3
$\begingroup$

Let $x$ and $y$ be two elements of $S_n$. Let $U(x,y)$ be the intersection $(BxB \cap B_{-} y B)/B$ inside the flag variety. Here $B$ and $B_{-}$ are the groups of upper and lower-triangular matrices respectively. Kazhdan and Lusztig define the $R$ polynomial $R_{x,y}(q)$. As explained in Theorem 1.3 of Deodhar, one such description is that $$R_{x,y}(q) = \# U(x,y)(\mathbb{F}_q).$$

Now, $U(x,y)$ is an affine variety, so I don't get to know that the weight filtration matches the cohomological filtration. But, looking at examples, it looks like the coefficient of $(-1)^{\ell(x) - \ell(y) - i } q^i$ in $R_{x,y}(q)$ is the Betti number $\dim H^i(U(x,y))$. For example, with $n=2$, $x=(21)$ and $y$ being the identity, then $U(x,y)$ is the torus $\mathbb{A}^1 \setminus \{ 0 \}$, it has $q-1$ points, and its betti numbers are $\dim H^0 = \dim H^1 = 1$.

This is probably something very standard, but I didn't find it the first few places I tried, and it seemed faster to ask here than to keep looking. Thanks!

$\endgroup$
5
  • $\begingroup$ As you probably know, $\\#U(x,y)(\mathbb{F}_q)$ the supertrace of the Frobenius acting on $H^*_c(U(x,y))$; one possibility is that the eigenvalue of Frobenius on $H^{2\ell(x)-2\ell(y)-i}_c(U(x,y))$ is $q^{\ell(x)-\ell(y)-i}$. This would happen if the pullback to the torus in the Deodhar decomposition is injective in cohomology. $\endgroup$
    – Ben Webster
    Commented Dec 19, 2011 at 17:51
  • $\begingroup$ Unfortunately, that pullback is not injective. I have an example in $GL_5$ with $R$-polynomial $q^6 - 4q^5 + 7q^4 - 8q^3 + 7q^2 - 4q +1$. If these are also betti numbers, then $\dim H^1(U)=4$ and $\dim H^2(U) = 7$. The image of $H^2(U)$ must land in the part of $H^2(T)$ generated by $H^1(U)$ (because pullback is a map of rings) and that part has dimension $\binom{4}{2} = 6$. $\endgroup$ Commented Dec 19, 2011 at 18:03
  • $\begingroup$ I seem to recall seeing some paper by someone somewhere in Europe whose name I didn't recognize giving a geometric interpretation of R-polynomials, possibly in connection with Bott-Samelson resolutions or buildings or both. Unfortunately I don't remember enough to track it down. $\endgroup$ Commented Dec 19, 2011 at 21:03
  • $\begingroup$ @Ben: Dear Ben, could you please let me knowthe fulln name of the Mathematician associated with $\text{Deodhar Decomposition}$. $\endgroup$
    – C.S.
    Commented Dec 26, 2011 at 15:08
  • $\begingroup$ Vinay Vithal Deodhar, according to genealogy.math.ndsu.nodak.edu/id.php?id=9927 . The decomposition Ben is referring to is from ams.org/mathscinet-getitem?mr=782232 $\endgroup$ Commented Dec 26, 2011 at 16:07

1 Answer 1

7
$\begingroup$

I think that this is one of these things that looks plausible in small examples but is false. For example, this would imply that the coefficients of R polynomials are alternating in $q$. This is implied by another conjecture called the Gabber-Joseph conjecture (roughly: coefficients of R-poynomials give dimensions of Ext groups between Verma modules), which is false. See "A counterexample to the Gabber-Joseph conjecture" by Brian Boe.

$\endgroup$
1
  • 3
    $\begingroup$ Besides correcting the spelling of Brian Boe's first name, I'll add a specific MathSciNet listing (it didn't get into arXiv): Boe, Brian D. (1-GA). A counterexample to the Gabber-Joseph conjecture. Kazhdan-Lusztig theory and related topics (Chicago, IL, 1989), 1–3, Contemp. Math., 139, Amer. Math. Soc., Providence, RI, 1992. I discussed this in Remark 8.11 of my 2008 book on the BGG category, hedging my bets at the end on whether there will be a nice interpretation of the R-polynomials. It still seems to be an open question. $\endgroup$ Commented Dec 19, 2011 at 21:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .