I have just learned here that we know numbers that are not periods; is it known meanwhile that the ring of periods is not a field? I know that it is conjectured that $1/\pi$ is not a period, but the existence of a period whose inverse is not a period seems to be still open. Is this correct?

More generally: is it believed that the unit group of the ring of periods is bigger than the nonzero algebraic numbers?

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    $\begingroup$ It would be extremely surprising if it was a field. D-finite functions satisfy all sorts of closure properties, but NOT inversion. And all periods that I know are actually evaluations of D-finite functions. $\endgroup$ Apr 6, 2010 at 18:13

2 Answers 2


I think the questions were about unconditional proofs or counter examples. I don't have an answer to any of those questions but I think it is still interesting to understand how the yoga of motives suggests natural answers to theses questions. Even though this may seem trivial to people familiar with the subject.

Let's work in the setting of Voevodsky's $\otimes$-triangulated categories $DM^{eff}(\mathbb{Q}):= DM_{gm}^{eff}(Spec(\mathbb{Q});\mathbb{Q}) \subset DM_{gm}(Spec(\mathbb{Q});\mathbb{Q}) =: DM(\mathbb{Q})$. Remember that the latter is obtained by formally inverting $\mathbb{Q}(1)$ and that it is a rigid $\otimes$-triangulated category.

Question 1: Is the ring of (effective) periods a field?

Following Beilinson's "Remarks on Grothendieck's standard conjecctures", let's assume

Motivic conjecture: There exists a non degenerate t-structure on Voevodsky's category $DM(\mathbb{Q}) := DM(Spec(\mathbb{Q});\mathbb{Q})$ and such that the Betti realization function $\omega_B: DM(\mathbb{Q}) \to D^bMod_f(\mathbb{Q})$ is a $t$-exact $\otimes$-functor.

This is an extremely strong conjecture as it implies the standard conjectures in characteristic 0.

Under this conjecture, the heart of the motvitic $t$-structure is a tannakian category $MM(\mathbb{Q})$. We have Betti and Rham realization functors $\omega_B,\omega_{dR}: MM(\mathbb{Q}) \rightrightarrows Mod_f(\mathbb{Q})$. And we can define $$ Per := Isom^\otimes(\omega_{dR},\omega_{B}) $$ This is a fpqc-torsor under the motivic Galois group $G_B := Aut^\otimes(\omega_B)$.

Define the algebra of motivic periods as the ring of regular functions on the Betti/de Rham torsor: $$ P_{mot} := \mathcal{O}(Per) $$ Integration of differential forms (or more generally the Riemann-Hilbert correspondance) defines an $\mathbb{C}$-point $$ Spec(\mathbb{C}) \longrightarrow Per $$ The image of the corresponding morphism $P_{mot} \to \mathbb{C}$ is the ring of periods $P$.

Note: I think this whole part is actually known unconditionnally in the setting of Nori's motives (see arXiv:1105.0865v4).

Period conjecture: The morphism $P_{mot} \to P$ is an isomorphism.

Now based on these tiny little conjectures we can say

Prop: $P_{mot}$ is not a field so $P$ isn't either.

Proof: Indeed in his comment G-torsor whose ring of regular functions is a field. @quasi-coherent explained how faithfull flatness would imply that if $P_{mot}$ were a field then it would be algebraic over $\mathbb{Q}$ which contradicts the fact that $2\pi i$ belongs to the image of $P_{mot}\to \mathbb{C}$.

Question 2 Is it true that $(P_{mot}^{eff})^\times = \overline{\mathbb{Q}}^\times$?

This post is getting too long already so I'll try and write down the rest later but the basic idea is that invertible effective motives are Artin motives. This can be proved in terms of weights or niveau (level).


Maybe I can sketch an argument for your first question.

Let $P$ be the ring of effective "formal" periods, generated by quadruples $[X,D,\omega,\gamma]$ consisting of a smooth projective $Q$-variety $X$, a normal crossing divisor $D$, top-form $\omega$, and singular cycle $\gamma$, as discussed in Kontsevich-Zagier.

Let $\omega: P \rightarrow C$ be the ring homomorphism obtained by integration, whose image is the ring of periods that you mention. You ask whether the image $\omega(P)$ could be a field.

If it were, then the induced map $Spec(C) \rightarrow Spec(P)$ would be a $C$-point of the scheme $Spec(P)$, whose image is a closed point of the scheme. As $P$ is the inductive limit of finitely-generated subrings $P_M$, we find that the induced map $Spec(C) \rightarrow Spec(P_M)$ has image a closed point of $Spec(P_M)$ for any motive $M$ which generates a sufficiently large ring of periods (or am I making a mistake about projective limits of affine varieties?).

This, incidentally, is opposite to the expectations of Grothendieck's period conjecture, which states that the image of $Spec(C) \rightarrow Spec(P_M)$ should be a generic point!

If the map $Spec(C) \rightarrow Spec(P_M)$ has image a closed point, then (since $Spec(P_M)$ is defined over $Q$), its image is a point defined over $\bar Q$. It follows that the dimension of the $Q$-Zariski closure of this point is zero.

But this dimension (zero) is equal (as remarked by Yves Andre in his paper "Galois Theory, Motives, and Transcendental Numbers") to the transcendence degree $TrDeg_Q[Per(M)]$, where $Per(M)$ is the set of periods of the motive $M$.

So, if one chooses a motive whose periods generate a transcendental extension of $Q$ (e.g., whose periods include $2 \pi i$), one should find a contradiction.

As for the second question... I'll try to say something when I'm not about to meet with students.

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    $\begingroup$ Something is wrong here. The only facts you are using about P is that it is the ascending union of countably many finitely generated $\mathbb{Q}$-algebras, and that has a map $\omega: P \to \mathbb{C}$ under which some elements have transcendental image. But these facts would be true of $P = \mathbb{Q}(t)$, with $\omega(t) = \pi$. $\endgroup$ Apr 6, 2010 at 20:39
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    $\begingroup$ I think David's right. I guess my problem is that a closed point in Spec(P) could get sent to a nonclosed point of Spec(P_M) for all M... I'll think about it some more, but for now I humbly retract my sketchy argument. $\endgroup$
    – Marty
    Apr 6, 2010 at 21:08

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