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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 (4),(5) and (7)? In which context are the expressions (4) and (5) 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.

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.

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} 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 (4),(5) and (7)? In which context are the expressions (4) and (5) 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.

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 (4),(5) and (7)? In which context are the expressions (4) and (5) 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.

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 $(t, \chi, \theta, \phi)$.

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} 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 (4),(5) and (7)? In which context are the expressions (4) and (5) 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.

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} 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 (4),(5) and (7)? In which context are the expressions (4) and (5) 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.