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When I read the paper "THE FARGUES–FONTAINE CURVE AND DIAMONDS" of Matthew Morrow, I have a question on page 11.

Question: The arthur said that the de Rham and crystalline period rings implicitly depended on having chosen an untilt of $F$, where $F=C_p^b$ is the tilt of the p-adic complex field $C_p$. I can't understand this. If the author just defines the $\infty$ point on Fontaine-Fargues curve to be the class $[C_p]$, why the author mention the period rings at the bottom of page11 when he defines the $\infty$ point?

Definition: The definitions of the de Rham and crystalline period rings I know are $B_{dR}:=Frac(lim W(R)[\frac{1}{p}]/{(ker\theta)}^n])$, $B_{cris}:=Frac(A_{cris}[\frac{1}{p}])$, $A_{cris}:=limA^0_{cris}/p^nA^0_{cris}$, and $A^0_{cris}$ is just the sub $W(R)$-module of $W(R)[\frac{1}{p}]$ generated by the $\frac{\xi^n}{n!}$ where $n$ takes all positive integers. These definitions come from Fontaine's readable book Theory of p-adic Galois Representations.

Thanks for any answers!

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    $\begingroup$ Doesn't the map $\theta$ (and hence $\xi$) depend on a choice of untilt? Also, I think $A^0_{cris}$ should be a subring of $B_{dR}$, not $W(R)[1/p]$, right? $\endgroup$ – Daniel Litt Jul 6 '19 at 15:07
  • $\begingroup$ @DanielLitt Thanks for your answer. The map $\theta$ is defined the natural extension of $W(R)\rightarrow O_C:(x_0,x_1,...,x_n,...)\rightarrow \sum p^nx_n^{(n)}$ where $x_n=(x_n^{(m)})$ and $x_n^{(m)}\in O_C$. And $A_{cris}^0\subseteq W(R)[\frac{1}{p}]\subseteq B_{dR}^+\subseteq B_{dR}$. The untilt of a perfectoid field $K$ is a pair $(L,r)$, where $L$ is a perfectoid field and $r:L^b\cong K$ is an isomorphism. But I still can't see why $\theta$ determines such a pair $(L,r)$. $\endgroup$ – user141691 Jul 6 '19 at 15:54
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    $\begingroup$ $C$ is a choice of untilt! $\endgroup$ – Daniel Litt Jul 6 '19 at 17:32
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As Daniel Litt says, the choice of $C$ is actually an untilt. The "classical" approach to period rings, which you might have in mind, was to start with a certain complete, algebraically closed field $C_p$, then to construct $R$ out of the quotient $O_{C_p}/p$ and out of its Witt vectors to construct the period rings.

The content of Morrow's Proposition 5.1 is that $R$ may arise from many other fields $C$ other than from your initially chosen $C_p$ and all these form, modulo $\varphi^\mathbb{Z}$, all the equivalence classes of untilts. But to produce $\theta$ you need a target, hence you need to pick one of these choices. By Theorem 2.3 ibid. this choice is a point on The Curve, call it $\infty$: the construction of the line bundle depends on this point. Morrow's remark that a choice has been made means the following: suppose $B_e$ could be constructed independently of the choice of an untilt. Then the construction in (6) would give the same result irrespectively of the chosen point, providing you with a canonical line bundle on it, or equivalently with a prefered point on $\mathbb{P}^1$, which is absurd. On the other hand, the two choices match with each other: a point on $\mathbb{P}^1$ (rather, on $X^{FF}$) call it $\infty$ and an untilt, for instance $C_p$.

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  • $\begingroup$ Thanks for your detailed answer. When I study p-adic Hodge, I haven't thought that we can replace $C_p$ with other complete algebraic closed non-archimedean field in the constructions of these period rings. One more question: Does the construction in (6) you mentioned mean the two hypotheses $Pic(X)\cong Z$ and $H^1(X,O_X(k))=0$ for all $k\geq0$ on page 11? $\endgroup$ – user141691 Jul 7 '19 at 13:45
  • $\begingroup$ Yea, especially in the isomorphism $\mathrm{Pic}(X)\cong\mathbb{Z}$. $\endgroup$ – Filippo Alberto Edoardo Jul 8 '19 at 16:12
  • $\begingroup$ Thank you very much ! $\endgroup$ – user141691 Jul 8 '19 at 16:38

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