# Algebraically generated groups
and their Lie algebras

Hanspeter Kraft and Mikhail Zaidenberg  Mathematisches Institut, Universität
Basel,
Rheinsprung 21, CH-4051 Basel, Switzerland<EMAIL_ADDRESS>Institut
Fourier, UMR 5582, Laboratoire de Mathématiques,
Université Grenoble Alpes, CS 40700, 38058 Grenoble cedex 9, France
<EMAIL_ADDRESS>
###### Abstract.

The automorphism group $\operatorname{Aut}(X)$ of an affine variety $X$ is an
ind-group. Its Lie algebra is canonically embedded into the Lie algebra
$\operatorname{Vec}(X)$ of vector fields on $X$. We study the relations
between closed subgroups $\mathfrak{G}$ of $\operatorname{Aut}(X)$ and the
corresponding Lie subalgebras of $\operatorname{Vec}(X)$.

We show that a subgroup $G\subseteq\operatorname{Aut}(X)$ generated by a
family of connected algebraic subgroups $G_{i}$ of $\operatorname{Aut}(X)$ is
algebraic if and only if the
$\operatorname{Lie}G_{i}\subseteq\operatorname{Vec}(X)$ generate a finite
dimensional Lie subalgebra of $\operatorname{Vec}(X)$.

Extending a result by Cohen-Draisma [CD03] we prove that a locally finite Lie
algebra $L\subseteq\operatorname{Vec}(X)$ generated by locally nilpotent
vector fields is algebraic, i.e. $L=\operatorname{Lie}G$ of some algebraic
subgroup $G\subseteq\operatorname{Aut}(X)$.

Along the same lines we prove that if a subgroup
$G\subseteq\operatorname{Aut}(X)$ generated by finitely many unipotent
algebraic groups is solvable, then it is a unipotent algebraic group.

Furthermore, we give an example of a free subgroup
$F\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$ generated by two algebraic
elements such that the Zariski closure $\overline{F}$ is a free product of two
nested commutative closed unipotent ind-subgroups.

To any ind-group $\mathfrak{G}$ one can associate a canonical ideal
$L_{\mathfrak{G}}\subseteq\operatorname{Lie}\mathfrak{G}$ generated by the Lie
algebras of algebraic subgroups of $\mathfrak{G}$, and another one associated
to the set of all algebraic subvarieties of $\mathfrak{G}$. We study
functorial properties of these ideals, ind-subgroups of finite codimension in
$\mathfrak{G}$ and the corresponding Lie subalgebras of
$\operatorname{Lie}\mathfrak{G}$.

###### Key words and phrases:

ind-group, automorphism group, Lie algebra, algebraically generated subgroup,
group action

###### 1991 Mathematics Subject Classification:

14J50 (primary), 14L40, 17B66, 22E65 (secondary)

A part of the paper was written during a stay of the second author at the Max
Planck Institute of Mathematics in Bonn. He thanks this institution for its
support and excellent working conditions.

###### Contents

  1. 1 Introduction
    1. 1.1 Notation
    2. 1.2 Ind-groups and vector fields
    3. 1.3 Algebraically generated groups
    4. 1.4 Locally finite subsets
    5. 1.5 Locally finite vector fields
    6. 1.6 The adjoint action on $\operatorname{Vec}(X)$
    7. 1.7 Toral subalgebras
    8. 1.8 Integration of Lie algebras
    9. 1.9 A free subgroup of $\operatorname{Aut}({\mathbb{A}}^{2})$ and its closure
    10. 1.10 Ind-groups and their Lie algebras
  2. 2 Algebraically generated subgroups
    1. 2.1 Orbits of algebraically generated subgroups
    2. 2.2 Vector fields associated to subvarieties of $\operatorname{Aut}(X)$
    3. 2.3 Lie algebras of algebraically generated groups
    4. 2.4 Ramanujam’s alternative
    5. 2.5 The finite dimensional case
    6. 2.6 Integration of locally finite Lie algebras
    7. 2.7 Solvable subgroups
    8. 2.8 Nested Lie algebras
  3. 3 An interesting example
    1. 3.1 Notation and first results
    2. 3.2 The Lie algebra of $\mathfrak{F}$
    3. 3.3 De Jonquière subgroups and amalgamated products
    4. 3.4 The fixed points under $S$
  4. 4 Homomorphisms of ind-groups and of their Lie algebras
    1. 4.1 A canonical Lie subalgebra of $\operatorname{Lie}\mathfrak{G}$
    2. 4.2 Surjective homomorphisms
    3. 4.3 Subgroups of finite codimension

##  1\. Introduction

###  1.1. Notation

Our base field ${\mathbb{K}}$ is algebraically closed and of characteristic
zero. ${\mathbb{A}}^{n}$ stands for affine $n$-space, and
${\mathbb{G}}_{a}:={\mathbb{K}}^{+}$ and ${\mathbb{G}}_{m}:={\mathbb{K}}^{*}$
denote the additive and the multiplicative group of ${\mathbb{K}}$. If $G$ is
an algebraic group, then a $G$-variety $X$ is an affine variety with an action
of $G$ such that the corresponding map $G\times X\to X$ is a morphism.

It is well known that for every $G$-variety $X$ there is a canonical (anti-)
homomorphism of Lie algebras
$\xi\colon\operatorname{Lie}G\to\operatorname{Vec}(X)$, where
$\operatorname{Vec}(X)$ stands for the Lie algebra of (algebraic) vector
fields on $X$. The construction of this homomorphism goes as follows. For any
$x\in X$ let $\mu_{x}\colon G\to X$ be the orbit map $g\mapsto gx$, and denote
by

$d\mu_{x}\colon\operatorname{Lie}G:=T_{e}G\to T_{x}X$

its differential in $e\in G$. If $A\in\operatorname{Lie}G$, then the
corresponding vector field $\xi_{A}$ is given by $\xi_{A}(x):=d\mu_{x}(A)\in
T_{x}X$. If $L\subseteq\operatorname{Vec}(X)$ denotes the image of
$\operatorname{Lie}G$, then one has $T_{x}Gx=L(x)$ for any $x\in X$. Moreover,
the differential $d\mu_{x}\colon T_{g}G\to T_{gx}Gx$ of the orbit map
$\mu_{x}$ is surjective in every $g\in G$. (For a reference one might look at
[Kra16, Appendix A.4].)

For a $G$-variety $X$ we have linear actions of $G$ on the coordinate ring
$\mathcal{O}(X)$ and on the vector fields $\operatorname{Vec}(X)$, and these
representations are locally finite and regular, i.e. every element is
contained in a finite dimensional $G$-invariant subspace and the linear action
of $G$ on any finite dimensional $G$-invariant subspace is regular. Moreover,
the homomorphism $\xi\colon\operatorname{Lie}G\to\operatorname{Vec}(X)$ is
$G$-equivariant, see [FK18, Sect. 7.3].

###  1.2. Ind-groups and vector fields

For an affine variety $X$ the group of regular automorphisms
$\operatorname{Aut}(X)$ is an affine ind-group. We refer to the paper [FK18]
for an introduction to ind-varieties and ind-groups and for basic concepts,
cf. [Sha81, Kum02]. For an ind-group $\mathfrak{G}$ the tangent space
$T_{e}\mathfrak{G}$ carries a natural structure of a Lie algebra which will be
denoted by $\operatorname{Lie}\mathfrak{G}$. In case of
$\operatorname{Aut}(X)$ there is a canonical embedding
$\xi\colon\operatorname{Lie}\operatorname{Aut}(X)\hookrightarrow\operatorname{Vec}(X)$
which is an anti-homomorphism of Lie algebras. It is constructed in a similar
way as explained above for an algebraic group $G$, see [FK18, Proposition
7.2.4].

The ind-groups and ind-varieties carry a natural (Zariski-) topology, called
the ind-topology. All topological notation in this paper will be with respect
to the ind-topology.

A subset $Y\subseteq\mathfrak{V}$ of an ind-variety
$\mathfrak{V}=\bigcup_{k}\mathfrak{V}_{k}$ is called algebraic if it is
locally closed and contained in $\mathfrak{V}_{k}$ for some $k$. If $X$ is a
$G$-variety, then the canonical map $G\to\operatorname{Aut}(X)$ is a
homomorphism of ind-groups, and the image of $G$ is closed and algebraic, see
[FK18, Proposition 2.7.1]. In particular, every algebraic subgroup
$G\subseteq\operatorname{Aut}(X)$ is closed. It follows that
$\operatorname{Lie}G\subseteq\operatorname{Lie}\operatorname{Aut}(X)$ is a Lie
subalgebra and that $\operatorname{Lie}G$ is canonically embedded into the
vector fields $\operatorname{Vec}(X)$.

Convention By abuse of notation, we will constantly identify, for an algebraic
group $G\subseteq\operatorname{Aut}(X)$, the Lie algebra $\operatorname{Lie}G$
with its image $\xi(\operatorname{Lie}G)\subseteq\operatorname{Vec}(X)$.

The main objective of these notes is to study the relations between the
automorphism group $\operatorname{Aut}(X)$ of an affine variety and its closed
subgroups on the one hand, and the corresponding Lie subalgebras of
$\operatorname{Vec}(X)$ on the other. Besides the introduction the paper
contains three different sections that we survey below. They are independent
and touch three different subjects. Section 2 is devoted to algebraically
generated subgroups of $\operatorname{Aut}(X)$ and their Lie algebras. Section
3 studies an example of an ind-subgroup of
$\operatorname{Aut}({\mathbb{A}}^{2})$ which is the closure of a nonabelian
free subgroup of $\operatorname{Aut}({\mathbb{A}}^{2})$, and the final Section
4 is about ind-subgroups of $\operatorname{Aut}(X)$, their Lie algebras and
homomorphisms.

###  1.3. Algebraically generated groups

A first result showing a very strong relation between the Lie algebra of an
ind-group and the group itself is the following, see [FK18, Proposition
2.2.1(3) and (4)].

######  Proposition 1.3.1.

Let $\mathfrak{G}$ be an ind-group. Then the connected component
$\mathfrak{G}^{\circ}$ is an algebraic group if and only if
$\operatorname{Lie}\mathfrak{G}$ is finite dimensional.

We will prove a similar statement for so-called algebraically generated
subgroups. Let $(G_{i})_{i\in I}$ be a family of connected algebraic subgroups
of $\operatorname{Aut}(X)$ where $X$ is an affine variety. The subgroup
$G\subseteq\operatorname{Aut}(X)$ generated by the $G_{i}$ is called
_algebraically generated_ ([AFK13]).

Let $L\subseteq\operatorname{Vec}(X)$ be the Lie subalgebra generated by the
Lie algebras $\operatorname{Lie}G_{i}$. We will see in Theorem 2.3.1 that $L$
is invariant under the action of $G$ and of the closure $\overline{G}$. This
implies that $L\subseteq\operatorname{Lie}\overline{G}$ is an ideal.

######  Question 1.

Do we have $L=\operatorname{Lie}\overline{G}$?

Our result in this direction is the following.

######  Theorem A.

Assume that $L$ is finite dimensional. Then $G$ is an algebraic subgroup of
$\operatorname{Aut}(X)$ and $\operatorname{Lie}G=L$.

We will prove a slightly more general result, see Theorems 2.3.1 and 2.5.1.
Our example in Section 3 gives an infinite dimensional $L$ where Question 1
above has a positive answer.

Theorem A above has several interesting application. An important one is the
following.

######  Theorem B.

Let $G\subseteq\operatorname{Aut}(X)$ be a subgroup generated by finitely many
unipotent algebraic subgroups. If $G$ is solvable, then $G$ is a unipotent
algebraic group.

The proof of Theorem B will be given in Section 2.7.

We say that $\operatorname{Aut}(X)$ verifies the Tits alternative if any
subgroup of $\operatorname{Aut}(X)$ is either virtually solvable, or it
contains a nonabelian free subgroup. Due to Tits’ original theorem, this
alternative holds for $\operatorname{GL}(n,{\mathbb{K}})$, hence for any
linear algebraic group. We refer to [AZ21] and the references therein for a
brief survey on the Tits alternative for automorphism groups.

An immediate consequence of Theorem B in this context is the following.

######  Corollary 1.3.2.

Let $G\subseteq\operatorname{Aut}(X)$ be a subgroup generated by a finite
family of unipotent algebraic groups. If the Tits alternative holds for
$\operatorname{Aut}(X)$, then either $G$ is a unipotent algebraic group or $G$
contains a nonabelian free subgroup.

######  Remark 1.3.3.

If $G$ is an algebraic group and $H_{1},\ldots,H_{m}\subseteq G$ a finite set
of connected closed subgroups, then the subgroup $H$ generated by the $H_{i}$
is closed. (In fact, there is a sequence $i_{1},i_{2},\ldots,i_{N}$ such that
the product $P:=H_{i_{1}}H_{i_{2}}\cdots H_{i_{N}}$ is dense (and
constructible) in $\overline{H}$. It then follows that $PP=\overline{H}$,
hence $H=\overline{H}$.)

As a consequence we see that a subgroup of a nested ind-group $\mathfrak{G}$
(see Definition 2.8.1) generated by finitely many connected algebraic groups
is an algebraic group.

Note that this statement does not hold for non-connected groups. E.g. the two
matrices $A:=\left[\begin{smallmatrix}0&-1\\\
1&\phantom{-}0\end{smallmatrix}\right]$ and
$B:=\left[\begin{smallmatrix}0&-\frac{1}{2}\\\
2&\phantom{-}0\end{smallmatrix}\right]$ have both order 4, but the product
$AB$ has infinite order. Thus the subgroup of
$\operatorname{SL}_{2}({\mathbb{K}})$ generated by the finite groups $\langle
A\rangle$ and $\langle B\rangle$ is not closed.

###  1.4. Locally finite subsets

Let $V$ be a vector space over ${\mathbb{K}}$. We denote by
$\operatorname{\mathcal{L}}(V)$ the algebra of linear endomorphisms of $V$. An
endomorphism $\lambda\in\operatorname{\mathcal{L}}(V)$ is called locally
finite if the linear span $\langle\lambda^{j}(v)\mid j\in{\mathbb{N}}\rangle$
is finite dimensional for any $v\in V$. It is called semisimple if there is a
basis of eigenvectors, and locally nilpotent if for any $v\in V$ there is an
$m\in{\mathbb{N}}$ such that $\lambda^{m}(v)=0$. Every locally finite
endomorphism $\lambda$ has a uniquely defined additive Jordan-decomposition
$\lambda=\lambda_{s}+\lambda_{n}$ where $\lambda_{s}$ is semisimple,
$\lambda_{n}$ locally nilpotent, and
$\lambda_{s}\circ\lambda_{n}=\lambda_{n}\circ\lambda_{s}$.

A subset $S\subseteq\operatorname{\mathcal{L}}(V)$ is called locally finite if
every $v\in V$ is contained in a finite dimensional subspace $W\subseteq V$
which is invariant under all elements from $S$. If
$S\subseteq\operatorname{GL}(V)$, then $S$ is locally finite if and only if
the group $\langle S\rangle$ generated by $S$ is locally finite.

Note that a locally finite subspace $A\subseteq\operatorname{\mathcal{L}}(V)$
is not necessarily finite dimensional. In fact, let $(e_{1},e_{2},\ldots)$ be
a basis of $V:={\mathbb{K}}^{\infty}$ and define
$\lambda_{k}\in\operatorname{\mathcal{L}}(V)$ by
$\lambda_{k}(e_{j}):=\delta_{kj}e_{j}$. Then
$A:=\bigoplus_{k}{\mathbb{K}}\lambda_{k}$ is an infinite dimensional locally
finite subspace.

######  Definition 1.4.1.

Let $X$ be an affine variety. A morphism $\varphi\colon X\to X$ is called
algebraic or locally finite if the induced endomorphism
$\varphi^{*}\in\operatorname{\mathcal{L}}(\mathcal{O}(X))$ is locally finite.
It is called semisimple if $\varphi^{*}$ is semisimple, and locally nilpotent
if $\varphi^{*}$ is locally nilpotent.

A subgroup $H\subseteq\operatorname{Aut}(X)$ is called locally finite if the
image of $H$ in $\operatorname{\mathcal{L}}(\mathcal{O}(X))$ is locally
finite.

Note that a subgroup $H\subseteq\operatorname{Aut}(X)$ is locally finite if
and only if the closure $\overline{H}\subseteq\operatorname{Aut}(X)$ is a
algebraic group. If $g\in\operatorname{Aut}(X)$ is an algebraic element, then
the closed commutative subgroup $\overline{\langle g\rangle}$ has the form
${\mathbb{G}}_{m}^{p}\times{\mathbb{G}}_{a}^{q}\times F$ where
$p\in{\mathbb{Z}}_{\geq 0}$, $q\in\\{0,1\\}$ and $F$ is a finite cyclic group.

There are many examples of subgroups $G\subseteq\operatorname{Aut}(X)$ that
are not locally finite while being generated by algebraic elements. We will
discuss such an example in Section 3. We also refer to the interesting
discussions in [FK18, 9.4.3–9.4.5] and [PR22] of subgroups
$G\subseteq\operatorname{Aut}(X)$ which consist of algebraic elements.

###  1.5. Locally finite vector fields

Recall that the vector fields $\operatorname{Vec}(X)$ are the derivations
$\operatorname{Der}(\mathcal{O}(X))$ of $\mathcal{O}(X)$. Since
$\operatorname{Der}(\mathcal{O}(X))\subseteq\operatorname{\mathcal{L}}(\mathcal{O}(X))$
we can talk about locally finite vector fields and locally finite subspaces of
$\operatorname{Vec}(X)$.

######  Example 1.5.1.

If $G\subseteq\operatorname{Aut}(X)$ is an algebraic subgroup, then its Lie
algebra $\operatorname{Lie}(G)\subseteq\operatorname{Vec}(X)$ is locally
finite. These Lie algebras are called algebraic.

In contrast to the general situation we have the following finiteness result.

######  Lemma 1.5.2.

Let $L\subseteq\operatorname{Vec}(X)$ be a locally finite subspace. Then $L$
is finite dimensional.

###### Proof.

If $f_{1},\ldots,f_{m}\in\mathcal{O}(X)$ is a set of generators, then every
vector field $\delta$ is determined by the values
$\delta(f_{1}),\ldots,\delta(f_{m})$. It follows that for any subspace
$L\subseteq\operatorname{Vec}(X)$ the linear map $L\to\mathcal{O}(X)^{m}$,
$\delta\mapsto(\delta(f_{1}),\ldots,\delta(f_{m}))$ is injective. If $L$ is
locally finite, then the image of this map is contained in the finite
dimensional subspace $\bigoplus_{i=1}^{m}Lf_{i}$, and the claim follows. ∎

The fact that a vector field is determined by its values on a finite
generating set of $\mathcal{O}(X)$ has the following important consequences.

######  Lemma 1.5.3.

Let $V\subseteq\mathcal{O}(X)$ be a finite dimensional subspace generating
$\mathcal{O}(X)$, and let $\delta\in\operatorname{Vec}(X)$ be a vector field.
If $V$ is $\delta$-invariant, then the following holds.

  1. (1)

$\delta$ is locally finite.

  2. (2)

$\delta$ is semisimple if and only if
$\delta|_{V}\in\operatorname{\mathfrak{g}l}(V)$ is semisimple.

  3. (3)

$\delta$ is locally nilpotent if and only if
$\delta|_{V}\in\operatorname{\mathfrak{g}l}(V)$ is nilpotent.

  4. (4)

If $\delta=\delta_{s}+\delta_{n}$ is the Jordan decomposition in
$\operatorname{\mathcal{L}}(\mathcal{O}(X))$, then
$\delta_{s},\delta_{n}\in\operatorname{Vec}(X)$.

  5. (5)

$V$ is $\delta_{s}$\- and $\delta_{n}$-invariant, and
$\delta|_{V}=\delta_{s}|_{V}+\delta_{n}|_{V}\in\operatorname{\mathfrak{g}l}(V)$
is the usual Jordan decomposition in $\operatorname{\mathfrak{g}l}(V)$.

###### Proof.

Denote by $V^{(n)}\subseteq\mathcal{O}(X)$ the linear subspace generated by
the $n$-fold products of elements from $V$. Then the following is obvious: (a)
$V^{(n)}$ is $\delta$-invariant; (b) if $\delta|_{V}$ is semisimple resp.
nilpotent, then so is $\delta|_{V^{(n)}}$. This proves (1)–(3). The claim (4)
follows from [FK18, Proposition 7.6.1], and (5) follows from (2) and (3). ∎

######  Question 2.

Let $L_{i}\subseteq\operatorname{Vec}(X)$ ($i\in I$) be a family of locally
finite Lie subalgebras, and denote by $L\subseteq\operatorname{Vec}(X)$ the
Lie subalgebra generated by the $L_{i}$. Is $L$ locally finite in case $L$ is
finite dimensional?

In this direction we have the following consequence of Theorem A.

######  Corollary.

Let $L\subseteq\operatorname{Vec}(X)$ be the Lie subalgebra generated by a
family of locally finite Lie subalgebras
$L_{i}\subseteq\operatorname{Vec}(X)$, $i\in I$. Assume that each $L_{i}$ is
algebraic, i.e. $L_{i}=\operatorname{Lie}G_{i}$ for some algebraic group
$G_{i}\subseteq\operatorname{Aut}(X)$. Then $L$ is locally finite if and only
if $L$ is finite dimensional. In this case the subgroup $G$ generated by the
$G_{i}$ is algebraic and $L=\operatorname{Lie}G$. In particular, $L$ is
algebraic.

###### Proof.

If $L$ is locally finite, then it is finite dimensional, by Lemma 1.5.2. If
$L$ is finite dimensional, then the claim follows from Theorem A. ∎

######  Remark 1.5.4.

For a locally finite vector field $\delta\in\operatorname{Vec}(X)$ there is a
uniquely determined minimal algebraic group $H\subseteq\operatorname{Aut}(X)$
such that $\delta\in\operatorname{Lie}H$. Moreover, $H$ is commutative and
connected, and $H$ is a torus if $\delta$ is semisimple and
$H\simeq{\mathbb{G}}_{a}$ if $\delta$ is locally nilpotent and non-zero, see
[FK18, Proposition 7.6.1].

###  1.6. The adjoint action on $\operatorname{Vec}(X)$

For any vector field $\delta$ we have the adjoint action
$\operatorname{ad}\delta$ on $\operatorname{Vec}(X)$ defined in the usual way:

$\operatorname{ad}\delta(\eta):=[\delta,\eta]=\delta\circ\eta-\eta\circ\delta.$

######  Lemma 1.6.1.

Let $\delta\in\operatorname{Vec}(X)$ be a locally finite vector field with
Jordan decomposition $\delta=\delta_{s}+\delta_{n}$. Then
$\operatorname{ad}\delta$ acting on $\operatorname{Vec}(X)$ is locally finite,
and
$\operatorname{ad}\delta=\operatorname{ad}\delta_{s}+\operatorname{ad}\delta_{n}$
is the Jordan decomposition.

###### Proof.

Let $V\subseteq\mathcal{O}(X)$ be a finite dimensional $\delta$-invariant
subspace which generates $\mathcal{O}(X)$. Then we have an inclusion

$\operatorname{Vec}(X)\hookrightarrow\operatorname{\mathcal{L}}(V,\mathcal{O}(X)),\quad\mu\mapsto\mu|_{V}.$

Given $\mu\in\operatorname{Vec}(X)$ we choose a finite dimensional
$\delta$-invariant subspace $W$ which contains $\mu(V)$. In
$\operatorname{\mathcal{L}}(\mathcal{O}(X))$ we have

$(\operatorname{ad}\delta)^{m}\mu=\sum_{i=0}^{m}n_{i}\delta^{m-i}\mu\delta^{i}$

with suitable $n_{i}\in{\mathbb{Z}}$. Hence $(\operatorname{ad}\delta)^{m}\mu$
sends $V$ to $W$ for all $m\geq 0$, and so the linear span
$\langle(\operatorname{ad}\delta)^{m}\mu\mid m\geq
0\rangle\subseteq\operatorname{Vec}(X)$ is finite dimensional. This shows that
$\operatorname{ad}\delta$ is locally finite.

If $\delta$ is locally nilpotent, then $(\delta|_{V})^{m}=0$ and
$(\delta|_{W})^{m}=0$ for a suitable $m>0$. The above formula implies that
$(\operatorname{ad}\delta)^{2m-1}\mu=0$, hence $\operatorname{ad}\delta$ is
locally nilpotent.

Next, assume that $\delta$ is semisimple. Then we have an eigenspace
decompositions $V=\bigoplus_{a}V_{a}$ and
$\mathcal{O}(X)=\bigoplus_{b}\mathcal{O}(X)_{b}$. As a consequence, we get

$\operatorname{\mathcal{L}}(V,\mathcal{O}(X))=\bigoplus_{a,b}\operatorname{\mathcal{L}}(V_{a},\mathcal{O}(X)_{b})$

which is an eigenspace decomposition for $\operatorname{ad}\delta$, and so
$\operatorname{ad}\delta$ is semisimple.

Finally, $\operatorname{ad}\delta_{s}$ and $\operatorname{ad}\delta_{n}$
commute, because $[\delta_{s},\delta_{n}]=0$, and thus
$\operatorname{ad}\delta=\operatorname{ad}\delta_{s}+\operatorname{ad}\delta_{n}$
is the Jordan decomposition. ∎

###  1.7. Toral subalgebras

######  Definition 1.7.1.

A finite dimensional Lie subalgebra $S\subseteq\operatorname{Vec}(X)$ is
called toral if it consists of semisimple elements. It follows that $S$ is
commutative [Hum72, 8.1, Lemma] and thus locally finite. In particular, we
obtain a weight decomposition $\mathcal{O}(X)=\bigoplus_{\alpha\in
S^{*}}\mathcal{O}(X)_{\alpha}$ where

$\mathcal{O}(X)_{\alpha}:=\\{f\in\mathcal{O}(X)\mid\tau f=\alpha(\tau)\cdot
f\text{ for }\tau\in S\\}.$

If $T\subseteq\operatorname{Aut}(X)$ is a torus, then
$\operatorname{Lie}T\subseteq\operatorname{Vec}(X)$ is a toral subalgebra, but
it is not true that every toral subalgebra is of this form. In fact, if
$T\subseteq\operatorname{Aut}(X)$ is a two dimensional torus, then the one
dimensional tori $S\subseteq T$ are the kernels of certain characters $\chi\in
X(T)\simeq{\mathbb{Z}}^{2}$ whereas every one dimensional subspace of
$\operatorname{Lie}T$ is a toral subalgebra. The best we can get is the
following result.

######  Lemma 1.7.2.

For a toral Lie subalgebra $S\subseteq\operatorname{Vec}(X)$ there exists a
uniquely defined smallest torus $T\subseteq\operatorname{Aut}(X)$ such that
$\operatorname{Lie}T\supseteq S$. Moreover, if a subspace
$M\subseteq\operatorname{Vec}(X)$ is invariant under the adjoint action of
$S$, then $M$ is invariant under $T$.

###### Proof.

(a) One easily verifies that the weight decomposition of $\mathcal{O}(X)$ has
the following properties, cf. [FK18, Proposition 7.6.1 and its proof]:

  1. $\bullet$

$\mathcal{O}(X)_{\alpha}\cdot\mathcal{O}(X)_{\beta}\subseteq\mathcal{O}(X)_{\alpha+\beta}$;

  2. $\bullet$

the set of weights $\Lambda:=\\{\gamma\mid\mathcal{O}(X)_{\gamma}\neq
0\\}\subseteq S^{*}$ generates a free subgroup
${\mathbb{Z}}\Lambda=\bigoplus_{i=1}^{k}{\mathbb{Z}}\mu_{i}$.

This defines a faithful action of the torus $T:=({\mathbb{G}}_{m})^{k}$ on
$\mathcal{O}(X)$ in the following way: For
$\alpha=\sum_{i}n_{i}\mu_{i}\in\Lambda$, $f\in\mathcal{O}(X)_{\alpha}$ and
$t=(t_{1},\ldots,t_{k})\in T$ we put

$t\,f:=t_{1}^{n_{1}}\cdots t_{k}^{n_{k}}\cdot f.$

The corresponding action of
$c=(c_{1},\ldots,c_{k})\in\operatorname{Lie}T={\mathbb{K}}^{m}$ is then given
by the vector field $\delta_{c}$ where

$\delta_{c}|_{\mathcal{O}(X)_{\alpha}}=\text{scalar multiplication with
}\textstyle\sum_{i}n_{i}c_{i}\text{ for
}\alpha=\sum_{i}n_{i}\mu_{i}\in\Lambda.$

It follows that the image of $\operatorname{Lie}T$ contains $S$ and that every
torus with this property contains $T$, see [FK18, proof of Proposition
7.6.1(2)].

(b) With respect to the adjoint action we also have a weight space
decomposition $\operatorname{Vec}(X)=\bigoplus_{\gamma\in
S^{*}}\operatorname{Vec}(X)_{\gamma}$, see Lemma 1.6.1. It has the property
that for $\delta\in\operatorname{Vec}(X)_{\gamma}$ we get
$\delta(\mathcal{O}(X)_{\alpha})\subseteq\mathcal{O}(X)_{\alpha+\gamma}$. It
follows that if $\operatorname{Vec}(X)_{\gamma}\neq 0$, then
$\gamma\in\Lambda$. Hence, the subspaces $\operatorname{Vec}(X)_{\gamma}$ are
weight spaces of the action of $T$ on $\operatorname{Vec}(X)$, and thus every
$S$-invariant subspace of $\operatorname{Vec}(X)$ is invariant under $T$. ∎

###  1.8. Integration of Lie algebras

The first part of the following result is due to Cohen and Draisma, see [CD03,
Theorem 1].

######  Theorem C.

Let $L\subseteq\operatorname{Vec}(X)$ be a Lie subalgebra. Then
$L\subseteq\operatorname{Lie}G$ for some algebraic subgroup
$G\subseteq\operatorname{Aut}(X)$ if and only if $L$ is locally finite. If $L$
is locally finite and generated by locally nilpotent elements, then $L$ is
algebraic, i.e. there is an algebraic group $G\subseteq\operatorname{Aut}(X)$
such that $L=\operatorname{Lie}G$.

The proof will be given in Section 2.6, see Theorem 2.6.1. A weaker form of
Question 2 is the following.

######  Question 3.

Let $\xi,\eta\in\operatorname{Vec}(X)$ be locally finite vector fields. If the
Lie subalgebra $L:=\langle\xi,\eta\rangle_{\text{\tiny Lie}}$ generated by
$\xi$ and $\eta$ is finite dimensional, does it follow that $L$ is locally
finite?

From Theorem C we get the following result in this direction.

######  Corollary.

Let $\\{\eta_{i}\mid i\in I\\}$ be a family of locally nilpotent vector
fields, and denote by $L:=\langle\eta_{i}\mid i\in I\rangle_{\text{\tiny
Lie}}$ the Lie subalgebra generated by the $\eta_{i}$. If $L$ is finite
dimensional, then $L$ is locally finite and algebraic.

###  1.9. A free subgroup of $\operatorname{Aut}({\mathbb{A}}^{2})$ and its
closure

In Section 3 we provide an example of a nonabelian free subgroup
$F\simeq\mathbb{F}_{2}$ of $\operatorname{Aut}({\mathbb{A}}^{2})$ generated by
two algebraic elements whose product is not algebraic. In particular, $F$ is
not locally finite. We compute its closure
$\mathfrak{F}:=\overline{F}\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$ and
describe the Lie algebra $\operatorname{Lie}\mathfrak{F}$. It occurs that
$\mathfrak{F}$ coincides with the double centralizer of $F$ (Lemma 3.1.1) and
is a free product of two abelian nested unipotent ind-subgroups $\mathfrak{J}$
and $\mathfrak{J}^{-}$ (Theorem 3.4.3(3)). Furthermore, any algebraic subgroup
$G$ of $\mathfrak{F}$ is abelian and unipotent and is conjugate to a subgroup
of $\mathfrak{J}$ or $\mathfrak{J}^{-}$ (Theorem 3.4.3(4)).

The construction shows that $\mathfrak{F}$ contains two algebraic subgroups
$U$ and $V$, both isomorphic to ${\mathbb{G}}_{a}$, with the following
properties:

$\mathfrak{F}=\overline{\langle
U,V\rangle}\quad\text{and}\quad\operatorname{Lie}\mathfrak{F}=\langle\operatorname{Lie}U,\operatorname{Lie}V\rangle_{\text{\tiny
Lie}},$

see Lemma 3.2.1. This is an instance where Question 1 has a positive answer.

Yet another instance is the following. Consider the ind-group
$\mathfrak{G}:=\operatorname{Aut}_{\mathcal{K}}(\mathcal{K}[x,y])$ where
$\mathcal{K}:={\mathbb{K}}[z]$, and let
$\mathfrak{G}^{t}\subseteq\mathfrak{G}$ be the subgroup consisting of tame
automorphisms of ${\mathbb{K}}[x,y,z]$. Then $\mathfrak{G}$ is connected and
closed in $\operatorname{Aut}_{\mathbb{K}}({\mathbb{K}}[x,y,z])$,
$\mathfrak{G}^{t}\subsetneqq\mathfrak{G}$ is closed and
$\operatorname{Lie}\mathcal{G}^{t}=\operatorname{Lie}\mathcal{G}$, see [FK18,
Theorem 17.3.1]. The group $\mathcal{G}^{t}$ is generated by the linear and de
Jonqières transformations which fix $z$. Hence $\mathcal{G}^{t}$ is
algebraically generated. By the Umirbaev-Shestakov Theorem, the one
dimensional unipotent subgroup $U$ of $\mathfrak{G}$ corresponding to the
locally nilpotent Nagata-derivation is not contained in $\mathfrak{G}^{t}$.
However, since the algebraically generated group
$\mathfrak{H}:=\langle\mathfrak{G}^{t},U\rangle$ is contained in
$\mathfrak{G}$ and
$\operatorname{Lie}U\subseteq\operatorname{Lie}\mathfrak{G}=\operatorname{Lie}\mathfrak{G}^{t}$
we have

$\operatorname{Lie}\overline{\mathfrak{H}}=L_{\mathfrak{H}}:=\langle\operatorname{Lie}\mathcal{G}^{t},\operatorname{Lie}U\rangle.$

###  1.10. Ind-groups and their Lie algebras

The theory of algebraic group actions on affine varieties can be generalized,
to some extent, to ind-groups acting on affine ind-varieties, see [FK18]. If
$\mathfrak{G}$ is an ind-group, then a $\mathfrak{G}$-variety $X$ is a variety
with a $\mathfrak{G}$-action such that the map $\mathfrak{G}\times X\to X$ is
a morphism of ind-varieties. For every ind-group $\mathfrak{G}$ the tangent
space $T_{e}\mathfrak{G}$ at the unit element $e\in\mathfrak{G}$ is well-
defined and carries a natural structure of a Lie algebra, denoted by
$\operatorname{Lie}\mathfrak{G}$. For every $\mathfrak{G}$-variety $X$ there
is a canonical homomorphism of Lie algebras
$\xi\colon\operatorname{Lie}\mathfrak{G}\to\operatorname{Vec}(X)$ constructed
as in the algebraic case. Again, $\mathfrak{G}$ acts linearly on both,
$\operatorname{Lie}\mathfrak{G}$ and $\operatorname{Vec}(X)$, and the
homomorphism $\xi\colon\operatorname{Lie}\mathfrak{G}\to\operatorname{Vec}(X)$
is $\mathfrak{G}$-equivariant.

A subset $M\subseteq\mathfrak{X}$ of an ind-variety $\mathfrak{X}$ is called
path-connected if for any two points $x,y\in M$ there is an irreducible
variety $Y$ and a morphism $\varphi\colon Y\to\mathfrak{X}$ such that
$x,y\in\varphi(Y)\subseteq\mathfrak{X}$, cf. [Ram64, p. 26]. One can always
assume that $Y$ is an irreducible curve, because any two points in an
irreducible variety are contained in an irreducible curve, cf. [FK18, Section
1.6].

In contrast to the finite-dimensional case, a strict closed subgroup of a
path-connected ind-group may have the same Lie algebra, see [FK18, Theorem
17.3.1]; cf. the example in Section 1.9. This disproves [Sha81, Theorems 1 and
2]. However, a homomorphism of a path-connected ind-group to an ind-group is
uniquely determined by the induced homomorphism of their Lie algebras, see
[FK18, Proposition 7.4.7]. In Section 4 we study surjective homomorphisms of
ind-groups and of their Lie algebras, and subgroups of finite codimension of
ind-groups.

##  2\. Algebraically generated subgroups

###  2.1. Orbits of algebraically generated subgroups

Let $\mathfrak{G}=\bigcup_{k}\mathfrak{G}_{k}$ be an ind-group. The following
concept was introduced in [AFK13].

######  Definition 2.1.1.

A subgroup $G\subseteq\mathfrak{G}$ is called algebraically generated if there
is a family $(G_{i})_{i\in I}$ of connected algebraic subgroups
$G_{i}\subseteq\mathfrak{G}$ generating $G$ as an abstract group: $G=\langle
G_{i}\mid i\in I\rangle$.

It is clear that an algebraically generated subgroup is path-connected, hence
connected. The following result can be found in [AFK13, Prop. 1.2 and 1.3].
For the reader’s convenience we provide a short argument.

######  Proposition 2.1.2.

Let $G=\langle G_{i}\mid i\in I\rangle\subseteq\operatorname{Aut}(X)$ be an
algebraically generated subgroup. For every $x\in X$ the $G$-orbit
$Gx\subseteq X$ is open in its closure $\overline{Gx}$. Moreover, there is a
finite sequence $j_{1},\ldots,j_{n}\in I$ such that $Gx=G_{j_{1}}\cdots
G_{j_{n}}x$ for all $x\in X$.

###### Proof.

(a) Choose a sequence $i_{1},\ldots,i_{m}$ such that
$\dim\overline{G_{i_{1}}\cdots G_{i_{m}}x}$ is maximal. Since
$\overline{G_{i_{1}}\cdots G_{i_{m}}x}$ is irreducible, it follows that
$G_{j_{1}}\cdots G_{j_{\ell}}x\subseteq\overline{G_{i_{1}}\cdots G_{i_{m}}x}$
for every sequence $j_{1},\ldots,j_{\ell}\in I$. Hence
$\overline{G_{i_{1}}\cdots G_{i_{m}}x}=\overline{Gx}$, and there is an open
dense set $U\subseteq\overline{Gx}$ contained in $G_{i_{1}}\cdots G_{i_{m}}x$.
This implies that $Gx=GU$ is open in $\overline{Gx}$.

(b) For the second claim we first remark that for every $x\in X$ with $\dim
Gx$ maximal there is an open set $U_{x}\subseteq X$ containing $x$ and a
sequence $\boldsymbol{i}_{x}=(i_{1},\ldots,i_{m})$, depending on $x$, such
that $\overline{G_{i_{1}}\cdots G_{i_{m}}y}=\overline{Gy}$ for all $y\in
U_{x}$. In fact, if $d:=\dim Gx$ is maximal and
$Gx\subseteq\overline{G_{i_{1}}\cdots G_{i_{m}}x}$, then
$\dim\overline{G_{i_{1}}\cdots G_{i_{m}}y}=d$ for all $y$ in an open
neighborhood of $x$. Now the union $U:=\bigcup_{x}U_{x}$ is a $G$-invariant
open set, hence covered by finitely many $U_{x_{j}}$. By joining the finitely
many sequences $\boldsymbol{i}_{x_{j}}$ corresponding to the points $x_{j}$,
we obtain a sequence $(j_{1},\ldots,j_{\ell})$ such that
$\overline{G_{j_{1}}\cdots G_{j_{\ell}}y}=\overline{Gy}$ for all $y\in U$. A
standard argument implies that $G_{j_{\ell}}G_{j_{\ell-1}}\cdots
G_{j_{2}}G_{j_{1}}G_{j_{2}}\cdots G_{j_{\ell}}y=Gy$ for all $y\in U$. Since
$X\setminus U$ is a closed $G$-invariant subvariety of smaller dimension, the
claim follows by induction. ∎

######  Remark 2.1.3.

A similar result holds for the action of a connected ind-group $\mathfrak{G}$
on an affine variety $X$, see [FK18, Proposition 7.1.2].

###  2.2. Vector fields associated to subvarieties of $\operatorname{Aut}(X)$

The construction of the embedding
$\xi\colon\operatorname{Lie}G\hookrightarrow\operatorname{Vec}(X)$ for an
algebraic subgroup $G\subseteq\operatorname{Aut}(X)$ (see 1.1) has the
following generalization, cf. [FK18, Section 3.2]. Consider the orbit map
$\mu_{x}\colon\operatorname{Aut}(X)\to X$, $g\mapsto gx$, which is an ind-
morphism, and its differential in $g\in\operatorname{Aut}(X)$:

$(d\mu_{x})_{g}\colon T_{g}\operatorname{Aut}(X)\to T_{gx}X.$

This defines, for any $v\in T_{g}\operatorname{Aut}(X)$, a vector field
$\xi^{(g)}_{v}\in\operatorname{Vec}(X)$ by

$(\xi^{(g)}_{v})(x):=(d\mu_{g^{-1}x})_{g}v\in T_{x}X,$

and thus a linear map $\xi^{(g)}\colon
T_{g}\operatorname{Aut}(X)\to\operatorname{Vec}(X)$.

######  Lemma 2.2.1 ([FK18, Proposition 3.2.4]).

The map $\xi^{(g)}\colon T_{g}\operatorname{Aut}(X)\to\operatorname{Vec}(X)$
is injective for every $g\in\operatorname{Aut}(X)$.

Let $Z$ be an affine variety and $\rho\colon Z\to\operatorname{Aut}(X)$ a
morphism. Composing the differential $d\rho\colon T_{z}Z\to
T_{\rho(z)}\operatorname{Aut}(X)$ with $\xi^{(\rho(x))}$ we obtain a linear
map $\xi^{(\rho)}=\xi^{(\rho,z)}\colon T_{z}Z\to\operatorname{Vec}(X)$. In the
following statement we use the obvious action of $\operatorname{Aut}(X)$ on
the vector fields $\operatorname{Vec}(X)$.

######  Lemma 2.2.2.

  1. (1)

For $g^{\prime},g^{\prime\prime}\in\operatorname{Aut}(X)$ set
$\tilde{\rho}\colon Z\to\operatorname{Aut}(X)$, $g\mapsto
g^{\prime}gg^{\prime\prime}$. Then
$\xi^{(\tilde{\rho})}=g^{\prime}\cdot\xi^{(\rho)}$.

  2. (2)

For a closed subvariety $Z\subseteq\operatorname{Aut}(X)$ the map
$\xi^{(z)}\colon T_{z}Z\to\operatorname{Vec}(X)$ is injective for any $z\in
Z$.

###### Proof.

(1) Replace the orbit map $\mu_{x}$ by $\tilde{\mu}_{x}:g\mapsto
g^{\prime}gg^{\prime\prime}x$, i.e.
$\tilde{\mu}_{x}=g^{\prime}\mu_{g^{\prime\prime}x}$. Then
$d\tilde{\mu}_{x}=g^{\prime}d\mu_{g^{\prime\prime}x}$. By definition,
$\xi^{(g^{\prime\prime}x)}=\xi^{(x)}$ and so
$\xi^{(\tilde{\rho})}=g^{\prime}\cdot\xi^{(\rho)}$.

(2) This follows from Lemma 2.2.1. ∎

######  Remark 2.2.3.

For an algebraic subgroup $G\subseteq\operatorname{Aut}(X)$ we obtain linear
maps $\xi^{(g)}\colon T_{g}G\to\operatorname{Vec}(X)$ for any $g\in G$. All
these maps have the same image, namely
$L:=\xi^{(e)}(\operatorname{Lie}G)\subseteq\operatorname{Vec}(X)$.

The following result is crucial and will be used at several places along the
paper.

######  Proposition 2.2.4.

Let $G_{1},\ldots,G_{m}\subseteq\operatorname{Aut}(X)$ be connected algebraic
subgroups, and consider the multiplication map

$\rho\colon G_{1}\times\cdots\times G_{m}\to\operatorname{Aut}(X).$

If $L_{i}\subseteq\operatorname{Vec}(X)$ denotes the image of
$\operatorname{Lie}G_{i}$, then the image of

$\xi^{(\rho,\mathbf{g})}=\xi^{(\rho,(g_{1},\ldots,g_{m}))}\colon
T_{g_{1}}G_{1}\oplus\cdots\oplus T_{g_{m}}G_{m}\to\operatorname{Vec}(X)$

is given by $\sum_{i}h_{i}\cdot L_{i}$ where $h_{1}=e$ and $h_{i}:=g_{1}\cdots
g_{i-1}$ for $i>1$.

###### Proof.

The linear map $\xi^{(\rho,\mathbf{g})}$ restricted to the summand
$T_{g_{i}}G_{i}$ is equal to $\xi^{(\rho_{i},g_{i})}$ where $\rho_{i}\colon
G_{i}\to\operatorname{Aut}(X)$ is the map $g\mapsto g_{1}\cdots
g_{i-1}gg_{i+1}\cdots g_{m}=h_{i}gg_{i+1}\cdots g_{m}$. Now Lemma 2.2.2(1)
shows that $\xi^{(\rho,\mathbf{g})}(T_{g_{i}}(G_{i}))=h_{i}\cdot L_{i}$, and
the claim follows. ∎

######  Corollary 2.2.5.

With the notation above consider the “orbit map” in $x\in X$,

$\rho_{x}\colon G_{1}\times\cdots\times G_{m}\to
X,\quad\mathbf{g}=(g_{1},\ldots,g_{m})\mapsto\mathbf{g}x:=g_{1}\cdots g_{m}x.$

Then the image of the differential $(d\rho_{x})_{\mathbf{g}}\colon
T_{g_{1}}G_{1}\oplus\cdots\oplus T_{g_{m}}G_{m}\to T_{\mathbf{g}x}X$ is given
by $\sum_{i}h_{i}\cdot L_{i}(g_{i}\cdots g_{m}x)$.

###### Proof.

By definition,
$(d\rho_{x})_{\mathbf{g}}=\operatorname{ev}_{\mathbf{g}x}\circ\xi^{(\rho,\mathbf{g})}$
where $\operatorname{ev}_{y}\colon\operatorname{Vec}(X)\to T_{y}X$ is the
evaluation map $\xi\mapsto\xi(y)$. Thus, by Proposition 2.2.4, the image of
$(d\rho_{x})_{\mathbf{g}}$ is given by

$\operatorname{ev}_{\mathbf{g}x}(\sum_{i}h_{i}L_{i})=\sum_{i}(h_{i}L_{i})(\mathbf{g}x)=\sum_{i}h_{i}L_{i}(g_{i}\cdots
g_{m}x).$

∎

###  2.3. Lie algebras of algebraically generated groups

As in 2.1, consider an algebraically generated subgroup $G=\langle G_{i}\mid
i\in I\rangle\subseteq\operatorname{Aut}(X)$. Set
$L_{i}:=\operatorname{Lie}G_{i}\subseteq\operatorname{Vec}(X)$, and let
$L:=\langle L_{i}\mid i\in I\rangle_{\text{\rm
Lie}}\subseteq\operatorname{Vec}(X)$ be the Lie subalgebra generated by the
$L_{i}$. If $Y\subseteq X$ is a $G$-invariant closed subset, then
$L_{i}|_{Y}\subseteq\operatorname{Vec}(Y)$ for all $i$ and so
$L|_{Y}\subseteq\operatorname{Vec}(Y)$. Choosing $Y:=\overline{Gx}$ we get
$L(x)\subseteq T_{x}Gx$ for any $x\in X$.

######  Theorem 2.3.1.

Let $G=\langle G_{i}\mid i\in I\rangle\subseteq\operatorname{Aut}(X)$ be an
algebraically generated subgroup, and define
$L:=\langle\operatorname{Lie}G_{i}\mid i\in I\rangle_{\text{\rm
Lie}}\subseteq\operatorname{Vec}(X)$ to be the Lie algebra generated by the
$\operatorname{Lie}G_{i}$. Then

  1. (1)

$L$ is stable under the action of $G$ and of
$\overline{G}\subseteq\operatorname{Aut}(X)$. In particular, $L$ is an ideal
in $\operatorname{Lie}\overline{G}$.

  2. (2)

We have $T_{x}Gx=L(x)$ for all $x\in X$. In particular, any vector field in
$L$ is tangent to the $G$-orbits.

  3. (3)

If $Z\subseteq\operatorname{Aut}(X)$ is a closed algebraic subset contained in
$G$, then the image of $T_{z}Z$ in $\operatorname{Vec}(X)$ is contained in $L$
for any $z\in Z$.

  4. (4)

If $H\subseteq\operatorname{Aut}(X)$ is an algebraic subgroup contained in
$G$, then $\operatorname{Lie}H\subseteq L$. In particular, $L$ depends only on
$G$ and not on the generating subgroups $G_{i}$.

###### Proof of Theorem 2.3.1.

(1) It suffices to show that $L$ is invariant under all the $G_{i}$. Since
$G_{i}$ is connected, we are reduced to prove that $L$ is invariant under the
action of $\operatorname{Lie}G_{i}$, i.e. under the adjoint action of
$L_{i}:=\operatorname{Lie}G_{i}$ (see [FK18, Proposition 6.3.4]) which holds
by construction.

(2) We already know that $L(x)\subseteq T_{x}Gx$. Now we choose a surjective
“orbit map”

$\mu_{x}\colon G_{j_{1}}\times\cdots\times G_{j_{n}}\to Gx\subseteq X,$

see Proposition 2.1.2. It follows that for a suitable
$\mathbf{g}=(g_{1},\ldots,g_{n})$ the differential

$(d\mu_{x})_{\mathbf{g}}\colon T_{g_{1}}G_{j_{1}}\oplus\cdots\oplus
T_{g_{n}}G_{j_{n}}\to T_{\mathbf{g}x}Gx$

is also surjective. Thus we get
$T_{\mathbf{g}x}Gx=\sum_{j}h_{j}L_{j}(g_{j}\cdots g_{n}x)$, by Corollary
2.2.5. Since $L$ is invariant under the action of $G$, we finally get

$h_{j}L_{j}(g_{j}\cdots g_{n}x)\subseteq h_{j}L(g_{j}\cdots
g_{n}x)=L(\mathbf{g}x),$

and the claim follows.

(3) Let $Z\subseteq\operatorname{Aut}(X)$ be a closed algebraic subset
contained in $G$. We first claim that $Z\subseteq G_{i_{1}}\cdots G_{i_{m}}$
for a suitable sequence $(i_{1},\ldots,i_{m})$. In fact, there is a sequence
such that the intersection $Z\cap G_{i_{1}}\cdots G_{i_{m}}$ is dense in $Z$,
hence contains a dense open set $U\subseteq Z$. Since $\dim(Z\setminus U)<\dim
Z$ the claim follows by induction on $\dim Z$.

Applying Proposition 2.2.4 to the multiplication map $\rho\colon
G_{i_{1}}\times\cdots\times G_{i_{m}}\to\operatorname{Aut}(X)$ and using
$\xi^{(\rho,\mathbf{g})}$ where $\mathbf{g}=(g_{1},\ldots,g_{m})$ and
$g_{1}\cdots g_{m}=z$ it follows that the image of $T_{z}Z$ in
$\operatorname{Vec}(X)$ is contained in $\sum_{i}h_{i}L_{i}\subseteq L$, since
$L$ is stable under the action of $G$, by (1).

(4) This is an immediate consequence of (3). ∎

###  2.4. Ramanujam’s alternative

The following result, due to Ramanujam, will be used in the proof of our
Theorem 2.5.1 below.

######  Proposition 2.4.1 ([Ram64, Theorem on p. 26]).

Let $G\subseteq\operatorname{Aut}(X)$ be a path-connected subgroup. Then one
of the following holds:

  1. (i)

$G$ is an algebraic subgroup;

  2. (ii)

$G$ contains algebraic subsets of arbitrary large dimension.

###### Proof.

Assume that we are not in case (ii), and let $Y\subseteq G$ be an irreducible
algebraic subset of maximal dimension $d$. We may assume that $e\in Y$. We
claim that $G\subseteq\overline{Y}=\overline{G}$. In fact, for any $g\in G$ we
can find an irreducible algebraic subset $S\subseteq G$ containing $e$ and
$g$. Then $\overline{S\cdot Y}$ is irreducible and meets $G$ in a dense open
set, hence $\overline{S\cdot Y}=\overline{Y}$. Since $g\in S\cdot Y$, the
claim follows.

As a consequence, $\overline{G}\subseteq\operatorname{Aut}(X)$ is an algebraic
subgroup. Since $Y\subseteq G$ is open and dense in $\overline{G}$ it follows
that $\overline{G}=Y^{-1}\cdot Y\subseteq G$, and so (i) holds. ∎

###  2.5. The finite dimensional case

Assume that $G=\langle G_{i}\mid i\in I\rangle$ is finite dimensional, that
is, an algebraic subgroup of $\operatorname{Aut}(X)$, see Proposition 2.4.1.
In this case, $L$ is finite dimensional and equal to $\operatorname{Lie}G$, by
Theorem 2.3.1(4). The next result shows that the converse is true as well. It
also proves Theorem A from Section 1.3 of the introduction.

######  Theorem 2.5.1.

Let $G=\langle G_{i}\mid i\in I\rangle\subseteq\operatorname{Aut}(X)$ be an
algebraically generated subgroup. Assume that the Lie algebra
$L\subseteq\operatorname{Vec}(X)$ generated by the $\operatorname{Lie}G_{i}$
is finite dimensional. Then $G$ is an algebraic subgroup with
$\operatorname{Lie}G=L$. In particular, $L$ is locally finite and algebraic.

###### Proof.

Let $Y\subseteq G$ be an irreducible algebraic subset. As in the proof of
Theorem 2.3.1(3) there is a sequence $(i_{1},\ldots,i_{m})$ such that
$G_{i_{1}}\cdots G_{i_{m}}$ contains $Y$. Thus the image of $T_{y}Y$ in
$\operatorname{Vec}(X)$ is contained in $L$. Since the map $\xi^{(g)}\colon
T_{g}\operatorname{Aut}(X)\to\operatorname{Vec}(X)$ is injective (Lemma 2.2.1)
we get $\dim Y\leq\dim L$. Now the claim follows from Ramanujam’ alternative
(Proposition 2.4.1). ∎

###  2.6. Integration of locally finite Lie algebras

The next result is our Theorem C from the introduction, see Section 1.8.

######  Theorem 2.6.1.

Let $L\subseteq\operatorname{Vec}(X)$ be a Lie subalgebra.

  1. (1)

If $L$ is finite dimensional and generated by locally nilpotent vector fields,
then $L$ is algebraic, i.e. there is a connected algebraic subgroup
$G\subseteq\operatorname{Aut}(X)$ such that $L=\operatorname{Lie}G$.

  2. (2)

If $L$ is locally finite, then there exists a connected algebraic subgroup
$G\subseteq\operatorname{Aut}(X)$ such that $L\subseteq\operatorname{Lie}G$.

###### Proof of (1).

Assume that $L$ is finite dimensional and generated by locally nilpotent
vector fields $\eta_{1},\eta_{2},\ldots,\eta_{m}$. By [FK18, Proposition
7.6.1] there are (uniquely determined) algebraic subgroups
$U_{i}\subseteq\operatorname{Aut}(X)$, $U_{i}\simeq{\mathbb{G}}_{a}$, such
that $\operatorname{Lie}U_{i}={\mathbb{K}}\eta_{i}$. Now Theorem 2.5.1 implies
that $G:=\langle U_{i}\mid i=1,\ldots,m\rangle\subseteq\operatorname{Aut}(X)$
is an algebraic group with $\operatorname{Lie}G=L$. ∎

######  Example 2.6.2.

Assume that $L\subseteq\operatorname{Vec}(X)$ is a locally finite semisimple
Lie subalgebra. Then there is a semisimple group
$G\subseteq\operatorname{Aut}(X)$ such that $L=\xi(\operatorname{Lie}G)$. In
particular, $L$ is algebraic.

In fact, the semisimple Lie algebra $L$ is generated by the root subspaces
with respect to some Cartan subalgebra (see [Hum72, Proposition 8.4(f)]),
hence by nilpotent elements. Lemma 1.5.3(3) shows that these elements are
locally nilpotent in $\operatorname{Vec}(X)$, and thus $L$ is generated by
locally nilpotent vector fields.

The proof of (2) needs some preparation. Let $L=L_{0}\ltimes R$ be the Levi
decomposition where $L_{0}$ is semisimple and $R:=\operatorname{rad}L$ is the
solvable radical (see [Pro07, 10.1.6, pp. 305–306]. The example above shows
that $L_{0}=\operatorname{Lie}G_{0}$ for a semisimple subgroup
$G_{0}\subseteq\operatorname{Aut}(X)$. It remains to find a solvable subgroup
$Q\subseteq\operatorname{Aut}(X)$ normalized by $G_{0}$ such that
$\operatorname{Lie}Q\supseteq R$. Then the product $G:=G_{0}Q$ has the
required property.

######  Lemma 2.6.3.

Let $M\subseteq\operatorname{Vec}(X)$ be a locally finite solvable Lie
subalgebra. Assume that $M$ contains with every element $\mu$ its semisimple
and locally nilpotent parts $\mu_{s}$ and $\mu_{n}$. Then the locally
nilpotent elements from $M$ form a Lie subalgebra $M_{n}$, and $M=S\ltimes
M_{n}$ where $S$ is a maximal toral subalgebra of $M$.

###### Proof.

Since $M$ is locally finite we may assume that
$M\subseteq\operatorname{\mathfrak{g}l}(V)$ where $V\subseteq\mathcal{O}(X)$
is a finite dimensional subspace generating $\mathcal{O}(X)$. Since $M$ is
solvable there is a basis of $V$ such that $M$ is contained in the upper
triangular matrices
$\mathfrak{b}_{d}\subseteq\operatorname{\mathfrak{g}l}_{d}$ where $d:=\dim V$
(Lie’s Theorem). Clearly, the nilpotent elements from $M$ belong to the
subalgebra $\mathfrak{n}_{d}\subseteq\mathfrak{b}_{d}$ of upper triangular
matrices with zeros along the diagonal, and so $M_{n}:=M\cap\mathfrak{n}_{d}$
is the ideal of $M$ consisting of the nilpotent elements. By Lemma 1.5.3 every
nilpotent element of $M$ is also locally nilpotent viewed as a vector field on
$X$, and vice versa. If $M=M_{n}$ then $S=0$ and $M=S\ltimes M_{n}$, as
desired.

Assume now that $M$ contains a semisimple element, and choose a maximal toral
subalgebra $S\subseteq M$. Now consider the corresponding weight space
decomposition $M=\bigoplus_{\alpha\in S^{*}}M_{\alpha}$ where

$M_{\alpha}:=\\{\mu\in
M\mid\operatorname{ad}\tau(\mu)=[\tau,\mu]=\alpha(\tau)\cdot\mu\text{ for
}\tau\in S\\}.$

Similarly, we get a decomposition
$\operatorname{\mathfrak{g}l}_{d}=\bigoplus_{\alpha\in
S^{*}}(\operatorname{\mathfrak{g}l}_{d})_{\alpha}$. Using the identity
$[s,a\cdot b]=[s,a]\cdot b+a\cdot[s,b]$ we get
$(\operatorname{\mathfrak{g}l}_{d})_{\alpha}\cdot(\operatorname{\mathfrak{g}l}_{d})_{\beta}\subseteq(\operatorname{\mathfrak{g}l}_{d})_{\alpha+\beta}$
which implies that $M_{\alpha}$ consists of nilpotent elements for $\alpha\neq
0$. Hence, $\bigoplus_{\alpha\neq 0}M_{\alpha}\subseteq\mathfrak{n}_{d}$ is a
solvable subalgebra consisting of nilpotent elements, while
$M_{0}=\operatorname{Cent}_{M}(S)\supseteq S$.

Let $\mu\in M_{0}$ have Jordan decomposition $\mu=\mu_{s}+\mu_{n}$. Since $S$
commutes with $\mu$ it also commutes with $\mu_{s}$ and $\mu_{n}$, and so
$\mu_{s}\in M_{0}$ and $\mu_{n}\in M_{0}\cap\mathfrak{n}_{d}$. Since $S$ is a
maximal toral subalgebra, we have $\mu_{s}\in S$, and so
$M_{0}=S\oplus(M_{0}\cap\mathfrak{n}_{d})$. As a consequence we see that
$M_{n}=(M_{0}\cap\mathfrak{n}_{d})\oplus\bigoplus_{\alpha\neq 0}M_{\alpha}$ is
a subalgebra consisting of nilpotent elements, and that $M=S\ltimes M_{n}$. ∎

The assumption in the lemma above leads to the following definition. A locally
finite subspace $W\subseteq\operatorname{Vec}(X)$ is called Jordan-saturated
(shortly $J$-saturated) if it contains with every element $\eta$ the
semisimple and the locally nilpotent parts $\eta_{s}$ and $\eta_{n}$. Clearly,
for any algebraic subgroup $G\subseteq\operatorname{Aut}(X)$ the Lie algebra
$\operatorname{Lie}G\subseteq\operatorname{Vec}(X)$ is $J$-saturated.

######  Lemma 2.6.4.

Let $R\subseteq\operatorname{Vec}(X)$ be a locally finite and solvable Lie
subalgebra. Then the smallest $J$-saturated Lie subalgebra
$\tilde{R}\subseteq\operatorname{Vec}(X)$ containing $R$ has the following
properties:

  1. (1)

$\tilde{R}$ is locally finite and solvable.

  2. (2)

If a subspace $V\subseteq\mathcal{O}(X)$ is $R$-invariant, then it is
invariant under $\tilde{R}$.

  3. (3)

If an automorphisms $\varphi\in\operatorname{Aut}(X)$ normalizes $R$, then it
normalizes $\tilde{R}$.

Notice that any algebraic Lie subalgebra of $\operatorname{Vec}(X)$ which
contains $R$ also contains $\tilde{R}$. Hence $\tilde{R}$ is the smallest
algebraic Lie subalgebra which contains $R$.

###### Proof.

As before, we choose a finite dimensional $R$-invariant subspace
$V\subseteq\mathcal{O}(X)$ which generates $\mathcal{O}(X)$. Choosing a
suitable basis of $V$ we may assume that the image of $R$ in
$\operatorname{\mathfrak{g}l}(V)=\operatorname{\mathfrak{g}l}_{d}$ is
contained in the upper triangular matrices $\mathfrak{b}_{d}$.

If $\eta=\eta_{s}+\eta_{n}$ is the Jordan-decomposition of an element $\eta\in
R$, then every finite dimensional $R$-invariant subspace
$V\subseteq\mathcal{O}(X)$ is invariant under $\eta_{s}$ and $\eta_{n}$, hence
invariant under the Lie subalgebra $R_{1}$ generated by
$R+{\mathbb{K}}\eta_{n}$. Thus $R_{1}$ is locally finite. The Jordan-
decomposition of $\eta$ carries over to the image of $\eta$ in
$\operatorname{\mathfrak{g}l}(V)=\operatorname{\mathfrak{g}l}_{d}$. Hence the
images of $\eta_{s}$ and $\eta_{n}$ also belong to $\mathfrak{b}_{d}$ which
shows that the image of $R_{1}$ is contained in $\mathfrak{b}_{d}$.

This procedure of adding the locally nilpotent part and forming the Lie
algebra has to end with a locally finite and solvable Lie subalgebra
$\tilde{R}$ containing $R$ with the properties (1) and (2).

Finally, assume that $\varphi\in\operatorname{Aut}(X)$ stabilizes $R$, and let
$\eta\in R$. Then $\varphi(\eta)=\varphi(\eta_{s})+\varphi(\eta_{n})$ is the
Jordan decomposition of $\varphi(\eta)\in R$, hence $\varphi(\eta_{s})$ and
$\varphi(\eta_{n})$ belong to $\tilde{R}$. Now it follows from the
construction that $\tilde{R}$ is invariant under $\varphi$. ∎

###### Proof of Theorem 2.6.1(2).

Let $L=L_{0}\ltimes R$ be the Levi decomposition where $L_{0}$ is semisimple
and $R$ is the solvable radical of $L$. By part (1) and Example 2.6.2 there is
a semisimple group $G_{0}\subseteq\operatorname{Aut}(X)$ with
$\operatorname{Lie}G_{0}=L_{0}$. Clearly, $R$ is $G_{0}$-invariant, as well as
$\tilde{R}$, the $J$-saturation of $R$, see Lemma 2.6.4. By Lemma 2.6.3 we
have $\tilde{R}=S\ltimes\tilde{R}_{n}$ where $S$ is toral and $\tilde{R}_{n}$
is solvable and consists of locally nilpotent elements. Thus, again by part
(1), there is a unipotent group $U\subseteq\operatorname{Aut}(X)$ such that
$\operatorname{Lie}U=\tilde{R}_{n}$.

It follows from Lemma 1.7.2 that there is a well-defined minimal torus
$T\subseteq\operatorname{Aut}(X)$ such that $\operatorname{Lie}T\supseteq S$,
and that $T$ normalizes $R_{n}$. By Lemmas 1.7.2 and 2.6.4, $T$ normalizes $U$
and $\tilde{R}_{n}$, and so the product
$Q:=TU=UT\subseteq\operatorname{Aut}(X)$ is a connected solvable subgroup with
$\operatorname{Lie}Q\supseteq\tilde{R}$. By construction, $Q$ is the smallest
subgroup with this property. It remains to show that $G_{0}$ normalizes $Q$.
In fact, $gQg^{-1}$ is a closed subgroup for any $g\in G_{0}$, and
$\operatorname{Lie}gQg^{-1}\supseteq g\tilde{R}g^{-1}=\tilde{R}$. ∎

###  2.7. Solvable subgroups

This section is devoted to the proof of our Theorem B.

######  Theorem 2.7.1.

Let $G\subseteq\operatorname{Aut}(X)$ be a subgroup generated by a finite
collection of unipotent algebraic subgroups. If $G$ is solvable, then $G$ is a
unipotent algebraic group.

The proof needs some preparation.

######  Lemma 2.7.2.

Let $G\subseteq\mathfrak{G}$ be a subgroup of an ind-group $\mathfrak{G}$.

  1. (1)

We have $(\overline{G},\overline{G})\subseteq\overline{(G,G)}$.

  2. (2)

If $G$ is solvable, then so is $\overline{G}$.

###### Proof.

(1) Let $\gamma\colon\mathfrak{G}\times\mathfrak{G}\to\mathfrak{G}$ be the
commutator map, $(g,h)\mapsto ghg^{-1}h^{-1}$. Then
$\gamma(\overline{G}\times\overline{G})\subseteq\overline{\gamma(G\times G)}$,
and so $(\overline{G},\overline{G})\subseteq\langle\overline{\gamma(G\times
G)}\rangle\subseteq\overline{(G,G)}$.

(2) From the derived series

$G^{(0)}:=G\supseteq G^{(1)}:=(G,G)\supseteq
G^{(2)}:=(G^{(1)},G^{(1)})\supseteq\cdots\supseteq G^{(n)}=\\{e\\}$

we find the series of closed subgroups of $\mathfrak{G}$:

$\overline{G}\supseteq\overline{G^{(1)}}=\overline{(G,G)}\supseteq\overline{G^{(2)}}=\overline{(G^{(1)},G^{(1)})}\supseteq\cdots\supseteq\overline{G^{(n)}}=\\{e\\}.$

It follows from (1) that the factor groups
$\overline{G^{(i)}}/\overline{G^{(i+1)}}$ are commutative, and the claim
follows. ∎

######  Lemma 2.7.3.

Let $\mathfrak{G}$ be an ind-group. Then
$\operatorname{Lie}\overline{(\mathfrak{G},\mathfrak{G})}\supseteq[\operatorname{Lie}\mathfrak{G},\operatorname{Lie}\mathfrak{G}]$.
In particular, if $\mathfrak{G}$ is solvable, then so is
$\operatorname{Lie}\mathfrak{G}$.

###### Proof.

(Cf. [FK18, Section 7.5])

(a) For a fixed $h\in\mathfrak{G}$ consider the morphism
$\gamma_{h}\colon\mathfrak{G}\to\mathfrak{G}$,
$\gamma_{h}(g):=hgh^{-1}g^{-1}$. Then the differential
$d\gamma_{h}\colon\operatorname{Lie}\mathfrak{G}\to\operatorname{Lie}\mathfrak{G}$
is given by $d\gamma_{h}:=\operatorname{Ad}(h)-\operatorname{id}$. Moreover,
the image of $d\gamma_{h}$ is contained in
$\operatorname{Lie}\overline{(\mathfrak{G},\mathfrak{G})}$.

Now fix $A\in\operatorname{Lie}\mathfrak{G}$ and consider the morphism
$\alpha_{A}\colon\mathfrak{G}\to\operatorname{Lie}\mathfrak{G}$,
$\alpha_{A}(h):=\operatorname{Ad}(h)A-A$. The differential of this map is
$-\operatorname{ad}(A)$. By (a), the image of $\alpha_{A}$ and hence of its
differential $d\alpha_{A}$ belongs to
$\operatorname{Lie}\overline{(\mathfrak{G},\mathfrak{G})}$ which implies that
$\operatorname{Lie}\overline{(\mathfrak{G},\mathfrak{G})}$ contains
$[\operatorname{Lie}\mathfrak{G},\operatorname{Lie}\mathfrak{G}]$.

(b) Now assume that $\mathfrak{G}$ is solvable. From the derived series for
$\mathfrak{G}$ we get the series

$\mathfrak{G}^{(0)}=\mathfrak{G}\supseteq\overline{\mathfrak{G}^{(1)}}\supseteq\overline{\mathfrak{G}^{(2)}}\supseteq\cdots\supseteq\overline{\mathfrak{G}^{(n)}}=\\{e\\}.$

of closed subgroups with the property that the factor groups are all
commutative (see Lemma 2.7.2 above and its proof). From (a) we get

$\operatorname{Lie}\overline{\mathfrak{G}^{(i+1)}}=\operatorname{Lie}\overline{(\mathfrak{G}^{(i)},\mathfrak{G}^{(i)})}=\operatorname{Lie}\overline{(\overline{\mathfrak{G}^{(i)}},\overline{\mathfrak{G}^{(i)}})}\supseteq[\operatorname{Lie}\overline{\mathfrak{G}^{(i)}},\operatorname{Lie}\overline{\mathfrak{G}^{(i)}}].$

Thus the Lie algebra
$\operatorname{Lie}\overline{\mathfrak{G}^{(i)}}/\operatorname{Lie}\overline{\mathfrak{G}^{(i+1)}}$
is commutative which implies that $\operatorname{Lie}\mathfrak{G}$ is
solvable. ∎

######  Lemma 2.7.4.

Let $L$ be a finitely generated Lie algebra. If $L$ is solvable, then $L$ is
finite dimensional.

###### Proof.

In the derived series

$L^{(0)}:=L\supseteq L^{(1)}:=[L^{(0)},L^{(0)}]\supseteq
L^{(2)}:=[L^{(1)},L^{(1)}]\supseteq\cdots\supseteq L^{(n)}=\\{0\\}$

every member $L^{(i)}$ is finitely generated. Therefore $L^{(i)}/L^{(i+1)}$ is
a commutative and finitely generated Lie algebra, hence finite dimensional,
and the claim follows. ∎

###### Proof of Theorem 2.7.1.

We can assume that $G$ is generated by one dimensional unipotent subgroups
$U_{i}\subseteq\operatorname{Aut}(X)$, $i=1,\ldots,m$. Then the image of
$\operatorname{Lie}U_{i}$ in $\operatorname{Vec}(X)$ is of the form
${\mathbb{K}}\eta_{i}$ with a locally nilpotent vector field $\eta_{i}$.
Denote by $L\subseteq\operatorname{Vec}(X)$ the Lie algebra generated by the
$\eta_{i}$, $i=1,\ldots,m$. Since the $U_{i}\subseteq G\subseteq\overline{G}$
we see that $L\subseteq\operatorname{Lie}\overline{G}$.

By Lemma 2.7.2 the closure $\overline{G}$ is solvable, hence
$\operatorname{Lie}\overline{G}$ is solvable, by Lemma 2.7.3. As a
consequence, $L$ is solvable, hence finite dimensional by Lemma 2.7.4. Now
Theorem 2.6.1(1) implies that $G$ is a connected algebraic group with
$\operatorname{Lie}G=L$. In particular, $L$ is locally finite. Since $L$ is
generated by locally nilpotent elements, it consists of locally nilpotent
elements (cf. first part of the proof of Lemma 2.6.3). Thus $G$ is a unipotent
algebraic group. ∎

###  2.8. Nested Lie algebras

For the following concept see [FK18, Section 9.4] and [KPZ17].

######  Definition 2.8.1.

A subgroup $G$ of an ind-group $\mathfrak{G}$ is called nested if there exists
a sequence $G_{1}\subseteq G_{2}\subseteq\cdots\subseteq\mathfrak{G}$ of
algebraic subgroups of $\mathfrak{G}$ such that $G=\bigcup_{i}G_{i}$.

A Lie algebra $L\subseteq\operatorname{Vec}(X)$ is called nested if there
exists a sequence $L_{1}\subseteq L_{2}\subseteq\cdots\subseteq L$ of finite
dimensional Lie algebras $L_{i}$ such that $L=\bigcup_{i}L_{i}$.

Note that any two filtrations of $G$ by algebraic subgroups are compatible. In
fact, if $G=\bigcup_{j}H_{j}$ is another filtration by algebraic subgroups,
then $G_{i}=\bigcup_{j}(G_{i}\cap H_{j})$ is an ascending union of closed
subgroups, hence $G_{i}=G_{i}\cap H_{j}$ for some $j$ and so $G_{i}\subseteq
H_{j}$.

This implies that a nested subgroup $G=\bigcup G_{i}\subseteq\mathfrak{G}$ has
a natural Lie algebra defined by
$\operatorname{Lie}G=\bigcup_{i}\operatorname{Lie}G_{i}$ which is nested and
does not depend on the filtration of $G$ by algebraic subgroups.

A nested subgroup $G=\bigcup_{i}G_{i}\subseteq\mathfrak{G}$ has a natural
structure of an ind-group, and the inclusion $G\hookrightarrow\mathfrak{G}$ is
a homomorphism of ind-groups. However, this is not necessarily a closed
immersion as the following example shows.

######  Example 2.8.2.

The subgroup $\mu_{2}\subseteq{\mathbb{G}}_{m}$ of elements of order a power
of $2$ is nested by the finite subgroups
$\simeq{\mathbb{Z}}/2^{n}{\mathbb{Z}}$. But it is not closed.

######  Lemma 2.8.3.

Assume that the nested subgroup $G\subseteq\mathfrak{G}$ is path-connected.
Then $G$ admits an ascending filtration by connected algebraic subgroups.

###### Proof.

If $G=\bigcup_{i}G_{i}$ is path connected, then $G=\bigcup_{i}G_{i}^{\circ}$.
∎

######  Question 4.

Let $G\subseteq\operatorname{Aut}(X)$ be a path-connected nested subgroup.
Does it follow that $G$ is closed?

In order to give a positive answer to this question one has to show that for
every algebraic subset $A\subseteq\operatorname{Aut}(X)$ the intersection
$A\cap G$ is contained in an algebraic group.

Clearly, if $G\subseteq\operatorname{Aut}(X)$ is a nested subgroup, then
$\operatorname{Lie}G\subseteq\operatorname{Vec}(X)$ is a nested Lie algebra
filtered by locally finite subalgebras. The converse of this is also true.

######  Proposition 2.8.4.

Let $L=\bigcup_{i}L_{i}\subseteq\operatorname{Vec}(X)$ be a nested Lie
subalgebra where all $L_{i}$ are locally finite. Then there exists a nested
subgroup $G\subseteq\operatorname{Aut}(X)$ such that
$\operatorname{Lie}G\subseteq\operatorname{Vec}(X)$ contains $L$. Moreover,
every closed ind-subgroup $\mathfrak{G}\subseteq\operatorname{Aut}(X)$ such
that $\operatorname{Lie}\mathfrak{G}\supseteq L$ contains $G$.

###### Proof.

Let $G_{i}\subseteq\operatorname{Aut}(X)$ be the minimal algebraic subgroups
with $\operatorname{Lie}G_{i}\supseteq L_{i}$ (Theorem 2.6.1). Then
$G_{i}\subseteq G_{i+1}$ for all $i$ and so $G:=\bigcup_{i}G_{i}$ is a nested
subgroup with the required properties.

Assume that $\operatorname{Lie}\mathfrak{G}\supseteq L$ and define
$G_{i}^{\prime}:=\mathfrak{G}\cap G_{i}$. Then
$\operatorname{Lie}G_{i}^{\prime}=\operatorname{Lie}\mathfrak{G}\cap\operatorname{Lie}G_{i}\supseteq
L_{i}$, hence $G_{i}^{\prime}=G_{i}$ and the claim follows. ∎

##  3\. An interesting example

In this section we study the subgroup
$F\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$ generated by
$\mathbf{u}:=(x+y^{2},y)$ and $\mathbf{v}:=(x,y+x^{2})$. We will show that $F$
is a free group in two generators and we will describe the closure
$\mathfrak{F}:=\overline{F}$ and its Lie algebra
$\operatorname{Lie}\mathfrak{F}$. It turns out that the following holds (see
Theorem 3.4.3).

  1. $\bullet$

$\mathfrak{F}$ is a free product of two nested abelian closed unipotent ind-
subgroups $\mathfrak{J}$ and
$\mathfrak{J}^{-}\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$. In
particular, $\mathfrak{F}$ is torsion free.

  2. $\bullet$

$\operatorname{Lie}\mathfrak{F}$ is generated, as a Lie algebra, by
$\operatorname{Lie}U$ and $\operatorname{Lie}V$ where
$U:=\overline{\langle\mathbf{u}\rangle}\simeq{\mathbb{G}}_{a}$ and
$V:=\overline{\langle\mathbf{v}\rangle}\simeq{\mathbb{G}}_{a}$.

  3. $\bullet$

Any algebraic subgroup of $\mathfrak{F}$ is abelian and unipotent and is
conjugate to a subgroup of $\mathfrak{J}$ or $\mathfrak{J}^{-}$.

###  3.1. Notation and first results

Let

$T:=\\{(ax,by)\mid
a,b\in{\mathbb{K}}^{*}\\}\subseteq\operatorname{GL}_{2}({\mathbb{K}})\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$

denote the standard 2-dimensional torus, and let

$\operatorname{SAut}({\mathbb{A}}^{2}):=\\{\mathbf{g}\in\operatorname{Aut}(X)\mid\operatorname{jac}(\mathbf{g})=1\\}$

where the Jacobian of $\mathbf{g}=(f,h)$ is defined in the usual way:
$\operatorname{jac}(\mathbf{g})=\det\left[\begin{smallmatrix}\frac{\partial
f}{\partial x}&\frac{\partial f}{\partial y}\\\ \frac{\partial h}{\partial
x}&\frac{\partial h}{\partial y}\end{smallmatrix}\right]$. Furthermore, we set

$\displaystyle\operatorname{Aut}_{0}({\mathbb{A}}^{2}):=\\{\mathbf{g}\in\operatorname{Aut}({\mathbb{A}}^{2})\mid\mathbf{g}(0)=0\\},$
$\displaystyle\mathfrak{A}:=\\{\mathbf{g}\in\operatorname{Aut}_{0}({\mathbb{A}}^{2})\mid
d\mathbf{g}_{0}=\operatorname{id}\\}.$

Note that $\mathbf{g}\in\operatorname{Aut}({\mathbb{A}}^{2})$ belongs to
$\mathfrak{A}$ if and only if the linear part of $\mathbf{g}=(f,h)$ is equal
to $(x,y)$. $\mathfrak{A}$ is a closed path-connected ind-subgroup of
$\operatorname{Aut}({\mathbb{A}}^{2})$ which contains $\mathbf{u}=(x+y^{2},y)$
and $\mathbf{v}=(x,y+x^{2})$, hence $\mathfrak{F}\subseteq\mathfrak{A}$.

Let $S:=\\{(\zeta
x,\zeta^{2}y)\mid\zeta^{3}=1\\}\subseteq\operatorname{SL_{2}}\subseteq\operatorname{SAut}({\mathbb{A}}^{2})$
be the cyclic diagonal subgroup of order 3. The elements of $S$ commute with
$\mathbf{u}$ and $\mathbf{v}$ and thus with all elements from $F$ and hence
from $\mathfrak{F}=\overline{F}$.

######  Lemma 3.1.1.

One has $S=\operatorname{Cent}_{\operatorname{Aut}({\mathbb{A}}^{2})}(F)$.

###### Proof.

Note that any $\mathbf{g}\in\operatorname{Aut}({\mathbb{A}}^{2})$ commuting
with $\mathbf{u}$ and $\mathbf{v}$, and therefore with $U$ and $V$, fixes the
origin which is the unique common fixed point of $U$ and $V$. It also leaves
invariant the set of orbits of $U$ and of $V$, i.e. the pencils of horizontal
and vertical lines. Hence, $\mathbf{g}\in T$, and an easy calculation shows
that, in fact, $\mathbf{g}\in S$. ∎

Considering the action of $S$ on $\operatorname{Aut}({\mathbb{A}}^{2})$ by
conjugation, one sees that $\operatorname{SAut}({\mathbb{A}}^{2})$ and
$\mathfrak{A}$ are stable under this action, hence

$\mathfrak{F}\subseteq\mathfrak{A}^{S}:=\\{\mathbf{g}\in\mathfrak{A}\mid\mathbf{g}\circ
s=s\circ\mathbf{g}\text{ for all }s\in S\\}.$

Moreover, $S$ also acts on the Lie algebras
$\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{2})$ and
$\operatorname{Lie}\mathfrak{A}$, and so

(1)
$\operatorname{Lie}\mathfrak{F}\subseteq\operatorname{Lie}\mathfrak{A}^{S}\subseteq(\operatorname{Lie}\mathfrak{A})^{S}.$

###  3.2. The Lie algebra of $\mathfrak{F}$

For the Lie algebras of $\operatorname{Aut}({\mathbb{A}}^{2})$ and
$\operatorname{SAut}({\mathbb{A}}^{2})$ we have the following description, see
[FK18, Lemma 15.7.1]:

  1. $\bullet$

$\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{2})\xrightarrow{\sim}\operatorname{Vec}^{c}({\mathbb{A}}^{2}):=\\{\delta\in\operatorname{Vec}({\mathbb{A}}^{2})\mid\operatorname{div}\delta\in{\mathbb{K}}\\}$,

  2. $\bullet$

$\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{2})\xrightarrow{\sim}\operatorname{Vec}^{0}({\mathbb{A}}^{2}):=\\{\delta\in\operatorname{Vec}({\mathbb{A}}^{2})\mid\operatorname{div}\delta=0\\}$,

  3. $\bullet$

$\operatorname{Lie}\mathfrak{A}\xrightarrow{\sim}\operatorname{Vec}^{0}_{0}({\mathbb{A}}^{2}):=\\{\delta\in\operatorname{Vec}^{0}({\mathbb{A}}^{2})\mid\delta(0)=0\\}$

where the divergence of a vector field $\delta=f\frac{\partial}{\partial
x}+h\frac{\partial}{\partial y}$ is defined as
$\operatorname{div}\delta=\frac{\partial f}{\partial x}+\frac{\partial
h}{\partial y}$.

Now consider the following vector fields from
$\operatorname{Vec}^{0}({\mathbb{A}}^{2})$,

$\partial_{i,j}:=(j+1)x^{i+1}y^{j}\frac{\partial}{\partial
x}-(i+1)x^{i}y^{j+1}\frac{\partial}{\partial y}$

where

$(i,j)\in\Lambda:=\\{(k,l)\in{\mathbb{Z}}^{2}\,\mid
k,l\geq-1,(k,l)\neq(-1,-1)\\}.$

The torus $T$ acts by conjugation on $\operatorname{Aut}({\mathbb{A}}^{2})$
and on $\operatorname{Vec}({\mathbb{A}}^{2})$, and the $\partial_{i,j}$ are
eigenvectors of weight $(i,j)$. Moreover, the action of $T$ on
$\operatorname{Vec}^{0}({\mathbb{A}}^{2})$ is multiplicity-free, and we have
the weight decompositions

$\operatorname{Vec}^{0}({\mathbb{A}}^{2})=\bigoplus_{(i,j)\in\Lambda}{\mathbb{K}}\partial_{i,j}\text{
\ and \
}\operatorname{Vec}^{0}_{0}({\mathbb{A}}^{2})=\bigoplus_{(i,j)\in\Lambda,i+j>0}{\mathbb{K}}\partial_{i,j}.$

Note that $\partial_{i,j}$ is fixed by the finite subgroup $S\subseteq T$ if
and only if $i-j\equiv 0\mod 3$. Setting $\Lambda_{0}:=\\{(k,l)\in\Lambda\mid
k-l\equiv 0\mod 3,(k,l)\neq(0,0)\\}$ we get

(2)
$\operatorname{Lie}\mathfrak{A}^{S}\subseteq(\operatorname{Lie}\mathfrak{A})^{S}=\bigoplus_{(i,j)\in\Lambda_{0}}{\mathbb{K}}\partial_{i,j}.$

The next lemma shows that
$\operatorname{Lie}\mathfrak{F}=\operatorname{Lie}\mathfrak{A}^{S}=(\operatorname{Lie}\mathfrak{A})^{S}$.
In contrast to the situation of algebraic groups, this does not automatically
imply that $\mathfrak{F}=\mathfrak{A}^{S}$. In fact, Furter-Kraft give an
example of a closed ind-subgroup of a path-connected ind-group which has the
same Lie algebra, but is strictly smaller (see [FK18, Theorem 17.3.1]). In
order to show that $\mathfrak{F}=\mathfrak{A}^{S}$ (see Theorem 3.4.3) we
therefore need an additional argument.

######  Lemma 3.2.1.

The Lie algebra $(\operatorname{Lie}\mathfrak{A})^{S}$ is generated by
$\operatorname{Lie}U$ and $\operatorname{Lie}V$, hence
$\operatorname{Lie}\mathfrak{F}=\operatorname{Lie}\mathfrak{A}^{S}=(\operatorname{Lie}\mathfrak{A})^{S}$.

###### Proof.

We have $\operatorname{Lie}U={\mathbb{K}}\partial_{-1,2}$ and
$\operatorname{Lie}V={\mathbb{K}}\partial_{2,-1}$. An easy calculation shows
that

$\operatorname{ad}\partial_{-1,2}(\partial_{i,j})=a\,\partial_{i-1,j+2}\,(i\geq
0)\text{ and
}\operatorname{ad}\partial_{2,-1}(\partial_{i,j})=b\,\partial_{i+2,j-1}\,(j\geq
0)$

with non-zero constants $a,b\in{\mathbb{K}}$. This implies that the Lie
algebra generated by $\partial_{-1,2}$ and $\partial_{2,-1}$ contains all
$\partial_{i,j}$ with $(i,j)\in\Lambda_{0}$, hence
$\langle\operatorname{Lie}U,\operatorname{Lie}V\rangle_{\text{Lie}}\supseteq(\operatorname{Lie}\mathfrak{A})^{S}$,
see formula (2). Since
$\operatorname{Lie}U,\operatorname{Lie}V\subseteq\operatorname{Lie}\mathfrak{F}$
the claim follows from the inclusions in (1). ∎

It is easy to see that the vector field $\partial_{i,j}$ is locally nilpotent
if and only if $i=-1$ or $j=-1$. In these cases, there are unique algebraic
subgroups $U_{-1,k}\simeq{\mathbb{G}}_{a}$ and
$U_{k,-1}\simeq{\mathbb{G}}_{a}$ with
$\operatorname{Lie}U_{-1,k}={\mathbb{K}}\partial_{-1,k}$ and
$\operatorname{Lie}U_{k,-1}={\mathbb{K}}\partial_{k,-1}$. Note that
$U_{-1,2}=U=\overline{\langle\mathbf{u}\rangle}$ and
$U_{2,-1}=V=\overline{\langle\mathbf{v}\rangle}$. Moreover, the subgroups
$U_{-1,k}$ and $U_{-1,l}$ commute as well as the vector fields
$\partial_{-1,k}$ and $\partial_{-1,l}$.

###  3.3. De Jonquière subgroups and amalgamated products

This and the subsequent sections are largely inspired by [AZ13, Section 4].
Consider the de Jonquière group

$\operatorname{Jonq}:=\\{(ax+h(y),cy+d)\mid
a,c\in{\mathbb{K}}^{*},d\in{\mathbb{K}},h\in{\mathbb{K}}[y]\\}$

and its subgroups

$\displaystyle\operatorname{Jonq}_{0}:=\\{(ax+yf(y),by)\mid
a,b\in{\mathbb{K}}^{*},f\in{\mathbb{K}}[y]\\}=\operatorname{Jonq}\cap\operatorname{Aut}_{0}({\mathbb{A}}^{2}),$
$\displaystyle\mathfrak{J}_{0}:=\\{(x+yf(y),y)\mid
f\in{\mathbb{K}}[y]\\}\subseteq\operatorname{Jonq}_{0}.$

$\mathfrak{J}_{0}$ is a commutative closed unipotent ind-subgroup, isomorphic
to ${\mathbb{K}}[y]^{+}$. It admits a filtration by closed unipotent subgroups
$\prod_{i=1}^{d}U_{-1,i}\simeq{\mathbb{G}}_{a}^{d}$, $d\geq 1$. Note that
$\operatorname{Jonq}_{0}=\mathfrak{J}_{0}\rtimes T$.

######  Lemma 3.3.1.

The group $\mathfrak{A}^{S}$ is the free product
$(\mathfrak{J}_{0})^{S}*(\mathfrak{J}_{0}^{-})^{S}$ where
$\mathfrak{J}_{0}^{-}:=\tau\circ\mathfrak{J}_{0}\circ\tau$ and
$\tau:=[\begin{smallmatrix}0&1\\\ 1&0\end{smallmatrix}]$.

###### Proof.

(1) The amalgamated product structure
$\operatorname{Aut}({\mathbb{A}}^{2})=\operatorname{Jonq}*_{\mathfrak{B}}\operatorname{Aff}_{2}$
where $\mathfrak{B}:=\operatorname{Jonq}\cap\operatorname{Aff}_{2}$ implies
that every element
$\mathbf{g}\in\operatorname{Aut}({\mathbb{A}}^{2})\setminus\mathfrak{B}$ can
be expressed in one of the following 4 forms, called the type of $\mathbf{g}$,

$\displaystyle\mathbf{g}=j_{1}a_{1}j_{2}\cdots
j_{n},\qquad\mathbf{g}=j_{1}a_{1}j_{2}\cdots j_{n}a_{n+1},$
$\displaystyle\mathbf{g}=a_{1}j_{1}a_{2}\cdots
j_{n},\qquad\mathbf{g}=a_{1}j_{1}a_{2}\cdots j_{n}a_{n+1}$

where $j_{k}\in\operatorname{Jonq}\setminus\mathfrak{B}$ and
$a_{k}\in\operatorname{Aff}_{2}\setminus\mathfrak{B}$. The length of the
expression, its type and the degrees of the $j_{k}$ are uniquely determined by
$\mathbf{g}$.

(2) Let now $\mathbf{g}\in\operatorname{Aut}_{0}({\mathbb{A}}^{2})$. Using
$\operatorname{Jonq}=\operatorname{Jonq}_{0}\rtimes\operatorname{Trans}_{2}$
and
$\operatorname{Aff}_{2}=\operatorname{GL}_{2}\rtimes\operatorname{Trans}_{2}$
where $\operatorname{Trans}_{2}\simeq{\mathbb{G}}_{a}^{2}$ is the group of
translations, one can assume that $j_{k}\in\operatorname{Jonq}_{0}\setminus B$
and $a_{k}\in\operatorname{GL}_{2}\setminus B$ where
$B\subseteq\operatorname{GL}_{2}$ is the Borel subgroup of upper triangular
matrices. Since $\operatorname{GL}_{2}\setminus B=U\tau B=B\tau U$ where
$U\subseteq B$ is the unipotent radical, we see that any
$\mathbf{g}\in\operatorname{Aut}_{0}({\mathbb{A}}^{2})$ has a presentation in
one of the 4 forms

$\displaystyle\mathbf{g}=j_{1}\tau j_{2}\cdots\tau
j_{n},\qquad\mathbf{g}=j_{1}\tau j_{2}\cdots j_{n}\tau,$
$\displaystyle\mathbf{g}=\tau j_{1}\tau\cdots\tau j_{n},\qquad\mathbf{g}=\tau
j_{1}\tau\cdots\tau j_{n}\tau$

where $j_{k}\in\operatorname{Jonq}_{0}\setminus B$. Again, the length, the
type and the degrees of the $j_{k}$ are uniquely determined by $\mathbf{g}$.

(3) Now we remark that $\operatorname{Jonq}_{0}=\mathfrak{J}_{0}\rtimes T$ and
$T\tau=\tau T$. It follows that we can reach one of the following forms

$\displaystyle\mathbf{g}=j_{1}\tau j_{2}\cdots\tau
j_{n}t,\qquad\mathbf{g}=j_{1}\tau j_{2}\cdots j_{n}\tau t,$
$\displaystyle\mathbf{g}=\tau j_{1}\tau\cdots\tau j_{n}t,\qquad\mathbf{g}=\tau
j_{1}\tau\cdots\tau j_{n}\tau t$

where $j_{k}\in\mathfrak{J}_{0}$ and $t\in T$. We claim that this form is
uniquely determined by $\mathbf{g}$. Since the type and the length are given
by $\mathbf{g}$, it suffices to compare the similar decompositions

$\mathbf{g}=j_{1}\tau j_{2}\cdots j_{n}t=j_{1}^{\prime}\tau
j_{2}^{\prime}\cdots j_{n}^{\prime}t^{\prime}\text{ and }\mathbf{g}=j_{1}\tau
j_{2}\cdots j_{n}\tau t=j_{1}^{\prime}\tau j_{2}^{\prime}\cdots
j_{n}^{\prime}\tau t^{\prime}.$

It follows that

$\tau j_{2}\cdots j_{n}t=j_{1}^{-1}j_{1}^{\prime}\tau j_{2}^{\prime}\cdots
j_{n}^{\prime}t^{\prime}\text{ \ resp. \ }\tau j_{2}\cdots j_{n}\tau
t=j_{1}^{-1}j_{1}^{\prime}\tau j_{2}^{\prime}\cdots j_{n}^{\prime}\tau
t^{\prime},$

hence $j_{1}^{\prime}=j_{1}$. Now the claim follows by induction.

(4) Using the equalities $st=ts$ and $s\tau s=\tau$ for $s\in S$ and $t\in T$
we find for an $S$-invariant
$\mathbf{g}\in\operatorname{Aut}_{0}({\mathbb{A}}^{2})$ of the form
$\mathbf{g}=j_{1}\tau j_{2}\cdots j_{n}t$ as in (3) above that

$\mathbf{g}=s\mathbf{g}s^{-1}=\begin{cases}(sj_{1}s^{-1})\tau(s^{-1}j_{2}s)\tau(sj_{3}s^{-1})\cdots(sj_{n}s^{-1})t&\text{if
$n$ is odd,}\\\
(sj_{1}s^{-1})\tau(s^{-1}j_{2}s)\tau(sj_{3}s^{-1})\cdots(s^{-1}j_{n}s^{-1})t&\text{if
$n$ is even.}\end{cases}$

In the first case, we get that each $j_{k}$ is $S$-invariant. In the second
case, we obtain $sj_{n}s=j_{n}$, a contradiction. Looking at the other forms
we see that for an $S$-invariant $\mathbf{g}$ the number of $\tau$’s must be
even and each $j_{k}$ is $S$-invariant, i.e.
$j_{k}\in(\mathfrak{J}_{0})^{S}\subseteq\mathfrak{A}^{S}$.

(5) Finally, assume that $\mathbf{g}\in\mathfrak{A}^{S}$. If $\mathbf{g}$ is
given in one of the forms of (3), then $t=\operatorname{id}$, because
$d\mathbf{g}_{0}=\operatorname{id}$ and $(dj_{k})_{0}=\operatorname{id}$ for
$j_{k}\in(\mathfrak{J}_{0})^{S}$. Since the number of $\tau$’s is even, we see
that such an automorphism $\mathbf{g}$ is a product of the form $\cdots
j_{k}j_{k}^{\prime}j_{k+1}j_{k+1}^{\prime}\cdots$ where
$j_{k}\in(\mathfrak{J}_{0})^{S}$ and
$j_{k}^{\prime}\in(\mathfrak{J}_{0}^{-})^{S}=(\tau\mathfrak{J}_{0}\tau)^{S}=\tau(\mathfrak{J}_{0})^{S}\tau$.
∎

###  3.4. The fixed points under $S$

Recall that $\mathfrak{F}\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$
denotes the closure of the free group $F$ generated by $\mathbf{u}$ and
$\mathbf{v}$.

######  Proposition 3.4.1.

$(\mathfrak{J}_{0})^{S}\subseteq\mathfrak{F}$. In particular,
$\mathfrak{F}=\mathfrak{A}^{S}=(\mathfrak{J}_{0})^{S}*(\mathfrak{J}_{0}^{-})^{S}$.

The proof needs some preparation. Set

$\varphi:=\mathbf{u}\circ\mathbf{v}=(x+y^{2},y)\circ(x,y+x^{2})=\\\
=(x+y^{2}+2x^{2}y+y^{4},y+x^{2})\in F=\langle\mathbf{u},\mathbf{v}\rangle$

and consider the powers $\varphi^{k}=(f_{k},h_{k})$. It follows that
$f_{k}=x+\text{terms of degree $\geq 2$}$ and $h_{k}=y+\text{terms of degree
$\geq 2$}$, and that all coefficients are non-negative integers.

######  Lemma 3.4.2.

The polynomial $f_{k}$ contains the term $cy^{3k-1}$ with $c\neq 0$.

###### Proof.

By definition, we have $f_{k+1}=f_{k}+h_{k}^{2}+2f_{k}^{2}h_{k}+h_{k}^{4}$.
Since all monomials in $f_{k}$ and $h_{k}$ have non-negative coefficients,
there is no cancellation of terms. Now assume that $f_{k}$ contains the terms
$y^{2}$ and $y^{3k-1}$. Then in the expression of $f_{k+1}$ the summand
$f_{k}$ contains the term $y^{2}$ and the summand $f_{k}^{2}h_{k}$ contains
$y^{2}\cdot y^{3k-1}\cdot y=y^{3k+2}$. Thus the claim follows by induction. ∎

###### Proof of Proposition 3.4.1.

We have seen in Section 3.3 that $\mathfrak{J}_{0}\simeq{\mathbb{K}}[y]^{+}$
admits a filtration by the closed unipotent subgroups
$\prod_{i=2}^{d}U_{-1,i}$, $d\geq 2$. It follows that
$\mathfrak{J}_{0}^{S}=\\{(x+y^{2}f(y^{3}),y)\mid
f\in{\mathbb{K}}[y]\\}\xrightarrow{\sim}{\mathbb{K}}[y]^{+}$, and that
$\mathfrak{J}_{0}^{S}$ admits a filtration by the closed unipotent subgroups
$\prod_{j=1}^{k}U_{-1,3j-1}$. Note that this product contains any automorphism
of ${\mathbb{A}}^{2}$ of the form $(x+\sum_{j=1}^{k}c_{j}y^{3j-1},y)$. We now
show by induction that $U_{-1,3j-1}\subseteq\mathfrak{F}$.

Assume that this holds for $j\leq k$ and consider the product $\psi$ of
$\varphi^{k+1}$ with an element
$\mathbf{g}\in\prod_{j=1}^{k}U_{-1,3j-1}\subseteq\mathfrak{F}$:

$\psi:=\varphi^{k+1}\circ\mathbf{g}=(f_{k+1},h_{k+1})\circ(x+\sum_{j=1}^{k}c_{j}y^{3j-1},y)=\\\
(f_{k+1}(x+\sum c_{j}y^{3j-1},y),h_{k+1}(x+\sum c_{j}y^{3j-1},y))=(f,h).$

Since the lowest degree term of $f_{k+1}(0,y)$ is $cy^{3k+2}$ we can choose
the coefficients $c_{j}$ is such a way that the lowest degree term in $f(0,y)$
is $cy^{3k+2}$. But then

$(t^{-3k-2}x,t^{-1}y)\circ(f,h)\circ(t^{3k+2}x,ty)\overset{t\to
0}{\longrightarrow}(x+cy^{3k+2},y)\in U_{-1,3k+2}.$

Since $\mathfrak{F}$ is closed and stable under the action of the torus $T$ we
get $U_{-1,3k+2}\subseteq\mathfrak{F}$, and the claim follows. ∎

Summing up we have the following result.

######  Theorem 3.4.3.

Let $F\subseteq\operatorname{Aut}({\mathbb{A}}^{2})$ be the subgroup generated
by $\mathbf{u}:=(x+y^{2},y)$ and $\mathbf{v}:=(x,y+x^{2})$, and denote by
$\mathfrak{F}:=\overline{F}$ its closure in
$\operatorname{Aut}({\mathbb{A}}^{2})$. Furthermore, define
$\mathfrak{A}:=\\{\mathbf{g}\in\operatorname{Aut}({\mathbb{A}}^{2})\mid\mathbf{g}(0)=0,d\mathbf{g}_{0}=\operatorname{id}\\}$.

  1. (1)

$F$ is a free group in the two generators $\mathbf{u},\mathbf{v}$, containing
non-algebraic elements.

  2. (2)

$\mathfrak{F}=\mathfrak{A}^{S}=\operatorname{Cent}_{\mathfrak{A}}(S)$ where
$S:=\\{(\zeta x,\zeta^{2}y)\mid\zeta^{3}=1\\}\subseteq T$ is the centralizer
of $F$ in $\operatorname{Aut}({\mathbb{A}}^{2})$.

  3. (3)

$\mathfrak{F}$ is a free product $\mathfrak{J}*\mathfrak{J}^{-}$ where
$\mathfrak{J}:=\\{(x+y^{2}f(y^{3}),y)\mid
f\in{\mathbb{K}}[y]\\}=(\mathfrak{J}_{0})^{S}$ and
$\mathfrak{J}^{-}=\\{(x,y+x^{2}f(x^{3}))\mid
f\in{\mathbb{K}}[x]\\}=\tau\mathfrak{J}\tau$. In particular, $\mathfrak{J}$ is
a commutative nested unipotent closed ind-subgroup of
$\operatorname{Aut}({\mathbb{A}}^{2})$ isomorphic to ${\mathbb{K}}[y]^{+}$.

  4. (4)

Every algebraic subgroup of $\mathfrak{F}$ is commutative and unipotent and is
conjugate to a subgroup of $\mathfrak{J}$ or of $\mathfrak{J}^{-}$.

###### Proof.

(1) The first part is clear. For the second we notice that
$\deg(\mathbf{u}\mathbf{v})^{k}$ tends to infinity for $k\to\infty$. Hence,
$\mathbf{u}\mathbf{v}\in\mathfrak{F}$ is not an algebraic element.

(2) & (3) This is Proposition 3.4.1 and Lemma 3.1.1

(4) This follows from a famous result of Serre’s, see [Ser03, Theorem 8],
because an algebraic subgroup of a free product
$\mathfrak{J}*\mathfrak{J}^{-}$ is of bounded length, hence conjugate to a
subgroup of one of the factors. ∎

######  Remark 3.4.4.

According to Lemma 3.2.1 the subgroup
$\mathfrak{F}\subseteq\operatorname{Aut}(X)$ is the closure of the subgroup
generated by $U$ and $V$, and $\operatorname{Lie}\mathfrak{F}$ is the Lie
algebra generated by $\operatorname{Lie}U$ and $\operatorname{Lie}V$. Thus our
example is a positive instance of our Question 1.

##  4\. Homomorphisms of ind-groups and of their Lie algebras

###  4.1. A canonical Lie subalgebra of $\operatorname{Lie}\mathfrak{G}$

Let $\mathfrak{G}=\bigcup_{d}\mathfrak{G}_{d}$ be an ind-group. For
$g\in\mathfrak{G}$ denote by
$\lambda_{g},\rho_{g}\colon\mathfrak{G}\xrightarrow{\sim}\mathfrak{G}$ the
left- and the right-multiplications by $g$, respectively. If
$Y\subseteq\mathfrak{G}$ is an algebraic subset and $g\in Y$, then
$\rho_{g^{-1}}$ defines an isomorphism $Y\xrightarrow{\sim}Yg^{-1}$ whose
differential in $g$ is a linear isomorphism
$T_{g}\,Y\xrightarrow{\sim}T_{e}\,Yg^{-1}\subseteq\operatorname{Lie}\mathfrak{G}$.
Define

$M_{Y}:=\operatorname{span}_{\mathbb{K}}\left(T_{e}Yg^{-1}=d\rho_{g^{-1}}(T_{g}Y)\mid
g\in Y\right)\subseteq\operatorname{Lie}\mathfrak{G}.$

More generally, if $\varphi\colon X\to\mathfrak{G}$ is a morphism where $X$ is
an algebraic variety, then we define

$M_{\varphi}:=\operatorname{span}_{\mathbb{K}}\left(d\rho_{\varphi(x)^{-1}}(\operatorname{Im}d\varphi_{x})\mid
x\in X\right)\subseteq\operatorname{Lie}\mathfrak{G}.$

######  Lemma 4.1.1.

Let $Y\subseteq\mathfrak{G}$ be an algebraic subset, and let $\varphi\colon
X\to\mathfrak{G}$ be a morphism, where $X$ is an algebraic variety. Then the
following holds.

  * (a)

$\dim M_{Y}<\infty$ and $\dim M_{\varphi}<\infty$.

  * (b)

For $g\in\mathfrak{G}$ we have $M_{Yg}=M_{Y}$ and
$M_{gY}=\operatorname{Ad}g\,(M_{Y})$.

  * (c)

For $g\in\mathfrak{G}$ we have $M_{\rho_{g}\circ\varphi}=M_{\varphi}$ and
$M_{\lambda_{g}\circ\varphi}=\operatorname{Ad}g\,(M_{\varphi})$.

  * (d)

If $A\subseteq X$ is a closed subset, then $M_{\varphi|_{A}}\subseteq
M_{\varphi}$.

  * (e)

If $Y$ is smooth and $U\subseteq Y$ open and dense, then $M_{U}=M_{Y}$.

  * (f)

If $Y=\overline{\varphi(X)}$, then $M_{Y_{\text{\it reg}}}\subseteq
M_{\varphi}\subseteq M_{Y}$.

###### Proof.

(a) Consider the closed subset $\tilde{Y}:=\overline{\\{hg^{-1}\mid h,g\in
Y\\}}$. Then $\rho_{g^{-1}}(Y)=Yg^{-1}\subseteq\tilde{Y}$ for all $g\in Y$,
and so $M_{Y}\subseteq T_{e}\tilde{Y}$. For $Y:=\overline{\varphi(X)}$ we
obviously have $M_{\varphi}\subseteq M_{Y}$ and the both claims follow.

(b) For $Z:=Yg$, $y\in Y$, $z:=yg\in Z$ we have $Zz^{-1}=Yy^{-1}$, hence
$M_{Yg}=M_{Y}$. For $Z:=gY$, $y\in Y$, $z:=gy$ we get
$Zz^{-1}=g(Yy^{-1})g^{-1}$, hence
$T_{e}Zz^{-1}=T_{e}g(Yy^{-1})g^{-1}=\operatorname{Ad}g\,(T_{e}Yy^{-1})$, and
so $M_{gY}=\operatorname{Ad}g\,(M_{Y})$.

(c) This is similar to (b).

(d) This is clear.

(e) If $V$ is a finite dimensional vector space and $(U_{y})_{y\in Y}$ a
family of $d$-dimensional subspaces parametrized by a variety $Y$, i.e. given
by a morphism of $Y$ to the Grassmannian, then $\langle U_{y}\mid y\in
Y\rangle=\langle U_{y}\mid y\in Y^{\prime}\rangle$ for any dense subset
$Y^{\prime}\subseteq Y$.

(f) There is an open dense set $X^{\prime}\subseteq X_{\text{\it reg}}$ where
$\varphi$ is smooth. This implies that
$\operatorname{Im}d\varphi_{x}=T_{\varphi(x)}Y$ for all $x\in X^{\prime}$.
Since the image $\varphi(X^{\prime})$ is open and dense in $Y_{\text{\it
reg}}$ we obtain from (d) and (e) that $M_{Y_{\text{\it
reg}}}=M_{\varphi(X^{\prime})}\subseteq M_{\varphi}$. The second inclusion is
obvious. ∎

######  Definition 4.1.2.

For an ind-group $\mathfrak{G}$ we define

$L_{\mathfrak{G}}:=\operatorname{span}_{\mathbb{K}}\left(M_{Y_{\text{\it
reg}}}\mid Y\subseteq\mathfrak{G}\text{ an irreducible closed algebraic
subset}\right)$

where $Y_{\text{\it reg}}\subseteq Y$ is the smooth locus. Given a
homomorphism $\Phi\colon\mathfrak{H}\to\mathfrak{G}$ of ind-groups, we define
$L_{\Phi}:=d\Phi(L_{\mathfrak{H}})\subseteq L_{\mathfrak{G}}$.

The next proposition shows that
$L_{\mathfrak{G}}\subseteq\operatorname{Lie}\mathfrak{G}$ is a canonical Lie
subalgebra.

######  Proposition 4.1.3.

  1. (a)

$L_{\mathfrak{G}}\subseteq\operatorname{Lie}\mathfrak{G}$ is stable under the
adjoint action of $\mathfrak{G}$, i.e. $L_{\mathfrak{G}}$ is an ideal in
$\operatorname{Lie}\mathfrak{G}$.

  2. (b)

For every algebraic subgroup $G\subseteq\mathfrak{G}$ we have
$\operatorname{Lie}G\subseteq L_{\mathfrak{G}}$.

  3. (c)

If $\mathfrak{H}\subseteq\mathfrak{G}$ is a closed subgroup then
$L_{\mathfrak{H}}\subseteq L_{\mathfrak{G}}$.

###### Proof.

(a) We have $\operatorname{Ad}h(M_{Y})=M_{hYh^{-1}}$. In fact,

$\operatorname{Ad}h(T_{e}Yg^{-1})=T_{e}(hYg^{-1}h^{-1})=T_{e}(hYh^{-1})g_{0}^{-1}$

where $g_{0}=hgh^{-1}$, and the claim follows.

(b) For an algebraic group $G\subseteq\mathfrak{G}$ we have
$d\rho_{g^{-1}}(T_{g}G)=\operatorname{Lie}G$ (cf. Remark 2.2.3), and so
$M_{G}=\operatorname{Lie}G$.

(c) This is clear. ∎

Recall that

$\operatorname{SAut}({\mathbb{A}}^{n}):=\\{\alpha\in\operatorname{Aut}({\mathbb{A}}^{n})\,|\,{\rm
jac}(\alpha)=1\\}$

is a normal subgroup of $\operatorname{Aut}({\mathbb{A}}^{n})$ with Lie
algebra $\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{n})$ being the
algebra $\operatorname{Vec}^{0}({\mathbb{A}}^{n})$ of polynomial vector fields
on ${\mathbb{A}}^{n}$ with zero divergence (see Section 3.2).

######  Corollary 4.1.4.

We have
$L_{\operatorname{SAut}({\mathbb{A}}^{n})}=\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{n})$
and
$L_{\operatorname{Aut}({\mathbb{A}}^{n})}=\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{n})$.

###### Proof.

By [Sha81, Lemma 3], the Lie algebra

$\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{n})=[\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{n}),\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{n})]$

is simple. It follows from Proposition 4.1.3(a) that
$\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{n})\subseteq
L_{\operatorname{Aut}({\mathbb{A}}^{n})}$. Now the claim is a consequence of
Proposition 4.1.3(b) due to the fact that

$\operatorname{Lie}\operatorname{Aut}({\mathbb{A}}^{n})=\operatorname{Lie}\operatorname{SAut}({\mathbb{A}}^{n})\oplus\operatorname{Lie}{\mathbb{K}}^{*},$

where ${\mathbb{K}}^{*}\subseteq\operatorname{Aut}({\mathbb{A}}^{n})$ is
realized as the subgroup of the scalar multiplications
$\lambda\cdot\operatorname{id}$, $\lambda\in{\mathbb{K}}^{*}$, cf. [FK18,
Prop. 15.7.2]. ∎

###  4.2. Surjective homomorphisms

Let $\Phi\colon\mathfrak{H}\to\mathfrak{G}$ be a homomorphism of ind-groups.
If $\Phi$ is surjective one expects that the induced homomorphism
$d\Phi\colon\operatorname{Lie}\mathfrak{H}\to\operatorname{Lie}\mathfrak{G}$
is also surjective.

######  Proposition 4.2.1.

Let $\Phi\colon\mathfrak{H}\to\mathfrak{G}$ be a surjective homomorphism of
ind-groups.

  * (a)

$M_{\Phi}$ is stable under the adjoint action of $\mathfrak{G}$, i.e.
$M_{\Phi}\subseteq\operatorname{Lie}\mathfrak{G}$ is an ideal.

  * (b)

If $H\subseteq\mathfrak{H}$ is an algebraic subgroup, then
$M_{\Phi}\supseteq\operatorname{Lie}\Phi(H)$.

###### Proof.

(a) For $h\in\mathfrak{H}$ and $g:=\Phi(h)\in\mathfrak{G}$ we get from Lemma
4.1.1(c) that $M_{\Phi|_{hX}}=\operatorname{Ad}g\,(M_{\Phi|_{X}})$ for any
closed algebraic subset $X\subseteq\mathfrak{H}$.

(b) This is clear, since $\Phi(H)\subseteq\mathfrak{G}$ is an algebraic
subgroup, and therefore
$d\Phi|_{H}\colon\operatorname{Lie}H\to\operatorname{Lie}\Phi(H)$ is
surjective. ∎

######  Question 5.

Assume we have a surjective homomorphism $\mathfrak{G}\to G$ where
$\mathfrak{G}$ is a path-connected ind-group and $G$ is an algebraic group.
Can one show that $\mathfrak{G}$ contains an algebraic subgroup which is sent
surjectively onto $G$?

###  4.3. Subgroups of finite codimension

Let $\mathfrak{G}=\bigcup_{d}\mathfrak{G}_{d}$ be an ind-group.

######  Definition 4.3.1.

We say that a closed subgroup $\mathfrak{H}\subseteq\mathfrak{G}$ is of finite
codimension if $\mathfrak{G}_{d}\cdot\mathfrak{H}=\mathfrak{G}$ for some
$d\in{\mathbb{N}}$.

For instance, if $\mathfrak{G}$ is connected and acts on an algebraic variety
$X$, then the stabilizer $\mathfrak{G}_{x}$ of any point $x\in X$ is of finite
codimension. In fact, the orbit $\mathfrak{G}x$ is an irreducible subvariety
of $X$ by [FK18, Prop. 7.1.2(1)], and so, $\mathfrak{G}x=\mathfrak{G}_{d}x$
for some $d$. The latter implies
$\mathfrak{G}=\mathfrak{G}_{d}\cdot\mathfrak{G}_{x}$. We do not know if every
subgroup $\mathfrak{H}$ of finite codimension of a connected ind-group
$\mathfrak{G}$ is the stabilizer of a point of a $\mathfrak{G}$-variety, i.e.
if $\mathfrak{G}/\mathfrak{H}$ has a canonical structure of an algebraic
variety.

######  Proposition 4.3.2.

Let $\mathfrak{G}$ be an ind-group and $\mathfrak{H}\subseteq\mathfrak{G}$ a
closed subgroup. Assume that there is a closed irreducible algebraic subset
$Z\subseteq\mathfrak{G}$ such that $Z\cdot\mathfrak{H}=\mathfrak{G}$. Then
$L_{\mathfrak{G}}=M_{Z_{\text{\it reg}}}+L_{\mathfrak{H}}$. In particular,
$L_{\mathfrak{H}}$ has finite codimension in $L_{\mathfrak{G}}$.

###### Proof.

Let $Y\subseteq\mathfrak{G}$ be a closed irreducible algebraic subset. Define
$\tilde{Y}:=\overline{Z\cdot Y}$. Then $\operatorname{Ad}z\,(M_{Y_{\text{\it
reg}})}\subseteq M_{\tilde{Y}_{\text{\it reg}}}$ for all $z$ in an open dense
subset of $Z_{\text{\it reg}}$. In fact, for all $z$ in an open dense subset
of $Z$ we have $zY_{\text{\it reg}}\cap\tilde{Y}_{\text{\it
reg}}\neq\emptyset$, hence $M_{zY_{\text{\it reg}}}\subseteq
M_{\tilde{Y}_{\text{\it reg}}}$.

Thus, in Definition 4.1.2 of $L_{\mathfrak{G}}$ we may assume that the
irreducible algebraic subsets are of the form $\overline{Z\cdot Y}$.

For any irreducible closed algebraic subset $Y\subseteq\mathfrak{G}$ of the
form above we can find an irreducible closed algebraic subset
$X\subseteq\mathfrak{H}$ such the $\overline{Z\cdot X}=Y$. Consider the map
$\eta\colon Z\times X\to Y$, $(z,h)\mapsto z\cdot h$. Then $\eta$ is smooth on
an open dense subset $U\subseteq Z_{\text{\it reg}}\times X_{\text{\it reg}}$.
In particular, $\eta(U)$ is open and dense in $Y_{\text{\it reg}}$. Now for
every $g=z\cdot h\in\varphi(U)$ where $(z,h)\in U$ we have the following
commutative diagram:

$\begin{CD}Z\times X@>{}>{\eta}>Y\\\
@V{\rho_{z^{-1}}\times\operatorname{Int}z\circ\rho_{h^{-1}}}V{\simeq}V@V{\simeq}V{\rho_{g^{-1}}}V\\\
Zz^{-1}\times z(Xh^{-1})z^{-1}@>{}>{\eta}>Yg^{-1}\end{CD}$

On the tangent spaces this induces the commutative diagram

$\begin{CD}T_{z}Z\oplus T_{h}X@>{}>{d\eta}>T_{g}Y\\\
@V{d\rho_{z^{-1}}\times\operatorname{Ad}z\circ
d\rho_{h^{-1}}}V{\simeq}V@V{\simeq}V{d\rho_{g^{-1}}}V\\\ T_{e}(Zz^{-1})\oplus
T_{e}(z(Xh^{-1})z^{-1})@>{}>{d\eta}>T_{e}(Yg^{-1})\end{CD}$

By assumption, $d\eta$ is surjective, hence

$T_{e}(Yg^{-1})=T_{e}(Zz^{-1})+\operatorname{Ad}z(T_{e}(Xh^{-1}))\subseteq
M_{Z_{\text{\it reg}}}+M_{zXz^{-1}}\subseteq M_{Z_{\text{\it
reg}}}+L_{\mathfrak{H}}.$

This holds for every $g\in\varphi(U)$ which is a dense subset of $Y_{\text{\it
reg}}$. Thus, by Lemma 4.1.1(e), we finally get $M_{Y_{\text{\it
reg}}}\subseteq M_{Z_{\text{\it reg}}}+L_{\mathfrak{H}}$, hence the claim. ∎

######  Corollary 4.3.3.

Let $X$ be an affine variety, and let
$\mathfrak{H}\subseteq\operatorname{Aut}(X)$ be a closed ind-subgroup. If
$\mathfrak{H}$ has finite codimension, then
$\operatorname{Lie}\mathfrak{H}\subseteq\operatorname{Lie}\operatorname{Aut}(X)$
has finite codimension, too.

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