Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question: What is the number of two-dimensional irreducible representations of a finite group ? How it can be expressed in groups-theoretic terms ? (Number of 1-dimensional irreps is |G/[G,G]| ).

The question is somewhat naive and actually I heard it from our teachers when I was an undergrad many years ago, it was always outspoken with some kind of mysterious flavour - "nobody knows, but may be...".


Some analogies:

There is a paper by V. Drinfeld 1981, which title is "Number of two-dimensional irreducible representations of the fundamental group of a curve over a finite field".

The main theorem express the number via the zeta function of the curve over F_q. (Russian pdf is for free - main formula can be seen from there). Of course, it is very specific class of the groups, however may be something can be done ?

Another analogy which comes to mind is related to topological quantum field theories, quantization of Wess-Zumino and Chern-Simons models. One consider the moduli space of d-dimensional representatation of the fundamental group. It is natarally symplectic manifold and its VOLUME is somewhat an anologue of the "number" of irreps for finite group. The volume can be calculated and is related to the famous Verlinde formula.

So, it is of course, both considerations are related to fundamental (=Galois) groups of CURVES.

Question: WHY? What makes fundamental (=Galois) groups of curves so specific ? Can it be somehow generalized to other classes of curves ?

share|improve this question
1  
Which field are you working over? I.e., you want to count homomorphisms to $GL_2(K)$ for which fields $K$? –  Ian Agol Sep 21 '12 at 17:46
    
@Agol complex numbers, any other field is also welcome. –  Alexander Chervov Sep 21 '12 at 18:38
1  
The case of $K=F_2$ is slightly nontrivial. $GL(2,F_2) \cong S_3$, and any nontrivial homomorphism is irreducible, although you want to count homomorphisms up to conjugacy. Transitive homomorphisms correspond to subgroups of $G$ of index $3$ and nontransitive nontrivial homomorphisms correspond to subgroups of index $2$. –  Douglas Zare Sep 21 '12 at 22:36
    
@Douglas Zare thank you ! –  Alexander Chervov Sep 22 '12 at 13:32
add comment

1 Answer 1

up vote 17 down vote accepted

This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of $2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then $G = Z(G)E,$ where $ E \cong {\rm SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary icosahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup). Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only possible non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated if it appears. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.

share|improve this answer
    
Thank you very much as always ! –  Alexander Chervov Sep 22 '12 at 13:36
add comment

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.