Let $A$ be a complex Banach algebra with identity 1. Define the exponential spectrum $e(x)$ of an element $x\in A$ by $$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G\_1(A)\},$$ where $G\_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity.



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> Is it true that $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$?
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Equivalently, is it true that 
$1-ab$ is in $G\_1(A)$ if and only if $1-ba$ is in $G\_1(A)$, for all $a,b \in A$?


Note: The usual spectrum has this property.


Just an additional note:


We have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if


1) The group of invertibles of $A$ is connected, because then the exponential spectrum of any element is just the usual spectrum of that element.


2) The set $Z(A)G(A) = \{ab: a \in Z(A), b\in G(A)\}$ is dense in $A$, where $Z(A)$ is the center of $A$. (One can prove this). In particular, we have $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$ if the invertibles are dense in $A$.


3) $A$ is commutative, clearly.


But what about other Banach algebras? Can someone provide a counterexample?



In a Banach algebra, do ab and ba have almost the same exponential spectrum?
Just to update this: a negative solution was recently given by Klaja and Ransford. See [arXiv 1510.08109.](http://arxiv.org/abs/1510.08109)


