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 $M(G)$?

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    $\begingroup$ It seems to me that (in a general dimension $d$), life is easier if you have a homomorphism from $G$ to ${\rm PSL}(d,K)$ 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. $\endgroup$ Jun 2, 2011 at 17:30
  • $\begingroup$ For your added question, I just noticed a cheap answer in Machi's new group theory text: take $N=|G|$ and let $\rho$ be the regular representation, twisted by $\alpha \in Z^2(G,K^\times)$, that is $e_g \cdot e_h = \alpha(g,h) e_{gh}$ where $\{e_g : g \in G\}$ is a basis of $K^G$. Then $\rho$ is a projective representation inducing $\bar \alpha \in H^2(G,K^\times)$. $${}$$ I've been wondering about associated covering groups in math.stackexchange.com/questions/423814/… $\endgroup$ Jun 18, 2013 at 18:17

1 Answer 1


As far as 1) is concerned :

Exercise 6.6.5 in Weibel's Introduction to Homological Algebra book :

For any field k and any n, let $\gamma$ denote the class in $H^2(PGL(n,k), k^*)$

corresponding to the extension $$1 \rightarrow k^* \rightarrow GL(n,k) \rightarrow PGL(n,k) \rightarrow 1$$ If $\rho : G \rightarrow PGL(n,k)$ is a projective representation, show that $\rho$ lifts to a linear representation $G \rightarrow GL(n,k)$ if and only if $\rho^{\ast}(\gamma) = 0$ in $H^2(G,k^{\ast})$. Here, $\rho^{\ast}$ is the obvious map from $H^2(PGL(n,k), k^{\ast})$ to $H^2(G, k^*)$ induced by $G \rightarrow PGL(n,k)$.

  • $\begingroup$ No problem. By the way, I should have mentioned : to prove this exercise, one considers the pullback of the two morphisms $G \rightarrow PGL(n,k)$ and $GL(n,k) \rightarrow PGL(n,k)$. The details are in exercise 6.6.4, the one right before 6.6.5 :-) $\endgroup$ Jun 2, 2011 at 17:58

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