# Derivation of $x \partial_y - y\partial_x = \partial_{\phi}$

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?

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

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

Insert that in your chain rule and you get it.

• This means that $x^{2}+y^{2}=1$ but I don’t get it ?
– Eli
Dec 11, 2019 at 8:42
• 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. Dec 12, 2019 at 17:06
• 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? 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 $$\partial_\phi = -\sin\theta\sin\phi \partial_x+\sin\theta\cos\phi \partial_y = -y\partial_x+x\partial_y.$$