15
$\begingroup$

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?

$\endgroup$
20
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
25
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ $k$ cyclotomic (including $k=\mathbf{Q}$) doesn't mean that eigenvalues have finite order... $\endgroup$ – YCor Apr 2 '19 at 14:22
  • 2
    $\begingroup$ 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. $\endgroup$ – YCor Apr 2 '19 at 14:26
  • $\begingroup$ @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. $\endgroup$ – Geoff Robinson Apr 2 '19 at 14:42
  • $\begingroup$ 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). $\endgroup$ – YCor Apr 2 '19 at 15:03
  • $\begingroup$ 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. $\endgroup$ – Geoff Robinson Apr 2 '19 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.