I have a question. I seen that Cauchy Schwarz inequality is not valide in the case of pseudo riemannian metric because it is not positive ( or negative) define, I would like to know if there is special cases where this inequality holds, for exemple for spacelike, timelike, .. cases. Thank you
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Ok thank you, but I don't see exactly how to prove it, so if you have a proof or a link to read more about this case. – zaar1992 Mar 22 '14 at 20:53
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zaar1992: sorry, my comment was incorrect. I have deleted it. – Gil Bor Mar 22 '14 at 21:01
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This is a question on linear algebra which can be discussed by just looking at vector spaces with a symmetric bilinear form $g$ (i.e. the nonlinear structure of a manifold is not relevant).
Of course you have the usual Schwarz inequality on subspaces $V$ where the metric tensor is positive (i.e. the restriction of the tensor to $V\otimes V$). The proof is just the same as in the Euclidean case.
In Lorentz space you have in fact the backwards Schwartz inequality: if $v, w$ are timelike vectors in a Lorentz space, then $|\langle v, w\rangle| \ge |v| |w|$ with equaility iff they are collinear.
See, e.g., Barrett O'Neill, Semi-Riemannian Geometry, page 141-144.
Thomas
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