## 14 September 2007

### multiple zeta values

Today I learned about Multiple Zeta Values in my department's graduate student "pizza" seminar. The speaker was Sarah Carr; the link is to her notes. Some of what follows basically rehashes definitions and conjectures from her notes; some of this is my own thoughts. In the future, you might expect to find math that I don't really understand but am pretending to on Fridays, because that's when this seminar meets.

One can define the "multiple zeta value" of a sequence of positive integers (k1, ..., kd), with k1 ≥ 2 (a condition which is needed for convergence), in the following way:

The ordinary Riemann zeta function is just the special case where d=1.

It turns out that these obey certain nice relations which can be found basically by just looking at the sums, for example

ζ(a) ζ(b) = ζ(a, b) + ζ(b, a) + ζ(a+b).

This allows one to compute some of these values; for example, if a = b = 2, we get

ζ(2)2 = 2 ζ(2, 2) + ζ(4)

and using a certain well-known results of Euler, namely that ζ(2) = π2/6 and ζ(4) = π4/90, we get ζ(2,2) = π4/120. Of course, one doesn't want to write ζ over and over again when studying these things, so we'd write something like

(a) * (b) = (a, b) + (b, a) + (a+b)

and this "*" is an example what's called the "stuffle product". (I swear I'm not making this name up!) You can read Carr's notes for the definition in general.

There's a natural way in which we can view sequences of integers as sequences of 0's and 1's; namely, replace each occurence of a by a-1 0's followed by a 1, so, for example, the sequence (2, 3) becomes (0, 1, 0, 0, 1). On these sequences one can define a relation called the "shuffle product", on which one has, for example,

(0, 1) Ш (0, 1) = 2(0, 1, 0, 1) + 4(0, 0, 1, 1)

or, in the original notation,

(2) Ш (2) = 2(2, 2) + 4(3, 1).

Sticking the ζs back in and turning Ш into multiplication is allowed; this is a result of Kontsevich. The proof hinges on a representation of ζ values as integrals, which is pretty natural; the integrals in question have nice power series expansions that coincide with the definition of the ζ values. Doing this, you get

ζ(2)2 = 2ζ(2,2) + 4ζ(3,1)

from which we conclude that ζ(3,1) = ζ(4)/4 = π4/360.

It turns out, though, that the relations one gets from considering the * and Ш operations are graded, in the sense that given a relation among the ζ values, the sum of the arguments in each term of that relation (the "weight" of each term) will be the same. For example, in the relation

ζ(2)2 = 2ζ(2,2) + 4ζ(3,1)

the three terms have weight 2+2, 2+2, and 3+1 respectively. It's conjectured that all the relations among ζ values come from * and Ш, from which it would follow are no relations among ζ values of different weight; this would mean that all ζ values are transcendental. Since putting a sequence which sums to m and one which sums to n into either * or Ш gives a sequence which sums to m+n, this would mean that the ζ values form a graded algebra.

I also don't know how many relations there are among ζ values of the same weight; one might hope that there are enough that we can find all the ζ values of even weight exactly by purely algebraic means, given that we know ζ(2n)? (In particular, this would imply that ζ of any sequence summing to 2n is π2n times some rational number; above, we see that ζ(4), ζ(3,1) and ζ(2,2) are all rational multiples of π4. But I don't have too much hope for that conjecture, because I can't even find ζ(2,1,1) that way! (I think that my inability to do this would follow from a conjecture of Zagier mentioned in Carr's notes, on the dimension of the grade-n part of the algebra of ζ values, but I don't trust myself.)

CORRECTION, Monday, September 17: There's a missing relation that I didn't know about. See this post.

#### 1 comment:

Michael said...

Have you read any Tom Pynchon especially his last book?