# Do Richardson varieties have rational singularities in arbitrary characteristic?

The title basically asks the question. I'll review the relevant terminology and explain what I have and haven't found in the literature.

Let $G$ be a reductive group. Let $v \leq w$ be elements of the Weyl group, with $X_v$ and $X^w$ the corresponding Schubert and opposite Schubert. Let $R_v^w = X_v \cap X^w$. This is called a Richardson variety. A variety $R$ is said to have rational singularities if there is a resolution of singularities $Z \to R$ which has certain nice cohomological properties.

I know of two references in the literature for the fact that Richardson varieties have rational singularities: Theorem 4.2.1 in Michel Brion's "Lectures on the Geometry of Flag Varieties" and Lemma 2 in his "Positivity in the Grothendieck Group of Flag Varieties". Both of these essentially give the same proof, though in different language. In the notation of "Lectures", they construct a variety $Z_{\underline{v}}^{\underline{w}}$ with a map to $R_v^w$ and proof that it has the correct cohomological properties. This part of the proof works in any characteristic.

However, in order to show that $Z_{\underline{v}}^{\underline{w}}$ is smooth, they appeal to Kleiman transversality, or to generic smoothness. These arguments only work in characteristic zero.

Does anyone know a reference which addresses this? (I have e-mailed Brion, and also Kumar, and am waiting to hear back, but I figured someone here might know this.)

• David, just curious, how do you define rational singularities in characteristic $p > 0$. – Karl Schwede Aug 18 '11 at 21:04
• Ah, a good question, I recently got this wrong. The right definition (see for example Brion and Kumar's book, section 3.4) is that there is a proper birational map $f: X \to Y$, with $X$ nonsingular, $f_{\ast} \mathcal{O}_X =\mathcal{O}_Y$ and $R^i f_{\ast} \mathcal{O}_X = R^i f_{\ast} \omega_X=0$ for $i>0$. – David E Speyer Aug 18 '11 at 21:10
• Does Frobenius splitting say something about rational singularities? – Alexander Woo Aug 19 '11 at 2:43
• It can be used to prove things like that, but for this particular question, I'm curious what Shrawan Kumar says. Certain Frobenius splittings (ie, global F-regularity) is enough to prove $f_* \omega_X = \omega_Y$ but it doesn't give you the higher $R^i f_* \omega_X = 0$ without knowing the existence of a resolution with a certain form. Perhaps I should point out that anything that is Frobenius split in char. $p > 0$ has something like Du Bois singularities (which can be viewed as a limit of rational singularities). – Karl Schwede Aug 19 '11 at 14:30