From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.physics.research,sci.physics,sci.math
Subject: Re: This Week's Finds in Mathematical Physics (Week 64)
Date: 1 Oct 1995 06:07:47 GMT

In article <DFqnw2.Apn@murdoch.acc.virginia.edu>,
Joerg Knappen <knappen@vkpmzd.kph.Uni-Mainz.DE> wrote:
>Seeing matrices with quaternionic or even octonionic entries,
>I have the question how the multiplication of those matrices can be
>consitently defined, and how one defines determinants of such beasts.

Products are defined in the usual way: (AB)_{ij}=Sum(a_{ik}b_{kj}).
Of course over non-associative rings you can't expect associative products
among their matrices.

For determinants, I guess the usual trick for associative algebras  A  is
to embed them in matrix rings over a commutative subfield  B  in the center
of  A  (that is,  B \subseteq A \subseteq M_n(B).) Then a matrix ring over
A, say  M_m(A),  may be embedded in  M_m( M_n ( B )),  which is naturally
isomorphic to  M_(mn) ( B ). Then everything in  M_m(A)  may be assigned
a determinant, which will be an element of  B  (not, unfortunately, of  A).
Roughly speaking, the determinant so defined is the norm_(A->B) of what
the determinant ought to be if you _could_ define the determinant as
an element of  A, but the problem is that there is no way to assign
such a determinant which preserves the product rule. There are no
semigroup homomorphisms  M_m(A) -> A, so you have to ask just what
properties you intended this 'determinant' to retain.

A related question is to ask about the existence of traces, even for
infinite dimensional algebras. About this question all I can recall is
that the issue arose in some K-theory and cyclic-homology discussions,
my notes from which have vanished without a, um, trace.

dave




