Locally constant sheaf but not constant

Question

I am searching for a locally constant sheaf which is not constant. Locally constant means that there exists an open covering of the total space such that the sheaf restricted to each open set in the cover is isomorphic to a constant sheaf. But the sheaf should not be constant throughout X. Does there exist such an example. Please help.

Answer 1

If $X$ is locally connected, locally constant sheaves are (up to isomorphism) exactly the sheaves of sections of covering spaces $\pi:Y\to X$.
Such a locally constant sheaf is a constant sheaf if and only the covering $\pi$ is trivial.
So any non trivial covering will give you a non-constant but locally constant sheaf.
The simplest example is the sheaf of sections of the two sheeted non trivial covering $\mathbb C^*\to \mathbb C^*:z\mapsto z^2$ or its restriction to the unit circle $S^1\to S^1: e^{i\theta}\mapsto e^{2i\theta}$

Answer 2

Suppose that $X$ is connected (and is reasonable enough that the usual theory of covering spaces applies), and choose a base-point $x \in X$. Then giving a locally constant sheaf (of abelian groups, say) on $X$ is the same as giving an abelian group $M$ with an action of $\pi_1(X,x)$.

The group $M$ arises as the stalk of the locally constant sheaf at $x$, and the action of $\pi_1(X,x)$ is obtained by choosing an element $m \in M$, think of $m$ as a section in a n.h. of $x$, and then move the section around loops based at $x$ by using the locally constant structure. (This is called the monodromy action of $\pi_1(X,x)$ on the stalk $M$.)

Note: this is a rephrasing of Georges Elencwajg's answer, using the language of $\pi_1$ rather than the language of covering spaces (although the latter is implicit in my answer).

Answer 3

There even exist examples where $X$ is connected. For instance, consider the covering map $\mathbf R\to S^1$ given by the exponential. The sheaf of continuous sections of this map is a locally constant sheaf of sets on $S^1$, but it is not constant because it has no global section.

Answer 4

Hint: Take $X$ to be the disjoint union of two spaces and take two different constant sheafs, one on each piece.

Answer 5

Here's a concrete example from differential geometry: pick a flat Riemannian manifold $M$ whose tangent bundle is not isomorphic to a trivial bundle. For example, take $M$ to be the Mobius strip. Then the kernel of the Levi-Civita connection $d^{\nabla}: \mathcal{T}_M \to \Omega_M^{1} \otimes \mathcal{T}_M$ is the sheaf of parallel vector fields, which is a locally constant sheaf. It cannot be a constant sheaf, since the tangent bundle does not admit a global frame.