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Suppose $G$ is a connected semisimple linear algebraic group with Lie algebra $\mathfrak{g}$ and $X$ is a homogeneous $G$-space with isotropy subgroup $H$ (associated Lie algebra $\mathfrak{h}$) that is normalised by a parabolic subgroup $P$. Write $A=P/H$.

Working over $\mathbb{C}$, Borho and Brylinski explain in Proposition 2.8 of "Differential Operators on Homogeneous Spaces I" how to use the moment map $T^\ast X\rightarrow \mathfrak{g}^\ast$ to induce a map from $\pi\colon T^\ast X/A\rightarrow \mathfrak{g}^\ast$. We may also understand this map as a map from the vector bundle $G\times^P\mathfrak{h}^\perp$ to $\mathfrak{g}^\ast$. It is a generalisation of the Springer resolution (which arises in case $H=P$ is a Borel).

The same construction can be made over an algebraically closed field of characterstic $p>0$. My question is what is known about the dimensions of the fibres of $\pi$ in this characteristic $p$ case? I am most interested in knowing the largest possible dimension of a fibre over a non-zero point in $\mathfrak{g}^\ast$ in the case where $P$ is a Borel and $H$ is its unipotent radical but more general results where $P$ is any parobolic and $H$ is its unipotent radical are also of interest. I am not interested in the case $H=P$.

Added following request for motivation and examples:

My motivation comes from a project aiming to understand the possible dimensions (that is the canonical dimension defined for Auslander-Gorenstein rings that corresponds intuitively but not precisely to GK-dimension for almost commutative algebras) of simple modules for the localisation of Iwasawa algebras $\mathbb{Z}_p[[G]]$ at the m.c. set generated by $p$ where $G$ is a compact $p$-adic Lie group of semisimple type. At the moment the answer we have is a simple function of the dimension of these fibres when the group $G$ in the question is the associated Lie group over the algebraic closure of $\mathbb{F}_p$. It would be good to give precise values.

We believe that it suffices to know the answer when $P$ is a Borel and $H$ is its unipotent radical which is why I am most interested in this case.

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Tiny edit. Aside from that, some hint about motivation would be helpful as well as maybe a small example. –  Jim Humphreys Sep 16 '10 at 22:16

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