Orthogonal Complex Hyperbolic Arrangements
with J. Carlson and D. Toledo; Symposium in Honor of C. H. Clemens, Contemp. Math. 312 (2002) 1-8

In the proof of lemma 3.1 we make an assertion which is true but not sufficiently justified in the paper. The setting is that Gamma is a lattice in PU(n,1) for n>1, carrying a hyperplane arrangement to itself, Hi is one of these hyperplanes, and Gammai is its stabilizer. In the paper we use without comment the fact that Gammai is a lattice in the automorphism group PU(n-1,1) of Hi.

There may be a simple way to see this, but we don't know of one. The truth of the assertion follows from very general theorems in ergodic theory. See M. Ratner's survey, Invariant measures and orbit closures for unipotent actions on homogeneous spaces, Geom. Funct. Anal. 4 (1994), 236-257, and its references, for background. Briefly, the first few pages of this article, up through conjecture 3 (which is proven later), imply the following:

Theorem: Suppose X is a symmetric space, Gamma is a lattice in Aut(X), and Y is a symmetric subspace of X corresponding to a subgroup of Aut(X) that is generated by unipotent elements. Then the closure of the image of Y in Gamma\X is the quotient of a symmetric subspace Z of X modulo a lattice in its stabilizer.

In our case, we take X to be complex hyperbolic n-space and Y to be the hyperplane Hi. The hypothesis n>1 implies that PU(n-1,1) is generated by unipotent elements. So we apply the theorem; since Y is a hyperplane, Z is either a hyperplane or all of CHn. The latter case is impossible because the arrangement is not dense in CHn. In the former case it is easy to see that Z is equivalent to Hi under Gamma, and it follows from the theorem that Gammai is a lattice in Aut(Hi), as desired.

The story behind the omission of this detail is that an earlier version of the paper treated a special class of hyperplane arrangements, rather than the more general version given in the published paper. For these special arrangements, the fact that Gammai is a lattice in PU(n-1,1) was obvious.