# I just can’t resist: there are infinitely many pairs of primes at most 59470640 apart

Everyone by now has heard about Zhang’s landmark result showing that there are infinitely many pairs of primes at most 70000000 apart.

His core result is that if a set of 3.5 * 10^6 (corrected, thanks to comment #2) numbers $H$ is admissible (see below), then there are infinitely many $n$ so that $n+H$ contains at least two primes. He then easily constructs an admissible set wherein the largest difference is 7 * 10^7, obtaining the stated result.

A set $H$ is admissible if there is no prime $p$ so $H \pmod p$ occupies every residue class. For a given $H$ this is clearly a checkable condition; there’s no need to look at primes larger than $|H|$.

(While Zhang went for a nice round number, Mark Lewko found his argument in fact gives 63374611, if you’re being overly specific about these things, which we are right now. :-)

In a short note on the arXiv yesterday, Tim Trudgian (whose office is not far from mine) pointed out another way to build an admissible set, giving a smaller largest difference, obtaining the result that there are infinitely many pairs of primes at most 59874594 apart. He considers sets of the form $H_m = {p_{m+1}, \ldots, p_{m+k_0}}$ (where $k_0$ is Zhang’s constant 3.5 * 10^7). These aren’t necessarily admissible, but they are for some values of $m$, and both Zhang and Tim noticed certain values for which this is easy to prove. Zhang used $H_m$ with $m=k_0$, while Tim’s observation is that $m_0=250150=\pi(k_0)$ also works. (Comment #2 below points out this isn’t right, and Zhang also intended $m=\pi(k_0)$, and the slack in his estimate is coming from just looking at the largest element of $H_m$, rather than the largest difference.) Thus the bound in his result is $p_{m_0+k_0}-p_{m_0+1} = 59874594$.

It turns out that checking admissibility for a given $H_m$ isn’t that hard; it takes about an hour to check a single value for $m ~ m_0$ (but if you find a prime witnessing $H_m$ not being admissible, it very often gives you a fast proof that $H_{m+1}$ is not admissible either, so searching is much faster).

I haven’t looked exhaustively, but one can check that $m_1 = m_0 / 2 = 125075$ gives an admissible $H_m$, and hence there are infinitely many pairs of primes at most $p_{m_1 + k_0} - p_{m_1 + 1} = 59470640$. (Sadly, it’s impossible to get below 59 million with this trick; no $m$ below 27000 works; all witnessed by $p=182887$ or $378071$.)

I just couldn’t resist momentarily “claiming the crown” for the smallest upper bound on gap size. :-) Of course the actual progress, that’s surely coming soon from people who actually understand Zhang’s work, is going to be in reducing his 3.5 * 10^6. You can read more about prospects for that in the answers to this MathOverflow question.

# ABC conjecture rumor

There is a rumor circulating here in Japan, to the effect that S. Mochizuki has proved the ABC conjecture. My understanding is that blogs are for spreading such things.

Apparently, he had predicted some years ago that he would finish a proof in 2012, so I suppose this is an “on-time delivery”. It is certainly no secret that his research program has been aiming at the conjecture for several years.

Our very own Noah Snyder did some original work on the function field version of this conjecture, when he was a high-school student.

Update (Sept 4, 2012): This rumor seems to be true! You can find the four “Inter-universal Teichmuller Theory” papers on the very bottom of his papers page.

# Fun with y^2=x^p-x

Here’s a basic example that comes up if you work with elliptic curves: Let $p$ be a prime which is $3 \mod 4$. Let $E$ be the elliptic curve $v^2=u^3-u$ over a field of characteristic $p$. Then $E$ has an endomorphism $F(u,v) = (u^p, v^p)$. It turns out that, in the group law on $E$, we have $F^2 = [-p]$. That is to say, $F(F(u,v))$ plus $p$ copies of $(u,v)$ is trivial.

I remember when I learned this trying to check it by hand, and being astonished at how out of reach the computation was. There are nice proofs using higher theory, but shouldn’t you just be able to write down an equation which had a pole at $F(F(u,v))$ and vanished to order $p$ at $(u,v)$?

There is a nice way to check the prime $3$ by hand. I’ll use $\equiv$ for equivalence in the group law of $E$. Remember that the group law on $E$ has $-(u,v) \equiv (u,-v)$ and has $(u_1,v_1)+(u_2,v_2)+(u_3,v_3) \equiv 0$ whenever $(u_1, v_1)$, $(u_2, v_2)$ and $(u_3, v_3)$ are collinear.

We first show that

$\displaystyle{ F(u,v) \equiv (u-1, v) - (u+1, v) \quad (\dagger)}$

Proof of $(\dagger)$: We want to show that $F(u,v)$, $(u+1,-v)$ and $(u-1,v)$ add up to zero in the group law of $E$. In other words, we want to show that these points are collinear. We just check:

$\displaystyle{ \det \begin{pmatrix} 1 & u^3 & v^3 \\ 1 & u-1 & -v \\ 1 & u+1 & v \end{pmatrix} = 2 v (v^2-u^3+u) = 0}$

as desired. $\square$.

Use of $(\dagger)$: Let $(u_0, v_0)$ be a point on $E$. Applying $F$ twice, we get

$\displaystyle{ F^2(u_0,v_0) \equiv F \left( (u_0-1,v_0) - (u_0+1,v_0) \right)}$

$\displaystyle{ \equiv (u_0-2,v_0) - 2 (u_0,v_0) + (u_0+2,v_0)}$.

Now, the horizontal line $v=v_0$ crosses $E$ at three points: $(u_0, v_0)$, $(u_0-2, v_0)$ and $(u_0+2, v_0)$. (Of course, $u_0 -2 =u_0+1$, since we are in characteristic three.) So $(u_0-2, v_0) + (u_0, v_0) + (u_0+1, v_0) \equiv 0$ and we have

$\displaystyle{F^2(u_0, v_0) \equiv -3 (u_0, v_0)}$

as desired. $\square$.

I was reminded of this last year when Jared Weinstein visited Michigan and told me a stronger statement: In the Jacobian of $y^2 = x^p-x$, we have $F^2 = [(-1)^{(p-1)/2} p]$, where $F$ is once again the automorphism $F(x,y) = (x^p, y^p)$.

Let me first note why this is related to the discussion of the elliptic curve above. (Please don’t run away just because that sentence contained the word Jacobian! It’s really a very concrete thing. I’ll explain more below.) Letting $C$ be the curve $y^2 = x^p-x$, and letting $p$ be $3 \mod 4$, we have a map $C \to E$ sending $(x,y) \mapsto (y x^{(p-3)/4}, x^{(p-1)/2})$, and this map commutes with $F$. I’m going to gloss over why checking $F^2 = [(-1)^{(p-1)/2} p]$ on $C$ will also check it on $E$, because I want to get on to playing with the curve $C$, but it does.

So, after talking to Jared, I was really curious why $F^2$ acted so nicely on the Jacobian of $C$. There are some nice conceptual proofs but, again, I wanted to actually see it. Now I do.

# Rationality of the zeta function mod p

Here’s a neat argument about counting points that you could present at the end of a second course in number theory. I’m sure it’s not original, but, hey, that’s what blogs are for!

Let $X$ be a smooth hypersurface in $\mathbb{P}^{n}$, over the field $\mathbb{F}_p$ with $p$ elements. The Weil conjectures are conjectures about the number of points of $X$ over $\mathbb{F}_{p^k}$. Specifically, they say that there should be some matrix $A$ such that

$\displaystyle{ \# X(\mathbb{F}_{p^k}) = 1+p^k+p^{2k} + \cdots + p^{(n-1)k} + (-1)^{n-1} \mathrm{Tr} (A^k),}$

and that the eigenvalues of $A$ should be algebraic integers of norm $p^{(n-1)/2}$.
Here I am using the Lefschetz hyperplane theorem to know what $H_{et}^i(X)$ is for $i \neq n-1$.

This is, of course, a famously hard theorem. The claim about the eigenvalues is the hardest part, but simply the existence of a matrix for which this formula holds is already quite hard; the first proof was due to Dwork.

What I am going to show you is that there is a much easier proof of the above formula modulo $p$; a proof of the sort that could be appear in Ireland and Rosen. Many of the terms above disappear mod $p$, so our goal is just to show that there is some matrix $B$ such that

$\displaystyle{ \# X(\mathbb{F}_{p^k}) \equiv 1 + (-1)^{n-1} \mathrm{Tr} (B^k) \mod p.}$

# Farey fractions, Ford circles, and SL_2.

The topic of this post came up during a conversation with some physicists about the fractional quantum Hall effect (which is quite fascinating, but I don’t feel particularly qualified to discuss).  I have decided to set it down here in the hope that, as long as I have an internet-capable device with me, I won’t have to rederive it in front of people again.  Some of this material appears in Apostol’s Modular functions and Dirichlet series in number theory and Conway’s The sensual form. I’d be happy to hear about other good treatments.

# The Brauer Groupoid

Recall that the Brauer group of a field k consists of central simple algebras over k up to Morita equivalence with the group operation given by tensor product. For example, the Brauer group of the real numbers is Z/2 because the only central simple algebras are matrix algebras over $\mathbb{R}$ or matrix algebras over the quaternions $\mathbb{H}$, and $\mathbb{H} \otimes \mathbb{H} \cong M_4(\mathbb{R})$. It is a well-known and fundamental fact that the Brauer group is isomorphic to the second Galois cohomology $H^2(\text{Gal}(\bar{k}/k), \bar{k}^*)$ where $\bar{k}$ is the seperable closure of k.

What I’d like to explain in this post is a follow-your-nose proof of this isomorphism which comes from thinking about fusion categories. Namely, attached to any fusion category there is a very natural object called Brauer-Picard groupoid (introduced by Etingof-Nikshych-Ostrik). For the special case of the fusion category of vector spaces over k the Brauer-Picard groupoid has a point for every seperable extension of k and the group of automorphisms of the point k gives exactly the Brauer group. However, one can also look at the group of automorphisms of other points, in particular the point $\bar{k}$. The group of automorphisms of that point is instead naturally isomorphic to the Galois cohomology $H^2(\text{Gal}(\bar{k}/k), \bar{k}^*)$. Since the groupoid is connected we see that the Brauer group coincides with the Galois cohomology. In fact, there’s a natural choice of arrow from $k$ to $\bar{k}$ and so a natural choice of isomorphism between the two groups.

This example came up in work in progress with Pinhas Grossman where we compute the Brauer-Picard groupoid of the fusion categories which come from the Haagerup subfactor. As we’ll see the automorphism group of a point in the Brauer-Picard groupoid has a subgroup consisting of certain “outer automorphisms.” I wanted to have a good example in hand where the outer automorphism group of different points were different in order to test certain lemmas. The example in this post is as extreme as things can get, for $k$ there are no nontrivial outer automorphisms, while for $\bar{k}$ the whole group consists of outer automorphisms.

# Divergent sums and the class number formula

Over on MathOverflow, we’ve had a bunch of discussions about the class number formula. In particular, Keith Conrad pointed out a paper of Orde which gives a beautiful nonsense proof of the class number formula. This post is my attempt to understand why Orde’s argument works. Specifically, I am going to use an idea which I learned from Terry Tao’s blog: Arguments about divergent sums are often really arguments about the constant term in asymptotic expressions for smoothed sums.

This post assumes familiarity with the concepts and notations of a first course in algebraic number theory.

# Lattices and their invariants

This post started out as an exposition on the monster Lie algebra, but it grew out of control, so I’m hacking off a chunk. Here, I’ll describe the basics of integer lattices.

Lattices show up in many mathematical contexts, some of which may be unexpected to the uninitiated. These contexts include the study of optimal periodic sphere-packings, the topology of 4-manifolds (where lattices give a full classification in the simply connected case), algebraic number theory, finite group theory, and theoretical high-energy physics. I will say almost nothing about these applications, though.

# The Weil Conjectures: The approach via the Standard Conjectures

The aim of this post is to outline a strategy for proving the Weil conjectures, proposed by Grothendieck and others. This strategy is incomplete; at various stages, we will need to assume conjectures which are still open today.

Our aim is to prove:

Theorem:
Let $X$ be a smooth projective variety, over a field of any characteristic. Let $H^*$ be a “reasonable” cohomology theory. Let $\omega \in H^2(X)$ be the hyperplane class for a projective embedding of $X$. Let $F$ be an automorphism of $X$, such that $F^* \omega = q \omega$. Then the eigenvalues of $F^*: H^r(X) \to H^r(X)$ are algebraic numbers and, when interpreted as elements of $\mathbb{C}$, have norm $q^{r/2}$.

In previous posts, we gave for a proof in characteristic zero and a proof in the case that $X$ is a curve. I also explained why I need to say how I am embedding these eigenvalues into $\mathbb{C}$. Our proof requires all the ideas of these previous posts, plus some new ones.
What’s so special about $(\sqrt{3} + \sqrt{7})/2$?