# The Unapologetic Mathematician

## 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

1. [...] Linear Spaces III We’re on a roll with our discussion of ordered linear spaces. So I want to continue past just describing these [...]

Pingback by Ordered Linear Spaces III « The Unapologetic Mathematician | September 24, 2007 | Reply

2. isn’t this absence of duals connected to the fact that not every positive polynomial is a sum of squares?

Comment by Dima | September 29, 2007 | Reply

3. It might be, though I must admit I’m not sure what order you’re using to define positive polynomials. It’s tough to get anything done without definitions we all know.

Comment by John Armstrong | September 29, 2007 | Reply

4. the “classical definition” of a positive polynomial would be a polynomial $f\in R[x_1,...,x_n]$, for $R$ a real closed field, that satisfies $f(x)>0$ for all $x\in R^n$.
(respectively, nonnegative polynomial would satisfy instead $f(x)\geq 0$)

More generally, people talk about positivity on a cone in $R^n$ (so, replace $R^n$ above with the cone), or even on a (semi)algebraic/analytic/etc set.

PS. I must admit I’ve lost the thread of your post somewhere halfway, so my remark is just a wild guess…

Comment by Dima | September 29, 2007 | Reply

5. Well, let’s see if this case is one of the things I’m talking about. First of all the vector space of polynomials is infinite-dimensional, so it’s not strictly what I’m covering here. However, the infinite-dimensional case is a natural generalization. It’s easy enough to verify that such positive polynomials do satisfy the other definitions for a cone, except for maybe the “generating” hypothesis.

The problem here is that I don’t know what the natural generalization of this condition is. If we just say “positive basis”, then we might run into trouble in ordered Hilbert spaces. But for the moment, can we find a basis of the space of all polynomials consisting of only positive polynomials? I don’t know offhand.

Comment by John Armstrong | September 29, 2007 | Reply

6. But for the moment, can we find a basis of the space of all polynomials consisting of only positive polynomials?
Yes, that’s certainly possible.

My remark was probably prompted by appearance of tensors in your post, and you talking about “tensor positivity”.

PS. there is a “practical” theory about what duality in the context of polynomials positive on a semialgebraic sets should look like. Somehow mysteriously (to me at least), it involves measures and their moments…

Comment by Dima | September 29, 2007 | Reply