Sequential Zeta Values - United States Naval AcademySZVs Product Structure Introduction cont’d The H-series (3), unlike MZVs, don’t have the property that the product of two such

SequentialZeta Values

ME Hoffman

Outline

Introduction

Proof of theSum Theoremfor H-series

IteratedIntegrals


SumConjecture forSZVs

ProductStructure

Sequential Zeta Values

Michael E. Hoffman

U. S. Naval Academy

Number Theory TalkMax-Planck-Institut für Mathematik, Bonn

17 June 2015

ME Hoffman Sequential Zeta Values


ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Outline

1 Introduction

2 Proof of the Sum Theorem for H-series

3 Iterated Integrals

4 Sequential Zeta Values

5 Sum Conjecture for SZVs

6 Product Structure



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Introduction

For positive integers a1, . . . , ak with a1 > 1 we define thecorresponding multiple zeta value (MZV) by

ζ(a1, a2, . . . , ak) =∑

n1>n2>···>nk≥1

1

na11 na22 · · · n

akk

. (1)

One calls k the depth and a1 + · · ·+ ak the weight. Euleralready studied the cases of depth 1 and depth 2, but arguablythe present era of MZVs of general depth began with the proofof the “sum theorem”∑

a1+···+ak=n, a1>1, ai≥1ζ(a1, . . . , ak) = ζ(n). (2)

This was proved by Euler for depth 2, by C. Moen for depth 3,and by A. Granville and D. Zagier for general depth.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Introduction cont’d

Recently Moen and I proved a rather different sum theorem.Define, for nonnegative integers a1, . . . , ak witha1 + · · ·+ ak ≥ 2, the series

ζ(a1|a2| · · · |ak) =∞∑n=1

1

(n + k − 1)a1(n + k − 2)a2 · · · nak. (3)

(In our paper these are called “H-series”). Note that this is asingle sum, in contrast to the k-fold sum (1). Then our result(Integers, 2014) is∑

a1+···+ak=n,ai≥0ζ(a1|a2| · · · |ak) = kζ(n) (4)

for n ≥ 2.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Introduction cont’d

The H-series (3), unlike MZVs, don’t have the property thatthe product of two such series is a finite sum of series of thesame kind. After some experimentation I arrived at a definitionof “sequential zeta values” (SZVs) which has the property thatthe product of two SZVs is a finite sum of SZVs. Further, thereis a plausible “sum conjecture” for SZVs that includes thetheorems (2) and (4) as special cases. This gives one hope thatSVZs are interesting mathematical objects. But so far we don’thave plausible number-theoretic or geometric interpretations forSZVs (though most can be expressed as iterated integrals).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Lemmas about H-series

We begin with a look at the proof of the sum theorem (4) asgiven in our paper; it relies on a chain of lemmas about theH-series. The first, which is entirely trivial, is that

m−1∑k=1

ζ(a1| · · · |ai−1|k |ai+1| · · · |aj−1|m − k |aj+1| · · · |an) =

1

j − i[ζ(a1| · · · |ai−1|0|ai+1| · · · |aj−1|m − 1|aj+1| · · · |an)

− ζ(a1| · · · |ai−1|m − 1|ai+1| · · · |aj−1|0|aj+1| · · · |an)]

for 1 ≤ i < j ≤ n.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Lemmas about H-series cont’d

From this it follows that∑ai0+···+aik=m

ζ(a1|a2| · · · |an) =

k∑j=1

(−1)j−1H(m−k)i0,ij−1(ij − i0) · · · (ij − ij−1)(ij+1 − ij) · · · (ik − ij)

(5)

for any fixed sequence 1 ≤ i0 < i1 < · · · < ik ≤ n, where

H(m)p,q =

q∑j=p

1

jm.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure


From the formula (5) it follows that the sum C (k, n;m) of allζ(a1|a2| · · · |an) with with exactly k + 1 of the ai nonzero anda1 + · · ·+ an = m can be written in the form

n−1∑j=1

c(n)k,j

jm−k

for c(n)k,j ∈ Q. It is an easy observation that the rational

numbers c(n)k,j have the symmetry/antisymmetry property

c(n)k,n−j = (−1)

k−1c(n)k,j .



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure


It is also easy to show from equation (5) and thesymmetry/antisymmetry property that

c(n)k,1 = (−1)

k−1c(n)k,n−1 =

1

(n − 1)!

[n

k + 1

],

where[nk

]is the Stirling number of the first kind, i.e., the

number of permutations of {1, 2, . . . , n} with exactly k disjointcycles. The last (rather tricky) lemma is that

c(n)k,j =

j∑q=1

q∑p=1

(−1)p−1[qp

][n+1−qk+2−p

](q − 1)!(n − q)!

for 1 ≤ k, j ≤ n − 1.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Proof of the Sum Theorem for H-series

Now

∑a1+···+an=m, ai≥0

ζ(a1|a2| · · · |an) =n−1∑k=0

C (k, n;m);

recall C (k , n;m) is the sum of those terms with nonzero entriesin exactly k + 1 positions. We have

C (0, n;m) = ζ(0| · · · |0|m)+ζ(0| · · · |m|0)+· · ·+ζ(m|0| · · · |0)

= ζ(m) + ζ(m)− 1 + · · ·+ ζ(m)− 1− 12m− · · · − 1

(n − 1)m

= nζ(m)−n−1∑j=1

n − jjm



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Proof of the Sum Theorem for H-series cont’d

That the remaining terms C (k, n;m) cancel the negative termsabove, i.e., that

n−1∑k=1

C (k , n;m) =n−1∑k=1

n−1∑j=1

c(n)k,j

jm−k=

n−1∑j=1

n − jjm

,

follows from the formula expressing c(n)k,j in terms of Stirling

numbers. It follows that∑a1+···+an=m, ai≥0

ζ(a1|a2| · · · |an) = nζ(m).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Iterated Integral Representation

For the multiple zeta value ζ(a1, . . . , ak) we have thewell-known iterated integral representation

ζ(a1, a2, . . . , ak) =

∫ 10ωa1−10 ω1ω

a2−10 ω1 · · ·ω

ak−10 ω1

where

ω0 =dt

t, ω1 =

dt

1− t.

Provided all the ai are positive, there is a similar representationfor the H-series ζ(a1| · · · |ak), i.e.,

ζ(a1|a2| · · · |ak) =∫ 10ωa1−10 dtω

a2−10 dt · · ·ω

ak−10 ω1.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Iterated Integrals cont’d

We can also represent H-series in the which the sequence haszeroes in positions other than the first or last as iteratedintegrals, e.g.,

ζ(2|0|1) =∫ 10ω0tdtω1

ζ(1|0|0|1|1) =∫ 10

t2dtdtω1

ζ(1|0|2|0|0|2) =∫ 10

tdtω0t2dtω0ω1.

(Initial zeroes don’t affect the value of the series, but trailingzeroes do.)



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Definition of SZVs

The H-series do not form a ring. But there is a larger class ofseries that does form a ring, and includes both MZVs andH-series as special cases. We call these sequential zeta values.Let J1, J2, . . . , Jk be sequences of nonnegative integers, say

Ji = (ai ,1|ai ,2| · · · |ai ,li )

and set `(Ji ) = li , |Ji | = ai ,1 + · · ·+ ai ,li . Then the sequentialzeta value ζ(J1, J2, . . . , Jk) is∑

n1>l1n2>l2 ···>lk−1nk>lk 0

1∏ki=1

∏ai,jj=1(n − j + 1)ai,j

where a >k b means a− b ≥ k . We call |J| =∑k

i=1 |Ji | theweight of the SZV, and (l1, l2, . . . , lk) its shape.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Convergence of SZVs

It is immediate from the definition that

ζ(J1, J2 . . . , Jk) ≤ ζ(|J1|, |J2|, . . . , |Jk |).

Hence ζ(J1, J2, . . . , Jk) converges provided |J1| > 1 and|Ji | ≥ 1 for i ≥ 2. In fact, the following is true.

Proposition

The SZV ζ(J1, J2, . . . , Jk) converges provided |J1| > 1, |Ji | ≥ 1for 1 < i < k, and |J1|+ |J2|+ · · ·+ |Jk | > k.

The only case requiring examination is if |Jk | = 0. In this caseζ(J1, . . . , Jk) can be written∑

n1>l1 ···nk−1>lk−1>lk

nk−1 − lk∏k−1i=1 n

ai,1i (ni − 1)ai,2 · · · (ni − li + 1)ai ,li

,



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Convergence of SZVs cont’d

which is

ζ(J1, . . . , ak−1,1| · · · |ak−1,lk−1 − 1)− ζ(J1, . . . , Jk−1).

This is a sum of convergent SZVs if |Jk−1| > 1. Otherwise thefirst term has last sequence 0, and we can iterate. The onlycase that would produce a non-convergent SZV would be|J1| = 2, |J2| = · · · = |Jk−1| = 1, and |Jk | = 0, but this isexcluded by the hypothesis.We note that a leading 0 in J1 can simply be omitted withoutaffecting the value of the series; so henceforth we assume thatJ1 starts with a nonzero integer. If also |J1| > 1, Ji has notrailing 0 for i < k, |Ji | ≥ 1 for 1 < i < k , and|J1|+ · · ·+ |Jk | > k , we call (J1, . . . , Jk) admissible.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Properties of SZVs

SZVs have some properties from partial fractions. Since

1

(n − p)(n − q)=

1

q − p

(1

n − p− 1

n − q

)it follows that

ζ(J1, . . . , ai ,1| · · · |ai ,q| · · · |ai ,p| · · · |ai ,li , . . . , Jk) =1

q − pζ(J1, . . . , ai ,1| · · · |ai ,p − 1| · · · |ai ,q| · · · |ai ,li , . . . , Jk)

− 1q − p

ζ(J1, . . . , ai ,1| · · · |ai ,p| · · · |ai ,q − 1| · · · |ai ,li , . . . , Jk)

whenever ai ,p, ai ,q > 0.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Properties of SZVs cont’d

Recall that we assume J1 has no leading 0. A trailing 0 in Jk israther complicated to describe in general; one has, e.g.,

ζ(2, 1, 1|0) =∑

i>j>k≥2

1

i2jk=

∑i>j>k≥1

1

i2jk−∑i>j≥2

1

i2j=

ζ(2, 1, 1)−∑i>j≥1

1

i2j+∑i≥2

1

i2= ζ(2, 1, 1)− ζ(2, 1) + ζ(2)− 1.

All other leading and trailing zeroes can be disposed of via

ζ(J1, . . . , Ji |0, Ji+1, . . . , Jk) = ζ(J1, . . . , Ji , 0|Ji+1, . . . , Jk) =ζ(J1, . . . , Ji , Ji+1, . . . , Jk)− ζ(J1, . . . , Ji |Ji+1, . . . , Jk).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Strings of 1’s

By the preceding results we have for p > 1

ζ(1| · · · |1︸︷︷︸p

, J2, . . . , Jk) =1

p − 1ζ(1| · · · |1︸︷︷︸

p−1

|J2, J3, . . . , Jk).

Using this one can easily give a formula for ζ(J1, . . . , Jk) wheneach Ji is a string of 1’s:

1

(|J1| − 1)(|J1|+ |J2| − 2) · · · (|J1|+ · · ·+ |Jk | − k)

× 1(|J1|+ · · ·+ |Jk | − k)!

.



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Other Strings

A similar result is

ζ(1| 0| · · · |0︸︷︷︸p

|1, J2, . . . , Jk) =

1

p + 1

p∑j=0

ζ(1| 0| · · · |0︸︷︷︸j

|J2, J3, . . . , Jk).

For example,

ζ(1|0|0|1, 2, 1) = 13

[ζ(1|2, 1) + ζ(1|0|2, 1) + ζ(1|0|0|2, 1)] .



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Iterated Integrals Again

Provided that the last entry of Jk is nonzero, the SZVζ(J1, . . . , Jk) can be represented as an iterated integral. Forexample,

ζ(3|1, 2) =∑i ,j≥1

1

(i + j + 1)3(i + j)j2=

∫ 10ω20dtω1ω0ω1

ζ(2, 1|0|1) =∑i ,j≥1

1

(i + j + 2)2(i + 2)i=

∫ 10ω0ω1tdtω1



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Sum Conjecture

An MZV is a sequential zeta value of shape (1, 1, . . . , 1), whilean H-series is a sequential zeta value of shape (k). It turns outthat the sum theorems (2) and (4) can be put into a commonform for SZVs. Recall that a sequence (J1, . . . , Jk) isadmissible if J1 does not start with 0, |J1| > 1, Ji has notrailing zero if i < k , |Ji | ≥ 1 for 1 < i < k, and|J1|+ |J2|+ · · ·+ |Jk | > k . For a given sequence S , let Am(S)be the set of admissible sequences of shape S and weight m;the length `(S) of a shape S = (l1, . . . , lk) is k .

Conjecture

For m > `(S), ∑p∈Am(S)

ζ(p) = ζ(m − `(S) + 1).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Sum Conjecture cont’d

It is simple to put the sum theorem (4) for H-series into thisform: in the identity∑

a1+···+ak=n, ai≥0ζ(a1| · · · |ak) = kζ(n),

the inadmissible terms on the right-hand side are those witha1 = 0. These add up to∑

a2+···+ak=n, ai≥0ζ(0|a2| · · · |ak) = (k − 1)ζ(n)

since ζ(0|a2| · · · |ak) = ζ(a2| · · · |ak); hence∑a1+···+ak=n, a1≥1, ai≥0

ζ(a1| . . . |ak) = ζ(n).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Sum Conjecture cont’d

To put the MZV sum theorem (2) in this form is a little moreinvolved; to avoid notational complications we treat the specialcase depth 3, weight 5. The MZV sum theorem says

ζ(3, 1, 1) + ζ(2, 1, 2) + ζ(2, 2, 1) = ζ(5) (6)

There are additional admissible triple sums

ζ(3, 2, 0) = ζ(3, 1)− ζ(3, 2)ζ(2, 3, 0) = ζ(2, 2)− ζ(2, 3)ζ(4, 1, 0) = ζ(3)− ζ(4)− ζ(4, 1).

When these are added to the left-hand side of equation (6), theresult is

ζ(3, 1, 1) + ζ(2, 1, 2) + ζ(2, 2, 1) + ζ(3, 1) + ζ(2, 2)

− ζ(3, 2)− ζ(2, 3)− ζ(4, 1)− ζ(4) + ζ(3) = ζ(5− 3 + 1).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Known Cases of Sum Conjecture

So the sum conjecture holds for shape (k) and for shape(1, . . . , 1). So far we have only been able to prove that the sumconjecture holds for a few other shapes. These include allshapes of the form (k ,m) for k ≥ 2. Shapes of the form (1,m)seem to be harder; the conjecture has been proved for shapes(1, 2) and (1, 3). We give the proof for shape (1, 2).

Proposition

If m ≥ 3, then ∑p∈Am(1,2)

ζ(p) = ζ(m − 1).



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Known Cases of Sum Conjecture cont’d

ζ(m−1)−2ζ(m)−1+m−1∑i=2

[2ζ(i ,m− i)−ζ(i |m− i)−ζ(i)+1]

= ζ(m − 1) + m − 2−m−2∑i=2

ζ(i)− ζ(1|m − 2)

using the sum theorem of MZVs. The second sum in (7) gives

m−2∑i=2

[ζ(i)− ζ(i |m − i − 1) + ζ(i)− 1] =

m−2∑i=2

ζ(i)− ζ(1|m − 3) + ζ(m − 2)− (m − 2)

and the conclusion follows.ME Hoffman Sequential Zeta Values


ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Product of SZVs

The product of two SZVs is a finite sum of SZVs, but theproduct structure (unlike that for MZVs) does not fall into theframework of quasi-shuffle products. As with the product ofMZVs, all shuffles appear in the product, together with termsin which sequences combine; a new feature is that a singlesequence in one factor can “paste together” two or moresequences in the other. Here are some examples.

ζ(2)ζ(2|1) = ζ(2, 2|1) + ζ(2|1, 2) + ζ(4|1) + ζ(2|3)ζ(2, 1)ζ(2|1) = ζ(2, 1, 2|1) + ζ(2, 2|1, 1) + ζ(2|1, 2, 1)

+ζ(2, 3|1) + ζ(2, 2|2) + ζ(4|1, 1) + ζ(2|3, 1) + ζ(4|2)ζ(2|1)2 = 2ζ(2|1, 2|1) + 2ζ(2|3|1) + ζ(4|2)



ME Hoffman

Outline

Introduction


IteratedIntegrals



ProductStructure

Product Structure cont’d

Here is another example.

ζ(1|1)ζ(2|1) = ζ(2|1, 1|1) + ζ(1|1, 2|1) + ζ(1|3|1)+ ζ(2|2|1) + ζ(3|2).

Since ζ(1|1) = 1, the left-hand side is simply ζ(2|1) = 2− ζ(2).That right-hand side equals this follows from earlier results:

ζ(2|1, 1|1) = ζ(1|1|1) + ζ(2|1)− ζ(2) + 1 + ζ(2|1|1)ζ(1|1, 2|1) = ζ(2|1)− ζ(1|1|1)ζ(1|3|1) = ζ(3|1)− ζ(1|2|1)ζ(2|2|1) = ζ(1|2|1)− ζ(2|1|1)ζ(3|2) = ζ(2)− 1− ζ(2|1)− ζ(3|1)


IntroductionProof of the Sum Theorem for H-seriesIterated IntegralsSequential Zeta ValuesSum Conjecture for SZVsProduct Structure

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Sequential Zeta Values - United States Naval AcademySZVs Product Structure Introduction cont’d The H-series (3), unlike MZVs, don’t have the property that the product of two such