# 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.

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?

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question poseé par TJR, n'est ce pas? It certainly seems to be quite subtle, but perhaps my intuition is faulty. If I recall correctly, one can prove that this works when A is the Calkin algebra, by using properties of the Fredholm index. I'll have to check this though. –  Yemon Choi Oct 25 '09 at 5:48
This sounds like a very interesting question. For an arbitrary Banach algebra, I would have thought that the answer is no. Have to think of a counterexample... –  George Lowther Oct 29 '09 at 23:43
This probably ought to have a "functional-analysis" tag on it, if anyone with the power & inclination to bestow such is reading –  Yemon Choi Oct 30 '09 at 9:48
So, anyone has comments / reading suggestions / ideas? –  Malik Younsi Nov 16 '09 at 17:03
I edited to add LaTeX, to merge the additional note from the closed duplicate, and to make the title (an imprecise version of) the question. (I now realize that the additional note also appears in the answer below, but I think that it will be more useful in the question.) –  Jonas Meyer Apr 6 '10 at 2:57