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I previously asked the following question of MathOverflow: Showing that $\phi$ is a Jordan morphism in which I was asking assistance with proving the following statement made in the introduction Spectrum preservig linear mappings in Banach algerbas by B. Aupetit and H. du T. Mouton (1994).

If $\phi$ is a linear mapping from a Banach algebra $A$ into another one $B$ such that $\phi(1)=1$ and $\phi(x)^{-1}=\phi(x^{-1})$ for $x$ invertible, then using exponentials it is easy to prove that $\phi$ is a Jordan morphism.

However, while reading the paper Spectral Characterization of Idempotents and Invertibility Preserving Linear Maps by M. Brešar and P. Šemrl (1999), I came across the following:

Let $A$ and $B$ be semi-simple Banach algebras and $\phi:A \to B$ a unital bijective linear map preserving invertibility. It is then conjectured that $\phi$ must be a Jordan isomorphism.

My question: When comparing the statements of the two papers, is the condition that $A$ and $B$ are semi-simple necessary to ensure that $\phi$ is Jordan?

I understand that the bijective condition of the latter statement, simply changes the (homo)morphism of the first statement into a (iso)morphism. I am simply interested in seeing how the condition that $A$ and $B$ being semi-simple comes into play.

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I don't know the sharpest examples to answer your question, but the basic "idea" is that the conjecture you mention (due I think to Kaplansky?) is only reasonable when the Jacobson radical is trivial, because it is the part which spectral theory cannot detect.

To give a more concrete illustration: suppose $A$ is a unital CBA whose Jacobson radical $J$ has codimension $1$. (Take your favourite radical CBA and adjoin an identity). Now every element of the coset $1+J$ is invertible. So if you take any linear bijection $\theta:J\to J$ and extend it to a linear bijection $\phi:A\to A$ in the obvious way, then $\phi$ will be invertibility-preserving; provided $\dim(J)\geq 2$ it should be possible to cook up $\theta$ so that $\phi$ is not a homomorphism.

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  • $\begingroup$ Thank you very much for your answer once again! Apologies for taking so long to accept - I did not realize that I haven't done so yet. I've not yet made an exact example of your answer, but just wondered - your answer acts as a (very nicely written, I might add) guide to constructing a linear mapping that is not a homomorphism. I'm wondering, however, if this will be sufficient to find an example which is not Jordan - i.e. if this will be sufficient to show that we do in fact require nontrivial Jacobson radical to ensure that $\phi$ (satisfying the hypothesis), is Jordan. $\endgroup$
    – user860374
    Commented Apr 6, 2017 at 6:27
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    $\begingroup$ Sorry, I should have clarified: Jordan homomorphisms between two commutative algebras are algebra homomorphisms (as long as we are working over a field of characteristic $\neq 2$ ). $\endgroup$
    – Yemon Choi
    Commented Apr 6, 2017 at 16:35

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