Problem to conceptualize $\frac{\partial }{\partial \theta}$ and $\frac{\partial }{\partial \varphi}$.

Question

I have some little problem to give a conception to $\frac{\partial }{\partial \theta}$ and $\frac{\partial }{\partial \varphi}$ on manifold (like $\frac{\partial }{\partial x}$ as well). For example, on the sphere, since $$(x,y,z)=(\sin \varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi),$$ we get $$\frac{\partial }{\partial \theta}=(-\sin\varphi\sin\theta,\sin\varphi\cos\theta,0)\quad \text{and}\quad \frac{\partial }{\partial \varphi}=(\cos\varphi\cos\theta,\cos\varphi\sin\theta,-\sin\varphi).$$

What is written in my book, it's that is a basis of the tangent space of the sphere (ok, why not... I can imagine that $T_p(\mathbb S^2)$ is a vector space of dimension 2, and locally homeomorphic to $\mathbb R^2$, and above we have a basis). But now, why the notation $\frac{\partial }{\partial \theta}$ and $\frac{\partial }{\partial \varphi}$ ? What would be for example $\frac{\partial f}{\partial \theta}$ of a function $f=f(\theta,\varphi)$ or as well $\frac{\partial f}{\partial \varphi}$ ? How would you compute these derivative on the $\mathbb S^2$ ?

Answer 1

Consider a $2$-sphere with coordinates given by latitude ($\phi$) and longitude ($\theta$). Except at the poles, this gives a local $2$ dimensional coordinate system (each point has many coordinates, but that won't matter).

Through each point, there is a unique curve on which longitude is constant, and through each point except the poles there is a nontrivial curve on which latitude is constant.

In general, partial derivatives of a function $f$ can be viewed as derivatives along curves in which many coordinates are constant, with respect to the change in the coordinate named in the partial derivative. In particular, the partial derivatives $\partial f /\partial \phi$ and $\partial f/ \partial \theta$ can be viewed as derivatives of $f$ along the curves from the previous paragraph. These derivatives can be computed, among other ways, by parameterizing the curves and using the usual formulas for derivatives of a function along parameterized curves.

If you want to get a basis for the tangent space at each point other than the poles, you can use the unit tangent vectors of the curves from the second paragraph, which are basically what you computed in the question.

Answer 2

It is a general fact that a vector field on a manifold $M$ is exactly the same thing as a derivation of the algebra $C^\infty(M)$, i.e. a linear map $\delta : C^\infty(M) \to C^\infty(M)$ that satisfies: $$\delta(fg) = \delta(f) g + f \delta(g).$$

Indeed if $X$ is a vector field and $f \in C^\infty(M)$, then $X \cdot f \in C^\infty(M)$ is defined by partial derivative along $X$, by the following formula that makes sense in a chart: $$(X \cdot f)(x) = \lim_{t \to 0} \frac{1}{t} (f(x+tX_x) - f(x))$$

One instance of derivation that you're very familiar with is differentiation of smooth functions in $C^\infty(\mathbb{R})$, given by $f \mapsto f'$ (that satisfies the well-known law $(fg)' = f'g+fg'$. Well as it turns out, the corresponding vector field on $\mathbb{R}$ is exactly the one usually denoted by $\frac{\partial}{\partial x}$. This is not a coincidence!

The correspondence between vector fields and derivations goes like this. Say you have a chart $\phi : U \to M$, where $U \subset \mathbb{R}^n$. Using $\phi$, you have an isomorphism of algebras between $C^\infty(\phi(U))$ and $C^\infty(U)$

Then you can define a derivation $\frac{\partial}{\partial x_i}$ on $C^\infty(U)$, simply using the standard formulas for an open subset of $\mathbb{R}^n$. Using $\phi$ this gives you a derivation on $C^\infty(\phi(U))$, which you can then use to define a vector field on $\phi(U)$, still called $\frac{\partial}{\partial x_i}$ (with the hidden $\phi$ behind the scenes).

This is exactly what you have in your question. When you restrict the domain of $(\theta,\phi)$, you get a chart from an open subset $U \subset \mathbb{R}^2$ to an open subset $V \subset \mathbb{S}^2$. Then you can use the formulas above to define a derivation on $C^\infty(V)$. And now it turns out that no matter how you chose $U$ to restrict $(\theta,\phi)$, this gives the same derivation! So you get a well-defined global derivation on $C^\infty(\mathbb{S}^2)$, which is given in local coordinates by differentiating wrt $\theta$.

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tl;dr Differentiating wrt a variable is something that is only valid in local coordinates. This gives a vector field, and sometimes that vector field can be extended to all of $M$, giving something that we still call $\partial/\partial x$ even though the formula for differentiating wrt is only valid in a given chart.