MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p:E\to B$ be a continuous map of topological spaces and set $F_x=p^{-1}(x)$ for an $x\in B$. Take a commutative ring $A$ and assume for simplicity that each $H^\*(F_x,A)$ is a free $A$-module. Let $a_1,a_2,\ldots \in H^\*(E,A)$ be classes that give a basis of $H^\*(F_x,A)$ when restricted to any $F_x$. Assume that the direct image $R^0p_\ast \underline{A}_E$ of the constant sheaf on $E$ is constant. The Leray-Hirsch principle says that $H^\*(E,A)$ is a free $H^\*(B,A)$-module generated by the $a_i$'s.

I would like to ask if anyone knows a reference for a similar result for étale cohomology. Ideally I would like to have a statement for $E,B$ varieties over an algebraically closed field $k$ and finite coefficients of order prime to $char (k)$.

share|cite|improve this question
up vote 11 down vote accepted

[[ I have added a discussion of when $p$ is smooth or has quotient singularities. ]] [[ I added a discussion on the cohomology of $[X/G]$. ]]

The étale case follows in a way that is altogether analogous to the topological case. Let me give a proof that gives a teeny bit of extra information. I assume that $\alpha_i$ is homogenous (with respect to cohomological degree) of degree $d_i$. Then $\alpha_i$ gives a map in the derived category $A[-d_i]\to Rp_\ast A$ and combining them a map $\bigoplus_iA[-d_i]\to Rp_\ast A$. If we can show that this map is an isomorphism then we get an isomorphism $\bigoplus_iR\Gamma(B,A)[-d_i]\to R\Gamma(E,A)$ which on taking cohomology gives the L-H theorem. That this is an isomorphism can be checked fibrewise and if the natural map $(R^ip_\ast A)_x\to H^i(F_x,A)$ is an isomorphism for all geometric points $x$ we are through.

This condition is true under one of the following conditions:

  • $p$ is proper (by the proper base change theorem).
  • $p$ is locally trivial by the Künneth formula.
  • The second case covers the case of a $G$-torsor. If $G$ acts on $X$ with finite stabiliser scheme (a condition slightly stronger than having finite stabilisers but which guarantees that $X/G$ exists) and the orders of the stabilisers are invertible in the ring of coefficients it is still true. This can be seen by looking at the stack quotient $X\to[X/G]$ which is a $G$-torsor (though with base a stack) and at $[X/G]\to X/G$ which induces an isomorphism in cohomology fulfilling the condition. It should also be possible to do directly imitating Deligne's proof (in SGA 4 1/2 I think) that the cohomology of $G$ is indendent of the characteristic (it use the sequence of fibrations $G\to G/U\to G/B$ where $B$ is a Borel subgroup and $U$ its unipotent radical).
    Addendum: Here are, as requested below by algori, some details on the fact that $\pi\colon[X/G]\to X/G$ induces an isomorphism for coefficients $A$ for which the order of the (group of connected components of the) stabilisers are invertible. (This of course is well-known, so well-known in fact that I don't know if there is a proper reference for it.) I will not use that the stack is a global quotient so we may as well consider $\pi\colon\mathcal X\to X$ where $\mathcal X$ is a stack with finite stabiliser scheme and $X$ is its spatial quotient. For simplicity I will assume that the automorphism groups are reduced (i.e., that $\mathcal X$ is a Deligne-Mumford stack). The general case can be proved along the same lines but would be longer and more technical. What we are going to show is that $R\pi_*A=A$. As the construction of the spatial quotient commutes with étale localisation on $X$ we may assume that $X$ is local strictly Henselian and then by the local structure theory of DM-stacks (to be found for instance in Laumon-Moret-Bailly) $\mathcal X$ has the form $[Y/G]$, where $G$ is a finite group which can be assumed to be the stabiliser of a point of $Y$ and hence has order invertible in $A$ and $Y$ is also local strictly Henselian. Now using the usual simplicial resolution $T_n=G^n\times Y$ of $[Y/G]$ we get that $H^*([Y/G],A)=H^*(G,A)=A$ as $H^*(T_n,A)=A^G$.
  • $p$ is smooth. This is proved as follows: For any $A$-complex $K$ on $B$ we get a map $\bigoplus_iK[-d_i]\to Rp_\ast p^\ast K$ which we want to show is an isomorphism. This map is functorial so in particular if is an isomorphism for two complexes in a distinguished triangle it is so for the third. By Noetherian induction (assuming for simplicity $B$ is Noetherian) we may assume that $B$ is local Henselian and that the statement is true for $B$ replaced by the complement $U$ of the closed point. We start by showing that that implies that if $K$ is an $A$-complex on $U$ and if $j\colon U\to B$ is the inclusion, then the result is true for $Rj_\ast K$. Indeed, this follows directly from smooth base change which implies that $p^\ast Rj_\ast K=Rj'_\ast p^\ast K$, where $j'\colon p^{-1}U\to E$ is the inclusion, and then the result follows from the induction hypothesis. On the other hand, if $K$ is supported on the closed point $x$ of $B$, then the map is $\bigoplus_iK_x[-d_i]\to R\Gamma(E_x,K_x)$ which is an isomorphism as it is for $K=A$. Now, the mapping cone of $K\to Rj_\ast j^\ast K$ has support at $x$ so the statement follows.
  • The only thing that is used about a smooth map is that $p$ is universally locally acyclic for the torsion primes of $A$ (SGA IV: Exp. XVI, Thm. 1.1). Unless I am mistaken this is true for $p$ that locally are of the form $E\times_GU\to B\times_GU$, where $E\to B$ is a smooth $G$-map, $U$ a $G$-scheme and $|G|$ is invertible in $A$.

Another way of dealing with the $(G,X)$ case which I think should be more efficient and general is, following Deligne, to split $X \to X/G$ up into $X\to X\times_GG/U\to X\times_GG/B\to X/G$ (where $U$ is the unipotent radical of a Borel subgroup $B$). Then $X\times_GG/B\to X/G$ is proper and in fact the Leray-Hirsch argument applies provided a large enough integer is invertible in the coefficients (it is in general not enough to invert the $|H|$ but one also needs to invert some primes intrinsincally defined by $G$) and $X\to X\times_GG/U$ has more or less affine spaces as fibres and induces an isomorphism if the $|H|$ are invertible. Finally $X\times_GG/U\to X\times_GG/B$ is essentially a torus bundle and the cohomology of $X\times_GG/U$ can be analysed in terms of the cohomology of $X\times_GG/B$ and the characteristic classes of the bundle.

share|cite|improve this answer
Thanks, Thorsten! A couple of remarks: 1. How exactly does one show in the etale case that the derived direct image decomposes as a direct sum of shifted constant sheaves? 2. I really would like to avoid assuming $p$ a fibration in the \'etale topology. – algori Jun 23 '10 at 13:47
1. This is what my argument was supposed to show, I can expand if you point out what part is unclear to you. 2. As I said if $p$ is proper you are always OK otherwise I might be a little bit tricky however there are quite a few situations when it is still OK. On the other hand most of the classical cases where one applies LH have algebraic analogues which are fibrations in the étale topology so I would need more specificity to say more. – Torsten Ekedahl Jun 23 '10 at 14:56
Torsten -- an typical example would be a smooth affine variety on which a reductive group acts with finite stabilizers. Then the geometric quotient exists and we can then consider the map into the quotient. As far as I understand this is not a fibration (or is it?) In any case the map is not proper. However in some cases one can construct the "Leray-Hirsch" cohomology classes. – algori Jun 23 '10 at 16:07
OK, I've added a section which should cover that case. – Torsten Ekedahl Jun 23 '10 at 20:16
Dear Torsten, I was wondering if you might help me understand your comment about the fibers of p_*A when p is smooth, since I'm having trouble picturing what you're saying. To my eye, your claim implies that the pushforwards of the constant sheaf along A^2 --> A^1 and A^2 \ 0 --> A^1 will agree, which doesn't square. What am I missing? – Dustin Clausen Jun 23 '10 at 21:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.