56 lines
2.4 KiB
TeX
56 lines
2.4 KiB
TeX
\section{Unitary Elements}
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\label{section:unitary-c-star}
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\begin{definition}[Unitary]
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\label{definition:unitary-element}
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Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$.
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\end{definition}
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\begin{lemma}
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\label{lemma:unitary-unit}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$.
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\end{lemma}
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\begin{proof}
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$\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$.
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\end{proof}
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\begin{definition}[Unitarily Equivalent]
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\label{definition:unitary-equivalent}
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Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$.
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\end{definition}
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\begin{lemma}
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\label{lemma:unitary-equivalent-same-stuff}
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Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then:
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\begin{enumerate}
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\item $\norm{x}_A = \norm{y}_A$.
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\item $\sigma_A(x) = \sigma_A(y)$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): Let $u \in A$ be unitary such that $x = uyu^*$, then $\norm{x}_A \le \norm{u}_A\norm{x}_A\normn{u^*}_A$. By \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, so $\norm{x}_A \le \norm{y}_A$. As the argument is symmetric, $\norm{x}_A = \norm{y}_A$.
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(2): Let $\lambda \in \complex$, then
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\[
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u(y - \lambda)u^* = uyu^* - \lambda uu^* = uyu^* - \lambda = x - \lambda
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\]
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Since $u, u^* \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_A(x) = \sigma_A(y)$.
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\end{proof}
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\begin{proposition}
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\label{proposition:unitary-spectrum}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$.
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 8.2]{Zhu}}}. ]
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By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus
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\[
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\bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)}
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\]
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by (4) of \autoref{proposition:c-star-algebra-gymnastics}. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_A(x) \subset \partial B_\complex(0, 1)$.
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\end{proof}
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