Symmetry of generalized group algebras

Abstract

A Banach* -algebra is symmetric if elementsf* *f have nonnegative spectrum. This is the case if and only if hermitian elements have real spectra [5]. We first observe that it suffices to show that A is symmetric iff B1(G, A) is symmetric, because B1(G, A) symmetric ==>BP(G, A) symmetric for any p, 1 ?p A symmetric ==>B 1(G, A) symmetric. The proof of the first of these implications will be accomplished by showing that if tEBP(G, A)CB'(G, A) then the spectrum of t in BP(G, A), op(t) equals the spectrum of t in B'(G, A), ol(t). Since BP(G, A)CB1(G, A), clearly a1(t)C p(t). On the other hand, BP(G, A) is an ideal in B'(G, A) [6]. Recall that Of X(EII(t) iff t/X has a quasiinverse, yx, say, in B1(G, A) [4, p. 28]. t/X and yx satisfy the relationship