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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 $G$ ?

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  • $\begingroup$ 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. $\endgroup$ Nov 14, 2017 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?) $\endgroup$ Nov 15, 2017 at 0:29
  • $\begingroup$ Yes, $X$ is a variety over an algebraically closed field of characteristic 0. $\endgroup$
    – user114666
    Dec 16, 2017 at 17:32

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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 $G=F_4$.

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