The first non-trivial example of a uniform algebra which comes to mind is the disc algebra $A(\mathbb{D})$. In a similar manner one can define its relatives $P(U)$ and $R(U)$, where $U$ is any region on the complex plane and $P$, $R$ stand for the closures, of respectively polynomials and rational functions on $U$.

What if we go beyond the case when spectrum is contained in the complex plane? Rudin proved, that if $K$ is scattered, then there are no non-trivial uniform subalgebras of $C(K)$.

Are there any classical or canonical examples of uniform algebras with non-metrizable spectra or is the theory of uniform algebras just a daughter of complex analysis?

Forgive me if this question seems to be to vague.

Addendum: A uniform algebra is a closed subalgebra of a commutative C*-algebra $C(K)$ which contains constant functions and separates points in $K$.

  • $\begingroup$ It might help if you include the definition of a uniform algebra. Just click the "edit" button to add it. $\endgroup$ – MTS Mar 13 '12 at 20:31
  • $\begingroup$ Contra MTS, I think that people who might be interested in this question, or able to answer it, know full well what a uniform algebra is. No harm done, in any case. $\endgroup$ – Yemon Choi Mar 13 '12 at 22:00
  • $\begingroup$ Yemon, sure. I mean that all of them are $C(\Omega)$s. $\endgroup$ – Alex Ortega Mar 13 '12 at 22:14

Here are some scattered thoughts: I'm not sure if they really answer what you seem to be driving at.

Complex analysis still gives you examples with nasty maximal ideal spaces, e.g. $H^\infty(\Omega)$. Even when $\Omega$ is the open unit disc, the spectrum is non-metrizable and not at all straightforward to understand (there ought to be some discussion in Gamelin's book).

$L^\infty$ has non-metrizable spectrum, although its structure is in some sense not as mysterious as that of $H^\infty$.

Note that every Banach space $E$ embeds as a closed, complemented subspace of a uniform algebra. Namely, take the canonical map from $E$ into $C(B)$, where $B$ is the closed unit ball of $E^*$ equipped with the weak$*$-topology, and let A be the closed subalgebra generated by the image of $E$. (This is a theorem of Milne; I learned of it from an article of Gamelin and Kislyakov, of which a preprint version can be found online.)

  • $\begingroup$ I think you meant "every Banach space E embeds as a closed COMPLEMENTED subspace of a uniform algebra." $\endgroup$ – Bill Johnson Mar 13 '12 at 23:25

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