Coxeter Arrangements and an Identity - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:58:57Z http://mathoverflow.net/feeds/question/6343 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6343/coxeter-arrangements-and-an-identity Coxeter Arrangements and an Identity Greg Muller 2009-11-21T02:21:40Z 2009-11-21T03:42:53Z <p>Let ${A_i}$ be a collection of $m$ hyperplanes in $\mathbb{C}^n$ which all pass through the origin (a <strong>central hyperplane arrangement</strong>). Such an arrangement is called <strong>Coxeter</strong> if reflecting across any hyperplane in ${A_i}$ sends the arrangement to itself (and so the reflections automatically will generate a Coxeter group).</p> <p>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 $$\gamma(v) = \sum_{1\leq i&lt; j\leq m}(n_i,n_j)\left(\prod_{k\neq i,j} (n_k,v) \right)$$ 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 <strong>puzzling</strong> if $\gamma(v)=0$ for all $v$.</p> <p>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.</p> <p><strong>Conjecture:</strong> The puzzling arrangements are exactly the Coxeter arrangements.</p> <p>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.</p> <p>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.</p> <p><hr /></p> <p><strong>Edit:</strong> 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</p> <p>$$\gamma = \left( \sum_i (n_i^*)^{-2}(n_i^*d_i-(n_i,n_i))\right) \prod_j n_j$$ </p> <p>Therefore, the 'puzzling' condition is then a statement about the function $\prod_jn_j$ being killed by a particular rational differential operator.</p> <p><hr /></p> <p><strong>Edit 2:</strong> Fixed the definition of Coxeter arrangements... I want all the reflections, not just a generating set.</p> http://mathoverflow.net/questions/6343/coxeter-arrangements-and-an-identity/6344#6344 Answer by David Speyer for Coxeter Arrangements and an Identity David Speyer 2009-11-21T02:56:52Z 2009-11-21T03:42:53Z <p><strong>Proof that Coxeter arrangements obey this identity:</strong> 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 $$\sum_{n_i \neq n_j,\ n_i, n_j \in H} \langle n_i, n_j \rangle \prod_{n_k \in H,\ n_k \neq n_i, n_j} \langle n_k, v \rangle.$$</p> <p>This is the two dimensional example you've already done.</p> <p><strong>Proof that only Coxeter arrangements obey this identity:</strong> 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.</p> <p>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 $I^{r-1}$. </p> <p>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.</p> <p>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.</p> <p><strong>Warning:</strong> I have not, myself, checked the two dimensional case which I am relying on.</p>