Why the two methods give correct answer for the killing fields?

Question

Consider the embedding of the $\mathbb{S}^2$ in $\mathbb{R}^3$, \begin{align} x & = r \sin \theta \cos \phi,\tag1 \\ y & = r \sin \theta \sin \phi,\tag2 \\ z & = r \cos \theta,\tag3 \end{align} with $r$ fixed at $r=1$.

It has $SO(3)$ symmetry. So now I want to express the generators of $SO(3)$, \begin{equation} L^{X^iX^j} = X^i \frac{\partial}{\partial X^j} - X^j \frac{\partial}{\partial X^i}\tag4 \end{equation} in the terms of the the intrinsic co-ordinates $(r, \theta, \phi)$.

Method 1 ( Similar to what we do when finding the angular momentum operators in spherical polar co-ordinates but Which according the answers and comments here is a miscalculation):

Consider $L^{YZ}=y\frac{\partial}{\partial z} - z \frac{\partial}{\partial y}$ (Rotation in Y,Z co-ordinates)

we express $\partial_z$ and $\partial_y$ in terms of $\partial_\theta$, $\partial_\phi$, using equations (1), (2) and (3) and the chain rule and plug in the equation for the generator. We first treat r as a variable. \begin{equation} \frac{\partial}{\partial z} = \cos \theta \, \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta}. \end{equation}

\begin{equation} \frac{\partial}{\partial y} = \sin \theta \sin \phi \, \frac{\partial}{\partial r} + \frac{\cos \theta \sin \phi}{r} \, \frac{\partial}{\partial \theta} + \frac{\cos \phi}{r \sin \theta} \, \frac{\partial}{\partial \phi}. \end{equation}

with r fixed at $r=1$, we have, \begin{align} \frac{\partial}{\partial z} &= - \sin \theta\, \frac{\partial}{\partial \theta},\tag5 \\ \frac{\partial}{\partial y} &= \cos \theta \sin \phi \, \frac{\partial}{\partial \theta} + \frac{\cos \phi}{\sin \theta} \, \frac{\partial}{\partial \phi}. \tag6 \end{align} Gives the correct answer, \begin{equation}L^{YZ}=\cot \theta \cos \phi \, \frac{\partial}{\partial \phi} + \sin \phi \, \frac{\partial}{\partial \theta}\end{equation}.

Method 2: Working with covectors and the going to dual basis. Please check the answer here We get the expressions for the dual basis, \begin{equation} \begin{bmatrix} \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} = \begin{bmatrix} 0 & \csc\theta\sec\phi \\ -\csc\theta & \cot\theta\csc\theta\tan\phi \end{bmatrix} \begin{bmatrix} \frac{\partial}{\partial\theta} \\ \frac{\partial}{\partial\phi} \end{bmatrix}\tag7 \end{equation} We get the correct expression for $L^{YZ}$.

Method 3 (My attempt a the justification):

Consider the map $f: (x, y, z) \rightarrow (\theta, \phi)$ defined by: \begin{align} \theta &= \cos^{-1}(z), \\ \phi &= \tan^{-1}\left(\frac{y}{x}\right). \end{align} This correctly maps only a subregion. The Jacobian matrix of this transformation is given by: \begin{align} J = \begin{pmatrix} \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} & \frac{\partial \theta}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{pmatrix} = \begin{pmatrix} 0 & 0 & -\frac{1}{\sqrt{1 - z^2}} \\ -\frac{y}{x^2 + y^2} & \frac{x}{x^2 + y^2} & 0 \end{pmatrix}. \end{align} The push-forward of the vector field $(0, z, -y)$, which is $L^x$ under the map $f$ is: \begin{align} f_*(0, z, -y) = \begin{pmatrix} \frac{y}{\sqrt{1 - z^2}} \\ \frac{xz}{x^2 + y^2} \end{pmatrix}. \end{align}

Now if we put, \begin{align} x & = \sin \theta \cos \phi, \\ y & = \sin \theta \sin \phi, \\ z & = \cos \theta, \end{align} we get the killing vector field in the $\partial_{\theta}, \partial_{\phi}$ basis as, $(\sin\phi,\cot\theta\cos\phi)$.

Question 1 : Method 2 seems to be the flawless way of handling the situation when working with general manifolds, but why do the other methods give the correct answer for the Killing Fields?

Question 2 : What is the interpretation of the different expressions for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial y}$ that we get from equations (5),(6) and (7)? In which context are the expressions (5) and (6) meaningful when we are talking about mapping between two manifolds?

Method 2 is standard in quantum mechanics but apparently it seems that there is more to it when we are working with fields on manifolds.

Answer 1

Your issue is that you are mixing up the vector basis with the corresponding dual basis. Essentially the dual basis is for forms while the former is for vectors, and they can be related to each other by the metric. For Cartesian coordinates, the metric is trivially the kronecker delta, so the form and vector bases match. However, when expressing things in terms of 2-sphere coordinates you use here, they are different, and thus you are comparing different quantities. $\mathrm dx^a$ is the form basis while $\frac{\partial}{\partial x^a}:=\partial_a$ is the vector basis. They are related by a completeness relation $\delta^a_b=\mathrm dx^a\partial_b=\frac{\partial x^a}{\partial x^b}$. In what you’ve written, (7) is correct while (6) is actually for $\mathrm dy$ instead of $\partial_y$. To get one from the other you must raise or lower indices with the metric.