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Let G be a semisimple Lie group and let k be some number field. Let A be a finite index subgroup of some group B which is discrete and Zariski dense in G. Suppose A is in G(k). Then can we expect that B is in G(k') where k' is a finite Galois extension of k? When can we expect k'=k?

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up vote 7 down vote accepted

I'll prove this when $G=SL(n,\mathbb{C})$. So assume $B\leq SL(n,\mathbb{C})$ is discrete, and $A\leq B$ is a finite-index subgroup. Let $A'\lhd A$ such that $A'\lhd B$ is the core, which is also finite index (by considering the kernel of the action on cosets of $A$ in $B$). Suppose that $A\subset SL(n,k)$ where $k$ is a number field, and assume that $k$ is the smallest field containing the entries of $A$. Since $B$ is Zariski dense in $SL(n)$, $A'$ is also Zariski dense. Take a maximal collection $a_1,\ldots, a_m\in A'$ of linearly independent elements. Since $A$ is finite index in $B$, the field generated over $\mathbb{Q}$ by the entries of $B$ will be spanned over $k$ by the entries of finitely many coset representatives of $A$ in $B$, and therefore will be a finitely generated extension field $k'$ of $k$. Consider the variety $X$ of representations $\rho:B\to SL(n,\mathbb{C})$ such that $\rho_{| A}=Id$. This is a variety again, because it is determined by finitely many matrices which are coset representatives of $A$ in $B$. Considering the fact that $\rho(b)^{-1} a_i \rho(b)=\rho(b^{-1}a_i b)=b^{-1}a_i b\in A'$, we see that $b\rho(b)^{-1}\in Z(a_i)$. But since the $a_i$ are linearly independent, we see that $b \rho(b)^{-1}\in Z(A')=Z(SL(n,k))$ since $A'$ is Zariski dense. Thus, we see that there are only finitely many possibilities for $\rho(b)$ for each $b$, and therefore the variety $X$ is finite. This in turn implies that $k'$ is a finite extension of $k$, since a transcendental extension would give rise to a variety $X$ of dimension $\geq 1$, since you could vary the transcendental generators at will.

Addendum: One may actually show that $k'$ is a Galois extension of $k$. Since $Z(SL(n,\mathbb{C}))=\{ \zeta I: \zeta^n=1\}$, we see that $\rho(b)=\zeta b$, for some $\zeta \in k', \zeta^n=1$. Thus, for different Galois embeddings $\sigma:k'\to \mathbb{C}$ which restrict to the identity on $k$, the entries of $\sigma(b)$ will change by multiplication by a cyclotomic number. So we conclude $b_{ij}^n \in k$ for each matrix entry $b_{ij}$ of $b$. This implies that $k'$ is a Galois extension of $k$ obtained by adding finitely many $n$-th roots to $k$.

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Nice argument. Note that "discrete" is not used, so everything is entirely algebraic. I think the key point is this: Let $G$ be an algebraic group over a field $k$, $\Omega$ an alg. closed extension of $k$. Assume $A\vartriangleleft B$ are subgroups of $G(\Omega)$ with $A\subset G(k)$, and that $A$ has finite centralizer in $G$ (e.g. $A$ is Zariski-dense and $G$ has finite center). Then $B\subset G(\overline{k})$. Proof: if $b\in B$ and $\rho$ is a $k$-automorphism of $\Omega$, then Agol's computation shows that $\rho(b)b^{-1}$ centralizes $A$, hence the "Galois" orbit of $b$ is finite. – Laurent Moret-Bailly Oct 8 '10 at 11:45

Agol's argument is really nice. For what it is worth, I would like to add a few comments. Suppose we assume that $G$ is absolutely simple (in particular, $G$ is centreless). Then Agol's argument shows that $k'=k$. Secondly, one can replace $G$ by a conjugate in its adjoint representation and assume that the smaller Zariski dense subgroup $A$ is contained in $G(k_0)$ where $k_0$ is the "smallest possible field", namely the field generated by the traces of elements of $A$ in the adjoint representation. All these are proved in an old paper of Vinberg.

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