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Nov 1 at 2:55 comment added user42804 Dear Prof. Knop, why is $SL(2)$ has wonderful compactification but $SL(3)$ does not?
Mar 18, 2019 at 19:22 comment added Friedrich Knop 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.
Mar 18, 2019 at 16:38 history edited LSpice CC BY-SA 4.0
Added link to De Concini and Procesi article; pointed to Pezzini article by DOI
Feb 25, 2018 at 15:43 comment added Allen Knutson Isn't de Concini-Procesi only for $G$ adjoint (so not $SL(n)$)?
Feb 4, 2018 at 22:40 vote accept CommunityBot
Feb 4, 2018 at 16:35 comment added YCor @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 $(A,B)\mapsto (A,(B^t)^{-1})$.
Feb 4, 2018 at 12:00 comment added user117617 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 $AMB^{-1}$?
Feb 4, 2018 at 1:55 history edited Michael Joyce CC BY-SA 3.0
deleted 17 characters in body
Feb 4, 2018 at 1:31 history answered Michael Joyce CC BY-SA 3.0