Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The enumeration of finite subgroups of $\operatorname{PGL}_2(\mathbb{C})$ is one of the classic classification problems: mathematicians in many fields know well that the answer is cyclic groups, dihedral groups, $A_4$, $S_4$ and $A_5$. Moreover each of these groups occurs exactly once, up to conjugacy.

I hadn't thought about this classification very much for many years (if at all), but recently I have had some impetus to do so. In particular I am currently reading this wonderful note of A. Beauville, which treats the classification of finite subgroups of $\operatorname{PGL}_2(K)$ for any field $K$ of characteristic $0$. Here the new work is not so much figuring out exactly which groups can be realized over a given $K$, but rather working out the classification up to $K$-rational conjugacy, which turns out to be an interesting problem in Galois cohomology.

In some recent ruminations I have been thinking about Galois cohomology of finite subgroups of $\operatorname{PGL}_N(K)$, but it occurs to me that I don't know anything concrete past $N = 2$. Well, I can see that every finite group occurs as a subgroup of $\operatorname{PGL}_N(K)$ for any field $K$ and sufficiently large $N$, so obviously one can only ask for so much.

So...what about $\operatorname{PGL}_3(\mathbb{C})$? In particular:

1) What are the finite subgroups of $\operatorname{PGL}_3(\mathbb{C})$.

2) Is it still the case that if a finite group can be embedded in $\operatorname{PGL}_3(\mathbb{C})$, the embedding is unique up to conjugacy? If so, is there a general principle at work here?

Note that Beauville gives proofs of everything in the $\operatorname{PGL}_2$ case. But much of his classification arguments turn on the "accidental isomorphism" $\operatorname{PGL}_2 \cong SO(q)$, where $q = x^2 + yz$. Perhaps if these arguments generalized to $\operatorname{PGL}_N$ in some well-known way, he would have given proofs which generalize as well...

share|improve this question

2 Answers 2

up vote 10 down vote accepted

As I recall, the finite subgroups of $PGL_N(\mathbb C)$ are classified for $N \le 7$. See Blitchfeldt's "Finite collineation groups" (1917) for $N=3, 4$. (Beware: the terminology is quite old---for instance, isomorphic only means something like isomorphic modulo a normal (or maybe central) subgroup in modern notation.)

Feit (The current situation in the theory of finite simple groups. Actes du Congr`es International des Math'ematiciens, Nice 1970) gives a list of maximal finite subgroups in these cases, though there are some cases that he missed in higher dimensions, but I don't have references with me now.

Incidentally, in my thesis, I tried classifying the finite solvable subgroups of $GSp_4(\mathbb C)$ in a more simple manner than Blitchfeldt, though at one crucial point, I couldn't get around using some of his work. Still, it might be useful (and easier to read than Blitchfeldt) if you want to know some basic techniques in these classification questions.

EDIT: It turns out I wrote down the list for $PGL_3(\mathbb C)$ in my thesis. (I knew I had it written down somewhere!)

http://etd.caltech.edu/etd/available/etd-05272004-224316/unrestricted/Thesis.pdf

See chapter 8 for the list. Cf. Appendix B for the GAP notation for the less well known of these groups.

share|improve this answer
    
thanks a lot for this answer. I hate to be that lazy student, but -- is there any chance that you could point me at a webpage that has a complete list?? –  Pete L. Clark Jun 3 '11 at 3:49
    
Pete, do you mean for N=3 or larger N also? I don't know of a webpage with a list for either, but at one point I wrote some things down about the PGL(3) case. If I can find my notes on that, I'll send them to you. I'm currently away for the summer, but I'll try searching my computer later today. (However, I'm not even sure if I typed them up!) –  Kimball Jun 3 '11 at 4:14
    
let's say $N = 3$ for now. Thanks for anything you can provide. –  Pete L. Clark Jun 3 '11 at 6:01
    
Sorry, I don't seem to have anything on my laptop. But if I find Blitchfeldt in the library here, I'll take another look. However I did discover that Yau and Yu's MAMS volume, Gorenstein quotient singularities in dimension three, (which is also not available online to me) contains the classification for SL(3,C) and this may be easier to read than Blitchfeldt or Feit. –  Kimball Jun 3 '11 at 12:51

On mathoverflow a similar question was asked. You can take the list for SU3 and mod out by the centers to get the list for PGL3. Several references for SU3 are given (in more than one answer), and some discussion of SU(n) is also given.

For instance A6 is in PGL3 because the Valentiner group 3.A6 is in SL3. I believe one of the papers referenced even goes so far as to list the groups by their image in PGL3. I remembered the discussion since that is where I learned the name of 3.A6, but I guess noone mentioned it at the time (so I am mentioning it here. VALENTINER!).

share|improve this answer
    
Thanks for your response. Two questions: (i) okay, I see maps $SU_3 \rightarrow GL_3 \rightarrow PGL_3$, so if I know the finite subgroups of $SU_3$ then I will get some finite subgroups of $PGL_3$. But why do I get all of them this way? (ii) What about the uniqueness of the embedding? –  Pete L. Clark Jun 3 '11 at 1:47
    
@Pete: (This answers (i)). Every finite group acts by isometries in some Hermitian inner product. This can be seen by picking an arbitrary Hermitian inner product and averaging it over the group action. Thus, every finite subgroup of $Gl_3$ is conjugate to one in $SU_3$. –  Jason DeVito Jun 3 '11 at 2:00
    
Uniqueness is a good question; I don't know. It won't hold for general n, I don't think. I think though the SU3 to GL3 to PGL3 is likely not to change any conjugacy; otherwise I lose a little faith in "representations are isomorphic iff their characters are equal" (which I think is still true for projective characters). –  Jack Schmidt Jun 3 '11 at 2:10
1  
@Jason: I know that every finite (indeed compact) subgroup of $GL_N(\mathbb{C})$ preserves a Hermitian form and thus is contained in a unitary group and all unitary groups are conjugate. This gets me down to $U(N)$. But it is clear that in general one cannot replace $U(N)$ with $SU(N)$ -- e.g. try $N = 1$. So I'm still not quite understanding this answer. –  Pete L. Clark Jun 3 '11 at 3:44
    
@Pete: Good point. In fact, for all $N$, there are finite groups which embed into $U(N)$ but not $SU(N)$, e.g., the $N$-fold direct sum of $\mathbb{Z}/2\mathbb{Z}$ embeds into $U(N)$ but not into $SU(N)$. –  Jason DeVito Jun 3 '11 at 4:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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