The topological center of the semigroup of free ultrafilters

Abstract

ABSTRACT. For any group G, it is proved that the topological center of the semigroup of free ultrafilters on the group G is empty.

Slowly oscillating functions in semigroup compactifications and convolution algebras

“…We may consider x as an ultrafilter on G and as such it is κ-complete. It is straightforward to show that the multiplication in βG is jointly continuous at (y, x) for every y in βG with ρ(y) < κ (see [33,Lemma 3]). Pick y from V such that ρ(y) < κ.…”

“…(v) In 1998, Protasov proved that the topological centre of the remainder βG \ G is empty [33]. His proof is very combinatorial, and relies partly on P-points (or more precisely on non-P-points) in βG for countable groups.…”

Banach algebras on semigroups and on their compactifications

“…Then one can ask whether, for u ∈ βS, the continuity of L u : βS → βS at v forces u to belong to S. (This is a stronger hypothesis than that {v} is determining for the left topological centre of βS.) Such a point v has been exhibited by Protasov [117,Theorem 1] in the case where S is a countable group.…”