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

It follows from the theory of Schur multiplier that any $n$-dimensional projective representation $\theta : G\to PGL(n,\mathbb{R})$ of a finite group $G$ is either an ordinary representation of $G$, i.e. $\theta : G\to GL(n,\mathbb{R})$, or lifts to an ordinary representation $\theta' : 2.G\to GL(n,\mathbb{R})$ of a double cover $2.G$ of $G$.

A direct reference to this fact would be very useful.

Is there a more direct way to see this, preferably suitable for non-algebraist readers? The quickest route I know is to mimick the usual proof that the $|G|$-th power of the cocycle is trivial, as in e.g. Theorem 11.15 in [1].

[1]: I.M.Isaacs, Character Theory of Finite Groups, Dover 1994.

share|improve this question
    
Concerning sources, I suspect most mathematicians who deal with this kind of question aim for more generality than you want. From what I know of the books on mathematical physics, they might take a narrower and more down-to-earth viewpoint on projective representations. (And some of the natural applications occur in physics.) Still, some formalism is bound to intrude, no matter how you do things. –  Jim Humphreys Sep 8 '12 at 18:23

1 Answer 1

I think a proof is also in Curtis and Reiner (Wiley, 1962). Not sure whether it counts as a non-algebraic proof, but if you think of the projective representation as a map $\sigma$ from $G$ to ${\rm GL}(n, \mathbb{R})$, defined only up to scalars, and for each $g \in G,$ and you make a particular choice of $g\sigma$ for each $g \in G,$ you can if necessary replace it by a (real) scalar multiple so that $d(g) = {\rm det}(g\sigma) \in \{1,-1\}$ for all $g \in G.$ Then the double cover you need (if you need one at all) is $\hat{G} = \{(g, d(g)):g \in G \}$ with the multiplication of the second component forced by making $\sigma$ a genuine representatin of the new group ${\hat G}.$

share|improve this answer
    
This does not seem to explain why there cannot be a nonsplit extension of a bigger, perhaps torsion-free, subgroup of $\mathbb{R}^*$ by $G$. –  Dima Pasechnik Sep 8 '12 at 9:23
    
@Dima Pasechnik: ? There are nonsplit extensions of (torsion-free) subgroups of $\mathbb{R}^*$ by $G$, for example $$1\to \langle 4 \rangle \to \langle 2 \rangle \to C_2 \to 1 $$ for $G=C_2$ cyclic of order $2$. –  Frieder Ladisch Sep 8 '12 at 13:29
    
@F.Ladisch, indeed, I was too quick and imprecise here. I meant to ask how this would imply the original statement, i.e. that any group with the universal lifting property for $\mathbb{R}$-projective representations of $G$ is finite (and thus either $G$ or $2.G$). –  Dima Pasechnik Sep 9 '12 at 5:25
    
@Dima: I don't see your objection to my argument. It lifts the projective representation of $G$ to a genuine representation of a double cover. It is only with perfect groups that one can speak about "the" double cover in any case. In that case, if there is a genuine double cover, it can be taken to be perfect, and the problems with superfluous scalars disappear. By the way, I think you should deal with absolutely irreducible groups in this problem. –  Geoff Robinson Oct 7 '12 at 11:51

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.