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edited Mar 16, 2021 at 16:21

Now, given a subgroup $\Lambda^\prime \le \Lambda$ (say, torsion-free), we have that $\Gamma^\prime = \Lambda^\prime \cap \Gamma$ is the stabilizer of $\mathbb{H}^3$ in $\mathbb{B}^3$, so we have a $\pi_1$-injective inclusion: $$ X^\prime = \Lambda^\prime \backslash \mathbb{H}^3 \looparrowright \Lambda^\prime \backslash \mathbb{B}^3 = Y^\prime$$$$ X^\prime = \Gamma^\prime \backslash \mathbb{H}^3 \looparrowright \Lambda^\prime \backslash \mathbb{B}^3 = Y^\prime$$ Since $\Lambda^\prime$ is torsion-free, so is $\Gamma^\prime$, so $X^\prime$ is a hyperbolic $3$-manifold. It is well-known (for example, see Goldman's book Complex Hyperbolic Geometry) that there is a $\mathrm{PU}(3,1)$-invariant $2$-form on $\mathbb{B}^3$ that descends to $Y^\prime$. Assuming that $Y^\prime$ is compact (again, equivalently, that $K \neq \mathbb{Q}$), this turns $Y^\prime$ into a smooth compact Kähler $3$-fold (in fact, a smooth projective $3$-fold). In the noncompact case, it is quasiprojective and admits a "nice" compactification, but I'll stick to the compact case for simplicity.

One can find, for example, hyperbolic knot and link complements that cannot be a totally geodesic submanifold of any complex hyperbolic $n$-manifold of any dimension $n$. The point is that there are knot and link complements for which the holonomy for the complete hyperbolic structure has elements whose trace istraces are not an algebraic integer (only an algebraic number). As Ian describes, this is an obstruction to being a totally geodesic submanifold of a complex hyperbolic manifold.

Now, given a subgroup $\Lambda^\prime \le \Lambda$ (say, torsion-free), we have that $\Gamma^\prime = \Lambda^\prime \cap \Gamma$ is the stabilizer of $\mathbb{H}^3$ in $\mathbb{B}^3$, so we have a $\pi_1$-injective inclusion: $$ X^\prime = \Lambda^\prime \backslash \mathbb{H}^3 \looparrowright \Lambda^\prime \backslash \mathbb{B}^3 = Y^\prime$$ Since $\Lambda^\prime$ is torsion-free, so is $\Gamma^\prime$, so $X^\prime$ is a hyperbolic $3$-manifold. It is well-known (for example, see Goldman's book Complex Hyperbolic Geometry) that there is a $\mathrm{PU}(3,1)$-invariant $2$-form on $\mathbb{B}^3$ that descends to $Y^\prime$. Assuming that $Y^\prime$ is compact (again, equivalently, that $K \neq \mathbb{Q}$), this turns $Y^\prime$ into a smooth compact Kähler $3$-fold (in fact, a smooth projective $3$-fold). In the noncompact case, it is quasiprojective and admits a "nice" compactification, but I'll stick to the compact case for simplicity.

One can find, for example, hyperbolic knot and link complements that cannot be a totally geodesic submanifold of any complex hyperbolic $n$-manifold of any dimension $n$. The point is that there are knot and link complements for which the holonomy for the complete hyperbolic structure has elements whose trace is not an algebraic integer (only an algebraic number). As Ian describes, this is an obstruction to being a totally geodesic submanifold of a complex hyperbolic manifold.

Now, given a subgroup $\Lambda^\prime \le \Lambda$ (say, torsion-free), we have that $\Gamma^\prime = \Lambda^\prime \cap \Gamma$ is the stabilizer of $\mathbb{H}^3$ in $\mathbb{B}^3$, so we have a $\pi_1$-injective inclusion: $$ X^\prime = \Gamma^\prime \backslash \mathbb{H}^3 \looparrowright \Lambda^\prime \backslash \mathbb{B}^3 = Y^\prime$$ Since $\Lambda^\prime$ is torsion-free, so is $\Gamma^\prime$, so $X^\prime$ is a hyperbolic $3$-manifold. It is well-known (for example, see Goldman's book Complex Hyperbolic Geometry) that there is a $\mathrm{PU}(3,1)$-invariant $2$-form on $\mathbb{B}^3$ that descends to $Y^\prime$. Assuming that $Y^\prime$ is compact (again, equivalently, that $K \neq \mathbb{Q}$), this turns $Y^\prime$ into a smooth compact Kähler $3$-fold (in fact, a smooth projective $3$-fold). In the noncompact case, it is quasiprojective and admits a "nice" compactification, but I'll stick to the compact case for simplicity.

One can find, for example, hyperbolic knot and link complements that cannot be a totally geodesic submanifold of any complex hyperbolic $n$-manifold of any dimension $n$. The point is that there are knot and link complements for which the holonomy for the complete hyperbolic structure has elements whose traces are not an algebraic integer (only an algebraic number). As Ian describes, this is an obstruction to being a totally geodesic submanifold of a complex hyperbolic manifold.

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created Mar 16, 2021 at 16:08

Toffee

Here is one general construction that give you infinitely many (closed or noncompact, finite volume) hyperbolic 3-manifolds as $\pi_1$-injective immersed submanifolds of Kähler 3-folds. They will be immersed totally geodesic submanifolds of complex hyperbolic $3$-manifolds.

In fact, for certain closed hyperbolic $3$-manifolds, I will build infinitely many homeomorphism classes of Kähler $3$-folds that contain it as a $\pi_1$-injective submanifold. Moreover, each hyperbolic $3$-manifold is a component of the fixed-set of an antiholomorphic involution on the $3$-fold (it might be interesting to further refine this to make sure it's precisely the fixed set - this can realize hyperbolic $3$-manifolds as real projective varieties of explicitly computable dimension/degree).

Consider the hyperbolic 3-manifolds of "simplest type". Fix a totally real number field $K$ with ring of integers $\mathcal{O}_K$ and a quadratic form $q$ on $K^4$ so that $q$ has signature $(3,1)$ at one real embedding of $K$ and is definite at the other embeddings. Then $$\Gamma = \mathrm{PO}(q, \mathcal{O}_K)$$ defines an arithmetic lattice in $\mathrm{PO}(3,1) \ge \mathrm{Isom}^+(\mathbb{H}^3)$ and $\Gamma \backslash \mathbb{H}^3$ is a finite volume hyperbolic $3$-orbifold ($\Gamma$ typically has torsion, e.g., the reflection through a hyperplane). For all but finitely many ideals $\mathcal{I} \subseteq \mathcal{O}_K$, the finite index subgroup $\Gamma(\mathcal{I})$ consisting of the elements that are congruent to the identity modulo $\mathcal{I}$ is torsion-free, hence $X_{\mathcal{I}} = \Gamma(\mathcal{I}) \backslash \mathbb{H}^3$ is a hyperbolic $3$-manifold. Finally, $X_{\mathcal{I}}$ is compact if and only if $K \neq \mathbb{Q}$. See the book The Arithmetic of Hyperbolic 3-manifolds by Maclachlan and Reid (particularly Chapters 9 and 10) for more on all of this.

Now, let $L / K$ be a totally imaginary quadratic extension. Notice that there are infinitely many choices of $L$ (e.g., there are infinitely many imaginary quadratic extensions of $\mathbb{Q}$). Every real embedding of $K$ extends to a unique complex conjugate pair of complex embeddings of $L$, and the nontrivial element of $\mathrm{Gal}(L/K)$ always extends to complex conjugation on $\mathbb{C} / \mathbb{R}$. This allows us to consider our quadratic form $q$ as a hermitian form on $L^4$, and we obtain an analogous lattice $$ \Lambda = \mathrm{PU}(q, \mathcal{O}_L) $$ in $\mathrm{PU}(3,1)$. The associated symmetric space is the unit ball $\mathbb{B}^3$ in $\mathbb{C}^3$ with its Bergman metric. There is an antiholomorphic involution of $\mathbb{B}^3$ with fixed point set the ball model of $\mathbb{H}^3$, and this is compatible with the natural inclusion $\Gamma \hookrightarrow \Lambda$.

Now, given a subgroup $\Lambda^\prime \le \Lambda$ (say, torsion-free), we have that $\Gamma^\prime = \Lambda^\prime \cap \Gamma$ is the stabilizer of $\mathbb{H}^3$ in $\mathbb{B}^3$, so we have a $\pi_1$-injective inclusion: $$ X^\prime = \Lambda^\prime \backslash \mathbb{H}^3 \looparrowright \Lambda^\prime \backslash \mathbb{B}^3 = Y^\prime$$ Since $\Lambda^\prime$ is torsion-free, so is $\Gamma^\prime$, so $X^\prime$ is a hyperbolic $3$-manifold. It is well-known (for example, see Goldman's book Complex Hyperbolic Geometry) that there is a $\mathrm{PU}(3,1)$-invariant $2$-form on $\mathbb{B}^3$ that descends to $Y^\prime$. Assuming that $Y^\prime$ is compact (again, equivalently, that $K \neq \mathbb{Q}$), this turns $Y^\prime$ into a smooth compact Kähler $3$-fold (in fact, a smooth projective $3$-fold). In the noncompact case, it is quasiprojective and admits a "nice" compactification, but I'll stick to the compact case for simplicity.

One can moreover ask that $\Lambda^\prime$ be normalized by the element $\iota \in \mathrm{Isom}(\mathbb{B}^3)$ that is the antiholomorphic involution with fixed set $\mathbb{H}^3$. This implies that $\iota$ descends to an antiholomorphic involution of $Y^\prime$ and that $X^\prime$ is contained in the fixed set of this involution.

Question 1: Which finite index subgroups of $\Gamma$ can be realized as such a $\Gamma^\prime$?

In general I don't know!

It isn't hard to use ideals $\mathcal{J} \subseteq \mathcal{O}_L$ so that $\mathcal{J} \cap \mathcal{O}_K = \mathcal{I}$ to get $X_{\mathcal{I}} \looparrowright Y_{\mathcal{J}}$. This at least gives infinitely many distinct $\Gamma^\prime \le \Gamma$, hence infintely many distinct hyperbolic $3$-manifolds. In fact, as we vary the form $q$ and the number field $K$, we get infinitely many distinct commensurability classes of hyperbolic $3$-manifolds, each of which contain infinitely many distinct manifolds that immerse as a $\pi_1$-injective submanifold of a Kähler $3$-fold. One can use elementary number theory to produce examples with antiholomorphic involutions fixing $X_{\mathcal{I}}$.
One can fix $K$ and $q$ and vary $L$ to instead fix the hyperbolic $3$-manifold, say some $X_{\mathcal{I}}$ for a fixed $\mathcal{I}$, and immerse it in infinitely many distinct commensurability classes of Kähler $3$-folds.
Alternately, once we have that $\Gamma^\prime$ is the stabilizer of $\mathbb{H}^3$ in $\Lambda^\prime$ as above, then $\Gamma^\prime$ is separable in $\Lambda^\prime$, by an argument due to Darren Long (written up in this setting by Nicolas Bergeron in Premier nombre de Betti et spectre du laplacien de certaines variétés hyperboliques, Enseign. Math. (2) 46 (2000), no. 1-2, 109–137). This means that: $$\Lambda^\prime = \underset{[\Gamma^\prime : \Delta] < \infty}{\bigcup_{\Lambda^\prime \le \Delta \le \Gamma^\prime}} \Delta $$ Thus we immerse (in fact, eventually embed) $X^\prime$ in the infinitely many distinct (commensurable) $3$-folds $\Delta \backslash \mathbb{B}^3$.

Something that seems pretty interesting is whether or not every finite index $\Gamma^\prime \le \Gamma$ can be realized as $\Lambda^\prime \cap \Gamma$ for some finite index $\Lambda^\prime \le \Lambda$. This would prove that every hyperbolic $3$-manifold covering our original hyperbolic $3$-orbifold $X$ admits such an embedding. I don't know if this is true.

One way to prove this is to show that the profinite topology on $\Lambda$ induces the full profinite topology on $\Gamma$. This is more or less immediate from the definition of the profinite topology as generated by finite index subgroups. One wants to show that every finite index subgroup of $\Gamma$ is separable in $\Lambda$ in the above sense. Long's argument only proves separability of the full stabilizer of $\mathbb{H}^3$ in $\Lambda^\prime$, not finite index subgroups of the stabilizer.

Cautionary example. The natural embedding $\mathrm{SL}_2(\mathbb{Z}) \le \mathrm{SL}_3(\mathbb{Z})$ gives an immersion: $$ X = \mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}(2) \looparrowright \mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}(3) = Y $$ However, one can prove, using the fact that $\mathrm{SL}_3(\mathbb{Z})$ has the congruence subgroup property but $\mathrm{SL}_2(\mathbb{Z})$ does not, that only very special fininte covers of $X = \mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}^2$ embed as a totally geodesic submanifold in some finite cover of $Y$. However, our lattices $\Lambda \le \mathrm{PU}(3,1)$ are known not to have the congruence subgroup property, so this isn't an obstruction.

One way to show that $\Lambda$ induces the full profinite topology on $\Gamma$ is to prove that there is some pair $\Gamma^\prime \le \Lambda^\prime$ as above so that $Y^\prime$ admits a retraction onto $X^\prime$. However, this is impossible in our setting by work of Carlson and Toledo (Harmonic mappings of Kähler manifolds to locally symmetric spaces, Inst. Hautes Études Sci. Publ. Math. No. 69 (1989), 173–201). This was also written up in a paper of Long and Reid (Subgroup separability and virtual retractions of groups, Topology 47 (2008), no. 3, 137–159).

Why else might you care about this separability problem? If there were some $\Gamma^\prime$ in $\Gamma$ that were not separable in $\Lambda$, then Agol, Groves, and Manning would tell you that there is a Gromov hyperbolic group that is not residually finite. See Residual finiteness, QCERF and fillings of hyperbolic groups, Geom. Topol. 13 (2009), no. 2, 1043–1073.

Question 2: Can every hyperbolic $3$-manifold be realized as a totally geodesic submanifold of a complex hyperbolic $3$-manifold as above?

No! See this post by Ian Agol: Complexifications of hyperbolic manifolds

One can find, for example, hyperbolic knot and link complements that cannot be a totally geodesic submanifold of any complex hyperbolic $n$-manifold of any dimension $n$. The point is that there are knot and link complements for which the holonomy for the complete hyperbolic structure has elements whose trace is not an algebraic integer (only an algebraic number). As Ian describes, this is an obstruction to being a totally geodesic submanifold of a complex hyperbolic manifold.