Can you describe the image of the exponential map $B(H)\to B(H)$?

Question

James Tener asks at the 20-questions seminar:

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

Answer 1

The commenters on the scratchpad wiki are very knowledgeable; I'm just expanding on Anonymous's answer.

The exponential map is not surjective. If you look at Halmos's paper 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 $\int |f(z)|^2\mathrm{d}z\wedge\mathrm{d}\bar z$ is finite. The operator to consider 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.

Answer 2

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?

Answer 3

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.

Answer 4

I was going to post this as a comment to David Speyer's Oct 12 13:23 post but it became too long.

I only read the abstract of the paper linked to in that answer so far. It says that they characterize the interior of the image as follows: an invertible operator A is in the interior of the image of the exponential map if and only if 0 is in the unbounded component of the complement of the spectrum of A. (The latter property is equivalent to the existence of an analytic logarithm with domain containing the spectrum of A.) A consequence of this characterization of the interior is the following class of operators contained in the image and in the boundary of the image: every normal operator N with 0 in a bounded component of the complement of the spectrum of N. That is, every normal invertible operator is in the image (take a Borel logarithm on the spectrum and apply "functional calculus" for normal operators), but many of them are also in the boundary of the image, including every unitary operator whose spectrum is the entire circle (e.g. a bilateral shift). I don't know why, but I was surprised to learn that the image isn't open.