I've been trying to understand geodesics of hyperbolic space $H^n$, and found a very similar question and a helpful answer here: Geodesics and Distance in Hyperbolic Space
However, I found it difficult to prove one point in the 2nd step of the above-mentioned page: "Show there is an element of $O_+(1,n)$ fixing $P$ and acting non-trivially on each point of $\mathbb{R}^{n+1}\backslash P$."
How can I find such $A \in O_+(1,n)$? Thanks in advance.
For $p \in H^n$ the vector $p=\gamma$ is normal to $T_p H^n$. We also have $(\gamma,\gamma)=-1$. Now choose $v \in P\backslash span(\gamma)$ and define $\alpha'=v+(v,\gamma)\gamma$. Now define $\alpha=\frac{\alpha'}{(\alpha',\alpha')}$. By this construction, we have $\alpha \perp \gamma$. This means $\alpha \in T_p H^n_k$. Extend $\alpha$ to an ordered orthonormal basis $(\alpha,v_2, \dots,v_n)$ of $T_p H^n_k$. We have that $B=(\gamma, \alpha,v_2, \dots,v_n)$ is an ordered orthonormal basis for $L^{n+1}$, that is $\mathbb{R}^{n+1}$ with the lorenz metric . Let $[I]^B_E$ change of basis matrix from $B$ to the standard basis $E$. We have from the construction: $$([I]^B_E)^T \eta [I]^B_E=\eta ,$$ where $\eta$ is the matrix representation of the Lorentz metric on the standard basis. This means that $[I]^B_E \in O(1,n)$. We know that the first column in $[I]^B_E$ is $\gamma$. The first coordinate of $\gamma$ is positive because $ \gamma =p \in H^n_k$ so $[I]^B_E \in O_+(1,n)$. Now define: $$A=\begin{pmatrix} 1 & 0 & 0 & \dots & 0\\ 0 & 1 & 0 & \dots & \vdots \\ 0 & 0 &-1 & \ddots & \vdots \\ \vdots& \vdots &\ddots &\ddots & 0 \\ 0 & \dots & \dots & 0 & -1 \end{pmatrix}$$ It is clear that we have: $$ A^t \eta A= \eta$$ So $A \in O_+(1,n)$. As $O_+(1,n)$ is a subgroup, we have that $[I]^E_B \in O_+(1,n)$, because it's an inverse of a subgroup element. We also have $[I]^B_E A [I]^E_B \in O_+(1,n)$. Define $M=[I]^B_E A [I]^E_B$. The construction of $M$ guarantees that it preserves $P$, and acts non-trivially on each point of $\mathbb{R}^{n+1}\backslash P$.
