The Unapologetic Mathematician

Mathematics for the interested outsider

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.

September 25, 2007 - Posted by | Category theory

1 Comment »

  1. 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 | Reply


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