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An exercise in "Riemannian Geometry" by Gallot, Hulin, Lafontaine (p. 53):

[Check that] For instance, there is no Lorentzian metric on the sphere $S^2$.

I am aware of this question and also of this discussion, but I would like to use an "elementary" partition of unity argument to prove the thesis (since in the book the previous quote links to this type of proof of existence for Riemannian metrics in order to obtain the result).

We can mimic the proof and construct a smooth tensor field with partitions of unity and we know that the sum does not need to be of given signature even if each (local) term of the sum is, but how this relates to impossibility? I guess that hairy ball theorem should appear somewhere.

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    You do not need a partition of unity. Just construct a continuous line field on the surface using the null cone of the Lorentzian metric. – 2017-02-11
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    Ok, but how can I assure that choosing a vector from the null cone at every point leads to a continuous line field? – 2017-02-11

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A lorentzian metric is of signature $(1,1)$. Using an auxiliary Riemannian structure this would implies that the tangent bundle of the sphere splits as the direct sum of two line bundles. A line bundle on the sphere is trivial, so it is impossible as for instance the Euler characteristic of the sphere is not 0.

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    Is "a line bundle on the sphere is trivial" something that I am supposed to know at the beginning of a basic Riemannian geometry course? I have never heard about that, at least in this form. Could you please provide further informations? – 2017-02-11
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    Afetr the choice of a riemanian metric the lorentzain metric is a symmetic matrix with one positive one negative eigenvalue. Consider the vector filed of norm one with positive eigenvalue.... – 2017-02-11
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    Ok, I think I can get it; just a little detail: the continuity of this vector field follows from the continuity (actually smoothness) of the positive sub-bundle and/or from the continuity of the Riemannian metric norm one condition? – 2017-02-12
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    The local continiuty of "this vector" field follws from the continuyty of the map which associate to a $(2,2)$ symmetric matrix $S$ with negative determinant a eigenvector with positive eigenvalue and norm one. The difficulty is that you have two such maps $V(S)$ and $-V(S)$. But globally one need to use the simple connectivity of the 2-sphere to define this vector... – 2017-02-12
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    Why simple connectivity and not connectivity? The continuity of the map(s) you mentioned and the connectedness of the sphere should allow the global extension, just like the definition of the pseudo-Riemannian metric signature does not depend on the point if the manifold is connected. What am I missing here? – 2017-02-12
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    The 2 vector fields V, -V (which are opposite) are not globally defined..Replace the sphere by the projective space. How is your vector field defined ? Think of a Mobius strip and V is a normal field of the soul of this strip, something like this – 2017-02-12
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    Sorry, I cannot get the difference with projective space case. The example of Moebius strip makes me think to orientability, but I am not able to realize how this could relate to global continuous extension - but I know that simply conn. manifolds are orientable. – 2017-02-12
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    It is easy to construct a Lorentz metric $l$ on a Mobius strip with no vector field V such that $l(V,V)>0$ – 2017-02-12