From: jriou@clipper.ens.fr (Joel Riou) Subject: Re: Automorphis group of GL_n(K) Date: 13 Feb 2001 17:16:00 GMT Newsgroups: sci.math tim_brooks@my-deja.com, dans le message (sci.math:393299), a �crit : > Let K be a field and n>=2. > What is the automorphism group of GL_n(K) ? If you want to know the automorphisms of GL_n(K) as an abstract group, it might be strange because, for instance any automorphism of K as a field would provide an automorphism of GL_n(K). But, if you want to know the group of automorphism of GL_n(K) as an algebraic group defined over K then the answer is more interesting (as far as i am concerned). If we denote by G the group of automorphisms of GL_n(K) (as an algebraic group), I claim that we have the following split exact sequence of groups : 1 -> Gl_n(k)/diagonal invertible matrices -> G -> Z/2Z -> 1 (where the non trivial element of Z/2Z act on GL_n(k)/k^* by the formula g -> transpose(g^-1)). In fact, any automorphism of GL_n(K) would be an interior automorphism (the conjugation by an element of GL_n(K)) or the composition of an interior automorphism and the automorphism g -> transpose(g^-1). Proof : let A denote the coordinate ring of M_n(K) (i.e. A=k[X_ij,1<=i,j<=n]), and B=A[Det^-1], the ring of regular functions on GL_n(K). Let Phi : B -> B by an automorphism of K-algebra which induces an automorphism of GL_n(K) (F:Gl_n(K)->GL_n(K)) as an algebraic group. Consider the regular function of GL_n(K) defined by f(x)=det(F(x)), f is clearly a morphism of algebraic groups Gl_n(K) -> k^*, and it is not hard to show that f(=Phi(Det)) must be of the form Det^n with n a nonzero integer. If you use this result with F and F^-1, you can prove that Phi(Det)=Det or Phi(Det)=Det^-1. For the rest of the discussion, we may suppose that Phi(Det)=Det (swap F and F'=( g -> transpose(F(g^-1)))). Then, we can prove that the automorphism Phi: A[Det^-1] -> A[Det^-1] is induced by an automorphism of A (it just use that fact the the ring K[X_ij] is factorial). So we have a morphism of algebraic varieties G : M_n(K) -> M_n(K) whose restriction to GL_n(K) is the automorphism F. So G is multiplicative on the Zariski-opened dense subset Gl_n(K), so G is multiplicative (i.e. G(AB)=G(A)G(B)). The center of Gl_n(K) must be mapped with the center of Gl_n(K) by F, so G induces an automorphism of the subvariety D of diagonal matrices in M_n(K). D is canonically isomorphic with the variety K. Furthermore, the map D -> D (restriction of G) is multiplicative, so it is easy to see that it must be the identity of D. But now, the relation G(AB)=G(A)G(B) with A diagonal means that G is homogenous of degre 1. Then G induces an automorphism of M_n(K) as a K-algebra. It is a good exercice to show that all automorphisms of M_n(K) as a K-algebra are of the form M -> gMg^-1 where g belongs to GL_n(K), q.e.d. -- Jo�l Riou - Joel.Riou@ens.fr