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I will preface this by saying that while I am familiar with the general theory of (semi)-Riemannian manifolds, I am a complete novice when it comes to the specifics of hyperbolic manifolds (I am using them for examples when computing cohomology with local coefficients). I was originally trained as a physicist, so I am also not very familiar with field theory (which comes up in the construction). So apologies in advance if there are some "dumb" questions here.

The purpose of my question is to understand the following construction referred to in the paper Local rigidity of hyperbolic manifolds with geodesic boundary by Kerckhoff and Storm, where they state:

"For each dimension $n>2$ [Gromov and Thurston] construct an infinite number of closed hyperbolic n-manifolds $V$ with the following properties: $V$ has a codimension 1 embedded, totally geodesic submanifold $M$ which itself has a codimension 1 embedded, totally geodesic submanifold $P$ (so $P$ is codimension 2 in $V$)."

The Gromov & Thurston paper being referred to is Pinching Constants for Hyperbolic Manifolds. It appears that the brief Section 1 contains the relevant construction, which I have snipped from the linked pdf:enter image description here

I am unfortunately confused about "why" this construction gives us a hyperbolic manifold containing a codimension 1 totally geodesic hypersurface itself containing a totally geodesic codimension 1 hypersurface. To be more specific, why use this $\Phi_n$ instead of the usual Lorentzian quadratic form, and why finite index subgroups of this particular $\Gamma_n$?

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    $\begingroup$ The usual Lorentz form gives a non-compact but finite volume manifold. The one involving $\sqrt{2}$ gives a compact hyperbolic manifold. The other conditions remain the same> BTW, are the letters Tha-Tha in Kannada? $\endgroup$
    – Venkataramana
    Commented Oct 11, 2018 at 12:44
  • $\begingroup$ Thanks for your comment! Unfortunately I'm still not sure "why" the $\sqrt{2}$ gives the desired result; E.g. does it matter that $2$ is prime? Would $\sqrt{7}$ or $\sqrt{21}$ give the same result? Also, I'm not sure what you mean by "are the letters Tha-Tha in Kannada"? $\endgroup$
    – ಠ_ಠ
    Commented Oct 11, 2018 at 20:15
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    $\begingroup$ in your username, the letters seem to be in Kannada. $sqrt 2$ is used so that there are two embeddings of the field ${\mathbb Q}(\sqrt{2})$ into $\mathbb R$ one of which gives the signature (n,1) for the quadratic form and the other gives (n+1,0) ,i.e. positive definite. Under these conditions, it can be proved that the manifold is compact. This works equally well with $\sqrt{7}$ or $\sqrt{21}$. $\endgroup$
    – Venkataramana
    Commented Oct 12, 2018 at 2:54
  • $\begingroup$ @Venkataramana Ah yep it's "tha" in Kannada. It's used to make the "look of disapproval", which is my username. And thanks for your further explanations. So it seems to me from your comment and Autumn Kent's answer that the only thing that matters for constructing this sort of hyperbolic manifold with totally geodesic hypersurface itself containing a totally geodesic hypersurface is that we pick $\Gamma_n$ to be a cocompact group containing the involution $x_1 \mapsto -x_1$. Is this correct? $\endgroup$
    – ಠ_ಠ
    Commented Oct 24, 2018 at 22:44
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    $\begingroup$ That is sort of right. What matters is that you have a quadratic form $Q$ over a totally real number field which has signature (n,1) is one real embedding but is definite in all others. Then you get a cocompact lattice in $SO(n,1)$ with the associated compact hyperbolic manifold which contains a totally geodesic hypersurface whcich contains a totally geodesic hypersurface. You may look at old papers of John Millson and Raghunathan which consider such manifolds. $\endgroup$
    – Venkataramana
    Commented Oct 25, 2018 at 2:15

1 Answer 1

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The fixed point set of the involution in the universal cover is a totally geodesic copy of hyperbolic space of codimension 1. So that gives you the first totally geodesic hypersurface in some finite cover of the given thing. And since the hypersurface is the same type of lattice, there is a finite cover of it that has a totally geodesic hypersurface by the same argument. There is a single finite cover of the original thing doing both at once.

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  • $\begingroup$ Thanks for your answer! Is it obvious why the particular choice of group $\Gamma_n$ allows us to do this? Also, for the involution, I assume that the authors are using the hyperboloid model of hyperbolic space, since in this case geodesics in $\mathbb{H}^n$ correspond to hyperplanes like $\{x \in \mathbb{R}^{n+1} \mid x_1 =0\} in $\mathbb{R}^{n+1}$? $\endgroup$
    – ಠ_ಠ
    Commented Oct 11, 2018 at 20:44
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    $\begingroup$ I think the $\Gamma_n$ is just chosen so that the involution is clear and that when you restrict to the $x_0 =0$ subspace you have another involution of the same kind. $\endgroup$
    – Autumn Kent
    Commented Oct 14, 2018 at 4:18
  • $\begingroup$ So instead of using $\Gamma_n$, can we just choose any cocompact group containing this involution? $\endgroup$
    – ಠ_ಠ
    Commented Oct 24, 2018 at 22:39
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    $\begingroup$ Well the involution isn't in your group, it's just that the involution exists and preserves the form. And then it's important that the hypersurface also has an involution. In this case that's taken care of because the subgroup stabilizing the hyperplane is the same type of group that you started with (you start with $\Gamma_n$, and the subgroup is like $\Gamma_{n-1}$). $\endgroup$
    – Autumn Kent
    Commented Oct 25, 2018 at 1:25

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