In what circumstances does a given group carry a Polish group topology? (A metric, separable, complete topology is called Polish, and a topology is a group if multiplication and inverse are continuous.) Two natural interpretations of this question can be considered.

1. We consider a group equipped only with its algebraic structure. This algebraic structure may or may not be compatible with the possibility of defining a Polish group topology on the group.

2. The group is endowed with a σ-algebra Σ of subsets, and the multiplication (regarded as function of two variables) and inverse operations axe assumed to be measurable with respect to Σ, that is, preimages of sets from Σ are in the product σ-algebra Σ × Σ in case of multiplication and in Σ in case of inverse. Now, we would like to investigate the possibility of putting a Polish group topology on the group whose Borel sets coincide with Σ.

Problem 2 is more “canonical” than 1 in the following sense. A group can carry many Polish group topologies compatible with its algebraic structure. Take for example the reals, R, with their natural topology and Rw with the product topology. Both these groups are topological groups with their respective Polish topologies. Now, R and R and Rω are isomorphic as groups (both of them are linear spaces over the rationals, Q, with bases of cardinality continuum) but the topologies are clearly different (R is locally compact while Rw is not).