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Let $A$ be a unital banach algebra generated by two elements $1$ and $x$. Then it seems $\sigma(x)$ cannot have holes. At least this is true in the case for disk algebras.

This amounts to prove that $\mathbb{C}\backslash\sigma(x)$ is connected. I tried something but failed.

Can somebody give a hint? Thanks!

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There is a generalization of this in A. Browder's "Cohomology of maximal ideal spaces". –  Jonas Meyer Dec 4 '12 at 3:05
    
At least when the sets are well-behaved, this is a generalization in view of Alexander's duality, right? This is extremely cool! –  levap Dec 4 '12 at 12:26
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up vote 6 down vote accepted

Denote by $\mathrm{sp}(A)$ the Gelfand spectrum of $A$. First, show that the Gelfrand transform $\hat{x} : \mathrm{sp}(A) \rightarrow \sigma(x)$ of $x$ is in fact a homeomorphism of topological spaces. This also implies that you can identify $C(\sigma(x))$ with $C(\mathrm{sp}(A))$ with the pullback.

Now, given $\lambda \notin \sigma(x)$, you know that $(x - \lambda)$ is invertible and so $(x - \lambda)^{-1} \in A$. Because $A$ is generated by a single element $x$, this implies that there are polynomials $p_n(z)$ such that $p_n(x) \to (x - \lambda)^{-1}$, inside $A$. Taking the Gelfand transform, you also have $\widehat{p_n(x)} \to \widehat{\frac{1}{x - \lambda}}$ inside $C(\mathrm{sp}(A))$. Using the identification above, you obtain in fact a sequence of complex polynomials $p_n(z)$ that converges uniformly over $\sigma(x)$ to the holomorphic function $\frac{1}{z - \lambda}$.

Assume that $\mathbb{C} \setminus \sigma(x)$ is not connected and decompose it as $$ \mathbb{C} \setminus \sigma(x) = \Omega_{\infty} \sqcup \Omega $$ where $\Omega_{\infty}$ is the open unbounded connected component and $\Omega \neq \emptyset$ is an open bounded (not necessarily connected) set which is the union of all other connected components of $\mathbb{C} \setminus \sigma(x)$. Define $K = \sigma(x) \sqcup \Omega$ (the spectrum together with the holes). Then, as $\partial{\Omega} \subset \sigma(x)$, the set $K$ is closed and bounded, hence compact and we have $\partial{K} = \partial{\sigma(x)}$. This, together with the maximum modulus principle, implies that for any polynomial $p(z)$ we have $$ \max_{z \in \sigma(x)} |p(z)| = \max_{z \in K} |p(z)|. $$

Let $\lambda \in \Omega$. Take the sequence $p_n(z) \to \frac{1}{z - \lambda}$ we have constructed above that converges uniformly over $\sigma(x)$. Let $m = \max_{z \in \sigma(x)} |{z - \lambda}|$. Then, we have $n$ such that $$ \max_{z \in \sigma(x)} |p_n(z) - \frac{1}{z - \lambda}| < \frac{1}{m}. $$ But this, together with the maximum modulus principle, implies that $$ \max_{z \in \sigma(x)} |(z - \lambda) p_n(z) - 1| = \max_{z \in K} |(z - \lambda) p_n(z) - 1| < 1.$$ Plugging in $z = \lambda$ we have $1 < 1$ which contradicts our decomposition.


Let us put some context to this argument. In general, given a compact subset $K \subset \mathbb{C}$, the complex polynomials are not going to be dense in $C(K)$. For one thing, if a function is a uniform limit of polynomials on $K$, then on $K^{\circ}$ (if it is non-empty), it will be holomorphic. But there are further topological restrictions. If $K$ is a compact set "with holes", then if you take any holomorphic function with a pole inside a hole, you won't be able to approximate it uniformly on $K$ with polynomials. On the other hand, if $K$ doesn't have a hole, then any function which is continuous on $K$ and holomorphic on $K^{\circ}$ can be approximated by polynomials. This is Mergelyan's theorem, see Chapter 20 of Rudin's "Real and Complex Analysis". What we have done here is to show that if there are holes, and if $\lambda$ is inside such a hole, then because of the special structure of $A$, it implies that the function $\frac{1}{z - \lambda}$ can be approximated uniformly by polynomials, which implies that in fact there cannot be any holes.

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Thanks! Very clear answer! Also I guess the picture is the following: You put $A$ as a larger algebra $C(K)$, and you realized the spectrum of $x$ in this larger algebra has no holes, and hence the spectrum in the original algebra has no holes. –  Hui Yu Dec 4 '12 at 4:29
    
Generally, $A$ is not $C(K)$ or $C(\mathrm{sp}(A))$, nor even a subalgebra as the Gelfand transform can fail to be injective and surjective. Even your case supports this - if you take $x$ to be a $2 \times 2$ nilpotent matrix, the Gelfand transform is onto but has a one dimensional kernel, so you can't identify $A$ with $C(\mathrm{sp}(A))$. –  levap Dec 4 '12 at 8:47
    
Well, the crucial thing here is $\partial K\subset \partial \sigma(x)$ so the analytic functions on $\sigma(x)$ can be identified with a subalgebra of analytic functions on $K$. –  Hui Yu Dec 4 '12 at 15:53
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