# Finitely generated matrix groups whose eigenvalues are all algebraic

Let $$G$$ be a finitely generated subgroup of $$GL(n,\mathbb{C})$$. Assume that there exists a number field $$k$$ (i.e. a finite extension of $$\mathbb{Q}$$) such that for all $$g \in G$$, the eigenvalues of $$g$$ all lie in $$k$$. This implies that $$g$$ is conjugate to an element of $$GL(n,k)$$.

Question: must it be the case that some conjugate of $$G$$ lies in $$GL(n,k)$$? Or at least $$GL(n,k')$$ for some finite extension $$k'$$ of $$k$$? If this is not true, what kinds of assumptions can I put on $$G$$ to ensure that it is?

## 3 Answers

At the positive side, if $$G$$ acts irreducibly on $$\mathbf{C}^n$$ and $$k$$ is an arbitrary subfield of $$\mathbf{C}$$, then the answer is yes (allowing some field extension $$k'$$ of degree dividing $$n$$). This even works assuming that $$G$$ is a multiplicative submonoid of $$M_n(\mathbf{C})$$ (keeping the irreducibility assumption).

See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)

Robert Israel's simple example shows that some assumption such as irreducibility has to be done.

Here's one easy example. Let $$G$$ be generated by $$\pmatrix{1 & x\cr 0 & 1}$$ for $$x$$ in some finite set $$X$$ of complex numbers. All eigenvalues are $$1$$, so we can take $$k = \mathbb Q$$. If $$G$$ is conjugate by $$S$$ to a subgroup of $$GL(2,\mathbb Q)$$, then the members of $$X$$ are in the field generated by the matrix elements of $$S$$, and we can choose $$X$$ so that this is impossible (e.g. take more than $$4$$ numbers that are algebraically independent).

Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $$G$$ of $${\rm GL}(n,\mathbb{C})$$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.

Hence the answer to your question is "yes" , if every eigenvalue of every element of $$G$$ is a root of unity and every element of $$G$$ is semisimple.

In that case, once we know that $$G$$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $$X$$ of $${\rm GL}(n,\mathbb{C})$$ is conjugate within $${\rm GL}(n,\mathbb{C})$$ to a subgroup of $${\rm GL}(n,\mathbb{Q}[\omega]),$$ where $$\omega$$ is a primitive complex $$|G|$$-th root of unity.

• $k$ cyclotomic (including $k=\mathbf{Q}$) doesn't mean that eigenvalues have finite order... – YCor Apr 2 '19 at 14:22
• and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element. – YCor Apr 2 '19 at 14:26
• @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either. – Geoff Robinson Apr 2 '19 at 14:42
• If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $\mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics). – YCor Apr 2 '19 at 15:03
• That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite. – Geoff Robinson Apr 2 '19 at 16:25