# Is it enough to give a time-orientation to define a spin structure?

Maybe I got it wrong and my question doesn't make sense, excuse me if that's the case. For a smooth Lorentz 4-manifold $$(M, g)$$ with signature $$(- + + +)$$ is it enough to give a time-orientation to define a spin structure? Or is spatial orientation necessary too?

More precisely, let $$\pi_{O}: F(M) \rightarrow M$$ be the orthonormal frame bundle (a principal $$O(1, 3)$$-bundle). So, if I choose the set $$L$$ consisting only of the orthonormal bases such that $$e_{0}$$ is a timelike vector ($$g(e_{0}, e_{0}) < 0$$), future-pointing (in the class designated as future) and $$e_{1}, e_{2}, e_{3}$$ are spacelike vectors, then do I have a principal $$SO(1, 3)$$-bundle $$(L, SO(1, 3), \pi_{SO})$$?

$$SO(1, 3) = \{A \in O(1, 3): det(A) = 1\ and\ A_{0}^{0} \geq 1\},$$ where $$Ae_{0} = (A_{0}^{0}, A_{1}^{1}, A_{2}^{2}, A_{3}^{3})^{T}$$. And the spin structure comes from the double covering $$\Lambda: SL(2, \mathbb{C}) \rightarrow SO(1, 3)$$.

My doubt is because the time-orientation is not a common orientation for manifolds (in the sense of the Jacobian sign), as there are examples of non-orientable manifolds that are time-orientable. Appreciate.

• @QGravity A reference is already given in my answer (i.e. the proof of this should be in Baum's book). But I'm not sure why you think the orientability is the non-obvious part here: Orientability means you can actually reduce the $\mathrm{O}(1,3)$-bundle of Lorentz transformations to the $\mathrm{SO}^+(1,3)$-bundle of orthochronous proper Lorentz transformations. Since a spin structure is by definition a lift of the $\mathrm{SO}^+(1,3)$ structure to a $\mathrm{Spin}(1,3)$-bundle, I'm not sure how you want to even define the notion of spin structure for non-orientable manifolds. Commented Apr 14, 2023 at 13:44
• I understand the orientability; What I do not understand is time-orientability. Orientability means $w_1(TM)=0$, where $w_1$ is the first Stiefel-Whitney class of the manifold $M$. Now on a $d$-dimensional Lorentzian manifold $M$, the tangent bundle decomposes as $TM=TM_T\oplus TM_S$, where $TM_T$ and $TM_S$ are rank-$1$ and rank-$(d-1)$ sub-bundles, respectively. Then, $w_1(TM)=0$ only implies $w_1(TM_T)+w_1(TM_S)=0$, and not $w_1(TM_T)=0$, which means time-orientability. For example a manifold could be time-orientable but not necessarily orientable. Commented Apr 14, 2023 at 14:11