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Let $A$ be a $C^*$-algebra or some norm-closed algebra of operators on a Hilbert space.

In the old paper

Hille, E. On Roots and Logarithms of Elements of a Complex Banach Algebra, Math. Annalen, Bd. 136, S. 46--.57 (1958)

the question is studied, which elements $x \in A$ have the property that the exponential function is open at $x$ (i.e. every neighborhood of $x$ maps to a neighborhood of $\exp(x)$). Under some conditions on the spectrum of $x$, Hille shows that the answer is positive.

Question: Is there an example of a unital $C^*$-algebra (or operator algebra) with the property that the exponential function is not open?

More specifically: Is the exponential function open for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$?

Is there an example of a Banach algebra where the exponential function is not open?

EDIT: Jonas Meyer has clarified that $B(H)$ is a counterexample. It remains unclear what happens in general, or whether there is any description of the class of Banach algebras, where the exponential map is open. In particular, it remains unclear for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$.

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  • $\begingroup$ You say that "Hille shows that the answer is positive" yet apparently the question is of the form "which elements...?". $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 5, 2010 at 16:14
  • $\begingroup$ Right, under some conditions on the element $x$, Hille can show that the exponential map is open at $x$. His conditions are sufficient but probably not necessary. If the answer to the Question is negative, one should ask about a precise characterization of the elements where $\exp$ is open. $\endgroup$
    – Andreas Thom
    Commented Aug 5, 2010 at 16:19

1 Answer 1

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The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the image but not in the interior of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.

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  • $\begingroup$ It remains interesting to ask the question for other $C^*$-algebras or operator algebras, such as the algebra of upper triangular operators on $\ell^2 {\mathbb N}$. $\endgroup$
    – Andreas Thom
    Commented Aug 5, 2010 at 20:22
  • $\begingroup$ You're welcome. I agree. I didn't answer one of your questions, and you may "unaccept" this answer if you want. (If no answer is "accepted", then it may be easier for people to quickly see that part of the question remains unanswered. But that's up to you.) $\endgroup$
    – Jonas Meyer
    Commented Aug 5, 2010 at 20:59

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