most recent 30 from http://mathoverflow.net 2013-05-18T19:53:32Z http://mathoverflow.net/feeds/user/25091 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106092/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/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order Vlad Firoiu 2012-07-13T01:06:02Z 2013-02-10T23:40:57Z

<p>I have read in a number of places that the lower Bruhat interval $[e, w]$ is rank-symmetric if and only if the KL-polynomial $P_{e, w}(q) = 1$. All of the proofs I've come across use "rationally smooth Schubert varieties", which I don't really understand.</p> <p>The KL polynomials can be defined purely in terms of the Iwahori-Hecke algebra of the Coxeter group, and satisfy a number of identities involving sums over Bruhat intervals. I would like to know then if there is a more direct way to prove that $[e, w]$ is rank symmetric iff $P_{e, w}(q) = 1$, using only the Hecke algebra (and Bruhat order).</p>

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order/102758#102758 Vlad Firoiu 2012-07-20T17:57:44Z 2012-07-20T17:57:44Z

<p>"Singular Loci of Schubert Varieties" by Billey and Lakshmibai is by far the best reference I've found so far for this question. Chapter 6 in particular deals with the combinatorial consequences of $P_{x, w}(q) = 1$.</p>

http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements/106121#106121 Vlad Firoiu 2012-09-09T20:12:02Z 2012-09-09T20:12:02Z

Well, you say &quot;the unique region $R$ having the property that the dot product $c\cdot x$ is maximized on $R$ at $v$&quot;. I'm not sure why this region would be unique.

http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements/106121#106121 Vlad Firoiu 2012-09-07T04:03:21Z 2012-09-07T04:03:21Z

Right, by 1-dimensional I meant single points (which of course are actually 0-dimensional). As for your conjectured solution, I don't think that you can uniquely assign to each vertex a region... In any case I'm stumped, and will ask Prof. Stanley if he has any ideas.

http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements/106121#106121 Vlad Firoiu 2012-09-01T23:42:33Z 2012-09-01T23:42:33Z

Thank you for all of the references! I'm currently trying to understand your conjectured solution, but I'm not sure what you mean by &quot;vertex&quot; of the arrangement... perhaps the 1-dimensional elements of $L(\mathcal{A})$?

http://mathoverflow.net/questions/106092/regions-of-hyperplane-arrangements Vlad Firoiu 2012-09-01T04:15:04Z 2012-09-01T04:15:04Z

Also relevant is Zaslavsky's Theorem, which gives an explicit formula for $R(\mathcal{A})$ in terms of the Mobius function on $L(\mathcal{A})$: $$R(\mathcal{A}) = \sum_{x\in L(\mathcal{A})} (-1)^{r(x)}\mu(\hat{0}, x)$$ I'm not sure how to use this though...

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order/102758#102758 Vlad Firoiu 2012-07-24T19:54:38Z 2012-07-24T19:54:38Z

The book was not one of my original sources. While I don't have an outright answer, it does says many interesting things regarding my original question. For example, another equivalent condition to $P_{e, w}(q) = 1$ is that $|T \bigcap [e, w]| = l(w)$, which is the equality case of Deodhar's inequality. This kind of statement is the kind of thing I was looking for, and should be very useful in my research.

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order Vlad Firoiu 2012-07-14T17:37:40Z 2012-07-14T17:37:40Z

@Christian: I'm using a C++ program called &quot;coxeter&quot;, written a while ago by Fokko du Cloux (<a href="http://math.univ-lyon1.fr/~ducloux/coxeter/coxeter3/english/coxeter3_e.html" rel="nofollow">math.univ-lyon1.fr/~ducloux/coxeter/coxeter3/&hellip;</a>). The code is a bit outdated (doesn't use standard C++ data structures) but otherwise pretty clean and fast.

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order Vlad Firoiu 2012-07-13T21:29:27Z 2012-07-13T21:29:27Z

Ah, so Jim Humphreys (mathoverflow user) = James Humphreys (author)? In that case, there probably isn't a known Coxeter-theoretic proof. I was hoping that such a proof would help with some related research I'm doing; maybe I'll switch to trying to find a proof myself.

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order Vlad Firoiu 2012-07-13T17:13:53Z 2012-07-13T17:13:53Z

@Qiaochu: I've checked it by computer for F4, H3, H4, A1-7, B1-6, D1-6, E6, E7. Not sure what else I'd check, aside from larger groups (which tend to take forever and cause memory overflows). @Jim: Most of what I know about Coxeter groups and Bruhat order is from Bjorner and Brenti's &quot;Combinatorics of Coxeter Groups&quot;, and my main source for Hecke Algebra and KL-polynomials is James Humphreys' &quot;Reflection Groups and Coxeter Groups&quot;.

http://mathoverflow.net/questions/102087/kazhdan-luzstig-polynomials-and-lower-intervals-in-the-bruhat-order/102124#102124 Vlad Firoiu 2012-07-13T17:00:09Z 2012-07-13T17:00:09Z

Running a few quick tests shows that the statement holds in A7, B6, D6, H4, E6, and E7. Larger groups overflow my memory :(