Let $R$ be a topological ring of characteristic zero. Assume that $R$ is commutative. Let $G :R \times R \to R$ be a continuous function satisfying $G(G(x,y),z)=G(x,G(y,z))$ and $G(0,x)=x$, for all $x,y,z \in R$.

If we assume that $G$ is a convergent power series then $G$ is also a formal group law and the Honda's argument(cf. Hazewinkel, "Formal Groups and Applications", Chapter 1, 5.8, p.37) proves that G admits a logarithm, i.e. $G(x,y)=h^{-1}(h(x)+h(y))$.

Question : Does $G$ admit a logarithm if it is only continuous?

EDIT Assuming that $G$ is continuous, what are the additional assumptions that should be made on $R$ and $G$ to guarantee the existence of logarithms?

  • 3
    $\begingroup$ But if $G$ is only continuous then the ring structure plays no rôle at all $\endgroup$ – მამუკა ჯიბლაძე Aug 9 '20 at 18:18
  • 3
    $\begingroup$ As @მამუკაჯიბლაძე says: If $R $ has the discrete topology, then continuity means nothing, and we are just requiring that $G $ is a binary operation that makes $R $ a semigroup with left identity $0$. Such operations exist in abundance; not all of them are groups, so not all of them have logarithms. $\endgroup$ – darij grinberg Aug 9 '20 at 18:22
  • $\begingroup$ What about $R=Q_p$? $\endgroup$ – A413 Aug 9 '20 at 18:40
  • 1
    $\begingroup$ For $Q_p$ you get arbitrary topological semigroup structure on the Cantor set with a point removed. $\endgroup$ – მამუკა ჯიბლაძე Aug 9 '20 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.