# ordered vector space

Let $k$ be an ordered field. An ordered vector space over $k$ is a vector space $V$ that is also a poset at the same time, such that the following conditions are satisfied

1. 1.

for any $u,v,w\in V$, if $u\leq v$ then $u+w\leq v+w$,

2. 2.

if $0\leq u\in V$ and any $0<\lambda\in k$, then $0\leq\lambda u$.

Here is a property that can be immediately verified: $u\leq v$ iff $\lambda u\leq\lambda v$ for any $0<\lambda$.

Also, note that $0$ is interpreted as the zero vector of $V$, not the bottom element of the poset $V$. In fact, $V$ is both topless and bottomless: for if $\bot$ is the bottom of $V$, then $\bot\leq 0$, or $2\bot\leq\bot$, which implies $2\bot=\bot$ or $\bot=0$. This means that $0\leq v$ for all $v\in V$. But if $v\neq 0$, then $0 or $-v<0$, a contradiction. $V$ is topless follows from the implication that if $\bot$ exists, then $\top=-\bot$ is the top.

For example, any finite dimensional vector space over $\mathbb{R}$, and more generally, any (vector) space of real-valued functions on a given set $S$, is an ordered vector space. The natural ordering is defined by $f\leq g$ iff $f(x)\leq g(x)$ for every $x\in S$.

Properties. Let $V$ be an ordered vector space and $u,v\in V$. Suppose $u\vee v$ exists. Then

1. 1.

$(u+w)\vee(v+w)$ exists and $(u+w)\vee(v+w)=(u\vee v)+w$ for any vector $w$.

###### Proof.

Let $s=(u\vee v)+w$. Then $u+w\leq s$ and $v+w\leq s$. For any upper bound $t$ of $u+w$ and $v+w$, we have $u\leq t-w$ and $v\leq t-w$. So $u\vee v\leq t-w$, or $(u\vee v)+w\leq t$. So $s$ is the least upper bound of $u+w$ and $v+w$. ∎

2. 2.

$u\wedge v$ exists and $u\wedge v=(u+v)-(u\vee v)$.

###### Proof.

Let $s=(u+v)-(u\vee v)$. Since $u\leq u\vee v$, $-(u\vee v)\leq-u$, so $s\leq v$. Similarly $s\leq u$, so $s$ is a lower bound of $u$ and $v$. If $t\leq u$ and $t\leq v$, then $-u\leq-t$ and $-v\leq-t$, or $v\leq(u+v)-t$ and $u\leq(u+v)-t$, or $u\vee v\leq(u+v)-t$, or $t\leq(u+v)-(u\vee v)=s$. Hence $s$ the greatest lower bound of $u$ and $v$. ∎

3. 3.

$\lambda u\vee\lambda v$ exists for any scalar $\lambda\in k$, and

1. (a)

if $\lambda\geq 0$, then $\lambda u\vee\lambda v=\lambda(u\vee v)$

2. (b)

if $\lambda\leq 0$, then $\lambda u\vee\lambda v=\lambda(u\wedge v)$

3. (c)

if $u\neq v$, then the converse holds for (a) and (b).

###### Proof.

Assume $\lambda\neq 0$ (clear otherwise). (a). If $\lambda>0$, $u\leq u\vee v$ implies $\lambda u\leq\lambda(u\vee v)$. Similarly, $\lambda v\leq\lambda(u\vee v)$. If $\lambda u\leq t$ and $\lambda v\leq t$, then $u\leq\lambda^{-1}t$ and $v\leq\lambda^{-1}t$, hence $u\vee v\leq\lambda^{-1}t$, or $\lambda(u\vee v)\leq t$. Proof of (b) is similar to (a). (c). Suppose $\lambda u\vee\lambda v=\lambda(u\vee v)$ and $\lambda<0$. Set $\gamma=-\lambda$. Then $\lambda u\vee\lambda v=\lambda(u\vee v)=-\gamma(u\vee v)=-(\gamma(u\vee v))=-(% \gamma u\vee\gamma v)=-((-\lambda u)\vee(-\lambda v))=-(-(\lambda v\wedge% \lambda u))=\lambda v\wedge\lambda u$. This implies $\lambda u=\lambda v$, or $u=v$, a contradiction. ∎

Remarks.

• Since an ordered vector space is just an abelian po-group under $+$, the first two properties above can be easily generalized to a po-group. For this generalization, see this entry (http://planetmath.org/DistributivityInPoGroups).

• A vector space $V$ over $\mathbb{C}$ is said to be ordered if $W$ is an ordered vector space over $\mathbb{R}$, where $V=W\oplus iW$ ($V$ is the complexification of $W$).

• For any ordered vector space $V$, the set $V^{+}:=\{v\in V\mid 0\leq v\}$ is called the positive cone of $V$. $V^{+}$ is clearly a convex set. Also, since for any $\lambda>0$, $\lambda V^{+}\subseteq V^{+}$, so $V^{+}$ is a convex cone. In addition, since $V^{+}-\{0\}$ remains a cone, and $V^{+}\cap(-V^{+})=\{0\}$, $V^{+}$ is a proper cone.

• Given any vector space, a proper cone $P\subseteq V$ defiens a partial ordering on $V$, given by $u\leq v$ if $v-u\in P$. It is not hard to see that the partial ordering so defined makes $V$ into an ordered vector space.

• So, there is a one-to-one correspondence between proper cones of $V$ and partial orderings on $V$ making $V$ an ordered vector space.

Title ordered vector space OrderedVectorSpace 2013-03-22 16:37:24 2013-03-22 16:37:24 CWoo (3771) CWoo (3771) 20 CWoo (3771) Definition msc 46A40 msc 06F20 ordered linear space TopologicalLattice positive cone