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Perhaps the first example in Teichmuller theory is the following proposition:

Proposition: Let $1<r<R$. Then two annular region $U_r=\{z\in \mathbb{C}\bigm|1<|z|<r\}$ and $U_R=\{z\in \mathbb{C}\bigm|1<|z|<R\}$ are not bio holomorphic.

We consider the following two Banach algebras:

$A_r=\{ f:\overline{U_r}\to \mathbb{C}\bigm|f\quad \text{is continuous and is holomorphic in the interior of this closed annular region}\}$

$A_R=\{ f:\overline{U_R}\to \mathbb{C}\bigm|f\quad \text{is continuous and is holomorphic in the interior of this closed annular region}\}$

Are these two complex Banach algebeas isomorphic algebras?

Note: In the literature are there some researchs which apply operator algebraic methods for consideration of some problems in Teichmuller theory?

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    $\begingroup$ An interesting question, but are you sure it has to do anything with noncommutativity? $\endgroup$
    – მამუკა ჯიბლაძე
    Jun 14, 2021 at 22:35
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    $\begingroup$ The spectrum of $A_r$ is $\overline{U_r}$, isn't it? Hence any isomorphism gives rise to a homeomorphism, which must be holomorphic. $\endgroup$
    – Narutaka OZAWA
    Jun 15, 2021 at 0:30
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    $\begingroup$ @NarutakaOZAWA is right: I suspect something like this result must be mentioned in one of the classical books on function algebras, such as Gamelin's book on Uniform Algebras, or maybe Stout's book. This is Gelfand theory for Banach algebras and has nothing to do with "noncommutative geometry" $\endgroup$
    – Yemon Choi
    Jun 15, 2021 at 4:13
  • $\begingroup$ @NarutakaOZAWA I am familiar with the Gelfand theory. so every isomorphism gives a homemorphism between maximal ideak spaces. But what is the reason this homeomorphisms must be holomorphic in the interior? $\endgroup$
    – Ali Taghavi
    Jun 15, 2021 at 10:43
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    $\begingroup$ If you have a more general question concerning noncommutative algebras which could provide some analogue of complex function theory or Teichmueller theory, then you could post that as a separate question. My point is that the particular question you asked has an easy answer which is nothing to do with "noncommutative X" $\endgroup$
    – Yemon Choi
    Jun 15, 2021 at 13:55

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