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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 $G$. If $X(G)$ 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 the following 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
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