Ordered Linear Spaces III

[UPDATE]: This whole post is badly-founded as it stands, because on further reflection it seems that does not have duals. See the post in the first link and its update for an explanation.

We’re on a roll with our discussion of ordered linear spaces. So I want to continue past just describing these things and prove a very interesting theorem that Howard and I hit upon. It may have been known before, but what I’ll say here is original work to the two of us.

Whenever we have a (symmetric) monoidal closed category with duals, we have an arrow . Indeed, we have an arrow

where we use symmetry on the left, the counit of the dual in the middle, and the left unit isomorphism on the right. By the closure adjunction, this arrow from to corresponds to an arrow from to .

Sometimes this arrow is an isomorphism. In the category of finite-dimensional vector spaces over a field, for instance, this property holds. It’s nice when this happens because then we can get a good understanding of every morphism in . But it turns out that it doesn’t hold in , even though ordered linear spaces are built on finite-dimensional vector spaces. And the exact way it fails is very interesting.

Let’s take a positive element of . This is a finite sum , where the are positive vectors in and the are positive linear functionals on . I’m writing the index as a superscript for a technical reason some of you might know, but it’s not important if you don’t. Anyhow, recall that a linear functional is positive if for all . Then if we feed a positive element of , we evaluate to find , which is a linear combination of positive vectors in with positive coefficients, and thus is positive. So corresponds to a positive map as it should. This is just the arrow described above, now in the special case of .

But let’s see what this means for the image of the function described by . The vectors are all inside the positive cone in , and so they span a classical subcone of . Then the image of must be inside this subcone. This imposes an extremely strong condition on and , which turns out to also be sufficient: if and only if the positive image of every positive function from to is contained in a classical subcone!

I think that we can push on from here to show that this implies that either or is classical, but I don’t know of a proof of this conjecture yet.

Okay, so we don’t have duals. So this day’s not a total loss I’ll throw this out to the group:

The construction we gave obviously exists and is natural in some sense. We define a “dual” cone as before. That is, a linear functional is considered positive if it sends the whole positive cone from to the positive ray in . Now the question is, in what sense is this a “dual”?

Advertisements