Let $X$ be a proper scheme over a base field $k$ (one could consider more general settings, but I am primarly interested in a "geometric" situation with $k$ being algebraically closed). Then the Picard functor of $X$ is representable by the Picard scheme $Pic(X)$ of $X$, whose set of $k$-points is the Picard group $H^1 (X, \mathcal{O}_X^*)$.

A natural generalization is to replace $H^1$ by $H^i$ for $i \geq 2$. For example, for $i=2$, we obtain the Brauer functor, whose set of of $k$-points is the Brauer group $H^2 (X, \mathcal{O}_X^*)$. Given the fact that I have never seen general existence results for a "Brauer scheme" and given some facts I learned from some people, it seems that the Brauer functor is not representable. So my first question is:

1) What is known about the representability of the Brauer functor? Is it representable in some cases, under which conditions?

When I asked someone who might know the answer, he told me that the non-representability of the Brauer group is something related to the result due to Mumford that the Chow group of 0-cycles of some surfaces is "too big" but I don't really understand the relation. So:

2) What is the obstruction to the representability of the Brauer functor? What is the relation with the size of some Chow groups?

I am interested in the questions 1) and 2) for any $i \geq 2$ and not just the Brauer case $i=2$. If the $i \geq 2$ case is not representable in general but the $i=1$ Picard case is representable, it is natural to ask:

3) What is the difference between the cases $i=1$ and $i \geq 2$ ? What is the "miracle" which does that the potential "bad things" happening for $i \geq 2$ do not happen for $i=1$?


1 Answer 1


Suppose that $Br(X)$ is representable in the following sense: there exists a $k$-scheme $B_X$ such that for each $k$-scheme $S$ there is a natural bijection $B_X(S)=Br(X_S)$, or perhaps we should rigidify by asking for $B_X(S)=Br(X_S)/Br(S)$. In any case, since $B_X(k)\rightarrow B_X(l)$ is injective for all field extensions $k\rightarrow l$, we see that $Br(X)\rightarrow Br(X_l)$ or $Br(X)/Br(k)\rightarrow Br(X_l)/Br(l)$ would have to be injective. There is no reason for this to be true in general. For instance, take an elliptic curve $E$ defined over $\mathbb{Q}$. Then, there is a short exact sequence $$0\rightarrow Br(\mathbb{Q})\rightarrow Br(E)\rightarrow H^1_{et}(k,E)\rightarrow 0.$$ Taking the extension $\mathbb{Q}\rightarrow\overline{\mathbb{Q}}$ in the argument above we have $Br(E)/Br(\mathbb{Q})$ is non-zero (it's in fact typically very big), while $Br(E_{\overline{\mathbb{Q}}})=0$.

The basic issue is that for $i>1$, the assignment $S\mapsto H^i_{et}(X_S,\mathbb{G}_m)$ is just not a sheaf in the étale topology over $Spec\, k$. One can go in a different direction and look at the Picard stack $Pic=K(\mathbb{G}_m,1)$ and its classifing stack, $BPic=K(\mathbb{G}_m,2)$. This is a higher stack in the sense of Simpson, which in my indexing scheme (which I hope agrees with others') would be a $2$-stack. It precisely represents the Brauer group, but sections $BPic(X)$ also have higher homotopy groups: $$\pi_iBPic(X)=\begin{cases} H^2_{et}(X,\mathbb{G}_m)&\text{if $i=0$,}\\Pic(X)&\text{if $i=1$,}\\\mathbb{G}_m(X)&\text{if $i=2$}\\0&\text{otherwise.}\end{cases}$$ This story appears in Toën's paper on derived Azumaya algebras.

  • $\begingroup$ I have question. I don't know too much about topology and higher stacks, but when I see K(G,n) I expect to find only one homotopy group. Instead, you are saying that BPic(X) is more like $K(G_m, \leq 2)$ rather than $K(G_m,2)$. No? Is there a moral explanation of this? $\endgroup$ Oct 29, 2014 at 15:54
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    $\begingroup$ You're absolutely correct that in topological spaces a K(G,n) has only one homotopy group. But, now the K(G,n)s I'm talking about are étale sheaves of spaces in some sense. They have only a single homotopy sheaf, but the global sections are more complicated. In general, $\pi_i(\Gamma(X,K(G,n)))\cong H^{n-i}(X,G)$ for $0\leq i\leq n$ and $0$ otherwise. $\endgroup$ Oct 29, 2014 at 16:56
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    $\begingroup$ To say it more explicitly, the topological analogue of $B \text{Pic}(X)$ is not $B^2 G$ but the space of maps from $X$ to $B^2 G$. $\endgroup$ Oct 30, 2014 at 8:22
  • $\begingroup$ You're absolutely right. $\endgroup$ Oct 30, 2014 at 14:44
  • $\begingroup$ @BenjaminAntieau I am a little bit confused, suppose $l/k$ is an extension, should the map of Brauer groups be of the reverse direction $\mathrm{Br}(X_l)\to\mathrm{Br}(X_k)$ by pullback? Also can you explain a little bit why injectivity is expected if a functor is representable? Thank you! $\endgroup$
    – Qixiao
    Oct 2, 2016 at 1:22

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