# Expected Mordell-Weil rank heuristics through Sato-Tate, Birch and
Swinnerton-Dyer conjectures

Dinesh S. Thakur  Department of Mathematics, University of Rochester,
Rochester, NY 14627<EMAIL_ADDRESS>
###### Abstract.

We present an heuristic argument for the prediction of expected Mordell-Weil
rank of elliptic curves over number fields, using Birch and Swinnerton-Dyer’s
original conjecture and Sato-Tate conjectures. We do calculations in some
cases and raise questions about their relations, if any, with the predictions
of various average rank models that have been considered.

For an elliptic curve $E$ over the rational number field $\mathbb{Q}$, write
$N_{p}$ for the cardinality of $E(\mathbb{F}_{p})$, and let
$P_{x}=\prod_{p\leq x}N_{p}/p$.

Then the strong original form of Birch and Swinnerton-Dyer (BSD) conjecture
[BSD63-65] says that there is a positive constant $C=C(E)$ such that $P_{x}$
is asymptotic to $C(\log x)^{r}$, as $x$ tends to infinity, where $r$ is the
(Mordell-Weil) rank of $E$, i.e., the rank of the finitely generated group
$E(\mathbb{Q})$.

Goldfeld [Goldfeld82] showed that this implies the usual $L$-function form of
this conjecture as well as the Riemann hypothesis for this function. The
finite field-function field Riemann Hypothesis, i.e., Hasse bounds imply much
weaker result $\log(P_{x})=O(x^{1/2+\epsilon})$, for any $\epsilon>0$, instead
of the asymptotics $r\log\log(x)+B$ expected from above. But they also imply
that we can write $N_{p}=p+1-2\sqrt{p}\cos(\theta_{p})$,
$\theta_{p}\in[0,\pi]$

The distribution of $\theta_{p}$ is predicted by the Sato-Tate (ST) conjecture
[Tate63] as equidistribution wrt. $(2/\pi)\sin^{2}(\theta)d\theta$, for $E$
with no complex multiplications (non-CM). This is now a theorem [BGHT11]
([Taylor08] for such $E$ with non-integral $j$-invariant.)

Now let us first calculate the expected value of $\log(P_{x})$ under the Sato-
Tate distribution on $\theta_{p}$ for non-CM elliptic curve. We write
$\mathbb{E}$ for the expectation.

Using $\log(1+y)=y-y^{2}/2+\cdots$, and $\sum_{p\leq
x}1/p=\log\log(x)+B+O(1/\log(x))$, for some constant $B$, we get

$\displaystyle\mathbb{E}(\log(P_{x}))$
$\displaystyle=\sum\mathbb{E}(\log(1+1/p-2\cos(\theta_{p})/\sqrt{p}))=\log\log(x)-\frac{1}{2}\log\log(x)+O(1)$
$\displaystyle=\frac{1}{2}\log\log(x)+O(1),$

using $\int_{0}^{\pi}2\cos(\theta)\sin^{2}(\theta)d\theta=0$ and
$\int_{0}^{\pi}2\cos^{2}(\theta)\sin^{2}(\theta)d\theta=\pi/4$ to get the
first two terms from the first 2 terms of the Taylor expansion, and putting
the error from the remaining convergent series in the big O term.

Comparison with the strong original BSD conjecture asymptotics of
$r\log\log(x)$ implies heuristically that the expected value of the Mordell-
Weil rank of $E(\mathbb{Q})$ in non-CM case is $1/2$.

Remarks (1) Note that we have not taken averages or expectations of actual
ranks under any family, but just matched the asymptotics of the expectation of
the quantity $\log(P_{x})$ under ST for non-CM case with the asymptotics
conjectured (a conjecture is, in fact, basically just an ‘expectation’!) and
called the result heuristic expected value just because it comes from such a
matching (and [Birch68] connection with the averages mentioned below).
Interestingly, this seems to fit well with the predictions (and some evidence
and bounds in some cases) by Dorian Goldfeld [Goldfeld79, Conjecture B] (for
quadratic twists) and others (e.g., [PR11, Conjecture 1.2] over global fields)
for the average rank as $1/2$ in various ‘averages scenarios’ when you take
averages by ordering the elliptic curves by the heights (normalized
coefficients sizes) or use families of quadratic twists ordered by the twist
parameter. (I do not know of any predictions (or rather results?) specifically
mentioning ordering by minimal discriminants or conductors, apart from Katz-
Sarnak function field work[KS99]). We refer to [BS15, PPVW19, Poonen18,
Poonen13] for some results and surveys. Is this a coincidence or is there a
better explanation? In the Sato-Tate original case [Tate63], one fixes $E$ and
averages over primes, but as Birch [Birch68] proved, one gets the same
probability distribution when we fix $p$ and average over curves $E$ over
$\mathbb{F}_{p}$. So there might be an explanation of the coincidence.

Andrew Sutherland suggested that probably the ordering of $E$’s over
$\mathbb{Q}$ by naive height has the property that for any fixed $p$ and all
sufficient large $M$ the first $M$ of the $E$’s have reductions mod $p$ that
are equidistributed over the isomorphism classes of $E$ over $\mathbb{F}_{p}$,
and Birch might apply to connect with actual averages.

Peter Sarnak suggested that going from one curve (as in the expectation
calculation) to a family, a serious complicated issue is how big the parameter
$x$ is compared to the conductors of the family one is supposed to be
averaging over.

(2) Similar calculation gives the same expectation $1/2$ for non-CM elliptic
curves over any number field, by exactly the same argument and using that sum
of the reciprocals of the norms of primes (up to $x$) still grows like
$\log\log x$. Only the constants seem to depend on the number field. In
contrast to [PR11, Conjecture 1.2] for global fields, Dorian Goldfeld
suggested to the author that $1/2$ may not now be expected to be the average
rank for the elliptic curves over number fields with many complex places,
since the discrepancy is known [DD09] for some quadratic twist families. The
‘expectations’ and predictions by experts of the rank distributions have
changed a few times over the time. The ‘Infinite family Rank at most 21’
heuristics of Granville [W15] and [PPVW19] for elliptic curves over
$\mathbb{Q}$ also seems to go wrong when generalized naively for general
number fields, as pointed out in [PPVW19, Sec. 12], [Poonen18, Sec. 4].

Next we look at CM $E/Q$, a similar calculation now with the predicted uniform
distribution of $\theta_{p}$ for ordinary primes and $a_{p}=0$ for super-
singular p gives

$\mathbb{E}(\log(P_{x}))=1/2\log\log(x)+O(1)$, now $1/2$ coming from the
density of super-singular primes, the ordinary primes contribution being O(1)
term: as $(1/2)\log\log(x)$ from the $y$ term (since
$\int_{0}^{\pi}\cos(\theta)d\theta=0$) cancels from the $(1/2)\log\log(x)$
from $-y^{2}/2$ term (since $\int_{0}^{\pi}2\cos^{2}(\theta)d\theta=\pi$.)

Again, this heuristically gives the expected rank as 1/2. This argument shows
that the expected rank is 1/2 for CM $E$ over field not containing the CM
field and 0 for CM $E$ over field containing the CM field.

I do not know whether these are the ‘expected’ averages in these cases. In
fact, I could not find any published predictions. I would love to know
references, if any.

Let us now calculate higher moments $\mathbb{E}((\log P_{x})^{n})$’s.

Write $x_{i}$ for $1/p_{i}-2\cos(\theta_{p})/\sqrt{p_{i}}$, where $p_{i}$ is
$i$-th prime. Then $\mathbb{E}(x_{i})=1/p_{i}$ and
$\mathbb{E}(x_{i}^{2})=1/p_{i}$, so that
$\sum\mathbb{E}(x_{i})=\log\log(x)+O(1)$ and
$\sum\mathbb{E}(x_{i}^{2}/2)=(1/2)\log\log(x)+O(1)$ by the calculations done
above. Since third and higher powers of $x_{i}$ can be ignored as they give
convergent sums, we have, up to O(1), $\mathbb{E}((\log
P_{x})^{n})=\mathbb{E}((\sum\log(1+x_{i}))^{n})=\mathbb{E}(\sum\prod_{i=1}^{n}(x_{i}-x_{i}^{2}/2))$
since terms with some repeated indices $i_{1}=i_{2}$ again contribute bounded
quantities as they involve at least $3/2$ powers of primes in denominator.

Let us do $n=2$ case first. Then up to O(1) again, with $i\neq j$ below, we
have

$\displaystyle\mathbb{E}((\log(P_{x}))^{2})$
$\displaystyle=\mathbb{E}(\sum\log(1+x_{i})\log(1+x_{j}))=\sum\mathbb{E}((x_{i}-x_{i}^{2}/2)(x_{j}-x_{j}^{2}/2)$
$\displaystyle=\sum\mathbb{E}(x_{i}x_{j}-x_{i}x_{j}^{2}/2-x_{i}^{2}x_{j}/2+x_{i}^{2}x_{j}^{2}/4)$
$\displaystyle=\sum[1/(p_{i}p_{j})-1/(2p_{i}p_{j})-1/(2p_{j}p_{i})+1/(4p_{i}p_{j})]$
$\displaystyle=\sum 1/(4p_{i}p_{j})=(\log\log(x))^{2}/4$

For general $n$, we get, by the simplification above that

$\mathbb{E}((\log P_{x})^{n})=\sum\mathbb{E}((\prod
x_{i})(\prod(-x_{j}^{2}/2)=\sum\prod\mathbb{E}(x_{i})\mathbb{E}(-x_{j}^{2}/2)$,
where we take k of $x_{i}$ terms and $n-k$ of $-x_{j}^{2}/2$ terms, so we get

$(\sum_{k=0}^{n}(-1)^{k}{n\choose
k}/2^{k})(\log\log(x))^{n}=(1-1/2)^{n}(\log\log(x))^{n}=(\log\log(x))^{n}/2^{n}$

Remarks The expectations (in the generic non-CM case that we only worked out
so far) now seem ‘unexpected’: $n$-th moment is basically $(\log\log
x)/2^{n}$, rather than $(\log\log x)/2$, which would be ‘expected’, if it is
supposed to be the just average of $r^{n}$, since half are expected to be of
rank 0 and half of rank 1, in the widely believed conjecture. Are the
expectations not supposed to be reasonable averages or is there some other
good explanation for higher moments discrepancy with average considerations?

Finally, we mention another (unrelated to above) observation regarding the
rank conjectures. In a beautiful work [Poonen02], Bjorn Poonen showed (among
other things) that Hilbert’s tenth problem over ${\cal O}_{K}$ is undecidable
for every number field $K$, if for every number field $K$, there exists
elliptic curve $E$ over $\mathbb{Q}$ with MW rank of $E$ over $\mathbb{Q}$ and
over $K$ being one. But the optimistic conjecture [Poonen02, 2.6] that the if
part is always true is incompatible with the BSD conjecture, since for some
number fields K, e.g., $K=\mathbb{Q}(\sqrt{-1},\sqrt{17})$ it is known (by
Karl Rubin see e.g., [Dokchitser13, Sec. 8.1]) that the analytic rank of $E$
over $K$ is always even for all $E$ over $\mathbb{Q}$, by root number
calculations. Bjorn Poonen tells me that he and Alexandra Shlapentokh proved
later that for this application to Hilbert’s tenth problem, it is enough to
know that for every quadratic extension of number fields $L/K$, there is an
elliptic curve $E$ of a positive rank over $K$ (not necessarily 1) that has
the same rank over $L$. In fact, Karl Rubin and Barry Mazur [MR10] have proved
that such $E$ exists, assuming that the Tate-Shafarevich group for elliptic
curves is always finite.

Acknowledgments I thank Dorian Goldfeld, Peter Sarnak, Bjorn Poonen
(especially for pointing out a wrong comment), Andrew Sutherland for their
comments.

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