# Manifolds that admit Lorentzian metrics?

John Lee says in "Riemannian Manifolds: An Introduction to Curvature":

With some more sophisticated tools from algebraic topology, it can be shown that every noncompact connected smooth manifold admits a Lorentz metric, and a compact connected smooth manifold admits a Lorentz metric if and only if its Euler characteristic is zero (see [O’N83, p. 149]).

The reference is

[O’N83] Barrett O’Neill, Semi-Riemannian Geometry with Applications to General Relativity, Academic Press, New York, 1983.

But I can't get access to this book. Is there another more available reference? It could be lecture notes, does not have to be a published text.

Thank you.

• O'Neill's text is available on libgen. – Jackozee Hakkiuz Apr 23 at 8:37

I don't know of another reference, but the proof is not too hard once you have the right tools, there are just a lot of them.

Let $$\dim M = n$$. We can reduce the structure group of $$TM$$ to $$O(n)$$ by choosing a Riemannian metric. If $$TM$$ admits an indefinite metric with signature $$(p, q)$$, then the structure group of $$TM$$ reduces to $$O(p, q)$$, which is an indefinite orthogonal group. Now $$O(p)\times O(q)$$ is a maximal compact subgroup of $$O(p, q)$$ and hence a deformation retract of it, so the structure group of $$TM$$ reduces further to $$O(p)\times O(q)$$. Therefore, there are vector bundles $$E$$ and $$F$$ with $$\operatorname{rank} E = p$$ and $$\operatorname{rank} F = q$$ such that $$TM \cong E\oplus F$$. Conversely, if $$TM \cong E\oplus F$$ with $$\operatorname{rank} E = p$$ and $$\operatorname{rank} F = q$$, then $$TM$$ admits an indefinite metric with signature $$(p, q)$$, for example $$g = g_E - g_F$$ where $$g_E$$ and $$g_F$$ are Riemannian metrics on $$E$$ and $$F$$ respectively.

The case of Lorentzian metrics corresponds to $$q = 1$$; some people would instead say $$p = 1$$, but it doesn't matter, it is equivalent. The above shows that $$TM$$ admits a Lorentzian metric if and only if $$TM \cong E\oplus L$$ where $$\operatorname{rank}(E) = n - 1$$ and $$L$$ is a real line bundle.

If $$M$$ is not closed, it follows from obstruction theory that $$TM$$ admits a nowhere zero section and hence $$TM \cong E\oplus\varepsilon^1$$ where $$\varepsilon^1$$ denotes the trivial real line bundle. Therefore $$M$$ admits a Lorentzian metric.

If $$M$$ is closed and orientable, there is a double cover$${}^*$$ $$p : M' \to M$$ associated to $$L$$ such that $$p^*L \cong \varepsilon^1$$ and hence $$TM' \cong p^*TM \cong p^*E\oplus p^*L \cong p^*E\oplus\varepsilon^1$$. As $$M'$$ is orientable, we see that $$\chi(M') = 0$$ by the Poincare-Hopf Theorem, but $$\chi(M') = 2\chi(M)$$, so $$\chi(M) = 0$$.

If $$M$$ is closed and non-orientable, let $$\pi : \widetilde{M} \to M$$ be the orientable double cover. Then $$T\widetilde{M} \cong \pi^*TM \cong \pi^*E\oplus\pi^*L$$. Applying the argument in the previous paragraph with $$L$$ replaced by $$\pi^*L$$, we see that $$\chi(\widetilde{M}) = 0$$, but $$\chi(\widetilde{M}) = 2\chi(M)$$, so $$\chi(M) = 0$$.

$${}^*$$Explicitly, $$M' = S(L)$$, the sphere bundle of $$L$$ with respect to a Riemannian metric, and $$p$$ is just the restriction of the projection $$L \to M$$ to $$S(L)$$. The claim follows once you know that the pullback of a vector bundle by its own projection is trivial and the normal bundle of the sphere bundle in a vector bundle is trivial; see here for the latter claim.