It is asserted in Definition 4.2 in [1] that the random operators U(t) defined there are unitary. As was pointed out to the author by Shizan Fang, it is clear that U(t) is an isometry but it is not obvious that U(t) is surjective. The purpose of this note is to fill this gap.
1998 Academic Press 1. INITIAL COMMENTS I would first like to point out that, even without verifying the surjectivity of U(t) defined in Definition 4.2 in [1], all of the results and all but one proof in [1] would still be valid. Indeed, the only place where the surjectivity of U(t) was used, other than for notational simplicity, was in the first proof of Theorem 4.14 in [1]. Nevertheless, Theorem 4.14 is still valid because of Theorem 6.2; see Remark 4.15 in [1]. The only notational changes that would need to be made are: (1) replace the orthogonal group O(H 0 (g)) on H 0 (g) by the set ISO(H 0 (g)) of isometries on H 0 (g) and (2) interpret U(t) H 4 (t) as
In the next section we will give a more satisfying remedy to the gap in Definition 4.2 in [1], namely the fact that U(t) is unitary.
2. A PROOF THAT U(T ) IS UNITARY
The reader is referred to [1] for the notation and definitions used in this corrigendum. Recall that S 0 /H 0 (g) is an orthonormal basis for H 0 (g) and article no. FU983244