6
$\begingroup$

In Bourbaki's General Topology, Chapitre III §6 Exercise 11, they define a Gelfand Ring as a topological ring $A$ such that

  1. The set $A^*$ ($=A^{-1}$) of invertibles is open.
  2. The uniform structure induced on $A^*$ by the given one on $A$ is compatible with the group structure.
Of course, the product being continuous, this amounts to only ask that $x\mapsto x^{-1}$ be continuous. My question is (as commutativity is not required):

Is there a link between this notion and the one commonly accepted in commutative algebra? (see https://en.wikipedia.org/wiki/Gelfand_ring)
Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ It looks more like a generalization of Banach algebras — for which using Gelfand's name seems quite natural. $\endgroup$
    – abx
    May 17, 2021 at 10:15
  • $\begingroup$ Right, note that although these (sometimes noncommutative) Gelfand rings form a category, it is not closed by projective limits. For example, the space of analytic functions $\mathcal{H}(\Omega)$ over a domain $\Omega\subset\mathbb{C}$ is a limit of Banach algebras but, in general, not a Gelfand ring (if I am not mistaken). $\endgroup$
    – Duchamp Gérard H. E.
    May 17, 2021 at 12:34

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.