Orientations and Covering Maps

Proof.

A covering map is a local diffeomorphism, so an orientation of \(M\) pulls back to an orientation of \(E\). Hence \(E\) is orientable.

Proof.

Let \(\mathcal{O}_E\) denote the given orientation on \(E\).

Suppose first that \(M\) is orientable, and let \(q\) be an arbitrary point of \(E\). Because \(M\) is connected it has exactly two orientations, and one of them, call it \(\mathcal{O}_M\), has the property that \(d\pi_q : T_q E \to T_{\pi(q)} M\) is orientation-preserving. The pullback orientation \(\pi^{*}\mathcal{O}_M\) then agrees with the given orientation at \(q\), so by connectedness it equals \(\mathcal{O}_E\) everywhere. Now let \(\varphi \in \mathrm{Aut}_\pi(E)\). The identity \(\pi \circ \varphi = \pi\) gives \[ \varphi^{*}\mathcal{O}_E = \varphi^{*}\bigl(\pi^{*}\mathcal{O}_M\bigr) = (\pi \circ \varphi)^{*}\mathcal{O}_M = \pi^{*}\mathcal{O}_M = \mathcal{O}_E, \] so \(\varphi\) is orientation-preserving. Hence the action of \(\mathrm{Aut}_\pi(E)\) preserves orientation.

Conversely, suppose the action of \(\mathrm{Aut}_\pi(E)\) is orientation-preserving, and let \(p \in M\). Choose a connected evenly covered neighbourhood \(U\) of \(p\); there is a smooth local section \(\sigma : U \to E\), which induces the pullback orientation \(\sigma^{*}\mathcal{O}_E\) on \(U\). Suppose \(\sigma_1 : U \to E\) is any other smooth local section over \(U\). Because \(\pi\) is a normal cover, \(\mathrm{Aut}_\pi(E)\) acts transitively on each fiber, so there is a deck transformation \(\varphi\) with \(\sigma_1(p) = \varphi\bigl(\sigma(p)\bigr)\). Then \(\varphi \circ \sigma\) is a local section agreeing with \(\sigma_1\) at \(p\), and since local sections of a covering are determined by one value on a connected domain, \(\sigma_1 = \varphi \circ \sigma\) on all of \(U\). Because \(\varphi\) is orientation-preserving, \[ \sigma_1^{*}\mathcal{O}_E = \sigma^{*}\varphi^{*}\mathcal{O}_E = \sigma^{*}\mathcal{O}_E, \] so the orientations induced by \(\sigma\) and \(\sigma_1\) coincide. We may therefore define an orientation \(\mathcal{O}_M\) on \(M\) by declaring it on each evenly covered open set to be the pullback induced by any local section; the computation just made shows these agree on overlaps, so \(\mathcal{O}_M\) is a well-defined orientation. Hence \(M\) is orientable.

Proof.

The smooth covering map \(q : S^n \to \mathbb{RP}^n\) is a two-sheeted normal cover whose only nontrivial deck transformation is the antipodal map \(\alpha(x) = -x\). Give \(S^n\) its standard orientation, determined by the outward unit normal \(N_x = x\): an ordered basis \((E_1, \dots, E_n)\) of \(T_x S^n\) is positively oriented exactly when \((x, E_1, \dots, E_n)\) is positively oriented in \(\mathbb{R}^{n+1}\). The antipodal map has differential \(-\,\mathrm{Id}\) on \(\mathbb{R}^{n+1}\), so it sends such a basis to \((-E_1, \dots, -E_n)\) at the point \(-x\), where the outward normal is \(N_{-x} = -x\). Comparing orientations, \[ \bigl(N_{-x}, -E_1, \dots, -E_n\bigr) = \bigl(-x, -E_1, \dots, -E_n\bigr) = (-1)^{n+1}\bigl(x, E_1, \dots, E_n\bigr) \] as a relation between signed bases of \(\mathbb{R}^{n+1}\), since negating all \(n+1\) vectors contributes the factor \((-1)^{n+1}\). Thus \(\alpha\) carries the induced orientation of \(T_x S^n\) to \((-1)^{n+1}\) times the induced orientation of \(T_{-x} S^n\), so \(\alpha\) is orientation-preserving precisely when \((-1)^{n+1} = 1\), that is, when \(n\) is odd. By the criterion for the base of a normal cover, \(\mathbb{RP}^n\) is orientable if and only if the deck action is orientation-preserving, which happens if and only if \(n\) is odd.

Proof.

Realize the open Möbius band as the quotient of \(\mathbb{R}^2\) by the smooth, free, and properly discontinuous action of \(\mathbb{Z}\) given by \[ n \cdot (x, y) = \bigl(x + n,\, (-1)^n y\bigr). \] The quotient map \(q : \mathbb{R}^2 \to E\) onto the quotient \(E\) is a smooth normal covering, and the deck group is the image of this \(\mathbb{Z}\)-action. Give \(\mathbb{R}^2\) its standard orientation, with orientation form \(dx \wedge dy\). The generator \((x, y) \mapsto (x + 1, -y)\) pulls this form back to \(dx \wedge d(-y) = -\,dx \wedge dy\), so it reverses orientation. The deck action is therefore not orientation-preserving, and by the criterion for the base of a normal cover, \(E\) is not orientable. The closed Möbius band, being the image of a strip \(\mathbb{R} \times [-r, r]\) under the same covering, fails to be orientable by the identical argument applied to the restricted cover.

Proof.

If \(M\) is orientable, then by the characterization of evenly covered sets, \(M\) itself is evenly covered by \(\widehat\pi\); hence \(\widehat M\) is the disjoint union of two open sets, each mapped homeomorphically onto \(M\), and each restriction of \(\widehat\pi\) is a diffeomorphism. The two components correspond to the two orientations of \(M\).

Now suppose \(M\) is nonorientable. Let \(W\) be a component of \(\widehat M\). The restriction of \(\widehat\pi\) to \(W\) is a covering map onto \(M\), so all its fibers have the same cardinality, which must be \(1\) or \(2\) since the fibers of \(\widehat\pi\) have cardinality \(2\). If the cardinality were \(1\), then \(\widehat\pi|_W\) would be an injective smooth covering map, hence a diffeomorphism, and its inverse would be a global smooth section of \(\widehat\pi\), inducing an orientation on \(M\)—contradicting nonorientability. Therefore the cardinality is \(2\), which forces \(W = \widehat M\), so \(\widehat M\) is connected; and since \(\widehat\pi\) is a local diffeomorphism with two-element fibers, it is a two-sheeted smooth covering map.

Proof Sketch.

Define \(\varphi\) by sending \(\tilde q \in \widetilde M\) to the pair \(\bigl(\widetilde\pi(\tilde q),\ (d\widetilde\pi_{\tilde q})_{*}\mathcal{O}^{\widetilde M}_{\tilde q}\bigr)\), where \(\mathcal{O}^{\widetilde M}\) is the given orientation of \(\widetilde M\) and \((d\widetilde\pi_{\tilde q})_{*}\) pushes it forward through the isomorphism \(d\widetilde\pi_{\tilde q}\) (an isomorphism because \(\widetilde\pi\), being a covering map, is a local diffeomorphism). By construction \(\widehat\pi \circ \varphi = \widetilde\pi\), so \(\varphi\) maps each fiber of \(\widetilde\pi\) into the corresponding fiber of \(\widehat\pi\). Since both \(\widetilde\pi\) and \(\widehat\pi\) are local diffeomorphisms, locally \(\varphi = \widehat\pi^{-1} \circ \widetilde\pi\) for suitable local sections, so \(\varphi\) is smooth and is itself a local diffeomorphism.

It remains to see that \(\varphi\) is bijective on each two-element fiber. Fix \(p \in M\); both \(\widetilde\pi^{-1}(p)\) and \(\widehat\pi^{-1}(p)\) have two elements, the latter being the two orientations of \(T_pM\). Since \(M\) is nonorientable and connected, its orientation covering \(\widehat M\) is connected by the Orientation Covering Theorem, and the same argument shows \(\widetilde M\) is connected: a disconnected oriented two-sheeted cover would split into two sheets each mapping diffeomorphically onto \(M\), whose inverse would be a global section and would orient \(M\). A connected two-sheeted cover is normal, with deck group of order two generated by the nontrivial automorphism \(\tau\) interchanging the two points of each fiber. By the criterion for the base of a normal cover, \(M\) nonorientable forces the deck action to be orientation-reversing, so \(\tau\) reverses the orientation of \(\widetilde M\). The two points \(\tilde q\) and \(\tau(\tilde q)\) of a fiber therefore push their orientations forward to the two opposite orientations of \(T_pM\); hence \(\varphi\) sends the two points of \(\widetilde\pi^{-1}(p)\) to the two distinct points of \(\widehat\pi^{-1}(p)\), so it is fiberwise bijective. A bijective local diffeomorphism is a diffeomorphism.

That \(\varphi\) is orientation-preserving is immediate from its definition: it carries \(\mathcal{O}^{\widetilde M}_{\tilde q}\) to the tautological orientation \((d\widetilde\pi_{\tilde q})_{*}\mathcal{O}^{\widetilde M}_{\tilde q}\) at \(\varphi(\tilde q)\), which is precisely how \(\widehat M\) is oriented. For uniqueness, suppose \(\varphi'\) is another orientation-preserving diffeomorphism over \(M\). At any point \(\tilde q\), both \(\varphi(\tilde q)\) and \(\varphi'(\tilde q)\) lie in the same two-element fiber \(\widehat\pi^{-1}(\widetilde\pi(\tilde q))\), and both must carry the orientation of \(\widetilde M\) to the same orientation of \(T_{\widetilde\pi(\tilde q)}M\), since each is orientation-preserving over \(M\); the two fiber points being distinguished exactly by their orientation, \(\varphi(\tilde q) = \varphi'(\tilde q)\). Thus \(\varphi\) is unique.

Proof.

Suppose \(M\) is not orientable. By the Orientation Covering Theorem its orientation covering \(\widehat\pi : \widehat M \to M\) is then a connected two-sheeted covering. A connected two-sheeted covering of \(M\) corresponds, under the covering-space dictionary, to an index-two subgroup of the fundamental group of \(M\), namely the image of the fundamental group of the cover. Thus nonorientability produces a subgroup of index two. Contrapositively, if the fundamental group of \(M\) has no subgroup of index two, then \(M\) is orientable. A simply connected manifold has trivial fundamental group, which has no subgroup of index two, so it is orientable.