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# Zhang’s theorem on bounded prime gaps

This post is based on Ben Green’s talk at YRM 2013, which neatly summarised Yitang Zhang’s proof of the following long-sought result.

Theorem (Zhang, 2013). There exists $H$ such that there are infinitely many pairs $p' > p$ of primes such that $p'-p \le H$.

Zhang gave $H \le 70,000,000$. The twin prime conjecture claims that $H=2.$ There’s an ongoing Polymath8 project dedicated to lowering $H$, with the currently accepted record being $12,006$.

Zhang’s idea was to marry two large bodies of work. The first, initiated by Selberg in the 1940s and carried on by Goldston-Pintz-Yildirim in 2005, investigates the link between the distribution of primes in arithmetic progression and bounded gaps. The second, whose numerous contributors included BombieriA.I. Vinogradov, Linnik, Heath-Brown, Iwaniec, Friedlander and Fouvry, studies how primes behave in APs.

When studying primes via sieves, it is convenient to use the von Mangoldt function $\Lambda$. The prime number theorem is equivalent to

$\sum_{x \le X} \Lambda(x) \sim X$.

For primes in AP: we expect that

$\sum_{X \ge x \equiv c \mod d} \Lambda(x) \simeq X/\varphi(d)$          (1)

whenever $d < X^{1-\varepsilon}$ and $(c,d)=1$. This is only known for $d \le \log^A X$, where $A$ is any positive number. Roughly speaking GRH is equivalent to (1) holding for $d < X^{1/2 - \varepsilon}$. Powerful work of Bombieri and Vinogradov from the 1960s establishes (1) for almost all $d.

Then in 2005, Goldston, Pintz and Yildirim showed that if (1) holds for almost all $d < X^\theta$ (some $\theta > 1/2$ then $H \sim (\theta-1/2)^{-3/2}$. Motohashi and Pintz adapted this to show that it suffices for almost all smooth $d < X^\theta$ to satisfy (1); Zhang appears to have done this independently.

In the 1980s, Bombieri, Fouvry, Friendlander and Iwaniec demonstrated (1) for almost all smooth $d < X^{4/7}$, but this required $c$ to be fixed. Zhang removed the technical condition of $c$ needing to be fixed.

To tackle the resulting Kloosterman sums BFFI used estimates from automorphic forms, together with Weil bounds, whereas Zhang used only the latter. We expect square root cancellation of these sums, as if primes were distributed randomly, and indeed for prime $p$ this is obtained via Weil bounds:

$\sum_{n \mod p} e((an+b\bar n)/p) \ll 2 \sqrt p$.

Crucially, if $p$ is composite then we can surpass square root cancellation just slightly.

The remainder of the talk elaborates on GPY. Let EH($\theta$) be the assertion that (1) holds whenever $d \le X^\theta$. We briefly recap. The Elliott-Halberstam conjecture is that EH($\theta$) holds for any $\theta <1$. The Bombieri-Vinogradov theorem is that EH($\theta$) holds whenever $\theta <1/2$. GPY show that if EH($\theta$) holds for some $\theta >1/2$ then $H < \infty$. So GPY fell just short of proving bounded prime gaps. GPY also showed, conditionally on EH, that $H \le 16$.

A set $\mathcal{H} = \{h_1, \ldots, h_k\}$ is admissible if, modulo any prime $p$, the set $\mathcal{H}$ misses a residue class modulo $p$. This condition is necessary for it to be possible that for infinitely many $n$, two of $n+h_1, \ldots, n+h_k$ are prime.

Theorem (GPY, 2005). Assume EH($\theta$) for some $\theta > 1/2$, and let $\{h_1,\ldots,h_k\}$ be admissible, where $k \ge k_0(\theta)$. Then, for infinitely many $n$, two of $n+h_1, \ldots, n+h_k$ are prime.

With Zhang’s work, this shows that improvements in $k_0$ lead to improvements in $H$; roughly speaking $H \sim k_0 \log k_0$. In fact it is clear that $H$ is bounded above by the diameter of any admissible $k_0$-tuple. Consequently, a lot of effort is being devoted towards the task of finding narrow admissible tuples.

We conclude with a sketch of the proof of the above Theorem of GPY. Define

$T = \frac{\sum_{n \le X} (\Lambda(n+h_1) + \ldots + \Lambda(n+h_k)) \nu(n)} { \sum_{n \le X} \nu(n)}$,

where $\nu(n)$ is the indicator function of $(n+h_1) \ldots (n+h_k)$ being almost prime. The idea is that if $T > \log X$ then, reasonably often, two of $n+h_1, \ldots, n+h_k$ would be prime. It remains to estimate the numerator and denominator of $T$. To this end, we introduce Selberg’s functions $\nu$ which, with judicious weights, approximate our previous $\nu$. Fix $D$ with $1 < D \ll X$, and put $\nu = \lambda^2$, where

$\lambda(n) = \sum_{D \ge d | n} \lambda_d$,

where the $\lambda_d$ are real weights with $\lambda_1 = 1$. Note that $\nu$ majorises the indicator function of the primes, for if $n$ is prime and $D < n \le X$ then $\nu(n) =1$.

First consider the denominator of $T$. Our weights can be chosen so that, as well as the two notions of $\nu$ being roughly compatible, we can show that

$\sum_{n \le X} \nu(n) \le \frac{2X}{\log X}$.          (2)

The derivation of (2) proceeds as follows. Note that

$\sum_{n \le X} \nu(n) = \sum_{d,d' \le D} \lambda_d \lambda_{d'} \sum_{[d,d'] | n \le X }1$,

where $[d,d']$ denotes the least common multiple. If $D^2 < X$ then the inner sum is roughly $X / [d,d']$, so

$\sum_{n \le X} \nu(n)\simeq X\sum_{d,d' \le D}\frac{\lambda_d\lambda_{d'}}{[d,d']}$.

The right hand sum is a quadratic form in $(\lambda(1), \ldots, \lambda(D))$, which can be minimised using general theory of the Selberg sieve, leading to the inequality (2).

The numerator of $T$ is amenable to a similar treatment. The attendant sum

$\sum_{[d,d']| n \le X} \Lambda(n+h)$

can be controlled providing that (1) holds with $[d,d']$ in place of $d$. A painful calculation gives

$T \ge \frac{2\theta}{1+1/\sqrt{k}}\log X$,

and this exceeds $\log X$ (for large $k$) if $\theta > 1/2$.

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# The j-invariant of CM elliptic curves

In this post I want to sketch the proof that the j-invariant of an elliptic curve over $\mathbb{C}$ with complex multiplication is an algebraic number. With more work, one can show that it is in fact an algebraic integer which underpins the famous observation that $e^{\pi\sqrt{163}}$ is remarkably close to an integer ( to 40 significant digits $e^{\pi\sqrt{163}}=262537412640768743.999999999999250072597$).

We start with some background about elliptic curves over $\mathbb{C}$.

A lattice $\Lambda \subseteq \mathbb{C}$ is a rank 2 discrete subgroup $\Lambda =\mathbb{Z}w_1 \oplus \mathbb{Z} w_2$ for $w_1, w_2$ $\mathbb{R}$-linearly independent. The quotient $\mathbb{C}/\Lambda$ carries a natural group structure and the Weierstrass $\wp$– function

$\wp(z;\Lambda)=\frac{1}{z^2} +\sum_{w\in\Lambda-{0}}\frac{1}{(z-w)^2}-\frac{1}{w^2}$

gives a group isomorphism $\mathbb{C}/\Lambda\longrightarrow E_{\Lambda}(\mathbb{C})$ sending $z\in\mathbb{C}$ to $(\wp(z;\Lambda),\wp'(z;\Lambda))$ where $E(\mathbb{C})$ denotes the $\mathbb{C}$-points of the elliptic curve

$E_{\Lambda} : y^2=4x^3-60G_{4}(\Lambda)x-140G_{6}(\Lambda)$

for $G_{2k}(\Lambda)=\sum_{w\in\Lambda-{0}}w^{-2k}$ the Eisenstein series of weight $2k$. To eliminate unwanted constants later we define $g_{2}(\Lambda)=60G_4(\Lambda)$ and $g_3(\Lambda)=140G_6(\Lambda)$.

The uniformisation theorem  shows that every elliptic curve $E/\mathbb{C}$ is isomorphic to $E_{\Lambda}$ for some $\Lambda\subseteq\mathbb{C}$.

For an elliptic curve given by the Weierstrass equation

$E: y^2=4x^3+Ax+B$

the discriminant of E is defined to be the quantity $\Delta(E)=-(A^3+27B^2)$ (more or less the discriminant of the RHS of the equation defining E) and the j-invariant of E to be the quantity $j(E)=-1728\frac{A^3}{\Delta}$. As one might hope from the name, if E’ is isomorphic to E and is itself given by a Weierstrass equation of the form above then we have $j(E')=j(E)$ (as any elliptic curve $E$ over $\mathbb{C}$ can be put in Weierstrass form we will refer to the j-invariant of E without first specifying an equation). In light of the correspondence between elliptic curves over $\mathbb{C}$ and lattices, we can think of j and $\Delta$ as functions on lattices. We find $\Delta(\Lambda)=g_2(\Lambda)^3-27g_3(\Lambda)^2$ and $j(\Lambda)=1728\frac{g_2(\Lambda)^3}{\Delta(\Lambda)}$.

Now suppose $\Lambda_{1}$ and $\Lambda_{2}$ are lattices and $\alpha\Lambda_{1}\subseteq\Lambda_{2}$ for some $\alpha\in\mathbb{C}$. Then multiplication by $\alpha$ induces a homomorphism $\mathbb{C}/\Lambda_{1}\longrightarrow\mathbb{C}/\Lambda_{2}$ and it turns out that this gives rise to an isogeny (morphism of algebraic varieties which is also a group homomorphism) $\phi_{\alpha}:E_{\Lambda_{1}}\longrightarrow E_{\Lambda_{2}}$ and that every isogeny arises in this way. In particular, elliptic curves $E_{\Lambda_{1}}$ and $E_{\Lambda_{2}}$ are isomorphic if and only if $\Lambda_{1}$ and $\Lambda_{2}$ are homothetic, i.e. $\Lambda_{1}=\alpha\Lambda_{2}$ for some $\alpha\in\mathbb{C}^{\times}$. Moreover, the endomorphism ring of  $E_{\Lambda}$ (isogenies $E_{\Lambda}\rightarrow E_{\Lambda}$ with addition and composition) is isomorphic to the ring

$R=\{\alpha\in\mathbb{C} : \alpha\Lambda\subseteq\Lambda\}$

Note that any lattice is homothetic to one of the form $\mathbb{Z}\oplus\tau\mathbb{Z}$ for some $\tau\in\mathbb{C}$. In this case, it is a simple check to see that either $R=\mathbb{Z}$ or $K=\mathbb{Q}(\tau)$ is an imaginary quadratic extension of $\mathbb{Q}$ and R is an order in K (a subring S of K, finitely generated as a $\mathbb{Z}$-module such that $S\otimes\mathbb{Q}=K$, for example the ring of integers $\mathcal{O}_{K}$). In the latter case we say that $E_{\Lambda}$ has complex multiplication (by K or R).

If we start with an imaginary quadratic field K then it is easy to construct an elliptic curve with complex multiplication by K. Indeed, let $\mathfrak{a}$ be any non-zero fractional ideal of K. After embedding K in $\mathbb{C}$$\mathfrak{a}$ becomes a lattice and we may consider the elliptic curve $E_{\mathfrak{a}}$ corresponding to $\mathbb{C}/\mathfrak{a}$. By definition, $\mathfrak{a}$ is an $\mathcal{O}_{K}$-module so $\mathcal{O}_{K}\subseteq \text{End}(E)$. Conversely, any order of K is a subring of $\mathcal{O}_{K}$ so actually $\text{End}(E)=\mathcal{O}_{K}$. Moreover, if two fractional ideals $\mathfrak{a}$ and $\mathfrak{b}$ have the same image in the class group $\text{cl}(\mathcal{O}_{K})$ of K then $\mathfrak{a}=\alpha\mathfrak{b}$ for some $\alpha\in K^{\times}$ and so $E_\mathfrak{a}$ and $E_{\mathfrak{b}}$ are isomorphic. This lets us define a map

$\text{cl}(\mathcal{O}_{K})\longrightarrow\mathcal{EL}(\mathcal{O}_{K})=\{\text{elliptic curves with CM by } \mathcal{O}_{K}\}/\text{iso}$

sending $\mathfrak{a}$ to $E_{\mathfrak{a}}$ which is easily seen to be injective. Moreover, if $E_{\Lambda}$ has complex multiplication by $\mathcal{O}_{K}$ for some $\Lambda=\mathbb{Z}\oplus\tau\mathbb{Z}$ then $\tau\in\mathcal{O}_{K}$ whence $\Lambda$ is a fractional ideal of $\mathcal{O}_{K}$. We have therefore established a bijection between $\text{cl}(\mathcal{O}_{K})$ and $\mathcal{EL}(\mathcal{O}_{K})$ and so, crucially, the latter set is finite.

We are now in a position to complete the proof that the j-invariant of an elliptic curve with complex multiplication  is an algebraic number but let’s first pause to consider the implications of this result. Since the j-invariant is a well defined function on isomorphism classes of elliptic curves it may also be viewed as a function on lattices up to homothety. That is, we have $j(\alpha\Lambda)=j(\Lambda)$ for all $\alpha\in\mathbb{C}^{\times}$. If we define the function $j(\tau)$ on the upper half plane $\mathbb{H}$ by $j(\tau)=j(\mathbb{Z}\oplus \tau\mathbb{Z})$ then $j(\tau)$ is a modular function of weight $0$ for $SL(2,\mathbb{Z})$. Its q-expansion begins

$j(\tau)=\frac{1}{q}+744+196884q+21493760q^2+...$

where $q=e^{2\pi i\tau}$ and so we are asserting that this function, when evaluated at any element of $\mathbb{H}$ that lies in an imaginary quadratic field, yields an algebraic number.  This  seemed truly remarkable to me when I first saw it. It is also worth remarking that the function $\Delta(\tau)$ defined in the obvious way is, in fact, a weight 12 modular form and famously has the product expansion

$\Delta(\tau)=(2\pi)^{12}q\prod_{n\geq1}(1-q^n)^{24}$

Continuing with the proof, fix E an elliptic curve with complex multiplication by $\mathcal{O}_{K}$ (the general case where $\mathcal{O}_K$  is replaced by an arbitrary order in K is more complicated and I will omit it for simplicity) and j-invariant $j(E)$. It suffices to show that the set $\{\sigma(j(E)) : \sigma \in \text{Aut}(\mathbb{C}) \}$ is finite. Indeed, if $x_{1}\in\mathbb{C}$ is transcendental over $\mathbb{Q}$ then we may find $x_{2},...,x_{n}\in\mathbb{C}$ with $L=\mathbb{Q}(x_{1},...,x_{n})$ isomorphic to the function field over $\mathbb{Q}$ in n indeterminates. Note that $x_{1}$  has (at least) n distinct images under automorphisms of L (given by permuting the generators) and since $\mathbb{C}$ is algebraically closed these extend to automorphisms of $\mathbb{C}$. As we may do this for every n we see that the set $\{\sigma(x_{1}) : \sigma \in \text{Aut}(\mathbb{C}) \}$ is infinite. Note conversely that if $x_{1}$ is algebraic over $\mathbb{Q}$ then any automorphism of $\mathbb{C}$ must send $x_{1}$ to some root of the minimal polynomial of $x_{1}$ over $\mathbb{Q}$ and so $\{\sigma(x_{1}) : \sigma \in \text{Aut}(\mathbb{C}) \}$ is finite and has order equal to the degree of $\mathbb{Q}(x_{1})$ over $\mathbb{Q}$.

Fix now $\sigma\in\text{Aut}(\mathbb{C})$ and define $E^{\sigma}$ to be the elliptic curve obtained from E by acting on the coefficients of E. That is, if E is the elliptic curve $y^2=4x^3-60G_{4}(\Lambda)x-140G_{6}(\Lambda)$ for some lattice $\Lambda$ then $E^{\sigma}$ is the elliptic curve $y^2=4x^3-60\sigma(G_{4}(\Lambda))x-140\sigma(G_{6}(\Lambda))$ . In particular, we have $j(E^{\sigma})=\sigma(j(E))$. If $\phi$ is an endomorphism of E then $\phi^{\sigma}$, obtained from $\phi$ by applying $\sigma$ to the equations defining $\phi$ (recall that $\phi$ is a morphism of algebraic varieties), is an endomorphism of $E^{\sigma}$. So $\text{End}(E)\cong\text{End}(E^{\sigma})$ and in particular, $E^{\sigma}$ also has complex multiplication by $\mathcal{O}_{K}$. The upshot of this is that $\sigma(j(E))=j(E')$ for some $E'\in\mathcal{EL}(\mathcal{O}_{K})$ and we have already seen that the set $\mathcal{EL}(\mathcal{O}_{K})$ is finite, thus $j(E)$ is an algebraic number.

In fact, the proof tells us more, namely that the degree of $\mathbb{Q}(j(E))$ over $\mathbb{Q}$ is at most the class number of K. In particular, if $\tau\in\mathbb{H}$ lies in an imaginary quadratic field K of class number one and the corresponding lattice is a fractional ideal of K, then $j(\tau)$ will actually be rational. As mentioned at the beginning, one may show that the j-invariant of a CM elliptic curve is in fact an algebraic integer so $j(\tau)$ will actually be in $\mathbb{Z}$ in this case. Choosing $\tau$ to be $\frac{1+\sqrt{-163}}{2}$, a generator of the ring of integers of the class number one field $\mathbb{Q}(\sqrt{-163})$ and substituting into the q-expansion for j, we have

$\frac{1}{q}+744+196884q+21493760q^2+...\in\mathbb{Z}$

where $q=-e^{-\pi\sqrt{163}}$. Since q is very small ($\approx 10^{-18}$) this predicts that $\frac{-1}{q}=e^{\pi\sqrt{163}}$ should be very close to an integer, as indeed is the case.

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