[proofplan]
We compute the curvature of the global Hopf connection by differentiating the defining $1$-form $\omega=\bar z_1\,dz_1+\bar z_2\,dz_2$. Then we pull $\omega$ back along the standard local section over the affine chart $z_2\neq 0$ and simplify the resulting expression, using the normalization factor $(1+|w|^2)^{-1/2}$. Finally, we differentiate the local connection form directly; since the structure group $U(1)$ is abelian, the local curvature is just $dA$.
[/proofplan]
[step:Differentiate the global Hopf connection form]
Let $z_1,z_2:S^3\to\mathbb{C}$ denote the coordinate functions obtained by restricting the standard coordinates on $\mathbb{C}^2$ to $S^3$. The Hopf connection form is
\begin{align*}
\omega=\bar z_1\,dz_1+\bar z_2\,dz_2.
\end{align*}
Using the [exterior derivative](/theorems/1525) rule $d(f\alpha)=df\wedge\alpha+f\,d\alpha$ for a smooth function $f$ and a differential form $\alpha$, and using $d(dz_1)=d(dz_2)=0$, we obtain
\begin{align*}
d\omega=d(\bar z_1\,dz_1)+d(\bar z_2\,dz_2)=d\bar z_1\wedge dz_1+d\bar z_2\wedge dz_2.
\end{align*}
For a principal $U(1)$-connection, the Lie algebra $\mathfrak{u}(1)$ is abelian, so the bracket contribution in the curvature formula vanishes. Thus the curvature $2$-form of the global connection is
\begin{align*}
\Omega=d\omega=d\bar z_1\wedge dz_1+d\bar z_2\wedge dz_2.
\end{align*}
[/step]
[step:Pull the Hopf connection back along the affine local section]
Define the positive smooth function
\begin{align*}
\rho:\mathbb{C}\to(0,\infty),\qquad w\mapsto 1+|w|^2.
\end{align*}
On the affine chart $U_2$, the standard local section is
\begin{align*}
s:\mathbb{C}\to S^3,\qquad w\mapsto \rho(w)^{-1/2}(w,1).
\end{align*}
Write its coordinate functions as
\begin{align*}
s_1:\mathbb{C}\to\mathbb{C},\qquad w\mapsto \rho(w)^{-1/2}w
\end{align*}
and
\begin{align*}
s_2:\mathbb{C}\to\mathbb{C},\qquad w\mapsto \rho(w)^{-1/2}.
\end{align*}
The pullback local connection form is
\begin{align*}
A=s^*\omega=\overline{s_1}\,ds_1+\overline{s_2}\,ds_2.
\end{align*}
Since $\rho$ is real-valued, $\overline{s_1}=\rho^{-1/2}\bar w$ and $\overline{s_2}=\rho^{-1/2}$. Differentiating the two coordinate functions gives
\begin{align*}
ds_1=\rho^{-1/2}\,dw+w\,d(\rho^{-1/2})
\end{align*}
and
\begin{align*}
ds_2=d(\rho^{-1/2}).
\end{align*}
Therefore
\begin{align*}
A=\rho^{-1}\bar w\,dw+\rho^{-1/2}(\bar w w+1)\,d(\rho^{-1/2}).
\end{align*}
Because $\bar w w+1=\rho$, this becomes
\begin{align*}
A=\rho^{-1}\bar w\,dw+\rho^{1/2}\,d(\rho^{-1/2}).
\end{align*}
The derivative of $\rho^{-1/2}$ is
\begin{align*}
d(\rho^{-1/2})=-\frac{1}{2}\rho^{-3/2}\,d\rho.
\end{align*}
Since $d\rho=d(w\bar w+1)=\bar w\,dw+w\,d\bar w$, we get
\begin{align*}
\rho^{1/2}\,d(\rho^{-1/2})=-\frac{1}{2}\rho^{-1}(\bar w\,dw+w\,d\bar w).
\end{align*}
Substituting this into the expression for $A$ gives
\begin{align*}
A=\rho^{-1}\bar w\,dw-\frac{1}{2}\rho^{-1}(\bar w\,dw+w\,d\bar w)=\frac{1}{2}\rho^{-1}(\bar w\,dw-w\,d\bar w).
\end{align*}
Since $\rho=1+|w|^2$, this is exactly
\begin{align*}
A=\frac{1}{2}\frac{\bar w\,dw-w\,d\bar w}{1+|w|^2}.
\end{align*}
[guided]
The local connection form is defined by pulling the global connection form back along a local section. Here the section is
\begin{align*}
s:\mathbb{C}\to S^3,\qquad w\mapsto \frac{(w,1)}{(1+|w|^2)^{1/2}}.
\end{align*}
Introduce
\begin{align*}
\rho:\mathbb{C}\to(0,\infty),\qquad w\mapsto 1+|w|^2.
\end{align*}
Then $s(w)=\rho(w)^{-1/2}(w,1)$. The two coordinate functions of $s$ are
\begin{align*}
s_1:\mathbb{C}\to\mathbb{C},\qquad w\mapsto \rho(w)^{-1/2}w
\end{align*}
and
\begin{align*}
s_2:\mathbb{C}\to\mathbb{C},\qquad w\mapsto \rho(w)^{-1/2}.
\end{align*}
Since $\omega=\bar z_1\,dz_1+\bar z_2\,dz_2$, pulling back by $s$ gives
\begin{align*}
A=s^*\omega=\overline{s_1}\,ds_1+\overline{s_2}\,ds_2.
\end{align*}
Now compute each term. Because $\rho$ is real-valued, complex conjugation gives $\overline{s_1}=\rho^{-1/2}\bar w$ and $\overline{s_2}=\rho^{-1/2}$. By the product rule,
\begin{align*}
ds_1=d(\rho^{-1/2}w)=\rho^{-1/2}\,dw+w\,d(\rho^{-1/2})
\end{align*}
and
\begin{align*}
ds_2=d(\rho^{-1/2}).
\end{align*}
Substituting these into $A$ gives
\begin{align*}
A=\rho^{-1}\bar w\,dw+\rho^{-1/2}\bar w w\,d(\rho^{-1/2})+\rho^{-1/2}\,d(\rho^{-1/2}).
\end{align*}
The last two terms combine because $\bar w w+1=|w|^2+1=\rho$, so
\begin{align*}
A=\rho^{-1}\bar w\,dw+\rho^{1/2}\,d(\rho^{-1/2}).
\end{align*}
The remaining term is the contribution from differentiating the normalization factor. Since the real function $t\mapsto t^{-1/2}$ has derivative $-\frac12 t^{-3/2}$, the chain rule gives
\begin{align*}
d(\rho^{-1/2})=-\frac{1}{2}\rho^{-3/2}\,d\rho.
\end{align*}
Also,
\begin{align*}
d\rho=d(1+w\bar w)=\bar w\,dw+w\,d\bar w.
\end{align*}
Therefore
\begin{align*}
\rho^{1/2}\,d(\rho^{-1/2})=-\frac{1}{2}\rho^{-1}(\bar w\,dw+w\,d\bar w).
\end{align*}
Putting this back into the formula for $A$ yields
\begin{align*}
A=\rho^{-1}\bar w\,dw-\frac{1}{2}\rho^{-1}(\bar w\,dw+w\,d\bar w)=\frac{1}{2}\rho^{-1}(\bar w\,dw-w\,d\bar w).
\end{align*}
Replacing $\rho$ by $1+|w|^2$ gives the desired local connection form:
\begin{align*}
A=\frac{1}{2}\frac{\bar w\,dw-w\,d\bar w}{1+|w|^2}.
\end{align*}
[/guided]
[/step]
[step:Differentiate the local connection form]
Keep
\begin{align*}
\rho:\mathbb{C}\to(0,\infty),\qquad w\mapsto 1+|w|^2,
\end{align*}
and define the $1$-form
\begin{align*}
\alpha=\bar w\,dw-w\,d\bar w.
\end{align*}
Then
\begin{align*}
A=\frac{1}{2}\rho^{-1}\alpha.
\end{align*}
Since $d(dw)=d(d\bar w)=0$, the exterior derivative of $\alpha$ is
\begin{align*}
d\alpha=d\bar w\wedge dw-dw\wedge d\bar w=2\,d\bar w\wedge dw.
\end{align*}
Also
\begin{align*}
d(\rho^{-1})=-\rho^{-2}d\rho=-\rho^{-2}(\bar w\,dw+w\,d\bar w).
\end{align*}
Using $d(f\alpha)=df\wedge\alpha+f\,d\alpha$, we obtain
\begin{align*}
dA=\frac{1}{2}d(\rho^{-1})\wedge\alpha+\frac{1}{2}\rho^{-1}d\alpha.
\end{align*}
The wedge product in the first term is
\begin{align*}
d\rho\wedge\alpha=(\bar w\,dw+w\,d\bar w)\wedge(\bar w\,dw-w\,d\bar w)=2|w|^2\,d\bar w\wedge dw.
\end{align*}
Therefore
\begin{align*}
dA=-\rho^{-2}|w|^2\,d\bar w\wedge dw+\rho^{-1}d\bar w\wedge dw.
\end{align*}
Combining the scalar coefficients gives
\begin{align*}
dA=(\rho^{-1}-|w|^2\rho^{-2})\,d\bar w\wedge dw=\rho^{-2}(\rho-|w|^2)\,d\bar w\wedge dw.
\end{align*}
Since $\rho=1+|w|^2$, we have $\rho-|w|^2=1$, and hence
\begin{align*}
F_A=dA=\rho^{-2}d\bar w\wedge dw=\frac{d\bar w\wedge dw}{(1+|w|^2)^2}.
\end{align*}
[/step]
[step:Identify the local curvature with the pullback of the global curvature]
Since exterior differentiation commutes with pullback, the same form is obtained by pulling back the global curvature:
\begin{align*}
s^*\Omega=s^*(d\omega)=d(s^*\omega)=dA=F_A.
\end{align*}
Thus the global curvature and the affine-coordinate curvature are precisely the displayed forms, completing the proof.
[/step]
