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

Let A$A$ be a complex Banach Algebraalgebra with identity 1. Define the exponential spectrum $e(x)$ of an element x in A$x\in A$ by

e(x)= {complex lambda's such that x-lamba*1 is not in G_1(A)},

$$e(x)= \{\lambda\in\mathbb{C}: x-\lambda1 \notin G_1(A)\},$$ where G_1(A)$G_1(A)$ is the connected component of the group of invertibles $G(A)$ that contains the identity 1.

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

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$1-ab$ is in G_1(A)$G_1(A)$ if and only if 1-ba$1-ba$ is in G_1(A)$G_1(A)$, for all a,b in A$a,b \in A$?

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

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

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 $A$.

$A$ is commutative, clearly.

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

Question about the exponential spectrum in a Banach algebra

Let A be a complex Banach Algebra with identity 1. Define the exponential spectrum of an element x in A by

e(x)= {complex lambda's such that x-lamba*1 is not in G_1(A)},

where G_1(A) is the connected component of the invertibles that contains the identity 1.

Is it true that e(ab) U {0} = e(ba) U {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.

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

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

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

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 $A$.