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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 $\frac{\partial}{\partial \theta}$, $\frac{\partial}{\partial \phi}$?

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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.

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  • $\begingroup$ 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? $\endgroup$ – jane Jan 20 at 9:34
  • $\begingroup$ @jane, yes, the resulting vector is the representation of $\mathbf e_1=\frac\partial{\partial\theta}$ in the embedding given by $f(\theta,\phi)=(x(\theta,\phi), y(\theta,\phi), z(\theta,\phi))$. $\endgroup$ – Klaas van Aarsen Jan 20 at 10:46

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