# vector field of a sphere

I have a problem to understand vector fields on manifolds. Theoretically I get it, but I cannot find a concrete example with numbers.

For example, I have the $$S^2$$ sphere and I take the classic polar parametriazation ($$\theta$$,$$\phi$$)$$\to$$ r($$\theta$$,$$\phi$$) in order to get the vector fields r$$\theta$$,r$$\phi$$.What are those vector fields?Are they a base of the tangent space?Obviously they are, but why?

Are they the same with the canonical base $$\frac{\partial}{\partial \theta}$$, $$\frac{\partial}{\partial \phi}$$?

## 1 Answer

The basis of the tangent space is indeed $$\frac\partial{\partial\theta},\frac\partial{\partial\phi}$$. This is an abstracted form, which is also sometimes written as $$\partial_\theta,\partial_\phi$$.

We can associate the points on the sphere with the coordinates in the map $$(\theta,\phi)$$ and then 'read' the basis vectors 'through the coordinate map'. If we do, we get: $$\frac\partial{\partial\theta}(\theta,\phi)=(1,0)\quad\text{and}\quad\frac\partial{\partial\phi}(\theta,\phi)=(0,1)$$ So we see that they span indeed a 2-dimensional tangent space. The differential map of the coordinate map transforms them to the unit vectors in its coordinate space.

For comparison, if we look at the embedding of $$S^2$$ in $$\mathbb R^3$$ with its canonical cartesian coordinate map $$(x,y,z)$$, we get: $$\frac\partial{\partial\theta}(\cos\phi\sin\theta,\ \sin\phi\sin\theta,\ \cos\theta)=(\cos\phi\cos\theta,\ \cos\phi\cos\theta,\ -\sin\theta)$$ And if we 'read through the cartesian coordinate map' $$(x,y)$$ of $$S^2$$, we get: $$\frac\partial{\partial\theta}(\cos\phi\sin\theta,\ \sin\phi\sin\theta)=(\cos\phi\cos\theta,\ \cos\phi\cos\theta)$$ which is the representation of $$\frac\partial{\partial\theta}$$ in the cartesian coordinate map.

• I am a bit confused now.Is that wright: r$\theta$:$R^2$ $\to$ $R^3$ as ($\theta$,$\phi$) $\to$ ($\frac{\partial x}{\partial \theta}$,$\frac{\partial y}{\partial \theta}$,$\frac{\partial z}{\partial \theta}$) and this is the same as e1? – jane Jan 20 at 9:34
• @jane, yes, the resulting vector is the representation of $\mathbf e_1=\frac\partial{\partial\theta}$ in the embedding given by $f(\theta,\phi)=(x(\theta,\phi), y(\theta,\phi), z(\theta,\phi))$. – Klaas van Aarsen Jan 20 at 10:46