## Problems with Ordered Linear Spaces — Solved!

Okay, there’s some sort of problem with these things. I defined the category , showed some properties, and tried to prove a theorem. Here I want to collect what I’m sure about and what I’m not sure about.

The biproduct, exponential, and monoidal product structures are all correct. We have *something* natural defined by , which seems sort of like a categorical dual, but maybe not.

What I tried to prove is that in general is not isomorphic to . But as Todd pointed out, if is a dual then we have an adjunction

but the definition of an exponential is an adjunction

and functors with the same left adjoint are naturally isomorphic.

So, if is a categorical dual, then is naturally isomorphic to , and there’s something wrong with the proof I offered that it isn’t.

On the other hand, if is *not* naturally isomorphic to , then is not a categorical dual. I haven’t found an example of a cone that has no positive basis with a positive dual basis, and I’m losing faith in the example I thought I’d sketched that even had one positive basis whose dual basis was not positive.

So one of these “proofs” is wrong — which is it?

*[UPDATE]:* Well, I figured it out. There’s a problem with the duals.

Start with any cone , and pick a positive basis in the cone. Now we can simplify our lives by moving to a cone with a particularly nice positive basis. We know that there’s a linear isomorphism taking the vector space to the free vector space , and taking our chosen basis of to the canonical basis of . We induce a cone on by saying if and only if . So without loss of generality our positive basis consists of the “coordinate vectors” , and the dual basis consists of the coordinate projections.

Now these vectors cut out a classical subcone of , and the positive vectors are exactly those in the positive orthant. If is not exactly this classical cone, then there exists some positive vector in outside the positive orthant. That is, some component of will be negative. But then that coordinate projector will send to a negative number, and so cannot be a positive linear functional. Thus the dual basis of a positive basis can only itself be positive if is classical.

So what does this mean? Just that is only the categorical dual to when is classical. Thus only in this case do we have an adjunction , and so only in this case do we find naturally in .

What happens for non-classical cones? Well, *is* a right adjoint, and so its evaluation transformation is couniversal. Since we also have a natural transformation (evaluate the second factor on the first) it must factor through the couniversal. That is, there is an arrow , which is the one we earlier calculated explicitly. But it’s not generally an isomorphism.

This is good! I’d be interested to see what you get going in the infinite-dimensional case; it looks like ordered linear spaces of arbitrary dimension form a complete cocomplete symmetric monoidal closed category, and that you’ve just shown that the finitely generated projectives in this category are free (by finitely generated projective A, I mean A \otimes (-) preserves arbitrary limits, and hence this time has a left adjoint A* \otimes (-) by an adjoint functor theorem; obviously we have here an ambidextrous adjunction by symmetry of the tensor).

I imagine of course you’d also be introducing suitable topologies in that setting…

Comment by Todd Trimble | September 26, 2007 |