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Let G be the Minkovski quadratic form on $\mathbf{R}^{n+1}$ : $G(x,x)=-x_{0}^{2}+x_{1}^{2}+ \cdots +x_{n}^{2}$. Consider the (half) hyperboloid in $\mathbf{R}^{n+1}$ : $H=\{ x \in \mathbf{R}^{n+1} : G(x,x)=-1 \quad and \quad x_{0} >0\}$. For $p \in H$, the tangent space $T_{p}H$ can be considered a hyperplane of $\mathbf{R}^{n+1}$ and we define $g: T_{p}H \times T_{p}H \rightarrow \mathbf{R}$ by $g (x,y)=-x_{0}y_{0}+x_{1}y_{1}+\cdots +x_{n}y_{n}$. How to show that $(H,g)$ is a Riemannian manifold?

The only problem here is to show the positivity of $g$, $\quad i.e. \forall v \in T_{p}H$, $v \not= 0$, $g(v,v)>0.$

Could anybody help on this? Thank you!

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up vote 2 down vote accepted

Describe the $n-$dimensional $T_p(H)$ explicitly as a subspace of $\mathbb R^{n+1}$ $$ T_p(H)= \{(-p_0,p_1,\dots,p_n)\}^\top $$ and $-p_0^2+p_1^2+\dots+p_n^2=-1$. For a tangent vector $v\neq 0$ by definition $$ -p_0v_0+p_1v_1+\dots+p_n v_n=0 $$ And since $p_0>0$ (which implies $p_0\geq1$) you get: $$ v_0={1\over p_0}(p_1v_1+\dots+p_nv_n) \Rightarrow v_0^2={1\over p_0^2}(p_1v_1+\dots+p_nv_n)^2 $$ Apply standard Cauchy-Schwarz: $$ (p_1v_1+\dots+p_nv_n)^2 \leq (p_1^2+\dots+p_n^2)(v_1^2+\dots+v_n^2)=(p_0^2-1)(v_1^2+\dots+v_n^2) $$ and hence $$ v_0^2\leq\left(1-\left({1\over p_0^2}\right)\right)(v_1^2+\dots+v_n^2) \leq v_1^2+\dots+v_n^2 $$

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Thank you so much! – user7762 Mar 21 '12 at 15:20

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