ar X iv : m at h / 06 06 59 4 v 1 [ m at h . O A ] 2 3 Ju n 20 06 THE CALKIN ALGEBRA HAS OUTER AUTOMORPHISMS

Assuming the continuum hypothesis, we show that the Calkin algebra has 2 ℵ 1 outer automorphisms. Let H be a separable infinite dimensional Hilbert space, let L(H) be the algebra of bounded operators on H, let K(H) be the algebra of compact operators on H, and let Q = L(H)/K(H) be the Calkin algebra. A long-standing problem asks whether every automorphism of Q is inner, that is, of the form x → u * xu for some unitary u ∈ Q. The earliest references we have found to this problem are implicitly in [4] (see page 126) and explicitly in [5] (see Remark 1.6(2)). The automorphisms whose existence we prove are, however, implemented by a unitary on every separable subalgebra of the Calkin algebra, and are thus not interesting from the point of view of extensions. In particular, it remains open whether there exists an automorphism of the Calkin algebra which sends the image of the unilateral shift to its adjoint. It is of course well known that, for any Hilbert space H, all automorphisms of L(H) are inner. Corollary 8.8 of [6] provides a factor of type II 1 with separable predual such that all automorphisms are inner. (In this context, recall that, by Corollary 5.13 of [12], all C* automorphisms of a von Neumann algebra are in fact von Neumann algebra automorphisms.) If A is a separable simple C*-algebra such that every automorphism of A is inner (in the multiplier algebra if A is not unital), then A must be isomorphic to the algebra of compact operators on some Hilbert space. This is Corollary 3 of [9], but, as pointed out by George Elliott, can also be derived from the existence of nontrivial central sequences [1]. Without simplicity, this is not true. In fact, it fails even for commutative C*-algebras. There exists a compact metric space X, with more than one point, such that the only homeomorphism h : X → X is the identity. We are grateful to Greg Kuperberg for providing this example. We don't know a reference, but the construction of such a subset of the plane is easy to outline. Start with a line segment in the plane. Attach another line segment to the midpoint. Now there are three line segments, the new one and two halves of the original. Attach two more line segments to one midpoint, three more to another, and four …