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Let $d\in \mathbb{Z}_{\ge 1}$, let $\sigma = (H_i)_{i\in \mathcal{I}}$ be a finite hyperplane arrangement in $\mathbb{R}^d$, where $H_i\subset \mathbb{R}^d$ is a hyperplane for $i\in \mathcal{I}$ (the hyperplanes may repeat), and let $\mathcal{A} = \pi_0(\mathbb{R}^d\setminus \bigcup_{i\in \mathcal{I}}H_i)$ be the set of chambers of the arrangement $\sigma$. For any pair of alcoves $a, b\in \mathcal{A}$, let $n_{a, b}\in \mathbb{Z}_{\ge 0}$ be the number of hyperplanes in $\sigma$ which seperate $a$ and $b$ (i.e. $n_{a, b}$ is the number of those $H\in \sigma$ such that $a$ and $b$ sit in different connected components of $\mathbb{R}^d \setminus H$).

Consider the following $\cal{A}\times \cal{A}$ matrix with coefficients in $\mathbb{Q}(t)\cap\mathbb{Z}[\![ t]\!]$:

$$ M = (M_{a,b})_{a,b\in \cal{A}(\sigma)},\quad M_{a,b} = \frac{t^{n_{a,b}}}{(1 - t^2)^d}. $$

Obviously, $M$ is symmetric and congruent to the identity matrix modulo $t$.

I'm wondering under what conditions on $\sigma$ are the coefficients of the inverse matrix $M^{-1}$ in $\mathbb{Z}[t]$.

Question: Is the following true?

The coefficients of $M^{-1}$ are polynomials in $t$ iff $\sigma$ is generic, i.e. the hyperplanes in $\sigma$ don't repeat and they intersect transversally.

Examples

  1. Consider $\mathbb{R}^1$ with coordinate $x$ and $\sigma = (\{x=i\})_{i = 0}^{r-1}$ for some $r\in \mathbb{Z}_{\ge 0}$. The arrangement $\sigma$ is generic and its $r$ hyperplanes divide $\mathbb{R}^1$ into $r+1$ open intervals. Then matrix $M$ is $(r+1)\times (r+1)$ and is given by $$ M = (1-t^2)^{-1}\begin{pmatrix}1 &t & t^2 & \cdots & t^{r} \\ t &1 & t &\cdots & t^{r-1} \\ \vdots & \vdots & \vdots & & \vdots \\ t^{r} & t^{r-1} & t^{r-2} & \cdots &1\end{pmatrix}. $$ We have in this case $$ M^{-1} = \begin{pmatrix}1 &-t & 0 & \cdots & 0& 0 \\ -t & 1+t^2 & -t &\cdots & 0 & 0 \\ 0 & -t & 1+t^2 &\cdots & 0& 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 + t^2 &-t \\ 0 & 0 & 0 & \cdots & -t &1\end{pmatrix}. $$
  2. Consider $\mathbb{R}^2$ with coordinates $x,y$ and $\sigma = (\{x=0\}, \{y=0\},\{x+y=0\})$. These three lines are concurrent at $0$ and divide $\mathbb{R}^2$ into 6 alcoves. Since the lines don't interesect transversally at $0$, the arrangement $\sigma$ is not generic. With a suitable order of the chambers we can write $M$ as a circular matrix $$ M = (1-t^2)^{-2}\begin{pmatrix}1 & t & t^2 & t^3 & t^2 & t \\ t & 1 & t & t^2 & t^3 & t^2 \\ t^2 & t & 1 & t & t^2 & t^3 \\ t^3 & t^2 & t & 1 & t & t^2 \\ t^2& t^3 & t^2 & t & 1 & t\\ t& t^2& t^3 & t^2 & t & 1 \end{pmatrix} $$ and the inverse $$ M^{-1} = (1 + t^2 + t^4)^{-1}\begin{pmatrix} 1 + t^2 & -t & -t^4& t^3 + t^5 & -t^4 & -t \\ -t &1 + t^2 & -t & -t^4& t^3 + t^5 & -t^4 \\ -t^4 & -t &1 + t^2 & -t & -t^4& t^3 + t^5 \\ t^3 + t^5& -t^4 & -t &1 + t^2 & -t & -t^4 \\ -t^4& t^3 + t^5& -t^4 & -t &1 + t^2 & -t \\ -t& -t^4& t^3 + t^5& -t^4 & -t &1 + t^2 \end{pmatrix} $$ is not polynomial, since it has the denominator $1 + t^2 + t^4$.
  3. Consider a slight modification of example 2 : $\mathbb{R}^2$ with coordinates $x,y$ and $\sigma = (\{x=0\}, \{y=0\},\{x+y=1\})$ (one translates the line $x+y = 0$ in order to "resolve" the concurrence). There is one extra chamber, and with a suitable order of chambers the matrix $M$ is written as $$ M = (1-t^2)^{-2}\begin{pmatrix}1 & t & t^2 & t^3 & t^2 & t & t^2 \\ t & 1 & t & t^2 & t^3 & t^2 & t \\ t^2 & t & 1 & t & t^2 & t^3 & t^2 \\ t^3 & t^2 & t & 1 & t & t^2 & t \\ t^2& t^3 & t^2 & t & 1 & t & t^2 \\ t& t^2& t^3 & t^2 & t & 1 & t \\ t^2 & t & t^2 & t & t^2 & t & 1 \end{pmatrix}. $$ It has an extra line and an extra column compared to example 2. In this case $\sigma$ is generic and the inverse matrix is given by $$ M^{-1} = \begin{pmatrix} 1 & -t & 0 & 0 & 0 & -t& t^2 \\ -t & 1+t^2 & -t & t^2 & 0 & t^2& -t -t^3 \\ 0 & -t & 1 & -t & 0 & 0 & t^2 \\ 0 & t^2 & -t & 1+t^2 & -t & t^2 & -t -t^3 \\ 0 & 0 & 0& -t & 1 & -t & t^2 \\ -t & t^2 & 0 & t^2 & -t & 1+t^2 & -t -t^3 \\ t^2 & -t - t^3 & t^2 & -t-t^3 & t^2 & -t-t^3 & 1 + t^2 + t^4 \end{pmatrix}, $$ which is polynomial.

Notes: SamHopkins suggests that $M$ without the denominator is the specialisation of the Varchenko matrix by setting $a_H = t$ for all $H\in \sigma$. It seems that the determinant formula of Varchenko can be useful. However, without the equal-weight specialisation $a_H = t$, the degree of the denominators of the inverse Varchenko matrix can be large (up to the degree of $\det M$) even when $\sigma$ is generic. One will need a formula for the minors of the Varchenko matrix to answer the question.

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    $\begingroup$ Is this related to the Varchenko matrix of a hyperplane arrangement? $\endgroup$ Dec 3 '20 at 18:05
  • $\begingroup$ @SamHopkins I don't know what it is. I'll look at it, thanks. I came up with this matrix from a problem of homological algebra. The power series in $t$ is the graded dimension of graded hom-spaces between projective modules. $\endgroup$
    – Wille Liou
    Dec 3 '20 at 18:07
  • $\begingroup$ The Varchenko matrix $V$ is the (in your notation) $\#\mathcal{A}\times\#\mathcal{A}$ matrix whose $a,b$ entry is $\prod x_{H}$ where the product is over all hyperplanes separating regions $a$ and $b$, and the $x_{H}$ are formal variables corresponding to the hyperplanes. Varchenko's theorem gives a product formula for the determinant of $V$: $\det V = \prod_{M \in \mathcal{L}(\sigma)} (1-x_M^2)^{n(M)p(M)}$, where $\mathcal{L}(\sigma)$ is the intersection poset of the arrangement and $x_M = \prod_{M \subseteq H} x_H$, and $n(M)$ and $p(M)$ are certain numbers attached to $M$. $\endgroup$ Dec 3 '20 at 18:24
  • $\begingroup$ See Varchenko, "Bilinear Form of Real Configuration of Hyperplanes ", sciencedirect.com/science/article/pii/S0001870883710030 $\endgroup$ Dec 3 '20 at 18:25
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    $\begingroup$ The paper arxiv.org/abs/1511.02923 may be relevant. $\endgroup$ Dec 4 '20 at 1:55

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