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Minkowski space time is defined in terms of a flat pseudo-Riemannian manifold. I have wondered if it can be redefined as Riamannian manifold and in the case what type of curvature would there appear.

Formally:

Let M be a semi-Riemannian manifold of dimension 4, corresponding to the Minkowski space, and let g be the metric tensor (non positive definite), T be the Riemann curvature tensor and P a generic point of M.

Question 1

Which (if any) of the following is true?

a. at every P there exists one system of coordinates for which the metric g becomes Riemannian (positive definite) in a ball of radius R non infinitesimal centred in P

b. there exists one system of coordinates for which g is Riemannian (positive definite) at all P of M

Comment: in words, can we, with a change of coordinates, get rid of semi-Riemannianity – either in finite region or globally?

If this is the case, how do we pay it in terms of curvature? This the next question:

Question 2

c. if previous a) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

d. if previous b) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

Thanks a lot

massimo
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2 Answers2

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We interpret OP's question (v3) as essentially asking

Can a Lorentzian manifold (with Minkowski signature) by coordinate transformations be redefined as a Riemannian manifold (with Euclidean signature)?

The answer is No since the metric signature of a pseudo-Riemannian manifold is invariant under general coordinate transformations. This follows e.g. from Sylvester's law of inertia. Recall that the metric tensor in any coordinate system is a real symmetric matrix, and therefore diagonalizable.

Qmechanic
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As mentioned by Qmechanic, the answer to your questions is no.

However, assuming space-time is oriented, we have the following:

For any pseudo-Riemannian metric $g$, there exists a normalized time-like one-form $h^0$ and a Riemannian metric $g^R$ so that $$ g = 2h^0\otimes h^0 - g^R $$

This yields a locally Euclidean topology compatible with the manifold topology.

Christoph
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