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Given $\mathbb{R}^4$, we define the Minkowski inner product on it by $$ \langle v,w \rangle = -v_1w_1 + v_2w_2 + v_3w_3 + v_4w_4$$ We say a vector is spacelike if $ \langle v,v\rangle >0 $, and it is timelike if $ \langle v,v \rangle < 0 $.

How can I show that if $v$ is timelike and $ \langle v,w \rangle = 0$ , then $w$ is either the zero vector or spacelike? I've tried to use the polarization identity, but don't have any information regarding the $\langle v+w,v+w \rangle$ term in the identity.

Context: I'm reading a book on Riemannian geometry, and the book gives a proof of a more general result: if $z$ is timelike, then its perpendicular subspace $z^\perp$ is spacelike. It does so using arguments regarding the degeneracy index of the subspace, which I don't fully understand. Since the statement above seems fairly elementary, I was wondering whether it would be possible to give an elementary proof of it as well.

Any help is appreciated!

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Do you know Sylvester's Law? –  Zach L. Dec 13 '12 at 15:39
Unfortunately I do not. Assuming I did, how could I prove the statement? –  Jonas Dec 13 '12 at 15:48
First, I don't think your statement is quite right. There are null vectors $w$ that are not zero and not timelike that satisfy $\langle v, w \rangle = 0$ for all vectors. But, if you could use Sylvester's Law, and were instead trying to show $\langle v,w \rangle = 0$ implies null or spacelike, then assume such a timelike $w$ exists. $w$ would then have to be timelike, giving two orthgonal timelike vectors. Extend these to an orthogonal basis of the space. This basis would then give signature $(n-2,2)$, while we know by Sylvester's law the signature is unique and should be $(n-1,1)$. –  Zach L. Dec 13 '12 at 15:54
@ZachL.: The Minkowski inner product is non-degenerate, so there is no null vector. –  23rd Dec 13 '12 at 15:59
@ richard: Of course. Not sure what I was thinking. It doesn't seem I can edit my comment, though. –  Zach L. Dec 13 '12 at 16:10

1 Answer 1

up vote 1 down vote accepted

Let $\langle v,v\rangle=-\lambda^2$. Normalize it by $\frac1\lambda$, we get $\langle v,v\rangle=-1$. Hence we can extend $\{v\}$ to a "orthonormal" basis $\{v,\,u_1,u_2,u_3\}$ of $\mathbb{R}^4$ such that $\langle u_i, u_i\rangle=1$ and $\langle v, u_i\rangle=\langle u_i, u_j\rangle=0$ for every $i\not=j$ (see here for the reason.) Now the rest is trivial.

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