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I find the definition of constructible $\bar{\mathbb Q}_l$-sheaves (or their derived category) on a variety of positive characteristic quite involved and ad Hoc. Roughly it goes as follows:

First one defines constructible sheaves modules over torsion rings in the naive way, then over finite $\mathbb Z_l$ extensions as certain projektive systems of sheaves with bigger and bigger torsion coefficients. To pass from finite $\mathbb Z_l$ extensions to finite $\mathbb Q_l$ extensions one tensors the hom spaces in our category with the quotient field. Finally $\bar{\mathbb Q}_l$ sheaves are certain inductive systems of sheaves over finite $\mathbb Q_l$ extensions.

Is there a good reason a priori to make the definition as it is? Why are more naive definitions flawed?

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What is a more naive definition? – Moosbrugger Dec 15 '11 at 22:13
One possible naive definition might include the constant sheaves $\mathbb{Z}_\ell$ etc. But I believe that $H^1(\mathbb{G}_m,\mathbb{Z}_\ell)=0$ with this choice, which would be the wrong answer: it should be $\mathbb{Z}_\ell$ like the circle. – Donu Arapura Dec 15 '11 at 23:26
@Donu: Indeed, but then the question really is: why $\ell$-adic sheaves at all, instead of some other coefficients? (And this is more or less the question Ryan has answered). @Jan: A good way to get a feel for this definition is to work out what it means for the spectrum of a field, to see that it's the same (modulo equivalence) as continuous representations of the Galois group of the field into a finite extension of $\mathbb{Q}_{\ell}$. – Moosbrugger Dec 16 '11 at 0:29
With the naive definition, $H^{1}(X,\mathbb{Z}_{\ell))=\Hom_{\text{continuous}}(\pi_{1}(X),\mathbb{Z}_{\ell‌​})$ with the discrete topology on $\mathbb{Z}_{\ell}$. This is generally zero, because (for nice schemes) $\pi_{1}(X)$ is profinite. With the nonnaive definition, it is Hom with the natural $\ell$-adic topology on $\mathbb{Z}_{\ell}$, which is what you want. – anon Dec 16 '11 at 4:54
Thanks Donu, Moosbrugger and anon these comments are useful to me! – Jan Weidner Dec 16 '11 at 14:18
up vote 21 down vote accepted

I believe the goal is to force the category of l-adic sheaves to keep some reasonable connection to geometry. No doubt someone more knowledgeable can say more about that, but I think if you start by understanding that torsion local systems are, by definition, representable by etale covers by finite relative group schemes, you will already realize why just translating the words into the full l-adic setting won't work. Namely, it is impossible to have an etale covering space that actually represents even the constant sheaf on the l-adic integers. Even at a lower level, you will find that a local system for $\mathbb{Z}/\ell^n \mathbb{Z}$ gets trivialized over progressively coarser covers as you reduce it mod powers of l, so you'd expect that an actual l-adic "local system" is only trivialized on the "intersection" of arbitrarily fine covers. This is not a cover; it is an inverse system of covers.

Why keep the connection to geometry? Again, not an expert, but I believe the biggest draw of etale cohomology is that it has a natural action of the etale fundamental group, which in the number-theoretic setting means some Galois group. For torsion local systems it's obvious that the fundamental group acts, since it is by definition constructed from automorphisms of finite etale covers, and so by actually building up an l-adic local system from torsion ones you retain the action. Furthermore, since the fundamental group itself has the inverse limit (profinite) topology, the representations you get are continuous.

Passing to fields: why tensor with $\mathbb{Q}_\ell$? Because first of all, this is what happens to module homomorphisms when you tensor with the fraction field. Why should we expect that any $\mathbb{Q}_\ell$ sheaf is like a $\mathbb{Z}_\ell$ sheaf tensored with the fraction field? Because, effectively, we have built finite generation into the construction, so we can always "find generators" with the largest possible denominators.

Finally, why take the limit over all finite extensions of $\mathbb{Q}_\ell$? Because again, by "finite generation" any "module" over the algebraic closure should be expected to have all its "structure coefficients" defined over some finite extension.

The construction is not at all ad-hoc; the idea is to start with a totally reasonable geometric definition and then apply standard algebraic tools (limits and colimits, though of categories rather than objects) to obtain a category still with good geometric properties but now also with good algebraic properties (defined over an algebraically closed field of characteristic zero).

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Thanks for explaining/motivating each step of the construction, this made things clearer for me! – Jan Weidner Dec 16 '11 at 14:19

One reason why one needs a large coefficient group like $\overline{\mathbb{Q}_\ell}$ instead of $\mathbb{Q}$ is that a reasonable cohomology theory should be a functor. Sometimes varieties over finite fields admit morphisms that cannot be captured by those smaller coefficients.

The standard example due to Serre is that of a supersingular elliptic curve $E$ over a finite field. $E$ has an endomorphism ring that is a quaternion order over $\mathbb{Z}$, but elliptic curves over $\mathbb{C}$ have 2-dimensional $H^1$. If you want the cohomology theory to behave like de Rham or singular cohomology, you need an action of the quaternion order on a two-dimensional vector space such that pullback along the multiplication-by-$n$ isogeny acts as multiplication by $n$. The only way to get a quaternion algebra to act on a 2-dimensional vector space is to make sure the quaternion algebra splits over the coefficient field of the vector space, i.e., after taking a tensor product, it becomes isomorphic to a ring of $2 \times 2$ matrices. The splitting is guaranteed over algebraically closed fields, but in this case, it fails to happen over $\mathbb{Q}$, or even $\mathbb{Q}_p$, where $p$ is the characteristic of the field of definition.

As it happens, you can get away with $\mathbb{Q}_\ell$ coefficients much of the time, but if you want to examine eigenvalues of the Frobenius acting on cohomology, it helps to have solutions to polynomial equations on hand.

As some comments on the question explained, one reason why $\mathbb{Q}_\ell$ coefficients are not given by the naïve construction is that the naïve construction will not yield a comparison with singular or de Rham cohomology. Torsion-free coefficients will not see finite covers, and finite covers are the essence of étale site.

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Thank you! Your answer is a nice complement to Ryans answer! – Jan Weidner Dec 16 '11 at 14:22

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