Martin Mathieu (Queen’s University Belfast)<\/p>\n
A contribution to Kaplansky’s problem<\/em><\/p>\n Abstract: A Jordan homomorphism between two unital, complex algebras A and B is a linear mapping T such that $T(x^2)=(Tx)^2$ for all $x\\in A$. Equivalently, T preserves the Jordan product $xy+yx$. Every surjective unital Jordan homomorphism preserves invertible elements. In 1970, Kaplansky asked whether the following converse is true: Suppose $T\\colon A\\to B$ is a unital surjective invertibility-preserving linear mapping between unital (Jacobson) semisimple Banach algebras A and B. Does it follow that T is a Jordan homomorphism?<\/p>\n In the past 50 years a lot of progress has been made towards a positive solution to Kaplansky’s problem, however, as it stands, it is still open. We will report on some recent joint work with Francois Schulz (University of Johannesburg, SA) which gives a positive answer if B is a C*-algebra with faithful tracial state. Until recently, the existence of traces had been a major obstacle to a solution. Moreover, in our approach, no assumption on the existence of projections (such as real rank zero) is necessary.<\/p>\n I will further discuss a sharpening of Kaplansky’s problem in which the assumption on T is reduced to the preservation of the spectral radius only (a spectral isometry).<\/p>\n","protected":false},"excerpt":{"rendered":" Martin Mathieu (Queen’s University Belfast) A contribution to Kaplansky’s problem Abstract: A Jordan homomorphism between two unital, complex algebras A and B is a linear mapping T such that $T(x^2)=(Tx)^2$ for all $x\\in A$. Equivalently, T preserves the Jordan product $xy+yx$. Every surjective unital Jordan homomorphism preserves invertible elements. In 1970, Kaplansky asked whether the … <\/p>\n