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Is it possible to choose a Lorentzian metric $g$ on a neighborhood of the origin in $\mathbb R^{1+n}$ so that the sectional curvature of all non-degenerate tangent timelike two planes at the origin is equal to -1 and the sectional curvature of all spacelike two planes is 1?

More generally, is it possible to choose $g$ so as to arbitrarily assign the sectional curvatures of all the non-degenerate tangent two-planes at the origin?

Thanks

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To your second question, the answer is no. Using polarization, t)he sectional curvature determines the Riemann curvature tensor so if we know the sectional curvature on "enough" two planes, it is uniquely determined. However, every algebraic curvature tensor has a geometric realization, so to determine whether a particular sectional curvature is possible, it's just a matter of verifying whether your desired sectional curvatures give rise to a valid algebraic curvature tensor.

For your first question, a Lorentzian metric has signature $(n,1)$. As such, it's not clear to me what you mean by timelike planes. Perhaps you are interested in the more general pseudo-Riemannian case?

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  • $\begingroup$ Here By timelike two plane I mean that the plane is spanned by two vectors $X,Y$ and that $g(X,X)g(Y,Y)-g(X,Y)^2<0$. $\endgroup$
    – Ali
    Commented Sep 17, 2020 at 4:26
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    $\begingroup$ I see. In this case I believe the answer to your first question is no as well. At any point, the sectional curvature should depend analytically on the choice of plane (at least for non-lightlike planes), so I don't think it's possible for it to be locally but not globally constant. $\endgroup$
    – Gabe K
    Commented Sep 17, 2020 at 10:29

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