cdot \e$ beside
$\OO(\AA^3)^{\bu'} \cdot \bu'$.
The difficulty lies in proving that $\Cent(\bu)_u$ is a group, generated by
$\OO(\AA^3)^{\langle \bu, \e \rangle} \cdot \e$ and $\OO(\AA^3)^{\bu'} \cdot \bu'$.

\subsection{Applications}

We present two applications of the structure theorems.
The first one concerns abstract automorphisms of $\Cent(\bu)$.

\begin{mainprop}[cf. Proposition~\ref{char.prop}]
		Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent, not a translation and
		let $\bu = d \cdot \bu'$ be a standard decomposition. Then,
		the subgroup
		$\OO(\AA^3)^{\bu'} \cdot \bu' \subseteq \Cent(\bu)$ is
		characteristic, i.e. it
		is invariant  under all abstract
		group automorphisms of $\Cent(\bu)$.
\end{mainprop}

The second application concerns the plinth divisor $\Gamma$.
The reduced scheme $\Gamma_{\red}$
has the following geometric description: The complement of $\Gamma_{\red}$
is the maximal open subset of $\AA^3 \aquot \GGa$
such that the algebraic
quotient $\pi \colon \AA^3 \to \AA^3 \aquot \GGa$ is a locally trivial principal
$\GGa$-bundle over it (see \cite[Proposition~5.4]{DuKr2014Invariants-and-Sep}).
So far - to the author's knowledge -
there is no geometric description of the \emph{scheme} $\Gamma$.
But in the case when $\Gamma$ is a non-empty fence
and $\bu$ is irreducible we can give one.

\begin{mainprop}[cf. Proposition~\ref{applicationfence.prop}]
	Let $\bu \in \Aut(\AA^3)$ be unipotent and irreducible, and assume that
	$\Gamma$ is a non-empty fence.
	Then $\Gamma \subseteq \AA^3 \aquot \GGa$
	is the largest closed subscheme fixed by $\Cent(\bu)_u$.
\end{mainprop}

\section{\texorpdfstring{Automorphisms of $\AA^2$ that preserve a divisor}
	    {Automorphisms of A2 that preserve a divisor}}

Let $\Gamma \subseteq \CC^2$ be an effective divisor. We have an exact
sequence
\[
	1 \to \Iner(\AA^2, \Gamma) \hookrightarrow \Aut(\AA^2, \Gamma)
	\rightarrow \Aut(\Gamma)
\]
(see Definition~\ref{def:RelativeAuto}). The next result uses heavily the main
result in \cite{BlSt2013Automorphisms-of-t}.

\begin{prop} $\textrm{ }$
	\lab{curves.prop}
	Let $\Gamma$ be a non-trivial effective divisor of $\AA^2$. Then
	\begin{enumerate}[i)]

		\item The subgroup $\Aut(\AA^2, \Gamma) \subseteq \Aut(\AA^2)$
		is closed and all elements of $\Aut(\AA^2, \Gamma)$ are algebraic.

		\item The following statements are equivalent
		\begin{enumerate}[a)]
			\item $\Gamma$ is a fence
			\item $\Aut(\AA^2, \Gamma)$ contains unipotent automorphisms
				$\neq \bid$.
			\item $\Iner(\AA^2, \Gamma) \neq \{ \bid \}$
			\item $\Aut(\AA^2, \Gamma)$ is not an algebraic group
		\end{enumerate}
	\end{enumerate}
\end{prop}

For the proof of this proposition we recall some facts about $\Aut(\AA^2)$.
The next result is a direct consequence of
\cite[Theorem~1]{BlSt2013Automorphisms-of-t}.

\begin{thm}
	\lab{algcurve.thm}
	An automorphism of $\AA^2$ that preserves an algebraic curve is
	conjugate to a triangular automorphism.
\end{thm}

In the next result we prove a slightly more general version of the
\name{Abhyankar-Moh-Suzuki}-Theorem which says that all closed
embeddings $\AAone \hookrightarrow \AA^2$ are equivalent up
to automorphisms of $\AA^2$
(see \cite{AbMo1975Embeddings-of-the-}).

\begin{prop}
	\lab{genAMS.prop}
	Let $\Gamma$ be a fence in $\AA^2$ and let
	$F \subseteq \CC$ be a closed 0-dimensional subscheme
	such that $\Gamma \simeq F \times \CC$. Then there exists an automorphism
	of $\CC^2$ that maps $\Gamma$ onto $F \times \CC$ (scheme-theoretically).
\end{prop}

\begin{proof}
	Clearly, we can assume that $\Gamma \neq \varnothing$.
	Moreover, we can easily reduce to the case, where $\Gamma$
	is a reduced scheme.
	Let $\Gamma_i$, $i \in I$ be the irreducible components of $\Gamma$.
	Let $i_0 \in I$ be fixed.
	By the \name{Abhyankar-Moh-Suzuki}-Theorem, there exists
	a trivial $\AAone$-bundle $f \colon \AA^2 \twoheadrightarrow  \AAone$
	such that $\Gamma_{i_0}$ is a fiber of $f$. Now, if the restriction
	$f |_{\Gamma_i} \colon \Gamma_i \to \AAone$ is non-constant, then it
	is surjective, since $\Gamma_i \simeq \AAone$. But this implies that
	$\Gamma_i \cap \Gamma_{i_0} \neq \varnothing$, a contradiction.
	Thus every $\Gamma_i$ is a fiber of $f$. This implies the proposition.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{curves.prop}] $\textrm{}$
	\begin{enumerate}[i)]

	\item
	Assume that $\Gamma = \div(a)$ for some non-zero
	$a \in \OO(\CC^2) = \CC[x, y]$.
	For $(f_1, f_2) \in \CC[x, y]^2$ we denote by $a_{ij}(f_1, f_2)$ the coefficient of
	the monomial $x^i y^j$ in the polynomial $a(f_1, f_2)$.
	The subgroup $\Aut(\AA^2, \Gamma)$ of $\Aut(\AA^2)$
	is defined by the equations
	\[
		a_{ij}(f_1, f_2) a_{kl}(x, y) = a_{kl}(f_1, f_2) a_{ij}(x, y) \quad
		\textrm{for all pairs $(i, j)$, $(k, l)$} \, .
	\]This proves the first statement.

	Let $\g \in \Aut(\AA^2, \Gamma)$. By Theorem~\ref{algcurve.thm},
	$\g$ is conjugate to a triangular automorphism and hence
	$\g$ is algebraic. This proves the second statement.

	\item
	$a) \Rightarrow b)$: This follows immediately form
	Proposition~\ref{genAMS.prop}.

	$b) \Rightarrow c)$:
	Let $\g \in \Aut(\AA^2, \Gamma)$ be a unipotent
	automorphism $\neq \bid$.
	Choose some $a \in \OO(\CC^2)$ such that $\Gamma = \div(a)$.
	As $\g$ preserves $\Gamma$, it follows that $a$ is a semi-invariant
	for the $\GGa$-action on $\AA^2$ induced by $\g$. Since $\GGa$
	has no non-trivial character, $a$ is an invariant. Hence,
	$\bid \neq a \cdot \g \in \Iner(\AA^2, \Gamma)$.

	$c) \Rightarrow d)$: Let $\g \in \Iner(\AA^2, \Gamma)$ with $\g \neq \bid$.
	By Theorem~\ref{algcurve.thm}, $\g$ preserves a trivial $\AAone$-bundle
	$f \colon \AA^2 \twoheadrightarrow \AAone$.
	Let $\Gamma_i$, $i \in I$ be the irreducible components of
	the reduced scheme $\Gamma_{\red}$.
	If every $\Gamma_i$ lies in a fiber of $f$, then $\Gamma$ is a fence and
	thus $\Aut(\AA^2, \Gamma)$ is not an algebraic group. Therefore
	we can assume that $f(\Gamma_i) \subseteq \AAone$ is dense
	for some $i$. As $\g$ is the identity on $\Gamma_i$, it follows that
	$\g$ maps each fiber on itself. Hence, there exists $\alpha \in \CC^\ast$
	and a polynomial $b(y)$ such that for each $y \in \AAone$ the restriction
	of $\g$ to the fiber $f^{-1}(y)$ is given by
	\[
		\g_y \colon \AAone \to \AAone \, , \quad
		x \mapsto \alpha x + b(y) \, .
	\]
	As $\g$ is the identity on $\Gamma_i$, it follows that $\g_y$ has a fixed point
	for all $y \in f(\Gamma_i)$.
	If $\alpha = 1$, then $\g_y$ is the identity map for all
	$y \in f(\Gamma_i) \subseteq \AAone$.
	Since $f(\Gamma_i)$ is dense in $\AAone$ we get a
	contradiction to the fact that
	$\g \neq \bid$.  Thus, $\alpha \neq 1$. But this implies that
	$\g_y$ has exactly one fixed point for each $y \in \AAone$.
	Thus, $\Gamma_i = V(ax + b(y)-x) \simeq \AAone$ and it is the only
	irreducible component of $\Gamma_{\red}$.
	Therefore, $\Gamma$ is again a fence
	and $\Aut(\AA^2, \Gamma)$ is not an algebraic group.

	$d) \Rightarrow a)$: Assume that $\Gamma$ is not a fence.
	By \cite[Theorem~1]{BlSt2013Automorphisms-of-t},
	$\Aut(\AA^2, \Gamma_{\red})$ is an algebraic group. Now,
	$\Aut(\AA^2, \Gamma)$ is an algebraic group as well, since
	it is a closed subgroup of
	$\Aut(\AA^2, \Gamma_{\red})$.
\end{enumerate}

\end{proof}

\section{Some basic properties of locally nilpotent derivations}
Let $A$ be a $\CC$-algebra and assume it is a unique factorization domain (UFD).
Let $B$ be a locally nilpotent derivation of $A$.
We call $B$ \emph{irreducible}, if $B \neq 0$ and the following holds:
if $B = f B'$ for some locally nilpotent derivation $B'$ and some $f \in \ker B$, then
$f \in A^\ast$ where $A^\ast$ denotes the subgroup of units of $A$.
Remark, if $A = \OO(\CC^n)$, then a unipotent automorphism
$\bu = \Exp(D)$ is irreducible if and only if $D$ is irreducible.

We list some basic facts about locally nilpotent derivations,
that we will use constantly (see \cite{Fr2006Algebraic-theory-o} for proofs).

\begin{lem} Let $A$ be a $\CC$-algebra and assume it is a UFD, and let $B$
be a locally nilpotent derivation of $A$. Then

\begin{enumerate}[i)]
	\item The units of $A$ lie in $\ker B$.
		In particular, $\CC \subseteq \ker B$.
	\item The kernel $\ker B$ is factorially closed in $A$, i.e. if
		$f, g \in A$ such that $fg \in \ker B$, then $f, g \in \ker B$.
	\item If $S \subseteq \ker B$ is a multiplicative system, then
		$B$ extends uniquely to a locally nilpotent derivation of the localization
		$A_S$.
	\item If $B \neq 0$, then there exists
		$f \in A$ such that $B(f) \in \ker B$ and $B(f) \neq 0$.
	\item If $s \in A$ such that $B(s) = 1$, then
		$A$ is a polynomial ring in $s$ over
		$\ker B$ and $B = \partial / \partial s$.
	\item For $f \in A$, the derivation $f B$ is locally nilpotent if and only if
		$f \in \ker B$.
	\item If $B$ is irreducible and $E$ is another locally nilpotent derivation
		of $A$ such that $E(\ker B) = 0$, then there exists
		$f \in \ker B$ such that $E = f B$.
	\item If $B \neq 0$, then there exists a unique irreducible
		locally nilpotent derivation $B'$
		(up to multiplication by some element of $A^\ast$) such
		that $\ker B = \ker B'$.
	\item If $B(f) \in f A$, then $B(f) = 0$.
	\item The exponential $\exp(B) = \sum_{i = 0}^{\infty} B^i / {i!}$
		is a $\CC$-algebra automorphism of $A$ and the map $\exp$ defines
		an injection from the set of locally nilpotent derivations of $A$
		to the set of $\CC$-algebra automorphisms of $A$.
\end{enumerate}
\end{lem}

\section{\texorpdfstring{Structure theorems for $\Cent(\bu)$}
	{Structure theorems for Cent(u)}}
\lab{Cent.sec}

\subsection{\texorpdfstring{The first unipotent subgroup in $\Cent(\bu)$}
		 {The first unipotent subgroup in Cent(u)}}
\label{The_first_subgroup.subsec}
Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent and let
$\bu = d \cdot \bu'$ be a standard
decomposition.
There exists an obvious subgroup of unipotent automorphisms in $\Cent(\bu)$:
The modifications of $\bu'$, i.e. the subgroup $\OO(\AA^3)^{\bu'} \cdot \bu'$.
This subgroup has another characterization:

\begin{prop}
	\lab{firstfam.prop}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent with standard decomposition
	$\bu = d \cdot \bu'$.
	The subgroup $\OO(\AA^3)^{\bu'} \cdot \bu'$
	consists of those automorphisms of $\AA^3$ that commute with
	$\bu$ and that induce the identity on $\AA^3 \aquot \GGa$,
	i.e. the sequence
	\[
		1 \to \OO(\AA^3)^{\bu'} \cdot \bu'
		\hookrightarrow \Cent(\bu)
		\stackrel{p}{\to} \Aut(\AA^3 \aquot \GGa, \Gamma, \Gamma')
	\]
	is exact, where $\Gamma$, $\Gamma'$ denote the plinth divisors of
	$\bu$, $\bu'$ respectively.
	Moreover, the homomorphisms in the sequence above are
	homomorphisms of ind-groups.
\end{prop}

This result is an immediate consequence of Remark~\ref{indhomo.rem}
and Lemma~\ref{identity.lem}.

\begin{rem}
	\label{indhomo.rem}
	Choose generators $v_1, v_2$ of the polynomial ring
	$\OO(\AA^3)^{\bu}$ and choose a $\CC$-linear retraction
	$rÊ\colon \OO(\AA^3) \twoheadrightarrow \OO(\AA^3)^{\bu}$.
	The map $p \colon \Cent(\bu) \to \Aut(\AA^3 \aquot \GGa, \Gamma)$
	is a morphism of ind-varieties due to the following commutative
	diagram
	\[
		\xymatrix{
			\End(\AA^3) \ar[rrr]^-
			{\g \mapsto (\g^\ast(v_1), \g^\ast(v_2))}_-{\textrm{morph.}}
			&&& \OO(\AA^3)^2
			\ar@{->>}[r]^-{r \times r}_-{\textrm{lin.}} & (\OO(\AA^3)^{\bu})^2 \\
			\Cent(\bu) \ar@{^(->}[u]_-{\textrm{loc. closed}}
			\ar[rrrr]^-{p} &&&& \Aut(\AA^3 \aquot \GGa, \Gamma, \Gamma')
			\ar@{^(->}[u]^-{\textrm{loc. closed}} \, .
		}
	\]
\end{rem}

\begin{lem}
	\lab{identity.lem}
	Let $A$ be a $\CC$-algebra and assume it is a UFD, let
	$B, B'$ be non-zero locally nilpotent derivations of $A$ such that $B'$
	is irreducible and $\ker B = \ker B'$.
	If $\varphi \colon A \to A$ is a $\CC$-algebra automorphism, then we have
	\[
		\varphi |_{\ker B} = \id \ \textrm{and} \
		\varphi \circ B = B \circ \varphi \quad \textrm{if and only if} \quad
		\varphi = \exp(fB') \, , \ f \in \ker B \, .
	\]
\end{lem}

\begin{proof}
	Assume that $\varphi |_{\ker B}$ is the identity and $\varphi$
	commutes with $B$. There exists
	$0 \neq d \in \ker B$, such that $A_d = \ker(B)_d[s]$ is a polynomial ring
	in an element $s \in A_d$ and $B(s) = 1$, if we extend $B$ to $A_d$.
	Since $\varphi$ commutes with $B$ there exists
	$g \in \ker(B)_d$ such that the extension $\tilde{\varphi}$ to $A_d$
	of $\varphi$ satisfies $\tilde{\varphi}(s) = s + g$.
	Now, we have $\varphi = \exp(g B) |_A$.
	A density argument shows that
	$\exp(t gB)(A) \subseteq A$ for all $t \in \CC$. Since
	\[
		gB = \frac{\exp(t gB) - \id}{t} \Big| _{t =0} \, ,
	\]
	we have $gB(A) \subseteq A$. Hence $gB$ is a locally nilpotent derivation
	of $A$ that vanishes on $\ker B = \ker B'$. Thus, $gB = f B'$
	for some $f \in \ker B$. The converse is clear.
\end{proof}

\begin{rem}
	\label{ActionByConj.rem}
	By Proposition~\ref{firstfam.prop}, $\Cent(\bu)$ normalizes
	$\OO(\CC^3)^{\bu'} \cdot \bu'$ and one can easily see, that the action is
	given by
	\[
		\g^{-1} \circ f \cdot \bu' \circ \g = \mu(\g) \g^\ast(f) \cdot \bu' \, , \quad
		\g \in \Cent(\bu) \, , \ f \in \OO(\CC^3)^{\bu'}
	\]
	where $\mu \colon \Cent(\bu) \to \CC^\ast$ is the homomorphism
	given by $\mu(\g) d = \g^{\ast}(d)$.
\end{rem}

\subsection{\texorpdfstring{Centralizer of a modified
	    translation in $\Aut(\AA^3)$}
	    {Centralizer of a modified translation in Aut(C3)}}
\label{modtranslation.sec}

\begin{prop}
	\lab{modtranslation.prop}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be a modified translation
	with standard decomposition $\bu = d \cdot \bu'$.
	Denote by $\Gamma$ the plinth divisor of $\bu$. Then
	\[
			1 \to \OO(\AA^3)^{\bu'} \cdot \bu'
			\hookrightarrow \Cent(\bu) \stackrel{p}{\to}
			\Aut(\AA^3 \aquot \GGa, \Gamma) \to 1
	\]
	is a split short exact sequence of ind-groups.
	Moreover there exists a closed subgroup of $\Cent(\bu)$
	that is mapped via $p$ isomorphically onto
	$\Aut(\AA^3 \aquot \GGa, \Gamma)$.
\end{prop}

\begin{rem}
	\label{AlgebraicElements.rem}
	Under the assumptions of Proposition~\ref{modtranslation.prop},
	$\Cent(\bu)$ consists only of algebraic elements, provided that
	$\Gamma$ is non-empty. Indeed, let $H \subseteq \Cent(\bu)$
	be a closed subgroup, such that $p$ induces an isomorphism
	$H \simeq \Aut(\CC^3 \aquot \CC^+, \Gamma)$. Let
	$(f \cdot \bu', \h) \in \OO(\CC^3)^{\bu'} \cdot \bu' \rtimes H \simeq \Cent(\bu)$.
	By Proposition~\ref{curves.prop} i),
	$R = \overline{\langle \h \rangle}$ is an algebraic subgroup of $H$.
	Hence
	\[
		W = \Span \{ \, \br^\ast(f) \ | \ \br \in R \, \}
	\]
	is a finite dimensional subspace of $\OO(\CC^3)^{\bu'}$ and
	$W \cdot \bu' \rtimes R \subseteq \OO(\CC^3)^{\bu'} \cdot \bu' \rtimes H$
	is an algebraic subgroup  that contains $(f \cdot \bu', \h)$
	(see Remark~\ref{ActionByConj.rem}).
\end{rem}

\begin{proof}[Proof of Proposition~\ref{modtranslation.prop}]

	By Proposition~\ref{firstfam.prop}, the sequence is left
	exact.
	By assumption, there exist coordinates $(x, y, z)$ on $\CC^3$
	such that $d \in \CC[y, z]$ and $\bu = (x+d, y, z)$. Moreover,
	we can identify the algebraic
	quotient $\pi \colon \AA^3 \to \AA^2$ with the map $(x, y, z) \mapsto (y, z)$
	and $\Gamma = \div(d)$.
	Let $\f \in \Aut(\AA^2, \Gamma)$. Then $\f^\ast(d) = \lambda(\f) d$
	for some $\lambda(\f) \in \CC^\ast$.
	One can see that
	$\Aut(\AA^2, \Gamma) \to \CC^\ast$, $\f \mapsto \lambda(\f)$
	is a homomorphism of ind-groups. Thus
	\[
		H = \{ \, \sigma \in \Aut(\CC^3) \ | \ \sigma(x, y, z) =
		(\lambda x, \f(y, z)) \ \textrm{with} \ \f \in
		\Aut(\AA^2, \Gamma) \ \textrm{and} \ \lambda(\f) = \lambda
		\, \}
	\]
	is a closed subgroup of $\Cent(\bu)$
	(note that the subgroup $\Aut(\AA^2, \Gamma) \subseteq \Aut(\AA^2)$
	is closed by Proposition~\ref{curves.prop}) and
	$p|_H \colon H \to \Aut(\AA^2, \Gamma)$ is an isomorphism of ind-groups.
\end{proof}

\begin{prop}
	Let $\bu \in \Aut(\CC^3)$ be unipotent. Then
	$\Cent(\bu)$ contains $(\CC^\ast)^2$ as an algebraic subgroup
	if and only if $\bu$ is a modified translation and the plinth divisor
	$\Gamma$ is given by $v^i w^j$ for some coordinates $(v, w)$
	of $\CC^3 \aquot \CC^+ \simeq \CC^2$.
\end{prop}

\begin{proof}
	Assume that $\Cent(\bu)$ contains an algebraic subgroup
	$T \simeq (\CC^\ast)^2$.
	By Proposition~\ref{modtranslation.prop} it follows
	that $T$ acts faithfully on $\CC^3 \aquot \CC^+$ and leaves the plinth
	divisor $\Gamma$ invariant. Hence, it follows from
	\cite{Bi1966Remarks-on-the-act} that
	there exist coordinates $(v, w)$ on $\CC^2 \simeq \CC^{3} \aquot \CC^+$
	such that $\Gamma$ is given by $v^i w^j$ for some integers $i, j$.
	By \cite{Bi1967Remarks-on-the-act} there exist coordinates
	$(x_1, x_2, x_3)$ of $\CC^3$ such that the action of $T$ is diagonal
	with respect to these coordinates. Hence there exist characters
	$\lambda_1$, $\lambda_2$, $\lambda_3$ of $T$ such that
	$\bt(x_1, x_2, x_3) =
	(\lambda_1(\bt)x_1, \lambda_2(\bt)x_2, \lambda_3(\bt)x_3)$ for all
	$\bt \in T$. Let $\bu = \Exp(D)$. By assumption we have for all
	$\bt \in T$ and $i = 1, 2, 3$
	\begin{equation}
		\label{Comm.eq}
		D(x_i) \circ \bt = \lambda_i(\bt) D(x_i).
	\end{equation}
	As the action of $T$ on $\CC^3$
	is faithful, the subgroup spanned by $\lambda_1, \lambda_2, \lambda_3$
	inside the characters of $T$ has rank $2$.
	Assume first that the $\lambda_i$ are pairwise different.
	Then there exist at least two different indices $k_1, k_2 \in \{1, 2, 3 \}$
	such that $\lambda_{k_i}$ lies not in the monoid
	spanned by $\{ \, \lambda_l \ | \ l \neq k_i \, \}$. By symmetry we can
	assume $k_1 = 1$, $k_2 = 2$.
	This implies that $D(x_i) \in x_i \CC[x_1, x_2, x_3]$ for $i = 1, 2$.
	Since $D$ is locally nilpotent we have $D(x_1)=D(x_2) = 0$.
	Hence, $\bu$ is a modified translation.
	Assume now that $\lambda_1 = \lambda_2 \neq \lambda_3$
	(the other cases follow by symmetry). Thus,
	$\lambda_3$ does not lie in the monoid spanned by $\lambda_1$
	and $\lambda_2$. Hence we get $D(x_3) \in x_3 \CC[x_1, x_2, x_3]$ and
	$D(x_1), D(x_2) \in \CC x_1 \oplus \CC x_2$. Since $D$ is locally nilpotent
	it follows that $D(x_3) = 0$ and the linear endomorphism
	$D |_{\CC x_1 \oplus \CC x_2}$ is nilpotent. This implies that $\bu$ is
	a modified translation.

	The converse of the statement is clear.
\end{proof}

\subsection{\texorpdfstring{The second unipotent subgroup in $\Cent(\bu)$}
	          {The second unipotent subgroup in Cent(u)}}
\label{The_second_subgroup.subsec}
Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent with standard decomposition
$\bu = d \cdot \bu'$.
Throughout this subsection we assume that the plinth divisor $\Gamma = \div(a)$
of $\bu$ is a fence. There exists another subgroup of unipotent automorphisms
inside $\Cent(\bu)$ in addition to $\OO(\AA^3)^{\bu'} \cdot \bu'$, that we describe
in this subsection.
\begin{lem}
	\lab{fact.lem}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent. If the
	plinth divisor $\Gamma = \div(a)$ is a fence,
	then there exists a variable $z$ of $\OO(\AA^3)^{\bu} = \OO(\CC^2)$
	such that $a \in \CC[z]$ and any such $z$ is a variable of
	$\OO(\AA^3)$.

\end{lem}

\begin{proof}
	By Proposition~\ref{genAMS.prop} there exists a coordinate system
	$(z, w)$ of $\AA^2$ such that the embedding
	$\div(a) = \Gamma \subseteq \AA^2$ is given by the standard embedding
	$F \times \AAone \subseteq \AA^2$ for some 0-dimensional
	closed subscheme $F$ of $\AAone$. Thus $a \in \CC[z]$.
	Since $\pi$ is a trivial $\AAone$-bundle over $\AA^2 \setminus \Gamma$,
	it follows that only finitely many fibers of
	$z \colon \AA^3 \to \AAone$ are non-isomorphic
	to $\AA^2$. Thus $z$ is a variable of $\OO(\CC^3)$, according to
	\name{Kaliman}'s Theorem \cite[Theorem~3]{DaKa2009A-note-on-locally-}
	(cf. also \cite[Theorem~3.1]{BaElEl2008Triangulable-local}).
\end{proof}

\begin{rem}
	\lab{rank2.rem}
	If $\bu = \Exp(D)$ is irreducible, then
	$\Gamma$ is a fence
	if and only if $\rank D \leq 2$ (i.e. there exists a variable $z$ of $\OO(\CC^3)$
	that lies in $\ker D$). This follows from the lemma above
	and from
	\cite[Theorem~2.4, Proposition~2.3]{DaFr1998Locally-nilpotent-}.
\end{rem}

\begin{defn}
	Let $A$ be a UFD and let $P \in A[x, y]$. We denote
	\[
		\Delta_P = -P_y \frac{\partial}{\partial x}
				  + P_x \frac{\partial}{\partial y}
	\]
	where $P_x$ and $P_y$ denote the partial derivatives of $P$ with respect
	to $x$ and $y$ respectively. Obviously, $\Delta_P$ is an
	$A$-derivation of $A[x, y]$ and $\Delta_P(P) = 0$.
\end{defn}

Let $D$, $D'$ be locally nilpotent derivations of $\OO(\AA^3)$
such that $\bu = \Exp(D)$ and $\bu' = \Exp(D')$.
Let $z \in \ker D$ be a variable such that $a \in \CC[z]$ and let
$(x, y, z)$ be a coordinate system of $\OO(\CC^3)$ (see Lemma~\ref{fact.lem}).
Let $A = \CC[z]$.
It follows now from \cite[Theorem~2.4]{DaFr1998Locally-nilpotent-} that there exists
$P \in A[x, y]$ such that
\[
	D' = \Delta_P \quad \textrm{and} \quad \ker D = \ker D' = \CC[z, P] \, .
\]
Obviously, $d$ divides $a$ in $\CC[z]$.
Let $a = d a'$. An easy calculation shows that $\div(a')$ is the plinth
divisor of $\bu'$ and that for all $Q \in A[x, y]$ we have
\begin{equation}
	\label{D'.eq}
	D(Q) = a \quad \textrm{if and only if} \quad D'(Q) = a' \, .
\end{equation}
By assumption $\Gamma = \div(a)$ is a fence and thus $a, a' \neq 0$.

\begin{lem}
	\lab{AdmissibleCompl.lem}
	Let $A = \CC[z]$. If $Q \in A[x, y]$ such that $D(Q) = a$, then $E = \Delta_Q$
	is an irreducible locally nilpotent derivation.
	Moreover, $E$ commutes with $D$.
\end{lem}

\begin{proof}
	Let $K$ be the quotient field of $A$.
	The extension of $D'$ to $K[x, y]$ satisfies $D'(Q/a')= 1$.
	Thus $K[x, y] = K[P, Q]$. $E$ is non-zero, since $E(P) = -a' \neq 0$.
	If we extend $E$ to a
	derivation of $K[x, y]$ one easily sees that
	$E$ is locally nilpotent. Thus $E$ is a non-zero locally nilpotent derivation of
	$A[x, y]$.

	By \cite[Theorem~2.4, Proposition~2.3]{DaFr1998Locally-nilpotent-})
	there exists $S \in A[x, y]$ and $0 \neq h \in A[P]$ such that
	$E = h \Delta_{S}$ and $\Delta_{S}$ is irreducible. Thus
	$-a' = E(P) = h \Delta_{S}(P) = - h \Delta_P(S)$. Hence
	$\Delta_P(S)$ lies in the plinth ideal of $\Delta_P$ and thus
	$\Delta_P(S)$ is a multiple of $a'$. This implies that
	$h \in \CC^\ast$ and proves that $E$ is irreducible.

	If we extend $\Delta_P$ and $\Delta_Q$ to $K[x, y] = K[P, Q]$,
	we get $\Delta_P = a' (\partial / \partial Q)$ and
	$\Delta_Q = -a' (\partial / \partial P)$.
	Thus $E$ commutes with $D'$. Since  $d \in \CC[z]$,
	$E$ and $D = d D'$ commute.
\end{proof}

\begin{defn}
For any $Q \in \OO(\CC^3)$ with $D(Q) = a$ we call
\[
	\e = \Exp(E) = \Exp(\Delta_Q)
\]
an \emph{admissible complement to $\bu$}.
\end{defn}

By \eqref{D'.eq}, we get that $\e$ is an admissible complement to $\bu$
if and only if $\e$ is an admissible complement to $\bu'$.
It follows from Lemma~\ref{AdmissibleCompl.lem} that
$\OO(\AA^3)^{\langle \e, \bu \rangle} \cdot \e$ is a subgroup of
unipotent automorphisms inside $\Cent(\bu)$.

\begin{rem}Ê$\textrm{}$
	\label{geom.rem}
	We have $\AA^2 \setminus \Gamma = U \times \AAone$
	for some non-empty open subset $U \subseteq \AAone$.
	The restriction of $\bu$ and of $\e$ to the open subset
	\[
		\pi^{-1}(\AA^2 \setminus \Gamma) =
		(U \times \AAone) \times \AAone = \Spec( \CC[z]_a[P, Q])
	\]
	are given by $(u, v, w) \mapsto (u, v, w + 1)$ and
	$(u, v, w) \mapsto (u, v + 1, w)$ respectively, where $(u, v, w)$
	is the coordinate system $(z, -P/a', Q/a)$.
\end{rem}

\subsection{\texorpdfstring{The property (Sat)}
		 {The property (Sat)}}
We introduce in this subsection a property for a subset
$S \subseteq \Aut(\AA^n)$ and
we will show that $\Cent(\bv)$ satisfies this property for any
unipotent automorphism $\bv \in \Aut(\AA^n)$.
This property will then play a key role when we describe the
set of unipotent elements inside the centralizer. One can think of this property
as a saturation feature on the unipotent elements in $S$.

\begin{defn}
	Let $S \subseteq \Aut(\AA^n)$ be a subset.
	We say that $S$ has the \emph{property (Sat)}
	if for all unipotent $\w \in \Aut(\AA^n)$
	and for all $0 \neq f \in \OO(\AA^n)^{\w}$ we have
	\[
		\tag{Sat}
		f \cdot \w \in S \quad \Longrightarrow \quad \w \in S \, .
	\]
\end{defn}

\begin{prop}
	\lab{propertyR.prop}
	If $\bv \in \Aut(\AA^n)$ is unipotent, then
	the subgroup $\Cent(\bv) \subseteq \Aut(\AA^n)$ satisfies the property (Sat).
\end{prop}

\begin{proof}
	Let $\bv = \Exp(B)$ and let $\w = \Exp(F)$.
	Assume that
	$f \cdot \w$ commutes with $\bv$ for some $\w$-invariant
	$0 \neq f \in \OO(\AA^n)$. If $\bv = \bid$ or $\w = \bid$,
	then (Sat) is obviously satisfied. Thus we assume $\bv \neq \bid \neq \w$.
	For the Lie-bracket we have
	\begin{equation}
		\label{Lie-bracket.eq}
		0 = [fF, B] = f [F, B] - B(f) F \, .
	\end{equation}
	Thus, it is enough to prove that $B(f) = 0$.

	First, assume that $F$ is irreducible.
	By \eqref{Lie-bracket.eq}, it follows that $f$ divides $B(f) F(g)$
	for all $g \in \OO(\AA^n)$. As $F$ is irreducible, it follows that
	$f$ divides $B(f)$.
	Since $B$ is locally nilpotent, it follows that $B(f) = 0$.

	Now, let $F = f' F'$ for some irreducible $F'$. Thus, $f f' F'$ commutes with
	$B$ and by the argument above, $B(ff') = 0$.
	Since $\ker B$ is factorially closed in $\OO(\AA^n)$, we have $B(f) =0$.
\end{proof}

\subsection{\texorpdfstring{The subgroup $N \subseteq \Cent(\bu)$}
		 {The subgroup N in Cent(u)}}
Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent.
We define in this subsection a subgroup $N$ of $\Cent(\bu)$
and we gather some facts about this group.
In the next subsection, we will prove that $N$ is exactly the set of unipotent
automorphisms $\Cent(\bu)$ if $\bu$ is not a translation.

\begin{defn} Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent
with standard decomposition $\bu = d \cdot \bu'$ and let
$\Gamma$ be the plinth divisor of $\bu$. Let
\[
	 N = N(\bu) = \left\{
			\begin{array}{rl}
				\OO(\AA^3)^{\langle \e, \bu' \rangle} \cdot \e
				\circ \OO(\AA^3)^{\bu'} \cdot \bu' &
				\textrm{if $\Gamma$ is a fence} \\
				\OO(\AA^3)^{\bu'} \cdot \bu' & \textrm{otherwise.}
			\end{array}
		     \right.
\]
where $\e$ is an admissible complement to $\bu$
(cf. subsec.~\ref{The_second_subgroup.subsec}). Moreover, let
\[
	M = M(\bu) = \left\{
			\begin{array}{rl}
				(\ker E \cap \ker D') E + \ker(D') D' &
				\textrm{if $\Gamma$ is a fence,} \\
				\ker(D') D' & \textrm{otherwise.}
			\end{array}
		     \right.
\]
where $\bu' = \Exp(D')$ and
$\e = \Exp(E)$ (cf. Lemma~\ref{AdmissibleCompl.lem}).
\end{defn}

\begin{prop}
	\lab{properties.prop}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent
	and assume it is not a translation. Then:
	\begin{enumerate}[i)]
		\item $N$ consists of
		unipotent automorphisms and we have $N = \Exp(M)$.

		\item $N$
		normalizes $\OO(\AA^3)^{\bu'} \cdot \bu'$
		and we have for all $\g \in N$ and for all
		$f \in \OO(\AA^3)^{\bu'}$
		\[
			\g^{-1} \circ f \cdot \bu' \circ \g = \g^\ast(f) \cdot \bu' \, .
		\]

		\item $N$ is a closed normal subgroup of $\Cent(\bu)$ that fits into
		the following split short exact sequence of ind-groups
		\[
				1 \to \OO(\AA^3)^{\bu'} \cdot \bu'
				\hookrightarrow N
				\stackrel{p|_N}{\longrightarrow}
				\Aut(\AA^2, \Gamma) \cap
				\Iner(\AA^2, \Gamma')_u
				\to 1 \, ,
		\]
		where $\Gamma'$ is the plinth divisor of $\bu'$.
		If $\Gamma$ is a fence, then the restriction of
		$p$ to ${\OO(\AA^3)^{\langle \bu', \e \rangle} \cdot \e}$
		is an isomorphism of ind-groups.
		In particular, $N$ is independent of the choice of $\e$.

		\item $N \subseteq \Aut(\AA^3)$
		satisfies the property (Sat).
	\end{enumerate}
\end{prop}

\begin{proof}
	Assume first that
	$\Gamma$ is not a fence. Then i) ii) and iv) are clear, iii) follows from
	Proposition~\ref{curves.prop}. Thus we can assume that
	$\Gamma$ is a fence.
	\begin{enumerate}[i)]
		\item Let $hE + fD' \in M$. By induction on $l \geq 1$ one sees that
		$(hE + fD')^l$ is a sum of terms of the form $g E^i (D')^j$
		where $g \in \ker D'$. From this fact, one can deduce that $hE + fD'$
		is locally nilpotent and hence $M$ consists only of locally nilpotent
		derivations.

		For all $f \in \ker D'$ and $h \in \ker D' \cap \ker E$ and $q \geq 0$
		we have
		\[
			fD' \ad(hE)^{q}
			= (-1)^q h^qE^q(f) D'
		\]
		where $A\ad(B) = [A, B]$.
		With the aid of this formula, an application of the
		Baker-Campbell-Hausdorff formula yields
		$\Exp(hE) \circ \Exp(fD') \in M$
		(see \cite[Proposition 1, \S 5, chp. V]{Ja1962Lie-algebras}).
		Hence $N \subseteq \Exp(M)$ which shows in particular, that
		$N$ consists of unipotent automorphisms.
		Moreover,
		$\exp hE$ and $\exp(fD' + hE)$ coincide
		on $\ker D'$.
		Lemma~\ref{identity.lem} implies
		$(\Exp hE)^{-1} \circ \Exp(fD' + hE) = \Exp(g D')$ for some
		$g \in \ker D'$ and thus $\Exp(M) \subseteq N$.

		\item This follows from Remark~\ref{ActionByConj.rem}.

		\item
		One can check that
		$N = p^{-1}(\Aut(\AA^2, \Gamma) \cap \Iner(\AA^2, \Gamma')_u)$ by
		using Proposition~\ref{firstfam.prop}.
		Since $\Aut(\AA^2, \Gamma) \cap \Iner(\AA^2, \Gamma')_u$ is a
		closed normal subgroup of
		$\Aut(\AA^2, \Gamma, \Gamma')$ it follows that $N$ is a closed normal
		subgroup of $\Cent(\bu)$.

		It is enough to show that the homomorphism
		$\OO(\AA^3)^{\langle \e, \bu' \rangle} \cdot \e \to
		\Aut(\AA^2, \Gamma) \cap \Iner(\AA^2, \Gamma')_u$ (induced by $p$)
		is an isomorphism of ind-groups. Injectivity follows from the fact that
		$\OO(\AA^3)^{\langle \e, \bu' \rangle} \cdot \e \cap
		\OO(\AA^3)^{\bu'} \cdot \bu' = \{ \bid \}$
		and surjectivity follows from a straightforward calculation, by using that
		$\Gamma$ is non-empty. The inverse map is clearly a morphism.

		\item Let $0 \neq hE + fD' \in M$. It is enough to prove that
		\begin{equation}
			\label{triangle}
			\gcd(h, f) = 1 \quad \Longrightarrow \quad
			\textrm{$hE + fD'$ is irreducible}
		\end{equation}
		where the greatest common divisor
		is taken in the polynomial ring $\ker D' = \CC[z, P]$
		(we use the notation of subsec.~\ref{The_second_subgroup.subsec}).
		Indeed, let $g B = hE + fD' \in M$ for some
		locally nilpotent derivation
		$B \neq 0$ and some $0 \neq g \in \ker B$ and let $h = \gcd(h, f) h_0$,
		$f = \gcd(h, f) f_0$. Thus $B$ vanishes on $\ker(h_0 E + f_0 D')$ and
		since $h_0 E + f_0 D'$ is irreducible, there exists
		$b \in \ker(h_0 E + f_0 D')$
		such that $B = b (h_0 E + f_0 D')$.
		This implies $g b = \gcd(h, f) \in \CC[z]$
		and therefore $b \in \CC[z]$. This shows that $B \in M$.

		Let us prove \eqref{triangle}.
		Since $E$ and $D'$ are irreducible
		(see Lemma~\ref{AdmissibleCompl.lem}) we can
		assume that $h$ and $f$ both are non-zero. A calculation shows
		\[
			h E + f D' = \Delta_{F} \, , \quad
			F = hQ + fP -Ê\int \left( \frac{\partial f}{\partial P} P \right) \ud P
		\]
		where the integration is taken inside the polynomial ring
		$\ker D' =\CC[z, P]$ and $\Delta_F$
		is taken with respect to $A[x, y]$ where $A = \CC[z]$.
		Let $f = \sum_{i=0}^n f_i(z) P^i$. Thus we have
		\[
			f P - \int \left( \frac{\partial f}{\partial P} P \right) \ud P =
			\sum_{i=0}^n f_i(z) \left( 1 - \frac{i}{i+1} \right) P^{i+1} \, .
		\]
		Denote this last polynomial by $G \in \CC[z, P]$.

		Now, assume towards a contradiction that $hE + fD'$ is not irreducible.
		Hence, we have $hE + fD' = bB$ for some locally
		nilpotent derivation $B$ and some
		non-constant $b \in \ker B$. By plugging in
		$P$ and $Q$ in $hE + fD' = bB$
		and using the fact that $\gcd(h, f) = 1$
		we see that $b$ divides $a'$
		(recall that $D'(Q) = a'$ and $E(P) = -a'$).
		Hence there exists a root $z_0$ of $a'$ such that the induced
		derivation of $\Delta_F = hE + fD'$ on
		$\CC[x, y, z] / (z-z_0) \simeq \CC[x, y]$
		vanishes. Thus, there exists a constant $c \in \CC$ such that
		\begin{equation}
			\label{square}
			h(z_0) Q(x, y, z_0) +
			\sum_{i = 0}^n
			f_i(z_0) \left( 1 - \frac{i}{i+1} \right) P^{i+1}(x, y, z_0) =  c \, .
		\end{equation}
		The polynomial $P(x, y, z_0) \in \CC[x, y]$ is non-constant,
		since otherwise $\bu = \Exp(\Delta_P)$ would have a
		two-dimensional fixed point set,
		contradicting the irreducibility
		(cf. \cite[2.10]{Da2007On-polynomials-in-}).
		If $h(z_0) = 0$, then we have $f(z_0, P) = 0$ by \eqref{square}.
		Hence $\gcd(h, f) \neq 1$, a contradiction. Thus we can assume
		$h(z_0) \neq 0$. It follows that
		$Q + h(z_0)^{-1} (G(z, P) - c)$ is divisible by $z - z_0$
		inside $\OO(\CC^3)$. Thus,
		\[
			D' \left( \frac{Q + h(z_0)^{-1}(G(z, P)-c)}{z-z_0} \right) =
			\frac{a'}{z-z_0} \, .
		\]
		But this contradicts the fact, that $a'$ is a generator
		of the plinth ideal of $D'$.
	\end{enumerate}
\end{proof}

\subsection{\texorpdfstring{The group $\Cent(\bu)$ as a semi-direct product}
		{The group Cent(u) as a semi-direct product}}
In this subsection, we prove our first main result: There exists an algebraic
subgroup $R \subseteq \Cent(\bu)$ such that $\Cent(\bu)$
is the semi-direct product of $N$ with $R$, if $\bu$ is not a translation.

\begin{thm}
	\lab{main.thm1}
	Let $\bu \in \Aut(\AA^3)$ be unipotent and assume that
	$\bu$ is not a translation. Then the subgroup
	$N \subseteq \Cent(\bu)$ is closed and normal, and
	there exists an algebraic subgroup $R \subseteq \Cent(\bu)$
	such that $\Cent(\bu) \simeq N \rtimes R$
	as ind-groups. Moreover, all elements of $\Cent(\bu)$ are algebraic.
\end{thm}

We prove the result for modified translations and reduce the general case to it.

\begin{proof}[Proof for a modified translation]
	Let $\bu = d \cdot \bu'$ be a standard decomposition. There exists
	a coordinate system $(x, y, z)$ such that
	$\bu'(x, y, z) = (x + 1, y, z)$ and
	$d \in \CC[y, z] \setminus \CC$.

	If $\Gamma$ is not a fence, then
	it follows from Proposition~\ref{curves.prop}
	that $\Aut(\AA^2, \Gamma)$ is an algebraic group. By
	Proposition~\ref{modtranslation.prop} there exists a
	closed subgroup $R$ of $\Cent(\bu)$ that is mapped via
	$p \colon \Cent(\bu) \twoheadrightarrow \Aut(\AA^2, \Gamma)$
	isomorphically onto $\Aut(\AA^2, \Gamma)$ and
	$\Cent(\bu) \simeq N \rtimes R$.

	Now, assume that $\Gamma = \div(a)$ is a non-empty fence.
	By Proposition~\ref{genAMS.prop} there exist coordinates $(y, z)$ of
	$\AA^2 = \AA^3 \aquot \GGa$ such that $a \in \CC[z]$.
	Thus we have a split short exact sequence of ind-groups
	\[
		1 \to \Aut(\AA^2, \Gamma)_u
		   \hookrightarrow \Aut(\AA^2, \Gamma)
		   \stackrel{q}{\to}
		   \CC^{\ast} \times \Aut(\AAone, V(a))
		   \to 1
	\]
	where $q$ sends an automorphism
	$(y, z) \mapsto (\lambda y + h, \alpha z + \beta)$ to
	$(\lambda, z \mapsto \alpha z + \beta)$.
	Let $R$ be the algebraic group $\CC^{\ast} \times \Aut(\AAone, V(a))$.
	Since $N$ is generated by $\OO(\AA^3)^{\bu'} \cdot \bu'$
	and $\OO(\AA^3)^{\langle \e, \bu' \rangle}
	\cdot \e \simeq \Aut(\AA^2, \Gamma)_u$
	(see Proposition~\ref{properties.prop}),
	we have the desired split short exact sequence of ind-groups
	\[
		1 \to N \hookrightarrow
		\Cent(\bu) \stackrel{q \circ p}{\longrightarrow}
		R \to 1 \, .
	\]
	By Remark~\ref{AlgebraicElements.rem}, every element of
	$\Cent(\bu)$ is algebraic.
\end{proof}

\begin{proof}[Proof in the general case]
	Let $\OO(\AA^3)^{\bu} = \CC[\tilde{y}, \tilde{z}]$ and let
	$\tilde{x} \in \OO(\CC^3)$
	such that $\bu^\ast(\tilde{x}) = \tilde{x} + a$, where
	$\Gamma = \div(a)$.
	Let $\tilde{\bu}  \in \Aut(\AA^3)$ be given by
	$\tilde{\bu}(\tilde{x}, \tilde{y}, \tilde{z}) = (\tilde{x} + a, \tilde{y}, \tilde{z})$
	where we interpret
	$a$ as a polynomial in $\tilde{y}$ and $\tilde{z}$.
	The morphism $\AA^3 \to \AA^3$
	induced by the inclusion
	$\CC[\tilde{x}, \tilde{y}, \tilde{z}] \subseteq \OO(\AA^3)$
	is birational and thus we get an injective group homomorphism
	\[
		\eta \colon \Cent(\bu) \longrightarrow \Cent(\tilde{\bu}) \, .
	\]
	In fact, $\eta$ is a homomorphism of ind-groups, due to the following
	commutative diagram, where
	$r \colon \OO(\CC^3) \twoheadrightarrow \CC[\tilde{x}, \tilde{y}, \tilde{z}]$
	is a $\CC$-linear retraction
	\[
		\xymatrix{
			\OO(\CC^3)^3 \ar[rrrr]^-
			{\g \mapsto (\g^\ast(\tilde{x}), \g^\ast(\tilde{y}),
			\g^\ast(\tilde{z}))}_-{\textrm{morph.}}
			&&&& \OO(\CC^3)^3
			\ar@{->>}[rr]^-{r \times r \times r}_-{\textrm{lin.}}
			&& \CC[\tilde{x}, \tilde{y}, \tilde{z}]^3 \\
			\Cent(\bu) \ar@{^(->}[u]_-{\textrm{loc. closed}}
			\ar[rrrrrr]^-{\eta} &&&&&& \Cent(\tilde{\bu})
			\ar@{^(->}[u]^-{\textrm{loc. closed}}
		}
	\]
	According to the first case, $\Cent(\tilde{\bu})$ is the semi-direct
	product of $N(\tilde{\bu})$ with some algebraic subgroup
	$\tilde{R} \subseteq \Cent(\tilde{\bu})$. Let $H \subseteq \tilde{R}$ be
	an algebraic subgroup. We claim that
	$\eta^{-1}(H) \subseteq \Cent(\bu)$ is an algebraic subgroup.
	Since $\eta \colon \Cent(\bu) \to \Cent(\tilde{\bu})$ is a
	homomorphism of ind-groups, it
	follows that $\eta^{-1}(H)$ is a closed subgroup. As $H$ is algebraic
	and thus acts locally finite on $\AA^3$, it follows
	that $\eta^{-1}(H)$ acts also locally finite on $\AA^3$ by
	\cite[Lemma~3.6]{KrSt2013On-Automorphisms-o}. This implies the claim.

	According to the claim all elements of $\Cent(\bu)$ are algebraic
	and $R = \eta^{-1}(\tilde{R})$ is algebraic as well.
	Since $\eta$ is an injective homomorphism of ind-groups we have the
	following commutative diagram
	\[
		\xymatrix{
			1 \ar[r] & N(\tilde{\bu}) \ar@{^(->}[r]
			& \Cent(\tilde{\bu})
			\ar@{->>}[r] & \tilde{R} \ar[r] & 1 \\
			1 \ar[r] & N(\bu) \ar[u]^-{\textrm{iso. of groups}}
			\ar@{^(->}[r] &
			\Cent(\bu) \ar[u]_{\eta} \ar@{->>}[r] & R
			\ar@{^(->}[u]_-{\textrm{cl. embedd.}} \ar[r] & 1
		}
	\]
	As the first column is a split short exact sequence of
	ind-groups, the second coloumn is also a split short exact sequence of
	ind-groups. This proves the theorem.
\end{proof}

\subsection{\texorpdfstring{The unipotent elements of $\Cent(\bu)$}
		 {The unipotent elements of Cent(u)}}
The goal of this subsection is to prove our second main result:
The unipotent elements of $\Cent(\bu)$ are exactly $N$ provided $\bu$
is not a translation.
As we know from Proposition~\ref{properties.prop} the set $N$
satisfies the property (Sat). This
will be a key ingredient in the proof.

\begin{thm}
   	\lab{main.thm2}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent and assume
	it is not a translation. Then the
	set of unipotent elements of $\Cent(\bu)$ is equal to $N$.
\end{thm}

\begin{proof}
	Let $\bu = d \cdot \bu'$ be a standard decomposition.
	Let $\g \in \Cent(\bu)$ be a unipotent automorphism with $\g \neq \bid$.
	If $\Gamma$ is not a fence, then
	$\Aut(\AA^2, \Gamma)$ contains no unipotent
	automorphism $\neq \bid$ (see Proposition~\ref{curves.prop}).
	By Proposition~\ref{firstfam.prop} it follows that
	$\g \in \OO(\AA^3)^{\bu'} \cdot \bu' = N$.

	Hence we can assume that $\Gamma = \div(a)$ is a fence.
	Let $z \in \OO(\AA^3)^{\bu}$ be a variable of $\OO(\CC^3)$
	such that $a \in \CC[z]$ (see Lemma~\ref{fact.lem}).
	If $\OO(\AA^3)^{\g} = \OO(\AA^3)^{\bu}$, then
	$\g$ is a modification of $\bu'$ and therefore $\g \in N$.
	Now, assume $\OO(\AA^3)^{\g} \neq \OO(\AA^3)^{\bu}$.
	Thus $\g$ is not a modification of $\bu'$ and hence
	$\bid \neq p(\g) \in \Aut(\AA^2, \Gamma)$. Since $p(\g)$ is unipotent and
	$0 \neq a \in \CC[z]$, it follows that $z \in \OO(\AA^3)^{\g}$ and thus
	$\OO(\AA^3)^{\langle \g, \bu \rangle}$ is an $\infty$-dimensional
	$\CC$-vector space.
	By Theorem~\ref{main.thm1} there exists
	an algebraic subgroup $R \subseteq \Cent(\bu)$
	and a split short exact sequence of ind-groups
	\[
		1 \to N \hookrightarrow \Cent(\bu) \stackrel{r}{\twoheadrightarrow}
		R \to 1 \, .
	\]
	If $\g \notin N$, then $\OO(\AA^3)^{\langle \g, \bu \rangle} \cdot \g \cap N = 1$,
	since $N$ satisfies the property (Sat).
	We get an injection
	$\OO(\AA^3)^{\langle \g, \bu \rangle} \to \Cent(\bu)
	\twoheadrightarrow R$, $h \mapsto r(h \cdot \g)$.
	Choose any filtration by finite dimensional $\CC$-subspaces
	to turn $\OO(\AA^3)^{\langle \g, \bu \rangle}$
	into an ind-group. It follows that
	$\OO(\AA^3)^{\langle \g, \bu \rangle} \to R$ is an injective
	homomorphism of ind-groups.
	But this implies that $R$ has algebraic subgroups
	of arbitrary high dimension, which is absurd.
	This finishes the proof of the theorem.
\end{proof}

If we endow $\Cent(\bu) / N$ with the algebraic group structure induced by
the semi-direct product decomposition coming from Theorem~\ref{main.thm1},
then we get immediately the following
corollary from Theorem~\ref{main.thm2}.

\begin{cor}
	\lab{mainthm2.cor}
	If $\bu \in \Aut(\AA^3)$ is unipotent and not a translation,
	then the algebraic group $\Cent(\bu) / N$ consists only of semi-simple
	elements. In particular, the connected component of the neutral element
	in $\Cent(\bu) / N$ is a torus.
\end{cor}

\section{\texorpdfstring{Applications}{Applications}}

\begin{prop}
	\label{applicationfence.prop}
	Let $\bu \in \Aut(\AA^3)$ be unipotent and irreducible, and assume that
	$\Gamma$ is a non-empty fence.
	Then $\Gamma \subseteq \AA^3 \aquot \GGa$
	is the largest closed subscheme fixed by $\Cent(\bu)_u$.
\end{prop}

\begin{proof}
	By Proposition~\ref{properties.prop} and Theorem~\ref{main.thm2},
	we get $p(\Cent(\bu)_u) = \Iner(\CC^2, \Gamma)_u$, where
	$p \colon \Cent(\bu) \to \Aut(\CC^2, \Gamma)$ is the canonical morphism.
	Thus $\Gamma$ is fixed by the action of $\Cent(\bu)_u$. Let
	$X \subseteq \CC^2$ be a closed subscheme that is fixed under
	$\Cent(\bu)_u$ and assume that $X$ contains $\Gamma$. Moreover, let
	$I(X) \subseteq \OO(\CC^2)$ be the vanishing ideal of $X$ and let
	$\Gamma = \div(a)$. By Proposition~\ref{genAMS.prop} there exist
	coordinates $(z, w)$ of $\CC^2$ such that $a \in \CC[z]$.
	Let $\sigma \in \Iner(\CC^2, \Gamma)_u$ be given by
	$\sigma(z, w) = (z, w + a)$.
	By assumption, we get $a = \sigma^\ast(w)-w \in I(X)$. But this implies
	that $X$ is a closed subscheme of $\Gamma$ and hence $X = \Gamma$.
\end{proof}

\begin{prop}
	\lab{char.prop}
	Let $\bid \neq \bu \in \Aut(\AA^3)$ be unipotent, not a translation, and let
	$\bu = d \cdot \bu'$ be a standard decomposition. Then the subgroup
	$\OO(\AA^3)^{\bu'} \cdot \bu'$ of $\Cent(\bu)$ is characteristic, i.e.
	$\OO(\AA^3)^{\bu'} \cdot \bu'$ is invariant under all abstract group
	automorphisms of $\Cent(\bu)$.
\end{prop}

\begin{lem}
	\label{DiagonalAction.lem}
	Let $T$ be a torus acting on $\CC^2$. Assume that there exist coordinates
	$(z, w)$ of $\CC^2$ such that $z$ is a semi-invariant. Then there exists
	$r \in \CC[z]$ such that the action of $T$
	is diagonal with respect to the coordinate system $(z, w+r)$.
\end{lem}

\begin{proof}[Proof of Lemma~\ref{DiagonalAction.lem}]
	By assumption, $\pi \colon \AA^2 \to \AAone$, $(z, w) \to z$
	is $T$-equivariant  with respect to a suitable $T$-action on $\AAone$. Due to
	\cite[Proposition~1]{KrKu1996Equivariant-affine}, every lift
	of a $T$-action on $\AAone$ to $\AA^2$ (with respect to $\pi$)
	is equivalent to a trivial lift (with respect to $\pi$). Thus, there
	exists $r \in \CC[z]$ such that $w + r$ is a semi-invariant with respect
	to the action of $T$. This finishes the proof.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{char.prop}]
Let $R \subseteq \Cent(\bu)$ be an algebraic subgroup such that
$\Cent(\bu) = R \ltimes N$ (see Theorem~\ref{main.thm1}).
Let $T = R^0$ be the connected component of the neutral element in $R$.
By Corollary~\ref{mainthm2.cor}, it is a torus (possibly $\dim T = 0$).

It follows from Theorem~\ref{main.thm1} that
$T \ltimes N \subseteq \Cent(\bu)$ is a subgroup of finite index.
Moreover, $T \ltimes N$ has no proper subgroup of finite index, as this group
is generated by groups that have no proper subgroup of finite index.
This implies that $T \ltimes N$ is a characteristic subgroup of $\Cent(\bu)$.
Let $G = T \ltimes N$
and let $H = \OO(\AA^3)^{\bu'} \cdot \bu' \subseteq N$.
It is now enough to prove, that $H$ is characteristic in $G$.
We divide the proof now in two cases.

\textbf{$\Gamma$ is not a fence:}
	If $\dim T = 0$, then we have $H = G$. So assume $\dim T > 0$.
	There exist coordinates $(v_1, v_2)$ of $\CC^3 \aquot \CC^+$
	such that the action of $T$ on $\CC^3 \aquot \CC^+$ is diagonal
	with respect to $(v_1, v_2)$ (see \cite{Ka1979Automorphism-group}).
	Let $\rho_1$ and $\rho_2$ be the
	characters of $T$ such that $\bt^\ast(v_i) = \rho_i(\bt) v_i$
	for all $\bt \in T$. Let $\bt \circ f \cdot \bu' \in \Cent_G(G^{(1)})$, where
	$G^{(1)} = [G, G]$ denotes the first derived group.
	A calculation shows for $i =1,2$ and for all $k \geq 0$
	\[
		\bid = [\bt \circ f \cdot \bu', [\bt^{-1}, v_i^k \cdot \bu']]
		= (1-\mu(\bt^{-1})\rho_i(\bt^{-1})^k)(1-\mu(\bt)\rho_i(\bt)^k)v_i^k \cdot \bu'
	\]
	where $\mu \colon T \to \CC^\ast$ is the character defined
	by $\mu(\bt) d = \bt^{\ast}(d)$ (see Remark~\ref{ActionByConj.rem}).
	Thus $\rho_i(\bt) = 1$ for $i =1,2$.
	As the action of $T$ on $\CC^3 \aquot \CC^+$
	is faithful, it follows that $\bt = 1$. Hence $H = \Cent_G(G^{(1)})$.

\textbf{$\Gamma$ is a fence:}
	Let $(z, P)$ be a coordinate system of $\AA^2 = \AA^3 \aquot \GGa$
	such that $\Gamma \subseteq \AA^2$ is given by the standard embedding
	$F \times \AAone \subseteq \AA^2$ for some 0-dimensional
	closed subscheme $F \subseteq \AAone$ and
	$\bu' = \Exp(\Delta_P)$ (see subsec.~\ref{The_second_subgroup.subsec}).
	As the torus $T$ leaves $\Gamma = V(a) \subseteq \CC^2$
	invariant and since $a \in \CC[z]$, there exists $q \in \CC$, such
	that $z+q$ is a semi-invariant for the action of $T$. By replacing
	$z+q$ with $z$, we can assume that $z$ is a semi-invariant.
	Moreover, by replacing $P$ with a suitable $P+r$ for some $r \in \CC[z]$
	we can assume that the action of $T$ with respect to $(z, P)$ is diagonal
	(see Lemma~\ref{DiagonalAction.lem}).
	Moreover, we denote by $\e$ an admissible complement to $\bu$.

	Assume first $\dim T = 0$. Let
	$h \cdot \e \circ f \cdot \bu' \in \Cent_G(G^{(1)})$.
	A calculation shows that
	\[
		\bid =
		[h \cdot \e \circ f \cdot \bu', [P^2 \cdot \bu', \e]] =
		-2h(a')^2 \cdot \bu'
	\]
	where $a' \in \CC[z]$ such that $\Gamma' = \div(a')$.
	Hence $h = 0$ and therefore $H = \Cent_G(G^{(1)})$.

	Assume now, $\dim T > 0$. Let $A = \CC[z]$ and let $A \ltimes A[P]$ be the
	semi-direct product defined by
	\begin{equation}
		\label{prod.eq}
		(h, f) \cdot (\bar{h}, \bar{f}) = (h + \bar{h}, f(P - \bar{h}a') + \bar{f})
	\end{equation}
	From Proposition~\ref{properties.prop} it follows that
	\[
		A \ltimes A[P] \stackrel{\sim}{\longrightarrow} N(\bu) \, , \quad
		(h, f) \longmapsto h \cdot \e \circ f \cdot \bu'
	\]
	is an isomorphism of groups. Under this isomorphism
	the subgroup $A[P]$ is sent onto $H$.
	It follows from Lemma~\ref{pres.lem} that $H = \Cent_G G^{(2)}$.
\end{proof}

\begin{lem}
	\lab{pres.lem}
	Let $A \ltimes A[P]$ be defined as in \eqref{prod.eq} and let
	$G = T \ltimes (A \ltimes A[P])$ where $T$ is a
	torus with $\dim T > 0$. Assume that
	$A[P] \subseteq G$ is a normal subgroup, that
	the action of $T$ by conjugation on $A[P]$ is given by
	$\lambda^{-1} f \lambda =
	\mu(\lambda)f(\rho_1(\lambda)z, \rho_2(\lambda)P)$
	for some characters $\mu$, $\rho_1$, $\rho_2$ and that
	$\ker \rho_1 \cap \ker \rho_2 = \{ 1 \}$.
	Furthermore we assume that the action
	of $T$ by conjugation on the quotient $G / A[P]$
	is non-trivial and the product in $G$ satisfies
	$(\lambda, 0, 0) \cdot (1, h, f) = (\lambda, h, f)$.
	Then $A[P] = \Cent_G G^{(2)}$.
\end{lem}

\begin{proof}[Proof of Lemma~\ref{pres.lem}]
	As the action by conjuagtion of $T$ on $G / A[P]$ is non-trivial, it follows
	that the first derived subgroup $G^{(1)}$ is not contained in
	$A[P]$. As $T$ is abelian it follows
	that $G^{(1)} \subseteq A \ltimes A[P]$ and as $A$ is abelian
	we conclude $G^{(2)} \subseteq A[P]$.
	Thus there exists $(1, h_0, f_0) \in G^{(1)}$ with $h_0 \neq 0$.
	As $A[P]$ is abelian, we get $A[P] \subseteq \Cent_G G^{(2)}$.
	Now, we prove $\Cent_G G^{(2)} \subseteq A[P]$. We have
	\[
		(1, 0, q - q(P + h_0a')) = [(1, 0, q), (1, h_0, f_0)] \in G^{(2)}
		\quad \textrm{for all $(1, 0, q) \in G^{(1)}$} \, .
	\]
	Moreover, $(1, 0, z^i P^j) \in G^{(1)}$
	for all $(i, j) \in \NN_0^2$ such that $\mu \rho_1^i \rho_2^j$ is not the trivial
	character, as we have $(1, 0, z^i P^j) = [(\lambda, 0, 0), (1, 0, z^i P^j)]$
	for some well chosen $\lambda \in T$. For all $j \geq 0$,
	the character $\mu \rho_1^i \rho_2^j$ is non-trivial,
	provided that $i$ is large enough. For all $(1, 0, f) \in A[P]$ we have
	\begin{eqnarray*}
		\Cent_G(1, 0, f) & = &
		\{ \, (\bar{\lambda}, \bar{h}, \bar{f}) \in G \ | \
		      (\bar{\lambda}, \bar{h}, \bar{f})^{-1} \cdot (1, 0, f) \cdot
		      (\bar{\lambda}, \bar{h}, \bar{f}) = (1, 0, f)  \, \} \\
		&=& \{ \, (\bar{\lambda}, \bar{h}, \bar{f}) \in G \ | \ f =
		(\bar{\lambda}^{-1}f\bar{\lambda})(P - \bar{h}a') \, \}
	\end{eqnarray*}
	Let $(\bar{\lambda}, \bar{h}, \bar{f}) \in \Cent_G G^{(2)}$.
	Since
	$(\bar{\lambda}, \bar{h}, \bar{f}) \in \Cent_G(1, 0, -z^i h_0 a')$
	for $i$ sufficiently large, we get $\bar{\lambda} \in \ker \rho_1$.
	Moreover,
	$(\bar{\lambda}, \bar{h}, \bar{f}) \in \Cent_G(1, 0, -z^i h_0 a' (2P + h_0 a'))$
	for sufficiently large $i$. This implies
	$\rho_2(\bar{\lambda}) = \mu(\bar{\lambda})^{-1}$ and
	$\bar{h} = ((\mu(\bar{\lambda})-1)/2)h_0$. Since
	$(\bar{\lambda}, \bar{h}, \bar{f}) \in
	\Cent_G(1, 0, -z^i h_0 a' (3P^2+3Ph_0a'+(h_0 a')^2))$ it follows that
	$\rho_2(\bar{\lambda})^2 = \mu(\bar{\lambda})^{-1}$.
	Therefore $\bar{\lambda} \in \ker \rho_2$, $\bar{h} = 0$. Hence we have
	$(\bar{\lambda}, \bar{h}, \bar{f}) = (1, 0, \bar{f}) \in A[P]$ and this proves
	$\Cent_G G^{(2)} \subseteq A[P]$.
\end{proof}

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TITLE: Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx = \frac 14 + \frac {\pi}8$
QUESTION [4 upvotes]: Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx=\frac 14 + \frac {\pi}8$

My attempt (integration by parts):
\begin{align}
I_n & = \int_0^1 \frac 1{(1+x^2)^n}\,dx = \left. \frac {x}{(1+x^2)^n} \right|_0^1+2n\int_0^1 \frac {x^2+1-1}{(1+x^2)^{n+1}}\,dx \\[10pt]
& =\frac 1{2^n}+2n \times I_n-2n\times I_{n+1}\implies I_{n+1}= \frac 1{2^{n+1}n}+\frac {2n-1}{2n}I_n.
\end{align}
where $I_1=\frac {\pi}{4}.$
Finding out that $I_2=\frac {\pi}{8}+\frac 14.$ But how do I find out that this is the only solution?

REPLY [3 votes]: Hint. The sequence $I_n$ is strictly decreasing because for $x\in (0,1]$,
$$0<\frac {1}{1+x^2} <1.$$
.

REPLY [2 votes]: Note that the sequence $(I_n)$ is decreasing, hence there is at most one value of $n$ such that $I_n=\frac{1}{4}+\frac{\pi}{8}$.


text:
\subsection{Input for tests: initial condition and wind field}
\label{A:Tests:Wind,Shape}

\begin{table}[h!]
\centering
\caption{Initial scalar fields}\label{tab:InitCond}
\begin{adjustbox}{width=1\textwidth}
	\begin{tabular}{|l|l|c|c|}
	\hline
	 \textbf{No} &  \textbf{Initial scalar field} & \textbf{Parameters} & \textbf{Source}	\\
	\hline
	1 & Cosine bell &  & \\
 	  &  $\begin{array}{lcl}
             q(\lambda,\theta, t_0)=\begin{cases}
	                         \frac{h_{max}}{2}(1+\cos\frac{\pi r}{\tilde{r}}), & \text{if } r<\tilde{r}\\
	                         0,              & \text{otherwise}
	                           \end{cases}
             \end{array}$
          & \makecell{ $r=arccos(\sin\theta_c\sin\theta+\cos\theta_c\cos\theta\cos(\lambda-\lambda_c))$
	\\$h_{max}=1$ \ \  $\tilde{r}=1/3$ \ \ $(\lambda_c,\theta_c)=(3\pi/2,0)$ }
          &  \cite{williamson1992standard}  \\
	\hline
	2 & Slotted cylinder &  & \\
	  & $\begin{array}{lcl}
	q(t_0,\lambda, \theta)=\begin{cases}
	c, & \text{if } r\leq\tilde{r}, |\lambda-\lambda_c|\geq\frac{\tilde{r}}{6}, \\
	c, & \text{if } r\leq\tilde{r}, |\lambda-\lambda_c|<\frac{\tilde{r}}{6},\ \  \theta-\theta_c<\frac{2}{3}\tilde{r}, \\
	b,              & \text{otherwise}
	\end{cases}
	\end{array}$ & \makecell{ $r=arccos(\sin\theta_c\sin\theta+\cos\theta_c\cos\theta\cos(\lambda-\lambda_c))$
	               \\ $\tilde{r}=1/2$ \\ $c=1$ \ \ $b=0$ \ \ $(\lambda_c,\theta_c)=(3\pi/2,0)$} & \cite{zalesak1979fully}  \\
	\hline
	3 & Vortex &  & \\
	  & $\begin{array}{lcl}
	         q(t_0,\lambda^{'}, \theta^{'})=1-tanh\big[\frac{\tilde{\rho}}{\gamma}\sin(\lambda^{'}-\omega(\theta^{'}) t)\big]
	     \end{array}$
	  & \makecell{ $\lambda^{'}(\lambda,\theta)=\text{arctan}\big[\frac{\cos\theta\sin(\lambda-\lambda_p)}
	                                                                   {\cos\theta\sin\theta_p\cos(\lambda-\lambda_p)-\cos\theta_p\sin\theta}\big]$,\\
	               $\theta^{'}(\lambda,\theta)=\text{arcsin}[\sin\theta\sin\theta_p+\cos\theta\cos\theta_p\cos(\lambda-\lambda_p)]$,\\
	               $\begin{array}{lcl}
	                  \omega(\theta^{'})=\begin{cases}
                                                V/(R\tilde{\rho}), & \text{if } \tilde{\rho}\neq0, \\
                                                0, & \text{otherwise}.
                                             \end{cases}
	                \end{array}$, \\
	               $V=v_0\frac{3\sqrt{3}}{2}sech^2(\tilde{\rho})tanh(\tilde{\rho})$ \\
	               $v_0=2\pi R/T$ \ \ $\tilde{\rho}=\tilde{\rho_0}\cos\theta^{'}$ \\
	               $R=6.371229\times10^6 [m]$ \ \  $T=1036800[s]$  \\
	               $(\lambda_p,\theta_p)=(\pi-0.8+\pi/4,\pi/4.8)$ \\
	               $\gamma=5$ \ \ $\tilde{\rho_0}=3$ }
	  & \cite{nair2002mass}  \\
	\hline
	4 & Two cosine bells &  & \\
 	  & $\begin{array}{lcl}
         q(t_0,\lambda,\theta)=\begin{cases}
         b+ch_1(\lambda, \theta), & \text{if } r_1<\tilde{r}, \\
         b+ch_2(\lambda, \theta), & \text{if } r_2<\tilde{r}, \\
         b, & \text{otherwise,}
         \end{cases}\end{array}$ & \makecell{ $\begin{array} {lcl}
	h_i(\lambda, \theta)=\begin{cases}
	\frac{h_{max}}{2}(1+\cos\frac{\pi r}{\tilde{r}}), & \text{if } r_i<\tilde{r}\\
	0,              & \text{otherwise}
	\end{cases}
	\end{array}$  \\ $r_i=arccos(\sin\theta_{c,i}\sin\theta+\cos\theta_{c,i}\cos\theta\cos(\lambda-\lambda_{c,i}))$
	\\$i=1,2$ \ \ $h_{max}=1$ $\tilde{r}=1/2$ \ \ $c=1$ \ \ $b=0$ \\
	$(\lambda_{c,1},\theta_{c,1})=(3\pi/4,0)$ \ \ $(\lambda_{c,2},\theta_{c,2})=(5\pi/4,0)$ }
          &  \cite{williamson1992standard}  \\
 	\hline
 	5 & Two slotted cylinders &  & \\
 	  & $\begin{array}{lcl}
 	     q(t_0,\lambda, \theta)=\begin{cases}
c, & \text{if } r_i\leq\tilde{r}, |\lambda-\lambda_{c,i}|\geq\frac{\tilde{r}}{6} \ \ \text{for } i=1,2, \\
c, & \text{if } r_1\leq\tilde{r}, |\lambda-\lambda_{c,1}|<\frac{\tilde{r}}{6},\ \  \theta-\theta_{c,1}<-\frac{5}{12}\tilde{r}, \\
c, & \text{if } r_2\leq\tilde{r}, |\lambda-\lambda_{c,2}|<\frac{\tilde{r}}{6},\ \  \theta-\theta_{c,2}>\frac{5}{12}\tilde{r}, \\
b,              & \text{otherwise},
\end{cases}
 	    \end{array}$ &\makecell{$r_i=arccos(\sin\theta_{c,i}\sin\theta+\cos\theta_{c,i}\cos\theta\cos(\lambda-\lambda_{c,i}))$ \\
 	    $i=1,2$ \ \ $\tilde{r}=1/2$ \ \ $c=1$ \ \ $b=0$ \\
 	    $(\lambda_{c,1},\theta_{c,1})=(3\pi/4,0)$ \ \ $(\lambda_{c,2},\theta_{c,2})=(5\pi/4,0)$ }
 	    & \cite{nair2010class} \\
 	    \hline
	\end{tabular}
\end{adjustbox}

\end{table}

\begin{table}[h!]
\centering
\caption{Velocity vector}\label{tab:Velocity}
\begin{adjustbox}{width=1\textwidth}
	\begin{tabular}{|l|l|c|c|}
	\hline
	\textbf{No} & \textbf{Velocity vector} & \textbf{Parameters} & \textbf{Source}	\\
	\hline
	1 & \multicolumn{2}{l|}{Solid body rotation: $\nabla\cdot \vec{v}=0$} &  \\
	  & \multicolumn{2}{l|}{Exact solution: after complete revolution, initial field reaches starting position} &  \\
        \hline
	  & $\begin{array} {lcl}
	v_\lambda(\lambda,\theta)=u_0(\cos\theta\cos\alpha+\sin\theta\cos\lambda\sin\alpha),\\
	v_\theta(\lambda,\theta)=-u_0\sin\lambda\sin\alpha
	\end{array}$  & \makecell{ $u_0=2\pi R/T$ \\ $R=6.371229\times10^6 [m]$ \ \ $T=1036800[s]$ \\ $\alpha=0^\circ$ \ \  $0\leq t\leq T$ } & \cite{williamson1992standard}  \\
	\hline
	2 & \multicolumn{2}{l|}{Deformational flow: $\nabla\cdot \vec{v}=0$} &  \\
	  & \multicolumn{2}{l|}{Exact solution: after complete revolution, initial field reaches starting position} &  \\
        \hline
	  & $\begin{array}{lcl}
	v_\lambda(\lambda,\theta, t)=k\sin^2(\lambda/2)\sin(2\theta)\cos(\pi t/T),\\
	v_\theta(\lambda,\theta, t)=\frac{k}{2}\sin\lambda\cos\theta\cos(\pi t/T)
	\end{array}$ &  $k=2.4$  \ \ $0\leq t\leq T$ & \cite{nair2010class}  \\
	\hline
	3 & \multicolumn{2}{l|}{Deformational flow: $\nabla\cdot \vec{v}\neq0$} &  \\
	  & \multicolumn{2}{l|}{Exact solution: after complete revolution, initial field reaches starting position} &  \\
        \hline
	  & $\begin{array}{lcl}
	v_\lambda(\lambda,\theta, t)=-k\sin^2(\lambda/2)\sin(2\theta)\cos^2\theta\cos(\pi t/T),\\
	v_\theta(\lambda,\theta, t)=\frac{k}{2}\sin\lambda\cos^3\theta\cos(\pi t/T)
	\end{array}$ & $k=1$ \ \ $0\leq t\leq T$  & \cite{nair2010class}  \\
	\hline
	4 & \multicolumn{2}{l|}{Moving vortices: $\nabla\cdot \vec{v}=0$} &  \\
	  & \multicolumn{2}{l|}{Exact solution: Forward- $q(t,\lambda,\theta)=1-tanh[\frac{\tilde{\rho}}{\gamma}\sin(\lambda^{'}-\omega t)]$,
	                                       Backward- $q(t,\lambda,\theta)=1-tanh[\frac{\tilde{\rho}}{\gamma}\sin(\lambda^{'}+\omega t)]$ } &  \\
        \hline
	  & $\begin{array}{lcl}
	     v_\lambda(t,\lambda,\theta)=u_0(\cos\theta\cos\alpha+\sin\theta\cos\lambda\sin\alpha)+\\
	                                 R\omega(\theta^{'})(\sin\theta_c\cos\theta-\cos\theta_c\cos(\lambda-\lambda_c)\sin\theta), \\
	     v_\theta(t,\lambda,\theta)=-u_0\sin\lambda\sin\alpha,+R\omega(\theta^{'})(\cos\theta_c\sin(\lambda-\lambda_c)).
	     \end{array}$
	 & \makecell{ $V=v_0\frac{3\sqrt{3}}{2}sech^2(\tilde{\rho})tanh(\tilde{\rho})$ \\
	              $v_0=2\pi R/T$\ \ $\tilde{\rho}=\tilde{\rho_0}\cos\theta^{'}$ \ \ $\tilde{\rho_0}=3$\\
	              $R=6.371229\times10^6 [m]$ \ \ $T=1036800[s]$ \\
                      $\begin{array}{lcl}
                         \omega(\theta^{'})=\begin{cases}
                                              V/(R\tilde{\rho}), & \text{if } \tilde{\rho}\neq0, \\
                                              0, & \text{otherwise}.
                                            \end{cases}
                       \end{array}$  \\
                      $(\lambda_c,\theta_c)=(\lambda_p+\omega t_n, \theta_p)$ \\
                      $(\lambda_p,\theta_p)=(\pi-0.8+\pi/4,\pi/4.8)$ \\
                      $t_n=n\Delta t$ \ \  $\Delta t=600[s]$ \ \
                      $0\leq t\leq T$ }
         & \cite{nair2008moving}  \\
	\hline
	\end{tabular}
\end{adjustbox}

	\label{tab2}
\end{table}

config: arxiv


TITLE: Can we use mathematical induction within a proof by mathematical induction?
QUESTION [3 upvotes]: My question is short and concise. It is that while proving a theorem or a result by induction can we use induction in between the proof on an intermediate expression.
I've encountered this many times that while proving by induction sometimes we land to an expression wherein I think proving that expression using induction can be helpful rather than carrying out some difficult manipulations.
I think that doing this is logically justified because the intermediate expression which we receive can also be asked to be proved as an independent result. And many a times while manipulating those lengthy expression we also use some of the properties which are themselves proven using induction or any other method of proof.
So is this process justified. One more thing is that if it is justified and we can use induction in between a proof by induction, then it follows that we can do that multiple times and may be we end up at a greatly simplified expression.
Please and let me know that whether this is worth doing?
As an example I just tried this (which I have still not proved) - $$(n+1)^{2n+1}< n^n(n+2)^{n+1} $$ After a series of steps within inductive step I reached to this which I have to prove- $$\left(\frac{k}{k+1}\right)^k>\left(\frac{k+2}{k+3}\right)^{k+2}$$
May be the manipulations are simple enough which I can not figure out(or don't want to) but I think I can use induction to prove this result ( which may make this a bit more complex, anyways may be if applied somewhere else it would make that simpler )
Thanks.

REPLY [3 votes]: Yes, it is ok to do an inductive proof to prove something that you need as part of another inductive proof.  For example, starting with the Peano Axioms, you can prove commutativity of addition ($\forall x \forall y \ x+y=y+x$) by induction on $x$, and you can prove the base ($\forall y \ 0+y=y+0$) by induction on $y$.


TITLE: Give an explicit definition for the different homomorphisms $\psi_1 ,\psi_2:\mathbb{Z}_{5}\to\mathbb{Z}_{5}$.
QUESTION [2 upvotes]: Can anyone please help me define clearly using the definition for a group homomorphism that for the given two homomorphisms it holds for all $g_1,g_2\in G$?
We know that the definition gives that there exists two groups, $G$ and $H$ and a function  $\psi_1:G\to H$. This is then called a homomorphism if the equality $\psi_1(g_1+g_2)=\psi_1(g_1)+\psi_1(g_2)$ holds for all $g_1,g_2\in G$.
So as a start,we take the first function $\psi_1:\mathbb{Z}_{5}\to\mathbb{Z}_{5}$. If we let $x,y\in\mathbb{Z}_{5}$, then $\psi_1(x+y)=\psi_1(x)+\psi_1(y)$?
Please brief me out on how to do this problem.

REPLY [3 votes]: The group $\mathbb{Z}_5$ is an additive cyclic group: $(\mathbb{Z}_5, +)$, so the homomorphism property for $\psi_i: \mathbb{Z}_5 \to \mathbb{Z}_5$ is given by
$$\psi_i(x + y) = \psi_i(x) + \psi_i(y)$$
For all homomorphisms, $\psi_i(0) = 0$ (a homomorphism maps identity to identity).
Then choose $\psi_i(1) = x \in \mathbb{Z}_5, x \neq 0$, the the homomorphism property guarantees that this assignment determines the assignment of all elements in $\mathbb{Z}_5$
For each element $x \in \mathbb{Z}_5$, $x \neq 0$, there exists a unique homomorphism determined by $\psi_i(1) = x$.
Hence, there are four such unique homomorphisms corresponding to $\psi_i(1) = i$, $i \in \{ 1, 2, 3, 4\}$


TITLE: Applying Compactness Theorem in Predicate Logic
QUESTION [0 upvotes]: $A$ is a first-order sentence over the language $L = [; R, =]$ where $R$ is a binary predicate symbol. Suppose that for each $n \geq 3$, $A$ has a model consisting of a directed cycle with n nodes, where $R$ represents the edge relation of a directed graph. Prove that $A$ has a model whose universe $M$ includes an infinite path; i.e. a set of distinct elements $v_0, v_1, ...$ such that $R^M(v_i, v_{i+1})$ holds for all $i \geq 0$

REPLY [1 votes]: Let $\Sigma$ be a set of sentence of the form
$$
\sigma_n = [\forall x_1 \forall x_2\cdots \forall x_n : R(x_1,x_2)\land \cdots \land R(x_{n-1},x_n) \to \exists y R(x_n,y)\land \bigwedge_{1\le i\le n} x_i\neq y].
$$
and let define $\tau=[\exists x\exists y :R(x,y)\land x\neq y]$.
Since $\{A,\tau\}\cup \{\sigma_1,\sigma_2,\cdots , \sigma_n\}$ has a model (consider the model of $A$ consisting of a directed cycle with $n+1$ nodes), there is a model of $\{A,\tau\}\cup \Sigma$, by compactness theorem.
Let $\mathfrak{A}$ be a model of $\{A,\tau\}\cup \Sigma$. Since $\mathfrak{A}$ satisfies $\tau$, there is a distinct $c_0,c_1\in \mathfrak{A}$ satisfy that $(c_0,c_1)\in R^\mathfrak{A}$. If there is a distinct $c_0,c_1,\cdots,c_{n-1}\in \mathfrak{A}$ satisfy
$$(c_0,c_1),(c_1,c_2),\cdots,(c_{n-2},c_{n-1})\in R^\mathfrak{A},$$
then by sentence $\sigma$ there is $c_n$ that distinct from $c_0,c_1,\cdots,c_{n-1}$ and satisfy $(c_{n-1},c_n)\in R^\mathfrak{A}$. So the set $\{c_n:n=0,1,2,\cdots\}$ is an infinite path over $\mathfrak{A}$.


TITLE: Riemannian holonomy of a covering
QUESTION [5 upvotes]: Suppose I have a connected Riemannian manifold $X$ and a covering $\pi:Y\to X$ with the pulled back metric on $Y$, making $\pi$ into a local isometry between Riemannian manifolds.
Suppose we have a closed loop in $Y$ based at a point $y\in Y$. This will give rise to an element of the holonomy group $\mathrm{Hol}_y(Y)$, which after we fix a basis of $T_yY$ becomes an element of $\mathrm{GL}(n)$ (or $\mathrm O(n)$ really).
If we project the path to $X$, we get a closed loop again, hence an element of $\mathrm{Hol}_xX$, where $x = \pi(y)$, and by using $\pi$ to map the basis of $T_yY$ to one of $T_xX$ this becomes an element of $\mathrm{GL}(n)$ as well, and we can talk about equality in a sensible way.
Since all along the paths the geometry is identical (around every point there is a neighbourhood on which $\pi$ induces an isometry) it seems clear that they must induce the same element in holonomy.
This would mean that the holonomy of $Y$ is a subgroup of the holonomy of $X$.
This inclusion can be strict, if $X$ is not orientable for example, its holonomy group might be $\mathrm O(n)$, but if we take $Y$ to be its oriented double cover its holonomy must be contained in $\mathrm{SO}(n)$.
Can anything more precise be said about the relation between the holonomy groups of $X$ and of $Y$? I would expect the fundamental group or rather the group of deck transformations to play a role, but I don't know in what form.
Thanks!

REPLY [1 votes]: I think Wolfgang Globke's answer could be generalised in the following way. Fix a point $x_0\in X$. The cover $\pi\colon Y\to X$ corresponds to a subgroup $H\subset \pi_1(Y,x_0)$.
A loop $\gamma$ in $Y$ based in a point $y_0\in\pi^{-1}(x_0)$ is null-homotopic if and only if $\pi(\gamma)$ lies in $H$. Define $$\mathrm{Hol}^H(X,x_0)=\{\mathrm{hol}_{x_0}(\gamma)\mid [\gamma]\in H\}\subset \mathrm{Hol}(X,x_0).$$ Then analogous to Helga Baum's proof we can show
$$\mathrm{Hol}(Y,y_0)=\mathrm{Hol}^H(X,x_0).$$
And just as before we obtain a surjection
$$H\to \mathrm{Hol}(X,x_0)/{\mathrm{Hol}^H(X,x_0)}=\mathrm{Hol}(X,x_0)/{\mathrm{Hol}(Y,y_0)}.$$