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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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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. – Jacques Carette Apr 6 2010 at 18:13

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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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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 $\omega(t) = \pi$. – David Speyer Apr 6 2010 at 20:39
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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. – Marty Apr 6 2010 at 21:08

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