Harari 2: Weak approximation on algebraic varieties (cohomology)

Let $ X$ be a smooth, geometrically integral variety over $ k$ (a number field), and suppose that $ X$ is projective. We denote by $ \overline{X(k)}$ the closure of $ X(k)$ in $ \prod_{v \in \Omega_k} X(k_v)=X(\mathbf A_k)$.

Here our aim is to: (i) explain the counterexamples to weak approximation; (ii) find `intermediate' sets $ E$ between $ \overline{X(k)}$ and $ X(\mathbf A_k)$; (iii) in some cases, prove that $ E=\overline{X(k)}$.


General setting

Let $ G/k$ be an algebraic group (usually linear, but not necessarily connected, e.g. $ G$ finite). If $ G$ is commutative: define $ H^i(X,G)$ the étale cohomology groups ($ i=1,2$; the cohomological dimension of a number field forgetting real places makes the higher cohomology groups uninteresting). In general, we have only the pointed set $ H^1(X,G)$ (defined by Cech cocycles for the étale topology). If $ X={\mathrm{Spec}}
k$, $ H^1(X,G)=H^1(\Gamma,G(\overline{k}))$, where $ \Gamma={\mathrm{Gal}}(\overline{k}/k)$. If $ G$ is linear, $ H^1(X,G)$ corresponds to $ G$-torsors over $ X$ up to isomorphism.

Take $ f \in H^i(X,G)$, define

$\displaystyle X(\mathbf A_k)^f=\left\{(M_v) \in X(\mathbf A_k):(f(M_v)) \in {\mathrm{img}}(H^i(k,G) \to
\textstyle{\prod}_v H^i(k_v,G))\right\}. $

Obviously $ X(k) \subset X(\mathbf A_k)^f$. We will see that in many cases $ \overline{X(k)} \subset X(\mathbf A_k)^f$

Example.

  1. $ {\mathrm{Br}}X=H^2(X,\mathbb{G}_m)$;

    $\displaystyle \overline{X(k)} \subset X(\mathbf A_k)^{{\mathrm{Br}}} = \bigcap_{f \in {\mathrm{Br}}X}
X(\mathbf A_k)^f. $

    (Indeed the Brauer group of the ring of integers of $ k_v$ is zero). $ X(\mathbf A_k)^{{\mathrm{Br}}}$ is the Brauer-Manin set of $ X$. Manin showed in 1970 that for a genus one curve with finite Tate-Shafarevich group, the condition $ X(\mathbf A_k)^{{\mathrm{Br}}} \neq \emptyset$ implies the existence of a rational point.

  2. Let $ f:Y \to X$ be a Galois, geometrically connected, nontrivial étale covering with group $ G$. Then $ f \in H^1(X,G)$, where $ G$ is considered as a constant group scheme. Then $ \overline{X(k)} \subset X(\mathbf A_k)^f$ (via Hermite's Theorem). It is possible to find $ (M_v) \not\in X(\mathbf A_k)^f$, which implies Minchev's result that $ X$ does not satisfy weak approximation.

Remark. If $ X$ is rational, then $ {\mathrm{Br}}X/{\mathrm{Br}}k=H^1(k,{\mathrm{Pic}}\overline{X})$ is finite, where $ \overline{X} = X \times_k \overline{k}$. Then $ X(\mathbf A_k)^{{\mathrm{Br}}}$ is `computable'.

Theorem. [H, Skorobogatov] If $ G$ is linear and $ f \in H^1(X,G)$, then $ \overline{X(k)} \subset X(\mathbf A_k)^f$ (and $ X(\mathbf A_k)^f$ is "computable").


Abelian descent theory

This was developed by Colliot-Thélène and Sansuc, and recently completed by Skorobogatov.

Theorem. Define

$\displaystyle X(\mathbf A_k)^{{\mathrm{Br}}_1}=\bigcap_{f \in {\mathrm{Br}}_1 X} X(\mathbf A_k)^f $

where $ {\mathrm{Br}}_1 X = \ker({\mathrm{Br}}X \to {\mathrm{Br}}\overline{X})$. Assume that $ X(\mathbf A_k)^{{\mathrm{Br}}_1} \neq \emptyset$. Then:
  1. We have

    $\displaystyle X(\mathbf A_k)^{{\mathrm{Br}}_1} = \bigcap_{\substack{ f \in H^1(X,S) \\
S\text{\textup{ of multiplicative type}}}} X(\mathbf A_k)^f. $

  2. Assume further that $ {\mathrm{Pic}}\overline{X}$ is of finite type, set $ S_0$ such that $ \widehat{S_0}={\mathrm{Pic}}\overline{X}$; then there exists a torsor $ f_0:Y \to X$ under $ S_0$ (a universal torsor, i.e. "as nontrivial as possible") such that

    $\displaystyle X(\mathbf A_k)^{{\mathrm{Br}}_1} = X(\mathbf A_k)^{f_0}. $

This Theorem is difficult, see Skorobogatov's book for a complete account on the subject. One of the ideas is to recover the Brauer group of $ X$ (mod. $ {\mathrm{Br}}k$) making cup-products $ [Y] \cup a$, where $ a \in H^1(k,\widehat S_0)$ and $ [Y]$ is the class of $ Y$ in $ H^1(X,S_0)$.

Now assume that $ X$ is a rational variety, so $ X(\mathbf A_k)^{{\mathrm{Br}}}=X(\mathbf A_k)^{{\mathrm{Br}}_1}$ (since $ {\mathrm{Br}}\overline{X}=0$). Assume $ X(\mathbf A_k)^{{\mathrm{Br}}} \neq \emptyset$. Consider a universal torsor $ f:Y
\xrightarrow{S_0} X$. If $ \sigma \in H^1(k,S_0)$, can define $ f^\sigma:Y^\sigma \to X$ where

$\displaystyle [Y^\sigma]=[Y]-\sigma \in H^1(X,S_0). $

Then

$\displaystyle X(\mathbf A_k)^f = \bigcup_{\sigma \in H^1(k,S_0)}
f^\sigma(Y^{\sigma}(\mathbf A_k)). $

If you can prove that the torsors $ Y^\sigma$ satisfy weak approximation, then $ \overline{X(k)}=X(\mathbf A_k)^f=X(\mathbf A_k)^{{\mathrm{Br}}}$, so the Brauer-Manin obstruction is the only one.

Example. There are many examples of this:

  1. Châtelet surface: $ y^2-az^2=P(x)$, $ a \in k^*/k^{*2}$, $ \deg P=4$. Colliot-Thélène, Sansuc, Swinnerton-Dyer showed that $ \overline{X(k)}=X(\mathbf A_k)^{{\mathrm{Br}}}$, so the Brauer-Manin obstruction is the only one. If $ P$ is irreducible, then $ {\mathrm{Br}}X/{\mathrm{Br}}k=0$, so $ X$ satisfies weak approximation.

    If $ P$ is reducible, we can have a counterexample to weak approximation, e.g. $ y^2-az^2=f_1(x)f_2(x)$, where $ \deg f_1=\deg f_2=2$, $ \gcd(f_1,f_2)=1$, in some cases there is an obstruction given by the Hilbert symbol $ f=(a,f_1)$.

  2. Conic bundles over $ \mathbb{P}^1$ with at most $ 5$ degenerate fibres. Results of Colliot-Thélène, Salberger, Skorobogatov covered at most $ 4$. In the case of $ 5$, the existence of a global rational point is easy to show, so the only problem is weak approximation, and which is due to Salberger, Skorobogatov 1993 (using descent and $ K$-theory).

Theorem. [Sansuc 1981] Let $ G$ be a linear connected algebraic group over $ k$, $ X$ a smooth compactification of $ G$, then the Brauer-Manin obstruction is the only one:

$\displaystyle \overline{X(k)} = X(\mathbf A_k)^{{\mathrm{Br}}}. $


Back to fibration methods

If $ p:X \to B$ is a fibration, we saw that if the base and the fibres satisfy weak approximation, under certain circumstances then $ X$ satisfies weak approximation.

Here we consider $ p:X \to \mathbb{P}^1$, a projective, surjective morphism (and the generic fibre $ X_\eta$ is smooth). Assume also that all fibres are geometrically integral (can do with all but one because of strong approximation on the affine line).


\begin{ques}
If $\overline{X_P(k)}=X_P(\mathbf A_k)^{{\mathrm{Br}}}$\ for almost...
... can you prove that $\overline{X(k)}=X(\mathbf A_k)^{{\mathrm{Br}}}$?
\end{ques}

Theorem. [H 1993, 1996] Yes, $ \overline{X(k)}=X(\mathbf A_k)^{{\mathrm{Br}}}$ if you assume that:

  1. $ {\mathrm{Pic}}\overline{X_\eta}$ is torsion-free, where $ \overline{X_\eta}=X_\eta \times_{K} \overline{K}$, $ K=k(\eta)$; e.g. $ X_\eta$ rational, or smooth complete intersection of dimension at least three.
  2. $ {\mathrm{Br}}\overline{X_\eta}$ is finite.

Two ideas:

  1. $ {\mathrm{Br}}X_\eta/{\mathrm{Br}}K \to {\mathrm{Br}}X_P/{\mathrm{Br}}k$ is an isomorphism for many $ k$-fibres $ X_P$ (`many' in the sense of Hilbert's irreducibility theorem).
  2. If $ \alpha_1,\dots,\alpha_r$ are elements of $ {\mathrm{Br}}X_\eta$, assume $ \alpha_i \in {\mathrm{Br}}U$, $ U \subset X$ an open subset. Use the `formal lemma': Take $ (M_v) \in X(\mathbf A_k)^{{\mathrm{Br}}}$, $ M_v \in U$, $ \Sigma_0$ a finite set of places; then there exists $ (P_v) \in X(\mathbf A_k)$, $ P_v \in U$, $ \Sigma
\supset \Sigma_0$ finite such that:
    1. $ P_v=M_v$ for $ v \in \Sigma_0$;
    2. $ \sum_{v \in \Sigma} j_v(\alpha_i(P_v))=0$ for $ 1 \leq i \leq r$, where $ j_v:{\mathrm{Br}}k_v \to \mathbb{Q}/\mathbb{Z}$ is the local invariant.

Applications: (i) Recover Sansuc's result just knowing the case of a torus; (ii) If you know that $ \overline{X(k)}=X(\mathbf A_k)^{{\mathrm{Br}}}$ for $ X$ a smooth cubic surface, then by induction the same holds for hypersurfaces, so if $ \dim X \geq 3$, then $ X$ satisfies weak approximation.


Nonabelian descent

If $ G/k$ is a finite but not commutative $ k$-group, it is possible that for $ f \in H^1(X,G)$, $ X(\mathbf A_k)^f \not\supset X(\mathbf A_k)^{{\mathrm{Br}}}$.

Theorem. [Skorobogatov 1997] There exists $ X/\mathbb{Q}$ a bi-elliptic surface such that $ X(\mathbb{Q})=\emptyset$, $ X(\mathbf A_\mathbb{Q})^{{\mathrm{Br}}} \neq \emptyset$.

Actually: $ X(\mathbf A_\mathbb{Q})^f=\emptyset$ for some $ f \in H^1(X,G)$, $ G(\overline{\mathbb{Q}})=(\mathbb{Z}/4\mathbb{Z})^2 \rtimes \mathbb{Z}/2\mathbb{Z}$.

There are similar statements for weak approximation (H 1998), e.g. take $ X/k$ any bi-elliptic surface, $ X(k) \neq \emptyset$, then $ \overline{X(k)} \subsetneq
X(\mathbf A_k)^{{\mathrm{Br}}}$.

Nevertheless the Brauer-Manin condition is quite strong, as shows the following result :

Theorem. [H 2001] We have:

  1. If $ G/k$ is a linear connected $ k$-group, $ f \in H^1(X,G)$, then

    $\displaystyle X(\mathbf A_k)^{{\mathrm{Br}}} \subset X(\mathbf A_k)^f. $

  2. If $ G$ is any commutative $ k$-group, $ f \in H^2(X,G)$, then

    $\displaystyle X(\mathbf A_k)^{{\mathrm{Br}}} \subset X(\mathbf A_k)^f. $

Open question : is the first part of this theorem still true for a $ G$ which is an extension of a finite abelian group by a connected linear group ? My guess is "no".




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