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

2. if $0\le u\in V$ and any , then $0\le \lambda u$.

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

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 $\perp$ is the bottom of $V$, then $\perp \le 0$, or $2\perp \le \perp$, which implies $2\perp =\perp$ or $\perp =0$. This means that $0\le v$ for all $v\in V$. But if $v\ne 0$, then or , a contradiction. $V$ is topless follows from the implication that if $\perp$ exists, then $\top =-\perp$ is the top.

For example, any finite dimensional vector space over $ℝ$, 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 $f\le g$ iff $f(x)\le g(x)$ 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. $(u+w)\vee (v+w)$ exists and $(u+w)\vee (v+w)=(u\vee v)+w$ for any vector $w$.

   Proof.

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

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

   Proof.

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

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

   1. if $\lambda \ge 0$, then $\lambda u\vee \lambda v=\lambda (u\vee v)$

   2. if $\lambda \le 0$, then $\lambda u\vee \lambda v=\lambda (u\wedge v)$

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

   Proof.

   Assume $\lambda \ne 0$ (clear otherwise). (a). If $\lambda >0$, $u\le u\vee v$ implies $\lambda u\le \lambda (u\vee v)$. Similarly, $\lambda v\le \lambda (u\vee v)$. If $\lambda u\le t$ and $\lambda v\le t$, then $u\le {\lambda }^{-1}t$ and $v\le {\lambda }^{-1}t$, hence $u\vee v\le {\lambda }^{-1}t$, or $\lambda (u\vee v)\le t$. Proof of (b) is similar to (a). (c). Suppose $\lambda u\vee \lambda v=\lambda (u\vee v)$ and . 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 $ℂ$ is said to be ordered if $W$ is an ordered vector space over $ℝ$, where $V=W\oplus iW$ ($V$ is the complexification of $W$).

• For any ordered vector space $V$, the set $V^{+}:=\{v\in V\mid 0\le v\}$ 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 $u\le v$ 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
Canonical name	OrderedVectorSpace
Date of creation	2013-03-22 16:37:24
Last modified on	2013-03-22 16:37:24
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	20
Author	CWoo (3771)
Entry type	Definition
Classification	msc 46A40
Classification	msc 06F20
Synonym	ordered linear space
Related topic	TopologicalLattice
Defines	positive cone