# The Unapologetic Mathematician

## Problems with Ordered Linear Spaces — Solved!

Okay, there’s some sort of problem with these things. I defined the category $\mathcal{O}rd\mathcal{L}in$, 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 $\hat{V}=\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$, which seems sort of like a categorical dual, but maybe not.

What I tried to prove is that in general $\hat{V}\otimes W$ is not isomorphic to $W^V$. But as Todd pointed out, if $\hat{V}$ is a dual then we have an adjunction
$V\otimes\underline{\hphantom{X}}\dashv\hat{V}\otimes\underline{\hphantom{X}}$
but the definition of an exponential is an adjunction
$V\otimes\underline{\hphantom{X}}\dashv(\underline{\hphantom{X}})^V$
and functors with the same left adjoint are naturally isomorphic.

So, if $\hat{V}$ is a categorical dual, then $W^V$ is naturally isomorphic to $\hat{V}\otimes W$, and there’s something wrong with the proof I offered that it isn’t.

On the other hand, if $\hat{V}\otimes W$ is not naturally isomorphic to $W^V$, then $\hat{V}$ 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 $V$, 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 $g$ taking the vector space $V$ to the free vector space $\mathbb{F}^n$, and taking our chosen basis of $V$ to the canonical basis of $\mathbb{F}^n$. We induce a cone on $\mathbb{F}^n$ by saying $(x_1,...,x_n)\geq0$ if and only if $g^{-1}(x_1,...,x_n)\geq_V0$. So without loss of generality our positive basis consists of the “coordinate vectors” $(0,...,0,1,0,...,0)$, and the dual basis consists of the coordinate projections.

Now these vectors cut out a classical subcone of $V$, and the positive vectors are exactly those in the positive orthant. If $V$ is not exactly this classical cone, then there exists some positive vector $v$ in $V$ outside the positive orthant. That is, some component of $v$ will be negative. But then that coordinate projector will send $v$ 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 $V$ is classical.

So what does this mean? Just that $\hat{V}=\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$ is only the categorical dual to $V$ when $V$ is classical. Thus only in this case do we have an adjunction $V\otimes\underline{\hphantom{X}}\dashv\hat{V}\otimes\underline{\hphantom{X}}$, and so only in this case do we find $\hat{V}\otimes W\cong W^V$ naturally in $W$.

What happens for non-classical cones? Well, $(\underline{\hphantom{X}})^V$ is a right adjoint, and so its evaluation transformation $\epsilon_W:V\otimes W^V\rightarrow W$ is couniversal. Since we also have a natural transformation $V\otimes\hat{V}\otimes W\rightarrow W$ (evaluate the second factor on the first) it must factor through the couniversal. That is, there is an arrow $\hat{V}\otimes W\rightarrow W^V$, which is the one we earlier calculated explicitly. But it’s not generally an isomorphism.