# Locally constant sheaves for the étale topology, lack of intuition about “étale-localness”

I have started studying some étale cohomology and I am trying to build up some intuition about the concept of local for the étale topology. I can understand some nice examples (like Kummer exact sequence) but I am still quite confused by some "easy" notions such as locally constant sheaves.

I believe that an étale sheaf which is étale locally isomorphic to the same constant sheaf should be also globally isomorphic to that constant sheaf if the isomorphisms verify some cocycle condition, but here is a toy example which seems to contradict this:

Let $k$ be a field, $n$ an integer invertible in $k$ and assume that $k$ does not contain all $n$-th roots of unity. Now consider the two following étale sheaves on $X=Spec\; k$:

• The sheaf of n-th roots of unity $\mu_n$;
• The constant sheaf $\mathbb Z/n \mathbb Z$.

They are not isomorphic since their sections on $Spec\; k$ are different, but they become isomorphic after some finite separable extension of scalars so they are isomorphic étale locally. To be precise, $U=Spec(k[T]/(T^n-1))$ is an étale cover of $X$ such that the pullbacks of the two sheaves are isomorphic.

Why are this two sheaves locally isomorphic but not isomorphic?
Is it normal that this isomorphism doesn't "patch"? (which would imply that the sheaves over the small étale site on $Spec\; k$ don't form a prestack)

If I try to think to all this "stalkwise", changing to the point of view of topoi, (I'm not very familiar with the theory of topoi so please correct me if I am writing nonsense) I believe that:
the topos of sheaves over $Spec\;k$ with the small étale site has enough points, a family of conservative points consisting of just one element (the étale local ring is some separable closure $k^{sep}$ of $k$); and on this local ring the two sheaves above coincide.
It should follow that as soon as we have a morphism of sheaves inducing this isomorphism on the stalk the two sheaves should be isomorphic, which is not the case.

Is it just because we don't have such a morphism or am I missing something more fundamental here?

• Even in the Zariski topology, sheaves which are locally isomorphic need not be globally isomorphic, e.g. locally free sheaves. This is a similar phenomenon here, or am I missing the point of your question? – Daniel Loughran Jan 18 '11 at 14:08
• The $\mu_n$ versus $\mathbb Z/n\mathbb Z$ example is a good one to understand. Your phrase <i>if the isomorphisms verify some cocycle condition</i> may be the key to dispelling whatever apparent contradiction is mixing you up: if you apply it to this example, I think you will find that the cocycle condition demands that the isomorphism must respect the action of the Galois group of that finite separable extension. – Tom Goodwillie Jan 18 '11 at 14:26
• Locally free/locally constant is not the same as "locally isomorphic to a free/constant sheaf". I would say that a better way to think of it is that "about every point, there exists a Zariski (resp. étale) open neighborhood such that the restriction of the sheaf to that neighborhood is free (resp. constant). The sense in which a local isomorphism induces a global isomorphism is as follows: If there exists a morphism of sheaves that induces isomorphisms locally, then it is an isomorphism. – Harry Gindi Jan 18 '11 at 14:35
• It's very easy (reading Atiyah-MacDonald or Hartshorne) to confuse oneself between "local" properties and "punctual (stalkwise)" properties. Usually, assuming some finiteness conditions, punctual properties may extend to a neighborhood of the point, but in general, this is not the case. For instance, it is definitely worth it to look at how EGA treats the general cases. – Harry Gindi Jan 18 '11 at 14:41
• Dear Harry, What do you mean by "Locally free/locally constant is not the same as "locally isomorphic to a free/constant sheaf" "? Regards, Matthew – Emerton Jan 18 '11 at 17:49

$Isom(F,G)$ is indeed an etale sheaf. If we take $F = \mathbb Z/n$ and $G = \mu_n$, then $G$ is a sheaf of $F$-modules, and so evaluation at the global section $1$ gives an isomorphism of sheaves $Hom(\mathbb Z/n,\mu_n) \cong \mu_n$, which identifies $Isom(\mathbb Z/n,\mu_n)$ with the subsheaf of $\mu_n$ whose sections are primitive $n$th roots of unity. Thus there is no global isomorphism precisely because (by assumption) there is no primitive $n$th root of $1$ in $k$.

Certainy if we take $l = k[X](X^n - 1)$ we can find a section of the $Isom$ sheaf over Spec $l$, but this section does not descend to a section over Spec $k$, because it does not satisfy the requisite gluing conditions on Spec $l \times$ Spec $l =$ Spec $l\otimes_k l$. (These gluing conditions amount to the Galois invariance that Tom Goodwillie refers to in his comment above.)

Perhaps the source of your confusion is that if $V$ is an open set of a topological space, then $V \cap V = V$, but in the etale site (in which generality intersection is replaced by fibre product), $V\times V$ is typically quite a bit larger than $V$.

• That's indeed what was confusing me, now everything is much more clear, thanks! – Lorenzo Jan 18 '11 at 19:04

The short answer is that locally isomorphic things needn't be globally isomorphic, and this isn't specific to the etale topology. Let me spell it out for locally constant sheaves of vector spaces on an ordinary (sufficiently nice) topological space $X$. Such sheaves correspond to representations of the fundamental group (see Why are local systems and representations of the fundamental group equivalent). Two locally constant sheaves $F$ and $G$ of the same rank are locally isomorphic, and in fact they pullback to to isomorphic sheaves on the universal cover $\tilde X\to X$. However, they won't be isomorphic unless the corresponding representations match. This is entirely analagous to the example of the nonisomorphic sheaves $\mathbb{Z}/n\mathbb{Z}$ and $\mu_n$ pulling to isomorphic sheaves on $Spec( k^{sep})$.

(As I was writing this, I realize that Emerton has already given an answer, but perhaps two is better than none.)

• Dear Donu, how do you feel about the following statement: Local isomorphisms between sheaves are global isomorphisms? – Harry Gindi Jan 18 '11 at 18:30
• Harry, Donu never used the expression "local isomorphism". He used the expression "locally isomorphic", which I take to mean "locally admitting an isomorphism" rather than "admitting a [perhaps globally defined map which is a] local isomorphism", although I admit that this sort of expression could be a trap for the unwary or the uninitiated. – Tom Goodwillie Jan 18 '11 at 19:12
• A different ambiguity in expressions containing "locally" has to do with the way that this word carries hidden quantifiers. When it is combined with another expression containing hidden quantifiers of its own, then there is a question of precedence. "This thing is locally isomorphic to one of those things" might mean (1) "this thing has a cover by objects U such that there is one of those things T such that U is isomorphic to T" or it might mean (2) "there is one of those things T such that this thing can has a cover by objects U isomorphic to T". – Tom Goodwillie Jan 18 '11 at 19:30
• You can get to (1) by way of "this thing can be covered by objects U such that U is isomorphic to one of those things". You can get to (2) by the equally reasonable "there is one of those things T such that this thing is locally isomorphic to T" – Tom Goodwillie Jan 18 '11 at 19:30
• Dear Harry, I meant my statement the way Tom interpreted it. As to how I feel about "local isomorphisms are...", I find it funnier than helpful. – Donu Arapura Jan 18 '11 at 19:32