projective representations of a finite group over reals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:34:08Z http://mathoverflow.net/feeds/question/106646 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106646/projective-representations-of-a-finite-group-over-reals projective representations of a finite group over reals Dima Pasechnik 2012-09-08T05:32:39Z 2012-09-08T07:55:44Z <p>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 $G$.</p> <p>A direct reference to this fact would be very useful.</p> <p>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]. </p> <p>[1]: I.M.Isaacs, Character Theory of Finite Groups, Dover 1994.</p> http://mathoverflow.net/questions/106646/projective-representations-of-a-finite-group-over-reals/106655#106655 Answer by Geoff Robinson for projective representations of a finite group over reals Geoff Robinson 2012-09-08T07:55:44Z 2012-09-08T07:55:44Z <p>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 <code>$d(g) = {\rm det}(g\sigma) \in \{1,-1\}$</code> for all $g \in G.$ Then the double cover you need (if you need one at all) is <code>$\hat{G} = \{(g, d(g)):g \in G \}$</code> with the multiplication of the second component forced by making $\sigma$ a genuine representatin of the new group ${\hat G}.$ </p>