Uniqueness of the wonderful compactification of a semi-simple group Let $G$ be a semi-simple group over an algebraically closed field of characteristic zero. In which cases there is a unique wonderful compactification of $G$ (modulo isomorphism)?
For instance, is the wonderful compactification of $SL(n,\mathbb{C})$ unique?
 A: I assume you mean the variety $G$ considered as a $G \times G$ variety via the action $(g,h) \cdot x = gxh^{-1}$, which is the standard interpretation in the literature. The variety $G$ is spherical as a $G \times G$ variety, meaning that it contains a dense $B \times B$-orbit (where $B$ is a Borel subgroup of $G$). This statement is the content of the well-known Bruhat decomposition of $G$.
There is a more general statement: If $X$ is a spherical $G$-variety, where $G$ is reductive algebraic group over an algebraically closed field of any characteristic, then if $X$ admits a wonderful compactification, it is unique up to $G$-isomorphism.
The variety $G$ does have a wonderful compactification, which is therefore unique up to $G \times G$-isomorphism. The existence of the wonderful compactification of $G$ goes back to the original work on wonderful compactifications, Complete symmetric varieties by De Concini and Procesi (MSN). I don't know where the first proof of uniqueness appears, but a good introduction to the subject that includes a full proof is Pezzini's survey article, Lectures on spherical and wonderful varieties.
Here's a sketch of the main ideas involved. Any wonderful compactification of $G$, which I will denote $Y$, contains a big cell $Y_0$ with the following properties:


*

*$Y_0$ is dense in $Y$

*$Y_0$ is $B \times B$-invariant and isomorphic to affine space

*$G \cdot Y_0 = Y$
So suppose $Y$ and $Y'$ were wonderful compactifications of $G$ with big cells $Y_0$ and $Y'_0$. Let $g, g' \in G$ and consider the unique birational $G \times G$-equivariant map $f: Y \dashrightarrow Y'$, i.e. $f(g_1 g g_2^{-1}) = g_1 g' g_2^{-1}$. This map induces an isomorphism $Y_0 \cong Y'_0$ because they have isomorphic coordinate rings (an argument is needed why the map can be extended to the big cell). Then $f$ extends to an isomorphism $Y \cong Y'$ since $(G \times G) \cdot Y_0 = Y$.
