Let $X$ be a complete finite-volume orientable hyperbolic $3$-manifold, and let $\Gamma$ be a Kleinian representation of $\pi_1(X)$. Let $K\Gamma:=\mathbb{Q}\big(\{\mathrm{tr}\mid\gamma\in\Gamma\}\big)$, i.e. its trace field. Let $\Gamma^{(2)}:=\langle\gamma^2\mid\gamma\in\Gamma\rangle$, the group generated by squares in $\Gamma$ and let $k\Gamma=K\Gamma^{(2)}$, i.e. its invariant trace field. Then these are number fields, $K\Gamma$ is a manifold invariant and $k\Gamma$ is a commensurability invariant.

I've been told that if $X$ contains a immersed closed totally-geodesic surface, then $K\Gamma$ is a degree $2$ extension of a real field $F$ (where $F$ is the trace field of the surface under a fixed embedding into $X$).

Is this correct? Do they really mean to say this of $k\Gamma$ rather than of $K\Gamma$? Are there some additional conditions I'm missing that do make this true for $K\Gamma$?

If this is true, would you please provide a reference? (I'm scouring through my old copy of Maclachlan and Reid and can't seem to find it.) If it isn't true, would you know a counterexample?

Edit: I think this is true if I add the requirement that $\Gamma$ be arithmetic. The only part of that I'm missing is that the trace field of the surface must be the full real part of the trace field of the $3$-manifold. I can't find a reference for that either though.

It's obvious (trivial even) in the arithmetic noncompact case.


The answer to this follows easily from section 9.5 of Machlachlan and Reid -- which somehow I managed not to see every other time I checked the book! The thing stated in my question is not true, but the following is true.

Suppose $X$ is arithmetic and contains an immersed closed totally-geodesic surface $S$. Then $\Gamma$ contains a non-elementary Fuchsian group $\Delta'$. Then $\Delta'$ is contained in an arithmetic Fuchsian group $\Delta$ which satisfies $[k\Gamma:k\Delta]=2$ and $k\Gamma\cap\mathbb{R}=k\Delta$.


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