# 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?

• 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. Oct 25, 2009 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... Oct 29, 2009 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 Oct 30, 2009 at 9:48
• So, anyone has comments / reading suggestions / ideas? Nov 16, 2009 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.) Apr 6, 2010 at 2:57

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

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.

• Yes, it does. This was my first post, I was looking for "functional analysis" and "Banach algebras" tags, but for some reason couldn't find them, so I picked those two tags. Oct 30, 2009 at 13:10
• Do you remember how the proof goes for the Calkin algebra? I'm not sure if the proof I had previously found works, there seems to be a gap, or an error in my notes. Nov 3, 2009 at 6:57
• I'm sorry Yemon, I hadn't seen your comment before. hmm I forgot some key steps in the argument, but I have a pdf with the proof somewhere. I'll find it and then I'll send it to you via ulaval. Nov 5, 2009 at 13:45
• Did you receive the proof, Yemon? Nov 16, 2009 at 17:03
• Yes, thanks. Unfortunately the proof seems to rely on subtle properties of the densely-defined trace on B(H) - it's not clear how one might generalise to other Banach algebras. Even the C*-case could be difficult. Nov 17, 2009 at 8:15