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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).

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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.

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    $\begingroup$ Do you mean that there exist real hyperbolic 3-manifolds whose $\pi_1$ has no embedding into $\mathrm{GL}_d(\mathbf{Z})$ for any $d$? I'm not sure this is what you're saying; I'm just trying to understand. I'd be happy if you slightly expand the argument. $\endgroup$
    – YCor
    Commented Feb 23, 2019 at 21:11
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    $\begingroup$ @YCor: No, just the discrete faithful rep. $\endgroup$
    – Ian Agol
    Commented Feb 23, 2019 at 21:40
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    $\begingroup$ I still don't understand. "the discrete faithful representation" is just one representation, but you want a property of all a class of representations... all discrete faithful representations valued in $SU(n,1)$ for all $n$? $\endgroup$
    – YCor
    Commented Feb 23, 2019 at 21:56
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    $\begingroup$ So if I understand correctly, you say that Esnault-Groechenig method proves that the structural representation of any $\pi_1$ of a (compact? closed?) hyperbolic complex 3-manifold has integral traces. By "structural" I mean the obvious one, in $SU(3,1)$... this certainly deserves a precise name. Second, if a compact real hyperbolic 3-manifold embeds geodesically into a compact complex one, then the restriction of the structural representation of the larger one restricts to to structural representation of the smaller one (this seems obvious but was initially unclear in my mind). $\endgroup$
    – YCor
    Commented Feb 23, 2019 at 22:54
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    $\begingroup$ I think, you are misapplying their result: They require $H^1(X, Ad\rho)=0$ where $\rho$ is a representation into $GL(n, {\mathbb C})$, while your representation is rigid in $SU(n,1)$. This is not the same. The simplest example is given by lattices with $b_1>0$: They are rigid in $SU(n,1)$ but not in $GL(n+1,C)$. Furthermore, my recollection from reading Deligne and Mostow (admittedly, long time ago) is that some of their non-arithmetic examples are non-integral. $\endgroup$
    – Moishe Kohan
    Commented Mar 16, 2021 at 16:37

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