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I have asked the following question on M.SE here, but I have not yet received a response. I do apologize of this is not the correct site to post it on - if so, please do let me know and I will remove the question if it is off-topic.

I am reading through a research paper (Spectrum preservig linear mappings in Banach algerbas by B. Aupetit and H. du T. Mouton) in which they say the following:

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.

How does one show the above statement true using exponentials?

My initial thought, is defining another mapping $\psi$ as some kind of "combination" of $\phi$ and the exponential, i.e. something in the line of

$$\psi(x) = \phi(\exp(x))$$ or $$\psi(x)=\exp(\phi(x))$$ or every $x \in A$.

Is this kind of reasoning correct? And, if so, can someone help guide me on the correct path? I am not looking for a complete proof of the result - only a little "nudge" in the right direction :).

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    $\begingroup$ I don't have an MSE account so I've posted an answer - hopefully I did not make a mistake. It turned out that in order to check my idea for a hint, I ended up with a more-or-less full proof. $\endgroup$
    – Yemon Choi
    Commented Feb 18, 2017 at 2:52

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A standard trick one sees in Banach algebra theory, relating additive and multiplicative structure, is to look at $\exp(\lambda a)$ for fixed $a\in A$ as a holomorphic function of $\lambda$, and then try to play games with power-series or Liouville's theorem or other complex-analytic techniques.

The given conditions on $\phi$ imply that $\phi(x)\phi(x^{-1})=1_B$ for all $x\in A$. In particular $$\phi(e^{\lambda a})\phi(e^{-\lambda a}) = 1_B$$

The LHS is a $B$-valued holomorphic function of $\lambda$, and we can work out the first few terms in the power-series expansion: $$ \begin{aligned} \phi(e^{\lambda a})\phi(e^{-\lambda a}) & = \phi( 1_A+ \lambda a + \frac{\lambda^2}{2}a^2 + \dots) \phi(1_A - \lambda a + \frac{\lambda^2}{2}a^2 + \dots ) \\ & = (\phi(1_A)+ \lambda\phi(a) + \frac{\lambda^2}{2} \phi(a^2) + \dots) (\phi(1_A) - \lambda\phi(a) + \frac{\lambda^2}{2} \phi(a^2) + \dots) \\ & = 1_B +\lambda^2 (\phi(a^2) - \phi(a)^2 ) + O(\lambda^3) \end{aligned} $$

From this we obtain $\phi(a^2)=\phi(a)^2$ for all $a\in A$, which is the Jordan homomorphism condition.

Actually, now that I have worked through this, one doesn't seem to need the full power of being holomorphic. You could use $\lambda\in{\bf R}$ with $\lambda\to 0$ to achieve the same effect.

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  • $\begingroup$ Thank you very much for your answer. And thank you for the mentioning of the fact that this is a "standard trick" - I am only in my first week as a Masters student (so still very new) and will now most certainly be keeping this trick in my back pocket for future use. Although I do understand your argument given, my only question is one that is mostly arithmetic based .. How exactly did you arrive at the last line of your calculations ($1_B + \lambda^2(\phi(a^2) - \phi(a)^2)$ ? I understand why you left the remainder as an $O(\lambda^3)$. $\endgroup$
    – user860374
    Commented Feb 18, 2017 at 11:14
  • $\begingroup$ Try expanding out the brackets and you will see that the linear terms (multiples of $\lambda$) cancel. You get three terms of order $\lambda^2$, two of which are equal to each other. The $1_B$ comes from the fact that $phi(1_A)=1_B$. $\endgroup$
    – Yemon Choi
    Commented Feb 18, 2017 at 15:43

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