Let $A$ be a $C^*$ algebra. Assume that the spectrum $Sp(a_1a_2\ldots a_{n-1}a_n)$ is unchanged as a set after a permutation of $a_i$'s. (unless possible emerge or removing 0 from the spectrum) Does this imply that $A$ is a commutative algebra? What about Banach algebra case?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
In the $C^\ast$-algebra case, yes, your condition implies $A$ is commutative. Indeed, assume to the contrary that $A$ is non-commutative. Then there exists a nonzero $u \in A$ s.t. $u^2 = 0$. (See, for example, the second proposition in this answer.) Let $a_1 = u$, $a_2 = u^\ast u$, $a_3 = u^\ast$. Then $a_1a_2a_3 = (uu^\ast)^2$ is a nonzero positive element, so $\text{Sp}(a_1a_2a_3)$ contains a positive real. However, $a_2a_1a_3 = u^\ast uuu^\ast = 0$, so $\text{Sp}(a_1a_2a_3) \setminus \{0\} \neq \text{Sp}(a_2a_1a_3) \setminus \{0\}$.
In the Banach algebra case, the answer is no. Indeed, a counterexample can be given by $A \subset M_2(\mathbb{C})$ consisting of upper triangular matrices. In that case $\pi: A \to \mathbb{C}^2$, defined by $\pi(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}) = (a, c)$, is an algebra homomorphism that preserves the spectrum (as $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ is invertible iff $a \neq 0$ and $c \neq 0$, so its spectrum is $\{a, c\}$). Thus,
$$\text{Sp}(a_1a_2 \cdots a_{n-1}a_n) = \text{Sp}_{\mathbb{C}^2}(\pi(a_1)\pi(a_2) \cdots \pi(a_{n-1})\pi(a_n))$$
is invariant under permutation of $a_i$.
