# Can you describe the image of the exponential map B(H)->B(H).

James Tener asks at the 20-questions seminar:

The exponential map exp:B(H)->B(H) is just defined by its Taylor series. Can you describe its image?

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The commenters on the wiki http://scratchpad.wikia.com/wiki/091006qa are very knowledgeable; I'm just expanding on Anonymous's answer.

The exponential map is not surjective. If you look at Halmos's paper http://www.ams.org/mathscinet-getitem?mr=53391 (link requires academic access) you will see a wide variety of invertible maps that are not exponentials. Here is the simplest example. Take 0 < u < v and let D be the annulus u < |z| < v in the complex plane. Our Hilbert space will be the space of analytic functions f on D such that \integral |f(z)|^2 is finite. The operator is multiplication by z.

To sketch Halmos' argument, H is complete because the property of being harmonic can be stated as a condition on integrals and thus passes through an L^2 limit. The logarithm of multiplication by z wants to be multiplication by \log z, but we can't define \log z on D without introducing a branch cut. Of course, D doesn't have to be this exact shape, any open region of the complex plane which has a function without a logarithm would work.

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I am now reading through http://www.ams.org/mathscinet-getitem?mr=870760 . This seems to be extremely relevant, and has many interesting references, but is tough going for me. Perhaps an analyst would like to take a crack at it?

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The exponential map is surjective on ${\rm GL}_2({\bf C})$ — look at Jordan normal form. Also, $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ is in $\exp({\rm SL}_2)$, as it is the exponential of $\begin{pmatrix} 0 & \pi \\ -\pi & 0 \end{pmatrix}.$ However, $\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$ is not in $\exp({\rm SL}_2({\bf C}))$.

Next, some information on the original question.

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