Let $\{A\_i\}$ be a collection of $m$ hyperplanes in $\mathbb{C}^n$ which all pass through the origin (a **central hyperplane arrangement**). Such an arrangement is called **Coxeter** if reflecting across any hyperplane in $\{A\_i\}$ sends the arrangement to itself (and so the reflections automatically will generate a Coxeter group).


Now, I will define a rather random-seeming condition on an arbitrary central arrangement. Choose a normal vector $n\_i$ to each $A\_i$. Consider the function on $\gamma$ on $\mathbb{C}^n$ given by
$$ \gamma(v) = \sum\_{1\leq i< j\leq m}(n\_i,n\_j)\left(\prod\_{k\neq i,j} (n\_k,v) \right) $$
The function $\gamma$ depends on the specific choice of normal vectors; however, whether $\gamma$ vanishes does not depend on the choice of normal vectors (since scaling a normal vector will scale the output). Call a central hyperplane arrangement **puzzling** if $\gamma(v)=0$ for all $v$.


The 'puzzling' condition came up in studying a very specific research problem. However, both the context and a day's worth of experimentation have lead to the following conjecture.


**Conjecture:** The puzzling arrangements are exactly the Coxeter arrangements.


It's worth noting that I can't show either direction of the conjecture. Brute force computation says that for $n=2$, the conjecture is true.


Just from the form, it kind of reminds me of a Weyl character formula-type identity, but I don't really know much about those. My hope is that this kind of identity is pretty well known to people who work with such things.




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**Edit:** There's another way of stating this identity, that's closer to the context in which I encountered it (differential operators). Let $n\_i^\*$ denote the function $(n\_i,-)$, and let $d\_i$ denote differentiation along the vector $n\_i$. Then


$$ \gamma = \left( \sum\_i (n\_i^\*)^{-2}(n\_i^\*d\_i-(n\_i,n\_i))\right) \prod\_j n\_j $$ 


Therefore, the 'puzzling' condition is then a statement about the function $\prod\_jn\_j$ being killed by a particular rational differential operator.




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**Edit 2:** Fixed the definition of Coxeter arrangements... I want all the reflections, not just a generating set.



Coxeter Arrangements and an Identity
**Proof that Coxeter arrangements obey this identity:** Group together summands according to the two-plane spanned by $n\_i$ and $n\_j$. For any two-plane $H$, every summand coming from that two-plane is divisible by $\prod\_{n\_k \not \in H} \langle n\_k, v \rangle$. Factoring out this common summand, the contribution from $H$ is
$$\sum\_{n\_i \neq n\_j,\ n\_i, n\_j \in H} \langle n\_i, n\_j \rangle \prod\_{n\_k \in H,\ n\_k \neq n\_i, n\_j} \langle n\_k, v \rangle.$$


This is the two dimensional example you've already done.


**Proof that only Coxeter arrangements obey this identity:**
Consider $H$, a two plane spanned by some $(n\_i, n\_j)$. Our first goal is to show that $H \cap \{ n\_k \}$ is a dihedral root system.


Let $r$ be the number of hyperplanes in your arrangement. Let $S$ be the ring of polynomial functions and let $I$ be the ideal generated by the functions $\langle n, \ \rangle$, for $n \in H$. Note that every term of your sum which does not come from $(n\_i, n\_j)$ with $n\_i$, $n\_j \in H$ lies in $I^{r-1}$. So, the sum of the terms with $n\_i$, $n\_j \in H$ must be zero modulo $I^{r-1}$. 


As before, all of those terms are divisible by $\prod\_{n\_k \not \in H} \langle n\_k, v \rangle$. This is not a zero divisor in $S/I^{r-1}$. So we can factor it out and deduce that
$$\sum\_{n\_i \neq n\_j,\ n\_i, n\_j \in H} \langle n\_i, n\_j \rangle \prod\_{n\_k \in H,\ n\_k \neq n\_i, n\_j} \langle n\_k, v \rangle \equiv 0 \ \mathrm{mod} \ I^{r-1}$$
But the left hand side is degree $r-2$, so it must be identically zero. By the two dimensional example which you have already done, this shows that $\{ n\_k: n\_k \in H \}$ is a root system.


So, for any $n\_i$ and $n\_j$, the set of $n\_k$ in the span of $(n\_i, n\_j)$ is a root system. In particular, the reflection of $n\_i$ by $n\_j$ is some $n\_k$. So your whole set of vectors is a root system.


**Warning:** I have not, myself, checked the two dimensional case which I am relying on.


