Orientation of disconnected manifolds

Question

Let $M$ be a smooth disconnected n-manifold. Can the orientation of $M$ have a different sign on different connected components and still be a consistent orientation? For example, consider the $0$-manifold $\partial [a, b] = \{a, b\}$. The induced orientation takes the sign $-1$ at $a$ and $+1$ at $b$. Can this still be considered as a consistently oriented manifold, but with differing signs on different connected components?

Answer 1

A smooth $n$-manifold $M$ need not be orientable. So let us additionally assume that it is orientable, i.e. admits an orientation. I do not know which definition of orientation you use, but let me assume an orientation of $M$ is a maximal subatlas $\omega$ of the differentiable structure on $M$ such that all transition functions $\tau : W \to W'$ between charts in $\omega$ have the property that all derivatives $d_x\tau : \mathbb R^n \to \mathbb R^n$, $x \in W$, have positive determinant.

The simplest case is that $M$ is connected. In that case it has exactly two distinct orientations $\omega_1, \omega_2$. You can say that these have different sign - in the sense that all transition functions $\tau : W \to W'$ from an chart in $\omega_1$ to a chart in $\omega_2$ have the property that all derivatives $d_x\tau$, $x \in W$, have negative determinant. However, for $n > 0$ this does not mean that one of $\omega_1, \omega_2$ is the positive orientation and the other the negative orientation. You can of course pick one of $\omega_1, \omega_2$ and declare it as the positive orientation, but this would be an arbitrary choice. There is no internal property of an orientation allowing it to be called positive. The case $n = 0$ is somewhat special. A connected $0$-manifold is a point, and an orientation of a point is the assigment of one of two values like "$+1,-1$" or "right, left" etc. If you decide for "$+1,-1$", then you can say that "$+1$" is the positive orientation. But be aware that this is again arbitrary, if you decide for "right, left" what would be the positive orientation?

If $M$ is not connected, then there are more than two orientations. In fact, each component can be orientated separately and independently from the other components. Thus, if the number of components is finite , say $k$, then we get $2^k$ orientations. If the number of components is infinite (countably infinite because $M$ has a countable basis), then we get uncountably many orientations.

In your example $\{a,b\}$ is a $0$-manifold with two components, thus we get four possible orientations.

Remark:

As you mentioned, you use John M. Lee's textbook "Introduction to Smooth manifolds". Reading the section Orientations of Manifolds carefully makes clear that a smooth manifold does not have something like an inherent positive or negative orientation. An orientation has to be chosen, and there is always more than one choice. But if you consider an oriented manifold (i.e. a pair $(M,\omega)$ consisting of a manifold $M$ and an orientation $\omega$) you may talk about positive and negative orientations relative to $\omega$. Here is a quotation with a few comments added:

Let $M$ be a smooth manifold. We define a pointwise orientation on $M$ to be a choice of orientation of each tangent space. By itself, this is not a very useful concept, because the orientations of nearby points may have no relation to each other. ... In order for orientations to have some relationship with the smooth structure, we need an extra condition to ensure that the orientations of nearby tangent spaces are consistent with each other.
Suppose $M$ is a smooth $n$-manifold with a given pointwise orientation. Recall that a local frame for $M $is an $n$-tuple of smooth vector fields $(E_1,\ldots,E_n)$ on an open set $U \subset M$ such that $(E_i\mid_p)$ forms a basis for $T_pM$ at each $p \in U$. We say that a local frame $(E_i)$ is (positively) oriented if $(E_1\mid_p,\ldots,E_n\mid_p)$ is a positively oriented basis for $T_pM$ at each point $p \in U$. [Added: This has to be understood relative to the given pointwise orientation, i.e. means that the equivalence class of ordered bases represented by $(E_1\mid_p,\ldots,E_n\mid_p)$ agress with the given orientation at $p$. In my opinion it is somewhat problematic to use the phrase "positively oriented" in this definition because it may mislead the reader to believe that this is an absolute property of $(E_1\mid_p,\ldots,E_n\mid_p)$. But of course this a matter of taste.] A negatively oriented frame is defined analogously. [Added: This means that a local frame is oriented if it is either positively or negatively oriented. Again this has to be understood relative to the given pointwise orientation; there is no absolute notion of being positively or negatively oriented. Note that if a local frame $(E_1,\ldots,E_n)$ is negatively oriented, then there also exists a positively oriented local frame on $U$; simply take $(E_2,E_1,E_3,\ldots,E_n)$. I would prefer to say that a local frame is consistent with the the given pointwise orientation instead of being oriented. Again a matter of taste.]
A pointwise orientation is said to be continuous if every point is in the domain of an oriented local frame. [Added: We may also require that all these local frames are positively oriented. In fact, if some of the local frames is negatively oriented, we can replace it by a positively oriented local frame.] An orientation of $M$ is a continuous pointwise orientation. An oriented manifold is a smooth manifold together with a choice of orientation. We say $M$ is orientable if there exists an orientation for it, and nonorientable if not.
If $M$ is $0$-dimensional, this definition just means that an orientation of $M$ is a choice of $\pm 1$ attached to each of its points. The local constancy condition is vacuous in this case, and the notion of oriented frames is not useful. Clearly every $0$-manifold is orientable.