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In Riemannian geometry, we have

Proposition Any Manifold has a Riemannian metric.

However, we cannot place the proof on pseudo-Riemannian situation because we do not hold the signature on manifold. So my question is

  • Is there always a pseudo-Riemannian metric on manifold, which satisfies signature $(s,v)$?

  • If not, what condition on manifold keeps it having the pseudo-Riemannian metric?

Any advice is helpful. Thx.

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A necessary and sufficient condition for a smooth $n$-manifold $X$ to admit a metric of signature $(r, s)$ is that the tangent bundle admits a direct sum decomposition into a bundle of rank $r$ and a bundle of rank $s$, or equivalently that the tangent bundle admits a reduction of the structure group to $GL_r \times GL_s$.

When either $r$ or $s$ is equal to $1$ this is equivalent to the existence of a nonvanishing vector field, which is automatic when $X$ is noncompact and governed by whether $\chi(X) = 0$ otherwise, by the Poincaré–Hopf theorem.

In general I think it's a difficult question to determine when this reduction is possible. There are necessary conditions coming from characteristic classes which show that, for example, an even-dimensional sphere $S^{2n}$ does not admit a metric of any indefinite signature. A sufficient condition is that the manifold smoothly fibers over an $r$-manifold (or an $s$-manifold).

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  • $\begingroup$ Can you please introduce a reference that explains these results? $\endgroup$ – QGravity May 3 '18 at 21:04
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This doesn't answer the general question, but Proposition 5.37 from O'Neill's Semi-Riemannian Geometry might be of interest:

Proposition. For a smooth manifold $M$ the following are equivalent:

(1) There exists a Lorentz metric on $M$.

(2) There exists a time-orientable Lorentz metric on $M$.

(3) There is a nonvanishing vector field on $M$.

(4) Either $M$ is noncompact, or $M$ is compact and Euler number $\chi(M)=0$.

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  • $\begingroup$ Can you please introduce a reference that explains these results? $\endgroup$ – QGravity May 3 '18 at 21:04

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