12
$\begingroup$

I am learning étale cohomology, and we defined the étale cohomological dimension of a scheme $X$ as the minimum of $n$ where $H_{ét} ^n (X,F)$ vanishes for all the torsion sheaves $F$, yet I don't get why we only restrict our view to torsion sheaves.

This seems to be a very stupid question, but I really have no idea why we only talk about torsion sheaves as while discuss étale cohomological dimensions. Is there some kind of fundamental reason or intuition behind this?

This may be too elementary, so if it is, please feel free to tell me.

Improve this question
$\endgroup$
5
  • $\begingroup$ Same question on MSE: math.stackexchange.com/questions/2295417/… $\endgroup$
    – Kirigaya Kazuto
    May 24, 2017 at 21:29
  • 6
    $\begingroup$ To the people downvoting, it is bad form to downvote a question without leaving a comment explaining why you downvoted! Personally I think this question is fine... (and I give you an upvote to help undo the -2) $\endgroup$
    – Stiofán Fordham
    May 24, 2017 at 22:34
  • $\begingroup$ Morally the reason we consider torsion sheaves is that the Galois group itself is profinite, so it only really makes sense to consider its cohomology with finite coefficients. Many related results illustrate this. The most basic is that the comparison theorem between topological and etale cohomology of proper complex varieties is only valid if we consider torsion coefficients. Functorially, considering homotopy types up to isomorphisms in finite cohomology is equivalent to profinite homotopy theory, and the profinite homotopy type of complex variety is functorial wrt Galois action. $\endgroup$
    – Anton Fetisov
    May 25, 2017 at 0:17
  • $\begingroup$ @sdf :I was not one of the downvoters, but I'd guess that the downvotes might well have been a response to the crossposting. $\endgroup$
    – Steven Landsburg
    May 25, 2017 at 4:21
  • $\begingroup$ @Anton Fetisov I don't entirely get the first sentence of your comment; can you evaluate on it, please? $\endgroup$
    – Kirigaya Kazuto
    May 25, 2017 at 13:35

1 Answer 1

Reset to default
12
$\begingroup$

I think it's a combination of two things:

First, we do etale cohomology with torsion sheaves only, because of pathologies with non-torsion sheaves, and so we care about the cohomological dimension with torsion sheaves only.

Second, the cohomological dimension is actually different if we include nontorsion sheaves. Thus we lose information about torsion sheaves if we include nontorsion sheaves.

Both of these can be demonstrated already with the multiplicative group $\mathbb G_m$. We expect $H^1$ to be one-dimensional by analogy with classical cohomology, but it is easy to calculate $H^1_{et}(\mathbb G_m, \mathbb Z)=0$ by the relationship with the fundamental group, say. So the nontorsion cohomology is bad for $\mathbb G_m$.

Next consider the exact sequence $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/n\mathbb Z \to 0$, which by the previously mentioned vanishing induces an exact sequence $0 \to H^1(\mathbb G_m, \mathbb Z/n\mathbb Z) \to H^2(\mathbb G_m, \mathbb Z) \to H^2(\mathbb G_m, \mathbb Z)$ which, because $H^1(\mathbb G_m , \mathbb Z/n)$ is non-trivial, implies $H^2(\mathbb G_m, \mathbb Z)$ is nontrivial, and hence the cohomlogical dimension allowing nontorsion sheaves is at least $2$, while the cohomological dimension with torsion sheaves only is $1$.

Because there is a distinction, we must pick either torsion or nontorsion, and we pick torsion for the better analogies with singular cohomology.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for your answer. So it seems that we only consider torsion coefficients because the étale setting makes sense (only) if we consider torsion coefficients, but I am curious about the fundamental reason behind it. Can you explain why this is as it is? $\endgroup$
    – Kirigaya Kazuto
    May 25, 2017 at 13:32
  • $\begingroup$ It seems that the comment above, by Anton Fetisov, sort of answers it, "Morally the reason we consider torsion sheaves is that the Galois group itself is profinite, so it only really makes sense to consider its cohomology with finite coefficients," but I don't get it entirely. $\endgroup$
    – Kirigaya Kazuto
    May 25, 2017 at 13:34
  • 2
    $\begingroup$ @KirigayaKazuto Sure. It's because we can only "see" a finite covering at a time, and hence only a finite amount of cohomology at a time. For instance $H^1(X, \mathbb Z/n)$ classifies $\mathbb Z/n$-torsors on $X$. On $\mathbb G_m$, such a torsor is given by the map $\mathbb G_m \to \mathbb G_m, z \mapsto z^n$. This map is algebraic, and in fact is etale. On the other hand $H^1(X, \mathbb Z)$ classifies $\mathbb Z$-torsors, and an example is the covering $\mathbb G_a \to \mathbb G_m , x\mapsto e^x$, which is not algebraic and so does not contribute to etale cohomology. $\endgroup$
    – Will Sawin
    May 25, 2017 at 13:35
  • $\begingroup$ Oh, great. That makes much more sense now. It seems that this was quite a stupid question afterall. Many thanks!! $\endgroup$
    – Kirigaya Kazuto
    May 25, 2017 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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