# Homomorphic images of algebraic groups

Uri Bader and Elyasheev Leibtag

###### Abstract.

We study topological group theoretic properties of algebraic groups over local
fields. In particular, we find conditions under which such groups have closed
images under arbitrary continuous homomorphisms into arbitrary topological
groups.

##  1\. Introduction

Let $k$ be a local field, and let ${\mathbf{G}}$ denote a linear algebraic
group over $k$. The group ${\mathbf{G}}(k)$ has a natural locally compact
topology, called _the $k$-point analytic topology_, making it a topological
group. In this paper, we are interested in properties of ${\mathbf{G}}(k)$
viewed as a topological group.

######  Convention.

Throughout this paper, unless explicitly specified otherwise, all topological
groups are assumed to be Hausdorff, hence Tychonoff.

A topological group that admits no weaker (Hausdorff) group topology is called
_minimal_. It is straightforward that every compact group is minimal.
Moreover, the center of a Polish minimal group must be compact, see Lemma
4.14. In [Omo66] and [Got73], H. Omori and M. Goto showed that for connected
real algebraic groups this necessary condition is in fact sufficient - a
connected real algebraic group with a compact center is minimal. The following
is an extension of the above result which applies to an arbitrary local field
of characteristic zero, see Theorem 7.1.

######  Theorem A.

Let $k$ be a local field of characteristic zero, and let ${\mathbf{G}}$ be a
connected $k$-algebraic group. Then the group ${\mathbf{G}}(k)$ is minimal if
and only if the center of ${\mathbf{G}}(k)$ is compact.

We note that the work of Omori and Goto uses the connectivity of the real
group $G$ in an essential way. Connectivity is a feature lacking over non-
Archimedian local fields which are totally disconnected. As such, our proof of
Theorem A is entirely different from theirs.

It is a standard fact that a Polish topological group is minimal if and only
if its image is closed under any _injective_ continuous group homomorphism
into any topological group, see Corollary 4.13 below. A topological group is
said to be _sealed_ if its image is closed under any (not necessarily
injective) continuous group homomorphism into any topological group. Clearly,
a Polish sealed group must have a compact center for all of its quotients.
Relying on the work of Omori and Goto, Mayer showed in [May97] that for
connected locally compact groups this necessary condition is in fact
sufficient - a connected locally compact group such that all of its quotients
have a compact center is sealed. The following is a generalization that
applies to connected algebraic groups over an arbitrary local field of
characteristic zero, see Theorem 8.1.

######  Theorem B.

Let $k$ be a local field of characteristic zero and let ${\mathbf{G}}$ be a
connected $k$ group with a $k$-Levi Decomposition,
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$. The following are
equivalent

  1. (1)

The group ${\mathbf{G}}(k)$ is sealed.

  2. (2)

For any normal unipotent $k$-algebraic subgroup ${\mathbf{N}}\lhd{\mathbf{G}}$
the center $Z(({\mathbf{G}}/{\mathbf{N}})(k))$ is compact.

  3. (3)

The center of the reductive group ${\mathbf{L}}$ is anisotropic (equivalently,
the center of ${\mathbf{L}}(k)$ is compact) and the action of ${\mathbf{L}}$
on $\mathcal{R}_{u}({\mathbf{G}})$ is with no non-trivial fixed points,

  4. (4)

The center of the reductive group ${\mathbf{L}}$ is anisotropic (equivalently,
the center of ${\mathbf{L}}(k)$ is compact) and the action of ${\mathbf{L}}$
on $\operatorname{Lie}(\mathcal{R}_{u}({\mathbf{G}}))$ is with no invariant
vectors.

###  1.1. Previous literature

As already mentioned, minimality questions for connected groups were
considered by Omori, Goto and Mayer, see [Got73], [Omo66] and [May97]. In the
special case of semisimple groups, Theorem A and B were established by Bader-
Gelander in [BG17] over an arbitrary local filed. In another direction, an
interesting recent work by Megrelishvili and Shlossberg gives minimality
criteria for groups of the form $\mathrm{SL}_{n}(F)$, where $F\leq k$ is an
arbitrary subfield, see [MS22], as well as [Shl22] concerning number theoretic
applications.

For an extensive survey on minimality and closed image properties we refer the
reader to [DM14],[Meg98], [DU98] and [Ban17].

Lately, Carter and Willis investigated the sealed property in the context of
locally compact totally disconnected groups arising as automorphism groups of
buildings, see [CW22].

Regarding infinite-dimensional isometry groups, Duchesne showed that the
Polish topology of the isometry group of the infinite-dimensional hyperbolic
space is minimal [Duc20].

In a recent work by Ghadernezhad and de la Nuez González, the authors
investigate minimality of the automorphism group of homogeneous structures,
see [GG19]. Moreover, in [dlNG22], de la Nuez González points out that the
compact-open topology on the homeomorphism group of a surface without boundary
is minimal.

###  1.2. A note on positive characteristic

In proving Theorems A and B, we use the existence of a Levi Decomposition over
$k$, and that the connected $k$-unipotent radical is $k$-split. Moreover, in
proving Theorem B, we use the fact that the Levi factor is completely
reducible. These properties are always satisfied in zero characteristic, but
do not generally hold over fields of positive characteristics, hence the zero
characteristic assumption we take in Theorems A and B. However, it seems that
even over a field of positive characteristic, these assumptions could be
removed in many cases, maybe in all cases. This direction of study will be
carried elsewhere. Yet, as a “proof of concept”, we show in Section 9 how the
tools developed in this paper may be used to reprove the aforementioned result
of Bader-Gelander, namely proving Theorems A and B in the special case of
semisimple groups over an arbitrary local field.

###  1.3. Description of the paper

In section 2 we introduce a baby case presenting the main techniques and ideas
that will be used throughout the paper. In particular, the impotent role
played by admitting a continuous linear action on a finite-dimensional
topological vector space. In section 3 we recall a few topological properties
of $k$-points of algebraic groups. We also provide auxiliary lemmas concerning
groups acting by automorphisms on locally compact abelian groups. Section 4 is
devoted to a brief survey of the minimal and sealed group properties, as well
as the relative properties for subgroups. The importance of these relative
properties is that in order to show minimality, we show relative minimality
and co-minimality of a subgroup, usually the normal subgroup under a semi-
direct product structure.

We point out that the standard topology on a finite-dimensional vector space
over a local field _is not_ minimal. In section 5 we show that the standard
topology on a finite-dimensional vector space over a local field _is_ minimal
amongst all group topologies admitting a continuous action by an adequate
multiplicative subgroup. Section 6 shows that if a vector space is a
representation space of an algebraic torus then the torus acts on the vector
space by an adequate multiplicative subgroup of the field, thus making the
vector space minimal with respect to a continuous torus action.

In section 7 we prove Theorem A. We first prove it for solvable groups, using
the linear action of the torus on the unipotent radical, and then treat the
general case by finding a co-compact solvable subgroup (the solvable radical
of a minimal parabolic) and showing that it is minimal. Section 8 is devoted
to the proof of Theorem B. As in the baby case of $\mathrm{PGL}_{2}(k)$, we
first show how minimality of simple groups implies the sealed property for
these groups. For general algebraic groups, we show by induction on the
nilpotency degree in the unipotent radical that the radical is relatively
sealed. In section 9 we show how the tools developed in this paper can be used
to prove previously known results for semisimple groups over local fields of
any characteristic.

###  1.4. Acknowledgments

We are grateful to Gil Goffer, Tal Cohen, and Yotam Hendel for many remarks
and suggestions for improvements. We would like to also thank Dikran Dikranjan
and Michael Megrelishvili for their informative comments and remarks.

Above all, we thank the members of the Midrasha on Groups at the Weizmann
Institute for their support, friendship, and professional encouragement.

##  2\. Baby case

In this section, we clarify some of the ideas involved in our proofs by
applying them to the special cases (already well-know)$A=k^{*}\ltimes k^{n}$
and ${\mathbf{G}}=\mathrm{PGL}_{2}$, see Propositions 2.9, 2.10 and 2.11. In
the subsequent sections, we will use the general results presented in this
section. Here $k$ is an arbitrary local field.

###  2.1. Some general facts

######  Theorem 2.1 (Banach, [Ban31]).

Every continuous surjective morphism between Polish groups is open.

######  Corollary 2.2.

Let $G$ be a Polish group with topology $\sigma$. If $G$ admits a weaker group
topology $\omega$, and a bijective $\omega$-continuous homomorphism $f\colon
G\to H$ onto a Polish group $H$, then $\omega=\sigma$.

###### Proof.

Since $\omega$ is weaker than $\sigma$, the map $f$ is $\sigma$-continuous as
well. By the Open Mapping Theorem 2.1, the map $f$ is a homeomorphism onto
$H$. For any open set $W\in\sigma$, $f(W)$ is open in $H$, since $f$ is
$\omega$-continuous, $f^{-1}(f(W))=W$ is open in $\omega$, thus
$\omega=\sigma$. ∎

######  Lemma 2.3 (Merson, Lemma 7.2.3 [DPS90]).

Let $G$ be a topological group with topology $\sigma$, and let
$\omega\leq\sigma$ a weaker group topology. If there exists a subgroup $H\leq
G$, such that both the subspace topologies coincide on $H$, and both quotient
topologies coincide on $G/H$, then $\omega=\sigma$.

###### Proof.

Let $S\in\sigma$ be a $\sigma$-open identity neighborhood. We will show that
there exist $W\in\omega$, a $\omega$-open identity neighborhood, with
$W\subset S$. Let $S^{\frac{1}{2}}\in\sigma$ be a symmetric identity
neighborhood such that $S^{\frac{1}{2}}S^{\frac{1}{2}}\subset S$. Since both
subspace topologies coincide on $H$, there exists $U\in\omega$ an open
identity neighborhood, with $U\cap H=S^{\frac{1}{2}}\cap H$. Let
$U^{\frac{1}{2}}\in\omega$ be a symmetric open identity neighborhood such that
$U^{\frac{1}{2}}U^{\frac{1}{2}}\subset U$. Since both quotient topologies
coincide, we have that the $\sigma$-open subset $(S^{\frac{1}{2}}\cap
U^{\frac{1}{2}})H$ is $\omega$-open as well. We set $W=(S^{\frac{1}{2}}\cap
U^{\frac{1}{2}})H\cap U^{\frac{1}{2}}\in\omega$. Fixing $y\in W$ we find $x\in
S^{\frac{1}{2}}\cap U^{\frac{1}{2}}$ and $h\in H$ such that $y=xh$. Then
$h=x^{-1}y\in U^{\frac{1}{2}}U^{\frac{1}{2}}\subset U$. So $h\in U\cap
H\subset S^{\frac{1}{2}}$, making $y=xh\in
S^{\frac{1}{2}}S^{\frac{1}{2}}\subset S$, as needed. ∎

A general strategy for proving minimality of a group $G$, using the Merson
Lemma, is by showing that there exists a subgroup $H\leq G$, such that for all
weaker group topologies, the subspace topology on $H$ and the quotient
topology on $G/H$ coincide with the a-piori stronger subspace and quotient
topologies.

###  2.2. Weaker additive group topologies on vector spaces

Let $k$ be a local field with absolute value $|\cdot|$, and let $V$ be a
$n$-dimensional $k$-vector space with norm $\|\cdot\|$. We consider the
topology on $k$ to be the locally compact field topology, and the topology on
$V$ to be the locally compact norm topology.

######  Lemma 2.4.

Any additive group topology on $V$ which is strictly weaker than the norm
topology does not contain bounded open sets.

###### Proof.

Let $\omega$ be a weaker topology on $V$ and assume that there exists a
bounded open $0_{V}$-neighborhood $W\in\omega$ with $\|W\|<\lambda$. We will
show that every norm-open ball contains an $\omega$-neighborhood, and hence
that the norm topology is equivalent to $\omega$. Let $B_{\varepsilon}\subset
V$ be the norm-open ball of radius $\varepsilon>0$. Denote by
$\overline{B_{\lambda}}$ the norm-closed ball of radius $\lambda$, and
$\operatorname{ann}(\varepsilon,\lambda)\coloneqq\overline{B}_{\lambda}\setminus
B_{\varepsilon}$. The annulus $\operatorname{ann}(\varepsilon,\lambda)$ is
compact in the norm topology, so it is compact with respect to a weaker
topology $\omega$ as well. As any group topology is Tychonoff, there exists a
$\omega$-open $0_{V}$-neighborhood $W^{\prime}$, such that
$W^{\prime}\cap\operatorname{ann}(\varepsilon,\lambda)=\emptyset$. Finally, we
get that the $\omega$-open set $W^{\prime}\cap W$ is contained in the open
ball $B_{\varepsilon}$, as needed. ∎

######  Lemma 2.5.

Let $B_{\lambda}$ be the open ball of radius $\lambda$ in $k$, and let
$m\colon B_{\lambda}\times V\to V$ be the scalar multiplication map. Then
there does not exist a strictly weaker additive group topology on $V$ such
that the multiplication map $m$, is a continuous map.

###### Proof.

As all norms on finite dimensional vector spaces induce the same topology, we
may assume without loss of generality that $\|\cdot\|$ is an
$\mathrm{L}^{\infty}$ norm. Assume by contradiction that $\omega$ is a
strictly weaker group topology on $V$ such that the scalar multiplication map
$m\colon B_{\lambda}\times V\to V$ is continuous with respect to $\omega$. The
unit sphere $S_{1}$ is norm-compact and therefore $\omega$-compact. Let
$W\in\omega$ be a $0_{V}$-neighborhood not intersecting the norm sphere
$S_{1}$. By continuity of $m$ there exist $W^{\prime}\in\omega$ a
$0_{V}$-neighborhood, and $\varepsilon<\lambda$ such that
$m(B_{\varepsilon}\times W^{\prime})\subset W$. By assumption, $\omega$ is
strictly weaker than the norm topology, hence by Lemma 2.4 $W^{\prime}$ is
unbounded. Let $w^{\prime}\in W^{\prime}$ be a vector such that
$\varepsilon^{-1}<\|w^{\prime}\|$. Since the norm is $\mathrm{L}^{\infty}$
there exists a scalar $x\in k$ such that $|x|=\|w^{\prime}\|^{-1}$. We get
that $(x,w^{\prime})\in B_{\varepsilon}\times W^{\prime}$, making
$m(x,w^{\prime})=xw^{\prime}\in S_{1}$ as
$\|xw^{\prime}\|=|x|\cdot\|w^{\prime}\|=1$, in contradiction to $W$ not
intersecting the unit sphere. ∎

In particular, we get as a corollary a fundamental theorem in functional
analysis.

######  Corollary 2.6 ([Bou87, Chapter I.§2.3 Theorem 2]).

Let $k$ be a local field. Any $n$-dimensional topological $k$-vector space $V$
is homeomorphic to the topological norm space $k^{n}$.

###### Proof.

Let $V$ be an $n$-dimensional topological vector space over a local field $k$.
This means that $V$ admits an additive group topology $\tau$, such that the
scalar multiplication map $m\colon k\times V\to V$ is continuous.
Let$\\{e_{i}\\}_{i=1}^{n}$ be a basis of $V$. For any $i=1,...,n$ the map

$m_{i}\colon k\to V\ \ m_{i}\colon k\mapsto k\cdot e_{i}$

is continuous. And since $\tau$ is an additive group topology we get a
continuous additive group homomorphism from the product $k^{n}$ onto $V$.

$\Psi\colon k^{n}\to V\ \
\Psi\colon(k_{1},...,k_{n})\mapsto\sum_{i=1}^{n}k_{i}\cdot e_{i}$

The continuity of $\Psi$ implies that the $\tau$ topology on $V$ is weaker
than the norm topology. By Lemma 2.5 above, $\tau$ is equal to the norm
topology, and therefore the topology is unique. ∎

###  2.3. Minimality of the homothety group

Equipped with the product topology, the continuous action of $k^{*}$ on $V$
makes $A=k^{*}\ltimes V$ a topological group. We call the group $A$ the
homothety group of $V$, and we call the standard product topology on $A$ the
analytic topology.

######  Lemma 2.7.

Let $\omega$ be a group topology on $A$ weaker than the analytic topology.
Then the restriction of $\omega$ to the subspace $V$ is equal to the analytic
norm topology.

###### Proof.

Let $\alpha$ be the action map of $k^{*}$ on $V$, that is,

$\alpha\colon k^{*}\times V\to V\times V,\ \ \alpha\colon(r,v)\mapsto(rv,v)$

Since $\omega$ is a group topology on $A$, the action by conjugation is
continuous with respect to the $\omega$-subspace topologies, thus the map
$\alpha$ is continuous with respect to the $\omega$-subspace topologies on
$k^{*}$ and on $V$. By strengthening the topology on the domain, the map
$\alpha$ is still continuous. So by letting $V$ admit the $\omega$-subspace
topology $\omega_{V}$, and embellishing $k^{*}$ with analytic topology, the
map $\alpha$ remains continuous. Since $\omega_{V}$ is an additive group
topology on $V$, the subtraction operation

$\mathrm{-}\colon V\times V\to V,\ \ \mathrm{-}\colon(v,w)\mapsto v-w$

is $\omega_{V}$-continuous. Observe that on the local field $k$, the addition
by $1$ map, $\mathrm{+_{1}}\colon B_{1}\to k^{*}$ mapping $a$ to $a+1$ is
continuous. We get that the map
$m\coloneqq\mathrm{-}\circ\alpha\circ(\mathrm{+_{1}}\times Id)$

$m\colon B_{1}\times V\to V\ \ m\colon(x,v)\mapsto(x+1)v-v=xv$

is a continuous multiplication by scalars in the unit ball. By Lemma 2.5
above, the restricted topology $\omega_{V}$ on $V$ is equal to the norm
topology. ∎

######  Lemma 2.8.

Let $\omega$ be a group topology on $A$, which is weaker than the analytic
topology. Then the $\omega$-quotient topology on $A/V$, $\omega_{A/V}$, is
equal to the analytic quotient topology on $A/V$, which is isomorphic to
$k^{*}$.

###### Proof.

The subgroup $V$ is normal in $A$, so the action by conjugation $A\times V\to
V$ is continuous with respect to the $\omega$-subspace topologies. Since $V$
is in the kernel of the action, we get continuous action of the quotient group
$A/V$ on $V$. As abstract groups $A/V$ is isomorphic to $k^{*}$ and the action
on $V$ is by scalar multiplication on the vector space $V$. By Lemma 2.7, the
$\omega$-subspace topology $\omega_{V}$, is equal to the norm topology on $V$.
We, therefore, get a continuous representation $\pi\colon
A/V\to\mathrm{GL}(V)$ onto the closed analytic subgroup of scalars,
$k^{*}\leq\mathrm{GL}(V)$. Thus $\pi$ is a continuous bijection onto a Polish
group, so by Corollary 2.2, the $\omega_{A/V}$-quotient topology is equal to
the Polish analytic topology on $A/V$. ∎

######  Proposition 2.9 (see [Meg98, Theorem 4.7(b)]).

The homothety group $k^{*}\ltimes V$, endowed with the analytic topology, is
minimal.

###### Proof.

Let $\omega$ be a group topology on $A=k^{*}\ltimes V$, weaker than the
analytic topology. By Lemma 2.7, the subspace topology $\omega_{V}$ is equal
to the analytic subspace topology, and by Lemma 2.8 the $\omega$-quotient
topology on $A/V$ is equal to the quotient of the analytic topology. Hence by
Lemma 2.3, $\omega$ is equal to the analytic topology. ∎

###  2.4. Minimality and sealedness of $\mathrm{PGL}_{2}(k)$

######  Proposition 2.10 ([BG17, Theorem 5.4]).

The group $\mathrm{PGL}_{2}(k)$, endowed with the analytic topology, is
minimal.

###### Proof.

Let $\omega$ be a group topology on $G=\mathrm{PGL}_{2}(k)$, weaker than the
analytic topology. The Borel subgroup $P$, is isomorphic to the homothety
group $k^{*}\ltimes k$, hence by Proposition 2.9, $P$ is minimal. Thus for any
weaker group topology $\omega$, the subspace topology on $P$ coincides with
the analytic topology. Moreover, since the analytic topology is Polish, $P$ is
$\omega$-closed. Hence the $\omega$-quotient topology on $G/P$ is Hausdorff
and since the analytic quotient topology on $G/P$ is compact it does not admit
any strictly weaker typology. By Merson’s Lemma 2.3 $\omega$ is equal to the
analytic topology on $G$. ∎

######  Proposition 2.11 ([BG17, Theorem 5.1]).

The group $\mathrm{PGL}_{2}(k)$, endowed with the analytic topology, is
sealed.

###### Proof.

Let $f\colon G=\mathrm{PGL}_{2}(k)\to H$ be a continuous homomorphism into a
topological group $H$. By [Mar91, Chapter I, Theorem (1.5.6)(i) and Theorem
(2.3.1)(b)], any non-trivial normal subgroup of $G$ is co-compact, thus if $f$
is not injective $f(G)$ is compact and obviously closed. We thus assume as we
may that $f$ is injective. By Proposition 2.10, the $f$-initial topology
$\tau^{f}$ must coincide with the analytic topology on $G$. Therefore, we get
that $f$ is a homeomorphism onto its image. Therefore the subspace topology on
$f(G)$ is Polish, hence closed, see [Mel16, Theorem 2.16]. ∎

##  3\. Preliminaries

###  3.1. Algebraic groups over local fields

We assume that the reader is familiar with the basic theory of algebraic
groups.

Throughout this section we let $k$ be a local field (of any characteristic),
with absolute value $|\cdot|$, and denote by $\bar{k}$ its algebraic closure.
Algebraic groups over $k$ will be identified with their $\bar{k}$ points and
will be denoted by boldface letters. Their $k$-points will be denoted by
corresponding Roman letters. We regard $G={\mathbf{G}}(k)$ as a topological
group equipped with a $k$-analytic structure.

As seen in the next two propositions, for a local field $k$, the geometric
properties of the group ${\mathbf{G}}$ interact nicely with the topological
$k$-point structure on $G$. Namely, under a $k$-morphism of algebraic groups
the image of the $k$-points is topologically closed, and geometrically
(Zariski) dense.

######  Proposition 3.1 ([Mar91, Chapter I, Proposition (2.5.3)(ii)]).

Let ${\mathbf{G}}$ be a connected $k$ group, then ${\mathbf{G}}(k)$ is Zariski
dense in ${\mathbf{G}}$.

######  Proposition 3.2.

Let ${\mathbf{G}}$, ${\mathbf{H}}$ be $k$-algebraic groups, and
$f\colon{\mathbf{G}}\to{\mathbf{H}}$ be a $k$-morphism. Then the image of the
$k$-points, $f(G)$, is a closed subgroup of $H$. Moreover, the kernel of $f$,
$\ker(f)$ is a $k$-algebraic group such that as $k$-analytic topological
spaces ${\mathbf{G}}(k)/{\ker(f)}(k)\cong f(G)$.

###### Proof.

This follows by [BDL17, Proposition 2.2], identifying ${\mathbf{H}}$ as a
$k$-variety with a ${\mathbf{G}}$ action given by left multiplication via $f$,
i.e. $g.h=f(g)h$. ∎

######  Remark 3.3.

Let ${\mathbf{G}}$, ${\mathbf{H}}$ and $f$ be as above. Note that $f(G)$ is
Zariski dense in the $k$-algebraic group $f({\mathbf{G}})$. But generally,
$f(G)$ is not onto the $k$-points $f({\mathbf{G}})(k)$. For example let
$k={\mathbb{R}}$, take ${\mathbf{G}}=\mathrm{SL}_{2}$ and let $f$ be the
central quotient map. Observe that
$f(G)=\mathrm{PSL}_{2}({\mathbb{R}})\subsetneq\mathrm{PGL}_{2}({\mathbb{R}})=f({\mathbf{G}})(k)$

As seen in the remark above, one should be careful when dealing with
$k$-points of quotient varieties, as these may not be $k$-points of a
$k$-algebraic quotient. We recall that in the case of a parabolic subgroup, or
in the case of split solvable groups, the quotient of the $k$-points is equal
to the $k$-points of the quotient group.

######  Definition 3.4 ([Bor12, Definition 15.1]).

A connected solvable $k$-group is called split if it admits a composition
series by $k$-subgroups with successive quotients $k$-isomorphic to
${\mathbf{G}}_{m}$ or ${\mathbf{G}}_{a}$.

######  Remark 3.5 ([Bor12, Theorem 15.4]).

If the field $k$ is perfect, then any connected unipotent group is $k$-split.

######  Proposition 3.6 ([Bor12, Corollary 15.7]).

Let ${\mathbf{G}}$ be a connected $k$-algebraic group and ${\mathbf{A}}$ a
normal connected $k$-split solvable subgroup. Then the $k$-point algebraic map
$\pi\colon G\to{\mathbf{G}}/{\mathbf{A}}(k)$ is surjective, i.e.
$G/A={\mathbf{G}}/{\mathbf{A}}(k)$.

######  Definition 3.7.

Let ${\mathbf{G}}$ be a connected $k$-algebraic group. A subgroup $P\leq G$ is
called parabolic if $P={\mathbf{P}}(k)$ for some $k$-parabolic subgroup
${\mathbf{P}}<{\mathbf{G}}$.

######  Lemma 3.8.

Let ${\mathbf{G}}$ be a connected $k$-algebraic group, and $P\leq G$ a
parabolic subgroup. Then the center of the parabolic subgroup, $Z(P)$, equals
the center of the group, $Z(G)$.

###### Proof.

Denote by ${\mathbf{P}}$ the Zariski closure of $P$. This is a parabolic
subgroup. Since $G$ is Zariski dense in ${\mathbf{G}}$, $Z(G)$ is central in
${\mathbf{G}}$, hence contained in ${\mathbf{P}}$. We conclude that
$Z(G)<Z(P)$. We will argue to show the other inclusion, $Z(P)<Z(G)$.

For any element in the center of $P$, $z\in Z(P)$, we may define a
$k$-morphism of algebraic varieties

$\mathbf{c}_{z}\colon{\mathbf{G}}\to{\mathbf{G}}\ \ g\mapsto[g,z]$

The map $\mathbf{c}_{z}$ is a $k$-morphism of algebraic varieties, as it is
defined by group operations in ${\mathbf{G}}$. Since $P$ is Zariski dense in
${\mathbf{P}}$, and $P\subset\mathbf{c}_{z}^{-1}(\\{e\\})$, we get that
${\mathbf{P}}\subset\mathbf{c}_{z}^{-1}(\\{e\\})$. Implying that for any
$g\in{\mathbf{G}}$, $p\in{\mathbf{P}}$,
$\mathbf{c}_{z}(gp)=\mathbf{c}_{z}(g)$. So the map $\mathbf{c}_{z}$ factors to
a $k$-morphism
$\hat{\mathbf{c}}_{z}\colon{\mathbf{G}}/{\mathbf{P}}\to{\mathbf{G}}$. Since
${\mathbf{P}}$ is parabolic, the quotient variety ${\mathbf{G}}/{\mathbf{P}}$
is a connected complete $k$-variety. Therefore the map $\hat{\mathbf{c}}_{z}$
must be constant. Thus $z\in Z({\mathbf{G}})$, and in particular in $Z(G)$, as
needed. ∎

####  3.1.1.  Algebraic linear action

Assume $k$ is of characteristic zero, and ${\mathbf{U}}$ is a connected
unipotent $k$-group. We have a $k$-isomorphism of algebraic varieties
${\mathbf{U}}\cong\operatorname{Lie}({\mathbf{U}})$. Thus letting
${\mathbf{G}}$ be a $k$-group acting $k$-algebraically by group automorphisms
on a connected abelian $k$-unipotent group, we tacitly assume that the $G$
action on $U$ is linear, where the linear structure is induced by the
$k$-isomorphism with the Lie algebra.

In the case that $k$ is of positive characteristic, auxiliary assumptions on
the $k$-groups ${\mathbf{G}}$ and ${\mathbf{U}}$ are needed, see [McN14] for a
detailed discussion.

###  3.2. Faithful actions on vector spaces

This subsection is devoted to proving proposition 3.11 bellow.

######  Lemma 3.9.

Let $L$ be a group acting faithfully by automorphisms on an abelian group $A$,
and denote $G=L\ltimes A$. Then any non-trivial normal subgroup $M\lhd G$
intersects $A$ non-trivially.

###### Proof.

Assume by contradiction that $M\cap A=\\{e\\}$. Observe that $[M,A]\subset
M\cap A=\\{e\\}$, let $m=l_{m}a_{m}$ be an element in $M$. For any $a\in A$,
$[a,m]=e$, so $a=mam^{-1}=l_{m}al_{m}^{-1}$. By the faithfulness of the
$L$-action, we get that $l=e$. Therefore $M\subset M\cap A=\\{e\\}$ is trivial
in contradiction to $M$ being non-trivial. ∎

######  Lemma 3.10.

Let $k$ be a local field, and $V$ be a finite-dimensional $k$-representation
of a group $L$. Let $N\leq V$ be an $L$ invariant additive subgroup. If the
$L$ representation $V$ is faithful and irreducible, then the $L$ action on the
quotient $V/N$ is faithful.

###### Proof.

By self-duality of local fields, the Pontryagin dual of the vector space $V$,
$\widehat{V}$, is a vector space isomorphic to $V$. The $L$ action on the
Pontryagin dual $\widehat{V}$ is defined by pre-composition and is linear
irreducible and faithful as well. Let $N^{0}=\left\\{\chi\in\widehat{V}:\
\forall n\in N\ \chi(n)=1\right\\}$ be the annihilator of the subgroup $N$.
The annihilator $N^{0}$ is $L$ invariant, so by irreducibility $N^{0}$
contains a linear basis of the dual space $\widehat{V}$. Therefore the $L$
action on $N^{0}$ is faithful making the $L$ action on the dual
$\widehat{N^{0}}\cong V/N$ faithful as well. ∎

######  Proposition 3.11.

Let $k$ be a local field, and $V$ be a finite-dimensional $k$-representation
of a group $L$ and set $G=L\ltimes V$. If the $L$ representation $V$ is
faithful and irreducible, then any normal subgroup $M\lhd G$ is either
contained in $V$ or contains $V$.

###### Proof.

Assume $V\nsubseteq M$, and denote $N=M\cap V$. The group $L$ acts by
automorphisms on $V/N$ and we may identify the quotient $G/N$ with $L\ltimes
V/N$. By lemma 3.10 the action of $L$ on $V/N$ is faithful. The normal
subgroup $M/N\lhd G/N$ intersects $V/N$ trivially, hence by lemma 3.9 $M/N$ is
trivial implying that $N=M$, thus $M\subseteq V$ as needed. ∎

######  Proposition 3.12.

Let $k={\mathbb{Q}}_{p}$ for some prime $p$, and let $V$ be an $n$-dimensional
$k$-vector space. Any additive subgroup $N\leq V$ which does not contain a
vector subspace is compact.

###### Proof.

We will show that $N$ is compact by showing its Pontryagin dual is discrete.
Observe that if $N^{0}$ is contained in a vector subspace
$W\subset\widehat{V}$, then $N$ contains the linear subspace
$W^{0}\coloneqq\\{v\in V:\forall\chi\in W\ \chi(v)=0\\}$. Since $N$ does not
contain a linear subspace we get that
$\operatorname{span}_{{\mathbb{Q}}_{p}}(N^{0})=\widehat{V}$. Hence, $N^{0}$
contains a basis, and since $N^{0}$ is ${\mathbb{Z}}$-invariant and closed it
is ${\mathbb{Z}}_{p}$-invariant and is, therefore, an open subset in
$\widehat{V}$. Therefore, $\widehat{V}/N^{0}$ is discrete and since
$\widehat{N}\cong\widehat{V}/N^{0}$, we get that $N$ is compact. ∎

##  4\. Minimal and Sealed groups

The purpose of this section is to survey two topological properties of groups,
called minimality and sealdness, as well as a relative version of these
properties. The objects studied in this article are $k$-points of algebraic
groups over a local field $k$. Therefore we will pay attention to the case
where $G$ is a locally compact Polish topological group. In order to keep this
article self-contained, we include proofs for the special case of locally
compact Polish groups. However, many of the lemmas are true for a more general
class of topological groups. For a full and comprehensive account of minimal
group topologies (as well as the relative notions for subgroups) and closed
image properties we refer the ready to [DM10], [DM14], [Meg98], [DU98] and
[Ban17].

###  4.1. Minimal topologies on groups

Let $G$ be a group, with a group topology $\tau$. That is $\tau$ is a
(Hausdorff) topology on $G$ such that the multiplication and inverse
operations are continuous. Any subgroup of a topological group is a
topological group with the subspace topology. We will denote the subspace
topology on $H\leq G$ by $\tau_{H}$, and denote the quotient topology on $G/H$
by $\tau_{G/H}$. Given two group topologies $\omega$ and $\tau$ on $G$, we say
that a group topology $\omega$ is weaker than $\tau$ if $\omega\leq\tau$, that
is, any $\omega$-open set is $\sigma$-open. We say $\omega$ is strictly weaker
than $\sigma$, if $\omega\leq\sigma$ and $\omega\neq\sigma$.

######  Definition 4.1.

A topological group $G$ is called minimal if it does not admit a strictly
weaker (Hausdorff) group topology.

######  Example 4.2.

Compact groups are minimal. Observe that the identity map between the strong
and weak topologies is a continuous and closed bijection.

######  Example 4.3.

The homothety group ${\mathbb{R}}^{*}\ltimes{\mathbb{R}}$ is minimal (See
proposition 2.9).

Example 4.3, is a well-known result (see [DS79]), the minimality of the
homothety group follows particularly from the work of Omori [Omo66, Theorem
1.1], which classifies all minimal real algebraic groups.

######  Example 4.4 (Omori).

Any real algebraic group with a compact center is minimal.

######  Definition 4.5.

Let $G$ and $H$ be topological groups, and $f\colon G\to H$ be an injective
continuous homomorphism. The topology on $G$ generated by inverse images of
open subsets in $H$ is called the $f$-initial topology and is denoted
$\tau^{f}$.

######  Remark 4.6.

The initial topology $\tau^{f}$ is a group topology weaker than the given
topology on $G$, and the map $f$ is open with respect to the initial topology.
Therefore, if $G$ is minimal then any injective continuous homomorphism is a
topological embedding onto the image.

######  Theorem 4.7 (Stoyanov, Prodanov[SP84]).

Let $G$ be a Polish abelian group. Then $G$ is minimal if and only if it is
compact.

###### Proof.

By the Stoyanov and Prodanov Theorem [DPS90, Theorem 2.7.7], any abelian
minimal group is pre-compact. Hence Polish abelian minimal groups are compact.
In this paper, we are interested in locally compact Polish groups, for sake of
completeness, we provide a short argument in the case $G$ is locally compact
abelian. Let $\beta\colon G\to\beta G$ be the Bohr compactification of the
locally compact abelian group $G$, see [Kat04, VII.5]. Since minimal groups
are homeomorphic to their images, we get that $G$ is homeomorphic to
$\beta(G)$, and since $G$ is locally compact, $\beta(G)$ is locally compact
and hence closed, implying that $G$ is compact. ∎

######  Lemma 4.8.

Let $G$ be a second countable topological group. For any $\omega$, a weaker
group topology on $G$ there exists a group topology
$\omega^{\prime}\leq\omega$, such that $\omega^{\prime}$ is second countable.

###### Proof.

Let $\sigma$ denote the given second countable topology on $\sigma$, and let
$\omega\leq\sigma$ be a weaker group topology. We will show that there exists
a second countable weaker group topology $\omega^{\prime}\leq\omega$.

Let $\\{W_{\alpha}\\}_{\alpha\in A}\in\omega$ be a system of open identity
neighborhoods. Since the group topology $\omega$ is Tychonoff,
$\cap_{\alpha\in A}\overline{W_{\alpha}}=\cap_{\alpha\in
A}W_{\alpha}=\\{e\\}$, where $\overline{W_{\alpha}}$ is the $\omega$-closure.
Observe that $\overline{W_{\alpha}}$ is $\sigma$-closed as well. Thus we
obtain an open cover $\bigcup_{\alpha\in A}(G\setminus
W_{\alpha})=G\setminus\\{e\\}$. The $\sigma$-subspace topology on
$G\setminus\\{e\\}$ is second countable and hence Lindelof, implying that
there exists a countable subcover $\\{W_{i}\\}_{i=1}^{\infty}$ such that
$\bigcup_{i=1}^{\infty}(G\setminus W_{i})=G\setminus\\{e\\}$, particularly
$\cap_{i=1}^{\infty}{W_{i}}=\\{e\\}$. By [BG17, Proposition 5.6] there exist a
weaker, a priori not necessarily Hausdorff, group topology
$\omega^{\prime}\leq\omega$, such that $\omega^{\prime}$ is second countable
and contains the collection $\\{W_{i}\\}_{i=1}^{\infty}$. But since
$\cap_{i=1}^{\infty}{W_{i}}=\\{e\\}$, the topology $\omega^{\prime}$ is indeed
Hausdorff and therefore $\omega^{\prime}$ is a second countable weaker group
topology. ∎

######  Corollary 4.9.

Let $G$ be a second countable group. If $G$ does not admit any weaker second-
countable group topologies then $G$ is minimal.

###### Proof.

Let $\sigma$ denote the given second countable topology on $G$, and let
$\omega\leq\sigma$ be a weaker group topology. By Lemma 4.8 there exist a
weaker group topology $\omega^{\prime}\leq\omega$, such that $\omega^{\prime}$
is second countable, therefore $\sigma=\omega^{\prime}$ implying that
$\sigma=\omega=\omega^{\prime}$, as needed. ∎

###  4.2. Relative minimality

The reader may benefit from keeping in mind the following example.

######  Example 4.10.

Consider the group ${\mathbb{R}}^{*}\ltimes{\mathbb{R}}$ with the standard
topology, and the subgroup
${\mathbb{R}}\leq{\mathbb{R}}^{*}\ltimes{\mathbb{R}}$. Although the group
${\mathbb{R}}$ is not minimal, it does have the following property (see Lemma
2.7): any weaker topology on ${\mathbb{R}}^{*}\ltimes{\mathbb{R}}$ agrees with
the standard topology on the subgroup ${\mathbb{R}}$.

######  Definition 4.11.

Let $G$ be a topological group, and let $U\leq G$ be a closed subgroup. We say
that $U$ is minimal in $G$ if all weaker group topologies on $G$ coincide on
the subgroup $U$.

In the context of Polish topological groups, minimality can be interpreted in
terms of a closed image property.

######  Lemma 4.12.

Let $G$ be a Polish topological group, and let $U\leq G$ be a closed subgroup.
Then $U$ is minimal in $G$ if and only if for any injective continuous
homomorphism $f\colon G\to H$ into a topological group $H$, the image $f(U)$
is closed in $H$.

###### Proof.

Assume that $U$ is minimal in $G$. Let $H$ be a topological group and $f\colon
G\to H$ be an injective continuous homomorphism with a dense image. The
$f$-initial topology $\tau^{f}$ is a weaker group topology on $G$, which must
coincide on $U$. Therefore, we get that the restriction $f_{U}\colon U\to H$
is an open bijection onto $f(U)$. Therefore the subspace topology on $f(U)$ is
Polish hence $f(U)$ is closed, see [Mel16, Theorem 2.16].

Now assume that the image of $U$ is closed under any injective map into a
topological group. Denote by $\sigma$ the given topology on $G$, and let
$\omega\leq\sigma$ be a weaker topology on $G$. Following Corollary 4.9 we
assume without loss of generality that $\omega$ is second countable.

Let $\iota\colon G\to\overline{G}$ be the completion of $G$ with respect to
the two-sided uniformity defined by the group topology $\omega$, see [Bou13,
Chapter III, §3, no. 4 ] and[Rai46], this completion is also known as the
Raikov-completion of the group.

Since $\omega$ is second countable, it follows by [Bou13, Chapter III, §3, no.
4, Proposition 7] that the group completion $\overline{G}$ is Polish. By
assumption, $\iota(U)$ is closed in $\overline{G}$, making $\iota(U)$ a Polish
group. The completion map $\iota$ is a homeomorphism onto the image, making
$\iota|_{U}\colon U\to\iota(U)$ a homeomorphism from $U$ to $\iota(U)$. Hence,
with respect to the $\omega$-subspace topology, $U$ is a Polish group.
Following the Banach Open Mapping Theorem 2.1, restricted to $U$, the identity
map onto the weaker topology is open, hence the Polish topology $\omega_{U}$
is equal to the original group topology on $U$. ∎

######  Corollary 4.13.

Let $G$ be a Polish group. Then $G$ is minimal if and only if the image of $G$
under any injective continuous homomorphism is closed.

######  Lemma 4.14 ([DPS90, Theorem 7.2.5]).

Let $G$ be a Polish group. If $G$ is minimal, then its center, $Z(G)$, is
compact.

###### Proof.

Denote by $Z=Z(G)$ the center of $G$. Let $H$ be a topological group and
$f\colon Z\to H$ a continuous injective homomorphism with a dense image. We
will show that $f(Z)$ is closed in $H$.

Since $Z$ is central, $\Delta_{Z}=\\{(z,f(z))\in G\times H:\ z\in Z\\}$, the
graph of the function $f$, is a closed normal subgroup of $G\times H$. Let
$\hat{f}\colon G\to(G\times H)/\Delta_{Z}$, be the quotient map restricted to
$G\times\\{e\\}$, $\hat{f}(g)=(g,e_{H})\Delta_{Z}$, and let $\delta\colon
H\to(G\times H)/\Delta_{Z}$ be the quotient map restricted to $\\{e\\}\times
H$, $\delta(h)=(e_{G},h)\Delta_{Z}$. Observe that $\hat{f}$ is injective, so
by minimality of $G$, $\hat{f}(G)$ is closed. From the following commuting
diagram,

${Z}$${H}$${G}$${(G\times
H)/\Delta_{Z}}$$\scriptstyle{f}$$\scriptstyle{\delta}$$\scriptstyle{\hat{f}}$

we get that $f(Z)=\delta^{-1}(\hat{f}(G))$, and as $\hat{f}$ and $\delta$ are
continuous, $f(Z)$ is a pre-image of the closed subset $\hat{f}(Z)$, hence
$f(Z)$ closed. We get that the center $Z$ is minimal therefore by Theorem 4.7
the group $Z$ is compact. ∎

######  Lemma 4.15.

Let $G$ be a Polish group, and let $A_{1},A_{2}\lhd G$ closed normal subgroups
that are minimal in $G$, and such that $A_{1}\cap A_{2}=\\{e\\}$. Then their
product $A_{1}A_{2}$ is minimal in $G$.

###### Proof.

Let $f\colon G\to H$ be an injective continuous homomorphism with a dense
image, we will show that $f(A_{1}A_{2})$ is closed. For any $N_{1},N_{2}\lhd
H$ closed normal subgroups. Letting $\rho\colon H\to H/N_{1}$ be the quotient
map, and $\iota\colon H/N_{1}\to H/N_{1}\times H/N_{2}$ be the embedding of
$H/N_{1}$ in the product $H/N_{1}\times H/N_{2}$. The maps $\rho$ and $\iota$
are continuous, making

$\psi\coloneqq\iota\circ\rho\colon H\to H/N_{1}\times H/N_{2},\ \
\psi:h\mapsto(hN_{1},eN_{2})$

a continuous map.

For $i=1,2$, the subgroup $A_{i}\lhd G$ is normal, so its image $f(A_{i})\lhd
f(G)$ is normal in the dense subgroup $f(G)$ since $f(A_{i})$ is closed in
$H$, we get that $f(A_{i})$ is a closed normal subgroup in $H$. Set
$N_{i}=f(A_{i})$, and denote by $\pi\colon H\to H/N_{1}\times H/N_{2}$ the
product of the quotient maps. Consider the continuous map

$f_{\times}\coloneqq\pi\circ f\colon G\to H/N_{1}\times H/N_{2},\ \
f_{\times}:g\mapsto(f(g)N_{1},f(g)N_{2})$

Since $A_{1}\cap A_{2}=\\{e\\}$, the map $f_{\times}$ is injective, by
relative minimality of $A_{2}$ in $G$, the subgroup $f_{\times}(A_{2})\leq
H/N_{1}\times H/N_{2}$ is closed. Therefore
$f(A_{1}A_{2})=\psi^{-1}(f_{\times}(A_{2}))$ is a preimage of a closed subset,
so it is closed. ∎

######  Corollary 4.16.

Let $G$ be a topological group, and let $A_{i}\lhd G$ $i\in\\{1,...,n\\}$ a
collection of normal subgroups that are minimal in $G$ such that $\cap
A_{i}=\\{e\\}$. Then the product $\prod_{i}A_{i}$ is minimal in $G$.

###  4.3. Co-minimality

Similarly to relative minimality, the reader may benefit from keeping in mind
the following example.

######  Example 4.17.

Consider the group ${\mathbb{R}}^{*}\ltimes{\mathbb{R}}$ with the standard
topology, and the subgroup
${\mathbb{R}}\leq{\mathbb{R}}^{*}\ltimes{\mathbb{R}}$. Although the group
${\mathbb{R}}^{*}$ is not minimal, it does have the following property (see
Lemma 2.8): any weaker topology on ${\mathbb{R}}^{*}\ltimes{\mathbb{R}}$
induces the same quotient topology on the quotient
${\mathbb{R}}^{*}\ltimes{\mathbb{R}}/{\mathbb{R}}$.

######  Definition 4.18.

Let $G$ be a topological group. A closed subgroup $U\leq G$ is said to be co-
minimal in $G$ if any weaker topology on $G$ induces the same quotient
topology on the quotient space $G/U$.

######  Example 4.19.

Any minimal co-compact subgroup is co-minimal.

######  Lemma 4.20.

Let $G$ be a topological group, and $U\leq G$ a subgroup. If $U$ is minimal in
$G$ and co-minimal in $G$, then $G$ is minimal.

###### Proof.

Let $\sigma$ denote the given group topology on $G$. For any
$\omega\leq\sigma$ weaker group topology, $\omega_{U}=\sigma_{U}$ and
$\omega_{G/U}=\sigma_{G/U}$, hence by Merson’s Lemma 2.3, $\omega=\sigma$. ∎

######  Corollary 4.21.

Let $G$ be a topological group, and $U\leq G$ be a compact open subgroup.
There does not exist a strictly weaker group topology $\omega$ on $G$, such
that $U$ is open in $\omega$.

###### Proof.

Denote by $\sigma$ the given group topology on $G$. For any $\omega\leq\sigma$
weaker group topology, by the compactness of $U$, $\omega_{U}=\sigma_{U}$. And
since $U$ is open with respect to both $\omega$ and $\sigma$, the quotient
topologies are both discrete, hence $\omega_{G/U}=\sigma_{G/U}$, hence by
Lemma 4.20, $\omega=\sigma$. ∎

###  4.4. Minimal properties of semi-direct products of groups

A group topology on a semi-direct product $L\ltimes U$ induces subspace
topologies on the subgroups $L$ and $U$, and $L$ acts continuously on $U$ by
conjugation. Conversely, for $L$ and $U$ topological groups with $L$ acting by
continuous automorphisms on $U$, the product topology is a topological group
structure on $L\ltimes U$. Here we will emphasize the relation between
topological conditions on the action of $L$ on $U$, and the minimality of $U$
as a subgroup of $L\ltimes U$.

######  Remark 4.22.

Assume $L$ acts continuously by automorphisms on $U$. If $C\leq L$ is a
subgroup in the kernel of the $L$ action then $(L\ltimes U)/C$ with the
quotient of the product topology is isomorphic to $L/C\ltimes U$ with the
product of the quotient topology. If $N\lhd U$ is an $L$-invariant closed
normal subgroup then $(L\ltimes U)/N$ with the quotient of the product
topology is isomorphic to $L\ltimes U/N$ with the product of the quotient
topology.

######  Remark 4.23.

Let $G$ be a topological group acting continuously on a topological space $X$.
Then $G$ acts continuously on $X$ with respect to any stronger topology on
$G$.

######  Lemma 4.24 ([DM10, Proposition 4.4]).

Let $L$ and $U$ be topological groups such that $L$ acts continuously by
automorphisms on $U$. Then $U$ does not admit a strictly weaker group topology
such that the action of $L$ on $U$ is continuous, if and only if $U$ is
minimal in $L\ltimes U$.

###### Proof.

Denote by $\sigma_{L}$ and $\sigma_{U}$ be the given group topologies on $L$
and $U$ respectively.

Assume $\omega$ is a group topology, weaker than the product group topology on
$L\ltimes U$. Since $\omega$ is a group topology, the subgroup $L$ with the
$\omega$-subspace topology, $\omega_{L}$, acts continuously by conjugation on
$U$ with respect to the $\omega$-subspace topology, $\omega_{U}$. By Remark
4.23, embellished with the stronger $\sigma_{L}$ group topology, the group $L$
acts continuously by conjugation on $U$ with respect to $\omega$-subspace
topology, $\omega_{U}$ on $U$ Therefore if $U$ does not admit a strictly
weaker group topology such that the action of $L$ on $U$ is continuous we get
that $\omega_{U}=\sigma_{U}$, hence $U$ is minimal in $L\ltimes U$.

Conversely, assume $U$ is minimal in $L\ltimes U$. If $U$ admits a weaker
group topology $\omega_{U}$, such that the action of $L$ on $U$ is continuous
is minimal in $G$. The product topology $\sigma_{L}\times\omega_{U}$ is a
group topology on $L\ltimes U$, and by relative minimality of $U$,
$\omega_{U}=\sigma_{U}$ as needed. ∎

######  Corollary 4.25.

Let $G=L\ltimes U$, and let $C\lhd L$ be a closed normal subgroup acting
trivially on $U$. Then $U$ is minimal in $G$, if and only if $U$ is minimal in
$G/C\simeq L/C\ltimes U$.

###### Proof.

Let $\pi\colon L\to L/C$ be the quotient map. Following the commutative
diagram,

${L\times U}$${L/C\times
U}$${U}$$\scriptstyle{\alpha}$$\scriptstyle{\pi}$$\scriptstyle{\alpha_{c}}$

and since $\pi$ is open and continuous, the map $\alpha$ is continuous if and
only if $\alpha_{c}$ is continuous. We get that $L$ acts continuously on $U$,
if and only if $L/C$ acts continuously on $U$. Hence by Lemma 4.24, $U$ is
minimal in $G$, if and only if $U$ is minimal in $G/C=L/C\ltimes U$. ∎

###  4.5. Sealed groups

We saw that Polish minimal groups have the property that any continuous
injective homomorphism mapping them to another topological group has a closed
image. In this subsection, we are interested in groups having a closed image
under any continuous homomorphism.

######  Definition 4.26.

Let $G$ be a topological group. We say that $G$ is sealed if for any
continuous homomorphism $f\colon G\to H$ into a topological group $H$, the
image $f(G)$ is closed.

######  Example 4.27.

Compact groups are sealed.

######  Example 4.28.

Simple Polish minimal groups are sealed, as any homomorphism from a simple
group is either trivial or injective.

Also here, the relative property is very useful.

######  Definition 4.29.

For a subgroup $U\leq G$, we say that $U$ is sealed in $G$, if for any
continuous homomorphism $f\colon G\to H$ into a topological group $H$, the
image $f(U)\leq H$ is closed.

######  Lemma 4.30 (Cf. [DU98, Proposition 4.2]).

Let $N\lhd G$ be a closed normal subgroup, with the quotient map $\pi\colon
G\to G/N$. If $N$ is sealed in $G$ and $U\leq G$ is a subgroup such that
$\pi(U)$ is sealed in $\pi(G)$, then $UN$ is sealed in $G$.

###### Proof.

Let $f\colon G\to H$ be a continuous homomorphism with a dense image. Since
the image of $N$ is closed, we get that $f(N)$ is a closed normal subgroup of
$H$. Therefore, the induced map $\hat{f}\colon G/N\to H/f(N)$ is a morphism of
topological groups. Since $\pi(U)$ is sealed in $\pi(G)$, we get that
$\hat{f}(\pi(U))$ is a closed subgroup in $H/f(N)$. Hence Its pre-image in
$H$, which is equal to $f(UN)$, is closed. ∎

######  Corollary 4.31.

Let $N\leq G$ be a co-compact subgroup. If $N$ is sealed, then $G$ is sealed.

###### Proof.

Since compact groups are sealed, the proof follows from Lemma 4.30 above, by
taking $U=G$. ∎

######  Corollary 4.32.

If $U_{1},U_{2}$ are normal subgroups that are sealed in $G$ then $U_{1}U_{2}$
is sealed in $G$.

###### Proof.

Let $\pi\colon G\to G/U_{2}$ be the quotient map. Given a topological group
$H$, and continuous homomorphism $f\colon G/U_{2}\to H$, the map
$f\circ\pi\colon G\to H$ is a continuous homomorphism. Since $U_{1}$ is sealed
in $G$, $f(\pi(U_{1}))$ is closed in $H$. Hence $\pi(U_{1})$ is sealed in
$\pi(G)$, so by Lemma 4.30, $U_{1}\cdot U_{2}$ is sealed in $G$. ∎

######  Corollary 4.33 ([DU98, Theorem 2.13]).

If $G_{1},G_{2}$ are sealed groups, then so is $G_{1}\times G_{2}$.

##  5\. Rigidity of scalar multiplication

Let $k$ be a local field with absolute value $|\cdot|$, and let $V$ be a
$n$-dimensional $k$-vector space with norm $\|\cdot\|$. We consider the
topology on $k$ to be the locally compact field topology, and the topology on
$V$ to be the locally compact norm topology.

Although the norm topology on $V$ is not minimal (see Theorem 4.7), we saw in
Lemma 2.5 that amongst all additive group topologies on $V$ admitting a
continuous scalar multiplication by elements in an open neighborhood of $0$,
the norm topology on $V$ is minimal.

The main goal of this section is to prove Lemma 5.1 below. The lemma asserts
that if the additive group $V$ admits a continuous scalar multiplication
restricted to an analytic multiplicative subgroup then the group topology on
$V$ is minimal.

######  Lemma 5.1.

Let $k_{0}$ be a local field such that $k$ is a finite filed extension of
$k_{0}$, and let $S\leq k^{*}$ be an infinite $k_{0}$-analytic subgroup. Then
$V$ does not admit an additive group topology, which is strictly weaker than
the norm topology, and such that the $S$ action on $V$, given by scalar
multiplication, is continuous.

Moreover, given an $S$-invariant compact subgroup $N\leq V$. The topological
group $V/N$ does not admit group topology that is strictly weaker than the
quotient topology induced from $V$, and such that the induced action of $S$ on
$V/N$ is continuous.

###  5.1. Moderated subsets

In order to prove Lemma 5.1 we will need the notion of a moderated set. We
will see in Lemma 5.8 that admitting a continuous multiplication over a
moderated subset guarantees that the analytic topology on the vector space is
minimal.

######  Definition 5.2.

Let $A\subset k$ be a subset containing $0$. We call the set $A$ moderated if
$A$ contains a null sequence $(a_{n})_{n\in{\mathbb{N}}}$, i.e.
$\lim_{n}\|a_{n}\|=0$, such that $1<|a_{n}|\cdot|a_{n+1}|^{-1}\leq\lambda$ for
some $\lambda\in{\mathbb{R}}^{+}$.

######  Example 5.3.

Any 0-neighborhood in $k$ is moderated.

######  Example 5.4.

Let $k={\mathbb{C}}$ and $A=S^{1}-1$ where $S^{1}$ is the unit circle in
${\mathbb{C}}$. For any value $\delta<1$, the set $A$ contains an element $a$
with absolute value $\delta$, so $A$ is moderated.

Example 5.4 can be viewed as a special case of Lemma 5.5 below, observing that
the set $S^{1}-1$ is an ${\mathbb{R}}$ analytic subset of ${\mathbb{C}}$.

Assume $k$ is a degree $m$ extension of a local field $k_{0}$. We may view
$k\simeq k_{0}^{m}$ as an $m$-dimensional $k_{0}$-analytic manifold. Lemma 5.5
asserts us that $k_{0}$-analytic subsets are moderated.

######  Lemma 5.5.

Assume that $k$ is a finite extension of a local field $k_{0}$, and let
$A\subset k$ be a positive dimensional $k_{0}$-analytic sub-manifold
containing $0$, then $A$ is a moderated subset.

###### Proof.

Since $A$ is a positive dimensional analytic sub-manifold, there exists an
analytic immersion $f\colon B\to A$ satisfying $f(0)=0$, where $B$ is the open
unit ball in $k_{0}$. We let $L$ be the derivative of $f$ at $0$ and denote
$a=\|L(1)\|$. We fix a non zero $x\in B$ and observe that for every $n$,
$\|L(x^{n})\|=|x^{n}|\cdot a$. We fix $\epsilon<a$ small enough so
$(a-\varepsilon)(a+\varepsilon)^{-1}>|x|$ and let
$\lambda=(a-\varepsilon)^{-1}(a+\varepsilon)|x|^{-1}$. By definition of the
differential, there exists $N\in{\mathbb{N}}$ such that for $n\geq N$,

$-\varepsilon<\frac{\|f(x^{n})\|}{|x^{n}|}-\frac{\|L(x^{n})\|}{|x^{n}|}<\varepsilon,$

thus

$0<a-\varepsilon<\frac{\|f(x^{n})\|}{|x^{n}|}<a+\varepsilon.$

We set $a_{n}=f(x^{n+N})\in A$ and note that for every $n$

$\frac{|a_{n}|}{|a_{n+1}|}=\frac{\|f(x^{n+N})\|}{|x^{n+N}|}\cdot\frac{|x^{n+N+1}|}{\|f(x^{n+N+1})\|}\cdot|x|^{-1},$

thus

$1<(a-\varepsilon)(a+\varepsilon)^{-1}|x|^{-1}<\frac{|a_{n}|}{|a_{n+1}|}<(a-\varepsilon)^{-1}(a+\varepsilon)|x|^{-1}=\lambda.$

This finishes the proof. ∎

######  Lemma 5.6.

Let $S\subset k$ be a subset such that the scalar multiplication map

$m_{0}\colon S\times V\to V,\ \ (s,v)\mapsto s\cdot v$

is continuous with respect to an additive group topology on $V$. Then the
scalar multiplication

$m_{1}\colon(S-1)\times V\to V,\ \ (s-1,v)\mapsto(s-1)\cdot v$

is continuous.

###### Proof.

The map $\mathrm{+_{1}}\colon k\to k$ sending $a$ to $a+1$ is continuous, maps
the subset $S-1$ to the subset $S$. By continuity of $m_{0}$ the map

$\alpha\colon S\times V\to V\times V,\ \ (s,v)\mapsto(s\cdot v,v)$

is continuous. The subtraction of vectors is a continuous operation on $V$.
Denoting the subtraction map by $\mathrm{-}\colon V\times V\to V$, we get
continuity of the the multiplication map,
$m_{1}\coloneqq\mathrm{-}\circ\alpha\circ(\mathrm{+_{1}}\times Id)$,

$m_{1}\colon(S-1)\times V\to V,\ \ (s-1,v)\mapsto s\cdot v-v=(s-1)\cdot v$

as needed. ∎

###  5.2. Weaker topologies on quotients of vector spaces

Similar to the assertion of Lemma 2.4, weaker topologies on quotients of
vector spaces can be classified as those having no bounded open subsets.

######  Lemma 5.7.

Let $N\leq V$ be a compact additive subgroup. Then any strictly weaker
topology on $V/N$ does not have any open sets with bounded pre-image in $V$.

###### Proof.

If $k$ is Archimedean, $V$ does not admit any non-trivial compact subgroups
and the claim follows from Lemma 2.4. So we may assume that $k$ is non-
Archimedean. Denote by $\pi\colon V\to V/N$ the quotient map. Assume by
contradiction that there exist a group topology $\omega$, which is a strictly
weaker topology on $V/N$ admitting a $0_{V/N}$-neighborhood $W\in\omega$, such
that $\|\pi^{-1}(W)\|<\lambda$. The norm closed ball,
$\overline{B}_{\lambda}\coloneqq\\{x\in V:\ |x|\leq\lambda\\}$, is a compact
subgroup of $V$ and therefore $\pi(\overline{B}_{\lambda})$ is compact
subgroup in $V/N$. Observe that $W\subset\pi(\overline{B}_{\lambda})$, thus
$\pi(\overline{B}_{\lambda})$ is $\omega$-open. Hence by Corollary 4.21 the
$\omega$ topology is equal to the quotient of the norm topology. ∎

We will see that, given a moderated subset in $k$, the norm topology is
minimal amongst the additive group topologies admitting a continuous
multiplication by this moderated set.

######  Lemma 5.8.

Let $A\subset k$ be a moderated subset, and let $m\colon A\times V\to V$ be
the scalar multiplication map. Then there does not exist an additive group
topology on $V$, which is strictly weaker than the norm topology. and such
that $m$ is a continuous map.

Moreover, given an $A$-invariant compact subgroup $N\leq V$. The topological
group $V/N$ does not admit a group topology that is strictly weaker than the
quotient topology induced from $V$, and such that the induced scalar
multiplication $m^{\prime}\colon A\times V/N\to V/N$ is a continuous map.

Reading the proof, the reader should have in mind the proof of Lemma 2.5,
which is a special case.

###### Proof.

We will show that any weaker topology on $V/N$ admitting a continuous
$A$-multiplication must have an open subset with a bounded pre-image in $V$.
The statement then follows from Lemma 5.7.

Let $\omega$ be a weaker topology on $V/N$ such that the map

$m^{\prime}\colon A\times V/N\to V/N,\ \ (a,v+N)\mapsto a\cdot v+N$

is continuous and assume by contradiction that all $\omega$ open set in $V/N$
have unbounded pre-images in $V$. Since $A$ is a moderated set there exists
$\lambda\in{\mathbb{R}}^{+}$ and a null sequence
$\\{a_{n}\\}_{n\in{\mathbb{N}}}\subset k$, with
$1<|a_{n}|\cdot|a_{n+1}|^{-1}\leq\lambda$. Let $\pi\colon V\to V/N$ be the
quotient map. Since $N$ is compact there exist $\beta\in{\mathbb{R}}^{+}$ such
that $\|N\|<\beta$. The annuls
$\operatorname{ann}{(\beta,\beta\lambda)}\coloneqq\overline{B}_{\beta\lambda}\setminus
B_{\beta}\subset V$ is a compact subset in the norm topology on $V$, and does
not intersect $N$. Therefore, the image of the annulus under $\pi$,
$\pi(\operatorname{ann}{(\beta,\beta\lambda)})$, is a compact set not
containing the identity $0_{V/N}$. Let $W$ be an $\omega$ open neighborhood
separating $0_{V/N}$ from $\pi(\operatorname{ann}{(\beta,\beta\lambda)})$. By
continuity of $m^{\prime}$, there exist a $\omega$ open $0_{V/N}$ neighborhood
$W^{\prime}$, and a open $0$ neighborhood $A_{0}\subset A$ such that
$m^{\prime}(A_{0}\times W^{\prime})\subset W$. Note that there exists $N$ such
that for every $n\geq N$, $a_{n}\in A_{0}$. Since we assumed by contradiction
that $\pi^{-1}(W^{\prime})$ is unbounded, there exists
$v\in\pi^{-1}(W^{\prime})$ such that

$\|v\|>\beta\cdot|a_{N}|^{-1}$

Since $|a_{n}|^{-1}$ is a strictly increasing sequence with
$\lim_{n}|a_{n}|^{-1}=\infty$, there exists $N<n_{0}$ such that

$\beta\cdot|a_{n_{0}}|^{-1}<\|v\|\leq\beta\cdot|a_{n_{0}+1}|^{-1},$

thus $a_{n_{0}}\cdot v\in\operatorname{ann}{(\beta,\beta\lambda)}$ since

$\beta=\beta\cdot|a_{n_{0}}|^{-1}\cdot|a_{n_{0}}|<\|a_{n_{0}}\cdot
v\|\leq\beta\cdot|a_{n_{0}+1}|^{-1}\cdot|a_{n_{0}}|\leq\beta\cdot\lambda$

Observe that $(a_{n_{0}},\pi(v))\in A_{0}\times W^{\prime}$. Since
$m^{\prime}(A_{0}\times W^{\prime})\subseteq W$, thus $W$ intersects the
annulus $\operatorname{ann}{(\beta,\beta\lambda)}$. This is a contradiction. ∎

We are now ready to prove the main lemma 5.1.

###### Proof of Lemma 5.1.

Following Lemma 5.6, we get a $S-1$ continuous multiplication on $V$.
Moreover, if $N\leq V$ in $S$-invariant it is $S-1$-invariant. By Lemma 5.5,
the $k_{0}$-analytic set $S-1$ is moderated. The lemma now follows from Lemma
5.8. ∎

##  6\. Linear algebraic action of Tori

We will be using the following consequence of Schur’s Lemma.

######  Lemma 6.1 (Schur’s lemma).

Let $(\pi,V)$ be a finite-dimensional irreducible $k$-representation of a
commutative group $T$. Then the division ring $\text{End}_{T}(V)$ is
commutative and it is a finite field extension of $k$. Moreover, as
$\text{End}_{T}(V)$-modules, $\text{End}_{T}(V)\cong V$.

###### Proof.

Since $T$ is commutative, for any $t\in T$ the linear operator $\pi(t)$ is
intertwining. We therefore get a map $\pi:T\to\text{End}_{T}(V)$.
Irreducibility of the $T$ representation implies that $V$ is irreducible as an
$\text{End}_{T}(V)$-module. Therefore $V$ must be a quotient of
$\text{End}_{T}(V)$, but since $\text{End}_{T}(V)$ is a division ring we get
that $V\cong\text{End}_{T}(V)$ as $\text{End}_{T}(V)$-modules. By irreducibly,
$\pi(T)$ spans $\text{End}_{T}(V)$, so the finite-dimensional division ring
$\text{End}_{T}(V)$ is commutative, and therefore a finite field extension of
$k$. ∎

In this section we let $k$ be a local field of characteristic zero and $V$ be
a finite-dimensional $k$-vector space. We let ${\mathbf{T}}$ be a
$k$-algebraic torus, and denote by $T={\mathbf{T}}(k)$ its $k$-points. We fix
a $k$-rational representation of algebraic groups $\pi\colon
T\to\mathrm{GL}(V)$. I.e. $\pi$ is the $k$-point map of a rational
$k$-representation. Note that by Zariski density of the $k$-points, if the
algebraic representation of ${\mathbf{T}}$ is irreducible then so is the
representation of its $k$-point $T$.

The topology assumed on the $k$-points of a $k$-algebraic group is the
$k$-analytic topology, which we will refer to as the analytic topology.

######  Lemma 6.2.

Assume that the representation $\pi$ is irreducible. Let $\alpha\colon T\times
V\to V$ denote the continuous action $\alpha(t,v)=\pi(t)v$. Then the additive
group $V$ does not admit a group topology strictly weaker than the norm
topology with respect to which $\alpha$ is a continuous action.

Moreover, suppose $N\leq V$ is a $T$-invariant compact subgroup. Then the
topological group $V/N$ does not admit a group topology that is strictly
weaker than the analytic topology and such that the induced action
$\alpha^{\prime}\colon T\times V/N\to V/N$ is a continuous map.

###### Proof.

By Schur’s Lemma above 6.1, $\pi(T)$ is a representation into
$K=\text{End}_{T}(V)$, for $K$ a finite field extension of $k$. The
representation map $\pi\colon T\to\mathrm{GL}(V)$ is a separable algebraic
morphism. Therefore, it follows by Proposition [Mar91, Chapter I, Proposition
(2.5.4)(ii)] that $\pi(T)$ is a $k$-analytic multiplicative subgroup of
$K^{*}$. By the Banach Open Mapping Theorem 2.1, $\pi$ is open onto its image
so by following diagram

${T\times V}$${\pi(T)\times
V}$${V}$$\scriptstyle{\alpha}$$\scriptstyle{\pi}$$\scriptstyle{m}$

the multiplication by $\pi(T)$ is continuous. The claim now follows directly
from Lemma 5.1. ∎

######  Corollary 6.3.

Assume that the $T$-representation $(\pi,V)$ has no invariant vectors. Let
$\alpha\colon T\times V\to V$ denote the continuous action
$\alpha(t,v)=\pi(t)v$. Then the additive group $V$ does not admit a group
topology strictly weaker than the norm topology, and such that $\alpha$ is a
continuous action.

Moreover, given a $T$-invariant compact subgroup $N\leq V$. The topological
group $V/N$ does not admit a group topology that is strictly weaker than the
analytic topology, and such that the induced action $\alpha^{\prime}\colon
T\times V/N\to V/N$ is a continuous map.

###### Proof.

By complete reducibility of tori, the vector space $V$ decomposes into
finitely many irreducible sub-representations $V=\oplus_{i=1}^{m}V_{i}$. Let
$N\leq V$ be a compact $T$-invariant subgroup, and let $\omega$ be a weaker
additive group topology on $V/N$ admitting a continuous $T$-action. We prove
by induction on the length of the representation, $m$, that $\omega$ is equal
to the analytic topology on $V/N$.

If $m=1$ the claim follows from Lemma 6.2 above. Assume $m>1$, and let
$V_{1}\leq V$ be an irreducible sub-representation. As analytic groups
$V_{1}/(N\cap V_{1})\cong V_{1}N/N$, therefore the $\omega$-subspace topology
on $V_{1}N/N$, induces a weaker group topology on $V_{1}/(V_{1}\cap N)$,
admitting a continuous $T$-action. Hence by Lemma 6.2 above, the
$\omega$-subspace topology on $V_{1}N/N$ is equal to the analytic topology. In
particular, $V_{1}N/N$ is an $\omega$-closed subgroup of $V/N$.

Let $\omega^{\prime}$ denote the $\omega$-quotient topology on
$(V/N)/(V_{1}N/N)$. Observe that as analytic groups $(V/N)/(V_{1}N/N)\cong
V/V_{1}N\cong(V/V_{1})/(V_{1}N/V_{1})$ are isomorphic, therefore
$\omega^{\prime}$ induces a weaker topology on $(V/V_{1})/(V_{1}N/V_{1})$
which admits a continuous $T$ action. Since $V_{1}N/V_{1}$ is a $T$-invariant
compact subset of $V/V_{1}$, which is of length $m-1$ and has no invariant
vectors, by induction hypothesis $\omega^{\prime}$ must be equal to the
analytic topology. Hence by applying Merson’s Lemma 2.3 to the subgroup
$V_{1}N/N$ we get that the topology $\omega$ is equal to the analytic topology
on $V/N$. ∎

######  Corollary 6.4.

Assume that the $T$-representation $(\pi,V)$ has no invariant vectors. Then
for any $T$-invariant compact subgroup $N\leq V$, the subgroup $V/N$ is
minimal in $T\ltimes V/N$.

###### Proof.

By Corollary 6.3, the topological group $V/N$ does not admit a group topology
that is strictly weaker than the analytic topology, and such that the induced
action $\alpha^{\prime}\colon T\times V/N\to V/N$ is a continuous map. Hence
by Lemma 4.24, $V/N$ is minimal in $T\ltimes V/N$. ∎

######  Lemma 6.5.

Let $C=\ker(\pi)$ be the kernel of the representation map. Then the
topological group $T/C$ does not admit a group topology that is strictly
weaker than the analytic topology, and such that the action of $T/C$ on $V$ is
continuous.

###### Proof.

Suppose $\omega$ is a weaker topology on $T/C$ such that the action on $V$ is
continuous. Since $\pi$ is a $k$-algebraic morphism, the image $\pi(T)$ is
closed in $\mathrm{GL}(V)$. Thus, the injective map $\hat{\pi}\colon
T/C\to\mathrm{GL}(V)$, is a bijection onto a Polish group. Hence by Corollary
2.2 the $\omega$-quotient topology on $T/C$ is equal to the quotient of the
analytic topology induced from $T$. ∎

##  7\. Minimal algebraic groups

The primary objective of this section is to prove the following.

######  Theorem 7.1.

Let $k$ be a local field of characteristic zero, and let ${\mathbf{G}}$ be a
connected $k$-algebraic group. Then the group $G={\mathbf{G}}(k)$ is minimal
if and only if the center of $G$ is compact.

The proof of Theorem 7.1 follows similar outlines to the proof Proposition
2.9. Proposition 7.5 establishes relative minimality of the unipotent radical,
generalizing Lemma 2.7, where relative minimality of the unipotent radical $V$
was shown. Lemma 7.6 provides co-minimality for the special case of solvable
groups, generalizing Lemma 2.8, which showed co-minimality of $V$. The Borel
subgroup $P\leq\mathrm{PGL}_{2}(k)$, which played a key role in the proof of
Proposition 2.10, is replaced here by a minimal parabolic subgroup.

Henceforth in this section, we let $k$ be a local field of characteristic
zero. As usual, algebraic groups over $k$ will be identified with their
$\bar{k}$ points and will be denoted by boldface letters. Their $k$-points
will be denoted by corresponding Roman letters. For a connected $k$-algebraic
group ${\mathbf{G}}$ we denote its unipotent radical by
$\mathcal{R}_{u}({\mathbf{G}})$, and the $k$-points of the unipotent radical
by $\mathcal{R}_{u}(G)$. A $k$-algebraic homomorphism is a homomorphism
$f\colon G\to H$ that is induced from a $k$-morphism
${\mathbf{G}}\to{\mathbf{H}}$ between algebraic groups, by restricting to its
$k$-points.

###  7.1. Relative minimality of the unipotent radical

We will show relative minimality of the unipotent radical $\mathcal{R}_{u}(G)$
in $G$. As a first step, we show that the center of the unipotent radical is
relatively minimal.

######  Lemma 7.2.

Let ${\mathbf{G}}$ be a connected $k$-algebraic group with unipotent radical
$\mathcal{R}_{u}({\mathbf{G}})$. If $G$ has a compact center, then the center
of the unipotent radical, $Z(\mathcal{R}_{u}(G))$, is minimal in $G$.

###### Proof.

We let ${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ be a Levi
decomposition of ${\mathbf{G}}$, thus ${\mathbf{L}}$ is a reductive $k$-group.
Denoting by ${\mathbf{V}}$ the center of the unipotent radical
$\mathcal{R}_{u}({\mathbf{G}})$, we get that
$V={\mathbf{V}}(k)=Z(\mathcal{R}_{u}(G))$, which is isomorphic to a finite
dimensional $k$-vector space, and the action of $L$ on $V$ is $k$\- linear,
see the discussion in §3.1.1.

Let $\omega$ be a weaker group topology on $G$. We will construct a bijective
$\omega_{V}$-continuous homomorphism from $V$ onto a Polish group, and
conclude by Corollary 2.2 that $V$ is minimal in $G$.

Fix a maximal $k$-torus ${\mathbf{T}}\leq{\mathbf{L}}$, and consider the
corresponding $k$-rational representation $\pi\colon T\to\mathrm{GL}(V)$.
Denote by $V^{T}=\\{v\in V:\pi(t)v=v\\}$ the subspace of $T$-invariant vectors
in $V$. Since the action of $T$ on $V$ is algebraic, the quotient space
$V_{T}=V/V_{T}$ is the $k$-points of the algebraic quotient
${\mathbf{V}}_{T}={\mathbf{V}}/{\mathbf{V}}^{\mathbf{T}}$. We get a
$k$-rational representation $\pi_{T}\colon T\to\mathrm{GL}(V_{T})$ such that
the $T$ action on $V_{T}$ is without invariant vectors.

The $\omega$-subspace topology on $T\ltimes V$ is a group topology, and with
respect to the $\omega$-subspace topologies, the torus $T$ acts continuously
on $V$. Since the action of $T$ is continuous, the invariant subspace $V^{T}$
is $\omega_{V}$-closed, making the $\omega$-quotient topology
$\omega_{V/V^{T}}$ on $V_{T}$ a group topology.

Remark 4.23 implies that endowing $T$ with the analytic topology, it still
acts continuously on $V$ equipped with the $\omega$-subspace topology,
$\omega_{V}$. Following Corollary 6.3, the additive group $V_{T}$ does not
admit a weaker topology such that the action of $T$ is continuous, hence the
$\omega$-quotient topology $\omega_{V/V^{T}}$ is equal to the analytic
topology.

Let $\\{{\mathbf{T}}_{i}\\}_{i=1}^{m}$ be a collection of maximal $k$-tori
that generate ${\mathbf{L}}$. Let $\Psi$ be the product of the quotient maps
from $V$ onto $V_{T_{i}}$. That is, $\Psi$ is a $k$-morphism of algebraic
groups

$\Psi\colon V\to\prod_{i=1}^{m}V_{T_{i}}.$

By Proposition 3.2, the image $\Psi(V)$ is a Polish subgroup.

For any $i=1,...,m$, the quotient map from $V$ onto $V_{T_{i}}$ is continuous
with respect to the $\omega$ induced topology. Since the $\omega$-quotient
typology, $\omega_{V/V^{T_{i}}}$, is equal to the analytic topology, the map
$\Psi$ is a $\omega_{V}$-continuous homomorphism onto the Polish group
$\Psi(V)$.

We are left to show that $\Psi$ is injective. The kernel
$\ker(\Psi)=\cap_{i=1}^{m}V^{T_{i}}$ is fixed by all ${\mathbf{T}}_{i}$, hence
is fixed by ${\mathbf{L}}$, and in particular by $L$. Since $\ker(\Psi)\leq V$
is central in $\mathcal{R}_{u}(G)$, it is a central subgroup of $G$. As $G$ is
assumed to have a compact center, $\ker(\Psi)$ must be compact, thus
$\ker(\Psi)$ must be trivial, as it is a vector subspace of $V$. We got that
the map $\Psi$ is indeed a $\omega_{V}$-continuous bijection from $V$ onto a
Polish space, as needed. ∎

Next, we observe that for any $k$-group, the relative minimality of the center
of its unipotent radical implies the relative minimality of the unipotent
radical itself.

######  Lemma 7.3.

Let ${\mathbf{U}}$ be a connected unipotent $k$-group. Let $\omega$ be a group
topology on $U$, weaker than the analytic topology. If the $\omega$-subspace
topology is equal to the analytic topology on $Z(U)$, then $\omega$ is equal
to the analytic topology on $U$.

###### Proof.

Let $\\{{\mathbf{Z}}_{i}\\}_{i=1}^{n}$ denote the upper central series of the
group ${\mathbf{U}}$. By Zariski-density, the collection
$Z_{i}={\mathbf{Z}}_{i}(k)$ is the upper central series for $U$. We show that
if the $\omega$-subspace topology on $Z_{i}$ is equal to the analytic topology
then the $\omega$-subspace topology on $Z_{i+1}$ is equal to the analytic
topology.

Indeed, assume $\omega_{Z_{i}}$ is equal to the analytic topology. Let
$\rho\colon{\mathbf{Z}}_{i}\to{\mathbf{Z}}_{i}/{\mathbf{Z}}_{i-1}$ be the
algebraic quotient map, for any $u\in{\mathbf{U}}$ define the $k$-morphism of
algebraic varieties

$\mathbf{c}_{u}\colon{\mathbf{Z}}_{i+1}\to{\mathbf{Z}}_{i},\
\mathbf{c}_{u}(h)=[h,u].$

Observe that the composition $\varphi_{u}=\rho\circ\mathbf{c}_{u}$ is a
homomorphism, as for any $x,y\in{\mathbf{Z}}_{i+1}$

$\displaystyle\varphi_{u}(xy)$ $\displaystyle=[xy,u]\ \ {\mathbf{Z}}_{i-1}$
$\displaystyle=xyuy^{-1}(u^{-1}u)x^{-1}u^{-1}\ \ {\mathbf{Z}}_{i-1}$
$\displaystyle=x[y,u]ux^{-1}u^{-1}\ \ {\mathbf{Z}}_{i-1}$
$\displaystyle=xux^{-1}u^{-1}[y,u]\ \ {\mathbf{Z}}_{i-1}$
$\displaystyle=\varphi_{u}(x)\varphi_{u}(y)$

So $\varphi_{u}$ is a $k$-morphism of algebraic groups, we will denote by
$\varphi^{k}_{u}\colon Z_{i+1}\to({\mathbf{Z}}_{i}/{\mathbf{Z}}_{i-1})(k)$ the
map between the $k$-points.

We claim that there exists a finite collection $\\{u_{j}\\}_{j=1}^{m}\in U$
such that $\cap_{j=1}^{m}\ker(\varphi^{k}_{u_{j}})=Z_{i}$.

First, we claim that it is enough to intersect over the Zariski-dense set $U$
i.e. that

$\bigcap_{u\in{\mathbf{U}}}\ker(\varphi_{u})=\bigcap_{u\in
U}\ker(\varphi_{u})$

Indeed, let $x\in\cap_{u\in U}\ker(\varphi_{u})$ then
$[x,U]\subset{\mathbf{Z}}_{i-1}$, by Zarsiki-density we get that
$[x,{\mathbf{U}}]\subset{\mathbf{Z}}_{i-1}$, implying that
$x\in\cap_{u\in{\mathbf{U}}}\ker(\varphi_{u})$. So
$\cap_{u\in{\mathbf{U}}}\ker(\varphi_{u})=\cap_{u\in
U}\ker(\varphi_{u})={\mathbf{Z}}_{i}$ as claimed.

Since any descending chain of Zariski closed subsets must eventually be
constant, there exists a finite collection $\\{u_{j}\\}_{j=1}^{m}\in U$ such
that $\cap_{j=1}^{m}\ker(\varphi_{u_{j}})={\mathbf{Z}}_{i}$. Since
$\ker(\varphi_{u_{j}})(k)=\ker(\varphi^{k}_{u_{j}})$, see Proposition 3.2, we
get that $\cap_{j=1}^{m}\ker(\varphi^{k}_{u_{j}})=Z_{i}$.

The product map

$\Psi=\prod_{i=j}^{m}\varphi^{k}_{u_{j}}\colon
Z_{i+1}\to({\mathbf{Z}}_{i}/{\mathbf{Z}}_{i-1})(k)^{m}$

is $k$-algebraic homomorphism, hence its image $\Psi(Z_{i+1})$ is a closed
subgroup with respect to the analytic topology on
$({\mathbf{Z}}_{i}/{\mathbf{Z}}_{i-1})(k)^{m}$. By construction,
$\ker(\Psi)=\cap_{j=1}^{m}\ker(\varphi_{u_{j}})=Z_{i}$, and so the map $\Psi$
factors through a $\omega$-quotient continuous bijection map from
$Z_{i+1}/Z_{i}$ onto the Polish group $\Psi(Z_{i+1})$. We get by Corollary
2.2, that the $\omega$-quotient topology on $Z_{i+1}/Z_{i}$ is equal to the
analytic topology, therefore, by applying Merson’s Lemma 2.3 to the subgroup
$Z_{i}\leq Z_{i+1}$ we get that the $\omega$-topology on $Z_{i+1}$ is equal to
the analytic topology, as needed.

By assumption, the $\omega$-subspace topology on $Z_{1}$ is equal to the
analytic topology, by applying the claim above inductively for $i=1,..,n$ we
get that the $\omega$ topology in $Z_{n}=U$ is equal to the analytic topology.
∎

######  Remark 7.4.

Note that the proof of Lemma 7.3 above, did not use any characteristic
assumptions on the local field $k$.

######  Proposition 7.5.

Let ${\mathbf{G}}$ be a connected $k$-algebraic group with unipotent radical
$\mathcal{R}_{u}({\mathbf{G}})$. If $G$ has a compact center, then the
unipotent radical $\mathcal{R}_{u}(G)$, is minimal in $G$.

###### Proof.

Suppose that $G$ has a compact center. By Lemma 7.2, the center of
$\mathcal{R}_{u}(G)$ is minimal, and we conclude by Lemma 7.3. ∎

###  7.2. Minimal conditions for solvable groups

In this subsection, we assume ${\mathbf{G}}$ is connected solvable
$k$-algebraic group. Recall that by structure theory of connected solvable
groups [Bor12, Theorem 10.6], ${\mathbf{G}}$ admits a Levi decomposition such
that the Levi subgroup is a maximal torus defined over $k$.

######  Lemma 7.6.

Let ${\mathbf{G}}$ be a connected solvable $k$-algebraic group, and let
${\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ be a $k$-Levi decomposition
of ${\mathbf{G}}$. Denote by $C=\\{t\in T:\forall u\in\mathcal{R}_{u}(G),\
[t,u]=e\\}$ the kernel of the conjugation action of $T$ on
$\mathcal{R}_{u}(G)$. Then there exist a finite dimensional $k$-vector space
$V$, and an algebraic $k$-rational representation $\pi\colon
G\to\mathrm{GL}(V)$, with kernel $\ker(\pi)=C\times\mathcal{R}_{u}(G)$ such
that:

  1. (a)

For any group topology $\omega$, weaker than the analytic topology on $G$, if
the $\omega$-subspace topology on $\mathcal{R}_{u}(G)$ equals the analytic
topology, then $\pi$ is $\omega$-continuous.

  2. (b)

For any closed subgroup $C_{0}\leq C$, and for any group topology
$\omega^{\prime}$, weaker than the analytic topology on $G/C_{0}$, if the
$\omega^{\prime}$-subspace topology on $\mathcal{R}_{u}(G)$ equals the
analytic topology, then the map induced by on the quotient $\hat{\pi}\colon
G/C_{0}\to\mathrm{GL}(V)$ is $\omega^{\prime}$-continuous.

###### Proof.

Let $\\{Z_{i}\\}_{i=1}^{m}$ be the upper central series of the nilpotent group
$\mathcal{R}_{u}(G)$. For any $i=1,...,m$, the subgroup $Z_{i}$ is normal in
$G$, and thus we have an algebraic action of $G$ on the finite dimensional
$k$-vector space $V_{i}=Z_{i}/Z_{i-1}$. We define the representation space $V$
by taking the direct sum of the central quotient spaces,
$V=\oplus_{i=1}^{m}V_{i}$, and thus attain a $k$-rational representation
$\pi\colon G\to\mathrm{GL}(V)$.

By construction, for any $i=1,...m$, $[\mathcal{R}_{u}(G),Z_{i}]\subset
Z_{i-1}$ hence $\mathcal{R}_{u}(G)\in\ker(\pi)$. By definition
$[C,Z_{i}]=\\{e\\}$, so $C\in\ker(\pi)$ as well, implying that
$C\mathcal{R}_{u}(G)\subset\ker(\pi)$. We will show that $\ker(\pi)\subset
C\mathcal{R}_{u}(G)$. Let $x\in\ker(\pi)$, $x=tu$ for some $t\in T$ and
$u\in\mathcal{R}_{u}(G)$, since $u\in\ker(\pi)$ we get that $t\in\ker(\pi)$.
We prove by induction on $i=1,...,m$ that $[t,Z_{i}]=\\{e\\}$. For $i=1$ the
assertion is obvious. Assume that $[t,Z_{i-1}]=\\{e\\}$, we will show that
$[t,Z_{i}]=\\{e\\}$. Fix $z_{i}\in Z_{i}$. Since $t$ is in the kernel of the
action, the unipotent element $[t,z_{i}]$ is contained in $Z_{i-1}$. By
induction assumption, the semisimple element $t$, and thus its inverse
$t^{-1}$, commutes with $[t,z_{i}]$. Observe that
$t^{-1}\cdot[t,z_{i}]=z_{i}tz_{i}^{-1}$ is semisimple, so by uniqueness of
Jordan decomposition $[t,z_{i}]=e$ as needed. We get that
$[t,Z_{m}]=[t,\mathcal{R}_{u}(G)]=\\{e\\}$, and hence $t\in C$, and $x\in
C\mathcal{R}_{u}(G)$. Thus $\ker(\pi)=C\mathcal{R}_{u}(G)$.

Since (a) is a special case of (b), obtained by choosing $C_{0}=\\{e\\}$, we
are left to prove (b). Let $C_{0}$ be a closed subgroup of $C$, and set
$G_{0}=G/C_{0}=T/C_{0}\ltimes\mathcal{R}_{u}(G)$. Let $\omega^{\prime}$ be a
weaker group on $G_{0}$, such that the $\omega^{\prime}$-subspace topology on
$\mathcal{R}_{u}(G)$ equals the analytic topology. The conjugation action of
$G_{0}$ on $Z_{i}$ is continuous with respect to the
$\omega^{\prime}$-subspace topologies. In turn, the induced action of $G_{0}$
on $Z_{i}/Z_{i-1}$ is continuous with respect to the $\omega^{\prime}$-induced
topologies. Since $Z_{i}\leq\mathcal{R}_{u}(G)$ the $\omega^{\prime}$-induced
topology in $Z_{i}$ is analytic and hence the $\omega^{\prime}$-induced
topology on $Z_{i}/Z_{i-1}$ is equal to the standard analytic norm topology on
the vector space. By the construction of the map $\pi$, the action of $G$ on
$V$ is via the conjugation action of $G$ on $Z_{i}$, which is
$\omega^{\prime}$ continuous. Therefore the action of $G/C_{0}$ on $V$ via the
conjugation action of $G_{0}$ on $Z_{i}$ is $\omega^{\prime}$ continuous. ∎

######  Corollary 7.7.

Let ${\mathbf{G}}$ be a connected solvable $k$-algebraic group, and let
${\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ be a $k$-Levi decomposition
of ${\mathbf{G}}$. Denote by $C=\\{t\in T:\forall u\in\mathcal{R}_{u}(G),\
[t,u]=e\\}$ the kernel of the conjugation action of $T$ on
$\mathcal{R}_{u}(G)$. Let $C_{0}\leq C$, be a closed co-compact subgroup of
$C$. Let $\omega$ be a group topology, weaker than the analytic topology on
$G/C_{0}$. If the $\omega$-subspace topology on $\mathcal{R}_{u}(G)\leq
G/C_{0}$ equals the analytic topology, then $\omega$ equals the analytic
topology on $G/C_{0}$.

###### Proof.

Denote $G_{0}=G/C_{0}=T/C_{0}\ltimes\mathcal{R}_{u}(G)$, and let
$A=C/C_{0}\times\mathcal{R}_{u}(G)\leq G_{0}$. By assumption, the
$\omega$-subspace topology on $\mathcal{R}_{u}(G)$ is equal to the analytic
topology, implying that $\mathcal{R}_{u}(G)$ is $\omega$-closed. Since
$C/C_{0}$ is compact the $\omega$-quotient topology on $A/\mathcal{R}_{u}(G)$
is equal to the analytic quotient topology, hence by Merson’s Lemma 2.3 the
$\omega$-subspace topology on $A$ is equal to the analytic topology. In
particular, we get that $A$ is $\omega$-closed and that the $\omega$-quotient
topology on $G_{0}/A$ is a group topology weaker than the analytic quotient
topology.

Following Lemma 7.6 above, there exists a $k$-rational representation
$\pi\colon G\to\mathrm{GL}(V)$, with kernel $C\times\mathcal{R}_{u}(G)$, such
that the representation map $\hat{\pi}\colon G_{0}\to\mathrm{GL}(V)$ is
$\omega$ continuous.

Since $\pi$ is algebraic the image $\pi(G)$ is closed, hence we get that
$\hat{\pi}$ is a $\omega$-continuous representation with kernel $A$. This
gives us a $\omega_{G_{0}/A}$ continuous injective map from $G_{0}/A$ onto the
Polish group $\pi(G)$. By Corollary 2.2 the $\omega$-quotient topology,
$\omega_{G_{0}/A}$ is equal to the analytic quotient topology on $G_{0}/A$.
Applying Merson’s Lemma 2.3 to the subspace $A$, we get that the $\omega$
topology on $G_{0}$ is equal to the analytic topology. ∎

######  Corollary 7.8.

Let ${\mathbf{G}}$ be a connected solvable $k$-algebraic group, and let
${\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ be a $k$-Levi decomposition
of ${\mathbf{G}}$. Denote by $C=\\{t\in T:\forall u\in\mathcal{R}_{u}(G),\
[t,u]=e\\}$ the kernel of the conjugation action of $T$ on
$\mathcal{R}_{u}(G)$. Let $\omega$ be a group topology, weaker than the
analytic topology on $G$. If $C$ is compact and the $\omega$-subspace topology
on $\mathcal{R}_{u}(G)$ equals the analytic topology, then $\omega$ is equal
to the analytic topology in $G$.

###### Proof.

Since $C$ is compact, this follows by Corollary 7.7 above, by taking
$C_{0}=\\{e\\}$. ∎

We can now show Theorem 7.1 for the special case of solvable groups.

######  Proposition 7.9.

Let ${\mathbf{G}}$ a connected solvable $k$-algebraic group. Then $G$ is
minimal if and only if the center of $G$ is compact.

Moreover, if the center of $G$ is compact then for any central subgroup
$C_{0}\leq G$, the group $G/C_{0}$ is minimal.

###### Proof.

If $G$ is minimal then by Lemma 4.14, its center is compact. Assume that the
center of $G$ is compact, and let
${\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ be a $k$-Levi decomposition
of ${\mathbf{G}}$. Since $G$ is Zariski dense in ${\mathbf{G}}$, $Z(G)\leq
Z({\mathbf{G}})$. Let $z\in Z(G)$ and write $z=tu$ for $t\in T$ and
$u\in\mathcal{R}_{u}(G)$, observe that $[t,u]=e$ hence the decomposition
$z=tu$ is a Jordan decomposition of $z\in Z({\mathbf{G}})$, implying that
$u\in Z(G)$, but since $Z(G)$ is compact $u=e$. Therefore $Z(G)\subset T$, and
$Z(G)=\\{t\in T:\forall u\in\mathcal{R}_{u}(G),\ [t,u]=e\\}$ is the kernel of
the conjugation action of $T$ on $\mathcal{R}_{u}(G)$.

Let $\omega$ be a group topology on $G$, weaker than the analytic topology.
Since $Z(G)$ is compact, by Proposition 7.5, $\mathcal{R}_{u}(G)$ is minimal
in $G$. We get that $\omega_{\mathcal{R}_{u}(G)}$ is equal to the analytic
topology, hence by Corollary 7.8, the topology $\omega$ is equal to the
analytic topology on $G$.

Let $C_{0}\leq Z(G)$ be a central subgroup. Since $\mathcal{R}_{u}(G)$ is
minimal in $G$, following Corollary 4.25, $\mathcal{R}_{u}(G)$ is minimal in
$G/C_{0}$. Hence by Corollary 7.7, $\omega^{\prime}$ is equal to the analytic
topology on $G/C_{0}$. ∎

###  7.3. Minimal algebraic groups

######  Lemma 7.10.

Let ${\mathbf{G}}$ be a connected reductive $k$-algebraic group, and let
$P\leq G$ be a minimal parabolic in $G$. If $G$ has a compact center, then $P$
is a minimal topological group.

Moreover, for any central $k$-algebraic subgroup ${\mathbf{Z}}_{0}\leq
Z({\mathbf{P}})$, the group $P/Z_{0}$ is minimal.

###### Proof.

By the structure of parabolic subgroups [Bor12, Proposition 20.6], the minimal
parabolic subgroup $P\leq G$ is equal to the semi-direct product
$\mathrm{C}_{G}(T)\ltimes\mathcal{R}_{u}(P)$, where ${\mathbf{T}}$ is a
maximal $k$-split torus, and $\mathrm{C}_{G}(T)$ is the centralizer of $T$ in
$G$.

Denote by ${\mathbf{A}}$ the solvable $k$-algebraic group
${\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$. We claim that $Z(A)<Z(P)$.
Indeed, for $a\in Z(A)$, since $T\subset A$, clearly $a\in\mathrm{C}_{G}(T)$
hence $a\in T$, so $[a,\mathrm{C}_{G}(T)]=\\{e\\}$ and thus $a\in Z(P)$.
Following Lemma 3.8, $Z(P)=Z(G)$, and $Z(G)$ is compact, therefore $Z(A)$ is
compact. Hence, by Proposition 7.9 the group $A$ is minimal.

Moreover, let $Z_{0}\leq Z(P)$ be a $k$-algebraic subgroup. As $Z(P)=Z(G)$,
the group $Z_{0}$ is central in $G$ and is therefore a central algebraic
torus. The group $A^{\prime}=TZ_{0}\ltimes\mathcal{R}_{u}(G)$ is equal to the
$k$-points of the $k$-algebraic group ${\mathbf{A}}{\mathbf{Z}}_{0}$. Since
$T\subset A^{\prime}$ for any $a^{\prime}\in Z(A^{\prime})$,
$a^{\prime}\in\mathrm{C}_{G}(T)$ hence $a\in TZ_{0}$ and therefore commutes
with $\mathrm{C}_{G}(T)$ as well, making $a\in Z(P)$. Hence $Z(A)$ is compact,
and by Proposition 7.9 the group $A^{\prime}/Z_{0}$ is minimal.

Following [Mar91, Chapter I, Proposition (2.3.6)], the anisotropic group
$\mathrm{C}_{G}(T)/T$ is compact. Therefore the solvable subgroup $A\leq P$ is
co-compact, and in particular, $A^{\prime}$ is co-compact in $P$. Hence $A$, (
$A^{\prime}/Z_{0}$), is a minimal co-compact subgroup of $P$, ($P/Z_{0}$) thus
by Lemma 4.20, $P$ ($P/Z_{0}$), is minimal. ∎

Since parabolic subgroups are co-compact, we get the minimality criteria for
reductive groups.

######  Proposition 7.11.

Let ${\mathbf{G}}$ be a connected reductive $k$-algebraic group. Then $G$ is
minimal if and only if the center of $G$ is compact.

Moreover, if the center of $G$ is compact then for any finite central subgroup
$Z_{0}\leq G$, the group $G/Z_{0}$ is minimal.

###### Proof.

If $G$ is minimal then, by Lemma 4.14, its center is compact. Assume that the
center of $G$ is compact. If $Z_{0}$ is a central subgroup of the reductive
group $G$, it is contained in any maximal torus, and in particular $Z_{0}\leq
P$. Let $\omega$ be a group topology on $G$ (or on $G/Z_{0}$), weaker than the
analytic topology. By Lemma 7.10, $\omega_{P}$ (respectively,
$\omega_{P/Z_{0}}$) is equal to the analytic topology. In particular $P$
(resp. $P/Z_{0}$), is $\omega$-closed. Since $P$ is co-compact in $G$, the
$\omega$-quotient topology is Hausdorff and equal to the analytic quotient
topology. Hence by Merson’s Lemma 2.3 $G$, ($G/Z_{0}$), is minimal. ∎

We are now ready to prove Theorem 7.1 which states that having a compact
center is equivalent to minimality for any connected algebraic group over a
local field of characteristic zero.

###### Proof of theorem 7.1.

If $G$ is minimal, then by Lemma 4.14, the center of $G$ is compact.

For the other direction, let
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$, be a Levi decomposition of
${\mathbf{G}}$, with ${\mathbf{L}}$ a reductive $k$-group. The central
subgroup, $Z({\mathbf{L}})^{\circ}$ is a $k$-torus [Bor12, Proposition 11.21],
and we denote by ${\mathbf{T}}\leq Z({\mathbf{L}})^{\circ}$ the maximal
$k$-split subtorus as in [Bor12, 8.15]. Set
${\mathbf{A}}={\mathbf{T}}\ltimes\mathcal{R}_{u}({\mathbf{G}})\leq{\mathbf{G}}$.
We will show that $A$ is minimal and co-minimal in $G$.

Let $\omega$ be a group topology on $G$, weaker than the analytic topology.
Let $C=\\{t\in T:\forall u\in\mathcal{R}_{u}(G),\ [t,u]=e\\}$, since $T\leq
Z(L)$, $C$ is a central subgroup of $G$ and hence compact. Moreover, by
Proposition 7.5, the $\omega$-subspace topology on $\mathcal{R}_{u}(G)$ is
equal to the analytic topology. Hence by Corollary 7.8, the $\omega$-subspace
topology on $A$ is equal to the analytic topology, i.e. $A$ is relatively
minimal in $G$.

In order to show co-minimality of $A$, we will show that the group $G/A$ is
minimal. Observe that since ${\mathbf{A}}$ is a $k$-split solvable group then
$({\mathbf{G}}/{\mathbf{A}})(k)\cong G/A$ therefore, we may regard $G/A$ as
$k$-points of the reductive $k$-algebraic group ${\mathbf{G}}/{\mathbf{A}}$.
We claim that the center of $L/T\cong G/A$ is compact.

Let $\pi\colon{\mathbf{L}}\to{\mathbf{L}}/{\mathbf{T}}$ be the $k$-algebraic
quotient map. We claim that
$\pi(Z({\mathbf{L}}))=Z({\mathbf{L}}/{\mathbf{T}})$, clearly
$\pi(Z({\mathbf{L}}))\subset Z({\mathbf{L}}/{\mathbf{T}})$. For the other
direction, let $s\in{\mathbf{L}}$ be an element that is mapped into
$Z({\mathbf{L}}/{\mathbf{T}})$, define the algebraic map

$\mathbf{c}_{s}\colon{\mathbf{L}}\to{\mathbf{L}},\ \ \mathbf{c}_{s}\colon
l\mapsto[s,l]$

Observe that
$\mathbf{c}_{s}({\mathbf{L}})\subset\mathcal{D}({\mathbf{L}})\cap{\mathbf{T}}$
and since ${\mathbf{L}}$ is connected, $\mathbf{c}_{s}({\mathbf{L}})$ is
connected. Following [Bor12, Proposition 14.2], the intersection
$\mathcal{D}({\mathbf{L}})\cap{\mathbf{T}}$ is finite, hence by connectedness
$\mathbf{c}_{s}({\mathbf{L}})=\\{e\\}$, so $s\in Z({\mathbf{L}})$ as needed.
By restricting the map $\pi$ to the central subgroup, we get a quotient map
$\pi\colon Z({\mathbf{L}})\to Z({\mathbf{L}}/{\mathbf{T}})$. By [Mar91,
Chapter I, Proposition (1.4.5)(i)], maximal $k$-split tori in
$Z({\mathbf{L}}/{\mathbf{T}})$ are images of maximal $k$-split tori in
$Z({\mathbf{L}})$. By the fact that $\pi$ maps ${\mathbf{T}}$ to $\\{e\\}$, we
get that $Z({\mathbf{L}}/{\mathbf{T}})$ has no maximal $k$-split torus,
implying that $Z(L/T)$ is compact. Hence by Proposition 7.11, the group $L/T$
is minimal. Finally, by applying Meson’s Lemma 2.3 to $A<G$, since the
$\omega$-subspace topology on $A$ is analytic and $G/A\cong L/T$ is minimal,
we get that the $\omega$ topology on $G$ is equal to the analytic topology on
$G$. ∎

##  8\. Sealed algebraic groups

We saw that for algebraic groups over local fields of characteristic zero,
having a compact center is a sufficient condition for minimality. Mayer showed
in [May97] that for connected locally compact groups this necessary condition
is in fact sufficient - a connected locally compact group such that all of its
quotients have compact centers is sealed. The following is a generalization
that applies to connected algebraic groups over an arbitrary local field of
characteristic zero.

######  Theorem 8.1.

Let $k$ be a local field of characteristic $0$ and let ${\mathbf{G}}$ be a
connected $k$-group with a $k$-Levi decomposition
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$. The following are
equivalent

  1. (1)

The group ${\mathbf{G}}(k)$ is sealed.

  2. (2)

For any normal unipotent $k$-algebraic subgroup ${\mathbf{N}}\lhd{\mathbf{G}}$
the center $Z(({\mathbf{G}}/{\mathbf{N}})(k))$ is compact.

  3. (3)

The center of the reductive group ${\mathbf{L}}$ is anisotropic (equivalently,
the center of ${\mathbf{L}}(k)$ is compact) and the action of ${\mathbf{L}}$
on $\mathcal{R}_{u}({\mathbf{G}})$ is with no non-trivial fixed points,

  4. (4)

The center of the reductive group ${\mathbf{L}}$ is anisotropic (equivalently,
the center of ${\mathbf{L}}(k)$ is compact) and the action of ${\mathbf{L}}$
on $\operatorname{Lie}(\mathcal{R}_{u}({\mathbf{G}}))$ is with no invariant
vectors.

######  Remark 8.2.

One can define condition (2’) to be the condition that the center of any
quotient group is compact, note that clearly (2’) implies (2). For connected
real algebraic groups, the equivalence of (2’) and (1) is shown in [May97,
Proposition 2.14] and the equivalence of (4) and (1) is shown in [May97,
Theorem 2.5]. In both cases the methods of proof rely on the result of Omori,
and they differ from the methods used here.

In this section we assume that $k$ is a local field of characteristic zero. As
usual, algebraic groups over $k$ will be identified with their $\bar{k}$
points and will be denoted by boldface letters. Their $k$-points will be
denoted by corresponding Roman letters.

We let ${\mathbf{G}}$ a connected $k$-algebraic group and denote its unipotent
radical by $\mathcal{R}_{u}({\mathbf{G}})$, the $k$-points of the unipotent
radical will be denoted by $\mathcal{R}_{u}(G)$. We recall that the group
$G^{+}$ is the subgroup of $G$ generated by the unipotent element in $G$, see
[Mar91, Chapter I, (1.5.2)].

###  8.1. Sealed reductive groups

In this subsection, we will consider the special case where ${\mathbf{G}}$ is
a connected reductive $k$-group.

Since the sealed property is closed under taking products and quotients and
has the three spaces property (see Lemma 4.30), the sealed property for
reductive groups follows by analyzing the case for almost $k$-simple groups.
The following Lemma is a version of Proposition 2.11.

######  Lemma 8.3.

Let ${\mathbf{G}}$ be a connected almost $k$-simple group. Then $G$ is sealed.

###### Proof.

Let $f\colon G\to H$, be a continuous homomorphism into a topological group
$H$. By [Mar91, Chapter I, Theorem (1.5.6)(i) and Theorem (2.3.1)(b)], any
non-trivial normal subgroup is either co-compact or central and finite. If the
kernel of $f$ is co-compact then the image $f(G)$ is compact and hence closed.
Thus we may assume that $\ker(f)\leq G$ is a finite central subgroup. Let
$\hat{f}\colon G/\ker(f)\to H$ be the injective map induced by $f$. By
Proposition 7.11 the group $G/\ker(f)$ is minimal hence
$f(G)=\hat{f}(G/\ker(f))$ is closed in $H$. ∎

We saw in Corollary 4.33, that a product of sealed groups is sealed. Thus, as
semisimple groups are a quotient of a product of $k$-almost simple groups, see
[Bor12, Theorem 22.10], we obtain the following theorem, which was proved by
Bader-Gelander in [BG17] using a different method.

######  Corollary 8.4 ([BG17, Theorem 5.1]).

Let ${\mathbf{G}}$ be a connected semisimple $k$-group, then $G$ is sealed.

######  Lemma 8.5.

Let ${\mathbf{G}}$ be a connected semisimple $k$-group, then $G^{+}$ is
sealed.

###### Proof.

Assume without loss of generality that ${\mathbf{G}}$ has no anisotropic
factors. Let $\tilde{{\mathbf{G}}}$ be the simply connected covering of
${\mathbf{G}}$. Following [Mar91, Chapter I, Theorem (2.3.1)(a)],
$\tilde{{\mathbf{G}}}(k)=\tilde{G}^{+}$, so by Corollary 8.4, $\tilde{G}^{+}$
is sealed. Since the quotient map,
$f\colon\tilde{{\mathbf{G}}}\to{\mathbf{G}}$ is a central $k$ isogeny, we have
by [Mar91, Chapter I, Proposition (1.5.5)] that
$f(\tilde{{\mathbf{G}}}(k)^{+})=G^{+}$, so $G^{+}$ is sealed as it is a
quotient of a sealed group. ∎

######  Corollary 8.6.

Let ${\mathbf{G}}$ be a connected semisimple $k$-group. Then any finite index
subgroup of $G$ is sealed.

###### Proof.

Let $H$ be a finite index subgroup of $G$. The subgroup $G^{+}$ is co-compact
and is contained in any subgroup of finite index (see [Mar91, Chapter I,
Corollary (1.5.7) and Theorem (2.3.1)(b)]). Since $G^{+}$ is sealed and co-
compact in $H$, by Corollary 4.31, $H$ is sealed. ∎

######  Proposition 8.7.

Let ${\mathbf{G}}$ be a connected reductive $k$ group, then $G$ is sealed if
and only if $G$ has a compact center.

###### Proof.

Following [Bor12, Proposition 14.2], the solvable radical
$\mathcal{R}_{s}({\mathbf{G}})=Z({\mathbf{G}})^{\circ}$ is a $k$-torus. Since
$Z(G)=Z({\mathbf{G}})(k)$ is compact we get that the solvable radical
$\mathcal{R}_{s}(G)$ is a sealed normal subgroup. The algebraic $k$-group
${\mathbf{G}}/\mathcal{R}_{s}({\mathbf{G}})$ is a connected semisimple
$k$-group, and the quotient group $G/\mathcal{R}_{s}(G)$ is isomorphic to a
finite index subgroup of ${\mathbf{G}}/\mathcal{R}_{s}({\mathbf{G}})$ [PRR93,
Proposition 6.14].

Hence by Corollary 8.6, $G/\mathcal{R}_{s}(G)$ is sealed. We get by the three
spaces property for sealed groups, Lemma 4.30, that the group $G$ is sealed. ∎

######  Lemma 8.8.

Let ${\mathbf{G}}$ be a connected reductive $k$ group such that $G$ has a
compact center. Then for any normal connected $k$ group
${\mathbf{N}}\lhd{\mathbf{G}}$, the center of $N$ is compact.

###### Proof.

Let ${\mathbf{Z}}\leq{\mathbf{G}}$ be the center of ${\mathbf{G}}$, the group
${\mathbf{G}}/{\mathbf{Z}}$ is semisimple and so the image of ${\mathbf{N}}$
under the central quotient map is a normal subgroup of a semisimple group so
is semisimple (see [Bor12, Proposition14.10]). Thus $Z(N/Z)$ is compact, hence
by compactness of $Z$, $Z(N)$ is compact. ∎

######  Corollary 8.9.

Let ${\mathbf{G}}$ be a connected reductive $k$-group with a compact center.
Then for any $k$-algebraic normal subgroup ${\mathbf{N}}$, the group
$N={\mathbf{N}}(k)$ is sealed.

###  8.2. Proof of Theorem 8.1

Ultimately, we will prove Theorem 8.1 by induction on the nilpotency degree of
the unipotent radical. First, we establish a few helpful results for the case
where the unipotent radical is abelian.

######  Lemma 8.10.

Let $k={\mathbb{Q}}_{p}$, and let ${\mathbf{G}}$ be a connected $k$-algebraic
group with a Levi decomposition
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ over $k$. Assume that
$\mathcal{R}_{u}({\mathbf{G}})$ is commutative, and that $L$ acts on
$\mathcal{R}_{u}(G)$ with no invariant vectors. Let $N\lhd\mathcal{R}_{u}(G)$
be an $L$ invariant compact open subgroup. Then $\mathcal{R}_{u}(G)/N$ is
minimal in $L\ltimes\mathcal{R}_{u}(G)/N$

###### Proof.

The abelian group $\mathcal{R}_{u}(G)$ is isomorphic to a finite-dimensional
$k$-vector space, which will be denoted $V=\mathcal{R}_{u}(G)$. Let
${\mathbf{T}}$ be a $k$-maximal torus in ${\mathbf{L}}$, let $(V/N)^{T}$ be
the subgroup of $T$-invariant elements in $V/N$, and denote by $(V/N)_{T}$ the
co-invariant quotient space. Clearly the set of $T$-invariant vectors in $V$,
$V^{T}$, is sent to the set of $T$-invariant vectors in the quotient space
$V/N$. Hence the quotient map $V\to(V/N)_{T}$ factors via the quotient space
$V_{T}$. Let $M_{T}\leq V_{T}$ be the kernel of the corresponding map
$V_{T}\to(V/N)_{T}$. As $N<V$ is open, $V_{T}/M_{T}\cong(V/N)_{T}$ is an
isomorphism of discrete groups. It follows by Corollary 6.3 that there is no
topology, weaker than the discrete on $V_{T}/M_{T}$ and correspondingly
$(V/N)_{T}$ under which the analytic group $T$ acts continuously.

Let $\omega$ be a group topology on $L\ltimes V/N$, weaker than the analytic
topology. With respect to the $\omega$-subspace topology, $L$ acts
continuously on $V/N$. Let $\omega^{\prime}$ denote the $\omega$-subspace
topology on $V/N$. Following Remark 4.23, furnished with the analytic
topology, the group $L$ acts continuously on $V/N$, the latter endowed with
the $\omega^{\prime}$ topology. We will show that $\omega^{\prime}$ is a
discrete group topology by showing the existence of a finite open subgroup.

The analytic subgroup $T\leq L$ acts continuously on $V/N$ with respect to the
$\omega^{\prime}$-topology, thus the $T$-invariant subgroup, $(V/N)^{T}$, is
$\omega^{\prime}$-closed. Therefore, the $\omega^{\prime}$-quotient topology
on $(V/N)_{T}$ is a group topology weaker than the analytic topology,
admitting a continuous $T$-action. By the considerations above, the
$\omega^{\prime}$-quotient topology on $(V/N)_{T}$ is the discrete topology.
Therefore, the subgroup $(V/N)^{T}$ is $\omega^{\prime}$-open in $V/N$.

Let $\\{{\mathbf{T}}_{i}\\}_{i=1}^{m}$ be a collection of maximal $k$-tori
that generate ${\mathbf{L}}$. We get that the subgroup
$A=\cap_{i=1}^{m}(V/N)^{T_{i}}$ is $\omega^{\prime}$-open in $V/N$. Let
$\pi\colon V\to V/N$ denote the projection map. Note that by construction

$\pi^{-1}(A)=\\{v\in V\ :\forall l\in L\ [l,v]\in N\\}$

We claim that $\pi^{-1}(A)$ does not contain a non-trivial
${\mathbb{Q}}_{p}$-vector subspace of $V$. Assume by contradiction that
$v\in\pi^{-1}(A)$ is a non trivial element contained in a vector subspace of
$\pi^{-1}(A)$. Since $Z(G)$ is compact, there exists $l\in L$ such that
$[l,v]\neq e$. By ${\mathbb{Q}}_{p}$-linearity, $\lambda[l,v]=[l,\lambda v]\in
N$, so by taking $\lambda\in{\mathbb{Q}}_{p}$ with a large absolute value we
get a contradiction to the the fact that $N$ is bounded. Since $\pi^{-1}(A)$
does not contain a ${\mathbb{Q}}_{p}$-vector subspace, by Proposition 3.12 it
is compact. Thus $A\subset V/N$ is finite.

Finally, we got that the set $A$ is a finite $\omega^{\prime}$-open subset in
$V/N$, hence the $\omega^{\prime}$-topology is equal to the analytic quotient
topology which is discrete. ∎

######  Lemma 8.11.

Let $k={\mathbb{Q}}_{p}$ or $k={\mathbb{R}}$. Let ${\mathbf{G}}$ be a
connected $k$-algebraic group with Levi decomposition
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ over $k$, such that
$\mathcal{R}_{u}({\mathbf{G}})$ is commutative. Assume that $Z(L)$ compact,
and that the action of $L$ on $\mathcal{R}_{u}(G)$ is with no invariant
vectors. Then $G$ is sealed.

###### Proof.

The abelian group $\mathcal{R}_{u}(G)$ is isomorphic to a finite-dimensional
$k$-vector space, which will be denoted $V=\mathcal{R}_{u}(G)$. By complete
reducibility of reductive groups, the vector space $V$ decomposes into
finitely many irreducible sub-representations $V=\oplus_{i=1}^{m}V_{i}$. We
prove by induction on the length of the representation $m$, that $G$ is
sealed.

Assume $m=1$. $G=L\ltimes V$, where the action of $L$ on $V$ is irreducible
and $k$-algebraic. Thus $C=\\{x\in L:\ \forall v\in V\ [x,v]=e\\}$, the kernel
of the action, is a normal $k$-subgroup of $L$. Thus $C$ is $k$-points of a
normal $k$-algebraic subgroup $\mathbf{C}\lhd{\mathbf{L}}$. Since
${\mathbf{L}}$ is reductive with a compact center, by Corollary 8.9, the
normal subgroup $C$ is sealed.

Let $H$ be a topological group and $f\colon G\to H$ a continuous homomorphism
with a dense image. The subgroup $f(C)$ is a closed normal subgroup of $H$,
making $H/f(C)$ a topological group.

Let $f_{c}\colon G/C\to H/f(C)$ be the factor map induced by $f$. Since
$G/C=L/C\ltimes V$ and $L/C$ acts on $V$ faithfully irreducible, by
Proposition 3.11 either $\ker(f_{c})\subset V$ or $V\subset\ker(f_{c})$. In
case $V\subset\ker(f_{c})$, the image of $f$ is equal to the image of $L$ and
since $L$ is sealed, $f(G)=f(L)$ is closed. Thus we assume as we may that
$\ker(f_{c})\leq V$. Since the action is irreducible, the $L$-invariant subset
$\ker(f_{c})$ does not contain a vector subspace. If $k={\mathbb{R}}$ then
$\ker(f_{c})$ must be discrete and since it is $L$ invariant it must be
trivial. In case $k={\mathbb{Q}}_{p}$, by Proposition 3.12, $\ker(f_{c})$ is
compact. Moreover, by irreducibility the subset $\ker(f_{c})$ contains a
basis, therefore $\ker(f_{c})$ is open. Following Lemma 8.10 we get that
$V/\ker(f_{c})$ is minimal in $L\ltimes V/\ker(f_{c})$, hence by Corollary
4.25, $V/\ker(f_{c})$ is minimal in $G/C$. Observe the following commutative
diagram

${G}$${H}$${L/C\ltimes V}$${H/f(C)}$${L/C\ltimes
V/\ker(f_{c})}$$\scriptstyle{f}$$\scriptstyle{f_{c}}$

By the relative minimality of $V/\ker(f_{c})$, the preimage of
$f_{c}(V/\ker(f_{c}))$ in $H$ is closed and is equal to $f(C\times V)$. The
group $C\times V$ is normal in $G$ hence $f(C\times V)$ is a closed normal
subgroup of $H$, making $H/f(C\times V)$ a topological group.

Let $f_{cv}\colon G/(C\times V)\to H/f(C\times V)$ be the corresponding factor
map induced by $f$. Following proposition 8.7, since $L$ has a compact center,
$L$ is sealed. Observe that $L/C\cong G/(C\times V)$. Consider the following
commutative diagram

${G}$${H}$${L/C}$${H/f(C\ltimes V)}$$\scriptstyle{f}$$\scriptstyle{f_{cv}}$

We get that $f_{cv}(G/(C\times V))$ is a closed subgroup in $H/f(C\times V)$
and therefore its preimage in $H$, which is equal to $f(G)$, is closed. This
finishes the case $m=1$.

Assume $m>1$, and let $V_{1}\leq V$ be an irreducible sub representation. By
the induction hypothesis, $L\times V_{1}$ and $L\times V/V_{1}$ are sealed.
Let $H$ be a topological group and $f\colon G\to H$ a continuous homomorphism
with a dense image. The subgroup $f(L\ltimes V_{1})$ is a closed normal
subgroup of $H$, making $H/f(L\ltimes V_{1})$ a topological group. Let
$f_{1}\colon G/V_{1}\to H/f(L\ltimes V_{1})$ be the corresponding factor map
induced by $f$. We get that $f_{1}(G/V_{1})$ is a closed subgroup in
$H/f(L\times V_{1})$ and therefore its preimage in $H$, which is equal to
$f(G)$, is closed. This finishes the proof. ∎

######  Proposition 8.12.

Let ${\mathbf{G}}$ be a connected $k$-algebraic group with Levi decomposition
${\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})$ over $k$, such that
$\mathcal{R}_{u}({\mathbf{G}})$ is commutative. Assume that $Z(L)$ compact,
and the action of $L$ on $\mathcal{R}_{u}(G)$ is with no invariant vectors,
then $G$ is sealed.

###### Proof.

As $k$ is a local field of characteristic zero, $k$ is a separable finite
field extension of $k_{0}={\mathbb{Q}}_{p}$ or $k_{0}={\mathbb{R}}$. Let
$\mathrm{R}_{k/k_{0}}{\mathbf{G}}$ be the restriction of scalars (from $k$ to
$k_{0}$) of the group ${\mathbf{G}}$. By functorial properties of the
restriction of scalars functor, [Mar91, Chapter I (1.7)] we get that
$\mathrm{R}_{k/k_{0}}{\mathbf{L}}\ltimes\mathrm{R}_{k/k_{0}}\mathcal{R}_{u}(G)$
is a $k_{0}$-Levi decomposition of $\mathrm{R}_{k/k_{0}}{\mathbf{G}}$.
Moreover $\mathrm{R}_{k/k_{0}}{\mathbf{G}}(k_{0})\cong{\mathbf{G}}(k)$. The
center of $\mathrm{R}_{k/k_{0}}{\mathbf{L}}(k_{0})=L$ is compact and the
action of $L$ on
$\mathrm{R}_{k/k_{0}}\mathcal{R}_{u}({\mathbf{G}})(k_{0})=\mathcal{R}_{u}(G)$
is with no invariant vectors, hence by Lemma 8.11 $G$, is sealed. ∎

###### Proof of Theorem 8.1.

Denote by $\\{{\mathbf{Z}}_{i}\\}_{i=1}^{m}$ the upper central series of
$\mathcal{R}_{u}({\mathbf{G}})$. By Zariski density of the $k$-points, the
collection $\\{Z_{i}\\}_{i=1}^{m}$ is the upper central series of
$\mathcal{R}_{u}(G)$.

Implication (1) to (2) is straightforward. Indeed, assume that $G$ is sealed
since $G$ is Polish and any quotient group is sealed and in particular
minimal. In particular, for any unipotent normal $k$-algebraic subgroup
${\mathbf{N}}\leq{\mathbf{G}}$ the group $G/N={\mathbf{G}}/{\mathbf{N}}(k)$
(see Proposition 3.6) is minimal and by Lemma 4.14, its center is compact.

Assume (2), observe that $L=G/\mathcal{R}_{u}(G)$ hence its center is compact.
We will show that the action of $L$ on $\mathcal{R}_{u}(G)$ is with no non-
trivial fixed points. Assume by contradiction that there exists a non-trivial
element $u\in\mathcal{R}_{u}(G)$ such that for any $l\in L$, $[l,u]=e$. Let
$i$ be the minimal index such that $u\in Z_{i}\setminus Z_{i-1}$. Observe that
${\mathbf{Z}}_{i-1}$ is a normal unipotent $k$-algebraic group hence by
assumption $Z(G/Z_{i-1})$ is compact. But then $uZ_{i-1}\in Z(G/Z_{i-1})$
which is a contradiction since compact algebraic groups have no unipotent
elements. Hence the action of $L$ on $\mathcal{R}_{u}(G)$ is with no non-
trivial fixed points and (3) holds.

By the Lie-correspondence for unipotent algebraic groups in characteristic
zero [Mar91, Chapter I, Proposition (1.3.1)], the logarithmic mapping
$\mathcal{R}_{u}({\mathbf{G}})\to\operatorname{Lie}(\mathcal{R}_{u}({\mathbf{G}})$
is an ${\mathbf{L}}$ equivariant bijection. Hence a non-trivial fixed point in
the unipotent radical will be mapped to a non-trivial invariant vector in the
Lie-Algebra, and vise versa, therefore (3) and (4) are equivalent.

We will prove that condition (3) implies (1) by induction on $n$, the
nilpotency degree of the unipotent radical $\mathcal{R}_{u}({\mathbf{G}})$. If
$n=0$, that is $\mathcal{R}_{u}({\mathbf{G}})=\\{e\\}$, then the group
${\mathbf{G}}$ is reductive and $G$ has compact center, it follows by
Proposition 8.7 that $G$ is sealed. If $n=1$, that is
$\mathcal{R}_{u}({\mathbf{G}})$ is abelian, it follows by Proposition 8.12
that $G$ is sealed. Assume that the nilpotency degree of
$\mathcal{R}_{u}({\mathbf{G}})$ is $n>1$. Set
${\mathbf{G}}_{1}={\mathbf{L}}\ltimes{\mathbf{Z}}_{1}$, and
${\mathbf{G}}^{1}={\mathbf{L}}\ltimes\mathcal{R}_{u}({\mathbf{G}})/{\mathbf{Z}}_{1}$.
We will show that condition (3) holds for the $k$-groups ${\mathbf{G}}_{1}$
and ${\mathbf{G}}^{1}$. Clearly $Z(L)$ is compact, and as
${\mathbf{Z}}_{1}\leq\mathcal{R}_{u}({\mathbf{G}})$, the action of
${\mathbf{L}}$ on ${\mathbf{Z}}_{1}$ is with non-trivial fixed point. We are
left to show that the ${\mathbf{L}}$-action on
$\mathcal{R}_{u}({\mathbf{G}})/{\mathbf{Z}}_{1}$ is with no non-trivial fixed
points.

Assume by contradiction that there exists a non-trivial fixed point in
$\mathcal{R}_{u}({\mathbf{G}})/{\mathbf{Z}}_{1}$, i.e. there exists an element
$u\notin{\mathbf{Z}}_{1}$ such that for any $l\in{\mathbf{L}}$,
$[l,u]\in{\mathbf{Z}}_{1}$. The commutation map

$\mathbf{c}_{u}\colon{\mathbf{L}}\to{\mathbf{Z}}_{1}\ \
\mathbf{c}_{u}(l)=[l,u]$

satisfies the following co-cycle relation

$c_{u}(xy)=xyuy^{-1}x^{-1}u^{-1}=x(yuy^{-1}u^{-1})x^{-1}(xux^{-1}u^{-1})=c_{u}(x)+x.c_{u}(y)$

with respect to the conjugation action of ${\mathbf{L}}$ on
${\mathbf{Z}}_{1}$. By [Mar91, Chapter VII, Lemma (5.22)] the first co-
homology of ${\mathbf{L}}$ is trivial, therefore there exist
$z\in{\mathbf{Z}}_{1}$ such that $\mathbf{c}_{u}(l)=l.z-z=[l,z]$. We get that
for any $l\in{\mathbf{L}}$

$\displaystyle e$ $\displaystyle=[l^{-1},u][l^{-1},z]^{-1}$
$\displaystyle=[l^{-1},u][z,l^{-1}]$
$\displaystyle=l^{-1}z^{-1}uzlu^{-1}zl^{-1}z^{-1}l$
$\displaystyle=l^{-1}z^{-1}[uz,l]z^{-1}l$ $\displaystyle=l^{-1}[uz,l]l$

hence for any $l\in{\mathbf{L}}$, $[uz,l]=e$ in contradiction to
$\mathcal{R}_{u}({\mathbf{G}})$ having no non-trivial fixed point. This shows
that the ${\mathbf{L}}$-action on
$\mathcal{R}_{u}({\mathbf{G}})/{\mathbf{Z}}_{1}$ is indeed with no non-trivial
fixed points.

Since the nilpotency degrees of $\mathcal{R}_{u}({\mathbf{G}}_{1})$ and of
$\mathcal{R}_{u}({\mathbf{G}}^{1})$ are strictly smaller then $n$, we get by
the induction hypothesis that $G_{1}$ and $G^{1}$ are sealed.

Finally, let $f\colon G\to H$, be a continuous homomorphism into a topological
group $H$ with a dense image. The normal subgroup $G_{1}\lhd G$ is sealed,
hence $f(G_{1})$ is a normal closed subgroup of $H$, making $H/f(G_{1})$ a
topological group. Since $Z_{1}\leq G_{1}$ the map $f$ factors through
$f_{1}\colon G/(Z_{1})\to H/f(G_{1})$. Observe that $G/Z_{1}\cong G^{1}$.
Consider the following commutative diagram

${G}$${H}$${G^{1}}$${H/f(G_{1})}$$\scriptstyle{f}$$\scriptstyle{f_{1}}$

The subgroup $f_{1}(G^{1})$ is a closed subgroup in $H/f(G_{1})$ and therefore
its preimage in $H$, which is equal to $f(G)$, is closed. This means that $G$
is sealed, and the proof is complete. ∎

##  9\. A note on positive characteristic

As a proof of concept, we will show how the tools developed in the previous
sections can be used to show the minimality of semisimple groups over local
fields of positive characteristics.

Similar to the proof of Proposition 7.11, we show that minimal-parabolic
subgroups are minimal as topological groups and conclude minimality by co-
compactness of parabolic subgroups. Throughout this section we let $k$ be a
local field of characteristic $p$.

######  Lemma 9.1.

Let ${\mathbf{G}}$ be a connected reductive $k$-group, and let $P\leq G$ be a
minimal parabolic in $G$. Then the unipotent radical
$\mathcal{R}_{u}({\mathbf{P}})$ is defined over $k$ and the group of
$k$-points $\mathcal{R}_{u}(P)$ is minimal in $P$.

###### Proof.

By the structure of parabolic subgroups [Bor12, Proposition 20.6], the minimal
parabolic subgroup ${\mathbf{P}}\leq{\mathbf{G}}$ is equal to the semi-direct
product
$\mathrm{C}_{\mathbf{G}}({\mathbf{T}})\ltimes\mathcal{R}_{u}({\mathbf{P}})$,
where ${\mathbf{T}}$ is a maximal $k$-split torus, and
$\mathrm{C}_{\mathbf{G}}({\mathbf{T}})$ is the centralizer of ${\mathbf{T}}$
in ${\mathbf{G}}$.

By [Mar91, Chapter I, Proposition (1.3.2)], the unipotent group
$\mathcal{R}_{u}({\mathbf{P}})$ is defined and split over $k$ and there exists
a ${\mathbf{T}}$-equivariant $k$-isomorphism
$\Phi\colon\mathcal{R}_{u}({\mathbf{G}})\to\operatorname{Lie}(\mathcal{R}_{u}({\mathbf{G}}))$.
Hence pulling the $k$-linear structure from the Lie-algebra action to the
center of the unipotent radical, we get that $Z=Z(\mathcal{R}_{u}(P))$ is
isomorphic to a finite-dimensional $k$-vector space admitting a $T$ linear
action. Since the torus is $k$-split, the linear space $Z$ admits a weight
space decomposition with respect to non-trivial characters $\chi_{i}\colon
T\to k^{*}$. This means $Z=\bigoplus_{i}V_{i}$ such that for any $v_{i}\in
V_{i}$ and $t\in T$, $t.v_{i}=\chi_{i}(t)\cdot v_{i}$. Following [Bor12,
Corollary 1. 8.11], for any $i$, there exists a co-character
$\lambda_{i}\colon k^{*}\to T$ such that the composition
$\psi_{m_{i}}=\chi_{i}\circ\lambda_{i}$ is the power map

$\psi_{m_{i}}\colon k^{*}\to k^{*},\ \ \psi_{m_{i}}\colon x\mapsto x^{m_{i}}$

Writing $m_{i}=p^{N_{i}}\cdot d_{i}$ with $N_{i}$ equal to the $p$-adic
valuation of $m_{i}$, we define the local field $k_{i}=k^{p^{N_{i}}}$. By
[Ser09, Part II, Chapter II, Inverse Function Theorem ] the map

$\psi_{d_{i}}\colon k^{*}\to k^{*},\ \ \psi_{d_{i}}\colon x\mapsto x^{d_{i}}$

is a local homeomorphism at $1_{k}$. Moreover, the Fourbenuius map

$\psi_{p^{N_{i}}}\colon k^{*}\to k_{i}^{*},\ \ \psi_{p^{N_{i}}}\colon x\mapsto
x^{p^{N_{i}}}$

is a homeomorphism from $k^{*}$ onto $k_{i}^{*}$.

We will show relative minimality of $\mathcal{R}_{u}(P)$ in
$A=T\ltimes\mathcal{R}_{u}(P)\leq P$ and thus obtain relative minimality of
$\mathcal{R}_{u}(P)$ in $P$. Let $\omega$ be a group topology on $A$ weaker
than the analytic topology. We claim that for any $i$, the subgroup $V_{i}$ is
minimal in $A$.

Following Remark 4.23, furnished with the analytic topology, the group $T$
acts continuously on $V_{i}$ endowed with the $\omega$-subspace topology. We
get that the action of $k^{*}$ on $V_{i}$ defined by $x.v=x^{m_{i}}v$ is
continuous, since it is the composition of the co-character $\lambda_{i}$ with
the $T$ action via the character $\chi_{i}$. Since $k$ is a finite filed
extension of $k_{i}$, $V_{i}$ is a finite-dimensional $k_{i}$-vector space.
Denote by $m\colon k_{i}\times V_{i}\to V_{i}$ the scalar multiplication. By
the following commutative diagram

${k^{*}\times V_{i}}$${k_{i}^{*}\times
V_{i}}$${V_{i}}$$\scriptstyle{\pi}$$\scriptstyle{\psi_{m_{i}}}$$\scriptstyle{m}$

there exists an open subgroup $1\in W\leq k^{*}$ such that
$\psi_{m_{i}}(W)\leq{k_{i}}^{*}$ is an analytic subgroup acting continuously
by multiplication on $V_{i}$. Thus by Lemma 5.1, the topology on $V_{i}$ is
equal to the analytic $k_{i}$-point topology which is equivalent to the
analytic $k$-point topology.

We got that for any $i$ the subgroup $V_{i}$ is minimal in $A$. Following
Corollary 4.16, the subgroup $Z=\oplus_{i}V_{i}$ is minimal in $A$. Therefore,
the $\omega$-subspace topology on $Z$ is equal to the analytic topology. Thus,
by Lemma 7.3 the $\omega$-subspace topology on $\mathcal{R}_{u}(P)$ is equal
to the analytic topology. ∎

######  Lemma 9.2.

Let ${\mathbf{G}}$ be a semisimple $k$ group, and let $P\leq G$ be a minimal
parabolic in $G$. Then for any central subgroup $Z_{0}$, the group $P/Z_{0}$
is minimal.

###### Proof.

Let $A=T\ltimes\mathcal{R}_{u}(P)$ be as in the proof of Lemma 9.1 above were
we established relative minimality of the unipotent radical
$\mathcal{R}_{u}(P)$. Using the same methods as in Lemma 7.6, the
${\mathbf{T}}$-equivariant $k$-morphism from $\mathcal{R}_{u}({\mathbf{P}})$
to its Lie algebra $\operatorname{Lie}(\mathcal{R}_{u}({\mathbf{P}}))$ fosters
a $k$-rational representation $\pi\colon A\to\mathrm{GL}(V)$ with kernel equal
to $C\ltimes\mathcal{R}_{u}(P)$, where $C$ is the kernel of the $T$ action.
The proof of the minimality of $P/Z_{0}$, now follows the same as in the proof
of Lemma 7.10. ∎

We, therefore, obtain the general result for semisimple groups over local
fields.

######  Theorem 9.3 (Bader-Gelander, Theorem 5.1 [BG17]).

Let $k$ be a local field, and ${\mathbf{G}}$ a semisimple $k$-group. Then
$G={\mathbf{G}}(k)$ is sealed.

###### Proof.

The theory developed by Borel and Tits on the structure of $G^{+}$ is valid in
positive characteristic as well. Therefore, the proof follows exactly as in
§8.1, where we first obtain the sealed property for almost $k$-simple groups
by minimality of $P/Z_{0}$ where $P$ is a minimal parabolic and $Z_{0}$ is a
central subgroup, see Lemma 8.3. Products and quotients of sealed groups are
sealed. As semisimple $k$-groups are a quotient of a product of $k$-almost
simple groups (see [Bor12, Theorem 22.10]), we get that the $k$-points of
semisimple $k$-group is sealed. ∎

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