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I am studying almost contact manifolds. Since every smooth manifold admits the Riemannian metric, with the help of this metric and contact structure one defines the contact metric manifold and then the Sasakian manifold. I am trying to study all above with semi-Riemannian metrics. So the first question comes is that.....

Does every smooth manifold admit proper semi-Riemannian metric?

Vipul Singh
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Aug 21 '22 at 21:20
  • Have you looked at the proof that all manifolds admit Riemannian metrics and tried to apply it? I don't know that much about semi-Riemannian metrics, but I have a hard time imagining what the important difference would be here - it's still supposed to be a partition of unity argument. – A. Thomas Yerger Aug 21 '22 at 22:45
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    The difference, @A.ThomasYerger, is that the space of indefinite inner products on a vector space is not convex. This leads to obstructions if the signature is indefinite. – Danu Aug 21 '22 at 22:49
  • Of course, if you want to be pedantic every smooth $n$-manifold admits metrics of signature $(n,0)$ and $(0,n)$. – Kajelad Aug 21 '22 at 22:52
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    For the nontrivial cases, see this question. – Kajelad Aug 21 '22 at 22:54

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No, and we can say more.

A smooth manifold $M$ admits a Lorenztian metric iff (a) it is noncompact and/or (b) its Euler characteristic, $\chi(M)$ vanishes (see O'Neill, $\S$5, Proposition 37). On the other hand, the Euler characteristic of every odd-dimensional manifold is zero. So, a smooth manifold $M$ admits no Lorenztian metric iff

  1. $M$ is compact,
  2. $\dim M$ is even, and
  3. $\chi(M) \neq 0$.

In particular, compact surfaces of nonzero Euler characteristic, e.g., the $2$-sphere, admit no semi-Riemannian metric of indefinite signature.

I don't know in general which smooth manifolds satisfying (1)-(3) of dimension $\geq 4$ admit a semi-Riemannian metric of indefinite (and hence necessarily non-Lorentzian) signature, but even-dimensional spheres, for example, admit no indefinite metric.

O'Neill, Semi-Riemannian Geometry, with Applications to General Relativity (1983).

See also: Manifolds that admit Lorentzian metrics?

Travis Willse
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  • Ok, It means an odd-dimensional manifold admits a Lorentz metric because its Euler characteristic vanishes. Does an odd-dimensional manifold admits a semi-Riemannian metric of any signature? – Vipul Singh Aug 22 '22 at 04:09
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    @Vipul: no, as described in the question linked by Kajelad in the comments there are obstructions coming from characteristic classes. For example, if a $5$-manifold admits a metric of signature $(2, 3)$ then its first Pontryagin class $p_1$ must vanish, but the first Pontryagin class of, say, $S^1 \times \mathbb{CP}^2$ doesn't vanish. – Qiaochu Yuan Aug 22 '22 at 16:36