I'm working through this example in Peter Olver's textbook Application of Lie Groups to Differential equations and I am having some trouble and was wondering if somebody could point out where I went wrong.
So in the first highlighted section, I am trying to understand why this coincides with the tangent space of the sphere. I know that the Hamiltonian vectors are not linearly independent but atleast of them are so they span a $2D$ subspace in the tangent space, however, when we look at the tangent space to the sphere using spherical coordinates I get that it is spanned by $$\frac{d}{d\theta} = -p\sin\theta\sin\phi \frac{d}{du^1} + p \cos\theta\sin\phi \frac{d}{du^2} = -u^2 \frac{d}{du^1} + u^1 \frac{d}{du^2}$$
$$\frac{d}{d\phi} = p\cos\theta\cos\phi \frac{d}{du^1} + p \sin\theta\cos\phi \frac{d}{du^2} - p\sin\phi \frac{d}{du^3}$$
Now the first one coincides with one of the Hamiltonian vectors however the second one I cannot seem to simplify or rewrite so that it looks like any of the other hamiltonian vectors or a linear combination of them.
As for the second highlited piece I wrote $ \phi = \arccos(\frac{u^3}{p})$ and $ \theta = \arctan(\frac{u^2}{u^1})$ by solving the spherical coordinates formulas and got the following by computing the gradient with respect to the coordinates $u = (u^1,u^2,u^3)$ $$\nabla_u(\theta) = (\frac{-u^2}{(u^1)^2 +(u^2)^2}, \frac{u^1}{(u^1)^2+(u^2)^2}, 0)$$ $$ \nabla_u(\phi) = (0,0,\frac{-1}{\sqrt{(u^1)^2+ (u^2)^2}}) $$
$$\nabla_u(\theta) \times \nabla_u(\phi) = (\frac{-u^1}{((u^1)^2 + (u^2)^2)^{3/2}}, \frac{-u^2}{((u^1)^2 + (u^2)^2)^{3/2}}, 0)$$ $$ u \cdot (\nabla_u(\theta) \times \nabla_u(\phi)) = (u^1,u^2,u^3) \cdot (\frac{-u^1}{((u^1)^2 + (u^2)^2)^{3/2}}, \frac{-u^2}{((u^1)^2 + (u^2)^2)^{3/2}}, 0) = \frac{-1}{\sqrt{(u^1)^2+(u^2)^2}}$$
Now changing that back into spherical coordinates I get $$ u \cdot (\nabla_u(\theta) \times \nabla_u(\phi)) = \frac{-1}{p\sin\phi}$$ $$ \implies - u \cdot (\nabla_u(\theta) \times \nabla_u(\phi)) = \frac{1}{p\sin\phi} $$
So I am off by a negative sign, I've done the calculations over and I still get the same thing I am not sure where I am going wrong unless this whole process of doing the gradient was wrong.
For your first question:
In the $(u_1,u_2,u_3)$ coordinates, you have expressed $\frac{d}{d\phi}$ as
$$ \frac{d}{d\phi} = \left( \rho \cos \theta \cos \phi, \; \rho \sin \theta \cos \phi, \; -\rho \sin \phi \right) $$
Using the coordinate-change formulas, you can write this in terms of the $u$'s as
$$ \frac{d}{d\phi} = \left( \frac{u_1u_3}{\sqrt{u_1^2+u_2^2}}, \; \frac{u_2u_3}{\sqrt{u_1^2+u_2^2}}, \; -\sqrt{u_1^2+u_2^2} \right) $$
You can check that this is expressible in terms of the $\hat{v}$'s by
$$ \frac{d}{d\phi} = \frac{u_2}{\sqrt{u_1^2+u_2^2}} \, \hat{v}_1 - \frac{u_1}{\sqrt{u_1^2+u_2^2}} \, \hat{v}_2 = \frac{1}{\sqrt{u_1^2+u_2^2}} \left( u_2 \hat{v}_1 - u_1 \hat{v}_2 \right) $$