# Schur multipliers over non-algebraically closed ground fields?

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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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. –  Geoff Robinson Jun 2 '11 at 17:30
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/… –  Jack Schmidt Jun 18 '13 at 18:17

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 $G \rightarrow PGL(n,k)$.

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I'm having trouble with the TeX in the above post for some reason...something screws up when I try to put some things on the same line towards the end of the post, and it doesn't come out right. –  Moshe Adrian Jun 2 '11 at 17:36
@Moshe: thanks for the reference. As I think I implied above, I am not exactly surprised by this, but it's very useful to have something in print... –  Pete L. Clark Jun 2 '11 at 17:51
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 :-) –  Moshe Adrian Jun 2 '11 at 17:58
@Moshe: the problem is almost certainly the asterisk *, which is trying to put things in italics. You can always escape the problem by protecting all math with backticks, but at least on some browsers this has the unwanted effect of changing the font-size of the math. You can also get into the habit of using \ast in place of *. –  Theo Johnson-Freyd Jun 2 '11 at 21:52
@Theo : Thanks, I fixed it! –  Moshe Adrian Jun 2 '11 at 22:25