The round metric on $S^2$ with radius $\rho$ is given by $g_{S^2(\rho)} = \rho^2 \cdot \left(\begin{matrix} \sin^2(\phi) & 0 \\ 0& 1 \end{matrix} \right)$. Here the top-left entry is given by $g_{\theta\theta}$ from viewing $(X,Y,Z) = (\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)$ of the inclusion of $S^2(\rho) \hookrightarrow \mathbb{R}^3$. Stereographic projection from the north pole $N=(0,0,1)$ to $\mathbb{R}^2 = \{Z=0\}$ has the form $\Psi: (X,Y,Z) \mapsto (x,y) = (R\cos\theta, R\sin\theta)$ where $R = \cot(\phi/2) = \frac{\rho\sin\phi}{1-\rho\cos\phi}$. Here, we think of $\theta$ and $\phi$ as parametrizing $S^2(r)$. I understand what it means for this matrix $g_{S^2(r)}$ to represent the round metric on $S^2(\rho)$, which is induced from the inclusion $\iota: S^2(\rho) \hookrightarrow \mathbb{R}^3$. Here, $\rho^2\sin^2(\phi) = ||\partial_\theta||^2_{g_{S^2(r)}}$ gives the value of the dot product of $\frac{\partial}{\partial \theta}$ with itself in $T_pS^2(\rho)$, where $p=(X,Y,Z)$ for $S^2(\rho)$ considered as a submanifold of $\mathbb{R}^3$. Similarly for $\rho^2 = || \frac{\partial}{\partial \psi}||^2_{g_{S^2(r)}}$, the value of $\partial_\psi$ dot product with itself in the round metric. If we view $\mathbb{R}^2$ as $\mathbb{C}$, then we can say that [here, I'll think of $\Psi$ as a map from $S^2(\rho) \rightarrow \mathbb{R}^2$ with $\theta$ and $\phi$ parametrizing $S^2(\rho)$] $$\Psi(\theta,\phi) = \cot(\phi/2) e^{i\theta} = z$$ is our complex coordinate.
Using the inverse of the stereographic projection map, $\Psi^{-1}: \mathbb{R}^2 \rightarrow S^2(\rho) \setminus \{N\}$, we can pull-back the round metric from $S^2(\rho)$, or, equivalently, can pull-back the flat metric from $\mathbb{R}^3$, back to $\mathbb{R}^2$. From John Lee's Riemannian Manifolds and other notes on the Fubini-Study metric on $\mathbb{P}^1_\mathbb{C}$, I gather that the pull-back metric should be given by $$ \frac{4}{(|z|^2 + 1)^2} [dx \otimes dx + dy \otimes dy].$$ My question is: What is the matrix representing this pull-back metric? Is it $ \frac{4}{(|z|^2 + 1)^2}$ times the $2 \times 2$ identity matrix if we view $\mathbb{R}^2$ again as a real manifold, or is it $\frac{4}{(|z|^2 + 1)^2}$ times the $1 \times 1$ identity matrix (aka a number) over $\mathbb{C}$?
Maybe it is reasonable to have the pullback of the round metric on $S^2(\rho)$ be a scalar multiple of the identity matrix, since $\theta \mapsto \theta + t$ represents an isometry of $S^2(\rho)$, so that $\frac{\partial}{\partial x}$ would be no longer than $\frac{\partial }{\partial y}$, and vice versa. [From the azimuthal symmetry of stereographic projection]. But if we forget the complex structure on $\mathbb{R}^2$ and instead of calling $z= Re^{i\theta}$, we use $(x,y) = (R\cos \theta, R\sin \theta)$ then I would think the $2\times 2$ matrix comes from the fact that $$ (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)} = (\psi^{-1})^*\overline{g}_{\mathbb{R}^3}$$ is determined by $$ \left(\begin{matrix} (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_x, \partial_x), & (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_x, \partial_y) \\ (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_y, \partial_x), & (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_y, \partial_y) \end{matrix} \right)$$
though when I try this I don't seem to get the $\frac{4}{(R^2+1)^2}\cdot \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right)$ that I would expect. Here the notation is that $\stackrel{\circ}{g}_{S^2(r)}$ is the round metric on $S^2(\rho)$ and $\overline{g}_{\mathbb{R}^3}$ is the flat metric on $\mathbb{R}^3$.
Update The answer I'm currently getting for $(\psi^{-1})^* \overline{g}_{\mathbb{R}^3}$ is $(\psi^{-1})^* \overline{g}_{\mathbb{R}^3}(\partial_i, \partial_j) = \left< C_i, C_j \right>$ for $C_i$ the ith column of the $3 \times 2$ Jacobian matrix $Jac \psi^{-1}$. This does not give that $<C_1, C_1> = \frac{4}{(|z|^2+1)^2}$, nor that $<C_1, C_2> = 0$.
Update 2 Eventually got it to work, some innocuous computational error was thwarting me previously.
They do give the same thing. That is,
$$(\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)} = (\psi^{-1})^*\overline{g}_{\mathbb{R}^3}$$ is determined by $$ \left(\begin{matrix} (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_x, \partial_x), & (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_x, \partial_y) \\ (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_y, \partial_x), & (\psi^{-1})^* \stackrel{\circ}{g}_{S^2(\rho)}(\partial_y, \partial_y) \end{matrix} \right).$$