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Consider the Lorentz-Minkowski space $E^n_1$, also known as $\mathbb{L}^n$. I want to prove that there are not lightlike linearly independent vectors $u, v, w \in E^n_1$ such that $u + v + w = 0$. How to do it? I'm still unfamiliar with the intuition behind such space.

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  • $\begingroup$ Does this space have 2 spacelike and one timelike variables? or what if not? $\endgroup$ – coffeemath Dec 8 '18 at 1:38
  • $\begingroup$ @coffeemath Those are all hypotheses I have. $\endgroup$ – user71487 Dec 8 '18 at 1:41
  • $\begingroup$ I don't see how this can be true... can't we take one of the vectors from the future light cone, say $\vec{u}$, and the other two to be $-\frac{1}{2}\vec{u}$? Don't we need to say that they are non-coplanar? $\endgroup$ – RandomMathGuy Dec 8 '18 at 1:41
  • $\begingroup$ @RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent. $\endgroup$ – Travis Willse Dec 8 '18 at 1:44
  • $\begingroup$ @Travis I should learn to read the titles more carefully! $\endgroup$ – RandomMathGuy Dec 8 '18 at 1:44
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Hint Suppose there were. Expand $$0 = [{\bf u} + {\bf v} + {\bf w}] \cdot [{\bf u} - ({\bf v} + {\bf w})]$$ to conclude that ${\bf v} \cdot {\bf w} = 0$.

Additional hint What is the matrix representation of the bilinear form $\cdot$ with respect to the basis $({\bf u}, {\bf v}, {\bf w})$?

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  • $\begingroup$ Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $\mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({\bf u}, {\bf v}, {\bf w})$ is not a basis of $\mathbb L^n$ for $n \neq 3$. $\endgroup$ – Travis Willse Dec 8 '18 at 1:48
  • $\begingroup$ Shouldn't I conclude ${\bf u} \cdot {\bf w} = 0$ instead? $\endgroup$ – user71487 Dec 8 '18 at 1:58
  • $\begingroup$ I don't see how---after all, the expression is symmetric in ${\bf v}$ and ${\bf w}$. $\endgroup$ – Travis Willse Dec 8 '18 at 2:02
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    $\begingroup$ Expanding gives $0 = {\bf u} \cdot {\bf u} - ({\bf v} + {\bf w}) \cdot ({\bf v} + {\bf w}) = -({\bf v} \cdot {\bf v} + 2 {\bf v} \cdot {\bf w} + {\bf w} \cdot {\bf w}) = -2 {\bf v} \cdot {\bf w}$. $\endgroup$ – Travis Willse Dec 8 '18 at 2:58
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    $\begingroup$ Not directly anyway. By symmetry of notation, we also have ${\bf u} \cdot {\bf v} = {\bf w} \cdot {\bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $\cdot$ is the zero bilinear form, a contradiction. $\endgroup$ – Travis Willse Dec 8 '18 at 3:00

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