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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.


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.


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

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I'm pretty sure Dunkl told me he had done a calculation equivalent to what David does below in one of his papers, but I don't remember where now. If you need a specific reference you might ask him. –  GS Jan 14 '10 at 10:32

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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.

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Hey, excellent! Well, it's also disappointing, since it proves a no-go theorem for something I was trying to do. This also doesn't satisfy the part of me that wants to know WHY this identity is true, but a proof is a proof. Thanks! –  Greg Muller Nov 21 '09 at 3:20
    
I could imagine a conceptual proof that Coxeter arrangements obey this condition. Let P be the polynomial defined by your formula, and let w be an element of the Coxeter group. Then $wP = (-1)^{\ell(w)} P$. On the other hand, P is in Sym^{n-2} V, where V is the reflection representation of W. If you could show that Sym^{n-2} V didn't contain any copies of the sign representation, that would be a proof. –  David Speyer Nov 21 '09 at 5:23
    
No ideas about a conceptual proof for the other direction. I'd be curious to see your proof in the rank 2 case. –  David Speyer Nov 21 '09 at 5:23
    
My proof in the rank 2 case was a brutish as you can imagine. I let the vectors be $(1,0)$, $(a_2,1)$,... $(a_n,1)$, and then computed the function $P$. From the condition that $P$ vanishes, it isn't hard to make the polynomial who's roots are the $a_i$, and then an inductive argument shows this is an even polynomial. Therefore, the root system is invariant under reflection. –  Greg Muller Nov 21 '09 at 6:51

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