# Lattice uniformities inducing unbounded convergence

Kevin Abela  Kevin Abela
Department of Mathematics
Faculty of Science
University of Malta
Msida MSD 2080, Malta<EMAIL_ADDRESS>,  Emmanuel Chetcuti  Emmanuel
Chetcuti, Department of Mathematics
Faculty of Science
University of Malta
Msida MSD 2080 Malta<EMAIL_ADDRESS>and  Hans Weber  Hans Weber
Dipartimento di Scienze Matematiche, Informatiche e Fisiche
Università degli Studi di Udine
I-33100 Udine
Italia<EMAIL_ADDRESS>
###### Abstract.

A net $(x_{\gamma})_{\gamma\in\Gamma}$ in a locally solid Riesz space
$(X,\tau)$ is said to be unbounded $\tau$-convergent to $x$ if
$|x_{\gamma}-x|\wedge u\mathop{\overset{\tau}{\longrightarrow}}0$ for all
$u\in X_{+}$. We recall that there is a locally solid linear topology
$\mathfrak{u}\tau$ on $X$ such that unbounded $\tau$-convergence coincides
with $\mathfrak{u}\tau$-convergence, and moreover, $\mathfrak{u}\tau$ is
characterised as the weakest locally solid linear topology which coincides
with $\tau$ on all order bounded subsets. It is with this motivation that we
introduce, for a uniform lattice $(L,u)$, the weakest lattice uniformity
$u^{\ast}$ on $L$ that coincides with $u$ on all the order bounded subsets of
$L$. It is shown that if $u$ is the uniformity induced by the topology of a
locally solid Riesz space $(X,\tau)$, then the $u^{*}$-topology coincides with
$\mathfrak{u}\tau$. This allows comparing results of this paper with earlier
results on unbounded $\tau$-convergence. It will be seen that despite the fact
that in the setup of uniform lattices most of the machinery used in the
techniques of [22] is lacking, the concept of ‘unbounded convergence’ well
fittingly generalizes to uniform lattices. We shall also answer Questions
2.13, 3.3, 5.10 of [22] and Question 18.51 of [21].

###### Key words and phrases:

uniform lattices, l-groups, order convergence, unbounded convergence, locally
solid topologies

###### 2000 Mathematics Subject Classification:

06B30, 06F30, 54H12, 46A40

##  1\. Introduction

Uniform lattices were introduced in [24, 26] as a generalization of
topological Boolean rings (= Boolean rings endowed with an FN-topology) and of
locally solid Riesz spaces or, more general, of locally solid $\ell$-groups. A
uniform lattice is a lattice with a lattice uniformity, i.e. a uniformity
making the lattice operations $\vee$ and $\wedge$ uniformly continuous. The
uniformity induced by a group topology $\tau$ on a Boolean ring is a lattice
uniformity iff $\tau$ is an FN-topology (in the sense of [8]); the right (left
or two-side) uniformity induced by a group topology $\tau$ on an $\ell$-group
is a lattice uniformity iff $\tau$ is locally solid (see [24, Propositions
1.1.7 and 1.1.8]).

Recently several authors have studied in a locally solid Riesz space
$(X,\tau)$ the concept of ‘unbounded $\tau$-convergence’ of a net
$(x_{\alpha})_{\alpha\in A}$ to $x$, i.e. $|x_{\alpha}-x|\wedge
u\mathop{\overset{\tau}{\longrightarrow}}0$ for all $u\in X_{+}$. ‘Unbounded
convergence’ was studied in [7, 13, 14] when $\tau$ is the norm topology on a
Banach lattice, and in [22, 6], more generally, when $\tau$ is a locally solid
linear topology on a Riesz space. Taylor [22, Theorem 2.3] showed that there
is a locally solid linear topology $\mathfrak{u}\tau$ on $X$ such that
unbounded $\tau$-convergence coincides with $\mathfrak{u}\tau$-convergence. It
can be shown that $\mathfrak{u}\tau$ is the weakest locally solid linear
topology which coincides with $\tau$ on all order bounded subsets (cf.
Corollary 4.3, Theorem 4.8).

This is the motivation to introduce here for a uniform lattice $(L,u)$ the
weakest lattice uniformity $u^{*}$ on $L$ which coincides with $u$ on all
order bounded subsets of $L$. The $u^{*}$-topology is then the weakest lattice
topology (i.e. making $\vee$ and $\wedge$ continuous) which coincides with the
$u$-topology on all order bounded subsets of $L$. If $u$ is the uniformity
induced by the topology of a locally solid Riesz space $(X,\tau)$, then the
$u^{*}$-topology coincides with $\mathfrak{u}\tau$ (see Theorem 4.9). This
allows comparing results of this paper with earlier results on unbounded
$\tau$-convergence. Apart from showing how the concept of ‘unbounded
convergence’ well fittingly generalizes to uniform lattices, we shall also
answer Questions 2.13, 3.3, 5.10 of [22] and Question 18.51 of [21].

The paper is organized as follows. In Section 2 we consider two modes of order
convergence (Definition 2.1) and prove a result (Theorem 2.4) that allows us
to answer [21, Question 18.51] and [22, Question 5.10] (see Corollaries 2.5
and 2.6). Section 3 is a summary of tools on uniform lattices that are most
relevant to our work. In Section 4 we extend the notion of unbounded
convergence to uniform lattices and show that our definition is a ‘faithful’
generalization to the one that has been studied in the existing literature on
locally solid Riesz spaces. Section 5 is devoted to the study of the
uniformity $u^{*}$ on sublattices. In particular, Theorems 5.7 and 5.8 answer
Questions 3.3 and 2.13 of [22].

##  2\. Order convergence, Order topology and Order continuous functions

Let $(P,\leq)$ be a partially ordered set. A subset $D$ of $P$ is _directed_
provided it is nonempty and every finite subset of $D$ has an upper bound in
$D$. Dually, a nonempty subset $F$ of $P$ is _filtered_ if every finite subset
of $F$ has a lower bound in $F$. If every subset of $P$ that is bounded from
above has a supremum and every subset of $P$ that is bounded from below has an
infimum, $P$ is said to be _Dedekind complete_ (or conditionally complete).
$P$ is _complete_ if every of its subsets has a supremum and infimum.

Let $(x_{\gamma})_{\gamma\in\Gamma}$ be a net in a set $X$. If
$\Gamma^{\prime}$ is some directed set and $\varphi:\Gamma^{\prime}\to\Gamma$
is increasing and final111i.e. for every $\gamma\in\Gamma$ there exists
$\gamma^{\prime}\in\Gamma^{\prime}$ such that
$\varphi(\gamma^{\prime})\geq\gamma$, then the net
$(x_{\varphi(\gamma^{\prime})})_{\gamma^{\prime}\in\Gamma^{\prime}}$ is called
a _subnet_ of $(x_{\gamma})_{\gamma\in\Gamma}$. Let $P$ be a general property
of nets. If there exists a $\gamma_{0}\in\Gamma$ such the subnet
$(x_{\gamma})_{\Gamma\ni\gamma\geq\gamma_{0}}$ has the property $P$ , then we
say that _$(x_{\gamma})_{\gamma\in\Gamma}$ satisfies $P$ eventually_. If
$(x_{\gamma})_{\gamma\in\Gamma}$ is increasing, its supremum exists and equals
$x$, we write $x_{\gamma}\uparrow x$. Dually, $x_{\gamma}\downarrow x$ means
that the net $(x_{\gamma})_{\gamma\in\Gamma}$ is decreasing with infimum equal
to $x$.

There are distinct definitions for order convergence that were used for
various applications. We shall here be concerned with the following two modes
of order convergence, namely $O_{1}$\- and $O_{2}$-convergence. For every
$a,b\in P$ define $[a,b]:=\\{x\in P:a\leq x\leq b\\}$.

######  Definition 2.1.

Let $(x_{\gamma})_{\gamma\in\Gamma}$ be a net and $x$ a point in a poset $P$.

  1. (i)

$(x_{\gamma})_{\gamma\in\Gamma}$ is said to _$O_{1}$ -converge_ to $x$ in $P$
if there exist two nets $(y_{\gamma})_{\gamma\in\Gamma}$,
$(z_{\gamma})_{\gamma\in\Gamma}$ in $P$ such that eventually $y_{\gamma}\leq
x_{\gamma}\leq z_{\gamma}$, $y_{\gamma}\uparrow x$ and $z_{\gamma}\downarrow
x$.

  2. (ii)

$(x_{\gamma})_{\gamma\in\Gamma}$ is said to _$O_{2}$ -converge_ to $x$ in $P$
if there exists a directed subset $M\subseteq P$, and a filtered subset
$N\subseteq P$, such that $\sup M=\inf N=x$, and for every $(m,n)\in M\times
N$ the net is eventually contained in $[m,n]$.

The notion of $O_{1}$-convergence can be traced back to Birkhoff [3] and
Kantorovich [15]. In contrast to $O_{1}$-convergence, in $O_{2}$-convergence,
the controlling nets are not indexed by the same directed set of the original
net. $O_{2}$-convergence is due to McShane [17]. It is easy to see for
$i\in\\{1,2\\}$ that:

  * •

For a net $(x_{\gamma})_{\gamma\in\Gamma}$, and point $x$,
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{1}$}}}}{\longrightarrow}}x\
\Rightarrow\
x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{2}$}}}}{\longrightarrow}}x$.

  * •

Every $O_{i}$-convergent net is eventually order bounded and every eventually
constant net is $O_{1}$-convergent to its eventual value.

  * •

Every subnet of an $O_{i}$-convergent net is $O_{i}$-convergent to the same
limit.

  * •

For a monotone net the two types of order convergence coincide.

  * •

When $P$ is a Dedekind complete lattice, the expression
‘$(x_{\gamma})_{\gamma\in\Gamma}$ $O_{i}$-converges to $x$’ conveys the
intuitive meaning222Note that for an $O_{i}$-convergent net in a Dedekind
complete lattice, the suprema/infima involved in (1) exist eventually.

(1)
$\liminf_{\gamma}x_{\gamma}=\sup_{\gamma\in\Gamma}\inf_{\gamma^{\prime}\geq\gamma}x_{\gamma^{\prime}}=x=\inf_{\gamma\in\Gamma}\sup_{\gamma^{\prime}\geq\gamma}x_{\gamma^{\prime}}=\limsup_{\gamma}x_{\gamma}.$

The following example shows that – even when the poset is a Boolean algebra or
a Riesz space – $O_{2}$-convergence does not imply $O_{1}$-convergence. This
example is credited to D. Fremlin [20, Ex. 2, pg. 140]. We include it for
completeness.

######  Example 2.2 ($O_{2}$-convergence is not the same as
$O_{1}$-convergence).

Let $X$ be an uncountable set and $\mathcal{A}$ the algebra of all finite and
cofinite subsets of $X$ (partially ordered by inclusion). Let
$x_{1},x_{2},x_{3},\dots$ be a sequence of distinct elements of $X$ and
$A_{n}:=\\{x_{n}\\}$. Then the sequence $(A_{n})_{n\in\mathbb{N}}$
O2-converges in $\mathcal{A}$ to $\emptyset$ since the filtered family of all
cofinite subsets of $X$ has infimum $\emptyset$, but is not O1-convergent
since any O1-convergent sequence in $\mathcal{A}$ is eventually constant.

Let $S(\mathcal{A}):=span\\{\chi_{A}:A\in\mathcal{A}\\}$ be the Riesz space of
$\mathbb{R}$-valued $\mathcal{A}$-simple functions. Then the sequence
$(\chi_{A_{n}})_{n\in\mathbb{N}}$ O2-converges to $0$, but is not
O1-convergent in $S(\mathcal{A})$.

A subset $X$ of $P$ is said to be _$O_{i}$ -closed_ if there is no net in $X$
that is $O_{i}$-converging to a point outside of $X$. In [1, Theorem 5] it was
shown that $X$ is $O_{1}$-closed iff it is $O_{2}$-closed. The collection of
all $O_{1}$-closed (= $O_{2}$-closed) subsets of $P$ forms the _order topology
$\tau_{O}(P)$_. It is easily seen that $[a,b]$ is $O_{1}$-closed for every
$a\leq b\in P$.

Motivated by the notion of _unbounded order convergence_ (uo-convergence for
short), studied for Riesz spaces in [10, 9, 11, 12], let us propose the
following definition.

######  Definition 2.3.

Let $P$ be a poset and let $F\subseteq P^{P}$. Let $i\in\\{1,2\\}$.

  1. (i)

The net $(x_{\gamma})_{\gamma\in\Gamma}$ is said to _$FO_{i}$ -converge to $x$
in $P$_ if $(f(x_{\gamma}))_{\gamma\in\Gamma}$ is $O_{i}$-convergent to $f(x)$
for every $f\in F$.

  2. (ii)

A subset $X\subseteq P$ is said to be _$FO_{i}$ -closed_ if there is no net in
$X$ that is $FO_{i}$-converging to a point outside of $X$.

Clearly,
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{$FO_{1}$}}}{\longrightarrow}}x\
\Rightarrow\
x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{$FO_{2}$}}}{\longrightarrow}}x$,
for every net $(x_{\gamma})_{\gamma\in\Gamma}$ and $x$ in the poset $P$, and
$F\subseteq P^{P}$. Although $FO_{1}$-convergence is different from
$FO_{2}$-convergence, we shall prove the following result that will play an
important role in order to answer [22, Question 5.10].

######  Theorem 2.4.

Let $P$ be a poset and $F\subseteq P^{P}$. Let
$(x_{\gamma})_{\gamma\in\Gamma}$ be a net in $P$ that $F{O}_{2}$-converges to
$x\in P$. Then $(x_{\gamma})_{\gamma\in\Gamma}$ has a subnet that
$FO_{1}$-converges to $x$.
In particular, if a net $(x_{\gamma})_{\gamma\in\Gamma}$ of $P$ is
${O}_{2}$-convergent to $x\in P$, then $(x_{\gamma})_{\gamma\in\Gamma}$ has a
subnet that $O_{1}$-converges to $x$.

###### Proof.

For every $f\in F$ let $M_{f}$ and $N_{f}$ be the directed, and respectively,
the filtered subsets of $P$ ‘witnessing’ the $O_{2}$-convergence of
$(f(x_{\gamma}))_{\gamma\in\Gamma}$ to $f(x)$, i.e. $M_{f}$ and $N_{f}$
satisfy: (i) $\sup M_{f}=\inf N_{f}=f(x)$, and (ii)
$(f(x_{\gamma}))_{\gamma\in\Gamma}$ is eventually in $[m,n]$ for every $m\in
M_{f}$ and $n\in N_{f}$.

For every finite subset $F_{0}$ of $F$, let $M_{F_{0}}$ denote the set of all
$g\in P^{F_{0}}$ satisfying $g(f)\in M_{f}$ for every $f\in F_{0}$; and
similarly, let $N_{F_{0}}$ denote the set of all $h\in P^{F_{0}}$ satisfying
$h(f)\in N_{f}$ for every $f\in F_{0}$. Define

$\Upsilon:=\\{(F_{0},g,h)\,:\,F_{0}\subseteq F,\,|F_{0}|<\infty,\,g\in
M_{F_{0}},\,h\in N_{F_{0}}\\}.$

It can easily be verified that the relation

$(F_{0},g,h)\leq(F_{0}^{\prime},g^{\prime},h^{\prime})\ \leftrightarrow\
F_{0}\subseteq F_{0}^{\prime}\ \mbox{and}\ g(f)\leq g^{\prime}(f),\ h(f)\geq
h^{\prime}(f)\ \forall f\in F_{0}$

induces a partial order on $\Upsilon$ and that $(\Upsilon,\leq)$ is a directed
set. Moreover, for every $(F_{0},g,h)\in\Upsilon$, there exists
$\gamma(F_{0},g,h)\in\Gamma$ such that $f(x_{\gamma})\in[g(f),h(f)]$ for every
$f\in F_{0}$ and for every $\gamma\geq\gamma(F_{0},g,h)$.

The product $\Upsilon\times\Gamma$ of the two directed sets $\Upsilon$ and
$\Gamma$ is again a directed set and the subset
$\Lambda\subseteq\Upsilon\times\Gamma$ consisting of all
$(F_{0},g,h,\gamma):=\bigl{(}(F_{0},g,h),\gamma\bigr{)}\in\Upsilon\times\Gamma$
satisfying $\gamma\geq\gamma(F_{0},g,h)$ is cofinal in $\Upsilon\times\Gamma$.
Hence, $\Lambda$ is a directed set when equipped with the partial order
inherited from $\Upsilon\times\Gamma$. One can easily verify that the
projection $\varphi:(F_{0},g,h,\gamma)\mapsto\gamma$ is isotone and final;
i.e. the net $\bigl{(}x_{\varphi(\lambda)}\bigr{)}_{\lambda\in\Lambda}$ is a
subnet of $(x_{\gamma})_{\gamma\in\Gamma}$.

We show that the net $(x_{\varphi(\lambda)})_{\lambda\in\Lambda}$ is
$FO_{1}$-convergent to $x$. Fix an arbitrary $f\in F$. Let
$\Lambda_{f}:=\\{(F_{0},g,h,\gamma)\in\Lambda:f\in F_{0}\\}$ and for every
$(F_{0},g,h,\gamma)\in\Lambda_{f}$ define $m(F_{0},g,h,\gamma):=g(f)$ and
$n(F_{0},g,h,\gamma):=h(f)$. By construction, the net
$(m(\lambda))_{\lambda\in\Lambda_{f}}$ is increasing, the net
$(n(\lambda))_{\lambda\in\Lambda_{f}}$ is decreasing and $m(\lambda)\leq
f(x(\lambda))\leq n(\lambda)$ for all $\lambda\in\Lambda_{f}$. Moreover,

$\displaystyle\sup\\{m(\lambda):\lambda\in\Lambda_{f}\\}=\sup M_{f}=f(x),$
$\displaystyle\inf\\{n(\lambda):\lambda\in\Lambda_{f}\\}=\inf N_{f}=f(x).$

Therefore $\bigl{(}f(x_{\varphi(\lambda)})\bigr{)}_{\lambda\in\Lambda_{f}}$ is
$O_{1}$-convergent to $f(x)$. Since
$\\{\lambda\in\Lambda:\lambda\geq\lambda_{0}\\}=\\{\lambda\in\Lambda_{f}:\lambda\geq\lambda_{0}\\}$
for $\lambda_{0}\in\Lambda_{f}$, also
$\bigl{(}f(x_{\varphi(\lambda)})\bigr{)}_{\lambda\in\Lambda}$ is
$O_{1}$-convergent to $f(x)$. This shows that
$(x_{\varphi(\lambda)})_{\lambda\in\Lambda}$ is $FO_{1}$-convergent to $x$. In
particular, by setting $F$ equal to the singleton set containing just the
identity function on $P$, one obtains that every net of $P$ that is
$O_{2}$-convergent to some point admits a subnet that is $O_{1}$-convergent to
the same point. ∎

######  Corollary 2.5.

Let $P$ be a poset and $F\subseteq P^{P}$.

  1. (i)

For $X\subseteq P$, denote by $\bar{X}^{FO_{i}}$ the set of $x\in P$ such that
there is a net in $X$ which $FO_{i}$-converges to $x$. Then
$\bar{X}^{FO_{1}}=\bar{X}^{FO_{2}}$.

  2. (ii)

A subset $X\subseteq P$ is $FO_{1}$-closed iff it is $FO_{2}$-closed. (In
particular, by setting $F$ equal to the singleton set containing the identity
map on $P$, one recovers [1, Theorem 6].)

  3. (iii)

The collection of all $FO_{1}$-closed (=$FO_{2}$-closed) subsets of $P$ forms
a topology on $L$. Denote this topology by $\tau_{FO}(P)$.

  4. (iv)

Let $(Y,\tau)$ be a topological space. For every $\varphi:P\to Y$ the
following assertions are equivalent:

    1. (a)

$\varphi$ is $\tau_{FO}(P)-\mathcal{\tau}$-continuous;

    2. (b)

If $(x_{\gamma})_{\gamma\in\Gamma}$ is $FO_{2}$-convergent to $x$ in $P$, then
$(\varphi(x_{\gamma}))_{\gamma\in\Gamma}$ is convergent to $\varphi(x)$ w.r.t.
$\tau$;

    3. (c)

If $(x_{\gamma})_{\gamma\in\Gamma}$ is $FO_{1}$-convergent to $x$ in $P$, then
$(\varphi(x_{\gamma}))_{\gamma\in\Gamma}$ is convergent to $\varphi(x)$ w.r.t.
$\tau$.

###### Proof.

(i) is a direct consequence of Theorem 2.4. (ii) follows from (i). The proof
of (iii) is routine. The implications (a)$\Rightarrow$(b)$\Rightarrow$(c) of
(iv) are trivial. To show (c)$\Rightarrow$(a) observe that if $X\subseteq Y$
is $\mathcal{T}$-closed, and $(x_{\gamma})_{\gamma\in\Gamma}$ is a net in
$\varphi^{-1}X$ that is $FO_{1}$-convergent to $x$, then $\varphi(x)\in X$ by
(c). So, $x\in\varphi^{-1}X$, i.e. $\varphi^{-1}X$ is $FO_{1}$-closed (i.e.
$\tau_{FO}(P)$-closed). ∎

The notion of unbounded order convergence on Riesz spaces has received a
considerable attention. For general results on this topic the reader is
referred to [10] and [11]. Let us recall that the notion of unbounded order
convergence is an abstraction of almost everywhere convergence in function
spaces and originally goes back to [18].

Unbounded order convergence can be generalized for lattices: Let $L$ be a
lattice and define $f_{s,t}$ on $L$ by $f_{s,t}(x):=(x\wedge t)\vee s$ for
$s,t,x\in L$. If $F=\\{f_{s,t}:s,t\in L\;,\;s\leq t\\}$, we write
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}O_{i}$}}}}{\longrightarrow}}x$
instead of
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$FO_{i}$}}}}{\longrightarrow}}x$
and $\mathfrak{u}O$-closed instead of $FO_{1}$-closed. In view of Proposition
2.8, $\mathfrak{u}O_{2}$-convergence in Riesz spaces coincides with unbounded
order convergence as defined e.g. in [10].

######  Corollary 2.6.

Let $\tau$ be a topology on a lattice $L$. Then the following conditions are
equivalent:

  * •

$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}O_{1}$}}}}{\longrightarrow}}x\
\Rightarrow\
x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\tau$}}}}{\longrightarrow}}x$,

  * •

$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}O_{2}$}}}}{\longrightarrow}}x\
\Rightarrow\
x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\tau$}}}}{\longrightarrow}}x$.

The fact that the two conditions of Corollary 2.6 (called in [22] the
$\mathfrak{u}O_{1}$-Lebesgue property and the $\mathfrak{u}O_{2}$-Lebesgue
property) are equivalent, answers [22, Question 5.10].

Let $A$ be a solid subgroup of a commutative $\ell$-group $(G,+)$. For
$i\in\\{1,2\\}$, the net $(x_{\gamma})_{\gamma\in\Gamma}$ is said to
$\mathfrak{u}_{A}O_{i}$-converge to $x$ in $G$ if $|x_{\gamma}-x|\wedge
a\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}x$
for every $a\in A_{+}$.

The proof of the following lemma is basically contained in the proof of [19,
Proposition 7.2]. It is also used in the proof of Theorem 4.9.

######  Lemma 2.7.

Let $(G,+)$ be a commutative $\ell$-group. For every $s,x,y\in G$ and $a\in
G_{+}$

$|f_{s,s+a}(x)-f_{s,s+a}(y)|\,\leq\,|x-y|\wedge
a\,=\,|f_{y-a,y}(x)-f_{y-a,y}(y)|+|f_{y,y+a}(x)-f_{y,y+a}(y)|.$

###### Proof.

To prove the inequality on the left-hand side observe that
$|f_{s,s+a}(x)-f_{s,s+a}(y)|\leq a$ follows since
$f_{s,s+a}(x),f_{s,s+a}(y)\in[s,s+a]$, and the inequality
$|f_{s,s+a}(x)-f_{s,s+a}(y)|\leq|x-y|$ follows by [4, p. 296].

Let us prove the equality on the right-hand side. First observe that

$\displaystyle f_{y,y+a}(x)-f_{y,y+a}(y)=$ $\displaystyle(x\wedge(y+a))\vee
y\,-\,(y\wedge(y+a))\vee y,$ $\displaystyle=$ $\displaystyle(x\wedge(y+a))\vee
y\,-\,y$ $\displaystyle=$ $\displaystyle((x-y)\wedge a)\vee 0$
$\displaystyle=$ $\displaystyle(x-y)^{+}\wedge a,$

and

$\displaystyle f_{y-a,y}(x)-f_{y-a,y}(y)=$ $\displaystyle(x\wedge
y)\vee(y-a)\,-\,y$ $\displaystyle=$ $\displaystyle((x-y)\wedge 0)\vee(-a)$
$\displaystyle=$ $\displaystyle(-(x-y)^{-})\vee(-a)$ $\displaystyle=$
$\displaystyle-((x-y)^{-}\wedge a).$

Therefore, since $(x-y)^{+}\wedge a$ and $(x-y)^{-}\wedge a$ are disjoint, we
obtain

$\displaystyle|f_{y,y+a}(x)-f_{y,y+a}(y)|+|f_{y-a,y}(x)-f_{y-a,y}(y)|=$
$\displaystyle((x-y)^{+}\wedge a)\,+\,((x-y)^{-}\wedge a)$ $\displaystyle=$
$\displaystyle((x-y)^{+}\wedge a)\,\vee\,((x-y)^{-}\wedge a)$ $\displaystyle=$
$\displaystyle|x-y|\wedge a.$

∎

Let us recall that in an $\ell$-group $O_{i}$-convergence is: (i) additive,
i.e. if
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}x$
and
$y_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}y$,
then
$x_{\gamma}+y_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}x+y$;
and (ii) ‘commutes with taking inverses’, i.e.
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}x$
iff
$-x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}-x$.
Making use of the decomposition $x=x^{+}-x^{-}$, one can easily deduce that
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}0$
iff
$|x_{\gamma}|\mathop{\overset{{\scriptscriptstyle\text{\text{$O_{i}$}}}}{\longrightarrow}}0$.
Hence, the following proposition immediately follows by Lemma 2.7.

######  Proposition 2.8.

Let $(G,+,\tau)$ be a commutative $\ell$-group and $A$ a solid subgroup of
$G$. Let

$F:=\\{f_{s,t}\,:\,s,t\in G,\,t-s\in A_{+}\\}.$

Then
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}_{A}O_{i}$}}}}{\longrightarrow}}x$
iff
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$FO_{i}$}}}}{\longrightarrow}}x$,
for every net $(x_{\gamma})_{\gamma\in\Gamma}$ and $x$ in $G$.

Proposition 2.8 shows that, for $A=G$, $\mathfrak{u}_{G}O_{i}$-convergence
coincides with $\mathfrak{u}O_{i}$-convergence. For $A=G$ and $i=2$,
Proposition 2.8 coincides with [19, Proposition 7.2].

######  Corollary 2.9.

Let $(G,+)$ be a commutative $\ell$-group . Let
$(x_{\gamma})_{\gamma\in\Gamma}$ be a net in $G$, $x\in G$ and
$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}O_{2}$}}}}{\longrightarrow}}x$.
Then $(x_{\gamma})_{\gamma\in\Gamma}$ has a subnet that
$\mathfrak{u}O_{1}$-converges to $x$.

###### Proof.

This follows at once by Proposition 2.8 and Theorem 2.4. ∎

######  Corollary 2.10.

Let $(G,+)$ be a commutative $\ell$-group and $A$ a solid subgroup of $G$. Let
$(Y,\tau)$ be a topological space and $\varphi$ a function from $G$ into $Y$.
Then the following conditions for $\varphi$ are equivalent:

  * •

$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}_{A}O_{1}$}}}}{\longrightarrow}}x\
\Rightarrow\
\varphi(x_{\gamma})\mathop{\overset{{\scriptscriptstyle\text{\text{$\tau$}}}}{\longrightarrow}}\varphi(x)$,

  * •

$x_{\gamma}\mathop{\overset{{\scriptscriptstyle\text{\text{$\mathfrak{u}_{A}O_{2}$}}}}{\longrightarrow}}x\
\Rightarrow\
\varphi(x_{\gamma})\mathop{\overset{{\scriptscriptstyle\text{\text{$\tau$}}}}{\longrightarrow}}\varphi(x)$.

###### Proof.

This follows by Proposition 2.8 and Corollary 2.5 ∎

In view of Proposition 2.8, Corollary 2.5(i) answers [21, Question 18.51].

##  3\. Lattice uniformities – setting up the basics

Let $L$ be a lattice and $\Delta:=\\{(x,x):x\in L\\}$ the diagonal of $L\times
L$. A subset $C$ of $L$ is called convex if $[a,b]\subseteq C$ whenever
$a,b\in C$ and $a\leq b.$ A lattice uniformity is a uniformity on a lattice
making the lattice operations $\vee$ and $\wedge$ uniformly continuous. A
uniform lattice is a lattice endowed with a lattice uniformity. One easily
verifies:

######  Proposition 3.1 ([24], Proposition 1.1.2).

Let $u$ be a uniformity on the lattice $L$. Then the following statements are
equivalent:

  1. (i)

$u$ is a lattice uniformity;

  2. (ii)

for every $U\in u$ there exists $V\in u$ satisfying $V\vee V\subseteq U$ and
$V\wedge V\subseteq U$;

  3. (iii)

for every $U\in u$ there exists $V\in u$ satisfying $V\vee\Delta\subseteq U$
and $V\wedge\Delta\subseteq U$.

It follows that the supremum of lattice uniformities on $L$ (built in the
complete lattice of all uniformities on $L$) is also a lattice uniformity. We
remark that – although it is more delicate to see – the analogous statement
holds true also for the infimum of lattice uniformities, (see [27, Remark
2.6]).

The topology induced by a lattice uniformity $u$ on $L$, the $u$-topology, is
a locally convex lattice topology, i.e. $\vee$ and $\wedge$ are continuous and
every point of $L$ has a neighbourhood base consisting of convex sets.

As mentioned in the introduction, uniform lattices generalize topological
Boolean rings on one-hand and locally solid $\ell$-groups (and therefore
locally solid Riesz spaces) on the other. More specifically, we recall that
(see [24, Proposition 1.1.7 & 1.1.8]) the uniformity induced by

  * •

an FN-topology (i.e. a locally convex group topology) on a Boolean ring, or

  * •

a locally solid group topology333A subset $A$ of an $\ell$-group $G$ is solid
if for every for every $a\in A$ and $x\in G$ satisfying $|x|\leq|a|$, it holds
that $x\in A$. A group topology on an $\ell$-group $G$ is locally solid if $0$
has a neighbourhood base consisting of solid sets.  on an $\ell$-group,

is a lattice uniformity.

For the lattice uniformity $u$ we let $N(u):=\cap_{U\in u}U$. We note that $u$
is Hausdorff iff $N(u)=\Delta$. We recall [24, Proposition 1.1.3] that if $u$
is a lattice uniformity on $L$, every $U\in u$ contains a $V\in u$ such that
the rectangle $[a\wedge b,a\vee b]^{2}\subseteq V$ for all pairs $(a,b)\in V$.

######  Proposition 3.2.

Let $u$ and $v$ be two lattice uniformities on $L$ that agree on every order
bounded subset of $L$. Then $N(u)=N(v)$. In particular, $u$ is Hausdorff iff
$v$ is Hausdorff.

###### Proof.

If $(a,b)\in N(u)$, then the rectangle $[a\wedge b,a\vee b]^{2}\subseteq N(u)$
and so, since the uniformity
${\left.\kern-1.2ptu\vphantom{\mid}\right|_{[a\wedge b,a\vee b]}}$ induced by
$u$ on $[a\wedge b,a\vee b]$ agrees with
${\left.\kern-1.2ptv\vphantom{\mid}\right|_{[a\wedge b,a\vee b]}}$, we get
$[a\wedge b,a\vee b]^{2}\subseteq N(v)$, i.e. $(a,b)\in N(v)$. This shows
$N(u)\subseteq N(v)$ and therefore $N(u)=N(v)$ by symmetry. ∎

A lattice uniformity $u$ on $L$ is said to be:

  * •

_order continuous (o.c.)_ if $(x_{\gamma})_{\gamma\in\Gamma}$ converges to $x$
in $(L,u)$ whenever $(x_{\gamma})_{\gamma\in\Gamma}$ is a monotone net in $L$
order converging to $x$;

  * •

_exhaustive_ if every monotone net in $L$ is Cauchy;

  * •

_locally exhaustive_ if ${\left.\kern-1.2ptu\vphantom{\mid}\right|_{[a,b]}}$
is exhaustive for every $a\leq b$ in $L$.

It is easy to see that a lattice uniformity $u$ is o.c. iff the $u$-topology
is coarser than the order topology $\tau_{O}(L)$: Let $u$ be o.c. and $C$ be
closed in $(L,u)$. If $(x_{\gamma})_{\gamma\in\Gamma}$ is a net in $C$, $x\in
L$ and $(y_{\gamma})_{\gamma\in\Gamma}$, $(z_{\gamma})_{\gamma\in\Gamma}$ are
nets in $L$ such that $y_{\gamma}\uparrow x$, $z_{\gamma}\downarrow x$ and
$y_{\gamma}\leq x_{\gamma}\leq z_{\gamma}$ for all $\gamma\in\Gamma$, then
both, $(y_{\gamma})$ and $(z_{\gamma})$, converge to $x$ w.r.t. $u$, therefore
$(x_{\gamma})$ converges to $x$ w.r.t. $u$ since the $u$-topology is locally
convex, hence $x\in C$, i.e. $C$ is order-closed. The other implication, that
$u$ is o.c. if the $u$-topology is coarser than $\tau_{O}(L)$, is obvious.

By [26, Proposition 6.1], $u$ is exhaustive iff every monotone sequence in $L$
is Cauchy.

######  Theorem 3.3 ([26], Proposition 6.3).

Let $(L,u)$ be a Hausdorff uniform lattice. Then the following conditions are
equivalent:

  1. (i)

$(L,u)$ is complete (as a uniform space) and $u$ is exhaustive;

  2. (ii)

$(L,\leq)$ is a complete lattice and $u$ is o.c..

######  Proposition 3.4 ([26], Proposition 6.14).

Let $S$ be a dense sublattice of a uniform lattice $(L,u)$. If $u|_{S}$ is
(locally) exhaustive, then $u$ is (locally) exhaustive, too.

We recall that every Hausdorff uniform lattice $(L,u)$ is a sublattice and a
dense subspace of a Hausdorff uniform lattice $(\tilde{L},\tilde{u})$, which
is complete as a uniform space (see [24, Proposition 1.3.1].
$(\tilde{L},\tilde{u})$ is called the completion of $(L,u)$.

From Theorem 3.3 and Proposition 3.4 immediately follows

######  Corollary 3.5.

Let $(L,u)$ be a (locally) exhaustive Hausdorff uniform lattice and
$(\tilde{L},\tilde{u})$ its completion. Then $(\tilde{L},\leq)$ is a
(Dedekind) complete lattice and $\tilde{u}$ is o.c..

In Theorem 3.7 we need for a uniform lattice $(L,u)$ the following Condition
(C) stronger than order continuity.

_Condition (C)_ : For every increasing (and resp. decreasing) order bounded
net $(x_{\gamma})_{\gamma\in\Gamma}$ in $L$ and every $U\in u$, there exists
an upper bound (resp. lower bound) $x\in L$ of
$(x_{\gamma})_{\gamma\in\Gamma}$, such that $(x_{\gamma},x)\in U$ eventually.

######  Proposition 3.6 ([26], Proposition 7.1.2).

  1. (i)

Any lattice uniformity satisfying Condition (C) is o.c. and locally
exhaustive.

  2. (ii)

If $u$ is a lattice uniformity on a Dedekind complete lattice, then $u$
satisfies Condition (C) iff $u$ is o.c..

######  Theorem 3.7 ([26], Corollary 7.2.4).

Let $u$ and $v$ be exhaustive lattice uniformities on the lattice $L$
satisfying the Condition (C). If $N(u)=N(v)$, then the $u$-topology coincides
with the $v$-topology.

Let us briefly consider locally solid $\ell$-groups (in particular, locally
solid Riesz spaces). Following [2] (for locally solid Riesz spaces) we call a
locally solid group topology a _Lebesgue topology_ (resp. _pre-Lebesgue
topology_) when the induced uniformity is o.c. (resp. locally exhaustive). In
[26, Proposition 7.1.5] it is shown that the uniformity induced by a locally
solid group topology on an Archimedean $\ell$-group satisfies Condition (C)
iff it is o.c.. Therefore, by Proposition 3.6 (a), if an Archimedean locally
solid $\ell$-group is Lebesgue, then it is pre-Lebesgue; this generalizes [2,
Theorem 3.23 on pg 87] where this implication is given for Archimedean locally
solid Riesz spaces.

The ‘uniqueness’ result in Theorem 3.7 implies that if $\tau$ and $\sigma$ are
Hausdorff Lebesgue locally solid group topologies on an Archimedean
$\ell$-group $G$, then $\sigma$ and $\tau$ induce the same topology on the
order bounded subsets of $G$. In particular, for locally solid Riesz spaces,
this gives [2, Theorem 4.22 on pg 106].

Sometimes, for example to apply Theorem 3.3, it is convenient to consider the
Hausdorff uniform lattice associated with a uniform lattice $(L,u)$. Let us
briefly outline this factorisation procedure.

######  Proposition 3.8 ([24], Proposition 1.2.4).

Let $(L,u)$ be a uniform lattice. Define the equivalence relation $\sim_{u}$
by setting $a\sim_{u}b$ iff $(a,b)\in N(u)$. Let $\hat{a}$ denote the
equivalence class containing the element $a\in L$. Then the operations
$\hat{a}\vee\hat{b}:=\widehat{a\vee b}$ and
$\hat{a}\wedge\hat{b}:=\widehat{a\wedge b}$ turn the quotient
$\hat{L}:=L/{\scriptscriptstyle\sim_{u}}$ into a lattice, and the family
$\widehat{u}:=\\{\widehat{U}:U\in u\\}$, where
$\widehat{U}:=\\{(\hat{a},\hat{b}):(a,b)\in U\\}$, is a Hausdorff lattice
uniformity on $\hat{L}$. For any closed subset $U$ of $(L,u)^{2}$ we have
$(a,b)\in U$ iff $(\hat{a},\hat{b})\in\hat{U}$.

Lattice uniformities can be generated by systems of particular
semimetrics444Semimetrics are assumed to be real-valued.. If $q>1$ and $D$ is
a system of semimetrics satisfying

$d(x\vee z,y\vee z)\leq q\cdot d(x,y)\quad\text{and}\quad d(x\wedge z,y\wedge
z)\leq q\cdot d(x,y)\quad\text{for every $x,y,z\in L$ and $d\in D$,}$

then the uniformity generated by $D$ on $L$ is a lattice uniformity. Vice
versa we have:

######  Proposition 3.9 ([25], Corollaries 1.4 and 1.6).

Let $(L,u)$ be a uniform lattice and $q>1$. Then $u$ is generated by a system
$D$ of semimetrics satisfying

$d(x\vee z,y\vee z)\leq q\cdot d(x,y)\quad\text{and}\quad d(x\wedge z,y\wedge
z)\leq q\cdot d(x,y)\quad\text{for every $x,y,z\in L$ and $d\in D$.}$

If $L$ is distributive, then one can here choose also $q=1$.

Let $(L,u)$ be a uniform lattice and $F$ a set of lattice homomorphisms
$f:L\rightarrow(L,u)$. The sets $U_{f}:=\\{(x,y)\in L^{2}:(f(x),f(y))\in
U\\}$, where $U\in u$ and $f\in F$, form a subbase of the initial uniformity
of $F$, i.e. of the coarsest uniformity on $L$ making $f$ uniformly continuous
for all $f\in F$ (see [5, Proposition 4 of II.2.3]). Using Proposition 3.1,
one sees that this uniformity is a lattice uniformity: If $U,V\in u$ with
$V\vee\Delta\subseteq U$ and $V\wedge\Delta\subseteq U$, then
$V_{f}\vee\Delta\subseteq U_{f}$ and $V_{f}\wedge\Delta\subseteq U_{f}$ for
all $f\in F$.

We are here interested in the initial uniformity w.r.t. functions of the type
$f_{a,b}(x):=(x\wedge b)\vee a$ or $g_{a,b}(x):=(x\vee a)\wedge b$.
Proposition 3.10 (a) is the reason that later on we only consider distributive
lattices. Proposition 3.10 (b) shows that it is indifferent whether we study
the initial uniformity w.r.t. functions of the type $f_{a,b}$ or of the type
$g_{a,b}$; moreover, we can hereby always assume that $a\leq b$.

######  Proposition 3.10.

For a lattice $L$ the following statements are equivalent:

  1. (i)

$L$ is distributive.

  2. (ii)

$f_{a,b}$ is a lattice homomorphism for all $a,b\in L$.

  3. (iii)

$g_{a,b}$ is a lattice homomorphism for all $a,b\in L$.

(b) If $L$ is distributive, then $f_{a,b}=f_{a,a\vee b}=g_{a,a\vee b}$ and
$g_{a,b}=g_{a\wedge b,b}=f_{a\wedge b,b}$ for every $a,b\in L$.

We omit the proof which is routine.

##  4\.  The uniformity $u^{*}$ inducing ‘unbounded $u$-convergence’.

In this section, let $u$ be a lattice uniformity on a distributive lattice
$L$.

For $a,b\in L$, let $u_{a,b}$ be the initial uniformity of $f_{a,b}$. Since
$f_{a,b}:(L,u)\rightarrow(L,u)$ is uniformly continuous, $u_{a,b}\subseteq u$
and, as explained before, $u_{a,b}$ is a lattice uniformity since $L$ is
assumed to be distributive. Moreover, $v_{a,b}\subseteq u_{a,b}$ if $v$ is a
lattice uniformity on $L$ coarser than $u$.

######  Proposition 4.1.

Let $a,b,c,d\in L$. Then

$u_{a\vee(b\wedge c),\,b\wedge d}\subseteq(u_{a,b})_{c,d}.$

In particular:

  1. (i)

If $a\leq c\leq d\leq b$, then $u_{c,d}\subseteq u_{a,b}$, and

  2. (ii)

$\left(u_{a,b}\right)_{a,b}=u_{a,b}$.

###### Proof.

First observe that for $x\in L$

$\displaystyle f_{a,b}\circ f_{c,d}(x)$ $\displaystyle=(((x\wedge d)\vee
c)\wedge b)\vee a$ $\displaystyle=(x\wedge d\wedge b)\vee(c\wedge b)\vee a$
$\displaystyle=(x\wedge(d\wedge b))\vee(a\vee(b\wedge c))=f_{a\vee(b\wedge
c),\,b\wedge d}(x),$

i.e. $f_{a,b}\circ f_{c,d}=f_{a\vee(b\wedge c),\,b\wedge d}$. The functions
$f_{c,d}:\left(L,(u_{a,b})_{c,d}\right)\to(L,u_{a,b})$ and
$f_{a,b}:(L,u_{a,b})\to(L,u)$ are uniformly continuous, and therefore so is
the function

$f_{a\vee(b\wedge c),\,b\wedge d}=f_{a,b}\circ
f_{c,d}:\left(L,\left(u_{a,b}\right)_{c,d}\right)\to(L,u).$

This shows that $u_{a\vee(b\wedge c),\,b\wedge d}\subseteq(u_{a,b})_{c,d}$.

In particular, if $a\leq c\leq d\leq b$, then $a\vee(b\wedge c)=c$ and
$b\wedge d=d$, and therefore
$u_{c,d}\subseteq\left(u_{a,b}\right)_{c,d}\subseteq u_{a,b}$, and by setting
$a=c$ and $b=d$ one also obtains $u_{a,b}=(u_{a,b})_{a,b}$. ∎

As observed before, it is enough to consider the uniformities $u_{a,b}$ for
$a\leq b$. Define $J(L):=\\{(a,b)\in L^{2}:a\leq b\\}$. If $\emptyset\neq
J\subseteq J(L)$ then the initial uniformity $u_{J}$ of $\\{f_{a,b}:(a,b)\in
J\\}$ is the uniformity having as a subbase the sets of the form

$U_{a,b}:=\\{(x,y)\in L^{2}:(f_{a,b}(x),f_{a,b}(y))\in U\\}\quad((a,b)\in J,\
U\in u).$

Note that $u_{J}$ is the supremum of the uniformities $u_{a,b}$ where
$(a,b)\in J$, and thus $u_{J}$ is again a lattice uniformity. Let
$u^{*}:=u_{J(L)}$.

######  Proposition 4.2.

Let $J$ and $J^{\prime}$ be non-empty subsets of $J(L)$.

  1. (i)

$v_{J}\subseteq u_{J}\subseteq u$ for every lattice uniformity $v$ coarser
than $u$.

  2. (ii)

If for every $(a,b)\in J$ there exists $(a^{\prime},b^{\prime})\in J^{\prime}$
satisfying $a^{\prime}\leq a\leq b\leq b^{\prime}$, then $u_{J}\subseteq
u_{J^{\prime}}$ and $(u_{J})_{J^{\prime}}=(u_{J^{\prime}})_{J}=u_{J}$.

  3. (iii)

$u_{J}$ is the weakest lattice uniformity which agrees with $u$ on $[a,b]$ for
all $(a,b)\in J$.

  4. (iv)

The $u_{J}$-topology is the weakest lattice topology which agrees with the
$u$-topology on $[a,b]$ for all $(a,b)\in J$.

###### Proof.

(i) is obvious.

(ii) The first assertion follows by (i) of Proposition 4.1 and by the fact
that $u_{J}$ is the supremum of the uniformities $u_{a,b}$ for all $(a,b)\in
J$.

For the second assertion, we first note that $(u_{J})_{J^{\prime}}\subseteq
u_{J}$ by (i). Using (i) and Proposition 4.1, we get
$u_{a,b}=(u_{a,b})_{a,b}\subseteq(u_{J})_{a,b}\subseteq(u_{J})_{J^{\prime}}$.
Hence $u_{J}\subseteq(u_{J})_{J^{\prime}}$ since $u_{J}$ is the supremum of
the uniformities $u_{a,b}$, $(a,b)\in J$. Similarly, one can prove
$u_{J}=(u_{J^{\prime}})_{J}$

(iii) Let us first show that $u_{J}$ agrees with $u$ on $[a,b]$ for all
$(a,b)\in J$. Let $(a,b)\in J$, $U\in u$ and $U_{a,b}:=\\{(x,y)\in
L^{2}:(f_{a,b}(x),f_{a,b}(y))\in U\\}$. Then
$U_{a,b}\cap[a,b]^{2}=U\cap[a,b]^{2}$ since $f_{a,b}(x)=x$ for $x\in[a,b]$.
Therefore $u_{a,b}$ and $u$ agree on $[a,b]$. It follows that $u_{J}$ and $u$
agree on $[a,b]$ since $u_{a,b}\subseteq u_{J}\subseteq u$. Now, let $v$ be a
lattice uniformity on $L$ which agrees with $u$ on $[a,b]$ for all $(a,b)\in
J$. We shall show that $u_{a,b}\subseteq v$ for $(a,b)\in J$. Since $v$ is a
lattice uniformity, $f_{a,b}:(L,v)\rightarrow(L,v)$ is uniformly continuous.
Since $f_{a,b}(L)\subseteq[a,b]$ and $v|_{[a,b]}=u|_{[a,b]}$, also
$f_{a,b}:(L,v)\rightarrow(L,u)$ is uniformly continuous. Therefore
$u_{a,b}\subseteq v$.

The proof of (iv) is similar to that of (iii). ∎

######  Corollary 4.3.

  1. (i)

$(u^{\ast})^{\ast}=u^{\ast}$.

  2. (ii)

$u^{*}$ is the weakest lattice uniformity which agrees with $u$ on any order
bounded subset of $L$.

  3. (iii)

The $u^{*}$-topology is the weakest lattice topology which agrees with the
$u$-topology on any order bounded subset of $L$.

In contrast to Corollary 4.3, in general, $u^{*}$ is not the weakest
uniformity which agrees with $u$ on every order bounded subset of $L$, and the
$u^{*}$-topology is not the weakest topology which agrees with the
$u$-topology on any order bounded subset of $L$:

######  Example 4.4.

Let $u$ be the usual uniformity on $\mathbb{R}$ induced by the absolute value.

  1. (a)

The infimum $v$ of all uniformities on $\mathbb{R}$ which agree with $u$ on
all bounded subsets of $\mathbb{R}$ is the trivial uniformity.

  2. (b)

The sets $U_{n}:=\\{(x,y)\in\mathbb{R}^{2}:|x-y|\leq\frac{1}{n}\text{ or
}x,y\geq n\text{ or }x,y\leq-n\\}$, $n\in\mathbb{N}$, form a base for $u^{*}$.

###### Proof.

(a) For $a\in\mathbb{R}$, and $n\in\mathbb{N}$ let $V_{n,a}$ be the subset of
$\mathbb{R}^{2}$ defined by

$(x,y)\in V_{n,a}\ \Longleftrightarrow\ \begin{cases}|x-y|\leq\frac{1}{n},\
\mbox{or}\\\ |x|,|y|\geq n,\ \mbox{or}\\\ |x|\geq n,|y-a|\leq\frac{1}{n},\
\mbox{or}\\\ |y|\geq n,|x-a|\leq\frac{1}{n}\,.\end{cases}$

The family $\\{V_{n,a}:n\in\mathbb{N}\\}$ form a base for a uniformity $v_{a}$
on $\mathbb{R}$ which agrees with $u$ on all bounded subsets of $\mathbb{R}$.
Then $n\rightarrow a$ $(v_{a})$ and therefore $n\rightarrow a$ ($v$) since
$v\subseteq v_{a}$. Let now $V\in v$ and $a,b\in\mathbb{R}$. Then
$n\rightarrow a$ ($v$) and $n\rightarrow b$ ($v$), hence $(a,b)\in V$. This
shows $V=\mathbb{R}^{2}.$

(b) immediately follows from the description of $u^{*}$ as initial uniformity.
∎

######  Remark 4.5.

Note that in Example 4.4, the $u$-topology coincides with the
$u^{*}$-topology; $u$ is complete whereas $u^{*}$ is not complete; the
completion of $(\mathbb{R},u^{*})$ is $[-\infty,+\infty]$ with the usual
compact uniformity.

######  Proposition 4.6.

Let $S$ be a sublattice of $L$ and $I:=J(S)=\\{(a,b)\in S^{2}:a\leq b\\}$.
Then $u_{I}$ is Hausdorff iff $u$ is Hausdorff and $x=\sup_{s\in S}s\wedge
x=\inf_{s\in S}s\vee x$ for every $x\in L$.

###### Proof.

Assume that $u_{I}$ is Hausdorff. Then also $u$ is Hausdorff since
$u_{I}\subseteq u$. Let $x\in L$; we show that $\sup_{s\in S}x\wedge s=x$. Let
$y\leq x$ with $x\wedge s\leq y$ for every $s\in S$. Then for every $(a,b)\in
I$ we have $x\wedge b=y\wedge b$, therefore $(x\wedge b)\vee a=(y\wedge b)\vee
a$ and hence $(x,y)\in N(u_{I})=\Delta$, i.e. $x=y$. The proof of $\inf_{s\in
S}x\vee s=x$ is similar.

Conversely, suppose that $u$ is Hausdorff and that $x=\sup_{s\in S}s\wedge
x=\inf_{s\in S}s\vee x$ for every $x\in L$. Let $(x,y)\in N(u_{I})$. Then,
since $u$ is Hausdorff, we get that $(x\wedge b)\vee a=(y\wedge b)\vee a$ for
every $(a,b)\in I$. Then,

$(x\wedge b)\vee a=(x\wedge(a\vee b))\vee a=(y\wedge(a\vee b))\vee a=(y\wedge
b)\vee a,$

for arbitrary $a$ and $b$ in $S$, and therefore

$x=\sup_{b\in S}x\wedge b=\sup_{b\in S}\inf_{a\in S}(x\wedge b)\vee
a=\sup_{b\in S}\inf_{a\in S}(y\wedge b)\vee a=\sup_{b\in S}y\wedge b=y.$

This shows that $u_{I}$ is Hausdorff. ∎

######  Corollary 4.7.

$u$ is Hausdorff iff $u^{\ast}$ is Hausdorff.

Corollary 4.7 also follows from Corollary 4.3 (ii) and Proposition 3.2.

We now consider the situation when $u$ is the uniformity induced by the
topology $\tau$ of a locally solid Riesz space, or more general, of a locally
solid commutative $\ell$-group. We will compare the $u_{J}$-topology and
$u^{*}$-topology, respectively, with $\mathfrak{u}_{A}\tau$ and
$\mathfrak{u}\tau$ introduced by Taylor [22]. The following theorem was
formulated by Taylor [22, Theorem 2.3 and §9] in the case that $\tau$ is a
locally solid linear topology on a Riesz space, but in [21] it is mentioned a
possible generalization for locally solid $\ell$-groups.

######  Theorem 4.8.

Let $(G,+,\tau)$ be a Hausdorff locally solid commutative $\ell$-group,
$\mathfrak{U}$ its $0$-neighbourhood system and $A$ a solid subgroup of $G$.
Then the sets

$\\{x\in G:|x|\wedge a\in U\\}\;(a\in A_{+},U\in\mathfrak{U})$

form a $0$-neighbourhood base for a locally solid group topology
$\mathfrak{u}_{A}\tau$ on $G$.

A net $(x_{\gamma})$ converges to $x$ w.r.t. $\mathfrak{u}_{A}\tau$ iff
$|x_{\gamma}-x|\wedge a\mathop{\overset{\tau}{\longrightarrow}}0$ for any
$a\in A_{+}$.

If $(G,\tau)$ is a locally solid Riesz space, then $(G,\mathfrak{u}_{A}\tau)$
is a locally solid Riesz space, too.

Following Taylor [22] we write $\mathfrak{u}\tau$ instead of
$\mathfrak{u}_{A}\tau$ if $A=G$. Convergence w.r.t. $\mathfrak{u}\tau$ is
called in [22] unbounded $\tau$-convergence. Moreover, in the terminology of
Taylor [22], $\tau$ is called unbounded iff $\mathfrak{u}\tau=\tau$.

######  Theorem 4.9.

Let $(G,+,\tau)$ be a Hausdorff locally solid commutative $\ell$-group and $A$
a solid subgroup of $G$. Let $u$ be the uniformity induced by $\tau$ and
$J=\\{(a,b)\in G^{2}:b-a\in A_{+}\\}$. Then the $u_{J}$-topology coincides
with $\mathfrak{u}_{A}\tau$. In particular, the $u^{*}$-topology coincides
with $\mathfrak{u}\tau$.

###### Proof.

Let $\mathfrak{U}$ be a $0$-neighbourhood base for $(G,\tau)$ consisting of
solid sets. For $s,t\in L$ with $t-s\in A_{+}$ and $U\in\mathfrak{U}$ let

$\hat{U}(x_{0},s,t)=\\{x\in G:f_{s,t}(x)-f_{s,t}(x_{0})\in U\\}.$

Then $\\{\hat{U}(x_{0},s,t)\,:\,U\in\mathfrak{U},\,t-s\in A_{+}\\}$ is a
neighbourhood subbase of $x_{0}$ w.r.t. the $u_{J}$-topology. We recall that
the family $\\{\tilde{U}(x_{0},a)\,:\,U\in\mathfrak{U},\,a\in A_{+}\\}$, where

$\tilde{U}(x_{0},a):=\\{x\in G:|x-x_{0}|\wedge a\in U\\},$

is a neighbourhood base of $x_{0}$ w.r.t. $\mathfrak{u}_{A}\tau$.

Let $U\in\mathfrak{U}$ and $s,t\in G$ with $a:=t-s\in A_{+}$. We show that
$\tilde{U}(x_{0},a)\subseteq\hat{U}(x_{0},s,t)$. Let $x\in\tilde{U}(x_{0},a)$.
By (ii) of Lemma 2.7, one obtains

$|f_{s,t}(x)-f_{s,t}(x_{0})|\leq|x-x_{0}|\wedge a\in U,$

i.e. $x\in\hat{U}(x_{0},s,t)$. This shows that the $u_{J}\mbox{-topology}$ is
coarser than $\mathfrak{u}_{A}\tau$.

Let us show the converse. Let $a\in A_{+}$ and $U\in\mathfrak{U}$. Choose
$V\in\mathfrak{U}$ with $V+V\subseteq U$, and set $r:=x_{0}-a$, $s:=x_{0}$ and
$t:=x_{0}+a$. We show that
$\hat{V}(x_{0},s,t)\cap\hat{V}(x_{0},r,s)\subseteq\tilde{U}(x_{0},a)$. For
$x\in\hat{V}(x_{0},s,t)\cap\hat{V}(x_{0},r,s)$ one has by Lemma 2.7

$|x-x_{0}|\wedge
a=|f_{r,s}(x)-f_{r,s}(x_{0})|\,+\,|f_{s,t}(x)-f_{s,t}(x_{0})|\in V+V\subseteq
U,$

hence $x\in\tilde{U}(x_{0},a)$. ∎

Under the assumption of Theorem 4.9, let $I:=\\{(-a,a):a\in A_{+}\\}$. Then
obviously $u_{I}\subseteq u_{J}$. The following example shows that the
$u_{I}$-topology can be strictly coarser the $u_{J}$-topology
(=$\mathfrak{u}_{A}\tau$).

######  Example 4.10.

Let $G=\mathbb{R}^{\mathbb{N}}$ and $\tau$ be the locally solid group topology
induced by $\|(x_{n})\|:=\sum_{n=1}^{\infty}|x_{n}|$. Let $A:=\ell_{\infty}$,
$I:=\\{(-a,a):a\in A_{+}\\}$ and $\sigma$ the $u_{I}$-topology. Then $\sigma$
is strictly coarser than $\mathfrak{u}_{A}\tau$.

###### Proof.

Let $x=(1,2,3,\dots)$, $e=(1,1,1,\dots)$, $y_{n}=\frac{1}{n}e$,
$x_{n}=x+y_{n}$.

We show that $x_{n}\rightarrow x$ ($\sigma$), but $x_{n}\nrightarrow x$
($\mathfrak{u}_{A}\tau$).

Let $a\in A$, $k\in\mathbb{N}$ with $e\leq a\leq ke$. Then $|(x_{n}\wedge
a)\vee(-a)-(x\wedge a)\vee(-a)|\leq\frac{1}{n}\chi_{[1,k]}$ where
$\chi_{[1,k]}$ denotes the characteristic function of $\\{1,2,\dots,k\\}$.
Therefore $\|(x_{n}\wedge a)\vee(-a)-(x\wedge
a)\vee(-a)\|\leq\frac{k}{n}\rightarrow 0$ ($n\rightarrow\infty$).

On the other hand, $|x_{n}-x|\wedge a=y_{n}$, $\|y_{n}\|=+\infty$ and
therefore $(x_{n})$ does not converge to $x$ w.r.t. $\mathfrak{u}_{A}\tau$. ∎

##  5\. The uniformity $u^{*}$ on sublattices

In this section, let $u$ be a lattice uniformity on a distributive lattice
$L$.

Let $D(L)$ denote the set of semimetrics $d$ on $L$ satisfying

$d(x\vee z,y\vee z)\leq d(x,y)\quad\text{and}\quad d(x\wedge z,y\wedge z)\leq
d(x,y)\quad\text{for every $x,y,z\in L$.}$

By Proposition 3.9, $u$ is generated by a subset of $D(L)$, and therefore by
the set $D_{u}$ of all $d\in D(L)$ which are uniformly continuous w.r.t. $u$.

######  Remark 5.1.

_Let $J\subseteq J(L)$. For every $(a,b)\in J$ and $d\in D(L)$ denote by
$d_{a,b}$ the semimetric on $L$ defined by
$d_{a,b}(x,y):=d(f_{a,b}(x),\,f_{a,b}(y))$. It is easy to see then that the
lattice uniformity $u_{J}$ is generated by the family $\\{d_{a,b}:d\in
D_{u},\,(a,b)\in J\\}$._

######  Lemma 5.2.

Let $d\in D(L)$.

  1. (i)

Let $P_{n}:L^{n}\to L$ be given inductively by $P_{1}(x):=x$ and

$P_{n}(x_{1},\dots,x_{n}):=\begin{cases}P_{n-1}(x_{1},\dots,x_{n-1})\,\vee\,x_{n},&\mbox{or
}{}\\\ P_{n-1}(x_{1},\dots,x_{n-1})\,\wedge\,x_{n}.&\end{cases}$

Then
$d(P_{n}(x_{1},\dots,x_{n}),P_{n}(y_{1},\dots,y_{n}))\leq\sum_{i=1}^{n}d(x_{i},y_{i})$.

  2. (ii)

$d(f_{a,b}(x),f_{a,b}(y))\leq d(f_{c,d}(x),f_{c,d}(y))+2d(a,c)+2d(b,d)$, for
every $a,b,c,d,x,y\in L$.

###### Proof.

The proof of (i) is by induction. The assertion is trivially true when $n=1$.
Suppose that
$d(P_{n-1}(x_{1},\dots,x_{n-1}),P_{n-1}(y_{1},\dots,y_{n-1}))\leq\sum_{i=1}^{n-1}d(x_{i},y_{i})$.
Let $\diamond\in\\{\wedge,\vee\\}$. Then,

$\displaystyle d(P_{n}(x_{1},\dots,x_{n}),P_{n}(y_{1},\dots,y_{n}))$
$\displaystyle=d(P_{n-1}(x_{1},\dots,x_{n-1})\diamond
x_{n},P_{n-1}(y_{1},\dots,y_{n-1})\diamond y_{n})$ $\displaystyle\leq
d(P_{n-1}(x_{1},\dots,x_{n-1})\diamond
x_{n},P_{n-1}(y_{1},\dots,y_{n-1})\diamond x_{n})$
$\displaystyle\qquad\qquad\qquad\qquad+d(P_{n-1}(y_{1},\dots,y_{n-1})\diamond
x_{n},P_{n-1}(y_{1},\dots,y_{n-1})\diamond y_{n})$ $\displaystyle\leq
d(P_{n-1}(x_{1},\dots,x_{n-1}),P_{n-1}(y_{1},\dots,y_{n-1}))+d(x_{n},y_{n})$
$\displaystyle\leq\sum_{i=1}^{n}d(x_{i},y_{i}).$

To prove (ii) one first writes

$d(f_{a,b}(x),f_{a,b}(y))\leq
d(f_{a,b}(x),f_{c,d}(x))+d(f_{c,d}(x),f_{c,d}(y))+d(f_{c,d}(y),f_{a,b}(y)),$

and then apply (i) to obtain

$d(f_{a,b}(x),f_{a,b}(y))\leq d(f_{c,d}(x),f_{c,d}(y))+2d(a,c)+2d(b,d).$

∎

######  Proposition 5.3.

Let $\emptyset\neq J\subseteq J(L)$ and let $\tilde{J}:=\bar{J}\cap J(L)$,
where $\bar{J}$ is the closure of $J$ in $(L\times L,u\times u)$. Then
$u_{J}=u_{\tilde{J}}$.

###### Proof.

The inclusion $u_{J}\subseteq u_{\tilde{J}}$ follows by Proposition 4.2(ii).
For the reverse inclusion, suppose that $W\in u_{\tilde{J}}$. We show that
there exists $U\in u_{J}$ such that $U\subseteq W$. In view of Remark 5.1
there exist $d\in D_{u}$, $(a,b)\in\tilde{J}$ and $\varepsilon>0$, such that
$(x,y)\in W$ for every $(x,y)\in L^{2}$ satisfying
$d_{a,b}(x,y)<5\varepsilon$. Let $(a^{\prime},b^{\prime})\in J$ with
$d(a,a^{\prime})<\varepsilon$ and $d(b,b^{\prime})<\varepsilon$. Then,
$U_{a^{\prime},b^{\prime}}:=\\{(x,y)\in
L^{2}:d_{a^{\prime},b^{\prime}}(x,y)<\varepsilon\\}\in u_{J}$ and by Lemma
5.2(ii)

$d_{a,b}(x,y)\leq
d_{a^{\prime},b^{\prime}}(x,y)+2d(a,a^{\prime})+2d(b,b^{\prime})<5\varepsilon,$

for every $(x,y)\in U_{a^{\prime},b^{\prime}}$. This implies that
$U_{a^{\prime},b^{\prime}}\subseteq W$. ∎

######  Corollary 5.4.

Let $S$ be a dense sublattice of $(L,u)$ and $v=u|_{S}$. Then
${\left.\kern-1.2ptu^{*}\vphantom{\mid}\right|_{S}}=v^{*}$.

###### Proof.

Let $J:=J(S)=\\{(a,b)\in S^{2}:a\leq b\\}$. First observe that
$J(L)\subseteq\tilde{J}$: If $(a,b)\in J(L)$ and $(x_{\alpha})$,
$(y_{\alpha})$ are nets in $S$ converging in $(L,u)$ to $a$ and $b$,
respectively, then $J\ni(x_{\alpha}\wedge y_{\alpha},x_{\alpha}\vee
y_{\alpha})\rightarrow(a,b)$ ($u$), hence $(a,b)\in\tilde{J}$.

Applying Proposition 5.3 we obtain $u^{*}=u_{J}$. Obviously
${\left.\kern-1.2ptu_{J}\vphantom{\mid}\right|_{S}}=v_{J}$.

Combining, we get
${\left.\kern-1.2ptu^{*}\vphantom{\mid}\right|_{S}}={\left.\kern-1.2ptu_{J}\vphantom{\mid}\right|_{S}}=v_{J}=v^{\ast}$.
∎

We now use besides Corollary 5.4 the following general observation:

######  Lemma 5.5 ([23], pg. 381).

Let $Y$ be a dense subspace of a uniform space $(X,v)$.

  1. (i)

Then $w\mapsto w|_{Y}$ defines a lattice isomorphism from the lattice of all
uniformities on $X$ coarser than $v$ onto the lattice of all uniformities on
$Y$ coarser than $v|_{Y}$.

  2. (ii)

Let $w$ be a uniformity on $X$ coarser than $v$ and $y_{0}\in Y$. Then

$\\{\bar{W}^{v}:W\text{ is a neighbourhood of }y_{0}\text{ in }(Y,w|_{Y})\\}$

is a neighbourhood base of $y_{0}$ in $(X,w)$; here $\bar{W}^{v}$ denotes the
closure of $W$ in $(X,v)$.

######  Theorem 5.6.

Let $S$ be a dense sublattice of $(L,u)$ and $v=u|_{S}$.

  1. (i)

Then $u^{*}=u$ iff $v^{*}=v$.

  2. (ii)

If the $u^{*}$-topology coincides with the $u$-topology, then $v^{*}$-topology
coincides with the $v$-topology.

###### Proof.

(i) By Lemma 5.5, $u^{*}=u$ iff $u^{*}|_{S}=u|_{S}$. Now observe that
$u|_{S}=v$ by definition and $u^{*}|_{S}=v^{*}$ by Corollary 5.4.

(ii) If the $u^{*}$-topology coincides with the $u$-topology, then the
$u^{*}|_{S}$-topology coincides with the $u|_{S}$-topology, i.e in view of
Corollary 5.4, the $v^{*}$-topology coincides with the $v$-topology. ∎

We don’t know whether in (ii) of Theorem 5.6 the converse implication is true.
Lemma 5.5 (ii) and Corollary 5.4 only imply that every $s\in S$ has the same
neighbourhood system in $(L,u^{*})$ and in $(L,u)$ if $v^{*}$-topology
coincides with the $v$-topology. But this is enough to answer the question of
Taylor (see [T19, Question 3.3]) whether the property ‘unbounded’ passes from
the Hausdorff locally solid Riesz space $(X,\tau)$ to its completion
$(\tilde{X},\tilde{\tau})$.

######  Theorem 5.7.

Let $H$ be a dense solid subgroup of a locally solid commutative $\ell$-group
$(G,\tau)$ and let $\sigma=\tau|_{H}$. Then $\mathfrak{u}\tau=\tau$ iff
$\mathfrak{u}\sigma=\sigma$.

###### Proof.

Let $u$ and $v$ be the uniformities induced, respectively, by $\tau$ and
$\sigma$. We use that the $u^{*}$-topology and the $v^{*}$-topology coincide,
respectively, with $\mathfrak{u}\tau$ and $\mathfrak{u}\sigma$ (see Theorem
4.9) and therefore $\mathfrak{u}\tau|_{H}=\mathfrak{u}\sigma$ (see Corollary
5.4).

If $\mathfrak{u}\tau=\tau$, then $\mathfrak{u}\sigma=\sigma$ by Theorem 5.6
(b). Vice versa, if $\mathfrak{u}\sigma=\sigma$, then $\mathfrak{u}\tau$ and
$\tau$ have the same $0$-neighbourhood system by Lemma 5.5(ii). Therefore
$\mathfrak{u}\tau=\tau$ since both, $\mathfrak{u}\tau$ and $\tau$, are group
topologies. ∎

The following Theorem generalizes [22, Proposition 2.12] and answers [22,
Question 2.13].

######  Theorem 5.8.

Let $S$ be a sublattice of the uniform lattice $(L,u)$. Then $S$ is
(sequentially) closed w.r.t. $u^{\ast}$ iff it is (sequentially) closed w.r.t.
$u$.

###### Proof.

For an infinite cardinal $\kappa$ let us say that a subset $U$ of a
topological space is $\kappa$-closed when $U$ contains the limit of every
convergent net $(x_{\alpha})_{\alpha\in A}$ where $|A|\leq\kappa$, that is
contained in $U$. We show that $S$ is $\kappa$-closed w.r.t. $u^{\ast}$ iff it
is $\kappa$-closed w.r.t. $u$. Since $u^{\ast}\subseteq u$, clearly every
$\kappa$-closed subset w.r.t. $u^{\ast}$ is $\kappa$-closed w.r.t. $u$. For
the converse, assume that $S$ is $\kappa$-closed w.r.t. $u$ and suppose that
$(s_{\alpha})_{\alpha\in A}$ is a net in $S$ such that $|A|\leq\kappa$ and
convergent to $x$ w.r.t. $u^{\ast}$. For every $s\in S$ and $z\in L$
satisfying $z\geq s$ we observe that $(s_{\alpha}\vee s)\wedge z\to(x\vee
s)\wedge z$ w.r.t. $u$. In particular, if we fix $\alpha^{\prime}\in A$ and
set $z:=s_{\alpha^{\prime}}\vee s$, we obtain – by the hypothesis that $S$ is
$\kappa$-closed w.r.t. $u$ – that $(x\vee s)\wedge(s_{\alpha^{\prime}}\vee
s)\in S$. Since $\alpha^{\prime}$ was arbitrary, and since $s_{\alpha}\to x$
w.r.t. $u^{\ast}$, this in turn implies that

$(x\vee s)\wedge(s_{\alpha}\vee s)\to(x\vee s)\wedge(x\vee s)=x\vee
s\qquad\text{w.r.t. $u$}.$

Using again the fact that $S$ is $\kappa$-closed w.r.t. $u$ we deduce that

(i) $x\vee s\in S$ for every $s\in S$.

(ii) Dually one obtains that $x\wedge s^{\prime}\in S$ for every
$s^{\prime}\in S$.

Let now $s\in S$. Then $s^{\prime}:=x\vee s\in S$ by (i) and $x=x\wedge
s^{\prime}\in S$ by (ii). ∎

######  Corollary 5.9.

Let $S$ be a sublattice of the uniform lattice $(L,u)$. Then the closures of
$S$ in $L$ w.r.t. $u$ and $u^{\ast}$ coincide.

By an obvious modification of the proof of Theorem 5.8 one obtains an
analogous result for unbounded order convergence, which generalizes [10,
Proposition 3.15].

######  Proposition 5.10.

Suppose that $x_{\alpha}\uparrow x$ implies $x_{\alpha}\wedge y\uparrow
x\wedge y$ and $x_{\alpha}\downarrow x$ implies $x_{\alpha}\vee y\downarrow
x\vee y$ for every net $(x_{\alpha})$ in $L$ and $x,y\in L$ (i.e. $L$ is
continuous in the sense of von Neumann – cf. [16, Definition 2.14]). Then a
sublattice $S$ of $L$ is $\mathfrak{u}O$-closed iff it is $O$-closed.

###### Proof.

$\Leftarrow$ can be proved as $\Leftarrow$ of Theorem 5.8 replacing
$u^{*}$-convergence by $\mathfrak{u}O$-convergence and $u$-convergence by
$O$-convergence. For $\Rightarrow$ observe that the additional assumption
exactly means that $O$-convergence implies $\mathfrak{u}O$-convergence.
Therefore any $\mathfrak{u}O$-closed subset of $L$ is $O$-closed. ∎

######  Proposition 5.11.

The following conditions are equivalent:

  1. (i)

$u$ is locally exhaustive;

  2. (ii)

$u^{\ast}$ is locally exhaustive;

  3. (iii)

$u^{\ast}$ is exhaustive.

###### Proof.

Since $u$ and $u^{*}$ coincide on bounded sets by Corollary 4.3, we have
(i)$\Leftrightarrow$(ii). (iii)$\Rightarrow$(ii) is obvious.

(i)$\Rightarrow$(iii): By definition, a net $(x_{\gamma})_{\gamma\in\Gamma}$
is $u^{\ast}$-Cauchy iff $(f_{a,b}(x_{\gamma}))_{\gamma\in\Gamma}$ is
$u$-Cauchy for every $(a,b)\in J(L)$. Let now $(x_{\gamma})_{\gamma\in\Gamma}$
be a monotone net in $L$ and $(a,b)\in J(L)$. Then
$(f_{a,b}(x_{\gamma}))_{\gamma\in\Gamma}$ is monotone and bounded, hence
$u$-Cauchy by (i), i.e. $(x_{\gamma})_{\gamma\in\Gamma}$ is $u^{\ast}$-Cauchy.
∎

######  Theorem 5.12.

Let $u$ and $v$ be Hausdorff lattice uniformities on $L$ satisfying Condition
(C). Then:

  1. (i)

the $u^{\ast}$-topology and the $v^{\ast}$-topology are equal;

  2. (ii)

$\overline{S}^{u}=\overline{S}^{v}$ for every sublattice $S$ of $L$.

###### Proof.

(i) First observe that the lattice uniformities $u^{\ast}$ and $v^{\ast}$ also
satisfy Condition (C) since $u^{*}\subseteq u$ and $v^{*}\subseteq v$.
Therefore $u^{\ast}$ and $v^{\ast}$ are locally exhaustive by Proposition 3.6,
hence exhaustive in view of Proposition 5.11. Moreover, Corollary 4.7 yields
that $u^{\ast}$ and $v^{\ast}$ are Hausdorff. So, (i) follows by Theorem 3.7.

(ii) follows from (i) and Corollary 5.9. ∎

Note that (ii) of the above theorem generalizes [22, Theorem 5.11].

The following lemma is needed in the proof of Theorem 5.14 to reduce it to the
Hausdorff case.

######  Lemma 5.13.

Let $(\hat{L},\hat{u})=(L/{\scriptscriptstyle\sim_{u}}\,,\,\widehat{u})$ be
the Hausdorff uniform lattice associated with $(L,u)$ according to Proposition
3.8. Then

$\left(\hat{L}\,,\,\widehat{u^{\ast}}\right)=\left(\hat{L}\,,\,(\widehat{u})^{\ast}\right).$

###### Proof.

Observe first that $N(u)=N(u^{\ast})$ by Proposition 3.2, i.e.
$L/{\scriptscriptstyle\sim_{u}}=L/{\scriptscriptstyle\sim_{u^{\ast}}}$. We
show that $(\widehat{u})^{\ast}=\widehat{(u^{\ast})}$. We shall make use of
the following fact that follows immediately by Proposition 3.8: If $v$ is a
lattice uniformity on $L$ satisfying $N(u)=N(v)$, then

$\widehat{V\cap[a,b]^{2}}=\widehat{V}\cap[\hat{a},\hat{b}]^{2},$

for every closed subset $V$ of $(L,v)^{2}$ and $a\leq b$ in $L$.

Hence, for every $a\leq b$ in $L$ we have

$\widehat{u^{\ast}}|_{[\hat{a},\,\hat{b}]}\,=\,\widehat{u^{\ast}|_{[a,\,b]}}\,=\,\widehat{u|_{[a,\,b]}}=\widehat{u}|_{[\hat{a},\,\hat{b}]}$

and therefore $\widehat{u^{\ast}}\supset\widehat{u}^{\ast}$. Conversely, if
$v$ is a lattice uniformity on $L$ with $N(v)=N(u)$ and
$\hat{v}=\widehat{u}^{\ast}$, then
$\widehat{v}|_{[\hat{a},\,\hat{b}]}=\widehat{u}|_{[\hat{a},\,\hat{b}]}$, and
therefore $v|_{[a,\,b]}=u|_{[a,\,b]}$ for every $a\leq b$. This implies that
$v\supset u^{\ast}$ and therefore
$\widehat{u}^{\ast}=\widehat{v}\supset\widehat{u^{\ast}}$. ∎

Let $u$ be Hausdorff and $(\tilde{L},\tilde{u})$ the completion of $(L,u)$. It
is easy to see that $\tilde{L}$ is again distributive, when $L$ is
distributive. Moreover, both properties of local exhaustivity and exhaustivity
pass directly to this completion by Proposition 3.4.

######  Theorem 5.14.

Let $(L,u)$ be a uniform lattice. If $u$ is exhaustive, then $u=u^{\ast}$.

###### Proof.

(i) Let us first suppose that $u$ is Hausdorff. Let $(\tilde{L},\tilde{u})$
denote the completion of $(L,u)$. Then $\tilde{L}$ is a complete lattice by
Corollary 3.5 and therefore obviously $\tilde{u}=(\tilde{u})^{\ast}$. On the
other hand,
${\left.\kern-1.2pt(\tilde{u})^{\ast}\vphantom{\mid}\right|_{S}}=u^{\ast}$ by
Corollary 5.4. Combining we get
$u={\left.\kern-1.2pt\tilde{u}\vphantom{\mid}\right|_{S}}={\left.\kern-1.2pt(\tilde{u})^{\ast}\vphantom{\mid}\right|_{S}}=u^{\ast}$.

(ii) In general, if $u$ is exhaustive, $\widehat{u}$ is Hausdorff and
exhaustive, and therefore $\widehat{u}=(\widehat{u})^{\ast}$ by (i).
Therefore, by Lemma 5.13, we obtain $\widehat{u}=\widehat{u^{\ast}}$. Since
$N(u)=N(u^{\ast})$, this implies555It is easy to see and follows from
Proposition 3.8 that if $u$ and $v$ are two lattice uniformities on $L$
satisfying $N(u)=N(v)$, then the induced uniformities $\widehat{u}$ and
$\widehat{v}$ on the common quotient
$L/{\scriptscriptstyle\sim_{u}}=L/{\scriptscriptstyle\sim_{v}}$ are equal iff
$u$ and $v$ are equal. that $u=u^{\ast}$. ∎

######  Corollary 5.15.

Let $(L,u)$ be a uniform lattice and let $u$ be locally exhaustive. Then
${\left.\kern-1.2ptu^{\ast}\vphantom{\mid}\right|_{S}}=({\left.\kern-1.2ptu\vphantom{\mid}\right|_{S}})^{\ast}$
for every sublattice $S$ of $L$.

###### Proof.

By Proposition 5.11 $u^{*}$ is exhaustive. Therefore $u^{*}|_{S}$ is
exhaustive, hence $(u^{*}|_{S})^{*}=u^{*}|_{S}$ by Theorem 5.14. Since
$u|_{S}\supseteq u^{*}|_{S}$, we have $(u|_{S})^{*}\supseteq(u^{*}|_{S})^{*}$.
Moreover, $u^{*}|_{S}\supseteq(u|_{S})^{*}$ since $u^{*}|_{S}$ agrees with
$u|_{S}$ on order-bounded subsets of $S$ (c.f. Proposition 4.2 (iii) ). We
have seen:

$(u|_{S})^{*}\supseteq(u^{*}|_{S})^{*}=u^{*}|_{S}\supseteq(u|_{S})^{*},$

${\left.\kern-1.2ptu^{\ast}\vphantom{\mid}\right|_{S}}=({\left.\kern-1.2ptu\vphantom{\mid}\right|_{S}})^{\ast}$.
∎

Since for locally solid Riesz spaces the Lebesgue property implies the pre-
Lebesgue property, Corollary 5.15 generalizes [22, Lemma 9.7] (therefore [13,
Corollary 4.6]).

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2022

[1,2]Dr Stefano Spessato ORCID: 0000-0003-0069-5853

[1]Università degli Studi ”Niccolò Cusano”, via don Carlo Gnocchi 3, Rome,
00166, Italy

# An equivalent definition for multiplicities of a quaternionic eigenvalue

<EMAIL_ADDRESS>*

###### Abstract

In this paper we introduce two definitions for algebraic and geometric
multiplicities of a quaternion right eigenvalue. This definitions are
equivalent to the classical ones. However, differently from the usual
definitions, the notions of root subspace and Jordan form of a quaternion
matrix are not required and all the properties can be proved easily.

###### keywords:

quaternion, right eigenvalue, geometric multiplicity, algebraic multiplicity

### Acknowledgments

I would like to thank the Unione Matematici Italiani (U.M.I.) for giving to me
the opportunity of held a talk during the First U.M.I. meeting of Ph.D.
students in May. Moreover I would also thank Alfredo Donno and Matteo Cavaleri
for their advice and their help.

Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

## Introduction

Spectral properties of matrices with quaternion entries are an important
research topic in last years. This happens for the large number of
applications of quaternions both in Pure and Applied Mathematics.
However, since the space of quaternions $\mathbb{H}$ is not a commutative
algebra, is quite difficult to obtain the quaternion version of the classical
theorems of Complex Linear Algebra. One of this result is the Quaternion
Spectral Theorem of Farenick and Pidkowich Spec . Sometimes it is also
difficult to define the equivalent notions: for example a quaternion matrix
has at least two kinds spectra, which are called the _right_ spectrum and the
_left_ spectrum. In particular, in this case, the most common choice is to
study the right spectrum of a quaternion matrix.
In this paper, after some basiscs of Quaternion Linear Algebra (Section 1), we
give two new definitions for the algebraic and geometric multiplicity (resp.
_a.m._ and _g.m._) of a right eigenvalue of a quaternion matrix (Section 2).
Moreover, always in the same Section, we also give the proofs of the main
properties of this multiplicities. In particular we prove

  * •

the invariance of the multiplicities under the similarity relation

  * •

the classical inequality $a.m.(q,A)\geq g.m.(q,A)$, where $q$ is a right-
eigenvalue of a quaternion matrix $A$,

  * •

the equality $a.m.(q,A)=g.m.(q,A)$, if $A$ is a normal matrix.

Finally, in Section 3, we prove that our definitions are equivalent to the
classical ones introduced in the book of Rodman Quat .
The advantage of our definitions is that they can be given without the notions
of _root subspace_ or _Jordan form_ of a quaternion matrix and so they can be
used also by people which are not in Quaternion Linear Algebra. Moreover also
the main properties can be proved without this tools.

##  1 Basics of Quaternion Linear Algebra

###  1.1 Quaternions

Let $\mathbb{H}$ be the unital associative $\mathbb{R}$-algebra whose
generators are $i$, $j$ and $k$ where

$i^{2}=j^{2}=k^{2}=ijk=-1.$ (1)

This algebra is called the algebra of real quaternions. Given $q=a+bi+cj+dk$
in $\mathbb{H}$ the imaginary part of $q$ is $Im(q)=bi+cj+dk$ and the real
part of $q$ is $Re(q)=a$. Then conjugate of a quaternion $q$ is the quaternion
$\overline{q}:=Re(q)-Im(q)$.
$\mathbb{H}$ is not a commutative algebra, indeed $ij=k$, but

$ji=ji(jj)(-1)=j(ij)j(-1)=jkj(-1)=ij(-1)=-k.$ (2)

Two quaternions $q_{1},q_{2}\in\mathbb{H}$ are _similar_ if there is $h$ in
$\mathbb{H}$ such that

$q_{1}=h^{-1}q_{2}h.$ (3)

If $q_{1}$ and $q_{2}$ are similar, then we write $q_{1}\sim q_{2}$ and we
denote by $[q_{1}]$ the class of $q$ in $\mathbb{H}$ such that $q\sim q_{1}$.
Observe that if $Im(q)=0$, then $[q]=\\{q\\}$, indeed $qh=hq$ for each $h$ in
$\mathbb{H}$. By Lemma 2.2 of the paper of Farenick and Pidkowich Spec  we
know that for each $q$ in $\mathbb{H}$ there is a unique $\lambda_{q}$ in
$\mathbb{C}$ such that $Im(\lambda_{q})\geq 0$ and

$[q]=[\lambda_{q}]=[\overline{\lambda_{q}}].$ (4)

###  1.2 Right spectrum of a quaternion matrix

Let us denote by $\mathbb{H}^{n}$ the space of column vectors of length $n$
with quaternionic entries. Let $\mathbb{H}^{m\times n}$ be the space of
$m\times n$ matrices with quaternionic entries. Given a matrix $A=(a_{ij})$ in
$\mathbb{H}^{m\times n}$ we define the conjugate transpose of $A$ as the
$n\times m$ matrix $A^{*}:=(\overline{a_{ji}})$. A square matrix $A$ is
hermitian if $A=A^{*}$ and it is normal if $AA^{*}=A^{*}A$.
For a square matrix $A$ in $\mathbb{H}^{n\times n}$, we say that
$q\in\mathbb{H}$ is a right eigenvalue if there is a $x\neq 0$ in
$\mathbb{H}^{n}$ such that $Ax=x\cdot q$. We say right spectrum of $A$ the set
$\sigma_{r}(A)$ of all right eigenvalues of $A$.
Notice that if $q$ is a right eigenvalue of $A$ and $q^{\prime}$ is an element
of $\mathbb{H}$ such that $q\sim q^{\prime}$ then also $q^{\prime}$ is a right
eigenvalue of $A$. Indeed if $q^{\prime}=sqs^{-1}$ and $Aw=w\cdot q$ then
$A(ws^{-1})=(ws^{-1})q^{\prime}$.
Observe that each quaternion $q=a+bi+cj+dk$ can be seen as $q=(a+ib)+(c+id)j$
where $a+ib$ and $c+id$ are complex numbers. In the same way every quaternion
matrix $A$ in $\mathbb{H}^{n\times n}$ can be decomposed as $A=A_{1}+A_{2}j$
where $A_{1}$ and $A_{2}$ are $n\times n$-complex matrices.
Then the complex adjoint matrix of $A$ is the matrix $f(A)$ in
$\mathbb{C}^{2n\times 2n}$ defined as

$f(A):=\begin{bmatrix}A_{1}&&A_{2}\\\
-\overline{A}_{2}&&\overline{A}_{1}\end{bmatrix}$ (5)

######  Proposition 1.1 (pag. 78 of Spec ).

For each $A,B$ in $\mathbb{H}^{n\times n}$ the following properties hold:

  * •

$f(Id_{n})=Id_{2n}$;

  * •

$f(A+B)=f(A)+f(B)$;

  * •

$f(A\cdot B)=f(A)\cdot f(B)\mathchar 24635\relax\;$

  * •

$f(A^{*})=f(A)^{*}$;

  * •

$f(A^{-1})=f(A)^{-1}$.

These properties are very useful in order to investigate the right spectrum of
a quaternion matrix. In particular we obtain the following result

######  Proposition 1.2.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$. Then

$Spec(f(A))=\\{\lambda\in\mathbb{C}|[\lambda]=[q]\mbox{ for each
}q\in\sigma_{r}(A)\\}.$ (6)

In particular, if $A=A^{*}$, then $\sigma_{r}(A)\subset\mathbb{R}$ and

$Spec(f(A))=\sigma_{r}(A).$ (7)

Proof: This is essentially the proof of points 1. and 2. of Lemma 2.5 of Spec
. Observe that if $A$ is a $1\times 1$-matrix, then
$f:\mathbb{H}\longrightarrow\mathbb{C}^{2\times 2}$. So, for each matrix
$A=(a_{rs})$ in $\mathbb{H}^{n\times n}$, we obtain a block matrix
$\pi(A):=(\pi(a_{rs}))=(f(a_{rs}))$ in $\mathbb{C}^{2n\times 2n}$.

######  Proposition 1.3.

For each matrix $A$ in $\mathbb{H}^{n\times n}$, the matrices $f(A)$ and
$\pi(A)$ are similar. This implies that

$Spec(f(A))=Spec(\pi(A))$ (8)

and, if we denote by $a.m.(\lambda,B)$ and $g.m.(\lambda,B)$ the algebraic
multiplicity and the geometric multiplicity of an eigenvalue $\lambda$ of a
matrix $B$, then

$a.m.(\lambda,\pi(A))=a.m.(\lambda,f(A))\mbox{ and
}g.m.(\lambda,\pi(A))=g.m.(\lambda,f(A)).$ (9)

Proof: The proof of this fact is similar to the the proof Theorem 5.9 of the
work of Donno, Cavaleri and S. Godsil . In their work, indeed, it is proved
that, given a quaternion matrix $A$, then $\pi(A)$ and $f(A)$ are similar as
complex matrices. So this fact implies also our statement. However, for sake
of completeness, we report the proof of the similarity of $f(A)$ and $\pi(A)$.
Notice that, for each entry $a_{rs}$, of $A$ there are two complex numbers
$a_{rs}^{[2]}$ and $a_{rs}^{[1]}$ such that

$a_{rs}=a_{rs}^{[1]}+a_{rs}^{[2]}j.$ (10)

Then, the matrix $\pi(A)$, is given by

$\pi(A)=\begin{bmatrix}a_{1,1}^{[1]}&a_{1,1}^{[2]}&...&a_{1,n}^{[1]}&a_{1,n}^{[2]}\\\
-\overline{a}_{1,1}^{[2]}&\overline{a}_{1,1}^{[1]}&...&-\overline{a}_{n,n}^{[2]}&\overline{a}_{n,n}^{[1]}\\\
\vdots&\vdots&...&\vdots&\vdots\\\
a_{n,1}^{[1]}&a_{n,1}^{[2]}&...&a_{n,n}^{[1]}&a_{n,n}^{[2]}\\\
-\overline{a}_{n,1}^{[2]}&\overline{a}_{n,1}^{[1]}&...&-\overline{a}_{n,n}^{[2]}&\overline{a}_{n,n}^{[1]}\\\
\end{bmatrix}$ (11)

Let us denote by $\tau:\\{1,...,2n\\}\longrightarrow\\{1,...,2n\\}$ the
permutation defined as

$\tau(r)=\begin{cases}n+k\mbox{ if }r=2k\\\ k+1\mbox{ if }r=2k+1\end{cases}$
(12)

Let $e_{b}$ be the $2n$-vector which has $1$ in the $b$-th entry and $0$
otherwise. Let us denote by $C_{\tau}$ the permutation $2n\times 2n$-matrix
such that the $b$-th column is $e_{\tau(b)}$ and let $R_{\tau}$ be its
transpose. Then

$f(A)=R_{\tau}\pi(A)C_{\tau}.$ (13)

Indeed

$\pi(A)C_{\tau}=\begin{bmatrix}a_{1,1}^{[1]}&a_{1,2}^{[1]}&...&a_{1,n}^{[1]}&a_{1,1}^{[2]}&...&a_{1,n-1}^{[2]}&a_{1,n}^{[2]}\\\
-\overline{a}_{1,1}^{[2]}&-\overline{a}_{1,2}^{[2]}&...&-\overline{a}_{1,n}^{[2]}&\overline{a}_{1,1}^{[1]}&...&\overline{a}_{1,n-1}^{[1]}&\overline{a}_{1,n}^{[1]}\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots&\vdots\\\
a_{n,1}^{[1]}&a_{n,2}^{[1]}&...&a_{n,n}^{[1]}&a_{n,1}^{[2]}&...&a_{n,n-1}^{[2]}&a_{n,n}^{[2]}\\\
-\overline{a}_{n,1}^{[2]}&-\overline{a}_{n,2}^{[2]}&...&-\overline{a}_{n,n}^{[2]}&\overline{a}_{n,1}^{[1]}&...&\overline{a}_{n,n-1}^{[1]}&\overline{a}_{n,n}^{[1]}\end{bmatrix}$
(14)

and

$R_{\tau}\pi(A)C_{\tau}=\begin{bmatrix}a_{1,1}^{[1]}&a_{1,2}^{[1]}&...&a_{1,n}^{[1]}&a_{1,1}^{[2]}&...&a_{1,n-1}^{[2]}&a_{1,n}^{[2]}\\\
a_{2,1}^{[1]}&a_{2,2}^{[1]}&...&a_{2,n}^{[1]}&a_{2,1}^{[2]}&...&a_{2,n-1}^{[2]}&a_{2,n}^{[2]}\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots&\vdots\\\
a_{n,1}^{[1]}&a_{n,2}^{[1]}&...&a_{n,n}^{[1]}&a_{n,1}^{[2]}&...&a_{n,n-1}^{[2]}&a_{n,n}^{[2]}\\\
-\overline{a}_{1,1}^{[2]}&-\overline{a}_{1,2}^{[2]}&...&-\overline{a}_{1,n}^{[2]}&\overline{a}_{1,1}^{[1]}&...&\overline{a}_{1,n-1}^{[1]}&\overline{a}_{1,n}^{[1]}\\\
\vdots&\vdots&&\vdots&\vdots&&\vdots&\vdots\\\
-\overline{a}_{n-1,1}^{[2]}&-\overline{a}_{n-1,2}^{[2]}&...&-\overline{a}_{n-1,n}^{[2]}&\overline{a}_{n-1,1}^{[1]}&...&\overline{a}_{n-1,n-1}^{[1]}&\overline{a}_{n-1,n}^{[1]}\\\
-\overline{a}_{n,1}^{[2]}&-\overline{a}_{n,2}^{[2]}&...&-\overline{a}_{n,n}^{[2]}&\overline{a}_{n,1}^{[1]}&...&\overline{a}_{n,n-1}^{[1]}&\overline{a}_{n,n}^{[1]}\end{bmatrix}$
(15)

Then if $A=(a_{rs})$, then

$R_{\tau}\pi(A)C_{\tau}=\begin{bmatrix}a_{rs}^{[1]}&a_{rs}^{[2]}\\\
-\overline{a}_{rs}^{[2]}&\overline{a}_{rs}^{[1]}\end{bmatrix}=f(A).$ (16)

Notice that, since $R_{\tau}$ is a permutation matrix, then
$R_{\tau}^{-1}=C_{\tau}$ and so $\pi(A)$ and $f(A)$ are similar. This means
that they have the same spectrum and the same multiplicities.

###  1.3 Schur’s triangularization theorem

We conclude this Section with a classical result on Quaternion Linear Algebra.
This is an immediate consequences of Fundamental Theorem of Algebra. A short
proof of this result can be found in the book of RotmanQuat

######  Lemma 1.4 (Theorem 5.3.5. of Quat ).

For every $A\in\mathbb{H}^{n\times n}$ there exists an eigenvalue which is a
complex number with nonnegative imaginary part.

As a consequence of this result, by applying the usual technique from Complex
Linear Algebra, it is possible to prove a quaternionic version of Schur’s
triangularization theorem.

######  Lemma 1.5 (Lemma 3.3 of Spec , Theorem 5.3.6 of Quat ).

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$. Then there are $U,T$ in
$\mathbb{H}^{n\times n}$ such that $U$ is unitary, $T$ is an upper triangular
matrix and

$T=U^{*}AU.$ (17)

Moreover the entries on the diagonal of $T$ are complex numbers $\mu_{i}$ such
that $Im(\mu_{i})\geq 0.$

###  1.4 Quaternionic dimension of a right $\mathbb{H}$-vector space

Our next step is to define the geometric multiplicity of a right eigenvalue
$q$ of a quaternionic matrix.

######  Definition 1.1.

An additive abelian group $V$ is a right $\mathbb{H}$-vector space if there is
a map $V\times\mathbb{H}\longrightarrow V$ under which the image of each pair
$(v,q)$ is denoted by $vq$, such that for all $q,q_{1},q_{2}$ in $\mathbb{H}$
and $v,v_{1},v_{2}$ in $V$:

  * •

$(v_{1}+v_{2})q=v_{1}q+v_{2}q$,

  * •

$v(q_{1}+q_{2})=vq_{1}+vq_{2}$,

  * •

$v(q_{1}q_{2})=(vq_{1})q_{2}$, and

  * •

$v1=v$.

######  Definition 1.2.

If $V$ is a nonzero right $\mathbb{H}$-vector space and if $X\subseteq V$ is a
nonempty subset, then:

  1. 1.

$X$ is a generating set for $V$ if for each $v$ in $V$,

$v=v_{1}q_{1}+...+v_{k}q_{k},$ (18)

where $v_{i}\in X$ and $q_{i}\in\mathbb{H}$ for each $i=1,...,k$,

  2. 2.

$X$ is an $\mathbb{H}$-independent set if, for any $k\in\mathbb{N}_{>0}$ and
for any distinct $v_{1},...,v_{k}$ in $V$ the equation

$v_{1}q_{1}+...+v_{k}q_{k}=0$ (19)

is satisfied only for $q_{1}=...=q_{k}=0$.

  3. 3.

$X$ is a basis for $V$ if $X$ is generating and $\mathbb{H}$-independent set.

######  Remark 1.

Every generating space $X$ of a right $\mathbb{H}$-vector space $V$ contains a
basis of $V$: the proof is exactly the same as for real vector spaces.

######  Theorem 1.6 (Theorem 4.2 of Spec ).

If a right $\mathbb{H}$-vector space $V$ has a generating set with $m$
elements, then $V$ has a basis with $n$ elements for some $n\leq m$. Moreover:

  1. 1.

every basis of $V$ has $n$ elements;

  2. 2.

if $1\leq k\leq n$ and if $v_{1},...,v_{k}$ in $V$ are
$\mathbb{H}$-independent, then there is a basis of $V$ that contains the
vectors $v_{1},...,v_{k}$.

######  Definition 1.3.

Let $V$ be a right $\mathbb{H}$-vector space and let $X$ be a basis for $V$.
Then the quaternionic dimension of $V$ is the cardinality of $X$. We denote it
by $dim_{\mathbb{H}}(V)$.

######  Remark 2.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$. Let $q$ be a right eigenvalue
of $A$. Let us denote by $V_{q}$ the right $q$-eigenspace, i.e. the subset of
$\mathbb{H}^{n}$ given by the right $q$-eigenvectors. Notice that $V_{q}$ is a
real vector space. Unfortunately it is not a right $\mathbb{H}$-vector space,
indeed if $v$ is a $q$-eigenvector and $q_{2}$ is an element of $\mathbb{H}$,
then $vq_{2}$ is a right $q_{2}^{-1}qq_{2}$-eigenvalue.

##  2 Multiplicities of a right eigenvalue

###  2.1 Algebraic multiplicity

Our next step is to define the algebraic multiplicity for a right eigenvalue
of a quaternion matrix.

######  Definition 2.1.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$ and let $q$ be a right
eigenvalue of $A$. Let $\lambda_{q}$ be the complex number such that
$Im(\lambda_{q})\geq 0$ and $[q]=[\lambda_{q}]$. Let $U$ and $T=U^{*}AU$ be
two matrices satisfying Lemma 1.5. Then the algebraic multiplicity  of $q$ (or
$a.m.(q,A)$) is the number of occurrences of $\lambda_{q}$ in the diagonal of
$T$.

######  Proposition 2.1.

The Definition 2.1 is well-given and it does not depend on the choice of $T$.
Moreover, if $q$ is a right eigenvalue of a quaternionic matrix $A$ such that
$[q]=[\lambda_{q}]$ where $\lambda_{q}\in\mathbb{C}$ and $Im(\lambda_{q})\geq
0$, then

$a.m.(q,A)=a.m.(\lambda_{q},f(A))=a.m.(\lambda_{q},\pi(A))$ (20)

if $Im(\lambda_{q})\neq 0$, and

$a.m.(q,A)=\frac{1}{2}a.m.(\lambda_{q},f(A))=\frac{1}{2}a.m.(\lambda_{q},\pi(A))$
(21)

if $\lambda_{q}$ is a real number.

Proof: Observe that $T=U^{*}AU$ implies that $\pi(T)$ and $\pi(A)$ are
similar, indeed $\pi(T)\sim f(T)=f(U)^{*}f(A)f(U)\sim f(A)\sim\pi(A)$. So, if
$[q]$ is a right eigenvalue of $A$, then $\lambda_{q}$ is an eigenvalue for
$\pi(A)$ and also for $\pi(T)$. This means that $\lambda_{q}$ occurs in the
diagonal of $\pi(T)$. Indeed $\pi(T)$ is a triangular matrix since the $i$-th
diagonal block of $\pi(T)$ is $\begin{bmatrix}\mu_{i}&0\\\
0&\overline{\mu_{i}}\end{bmatrix}$ where $\mu_{i}$ are the diagonal entries of
$T$. So $\lambda_{q}$ occurs also in the diagonal of $T$ and, in particular,

$a.m.(q,A)=a.m.(\lambda_{q},\pi(T))=a.m.(\lambda_{q},\pi(A))=a.m.(\lambda_{q},f(A))$
(22)

if $Im(\lambda)\neq 0$, and

$a.m.(q,A)=\frac{1}{2}a.m.(\lambda_{q},\pi(T))=\frac{1}{2}a.m.(\lambda,\pi(A))=\frac{1}{2}a.m.(\lambda,f(A))$
(23)

otherwise.
Finally, if there are some matrices $U_{1}$ and $T_{1}\neq T$ satisfying the
Lemma 1.5, then $T_{1}=U_{1}^{*}UTU^{*}U_{1}$. So $\pi(T_{1})$ and $\pi(T)$
are similar because $f(T_{1})$ and $f(T)$ are similar.
This proves that the definition of algebraic multiplicity is well-given.

###  2.2 Geometric multiplicity

Our next step is to define the geometric multiplicity of a right eigenvalue
$q$ of a quaternionic matrix.

######  Definition 2.2.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$ and let $q$ be a right
eigenvalue of $A$. Then the $[q]$-eigenspace is the subset of $\mathbb{H}^{n}$

$V_{[q]}:=\sum\limits_{\tilde{q}\in[q]}V_{\tilde{q}}$ (24)

which is the space of all finite sums of vectors $v_{1}+...+v_{k}$ where for
each $i$ the vector $v_{i}$ is in $V_{q_{i}}$ for some $q_{i}$.

######  Remark 3.

Notice that $V_{[q]}$ is a right $\mathbb{H}$-vector space. Moreover, for each
$\tilde{q}$ in $[q]$ is possible to find a basis of $V_{[q]}$ contained in
$V_{\tilde{q}}$. Indeed if $v$ is in $V_{\tilde{q}}$ where
$s\tilde{q}s^{-1}=q$ for some $s$ in $\mathbb{H}$, then $vs^{-1}$ is an
element of $V_{q}$. Then, since $V_{q}$ is a real vector space, then we can
find a basis111in the real vector spaces sense $\\{e_{i}\\}$ of $V_{q}$ and
some real coefficients $\\{a^{i}\\}$ such that
$vs^{-1}=e_{1}a^{1}+...+e_{k}a^{k}$ and so

$v=e_{1}(a^{1}s)+...+e_{k}(a^{k}s)$ (25)

This means that $\\{e_{i}\\}$ is a generating set for $V_{[q]}$ and so it
contains a basis for the right $\mathbb{H}$-vector space $V_{[q]}$.

######  Definition 2.3.

Let $q$ be a right eigenvalue of $A$ in $\mathbb{H}^{n\times n}$. The
geometric multiplicity of $q$ is

$g.m.(q,A):=dim_{\mathbb{H}}V_{[q]}.$ (26)

######  Remark 4.

Let $A$ and $D$ be two quaternionic matrix. Assume the existence of an
invertible matrix $U$ such that $D=U^{-1}AU$. Then the eigenvalues of $A$ and
of $D$ are the same and they also have the same geometric multiplicity.
Indeed, given an eigenvalue $q$ of $A$ and given $\\{x_{i}\\}$ a basis of the
$[q]$-eigenspace, then $\\{Ux_{i}\\}$ is a basis for the $[q]$-eigenspace of
$A$.

######  Proposition 2.2.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$ and let $q$ be a right
eigenvalue. Let $\lambda_{q}$ be the element of $\mathbb{C}$ such that
$Im(\lambda_{q})\geq 0$ and $[q]=[\lambda_{q}]$. Then, if $Im(\lambda_{q})\neq
0$

$g.m.(q,A)=g.m.(\lambda_{q},f(A))=g.m.(\lambda_{q},\pi(A))$ (27)

and, if $\lambda_{q}$ is real,

$g.m.(q,A)=\frac{1}{2}g.m.(\lambda_{q},f(A))=\frac{1}{2}g.m.(\lambda_{q},\pi(A)).$
(28)

Proof: Fix a right eigenvalue $q$ of $A$. Then choose a basis
$\\{v_{1},...,v_{d}\\}$ of $V_{[q]}$ which is contained in $V_{\lambda_{q}}$
and complete it in order to obtain a basis
$X=\\{v_{1},...,v_{d},x_{d+1},...,x_{n}\\}=:\\{s_{i}\\}$ of $\mathbb{H}^{n}$.
Let $\mathcal{C}_{X}$ be the matrix which has in the $i$-th column the vector
$s_{i}$ of $X$. Let $b_{i}$ be the vector in $\mathbb{H}^{n}$ which has the
$i$-th entry equal to $1$ and $0$ otherwise. Observe that $\mathcal{C}_{X}$ is
invertible and its inverse is given by the matrix which has in the $j$-th
column the components of $b_{i}$ with respect to $X$. Then, we define

$D_{\lambda}:=\mathcal{C}_{X}^{-1}\cdot A\cdot\mathcal{C}_{X}.$ (29)

The matrix $D_{\lambda}$ has the form

$D_{\lambda}=\begin{bmatrix}\lambda_{q}Id_{d}&B_{1}\\\ 0&B_{2}\end{bmatrix}$
(30)

The matrix $f(D_{\lambda})=f(\mathcal{C}_{X}^{-1})f(A)f(\mathcal{C}_{X})$ is
given by

$f(D_{\lambda}):=\begin{bmatrix}\lambda_{q}Id_{d}&\star&0&\star\\\
0&\star&0&\star\\\ 0&\star&\overline{\lambda}_{q}Id_{d}&\star\\\
0&\star&0&\star\end{bmatrix}.$ (31)

Since $f(A)$ is similar to $f(D_{\lambda})$, we obtain that

$g.m.(\lambda_{q},f(D_{\lambda}))=g.m.(\lambda_{q},f(A))\geq g.m.(q,A)$ (32)

if $Im(\lambda_{q})>0$ and

$g.m.(\lambda_{q},f(D_{\lambda}))=g.m.(\lambda_{q},f(A))\geq 2g.m.(q,A)$ (33)

if $\lambda_{q}$ is real. Indeed if $\lambda_{q}$ has $Im(\lambda_{q})>0$ then
the $\lambda_{q}$-eigenspace of $f(D_{\lambda})$ contains
$Span(e_{1},...,e_{d})$ where $e_{i}$ is the vector which has $1$ in the
$i$-the entry and zero otherwise. On the other hand, if $\lambda_{q}$ is real,
then the $\lambda_{q}$-eigenspace of $f(D_{\lambda})$ contains
$Span(e_{1},...,e_{d},e_{n+1},...,e_{n+d})$.
For the other inequality, assume that there is a vector $w=(w_{A},w_{B})$ in
$\mathbb{C}^{2n}$ such that $f(D_{\lambda})w=w\lambda_{q}$. By Lemma 2.4 of
Spec , we know that $y:=w_{A}-\overline{w}_{B}\cdot j$ in $\mathbb{H}^{n}$ is
a vector in $V_{\lambda_{q}}$. This means that, with respect to the basis $X$,

$\begin{split}y&=v_{1}a^{1}+...+v_{d}a^{d}+x_{d+1}\cdot 0+...+x_{n}\cdot 0\\\
&=v_{1}(w_{A}^{1}-\overline{w}_{B}^{1}j)+...+v_{d}(w_{A}^{d}-\overline{w}_{B}^{d}j)\\\
&+x_{d+1}\cdot(w_{A}^{d+1}-\overline{w}_{B}^{d+1}j)+...+x_{n}\cdot(w_{A}^{n}-\overline{w}_{B}^{n}j).\end{split}$
(34)

So, for each $r=d+1,...,n$, we obtain that $w_{A}^{r}=w_{B}^{r}=0$. This means
that $w$ is in $Span(e_{1},...,e_{d},e_{n+1},...,e_{n+d})$. This is sufficient
to conclude if $\lambda_{q}$ is real. If $\lambda_{q}$ is not real, observe
that $Span(e_{n+1},...,e_{n+d})$ is contained in the
$\overline{\lambda}$-eigenspace of $D_{\lambda}$ and so, in particular, $w$ is
in $Span(e_{1},...,e_{d})$.

###  2.3 Main properties of multiplicities

The first property of algebraic and geometric multiplicities is the classical
inequality between them.

######  Proposition 2.3.

Let $A\in\mathbb{H}^{n\times n}$. Then, if $q$ is a right eigenvalue of $A$,
then

$a.m.(q)\geq g.m.(q).$ (35)

The second one is about the invariance of multiplicities

######  Proposition 2.4.

Let $A$ and $B$ two matrices in $\mathbb{H}^{n\times n}$. Assume that they are
similar. Then their right spectrum coincides and if $q$ is an eigenvalue of
$A$ and $B$, then

$a.m.(q,A)=a.m.(q,B)\mbox{ and }g.m.(q,A)=g.m.(q,B).$ (36)

The following Corollary holds when $A$ is a normal matrix. In order to
introduce it, we need a result from Spec .

######  Theorem 2.5 (Theorem 3.3 of Spec ).

If $A\in\mathbb{H}^{n\times n}$ is normal, i.e. $AA^{*}=A^{*}A$ then there are
matrices $D$, $U$ in $\mathbb{H}^{n\times n}$ such that

  * •

$U$ is a unitary matrix, $D$ is a diagonal matrix, and $U^{*}AU=D$;

  * •

each diagonal entry of $D$ is a complex number $\lambda$ such that
$Im(\lambda)=0$;

  * •

$q\in\mathbb{H}^{n\times n}$ is a right eigenvalue of $A$ if and only if
$[q]=[\lambda]$ for some diagonal element $\lambda$ of $D$.

######  Proposition 2.6.

Let $A\in\mathbb{H}^{n\times n}$ be a normal matrix. Then, if $q$ is a right
eigenvalue of $A$, then

$a.m.(q)=g.m.(q).$ (37)

##  3 Equivalence of the definitions

As stated in the Introduction, the definitions given Section 2 are equivalent
to the definitions of Quat . In this final section we quickly introduce the
classical definitions and we show this equivalence.
In order to define the classical algebraic and the geometric multiplicity of a
right eigenvalue we need the notions of minimal polynomial $p^{A}$ related to
$A$ and of the root subspace $\mathcal{M}$ of $A$ corresponding to $p^{A}$.
Notice that $\mathbb{H}^{n\times n}$ is a finite dimensional, real vector
space. Then there are some polynomials $p(t)\in\mathbb{R}[t]$ such that
$p(A)=0$.

######  Definition 3.1.

The minimal polynomial of $A$ is the polynomial $p^{A}(t)\neq 0$ which is
monic and it has minimal degree.

######  Remark 5.

The minimal polynomial $p^{A}$ of a matrix $A\in\mathbb{H}^{n\times n}$ is
unique.

Assume that

$p^{A}(t)=p_{1}(t)^{m_{1}}\cdot...\cdot p_{k}(t)^{m_{k}}$ (38)

where $p_{j}(t)$ are distinct irreducible real polynomial which are monic.
Notice that for each eigenvalue $\lambda$ of $A$ there is exactly one
polynomial $p_{j}$ such that $\lambda$ is a root of $p_{j}$. Then the
following definitions make sense.

######  Definition 3.2.

A root subspace $\mathcal{M}_{j}$ of $A$ is the set

$\mathcal{M}_{j}:=\\{u\in\mathbb{H}^{n}|p_{j}(A)^{m_{j}}u=0\\}$ (39)

where $j=1,...,k$. The root subspace of the eingenvalue $\lambda$ is the root
subspace related to the polynomial $p_{j}$ which has $\lambda$ as a root.

######  Remark 6.

Given a right eigenvalue $q$ of a quaternion matrix $A$, then
$V_{[q]}\neq\mathcal{M}_{j}$, where $\mathcal{M}_{q}$ is the root subspace
related to $q$. An example is the following: notice that $0$ is an eigenvalue
of the matrix

$A=\begin{bmatrix}0&0&0&0\\\ 0&0&0&0\\\ 1&0&0&0\\\ 0&1&0&0\\\ \end{bmatrix}.$
(40)

Then the minimal polynomial of $A$ is $p^{A}(t)=t^{2}$ and so
$\mathcal{M}_{0}=ker(A^{2})=\mathbb{H}^{4}$. So
$dim_{\mathbb{H}}(\mathcal{M}_{0})=4$. On the other hand, we know that the
quaternionic dimension of $V_{[0]}$ is equal to $\frac{1}{2}g.m.(0,f(A))=2$.

######  Definition 3.3.

Let $\lambda$ be an eigenvalue of $A$. Let $p_{j}(t)$ be the unique polynomial
in Equality 38 such that $\lambda$ is a root of $p_{j}(t)$. Then the
quaternion geometric multiplicity of $\lambda$ (Definition in pag. 99 of Quat
) is the number the maximal number of $\mathbb{H}$-linearly independent
eigenvectors of $A$ in $\mathcal{M}_{j}$. We denote this number by
$q.g.m.(\lambda)$.
On the other hand, the quaternion algebraic multiplicity of $\lambda$
(Definition in pag. 99 of Quat ) is defined as the quaternion dimension of
$\mathcal{M}_{\lambda}$. We will denote this number by $q.a.m.(\lambda)$.

######  Remark 7.

In general $q.a.m.(\lambda)\geq q.g.m.(\lambda)$, indeed on
$\mathcal{M}_{\lambda}$ there could be some vectors $v$ which are not
eigenvectors of $A$.

The equivalence of the definitions is a Corollary of the existence and the
unicity 222unless to some permutation of blocks of a Jordan form for each
quaternion matrix $A$. In particular, on the diagonal of the Jordan form of a
quaternion matrix there are complex numbers $\lambda_{i}$ such that
$Im(\lambda_{i})\geq 0$. The proof of these facts is quite long and it is
proved in Section 5.6. of Quat  (Theorem 5.5.3., from p.102 to p.109).
In particular we need Theorem 5.5.5. of Quat .

######  Theorem 3.1 (Theorem 5.5.5., Quat ).

Let $A\in\mathbb{H}^{n\times n}$ and let $\lambda\in\sigma(A)$. Then:

  * •

the number of Jordan blocks in the Jordan form of $A$ corresponding to the
eigenvalues similar to $\lambda$ is equal to $q.g.m.(\lambda)$;

  * •

the sum of the sizes of Jordan blocks in the Jordan form of $A$ corresponding
to the eigenvalues similar to $\lambda$ is equal to $q.a.m.(\lambda)$.

######  Corollary 3.2.

Let $A$ be a matrix in $\mathbb{H}^{n\times n}$. Then, for each eigenvalue
$\lambda$ of $A$,

$a.m.(q,A)=q.a.m.(q,A)\mbox{ and }g.m.(q,A)=q.g.m.(q,A).$ (41)

Proof: Let $J$ be the Jordan form of $A$. Notice that $J$ and $A$ are similar.
Let $q$ be a right eigenvalue of $A$ and let $\lambda_{q}$ be the complex
number with $Im(\lambda_{q})\geq 0$ and $\lambda_{q}\sim q$. Then, if
$Im(\lambda_{q})\neq 0$

$q.g.m.(q,A)=q.g.m.(\lambda_{q},J)=g.m.(\lambda_{q},f(J))$ (42)

and, if $\lambda_{q}$ is real,

$q.g.m.(q,A)=q.g.m.(\lambda_{q},J)=\frac{1}{2}g.m.(\lambda_{q},f(J)).$ (43)

Moreover

$q.a.m.(q,A)=q.a.m.(\lambda_{q},J)=a.m.(\lambda_{q},f(J))$ (44)

and, if $\lambda_{q}$ is real,

$q.a.m.(q,A)=q.a.m.(\lambda_{q},J)=\frac{1}{2}a.m.(\lambda_{q},f(J)).$ (45)

Since $A$ and $J$ are similar, then we conclude by appliying Propositions 2.4,
2.2 and 2.1.

## References

  * (1) A. Donno, M. Cavaleri, S. Spessato, _Godsil-Mckay switching for gain graphs_ , 2022, preprint on arxiv: https://arxiv.org/pdf/2207.10986.pdf
  * (2) D. R. Farenick, B. A.F. Pidkowich, _The spectral theorem in quaternions_ , Linear Algebra Appl., 371 (2003) 75–102, MR 1997364
  * (3) L. Rodman, _Topics in Quaternion Linear Algebra_. Princeton University Press, 2014. JSTOR, http://www.jstor.org/stable/j.ctt6wpz0p.