For space-like linear independent vectors $x,y$, $V:=\text{span}\{x,y\}$ is space-like iff the Lorentz-orthogonals of $x$ and $y$ intersect

Question

A vector $x \in \mathbb{R}^{n+1}$ is called space-like if $\lVert x \rVert^2 > 0$ and time-like if $\lVert x \rVert^2 < 0$ with respect to the norm induced by the Lorentzian scalar product.

A subspace $V \subset \mathbb{R}^{n+1}$ is called time-like if it contains a time-like vector.
$V$ is called space-like if every nonzero vector in $V$ is space-like.

We set $H^n:= \{x \in \mathbb{R}^{n+1} | \lVert x \rVert^2 = -1\}$ (The hyperboloid model).
A hyperplane of $H^n$ is the intersection of $H^n$ with an n-dimensional time-like subspace of $\mathbb{R}^{n+1}$.

Let $x,y \in \mathbb{R}^{n+1}$ be linearly independent space-like vectors. Prove that:
The vector subspace $V:=\text{span}\{x,y\}$ is space-like if and only if
the hyperplanes $P$ and $Q$ of $H^n$ Lorentz orthogonal to $x$ and $y$ respectively, intersect.

My ideas so far:

1. I know that for a time like vector $v_1$ if $v_2$ is orthogonal to $v_1$, then $v_2$ is space-like.
2. I have the identity $V^{L} = \text{span}\{x\}^{L} \cap \text{span}\{y\}^{L}$. (Where by $^L$ I denote the Lorentz-orthogonal complement).
3. I know that $V$ space-like is equivalent to $|x \circ y| \leq \lVert x \rVert \lVert y \rVert$ for linearly independent space-like vectors $x,y$.

P.S.: This is also part of Theorem 3.2.6 in "Foundations of Hyperbolic manifolds" by John G. Ratcliffe.

Answer 1

So, we need to show: For linearly independent $x,y\in\mathbb{R}^{n+1}$ and for $P = \mathrm{span}\{x\}^L\cap H^n$, $Q = \mathrm{span}\{y\}^L\cap H^n$: $$ V = \mathrm{span}\{x,y\}\ \mathrm{is\ spacelike}\quad\Leftrightarrow\quad P\cap Q\neq\emptyset$$

Note that $$ P\cap Q = \mathrm{span}\{x\}^L \cap \mathrm{span}\{y\}^L \cap H^n = V^L\cap H^n$$ using your item 2.

Also note that $v\in H^n$ if and only if $\|v\|^2 = -1$ by definition, which implies in particular that $v$ is timelike if $v\in H^n$.

"$\Leftarrow$": Suppose $P\cap Q\neq\emptyset$, so there exists some $v\in P\cap Q$. Using the notes above, we know

• $v\in \mathrm{span}\{x\}^L$, i.e., $v\circ x = 0$,
• $v\in \mathrm{span}\{y\}^L$, i.e., $v\circ y = 0$, and
• $\|v\|^2 = -1$, i.e., $v$ is timelike.

Using your item 1., we also know that if there is some $0\neq w\in\mathbb{R}^{n+1}$ with $v\circ w = 0$, then $w$ is spacelike. Hence, for any vector $w=\alpha x + \beta y\in V$ with $\alpha,\beta\in\mathbb{R}$, we have $$ v\circ w = \alpha (v\circ x) + \beta (v\circ y) = \alpha\cdot0+\beta\cdot0=0.$$ thus, every non-zero vector in $V$ is spacelike, so $V$ is spacelike.

"$\Rightarrow$": This follows from the general

Claim: $V$ spacelike $\Rightarrow$ $V^L$ timelike

A proof for this can be found for example in this answer, using a bit different notation. Here is a proof using the notation at hand:

As $V$ is non-degenerate, $\mathbb{R}^{n+1}$ decomposes as $\mathbb{R}^{n+1} = V\oplus V^L$. Hence, any vector $z\in\mathbb{R}^{n+1}$ can be written as $z = v+w$ with $v\in V$ and $w\in V^L$. This yields $$ z\circ z = (v+w)\circ(v+w) = v\circ v + w\circ w.$$ As $v\in V$ and $V$ is spacelike, we have $v\circ v \geq 0$. Now assume that $V^L$ was not timelike, then we also had $w\circ w \geq 0$ as $V^L$ could not contain any timelike vector, implying $z\circ z \geq 0$. So, this would mean that there were no timelike vectors in $\mathbb{R}^{n+1}$ at all, which is a contradiction. So $V^L$ is timelike which proves the claim.

So, there is a timelike vector $0\neq v \in V^L$. But as $V^L$ is a subspace, also $\tilde{v} = v/\|v\|^2\in V^L$ and thus $\|\tilde{v}\|^2 = -1$, which implies $\tilde{v}\in H^n$ using the notes above. Hence, $\tilde{v}\in P\cap Q$, which proves $P\cap Q\neq\emptyset$.

Answer 2

$ \newcommand\R{\mathbb R} $

I'd like to provide a proof using geometric algebra since it is quite simple.

Notation:

• We have an orthonormal basis $\{e_i\}_{i=0}^n$ we have $e_0^2 = -1$ and $e_1^2 = e_2^2 = \cdots e_n^2 = 1$.
• $I = e_0e_1\cdots e_n$ is the unit pseudoscalar.
• $A \vee B = (AI^{-1})\wedge(BI^{-1})I$ is the regressive product.
• $\widetilde A$ is the reverse of $A$.
• $||A||^2 = \widetilde AA$.
• If $B$ is a blade, then $[B] = \{x \in \R^{n+1} \;:\; x\wedge B = 0\}$ is the linear subspace represented by $B$.

Given any blade $B$, it is easy to verify that $||B||^2 > 0$ if and only if $[B]$ is space-like, and $||B||^2 < 0$ if and only if $[B]$ is time-like. Note that in particular $||I||^2 < 0$ since $\R^{n+1}$ is time-like.

Let $x, y \in \R^{n+1}$ be linearly independent. Then the hyperplanes orthogonal to $x$ and $y$ are $[xI]$ and $[yI]$. Since $[xI] \not= [yI]$, we have $[(xI)\vee(yI)] = [xI]\cap[yI]$. But $$ X := (xI)\vee(yI) = (xII^{-1})\wedge(yII^{-1})I = x\wedge y\,I. $$

Now note that

• $[x\wedge y] = \mathrm{span}\{x, y\}$,
• $[xI]\cap[yI]\cap H^n = [X]\cap H^n \not= \varnothing$ if and only if $[X]$ is time-like,
• and $||X||^2 = ||x\wedge y||^2||I||^2 = -||x\wedge y||^2$

If $[x\wedge y]$ is space-like, then $||x\wedge y||^2 > 0$; hence $||X||^2 < 0$ and $[X]$ is time-like.

If $[X]$ is time-like, then $||X||^2 < 0$; hence $||x\wedge y||^2 > 0$ and $[x\wedge y]$ is space-like.