With: $$ \mathbb{H}^n=\left\{ (x_0,x_1,\dots, x_n)\in \mathbb{R}^{n+1}: \; x_0^2=1+x_1^2+\cdots +x_n^2,\; x_0>0\right\}. $$ and the form $$ \langle\langle(u_0,u_1,\dots, u_n),(v_0,v_1,\dots,v_n)\rangle\rangle=-u_0v_0+u_1v_1+\cdots+u_nv_n, $$ I have to prove that $$ \nabla_X Y =\overline{\nabla}_X Y -\langle\langle X,Y\rangle\rangle P, $$ where $\overline{\nabla}$ is the connection in $\mathbb{R}^{n+1}$ and $P(p)=p$.
I have tried what follows: I know that the normal vector to a point $p=(p_0,p_1,\dots,p_n)\in\mathbb{H}^n$ is $(-p_0,p_1,\dots,p_n)$ and also that ($\langle, \rangle$ is the usual product): $$ \nabla_X Y =\overline{\nabla}_X Y -\langle \overline{\nabla}_XY,N\rangle N. $$
Since $\langle \overline{\nabla}_XY,N\rangle=X(\langle Y,N\rangle)-\langle Y,\overline{\nabla}_X N\rangle=-\langle Y,\overline{\nabla}_X N\rangle$. Now $\overline{\nabla}_X N=\sum X(N_i)\frac{\partial}{\partial x_i}=X'$, where $X'$ is $X$, but with a change of sign in the first coordinate. Now $\langle Y,\overline{\nabla}_X N\rangle=\langle Y,X'\rangle=\langle \langle Y,X \rangle\rangle$. So I get $$ \nabla_X Y =\overline{\nabla}_X Y -\langle\langle X,Y\rangle\rangle N, $$ which is the same as above but with a change in a sign. Where is my mistake?
The problem was that: $$ \nabla_X Y \not=\overline{\nabla}_X Y -\langle \overline{\nabla}_XY,N\rangle N. $$ Thi is only valid when the metric of the subspace ($\mathbb{H}^n$ in this case) is the same of the bigger space ($\mathbb{R}^n$ here) and is the usual Euclidean metric. So we need to put the hyperbolic metric in $\mathbb{R}^n$. Since the Koszul formula is true in Semi-Riemannian geometry we prove that $(\overline{\nabla}_XY)^T=\nabla_XY\Rightarrow \nabla_X Y =\overline{\nabla}_X Y -\langle\langle \overline{\nabla}_XY,P\rangle\rangle \frac{P}{\langle\langle P,P\rangle\rangle}=\overline{\nabla}_X Y +\langle \overline{\nabla}_XY,N\rangle P$. Now the argument is the same and the result follows.