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I am reading a book that seems to claim the following. I suspect that there may be a misprint, or some assumptions missing. For $A=(x_1,y_1,z_1),B=(x_2,y_2,z_2)\in R^3$, define $$\langle A,B\rangle=-z_1z_2+x_1x_2+y_1y_2.$$ Suppose $C=(c_1,c_2,c_3)$, $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$ are such that $\langle C,C\rangle=-1,c_3\ge 1$,$\langle u,u\rangle=1$, $\langle C,u\rangle=0$, $\langle v,v\rangle=1$ and $\langle C,v\rangle=0$. The book claims that $$|\langle u,v\rangle|\le 1,$$ which does not seem to be correct.

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up vote 2 down vote accepted

The book is correct. Define $$V = \{x \in \mathbb{R}^3\ |\ \langle x,C \rangle = 0\}.$$ Thus $V$ is a two-dimensional subspace of $\mathbb{R}^3$. The key point here is the following exercise :

Exercise : The restriction of $\langle \cdot, \cdot \rangle$ to $V$ is positive-definite (ie it is an actual inner product!).

This exercise implies the fact from your book via an application of the Cauchy-Schwarz inequality to your vectors $u$ and $v$.

Here's a hint for the exercise : your quadratic form on $\mathbb{R}^3$ has signature $1$. What must the signature of the restriction of it to $V$ be?

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