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Section 6.12 A gallimaufry of applications (of Gross-Zagier) II (Alex)

Subsection 6.12.1 More on computation

A more advanced trick: We have the standard Heegner point outlined above, there are several speedups possible:

  1. Use Atkin-Lehner involutions to reduce the size of \(A\) in \((A,B,C)\text{.}\)
  2. Use faster algorithms for point counting, e.g. on CM curves we have a simpler expression for \(a_p\)'s which can be computed with Cornichias algorithm.
  3. Cremona-Silverman: Want to reduce the precision needed, how? What information do we know after running the method, an approximation of \(Q\approx P \in E(\RR)\text{.}\) We also know by Gross-Zagier
    \begin{equation*} \hat h(P)\text{.} \end{equation*}
    If
    \begin{equation*} x(P) = \frac{n}{d^2} \end{equation*}
    then
    \begin{equation*} 2 \log (d)=\widehat{h}(P)-\widehat{h}_{\infty}(P)-\sum_{p | N \atop {p \nmid d \atop p^{2} | \operatorname{disc}(E)}} \widehat{h}_{p}(P)\text{.} \end{equation*}
    Using \(Q\) we can compute
    \begin{equation*} \hat h_\infty (P)\text{.} \end{equation*}
    For each \(\hat h_p\) there are only finitely many possibilities, and so in total we have finitely many possible values of
    \begin{equation*} \widehat{h}(P)-\widehat{h}_{\infty}(P)-\sum_{p | N \atop {p \nmid d \atop p^{2} | \operatorname{disc}(E)}} \widehat{h}_{p}(P) \end{equation*}
    giving finitely many possible \(d\) values, from which
    \begin{equation*} numerator (x(P)) = round(x(P) d^2) \end{equation*}
    can be found. In all this allows us to work as we wanted, with an accuracy slightly more than half.

These speed ups are in PARI/GP and Remarks. as \(E\) in last times example is a CM curve, the computation time of the above example can be reduced from a couple of minutes to 7 seconds (factor of 20).

Gotta get height:

Example 6.12.1. A big example.

Let

\begin{equation*} E\colon y^2 = x^3 + \underbrace{2^{5} \cdot 3^{3} \cdot 5^{5} \cdot 7^{3} \cdot 11^{5} \cdot 13^{4}}_{=4259854045547100000} \end{equation*}

be a Mordell curve. We can compute that the analytic rank is 1 and that

\begin{equation*} L(E,1) \approx 28.43512495 \end{equation*}

We need an imaginary number field in which \(2,3,5,7,11,13\) all split, the smallest such is

\begin{equation*} K = \QQ(\sqrt{-1559})\text{.} \end{equation*}

So this \(D = D_K = -1559\) is the smallest possible Heegner discriminant followed by \(-2999,-3071,-5711,-6431,-6551,-8399,-8711,-9071,-9239\text{.}\) We can also compute that the twist \(E_D\) has analytic rank 0 and

\begin{equation*} L(E_D,1) \approx 0.34784 \end{equation*}

We can ask Pari/GP for a Heegner point and we obtain

\begin{equation*} P= (2673768366314301463225804132167786458743211296161665928714089443034915780624646102119694052638661139083614147052130407079631451208185292430763007319638022105825501399385761185566630883388394680375240310954529690758647194911112230527266639143771344809007916379476769393746579657952261851411688268671539912820072222763224027388823804772192104547803223898847305487037661307828748844799602593905135066916765339004127610241/14711766058713677663384943791747898516946843571332438782108054674511943585256352373257820891021633925632603928846607462754678923049647017632180676816464079929265454994902080754858496848491578669209122991110372607577751981118650453474461080043444438030297579085121835217242209089550661820106126782163015526489468149613211400477191028795214873959696995779238660836093943085146259968306971312571346641475392784100 : 138256584957437413549697979638187560540459292802582866047104251079871173001665466386110481033185180290975183359260945144859657131458366183036141245306630348292671101486345526463132664316934927829557544796041549654148869945306252397609306930530983551604764354393433405906021719607528424338682185259543058868693501829232087760637049066670461122157567207439393246729072214062375989571116431670132631712767339509965920423129094613543923419661857943413970492292794841174646487317425335124224816154049239888798416706708946385513743341645910420176133057435782209340663561195089350700342411480965751945453146090192411927796854564432961/56428332331562671615794363625451656023382470684636509020204202990180688177696621616970962649738545712542796767776584129330909917100366186924013354291641329738822701259890430806380424060475365542745244595324918343297529518263108027693527497847179482876257977247331950705161864795745543796461954042373098284699960118594562408063763160815610542420513488629513957169549566241725330887507912166227023322296844673603178204516200038314906488548391012692454233921581644355608983371337760875769764741031620675341450340457565846005848812354464193992153964872086150518588699778897808812845613993026309040085527357183723889000 : 1) \end{equation*}

this has naïve height \(2.67 \times 10^{417}\) (numerator of the \(x\)-coord) and (logarithmic) canonical height \(\approx 956.282209515622\text{.}\)

By saturating we find that in fact \(P = 2\cdot Q\) where \(Q\) has canonical height

\begin{equation*} 956.282209515622/4 \approx 239.070552378906\text{.} \end{equation*}

And

\begin{equation*} Q =(8868405892624209482831543296890050376591681305386772268723757017948071928478387800757391866726575892769529/12199065318293186020638466216615329130692308979647554152111639302853625564281014711694611565598225 : 835157422328503915572710156100004478252440344488413901374204905998433900560826559924171903629096959304261439875564863428904842353573598701320904439166984417667/42607871079159903759545082718655204271158039869031410587066955866681866212965161047419746875287012910440605581735106219437475107311541956522410375 : 1)\text{.} \end{equation*}

Subsection 6.12.2 Gauss's class number problem

Gauss was interested in binary quadratic forms and did a lot of computations with them. He in particular conjectured that

\begin{equation*} \begin{array}{|c|c|c|c|c|c|} \hline h(D) \amp {1} \amp {2} \amp {3} \amp {4} \amp {5} \\ \hline \# \text { of fields } \amp {9} \amp {18} \amp {16} \amp {54} \amp {25} \\ \hline \text { largest } {|D |}\amp{163} \amp {427} \amp {907} \amp {1555} \amp {2683} \\ \hline \end{array} \end{equation*}

The most famous instance of this being the first column. The Gauss class number 1 problem (Section 303 of his Disquisitiones Arithmeticae (1798)) there are only 9 imaginary quadratic fields of class number 1. This was first proved by Heegner, where he introduced analytic techniques into the field of elliptic curves, hence the name Heegner points. Warning: this is somewhat distinct and not exactly what we will mention now

The first instances of the small class number phenomenon go back to Euler who noted that

\begin{equation*} x^2 - x + 41 \end{equation*}

was prime for \(0 \le x \le 40\) (the maximum possible), Euler called such numbers lucky and could not find more.

We now know that this is due to \(\QQ(\sqrt{-163})\) having class number one, hence all small primes remaining inert. If a small prime split we would have an element of small prime norm, but the norm form shows this is not possible. Explicitly all primes less than

\begin{equation*} \frac{1 + |D_K|}4 \end{equation*}

are inert.

Even when \(h(D) \gt 1\) we still observe a similar phenomenon.

For \(\QQ(\sqrt{-427})\) as above we have of the primes up to \((1 + 427)/4 = 107\) only \(17,31,59,89,101\) are split (\(7,61\) ramified).

Why am I telling you this? Because it leads to the following theorem:

Here

\begin{equation*} L_{E / \QQ(\sqrt{D})}(s) \end{equation*}

has \(L_E\) as a factor so Goldfeld needed an elliptic curve \(E/\QQ\) with

\begin{equation*} \ord_{s=1} L_E(s) = 3 \end{equation*}

that is analytic rank 3, to obtain the order 4 vanishing of

\begin{equation*} L_{E / \QQ(\sqrt{D})}(s)\text{.} \end{equation*}

The bounds in this proof can be completely explicit, leading to lists of all imaginary quadratic fields with class number below \(100\text{.}\)

If \(\chi _D = \legendre{D}{\bullet}\) is the associated character to \(\QQ(\sqrt{D})\) of small class number we therefore have

\begin{equation*} L(s,\chi _D) = \prod_p \left(1-\frac{\chi _D(p)}{p^s}\right)^{-1} \end{equation*}
\begin{equation*} \sim \prod_p \left(1+\frac{1}{p^s}\right)^{-1} = \frac{\zeta (2s)}{\zeta (s)}\text{.} \end{equation*}

(mumble mumble approximate functional equation).

How to obtain an \(E\) with proven analytic rank 3.

Gross-Zagier showed that for

\begin{equation*} E = 37b3\colon y^2 = x^{3} + 10 x^{2} - 20 x + 8 \end{equation*}

of conductor 37 and rank 0, we can twist by \(d=-139\) to get a curve of conductor \(714877\text{.}\)

\begin{equation*} E_{-139}\colon -139 y^2 = x^{3} + 10 x^{2} - 20 x + 8 \end{equation*}

where

\begin{equation*} L_E(s,\chi _d) = L_{E_{d}}(s)\text{.} \end{equation*}

Doing a single Heegner point computation we find that \(P_d\) is zero and hence \(h(2P_d)= 0\text{.}\) Using Gross-Zagier we have

\begin{equation*} L_E(1)L'_{E_d}(1) = c \Omega_d \Omega h_{E_d}(2P_d) \end{equation*}

This implies that

\begin{equation*} L_{E_d}(1)L_E'(1) = 0 \end{equation*}

we have

\begin{equation*} L_{E_d}'(1) = 0 \end{equation*}

and as \(L_{E_d}(s)\) has odd functional equation we can calculate

\begin{equation*} L_{E_d}'''(1) \ne 0 \end{equation*}

hence the analytic rank is 3.

So \(E_{-139}\) can be used for Goldfeld's technique.

We now let \(E/\QQ\) be our twisted curve forgetting that it came from \(37b3\text{.}\)

The \(L\)-function of this curve has functional equation

\begin{equation*} \left(\frac{\sqrt{N}}{2 \pi}\right)^{1+s} \Gamma(1+s) L_{E}(1+s)=-\left(\frac{\sqrt{N}}{2 \pi}\right)^{1-s} \Gamma(1-s) L_{E}(1-s) \end{equation*}

if \(D\) is a fundamental discriminant of class number 1 with \(|D| \gt 163\) we can define

\begin{equation*} \Lambda_{D}(s)=\left(\frac{N|D|}{4 \pi^{2}}\right)^{s} \Gamma(1+s)^{2} L_{E}(s) L_{E}\left(s, \chi_{D}\right) \end{equation*}

so that

\begin{equation*} \Lambda_{D}(1+s)=w \cdot \Lambda_{D}(1-s) \end{equation*}

with \(w=\chi _D(-37\cdot 139^2) = 1\text{.}\)

The function

\begin{equation*} L_{E / \QQ(\sqrt{D})}(s) = L_E(s)L_E(s,\chi _D) \end{equation*}

therefore has a zero of even order at \(s=1\text{,}\) given that \(L_{E}(s)\) has an order 3 zero by construction

\begin{equation*} L_{E / \QQ(\sqrt{D})}(s) \end{equation*}

has an order 4 zero.

To give a flavour of the class number one problem assume \(D\) sufficiently large with \(h(D) = 1\) still, then consider

\begin{equation*} I_D = \frac{1}{2\pi i} \int_{2-i\infty }^{2+i\infty } \Lambda_D(1+s) \frac{\diff s}{s^3}\text{.} \end{equation*}
\begin{equation*} \begin{aligned} I_{D} \amp=\frac{1}{2 \pi i} \int_{-2-i \infty}^{-2+i \infty} \Lambda_{D}(1+s) \frac{d s}{s^{3}} \\ \amp=-\frac{1}{2 \pi i} \int_{2-i \infty}^{2+i \infty} \Lambda_{D}(1+s) \frac{d s}{s^{3}} \\ \amp=-I_{D} \end{aligned}\text{.} \end{equation*}

We now want to show that

\begin{equation*} I_D \ne 0 \end{equation*}

under the same assumptions on \(D\text{.}\)

Have euler products

\begin{equation*} \begin{array}{c}{L_{E}(s)=\prod_{p}\left(1-\frac{\alpha_{p}}{p^{s}}\right)^{-1}\left(1-\frac{\beta_{p}}{p^{s}}\right)^{-1}} \\ {L_{E}\left(s, \chi_{D}\right)=\prod_{p}\left(1-\frac{\alpha_{p} \chi_{D}(p)}{p^{s}}\right)^{-1}\left(1-\frac{\beta_{p} \chi_{D}(p)}{p^{s}}\right)^{-1}}\end{array} \end{equation*}

and once again many small primes splitting means that \(L_E(s)L_E(s,\chi _D)\) is analytically like

\begin{equation*} \phi(s):=\prod_{p}\left(1-\frac{\alpha_{p}^{2}}{p^{2 s}}\right)^{-1}\left(1-\frac{\beta_{p}^{2}}{p^{2 s}}\right)^{-1} \end{equation*}

Goldfeld then uses

\begin{equation*} I_{D}^{*}=\frac{1}{2 \pi i} \int_{2-i \infty}^{2+i \infty}\left(\frac{37 \cdot 139^{2}|D|}{4 \pi^{2}}\right)^{1+s} \Gamma(1+s)^{2} \phi(1+s) \frac{d s}{s^{3}} \end{equation*}

and

\begin{equation*} 0=I_{D}=I_{D}^{*}+\text { Error } \end{equation*}

to get the final contradiction.

Remark 6.12.4.

There is also work by Mestre and Buhler-Gross-Zagier on

\begin{equation*} y^{2}+y=x^{3}-7 x+6 \end{equation*}

the smallest conductor rank 3 curve 5077, where they verify BSD explicitly, giving one the first example in rank 3. (To this day, it is not possible, even in principle, to establish BSD for any curve of rank 4 or greater since there is no known method for rigorously establishing the value of the analytic rank when it is greater than 3.) Once again Gross-Zagier is used, if \(L'_E(1)\) is calculated to be small but possibly non-zero it must be a multiple of the height of a small point, but we can look and find no small points, hence we obtain vanishing of the derivative at \(s=1\text{.}\) This implies, for parity reasons that \(L_E(s)\) has analytic rank 3 or more. The third derivative can then be calculated and seen to be non-zero.

This smaller curve gives better bounds in Goldfelds method.

Using work of Oesterle they obtain

\begin{equation*} h(D) \gt \frac{1}{55} \log|D| \end{equation*}

for prime \(D\text{.}\)

It's super effective?!