On a $S^2$-sphere we can define the coordinates $$x = \sin(\theta)\cos(\phi)\\ y = \sin(\theta)\sin(\phi)\\z=\cos(\theta).$$ Then I want to show that $$x \partial_y - y\partial_x = \partial_{\phi}.$$

I know the possibility to use the chain rule $$\partial_{\phi} = \frac{\partial x}{\partial \phi} \partial_x+ \frac{\partial y}{\partial \phi}\partial_y$$ to prove this, but for that I have to know that the $x \partial_y - y\partial_x$ equals $\partial_{\phi}$. I wanted to try it without knowing it, so I thought if I put in $x$ and $y$ from the definition above and write $$\partial_x = \frac{\partial \theta}{\partial x} \partial_{\theta}+ \frac{\partial \phi}{\partial x}\partial_{\phi} = \left( \frac{\partial x}{\partial \theta} \right)^{-1} \partial_{\theta}+ \left( \frac{\partial x}{\partial \phi} \right)^{-1}\partial_{\phi} \\= \left( \cos(\theta) \cos(\phi) \right)^{-1}\partial_{\theta} + \left(- \sin(\theta) \sin(\phi) \right)^{-1}\partial_{\phi}$$ and analogous for $\partial_y$, but if I insert this in $x \partial_y - y\partial_x$ I don't get the right answer. What am I doing wrong?


2 Answers 2


$$\frac{\partial x}{\partial{\phi}}=-\sin(\theta)\sin(\phi)=-y$$

$$\frac{\partial y}{\partial{\phi}}=\sin(\theta)\cos(\phi)=x$$

Insert that in your chain rule and you get it.

  • $\begingroup$ This means that $x^{2}+y^{2}=1$ but I don’t get it ? $\endgroup$
    – Eli
    Dec 11, 2019 at 8:42
  • $\begingroup$ Thank you for your answer. I knew that, but I think I didn't explain my question precise enough. I wondered how I can get from a differential expression like $x \partial_y - y \partial_x$ in cartesian coordinates to an expression $a(\theta,\phi)\partial_{\theta}+b(\theta,\phi)\partial_{\phi}$ in general, because if I used the second method, that I described above, I get a wrong answer. And I don't know what's wrong with my method. $\endgroup$
    – Lukas
    Dec 12, 2019 at 17:06
  • $\begingroup$ But I thought about it again and I realized, that you can make an ansatz like $a(\theta, \phi)\partial_{\theta}+b(\theta,\phi)\partial_{\phi}$ and then express $\partial_{\theta}$ and $\partial_{\phi}$ via chain rule with $\partial_x$ and $\partial_y$. Then you can compare coefficients with the differential expression in cartesian coordinates and you get your answer. But this is a long calculation, does anyone of you know a faster way? $\endgroup$
    – Lukas
    Dec 12, 2019 at 17:11

I know the possibility to use the chain rule $$\partial_{\phi} = \frac{\partial x}{\partial \phi} \partial_x+ \frac{\partial y}{\partial \phi}\partial_y$$ to prove this, but for that I have to know that the $x\partial_y-y\partial_x$ equals $\partial_\phi$.

I don't get your point there. To use the chain rule you absolutely don't need to know this.

Let $M$ be a smooth manifold, like $\mathbb{R}^3$ or $S^2$. Let $(x,U)$ and $(y,V)$ be two coordinate systems on the open subsets $U$ and $V$ of $M$, respectively. Suppose further that the two overlap, $U\cap V\neq \emptyset$. Then in the overlap we can change coordinates. That means that if $p\in U\cap V$, then it has both $x(p)$ and $y(p)$ coordinates and these two can be related: $$x(p)=x(y^{-1}(y(p))=(x\circ y^{-1})(y(p)).$$

Now, in the overlap we have two partial derivative operations: partials with respect to $x^i$ and with respect to $y^j$. These operators are related through the chain rule: $$\dfrac{\partial}{\partial y^j}=\dfrac{\partial x^i}{\partial y^j}\dfrac{\partial}{\partial x^i},$$

and this is totally general, this can be found from the change of variables expression which in effects expresses each $x^i$ as some function of the $y^j$. For more details check "A Comprehensive Introduction to Differential Geometry Vol. 1" by Michael Spivak.

Now let's talk about $S^2$ in particular. First by definition $$S^2=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2+z^2=1\}.$$

One possible chart is obtained by solving $x^2+y^2+z^2=1$ in $z$ as $$z = \pm \sqrt{1-x^2-y^2}$$

Now take the plus sign. This gives the upper hemisphere $U$ and the coordinates are the $(x,y)$. Geometrically you project from the upper hemisphere onto the $xy$-plane.

Another possible chart is obtained by using angular coordinates. The points of $S^2$ are written as you have done in the question. The open set $V$ is obtained by removing the points at $\theta =0,2\pi$. These charts clearly overlap.

Moreover, the $(x,y)$ of the first chart is related to the $(\theta,\phi)$ of the second by the formulas you have, so $$\dfrac{\partial x}{\partial \phi} = -\sin\theta\sin\phi,\quad \dfrac{\partial y}{\partial \phi}=\sin\theta\cos\phi,$$

and the general chain rule gives $$\partial_\phi = -\sin\theta\sin\phi \partial_x+\sin\theta\cos\phi \partial_y = -y\partial_x+x\partial_y.$$


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