CRM Talk and Handout

Jon Borwein

July 10, 1997

Evaluations of multi-dimensional

polylogarithmic sums: a compendium of

results for arbitrary depths

Jonathan Borwein, Simon Fraser University

C E C M

Centre for Experimental &Constructive Mathematics

May 20, 1997 CRM Montreal

Atelier: Math�ematiques experimentales et

combinatoire

Joint work: with David Broadhurst (OU)

David Bradley (CECM)

Roland Girgensohn (Munich)

Petr Lisonek (CECM)

Papers at: http://www.cecm.sfu.ca/preprints/

1

ABSTRACT.This talk and papers concern

sums of the type

�(a; b; c) :=

1Xx=1

x�1Xy=1

y�1Xz=1

1

xaybzc;

and their n-variable analogues and polyloga-

rithmic extensions.

� For which integers a; b; c are these sums re-

ducible: expressible in terms of the simpler se-

ries

�(a; b) :=

1Xx=1

x�1Xy=1

1

xayband �(a) :=

1Xx=1

1

xa?

� Sums of this type are (triple, double or single)

Euler sums or Multi-dimensional zeta function

values (MZVs), as Euler was the �rst to study

them and, of course, the single Euler sums are

integral values of the Riemann zeta function.

2

�Multi-dimensional polylogarithmic sums, zeta

function values (MZVs) and their friends the

Euler/Zagier sums, seem to crop up ubiqui-

tiously in classical analysis, combinatorics, knot

theory, quantum �eld theory and elsewhere.

� I will touch on some of the recent results

(many heuristic) that my co-workers and I have

obtained through a mixture of introspection,

experimental mathematics, physics and mas-

sive high{performance computation. I will sketch

some of the experimentally discovered results

we have obtained and indicate how in some

cases experiment has also directed the proof.

� My starting point (Parseval):

1Xk=1

�1+

1

2+ � � �+

1

k

�2k�2 � 4:599873 � � �

�17

4�(4) =

17�4

360

3

� Also

1Xk=1

�1+

1

2+ � � �+

1

k

�3k�3 = �

33

16�(6)+2�2(3)

�

�

OUTLINE

Part 0. Preliminaries: Integer Relation Algorithms

Part I. Reducibility: Dimensional conjectures

Part II. Duality: Evaluations and computations

Part III. Proof of a conjecture due to Zagier

Part IV. More conjectured k-fold evaluations

Part V. A \tie-in" with knots

4

COMMENTS and PRELIMINARIES

� Complex analytic parameters are less central.

� Extensive use of \Integer Relation Algorithms":

PSLQ/LLL { or lattice basis reduction.

� Exclusion bounds are especially useful.

�Many proofs are out of reach { understanding

is not.

� Great test bed for \experimental methodol-

ogy":�

�

EZ{FACE

the remote interface we are building (Web,

Maple, C).

� Special thanks to OU { and MUN { comput-

ers.

5

PART 0:INTEGER RELATION DETECTION

� Let x = (x1; x2; � � � ; xn) be a vector of real

numbers. Then x is said to possess an integer

relation if there exist integers ai not all zero

such that

a1x1+ a2x2+ � � �+ anxn = 0

Problem: Find the integers ai if they exist. If

they do not, obtain a sequence of bounds on

the size of any possible integers ai.

� Euclid's algorithm gives solution for n = 2.

� Euler, Jacobi, Poincare, Minkowski, Perron,

Brun, Bernstein, and others tried to �nd a gen-

eral algorithm for n > 2.

� First general algorithm was discovered in 1977

by Ferguson and Forcade.

� Algorithms discovered since 1977: LLL, HJLS,

PSOS, PSLQ (1991).

6

Application to Number Theory

Suppose � can be computed to high precision.

Then form the n-long vector x = (1; �; �2; � � � ; �n�1)

and apply an integer relation algorithm.

� If a relation is found, the solution integers

ai are precisely the coe�cients of a poly-

nomial satis�ed by �.

� If no relation is found, bounds are obtained

within which no polynomial can exist.

It is also possible to explore whether � satis�es

an identify of the form

�p = 2a 3b 5c 7d �k el m � � �

by simply taking logarithms.

� g46 solves 1�5292x�7450x2�5292x3+x4 =

0.

7

AN EXEMPLARY APPLICATION

� Thanks to Ap�ery it is well known that

�(2) = 3

1Xk=1

1

k2�2kk

��(3) =

5

2

1Xk=1

(�1)k�1

k3�2kk

��(4) =

36

17

1Xk=1

1

k4�2kk

�

� These results suggest that

Z5 = �(5)=1Xk=1

(�1)k�1

k5�2kk

�might also be a simple rational or algebraic

number.

Integer relation result: If Z5 satis�es a poly-

nomial of degree 25 or less, then the Euclidean

norm of the coe�cients must exceed 2�1037.

8

Part I. Reducibility:Dimensional conjectures

�

�

EULER SUMS

�(i1; i2; : : : ; ik ; �1; �2; : : : ; �k) :=

Xn1>n2>:::>nk>0

�n11 �

n22 � � ��

nkk

ni11 n

i22 � � �n

ikk

� �i 2 f1;�1g de�ne Euler sums.

� General parameters yield Eulerian polyloga-

rithms.

�

�

MZVs (MULTIPLE ZETA VALUES)

�(i1; i2; : : : ; ik) :=X

n1>n2>:::>nk>0

1

ni11 n

i22 � � �n

ikk

� That is: �i � 1. As a �rst taste:

�(a; b) + �(b; a) = �(a)�(b)� �(a+ b)

reduces �(a; a).

9

A REDUCTION of EULER'S

� Maple can \prove" Euler's result: generat-

ingfunctionology yields

�(m;1) =(�1)m

(m� 1)!

Z 1

0

lnm�1(t) ln(1� t)

1� tdt

=(�1)m

2(m� 1)!

Z 1

0

(m� 1) lnm�2(t) ln2(1� t)

tdt

which yields a �-function derivative

�(m;1) =(�1)m

2(m� 2)!B(m�2)1 (0) (1)

where B1(y) :=@2

@x2�(x; y)

����x=1

.

10

Since

@2

@x2�(x; y) =

�(x; y)�((x)�(x+ y))2+ (0(x)�0(x+ y))

�;

we have a digamma representation via

B1(y) =1

y

�(� �(y+1))2+ (�(2)�0(y+1))

�:

� Indeed, we may implement (1) in Maple or

Mathematica very painlessly and discover its

Riemann �-function expression:

�(n;1) =1Xk=1

�1+

1

2+ � � �+

1

k

�(k+1)�n

=n�(n+1)

2�1

2

n�2Xk=1

�(n� k)�(k+1):

11

A KEY PROBLEM

� Find the dimension of aminimal generating

set for a (Q;+; �)-algebra that contains

� all Euler sums of weight n and depth k,

generated by Euler sums ... En;k

� all MZVs of weight n and depth k,

generated by Euler sums ... Mn;k

� all MZVs of weight n and depth k,

generated by MZVs ... Dn;k

12

CONJECTURED GENERATING FUNCTIONS

(Broadhurst{Kreimer, Zagier)

Yn�3

Yk�1

(1� xnyk)En;k?= 1�

x3y

(1� x2)(1� xy)

Yn�3

Yk�1

(1� xnyk)Mn;k?= 1�

x3y

1� x2

Yn�3

Yk�1

(1� xnyk)Dn;k?= 1�

x3y

1� x2

+x12y2(1� y2)

(1� x4)(1� x6)

For k = 2, n odd and k = 3, n even, the result

for Dn;k proven by \elementary methods" by

Borwein & Girgensohn.

� Dn;k has a disconcertingly complicated ratio-

nal generating function.

13

�

�

EXAMPLES of Reductions

MZV via MZVs

�(4;1;3) =

��(5;3)+ 7136�(8)� 5

2�(5)�(3)+ 1

2�(3)2�(2)

MZV via MZVs and Euler sums

�(4;2;4;2) =

�102427

�(�9;�3)� 2679915528

�(12)� 104027

�(9;3)

�763�(9)�(3)� 160

9�(7)�(5)+ 2�(6)�(3)2

+14�(5;3)�(4) + 70�(5)�(4)�(3)� 16�(3)4

� �(5;3); �(�9;�3) are irreducible

� Functional equations assist:

�(a; b; c) + �(a; c; b) + �(c; a; b) =

�(c)�(a; b)� �(a; b+ c)� �(a+ c; b)

14

En;k

k 1 2 3 4 5 6

n

3 1

4 1

5 1 1

6 1 1

7 1 2 1

8 2 2 1

9 1 3 3

10 2 5 3

11 1 5 7

12 3 8 9

13 1 7 14

14 3 14 20

15 1 9 25

16 4 20 42

17 1 12 42

18 4 30 75

19 1 15 66

20 5 40 132

15

Mn;k

k 1 2 3 4 5 6

n

3 1

4

5 1

6

7 1

8 1

9 1

10 1

11 1 1

12 2

13 1 2

14 2 1

15 1 3

16 3 2

17 1 5 1

18 3 5

19 1 7 3

20 4 8 1

16

Dn;k

k 1 2 3 4 5 6

n

3 1

4

5 1

6

7 1

8 1

9 1

10 1

11 1 1

12 1 1

13 1 2

14 2 1

15 1 2 1

16 2 3

17 1 4 2

18 2 5 1

19 1 5 5

20 3 7 3

17

� Generating functions con�rmed numerically�

in the following ranges:

En;k: (with REDUCE and PSLQ)

k = 2 and n � 44 ... k = 7 and n � 8

Mn;k: (with REDUCE and PSLQ)

k = 2 and n � 17 ... k = 7 and n � 20

Dn;k: (modulo a big prime)

�With REDUCE and FORTRAN at OU: (DEC�,

256Mb, 333Mhz)

k = 3 and n � 141 ... k = 7 and n � 21

� With FORTRAN at MUN:(DEC�, 4 � 1Gb,

400Mhz)

k = 3 and n � 161 ... k = 7 and n � 23

�A sizable subset is proven symbolically.Tools: partial fractions/functional equations/shu�es

18

ETINGHOF: \Let VN � C be the Q-vector

space generated by MZV's of weight N , and

WN be the subspace of VN of all elements rep-

resentable as rational polynomials of MZVs of

lower weights, with all terms have weight N .

Let DN = dimQ(VN=WN). You conjectured�

that DN = D�N where D�N are given by the for-

mula

1YN=1

(1� xN)D�

N = 1� x2 � x3:

This conjecture consists of 2 parts: the upper

bound DN � D�N , and the lower bound DN �

D�N . The upper bound involves proving that

�(2n+1) is irrational and is well beyond the range

of accessibility for today's number theory. On

the other hand, the lower bound reduces to a

purely algebraic problem, which can be solved

modulo:"

�A very special sub{case: DN =P

kDN;k.

19

Drinfeld (1991)-Deligne Conjecture.The

graded Lie algebra of Grothendieck-Teichmuller

has no more than one generator in odd de-

grees, and no generators in even degrees.

� Evaluate+PSLQ) exact reduction code has

been performed for:

1. all alternating sums to weight 9;

2. all MZV's to weight 14.

20

Part II. DualityEvaluations and computations

For non-negative integers s1; : : : ; sk, let

�a(s1; : : : ; sk) :=X

nj>nj+1>0

a�n1kY

j=1

n�sjj ; (2)

a special case of our multidimensional polylog-

arithm. Note that

�a(s) =Xn>0

1

anns= Lis(a

�1)

is the usual polylogarithm for s 2 N and jaj > 1.

� Put

� := �1 � := �2;

!a :=dx

x� a;

and

�(�1; : : : ; �k) :=X

nj>nj+1>0

kYj=1

�njj

nj;

a unit Euler sum.

21

� s := s1 + � � � + sk and r := r1 + � � � + rk,

denote weights of strings in N. Iterated integral

representations:�

�a(s1; : : : ; sk) = (�1)kZ 1

0

kYj=1

!sj�1

0 !a;

and dually

�a(s1; : : : ; sk) = (�1)s+kZ 1

0

1Yj=k

!1�a!sj�1

1 ;

follow on changing x 7! 1�x at each level. So:

(�1)k�a(s1+2; f1gr1; : : : ; sk +2; f1grk)

= (�1)rZ 1

0

kYj=1

!sj+1

0 wrj+1a ;

and dually

(�1)k�a(s1+2; f1gr1; : : : ; sk +2; f1grk)

= (�1)sZ 1

0

1Yj=k

!rj+1

1�a wsj+1

0 :

�Integrated over 0 � x1 � x2 � : : : � xs � 1.

22

� a := 1 gives the \duality for MZVs":

�(s1+2; f1gr1; : : : ; sk +2; f1grk)

= �(rk+2; f1gsk; : : : ; r1+ 2; f1gs1):

� a := 2 gives a corresponding \kappa-to-unit-

Euler" duality:

�(s1+2; f1gr1; : : : ; sk+2; f1grk)

= (�1)r+k�(1; f1grk;1; f1gsk; : : : ;1; f1gr11; f1gs1):

� A more general, less convenient, \kappa-to-

unit-Euler" duality similarly derivable is

�(s1; : : : ; sk) = (�1)k�(�1; �2=�1; �3=�2; : : : ; �s=�s�1);

where

[�1; : : : ; �s] = [�1; f1gsk�1; : : : ;�1; f1gs1�1]:

23

SOME �$ � DUALITY EXAMPLES

�(1) =Xn�1

1

n2n= � log(1=2)

=Xn�1

(�1)n+1

n= ��(1);

�(2) =Xn�1

1

n22n= Li2(1=2)

=Xn�1

(�1)n+1

n

n�1Xk=1

(�1)k

k= ��(1;1);

�(r+2) =Xn�1

1

nr+22n= Lir+2(1=2)

= ��(1;1; f1gr); (r � 0)

24

SOME MORE DUALITY EXAMPLES

�(f1gn) = (�1)n�(�1; f1gn�1) =(ln 2)n

n!

�(2; f1gn) = (�1)n+1�(�1; f1gn;�1)

�(f1gm+1;2; f1gn) = (�1)m+n

�(�1; f1gn; f�1g2; f1gm)

� �(1; n+2) = �(�1;�1; f1gn;�1)

and

�(1; n) =

Z 12

0

Lin(z)

1� zdz:

In particular,

�(1;2) =5

7Li2(

1

2)Li1(

1

2)�

2

7Li3(

1

2)+

5

21Li1(

1

2)3

and

�(1;3) = Li3(1

2)Li1(

1

2)�

1

2Li2(

1

2)2:

25

TWO �-REDUCTIONS

� Using integral ideas similar to below:

GOOD. Every MZV of depth N is a sum of

2N �'s of depth N . (Hence easily computed.)

BETTER: !0 = dx=x, !1 = �dx=(1� x),

�(s1; : : : ; sk) :=X

nj>nj+1>0

kYj=1

n�sjj

again has representation

�(s1; : : : ; sk) = (�1)kZ 1

0!s1�10 !1 : : : !

sk�10 !1:

� Domain, 1 > xj > xj+1 > 0, in n =Pj sj

variables, splits into n+1 parts: each being a

product of regions 1 > xj > xj+1 > �, for �rst

r variables, and � > xj > xj+1 > 0, for rest.

� xj ! 1�xj replaces integral of the former by

the latter type, with � replaced by �� := 1� �.

Thence:

26

� Let S(!0; !1) be the n{string !s1�10 !1 : : : !

sk�10 !1

specifying a MZV. Let Tr denote the substring

of the �rst r letters and Un�r the complemen-

tary substring, on the last n� r letters, so that

S = TrUn�r

for n � r � 0. Then

�(s1; : : : ; sk) =

Z 1

0S =

nXr=0

�

Z ��

0

eTr Z �

0Un�r

where e indicates reversal of letter order.� The alternate polylogarithm integral

�z(s1; : : : ; sk) =X

nj>nj+1>0

z�n1kY

j=1

n�sjj

=

Z z

0!s1�10 !1 : : : !

sk�10 !1

produces the MZV as the scalar product of two

vectors, composed of �z{values with z := p and

z := q, for any 0 < p < 1 and

1

p+1

q= 1:

27

�

�

ILLUSTRATION

� Usually set p = q = 2 (�2 = �)

� Thus, for any 1p+ 1

q= 1

�(2;1;2;1;1;1) = �p(2;1;2;1;1;1)

+�p(1;1;2;1;1;1)�q(1) + �p(1;2;1;1;1)�q(2)

+�p(2;1;1;1)�q(3) + �p(1;1;1;1)�q(1;3)

+�p(1;1;1)�q(2;3)+ �p(1;1)�q(3;3)

+�p(1)�q(4;3)+ �q(5;3) = �(5;3)

� Uses 2n �'s (no higher depth, same weight)

� This is also a duality result (q !1)!

� Di�erentiation ) �(0; f1gn) = �(f1gn).

� Applied to �(n + 2) this provides a lovely

closed form for �(2; f1gn).

28

Seq := proc(s, t) local k, n;

if 1 < s[1] then

[s[1] - 1, seq(s[k], k = 2 .. nops(s))], [1, op(t)]

else [seq(s[k], k = 2 .. nops(s))],

[t[1] + 1, seq(t[k], k = 2 .. nops(t))]

fi end

SEQ := proc(a) local w, k, s;

w := convert(a, `+`); s := a, [];

for k to w do s := Seq(s); print(s, k) od;

s[2] end

>SEQ([5,3]);

[4, 3], [1], 1

[3, 3], [1, 1], 2

[2, 3], [1, 1, 1], 3

[1, 3], [1, 1, 1, 1], 4

[3], [2, 1, 1, 1], 5

[2], [1, 2, 1, 1, 1], 6

[1], [1, 1, 2, 1, 1, 1], 7

[], [2, 1, 2, 1, 1, 1], 8

[2, 1, 2, 1, 1, 1]

29

� Time for D digits for MZV �(s1; : : : ; sk), of

weight n, is roughly c(n)D precision D multi-

plications (with c(n) / n, for large n, whatever

the depth, k).

� Idea extends somewhat to all Euler sums.

� 10;000 digits of �(5;3) in 7 hours 41 minutes

on Manyjars (CECM's 194 MHz R10000 SGI).

� Improved MPFUN: 47 minutes to get 20,000

digits, on Tempus (400 MHz Dec at MUN).

� 5000 digit PSLQ computations now feasi-

ble/in train.

30

Part III. Proofof a conjecture due to Zagier

� For r � 1 and n1; : : : ; nr � 1, again specialize

the polylogarithm to one variable�

L(n1; : : : ; nr;x) :=X

0<mr<:::<m1

xm1

mn11 : : :mnr

r:

� Thus

L(n;x) =x

1n+

x2

2n+

x3

3n+ � � �

is the classical polylogarithm, while

L(n;m; x) =1

1m

x2

2n+ (

1

1m+

1

2m)x3

3n

+(1

1m+

1

2m+

1

3m)x4

4n+ � � � ;

L(n;m; l; x) =1

1l

1

2m

x3

3n

+(1

1l

1

2m+

1

1l

1

3m+

1

2l

1

3m)x4

4n+ � � �

�L(n1; : : : ; nr; x) = �x�1(n1; : : : ; nr) to highlight depen-dence on x.

31

� Series converge absolutely for jxj < 1 (condi-

tionally on jxj= 1 unless n1 = 1 and x = 1).

� These polylogarithms

L(nr; : : : ; n1; x) =X

0<m1<:::<mr

xmr

mnrr : : :m

n11

;

are determined uniquely by

d

dxL(nr; : : : ; n1; x) =

1

xL(nr � 1; : : : ; n2; n1; x)

if nr � 2; while for nr = 1

d

dxL(nr; : : : ; n1; x) =

1

1� xL(nr�1; : : : ; n1;x)

with the initial conditions

L(nr; : : : ; n1; 0) = 0

for r � 1 and

L(;;x) � 1:

32

� Thus, if s := (s1; s2; : : : ; sN) and w :=Psi

then every periodic polylogarithm leads to a

function

Ls(x; t) :=Xn

L(fsgn;x)twn

which satis�es an algebraic ODE in x, or alter-

natively, nice recurrence relations.

� In the simplest case, with N = 1, the ODE

is DsF = tsF where

Ds :=

�(1� x)

d

dx

�1 �xd

dx

�s�1and the solution (by series) is a generalized

hypergeometric function:

Ls(x; t) := 1+Xn�1

xnts

ns

n�1Yk=1

1+

ts

ks

!;

as follows from considering Ds(xn).

33

� For N = 1 and negative integers, we similarly

obtain

L�s(x; t) := 1+Xn�1

(�x)nts

ns

n�1Yk=1

1+ (�1)k

ts

ks

!;

and in particular, L�1(2x�1; t) sats�es the hy-

pergeometric ODE.

� Indeed

L�1(1; t) =

1

�(1+ t=2;1=2� t=2):

34

� Let F(a; b; c; x) denote the hypergeometric

function. Then:

THEOREM.

1Pn=0

L(3;1;3;1; : : : ;3;1| {z }n�fold

; x) t4n =

F

t(1 + i)

2;�t(1 + i)

2; 1;x

!F

t(1� i)

2;�t(1� i)

2; 1; x

!

Proof. Both sides of the putative identity

start

1 +t4

8x2+

t4

18x3+

t8+44t4

1536x4+ � � �

and are annihilated by the operator

D31 :=

�(1� x)

d

dx

�2 �xd

dx

�2� t4

� Once discovered this can be checked vari-

ously in Maple!

35

THE MAPLE CODE

deq:=proc(F) D(D(F))+A*D(F)+B*F; end;

eqns:=

{(deq(H))(x),D((deq(H)))(x),D(D((deq(H))))(x)};

for p from 0 to 4 do

eqns:=subs((`@@`(D,p))(H)(x)=y[p],eqns); od:

yi_sol:=solve(eqns,{seq(y[i], i=0..4)});

id:=x->x; T:=x->t;

# The annihilator to be checked for any product

A31:=proc(F)

(1-id)*D((1-id)*D(id*D(id*D(F)))) - T^4*F; end;

Z:=expand(A31(F1*F2)(x));

for p from 0 to 4 do

Z:=subs((`@@`(D,p))(F1)(x)=y[1,p],

(`@@`(D,p))(F2)(x)=y[2,p],Z);

od:

36

# The annihilator to be checked for this product

a[1]:= x -> 1/x;

b[1]:= x -> I*t^2/2*1/(x*(1-x));

a[2]:= x -> 1/x;

b[2]:= x -> -I*t^2/2*1/(x*(1-x));

for i from 1 to 2 do

for o from 2 to 4 do

y[i,o]:=

subs(subs(A=a[i],B=b[i],

y[0]=y[i,0],y[1]=y[i,1],yi_sol),y[o]):

od:

od:

normal(Z);

# IS ZERO SHOWING THE 2_F_1 PRODUCT

IS ANNIHILATED

37

Corollary (Zagier).

�(3;1;3;1; : : : ;3;1| {z }n�fold

) =2�4n

(4n+2)!

Proof. We have

F(a;�a; 1; 1) =1

�(1� a)�(1 + a)=

sin�a

�a

and hence, setting x = 1,

F

t(1 + i)

2;�t(1 + i)

2; 1; 1

!F

t(1� i)

2;�t(1� i)

2; 1; 1

!

=2

�2t2sin

�1+ i

2�t

�sin

�1� i

2�t

�

=cosh�t� cos�t

�2t2=

1Xn=0

2�4nt4n

(4n+2)!

� Proof is Zagier's modi�cation of Broadhurst's

based on extensive empirical work by B3GL.

38

� Compare the much easier:

1Xn=0

L(f2gn;x) t2n = F (it;�it; 1;x)

and, more generally,

P1n=0L(fpgn; x) t

pn

= pFp�1(�!t;�!3t; : : : ;�!2p�1t; 1; : : : ;1; x)

where

!p = �1:

� The amazing factorizations in the result for

�(f3;1gn) and

�(f3;1gn) = 4�n�(f4gn) =1

2n+1�(f2;2gn)

beg the question \what other deep Clausen{

like hypergeometric factorizations lurk within?"

39

Part IV.More conjectured k-fold evaluations.

� as we saw iterated rational integral represen-

tations prove dual results� like

�(m1+2; f1gn1; : : : ;mp+2; f1gnp) =

�(np+2; f1gmp; : : : ; n1+2; f1gm1)

and the �rst self{dual case

�(f3gn) = �(f2;1gn)

is even easier; but

�(f�2;1gn) = 8�n�(f2;1gn)

remains mysterious.

�with lovely extensions for our polylogs that are also ex-

cellent computationally: especially those relating unit

Euler sums to polylogs at 1

2.

40

� A spectacular true super{Zagier identity is:

�(f2gm; f3; f2gm;1; f2gmgn)?=

2(m+1) � �4(m+1)n+2m

(2fm+1gf2n+1g)!

which we conjecture has a super factorization.

� Equally, �(2; f1;3gn)?=

4�nnX

k=0

(�1)k�(f4gn�k)f(4k+1) �(4k+2)

�4

kXj=1

�(4j � 1) �(4k � 4j +3)g

is probably provable by our present methods.

� Extensive computation suggests there are no

other such complete reductions. But:

�(2n;3;1)+ �(3;2n;1)+ �(3;1;2n) = �(2n+2) !

and much more � � �

41

�

�

A TASTE OF THINGS TO COME

� De�ne Z(m0; : : : ;m2n) :=

�(f2gm0;3; f2gm1

;1; f2gm2; : : : ;3; f2gm2n�1;1; f2gm2n) ;

with f2gmj inserted after the jth character of

the string f3;1gn. Then the super-Zagier equa-

tion reads

(2n+1)Z(fmg2n+1)?= Z(2n(m+1)+m) :

� We conjecture that for integers k;m;n � 0,

2nXr=0

Z(fmgr; k; fmg2n�r)?= Z(2n(m+1)+ k) ;

and this is proven for k := 1 and m := 0.

� That is

2nXr=0

Z(f0gr;1; f0g2n�r) =�4n+2

(4n+3)!:

42

Part V. A \tie-in"with knots (Kreimer)

MZVs ( FEYNMAN DIAGRAMS

) BRAIDS ) KNOTS

� In simplest case counter terms from renor-

malization in QFT provably associate odd �'s

and 2{odd torus knot (PICTURE):

torus(2;2n+1) � torus(2n+1;2), �(2n+1):

� QUESTION. Can one naturally make this

association directly? [The HOMFLY polyno-

mials and other invariants do not seem rich

enough.]

� B{K do not associate �(a; b) exclusively with

torus knots!

43

y �(5;3) (or the chosen irreducible at level 8)

goes with torus(4;3).

y �(7;3) (or the chosen irreducible at level 10)

goes with torus(5;3).

y �(9;3) and z(�9;�3) (or whichever pair of

irreducibles at level 12) attach a to pair of non-

torus knots.�

� Somewhat \accidently" that the knots at-

tached to z(a; b) are torus knots at weight 8

and 10. The general idea is that the irre-

ducibles f�(�odd0;�odd)jodd0 > odd > 1g at-

tach to the three-braid knots with braid words:

�1�odd0�22 �1�

odd2 . It just so happens that

�1�32�1�

32 = torus(4;3) = 819

�1�52�1�

32 = torus(5;3) = 10124

�See Physics Letters, B393 (1997), 403. This pa-per gives a tentative association of zetas with positiveknots.

44

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