total order on finite dimensional vector space over $\mathbb{R}$

Question

We know that, If we start with a basis of a finite dimensional vector space $V$ over $\mathbb{R}$ by using lexicographic ordering with respect to that basis, we have a total ordering $\lt$ on $V$ satisfying,

i) $v \in V, v_1 \lt v_2 \implies v_1 + v \lt v_2 + v$

ii) $ \alpha \in \mathbb{R}, v_1 \lt v_2 \implies $

a) $\alpha v_1 \lt \alpha v_2$ if $\alpha \gt 0$

b) $\alpha v_1 \gt \alpha v_2$ if $\alpha \lt 0 $

I just wondering, given a such a total ordering on $V$ , can we find a basis of $V$ with respect to which the lexicographic order will be the same as the one we start with?

is there any similar theory of infinite dimensional vector spaces?

Thanks in Advance

Answer 1

Let $<$ be such a total ordering on $V$ (which is, as in the question, finite dimensional). Then the set $C=\{v \in V: v\geq 0\}$ clearly is a convex cone. Since $C \cap (-C) =\{0\}$ and since $V=(-C) \cup C$ there is a $v_0 \in V$ which does not lie in the closure of $C$. Thus there is a nonzero linear form $l$ on $V$ such that $l \geq 0$ on $C$ (see below in "edit"). Now the zero set of $l$ is a hyperplane as Harald thought of:

Let $l(v)<0$ then $v \not\in C$ thus $v<0$.

Let $l(v)>0$ and assume $v \leq 0$ then $-v \in C$ which contradicts $l(-v)<0$.

edit: This follows from biduality of convex cones, see http://arxiv.org/abs/1006.4894 (section 2.1): Assume that $C^*=\{0\}$ (notation as in the paper), then we have $V=\{0\}^*=(C^*)^*=\textrm{cl}(C)$, but we have $V \neq \textrm{cl}(C)$ ($\textrm{cl}(C)$ is the closure of $C$).

I expect that you can find the proof of this biduality theorem (which uses the separation theorem that I mentioned in the first place) in a standard book about convexity, for example in Barvinok's "A First Course in Convexity".

Answer 2

In case of an infinite dimensional vector space there exists a basis that represents the order.

Nevertheless the separating “Hyperplane” may be the whole vector space itself. For example take $\mathbb{R}^{\mathbb N}$ with the natural orders on $\mathbb R$ and $\mathbb N$. In this vector space every vector is infinitesimal with respect to some other vector. If there were a hyperplane that separates the positive cone and the negative cone, then there would have been a vector that is not infinetesimal with respect to any other vectors.

This topic is discussed in this article.