Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The motivation for this question is to understand a recent theorem of Francis Brown which implies that all periods of mixed Tate motives over $\mathbb{Z}$ lie in $\mathcal{Z}[\frac{1}{2\pi i}]$, where $\mathcal{Z}$ is the $\mathbb{Q}$-span of the set of multiple zeta values (of positive integer arguments). My picture of mixed Tate motives is not very clear, and I would like to be able to relate their periods to something I understand better.

There is a survey article of Kontsevich and Zagier which defines a period as a complex number whose real and imaginary parts are given by convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ cut out by finitely many polynomial inequalities with rational coefficients.

What is the relationship between the set of periods of mixed Tate motives over $\mathbb{Z}$ and the set of periods in the sense of Kontsevich/Zagier? Does one of these sets contain the other?

I would be interested to see examples of periods of one kind which are not periods of the other.

share|cite|improve this question
Multiple zeta values can be defined by iterated integrals (see e.g. It follows that periods of mixed Tate motives are periods in the sense of Kontsevich and Zagier (maybe you need to invert $2\pi i$). – François Brunault Apr 7 '13 at 19:26
See also… for a more detailed explanation. – François Brunault Apr 7 '13 at 19:34
@François Brunault: But how do you invert $2\pi i $? Is this (now known to be) a period in the sense of Kontsevich--Zagier? In their paper they (if I remember well) mention specifically the (equivalent) question whether $1/\pi$ is a period or not. So, to me for the inclusion this is the point. – quid Apr 7 '13 at 21:13
@quid: You're right that it is only conjectured that $1/(2\pi i)$ is not a period. Anyway, it is necessary to invert $2\pi i$ to have a valid statement because the period of the Tate motive $\mathbf{Q}(n)$ is $(2\pi i)^n$. – François Brunault Apr 8 '13 at 6:45
This explains that it has become rather standard to consider the ring of extended periods $\widehat{\mathcal{P}} := \mathcal{P}[\frac{1}{2\pi i}]$. – François Brunault Apr 8 '13 at 6:47

1 Answer 1

I think the key of this issue is that recent papers, including those of Brown, routinely refer to "the algebra of periods of Kontsevich-Zagier", when they mean $\mathcal{P}[\frac{1}{2\pi i}]$. The reason is that the definition is more natural and general: it captures all periods of all mixed motives over $\mathbb{Q}$. The more classical periods of $\mathcal{P}$ defined by convergent integrals are usually called effective periods.

By analogy, the reason that you need $(2\pi i)^{-1}$ to get all periods coming from mixed motives is equivalent to having to invert $\mathbb{Z}(-1)$ to obtain an abelian category of Nori mixed motives $\mathrm{MM}(\mathbb{Q})$.

Let's denote the set of periods of mixed Tate motives over $\mathbb{Z}$ by $\mathcal{P}_\mathrm{MT}$, the effective (original) periods by $\mathcal{P}^+$, and the complete algebra of periods (i.e. $\mathcal{P}[\frac{1}{2\pi i}]$) by $\mathcal{P}_{KZ}$.

  • $\mathcal{P}_\mathrm{MT} \subseteq \mathcal{P}_{KZ}$

This is an easy consequence of Brown's theorem. All the periods in $\mathcal{P}_\mathrm{MT}$ are generated by multiple zeta values and $(2\pi i)^{-1}$. The latter is in $\mathcal{P}_{KZ}$ by definition, and the former by Chen integration (I think the first one to put this on writting was Don Zagier).

  • $\mathcal{P}_{KZ} \nsubseteq \mathcal{P}_\mathrm{MT}$

Special values of the L-function of a (not simple) Artin motive $M$ such that $L(M,s)$ is entire should do the trick. For an unconditional example you can use Dirichlet characters.

  • $\mathcal{P}_\mathrm{MT} \nsubseteq \mathcal{P}^+$ (open)

You would need to prove that $(2\pi i)^{-1} \notin \mathcal{P}^+$, but that of course is open.

On a sad side note, we can't even prove that $\mathcal{P}_{KZ} \nsubseteq \mathcal{P}^+$.

share|cite|improve this answer
Nori not Mori, I think. – Donu Arapura Oct 4 at 23:21
@DonuArapura Of course. Nice catch! – Myshkin Oct 5 at 0:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.