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What are the automorphisms of $SL_n$ as an algebraic variety?

In other words, let $k$ be an algebraically closed field of characteristic 0 (e.g., $k=\mathbb{C}$). Let $\tau$ be an automorphism of $SL_n$ regarded as an algebraic variety over $k$. Assume that $\tau$ takes the unit element $e$ of $G$ to itself. Is it true that $\tau$ is an automorphism of $SL_n$ as an algebraic group over $k$?

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    $\begingroup$ What about inversion (for $n>1$)? $\endgroup$
    – ACL
    Jan 15, 2013 at 16:27
  • $\begingroup$ @ACL: Thank you, Antoine. Are there any other automorphisms? $\endgroup$ Jan 15, 2013 at 16:33
  • $\begingroup$ @Mikhail: Your edit needs some more editing. Aside from this, what motivates the original question? $\endgroup$ Jan 15, 2013 at 17:04
  • $\begingroup$ @Jim: I have removed the edit. The original question was motivated by my previous question mathoverflow.net/questions/118356/… and a comment of Tom Goodwillie. I am trying to construct a finite subgroup $H\subset G=SL_{n,\mathbb{C}}$ and an automorphism $\sigma$ of $\mathbb{C}$ such that the $\mathbb{C}$-varieties $G/H$ and $\sigma(G/H)=G/\sigma H$ are not isomorphic. $\endgroup$ Jan 15, 2013 at 17:50
  • $\begingroup$ Note that the group generated by automorphisms, left translations and inversion is finite-dimensional (actually $2(n^2-1)$); while the example by Mariano gives a faithful action of an infinite dimensional abelian group. $\endgroup$
    – YCor
    Jan 15, 2013 at 18:10

2 Answers 2

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The coordinate ring when $n=2$ is $A=k[a,b,c,d]/(ad-bc-1)$.

If $f\in k[b,c]$, there is an automorphism $\phi:A\to A$ such that $\phi(a)=a+bf$, $\phi(c)=c+df$, $\phi(b)=b$ and $\phi(d)=d$.

One could conjecture that the automorphism group in this case is generated by $SL_2$, inversion and this sort of triangular automorphisms, much as in the Makar-Limanov–Jung–van der Kulk theorem for $k[x,y]$ (This is a very optimistic conjecture, though: this is a $3$-dimensional affine variety quite close to affine space and there are non-tame automorphisms of the latter...)

In general, I doubt we know the automorphism group.

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  • $\begingroup$ Of course, this trick works for all $n$. $\endgroup$ Jan 15, 2013 at 16:44
  • $\begingroup$ The polynomial f should be inside k[b, d]. $\endgroup$
    – Anonymous
    May 16, 2018 at 8:51
  • $\begingroup$ In fact, there are so-called "wild" automorphisms of k[a, b, c, d] / (ad-bc-1), i.e. automorphisms which do not belong to the subgroup generated by SL_2, inversion and the above mentioned "triangular" automorphisms. This is done in projecteuclid.org/euclid.jmsj/1359036456 $\endgroup$
    – Anonymous
    May 16, 2018 at 8:53
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The automorphism group is massive!

Flexible varieties and automorphism groups, I. Arzhantsev, H. Flenner, S. Kaliman, F. Kutzschebauch, M. Zaidenberg, http://arxiv.org/abs/1011.5375.

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