Of course not.

But after reading a bit, some points make me believe it should be:

Let $S$ be a nice$^{\*}$ surface defined over $Spec\ \mathbb{Z}$.

  1. The Brauer group $Br(S\otimes \bar{\mathbb{Q}})$ is an abelian divisible group,
  2. It is also a $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ module,
  3. For good primes there are reductions $Br(S)\rightarrow Br(S\otimes \mathbb{F}_q)$,
  4. These $Br(S\otimes \mathbb{F}_q)$ are finite,
  5. There is a formal Brauer group $\hat{Br}(S)$ of dimension 1,
  6. The coefficients of $\hat{Br}$, in suitable natural coordinates, relate to $|Br(S\otimes \mathbb{F}_q)|$.
  7. There are some examples where the associated L-function comes from a modular form (of weight 3). I'm not sure if this is conjectured (let alone known) in general.

Since the Brauer group observes many characteristics of an abelian variety (all properties) of dimension 1 (properties 5 and 7 [weight isn't two, but it's the right space]), my vague question is: how far is it from actually being a variety?

There are some easy examples of $S$ with $|Br(S\otimes \mathbb{F}_q)|$ varying between $1$ and $4(q-4)$, as $q$ varies over the primes. This is a clear point of departure from elliptic curves and varieties in general.

Maybe there's a family of natural galois-module homomorphisms into certain abelian varieties defined over $\mathbb{Q}$, commuting with the reduction maps and restriction (or some other appropriate term) to formal groups?

What's going on with these Brauer groups?

$^\*$ say a K3 surface. Something that (1) is true for (so not a rational surface) and (4) is proven for.

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    $\begingroup$ Virtual second +1 for 'of course not'. $\endgroup$ – David Roberts Jun 14 '11 at 23:43
  • $\begingroup$ nice question! the familiar analogy is that the Brauer group of a surface over a finite field is "like" the Tate-Shafarevitch group of an elliptic curve over a global field as, for example, in numdam.org/numdam-bin/fitem?id=CM_1979__38_2_163_0 $\endgroup$ – SGP Jun 15 '11 at 1:50
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    $\begingroup$ I believe that one difference between the $1$-dimensional formal Brauer group of a K3-surface and the $1$-dimensional formal group of an elliptic curve is that the heights of the former can be found in a greater range that those of the latter. I don't have a reference, and I'm remembering some talk from a long time ago, so I'm afraid I cannot be more explicit. But, perhaps this is related to the chromatic filtration in stable homotopy. Hopefully someone more knowledgeable than I can chime in here. $\endgroup$ – Benjamin Antieau Jun 23 '11 at 19:24

This is not a general answer to your question, but evidence of the intriguing connection between Brauer groups of surfaces and elliptic curves. Let $X$ be a K3 surface over the complex numbers $\mathbb{C}$. Then, the rank of $H^2(X,\mathbb{Z})$ is $22$, and the Hodge numbers are $h^{0,2}=h^{2,0}=1$ and $h^{1,1}=20$. If the Neron-Severi group of $X$ is as big as possible, namely if it has rank $20$, then the image of $H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)$ has rank $2$ and so is a lattice. Therefore, the cokernel of this map is an elliptic curve. But, since $H^3(X,\mathbb{Z})=0$, there is an exact sequence $$H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X^*)\rightarrow 0.$$ Therefore, $H^2(X,\mathcal{O}_X^*)$ is an elliptic curve $E$. The torsion of $H^2(X,\mathcal{O}_X^*)$ is therefore precisely the Brauer group of $X$, and these are precisely the torsion points of $E$.

One can produce similar examples using abelian surfaces. For instance, if $E$ is a CM elliptic curve, then $E\times E$ is an abelian surface such that $H^2(E\times E,\mathcal{O}^*)$ is isomorphic to $E$.

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    $\begingroup$ A long time ago (for me), I sketched an argument that on the product $E\times X$, where $X$ is a K3 surface with Neron-Severi group of rank $20$ and $E$ is the elliptic curve $H^2(X,\mathcal{O}_X^*)$ there is a universal gerbe, an element of $H^2(X\times E,\mathcal{O}_X^*)$ that restricts on a point of $E$ to the point it classifies in $H^2(X,\mathcal{O}_X^*)$. I called this the Poincare gerbe. Unfortunately, I have never written this up in detail, so I've never checked it carefully. $\endgroup$ – Benjamin Antieau Jun 15 '11 at 3:14
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    $\begingroup$ 1. If in the last sentence, instead of taking $E\times E$, we take $E\times E / <\zeta\times\zeta^{-1}>$, where $E$ has CM by either $\mathbb{Z}[\zeta_4]$ or by $\mathbb{Z}[\zeta_6]$), is the $H^2$ still isomorphic to $E$? 2a. What, if any, of this passes on to etale cohomology? 2b. Is there a sense in which we can algebraically, as opposed to analytically, understand "the torsion points of $E$"? Your answer has helped me try to comprehend what's going on, thanks. $\endgroup$ – Dror Speiser Jun 23 '11 at 18:38
  • $\begingroup$ If $A$ is an abelian surface which is isogenous to the product of two isogenous CM elliptic curves $E$ and $F$ (this is the case in your comment), then $H^2$ is another elliptic curve $G$ isogenous to $E$ and $F$. But, I don't think that it has to be equal. As for getting at these points algebraically, there is a surjection from $H^2(X_{et},\mu_{n})\rightarrow G[n]$. This surjection is not ever an isomorphism (since $A$ is algebraic). As for a reference for this, I believe I read about it in Shioda and Mitani's paper Singular abelian surfaces and binary quadratic forms. $\endgroup$ – Benjamin Antieau Jun 23 '11 at 19:21
  • $\begingroup$ Benjamin, I have a question regarding this really interesting construction. I believe that K3 surfaces with maximal Picard number are all related to CM elliptic curves. (E.g. taking X to be the diagonal quartic in $\mathbb{P}^3$ (over the complexes, say), I'd expect E to be the 'lemiscatic curve' $C : y^2 = x^3 - x$ which has CM by Z[i].) My question then is: can you construct a non-trivial endomorphism of $E = H^2(X,O_X)/\iota(H^2(X,\mathbb{Z}))$ from the data above? $\endgroup$ – RP_ May 1 '12 at 13:09
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    $\begingroup$ Hi René, I'm not sure how to see the endomorphism of $E$ directly from the K3 surface in these cases. Every K3 surface with Neron-Severi group of rank 20 is either a Kummer surface, or it has a symplectic involution (these are called Nikulin involutions I believe), and the quotient of X by this involution is a Kummer surface. Perhaps when X has a Nikulin involution, this leads to an extra automorphism of the elliptic curve E. When X is Kummer, its abelian cover is an abelian surface with Neron-Severi rank 6, which is a product of isogenous CM elliptic curves, by the Shioda-Mitani paper above. $\endgroup$ – Benjamin Antieau May 3 '12 at 18:29

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