A Metric, which is a (2,0) tensor on a manifold $M$ that is symmetric and non-degenerate.

A Riemannian Meric is a metric that is positive definite.

I'm reading proposition 13.3 from Lee's intro to smooth manifolds.

I understand that we can prove that every smooth manifold admits Riemannian Metric using partition of unity.

I'm just wondering is it possible to use that same proof to prove that every smooth manifold admits some other metric that is not positive definite?

I don't see how the proof relies on the positive definiteness.

  • $\begingroup$ The proof does rely on positive-definiteness: at one point you need to use that a linear combination of positive-definite inner products on a vector space with positive coefficientes is itself a positive-definite inner product, and that is not true if you remove the positive-definiteness. $\endgroup$ – Mariano Suárez-Álvarez Oct 23 '16 at 5:29
  • $\begingroup$ Yes, but the thing is cant you just construct some non positive definiteness metric locally and take linear combination of those to construct a global non-positive definite metric? $\endgroup$ – Phantom Oct 23 '16 at 6:55
  • $\begingroup$ My comment above gives the precise reason why you cannot! $\endgroup$ – Mariano Suárez-Álvarez Oct 23 '16 at 7:00
  • $\begingroup$ Are you saying that linear combination does not preserve other properties besides positive definiteness? $\endgroup$ – Phantom Oct 23 '16 at 7:11
  • 1
    $\begingroup$ Exactly. This is the content of my first comment :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 23 '16 at 9:22

You cannot, because there are differential-topological obstructions to the existence: not all manifolds admit metrics of a given signature.

You can see the discussion here for example. The simplest case is that of metrics of signature $(1,n-1)$: they exist iff the tangent bundle contains a $1$-dimensional subbundle, and that occurs iff the Euler class of the manifold is zero.


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