2
$\begingroup$

My question: is it true that we can define a spectrum of $\mathbb{C}$-algebra $A$ in such a way that it becomes a complex manifold with the algebra of holomorphic functions $A$? Maybe it will work if we restrict to algebras that already are an algebra of holomorphic functions for some manifold?

I'm aware of the construction of Gelfand spectrum but it gives us only continuous functions on the spectrum and not holomorphic functions...


It is a cross-post from math.exchange https://math.stackexchange.com/questions/3510203/holomorphic-spectrum

$\endgroup$
  • $\begingroup$ If you crosspost on mathstackexchange and mathoverflow, please say so in your post and provide a link (on both sites). $\endgroup$ – MaoWao Jan 17 at 10:56
  • $\begingroup$ @MaoWao thank you so much for explaining this rule! $\endgroup$ – Gregory G Jan 17 at 17:07
3
$\begingroup$

In some cases the spectrum of a commutative Banach algebra $\mathcal A$ may contain "analytic disks" on which the Gelfand transforms of members of $\mathcal A$ correspond to analytic functions. You might look at section 1.5 of Andrew Browder, "Introduction to Function Algebras".

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.