Equality characterization of the reverse Cauchy-Schwarz inequality in a Lorentzian manifold

Question

Let $(M, g)$ be a Lorentzian manifold (signature -++...) with a time orientation and suppose that $v, w \in T_pM$ are causal vectors that are in the same light cone(ie, both future-directed or past-directed). Denoting by $∥v∥_g = \sqrt{|g(v, v)|}$ the “norm” of $v$. Then the reverse C-S inequality holds with equality iff v and w are colinear: $−g(v, w) ≥ ∥v∥_g∥w∥_g \tag{*}$

My try:

Adapting of the proofs of the C-S inequality in the wikipedia page of the C-S inequality : finite dimensional real vector space case I get:

Since $v$ and $v $are causal, they are timelike or lightlike we have that $∥w∥_g^2=|g(w,w)|=-g(w,w)$ and $∥v∥_g^2=|g(v,v)|=-g(v,v)$ and since they are equally directed $g(v,w)\le 0$

Define the function $p:\Bbb R \to \Bbb R: t\to p(t):=g(tw+v,tw+v)$

$p(t):=g(tw+v,tw+v)=t^2g(w,w)+2g(v,w)+g(v,v)$

$=-t^2∥w∥_g^2+2g(v,w)-∥v∥_g^2 \le 0 \forall t \in \Bbb R$

Then the discriminant of this quadratic equation ( $g(w,w)\neq 0$)

$\Delta=4g(v,w)^2-4g(w,w)g(v,v)\le 0$

from which (*) holds.

For the case $g(w,w)=0$ the inequality is also trivially satified. This concludes the prove of $(*)$

To show the equality characterization:

$(*)\iff v$ and $w$ are colinear

I already proved that if $v$ and $w$ are colinear, $(*)$ holds. But I am having trouble proving the converse:

If$(*)$ holds, then $\Delta =0$ and $p(t)=(t∥w∥_g+∥v∥_g)^2,$

Let $t_0:=-\frac{∥v∥_g}{∥w∥_g}$

$0=p(t_0)=g(t_0∥w∥_g+∥v∥_g,t_0∥w∥_g+∥v∥_g)$

In the Riemannian case here one would use positive definiteness of $g$, ie. $g(x,x)=0 \implies x=0$, but in a Lorentzian manifold, that condition is dropped so I can't obtain that $ t_0∥w∥_g+∥v∥_g =0$ as I wished to conclude that the vectors are colinear.

a)How do I really prove $(*)\implies v $ and $w$ are colinear then?

By the way,

b)Is the prove of (*) OK?

Answer 1

To prove the reverse Cauchy-Schwarz inequality $|g(v,w)|\geq \|v\|_g\|w\|_g$, you actually don’t need to assume they lie in the same cone (but if you do, then the proportionality constant will be non-negative).

We’ll rely on some basic properties of Lorentzian inner product spaces, such as orthogonal complements of timelike vectors is a spacelike subspace, and that if we have two non-zero null vectors which are orthogonal then they must be a (non-zero) multiples of each other (otherwise you’d violate Lorentzian signature). Here’s the proof of the reverse Cauchy-Schwarz inequality:

• if $v=0$, then for all $w$, the claim is obvious. Likewise, if $w=0$, then again for all $v$, the claim is obvious. Henceforth we can assume $v\neq 0$ and $w\neq 0$.
• Suppose $v,w$ are both non-zero null vectors. Then, the inequality is obvious because we have $0$ on the RHS. We have equality if and only if $g(v,w)=0$, i.e $v,w$ are orthogonal. By the ‘fact’ I mentioned above (a good exercise for you to work out), it follows $v,w$ are multiples of each other.
• We’re finally left with the case where atleast one of the vectors is timelike, let’s say it is $v$. Notice the inequality is homogeneous, meaning if we scale $v$ by a positive number, the inequality still holds (and equality holds before scaling if and only if it holds after scaling). Hence, we may WLOG assume that $g(v,v)=-1$. By the link above, we have an orthogonal direct sum decomposition $V=\Bbb{R}v\oplus \{v\}^{\perp}$, with the second factor being a spacelike subspace. So, there exist unique $a\in\Bbb{R}$ and $s\in \{v\}^{\perp}$ (i.e either $0$ or spacelike vector) such that $w=av+s$. From this, we have \begin{align} -g(v,w)=-g(v,av+s)=-[a\cdot (-1)+0]=a. \end{align} Next, \begin{align} g(w,w)&=a^2g(v,v)+2ag(v,s)+g(s,s)=-[g(v,w)]^2+g(s,s), \end{align} so rearranging and keeping in mind that $w$ is causal, we get \begin{align} [g(v,w)]^2&=|g(w,w)|+g(s,s)\geq |g(w,w)|, \end{align} so taking square roots yields \begin{align} |g(v,w)|&\geq\|w\|_g=\|w\|_g\cdot\|v_g\|, \end{align} since $\|v\|_g=1$. Clearly, we have equality if and only if $g(s,s)=0$, which happens (since $s$ lies in the spacelike subspace $\{v\}^{\perp}$) if and only if $s=0$, i.e if and only if $w=av$.

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Here are some good-to-know consequences of the reverse Cauchy-Schwarz inequality:

Corollary

• Hyperbolic Angle: If $v,w$ are timelike vectors and lie in the same timecone, then there is a unique $\beta\in [0,\infty)$ such that $\langle v,w\rangle_g=-\|v\|_g\|w\|_g\cosh(\beta)$. The number $\beta$ is called the hyperbolic/boost angle between $v$ and $w$.

• Reversed Triangle Inequality: If $v,w$ are causal vectors which lie in the same cone, then $\|v+w\|_g\geq \|v\|_g+\|w\|_g$, with equality if and only if one is a non-negative multiple of the other.

To prove the claim about hyperbolic angles, notice that $\frac{\langle v,w\rangle_g}{-\|v\|_g\|w\|_g}=\frac{|\langle v,w\rangle_g|}{\|v\|\cdot\|w\|_g}$ (because both vectors are in the same cone; and the division is justified since they’re timelike vectors). This number lies in $[1,\infty)$ due to the reversed Cauchy-Schwarz inequality, so now the claim follows because $\cosh$ restricts to a bijection $[0,\infty)\to [1,\infty)$.

To prove the reversed triangle inequality (not to be confused with the reversed triangle inequality from real analysis), we have \begin{align} \|v+w\|_g^2&=|g(v,v)+2g(v,w)+g(w,w)|\\ &=-g(v,v)-2g(v,w)-g(w,w)\\ &=\|v\|_g^2+2|g(v,w)|+\|w\|_g^2\\ &\geq \|v\|_g^2+2\|v\|_g\|w\|_g+\|w\|_g^2\\ &=(\|v\|_g+\|w\|_g)^2. \end{align} In the second equal sign, what I used is that $g(v,v),g(w,w)\leq 0$ since they’re causal and $g(v,w)\leq 0$ since $v,w$ lie in the same cone; that’s why the absolute values simplify in that manner. Now, taking square roots proves the reversed triangle inequality. Finally, we have equality in the fourth line if and only if we have equality in the reversed Cauchy-Schwarz inequality, which happens if and only if $v,w$ are collinear; but now since they lie in the same cone, this multiple must be non-negative.

This reversal of the triangle inequality is of course the root of the infamous twin paradox of special relativity (not an actual paradox, so it’s better to call it the twin-phenomenon).