6
$\begingroup$

A spherical variety is a normal variety $X$ together with an action of a connected reductive affine algebraic group $G$, a Borel subgroup $B\subset G$, and a base point $x_0\in X$ such that the $B$-orbit of $x_0$ in $X$ is a dense open subset of $X$.

A wonderful variety is a smooth complete variety $X$ with the action of a semisimple simply connected group $G$ such that there is a point $x_0\in X$ with open $G$ orbit and such that the complement $X\setminus G\cdot x_0$ is a union of prime divisors $E_1,\cdots, E_t$ having simple normal crossing, and such that the closures of the $G$-orbits in $X$ are the intersections $\bigcap_{i\in I}E_i$ where $I$ is a subset of $\{1,\dots, t\}$.

Now, fix a connected reductive affine algebraic group $G$ and a Borel subgroup $B\subset G$. Could there exist two non isomorphic smooth complete varieties that are spherical with respect to $(G,B)$ and wonderful with respect to $G$ ?

$\endgroup$
  • $\begingroup$ wouldn't already G/P for two different parabolic subgroups P fit the bill? They are already complete, so there is no condition on boundary divisors. There are some easy non-isomorphic examples, e.g. Grassmannians Gr(k,n) for different k. $\endgroup$ – Konrad Voelkel Nov 14 '17 at 10:21
  • $\begingroup$ @Misa: I guess your question is posed over an algebraically closed field of characteristic 0 such as $\mathbb{C}$? (Does the characteristic matter?) $\endgroup$ – Jim Humphreys Nov 15 '17 at 0:29
  • $\begingroup$ Yes, $X$ is a variety over an algebraically closed field of characteristic 0. $\endgroup$ – J. Ross Dec 16 '17 at 17:32
8
$\begingroup$

The only groups which act on only one wonderful variety are tori (with $X$ being a point). All other admit at least $X=G/B$ and $X=G/G$.

If one fixes the open $G$-orbit then there is at most one wonderful completion (Luna-Vust, Luna).

It is known that the number of wonderful varieties for $G$ is finite (work of Alexeev-Brion). By now, they are all classified (work of mostly Luna, Losev, and Bravi-Pezzini). The result is way too complicated to describe it here. Bravi and Luna have written a nice survey on the case where $G=F_4$.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

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