The Unapologetic Mathematician

Mathematics for the interested outsider

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”?

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September 24, 2007 - Posted by | Category theory

3 Comments »

  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…

    Read the updates on this post and the previous one.

    Comment by John Armstrong | September 24, 2007 | Reply

  3. […] Well, as today’s main post sort of died, I really should post some good news. Jeff Morton has started a weblog! It’s a […]

    Pingback by New Blath! « The Unapologetic Mathematician | September 25, 2007 | Reply


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