# 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 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.

# What is a QFT?

I’m going to do a series on trying to get at what exactly a mathematician means when referring to a “quantum field theory” or QFT. This is pretty tough to get at since if you ask a physicist the answer is likely to be completely different than a mathematician. Also, people within disciplines often refer to different things when they mention it.

Problem: A mathematician needs to frame problems about QFT’s in terms that are familiar. A physicist must do the same likewise. When translating, these two frames are precisely the same (at least in theory, physicist have been able to do things that work in which the exact mathematical reason that they work is still a mystery and such…). This post is to try to bring some of these concepts to mathematical language, since even books with titles such as QFT for Mathematicians assume way too much knowledge of physics on the first page and don’t truly get the translation across.

First, let’s start a little more basic. What is a classical field theory? There are two main parts to this. There is the fiber bundle and the Lagrangian. I think we see why this is interesting to me now. I’m quite interested in algebraic topology, so the term fiber bundle gives good indication that we are headed that way.

For those not familiar with the fiber bundle, it really is quite a brilliant construction. It is so natural (although it might take a bit of time of staring at definitions to think so), yet powerful. A fiber bundle consists of a base space B and a total space E (topological of course). We have a “projection” type map $\pi: E\to B$ that is locally trivial. Now we also have the fiber space F. Here is where that locally trivial condition applies. What we mean is that for any $x\in B$, we can find a neighborhood of B, say U, such that $\pi^{-1}(U)$ is homeomorphic to $U\times F$ and so if we factor through that homeomorphism and then project, it is the same as just the $\pi$ map.

This is probably overwhelming at this point, but let’s look at how simple it really is through some examples. You should easily get the feel for it that way. The trivial bundle is to just taking $E=B\times F$ and the projection really is projecting $\pi: E\to B$ by $(b,f)\mapsto b$. This trivially satisfies the definitions, and the “fibers” just look like “slices” of E in that you get the whole of F but just a point of B ($\pi^{-1}(b)=\{b\}\times F$).

Another common example is to look at the tangent bundle (hmm “bundle” is in the name…) of a manifold. This has some more structure to it since it is a vector space as the fiber space and is hence known as a “vector bundle.” So in particular we could let our manifold be none other than $S^n$. Then our vectors we have in $\mathbb{R}^{n+1}$, so $E=\{(x,v)\in S^n\times\mathbb{R}^{n+1} : x\perp v\}$. Another vector bundle along the same lines is the normal bundle to a manifold. I don’t want to go too far off topic, but essentially fiber bundles tell us how much a space is “twisted.”

I know that is rather inadequate, but I do want to get to QFT’s, and I’m only on the first part of the definition of a CFT. One more quick thing that will be needed at some point in the future is called the section of a fiber bundle. Sadly global sections are not in general possible, but we can do it locally. Take an open set U in the base space B, then the section is a continuous map $f:U\to E$ such that $\pi(f(x))=x$ for every x in U. At this point we can start throwing around terms like sheaf, but we’ll move on.

So let’s sum up and reword with a view towards CFT. We mostly want things to be manifolds, since this is classical physics. So basically our fiber bundle will look locally like $F\times M\to M$. Now we say that a classical field is a section of that fiber bundle. Does this fit with our traditional view?

Classical mechanics is just taking our manifold (base space) to be time, i.e. $\mathbb{R}$ and the fiber space to be the space of all possible configurations of the mechanical system. So this works. Let’s get the Lagrangian into place now. Another classical field theory is statistical mechanics. We could just let our space be traditional 3-dimensional space $\mathbb{R}^3$ and the fiber space is $\mathbb{R}$. The field is just a real function on $\mathbb{R}^3$. Here we will have the Lagrangian as a function $L: Fields \times B\to \mathbb{R}$. Remember that fields in this case are functions on $\mathbb{R}^3$. The standard is $L(\phi)=\sum\left(\frac{\partial \phi}{\partial x_i}\right)^2+m^2\phi^2+\lambda\phi^4$.

I just realized that Lagrangians could probably use an entire post on their own. So don’t worry about what is written there. It is just a flavor of next post. I’ll finish up CFT’s next time with basically a post on Lagrangians and some more examples that should give a better feel.