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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?

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    $\begingroup$ Wonderful compactification is for an algebraic group with an action on a variety. When you just refer to $G$, what is the group and what is the variety? $G$ acting on itself by left translation? $G$ acting on itself by conjugation? $G\times G$ acting on $G$ by left-right translation? $\endgroup$
    – YCor
    Commented Feb 4, 2018 at 0:04
  • $\begingroup$ For the definition: en.wikipedia.org/wiki/Wonderful_compactification $\endgroup$
    – YCor
    Commented Feb 4, 2018 at 16:37

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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:

  1. $Y_0$ is dense in $Y$
  2. $Y_0$ is $B \times B$-invariant and isomorphic to affine space
  3. $G \cdot Y_0 = Y$

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 $(G \times G) \cdot Y_0 = Y$.

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  • $\begingroup$ Thanks a lot for the answer. Indeed I was considering $G = SL(n)$ and the action of $G\times G$ on $G$ given by $((A,B),M)\mapsto AMB^{t}$ (where $B^{t}$ is the transpose of $B$). Is it still fine if we consider this action instead of $AMB^{-1}$? $\endgroup$
    – user117617
    Commented Feb 4, 2018 at 12:00
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    $\begingroup$ @J_Cole Yes, because it's just another system of coordinates for the same action. More formally, let $H_1$ acts on $X,Y$ and $f:X\to Y$ is $H_1$-equivariant. If $H_2$ is another algebraic group and $u:G\to H$ an isomorphism, then $H_2$ acts on $X,Y$ through $u$ and $f$ is $H_2$-equivariant, and vice-versa. So $X\to Y$ is a wonderful compactification for the $H_1$-action iff it's a wonderful compactification for the $H_2$-action. Now apply this to $G=SL(n)$, $H_1=H_2=G\times G$ and $u$ the automorphism given by $(A,B)\mapsto (A,(B^t)^{-1})$. $\endgroup$
    – YCor
    Commented Feb 4, 2018 at 16:35
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    $\begingroup$ Isn't de Concini-Procesi only for $G$ adjoint (so not $SL(n)$)? $\endgroup$
    – Allen Knutson
    Commented Feb 25, 2018 at 15:43
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    $\begingroup$ Allen is right: $SL(n)$ has no wonderful compactification unless $n=2$. This is because wonderfulness entails smoothness. Nevertheless, any semisimple group has has a certain canonical compactification which is wonderful if it happens to be smooth. That's the case for all adjoint groups, for $SL(2)$ and maybe some others. $\endgroup$
    – Friedrich Knop
    Commented Mar 18, 2019 at 19:22
  • $\begingroup$ Dear Prof. Knop, why is $SL(2)$ has wonderful compactification but $SL(3)$ does not? $\endgroup$
    – user42804
    Commented Nov 1 at 2:55

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