# Thursday, September 02 ## Correspondence of Forms :::{.remark} Last time: standard/reduced complexes, forms, and $H^1$. A meta-definition for today: let $k, L \in \Field$ with $L\slice k$ finite and separable, and $X\slice k$ an object over $k$ (e.g. an algebraic variety, possibly with extra structure). A **form** of $X\slice k$ split by $L$ is an object $X' \slice k$ of the same class as $X$ such that $X_L \mapsvia{\sim} X_L'$. ::: :::{.theorem title="A meta-theorem"} The theorem was that there is a canonical bijection \[ \correspond{ \text{Forms of } X \\ \text{split by } L } &\mapstofrom H^1_\Gal(L\slice k; \Aut_k(X_L)) \] Note that we didn't assume the coefficients formed an abelian group, so we'll explain this today. It is true that $\Aut(X_L) \in \mods{\Gal(L\slice k)}$. We'll say that $X'$ is just **a form** of $X$ if there exists some $L'$ finite separable that splits $k$. In this case there is a correspondence \[ \correspond{ \text{Forms of }X } &\mapstofrom H^1_{\Gal}(L\slice k; \Aut_k(X_{k^s}) ) \] ::: :::{.proof title="A meta-proof"} What is the map? Given a form $X'$, we by definition have $F: X_L' \mapsvia{\sim} X_L$, and we want a map $\Gal(L\slice k) \to \Aut_k(X_L)$ such that $\delta f = 0$ for the differential in cohomology. Since $X'$ is defined over $k$, we have an action $\Gal(L\slice k)\actson X_L'$, i.e. a map $\Gal(L\slice k) \to \Aut(X_L')$, which we can compose with the given isomorphism to obtain \[ f: \Gal(L\slice k) \to \Aut(X_L')\to \Aut(X_L) .\] We have $f( \sigma \tau) = f(\sigma) f(\tau)^\sigma$. What happens if we change the isomorphism $F$ to some $F'$, changing by some $g\in \Aut(X_L)$ :::{.exercise title="?"} Here $f$ changes by a map of the form $\sigma \to g(g\inv)^\sigma$. ::: We'll write an inverse map using Galois descent. Given $f: \Gal(L\slice k)\to \Aut_k(X_L)$ with $f(\sigma \tau) = f( \sigma) f( \tau)^{ \sigma}$, we want to construct a form of $X$. Assume $X \in \Aff\Sch$, so $X = \spec(A)$ for some $A\in\kalg$, then define \[ X' \da \spec(A\tensor_k L)^{\Gal(L\slice k)} \] where the action is given by $f$. ::: :::{.remark} What is $\Aut_k(X\slice L)$ is nonabelian? Then we just make this proof a definition, and set \[ H^1(L\slice k ; G) \da \ts{f: \Gal(L\slice k)\to G \st f( \sigma \tau) = f( \sigma) f( \tau)^{ \sigma} }/ ( \sigma\to g(g\inv ) ^{ \sigma}) .\] Here the maps are of finite discrete groups. This is a pointed set, using the constant map as a basepoint. ::: ## Torsors :::{.definition title="Torsor"} Recall that for $G \in \Alg\Grp\slice k$, a **torsor** for $G$ (or a *principal homogeneous space*) is 1. A form of $G$ under the left action of $G$ on itself, i.e. a variety $X$ with a left $G\dash$action $G\times X\to X$ where $X_L \mapsvia{ \sim} G_L$ using the left-translation action. 2. A $G\dash$variety $X$ such that $G\times X \mapsvia{\sigma, \pi_2} X\times X$ is an isomorphism. ::: :::{.claim} Note that these are equivalent if $G$ is smooth, which for us will always happen in characteristic zero. ::: :::{.theorem title="?"} If $G$ is smooth, then $G\dash$torsors are canonically in bijection with $H^1(k; G(k^s))$, and $G\dash$torsors split by $L$ biject with $H^1(L\slice k; G(L))$. ::: :::{.exercise title="?"} Prove this! It suffices to show that $\Aut_{G_L}(G_L)\cong G_L$ as a $\Grp\Sch\slice{G_L}$. ::: ## Example: Kummer Theory :::{.example title="Kummer theory"} Suppose $\mu_p \subseteq k$, so $k$ contains all $p$th roots of unity. Then a $\mu_p\dash$torsor is the same as a $\ZZ/p$ Galois extension of $k$, where we allow $k^p =\mu_p$ itself. ::: :::{.theorem title="?"} There is a bijection \[ \correspond{ \ZZ/p \dash\text{extensions} } &\mapstofrom H^1(k; \mu_p) \] ::: :::{.proof title="?"} Use the SES \[ 1 \to \mu_p \to (k^s)\units \mapsvia{x\mapsto x^p} (k^s)\units \to 1 ,\] which yields a LES \[ 1 \to H^0(k; \mu_p) \to H^0(k; (k^s)\units) \mapsvia{x\mapsto x^p} H^0(k; (k^s)\units )\to H^1(k; \mu_p) \to H^1(k; (k^s)\units) ,\] and identifying terms yields \[ 0 \to k\units / (k\units)^p \to H^1(k; \mu_p) \to H^1(k; (k^s)\units) .\] ::: :::{.example title="?"} What is $H^1(k; (k^s)\units)$? Use that $L\units = \Aut(V\slice L)$ where $V$ is a 1-dimensional vector space over $L$. The claim is that by Galois descent, forms for a vector space split by $L$ are precisely vector spaces over $k$, which makes them all trivial. This in fact implies the more general fact that $H^1(k; \GL_n(k^s)) = 1$. ::: :::{.remark} Kummer theory gives us an explicit form of the map and identifying terms yields \[ 0 \to k\units / (k\units)^p \mapsvia{x\mapsto k[x^{1\over p} ]} H^1(k; \mu_p) \to H^1(k; (k^s)\units) .\] This can be found by unwinding the definition of the map from the snake lemma, or noting that the kernel of a map from the absolute Galois group cuts out exactly this field. ::: ## Geometry of Brauer Groups :::{.example title="of $H^1$"} $H^1(k; G)$ are forms of objects with automorphism groups $G$. - Vector spaces are obtained by taking $G = \GL_n$. - Forms of $\PP^n$, i.e. Severi-Brauer varieties, come from taking $G \da \PGL_{n+1}$. - For $G$ finite, a form of $G$ is an étale $k\dash$algebra (product of separable extensions of $k$ with total Galois group $G$). - For $G$ simple, these are Galois extensions with Galois group $G$. For $G\da \ZZ/p$, this is Kummer theory. - For $E$ an elliptic curve, all genus 1 curves are torsors for their Jacobian. So genus 1 curves $C$ with $\Jac(C) \cong E$ biject with $H^1(k; E(k^s))$. ::: :::{.remark} We'll now look at $H^2$, and there is a correspondence \includegraphics{figures/SectionsOne.pdf} Given a set-theoretic section $s: G\to G'$, we get a map \[ f_s: G\cartpower{2} &\to A \\ (g_1, g_2) &\mapsto s(g_1) s(g_2) s(g_1 g_2)\inv .\] Note that if $s$ is a group morphism, this is just the constant map. :::{.claim} One needs to show the following: 1. $\delta f_s = 0$, so one gets a cocycle. 2. Changing $s$ changes $f_s$ by a coboundary. 3. Make the inverse. ::: The group operation here is $G' \cdot G'' \da G' \fiberprod{G} G'' / A$, and the multiplication map is \[ (a_1, g_1) \cdot (a_2, g_2) \da (a_1 a_2 f_s(g_1, g_2), \, g_1 g_2) .\] ::: :::{.remark} Suppose $1\to Z\to H' \to H\to 1$ is a SES of groups with a $G\dash$action such that $Z$ is in the center of $H'$. Then there is a "LES" \begin{tikzcd} 1 \\ \\ {Z^G} && {(H')^G} && {H^G} \\ \\ {H^1(G; Z)} && {H^1(G; H')} && {H^1(G; H)} \\ \\ {H^2(G; Z)} \arrow[from=1-1, to=3-1] \arrow[from=3-1, to=3-3] \arrow[from=5-1, to=5-3] \arrow[from=3-3, to=3-5] \arrow["{\delta_0}"', from=3-5, to=5-1, in=180,out=0] \arrow[from=5-3, to=5-5] \arrow["{\delta_1}"', from=5-5, to=7-1, in=180, out=0] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwwLCIxIl0sWzAsMiwiWl5HIl0sWzIsMiwiKEgnKV5HIl0sWzQsMiwiSF5HIl0sWzAsNCwiSF4xKEc7IFopIl0sWzIsNCwiSF4xKEc7IEgnKSJdLFs0LDQsIkheMShHOyBIKSJdLFswLDYsIkheMihHOyBaKSJdLFswLDFdLFsxLDJdLFs0LDVdLFsyLDNdLFszLDQsIlxcZGVsdGFfMCIsMl0sWzUsNl0sWzYsNywiXFxkZWx0YV8xIiwyXV0=) Note that some terms here are only sets, so exactness means that differentials surject onto kernels, and $H^1(G; Z)\actson H^1(G;H')$ and $H^1(G; H)$ is the quotient by this action. ::: :::{.remark} ::: :::{.definition title="Brauer group"} Take $1\to \GG_m \to \GL_n \to \PGL_n \to 1$, then we get a map \[ H^1(k; \PGL_n(k^s)) \mapsvia{\iota_n} H^2(k, (k^s)\units) .\] Then define the **Brauer group** of $k$ to be \[ \Br(k) \da \Union_{n} \im(\iota_n) .\] ::: :::{.remark} Studying $H^2$ is hard in general, so this fact is the reason we can actually study Brauer groups. \todo[inline]{Something about Hilbert 90} This surjection gives us geometric objects to work with. We'll show this is a group next time, along with the following theorem: ::: :::{.theorem title="?"} \[ \Union_n \im(\iota_n) = H^2(k; (k^s)\units) .\] :::