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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Sep 8, 2012 at 18:23

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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 ${\hat G}.$

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  • $\begingroup$ This does not seem to explain why there cannot be a nonsplit extension of a bigger, perhaps torsion-free, subgroup of $\mathbb{R}^*$ by $G$. $\endgroup$
    – Dima Pasechnik
    Commented Sep 8, 2012 at 9:23
  • $\begingroup$ @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 $2$. $\endgroup$
    – Frieder Ladisch
    Commented Sep 8, 2012 at 13:29
  • $\begingroup$ @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 $2.G$). $\endgroup$
    – Dima Pasechnik
    Commented Sep 9, 2012 at 5:25
  • $\begingroup$ @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. $\endgroup$
    – Geoff Robinson
    Commented Oct 7, 2012 at 11:51

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