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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 ab is the same than the one of ba 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, though I dont remember the exact details about how to prove this.

(EDIT) Just an additional note :

We have e(ab) U {0} = e(ba) U {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) = {a*b: a in Z(A), b invertible} is dense in A, where Z(A) is the center of A (the set of elements of A that commute with every other element of A). (One can prove this). In particular, we have e(ab) U {0} = e(ba) U {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?

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 ab is the same than the one of ba 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.

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 ab is the same than the one of ba 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, though I dont remember the exact details about how to prove this.

(EDIT) Just an additional note :

We have e(ab) U {0} = e(ba) U {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) = {a*b: a in Z(A), b invertible} is dense in A, where Z(A) is the center of A (the set of elements of A that commute with every other element of A). (One can prove this). In particular, we have e(ab) U {0} = e(ba) U {0} for all a,b in A if the invertibles are dense in A.

3. A is commutative, clearly.

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 ab is the same than the one of ba 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, though I dont remember the exact details about how to prove this.

(EDIT) Just an additional note :

We have e(ab) U {0} = e(ba) U {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) = {a*b: a in Z(A), b invertible} is dense in A, where Z(A) is the center of A (the set of elements of A that commute with every other element of A). (One can prove this). In particular, we have e(ab) U {0} = e(ba) U {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?

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 ab is the same than the one of ba 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, though I dont remember the exact details about how to prove this.

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 ab is the same than the one of ba 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, though I dont remember the exact details about how to prove this.

(EDIT) Just an additional note :

We have e(ab) U {0} = e(ba) U {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) = {a*b: a in Z(A), b invertible} is dense in A, where Z(A) is the center of A (the set of elements of A that commute with every other element of A). (One can prove this). In particular, we have e(ab) U {0} = e(ba) U {0} for all a,b in A if the invertibles are dense in A.

3. A is commutative, clearly.