15
$\begingroup$

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.

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).

$\endgroup$
9
  • 1
    $\begingroup$ Have you tried looking for a counterexample in a non-Weyl group? $\endgroup$ Jul 13, 2012 at 1:55
  • $\begingroup$ @Vlad: It would help to add a reference or two (maybe the 2009 Annals paper by Bjorner-Ekedahl?) and/or a bit more background. There are other open problems about non-crystallographic Coxeter groups and KL polynomials, e.g., the KL conjecture that coefficients of the polynomials should be non-negative. In the crystallographic case they can be interpreted as dimensions of cohomology groups. Alvis checked by computer the $H_3, H_4$ groups, while Soergel has proposed an ambitious program to show non-negativity without algebraic geometry. Your problem also looks very hard. $\endgroup$ Jul 13, 2012 at 16:02
  • 1
    $\begingroup$ @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 "Combinatorics of Coxeter Groups", and my main source for Hecke Algebra and KL-polynomials is James Humphreys' "Reflection Groups and Coxeter Groups". $\endgroup$
    – xpilot
    Jul 13, 2012 at 17:13
  • 1
    $\begingroup$ @Vlad: To your question about Jim, yes they are the same person. To speculate, I doubt this is any easier than nonnegativity. Perhaps it might be possible to prove this assuming nonnegativity, but of course that is cheating since nonnegativity also requires geometry to prove. And, to give advice, I think you get further in this subject if you at least know enough of the algebraic geometry to understand the basic ideas coming from that side of the subject, even if you can't master the machinery well enough to use it to prove anything. The same can probably be said of the rep theory. $\endgroup$ Jul 13, 2012 at 23:05
  • 1
    $\begingroup$ @Christian: I'm using a C++ program called "coxeter", written a while ago by Fokko du Cloux (math.univ-lyon1.fr/~ducloux/coxeter/coxeter3/english/…). The code is a bit outdated (doesn't use standard C++ data structures) but otherwise pretty clean and fast. $\endgroup$
    – xpilot
    Jul 14, 2012 at 17:37

3 Answers 3

3
$\begingroup$

The statement does indeed hold in type H3; you can find the computation at http://sage.lacim.uqam.ca/home/pub/15/. The machine is still running (and I don't know how long it might take) doing type H4.

So, if no Coxeter theoretic proof is known, there is hope to find one...

$\endgroup$
3
  • $\begingroup$ @Christian: This kind of experimentation for finite Coxeter groups is certainly essential, though $H_4$ is much bigger than $H_3$. (Be patient.) $\endgroup$ Jul 13, 2012 at 16:06
  • 1
    $\begingroup$ Running a few quick tests shows that the statement holds in A7, B6, D6, H4, E6, and E7. Larger groups overflow my memory :( $\endgroup$
    – xpilot
    Jul 13, 2012 at 17:00
  • $\begingroup$ @Christian (and Vlad): Anyone concerned about Lie-theoretic computation should be aware of the software versions posted at the Atlas of Lie Groups site liegroups.org/software/index.html Fokko du Cloux was an essential part of the original team, but he died prematurely of ALS. $\endgroup$ Jul 16, 2012 at 15:01
2
$\begingroup$

The answer to the question is "yes", allowing for a generous interpretation of "direct way". This will follow from the recently posted work of Ben Elias and Geordie Williamson on non-negativity of coefficients of Kazhdan-Lusztig polynomials for an arbitrary Coxeter group here.

See the Update to my MO question here, which refers to the 1991 conference report by Jim Carrell (with Dale Peterson): in the first section, the equivalence you want is formulated for an arbitrary Coxeter group under the hypothesis that coefficients of relevant K-L polynomials are all non-negative. (This may be one of the sources you are referring to.)

At first I had overlooked this type of answer to my own question. (I'm still looking for other consequences of the non-negativity theorem, of course, but this one is interesting.) Note that for general Coxeter groups, one needs an approach which doesn't involve the geometry of Schubert varieties. What Elias and Williamson seem to do is avoid all that algebraic geometry by providing a sophisticated substitute.

$\endgroup$
0
$\begingroup$

"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 $P_{x, w}(q) = 1$.

$\endgroup$
2
  • $\begingroup$ @Vlad: I'm not sure whether this is given as another comment or as an answer to your question. Does the book do anything to answer that or is it one of the unnamed sources you started with? II realize the comments were already getting numerous.) $\endgroup$ Jul 20, 2012 at 19:23
  • $\begingroup$ 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. $\endgroup$
    – xpilot
    Jul 24, 2012 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.