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.