MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – Ben Webster Dec 19 '11 at 17:51
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 $\binom{4}{2} = 6$. – David Speyer Dec 19 '11 at 18:03
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. – Alexander Woo Dec 19 '11 at 21:03
@Ben: Dear Ben, could you please let me knowthe fulln name of the Mathematician associated with $\text{Deodhar Decomposition}$. – S.C. Dec 26 '11 at 15:08
Vinay Vithal Deodhar, according to . The decomposition Ben is referring to is from – David Speyer Dec 26 '11 at 16:07
up vote 5 down vote accepted

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.

share|cite|improve this answer
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. – Jim Humphreys Dec 19 '11 at 21:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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