## Derived Categories 5: Some Examples

Today we’ll just unravel some toy cases, because understanding these examples is really important, but they tend to be done in generality where the intuition goes away. The first fact is something that we’ll need to use to prove a structure theorem for objects in the derived category of a curve.

As usual, let’s fix ${X/k}$ a smooth, projective variety over a field and let ${D(X)}$ be the bounded derived category of coherent sheaves on ${X}$. In fact, our first fact will work for the derived category of any abelian category. It says that if we take an object (i.e. complex) ${A}$ whose cohomology vanishes for all ${n>i}$, then we can always put it into an exact triangle ${B\rightarrow A\rightarrow H^i(A)[-i]}$.

Let’s be extremely rough and vague to motivate this at first. Given any morphism ${B\stackrel{f}{\rightarrow}A}$ just from the axioms of a triangulated category we can always complete this to an exact triangle ${B\rightarrow A\rightarrow cone(f)}$. Roughly speaking the cone “behaves like a quotient.” I’ll remind you of the exact definition in a second. Our other motivating idea is that we always have a “truncation” process. I’ll chop the very last part of the complex off and then use the “inclusion” map. If the cone is like a quotient I’ll kill everything except that part that I chopped off, and I’ll just be left with the cohomology.

We won’t prove this in general, but we’ll do it in the case of a very small complex to reacquaint ourselves with the cone construction and quasi-isomorphisms. This is one of those cases where this toy case is actually no less general than the general proof when you see what is going on.

Recall that a morphism of complexes ${f: A^\bullet \rightarrow B^\bullet}$ is a collection of morphisms ${f^i: A^i\rightarrow B^i}$ commuting with the differentials: ${d_B\circ f^i=f^{i+1}\circ d_A}$. The complex denoted ${cone(f)=C^\bullet}$ is given by ${C^i=A^{i+1}\bigoplus B^i}$. The differentials are a little tricky, but sort of the only obvious thing that actually makes it into a complex: ${d_C^i=\left(\begin{matrix} -d_A & 0 \\ f^{i+1} & d_B\end{matrix}\right).}$

For simplicity, let’s just assume that our complex has the following form ${0\rightarrow A^0\rightarrow A^1\rightarrow A^2 \rightarrow A^3 \rightarrow 0}$. We’ll assume that cohomology at the second spot is the highest non-zero cohomology. We’ll truncate there and then form the cone. The truncated complex just looks like ${0\rightarrow A^0\rightarrow A^1\rightarrow \ker (d^2)\rightarrow 0}$. Our morphism, ${f}$, of complexes in this easy example at the three different types of places is either the identity morphism, inclusion morphism, or ${0}$ morphism.

The cone complex of this morphism is (I’ll be redundant and put in the zeros for clarity):

$\displaystyle 0\rightarrow A^0\oplus 0 \rightarrow A^1\oplus A^0\rightarrow ker(d^2)\oplus A^1 \rightarrow 0 \oplus A^2\rightarrow 0 \oplus A^3 \rightarrow 0$

One important thing to notice is that at ${A^0\oplus 0}$ is sitting in degree ${-1}$ by definition. By construction the last few maps are again just the original ${d}$ maps. Thus ${H^3(cone(f))=H^3(A^\bullet)=0}$ by assumption. Also, ${H^2(cone(f))=H^2(A^\bullet)}$. If we can show that all other cohomology vanishes, then we will have shown that ${cone(f)}$ is quasi-isomorphic to the single term complex ${H^2(A^\bullet)[-2]}$. This will show the lemma.

This part just follows by “fun with the cone construction.” Note that the first differential is ${(a,0)\mapsto (-d(a), a)}$. If this element is ${(0,0)}$, then ${a=0}$. Thus the kernel is trivial and we get ${H^{-1}(A^\bullet)=0}$. The next one is even more fun. The map is ${(a,b)\mapsto (-d(a), a+d(b))}$. The image of the previous is in the kernel of the next by just doing the maps successively ${(a,0)\mapsto (-d(a), a)\mapsto (d^2(a), -d(a)+d(a))=(0,0)}$. Now we see why that minus sign was important to make the cone a complex.

To check that the image is equal to the kernel, now suppose ${(a,b)}$ is in the kernel. In particular, this means ${a=-d(b)}$ by looking at the second term. The claim is that ${(a,b)}$ is the image of ${(b,0)}$. Well, ${(b,0)\mapsto (-d(b), b)=(a,b)}$. Thus ${H^0(cone(f))=0}$. And now you get the point. There is only one more to check, but it follows the same way.

Of course you could do this with cohomology bounded below with the other truncation functor where you replace the lowest term with a cokernel. In general, this shows an incredibly useful proposition: Given any complex in the bounded derived category of an abelian category there is a finite filtration by truncation ${0\rightarrow A_k \rightarrow A_{k+1} \rightarrow \cdots \rightarrow A_j\rightarrow A^\bullet}$ where the “quotient” (i.e. cone) at each step is ${A_{n-1}\rightarrow A_{n}\rightarrow H^{n}(A^\bullet)[-n]}$. This notation is meant to reflect that ${H^n(A^\bullet)=0}$ if ${n}$ is not in the range ${[k,j]}$.

Here is a remarkable consequence of this proposition. Let ${C/k}$ be a smooth projective curve. Any object ${\mathcal{F}^\bullet\in D(C)}$ is isomorphic to the direct sum of its cohomology sheaves ${\bigoplus_i \mathcal{H}^i(\mathcal{F}^\bullet)[-i]}$. In particular, every object in the derived category is a direct sum of coherent sheaves.

Now we see the power of being able to filtrate by cohomology, because it will allow us to make an induction argument. We prove this by inducting on the length of the complex. The base case is by definition. Suppose the result is true for a length ${k-1}$ complex. Suppose ${\mathcal{F}^\bullet}$ has length ${k}$. By shifting, we suppose ${\mathcal{H}^i(\mathcal{F}^\bullet)\neq 0}$ only in the range ${1\leq i \leq k}$. Consider the first step of the filtration ${\mathcal{E}^\bullet \rightarrow \mathcal{F}^\bullet \rightarrow \mathcal{H}^{k}(\mathcal{F}^\bullet)[-k]\rightarrow \mathcal{E}^\bullet [1]}$.

If this triangle splits we are done, because then the middle term will be a direct sum of the outer terms and by the inductive hypothesis ${\mathcal{E}^\bullet}$ splits as a direct sum of its cohomology. We may as well write it this way

$\displaystyle \mathcal{E}^\bullet\simeq \bigoplus_{i=1}^{k-1} \mathcal{H}^i(\mathcal{E}^\bullet)[-i].$

A sufficient condition to show the triangle splits is to check that ${Hom_{D(C)}(\mathcal{H}^{k}(\mathcal{F}^\bullet)[-k], \mathcal{E}^\bullet [1])=0}$. But by what we just wrote this is the same as

$\displaystyle Hom_{D(C)}(\mathcal{H}^k(\mathcal{F}^\bullet), \bigoplus_{i=1}^{k-1}\mathcal{H}^i(\mathcal{E}^\bullet)[k+1-i]) \simeq \bigoplus_{i=1}^{k-1}Ext_C^{k+1-i}(\mathcal{H}^k(\mathcal{F}^\bullet), \mathcal{H}^i(\mathcal{E}^\bullet))$

Since smooth curves have homological dimension ${1}$ and the exponent is always strictly larger than ${1}$ the Ext groups all vanish. This shows the triangle splits and hence any object in the derived category of a smooth curve splits as a sum of its cohomology sheaves. This exact same proof also shows that for any abelian category with homological dimension less than or equal to ${1}$ the objects of the derived category split as a sum of their cohomology.

## The Derived Category 4: A Nice Spanning Class

Recall that we are assuming that ${X/k}$ is a smooth projective variety. Let’s also say it is of dimension ${n}$. We’re going to be lazy (i.e. sane) and all functors will be derived when talking about the derived category even though the ${\mathbf{L}}$ and ${\mathbf{R}}$ will be omitted. Our derived category always has some (auto-) functors. For example, we definitely have the shift functor that comes with any triangulated category ${[k]}$ just by shifting where the sheaves occur in the complex.

Also, given any coherent sheaf we have the functor ${\mathcal{F}\otimes - : D(X)\rightarrow D(X)}$. In particular, we could tensor with the canonical bundle and shift by ${n}$. This functor is so useful it has a name and notation ${S_X(\mathcal{E})=\mathcal{E}\otimes \omega_X [n]}$ (again, ${\mathcal{E}}$ is any object of ${D(X)}$ and hence a complex even though I didn’t write ${\mathcal{E}^\bullet}$). We call this the Serre functor.

This name just comes from the fact that the generalized form of Serre duality for the derived category says that there is a functorial isomorphism

$\displaystyle \eta: Hom_{D(X)}(\mathcal{E}, \mathcal{F})\stackrel{\sim}{\rightarrow} Hom_{D(X)}(\mathcal{F}, S_X(\mathcal{E}))^*.$

Notice that if ${\mathcal{E}}$ and ${\mathcal{F}}$ are honest sheaves sitting in degree ${0}$ we can use that ${Ext^i(\mathcal{E}, \mathcal{F})=Hom_{D(X)}(\mathcal{E}, \mathcal{F}[i])}$ to derive the special case ${Ext^i(\mathcal{E}, \mathcal{F})\simeq Ext^{n-i}(\mathcal{F}, \mathcal{E}\otimes \omega_X)^*}$ which is the standard form of Serre duality given in classic texts like Hartshorne.

For the rest of today let’s look at a very important concept from triangulated categories. One might wonder how much we can know about certain triangulated categories just from knowing certain special classes of objects. A collection of objects ${\Omega}$ is called a spanning class for a triangulated category ${\mathcal{D}}$ if the following hold:

If ${Hom(A, B[i])=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

If ${Hom(B[i], A)=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

It is not in general true that these two conditions are equivalent, but it is easy to check that Serre duality for ${D(X)}$ will allow us to only have to check one of the conditions. The idea of spanning classes (which may not come up for awhile) is that you can check certain properties just on these objects to get properties on the whole category. For example, one can use this idea to prove necessary and sufficient conditions for a Fourier-Mukai transform to be fully-faithful.

Since our triangulated category ${D(X)}$ is somehow built out of ${X}$, to any (closed) point of ${X}$ we have a natural object associated to it that we’ll call ${k(x)}$. This is just the skyscraper sheaf at the point ${x}$. One hope would be that the set of objects of this form is a spanning class. This intuitively makes sense, because checking a property on this class in the derived category is sort of like checking a property on “points” of variety. It is indeed the case that this forms a spanning class.

Suppose ${\mathcal{F}}$ is a non-trivial object of ${D(X)}$. We’ll check the second condition. This says that we must produce some closed point ${x}$ and some integer ${i}$ so that ${Hom(\mathcal{F}, k(x)[i])\neq 0}$ (well, almost, we used Serre duality again to flip the i over to the other side). We will use the standard local-to-global spectral sequence

$\displaystyle E_2^{p,q}=Ext^p(\mathcal{H}^{-q}(\mathcal{E}), \mathcal{G})\Rightarrow Ext^{p+q}(\mathcal{E}, \mathcal{G}).$

If we plug in ${\mathcal{E}=\mathcal{F}}$ and ${\mathcal{G}=k(x)}$ we get

$\displaystyle E_2^{p,q}=Hom(\mathcal{H}^{-q}(\mathcal{F}), k(x)[p])\Rightarrow Hom(\mathcal{F}, k(x)[p+q]).$

Let ${m}$ be the maximal ${m}$ such that ${\mathcal{H}^m(\mathcal{F})\neq 0}$. The sheaf itself is assumed non-trivial, so there exists ${m}$ with that sheaf non-zero, but ${X}$ is regular so there are only finitely many non-zero and hence such an ${m}$ exists. We will now argue that ${E_2^{0, -m}=E_{\infty}^{0, -m}}$ by showing that all differentials with source and target ${E_r^{0,-m}}$ for any ${r}$ must be trivial.

On the one hand, ${E_2^{p,q}}$ is the ${p}$-th Ext group between coherent sheaves, so when ${p<0}$ it always vanishes. This means that any differential with target ${E_r^{0,-m}}$ must be trivial. On the other hand, our choice of ${m}$ maximal implies that any differential with source ${E_r^{0, -m}}$ is trivial.

Now ${\mathcal{H}^m(\mathcal{F})}$ is non-trivial, so in particular it has non-trivial support which is a closed set and hence contains some closed point ${x}$. This tells us that ${E_{\infty}^{0, -m}=E_2^{0, -m}=Hom(\mathcal{H}^m(\mathcal{F}), k(x))\neq 0}$. But this says that ${Hom(\mathcal{F}, k(x)[-m])\neq 0}$ which is what we set out to prove and hence the collection of skyscraper sheaves of closed points do form a spanning set.

## The Derived Category 3: Derived Functors

Let’s get back to some math. Today I’ll restart a series that I started forever ago on the derived category of a variety. I briefly described what the derived category of a variety is in order to give a very sketchy outline of what Homological Mirror Symmetry is. If you’ve forgotten the construction you can go read about it. I’ll do a one paragraph recap here, so if you don’t care about the details then that should suffice.

For us, in this series we will assume our varieties are all smooth, projective (and irreducible) over a field ${k}$. The notation ${D(X)}$ will mean the bounded derived category of coherent sheaves on ${X}$. It is no longer abelian, but it is a ${k}$-linear, triangulated category. One way to construct ${D(X)}$ is to form the category of complexes of coherent sheaves, then consider morphisms of complexes up to homotopy equivalence, then invert all quasi-isomorphisms. Thus a morphism ${\mathcal{A}^\bullet \rightarrow \mathcal{B}^\bullet}$ is a “hat.” There is a complex ${\mathcal{C}^\bullet}$ quasi-isomorphic to ${\mathcal{A}^\bullet}$ and a morphism in the homotopy category ${\mathcal{C}^\bullet\rightarrow \mathcal{B}^\bullet}$.

Recall how we form derived functors in classical-land. The most beloved derived functor in algebraic geometry is probably the global section functor, because the ${i}$-th right derived functor is just sheaf cohomology: ${R^i\Gamma (X, \mathcal{F})=H^i(X, \mathcal{F})}$. What do we do? We take our sheaf and replace it with an injective resolution (an exact sequence) ${0\rightarrow \mathcal{F}\rightarrow \mathcal{I}^\bullet}$. Then we take global sections of each term (and chop off the first guy) to get a complex which is possibly no longer exact. The ${i}$-th derived functor is now just cohomology at the ${i}$-th spot.

One of the beautiful things about the derived category is that we can keep track of all of this information all at once using a “total” derived functor. Secretly what is going on is that we started with an additive functor ${\Gamma: Coh(X)\rightarrow Vec(k)}$ to vector spaces over ${k}$ (in general, between any two abelian categories). When we first started talking about this we noted that the homotopy category of the full subcategory of injectives is isomorphic to the derived category: ${\mathcal{K}^+(\mathcal{I})\simeq D^+(X)}$. So step one of finding an injective resolution just amounts to replacing the complex ${\cdots \rightarrow 0 \rightarrow \mathcal{F}\rightarrow 0 \rightarrow \cdots}$ with a quasi-isomorphic complex of injectives (i.e. use this equivalence!).

In the homotopy category it is perfectly fine to take global sections of everything and get another complex. Then we just go back to the derived category by the universal quotient functor (of inverting quasi-isomorphisms). What did this do? Well, we may as well generalize. If I have a left exact functor between abelian categories (and I have enough injectives) ${F: \mathcal{A}\rightarrow \mathcal{B}}$, then I can make a total derived functor ${\mathbf{R}F: D^+(\mathcal{A})\rightarrow D^+(\mathcal{B})}$ by replacing a complex by a quasi-isomorphic complex of injectives and applying ${F}$ to everything (and strictly speaking passing back to the derived category).

It takes a complex to a complex, but it is keeping track of all the information of a classical derived functor, because it is literally the exact same process but we just omitted that last step of taking cohomology. So what we end up with is a complex with the property that ${\mathcal{H}^i(\mathbf{R}\Gamma(\mathcal{F}))=R^i\Gamma(\mathcal{F})=H^i(X, \mathcal{F})}$. Since we’re in our nice variety situation all higher cohomology vanishes so we can actually stay in the bounded derived categories.

Note that this isn’t merely a way to keep track of all the ${R^iF}$ at once. It wouldn’t be that useful if this was the only thing it was doing. If we have any of our common left exact functors ${F:Coh(X)\rightarrow Coh(Y)}$, we get a functor ${\mathbf{R}F: D(X)\rightarrow D(Y)}$. So we apply the derived functor to complexes and not just objects of ${Coh(X)}$! This is a vast generalization. A word of warning here. Strictly speaking we have to keep making use of two facts (presented previously).

First, the natural functor ${D(X)\rightarrow D^b(QCoh(X))}$ induces an equivalence between ${D(X)}$ and the full subcategory of complexes of quasi-coherent sheaves with coherent cohomology. This allows us to actually have enough injectives to form the resolutions. Second, we have enough conditions on ${X}$ so that we can do things in either the bounded below or bounded above versions of the derived category, but keep landing inside the bounded derived category since our resolutions may not necessarily be finite a priori.

The common functors to which I referred above are the pushforward of a map ${f: X\rightarrow Y}$. If I input a sheaf and take cohomology, then we recover the higher direct images ${\mathcal{H}^i(\mathbf{R}f_*(\mathcal{F}))=R^if_*(\mathcal{F})}$. We have ${\mathcal{H}om(\mathcal{F}, -): QCoh(X)\rightarrow QCoh(X)}$. Again, smoothness saves us and we get a functor on the bounded derived category with the property that ${\mathcal{H}^i(\mathbf{R}\mathcal{H}om(\mathcal{F}, \mathcal{G}))=\mathcal{E}xt^i(\mathcal{F}, \mathcal{G})}$. We could keep going, but the only other major one I foresee coming up in the near future is the derived tensor product. Of course this will be left derived and so we have to do the whole process above but for right exact functors.

The subtlety about reversing everything is that when you unravel the definitions you’ll find that ${\mathcal{H}^{-i}(\mathcal{F}\otimes^{\mathbf{L}}\mathcal{G})=\mathcal{T}or_i(\mathcal{F}, \mathcal{G})}$. A note to the detail oriented reader. I may forget to put bullets in the superscripts, but everything from here on out should be read as a complex of sheaves and not just a sheaf. I also may get lazy and leave off the R and L for right and left derived, but functors between derived categories will ALWAYS be derived.

Overall, this was just an annoying technical post I had to do. Next time I want to get to some actual geometry of the derived category!

## Mathematical Music Theory 4: Intro to Post-Tonal Methods

Alright. Let’s skip from first week of first year of music theory to something that probably won’t come up until a music theory elective in your third year (yes, we’re skipping two full years of theory here). What I’m going to describe is usually encountered in a class on “Post-tonal theory.” This can be misleading because for the most part it is an extremely useful mathematical way of thinking about music theory that doesn’t particularly have to do with atonal music or 12-tone serialism.

As we’ve already pointed out our Western 12 tone scale is essentially taking an octave and dividing it up into 12 parts. Since an octave (or 12 semitones up or down) gives the same note we can mathematically think of things more clearly by just labelling a C with 0, a C# with 1, a D with 2 and so on up to labelling a B with 11. When we back to C we “wrap around” and call it 0 again.

A great way to visualize this is to draw a 12-sided figure with all the side lengths the same (a regular dodecagon). Now if we take a C major chord: 0, 4, 7, then transposing it to a major chord 3 semitones up just amounts to adding every number by 3, i.e. 3, 7, 10. In fact, given any set of notes, we have the operation of transposition $T_n (i_1, i_2, \ldots, i_k)=(i_1+n, i_2+n, \ldots, i_k +n) \ \text{mod} \ 12$ where mod 12 means we add by wrapping around and consider 12=0, 13=1, 14=2, etc (because they are the same notes!!).

We can also do something called inversion. This just amounts to exactly inverting every interval. This amounts to negating every single number and then figuring out what this number is mod 12. So the inversion of the C major chord: [0, 4, 7] is [0, -4, -7]=[0, 8, 5] or if we really are considering “chords” then the order doesn’t matter so it is [0,5,8]. But this is just an f minor chord! We call this operation I for “inversion.” It can be visualized as a reflection of the dodecagon as follows (don’t make fun, I whipped this together using Google draw in a minute or so):

It is pretty clear that doing $T_n$ for all choices of n to [0,4,7] gives you all 12 majors chords and if you do both I and $T_n$ then you’ll get all 12 minor chords too. The operations of transpositions and inversions forms something called a group. In fact, visualizing with a regular 12-gon immediately tells us that the T/I group is what mathematicians call $D_{12}$ the Dihedral group of symmetries of the dodecagon. It has 24 elements.

We call an unordered collection of numbers between 0 and 12 a pitch class set, and we get that $D_{12}$ acts on the set of pitch class sets. We just proved that the orbit of [0,4,7] under this action consists of exactly the collection of major and minor triads. Note that none of the triads are sent to themselves, i.e. given a non-trivial symmetry/combination of transpositions and inversions we will always get a distinct new triad. Mathematicians might say this in a fancy way: the set of major and minor triads is a torsor under the T/I-action.

It turns out this is a “generic” phenomenon in the sense that choosing some random pitch class set you are likely (in that the probability is greater than 50%) to have chosen one that has this property. We could say that it has the property of having no T/I-symmetry. Conversely, we could call a k-chord (read: an unordered chord with k notes in it) T/I-symmetric if there is some choice of non-trivial transposition and inversion such that the chord is sent to itself.

Now even though these are more rare, it turns out that for any choice of k, there is always a k-chord with this property. These exist for rather silly reasons. For example, [0,1,2, ... , k] is always an example of such a chord (exercise: why?). For less trivial examples you could take the whole-tone scale [0,2,4,6,8,10]. If you do $T_2$ then you certainly get the whole tone scale back again. Inversion also fixes this 6-chord. This tells us that up to inversion and transposition there are only 2 distinct whole tone scales (if you want overkill then the subgroup generated by $T_2$ and I has 12 elements, so the Orbit-Stabilizer Theorem tells us this fact).

Here is an interesting question from pure music theory that to my knowledge is still open (although I suspect it is fairly easy to answer and if I spent time trying to figure out the answer in place of writing this post I’d have the answer). None of this was specific to dividing up an octave into 12 notes. Suppose you invent a tonal system with n notes instead. Then you’d have an action of $D_n$ on the k-chords. Is there a simple closed form formula for the number of k-chords that are T/I-symmetric? More importantly, for a given n, which k gives the most number of k-chords with T/I-symmetry.

I should point out that if you rule out the “silly examples” of T/I-symmetry given by a strictly chromatic scale, then there is actually utility in figuring this out. T/I-symmetry has played a great role in the history of composition. For example, the augmented triad, the French augmented sixth chord, the diminished seventh, the famous chord from Stravinsky’s Petrushka, the hexatonic scale, the whole tone scale, and the octatonic scale are all examples. So I think this is more than just a novelty problem.

## Brauer-Manin Obstruction

We will continue with the Brauer group after this post, but today let’s answer the question: Why should we care about the Brauer group? We gave one good reason back when talking about rational surfaces. It provided something that might be computable to detect whether or not our variety had good reduction.

Today let’s consider what is probably the most well-known use of the Brauer group. Suppose we have ${X/K}$ a (smooth, proper) variety over a number field. We want to know whether or not there are any ${K}$-rational points. Of course, classically if you hand me this variety using an equation, then the problem reduces to a Diophantine equation. So this problem is very old (3rd century … and one could argue we still don’t have a great handle on it).

Suppose ${Spec \ K\rightarrow X}$ is a rational point. Then we can use the embedding ${K\hookrightarrow K_v}$ into various completions of ${K}$ to get ${Spec \ K_v\rightarrow X}$ points over all the local fields. This isn’t saying much. It says that if we have a global point, then we have points locally everywhere. The interesting question is whether having points locally everywhere “glue” to give a global point. When a class of varieties always allows you to do this, then we say the varieties satisfy the Hasse principle.

This brings us to the content of today’s post. We will construct the Brauer-Manin obstruction to the Hasse principle. We’ve already considered the following pairing in a previous post. For any prime ${v}$, we get a pairing ${Br(X_{K_v})\times X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$ which is just given by pullback on étale cohomology using the map defining the point ${Spec \ K_v\rightarrow X_{K_v}}$ followed by the canonical identification ${Br(K_v)\stackrel{\sim}{\rightarrow}\mathbb{Q}/\mathbb{Z}}$. We could write ${(\alpha, x_v)=x_v^*(\alpha)}$ or more typically ${\alpha(x_v)}$.

We can package this into something global as follows. By restriction we have ${\displaystyle Br(X)\hookrightarrow \prod_v Br(X_{K_v})}$ which we will write ${\alpha\mapsto (\alpha_v)}$. Now we just sum to get ${\displaystyle Br(X)\times \prod_v X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$, so our global pairing is ${\displaystyle (\alpha, (x_v))\mapsto \sum_v \alpha_v(x_v)}$.

Recall that we have an exact sequence from the post on Brauer groups of fields ${0\rightarrow Br(K)\rightarrow \oplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0}$ by summing the local invariants. If our local points glue to give a global point ${x\mapsto (x_v)}$, then the pairing factors through this sequence and hence ${(\alpha, (x_v))=0}$ for any choice of ${\alpha}$.

This is our obstruction. The above argument says that if ${(\alpha, (x_v))\neq 0}$, then ${(x_v)}$ cannot possibly glue to give a rational point. We will give this set a name:

$\displaystyle X^{Br}=\{(x_v)\in \prod_v X(K_v) : (\alpha, x_v)=0 \ \text{for all} \ \alpha\in Br(K)\}$

We will call this the Brauer set of ${X}$ (earlier we called it being “Brauer equivalent to ${0}$” since we saw this type of condition was an equivalence relation on Chow groups). It is now immediate that ${X(K)\subset X^{Br}}$. We should think of this as the collection of local points that have some chance of coming from a global point. Now we have two obstructions to the existence of rational points. The first is that ${\prod_v X(K_v)\neq \emptyset}$ which is just the trivial condition that there are local points everywhere. The second is the Brauer-Manin obstruction which says that ${X^{Br}\neq \emptyset}$.

If the local points condition is necessary and sufficient for the existence of rational points, then of course that is exactly the Hasse principle, so we say the Hasse principle holds. If the Brauer-Manin condition (plus the local condition) is necessary and sufficient, then we say that the Brauer-Manin obstruction is the only obstruction to the existence of rational points. It would be fantastic if we could somehow figure out which varieties had this property.

Caution: It is not an open problem to determine whether or not the Brauer-Manin obstruction is the only obstruction. There are known examples where there is no B-M obstruction and yet there are still no rational points. As far as I can tell, it is conjectured, but still open that if the Tate-Shafarevich group of the Jacobian of a curve is finite, then the B-M obstruction is the only obstruction to the existence of rational points on curves over number fields. So even in such a special low-dimensional situation this is a very hard question.

## More Complicated Brauer Computations

Let’s wrap up some of our Brauer group loose ends today. We can push through the calculation of the Brauer groups of curves over some other fields using the same methods as the last post, but just a little more effort.

First, note that with absolutely no extra effort we can run the same argument as yesterday in the following situation. Suppose ${X}$ is a regular, integral, quasi-compact scheme of dimension ${1}$ with the property that all closed points ${v\in X}$ have perfect residue fields ${k(v)}$. Let ${g: \text{Spec} K \hookrightarrow X}$ be the inclusion of the generic point.

Running the Leray spectral sequence a little further than last time still gives us an inclusion, but we will usually want more information because ${Br(K)}$ may not be ${0}$. The low degree terms (plus the argument from last time) gives us a sequence:

$\displaystyle 0\rightarrow Br'(X)\rightarrow Br(K)\rightarrow \bigoplus_v Hom_{cont}(G_{k(v)}, \mathbb{Q}/\mathbb{Z})\rightarrow H^3(X, \mathbb{G}_m)\rightarrow \cdots$

This allows us to recover a result we already proved. In the special case that ${X=\text{Spec} A}$ where ${A}$ is a Henselian DVR with perfect residue field ${k}$, then the uniformizing parameter defines a splitting to get a split exact sequence

$\displaystyle 0\rightarrow Br(A)\rightarrow Br(K)\rightarrow Hom_{cont}(G_k, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Thus when ${A}$ is a strict local ring (e.g. ${\mathbb{Z}_p}$) we get an isomorphism ${Br(K)\rightarrow \mathbb{Q}/\mathbb{Z}}$ since ${Br(A)\simeq Br(k)=0}$ (since ${k}$ is ${C_1}$). In fact, going back to Brauer groups of fields, we had a lot of trouble trying to figure anything out about number fields. Now we may have a tool (although without class field theory it isn’t very useful, so we’ll skip this for now).

The last computation we’ll do today is to consider a smooth (projective) curve over a finite field ${C/k}$. Fix a separable closure ${k^s}$ and ${K}$ the function field. First, we could attempt to use Leray on the generic point, since we can use that ${H^3(K, \mathbb{G}_m)=0}$ to get some more information. Unfortunately without something else this isn’t enough to recover ${Br(C)}$ up to isomorphism.

Instead, consider the base change map ${f: C^s=C\otimes_k k^s\rightarrow C}$. We use the Hochschild-Serre spectral sequence ${H^p(G_k, H^q(C^s, \mathbb{G}_m))\Rightarrow H^{p+q}(C, \mathbb{G}_m)}$. The low degree terms give us

$\displaystyle 0\rightarrow Br(k)\rightarrow \ker (Br(C)\rightarrow Br(C^s))\rightarrow H^1(G_k, Pic(C^s))\rightarrow \cdots$

First, ${\ker( Br(C)\rightarrow Br(C^s))=Br(C)}$ by the last post. Next ${H^1(G_k, Pic^0(C^s))=0}$ by Lang’s theorem as stated in Mumford’s Abelian Varieties, so ${H^1(G_k, Pic(C^s))=0}$ as well. That tells us that ${Br(C)\simeq Br(k)=0}$ since ${k}$ is ${C_1}$. So even over finite fields (finite was really used and not just ${C_1}$ for Lang’s theorem) we get that smooth, projective curves have trivial Brauer group.

## Brauer Groups of Curves

Let ${C/k}$ be a smooth projective curve over an algebraically closed field. The main goal of today is to show that ${Br(C)=0}$. Both smooth and being over an algebraically closed field are crucial for this computation. The computation will run very similarly to the last post with basically one extra step.

We haven’t actually talked about the Brauer group for varieties, but there are again two definitions. One has to do with Azumaya algebras over ${\mathcal{O}_C}$ modulo Morita equivalence. The other is the cohomological Brauer group, ${Br'(C):=H^2(C, \mathbb{G}_m)}$. As already stated, it is a big open problem to determine when these are the same. We’ll continue to only consider situations where they are known to be the same and hence won’t cause any problems (or even require us to define rigorously the Azumaya algebra version).

First, note that if we look at the Leray spectral sequence with the inclusion of the generic point ${g:Spec(K)\hookrightarrow C}$ we get that ${R^1g_*\mathbb{G}_m=0}$ by Hilbert 90 again which tells us that ${0\rightarrow H^2(C, g_*\mathbb{G}_m)\hookrightarrow Br(K)}$. Now ${K}$ has transcendence degree ${1}$ over an algebraically closed field, so by Tsen’s theorem this is ${C_1}$. Thus the last post tells us that ${H^2(C, g_*\mathbb{G}_m)=0}$.

The new step is that we need to relate ${H^2(C, g_*\mathbb{G}_m)}$ to ${Br(C)}$. On the étale site of ${C}$ we have an exact sequence of sheaves

$\displaystyle 0\rightarrow \mathbb{G}_m\rightarrow g_*\mathbb{G}_m\rightarrow Div_C\rightarrow 0$

where ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$.
Taking the long exact sequence on cohomology we get

$\displaystyle \cdots \rightarrow H^1(C, Div_C)\rightarrow Br(C)\rightarrow H^2(C, g_*\mathbb{G}_m)\rightarrow \cdots .$

Thus it will complete the proof to show that ${H^1(C, Div_C)=0}$, since then ${Br(C)}$ will inject into ${0}$. Writing ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$ and using that cohomology commutes with direct sums we need only show that for some fixed closed point ${(i_v): Spec(k(v))\hookrightarrow C}$ that ${H^1(C, (i_v)_*\mathbb{Z})=0}$.

We use Leray again, but this time on ${i_v}$. For notational convenience, we’ll abuse notation and call both the map and the point ${v\in C}$. The low degree terms give us ${H^1(C, v_*\mathbb{Z})\hookrightarrow H^1(v, \mathbb{Z})}$. Using the Galois cohomology interpretation of étale cohomology of a point ${H^1(v,\mathbb{Z})\simeq Hom_{cont}(G_{k(v)}, \mathbb{Z})}$ (the homomorphisms are not twisted since the Galois action is trivial). Since ${G_{k(v)}}$ is profinite, the continuous image is compact and hence a finite subgroup of ${\mathbb{Z}}$. Thus ${H^1(C, v_*\mathbb{Z})=0}$ which implies ${H^1(C, Div_C)=0}$ which gives the result that ${Br(C)=0}$.

So again we see that even for a full curve being over an algebraically closed field is just too strong a condition to give anything interesting. This suggests that the Brauer group really is measuring some arithmetic properties of the curve. For example, we could ask whether or not good/bad reduction of the curve is related to the Brauer group, but this would require us to move into Brauer groups of surfaces (since the model will be a relative curve over a one-dimensional base).

Already for local fields or ${C_1}$ fields the question of determining ${Br(C)}$ is really interesting. The above argument merely tells us that ${Br(C)\hookrightarrow Br(K)}$ where ${K}$ is the function field, but this is true of all smooth, proper varieties and often doesn’t help much if the group is non-zero.

## Brauer Groups of Fields

Today we’ll talk about the basic theory of Brauer groups for certain types of fields. If the last post was too poorly written to comprehend, the only thing that will be used from it is that for fields we can refer to “the” Brauer group without any ambiguity because the cohomological definition and the Azumaya (central, simple) algebra definition are canonically isomorphic in this case.

Let’s just work our way from algebraically closed to furthest away from being algebraically closed. Thus, suppose ${K}$ is an algebraically closed field. The two ways to think about ${Br(K)}$ both tell us quickly that this is ${0}$. Cohomologically this is because ${G_K=1}$, so there are no non-trivial Galois cohomology classes. The slightly more interesting approach is that any central, simple algebra over ${K}$ is already split, i.e. a matrix algebra, so it is the zero class modulo the relation we defined last time.

I’m pretty sure I’ve blogged about this before, but there is a nice set of definitions that measures how “far away” from being algebraically closed you are. A field is called ${C_r}$ if for any ${d,n}$ such that ${n>d^r}$ any homogeneous polynomial (with ${K}$ coefficients) of degree ${d}$ in ${n}$ variables has a non-trivial solution.

Thus the condition ${C_0}$ just says that all polynomials have non-trivial solutions, i.e. ${K}$ is algebraically closed. The condition ${C_1}$ is usually called being quasi-algebraically closed. Examples include, but are not limited to finite fields and function fields of curves over algebraically closed fields. A more complicated example that may come up later is that the maximal, unramified extension of a complete, discretely valued field with perfect residue field is ${C_1}$.

A beautiful result is that if ${K}$ is ${C_1}$, then we still get that ${Br(K)=0}$. One could consider this result “classical” if done properly. First, by Artin-Wedderburn any finite dimensional, central, simple algebra has the form ${M_n(D)}$ where ${D}$ is a finite dimensional division algebra with center ${K}$. If you play around with norms (I swear I did this in a previous post somewhere that I can’t find!) you produce the right degree homogeneous polynomial and use the ${C_1}$ condition to conclude that ${D=K}$. Thus any central, simple algebra is already split giving ${Br(K)=0}$.

We might give up and think the Brauer group of any field is ${0}$, but this is not the case (exercise to test understanding: think of ${\mathbb{R}}$). Let’s move on to the easiest example we can think of for a non-${C_1}$ field: ${\mathbb{Q}_p}$ for some prime ${p}$. The computation we do will be totally general and will actually work to show what ${Br(K)}$ is for any ${K}$ that is complete with respect to some non-archimedean discrete valuation, and hence for ${K}$ a local field.

The trick is to use the valuation ring, ${R=\mathbb{Z}_p}$ to interpolate between the Brauer group of ${K}$ and the Brauer group of ${R/m=\mathbb{F}_p}$, a ${C_1}$ field! Since ${K}$ is the fraction field of ${R}$, the first thing we should check is the Leray spectral sequence at the generic point ${i:Spec(K)\hookrightarrow Spec(R)}$. This is given by ${E_2^{p,q}=H^p(Spec(R), R^qi_*\mathbb{G}_m)\Rightarrow H^{p+q}(G_K, (K^s)^\times)}$.

By Hilbert’s Theorem 90, we have ${R^1i_*\mathbb{G}_m=0}$. Recall that last time we said there is a canonical isomorphism ${Br(R)\rightarrow Br(\mathbb{F}_p)}$ given by specialization. This gives us a short exact sequence from the long exact sequence of low degree terms:

$\displaystyle 0\rightarrow Br(\mathbb{F}_p)\rightarrow Br(\mathbb{Q}_p)\rightarrow Hom(G_{\mathbb{F}_p}, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Now we use that ${Br(\mathbb{F}_p)=0}$ and ${G_{\mathbb{F}_p}\simeq \widehat{\mathbb{Z}}}$ to get that ${Br(\mathbb{Q}_p)\simeq \mathbb{Q}/\mathbb{Z}}$. As already mentioned, nothing in the above argument was specific to ${\mathbb{Q}_p}$. The same argument shows that any (strict) non-archimedean local field also has Brauer group ${\mathbb{Q}/\mathbb{Z}}$.

To get away from local fields, I’ll just end by pointing out that if you start with some global field ${K}$ you can try to use a local-to-global idea to get information about the global field. From class field theory we get an exact sequence

$\displaystyle 0\rightarrow Br(K)\rightarrow \bigoplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0,$

which eventually we may talk about. We know what all the maps are already from this and the previous post. The first is specialization (or corestriction from a few posts ago, or most usually this is called taking invariants). Then the second map is just summing since each term of the direct sum is a ${\mathbb{Q}/\mathbb{Z}}$.

Next time we’ll move on to Brauer groups of curves even though so much more can still be said about fields.

## Intro to Brauer Groups

I want to do a series on the basics of Brauer groups since they came up in the past few posts. Since I haven’t really talked about Galois cohomology anywhere, we’ll take a slightly nonstandard approach and view everything “geometrically” in terms of étale cohomology. Everything should be equivalent to the Galois cohomology approach, but this way will allow us to use the theory that is already developed elsewhere on the blog.

I apologize in advance for the sporadic nature of this post. I just need to get a few random things out there before really starting the series. There will be one or two posts on the Brauer group of a “point” which will just mean the usual Brauer group of a field (to be defined shortly). Then we’ll move on to the Brauer group of a curve, and maybe if I still feel like continuing the series of a surface.

Let ${K}$ be a field and ${K^s}$ a fixed separable closure. We will define ${Br(K)=H^2_{et}(Spec(K), \mathbb{G}_m)=H^2(Gal(K^s/K), (K^s)^\times)}$. This isn’t the usual definition and is often called the cohomological Brauer group. The usual definition is as follows. Let ${R}$ be a commutative, local, (unital) ring. An algebra ${A}$ over ${R}$ is called an Azumaya algebra if it is a free of finite rank ${R}$-module and ${A\otimes_R A^{op}\rightarrow End_{R-mod}(A)}$ sending ${a\otimes a'}$ to ${(x\mapsto axa')}$ is an isomorphism.

Define an equivalence relation on the collection of Azumaya algebras over ${R}$ by saying ${A}$ and ${A'}$ are similar if ${A\otimes_R M_n(R)\simeq A'\otimes_R M_{n'}(R)}$ for some ${n}$ and ${n'}$. The set of Azumaya algebras over ${R}$ modulo similarity form a group with multiplication given by tensor product. This is called the Brauer group of ${R}$ denoted ${Br(R)}$. Often times, when an author is being careful to distinguish, the cohomological Brauer group will be denoted with a prime: ${Br'(R)}$. It turns out that there is always an injection ${Br(R)\hookrightarrow Br'(R)}$.

One way to see this is that on the étale site of ${Spec(R)}$, the sequence of sheaves ${1\rightarrow \mathbb{G}_m\rightarrow GL_n\rightarrow PGL_n\rightarrow 1}$ is exact. It is a little tedious to check, but using a Čech cocycle argument (caution: a priori the cohomology “groups” are merely pointed sets) one can check that the injection from the associated long exact sequence ${H^1(Spec(R), PGL_n)/H^1(Spec(R), GL_n)\hookrightarrow Br'(R)}$ is the desired injection.

If we make the extra assumption that ${R}$ has dimension ${0}$ or ${1}$, then the natural map ${Br(R)\rightarrow Br'(R)}$ is an isomorphism. I’ll probably regret this later, but I’ll only prove the case of dimension ${0}$, since the point is to get to facts about Brauer groups of fields. If ${R}$ has dimension ${0}$, then it is a local Artin ring and hence Henselian.

One standard lemma to prove is that for local rings a cohomological Brauer class ${\gamma\in Br'(R)}$ comes from an Azumaya algebra if and only if there is a finite étale surjective map ${Y\rightarrow Spec(R)}$ such that ${\gamma}$ pulls back to ${0}$ in ${Br'(Y)}$. The easy direction is that if it comes from an Azumaya algebra, then any maximal étale subalgebra splits it (becomes the zero class after tensoring), so that is our finite étale surjective map. The other direction is harder.

Going back to the proof, since ${R}$ is Henselian, given any class ${\gamma\in H^2(Spec(R), \mathbb{G}_m)}$ a standard Čech cocycle argument shows that there is an étale covering ${(U_i\rightarrow Spec(R))}$ such that ${\gamma|_{U_i}=0}$. Choosing any ${U_i\rightarrow Spec(R)}$ we have a finite étale surjection that kills the class and hence it lifts by the previous lemma.

It is a major open question to find conditions to make ${Br(X)\rightarrow Br'(X)}$ surjective, so don’t jump to the conclusion that we only did the easy case, but it is always true. Now that we have that the Brauer group is the cohomological Brauer group we can convert the computation of ${Br(R)}$ for a Henselian local ring to a cohomological computation using the specialization map (pulling back to the closed point) ${Br(R)\rightarrow Br(k)}$ where ${k=R/m}$.

## L-series of CM Elliptic Curves

This will be the last post in the CM elliptic curve series. Last time we covered the main theorem of complex multiplication. Today we’ll very, very briefly sketch one amazing use of the main theorem. We’ll first talk about how to associate a Grössencharacter to a CM elliptic curve and then use this to better describe the ${L}$-series of an elliptic curve.

Here’s why this will be amazing. Awhile ago we talked about ${L}$-series of varieties and various modularity conjectures. One of the huge, major theorems of modern number theory (which wasn’t proved in full until 2003, and built on all of Wiles and Taylor’s results) is the so-called Modularity Theorem. It says that an elliptic curve ${E/\mathbb{Q}}$ is modular and hence its ${L}$-series has an analytic continuation to all of ${\mathbb{C}}$.

This is still open (as far as I know) for elliptic curves over an arbitrary number field, but today we’ll see that we can use the theory we’ve built to show that any CM elliptic curve over any number field has an ${L}$-series that analytically continues to the whole plane.

Fix ${E/L}$ an elliptic curve over a number field with CM by ${R_K}$, the ring of integers in a quadratic imaginary field ${K}$. We use the main theorem of CM to do the following. Fix an idele ${x\in \mathcal{J}_L}$ and let ${s=N_{L/K}(x)\in\mathcal{J}_L}$. There is a unque ${\alpha\in K^*}$ such that ${\alpha R_K=(s)}$ and for any fractional ${\frak{a}}$ in ${K}$ and any analytic iso ${f: \mathbb{C}/\frak{a}\stackrel{\sim}{\rightarrow} E(\mathbb{C})}$ we get a commutative diagram:

$\displaystyle \begin{matrix} K/\frak{a} & \rightarrow & K/\frak{a} \\ \downarrow & & \downarrow \\ E(L^{ab}) & \rightarrow & E(L^{ab}) \end{matrix}$

What this gives us is a map ${\alpha_{E/L}: \mathcal{J}_L\rightarrow K^*\subset\mathbb{C}^*}$. Recall that a Grössencharacter of a number field ${L}$ is such a map that is trivial on ${L^*}$. We can alter this to the map ${\Psi_{E/L}:\mathcal{J}_L\rightarrow \mathbb{C}^*}$ by ${\Psi_{E/L}(x)=\alpha_{E/L}(x)(N_{L/K}(x^{-1}))_\infty}$. It turns out this is our desired Grössencharacter.

Recall how we formed the L-series of a variety over ${\mathbb{Q}}$. There is nothing special about ${\mathbb{Q}}$ going on in that constuction and so the same thing can be done for any elliptic curve ${E/L}$ where ${L}$ is a number field. Basically you piece it together as a product over primes of some expression involving the characteristic polynomial of the Frobenius elements acting on the cohomology of the reductions of ${E}$ mod these primes.

Given a Grössencharacter ${\Psi: \mathcal{J}_L\rightarrow \mathbb{C}^*}$ we can define the Hecke ${L}$-series to be ${\displaystyle L(s,\Psi)=\prod_{\frak{p}}(1-\Psi(\frak{p})q_\frak{p}^{-s})^{-1}}$, where ${q_\frak{p}}$ is the size of the residue field at ${\frak{p}}$. Hecke proved that this ${L}$-series has an analytic continuation to the complex plane.

Duering proved that if ${E/L}$ is an elliptic curve with CM by ${R_K}$, then ${L(E/L, s)=L(s, \Psi_{E/L})L(s, \overline{\Psi_{E/L}})}$. In fact, even better is that if ${K}$ is not contained in ${L}$, then the ${L}$-series of the elliptic curve over ${L}$ is precisely the Hecke ${L}$-series of the Grössencharacter attached to the base-changed elliptic curve ${E/KL}$. In either case, we see that it is much easier to prove that the ${L}$-series of an elliptic curve with CM by the ring of integers in a quadratic imaginary field has an analytic continuation.