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Let $A$ be a C$^{*}$-algebra. Let $\hat{A}$ denote the set of all irreducible representations of $A$. Suppose $\pi\in\hat{A}$ has the following property: for all $a\in A$, the map from $\hat{A}\to\mathbb{R}$ given by $\tau\mapsto\|\tau(a)\|$ is continuous at $\pi$.

Let $e\in \pi(A)$ be a projection. Then there is a positive element $y\in A$ such that $\pi(y)=e$. We have that $\operatorname{sp}(\pi(y))\subseteq\{0,1\}$.

How can we show that there is a neighbourhood $V\subseteq\hat{A}$ of, $\pi$, such that for all $\tau\in V$, $\operatorname{sp}(\tau(y))\subseteq (-\frac{1}{4},\frac{1}{4})\cup(\frac{3}{4},\frac{5}{4})$?

This is part of a proof in a paper by Dixmier. He says it is a well known argument, but I don't believe I've seen it before. Any incite is appreciated. Thanks.

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Choose $V$ such that $|\|\tau(y)\|-\|\pi(y)\|\,|<\varepsilon$ for all $\tau\in V$. Then $$ \|\tau(y)^2-\tau(y)\|=\|\tau(y^2-y)\|<\varepsilon+\|\pi(y^2-y)\|=\varepsilon+\|e^2-e\|=\varepsilon. $$ The element $\tau(y)^2-\tau(y)$ is selfadjoint. For any $\lambda\in\sigma(\tau(y)^2-\tau(y))$, we have $|\lambda|<\varepsilon$. Also, $$ \sigma(\tau(y)^2-\tau(y))=\{\lambda^2-\lambda:\ \lambda\in\tau(y)\}. $$ Thus we have $|\lambda^2-\lambda|<\varepsilon$ for all $\lambda\in\sigma(\tau(y))\subset\mathbb R$. This forces $$\tag1 \frac{1-\sqrt{1+4\varepsilon}}2<\lambda<\frac{1-\sqrt{1-4\varepsilon}}2, $$ or $$\tag2 \frac{1+\sqrt{1-4\varepsilon}}2<\lambda<\frac{1+\sqrt{1+4\varepsilon}}2. $$ Now choose $\varepsilon$ small enough such that $(1)$ and $(2)$ imply $$ \lambda\in\left(-\tfrac14,\tfrac14\right)\cup\left(\tfrac34,\tfrac54\right) $$

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  • $\begingroup$ Thanks, Martin. I understand the spectrum calculation you are making, but I'm a bit confused as to how to apply it. $a=e=\pi(y)$ and $b=\tau(y)$ are elements in distinct C$^{*}$-algebras ($\pi(A)$ and $\tau(A)$) aren't they? So how are we to make sense of $e-\tau(y)$? $\endgroup$ – ervx May 20 at 14:19
  • $\begingroup$ You are right. It was much simpler than that. $\endgroup$ – Martin Argerami May 20 at 21:33
  • $\begingroup$ Thank you very much, Martin! I had never encountered an example of something worked out in this way before. $\endgroup$ – ervx May 21 at 12:57

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