# Thursday, October 28 :::{.remark} Let $k\in\Global\Field, G\in \Grp\Sch\slice k, T\in \torsors{G}\slice X$. We defined the Selmer group as \[ \Sel_{T\slice X} (k; G) \da \ts{ \tau \in H^1(k; G) \st \tau_{\kv} \in \im\qty{ X(\kv) \mapsvia{x\to T_x} H^1(\kv; G)} } .\] ::: :::{.theorem title="?"} For $G$ smooth affine and $X$ proper, $\Sel_{T\slice X}(k; G)$ is finite. ::: :::{.corollary title="?"} The following disjoint union is finite if $X$ is proper: \[ X(k) = \Disjoint_{\Sel_{T\slice X} (k; G)} \ts{x\in X(\kv) \st \tau_x \cong \tau} .\] ::: :::{.corollary title="Weak Mordell-Weil theorem"} For $A\in \Ab\Var\slice k$, which is a smooth proper group scheme, then $A(k)/mA(k)$ is finite for any $m\in \ZZ_{\geq 0}$. ::: :::{.remark} Note that Mordell-Weil is about finite generation of $A(k)$, which implies that the quotient is finite -- for counterexamples, take $\QQ$ and $\QQ/m\QQ = 0$. Existence of [[variations of Hodge structure]]: follows from [[nonabelian Hodge theory]]. ::: ## Proof of finiteness of Selmer Sets :::{.remark} Let $T$ be defined as multiplication by $n$ to get a SES \[ 0 \to A[n] \to A \mapsvia{x\mapsto nx} A \to 0 .\] Then we get a diagram \begin{tikzcd} {A(k)} && {H^1(k; A[n])} \\ \\ {A(k) / nA(k)} && {\Sel_{T\slice X}(k; A[n])} \arrow["{x\mapsto T_x}", from=1-1, to=1-3] \arrow[from=1-3, to=3-3] \arrow[from=1-1, to=3-1] \arrow[from=3-1, to=3-3] \arrow[dashed, hook, from=3-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJBKGspIl0sWzIsMCwiSF4xKGs7IEFbbl0pIl0sWzAsMiwiQShrKSAvIG5BKGspIl0sWzIsMiwiXFxTZWxfe1RcXHNsaWNlIFh9KGs7IEFbbl0pIl0sWzAsMSwieFxcbWFwc3RvIFRfeCJdLFsxLDNdLFswLDJdLFsyLDNdLFsyLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==) ::: :::{.remark} We'll separately handle the ramified and unramified cases. We'll need to set up some notation: let $k\in \Global\Field, S \subseteq \places(k)$ a finite subset of places, $\mcg\in \Grp\Sch\slice{\OO_{k, S}}$. There is a SES \[ 0 \to \mcg^0 \to \mcg \to \mcg/\mcf \to 0 ,\] where $\mcg_0$ is the component of the identity. Assume $\mcf$ is finite étale, which can always be achieved by enlarging $S$ if necessary. We'll use Roman letters to denote fibers, whence we get a SES over $k$: \[ 0 \to G^0 \to G\to F \to 0 .\] ::: :::{.definition title="?"} Define the set of torsors unramified away from $S$: \[ H^1_S(k; G) \da \ts{ \tau\in H^1(k; G) \st \tau_{\kv} \in\im\qty{H^1(\OO_{\kv};\mcg) \to H^1(k; \mcg) }\,\forall v\in S} .\] Note that this set depends on the model chosen for $\mcg$. ::: :::{.theorem title="?"} The unramified case of Selmer finiteness - $H_S^1(k; G) \to H_S^1(k; G) \to \prod_{v\in S} H^1(\kv, F)$ has finite fibers. - If $k\in \Number\Field$ then $H^1_S(k; G)$ is finite. ::: :::{.remark} This theorem is due to Lang, who was an AIDS denialist? **Yikes.** ::: :::{.remark} This $H^1_S(k; G)$ is an approximation to something slightly more natural, $H^1_\et(\OO_{k, S}; \mcg)$, the $G\dash$torsors over $\OO_{k, S}$. ::: :::{.slogan} Torsors for connected group schemes are not interesting? ::: :::{.theorem title="?"} Let $G\in \smooth\Alg\Grp\slice{\FF_q}$ be connected, then $H^1(\FF_q; G) = \ts{\pt}$. ::: :::{.exercise title="?"} Write down a variety over $\FF_q$ with no rational points! ::: :::{.proof title="?"} Let $T\in H^1(\FF_q; G)$, we want to show $T(\FF_q) \neq \emptyset$. Given a rational point, we want to show $G\cong T$. Take $G\times T\to T$ and for any $x\in T(\FF_q)$ take the map \[ G &\to T \\ x&\mapsto gx .\] :::{.exercise title="?"} Prove this is an isomorphism. Hint: use the following diagram: \begin{tikzcd} G && T && {G\times T} && {T\times T} \\ \\ && X &&& T \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=3-3] \arrow[from=1-5, to=3-6] \arrow[from=1-7, to=3-6] \arrow[from=3-3, to=3-6] \arrow[from=1-5, to=1-7] \arrow[from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJHIl0sWzIsMCwiVCJdLFsyLDIsIlgiXSxbNCwwLCJHXFx0aW1lcyBUIl0sWzYsMCwiVFxcdGltZXMgVCJdLFs1LDIsIlQiXSxbMCwxXSxbMSwyXSxbMyw1XSxbNCw1XSxbMiw1XSxbMyw0XSxbMCwyXV0=) ::: Let $t\in T(\fqbar)$, then any $t'\in T(\fqbar)$ satisfies $t'=gt$ for a unique $g\in G(\fqbar)$, since this is a principal homogeneous space (so the action is free and simply transitive). Consider Frobenius-fixed elements: take $\sigma\in\Gal(\fqbar \slice {\FF_q})$, then if $t' = (t')^{ \sigma}$ then $gt = g^{\sigma}t^{ \sigma}$ and $(g^{ \sigma})\inv \tau= t^{ \sigma}$. Now $t^{ \sigma} = bt$ for a unique $b\in G(\fqbar)$, so to solve this equation it suffices to show that every $b \in G(\fqbar)$ can be written in the form $b = (g^{ \sigma})\inv g$. :::{.claim} Consider the following funny action $G\actson G$ by $g\cdot t \da (g^{ \sigma})\inv t g$, a twisted conjugation. This acts transitively on $G(\fqbar)$ and thus has one orbit. ::: Why this implies the theorem: take $t=\id$, then any $b\in G(\fqbar)$ is in the orbit of $t$ and $b = (g^{\sigma})\inv g$. ::: :::{.example title="?"} For $G = \GG_a$, this yields $g\cdot t = tg - g^p$. For $G = \GG_m$ it yields $g\cdot t = t - g^{1-p}$. ::: :::{.observation} For fixed $t \in G(\fqbar)$, the following map is generically étale: \[ G_{\fqbar} &\to G_{\fqbar} \\ g &\mapsto (g^{ \sigma})\inv t g ,\] and thus the image contains an open subset of $G_{\fqbar}$. So any two orbits are open, and hence intersect since $G$ is connected. Since orbits are equal or disjoint, there is just one orbit. Proof: continued next time! ::: :::{.remark} Some asides on philosophy: hyperbolicity should be related to having rational points, and having "big" fundamental group and being general type. A recent theorem by Ellenberg-Lawrence-Venketesh: for $X\in \Var\slice k$ for $k\in \Number\Field$ and $\pi_1 X$ is big, then $\size X(\OO_{k, S})$ has height bounded above by $H$ which grows like $H^{\eps}$ for any $\eps$. > See paper: Here "big $\pi_1$" means that for $X_\CC$, there is a variation of Hodge structure on $X$ such that the map $X\to \mcd/\Gamma$ into the period domain is quasi-finite. :::