On the 3-sphere I have found the vector fields
$X_1=(-x_2,x_1,-x_4,x_3)$,
$X_2=(-x_3,x_4,x_1,-x_2)$,
$X_3=(-x_4,-x_3,x_2,x_1)$,
in the basis $\left\{\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\frac{\partial}{\partial x_3},\frac{\partial}{\partial x_4}\right\}$. Here I embedded $S^3$ in $\mathbb{R}^4$, so the $x_i$ are coordinates in $\mathbb{R}^4$ and $x_1^2+x_2^2+x_3^2+x_4^2=1$. The above vector fields form an orthogonal basis of the tangent space $T_x S^3$ for each $x\in S^3$.
The structure functions $C_{ij}^k(x)$ are then defined by
$[X_i,X_j]=\sum_{k=1}^3 C_{ij}^k(x)X_k$.
Computing them, we get for example $C^1_{2,3}=-2$ and $C^2_{2,3}=C^3_{2,3}=0$.
Now we can define the dual forms of the above vector fields by
$\omega_1=(-x_2,x_1,-x_4,x_3)$,
$\omega_2=(-x_3,x_4,x_1,-x_2)$,
$\omega_3=(-x_4,-x_3,x_2,x_1)$,
this time in the basis $\{dx_1,dx_2,dx_3,dx_4\}$. Then $\omega_i(X_j)=\delta_{ij}$. But then the structure functions are defined by
$d\omega_k=\sum_{i<j} C_{ij}^k(x)\omega_i\wedge\omega_j$.
But if I compute for example $d\omega_1$ I get $d\omega_1=2dx \wedge dy + 2 dz\wedge dt$, which is not equal to $C_{2,3}^1(x)\omega_2\wedge \omega_3=-2\omega_2\wedge \omega_3$! What am I doing wrong here?
I have the feeling the problem lies in the fact that the $x_i$ are not independent on $S^3$...