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Let $\Lambda$ be a lattice (i.e. $\Lambda \simeq \mathbb{Z}^n$) with a positive subcone $\Lambda^+$. Let $H: \Lambda^+ \rightarrow \mathbb{C}$ be a function such that $\forall\mu \in \Lambda^+$, $b_\mu(\lambda) := \frac{H(\lambda + \mu)}{H(\lambda)}$ is a polynomial in $\lambda$. To put it another way, let $b: \Lambda \rightarrow Rat(\Lambda)$ be a cocycle (in the group cohomology complex with $Rat(\Lambda)$ being rational functions on $\mathbb{C}\Lambda$ and the translation action) such that any element of the positive subcone is a polynomial. I think, but haven't figured out how to prove, the following:

$$b_\mu ~~\text{factors as}~~ b_\mu = \prod_i (\alpha_i + k_i), ~~\text{where}~~ \alpha_i \in \text{Hom}(\Lambda, \mathbb{C})$$

In fact, I have a bigger conjecture:

$H(\lambda)$ is a product of factorials of linear factors (i.e. factors of the form $\alpha_i + k_i$ with $\alpha_i \in \text{Hom}(\Lambda, \mathbb{Z})$ multiplied by an exponential function multiplied by a product of linear factors.)

A few examples: The most basic examples are products of factorials of linear factors; however, we can also consider the case $\Lambda = \mathbb{Z}, H(\lambda) = (3 \lambda)! (3 \lambda + 2)$.

I wasn't quite sure what to tag this with, so please feel free to tag it appropriately.

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  • $\begingroup$ I don't quite get it. You have a multivariate (if $\operatorname{rank}\Lambda>1$) polynomial; why should it split into linear factors? $\endgroup$ Nov 14, 2014 at 9:35
  • $\begingroup$ Also, what if $H(\lambda):=\prod_{j=1}^\lambda (j^2+1)$, for $\lambda\in\mathbb{Z}_+$ ? $\endgroup$ Nov 14, 2014 at 11:10
  • $\begingroup$ Is this related to the Weyl dimension formula? That is, in the end of properties on en.wikipedia.org/wiki/Schur_polynomial#Properties ? This formula is essentially a lattice point counting number. $\endgroup$ Nov 14, 2014 at 12:49
  • $\begingroup$ Alex: I don't just have one polynomial, I have many with a certain compatibility condition; I think that that compatibility condition implies factoring. $\endgroup$
    – user44191
    Nov 14, 2014 at 17:06
  • $\begingroup$ Pietro: any polynomial in one variable will split; for yours, $b_1(\mu) = (\lambda + 1)^2 + 1 = (\lambda + 1 + i)(\lambda + 1 - i)$. $\endgroup$
    – user44191
    Nov 14, 2014 at 17:09

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The positive answer to this question appears in the appendix to Sato, Shintani, and Muro's paper on b-functions (which were the source of this question in the first place).

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