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 , Joerg Knappen 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