p-adic Methods/L-functions

People

Salman Butt
Philip Candelas
Henri Cohen
Xenia de la Ossa
Fernando Rodriguez Villegas
Mark Watkins

Goals

Compute (conjectural) formulas for the conductor of the following Calabi-Yau manifold cut out by the equation

\sum_{i=1}^d x_i^d - d\psi \prod_{i=1}^d x_i=0.

For example, for d=3 we have the elliptic curve given in Weierstrass form by

y^2 -3\psi xy +9y= x^3 - 27(\psi^3+1)

with discriminant \Delta=(\psi^3-1)^3\cdot 3^9. We write \psi^3-1=3^a v with (3,v)=1 and conjecture that the formula for the conductor is

N = \left\{\begin{array}{ll} 3^3\cdot sqf(v) & a\leq 2 \\ 3^2\cdot sqf(v) & a\geq 3\end{array}\right.

where sqf(v) is the square-free part of v. This formula has been verified for all integral \psi\in[-10000,10000] not equal to 1 (though it was not investigated why a power of 3 is lost whenever \psi^3-1\equiv 0 \pmod{27}).

The goal is to fill in the following table of local factors of the conductor (for \psi\in\mathbb{Q}):

d	p\mid (\psi^d-1)	2	3	5
2
3
4
5

Such a formula will give us an idea (via the conductor) of the Siegel modular forms of interest to us.

Activities

Thursday, July 30

Fernando, inspired by DNA, made a conjecture for the L-function associated to \psi=\infty and, with some input from Mark, guessed the gamma factor was simply \Gamma(s/2). This allowed us to guess that the local factor for a prime occurring in the denominator of \psi^5-1 is 1-T. Henri then ran some numerical computations to determine the appropriate conductor as well determining the correct local factor for when 5 was in the denominator of \psi^5-1, leading to the following numerically verified table:

\psi	\psi^5-1	N	\varepsilon
1/2	2^-5.31	2³.31.5⁵	+
1/3	2.3^-5.11²	2.3³.11.5⁵	+
1/4	2^-10.3.11.31	2³.3.11.31.5^5	-
-1/3	2².3^-5.61	2.3³.61.5⁵	+
2/3	3^-5.211	3³.211.5⁵	+
5	2².11.71	2.11.71.5⁵	+
-5	2.3.521	2.3.521.5⁵	-
1/5	2².5^-5.11.71	2.11.71	-
-3/5	2³.5^-5.421	2.421	+

Considering the rational \psi's, we conjecture the following formula for the sign of the functional equation:

\varepsilon=\varepsilon_\psi = \left(\frac{*}{5}\right)\times \prod_{v_p(\psi^5-1)<0,p\neq 5} \left(\frac{5}{p}\right) \times \prod_{v_p(\psi^5-1)>0,p\neq 5}\left(-\left(\frac{5}{p}\right)\right)

where * is to be determined.

Wednesday, July 29

Henri and Salman developed code to test out conjectural conductors for varying \psi, allowing them to fill in the following table using numerical experiments

\psi	\psi^5-1	N	\varepsilon
-1	2	2.5⁵	+
2	31	31. 5⁵	-
-2	3.11	3.11.5⁵	-
3	2.11²	2.11.5⁵	-
-3	2².61	2.61.5⁵	-
4	3.11.31	3.11.31.5⁵	+
-4	5².41	41.5⁴	+
6	5².311	311.5⁴	+
-6	7.11.101	7.11.101.5⁵	+
7	2.3.2801	2.3.2801.5⁵	-
-7	2³.11.191	2.11.191.5⁵	+
8	3.7.151	7.31.151.5⁵	+
-8	3².11.331	3.11.331.5⁵	+
-15	2⁴.31.1531	2.31.1531.5⁵	+
1/3	2.3^-5.11²	?	?
1/11	2.5².11^-5.3221	?	?

We conjecture the sign of the functional equation is simply

\prod_{p\mid N, p\neq 5} \left(-\left(\frac{5}{p}\right)\right).

For integral \psi, we conjecture N is just 5^{5-i}\cdot sqf(\psi) where sqf(\psi) is just the squarefree part of \psi and i=1 if 5 divides \psi^5-1 and i=0 otherwise. The case of rational \psi is still left unresolved.

Tuesday, July 28

Following a suggestion of David Farmer, Henri and Salman wrote gp code to compute only the a coefficient out to 10⁵, which took 6.5 hours. At the same time, they optimized the code heavily by performing all the computations mod p^2 instead of p-adically, which reduced the computation of the a's from 6.5 hours to about 11 minutes.
Mark and Henri numerically verified independently that the gamma factors and conjectural conductor was correct for \psi=-1 (or was it \psi=2?).

Friday, July 24

Henri and Fernando worked out the formulas for the approximal functional equation in the case of our L-function in the most naive way (setting the smoothing function g in Mike's notation to be 1). Let W(s) encapsulate the gamma factors, conductor, and power of 2\pi for our L-function:

W(s) := \left(\frac{2\pi}{a}\right)^{-2s}\Gamma(s-1)\Gamma(s),

where our functional equation will send s to 4-s. Then the Mellin transform of W is given by the K-Bessel function:

\mathcal{M}^{-1}(W) = Y(x) = \frac{a}{\pi}x^{-1/2} K_1\left(\frac{4\pi}{a}x^{1/2}\right).

The incomplete Mellin transform is then

F(x,s) = \int_x^\infty t^s Y(t) \frac{dt}{t} = W(s) - \int_0^x t^s Y(t) \frac{dt}{t}.

Setting W(x,s)=\int_0^x t^s Y(t) \frac{dt}{t}, we find

W(x,s) = 8\left(\frac{a}{4\pi}\right)^{2s}\left[-\left(\frac{4\pi}{a}x^{1/2}\right)^{2s-2}K_0\left(\frac{4\pi}{a}x^{1/2}\right)+(2s-2)I\left(\frac{4\pi}{a}x^{1/2},2s-2\right)\right],

where

I(x,s) = \int_0^x t^sK_0(t)\frac{dt}{t}.

A little work shows that

I(x,s) = \left(\log 2 - \gamma -\log x\right) \sum_{k\geq 0} \frac{x^{2k+s}}{(2k+s)2^{2k}k!^2} + \sum_{k\geq 0} \left[\frac{x^{2k+s}}{(2k+s)2^{2k}k!^2}\left(H_k+\frac{1}{2k+s}\right)\right],

where \gamma is Euler's constant and H_k=1+\frac{1}{2}+\cdots+\frac{1}{k} is the harmonic series truncated at k. Putting all this together allows us to compute F(x,s), and using the coefficients of our L-function for the quintic surface, we aim to determine the unknown a (and hopefully the conductor) by varying x and s.

Thursday, July 23

Fernando presented a p-adic hypergeometric function (following Katz): Fix p a prime and set

H\left(\begin{array}{c}\alpha \\ \beta \end{array}\big|\lambda\right) = \frac{1}{(1-p)c_0\left(\begin{array}{c}\alpha\\\beta\end{array}\right)}\sum_{m=0}^{p-2} c_m\left(\begin{array}{c}\alpha \\ \beta\end{array}\right) Teich(\lambda)^{-m},

where \alpha,\beta are are sequences of rational numbers, \lambda is a parameter, and the c_m's are given below.

c_m\left(\begin{array}{c}\alpha\\\beta\end{array}\right) = \prod_{j=1}^r \frac{\Gamma_p\left(\left\{\alpha_j+\frac{m}{p-1}\right\}\right)}{\Gamma_p\left(\left\{1-\beta_j+\frac{m}{p-1}\right\}\right)}\times (-p)^{\mathcal{L}_0(\alpha,\beta,m)} \times p^{-\mathcal{L}_1(\beta,m)}\times \Gamma_p(1/2)^d,

where r is the length of \alpha and \beta, \Gamma_p is the p-adic gamma function, \{a\} is the fractional part of a,

\mathcal{L}_0(\alpha,\beta,m)=\sum_{j=1}^r \left(\left\{\alpha_j+\frac{m}{p-1}\right\}-\left\{1-\beta_j+\frac{m}{p-1}\right\}\right)+\frac{d}{2},

\mathcal{L}_1(\beta,m) = \#\left\{j\mid\left\{1-\beta_j+\frac{m}{p-1}\right\}=0\right\},

and

d=\#\{j\mid\beta_j=1\}.

Using this, we were able to show computationally that for w\neq 0,1,1/2

H\left(\begin{array}{cc}1/4 & 3/4 \\ 1 & 1\end{array}\big|w\right)^2 = p + H\left(\begin{array}{ccc}1/4 & 1/2 & 3/4 \\ 1 & 1 & 1\end{array}\big|4w(1-w)\right).

The left hand side of the above equation is the square of the trace of Frobenius on H^1 (of ?) while the right hand side is the trace of Frobenius on H^2 (of ?).

Wednesday, July 22

We pushed to the d=4 case (we switch from \psi to t):

S_t: x_1^4+x_2^4+x_3^4+x_4^4 - 4tx_1x_2x_3x_4=0.

This surface has periods corresponding to the hypergeometric sum

\,_3F_2\left(\begin{array}{c} \frac{1}{4} \, \frac{3}{4} \\ 1 \end{array}\bigg| 4^4\lambda\right)

where \lambda=1/(4t)^4. We have

\,_3F_2\left(\begin{array}{c} \frac{1}{4} \, \frac{3}{4} \\ 1 \end{array}\big| 4^4\lambda\right) = \,_2F_1\left(\begin{array}{c} \frac{1}{8} \, \frac{3}{8} \\ 1 \end{array}\big| 4^4\lambda\right)^2.

Using some relations on hypergeometric sums, we determined

\,_2F_1\left(\begin{array}{c} \frac{1}{8} \, \frac{3}{8} \\ 1 \end{array}\big| 4z(z-1)\right) = \,_2F_1\left(\begin{array}{c} \frac{1}{4} \, \frac{3}{4} \\ 1 \end{array}\big| z\right).

Moreover

_2F_1\left(\begin{array}{c} \frac{1}{4} \, \frac{3}{4} \\ 1 \end{array}\big| z\right) = \sum \frac{(4n)!}{(2n)!n!^2} \left(\frac{z}{4^3}\right)^n = \omega_E\left(\frac{z}{4^3}\right)

where \omega_E is the period of the elliptic curve given by the equation

E: x^2y+y^2+1=uxy,

and z=4^3/u^4. In Weierstrass form,

E: y^2-uxy=x^3+x

with discriminant u^4-4^3. Moreover, we have the following relations on our parameters:

\lambda=\frac{u^4-4^3}{u^8}, \qquad t^4-1 = \frac{(u^4-128)^2}{4^4(u^4-4^3)}.

We can make a further change of variables by setting u=1/v to get

E_v:y^2-xy=x^3+vx.

In this form we have

\sum_{n\geq 0} \frac{(4n)!}{n!^4} (v(1-4^3v))^n = \left(\sum_{n\geq 0} \frac{(4n)!}{(2n)!n!^2} v^n\right)^2.

We know hope to compute the conductor of the symmetric square of E with Mark's help.

Tuesday, July 21

Xenia (and Philip) introduced Calabi-Yau manifolds on a "baby-baby" level as a precursor/preview of the later workshop talk. They also discussed why Calabi-Yau manifolds are interesting to physicists and why they would be interested in the number of their \mathbb{F}_p-solutions (which are intimately related to the periods of the manifold, which are of prime importance to Calabi-Yau manifolds).
Philip discussed the complex structure and Kahler class of the moduli spaces of Calabi-Yau manifolds. He also discussed mirror symmetry and its implications for the associated zeta functions, as well as motivating the interest in Siegel modular forms.
The group set up a goal to determine the size of the conductor of the Siegel modular forms that may be related to Siegel modular forms. Henri ran some (very quick!) experiments in Pari/GP to see what happens in the elliptic curve case from which we settled on the formula above.

Monday, July 20

Fernando discussed some (number-theoretic) motivation for why we want to look at p-adic methods to compute L-functions. In particular, he gave an overview of how to count the number of points on the Legendre family of elliptic curves
E_\lambda: y^2=x(x-1)(x-\lambda)
using the p-adic gamma function. He also suggested an overarching strategy of computing traces of Frobenius using p-adic methods and without knowing anything about the geometric object on which Frobenius is acting.
Henri presented the p-adic gamma function, using a construction using the Hurwitz \zeta-function and Volkenborn integral that avoids viewing the gamma function initially as an interpolation. One can recover interpolation however with a little bit of care.