Section of the $\text{O}(1,n)$ bundle versus section of the Grassmann bundle

Question

A manifold in general relativity only admits a metric tensor on the condition that the Grassmann bundle admits a section (this is due to the bundle of metric tensors being $\approx \text {Gr}(1,n) \times \text{S}(n,0)$, with $\text{S}(n,0)$ the set of Riemannian metrics which always has a section on paracompact manifolds). This corresponds to all manifolds except compact manifolds with an Euler characteristic $\chi \neq 0$.

But there is another way to construct the metric tensor, from a frame field and the bundle metric on the tangent bundle. So I am assuming that a section of the orthonormal tangent bundle $\text{O}(1,n)$ should only exist under the same condition. What would be the proof of this, though?

Answer 1

If the frame bundle has a global section $\{\theta^j\}_{j=0}^n$, then you obtain a Lorentzian metric by $$g=-\theta^0\otimes\theta^0+\sum_{i=1}^n\theta^i\otimes\theta^i.$$ However, this is overkill. You could just select $\theta^0$, a Riemannian metric $h$ on $\mathscr M$, and let $X$ be the dual of $\theta^0$ with respect to $h$. Then the usual procedure gives a Lorentzian metric $$g(Y,Z)=h(Y,Z)-2\frac{h(X,Y)h(X,Z)}{h(X,X)},\quad Y,Z\in\Gamma(T\mathscr M).$$ But the condition that $\mathscr M$'s frame bundle admits a global section implies it is parallelizable, that is, $T\mathscr M\approx \mathscr M\times\Bbb R^n$. For $\mathscr M$ compact, this implies $\chi(\mathscr M)=0$ by Hopf's theorem. So you gain no new information by casting the problem in this light, because it's a much stronger condition.