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Let $A$ be a complex unital Banach algebra, and let $A^\times$ denote the set of (two-sidedly) invertible elements of $A$. Let $\exp : A \rightarrow A^\times$ denote the exponential map.

My question is, in general is the exponential map surjective onto $A^\times$?

What I understand so far is as follows:

  • If we replace "complex" with "real", then the answer becomes no. E.g. we can take $A$ to be $\mathbb{R}$ itself, or more generally the matrix algebra $\mathrm{Mat}_n(\mathbb{R})$ for any integer $n>0$. (reference)
  • The answer becomes "yes" in the case $A = \mathrm{Mat}_n(\mathbb{C})$, for integer $n>0$. (reference)

However I'm not sure how to generalize the 2nd bullet point, nor how to find a counter-example. Would anyone have any suggestions on how to think about this?

Arctic Char
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I.A.S. Tambe
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2 Answers2

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Take the algebra of continuous complex-valued functions on $S^1\subset\mathbb C$ (with the sup-norm). The identity function is not exp of any continuous function, so the answer is no.

In general, such questions are answered using the holomophic calculus for Banach algebras.

user8268
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The set of invertible elements $G(A)$ is an open subset of $A$. $\exp(A)$ is precisely the connected component of the unit $1\in A$. Thus, if $G(A)$ is connected, then $G(A) = \exp(A)$. Otherwise, define an equivalence relation on $G(A)$ by $$x\sim y \Leftrightarrow xy^{-1}\in\exp(A).$$ Let $\{x_{\alpha}\exp(A) : \alpha\in J\}$ denote the set of equivalence classes of $\sim$. These equivalence classes are precisely the (open) connected components of $G(A)$.

For the two examples you've provided:

  1. If $A=M(\mathbb{C}^n)$ then $G(A)$ is connected.
  2. If $A=M(\mathbb{R}^n)$ then $G(A)$ has two connected components; $\exp(A) = \{x\in A: \det(x)>0\}$ and $\{x\in A: \det(x)<0\}$
Onur Oktay
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