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Does the following statement:

"Let $G$ be a finitely generated group and let $X(G)$ be the $SL(2,\mathbb{C})$-character variety of $G$. Suppose $X(G)$ contains an irreducible component $X_0$ such that for every $g\in G$ and $\chi_{\rho}\in X_0$, $\chi_{\rho}(g)$ is an algebraic number. Then $\dim_{\mathbb{C}}X_0=0$."

contradict this next statement which is true by result of Thurston:

"The trace field of a finite covolume Kleinian group is a finite extension of $\mathbb{Q}$"

I am asking because if we take the figure-8 complement then it is a complete finite volume hyperbolic 3-manifold with finitely generated fundamental group. However the closure of the set of irreducible character has $\dim_{\mathbb{C}} =1 > 0$.

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The first statement is trivially true. The second result is essentially, due to Selberg, definitely not Thurston and follows easily from Mostow rigidity theorem. Why do you think there is a contradiction here? – Misha Jun 7 '13 at 22:03
    
I am confused about the dimension. If $dim_{\mathbb{C}} X_0>0$ for an irreducible component what conditions do we need to add in order to have some transcendental trace? What is the contrapositive of the first statement ? – christian Jun 7 '13 at 22:28
1  
christian: Consider a connected Riemann surface $Y$ and a nonconstant holomorphic map $\tau: Y\to {\mathbb C}$. Does it takes some transcendental values? Now, think about your question. – Misha Jun 7 '13 at 22:54
    
christian: Here is another thing you should consider. Suppose $\Gamma$ is a lattice in $SL(2,\C)$ and $\rho: \Gamma\to SL(2,C)$ is a representation. Is the image of $\Gamma$ also a lattice? – Misha Jun 7 '13 at 23:45

As Misha said in the comments, if the dimension of the $\mathrm{SL}(2,\mathbb{C})$-character variety is greater than 0, there will be characters with values that are not algebraic numbers. This is clear since traces of words in $G$ generate the coordinate ring of $X(G)$. In particular, let $t_{g_1}$,...,$t_{g_m}$ be a generating set for the coordinate ring, where $t_{g_i}(\chi_\rho)=\mathrm{tr}(\rho(g_i))$. Then the map $T:X(G)\to \mathbb{C}^m$ defined by $\chi_\rho\mapsto (...,t_{g_i}(\chi_\rho),...)$ is an algebraic embedding. In particular, notice that for any $g\in G$ that $t_g(\chi_\rho)=\mathrm{tr}(\rho(g))=\chi_\rho(g)$, and so the image of the characters coincides with the image of the coordinates.

And the second statement appears in Thurston's, The geometry and topology of 3-manifolds, Mimeographed lecture notes, Princeton University, 1978 as Proposition 6.7.4 according to Alan Reid's A note on trace-fields of Kleinian groups, Bull. London Math. Soc. 22 (1990), no. 4, 349–352, which is worth reading to see how to get an invariant of the commensurability class from the trace field. As Misha states it is a consequence of Mostow Rigidity.

Again as Misha noted in the comments, there is no contradiction between these two statements. The Kleinian group $\Gamma$ is a discrete subgroup of $\mathrm{PSL}(2,\mathbb{C})$ such that $H^3/\Gamma$ has finite volume, and the trace-field of $\Gamma$ is defined to be the field $\mathbb{Q}(\mathrm{tr}(\gamma)\ |\ \gamma \in \Gamma)$. On the other hand, coordinates in the character variety are coordinates on the full moduli space of (unimodular) representations of $G$. Not every point in the character variety corresponds to a discrete subgroup, let alone a hyperbolic structure.

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