TITLE: Is it possible to define orientability using orientation preserving loops?
QUESTION [7 upvotes]: Wikipedia says that the orientable double cover corresponds to the subgroup of orientation preserving loops in $\pi_1$ (which is of index 1 or 2 apparently).
My questions are:

What is an orientation preserving loop? 
Is it possible to use it to define orientability?

Many thanks

REPLY [7 votes]: An orientation preserving loop is one that preserves the orientation of objects that travel along it.
For example, it is conceivable that the universe is a non-orientable 3-manifold.  In this case, there would be certain trips you could take in a rocket ship starting at Earth so that when you returned you would be mirror flipped!  That is, your heart would be on your right (though you would insist that it was still on your left), and you would have trouble digesting most food.  The closed path that you take for such a trip is an orientation-reversing loop.
Every non-orientable manifold has such loops.  For example, a flatlander who lived on a Möbius strip would be reversed if he or she traveled once around the strip.  (See the video Wind and Mr. Ug on YouTube.)
There are many possible ways to define orientation-preserving and orientation-reversing loops rigorously  Let $M$ be an $m$-manifold, and let $\gamma\colon [0,1]\to M$ be a loop in $M$.

If $B^m$ denotes the unit $m$-ball in $\mathbb{R}^m$ with center point $p$, then exactly one of the following statements is true:

There exists an isotopy of embeddings $\Gamma_t\colon B^m \to M$, so that $\Gamma_t(p) = \gamma(t)$ for all $t\in[0,1]$ and $\Gamma_0 = \Gamma_1$.
There exists an isotopy of embeddings $\Gamma_t\colon B^m \to M$, so that $\Gamma_t(p) = \gamma(t)$ for all $t\in[0,1]$ and $\Gamma_1 = \Gamma_0\circ R$ for some reflection $R$ of $B^m$.

In the first case, the loop $\gamma$ is orientation preserving, and in the second case $\gamma$ is orientation reversing.  (This definition formalizes the idea of moving around a small object and then checking whether or not it was "flipped" by the motion.)
If $M$ is a surface and $\gamma$ is locally an embedding, then $\gamma$ is orientation preserving if and only if $\gamma$ extends to a local embedding of an annulus, and $\gamma$ is orientation reversing if and only if $\gamma$ extends to a local embedding of a Möbius strip.  This definition can be extended to higher dimensions by replacing the annulus by $B^{m-1} \times S^1$ and replacing the Möbius strip by an $m$-dimensional Möbius strip (i.e. the space obtained by gluing the top and bottom faces of $B^{m-1} \times [0,1]$ together via a reflection).
If $\gamma$ is a simple closed curve, then $\gamma$ is orientation-preserving if and only if $\gamma$ has an orientable neighborhood.  In particular, $\gamma$ is orientation preserving if and only if a regular neighborhood of $\gamma$ is orientable.
If $M$ is a differentiable manifold and $\gamma$ is a differentiable path, then $\gamma$ is orientation-preserving if and only if there is a continuous choice of a tangent frame (ordered basis of tangent vectors) along the curve $\gamma$, such that the frame at the beginning is the same as the frame at the end.
If $M$ is a differentiable $m$-manifold embedded (or immersed) in $\mathbb{R}^{m+1}$ and $\gamma$ is a differentiable path, then $\gamma$ is orientation preserving if and only if there is a continuous choice of a normal vector to $M$ along $\gamma$.  (If the loop $\gamma$ is orientation reversing, then the normal vector will "flip" when you go once around, e.g. on the Möbius strip.)

By the way, the answer to your second question is of course yes: a manifold is orientable if and only if every closed loop is orientation-preserving.