# 09/29 (Lecture 12) ## Brauer Groups Goal: for $C$ a smooth curve over $k=\bar k$, we've computed \[ H^i(C; \GG_m) = \begin{cases} \OO_C\units(C) & i=0\\ \Pic(C) & i=1 \\ 0 & i > 1 \end{cases} ,\] where the $i=1$ case followed from Hilbert 90, a special case of faithfully flat descent, but currently degrees $i > 1$ is a mystery and our goal will be to show that it vanishes in these degrees. Today we'll look at $i=2$, and recall that we've reduced this computation to understanding the Galois cohomology of the function field $H^i(k(C); \GG_m)$ and of the strict Henselization $H^i(k_{\bar x}; \GG_m)$ where $\bar x$ is a closed point and $K_{\bar x}$ is the strict Henselization of $\OO_{C, x}$. We'll try to understand the Galois cohomology of a field with coefficient in $\bar{k}\units$, or equivalently $\GG_m$ thought of as a sheaf on the étale site. We'll discuss $i=2$, and a general principle in group cohomology is that if one understands $i=1, 2$ then one can often understand all degrees. In general, $H^1$ has a geometric interpretation in terms of classifying torsors, which are relatively concrete objects. On the other hand, although $H^2$ will have some geometric meaning, it is much harder: it will classify certain stacks called **gerbes**. A miracle is that $H^2(\GG_m)$ has real meaning, and is very closely related the Brauer group and thus related to real "physical" objects, namely certain torsors. Recall that we defined the *cohomological Brauer group of $X$* (\cref{def:cohom_brauer_group}) as \[ \Br^{\coh }\da \Br'(X) \da H^i(X_\et; \GG_m)_{\tors } .\] We also started defining the Brauer group by considering \[ \Union_n \ts{\text{étale locally trivial } \PGL_n\dash\text{torsors} / X} \mapsvia{\delta} H^2(X_\et; \GG_m) ,\] and defining $\Br(X) \da \im f$ as a set, which is a reasonably concrete geometric object. This map came from a LES in cohomology, coming from a SES of sheaves, not all of which were abelian. The definition of $\delta$ was the boundary map of \[\label{eq:12_union_1} \Union_n H^1(X_\et; \PGL_n) \mapsvia{\delta} H^2(X_\et; \GG_m) \] arising from the SES of sheaves of groups on $X_\et$, \[ 1 \to \GG_m \mapsvia{} \GL_m \mapsvia{} \PGL_n \to 1 .\] We argued last time that left-exactness was not hard to see and that this was actually exact in the Zariski topology since $\GL_n\to \PGL_n$ is a $\GG_m\dash$torsor and thus Zariski locally trivial. We'll now discuss what the boundary map $\delta$ actually means, since we're considering a LES associated to a SES of potentially nonabelian groups.[^reference_verdier] [^reference_verdier]: Best reference: Giraud, "Cohomologie non Abelienne". :::{.remark} Making the LES here is a little subtle -- one generally gets a long exact sequence of *sets* here which terminates at the $H^2$ we're interested in, although one usually doesn't get a map of the form $H^1(C) \to H^2(B)$ for a SES $A \mapsvia{}B \mapsvia{}C$, you need more than that $A$ is abelian: it needs to be in the center). ::: We'll now try to make $\delta$ explicit in terms of Čech cohomology, which is the only way we have to make sense of the LHS set in \cref{eq:12_union_1}. We defined it to be the set of étale locally trivial $\PGL_n\dash$ torsors, which has a description in terms of $\Hc^1$: the boundary map doesn't usually make sense for a nonabelian group, but it does in very low degrees such as $i=1$. To make the boundary map, we need to implement the snake lemma. Start with $[T]\in H^i(X_\et; \PGL_n)$ a $\PGL_n\dash$ torsor split by some cover $U\mapsvia{}X$ since it is locally trivial. On $U\fiberprod{X} U$, the descent data is given by a section $\Gamma(U\cross_X U; \PGL_n)$ as a sheaf. This is because descent data is an isomorphism on this double intersection and an automorphism of $\PGL_n$ as a $\PGL_n\dash$torsor is the same as a section to the sheaf $\PGL_n$. Note that this descent data satisfies the cocycle condition, so we get an element in the Čech complex. How do we apply the boundary map to such an element? After potentially refining $U$ we can lift this descent data to a section $s\in \Gamma\qty{U\cross_{X}U,GL_{n}}$. This is possible since $\GL_n \surjects \PGL_n$ is a surjection of schemes -- note that there is some subtlety in why one can refine $U$ instead of $U\fiberprod{X} U$. After lifting descent data to a section of $\GL_n$, we now want to me an element of $H^2$. Note that $H^{2}\qty{\GG_{m}}$ is represented by a section to $\GG_{m}$ over $U\cross_{X}U\cross_{X}U$. We started with something satisfying the cocycle condition but lifted in an arbitrary way, so it may no longer satisfy the cocycle condition. We can measure the failure by considering \[ \pi_{23}^{*}\pi_{12}^{*}s \qty{\pi_{13}s}\inv \in \Gamma\qty{U\fiberpower{X}3, \GG_{m}} .\] A priori this is just a section to $\GL_{n}$ but we know that this goes to 1 in $\PGL_{n}$, meaning it must come from $\GG_{m}$. :::{.claim} This is a 2-cocycle representing an element of $H^{2}\qty{X_{\et}; \GG_{m}}$. ::: :::{.exercise title="?"} Prove the claim. In particular, check that $d$ of this cocycle is zero. ::: :::{.slogan} We've proved the following: the **Brauer class** of $T$ $\delta\qty{[T]}$ is the obstruction to lifting a $\PGL_{n}\dash$torsor $T$ to a $\GL_{n}\dash$torsor. If this class vanishes, a lift exists. This is what you might expect: the image of something coming from a boundary map is the obstruction to coming from the previous map. ::: We've just mapped from a set to a group, so we don't know that the image is a group yet, and we don't yet know that the image is in $\Br^{\coh}$ since we don't know if the image is torsion. ### Geometric Interpretations of $\PGL_{n}\dash$Torsors as Brauer Classes There is a general principle: suppose $T\in\Sh^{\set}\qty{X_{\et}}$ and $G = \ul{\Aut}\qty{T}$ as a sheaf, whose sections are given on an open $U$ by pulling back $T$ to $U$ and compute its automorphisms. In all of our examples, we'll think of $T$ as a scheme. There is a natural bijection \[ \correspond{\text{ Locally split } \\ G\dash\text{torsors}} &\mapstofrom \correspond{\text{ forms of } T} \\ \ul{\Isom}(F, T) &\mapsfrom F \\ \tau &\mapsto (\tau\times T)/G ,\] where a **form** is defined to be a sheaf on $X_{\et}$ locally isomorphic to $T$. Applied to $G = \GL_n$ and trivial vector bundles, this says that $\GL_n\dash$torsors are equivalent to vector bundles. The map here is given by sending a form $F$ to $\ul{\mathrm{Isom}}\qty{F, T}$. This is locally split since $F$ is locally trivial, i.e. locally isomorphic to $T$, and so base-changing to some cover where $F$ will make this torsor split. The reverse map is sending $\tau \to \qty{\tau\cross T}/G$, taking the sheaf quotient by a diagonal action. In good cases the quotient will be represented by a scheme, which isn't hard to see since $G$ acts simply transitively, so this essentially gets rid of $\tau$ and makes the fibers isomorphic to $T$ instead. :::{.example title="?"} $\GL_{n}\dash$torsors correspond to vector bundles. Note that this is **not** an equivalence of categories: there are maps of vector bundles which don't come from maps of torsors, e.g. any map that is not an isomorphism. Here the left/right arrows mean that there is a bijection between *sets*, up to isomorphism on each side. ::: :::{.example title="?"} Let $G=\PGL_{n}$, then what are objects with automorphism group $\Aut\qty{G} = \PGL_{n}$? An example that works here is $\Aut_{X}\qty{\PP_{X}^{n-1}} = \PGL_{n}$. ::: :::{.exercise title="?"} You may have seen this stated over an algebraically closed based field, but (nontrivially) this holds over a general base. Prove this. *(Hint: you may need to use the theorem on formal functions or formal GAGA.)* ::: :::{.corollary title="?"} There is a natural bijection \[ \correspond{\PGL_{n}\dash\text{torsors}} &\mapstofrom \correspond{\text{Forms of } \PP^{n-1}} \] These are referred to as **Severi-Brauer $X\dash$Schemes**.[^sev-brauer-var] [^sev-brauer-var]: If $x = \spec k$ is a point, then these are referred to as **Severi-Brauer Varieties**. ::: :::{.example title="?"} The algebra $M \da \Mat_{n\cross n}(A)$ over a ring $A$ has $\Aut(M) = \PGL_{n}$, which is usually referred to as the *Noether-Skolem* theorem. Thus there is a bijection \[ \correspond{\PGL_n\dash\text{torsors}} &\mapstofrom \correspond{\text{Forms of } \Mat_{n\cross n}(\OO_X\sumpower n) = \Endo_{\OO_{x}} (\OO_X\sumpower n )} ,\] and forms of this type are referred to as **Azumaya algebras**. Over a field these are central simple algebras, and it is a fact (which we may see) that over a field these are always division algebras. ::: :::{.question} How can we combine these to send forms of $\PP^{n-1}$ directly to an Azumaya algebra? ::: :::{.remark} Key takeaway: elements in the Brauer group (which we'll soon prove is a group) have concrete representatives: - Forms of $\PP^{n-1}$, which you can think of as étale locally trivial $\PP^{n-1}$ bundles, or - Forms of a matrix algebra, i.e. coherent sheaves of algebras which étale locally endomorphisms of the trivial bundle. ::: ## Twisted Sheaves People usually think about Brauer groups as one of these two objects. We'll discuss a third way. When we defined the boundary map, we took descent data from $\PGL_n$, i.e. elements of this group on double intersections, and we lifted those to $\GL_{n}$. :::{.definition title="Twisted Sheaves"} Let $U \to X$ be an étale cover and suppose $\alpha \in \Gamma(U\fiberprod{X}3, \GG_{m})$ represents a 2-cocycle $[\alpha] \in H^{s}i(X_{\et}; \GG_{m})$. An **$\alpha\dash$twisted sheaf** is $\mcf\in \QCoh(U)$, an isomorphism $\varphi:\pi_{1}^{*} \mathcal{F} \ms \pi_{2}^{*} \mathcal{F}$ (which looks like descent data) which satisfies the cocycle condition up to $\alpha$, i.e. \[ \pi_{23}^{*} \phi \circ \pi_{12}^{*}\phi = \alpha \pi_{13}^{*}\phi .\] ::: :::{.remark} Note that if $\alpha$ were the constant function 1, this would be the descent data for a quasicoherent sheaf, so we're twisting what it means to be a sheaf. This is exactly what we did when defining the boundary map; we started with a $\PGL_{n}\dash$torsor, lifted to a $\GL_n\dash$torsor -- which perhaps can not be done on $U$ since the torsor is split on $U$, but we can lift to a quasicoherent sheaf on $U$ -- and got isomorphisms between the $\GL_{n}\dash$torsors on the double overlaps. These failed to satisfy the cocycle condition, and the Brauer class measured this failure. So in defining $\delta$, our intermediate step was precisely an $\alpha\dash$twisted sheaf. ::: So we'll define the category $\QCoh(X, \alpha)$ whose objects are $\alpha\dash$twisted sheaves and whose morphisms are morphisms of sheaves on $U$ commuting with the pseudo-descent data $\varphi$. :::{.example title="?"} $\QCoh(X, 1)$ is canonically isomorphic to $\QCoh(X)$, which is the very famous theorem of **étale descent**. ::: ### Categorical Properties of $\QCoh(X, \alpha)$ :::{.proposition title="?"} Let $\alpha, \alpha'$ be 2-cocycles for $\GG_{m}$. Then 1. $[\alpha]\in\Br(X)$ (so its in the image the boundary map from $\PGL_{n}$) iff there exists an $\alpha$twisted vector bundle.[^proved_already] 2. $\QCoh(X,\alpha)$ is an abelian category (easy) and has enough injectives (hard) if $X$ is "nice". 3. There is a tensor functor \[ \tensor: \QCoh(X, \alpha) \cross \QCoh(X, \alpha') \mapsvia{} \QCoh(X, \alpha \cdot \alpha') ,\] where is $\alpha,\alpha'$ are defined on different covers than one passes to a common refinement, and a hom functor \[ \Hom: \QCoh(X, \alpha) \cross \QCoh(X, \alpha') \to \QCoh(X, \alpha' \alpha\inv) \] given by literally tensoring or homming sheaves together and use the induced pseudo-descent data. 4. There are functors \[ \Sym^{n}, \Wedge^{n} \mapsvia{} \QCoh(X, \alpha) \mapsvia{} \QCoh(X, n \alpha) .\] 5. There is a canonical equivalence $\QCoh(X, 1) \iso \QCoh(X)$. [^proved_already]: Note that we've proved this already since $\Br \subseteq \im \delta$, but part of our construction of $\delta$ involved constructing an $\alpha\dash$twisted sheaf out of a $\PGL_{n}\dash$torsor. ::: :::{.proof title="Sketch/Discussion"} For (1), if $\alpha\in \Br(X)$, then there is $\PGL_{n}\dash$torsor whose boundary is $\alpha$, and our construction of $\delta$ yielded an $\alpha\dash$twisted vector bundle. Given such a bundle, one can obtain a Brauer class by constructing a twisted form a $\PP^{n}$: take the projectivization. Then the quasi descent data becomes actual descent data since the failure was precisely in scalars, which you're now modding out by. For (3), just work out what happens to $\alpha, \alpha'$ when tensoring or homming. Similarly, all of the rest except for **(2)** follow from definitions. ::: :::{.exercise title="?"} Try to prove (3) and (4). ::: :::{.corollary title="?"} $\Br(X)$ is a group and not just a set. ::: :::{.proof title="?"} For the group operation, let $\mathcal{E}, \mathcal{E}'$ be $\alpha,\alpha'\dash$twisted vector bundles respectively. Note that $\mce \tensor \mce'$ is an $\alpha\cdot\alpha'\dash$twisted vector bundle and having a twisted vector bundle is the same as having a Brauer class. For inverses, suppose $\alpha$ is a Brauer class and $\mathcal{E}$ is a twisted vector bundle. To get an $\alpha\inv\dash$twisted vector bundle, one can just take the dual $\mathcal{E}\dual$. ::: ### Results On Twisted Bundles :::{.proposition title="?"} Suppose that $\alpha$ is a 2-cocycle for $\GG_{m}$, then $[\alpha]$ is trivial iff there exists an $\alpha\dash$twisted line bundle. ::: :::{.proof title="?"} $\implies$: take $\OO_X$, and use the following implicit lemma. :::{.lemma title="?"} If $[\alpha] = [\alpha']$, then $\QCoh(X,\alpha) \simeq \QCoh(X, \alpha')$, although this is not a canonical equivalence. ::: Now if $[\alpha] = 1$ then $\QCoh(X, \alpha) \iso \QCoh(X, 1) = \QCoh(X)$. $\impliedby$: Suppose you have an $\alpha\dash$twisted line bundle. Then writing down a line bundle is done by taking a cover, taking the trivial bundle on each open, and then the double intersections have gluing data which are sections of $\GG_m$. To prove $\alpha$ is a trivial Brauer class, note that the descent data (gluing data) is the same as a section $\beta\in \Gamma(U \cross_{X} U, \GG_{m})$. Then the Čech differential $\check{\delta}$ satisfies $\check{\delta}(\beta)=\alpha$, and thus by definition $\alpha = 0 \in H^{2}(\GG_{m})$. ::: :::{.exercise title="?"} Prove the lemma used in this proof. ::: :::{.corollary title="?"} Suppose $\mathcal{E}$ is an $\alpha\dash$twisted vector bundle of rank $n$. Then $[\alpha]\in H^{2}(X_{\et}; \GG_m)$ is $n\dash$torsion. ::: :::{.proof title="?"} We've just proved this for $n=1$, so for the general case, note that we can form a line bundle from a vector bundle by taking determinants. So consider $\Wedge^{n}\mathcal{E}$, which is a $\alpha^{n}\dash$twisted line bundle, and $\Wedge^n$ multiples a Brauer class by $n$. Now by the previous proposition, \[ 0 = [\alpha^{n}] = [\alpha]^{n} .\] ::: :::{.corollary title="?"} \[ \Br(X) \subseteq \Br'(X) \da H^{2}(X_{\et}; \GG_{m})_{\tors} .\] ::: ### Examples of Brauer Classes :::{.example title="?"} Let $X\da \ts{ x^{2}+y^{2}+z^{2} = 0}$ over $\RR$ is a smooth conic with no rational points. We know via stereographic projection that a smooth projective conic with a rational point would be isomorphic to $\PP^{1}$, so this becomes isomorphic to $\PP^{1}$ after base-changing to $\CC$ where it does have a rational point. So $X\fiberprod{\spec \RR} \spec \CC \cong \PP^{1}$, which implies that $X$ is a twisted form of $\PP^{1}_{\RR}$. In fact, $\delta\qty{[X]}$ generates $\Br(\RR) = \ZZ/2\ZZ$. ::: :::{.example title="?"} What is the generator of this group if we want an Azumaya algebra (or equivalently a CSA)? Take Hamilton's quaternions: these in fact form a division algebra and hence an Azumaya algebra representing the same element. ::: :::{.remark} Given an $\alpha\dash$twisted sheaf $\mathcal{E}$, taking the projectivization yields a Severi-Brauer. Since $\mathcal{E}$ was a vector bundle with descent data twisted by a scalar and projectivizing kills scalars, this yields honest descent data. However, descent is not effective for schemes. So to say that this is actually a variety instead of a sheaf representing this descent, an extra argument is needed: projective space is anti-canonically polarized. Descent data for a polarized variety -- a variety plus an ample line bundle -- is effective, and any descent data on $\PP^n$ is automatically polarized using $\omega_{\PP^n}\dual$. Taking $\Endo(\mathcal{E})$ yields an Azumaya algebra since taking $\Endo$ here kills the twisting cocycle and yields an honest sheaf of algebras. It's locally isomorphic to a matrix algebra since passing to any cover where $\mathcal{E}$ is trivialized yields a matrix algebra. Moreover, there is a way to go from Azumaya algebras to Severi-Brauers, by taking moduli of certain ideals in the algebra, so we have a diagram \begin{tikzcd} & \ts{\text{Severi-Brauers}} \\ \mathcal{E} \in \ts{\text{Twisted sheaves}} \ar[ru, "\PP(\wait)"] \ar[rd, "\Endo(\wait)"'] & \\ & \ts{\text{Azumaya algebras}} \ar[uu, dotted] \end{tikzcd} ::: :::{.example title="3"} An example from class field theory: \[ \Br(\QQ_{p}) = \QQ/\ZZ .\] ::: :::{.remark} Over a field, any 2-torsion Brauer class is represented by a quaternion algebra: using the usual formula where e.g. $i^{2} = -1$, you change $-1$ to some other numbers. In particular, the coset $[{1\over 2}] \in \QQ/\ZZ$ is represented by an algebra resembling the Hamilton quaternions. ::: :::{.example title="?"} There is a map \[ 0 \mapsvia{} \Br(\QQ) \mapsvia{} \bigoplus_{r\in \places{\QQ}} \Br(\QQ_{r}) \mapsvia{\epsilon} \QQ/\ZZ \mapsvia{} 0 ,\] where the first map is induced by pulling back via $\QQ\injects \QQpadic$, $\eps$ is given by adding, and $\places{\QQ}$ is the set of all places of $\QQ$, including $\infty$. Here the map is into the direct sum since the Brauer class is split for almost all places, which is a fun exercise. Thus there is an explicit description of $\Br(F)$ for $F=\QQ$ and in fact any number field. ::: :::{.exercise title="Fun!"} Use the Severi-Brauer interpretation to show that if $\alpha\in \Br(\QQ)$ then $\ro{\alpha}{\QQ_{\nu}} = 0$ for all places $\nu$. ::: :::{.remark} How to interpret multiplication: let $\mathcal{A}_{1}, \mathcal{A}_{2}$ be Azumaya algebras representing $\alpha_{1}, \alpha_{2}$, then $\mathcal{A}_{1} \tensor \mathcal{A}_{2}$ is an Azumaya algebra representing $\alpha_{1} \cdot \alpha_{2}$. This follows because being an Azumaya algebra is a local property, so one can just pass to a cover where this reduces to a fact that matrix algebras are closed under tensor products. Let $\mathcal{P}_1 \da \PP(\mathcal{E})$ and $\mathcal{P}_2 \da \PP(\mathcal{E}')$ be Severi-Brauers representing $\alpha_1, \alpha_2$ respectively. Then $\PP(\mathcal{E} \tensor \mathcal{E}')$ represents $\alpha_1 \alpha_2$. This is because Severi-Brauers come with a natural embedding into projective space given by the anticanonical line bundle, and these can be combined to get an explicit geometric description. ::: :::{.question} It's not easy to write down equations for Severi-Brauers! We can do this for curves (conics), but surfaces are difficult and we don't know how to do much for threefolds. ::: :::{.question title="The period-index question"} Given $\alpha\in\Br(X)$, what is the minimum rank or the $\gcd$ of the ranks of an $\alpha\dash$twisted vector bundle? This can be easily done for 2-torsion over a field, but this is generally a hard question. In general, if you have an $\alpha\dash$twisted bundle of rank $n$ then it's $n\dash$torsion, but the converse is known by examples not to be true. Asking for explicit (and small) representatives of these objects is generally a hard question. ::: Next time: use this theory to understand $H^i(k; \GG_m)$ for $k$ a field. We'll try to prove the following: :::{.theorem title="?"} \envlist 1. If $k(C)$ is the function field of a curve over $k = \bar k$, then $H^2(k(C); \GG_m) = 0$ and thus $\Br(k) = 0$. 2. If $k_{\bar x}$ is a strictly Henselian DVR, then $H^2(K_{\bar x}; \GG_m) = 0$. ::: This will be the key ingredient in computing the étale cohomology of curves.