I have to prove the next equality $$ \psi^*\left(\frac{xdx+ydy+zdz}{x^2+y^2+z^2}\right)=dt $$ wher $\psi^*$ is the pull-back of $$ \psi:S^2\times\mathbb{R}\rightarrow \mathbb{R}^3-\{(0,0,0)\}, \ $$ $$ \psi((v_1,v_2,v_3),t)=e^t(v_1,v_2,v_3) $$
Using the pullbacks properties, I have that $$ [\psi^*\left(\frac{xdx+ydy+zdz}{x^2+y^2+z^2}\right)](v_1,v_2,v_3,t)= $$ $$ [(\frac{x}{x^2+y^2+z^2}\circ\psi)d(x\circ\psi)](v_1,v_2,v_3,t)+[(\frac{y}{x^2+y^2+z^2}\circ\psi)d(y\circ\psi)](v_1,v_2,v_3,t)+[(\frac{z}{x^2+y^2+z^2}\circ\psi)d(z\circ\psi)](v_1,v_2,v_3,t)= $$ $$ dt+(v_1dv_1+v_2dv_2+v_3dv_3) $$ Now I should show that if I apply what I have there to a tangent vector I have the same result as applying $dt$ to the same vector, but I don't see how to apply the covectors to the vectors. It might be helpful to use that $i^*(xdx+ydy+zdz)=0$ where $i:S^2\rightarrow \mathbb{R}^3$ is the canonical inclusion. However, I get that $$ [i^*(xdx+ydy+zdz)](v_1,v_2,v_3)=v_1dv_1+v_2dv_2+v_3dv_3 $$ I don't know why that should be $0$, neither how to use it in the original problem.
Instead of $v_1,v_2,v_3,t$ you can just use $x,y,z,t$, then it gives
$$ \frac{x}{x^2+y^2+z^2}\circ \psi =\frac{e^t x}{e^{2t}(x^2+y^2+z^2)}=e^{-t}\frac{x}{x^2+y^2+z^2}=e^{-t}x $$
as $x^2+y^2+z^2=1$ in $S^2$, and
$$ d(x\circ \psi )=d(e^tx)=e^tdx+xe^t dt=e^t(dx+xdt) $$
Therefore
$$ \psi ^*\left(\frac{xdx+ydy+zdz}{x^2+y^2+z^2}\right)=\frac{xdx+ydy+zdz}{x^2+y^2+z^2}+dt=xdx+ydy+zdz+dt $$
So we only need to show that $xdx+ydy+zdz=0$ in $S^2$. However this follows from the fact that
$$ xdx+ydy+zdz=\frac1{2}d(x^2+y^2+z^2)=\frac12d(1)=0 $$
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