Christoffel symbols of Poincaré metric using orthonormal tetrad formalism I want to calculate the Christoffel symbols for the Poincaré metric using the orthonormal tetrad formalism.
$$ds^2 = y^{-2}dx^2 + y^{-2}dy^2.$$
I introduce a non coordinate orthonormal basis with one-forms
$$ds^2 = (\omega^{\hat{x}})^2 + (\omega^{\hat{y}})^2.$$
The new basis one forms can be identified as
$$\omega^{\hat{x}} = y^{-1}dx $$
$$\omega^{\hat{y}} = y^{-1}dy. $$
I calculate the exterior derivatives
$$d (\omega^{\hat{x}}) = d(y^{-1}dx) = - y^{-2} dy \wedge dx = -  \omega^{\hat{y}} \wedge \omega^{\hat{x}} = \omega^{\hat{x}} \wedge \omega^{\hat{y}}$$
$$d (\omega^{\hat{y}}) = d(y^{-1}dy) = 0$$
$$d (\omega^{\hat{x}}) = -\Gamma^{\hat{x}}_{\;\hat{i}} \wedge \omega^{\hat{i}} = -\Gamma^{\hat{x}}_{\;\hat{x}} \wedge \omega^{\hat{x}} - \Gamma^{\hat{x}}_{\;\hat{y}} \wedge \omega^{\hat{y}} $$
$$\Gamma^{\hat{x}}_{\;\hat{y} } = - \omega^{\hat{x}}, \,\,\,
\Gamma^{\hat{x}}_{\;\hat{x} } =  \omega^{\hat{y}}$$
$$\Gamma^{\hat{x}}_{\;\hat{y} \hat{x}} = -1, \,\,\, \Gamma^{\hat{x}}_{\;\hat{x} \hat{y}} = 1$$
Transformation matrices are:
$$
\Lambda^{\hat{a}}_{\;b}  = \begin{bmatrix}
    y^{-1}       & 0   \\
    0       & y^{-1}   \\
\end{bmatrix}
$$
$$
(\Lambda^{-1}) ^{a}_{\;\hat{b}}  = \begin{bmatrix}
    y       & 0   \\
    0       & y   \\
\end{bmatrix},
$$
hence
$$\Gamma^{x}_{\;yx} = (\Lambda^{-1}) ^{x}_{\;\hat{x}} \Gamma^{\hat{x}}_{\;\hat{y} \hat{x}} 
(\Lambda) ^{y}_{\;\hat{y}} (\Lambda) ^{x}_{\;\hat{x}} = -y^{-1} $$
The solution is correct, but for all other Christoffel symbols I get $0$.
Correct solutions should be
$$\Gamma^{x}_{xx} = 0, \ \ \ \Gamma^{x}_{yx} = \frac{-1}{y}, \ \ \ \Gamma^{y}_{xx} = \frac{1}{y}, \ \ \  \Gamma^{y}_{yy} = \frac{-1}{y}$$
How can I get
$$\Gamma^{y}_{xx} = \frac{1}{y}, \ \ \  \Gamma^{y}_{yy} = \frac{-1}{y}~?$$
The procedure I follow is based on Relativity Demistified from David McMahon (Chapter 5)
 A: $$ds^2 = y^{-2}dx^2 + y^{-2}dy^2.$$
I introduce a non coordinate orthonormal basis with one-forms
$$ds^2 = (\omega^{\hat{x}})^2 + (\omega^{\hat{y}})^2.$$
The new basis one forms can be identified as
$$\omega^{\hat{x}} = y^{-1}dx $$
$$\omega^{\hat{y}} = y^{-1}dy. $$
I calculate the exterior derivatives
$$d (\omega^{\hat{x}}) = d(y^{-1}dx) = - y^{-2} dy \wedge dx = -  \omega^{\hat{y}} \wedge \omega^{\hat{x}} = \omega^{\hat{x}} \wedge \omega^{\hat{y}}$$
$$d (\omega^{\hat{y}}) = d(y^{-1}dy) = 0$$
$$d (\omega^{\hat{x}}) = -\Gamma^{\hat{x}}_{\;\hat{i}} \wedge \omega^{\hat{i}} = -\Gamma^{\hat{x}}_{\;\hat{x}} \wedge \omega^{\hat{x}} - \Gamma^{\hat{x}}_{\;\hat{y}} \wedge \omega^{\hat{y}} $$
$$\Gamma^{\hat{x}}_{\;\hat{y} } = - \omega^{\hat{x}}, \,\,\,
\Gamma^{\hat{x}}_{\;\hat{x} } =  \omega^{\hat{y}}$$
$$\Gamma^{\hat{x}}_{\;\hat{y} \hat{x}} = -1, \,\,\, \Gamma^{\hat{x}}_{\;\hat{x} \hat{y}} = -1$$
$$\Gamma^{\hat{y}}_{\;\hat{y} } = - \omega^{\hat{y}}, \,\,\,
\Gamma^{\hat{y}}_{\;\hat{x} } =  \omega^{\hat{x}}$$
$$\Gamma^{\hat{y}}_{\;\hat{y} \hat{y}} = -1, \,\,\, \Gamma^{\hat{y}}_{\;\hat{x} \hat{x}} = 1$$
Transformation matrices are:
$$
\Lambda^{\hat{a}}_{\;b}  = \begin{bmatrix}
 y^{-1}       & 0   \\
 0       & y^{-1}   \\
\end{bmatrix}
$$
$$
(\Lambda^{-1}) ^{a}_{\;\hat{b}}  = \begin{bmatrix}
 y       & 0   \\
 0       & y   \\
\end{bmatrix},
$$
hence
$$\Gamma^{x}_{\;yx} = (\Lambda^{-1}) ^{x}_{\;\hat{x}} \Gamma^{\hat{x}}_{\;\hat{y} \hat{x}} 
(\Lambda) ^{y}_{\;\hat{y}} (\Lambda) ^{x}_{\;\hat{x}} = -y^{-1} $$
The other Christoffel symbols can be calculated in a similar way.
The solutions  are correct.
$$\Gamma^{x}_{xx} = 0, \ \ \ \Gamma^{x}_{yx} = \frac{-1}{y}, \ \ \ \Gamma^{y}_{xx} = \frac{1}{y}, \ \ \  \Gamma^{y}_{yy} = \frac{-1}{y}$$
