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 écrit :
> 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.

-- 
Joël Riou - Joel.Riou@ens.fr