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In a $C^*$ algebra why $\|a\|\not =\rho(a)$ for any $a$? Where $\rho(a)$ is the spectral radius.

It can be shown that the equality holds for self-adjoint elements. Then that can be used to show that the Gelfand transform is an isometry. Thus $\|a\|=\|\hat{a}\|=max \{|\phi(a)| \hspace{0.2cm}| \phi\in Spec(A) \}$. First equality follows from the Gelfand transformation being an isometry. The second follows form the definition of the norm on $C(Spec(A))$. But we know that $spec(a)=\{\phi(a) \hspace{0.2cm}| \phi\in Spec(A) \}$. thus $\rho(a)=\rho(\hat{a})$ and the result follows. My professor said this is incorrect. What I am doing wrong here?

Edit:

I am assuming the algebra is commutative

Sorfosh
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  • There certainly are C* algebras where $|a| = \rho(a)$ for all $a$ (e.g. continuous functions on a compact Hausdorff space). On the other hand, there are $2 \times 2$ matrices with $\rho(a) = 0$ and $|a| \ne 0$. – Robert Israel Feb 10 '21 at 21:40
  • @RobertIsrael I understand I am wrong. But where does my reasoning fail? ( I am assuming $C^*$ algebra is commutative btw) – Sorfosh Feb 10 '21 at 21:42
  • @RobertIsrael Also any commutative $C^$ algebra is isometrically isomorphic to a $C(X)$ where $X$ is a compact Hausdorff space, namely $X= Spec(A)$ – Sorfosh Feb 10 '21 at 21:43
  • It is true for commutative C*-algebras. – Eric Wofsey Feb 10 '21 at 21:44
  • On a commutative C* algebra, we do have $|a| = \rho(a)$. – Robert Israel Feb 10 '21 at 21:44
  • You are also assuming that the C*-algebra is unital. – J. De Ro Feb 10 '21 at 21:44
  • Ohhh, that clears everything up. Thank you very much everyone. Someone can post an answer or i can just close the question – Sorfosh Feb 10 '21 at 21:45

1 Answers1

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I think your proof is correct and your professor is wrong.

A faster proof is using the full power of Gelfand transformation. Indeed, without loss of generality we may assume that $A= C(X)$ where $X$ is a compact Hausdorff space. Then if $f \in C(X)$, then $\operatorname{spec}(f) = f(X)$ and thus $$\rho(f) =\max_x |f(x)| = \|f\|_\infty.$$

For non-commutative $C^*$-algebras, the statement is false. Consider $$A:=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\in M_2(\mathbb{C}).$$ Then $A^2=0$ and thus $\operatorname{spec}(A) = \{0\}$ but $\|A \| \neq 0$.

J. De Ro
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  • I think I miscommunicated with him, just as I did just now. I assumed the algebra is commutative, which is the key assumption. I think he thought i mean it is true for any $C*$ algebra which is false. – Sorfosh Feb 10 '21 at 21:46
  • @Sorfosh Yes that's right. For your convenience, I added some details to my answer. – J. De Ro Feb 10 '21 at 21:50