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Let $pr:X\to Z$ be a line bundle of (possibly) singular varieties (that could be reducible) over a characteristic $p$ field ($p$ could be zero); $U$ is the complement to the zero section $Z\to X$. Then for any $l\neq p$, $n>0$, there should exist a Gysin long exact sequence for the (etale or singular) cohomology $\dots \to H^i(X,\mathbb{Z}/l^n\mathbb{Z}) \to H^i(U,\mathbb{Z}/l^n\mathbb{Z})\to H^{i-1}(Z,\mathbb{Z}/l^n\mathbb{Z}(-1))\to \dots$ that is functorial in $pr$. Is this true?

Note that one can compute the cohomology of $U$ as the hypercohomology $H^*(Z,Rpr'_\ast\mathbb{Z}/l^n\mathbb{Z}_U)$, where $pr': U\to Z$ is the corresponding prinicple $G_m$-bundle. Hence the problem is to verify that $R^1pr'_*\mathbb{Z}/l^n\mathbb{Z}_U\cong \mathbb{Z}/l^n\mathbb{Z}_Z(1)$. Certainly, there is such an isomorphism for the trivial $G_m$-bundle $U\cong G_m\times Z$. Since this isomorphism does not seem to depend on the choice of a trivialization, one can 'glue' 'global' functorial isomorphisms from these 'local' isomorphisms. Yet I would prefer to have a reference for this result (or for some similar one).

Cf. https://mathoverflow.net/questions/89171/on-the-cohomology-of-g-m-bundles-and-purity-for-singular-varieties

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  • $\begingroup$ Do you mean $pr$ to be a projective bundle? $\endgroup$
    – Marc Hoyois
    Commented Mar 5, 2012 at 1:32
  • $\begingroup$ Actually, $pr$ is a line bundle. One can certainly compactify it to get a $P^1$-bundle. The problem is that the etale cohomology of a line bundle should be isomorphic to the one of the base (though I am not sure that I know the proof), and it seems to be more difficult to compute the cohomology of a $P^1$-bundle. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 5, 2012 at 5:55
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    $\begingroup$ Right, but then wouldn't this fail in the simplest case where $Z$ is a separably closed point and $X=\mathbb{A}^1$? For then the cohomology of $X$ and $Z$ is trivial but $U$ has the cohomology of a circle. The cohomology of a projective bundle, on the other hand, satisfies the same projective bundle formula as in topology (in complete generality; this is in SGA5, exposé VII, Theorem 2.2.1), so that $H^\ast(X,\Lambda)=H^\ast(Z,\Lambda)\oplus H^{\ast-2}(Z,\Lambda(-1))$ (if I get the gradings right). $\endgroup$
    – Marc Hoyois
    Commented Mar 5, 2012 at 6:38
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    $\begingroup$ Ah, and I just notice now that the Gysin exact sequence is proved 2 pages earlier, in Corollaire 1.5. It works for any rank $r$ vector bundle $X\to Z$, $Z$ any scheme, and reads $\to H^i(Z,L)\to H^i(U,q^\ast L)\to H^{i-2r+1}(Z,L(-r))\to H^{i+1}(Z,L)\to$. Here $q: U\to Z$ is the projection and $L$ is any $\mathbb{Z}/n$ module where $n$ is coprime to the residual characteristics of $Z$. $\endgroup$
    – Marc Hoyois
    Commented Mar 5, 2012 at 6:52
  • $\begingroup$ And in my previous comment I was confused about reduced vs nonreduced cohomology, please disregard the "counterexample". The gradings in the sequence in your original post seem incorrect though. $\endgroup$
    – Marc Hoyois
    Commented Mar 5, 2012 at 6:59

1 Answer 1

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Comment promoted to answer:

The following is Corollaire 1.5 in SGA5, Exposé VII:

Let $Z$ be any scheme, $p:X\to Z$ a rank $r$ vector bundle, $U=X-Z$, and $q:U\to Z$ the retsriction of $p$. Let $n$ be coprime to the residual characteristics of $Z$ and let $L$ be a sheaf of $\mathbb{Z}/n$-modules on $Z$. Then there is a long exact sequence

$$\to H^{\ast-2r}(Z,L(-r))\to H^\ast(Z,L)\to H^\ast(U,q^\ast L)\to H^{\ast+1-2r}(Z,L(-r))\to$$

That's how the corollary is stated but they also show that the sequence is naturally isomorphic to the more familar-looking sequence for cohomology with supports

$$\to H^{\ast}_Z(X,p^\ast L)\to H^\ast(X,p^\ast L)\to H^\ast(U,q^\ast L)\to H^{\ast+1}_Z(X,p^\ast L)\to$$

You should definitely have a look at the rest of exposé VII in SGA5, which proves a lot of related "expected" theorems for étale cohomology with very mild hypotheses (compared to what can be found elsewhere).

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  • $\begingroup$ Yes, you are quite right: I found two papers where some weaker results are proved. $\endgroup$
    – Mikhail Bondarko
    Commented Mar 6, 2012 at 9:37

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