Question about 1-forms on $S^3$ I am an undergrad student, and doing an introduction to smooth manifolds course. I came across the following question:
Let $\theta_1,\theta_2,\theta_3 \in \Omega^1(S^3)$ be 1-forms on $S^3\subset \Bbb{R}^4$ given by:
$\theta_1 := −y dx + x dy + t dz − z dt,\\
\theta_2 := −z dx − t dy + x dz + y dt,\\
\theta_3 := −t dx + z dy − y dz + x dt$
(where i use conventions, dx and the other forms on the right hand
side are the restrictions from $\Bbb{R}^4$ to $S^3$ just that we do not write $dx|_{S^3}$ etc all the
time).
now they ask me to show that:
$$d\theta_1=-2\theta_2\wedge\theta_3$$where $d: \Omega^k(S^3)\to\Omega^{k+1}(S^3)$ is DeRham differential and $\cdot \wedge \cdot: \Omega^k(S^3)\times \Omega^l(S^3) \to \Omega^{k+l}(S^3)$ is the wedge product.
I have worked on this for a while now but can't seem to figure it out. I feel like that if:
for $p \in S^3$, $p:=(x,y,z,t)$ and $w \in T_pS^3$, $w=(w_1,w_2,w_3,w_4)$, $$tw_3-yw_1+xw_2-zw_4=0$$
i would be much closer to the anwser, but i don't see why/if this is true.
a hint or something in the right way would be much appreciated.
 A: This approach that should work.
(1) Show that, for any $p = (x,y,z,t) \in S^3$,
$$ 
v_1 = -y\partial_x + x\partial_y + t \partial_z - z \partial_t, \\
v_2 = -z \partial_x - t\partial_y + x\partial_z + y \partial_t, \\
v_3 = -t \partial_x + z \partial_y - y \partial_z + x \partial_t $$
is a basis for the tangent space $T_p S^3$.
[First show that $v_1$, $v_2$ and $v_3$ really do lie tangent to $S^3$ at $p$. You can do this by checking that $v_1$, $v_2$ and $v_3$ are each orthogonal (w.r.t. the Euclidean inner product) to $v_0 = x\partial_x +y\partial_y + z \partial_z + t\partial_t$, which is the normal vector to $S^3$ in $\mathbb R^4$. Then show that $v_1$ and $v_2$ and $v_3$ are linearly independent in $T_p S^3$. You can do this by showing that $v_1, v_2, v_3 $ and $v_0$ are linearly independent in $T_p \mathbb R^4$, and this can be accomplished by showing that the determinant of the $4\times 4$ matrix whose entries are the components of $v_1$, $v_2$, $v_3$ and $v_0$ is equal to one, if $x^2 + y^2 + z^2 + t^2  = 1$.]
(2) Show that 
$$ \theta_i (v_j) = \delta_{ij}$$
and
$$ d \theta_i (v_j, v_k) = - 2 \epsilon_{ijk},$$
at all points on $S^3$.
[I'm viewing the 1-form $\theta_i$ as a linear map $T_p S^3 \to \mathbb R$, and I'm viewing the 2-form $d\theta_i $ as an antisymmetric bilinear map $T_p S^3 \times T_p S^3 \to \mathbb R$. In my notation,  $\delta_{ij}$ is the Kronecker delta  and $\epsilon_{ijk}$ is the Levi-Civita antisymmetric symbol.]
(3) To show that $d \theta_1 = -2 \theta_2 \wedge \theta_3$, it is sufficient to check  that $d \theta_1$ and $-2 \theta_2 \wedge \theta_3$ give the same results when acting on a basis of $T_p S^3$. In other words, in view of part (1), it's sufficient to check that
$$ d \theta_1 (v_2, v_3) = -2(\theta_2 \wedge \theta_3)(v_2, v_3) , $$
$$ d \theta_1 (v_3, v_1) = -2(\theta_2 \wedge \theta_3)(v_3, v_1) , $$
$$ d \theta_1 (v_1, v_2) = -2(\theta_2 \wedge \theta_3)(v_1, v_2) . $$
These equations should follow straight-forwardly from part (2).

Remark: For part (2), the most direct way to show that $d \theta_i (v_j, v_k) = - 2 \epsilon_{ijk}$ would be to simply compute $d \theta_i$. (For example, by direct computation, we have $d\theta_1  = 2dx \wedge dy - 2dz \wedge dt$, and so on.)
However, a more elegant way to do this would be to view $v_1, v_2 $ and $v_3$ as vector fields on $S^3$. Then we can use the standard result about the action of exterior derivatives on vector fields:
$$ d\theta_i (v_j, v_k) = v_j (\theta_i (v_k)) - v_k (\theta_i (v_j)) - \theta_i ([v_j, v_k]),$$
Here $[v_j, v_k]$ is the commutator (Lie bracket) of $v_j$ and $v_k$. The result then follows quite quickly once you realise that the Lie bracket is given by $$[v_j, v_k] = 2 \sum_{i=1}^3\epsilon_{ijk} v_i.$$
