# Grothendieck's question on the Brauer group for groups

Let $G$ be a group, and let $M(G)=H^2(G,\mathbb{C}^*)$ be the Schur multiplier of $G$. There is a group $Br(G)$ of complex projective representations of $G$ modulo those that can be lifted to linear representations. Projective representations of degree $n$ are classified by cohomology classes $H^1(G,PGL_n(\mathbb{C}))$. There are maps of pointed sets $$H^1(G,PGL_n(\mathbb{C}))\rightarrow M(G)$$ arising from the exact sequences of (trivial) $G$-modules $$1\rightarrow \mathbb{C}^*\rightarrow GL_n(\mathbb{C})\rightarrow PGL_n(\mathbb{C})\rightarrow 1.$$ Let $Br(G)$ be the subgroup of $M(G)$ generated by the images of these boundary maps for all $n$. (Of course, $Br(G)$ can be defined directly without using group cohomology.)

My question is: when is $Br(G)\rightarrow M(G)$ surjective? This is analogous to the question of Grothendieck on the difference between $Br(X)$ and $H^2(X_{et},\mathbb{G}_m)$ of schemes. If one works with the classifying space of $G$, then everything may be interpreted in terms of topological Azumaya algebras, and the question is precisely the same as Grothendieck's.

I know of three broad cases where this is true. First, when $G$ is a finite group, this follows fairly easily from group cohomology. Roughly speaking, one takes a cocycle representative $\alpha\in Z^2(G,\mathbb{C}^*)$. Then, one uses the cocycle to create a twisted group algebra $\mathbb{C}^{\alpha}[G]$. This algebra is a semi-simple algebra over the complex numbers, so it has various finite-dimensional representations. Any one of them gives a projective representation for $G$ with cohomology class $\alpha$. See for instance Karpilovsky's book Projective representations of finite groups.

Second, when $G$ is a group such that $BG$, the classifying space, has the homotopy type of a finite CW-complex, it follows from Serre's theorem that $Br=Br'$ for finite CW-complexes. See Grothendieck's Groupe de Brauer I.

Third, when $G$ is a good group in the sense of Serre, then $Br(G)=M(G)$. A group $G$ is good if it has the same cohomology as its finite completion. This result is proven in Schroer's paper Topological methods for complex analytic Brauer groups.

So, my real question is: does anyone know any more general results than these? For instance, is $Br(G)=M(G)$ for any finitely presented group?

• For non-linear groups, I would expect, by your definition, that Br(G)=0. But there is really no guarantee that M(G)=0. Jun 14, 2011 at 22:00
• @André: since the groups $PGL_n(\mathbb{C})$ are themselves linear, it is true that if $G$ is not linear, then there are no faithful projective representations of $G$. I do not know if this is enough to say $Br(G)=0$. It does say that if $G$ is a group such that no quotient is linear, then $Br(G)=0$. Jun 15, 2011 at 2:59

Note that it is not hard to construct examples where $Br(G)$ is not equal to $M(G)$. To see this, let $H_1=\mathbb{Z}/p\times\mathbb{Z}/p$, and let $\alpha_1\in H^2(H_1,\mathbb{C}^*)=\mathbb{Z}/p$ be a generator. The class $\alpha_1$ is represented by a projective representation $p_1:H_1\rightarrow PGL_p(\mathbb{C})$. Now, let $H_i=(H_1)^i$, and let $\alpha_i$ be the Brauer class of the projective representation $$p_i:H_i=(H_1)^i\xrightarrow{p_1\times\cdots\times p_1}(PGL_p(\mathbb{C}))^i\rightarrow PGL_{p^i},$$ where the final arrow is given by embedding $PGL_p(\mathbb{C})^i$ into $PGL_{p^i}$ via block matrices down the diagonal. Then, it is not hard to show that the index of $\alpha_i$ is $p^i$. Recall that the index of a class of $Br(G)$ is the least common divisor of all projective representations having that class.

Now, define $G=*_i H_i$, the free product of the $H_i$ for all positive integers $i$. The corresponding classifying space $BG$ is the wedge sum of the $BH_i$. Therefore, the collection of continuous pointed maps $$\alpha_i:BH_i\rightarrow K(\mathbb{C}^*,2)$$

induces a continuous pointed map $$\alpha:BG\rightarrow K(\mathbb{C}^*,2)$$such that the composition $$BH_i\rightarrow BG\rightarrow K(\mathbb{C}^*,2)$$ is $\alpha_i$.

It is easy to see that $ind(\alpha_i)$ divides $ind(\alpha)$ for all $i$. But, since $ind(\alpha_i)=p^i$, it follows that $ind(\alpha)$ is infinite, and hence that $\alpha\notin Br(G)$, as desired.

Obviously, $G$ is not finitely presented. In relation to André's comment above, it seems likely that it is not linear either.

• Obviously this example could be generalized to any sequence of groups $H_i$ with classes $\alpha_i\in Br(H_i)$ with bounded periods but unbounded indices. Jul 7, 2011 at 22:57
• I don't think that $B(\ast_i H_i) = \coprod_i BH_i$. For a start, classifying spaces are connected. Also, consider this statement in the case of $G = Z\ast Z$. This would imply that $BG = U(1) \coprod U(1)$, an absurdity. Are you thinking of the fact one can construct a functorial $B$ which preserved products (in the category of k-spaces)? Jul 7, 2011 at 23:32
• @David: oops. I meant pointed coproduct, so that we're taking a wedge product. Then this just follows from van Kampen. I will clarify that in the answer. Thanks for pointing it out. Jul 7, 2011 at 23:34