Singular cohomology and algebraic de Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q$ to $\mathbb Q$-vectors spaces. They come with the extra structure of Weil cohomology theories - grading, cup product, orientation map, and cycle class map. We can consider the space of all invertible natural transformations from singular cohomology to algebraic de Rham cohomology that respect this structure. For purely formal reasons, this is an affine scheme. The coordinates are the matrix coefficients of the natural transformation for each algebraic variety, and the relations are given by the various commutative diagrams that must hold. Conjecturally, this scheme is a torsor for the motivic Galois group of $\mathbb Q$.

de Rham's theorem, together with GAGA, gives us an isomorphism between these two functors when tensored with $\mathbb C$. In other words, it gives us a $\mathbb C$-point of this affine scheme.

Grothendieck's period conjecture says that this point is a generic point - so the ring of periods in $\mathbb C$ is equal to the whole ring of functions on this scheme.

On the other hand, consider the $p$-adic analogue. $p$-adic etale cohomology and algebraic De Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q_p$ to $\mathbb Q_p$-vector spaces. They are Weil cohomology theories.

We can consider the space of all isomorphisms between these two functors as an affine scheme over $\mathbb Q_p$.

$p$-adic Hodge theory gives a $B_{dR}$-valued point of this scheme.

However, it is not the generic point. The reason is that the isomorphism described by $p$-adic Hodge theory naturally factors through the category of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$-representations. Hence it lies in a closed subscheme that is a torsor under, not the full motivic Galois group, but the Zariski closure of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$ inside it. This group is much smaller because the Tate conjecture fails over $\mathbb Q_p$, and a map of Tannakian categories that is not a full functor corresponds to a map of Tannakian groups that is not surjective.

So the analogue of the period conjecture fails badly over $\mathbb Q_p$. This leads me to my question: Grothendieck's period conjecture essentially gives us a description of the scheme of isomorphisms betweeen singular and de Rham cohomology - it's the spectrum of the period ring. However, we do not have a similar picture of the scheme of isomorphisms between $p$-adic etale cohomology and de Rham cohomology. The period ring only gives us one point.

Can we explicitly describe any other points of this scheme of isomorphisms?

  • $\begingroup$ @Olivier I meant to write $B_{dR}$ instead of $\overline{\mathbb C}_p$. Given that, do I need to use a different scheme? $\endgroup$
    – Will Sawin
    Commented Mar 17, 2015 at 20:56
  • 1
    $\begingroup$ To get an isomorphism between algebraic de Rham cohomology and singular cohomology one only needs de Rham's theorem and Serre's GAGA, so no Hodge theory is required. $\endgroup$
    – naf
    Commented Mar 18, 2015 at 8:05
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    $\begingroup$ Will Sawin, first of all, thank you for this interesting (but hard) question. My impression is that part of the difficulty stems from the fact that your analogy is perhaps not so close: a closer analogue might be the collection of all comparison theorems between étale cohomology for each $p$ and de Rham cohomology for a variety over $\mathbb Q$ (rather than the single comparison theorem for a variety over $\mathbb Q_{p}$. In that case, one recovers I believe the full motivic group. As for your precise question, I guess one can take a variety over $\mathbb Q$, select a $p$ such that the Galois $\endgroup$
    – Olivier
    Commented Mar 18, 2015 at 19:33
  • $\begingroup$ representation is abnormally small (compared to the full motivic group) and try to work from there (one can find examples among abelian varieties). Another question is whether you should restrict to isomorphism respecting the filtration on both sides (as I think you should). $\endgroup$
    – Olivier
    Commented Mar 18, 2015 at 19:37
  • $\begingroup$ @AknazarKazhymurat Thanks! I have figured nothing out. $\endgroup$
    – Will Sawin
    Commented Mar 29, 2019 at 8:29


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