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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: http://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)?

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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... –  Nick Gill Apr 8 '13 at 15:35

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up vote 6 down vote accepted

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 $[G:Z(G)] = 216.$

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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.

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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 $\sqrt{-7}.$ –  Geoff Robinson Apr 9 '13 at 22:57
    
Yes you are right, I had miscalculated the character field! –  Derek Holt Apr 10 '13 at 8:00

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