A Mind for Madness

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Brauer-Manin Obstruction

We will continue with the Brauer group after this post, but today let’s answer the question: Why should we care about the Brauer group? We gave one good reason back when talking about rational surfaces. It provided something that might be computable to detect whether or not our variety had good reduction.

Today let’s consider what is probably the most well-known use of the Brauer group. Suppose we have {X/K} a (smooth, proper) variety over a number field. We want to know whether or not there are any {K}-rational points. Of course, classically if you hand me this variety using an equation, then the problem reduces to a Diophantine equation. So this problem is very old (3rd century … and one could argue we still don’t have a great handle on it).

Suppose {Spec \ K\rightarrow X} is a rational point. Then we can use the embedding {K\hookrightarrow K_v} into various completions of {K} to get {Spec \ K_v\rightarrow X} points over all the local fields. This isn’t saying much. It says that if we have a global point, then we have points locally everywhere. The interesting question is whether having points locally everywhere “glue” to give a global point. When a class of varieties always allows you to do this, then we say the varieties satisfy the Hasse principle.

This brings us to the content of today’s post. We will construct the Brauer-Manin obstruction to the Hasse principle. We’ve already considered the following pairing in a previous post. For any prime {v}, we get a pairing {Br(X_{K_v})\times X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}} which is just given by pullback on ├ętale cohomology using the map defining the point {Spec \ K_v\rightarrow X_{K_v}} followed by the canonical identification {Br(K_v)\stackrel{\sim}{\rightarrow}\mathbb{Q}/\mathbb{Z}}. We could write {(\alpha, x_v)=x_v^*(\alpha)} or more typically {\alpha(x_v)}.

We can package this into something global as follows. By restriction we have {\displaystyle Br(X)\hookrightarrow \prod_v Br(X_{K_v})} which we will write {\alpha\mapsto (\alpha_v)}. Now we just sum to get {\displaystyle Br(X)\times \prod_v X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}, so our global pairing is {\displaystyle (\alpha, (x_v))\mapsto \sum_v \alpha_v(x_v)}.

Recall that we have an exact sequence from the post on Brauer groups of fields {0\rightarrow Br(K)\rightarrow \oplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0} by summing the local invariants. If our local points glue to give a global point {x\mapsto (x_v)}, then the pairing factors through this sequence and hence {(\alpha, (x_v))=0} for any choice of {\alpha}.

This is our obstruction. The above argument says that if {(\alpha, (x_v))\neq 0}, then {(x_v)} cannot possibly glue to give a rational point. We will give this set a name:

\displaystyle X^{Br}=\{(x_v)\in \prod_v X(K_v) : (\alpha, x_v)=0 \ \text{for all} \ \alpha\in Br(K)\}

We will call this the Brauer set of {X} (earlier we called it being “Brauer equivalent to {0}” since we saw this type of condition was an equivalence relation on Chow groups). It is now immediate that {X(K)\subset X^{Br}}. We should think of this as the collection of local points that have some chance of coming from a global point. Now we have two obstructions to the existence of rational points. The first is that {\prod_v X(K_v)\neq \emptyset} which is just the trivial condition that there are local points everywhere. The second is the Brauer-Manin obstruction which says that {X^{Br}\neq \emptyset}.

If the local points condition is necessary and sufficient for the existence of rational points, then of course that is exactly the Hasse principle, so we say the Hasse principle holds. If the Brauer-Manin condition (plus the local condition) is necessary and sufficient, then we say that the Brauer-Manin obstruction is the only obstruction to the existence of rational points. It would be fantastic if we could somehow figure out which varieties had this property.

Caution: It is not an open problem to determine whether or not the Brauer-Manin obstruction is the only obstruction. There are known examples where there is no B-M obstruction and yet there are still no rational points. As far as I can tell, it is conjectured, but still open that if the Tate-Shafarevich group of the Jacobian of a curve is finite, then the B-M obstruction is the only obstruction to the existence of rational points on curves over number fields. So even in such a special low-dimensional situation this is a very hard question.

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