Timeline for Groupoid of points, shape and stratified shape of $\operatorname{Sh} (X_\text{pro-ét})$
Current License: CC BY-SA 4.0
10 events
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| Apr 27, 2022 at 18:54 | comment | added | user40276 | . So this additional path seems to be a possibly infinite chain of finite paths of spec./gen. | |
| Apr 27, 2022 at 18:53 | comment | added | user40276 | Thanks for the reply. By Galois, I've meant $G/N$ with $N$ normal. I took some time to look better at Raikov completions. Please, correct me if I'm screwing up anything. Suppose $G$ is Hausdorff with a basis of $1_G$ given open subgroups. The completion $G^*$ seems to contain exactly the same open sets as $G$ (?). So they are equal as locales. In your above example, one just need to (Raikov) complete (?), so additional points are parametrised by Cauchy filters $\{ g_U U \}$ (for the left uniformity) and each $G/U$ acts by paths of specialisations and generalisations.(...) | |
| Apr 20, 2022 at 13:44 | comment | added | M L | 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 | |
| Apr 20, 2022 at 13:36 | comment | added | M L | 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 | |
| Apr 19, 2022 at 23:21 | comment | added | user40276 | 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. | |
| Apr 19, 2022 at 22:54 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
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| Apr 19, 2022 at 22:49 | comment | added | M L | I'm sorry for the confusion, I was addressing point 2). I have just edited my answer. | |
| Apr 19, 2022 at 22:48 | history | edited | M L | CC BY-SA 4.0 |
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| Apr 19, 2022 at 21:57 | comment | added | user40276 | 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). | |
| Apr 19, 2022 at 21:17 | history | answered | M L | CC BY-SA 4.0 |