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?

  • $\begingroup$ Are your symmetric spaces required to be connected? If yes, then the answer is clearly negative. $\endgroup$ Mar 11 at 4:19
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    $\begingroup$ @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. $\endgroup$ Mar 12 at 0:22
  • $\begingroup$ 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. $\endgroup$ Mar 12 at 0:59
  • $\begingroup$ 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. $\endgroup$ Mar 12 at 17:08
  • $\begingroup$ No, that will not suffice either. The orbifold locus could be a 3-valent graph. $\endgroup$ Mar 12 at 18:17

1 Answer 1


Here is what I think is the correct setup:

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.

This picture serves a correction to the erroneous claim in the post "I believe it follows that..."

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.

  • $\begingroup$ I appreciate the elaboration and correction, thank you! $\endgroup$ Mar 14 at 17:32

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