In a Banach algebra, do ab and ba have almost the same exponential spectrum?

Question

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.

Is it true that $e(ab)\cup\{0\} = e(ba)\cup\{0\}$ for all $a,b \in A$?

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?

Answer 1

Just to update this: a negative solution was recently given by Klaja and Ransford. See arXiv 1510.08109.

Answer 2

Bonjour Yemon, Oui, c'est un problème proposé par TJR dans le cadre d'un projet de recherche d'été CRSNG.

It is indeed a very subtle question, I thought it might interest some people here. This problem appears to be strongly related to the topology of the group of invertible elements, which is difficult to study.

And yes, one can show that the exponential spectrum of a*b is the same than the one of b*a in the Calkin algebra. It follows from the fact that 1-ab is of Fredholm index zero if and only if 1-ba is of Fredholm index zero.