# Notion of positivity in $\mathbb{Q}(q)$

When dealing with an element $P \in \mathbb{Z}[q]$, a natural combinatorial question is to ask whether $P$ has non-negative coefficients. For instance, this is true if $P$ is a Kazhdan-Lusztig polynomial. I have recently been looking at certain elements of the field $\mathbb{Q}(q)$ that have a similar flavor to Kazhdan-Lusztig polynomials and have observed some positivity properties.

What is the correct definition of non-negativity in this ring?

At first I thought a natural candidate would be to say that $f \in \mathbb{Q}(q)$ is non-negative if $f = P/Q$, $P, Q \in \mathbb{Z}[q]$, $P$ and $Q$ have no common factors, and $P$ and $Q$ have non-negative coefficients. However, this is not a good definition because then the set of non-negative elements is not a semiring. For example, $[2],[3],$ and $\frac{1}{[6]}$ are all non-negative by this definition, but $\frac{[2][3]}{[6]} = \frac{q^2}{1-q^2+q^4}$ is not, where $[k] = \frac{q^k - q^{-k}}{q - q^{-1}}$.

Another possible definition of non-negativity would be $f \in \mathbb{Q}(q)$ is non-negative if $f(a)$ is non-negative for all non-negative real $a$. With this definition, the set of non-negative rational functions is a semiring. This is strictly weaker than the definition of non-negativity above.

Has anyone come across this definition of non-negativity before?

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The natural definition of non-negativity which occurs to me is that the coefficients in the Laurent expansion are all non-negative. – Qiaochu Yuan Jan 17 '11 at 22:41
(Also, minor nitpick: the coefficients of the KL-polynomials are known to be non-negative for Weyl groups and a few other Coxeter groups, but this is open for general Coxeter groups.) – Qiaochu Yuan Jan 17 '11 at 22:50
My first impulse is the same as Qiaochu's. Not only is this obviously a semi-ring, it also is a notion that naturally arises in categorifications. For example, the Hilbert series of a non-negatively graded ring is positive in this sense, but its denominator has a negative sign. – Ben Webster Jan 17 '11 at 22:56
On the other hand, clearly the right answer is "Whatever you can prove or conjecture in you situation." – Ben Webster Jan 17 '11 at 22:57