# The Unapologetic Mathematician

## Ordered Linear Spaces III

[UPDATE]: This whole post is badly-founded as it stands, because on further reflection it seems that $\mathcal{O}rd\mathcal{L}in$ 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 $A^*\otimes B\rightarrow B^A$. Indeed, we have an arrow
$A^*\otimes B\otimes A\cong A^*\otimes A\otimes B\rightarrow\mathbf{1}\otimes B\cong B$
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 $A^*\otimes B\otimes A$ to $B$ corresponds to an arrow from $A^*\otimes B$ to $B^A$.

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 $B^A$. But it turns out that it doesn’t hold in $\mathcal{O}rd\mathcal{L}in$, 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 $A^*\otimes B$. This is a finite sum $f=\sum_{i=1}^n\alpha^i\otimes b_i$, where the $b_i$ are positive vectors in $B$ and the $\alpha^i$ are positive linear functionals on $A$. 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 $\alpha_i(a)\geq_\mathbb{F}0$ for all $a\geq_A0$. Then if we feed $f$ a positive element of $A$, we evaluate to find $f(a)=\sum_{i=1}^n\alpha^i(a)b_i$, which is a linear combination of positive vectors in $B$ with positive coefficients, and thus is positive. So $f$ corresponds to a positive map as it should. This is just the arrow described above, now in the special case of $\mathcal{O}rd\mathcal{L}in$.

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

I think that we can push on from here to show that this implies that either $A$ or $B$ 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 $\hat{V}=\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$ as before. That is, a linear functional is considered positive if it sends the whole positive cone from $V$ to the positive ray in $\mathbb{F}$. Now the question is, in what sense is this a “dual”?

September 24, 2007 - Posted by | Category theory

1. I’m a little confused by something here. You say that the monoidal category OrdLin admits duals, meaning that for each object V there’s an object V* together with an adjunction

(V \otimes –) –| (V* \otimes –).

On the other hand, you’ve identified an exponential whereby

(V \otimes –) –| (–)^V.

We seem to get a natural isomorphism V* \otimes — ~ (–)^V since both functors have the same left adjoint. So I’m not understanding why V* \otimes W ~ W^V is an issue.

Comment by Todd Trimble | September 24, 2007 | Reply

2. Damn…