\input amstex
\documentstyle{amsppt}

\magnification=\magstephalf

\parskip=3pt 
\hsize=15true cm \vsize=22true cm \NoBlackBoxes \TagsOnRight \NoRunningHeads 
\UseAMSsymbols    \nologo

\def\R{\Bbb R} \def\Q{\Bbb Q} \def\C{\Bbb C} \def\Z{\Bbb Z} \def\p{\frak p}
\def\z{\zeta} \def\G{\Gamma} \def\U{\Cal O^\times} 

%\magnification=1200
\def \nr#1{{\noindent{\bf #1}\quad}}
\def \newline{{\par\noindent}}
\def \BZ{{\Bbb Z}}
\def \BN{{\Bbb N}}
\def \BQ{{\Bbb Q}}
\def \BA{{\Bbb A}}
\def \BR{{\Bbb R}}
\def \Sig{{\Cal S}}
\def \BC{{\Bbb C}}
\def \BX{{X}}
\def \CB{{\Cal B}}
\def \CO{{\Cal O}}
\def \SL{\text{SL}}
\def \Ker{\text{Ker}}
\def \Li{Li}
\def \Gal{\text{Gal}}
\def \rank{\text{rank}}
\def \sha{{\amalg\hskip -3.5pt\amalg}}
\def\Sh{\sha}
\def \isom {{\sim}}

\def \Qbar {{\overline{\Bbb Q}}}
\def \line{{\par\noindent}}
\def \Erd {{Erd\H os}}
\pageno=51

\topmatter\title Some problems related to polylogarithms\endtitle
\author D. Zagier\endauthor\endtopmatter\bigskip\bigskip


\vskip -1.2cm
\noindent
{\bf 1) \Erd-Stewart-Tijdeman for function fields }(cf~[2], pp.~390-391).
Let $F$ be a field, $S=\{p_1,\dots,p_s\}$ a set of primes.
We set
%$$\BX(S)=\{x\mid x,1-x \quad\text{are}\ \ S-\text{units}\}.$$
$$\BX(S)=\{x\mid \text{$x$ and $1-x$ are $S$-units}\}.$$

\line
{\sl Question} \ For a given value of $s=|S|$, how big can $|\BX(S)|$ be?
\medskip
\line
In the case $F=\BQ$ one knows that the number 
$N(s):=\max\limits_{|S|=s} |\BX(S)| $
is bounded above by $1000\cdot 50^s$ (Evertse) and below by
$e^{(2-o(1))\sqrt{s/\log s}}$ for $s$ large (\Erd-Stewart-Tijdeman),
and one expects that $N(s)=\exp(s^{\frac23+o(1)})\,.$  For number
fields, Evertse's result still holds (with an appropriately modified bound),
but the situation for the lower bound is less clear. (The proof 
of \Erd-Stewart-Tijdeman applies the pigeonhole principle to
the solutions of the equation $A+B=C$ in integers having only prime
factors from $S$, and seems to be specific to $\BQ$.) %\medskip\line
The really interesting question, however, is the function field case:
\medskip\line
{\sl Problem} \ Is there a lower bound for $|X(S)|$ in the case $F=\Q(t)$
which grows more than polynomially in $s$?
\line
A consequence of this would be the existence of non-trivial functional equations for 
polylogarithms at any level, since the number of conditions required to make an
element $\sum n_i\,[x_i]\in\Q[\BX(S)]$ belong to the $m$th Bloch group grows like $s^m$,
and an element of the $m$th Bloch group for $\Q(t)$ is a functional equation for $\Li_m(z)$.


\bigskip\noindent
{\bf 2) Ladders} (cf.~[2], pp.~387-389). For $\alpha\in \Qbar$, $\alpha$ not
a root of unity, define
$$\eqalign{\delta(\alpha)=\dim \Ker\bigg(\bigoplus_{i=0}^\infty \BZ\ &\to \ \Qbar^\times\bigg)\,,\cr
(c_i)_{i\ge 0} \ &\mapsto\ \alpha^{c_0}\prod_{i=1}^\infty(1-\alpha^i)^{c_i} \,.\cr}$$
It is known that this is always finite.
If $\ \delta(\alpha)\ge k \,[\BQ(\alpha)\colon\BQ]$ then one gets examples
of polylogarithmic relations (``ladders") up to order $m=2k+2$, the best example known 
being provided by Lehmer's algebraic number of degree 10, for which  $\delta(\alpha)$ is at least $\ge 73$ (Cohen-Lewin-Zagier, with 2 extra ones added by Niklasch), 
yielding polylogarithmic relations for all $\Li_m$ up to $m=16$.

\medskip\line
{\sl Problem} \ How big can $\delta(\alpha)$ be? Is there a global upper bound? Find examples with big
ratio ${\delta(\alpha)\over \deg(\alpha)}$, where $\deg(\alpha)$ denotes the degree of the minimal
polynomial of $\alpha$ over $\BQ$.
\bigskip\noindent
{\bf 3)  Rational conformal field theories.}
Let $A$ be a symmetric $n\times n-$matrix with rational coefficients. Then look
for $x\in \big(\BC-\{0,1\}\big)^n$ such that 
$$1-x\equiv x^A \quad \text{mod} \ (\BC^*)_{\text{tors}}\, ,$$
i.e. \ $1-x_i=\zeta_i\prod_j{x_j^{a_{ij}}}$ \ for all $i$, where $\zeta_i$ is a root of unity.
Since these are $n$ equations in $n$ unknowns, the  solution set is 
generically 0-dimensional, so that a typical solution $x=(x_1,\dots,x_n)$ belongs to $\Qbar^n$.
One immediately checks that $\sum_{i=1}^n{x_i\wedge(1-x_i)}=0$, so that
$\xi:=\sum_{i=1}^n{[x_i]}$ defines an element of the second Bloch group of $\Qbar$.
\medskip\line
{\sl Problem} \  Find (some, or all) matrices $A$ such that
\line i) { $\xi$ is zero in the Bloch group (i.e. $\xi$ is a linear combination of 5-term relations).}
\line
>From Borel's theorem and the known identification of the Bloch group and 
the Bloch-Wigner function 
$D$ with $K_3$ and the regulator, we see that i) is equivalent to 
$$\sum_{i=1}^n{D(x_i^\sigma)=0}\,,\qquad\forall\sigma\in \Gal(\Qbar/\BQ)\,.$$
Since $D\vert_\BR\equiv 0$, one has the related problem

\line i$'$) {find $A$ with solutions $(x_1,\dots,x_n)$ which are 
totally real algebraic numbers.}

\medskip\line
For instance, for $n=1$, $A=(a_{11})=({p\over q})$, the problem is to find
$p,q$ such that all roots of $x^p+x^q=1$ are real, and it is easily seen 
that the only solutions are \ $a_{11}=1,{1\over 2},2$.


\medskip\line
The problem arises in conformal field theory [1] where it is conjectured that the classification of so-called rational conformal field theories (which includes the famous 
ADE classification with Dynkin diagrams as a very special case) is {\sl equivalent} to the 
classification of matrices $A$ satisfying i) or i$'$).
%\medskip\line
A third property which is believed to be equivalent to these two is that
for some vector $b\in \BQ^n$ and number $c\in\BQ$ the series
$$f_{A,b,c}(q):=\sum_{r_1,\dots,r_n=0}^{\infty}{{q^{{1\over 2}r A r^t + b r^t + c}
\over (q)_{r_1}\cdots (q)_{r_n} }}\ ,\qquad (q)_r:=(1-q)(1-q^2)\cdots (1-q^r),$$
is a modular function. In the example $A=(1)$ 
%$({1\over 2})$ and $(2)$
above, the corresponding values of $(b,c)$ are
$(0,-\frac{1}{48})$, $(\frac12,\frac{1}{24})$, and $(-\frac12,\frac{1}{24})$, 
%and as a typical example we give the modular function for the first of these
the modularity of (for instance) the first of these following from the classical identity
$$f_{1,0,-\frac1{48}}(q)\,=\,\frac{\eta(\tau)^2}{\eta(\tau/2)\eta(2\tau)}\,,$$
where $\eta(\tau)$ denotes the Dedekind eta function and $q=e^{2\pi i \tau}$.
%For $A=(2)$, the values $b=0$ and $b=1$ yield the well-known Rogers-Ramanujan identities.
%For $A=(2)$, the pairs $(b,c)=(0,-\frac1{60})$ and $(1,\frac{11}{60})$ 
% come from the well-known Rogers-Ramanujan identities.
%For $A=(2)$ the modularity of $f_{A,b,c}$ for $(b,c)=(0,-\frac1{60})$ or $(1,\frac{11}{60})$ 
%follows from the well-known Rogers-Ramanujan identities.
For $A=(2)$ the function $f_{A,b,c}$ is modular for $(b,c)=(0,-\frac1{60})$ or $(1,\frac{11}{60})$ 
by virtue of the well-known Rogers-Ramanujan identities.

\bigskip\noindent
{\bf 4)  $K-$groups by hand.} Show ``by hand" that the second Bloch group of $\BQ$ is torsion
%$$ker\bigg( \BZ(\BQ)\to\Lambda^2(\BQ^*)\bigg) \,,$$
%$\delta(x)=\cases{0&,$x=0,1,$\cr x\wedge(1-x)&else\cr}$ is torsion,
i.e.~any linear combination $\sum{}n_i[x_i]$, ($n_i \in \BQ$, $x_i\in \BQ-\{0,1\}$) which satisfies
$$\sum{}n_i\big(x_i\wedge (1-x_i)\big)=0$$ 
is a rational combination of 5-term relations.
\line
(This corresponds to the fact that $K_3(\BQ)\otimes \BQ=0$. The analogous problem for the 
vanishing of $K_2(\BQ)\otimes \BQ$ would be to show that any element in $\bigwedge^2(\BQ^\times)$ 
is a rational linear combination of elements $x\wedge(1-x)$,  and this {\it does} have an 
elementary solution. See Milnor's book on algebraic $K-$theory, chap.11.) 


\bigskip\noindent \def\l{\;\longleftrightarrow\;}
{\bf 5) Enlarged Mordell-Weil group.} There is a well-known analogy
$$\aligned
 \text{(elliptic curve)}\quad E/\BQ &\l  F
    \quad \text{(number field)}\\
\text{(rational points)}\quad E(\BQ)&\l \CO_F^\times
    \quad \text{(unit group)}\\
\text{(Shafarevich group)}\quad\sha&\l Cl(F)\ 
    \quad\text{(class group)}\\
\text{(integral points)}\quad E(\BZ)&\l \BX(\emptyset)
    \quad \text{(exceptional units)}.\endaligned\tag*$$
(Here ``integral points" means integral points of an affine model of $E$
and ``exceptional units" are elements $u$ of $F$ such that both $u$ and $1-u$
are units.) This analogy is
most visible and best known in the context of the Birch-Swinnerton-Dyer 
conjecture, in which the rank, covolume, and torsion subgroup of $E(\Q)$ and 
the (conjecturally finite) order of $\Sh$ enter in the same way as the
corresponding invariants of $\U_F$ and the order of $\text{Cl}_F$ enter
the Dirichlet class number formula. 
\line
{\sl Problem} Can one find natural extensions of $E(\BQ)$, each of finite rank but with the ranks not bounded, which correspond under this analogy to the 
groups $\CO_{F,S}^\times$=$\{S$-units$\}$, $S$ a finite set of prime ideals
in $F$?
One aspect of the analogy $(*)$ is that the  4-term exact sequence
$$1\to\U_F\to F^\times\to\bigoplus_\p\Z\to\text{Cl}_F\to0 \tag1$$
(sum over prime ideals $\p$ of $\Cal O_F$), can be used as a model for
the sequence
$$0\longrightarrow E(\BQ) \longrightarrow E(\BA) \longrightarrow
\text {Form}( E/\BA,\BQ)
\longrightarrow \sha \longrightarrow 0 \,\tag2$$
where $\BA$ denotes the ring of adeles of $\BQ$ and $\text {Form}( E/\BA,\BQ)$
is the group of $\BQ$-forms of $E/\BA$, i.e.~
$\Q$-isomorphism classes of pairs $(C,\phi)$ where $C$ is a curve 
defined over $\BQ$ and $\phi$ an isomorphism $E\to C$ over $\BA$ (cf.~ [3]).
The groups $\CO_{F,S}^\times$ lie in the second group of (1) and eventually
exhaust it, so it might be reasonable to try to solve the problem by taking 
some subgroups of $E(\BA)$. What we {\sl don't} want is to enlarge $E(\Q)$
to $E(K)$, $[K:\BQ]< \infty$, eventually ending up with $E(\Qbar)$ rather 
than $E(\BA)$. That would be analogous to replacing $\CO_{F,S}^\times$
by the unit group of larger fields, rather than to non-units of the original
field $F$.  %\smallskip\line
The problem can therefore be seen as the elliptic analogue to that of 
finding the whole of a number field $F$ if we have been given only the 
units (say, numerically).

\smallskip
The connection (and potential application) to polylogarithms is as follows.
  % \subheading{Potential application and connection to polylogarithms}
Generalizing the Bloch-Wigner dilogarithm function 
$D(z)=\Im\bigl({Li}_2(z)+\log|z|\log(1-z)\bigr)$ one has
the elliptic dilogarithm function $D_E(z)=\sum_{n\in\Z}D(q^nz)$
for a point $z\in\C^\times/q^{Z}=E(\C)$ on an elliptic curve $E/\C$.
If $E$ is defined over $\Q$ and $\sum_i n_iP_i$ is a formal integral 
linear combinations of points of $E(\Q)$ satisfying certain auxiliary 
conditions (found experimentally by H\. Cohen and myself and then 
corrected and refined by Rolshausen, Wildeshaus, Goncharov and Levin), 
then $\sum_i n_iD_E(P_i)$ is conjecturally a rational multiple of the
value at $s=2$ of the Hasse-Weil zeta function at $s=2$, and there
are similar conjectural formulas, involving higher elliptic polylogarithms,
for other special values of $L$-series associated to $E$.  The problem
with these conjectures is that, unlike the situation for number fields
where there are (conjecturally, and also experimentally) 
always enough linear combination of elements of the field to
satisfy the conditions defining the higher Bloch groups, and hence to
give special values of the zeta function of the field in terms of
polylogarithms, here it very often happens that  
  % $E(\Q)$ is too small to contain any linear com
$E$ has too few rational points to yield linear combinations
satisfying all the requisite conditions.  For example, $E(\Q)$ may be
the trivial group, in which case one certainly cannot find examples,
or $E(\Q)$ may have rank 1 but with a generator whose multiples all 
contain different prime factors, in which case one will again not be
able to satisfy the needed conditions.
 % (In the rank 1 case one needs at leat $n+3$ rational points which
 % are $p$-integral for $p$ outside some some set of $n$ primes 
This is analogous to the situation we would have in the case of a number
field $F$ if we tried to find elements $\sum n_i[x_i]\in\Z[F]$ with
all $x_i$ and $1-x_i$ integral outside a given finite set of primes $S$:
the number of such $x_i$ is always finite, and if it happened to be too
small then we could not satisfy all the conditions defining the $m$th
Bloch group. (The number of such conditions grows roughly like $|S|^m$.)
The solution there would be obvious: simply replace the set $S$ by a
bigger one, thus  increasing both the rank of the finitely generated
group $\U_{F,S}$ and the cardinality of the finite set
$X(S)=\{x\in F\mid x,\,1-x\in\U_{F,S}\}$; doing this also
increases the number of conditions that have to be satisfied, but the
size of $X(S)$ grows faster and one eventually wins the race.  Here we
would like to do the same thing, but are stymied by the lack of a
natural extension of $E(\Q)$ analogous to the extension $\U_{F,S}$
of $\U_F$.  If such extensions exist, then it is reasonable to
expect that the elliptic dilogarithm function $D_E$ will be defined on
these groups also and that for sufficiently large extensions one will be
able to find as many combinations as one wants which belong to the image
of the Beilinson-Bloch regulator map and hence conjecturally give special
values of the $L$-series associated to $E$ and its symmetric powers.
\medskip
\Refs\nofrills{References}

\widestnumber\key{M}
\ref \key  1   \by Nahm, W., Recknagel, A.,Terhoeven, M., \paper Dilogarithm identities in conformal field theory,  \jour Modern Phys. Lett. A,
 8,  no. 19 \yr 1993 \pages 1835--1847\endref  

\ref \key  2   \by Zagier, D. \paper Special values and functional equations of
polylogarithms, {\rm Appendix A \ in } Structural properties of polylogarithms\publ (L.~Lewin ed.), AMS 
monographs\publaddr Providence \yr 1991\pages 377--400\endref

\ref \key   3   \by Zagier, D. \paper The Birch--Swinnerton-Dyer conjecture from a na\"\i ve point of view,
{\rm in } Arithmetic Algebraic Geometry
\publ (G.~v.d.Geer, F.~Oort, J.~Steenbrink, eds.), Prog. in Math. 89, Birkh\"auser \publaddr 
Boston \yr 1991\pages 377--389\endref


\bye



====



\subheading{Question}  Let $E$ be an elliptic curve over $\Q$. Can one extend
$E(\Q)$ to larger groups, of finite but unbounded ranks, in some sensible way?
This question is suggested by
the analogy between the Mordell-Weil group $E(\Q)$ of $E/\Q$
and the unit group $\U_F$ of a number field and of the Shafarevich group
$\Sh(E/\Q)$ with the class group $\text{Cl}_F$.  This analogy is of course
most visible and best known in the context of the Birch-Swinnerton-Dyer 
conjecture, in which the rank, covolume, and torsion subgroup of $E(\Q)$ and 
the (conjecturally finite) value of $|\Sh|$ enter in the same way as the
corresponding invariants of $\U_F$ and the order of $\text{Cl}_F$ enter
the Dirichlet class number formula.  It has other aspects, however.  For
instance, the finite set of elements $x\in\U_F$ for which $1-x$ also belongs
to $\U_F$, the so-called exceptional units of $F$, look in many ways like
the integral points on $E$, as was was noticed several years ago by
H\. Cohen and others.  In another direction, the groups $\U_F$ and 
$\text{Cl}_F$ fit naturally into a 4-term exact sequence
$$1\to\U_F\to F^\times\to\bigoplus_\p\Z\to\text{Cl}_F\to0 \tag*$$
(sum over prime ideals $\p$ of $\Cal O_F$), and there are similar
4-term exact sequences for $E(\Q)$ and $\Sh$, as pointed out in [Znaive].
In the number field case one can relax the condition ``$x$ is a unit" to
``$x$ is a $\p$-unit except at finitely many primes $p$", i.e., replace
$\U_F$ by the larger group $\U_{F,S}$ of $S$-units for some finite set
of primes $S$.  The groups $\U_{F,S}$ are still of finite rank, but their
ranks grow as $S$ does and they eventually fill up the whole group $F^\times$,
(= the second term of the sequence (*)). The question is whether there is
an analogous construction which works for $E(\Q)$.





It is well-known that, for each set of primes $S$ in $F$, $\rank\big(\CO_{F,S}^\times\big)=r_1+r_2+|S|-1$. So 
we can exhaust $F^\times=\bigcup_{S}\CO_S^\times$.
\line
Try to imitate the well-known exact sequence
$$0\longrightarrow \CO_F^\times \longrightarrow F^\times \longrightarrow \bigoplus_{\frac P}{\BZ}
\longrightarrow Cl_F \longrightarrow 0 $$
for elliptic curves, i.e.~try to find a certain series of ``extensions'' of $E(\BQ)$ which exhausts some arithmetic 
object such that
$$0\longrightarrow E(\BQ) \longrightarrow ?? \longrightarrow ???
\longrightarrow \sha \longrightarrow 0 \,.$$
Several candidates have been given in [Z].

?? should be connected to $E(\BR)\times\prod_p{E(\BQ_p)}$ or maybe only $E(\BR)$, the role of the
regulator should be played by the elliptic polylog.
\line
[Suppose someone has told you the units of a number field $\CO_F$ in
$\BC^\times$. Try to deduce 
$F^\times$...]
\medskip








\bye



\bigskip\noindent
{\bf 6) Uniqueness of Dirichlet series.} Assume a ``Gamma factor'' $\gamma(s)$ is given, together
with a weight $k\in \BN$ and effective constants $C_1,C_2$.
\medskip\noindent
{\sl Problem} \ Does there exist an $n_0$ such that any $\varphi(s)=\sum_{n\ge no}{a_n\over n^s}$
%with coefficients restricted by 
which fulfills (i) and (ii)
\item{(i)} $|a_n|\le C_1 n^{C_2}$, $a_n\in \BC$,
\item{(ii)} $\gamma(s) \varphi(s)$ is holomorphic and is invariant under $s\to k-s$,
has to vanish identically?
\medskip\noindent
Example: $\gamma(s)=(2\pi)^{-s}\Gamma(s)$ 
$$ f(z)=\sum_{n=1}^{\infty}a_n e^{2\pi i n z}\ \in S_k(\SL_2(\BZ))$$
a cusp form, then it is well known that $n_0={k\over 12}$ is enough.
\bigskip

\bye


\serieslength=80

try(a,b)=s=1; l=1+o(q)^80; for(r=1,20,l=l-l*q^(4*r);s=s+q^(a*r^2+b*r)/l); s

try(a,b)=s=1; l=1+o(q)^80; for(r=1,20,l=l-l*q^(2*r);s=s+q^(a*r^2+b*r)/l); s



try(a,b)=s=1; l=1+o(q)^80; for(r=1,3,l=l-l*q^(4*r);print(r,"  ",l,"   ",s);s=s+q^(a*r^2+b*r)/l); s


pr(f)=v=vector(60,n,0);for(n=1,60,v[n]=coeff(f,n);f=f*(1-q^n+o(q)^80)^v[n]); v



v=pr(try(1,0)*eta(q^4)/eta(q^10)); \
  for(n=1,60,print1(v[n]," ");if(n%10,,print("   ")))
1 -1 0 0 0 0 0 -1 1 0    
1 -1 0 0 0 0 0 -1 1 0    
1 -1 0 0 0 0 0 -1 1 0    
1 -1 0 0 0 0 0 -1 1 0    
1 -1 0 0 0 0 0 -1 1 0    
1 -1 0 0 0 0 0 -1 1 0    


? pr(try(2,0)/eta(q^4)^2*eta(q^8)*eta(q^2))
%129 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

? pr(try(2,2)*eta(q^4)/eta(q^8))
%147 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

===== own tries
try(a,b)=s=1; l=1+o(q)^80; for(r=1,20,l=l-l*q^(r);s=s+q^(a/2*r^2+b*r)/l); s





? try(2,0)                                                                    
%49 = 1 + q + q^2 + q^3 + 2*q^4 + 2*q^5 + 3*q^6 + 3*q^7 + 4*q^8 + 5*q^9 + 6*q^1
0 + 7*q^11 + 9*q^12 + 10*q^13 + 12*q^14 + 14*q^15 + 17*q^16 + 19*q^17 + 23*q^18
 + 26*q^19 + 31*q^20 

===
f(b,a)=s=1; ll=1+o(q)^30; l=1+o(q)^30;  for(r=0,20, ll=ll-ll*a*q^r);\
 for(j=1,20,l=l*(1-q^j)*(1-a*q^j);  s=s+b^j*q^(j*j)/l);ll*s

len=40;
f(a,b)=s=1+o(q^len);for(k=1,len,s= s+(a*b)^(k*(k-1)/2)*(a^k+b^k)); s

len=40;
f(a,b)=s=1+o(q^len);for(k=1,len,s= s+(a*b)^(k*(k-1)/2)*(a^k+b^k)); s




