# Tuesday, November 09

:::{.remark}
The descent obstruction does not suffice!
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:::{.theorem title="?"}
There exists a nice $X\in \Var\slice \QQ$ with $X(\AA)^{\et, \Br} \neq \emptyset$ but $X(\QQ) = \emptyset$.
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:::{.remark}
$X$ will be a quadric 3-fold bundle over a curve $C$ with $\size C(\QQ) = 1$ with 2 singular fibers.
By the Lefschetz hyperplane theorem, we'll be able to understand its $\pi_1$, even though this won't be a Serre fibration and thus we won't get a LES in homotopy.
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## Proof

:::{.remark}
First make $Y$, a quadric 3-fold bundle over $\PP^1\slice \QQ$ with 2 singular fibers.
Since we're fibering over $\PP^1$, we'll define this in two affine patches and then glue.
Define
\[
Y_t &\da \ts{ t(t-1) x_0^2 + x_1^2 + \cdots + x_4^2 = 0} \subseteq \AA^1\slice t \times\PP^1\slice{[x_0,\cdots, x_4]} \\
Y_T &\da \ts{ (1-T) X_0^2 + x_1^2 + \cdots + x_4^2 = 0} \subseteq \AA^1\slice T \times\PP^1\slice{[x_0,\cdots, x_4]}
.\]
Then define a gluing by
\[
t &\mapsto {1\over T} \\
x_i &\mapsto x_i, \qquad i=1,\cdots, 4\\
x_0 &\mapsto {T \over X_0}
.\]

Now check that $Y_0, Y_1$ are singular, and $Y_{\infty} = \ts{\sum x_i^2 = 0}$:

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> Something about $(\CC\units \times Q)/ \ts{\pm 1} \to \CC\units$?
> This makes the geometry in this situation easy.

Since $Y_\infty(\RR) = \emptyset$ implies $Y_\infty(Q) = \emptyset$.
Note that if $t=0, 1$ this is a point, for $0< t< 1$ this is an ellipsoid, and for $t>1$ or $t<0$ there are no real points.
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:::{.remark}
Issue: $\PP^1$ has lots of rational points, so we'll make a curve $C$ with a map $C\to \PP^1$ and pull back.
Let $C\slice \QQ$ be a smooth proper curve such that 

- $\size C(\QQ) = 1$, so write $c \da C(\QQ)$ as this single rational point.
  Write $U \subseteq C(\RR)$ as the connected component of the identity.
  The real points are compact 1-manifolds, so a disjoint union of circles whose number depends on $g(C)$.
  Fun question: what is $\pi_0 \mg(\RR)$?
- $C \mapsvia{f} \PP^1$ such that 
  - $f(c) = \infty$ such that
  - $f$ is étale at $c$ and also over $0, 1 \in \PP^1(\QQ)$.
  - $f(U) \ni 1$.

:::

:::{.example title="?"}
$y^2 = x^3-3$ is such a curve, see LMFDB.

- Probably the point is $\infty$!

- Pick a uniformizer at $\infty$, then compose with an affine map $z\mapsto az+b$ to obtain the 2nd condition.

- $f(U)$ contains an open subset of $\PP^1(\RR)$, using that $f$ is étale and thus a local homeomorphism, so apply the implicit/inverse function theorems.
  So compose with a generic $z\mapsto z+1$ such that $f(U) \ni 1$.
  This doesn't mess up the previous affine map, since it's generic.

Now define $X\da Y\fiberprod{\PP^1} C$ as the pullback.
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:::{.claim}
$X(\QQ) = \ts{ \infty }$.
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:::{.proof title="?"}
Note
\[
X(\QQ) \mapsvia{\pi} C(\QQ) 
,\]
and $\pi\inv(c) \cong Y_\infty$ which has no rational points.
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## Showing $X(\AA)^{\et, \Br}$


:::{.remark}
Next goal: show the étale-Brauer set is empty.
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:::{.example title="?"}
We can show $X(\AA)^{\Br} \neq \emptyset$.

1. An exercise using the Leray spectral sequence shows there is a surjection $\Br(C) \to \Br(X)$.

2. Observe that $x\in X(\AA)$ is also in $X(\AA)^{\Br}$ if
  - $\pi(x_{\infty}) \in C(\QQ_{v})$ is equal to $C$ for all finite $v$.
  - $\pi(x_{\infty}) \in U$.

3. Find such a $(x_v) \in X(\AA)$.
  - For $v$ finite, choose $x_v\in Y_\infty(\QQ_v)$.
    Then $\ts{ [x_0: \cdots : x_n] \st \sum x_i = 0 } \neq \emptyset$.
  - For $v$ infinite, choose any $c'$ lying over $1\in \PP^1$, so $c'\in U \intersect f\inv(1)$.
    Then $X_{c'} \cong Y_1$ but $Y_1(\RR) \neq \emptyset$ since it contains $p = [1:0:\cdots:0]$.

Why (2) is true: for $v$ finite, $\mathrm{inv}_v(x_v^* \alpha) = \mathrm{inv}_v(\pi(x_v)^* \alpha') = \mathrm{inv}_v(c^* \alpha')$ where \( \alpha\in \Br(X), \alpha= \pi^* \alpha', \alpha'\in \Br(C) \), and the same holds for $v$ replaced by $\infty$.
Use that this function is constant on $U$ since it's connected.
Now $\sum_v \mathrm{inv}_v( x_v^* \alpha) = \sum_v \mathrm{inv}_v(c^* \alpha') = 0$, which shows it's in the Brauer set.
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:::{.remark}
So the Brauer-Manin obstruction it not sufficient to obstruct rational points.
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## Étale-Brauer-Manin Set


:::{.remark}
Recall the definition:
\[
X(\AA)^{\et, \Br} \da 
\Intersect_{\substack{ 
T \in H^1(X; G) = \torsors{G} \\ G \text{ finite étale} 
}} 
\Union_{\tau \in H^1(k; G) } \im\qty{ T^\tau(\AA)^{\Br} \to X(\AA) }
.\]
Let $(x_v) \in X(\AA)$ as above.

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:::{.claim}
For each finite étale $G$ and each $T\in H^1(X; G)$ there exists a $\tau \in H^1(k; G)$ such that $(x_v)$ lifts to $T^\tau(\AA)^{\Br}$.
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:::{.proof title="?"}
In steps:

1. (SGA1) $T$ is a pullback of $\tilde T\in \torsors{G}\slice C$, using that $\pi_1 X = \pi_1 C$.
2. There exists a $\tau \in H^1(k; C)$ such that $c\in C(\QQ)$ lifts to $\tilde c\in \tilde T^\tau(\QQ)$.
3. For $v$ finite, use the diagram:

\begin{tikzcd}
	& {T^\tau(\QQ_v)} && {X(\QQ_v)} & {\ni x_v} \\
	\\
	{\tilde c\in} & {\tilde T^\tau(\QQ_v)} && {C(\QQ_v)}
	\arrow[from=3-2, to=3-4]
	\arrow[from=1-4, to=3-4]
	\arrow[from=1-2, to=1-4]
	\arrow[from=1-2, to=3-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwLCJUXlxcdGF1KFxcUVFfdikiXSxbMywwLCJYKFxcUVFfdikiXSxbMSwyLCJcXHRpbGRlIFReXFx0YXUoXFxRUV92KSJdLFszLDIsIkMoXFxRUV92KSJdLFswLDIsIlxcdGlsZGUgY1xcaW4iXSxbNCwwLCJcXG5pIHhfdiJdLFsyLDNdLFsxLDNdLFswLDFdLFswLDJdXQ==)

  Then $T^\tau_{\tilde c} \cong X_{c}$ implies $x_v$ lifts.

4. For $v$ infinite, we have $x_{\infty } \in Y_{1}(\RR) = X_{c'}(\RR)$.

  - Check that $c'$ lifts to $c'' \in \tilde T^\tau(\RR)$ in the same connected component of $\tilde c$.
    This is because $\tilde T^\tau(\RR) \to C(\RR)$ is étale, and thus locally a covering map and thus surjective on any nonempty component.
    Thus $T^\tau_{c''} \cong X_{c'}$ and thus $x_\infty$ lifts.


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