On the 3-sphere I have found the vector fields




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




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 $S^3$...

  • 1
    $\begingroup$ You should get equality when restricting all yr forms to $S^3$. That is, modulo $d(x^2)=2\sum x_idx_i$. $\endgroup$
    – Gil Bor
    Jan 30, 2014 at 5:03


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