CoCalc Public Fileswww / talks / harvard-talk-2004-12-08 / e2heights / cycloht.tex
Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
1\documentclass[11pt]{article}
2\bibliographystyle{amsalpha}
3\include{macros}
4\title{Computing $p$-adic Sigma Functions and Cyclotomic Heights}
5\author{Barry Mazur, William Stein, John Tate}
6\begin{document}
7\maketitle
8
9\section{Introduction}
10Fix an elliptic curve $E$ over $\Q$, given by a minimal
11Weierstrass equation
12\begin{equation}\label{eqn:weq}
13 y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6.
14\end{equation}
15Suppose $p$ is a prime
16of good ordinary reduction for $E$, so $$p \nmid N_E \cdot a_{E,p},$$
17where $N_E$ is the conductor of $E$ and $a_{E,p}=p+1-\#E(\F_p)$.
18
19\section{The Canonical $p$-adic sigma function}
20Let
21\begin{equation}\label{eqn:sigma}
22  \sigma(t) = t + s_1t^2 + s_2 t^3 + s_3 t^4 + \cdots
23       \in t (1+t\Z_p[[t]])
24\end{equation}
25be the canonical $p$-adic $\sigma$-function attached to $E$
26as in \cite{mazur-tate:sigma}.
27
28\subsection{Computing $\sigma(t)$}
29
30[[Describe the resultin algorithm here -- method to compute $\sigma$.]]
31
32
33Theorem 3.1 of \cite{mazur-tate:sigma} asserts that if $m\in \Z$ and
34$Q\in E^f(\overline{R})$, then
35\begin{equation}\label{eqn:sigma_mq}
36\sigma(m Q) = \sigma(Q)^{m^2} \cdot f_m(Q).
37\end{equation}
38We use this relation
39theorem to determine $s_1$ and $s_2$ in terms of the formal
40endomorphism $[p]$ and the $p$th division polynomial.  [[Idea -- maybe
41we could just use $m=2$ instead of $m=p$?!  It's probably easier to
42compute $[2]$ and $f_2$ than to compute $[p]$ and $f_p$; if not $m=2$,
43then $m=3$?]]  Then we use knowledge of $s_1$ and $s_2$ and the
44differential equation satisfied by $\sigma$ to determine the constant
45$c\in\Z_p$ such that $x(t) - c$ is the canonical eks'' (pronounced
46like the letter x'') function of
47\cite{mazur-tate:sigma}.  Finally, given $x(t) - c$, we find the
48unique series $\sigma(t) \in t(1+t\Z_p[[t]])$
49that satisfies the differential equation
50\begin{equation}\label{eqn:de1}
51  x(t) - c = -\frac{d}{\omega}\left(\frac{1}{\sigma}
52       \frac{d\sigma}{\omega}\right)
53\end{equation}
54of \cite[pg.~665]{mazur-tate:sigma}.
55
56Let $\omega = dx/(2y+a_1x + a_3)$ be the invariant differential
57on $E$.  Inverting the power series in equation 15 of
58\cite{tate:arithmetic}, we have
59$$60 \frac{dt}{\omega} = 611 - a_1 t - a_2 t^2 - 2a_3 t^3 + (-2 a_1 a_3 - 2 a_4)t^4 + \cdots 62 = \sum b_n t^n, 63$$
64where the $a_i$'s are as in (\ref{eqn:weq}).
65\comment{
66> K<a1,a2,a3,a4> := FieldOfFractions(PolynomialRing(Q,4));
67> R<t> := PowerSeriesRing(K);
68> f := 1 + a1*t + (a1^2+a2)*t^2 + (a1^3+2*a1*a2+2*a3)*t^3 + (a1^4+3*a1^2*a2+6*a1*a3+a2^2+2*a4)*t^4 + O(t^5);
69> 1/f;
701 - a1*t - a2*t^2 - 2*a3*t^3 + (-2*a1*a3 - 2*a4)*t^4 + O(t^5)
71}
72Using the chain rule, etc., and writing
73$\sigma(t) = \sum_{n\geq 1} s_{n-1} t^n$ (as in (\ref{eqn:sigma})),
74we see that
75the differential equation (\ref{eqn:de1}) written in terms of~$t$
76is
77\begin{equation}\label{eqn:de_big}
78\frac{d}{dt}\left(
79 \frac{\left(\sum_{n\geq 1} n s_{n-1} t^{n-1}\right) \cdot
80       \left(\sum_{n\geq 0} b_n t^n\right)}{\sum_{n\geq 1} s_{n-1} t^n}
81  \right)\cdot \sum_{n \geq 0} b_n t^n  = c - x(t).
82\end{equation}
83Here
84$$85 x(t) = \frac{1}{t^2} - \frac{a_1}{t} - a_2 - a_3 t - (a_4+a_1 a_3)t^2 + \cdots 86$$
87is the formal expansion of $x$ in terms of $t$.
88\comment{
89> K<a1,a2,a3,a4,s1,s2,s3,s4,s5> := FieldOfFractions(PolynomialRing(Q,9));
90> R<t> := LaurentSeriesRing(K);
91> w := 1 - a1*t - a2*t^2 - 2*a3*t^3 + (-2*a1*a3 - 2*a4)*t^4 + O(t^5);
92> sig := t + s1*t^2 + s2*t^3 + s3*t^4 + s4*t^5 + s5*t^6 + O(t^7);
93> sigprime := Derivative(sig);
94> Derivative(((sigprime*w)/sig))*w;
95-t^-2 + a1*t^-1 + -a1*s1 - s1^2 + 2*s2 + (a1^2*s1 + a1*a2 +
96    3*a1*s1^2 - 6*a1*s2 - 2*a2*s1 - 2*a3 + 2*s1^3 - 6*s1*s2 +
97    6*s3)*t + (-2*a1^2*s1^2 + 4*a1^2*s2 + 3*a1*a2*s1 - 5*a1*s1^3
98    + 15*a1*s1*s2 - 15*a1*s3 + a2^2 + 4*a2*s1^2 - 8*a2*s2 -
99    6*a3*s1 - 4*a4 - 3*s1^4 + 12*s1^2*s2 - 12*s1*s3 - 6*s2^2 +
100    12*s4)*t^2 + O(t^3)
101}
102Expanding the left hand of (\ref{eqn:de_big})
103we see that
104\begin{align*}
105&-\frac{1}{t^2} + \frac{a_1}{t} + (-a_1 s_1 - s_1^2 + 2s_2)
106      +\\
107&\quad (a_1^2 s_1 + a_1 a_2 +
108    3 a_1 s_1^2 - 6 a_1 s_2 - 2 a_2 s_1 - 2 a_3 + 2 s_1^3 - 6 s_1 s_2 +
109    6 s_3) t +\\
110&\quad (-2 a_1^2 s_1^2 + 4 a_1^2 s_2 + 3 a_1 a_2 s_1 - 5 a_1 s_1^3
111    + 15 a_1 s_1 s_2 - 15 a_1 s_3 + a_2^2 + 4 a_2 s_1^2 - \\
113    6 a_3 s_1 - 4 a4 - 3 s_1^4 + 12 s_1^2 s_2 - 12 s_1 s_3 - 6 s_2^2 +
114    12 s_4) t^2 + \cdots
115 \\
116&=
117c - \left(\frac{1}{t^2} - \frac{a_1}{t} + - a_2 - a_3 t - (a_4+a_1 a_3)t^2 + \cdots\right).
118\end{align*}
119Equating constant coefficients, we see that
120$$121 c = a_2 - a_1 s_1 - s_1^2 + 2s_2. 122$$
123We will not use the higher terms yet...
124
125Next we use the relation (\ref{eqn:sigma_mq}) to compute
126$s_1$ and $s_2$ in terms of $[p](t)$ and $f_p(t)$.
127Write the formal multiplication by $p$ as
128$$129 [p](t) = pt + r_2 t^2 + r_3 t^3 + \cdots \in \Z_p[[t]] 130$$
131and the formal division polynomial'' $f_p(t)$ as
132$$133 f_p(t) = pt^{1-p^2} + d_1 t^{2-p^2} + d_2 t^{3-p^2} + \cdots 134 \in (\Z_p[[t]])[1/t]. %\left[\frac{1}{t}\right]. 135$$
136The identity (\ref{eqn:sigma_mq})
137is
138$$139\sum_{n\geq 1} s_{n-1}\cdot 140\left(pt + r_2 t^2 + r_3 t^3 + \cdots \right)^n 141 =\qquad\qquad\qquad\qquad$$
142$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{p^2} \cdot 143 \left(pt^{1-p^2} + d_1 t^{2-p^2} + d_2t^{3-p^2} + \cdots\right). 144$$
145Multiply both sides by $t^{p^2-1}$, so that both sides are
146in $\Z_p[[t]]$.
147$$148\sum_{n\geq 1} s_{n-1}\cdot 149\left(pt + r_2 t^2 + r_3 t^3 + \cdots \right)^n\cdot t^{p^2-1} 150 =\qquad\qquad\qquad\qquad$$
151$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{p^2} \cdot 152 \left(p + d_1 t + d_2 t^2 + \cdots\right). 153$$
154
155
156[[Since we have this for all $m$, I'm going to replace $p$ by $2$ and
157see what happens.  Also, $f_2(t) = 2t^{1-2^2} + \cdots$ and
158$[2](t)=2t+r_2t^2+\cdots$.]]
159$$160t^3 \cdot \sum_{n\geq 1} s_{n-1}\cdot 161\left(2t + r_2 t^2 + r_3 t^3 + \cdots \right)^n 162 =\qquad\qquad\qquad\qquad$$
163$$\qquad\qquad\qquad\left(\sum_{n \geq 1} s_{n-1} t^n\right)^{4} \cdot 164 \left(2 + d_1 t + d_2 t^2 + \cdots\right). 165$$
166Expanding both sides we get
167$$168 2 t^4 + (4 s_1 + r_2) t^5 + (4 s_1 r_2 + 8 s_2 + r_3) t^6 + \cdots = 169\qquad\qquad\qquad 170$$
171$$172\qquad\qquad\qquad 1732 t^4 + (8 s_1 + d_1) t^5 + (12 s_1^2 + 4 s_1 d_1 + 8 s_2 + d_2) t^6 + \cdots. 174$$
175Equating coefficients of $t^5$ we see that
176$$177 4 s_1 + r_2 = 8 s_1 + d_1 178$$
179so
180$$181 s_1 = \frac{r_2 - d_1}{4}. 182$$
183Equating coefficients of $t^6$ shows that
184$$1854 s_1 r_2 + 8 s_2 + r_3 = 186 12 s_1^2 + 4 s_1 d_1 + 8 s_2 + d_2, 187$$
188so we get NO INFORMATION ABOUT $s_2$ from this.
189
190[[Observation: Actually one could compute $\sigma$ directly from
191equating coefficients!!!  See the commented MAGMA session below, where
192it appears to me that one can simply read off the other coefficients
193of $\sigma$ from (\ref{eqn:sigma_ma}) with $m=2$.]]
194
195However, we do get information from the coefficient of $t^7$.
196Taking differences, we have that
197\begin{verbatim}
1980 = (-4*s1 - d1 + r2)*t^5 + (-12*s1^2 - 4*s1*d1 + 4*s1*r2 - d2 +
199    r3)*t^6 + (-8*s1^3 - 6*s1^2*d1 - 24*s1*s2 - 4*s1*d2 + s1*r2^2
200    + 4*s1*r3 - 4*s2*d1 + 12*s2*r2 + 8*s3 - d3 + r4)*t^7 +
201    (-2*s1^4 - 4*s1^3*d1 - 24*s1^2*s2 - 6*s1^2*d2 - 12*s1*s2*d1 -
202    24*s1*s3 - 4*s1*d3 + 2*s1*r2*r3 + 4*s1*r4 - 12*s2^2 - 4*s2*d2
203    + 6*s2*r2^2 + 12*s2*r3 - 4*s3*d1 + 32*s3*r2 + 24*s4 - d4 +
204    r5)*t^8 + O(t^9)
205\end{verbatim}
206Thus
207$$208-8 s_1^3 - 6 s_1^2 d_1 - 24 s_1 s_2 - 4 s_1 d_2 + s_1 r_2^2 209 + 4 s_1 r_3 - 4 s_2 d_1 + 12 s_2 r_2 + 8 s_3 - d_3 + r_4 = 0. 210$$
211Unfortunately we have both $s_1$ and $s_3$ together
212in this formula, so this doesn't determine $s_2$.
213Using $m=-1$ we get
214\begin{verbatim}
2150 = (2*s1 - d1 + r2)*t^2 + (-s1*d1 - 2*s1*r2 - d2 + r3)*t^3 + (-s1*d2
216    + s1*r2^2 - 2*s1*r3 - s2*d1 + 3*s2*r2 + 2*s3 - d3 + r4)*t^4 +
217    (-s1*d3 + 2*s1*r2*r3 - 2*s1*r4 - s2*d2 - 3*s2*r2^2 + 3*s2*r3
218    - s3*d1 - 4*s3*r2 - d4 + r5)*t^5 + O(t^6).
219\end{verbatim}
220The coefficient of $t^4$ gives a different linear equation for $s_2$
221and $s_3$, and probably taking two of these together we'll be able to
222solve for both.
223
224For $m=3$ it's:
225\begin{verbatim}
226(-18*s1 - d1 + r2)*t^10 + (-108*s1^2 - 9*s1*d1 + 6*s1*r2 - d2 +
227    r3)*t^11 + (-252*s1^3 - 36*s1^2*d1 - 216*s1*s2 - 9*s1*d2 +
228    s1*r2^2 + 6*s1*r3 - 9*s2*d1 + 27*s2*r2 + 54*s3 - d3 +
229    r4)*t^12 + (-378*s1^4 - 84*s1^3*d1 - 756*s1^2*s2 - 36*s1^2*d2
230    - 72*s1*s2*d1 - 216*s1*s3 - 9*s1*d3 + 2*s1*r2*r3 + 6*s1*r4 -
231    108*s2^2 - 9*s2*d2 + 9*s2*r2^2 + 27*s2*r3 - 9*s3*d1 +
232    108*s3*r2 + 216*s4 - d4 + r5)*t^13 + O(t^14)
233\end{verbatim}
234
235\subsection{Computing formal multiplication by $p$}
236
237\subsection{Computing division polynomials}
238
239\section{The $p$-adic cyclotomic height}
240Let $E$ and $p$ be as in the introduction.
241
242Suppose $Q \in E(\Q)$ is a point that reduces to the point at infinity
243modulo $p$, and lies in the identity component modulo every bad prime.
244For example, if $M$ is the least common multiple of $\#E(\F_p)$ and
245the Tamagawa numbers $c_\ell$ of $E$ at bad primes, then every element
246of $M E(\Q)$ satisfies our hypothesis on $Q$.
247Since ...??, we can write
248$$249 Q=(x,y)=\left(\frac{a}{e^2},\frac{b}{e^3}\right), 250$$
251with $a,b,e\in\Z$ and $\gcd(a,e)=1$.  Let $t=-x/y$. Then the $p$-adic
252cyclotomic height $h_p(Q)$ is
253\begin{equation}\label{eqn:hpq}
254   h_p(Q) = \log_p\left(\frac{\sigma(t)}{e}\right).
255 \end{equation}
256[[I think exactly this formula appears on the upper right page of a preprint
257Tate was holding yesterday, which perhaps I don't have a copy.]]
258Here
259$$260 \log_p : \Q_p^* \to \Q_p 261$$
262is the the unique extension of the usual $p$-adic logarithm
263on $1+p\Z_p$ to a homomorphism from the multiplicative group
264of $\Q_p^*$ to the additive group of $\Q_p$.  Thus if $au \in \Q_p^*$
265with $u\in 1 + p\Z_p$, then $\log_p(au) = \log_p(u)$, and
266one computes $\log_p(u)$ using the formula
267$$268 \log_p(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots\right) 269$$
270with $x=1-u$.
271[[Again, there's a discussion of this in a paper Tate was holding,
272that I don't seem to have.]]
273
274The cyclotomic height $h_p(Q)$ extends uniquely to a $\Z_p$-valued
275function on $E(\Q)$ that satisfies $h_p(nQ) = n^2 Q$ for all
276integers $n$.   If $M$ is as above, then
277we can compute $h_p(P)$ for any $P\in E(\Q)$ by computing
278$$279 h_p(P) = \frac{h_p([M]P)}{M^2} = \frac{1}{M^2}\cdot 280\log_p\left(\frac{\sigma(t)}{e}\right), 281$$
282where $t=-x/y$ and $e$ are for $Q=[M]P=(x,y)=(a/e^2,b/e^3)$.
283
284
285
286\bibliography{biblio}
287\end{document}
288
289
290Using $m=2$ still,
291taking the difference of the two sides and setting the
292$s_{2n}=0$, since $\sigma$ is odd, we see that
293\begin{verbatim}
2940 = (-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
295    r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
296    - d4 + r5)*t^8 + O(t^9)
297\end{verbatim}
298\comment{
299> K<s1,s2,s3,s4,s5,d1,d2,d3,d4,d5,r2,r3,r4,r5> := FieldOfFractions(PolynomialRing(Q,14));
300> R<t> := LaurentSeriesRing(K);
301> sig := t + 0*t^2 + s2*t^3 + 0*t^4 + s4*t^5 + 0*t^6 + O(t^7);
302> m:=2; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6); divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
303(-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
304    r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
305    - d4 + r5)*t^8 + O(t^9)
306}
307The coefficient of $t^7$ determines $s_2$, so long as
308$r_2 - 3d_1\neq 0$.  Since $r_2 = d_1$, we're OK, at least as
309long as $d_1\neq 0$.
310So we get
311$$312 s_2 = \frac{d_3-r_4}{8d_1}. 313$$
314Now that we know $s_2$, the coefficient of $t^8$ determines
315$s_4$.  Etc., and it seems likely we can easily determine $\sigma(t)$
316from the formal $[2]$ and $f_2(t)$, without having to mess with
317any differential equations.   Surprising if right.
318
319Here is a relevant MAGMA session, which suggests that the
320information coming from (\ref{eqn:sigma_ma}) for different $m$ is
321equivalent:
322
323\begin{verbatim}
324> K<s1,s2,s3,s4,s5,d1,d2,d3,d4,d5,r2,r3,r4,r5> :=
325  FieldOfFractions(PolynomialRing(Q,14));
326> R<t> := LaurentSeriesRing(K);
327> sig := t + 0*t^2 + s2*t^3 + 0*t^4 + s4*t^5 + 0*t^6 + O(t^7);
328> m:=2; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
329  divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
330  t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
331(-d1 + r2)*t^5 + (-d2 + r3)*t^6 + (-4*s2*d1 + 12*s2*r2 - d3 +
332    r4)*t^7 + (-12*s2^2 - 4*s2*d2 + 6*s2*r2^2 + 12*s2*r3 + 24*s4
333    - d4 + r5)*t^8 + O(t^9)
334> m:=3; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
335  divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
336  t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
337(-d1 + r2)*t^10 + (-d2 + r3)*t^11 + (-9*s2*d1 + 27*s2*r2 - d3 +
338    r4)*t^12 + (-108*s2^2 - 9*s2*d2 + 9*s2*r2^2 + 27*s2*r3 +
339    216*s4 - d4 + r5)*t^13 + O(t^14)
340> m:=5; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
341  divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
342  t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
343  (-d1 + r2)*t^26 + (-d2 + r3)*t^27 + (-25*s2*d1 + 75*s2*r2 - d3 +
344    r4)*t^28 + (-1500*s2^2 - 25*s2*d2 + 15*s2*r2^2 + 75*s2*r3 +
345    3000*s4 - d4 + r5)*t^29 + O(t^30)
346> m:=1; mult := m*t + r2*t^2+r3*t^3+r4*t^4+r5*t^5 + O(t^6);
347  divpol := m+d1*t+d2*t^2+d3*t^3+d4*t^4+d5*t^5+O(t^6);
348  t^(m^2-1)*Evaluate(sig,mult) - sig^(m^2)*divpol;
349(-d1 + r2)*t^2 + (-d2 + r3)*t^3 + (-s2*d1 + 3*s2*r2 - d3 +
350    r4)*t^4 + (-s2*d2 + 3*s2*r2^2 + 3*s2*r3 - d4 + r5)*t^5 +
351    O(t^6)
352\end{verbatim}
353Thankfully, we can't use $m=1$ since $d_1=0$ there.
354