It is well-known that finite subgroups of $PGL_2(\mathbb{C})$ are cyclic groups, dihedral groups, A4, S4 and A5 and each of these groups occurs exactly once (up to conjugacy). These facts are classical.

If $K$ is an arbitrary field then theres are Beauville's notes about $PGL_2(K)$: http://arxiv.org/abs/0909.3942

In dimension $n=3$ there also classical results about subgroups of $PGL(3,\mathbb{C})$. One can also show the connection between subgroups of $SU(3)$ and $PGL(3,\mathbb{C})$. See this discussion for details: https://math.stackexchange.com/questions/42904/finite-subgroups-of-pgl3-c

So my questions is what can we say about finite subgroups of $PGL(3,K)$ where $K$ is not necessarily algebraicly closed? More concretely, knowing, for example, finite subgroups of $PGL(3,\mathbb{C})$ can one classify finite subgroups in $PGL(3,K)$ for any subfield $K\subset\mathbb{C}$ (up to conjugacy)?

  • 2
    $\begingroup$ Once you've got finite subgroups of $PGL(3,\mathbb{C})$ you should be able to look at their character table to see, for what fields $K$, $PGL(3,K)$ contains any given group. At least that's the way I'd start thinking about it... $\endgroup$ – Nick Gill Apr 8 '13 at 15:35

Edited in view of Derek Holt's comment on Schur indices: These things are well studied in the literature. You probably want to restrict to irreducible subgroups, and it's probably just as well to work with ${\rm GL}(3,K).$ In such a low dimensions, the Schur index usually will not play much of a role. The Schur index usually arises because it can happen that a complex irreducible character $\chi$ may take values in a field $K,$ but might not be afforded by a representation over $K.$ There is, however, a smallest integer $m_{K}(\chi)$ such that the character $m_{K}(\chi) \chi$ is afforded by a representation over $K$ and $m_{K}(\chi)$ divides $\chi(1).$ If $m_{K}(\chi) =3,$ then representation affording $\chi$ can only be realised over a degree $3$ extension of $K$.Except in degenerate cases, we won't have irreducible representations over $K$ which are not absolutely irreducible, since such a a representation would break up over some extension field of $K$ as a sum of Galois conjugate representations of the same degree. The finite irreducible subgroups of ${\rm GL}(3,\mathbb{C})$ have been known for a century or so. Such an imprimitive group has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{3}.$ The primitive ones may be rescaled so that all elements are unimodular, and once this is done, we obtain $G/Z(G)$ isomorphic to $A_{5}, A_{6},{\rm PSL}(2,7)$ or else $G$ is a solvable group with $G/O_{3}(G)$ isomorphic to ${\rm SL}(2,3)$ and $[G:Z(G)] = 216.$

| cite | improve this answer | |

Just to complicate things a little, it is possible for a 3-dimensional representation to have Schur index 3. For example, the group of order 63 with centre of order 3 and presentation

$\langle x,y \mid x^7= y^9=1, y^{-1}xy = x^2\rangle$

has four such 3-dimensional complex characters. Their character rings are all equal to the cyclotomic field $F$ of $21$st roots of 1, which is an extension of ${\mathbb Q}$ of degree 12.

I don't know how easy it is to decide whether this representation can be written over some specified subfield $K$ of ${\mathbb C}$. Certainly $K$ must be an extension of $F$ of degree at least 3. One possible $K$ of minimal degree is the field of $|G|$-th roots of 1, but there are probably others.

| cite | improve this answer | |
  • $\begingroup$ To amplify: The Fitting subgroup of your group is cyclic of order $21.$ Take a faithful 1-dimensional representation of that Fitting subgroup. It induces to an absolutely irreducible representation of degree 3. The induced representation is 3-dimensional, and is realized over the field generated by the $21$st-roots of unity, a degree $12$ extension of the rationals. The field generated by the character is actually a degree $4$ extension, generated by a primitive cube root of unity and $\sqrt{-7}.$ $\endgroup$ – Geoff Robinson Apr 9 '13 at 22:57
  • $\begingroup$ Yes you are right, I had miscalculated the character field! $\endgroup$ – Derek Holt Apr 10 '13 at 8:00

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.