Timeline for Grothendieck says: points are not mere points, but carry Galois group actions
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19 events
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| Jan 30, 2021 at 14:54 | comment | added | Will Sawin | A good analogy is that $\operatorname{Gal} ( \mathbb C(x))$ is the inverse limit of the profinite fundamental groups of Zariski open subsets of $\mathbb P^1$, and the complex points of the Zariski open subsets of $\mathbb P^1$ are the classifying spaces of their fundamental groups, so the classifying space of the Galois group is some kind of profinite completion of a inverse limit of complex points of Zariski open subsets of $\mathbb P^1$, which is exactly what we would want the homotopy type to be - take the sphere and remove all the points. | |
| Jan 30, 2021 at 14:51 | comment | added | Will Sawin | Since this question is bumped, I want to point out that the reason we should think of fields as $K(\pi,1)$s is not just that their etale covers are classified by $G$-sets for $G$ their Galois group but that their etale open sets are classified by $G$-sets, and every open set is a cover. This fact is trivial, its interpretation in terms of homotopy types is probably not. There is every reason to treat the $\operatorname{Spec}(\mathbb Q)$ as behaving like $K ( \operatorname{Gal} (\overline{\mathbb Q}/Q),1)$. | |
| Mar 19, 2016 at 19:29 | vote | accept | Arrow | ||
| Mar 18, 2016 at 16:24 | comment | added | Tom Church | You're probably right, I wasn't thinking about compact support or not, but the distinction is relevant here. It does still seem you need Artin-Verdier duality to promote simple connectivity to contractibility. I'm not at all a number theorist though! | |
| Mar 18, 2016 at 16:14 | comment | added | Qiaochu Yuan | @Tom: hmm, so at mathoverflow.net/questions/3103/… there's an argument that the higher etale homotopy groups of $\text{Spec } \mathbb{Z}$ vanishes. I was under the impression that $\text{Spec } \mathbb{Z}$ only looked $3$-dimensional if you look at cohomology with compact support...? | |
| Mar 18, 2016 at 16:00 | comment | added | Tom Church | No, as far as I understand, Artin-Verdier duality means that $\text{Spec}\mathbb{Z}$ has cohomological dimension at least 3 (in fact equal to 3, if 2-torsion is ignored). This kind of thing is precisely why I made my original comment. | |
| Mar 18, 2016 at 15:49 | comment | added | Qiaochu Yuan | @Tom: well, later in this answer I go on to say that $\text{Spec } k$ should look like $B \text{Gal}(k_s/k)$ (because its "universal cover" $\text{Spec } k_s$ really looks like a point), so in fact I do suggest this! As for the difference between ramified vs. unramified extensions, this should correspond to the difference between looking at etale extensions of $\mathbb{Q}$ vs. $\mathbb{Z}$, and isn't the etale homotopy type of $\text{Spec } \mathbb{Z}$ contractible? | |
| Mar 18, 2016 at 15:42 | comment | added | Tom Church | Another problem with this "suggestion": these covers are not actually covers, but ramified covers! So perhaps the claim should be that $\text{Gal}(\mathbb{Q}^{\text{ur}}/\mathbb{Q})=0$ suggests that $\text{Spec}(\mathbb{Q})$ is a point (but of course it still isn't). | |
| Mar 18, 2016 at 15:41 | comment | added | Tom Church | @QiaochuYuan: Sorry, I don't mean to pick on you too much; it's just that this particular misconception for this particular example is unusually persistent. I'm not sure why certain people continue to say "suggests that X is Y" when they mean "suggests that X and Y have the same fundamental group", especially since they don't extend the same suggestibility to other fields: I haven't heard anyone say this suggests $\text{Spec}(\mathbb{Q})$ should be a profinite $K(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}),1)$. | |
| Mar 18, 2016 at 15:31 | comment | added | Qiaochu Yuan | @Tom: hmm, I was hoping that the edited "this suggests" would be enough hedging for you. I am certainly not intending to imply that what I wrote is enough to conclude that the etale homotopy type is in fact a profinite circle, only that it suggests this hypothesis. My main motivation for saying "profinite circle" at all is that I wanted to give a concrete example where one can really have a topological picture in mind for what Spec of a field looks like, and it really does not look like a point. | |
| Mar 18, 2016 at 15:26 | comment | added | Tom Church | @QiaochuYuan: Thanks for the edit. But the claim that "you can think of $\text{Spec}\,\mathbb{F}_q$ as behaving like a 'profinite circle'" still seems to rely on the fact that $\pi^{et}_i(\text{Spec}\,\mathbb{F}_q)=0$ for $i\geq 2$, or at least that $H^i_{et}(\text{Spec}\,\mathbb{F}_q;\,\mathbb{Z}_{\ell})=0$ for $i\geq 2$. These facts are true, as far as I understand, but not trivial, and in any case are not related to the position of Grothendieck the OP asked about. | |
| Mar 18, 2016 at 15:03 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Mar 18, 2016 at 15:03 | comment | added | Qiaochu Yuan | @Tom: right, I made too strong a statement there. | |
| Mar 18, 2016 at 15:00 | comment | added | Tom Church | The fact that the finite covering theory of $\text{Spec}\,\mathbb{F}_q$ is the same as the finite covering theory of $S^1$ is certainly evidence that they have the same étale fundamental group. But in what sense could this be enough information to determine the étale homotopy type? After all, the finite covering theory of $\text{Spec}\,\mathbb{C}$ is the same as the finite covering theory of $S^{13}$, but that doesn't mean they have the same étale homotopy type. | |
| Mar 18, 2016 at 4:38 | comment | added | Theo Johnson-Freyd | So in the case I mentioned, which is you know the first case of an inseparable extension, the group you should work with is Hopf algebra $K[x]/(x^p)$ with $x$ primitive. (Linear dual of this is $\mathcal O(G)$.) Note that when $p=2$ this showed up in our discussion in your office. Anyway, I think this story is well-studied, just not by me. (I vaguely remember an article on differential Galois theory beginning with this case as motivation.) | |
| Mar 18, 2016 at 3:33 | comment | added | Qiaochu Yuan | @Theo: you can apply comonadic descent to the restriction-induction adjunction between $K$-modules and $L$-modules. You'll get that $K$-modules are $L \otimes_K L$-comodules in $L$-modules, where $L \otimes_K L$ is the comonad, explicitly described as a coalgebra in $(L, L)$-bimodules over $K$. If $K \to L$ is a finite Galois extension then $L \otimes_K L$ is a twisted version of the group coalgebra of the Galois group $G$. In your inseparable case it may be in some way related to the Lie algebra generated by that derivation; I haven't worked through any details here. | |
| Mar 18, 2016 at 3:10 | comment | added | Theo Johnson-Freyd | I meant to ask earlier today, but forgot --- do you have any insights into situations when the "galois group" behaves not as a profinite group but actually an algebraic group? Here's the case I have in mind. Suppose $K \to L$ is purely inseparable. Then there's a good sense in which $L$ is a torsor over $K$ for an algebraic group. Example: take $K = F_p(t)$ and $L = F_p(s)$ with $s^p=t$; then one should consider the "Galois group" exponentiating the derivation $\partial / \partial s$. | |
| Mar 17, 2016 at 21:23 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Mar 17, 2016 at 21:11 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |