Finiteness of X(k)/B for Rational Surfaces

Recall our setup. We start with a projective surface ${X/k}$ that becomes rational after some finite extension of scalars ${k'/k}$. Let ${C_0(X)}$ be the group of ${0}$-cycles of degree ${0}$. Last time we defined the Manin pairing ${(-,-): C_0(X)\times (Br(X)/Br(k))\rightarrow Br(k)}$ using the corestriction map ${(\sum n_ix_i, a)=\prod_i cor_{k(x_i)/k}(a(x_i))^{n_i}}$. Two rational points are called Brauer equivalent if ${(x-y, a)=1}$ for all ${a\in Br(X)}$, and denote the set of rational points up to Brauer equivalence by ${X(k)/B}$.

Now let ${N=NS(X_{k'})}$ be the Néron-Severi group of ${X_{k'}}$. It turns out we can factor the Manin pairing as follows:

$\displaystyle \begin{matrix} C_0(X)\times (Br(X)/Br(k)) & \longrightarrow & Br(k) \\ \downarrow & & \uparrow \\ A_0(X)\times H^1(G, N) & \longrightarrow & H^1(G, N\otimes \overline{k}^\times)\times H^1(G, N)\end{matrix}$

The goal of today is to say something about this factoring. Last time we wrote down the Hochschild-Serre spectral sequence and said the map ${Br(k)\rightarrow Br(X)}$ was just the quotient map ${E_2^{2,0}\rightarrow E_\infty^{2,0}}$ followed by the inclusion. Note that since all differentials are ${0}$ for all ${E_n^{1,1}}$ we get that it equals ${H^1(G, N)}$ and sits inside ${Br(X)}$. Thus we have a sequence ${Br(k)\rightarrow Br(X)\rightarrow H^1(G,N)}$ whose composition is ${0}$ and hence gives a map

$\displaystyle Br(X)/Br(k)\rightarrow H^1(G, N).$

This defines for us the left vertical map, since the left factor is just projection from all ${0}$-cycles of degree ${0}$ to ${0}$-cycles modulo rational equivalence of degree ${0}$. The right vertical map is just the one induced on group cohomology via the standard intersection pairing on the surface ${N\otimes \overline{k}^\times \times N\rightarrow \overline{k}^\times}$.

This leaves us with the bottom map. Call it ${\Phi \times id}$ where ${\Phi:A_0(X)\rightarrow H^1(G, N\otimes \overline{k}^\times )}$. It turns out that the majority of Bloch’s paper is merely defining this map and checking that the above diagram commutes, so we won’t get into that. It involves lots of K-theory which I’m not going to get into.

Supposing the above, the main theorem of the paper is that ${Im \Phi}$ is finite in the case of our hypotheses. We can check the nice corollary that ${X(k)/B}$ is finite. If ${X(k)}$ is empty we’re done, so fix some ${x_0\in X(k)}$. The proof is that we can make ${\Psi: X(k)\rightarrow H^1(G, N\otimes \overline{k}^\times)}$ by ${\Psi(x)=\Phi([x]-[x_0])}$. Since ${Im \Psi\subset Im \Phi}$, it must have finite cardinality. We need only check that distinct Brauer classes stay distinct to get the result, but this follows from commutativity of the diagram and the fact that Brauer classes are by definition distinguished under the Manin pairing.

It turns out that Manin had already proved that ${X(k)/B}$ is finite for cubic surfaces, so Bloch’s result extends this to any rational surface. As a consequence of the construction of ${\Phi}$, Bloch also gets the strange result that if ${X}$ is a conic bundle, i.e. ${X\rightarrow \mathbf{P}^1}$ has generic fiber a conic, and ${k}$ is local, then if ${X}$ has good reduction then ${|Im\Phi |=1}$. Thus at places of good reduction ${A_0(X)}$ is trivial.

Note how useful this is. For example, take some conic bundle over ${\mathbf{Q}_p}$. Good reduction means that there exists some proper, regular model ${\frak{X}/\mathbf{Z}_p}$ whose generic fiber is ${X}$ and whose special fiber is non-singular. It is hard to tell whether or not ${X}$ has good reduction, because you might accidentally be picking the wrong model. With this condition of Bloch, one can sometimes explicitly calculate some non-trivial element of ${A_0(X)}$ (Manin actually does this using the defining equation of a class of Châtalet surfaces!) which tells you ${X}$ has bad reduction.

To phrase this a different way, to test for honest bad reduction without some criterion requires a choice of model over ${\mathbf{Z}_p}$. There could be infinitely many distinct choices here, so it could be hard to tell if you’ve exhausted all possibilities. This criterion of Bloch says that no choice needs to be made. Bad reduction can be tested inherently from the variety over ${\mathbf{Q}_p}$.

Brauer Equivalence of Rational Points

Today I’m going to start a series on the arithmetic of rational surfaces. I feel like this theory is fairly unknown, but it provides such a wonderful source of examples that don’t exist in the curve case. Everything is so explicit with equations and concrete calculations.

The theorem I want to get to is in Bloch’s paper “On the Chow Groups of Certain Rational Surfaces.” It says that for a certain class of rational surfaces, good/bad reduction can be detected on the Chow group of ${0}$-cycles of degree ${0}$. We will also pull a lot from Manin’s book Cubic Surfaces and possibly from some papers of Colliet-Thélène.

Rather than prove this amazing result, I want to describe some of the constructions and ideas that go into it. Let’s get some terminology out of the way. For our purposes, a projective surface ${X/k}$ will be called rational if there exists some ${k'/k}$ for which the extension of scalars ${X_{k'}}$ is birational with ${\mathbf{P}^2_{k'}}$.

Fix an algebraic closure ${\overline{k}}$ and let ${G=Gal(\overline{k}/k)}$. Nothing in the immediate theory should require this, but since the goal is a theorem about reduction type, without loss of generality ${k}$ is either assumed global or complete, local, arising as the fraction field of some DVR of mixed characteristic. If you want to push the theory through for positive characteristic, you will need to at least alter all the ${\overline{k}}$ to ${k^{sep}}$.

Bloch’s proof involves playing around with Brauer groups and in particular realizing a certain pairing of Manin in a new way. Our goal for today will just be to work out some of the basic Brauer group theory. Recall that ${Br(X)=H^2(X, \mathbf{G}_m)}$. Unless otherwise stated, all cohomology will be étale or Galois (the distinction should be obvious from context and so subscripts will be omitted). This means that ${Br(k)=H^2(Spec(k), \mathbb{G}_m)=H^2(G, \overline{k}^\times)}$.

First, we’ll check that there is a map ${Br(k)\rightarrow Br(X)}$ in a sort of ridiculous way, because we’ll need to use the map involved at a later point. For étale cohomology, we have the convergent first-quadrant Hochschild-Serre spectral sequence ${E_2^{p,q}=H^p(G, H^q(X_{\overline{k}}, \mathbf{G}_m))\Rightarrow H^{p+q}(X, \mathbf{G}_m)}$. Thus writing ${H^2(X, \mathbf{G}_m)=Br(X)\simeq E_{\infty}^{0,2}\oplus E_{\infty}^{1,1}\oplus E_{\infty}^{2,0}}$ and using that ${E_{\infty}^{2,0}}$ is a quotient of ${E^{2,0}_2=Br(k)}$ we get a map via inclusion then projection.

Given any point ${x_i: Spec(k(x_i))\rightarrow X}$, we have a finite index subgroup ${G'=Gal(\overline{k}/k(x_i))}$ of ${G}$. By functoriality of Galois cohomology, this gives us two group homomorphisms. The restriction, ${res: Br(k)\rightarrow Br(k(x_i))}$ and corestriction (aka the transfer), ${cor_{k(x_i)/k}: Br(k(x_i))\rightarrow Br(k)}$. A well-known property of these two maps is that ${(cor)\circ (res)=[G: G']}$ is multiplication by the index (or by Galois theory in this case, ${[k(x_i):k]}$).

The last map we need before we can piece together Manin’s pairing is that we can pullback on cohomology given a point ${x_i^*: Br(X)\rightarrow Br(k(x_i))}$. We denote this by ${a(x_i)}$ (thought of as the Brauer class restricted to ${x_i}$). Define ${C_0(X)}$ to be the free abelian group generated by the points of ${X}$ (i.e. ${0}$-cycles) of degree ${0}$. Caution: This is not the Chow group or anything because we aren’t moding out by any sort of rational or algebraic equivalence yet. We only require degree ${0}$.

For those not used to the degree map when over non-algebraically closed fields, recall that ${deg(\sum n_ix_i)=\sum [k(x_i): k]n_i}$, so it is possible that ${2x_1-3x_2}$ is a degree ${0}$ cycle if the residue field extensions are ${3}$ and ${2}$ respectively.

The Manin pairing is ${(-,-): C_0(X)\times (Br(X)/Br(k))\rightarrow Br(k)}$ given by ${\displaystyle \left(\sum n_ix_i, a\right)=\prod_i cor_{k(x_i)/k}(a(x_i))^{n_i}}$. Certainly at the level of ${C_0(X)\times Br(X)}$ everything is well-defined by the above maps. There is some work in checking that we can pass to the quotient.

Here is why it works. If we take a class ${a\in Br(X)}$ that is in already in ${Br(k)}$, then ${\displaystyle Br(k)\rightarrow Br(X)\stackrel{x_i^*}{\rightarrow} Br(k(x_i))}$ is just ${res(a)}$ (this is somewhat non-obvious from our definition). Now given any element ${\sum n_ix_i\in C_0(X)}$ we need to check that it pairs with ${a}$ to ${1}$. We crucially use our arithmetic definition of degree ${0}$ here.

$\displaystyle \begin{matrix} (\sum n_ix_i, a) & = & \displaystyle \prod_i cor_{k(x_i)/k}(a(x_i))^{n_i} \\ & = & \displaystyle\prod_i (cor_{k(x_i)/k}\circ res)(a^{n_i}) \\ & = & \displaystyle\prod_i a^{[k(x_i):k]n_i} \\ & = & \displaystyle a^{\sum [k(x_i):k]n_i} \\ & = & a^0 =1 \end{matrix}$

This gives us a well-defined pairing which I’ll refer to as the Manin pairing (non-standard terminology to my knowledge). We say that two rational points ${x,y\in X(k)}$ are Brauer equivalent if ${(x-y, a)=1}$ for all ${a\in Br(X)}$. The key idea of the next post will be to rewrite this pairing in a way that allows us to check that for a smooth rational surface over a global field the set of rational points up to Brauer equivalence: ${X(k)/B}$ is finite. But more importantly, the set has only one element (i.e. all points are Brauer equivalent) at places of good reduction.

The Myth of a Close Election

Before presenting the argument for why the current US presidential race is not as close as it seems, let’s first get out of the way some of the major reasons why it seems close. First, there is the bias of the media. What is more exciting: watching a photo finish win of .01 seconds or watching someone take the lead early and never give up ground?

All of our news sources have a vested financial interest in portraying the presidential race as close. It keeps people coming to their webpages or turning on the news to see how things have changed. The myth was created because it sells. But this is far, far from the only reason someone would want to keep this myth alive.

If your candidate is the one that has the bad odds, then you will want to portray the race as close to keep up hope. No one wants to admit (or even believe) that they are going to lose. This is about as old and established a cognitive bias as you can find. What about if your candidate is the one that is ahead? Well, there is very good reason to portray the race as close in this case as well. If all the people that are going to vote think it is an easy win, the turnout may go down causing a sudden upset that shouldn’t have happened.

This gives good reason for even non-news sources to keep up the impression that it is a close race. In fact, I can’t think of any reason someone would not have some interest in skewing the numbers to make the race seem closer than it is. In our weird system of the electoral college, it is actually quite easy to keep this myth up. You just give standard national polling data. All of a sudden it looks like a dead even match. One day one candidate is up, the next day the other. Back and forth it goes. How exciting!

It turns out that the most careful analysis out there, Nate Silver at Fivethirtyeight, has as of yesterday a 79% chance of Obama winning and a 21% chance of Romney winning. Before discussing what this means, I’ll first point out that this is a true professional statistical analysis. He uses tons of polls in all of the states (sometimes 10 for a single state!) and not just one that suits his purpose. He takes into account noise and how historically accurate the polls have been at different times leading up to the election. It is a fully developed statistically model (as opposed to places like Real Clear Politics which takes straight from polls without filtering through a model).

He has used his methods in predicting sports and elections in the past and has an impeccable track record for accuracy. Now that that is out of the way, what do the numbers mean? Well, they mean what they say. A better tactic is to point out what they don’t mean. A 79% chance of a win is not a sure thing. In fact, people go to Vegas and play odds much, much (much, much) worse than 21% and win! Does this mean they had better odds than predicted? No. It means sometimes you win when you have .0001% chance at winning.

This is statistics we’re talking about, so there is never any sort of guarantee. If Romney wins, will Nate Silver be wrong? No! And that is the crucial point. If every time Silver gave 79/21 odds, the person with the 79% chance of winning won, then that would definitively prove Nate’s model incorrect. In order for the model to make accurate predictions it turns out that 21% of the time that he makes this 79/21 prediction, the person with the 21% chance of winning has to win.

You can go to his blog and check out his methods for yourself, but his track record should give us a bit of confidence that he knows what he’s doing. Now back to the original question: Is the US presidential race close? Armed with these stats, the answer is subjective and you can decide for yourself. To me it would be a lie to say that it is some sort of blowout, but under no stretch of the imagination is 4-1 odds close. I’m not a betting man, but I’d easily take the occasional 4-1 odds, and that says to me that it isn’t a very close race.