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In the proof of theorem 2.1.11 from C*-Algebras and Operator Theory, by Gerald Murphy the author is trying to prove $\sigma_B(u) = \sigma_A(u)$, where $B$ is a C*-subalgebra of $A$ and $u \in B$ is a unitary element.

The author starts the proof by assuming $B$ has a hermitian element $b$. He mentions that since $\sigma_A(b) \subset \mathbb{R}$, it has no holes. I understand why $\sigma_A(b) \subset \mathbb{R}$, but don't understand why it has no holes in $\mathbb{R}$.

After example 1.2.6, the author defines 'holes' of $K \subset \mathbb{C}$ as connected components of $\mathbb{C} \setminus K$.

Since $\sigma_A(b) \subset \mathbb{R}$, the statement that $\sigma_A(b)$ has no holes would imply $\sigma_A(b)$ is connected!?

codehumor
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  • Perhaps you meant a self adjoint element, as the spectrum of a unitary element is contained in the unit circle. – Ryszard Szwarc Jul 12 '23 at 17:15
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    What does "holes" mean here? In any case, it's certainly not true that the spectrum has to be connected. – Eric Wofsey Jul 12 '23 at 18:08
  • Perhaps the author means that for self-adjoint $b$, $\sigma_A(b)$ is simply-connected as a subspace of $\Bbb{C}$ - which is true even if it's not connected - as any loop into $\sigma_A(b)$ is contractible. I believe for any $b$ saying that $\sigma_A(b)$ is simply-connected as a subspace of $\Bbb{C}$ is equivalent to the resolvent of $b$ being a connected open subset of $\Bbb{C}$ - and this may be important when constructing a holomorphic function on the resolvent. – Chad K Jul 12 '23 at 18:41
  • See this – Ryszard Szwarc Jul 12 '23 at 20:28
  • The book OP is referring to is the one by Murphy, and in the proof of Theorem 2.1.11 the non-sensical statement about the spectrum having no holes appears exactly as OP has written (except OP has switched "hermitian" with "unitary"). Don't know how I never noticed this before, but it really makes no sense to me lol – s.harp Jul 12 '23 at 20:45
  • Well I would assume "holes" has some technical meaning defined earlier in the text that would make the statement make sense. I just don't know what that meaning is... – Eric Wofsey Jul 12 '23 at 21:05
  • Sorry for the typo about 'unital' amd 'hermitian', I've updated the question. Added the definition of 'holes' as the author defines it. – codehumor Jul 13 '23 at 06:08
  • I think i see it now. My problem was because I was thinking of $\sigma_A(b)$ as a subspace of $\mathbb{R}$. Viewing so doesn't make any sense even in the author's definition of holes. But considering $\sigma_A(b)$ as a subspace of $\mathbb{C}$ would make it have a connected resolvant. Which makes it having no bounded components. – codehumor Jul 13 '23 at 06:12

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The book in question defines "holes" as bounded connected components of $\mathbb C\setminus K$. So the claim is that for compact $K\subset \mathbb R$, the set $\mathbb C\setminus K$ has no bounded connected components (it would of course be non-sensical to claim that it has no connected components at all). As noted in the comments, it is crucial that we look at the complement in $\mathbb R$ and not in $\mathbb C$.

In fact, $\mathbb C\setminus K$ is always path connected in this situation. To see this, it suffices to prove that every $\lambda\in (\mathbb C\setminus K)\cap \mathbb R$ can be connected by a path in $\mathbb C\setminus K$ with every element $\mu$ of $\mathbb C\setminus \mathbb R$ - connecting arbitrary pairs of elements in $\mathbb C\setminus K$ can then be achieved by concatenating paths.

The piecewise linear curve $$ \gamma\colon [0,1]\to\mathbb C,\,\gamma(t)=\begin{cases}\lambda+2it{\operatorname{Im \mu}}&\text{if }0\leq t\leq 1/2\\ (2t-1)\mu+(2-2t)(\lambda+i{\operatorname{Im \mu}})&\text{if }1/2<t\leq 1\end{cases} $$ connects $\lambda$ and $\mu$. Moreover, $\gamma$ only intersects $\mathbb R$ at $t=0$ and $\gamma(0)=\lambda\in \mathbb C\setminus K$. Thus $\operatorname{im}\gamma$ lies inside $\mathbb C\setminus K$ as claimed.

MaoWao
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