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.


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 $\langle W, V \rangle = -1.$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.