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In Lorentz-Minkowski space $\mathbb{L}^3$, often denoted as $\mathbb{R}^{4}_1$ or $\mathbb{R}^{3,1}$, using the metric $\langle\cdot,\cdot\rangle=-dx_1^2+dx_2^2+dx_3^2+dx_4^2$, we consider a lightlike surface $M^2\subset\mathbb{L}^3$ and a vector field $V\in\mathfrak{X}(M)$ such that $V$ is lightlike, that is, $\langle V,V\rangle=0$ with $V\neq0$.

I've been reading an article in which it assumes that we can complete a basis of the tangent bundle $TM$ with another lightlike field $W\in\mathfrak{X}(M)$ such that, also, $\langle V,W\rangle=-1$.

How can we prove this existence? I'm pretty sure this is standard, as I suppose it's similar to assuming the existence of an orthonormal basis maybe, also it sounds reasonable considering that both $V,W$ need to be on the light cone which forms a $45º$ angle, if I recall correctly. Yet, I've tried to prove it to no avail.

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First let $U \in \mathfrak X(M)$ be any non-zero vector field completing $V$ to a frame (field of bases) of $TM$. (I'm assuming we're really talking about local frames here - depending on the topology of $M$ there may not be any global frame at all.) Since $M$ is lightlike, we know $U$ is lightlike.

Letting $U = (u, \mathbf u)$, $V=(v,\mathbf v)$ denote the decompositions into timelike and spacelike components (i.e. $u = dx_1(U), \mathbf u = (dx_2(U),dx_3(U),dx_4(U))$), note that the fact that $U,V$ are lightlike means exactly that $u^2 = |\mathbf u|^2, v^2 = |\mathbf v|^2.$ Since $U,V$ are non-vanishing, $u,v$ are also and thus we can assume without loss of generality that $u,v$ are positive.

This implies that $f := \langle U,V \rangle = -uv + \mathbf{u \cdot v} <0$ by the Cauchy-Schwarz inequality, since we know that $U,V$ (and thus $\mathbf{u},\mathbf v$) are linearly independent.

Now, we simply normalize $U$: let $W = -\frac{1}{f}U$ so that $\langle W, V \rangle = -1.$

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  • $\begingroup$ Thanks for the detailed answer, Anthony. $\endgroup$
    – F.Webber
    Dec 17 '17 at 1:40

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