# Polynomiality of the inverse of equally weighted Varchenko matrices attached to hyperplane arrangements

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.

• Is this related to the Varchenko matrix of a hyperplane arrangement? Dec 3 '20 at 18:05
• @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. Dec 3 '20 at 18:07
• 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 $M$. Dec 3 '20 at 18:24
• See Varchenko, "Bilinear Form of Real Configuration of Hyperplanes ", sciencedirect.com/science/article/pii/S0001870883710030 Dec 3 '20 at 18:25
• The paper arxiv.org/abs/1511.02923 may be relevant. Dec 4 '20 at 1:55