# (Time-)Orientability in the Language of Fiber Bundles

I'm currently studying spin geometry through Hamilton's book Mathematical Gauge Theory. At a given point, Hamilton considers a pseudo-Riemannian manifold, which I'll take to be Lorentzian in $$d=3+1$$ spacetime dimensions for concreteness and simplicity. Hamilton states that the bundle of frames can be seen as a principal $$O(3,1)$$-bundle. I disagree, as the bundle of orthonormal frames is technically a different bundle, but I can understand the basic construction (which is discussed, for example, in Wald's General Relativity). After this comment, Hamilton defines that a spacetime is orientable if the bundle of orthonormal frames can be reduced to a principal $$SO(3,1)$$-bundle under the embedding $$SO(3,1) \subset O(3,1)$$ and it is time-orientable if the bundle of orthonormal frames can be reduced to a principal $$O^+(3,1)$$-bundle under the embedding $$O^+(3,1) \subset O(3,1)$$. An orientable, time-orientable spacetime is a spacetime in which the bundle of orthonormal frames can be reduced to a principal $$SO^+(3,1)$$-bundle under the embedding $$SO^+(3,1) \subset O(3,1)$$.

In general relativity books, such as Wald's, orientability and time-orientability are defined differently. As far as I remember, a spacetime is orientable if one can choose a smooth volume form on the spacetime and it is time-orientable if one can choose a smooth everywhere-timelike vector field on the spacetime.

I assume these two definitions are equivalent (please correct me if I'm mistaken). My question is then the following: why are the existence of these structures (a volume element or a timelike vector field) necessary and sufficient for the existence of these bundle reductions? Which sorts of complications would arise if such structures were not available?

I’ll answer the orientability question in detail for vector bundles in general (you can phrase some of these at the level of the frame bundle of the vector bundle as well), and let you make the necessary modifications for the time-orientability case. It seems long, but really, the length is more so an explanation of notation, not fundamental ideas. The only real ideas used are a standard partition-of-unity trick to patch together local results, and the idea that a global nowhere-vanishing section of the top exterior power provides us with a ‘reference’ to compare stuff against.

Let $$(E,\pi,B)$$ be a rank $$k$$ vector bundle. You can think of this as a fiber bundle with typical fiber $$\Bbb{R}^k$$, and structure group $$\operatorname{GL}(k,\Bbb{R})$$. Then, the following statements are all equivalent, in which case we say the vector bundle $$(E,\pi,M)$$ is orientable (as a vector bundle)

1. $$E$$ has a vector bundle atlas with the local trivialization transitions consisting of linear maps of positive determinant (i.e there is a reduction of structure group from $$\operatorname{GL}(k,\Bbb{R})$$ to $$\operatorname{GL}^+(k,\Bbb{R})$$)
2. the structure group can be reduced from $$\operatorname{GL}(k,\Bbb{R})$$ to $$\operatorname{SL}(k,\Bbb{R})$$
3. there is a nowhere-vanishing section of $$\bigwedge^k(E)$$
4. $$\bigwedge^k(E)$$ is a trivializable vector bundle
5. $$\bigwedge^k(E^*)$$ is a trivializable vector bundle
6. there is a nowhere-vanishing section of $$\bigwedge^k(E^*)$$

The equivalence of (3) and (4) is easy: a vector bundle (in this case $$\bigwedge^k(E)$$) admits a nowhere-vanishing section if and only if it is trivializable. The equivalence of (4) and (5) follows from the canonical isomorphism $$\bigwedge^k(E^*)\cong [\bigwedge^k(E)]^*$$ and the easy fact that a vector bundle is trivializable if and only if the dual vector bundle is. Next, (5) and (6) are equivalent for the same reason (3) and (4) are.

Proof $$(2)\implies (1)$$

Obvious since $$\operatorname{SL}(k,\Bbb{R})\subset\operatorname{GL}^+(k,\Bbb{R})$$.

Proof $$(1)\implies (3)$$

Let $$\mathcal{V}=\{(U_{\alpha},\phi_{\alpha})\}_{\alpha\in A}$$ be a vector bundle atlas, i.e each $$U_{\alpha}\subset B$$ open, and $$\phi_{\alpha}:\pi^{-1}(U_{\alpha})\to U_{\alpha}\times\Bbb{R}^k$$ smooth, fiber-preserving and fiberwise linear. With some topological finessing, we may WLOG assume $$\{U_{\alpha}\}_{\alpha\in A}$$ is a locally-finite open cover of $$B$$. By considering $$s_{\alpha,i}(b)=\phi_{\alpha}^{-1}(b,e_i)$$, where $$e_i$$ is the $$i^{th}$$ standard basis vector of $$\Bbb{R}^k$$, we get a collection of $$k$$ local sections $$\{s_{\alpha,1},\dots,s_{\alpha,k}\}$$ of $$E$$ over $$U_{\alpha}$$. Thus far, it has just been notation. But now comes the construction: fix a smooth partition of unity $$\{f_{\alpha}\}_{\alpha\in A}$$ of $$B$$ subordinate to the open cover $$\{U_{\alpha}\}_{\alpha\in A}$$, and define for each index $$\alpha\in A$$ and point $$b\in B$$, \begin{align} \zeta_{\alpha}(b)&:= \begin{cases} f_{\alpha}(b)\cdot s_{\alpha,1}(b)\wedge\cdots\wedge s_{\alpha,k}(b)\quad&\text{if b\in U_{\alpha}}\\ 0&\text{else} \end{cases} \end{align} Then, because $$f_{\alpha}$$ has support contained inside $$U_{\alpha}$$, it follows that each $$\zeta_{\alpha}$$ is a smooth global section of $$\bigwedge^k(E)$$. Now, define $$\zeta:=\sum\limits_{\alpha\in A}\zeta_{\alpha}$$; this looks like an infinite series, but it is not; for each point $$b\in B$$, there is a neighbourhood $$V$$ which intersects only finitely many $$U_{\alpha}$$’s (due to local-finiteness); and so in this neighbourhood $$V$$, we have that $$\zeta|_V$$ is a finite sum of $$\zeta_{\alpha}$$’s, and thus $$\zeta|_V$$ is locally smooth, and since smoothness is a local property, we have that $$\zeta:B\to E$$ is a smooth global section of $$E$$. We now claim it is nowhere-vanishing. Fix any point $$b\in B$$. Let $$\alpha_1,\dots,\alpha_m$$ be those indices such that $$b\in U_{\alpha_j}$$. Then, \begin{align} \zeta(b)&=\sum_{\alpha}\zeta_{\alpha}(b)=\sum_{j=1}^mf_{\alpha_j}(b)\cdot s_{\alpha_j,1}(b)\wedge\cdots\wedge s_{\alpha_j,1}(b). \end{align} Now, observe that these $$s_{\alpha_j,i}(b)$$’s are sections corresponding to different local trivializations. But since the point $$b$$ lies common to all their domains, we can express each of these relative to a common local frame of sections, for example $$s_{\alpha_1,i}(b)$$. Recall now that \begin{align} s_{\alpha_j,1}(b)\wedge\cdots\wedge s_{\alpha_j,k}(b)&=(\text{some positive determinant})\cdot s_{\alpha_1,1}(b)\wedge\cdots\wedge s_{\alpha_1,k}(b) \end{align} Here, we have to consider the bundle transition between $$\phi_{\alpha_j}$$ and $$\phi_{\alpha_1}$$ over the fiber $$E_b$$, and consider its determinant (because that’s how wedge products behave under change of basis); let me call this determinant $$\delta_j(b)$$. This determinant is positive by hypothesis. So, we have \begin{align} \zeta(b)&=\left(\sum_{j=1}^mf_{\alpha_j}(b)\cdot \delta_j(b)\right)\cdot s_{\alpha_1,1}(b)\wedge\cdots\wedge s_{\alpha_1,k}(b). \end{align} In the brackets, we have each $$\delta_j(b)>0$$, and each $$f_{\alpha_j}(b)\geq 0$$ and also $$1=\sum_{\alpha}f_{\alpha}(b)=\sum_{j=1}^mf_{\alpha_j}(b)$$, so there is some $$f_{\alpha_j}(b)>0$$. So, the bracket is strictly positive, and so multiplying by the wedge (which is non-zero) gives that $$\zeta(b)\neq 0$$. Since $$b\in B$$ was arbitrary, we have that $$\zeta$$ is indeed a smooth nowhere-vanishing global section of $$\bigwedge^k(E)$$.

Proof $$(3)\implies (2)$$

Suppose now that we have a nowhere-vanishing smooth global section $$\zeta$$ of $$E$$. Fix a vector bundle atlas $$\mathcal{V}=\{(U_{\alpha},\phi_{\alpha})\}_{\alpha\in A}$$ such that each $$U_{\alpha}$$ is connected. Letting $$s_{\alpha,i}$$ be the local sections as defined above, note that there exist smooth functions $$h_{\alpha}:U_{\alpha}\to\Bbb{R}$$ such that \begin{align} \zeta|_{U_{\alpha}}&=h_{\alpha}\cdot s_{\alpha,1}\wedge\cdots\wedge s_{\alpha,k}. \end{align} Since $$\zeta$$ is nowhere-vanishing, it follows $$h_{\alpha}$$ is a nowhere-vanishing function; now define the sections \begin{align} t_{\alpha,1}:=h_{\alpha}\cdot s_{\alpha,1},\quad\text{and}\quad t_{\alpha,i}=s_{\alpha,i}\quad (2\leq i \leq k). \end{align} Then, $$\{t_{\alpha,i}\}_{i=1}^k$$ still form a local frame of sections (since $$h_{\alpha}$$ was nowhere-vanishing) thus gives us a local trivialization $$\psi_{\alpha}$$. Now for any two indices $$\alpha,\beta$$, on the intersection of domains, $$U_{\alpha}\cap U_{\beta}$$, we have (by definition of $$t$$’s) \begin{align} \zeta=t_{\alpha,1}\wedge\cdots\wedge t_{\alpha,k}=t_{\beta,1}\wedge\cdots\wedge t_{\beta,k}. \end{align} So, the transition from $$\psi_{\alpha}$$ to $$\psi_{\beta}$$ has determinant $$+1$$. Hence $$\mathcal{V}_{+1}:=\{(U_{\alpha},\psi_{\alpha})\}_{\alpha\in A}$$ is a vector bundle atlas with transitions lying in $$\operatorname{SL}(k,\Bbb{R})$$, and thus we have reduced the structure group.

To summarize, if you assume (1), then to get (3), you define the obvious local section of the top exterior power, and then you patch it together to get a smooth global section of the top exterior power. The positivity of the transition determinants implies this section is nowhere-vanishing (I bolded the places where this comes into play above). Next, to go from (3) to (2), we start with any vector bundle atlas, and by using $$\zeta$$ as our method of ‘normalizing’ we have absorbed the factor $$h_{\alpha}$$ into the first section $$s_{\alpha,1}$$ to get the $$t_{\alpha}$$’s; and these by construction have equal top-wedges, and thus the transition between them has determinant $$+1$$. Said again: the $$\zeta$$ serves as our ‘reference’ (it was arbitrary, but once we fix it, we have that reference which allows us to get the nowhere-vanishing $$h_{\alpha}$$’s and thus $$t_{\alpha}$$’s). Thus, we have the equivalence of all statements (1)-(6).

Suppose now that $$E$$ has a smooth bundle metric $$g$$ of signature $$(p,q)$$ (with $$p+q=k$$) (the existence of this bundle metric is equivalent to the existence of a reduction of the structure group to the indefinite orthogonal group $$\operatorname{O}(p,q)$$). In the Riemannian case $$(p,q)=(0,k)$$, this always exists (again, a partition-of-unity argument). Then statements (1)-(6) are equivalent to

1. there exists a reduction of the structure group from $$\operatorname{O}(p,q)$$ to $$\operatorname{SO}(p,q)$$.

Proof $$(7)\implies (1)$$

Trivial since by definition, elements of $$\text{SO}(p,q)$$ have positive determinant.

Proof $$(3)\implies (7)$$.

For this, we mimic the above proof of $$(3)\implies (2)$$: start with a vector bundle atlas $$\mathcal{V}_g$$ with transitions in $$\operatorname{O}(p,q)$$ (like I mentioned above, this is equivalent to the existence of the bundle metric $$g$$) and WLOG assume the $$U_{\alpha}\subset B$$ are all connected. In this case, the corresponding sections $$\{s_{\alpha,1},\dots, s_{\alpha,k}\}$$ are $$g$$-orthonormal. This time, we define $$t_{\alpha,1}:=\frac{h_{\alpha}}{|h_{\alpha}|}\cdot s_{\alpha,1}$$ (since $$h_{\alpha}$$ is nowhere-vanishing, this is still smooth), and $$\{t_{\alpha,i}\}_{i=1}^k$$ is still $$g$$-orthonormal since we only modified the first section by a sign. Now, we have on the intersection of two domains $$U_{\alpha}\cap U_{\beta}$$ that \begin{align} \zeta&=|h_{\alpha}| t_{\alpha,1}\wedge\cdots\wedge t_{\alpha,k}=|h_{\beta}|\cdot t_{\beta,1}\wedge \cdots\wedge t_{\beta,k}. \end{align} This shows the transition from the corresponding trivialization $$\psi_{\alpha}$$ to $$\psi_{\beta}$$ has positive determinant (it already landed in $$\operatorname{O}(p,q)$$, but now in particular we have shown it lands in $$\operatorname{SO}(p,q)$$). Thus we have found a reduction of structure group from $$\operatorname{O}(p,q)$$ to $$\operatorname{SO}(p,q)$$.

Thus, in the case that $$E$$ has a bundle metric $$g$$ of signature $$(p,q)$$, we have that statements $$(1)$$-$$(7)$$ are equivalent (and only $$(7)$$ makes reference to the metric).

The case of $$TM$$

Finally, suppose $$M$$ is a smooth manifold. We say $$M$$ is orientable as a manifold if the tangent bundle $$(TM,\pi,M)$$ as a vector bundle over $$M$$. So, when we take $$E=TM$$, the statement you see in Wald is $$(6)$$, i.e the existence of a smooth volume form on $$M$$. An analogue of statement $$(1)$$ is also one of the common ways of defining orientability of a manifold (require the existence of charts of $$M$$ whose transitions have positive Jacobian determinants… in this case rather than flipping a sign of the section $$s_1$$, you replace the sign of the coordinate function $$x^1$$ on the base manifold, if necessary).

So, once again, the reason why the existence of nowhere-vanishing sections of the top exterior power (or its dual) is so crucial is that it provides us a ‘reference’ to compare against, so that we can modify signs/normalize appropriately (i.e absorb $$h_{\alpha}$$ or $$\text{sign}(h_{\alpha})$$ into $$s_{\alpha,1}$$ to get $$t_{\alpha,1}$$). The reason for this necessity is that there is no notion of positivity/negativity of differential forms (or more generally the top exterior power $$\bigwedge^k(E)$$). We can only compare signs/magnitudes once we have a ‘reference’ quantity to measure against. It’s kind of like how for a hyperplane, there are two choices of unit normals, but there is no intrinsic way to say one is better than the other; but after you fix one, you can compare signs/magnitudes of normal vectors relative to that chosen one.

This need for having a ‘reference’ to compare (signs or normalizations) against, is also why we can only define integrals of differential forms on oriented manifolds. On unoriented/non-orientable manifolds, we would end up with inconsistencies when we try to piece together various chart-based definitions because we wouldn’t know whether to put a $$+$$ or $$-$$.