Why do I have an extra differential form in $S^3$

Question

I'm trying to get used to formalism of Riemannian Geometry so I wanted to obtain a base for the cotangent space of $S^3$. $$S^{3}=\left\{ \left(x,\,y,\,z,\,t\right)\in\mathbb{R}^{4}:\,\,x^{2}+y^{2}+z^{2}+t^{2}=1\right\}$$ I used the classic vector fieldsto parallelize the sphere $$X_{1}= y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}-t\frac{\partial}{\partial z}+z\frac{\partial}{\partial t},$$ $$ X_{2}= t\frac{\partial}{\partial x}-z\frac{\partial}{\partial y}+y\frac{\partial}{\partial z}-x\frac{\partial}{\partial t},$$ $$ X_{3}= -z\frac{\partial}{\partial x}-t\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}+y\frac{\partial}{\partial t},$$ $$ X_{4}= x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}+t\frac{\partial}{\partial t}.$$ Then I wanted to obtain the dual basis for the cotangent space so I inverted the matrix $$ R =\left(\begin{array}{cccc} y & -x & -t & z\\ t & -z & y & -x\\ -z & -t & x & y\\ x & y & z & t \end{array}\right),$$ And obtained the following forms $$ \omega^{1}= y\mbox{d}x-x\mbox{d}y-t\mbox{d}z+z\mbox{d}t,$$ $$ \omega^{2}= t\mbox{d}x-z\mbox{d}y+y\mbox{d}z-x\mbox{d}t,$$ $$ \omega^{3}= -z\mbox{d}x-t\mbox{d}y+x\mbox{d}z+y\mbox{d}t,$$ $$ \omega^{4}= x\mbox{d}x+y\mbox{d}y+z\mbox{d}z+t\mbox{d}t.$$ Then I considered the map F between $S^{3}$ and $\mathbb{R}^{4}$ $$F\left(\psi,\,\theta,\,\phi\right)=\left(\sin\psi\sin\theta\cos\phi,\sin\psi\sin\theta\sin\phi,\sin\psi\cos\theta,\cos\psi\right).$$ And found the pull-back of the forms $\omega^{1},\,\omega^{2},\,\omega^{3}$ $$\omega^{1}= \cos\theta\mbox{d}\psi-\sin\psi\cos\psi\sin\theta\mbox{d}\theta-\sin^{2}\psi\sin^{2}\theta\mbox{d}\phi,$$ $$ \omega^{2}= \sin\theta\cos\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\sin\phi-\cos\psi\cos\theta\cos\phi\right)\mbox{d}\theta -\sin\psi\sin\theta\left(\cos\psi\sin\phi+\sin\psi\cos\theta\cos\phi\right)\mbox{d}\phi,$$ $$ \omega^{3}= -\sin\theta\sin\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\cos\phi+\cos\psi\cos\theta\sin\phi\right)\mbox{d}\theta -\sin\psi\sin\theta\left(\cos\psi\cos\phi-\sin\psi\cos\theta\sin\phi\right)\mbox{d}\phi.$$ The problem is that I also have the pull-back of the last form $$\omega^{4}=\sin^{2}\theta\sin\psi\cos\psi \left(2\sin\phi\cos\phi-1\right)d\psi$$ Why should I exclude and keep just the other three? It is clear from $$ \omega^{4}= x\mbox{d}x+y\mbox{d}y+z\mbox{d}z+t\mbox{d}t.$$ that $\omega^{4}$ is normal to the sphere but why then once I pulled back don't I get 0 as I supposed? Was I seeing that wrong?

Answer 1

Since

$$\omega^4= x dx + ydy + zdz + wdw = \frac 12 d(x^2 + y^2 + z^2 + w^2),$$

we have

$$\omega^4 |_{\mathbb S^3} = \frac 12d[ (x^2 + y^2 + z^2 + w^2)|_{\mathbb S^3}] = \frac 12d (1) = 0.$$

I think there are some calculation mistakes. I calculated the $d\psi$ component of $\omega^4|_{\mathbb S^3}$ and get $0$ instead of your expression: Explicitly, since

$$\begin{split} \omega^4 |_{\mathbb S^3} &= \sin\psi \sin\theta \cos\phi d ( \sin\psi \sin\theta \cos\phi) + \sin\psi \sin\theta \sin\phi d ( \sin\psi \sin\theta \sin\phi) \\ &\ \ + \sin\psi \cos\theta d ( \sin\psi \cos\theta) + \cos \psi d (\cos \psi) \end{split}$$

So the $d\psi$-component is

$$\sin\psi \cos\psi \sin^2\theta \cos^2\phi + \sin\psi \cos\psi \sin^2\theta \sin^2\phi + \sin\psi \cos\psi \cos^2\theta -\cos\psi \sin\psi$$

and is zero.