Recently some arithmetic dynamicists came to town, bringing with them some interesting problems in arithmetic geometry.

I started thinking a bit about one of their problems, and it got me wondering about Schur multiplier groups over an arbitrary field. Traditionally, if $G$ is a group -- let us say it is finite -- then the **Schur multiplier group** $M(G)$ is $H^2(G,\mathbb{C}^{\times})$, i.e., group cohomology, with $\mathbb{C}^{\times}$ viewed as a trivial $G$-module. This group is also Brauer-like in that it measures obstructions to projective representations of $G$ -- i.e., homomorphisms $\rho: G \rightarrow \operatorname{PGL}_N(\mathbb{C})$ -- to be liftable to honest representations of $G$ -- i.e., homomorphisms $\tilde{\rho}: G \rightarrow \operatorname{GL}_N(\mathbb{C})$. It is not hard to see that you don't actually need to work over $\mathbb{C}$: if $\# G = n$, you can work over any field $K$ such that $K^{\times} = K^{\times n}$ and has primitive $n$th roots of unity.

But now suppose I have an arbitrary ground field $K$ and a homomorphism $\rho: G \rightarrow \operatorname{PGL}_N(K)$ which I am wondering lifts to a representation of $G$. What is the theory of this? Two basic questions:

1) Is it still true that the appropriate group to look at is $M_K(G) = H^2(G,K^{\times})$?

**Added**: Let me sharpen this question. The answer below shows that a projective representation gives rise to a class in $M_K(G)$ no matter what the ground field may be. But in the classical case the converse is also true: every element of $M_{\overline{K}}(G)$ arises in this way from a projective representation, uniquely up to projective equivalence. Does that still hold over an arbitrary ground field? I am a bit skeptical at the moment...

2) If the answer to 1) is *yes*, then it seems that the theory will have a much different flavor over an arbitrary field. (Here I say arbitrary but I am quite willing to assume for the moment that the characteristic of $K$ does not divide the order of $G$, so that we are in the setting of classical representation theory. This assumption will be in force in what follows.) For instance, if $G$ is cyclic of order $n$, then $M_K(G) \cong K^{\times}/K^{\times n}$. This means that over something like a number field there will be many projective representations of finite cyclic groups which do not lift. I think this is correct. In particular, I believe that for the cyclic group of order $2$, the map $G \rightarrow \operatorname{PGL}_2(K)$ associated with the order $2$ linear fractional transformation $z \mapsto \frac{\alpha}{z}$ is liftable to $\operatorname{GL}_2(K)$ iff $\alpha \in K^{\times 2}$. On the other hand I would like to deduce from the theory of "rational Schur multiplier groups" facts like the following: if $G$ is cyclic of odd order $n$ then every projective representation $\rho: G \rightarrow \operatorname{PGL}_2(K)$ lifts to a representation. (Again, in this case, if I am not mistaken, this can be shown by hand without much trouble, but I would like to see it come out of some general Schur-like theory.) In particular are there examples of computations of $M_K(G)$ in the literature for ~~simple~~ easy finite groups $G$, as there are for the usual $M(G)$?

`${\rm PSL}(d,K)$`

since you then get a factor set consisting of roots of unity in $K$. In other words, it may that`$K^{\times}/(K^{\times})^{d}$`

has special relevance for the problem. – Geoff Robinson Jun 2 '11 at 17:30