## Algebraically Determined Semidirect Products

Description:
Let G be a Polish group. We say that G is an algebraically determined Polish group if given any Polish group L and any algebraic isomorphism from L to G, then the algebraic isomorphism is a topological isomorphism. We will prove a general theorem that gives useful sufficient conditions for a semidirect product of two Polish groups to be algebraically determined. This will smooth the way for the proofs for some special groups. For example, let H be a separable Hilbert space and let G be a subset of the unitary group U(H) acting transitively on the unit sphere. Assume that -I in G and G is a Polish topological group in some topology such that H x G to H, (x,U) to U(x) is continuous, then H x G is a Polish topological group. Hence H x G is an algebraically determined Polish group. In addition, we apply the above the above result on the unitary group U(A) of a separable irreducible C*-algebra A with identity acting transitively on the unit sphere in a separable Hilbert space H and proved that the natural semidirect product H x U(A) is an algebraically determined Polish group. A similar theorem is true for the natural semidirect product R^{n} x G(n), where G(n) = GL(n,R), or GL^{+}(n,R), or SL(n,R), or |SL(n,R)|={A in GL(n,R) : |det(A)|=1}. On the other hand, it is known that the Heisenberg group H_{3}(R) , (R, +), (R{0}, x), and GL^{+}(n,R) are not algebraically determined Polish groups.

Date:
May 2011

Creator:
Jasim, We'am Muhammad