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$\DeclareMathOperator\Sh{Sh}\DeclareMathOperator\Pt{Pt}$Maybe this is well-known or even a stupid misunderstanding of something very basic. It's well-known that the groupoid of points (i.e., groupoid completion of the category of points) of $\Sh (X_\text{ét})$, $\Pt (X_\text{ét})$, is described by geometric points as objects and paths of étale specialisations and generalisations between them (SGA4). For the corresponding (pro-)étale $\infty$-topoi, $\Sh_{\infty} (X_\text{(pro-)ét})$, the $\infty$-groupoid of points is $1$-truncated since $\Sh_{\infty} (X_\text{(pro-)ét})$ is $1$-localic and, hence, it doesn't matter if one works with the respective $\infty$-topoi.

On the other side, $\Pt (X_\text{pro-ét})$ has geometric points as points, however such collection is not even conservative (see 61.18 Points of the pro-étale site).

Now, there's also the shape of $\Sh_{\infty} (X_\text{ét})$, $\Pi_{\infty} (X_\text{ét})$, and its profinite version, $\widehat{\Pi}_{\infty} (X_\text{ét})$, which coincides with the étale homotopy type of $X$. When $X$ is qcqs, it's also known that $\Pi_{\infty} (X_\text{ét})\cong \Pi_{\infty} (X_\text{pro-ét})$ (6.1.6 in Barwick, Glasman, and Haine - Exodromy).

Let $\pi^\text{BS}_1 (X, \overline{x})$ be defined as the automorphisms of a fiber functor from étale coverings satisfying the valuative criterion for properness (Bhatt and Scholze - The pro-étale topology for schemes). In $\Sh (X_\text{pro-ét})$, there are several points that are not geometric (as mentioned above) and, in fact, they are not even a conservative family. On the other side, $\pi_1 ({\Pi}_{\infty} (X_\text{ét}), \overline{x})$, when $X$ is connected locally Noetherian, has dense image in $\pi^{BS}_1 (X)$ (Rem 7.4.12 in Bhatt and Scholze - The pro-étale topology for schemes).

  1. What's the relation between $\pi^\text{BS}_1 (X, \overline{x})$ and $\Pt (X_\text{pro-ét})$? Is it just the connected component of $\overline{x}$?

  2. Is every element of $\pi^\text{BS}_1 (X, \overline{x})$ given by a path of specialisations and generalisations?

  3. What are the points and paths in $\Pt (X_\text{pro-ét})$? I think the points are given by connected affine $w$-contractible objects, i.e., connected $w$-strict local rings, i.e., a local ring having a section for every affine fpqc covering.

Now, let's purposefully complicate everything a little more. For $X$ qcqs, one can define a profinite stratified space or, equivalently (Hochster's duality), a spectral stratified $\infty$-topoi ${\widehat{\Pi}_{(\infty, 1)}^{X_\text{Zar}} (X_\text{(pro-)ét})}$ (Barwick, Glasman, and Haine - Exodromy). By taking the groupoid completion, one gets back $\Pi_{\infty} (X_\text{(pro-)et})$ (recall that, as mentioned above, changing from étale to pro-étale gives an equivalent shape) and, by taking the materialisation (I guess it means just taking the real limit in spaces of a profinite stratified space), one gets a $X_\text{Zar}$-stratified version of $\Pt (X_\text{(pro-)ét})$ before the groupoid completion.

  1. Do the materialisations coincide at first or after inverting all the maps?

  2. What's really the difference between ${\widehat{\Pi}_{(\infty, 1)}^{X_\text{Zar}} (X_\text{ét})}$ and ${\widehat{\Pi}_{(\infty, 1)}^{X_\text{Zar}} (X_\text{pro-ét})}$?

Maybe I've screwed up something in my assertions. If so, please, comment below.

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  • $\begingroup$ @PiotrAchinger I'm not well-versed too. I've included the reference (Remark 7.4.12 in BS). What's the étale $\pi_1$ in your comment? I think you mean profinite $\pi_1$. If so, you need to take the profinite completion of the shape or take the finite étale topoi. Am I misunderstanding anything? $\endgroup$
    – user40276
    Apr 19, 2022 at 20:44
  • $\begingroup$ Sorry, I deleted my comment because I could not edit it. Let me repeat here my comments so that the above makes sense: 1) the map with dense image goes from proetale $\pi_1$ to etale $\pi_1$; in fact there is a third prodiscrete group $\pi_1^{\rm SGA3}$ and maps $\pi_1^{\rm proet}\to \pi_1^{\rm SGA3}\to \pi_1^{\rm et}$ which are respectively the prodiscrete and profinite completion. 2) Artin-Mazur observed that $\pi_1^{\rm SGA3}$ is the $\pi_1$ of the etale homotopy type. Does that imply that $\Pi_\infty(X_{\rm et}$ and $\Pi_\infty(X_{\rm proet})$ are different bc they have different $\pi_1$? $\endgroup$
    – Piotr Achinger
    Apr 19, 2022 at 20:53
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    $\begingroup$ $$\begin{align} & pro-ét \\ {} \\ & \text{pro-ét} \end{align}$$ Above one sees what some parts of this posting looked like before and after the edits by LSpice. In particular, note that a hyphen looks different from a minus sign. $\endgroup$
    – Michael Hardy
    Apr 19, 2022 at 21:00
  • $\begingroup$ @PiotrAchinger I've edited the question and corrected a mistake. I said profinite étale $\pi_1$ when I've actually meant étale $\pi_1$ for the map with dense image. In order to get the usual SGA1 $\pi_1$ you need to take profinite completion of the shape. $\endgroup$
    – user40276
    Apr 19, 2022 at 21:02

1 Answer 1

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I think that the answer to 2) is "No". For example, take two genus > 0 smooth curves over an algebraically closed field and glue them at a closed point. Then their pro-étale fundamental group $\pi_1^\mathrm{BS}$ is the "Noohi completion" of the topological free product of the usual (profinite) étale fundamental groups of each of the curves $(\pi_1^\mathrm{ét}(C_1) *_{\mathrm{top}} \pi_1^\mathrm{ét}(C_2))^{\mathrm{Noohi}}$. I think that the sequences of specializations and generalizations should correspond to the elements of this (topological) free product, before taking the completions, and so should in general just have a dense image, and not give the entire group (the "free Noohi product" of two profinite groups is usually not profinite).

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  • $\begingroup$ Thanks for the answer. That might be stupid, but why, for a fixed geometric point, an automorphism of that is given by an element of $\pi_1^{BS}$? Are you giving a positive answer to 1) ? For the étale case a natural transformation between points is given by a specialisation. I'm not sure if this is true for the pro-étale site, though .$\pi_1^{BS}$ seems to have only paths of this form (all automorphisms of the fiber functor comes from a bunch of zig-zags of specialisations). $\endgroup$
    – user40276
    Apr 19, 2022 at 21:57
  • $\begingroup$ I'm sorry for the confusion, I was addressing point 2). I have just edited my answer. $\endgroup$
    – M L
    Apr 19, 2022 at 22:49
  • $\begingroup$ Ah!Ok! I've got it now. But can you explicitly give an example of such automorphism of the fiber functor even if it's a guess? This completion adds more Galois objects, but I don't know how to think about then. In the ordinary profinite case, each normal subgroup gives a Galois object which transport stuff by zig-zags of gen. and spec. Completion seems to extend this paths "transfinitely" (parametrising by a filter). Anyway, I don't even know the image I'm supposed to have of the automorphisms of a fixed Galois object corresponding to an open non-discrete subgroup of the identity. $\endgroup$
    – user40276
    Apr 19, 2022 at 23:21
  • $\begingroup$ I am not 100% sure, what do you mean by Galois objects. If you mean elements of the category $G-\mathrm{Set}$ (discrete sets with a continuous action by G), then for a (Hausdorff) topological group there is an equivalence $G-\mathrm{Set} \simeq G^{\mathrm{Noohi}}- \mathrm{Set}$. If you mean connected $G$-sets that are of the form $G/N$ for an open normal $N$, then remember that in general in a Noohi topological group open normal subgroups do not necessarily form a basis of $1_G$, i.e. not every covering has a Galois closure. tbc $\endgroup$
    – M L
    Apr 20, 2022 at 13:36
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    $\begingroup$ In explicit topological group theory language, taking Noohi completion of some group $G$ amounts to first weakening the topology of $G$ so that open subgroups form a basis of $1_G$ and then taking the Raikov completion of (the Hausdorff quotient of) the obtained group. So to make my answer above precise, one would have to first prove that indeed all specialization/generalization paths sit in this $G$ at hand (i.e. the free topological product of the two etale fundamental groups)and then argue that the Raikov completion introduces new elements. But I will try to think of a more explicit example $\endgroup$
    – M L
    Apr 20, 2022 at 13:44

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