3
$\begingroup$

Let $(W,S)$ be a Coxeter system with length function $\ell$ and $T=\bigcup_{w\in W}wSw^{-1}$.

Set

$N(u,v):=\{t\in T: u< tu \le v\}$, $\overline{\ell}(u,v):=|N(u,v)|$,

$\ell(u,v):=\ell(v)-\ell(u)$ and $df(u,v):=\overline{\ell}(u,v)-\ell(u,v)$.

Following [1], we call $df(u,v)$ the defect of the interval $[u,v]$.

By [2, Theorem C] and the fact that coefficients of Kazhdan-Lusztig polynomials are nonnegative, we have the following:

Let $x,w\in W$ and $x\le w$. Then we have the following equivalent statements:

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

  2. $P_{y,w}(q)=1$ for all $y\in [x,w]$.

  3. $df(y,w)=0$ for all $y\in [x,w]$.

Conjecture 1: $df(x,w)=0$ $\iff$ $df(y,w)=0$ for all $y\in [x,w]$.

There are two evidences to support my conjecture 1.

First, as noted in [1], there are similarities between $P_{x,w}(1)$ and $df(x,w)$ for $x\le w$.

  1. Suppose $ws<w$ and $rw<w$. We have $P_{xs,w}(1)=P_{x,w}(1)=P_{rx,w}(1)$ and $df(xs,w)=df(x,w)=df(rx,w)$.

  2. We have $P_{x,w}(1)\ge 1$ and $df(x,w)\ge 0$.

Now the equivalence (1)$\iff$(2) suggests that the conjecture may be true.

Apart from that, it is well-known that $[q^i](P_{x,w}(q))\ge [q^i](P_{y,w}(q))$ for all $y\in [x,w]$.

This implies that $P_{x,w}(1)\ge P_{y,w}(1)$ for all $y\in [x,w]$.

This suggests my second conjecture:

Conjecture 2: $df(x,w)\ge df(y,w)$ for all $y\in [x,w]$.

Clearly, conjecture 2 is true would imply conjecture 1 is true.

Second, the conjecture 1 is true for the case $W$ is a Weyl group of type $A$.

enter image description here

Let $e_x=x\cdot B$.

Theorem 1: $X_w$ is smooth at $e_x\iff \dim X_w=\dim T_{e_x} X_w$.

Proof of Theorem 1: See [3, Remark 13.1.4].

Theorem 2: $\dim X_w=\ell(w)$.

Proof of Theorem 2: See [3, Remark after Definition 10.8.1].

Theorem 3: $X_w$ is rationally smooth at $e_x\iff df(y,w)=0$ for all $y\in [x,w]$.

Proof of Theorem 3: See [4, Section 13.2].

Theorem 4: $X_w$ is smooth at $e_x \implies X_w$ is rationally smooth at $e_x$.

Proof of Theorem 4: See [4, Chapter 6].

Theorem 5: Suppose $W$ is a Weyl group of type A. Then $\dim T_{e_x} X_w=|\{r\in T: r x\le w\}|$.

Proof of Theorem 5:

By [3, Theorem 13.2.6], we get $\dim T_{e_x} X_w=|\{s_\alpha: \alpha\in x\Phi^+, s_\alpha x\le w\}|$.

It suffices to show $\{s_\alpha: \alpha\in x\Phi^+, s_\alpha x\le w\} =\{s_\alpha\in T: s_\alpha x\le w\}$. This follows from $\varphi:x\Phi^+\cong T$ defined by $\varphi(\alpha)=s_\alpha$.

Theorem 6: Suppose $W$ is a Weyl group of type $A$. Then $df(x,w)=0\implies df(y,w)=0$ for all $y\in [x,w]$.

Proof of Theorem 6:

Suppose $df(x,w)=0$. Then $|\{r\in T: x<rx\le w\}|=\ell(w)-\ell(x)$. Note that $\ell(x)=|\{r\in T: rx<x\}|$. This implies that $\ell(w)=|\{r\in T: rx\le w\}|$. Then $\dim X_{w}=\dim T_{e_x}X_w$ by Theorems 2,5. By Theorem 1, we get $X_w$ is smooth at $e_x$. By Theorem 4, we get $X_w$ is rationally smooth at $e_x$. By Theorem 3, we get $df(y,w)=0$ for all $y\in [x,w]$. The claim follows.

How to prove/disprove Conjectures 1 and 2?


[1]: Masato Kobayashi, Combinatorics on Bruhat Graphs and Kazhdan-Lusztig Polynomials, Pure Mathematical Sciences, Vol. 3, 2014, no. 1, 23 - 33

[2]: J. Carrell, The Bruhat graph of a Coxeter group, a conjecture of Deodhar and rational smoothness of Schubert varieties, Proc. Symp. Pure Math., 56 (1994), 53-61.

[3]: Lakshmibai, Venkatramani, and Justin Brown. Flag varieties: an interplay of geometry, combinatorics, and representation theory. Vol. 53. Springer, 2018.

[4]: Sarason, I. G., Billey, S., Sarason, S., & Lakshmibai, V. (2000). Singular loci of Schubert varieties (Vol. 182). Springer Science & Business Media.

$\endgroup$
1
  • $\begingroup$ For the record, reference [4] is authored only by Billey and Lakshmibai. I don't think anyone understands why various internet sources want to add Sarason as an author, but it's presumably related to misinterpreting Billey's first name "Sara" as an extra person. This is the first time I've seen Sarason listed as an author twice though... $\endgroup$ – Oliver Oct 10 '19 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.