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Grothendieck calls a "discretification" of a profinite group $\widehat G$, a discrete group $G$ whose profinite completion is isomorphic to $\widehat G$.

Does Grothendieck also define a notion of the "discretification" of a functor to profinite groups, and particulary that of the algebraic fundamental group functor?

The automorphisms of the geometric point act on the discretifications of the fundamental group functor; and thus one can ask:

are all discretifications of the the algebraic fundamental group functor related by automorphisms of the geometric point? To simplify the question, one may abelianise the functor and/or restrict it to a full subcategory of schemes $X\longrightarrow S$, $S=\mathrm{Spec}\ k$ a number field or $k=K$.

Here is the same question with some more notation and detail. Please excuse my poor notation: I am not very familiar with the language of schemes.

Say a functor $F$ to finitely generated Abelian groups is a discretification of a functor $F'$ to profinite groups iff for every $X$, $F(X)$ is a discretification of $F'(X)$. Equivalently, one is given an equivalence $\widehat F \Rightarrow F'$ (where $\widehat F(X)=\widehat{F(X)}$ is the profinite completion of $F(X)$.)

The algebraic fundamental group is, or at least defines, a functor $\pi_1^{alg}: \mathrm{Spec}\ K\text{\Schemes} \longrightarrow \text{ProfiniteGroups}$ from the category $\mathrm{Spec}\ K\text{\Schemes}= \mathrm{Mor} (\mathrm{Spec}\ K, \text{\Schemes})$ of Schemes cosliced over a geometric point $\mathrm{Spec}\ K$.

There is a canonical $\mathrm{Aut}(K/\mathbb{Q})$ action on $\mathrm{Spec}\ K\text{\Schemes}$ which extends to an action on the discretifications of this functor. Did Grothendicek conjecture anything about this action?

Is there a better way to state this question? E.g. using the fact that for K the field of algebraic numbers $\mathrm{Aut}(K/\mathbb{Q})=\pi_1^{alg}(\mathrm{Spec}\ \mathbb{Q}, \mathrm{Spec}\ K)$.

One can modify the question by replacing the category of Schemes by a subcategory of Schemes over $\mathrm{Spec}\ k$ or some other scheme, and then considering double cosets of $\mathrm{Aut}(K/\mathbb{Q})$ and $\mathrm{Aut}(k/\mathbb{Q})$ action; as an example, one may consider the full subcategory of etale covers of direct products of moduli spaces of curves....

Perhaps I should also add that probably I can prove some very partial positive results, for some very small and "abelian" subcategories of Schemes. That's the motivation for the question.

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  • $\begingroup$ Possibly related question (or not - I'm out of my depth): mathoverflow.net/questions/4972/… . $\endgroup$
    – HJRW
    Commented Aug 21, 2012 at 11:41
  • $\begingroup$ Thanks. Yes, this kind of question is probably indirectly related...But I do not see a direct relation. $\endgroup$
    – o a
    Commented Aug 21, 2012 at 12:33
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    $\begingroup$ Is it obvious whether there exists any discretification? $\endgroup$
    – Will Sawin
    Commented Aug 21, 2012 at 17:32
  • $\begingroup$ Will Savin: the topological fundamental group is intended to be an example. it works for K=C or K⊂C and an appropriate subcategory of Schemes. (Well, the topological fundamental group of a smooth projective complex algebraic variety is not necessarily residually finite, so you have to do something about that). In general, I do not know; perhaps my definition is too naive to work for general Schemes. Also, in char p, I do not know; but possibly it is not too hard to construct by hand. $\endgroup$
    – o a
    Commented Aug 21, 2012 at 19:04
  • $\begingroup$ I see. I thought it wouldn't work for $K\subset \mathbb C$ because of the classic example that shows the fundamental group of a complex manifold is not determined by its algebraic structure, but that is just an example of the $Aut(\mathbb C/\mathbb Q)$ action on discretization. In characteristic $p$, The Artin-Schreier sequence shows that $H^1(X,\mathbb Z/p)= Mor(\pi_1(X),\mathbb Z/p)$ is infinite-dimensional for a general scheme, so $\pi_1(x)$ is infinitely generated, which doesn't rule out a by-hand construction but makes it seem difficult. $\endgroup$
    – Will Sawin
    Commented Aug 21, 2012 at 20:43

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