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I am reviewing some things I studied over a year ago about the arithmetic of hyperbolic $3$-manifolds, and hoping for some help getting the rust out.

Let $\Gamma$ be a torsion-free Kleinian group, i.e. a discrete subgroup of $\mathrm{PSL}_2(\mathbb{C})$ with no elliptic elements. Suppose further that $\Gamma$ has finite co-volume, i.e. the manifold $\mathcal{H}^3/\Gamma$ given by the Möbius action of $\Gamma$ on the hyperbolic upper half-space $\mathcal{H}^3$ has finite volume. Mostow-Prasad rigidity tells us that if $\Gamma'\cong\Gamma$ is another finite-covolume torsion-free Kleinian group, then there exists some $\delta\in\mathrm{PSL}_2(\mathbb{C})$ such that this isomorphism is realized by $$\Gamma\rightarrow\Gamma', \quad \gamma\mapsto\delta\gamma\delta^{-1}.$$

The trace field of $\Gamma$ is defined as $\mathbb{Q}\big(\big\{\mathrm{tr}(\gamma)\mid\gamma\in\Gamma\big\}\big)$, and I'll denote this by $K\Gamma$. Since the trace is invariant under conjugation, it is invariant of the isomorphism class of $\Gamma$. A theorem of Neumann-Reid tells us that the trace field is a concrete number field, i.e. a finite extension of $\mathbb{Q}$ with a fixed embedding into $\mathbb{C}$.

Now suppose the trace field of $\Gamma$ has a complex embedding $\sigma$ in addition to the identity map and complex conjugation (this forces $\Gamma$ to be non-arithmetic), and define $$\sigma(\Gamma):= \left\{\begin{pmatrix} \sigma(a) & \sigma(b)\\ \sigma(c) & \sigma(d) \end{pmatrix}\;\middle|\; \begin{pmatrix} a & b\\ c & d \end{pmatrix}\in\Gamma\right\}.$$ This seems to give $\Gamma\cong\sigma(\Gamma)$ and $K\Gamma\neq K\sigma(\Gamma)$, but that would contradict the previous paragraph.

What am I missing?

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The answer is that any embeddings besides the identity and complex conjugation are not discrete. Thus $\sigma(\Gamma)$ as in the question is not actually a Kleinian group, so the rigidity theorem does not apply to it.

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  • $\begingroup$ Right.......... $\endgroup$ Nov 22, 2017 at 23:16

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