# 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?

• 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? – YCor Feb 4 '18 at 0:04
• For the definition: en.wikipedia.org/wiki/Wonderful_compactification – YCor Feb 4 '18 at 16:37

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

• 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 $AMB^{-1}$? – J_Cole Feb 4 '18 at 12:00
• @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 $(A,B)\mapsto (A,(B^t)^{-1})$. – YCor Feb 4 '18 at 16:35
• Isn't de Concini-Procesi only for $G$ adjoint (so not $SL(n)$)? – Allen Knutson Feb 25 '18 at 15:43
• 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. – Friedrich Knop Mar 18 at 19:22