# Inheritance of arithmeticity properties in orbifold strata

Suppose $$M = K\backslash G/\Gamma$$ is a quotient of a symmetric space by a lattice. I don't know all of the proper adjectives to apply here (e.g. what should be said about $$G$$ and so on), but I wouldn't be upset if we just took $$G=\operatorname{O}(1,n)$$ and $$K = \operatorname{O}(n)$$ so that $$M$$ is a finite volume hyperbolic orbifold.

Edit: as pointed out by @MoisheKohan in the comments, I really shouldn't be asking about the real hyperbolic setting. Let's instead take $$G=\operatorname{U}(1,n)$$ and $$K = \operatorname{U}(n)$$ so that $$M$$ is a finite volume complex hyperbolic orbifold.

Now suppose that $$M' \subset M$$ is a connected component of an orbifold locus. I believe it follows that $$M'$$ is also a lattice quotient (of smaller dimension), and feel free to correct me if I'm wrong about that. Now what can be said about the relationship between the arithmeticity of $$M$$ and $$M'$$? E.g. does the (non)arithmeticity of one imply the (non)arithmeticity of the other?

• Are your symmetric spaces required to be connected? If yes, then the answer is clearly negative. Mar 11 at 4:19
• @MoisheKohan could you elaborate a bit? (Non)arithmetic lattices have come up in the things I've been working on but almost as a side effect rather than the main focus. I'm not an expert. Mar 12 at 0:22
• OK, suppose you have a lattice $\Gamma_0< PSL(2,{\mathbb R})$ uniformizing a hyperelliptic hyperbolic surface $S$. Then $S$ admits a hyperelliptic involution $\tau$. Lifting $\tau$ to the hyperbolic plane, we obtain a lattice $\Gamma< PSL(2,{\mathbb R})$ which contains $\Gamma_0$ as an index 2 subgroup. The quotient $H^2/\Gamma$ is an orbifold $O$ with at least two singular points (projections to $O$ of the fixed-points of $\tau$). Hence, the orbifold locus of $O$ is disconnected, implying that it cannot be a lattice quotient (under the most common definition). Hence, your belief is wrong. Mar 12 at 0:59
• Ah I see. Let me modify my question to request that $M'$ be a connected component of an orbifold locus, that's what I really had in mind. Mar 12 at 17:08
• No, that will not suffice either. The orbifold locus could be a 3-valent graph. Mar 12 at 18:17

Let $$X$$ be a symmetric space of noncompact type, $$\Gamma$$ is a lattice in the isometry group of $$X$$. Then $$\Gamma$$ has finitely many $$\Gamma$$-conjugacy classes of finite subgroups. Let $$\Phi$$ be one of these finite subgroups (I will pick one from each conjugacy class). Then $$\Gamma$$ fixes a symmetric subspace $$Y\subset X$$. The (set-wise) $$\Gamma$$-stabilizer $$\Gamma_Y$$ of $$Y$$ is a sublattice in $$\Gamma$$: It acts on $$Y$$ as a lattice. The quotient-orbifold $$Y/\Gamma_Y$$ has a natural projection $$Q_Y$$ to the orbifold $$X/\Gamma$$, the projection map $$Y/\Gamma_Y\to X/\Gamma$$ is an isometric immersion in the sense of orbifolds. This defines a (finite) stratification of the orbifold locus of $$X/\Gamma$$: The strata are $$Q_Y$$'s.
Now, one can ask meaningful questions relating arithmeticity of $$\Gamma$$ and the sublattices $$\Gamma_Y$$. If memory serves me well, arithmeticity of $$\Gamma$$ implies arithmeticity of each $$\Gamma_Y$$. However, the converse is false: It is surely false for $$X={\mathbb H}^n$$: One can build examples in all dimensions using the Gromov--Piatetsky-Shapiro construction. I did not check Deligne-Mostov nonarithmetic examples, but, likely, you will find some arithmetic complex one-dimensional immersed totally geodesic suborbifolds. Since you like DM-examples, you should try finding such.