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I am studying Kerr Spacetime and I am not sure about something used in a proof I am trying to understand.

I am wondering, if you consider a 4-dimensional Lorentzian manifold $\mathcal{M}$ and $X_i \in \mathfrak{X}(\mathcal{M})$, $i=1,\dots 3$ spacelike. Consider the hypersurface $N$ where $t=t_0=const$ and suppose the vectors $X_i $ form a basis of $T_p(N)$ at each point. Does this make $T_p(N)$ (and therefore $N$) spacelike already? Because in the proof they say that the $X_i$ are also mutually orthogonal and I am not sure if this is actually needed here.

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    $\begingroup$ (1,0)=(1,2)+(1,-2). $\endgroup$
    – WillO
    May 11, 2019 at 13:59

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No, e.g. in Minkowski space we can reconstruct timelike vectors as linear combinations of spacelike vectors.

However, orthogonality of a spacelike basis is a sufficient condition to ensure that any linear combination is again spacelike.

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    $\begingroup$ A possibly useful necessary and sufficient condition for three vectors to span a spacelike 3-dimensional subspace of Minkowski space would be that the $3\times3$ matrix of their inner products has all its eigenvalues positive. $\endgroup$
    – Andreas Blass
    May 12, 2019 at 0:52

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