5
$\begingroup$

For any topological space $X$ we have a natural functor

$\text{Cov}_X \rightarrow \text{Fun}(\pi_1(X),\text{Set})$

from the category of coverings of $X$ to the category of functors $\pi_1(X) \rightarrow \text{Set}$ from the fundamental groupoid to sets, which to a covering $p: Y \rightarrow X$ associates the fiber functor. This functor is always faithful, and for nice spaces (locally path-connected) it is fully faithful. For even nicer spaces (locally path-connected and semi-locally simply connected) it is even an equivalence.

However for general spaces this shouldn't be true. In particular it should not be true for all totally disconnected spaces, because this would imply that every covering of a totally disconnected space with constant fiber is trivial, since the set-theoretic identity $X \rightarrow X$ induces an isomorphism between $\pi_1(X)$ and the fundamental groupoid of $X$ considered as a discrete topological space.

So my first question is: Does there exist a totally disconnected space which admits a non-trivial covering with constant fiber?

The isomorphism classes of coverings of $X$ of degree $n$ are classified by the first Cech-cohomology set $H^1(X,\underline{S_n})$ with values in the constant sheaf $\underline{S_n}$. Therefore all coverings of degree $n$ of $X$ are trivial precisely iff

$H^1(X,\underline{S}_n) = 1$

I was hoping one could construct an example for $n = 2$ and $X = \mathbb{Z}_3$ by showing that

$H^1(\mathbb{Z}_3,\underline{S_2}) \neq 1$

So my second question is: Does $H^1(\mathbb{Z}_3, \underline{S_2})$ vanish?

Identify $S_2$ with the group $\mu_2 = \{\pm 1\}$ and consider the exact sequence

$ 1 \rightarrow \underline{\mu_2} = \mathcal{C}_{\mu_2} \rightarrow \mathcal{C}_{\mathbb{C}^\times} \rightarrow \mathcal{C}_{\mathbb{C}^\times} \rightarrow 1 $

of sheaves of abelian groups induced by the sequence

$ 1 \rightarrow \mu_2 \rightarrow \mathbb{C}^\times \stackrel{z \mapsto z^2}{\rightarrow} \mathbb{C}^\times \rightarrow 1$

Here $\mathcal{C}_A$ for an abelian topological group $A$ denotes the sheaf of continuos functions with values in $A$. Taking the long exact sequence in Cech cohomology we see that we have an embedding

$ \text{coker}(\mathcal{C}_{\mathbb{C}^\times}(\mathbb{Z}_3) \stackrel{f(z) \mapsto f(z)^2}{\rightarrow} \mathcal{C}_{\mathbb{C}^\times}(\mathbb{Z}_3)) \hookrightarrow H^1(\mathbb{Z}_3,\underline{\mu_2})$

Therefore we could show non-vanishing by constructing a continuous function $\mathbb{Z}_3 \rightarrow \mathbb{C}^\times$ which does not admit a continous square root. I almost thought I had found such an example by mapping $\mathbb{Z}_3$ to a Sierpinski gasket in $\mathbb{C}^\times$ whose 'central hole' contains the origin, but it seems that one can actually construct a square root in this example.

My third (and final) question therefore is: Does there exist a continuous function $\mathbb{Z}_3 \rightarrow \mathbb{C}^\times$ which is not a square of another such function?

$\endgroup$
1
$\begingroup$

Here is a reference for the sheaf cohomology of totally disconnected spaces: R. Wiegand, 1969. This should answer your questions when the sheaf is abelian.

In short, the cohomological dimension of a locally compact totally disconnected space X is 0 iff X is a disjoint union of compact open sets. This gives $H^1({\bf Z}_3, S_2) = 0$. This also allows Wiegand to build several examples of such spaces with non-zero dimension, e.g. the space of ordinals less than the first uncountable ordinal $[0,\Omega[$. See the end of the article for other examples.

$\endgroup$
  • $\begingroup$ Thanks! The paper you mentioned at some point (thm. 4.1) uses the fact that taking sections $\Gamma(U,-)$ is exact for $U$ compact. Now that I think about it, this is actually obvious for locally profinite spaces, since we can make any finite covering by clopen sets into a disjoint one. My bad. Now I'm really dying to find out the answer to question #1. $\endgroup$ – Nicolas Schmidt Apr 15 '15 at 22:17
  • 1
    $\begingroup$ @Schmidt. I think that the example 5.3 in the paper answers your first question by the affirmative. That example has $H^1(X,Z_2)\neq 0$ but it is not paracompact and my Cech cohomology skills are a bit rusty so I can't be completely sure. $\endgroup$ – coudy Apr 15 '15 at 22:36
  • $\begingroup$ But we're safe, since the first Cech cohomology always coincides with sheaf cohomology (for instance [Hartshorne, Ex. III.4.4(c)]). Unfortunately the paper doesn't go into details here. Also I hope they use '$Z_2$' to denote the group with two elements ... $\endgroup$ – Nicolas Schmidt Apr 15 '15 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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