Regions of Hyperplane Arrangements - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:18:30Z http://mathoverflow.net/feeds/question/106092 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements Regions of Hyperplane Arrangements Vlad Firoiu 2012-09-01T04:10:02Z 2013-04-11T15:49:02Z <p>Let $\mathcal{A}$ be an arrangement of the hyperplanes $h_1, h_2, \ldots h_n$. $\mathcal{A}$ partitions the underlying space $V$ into connected regions, denoted by $R(\mathcal{A})$. I would like to enumerate the regions using the intersection lattice $L(\mathcal{A})$ of $\mathcal{A}$.</p> <p>Given a hyperplane $h \in \mathcal{A}$, we can define the following two arrangements:<br> $\mathcal{A}-h$ is the arrangement obtained by removing $h$.<br> $\mathcal{A}/h$ is the arrangement obtained by contracting to $h$; that is, the new underlying space is $h$, and the new hyperplanes are the intersections of the old hyperplanes with $h$.</p> <p>It is not hard to see that $|R(\mathcal{A})| = |R(\mathcal{A}-h)| + |R(\mathcal{A}/h)|$. Indeed, each region in $R(\mathcal{A}/h)$ corresponds to a region in $R(\mathcal{A}-h)$ which $h$ cuts in two.</p> <p>To review, $L(\mathcal{A})$ is the set of intersections of hyperplanes, ordered by reverse inclusion. It has bottom element $\hat{0} = V$, but only has a top element if all of the hyperplanes intersect at a point. Thus, joins (which are intersections) may fail to exist, while meets do always exist. Each element is the join of the hyperplanes below it. (For a better overview of this material, see www.math.rice.edu/~samans/ZaslavskyTheorem.pdf).</p> <p>For each $x\neq \hat{0}$, let $f(x)$ be the maximal $i$ such that $h_i \leq x$, and let $h(x) = h_{f(x)}$. Define an <em>increasing chain</em> in $L(\mathcal{A})$ to be a sequence $\hat{0} = x_0 \triangleleft x_1 \triangleleft \cdots \triangleleft x_m$ such that $f(x_i)$ is increasing for $i\geq 1$ ($\triangleleft$ denotes covering in the intersection lattice). Note that $x_i = x_{i-1} \lor h(x_i)$. Let $C(\mathcal{A})$ denote the set of all increasing chains.</p> <p>It is not too hard to see that $|C(\mathcal{A})| = |C(\mathcal{A} - h_1)| + |C(\mathcal{A}/h_1)|$, given an appropriate ordering of the atoms in $\mathcal{A}/h_1$. It then follows by induction that $|C(\mathcal{A})| = |R(\mathcal{A})|$ and that $|C(\mathcal{A})|$ does not depend on initial order of the hyperplanes.</p> <p>My question is then: does there exist a "natural" bijection between $R(\mathcal{A})$ and $C(\mathcal{A})$?</p> http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements/106121#106121 Answer by Patricia Hersh for Regions of Hyperplane Arrangements Patricia Hersh 2012-09-01T17:52:29Z 2012-09-01T19:15:24Z <p>Here's a conjectural solution, motivated by the idea of a "line shelling". Begin by choosing a "generic" nonzero vector $c$ in your ambient space, i.e. a vector such that it spans a line which crosses all of your hyperplanes. </p> <p>Next, associate each vertex $v$ in the hyperplane arrangement to the unique region $R$ having the property that the dot product $c\cdot x$ is maximized on $R$ at $v$. Now assign to each vertex $v$ the set of hyperplanes which bound this region $R$ and pass through $v$. Each such $R$ is then bijectively mapped to the increasing chain in the intersection poset labeled by this list of chosen hyperplanes arranged in ascending order. </p> <p>Similarly, label each of the remaining regions $R$ by its set of bounding hyperplanes $\mathcal{H}$ which have the property that for each chosen $\mathcal{H}$ there is another bounding hyperplane $\mathcal{H'}$ of $R$ and a line segment going in the direction $c$ which starts at a point $p'$ on $\mathcal{H'}$ and ends at a point $p$ on $\mathcal{H}$, with $c\cdot x$ increasing as we progress along the segment from $p'$ to $p$. Bijectively map each of these regions $R$ to the increasing chain in the intersection poset labeled by its chosen collection of hyperplanes, again listed in increasing order.</p> <p>It would be great if you or someone else wants to figure out whether this conjecture is correct. I wouldn't be surprised though if someone has thought about your question before -- your question certainly ties in with a lot of interesting work in the literature.</p> http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements/127260#127260 Answer by Michael Falk for Regions of Hyperplane Arrangements Michael Falk 2013-04-11T15:49:02Z 2013-04-11T15:49:02Z <p>There is an inductive construction of the bijection you seek in a paper of Ken Jewell and Peter Orlik, in the proceedings of Arrangements in Boston (1998), published in Topology and Its Applications. Your increasing chains are in bijection with nbc (= "no-broken-circuit) sets in the arrangement. (The nbc set associated with $x_0 &lt; \cdots &lt; x_m$ is {f(x_0), ... , f(x_n)}, identifying hyperplanes with their labels.) Jewell and Orlik assign a chamber to each nbc set, bijectively, in Lemma 3.14 of that paper. Their bijection is built up one hyperplane at a time, that is, by deletion-contraction. </p>