From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,sci.physics
Subject: Re: The square root of complex conjugation
Date: 9 Nov 1997 06:53:19 GMT

In article <640gqb$10c@gap.cco.caltech.edu>,
Toby Bartels <toby@ugcs.caltech.edu> wrote:
>C and I are the only multiplicative, additive, measurable operators
>*on the field of complex numbers*.
>But conjugation by any invertible matrix is such an operator
>*on the algebra of matrices*.
>I don't know if there are any others.

I haven't been following this thread so I can only hope this comment is
on-target, but:

All nonzero R-linear ring endomorphisms on the matrix ring  M_n(R)  are inner,
that is, of the form  f(x) = y^(-1) x y  for some invertible  y  in M_n(R).

You didn't actually specify "R-linear", above, but  R  is the center of this
matrix ring, and so is invariant under automorphisms anyway; if you are
interested in some class of maps which, when restricted to  R,  can only
be the identity map, then any such map which is multiplicative on
M_n(R)  is in fact R-linear anyway.

Incidentally, there is a related collection of theorems of the form,
	"If  f : GL(n,R) -> GL(n,R) is a [...] homomorphism, then it
	is of the form  f(x) = (det x)^s |det x|^t h(x)  for some real
	s  and  t,  where either h=identity or h(x)=(x^t)^(-1)."
but I don't think a very weak hypothesis is possible in [...] since
for example we have this endomorphism of  GL(2,R):
		[ 1 log(|det(x)|) ]
	x   ->	[                 ]
		[ 0      1        ]

Perhaps this family of results can be of some use for your question on
the algebras, assuming you're talking about multiplicative maps taking
identity to identity, and thus invertible elements to invertible elements.

dave