I am reading the following article of Berger, p8 and I don't understand the idea:

$C_p:=\widehat{\overline{\mathbb Q_p}}$ does not contain the periods

The text seem to reason as follows

(under some conditions) $$ H^0(K, C_p(\chi^{-1})) = \{x \in C_p \, : \, gx = \chi(\sigma)x \forall \sigma \in G_K\} =0 $$ for some character $\chi:G_K \rightarrow \mathbb Z_p^\times$, where this set is the "set of periods".


  • How does this relate to the classical of notion of periods/why is this set periods?

Details / explanations would be appreciated!

My thoughts: (Can ignore)

What I know: one formualtion of periods in the $\mathbb Q/\mathbb C$ setting is that that they are coefficients in the comparison iso. $$ C_{dR}: H^n_{dR}(X(\mathbb C), \mathbb Q) \otimes \mathbb C \simeq H^n_{Betti}(X(\mathbb C), \mathbb Q) \otimes_{\mathbb Q} \mathbb C $$ I believe our example here is consider $\mathbb G_{m,\mathbb Q_p}$.

What I don't see: how Galois groups even come into play.


1 Answer 1


Consider the Tate motive $\mathbb Q(1)$. Its de Rham realization is simply $\mathbb Q$ (with the filtration $F^{-1}\mathbb Q=\mathbb Q$ and $F^{0}=0$) and its Betti realization is $2\pi i\mathbb Q$. The comparison theorem you recalled works because after extension of scalars to $\mathbb C$, you can multiply by $2\pi i$.

However, the observation of Tate recalled by Laurent Berger that you mentioned above tells you that $\mathbb C_{p}(1)=\{0\}$. Thus $\mathbb C_{p}(1)$, which is the étale realization of $\mathbb Q(1)$ after extension of scalars to $\mathbb C_{p}$ is not isomorphic to $\mathbb C_{p}$, which is the de Rham realization of $\mathbb Q(1)$ after extension of scalars to $\mathbb C_{p}$. You would encounter similar problems if you were to believe that the period ring for the Betti-de Rham comparison isomorphism is $\mathbb R$ (in that case there would be no compatibility in the filtration and Hodge structure)

So one needs to extend to a ring which contains something like $2\pi i$ in the $p$-adic world, and that is the ring $B_{{dR}}$ of Jean-Marc Fontaine.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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