Abstract

We say that the completion theorem holds for a uniform space \((X,\fancyscript{U})\) if, for every uniformly continuous function \(\alpha :(X,\fancyscript{U})\rightarrow (X,\fancyscript{U})\), the Bowen entropy of \(\alpha \) coincides with the Bowen entropy of \(\widetilde{\alpha }\), the extension of \(\alpha \) to the completion \((\widetilde{X},\widetilde{\fancyscript{U}})\) of \((X,\fancyscript{U})\). We study the completion theorem in the realm of abelian topological groups. Namely, we prove that it fails to be true in a drastic way by showing that every (abstract) abelian group \(G\) can be endowed with a totally bounded group topology \(\tau \) such that the topological group \((G,\tau )\) has endomorphisms of zero entropy whose extension to the Raĭkov completion of \((G,\tau )\) has infinite entropy. Our proof uses the structure theorems for abelian groups, the properties of the Bohr topology and Pontryagin duality. The case of metrizable groups is also analyzed in the case of Bernoulli shifts of finite groups, dense subgroups of the circle \(\mathbb T\) and the reals \(\mathbb R\). The key tool is the so-called \(e\)-supporting family of an endomorphism \(\alpha \) of an abelian metrizable group \(G\) with respect to a neighborhood \(U\) of the neutral element of \(G\). Several open questions are proposed.

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