TITLE: Complexifications of hyperbolic manifolds
QUESTION [20 upvotes]: I'm wondering when a compact hyperbolic $n$-manifold ($n \geq 3$) can embed in a complex hyperbolic $n$-manifold as a real algebraic subvariety so that it is a component of the fixed point set of complex conjugation? 
I suspect it might to be possible to answer this for arithmetic hyperbolic manifolds by an arithmetic construction. But I'm particularly interested in the 3-dimensional case, where most manifolds are not arithmetic. 
One could vaguely hope to approach this using algebraic geometry. Make the manifold into a real algebraic variety, and then embed into a non-singular complex projective variety by resolving singularities. If this variety satisfies the Yau-Miyaoka inequality, then it is complex hyperbolic (see Theorem 1.3 of this paper for the statement and references). Obviously I have no idea how to achieve this though.
The restriction $n\geq 3$ is necessary, since in dimension 2 there are moduli spaces of hyperbolic structures, but compact complex hyperbolic surfaces are rigid and hence countable (the $1$-dimensional case is trivial).

REPLY [9 votes]: There exist hyperbolic 3-manifolds which cannot embed totally geodesically in complex hyperbolic manifolds, answering this
question in the negative.
Recently it was shown that complex hyperbolic manifolds have integral discrete faithful representations; in particular, the traces of the matrices in $SU(n,1)$ have integral traces.
See Theorem 1.3.1 of Baldi-Ullmo or Theorem 1.5(3) of Bader et. al..
This follows from a result of Esnault-Groechenig that cohomologically rigid representations of the fundamental group of smooth projective varieties must be integral.
Compact hyperbolic $n$-manifolds are projective varieties, and the discrete faithful representation into $SU(n,1)$ is unique up to conjugation and cohomologically rigid by Mostow rigidity. Hence this representation must have integral traces (this corollary was pointed out to me by David Fisher, but see the above citations for more details).
However, there are hyperbolic 3-manifolds such that the discrete faithful representation of the fundamental group into $SO(3,1)$ has non-integral traces, implying that they cannot embed isometrically in a complex hyperbolic 3-manifold. See Theorem 1.8.