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I have a basic question about the meaning of Tate twists in étale cohomology.

I want the understand a statement of the form $$H^1(U,\Lambda) \cong \Lambda(-1)$$ in which $U$ is the spectrum of a localization of a regular, strictly henselian local ring $A$ - I don't think giving more details would be useful - and $\Lambda = \mathbb{Z}/\ell^n \mathbb{Z}$ for some prime number $\ell$ which is invertible in $A$ (and some positive integer $n$).

I don't know how to "read" this statement. I'm well aware of the existence of the étale sheaves $\Lambda(1)$, $\Lambda(-1)$, ... and I know how they are defined. But $H^1(U,\Lambda)$ is a cohomology group (or $\Lambda$-module), not a sheaf. So, what is $\Lambda(-1)$ in the above equality? At first I thought that this would mean that the underlying group is precisely $\Lambda$ equipped with some Galois action - but of which Galois group? there is no obvious one - which explains the "(-1)". But I guess this is wrong...

Also, how is the above equality related to the statement $$H^1(U,\mu_{\ell^n}) = H^1(U,\Lambda(1)) \cong \Lambda?$$ This is a statement which I think I understand; $\Lambda(1)$ is a perfectly honest étale sheaf, you can consider étale cohomology with coefficients in this sheaf and get the group $\Lambda$ as your output. However, there must be some subtle differences between these two isomorphisms...?

Any help would be greatly appreciated.

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  • $\begingroup$ I have an answer in the case that there is some field $k$ in the background so that $A$ is a finite type $k$-algebra (maybe this is always the case for a strictly henselian ring? I haven't thought about that). $\endgroup$
    – Matt
    Sep 3, 2013 at 19:31
  • $\begingroup$ I don't see where such a field $k$ should come from. If you want the full reference, it is Lemma A.1 in the survey article "Réduction semi-stable des courbes d'après Artin, Deligne, Grothendieck, Mumford, Saito, Winters, ..." by Ahmed Abbes, which appears in Birkhauser, Progress in Mathematics, Volume 187. $\endgroup$
    – Evariste
    Sep 6, 2013 at 11:34
  • $\begingroup$ I don't have access to that paper right now, but from the title I'm pretty sure such a field exists. The setup should be something like $C/K$ is a proper smooth curve of genus $g\geq 2$. After that, everything should be done "over $K$". $\endgroup$
    – Matt
    Sep 6, 2013 at 14:15
  • $\begingroup$ I really don't think is the case. In Appendix A, Abbes proves the local description of vanishing cycles given in SGA 7, 1.3.3 - however, Lemma A.1 seems to be "100% base field free". $\endgroup$
    – Evariste
    Sep 8, 2013 at 21:31
  • $\begingroup$ Do you know how this typically works and are just confused about this situation with the ring? For example, for semi-stable reduction the setup is usually some proper model $\mathcal{C}\to Spec(A)$ with $A$ strictly henselian and having generic fiber $\pi: C\to K=Frac(A)$ a smooth genus $g\geq 2$ curve. The typical situation with $\ell$-adic cohomology would be to consider $H^1(C_{K^{sep}}, \Lambda)$. In this case there is a natural $Gal(K^{sep}/K)$ action on it. It comes from taking the stalk of $R^1\pi_*\Lambda$ at the chosen geometric point. $\endgroup$
    – Matt
    Sep 8, 2013 at 22:08

1 Answer 1

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You are right that $H^1(U, \Lambda)$ should be interpreted as a group plus a Galois action.

Notation: $U$ is the spectrum of the henselian local ring $(A, \mathfrak{m}, k)$.

Lemma: The category of étale covers of $U$ is naturally equivalent to the category of étale covers of $Spec(k)$.

The only place I've seen this spelled out in detail is in Lemma 10.143.8 of the Stacks project. The functor is the natural one given by moding out $S\mapsto S/\mathfrak{m}S$.

This induces an isomorphism in étale cohomology $$H^i(U,\Lambda)\stackrel{\sim}{\to}H^i(Spec(k), \Lambda).$$ I think this is the key confusing abuse of notation in the question.

Now we have a canonical identification by taking the stalk of $\Lambda$ at the geometric point with standard Galois cohomology $$H^i(Spec(k),\Lambda)\stackrel{\sim}{\to}H^i(Gal(k^{sep}/k), \Lambda).$$

Since this latter group is Galois cohomology, it has a natural Galois action on it (in fact it is the one I wrote in the comment and can already be read off without identifying the étale cohomology of $Spec(k)$ with Galois cohomology).

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  • $\begingroup$ But $k$ will be algebraically closed in the situation described above, so... I'm still confused! $\endgroup$
    – Evariste
    Sep 9, 2013 at 6:39
  • $\begingroup$ @Evariste Ah. But removing the divisor messes up the residue field. If you just take the standard power series ring over an algebraically closed field and remove the coordinate axes, then the residue field will be something like Laurent series. $\endgroup$
    – Matt
    Sep 9, 2013 at 14:39
  • $\begingroup$ Ok, thanks. I see the point now :-) $\endgroup$
    – Evariste
    Sep 9, 2013 at 15:39
  • $\begingroup$ One question remains: "Also, how is the above equality related to the statement..." $\endgroup$
    – Evariste
    Sep 9, 2013 at 19:17
  • $\begingroup$ @Evariste My guess is that the two should be equivalent, but it isn't obvious to me how to prove it. $\endgroup$
    – Matt
    Sep 10, 2013 at 4:04

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