# 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 | 3 Comments

## Ordered Linear Spaces II

Since I was a little slow posting things at the end of last week due to the conference, I’ll continue my discussion of ordered linear spaces with this observation: because each ordered linear space is a vector space with extra structure, the category $\mathcal{O}rd\mathcal{L}in$ inherits a lot from the category of vector spaces.

For one thing, given a pair of vector spaces ${V}$ and $W$, we can take their direct sum $V\oplus W$. Now if each of them has an identified cone of positive vectors, we can set up a cone on the direct sum by insisting that the structural maps $\pi_V$, $\pi_W$, $\iota_V$, and $\iota_W$ are positive. Let’s write these as logical statements and see what they imply:

• If $(v,w)\geq_{V\oplus W}0$ then $\pi_V(v,w)=v\geq_V0$.
• If $(v,w)\geq_{V\oplus W}0$ then $\pi_W(v,w)=w\geq_W0$.
• If $v\geq_V0$ then $\iota_V(v)=(v,0)\geq_{V\oplus W}0$.
• If $w\geq_W0$ then $\iota_W(w)=(0,w)\geq_{V\oplus W}0$.

The projections tell us that if a pair $(v,w)$ is positive in $V\oplus W$, then each of its components is positive in its respective vector space. On the other hand, the injctions tell us that if each component of a pair is positive, then each of their images in $V\oplus W$ must be positive, and so the sum of the images — the pair itself — must be positive. That is, a pair is positive if and only if each of its components is positive. This uniquely specifies the cone on the direct sum so as to make it a biproduct in $\mathcal{O}rd\mathcal{L}in$.

This category is also monoidal closed. There are various natural monoidal structures we could use, so we’ll start with the exponential this time. Now, in $\mathbf{FinVect}_\mathbb{F}$ we have an exponential — ${W^V}$ is the vector space of all $\mathbb{F}$-linear maps from ${V}$ to $W$. Is there a natural cone in this vector space? Indeed there is! It’s just the cone of all positive maps! That is, $f\geq_{W^V}0$ if and only if $v\geq_V0$ implies $f(v)\geq_W0$.

So what’s the tensor product? Well, we start with the vector space tensor product and try to find a cone. This should give an adjunction $\hom_{\mathcal{O}rd\mathcal{L}in}(U\otimes V,W)\cong\hom_{\mathcal{O}rd\mathcal{L}in}(U,W^V)$. So let’s read this as another logical statement. A linear map $\mu:U\rightarrow W^V$ is positive (and thus in $\hom_{\mathcal{O}rd\mathcal{L}in}(U,W^V)$) if
$u\geq_U0\Rightarrow\mu(u)\geq_{W^V}0$
Expanding this condition on $\mu (u)$, we get that $\mu$ is positive if
$(u\geq_U0)\&(v\geq_V0)\Rightarrow[\mu(u)](v)\geq_W0$
But $(u,v)\mapsto[\mu(u)](v)$ is the usual closure adjunction in the category of vector spaces, turning a function-valued function of one variable into a vector-valued function of two variables. And we want every positive map from $U\otimes V$ to $W$ to correspond to exactly one $\mu$ in just this way. Thus the cone on $U\otimes V$ that makes the tensor product for $\mathcal{O}rd\mathcal{L}in$ into a left adjoint to the exponential is that of all finite sums of tensor pairs of positive elements. That is, if $x=\sum_{i=1}^n u_i\otimes v_i$ with all the $u_i$ and $v_i$ positive in their respective cones. As an exercise, verify that this tensor product is also symmetric.

The monoidal identity has the base field $\mathbb{F}$ as its underlying vector space. For its cone, take the positive ray. It’s straightforward to check that $V\otimes\mathbb{F}\cong V$. As usual, we use the tensor identity to represent the “underlying set” functor. That is, we define the underlying set of a cone ${V}$ by ${\hom_{\mathcal{O}rd\mathcal{L}in}(\mathbb{F},V)}$. Such a positive map $f$ is a linear map from $\mathbb{F}$ to ${V}$ that picks out a positive point $f(1)\geq_V$, and there is exactly one such map for each positive point in ${V}$. That is, the underlying set is exactly the set of points in the positive cone of ${V}$. As a check, note that this means the underlying set of ${W^V}$ is the set $\hom_{\mathcal{O}rd\mathcal{L}in}(V,W)$.

As if that weren’t enough, $\mathcal{O}rd\mathcal{L}in$ has duals! Indeed, we have a cone in the dual vector space defined by ${\lambda\geq_{V^*}0}$ if and only if ${\lambda(v)\geq_\mathbb{F}0}$ for all $v\geq_V0$. Or in other words, $\lambda\in\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$. We just need natural maps $\epsilon_V:V\otimes V^*\rightarrow\mathbb{F}$ and $\eta:\mathbb{F}\rightarrow V^*\otimes V$ to make this really a categorical dual. The first of these is evaluation — $\epsilon(v,\lambda)=\lambda(v)$. The second one picks out an identified element in $V^*\otimes V$. How can we do this?

Well, in a vector space we can pick a basis $\{e_i\}$ of ${V}$ and get a dual basis $\{f_i\}$ of $V^*$ defined so that $f_j(e_i)$ is $1$ if $i=j$ and ${0}$ otherwise. Then we can sum up $\sum_{i=1}^nf_i\otimes e_i$. It’s well-known (though I haven’t shown it yet) that this sum doesn’t depend on which basis we picked! That is, it’s just a property of the vector space ${V}$.

Now if ${V}$ has a cone, we know we can find a positive basis. And then it turns out the dual basis will be positive in the dual cone. Putting these together, it turns out that the above element $\sum_{i=1}^nf_i\otimes e_i$ is always in the positive cone of $V^*\otimes V$, even if we didn’t start by picking a positive basis! All we need is the fact that this sum can be written as a sum of positive tensor pairs.

From here, it’s an easy calculation to verify that $\eta$ and $\epsilon$ satisfy the two required equations, making $V^*$ the dual of ${V}$.

[UPDATE]: Okay, that last bit doesn’t seem to work. The dual basis is not in general positive. That was a fact that I quoted from my conversation with Howard, so I think he made a mistake there. It’s my own fault for not verifying it, but now I’ve found an example where it fails. I’m working on finding an example where it fails for all positive bases. As it stands, $\mathcal{O}rd\mathcal{L}in$ does not in general have duals.

Them’s the breaks when you’re working on the edge of what you know.

September 24, 2007 Posted by | Category theory | 6 Comments