Construction of area 2-form on S^2 using Hopf Map given a complex scalar field  $\phi(x_1,x_2,x_3)$ which is a map $\phi:S^3 \mapsto S^2$ where $S^3 = R^3 \cup \{\infty\}$ and $S^2 = C^1 \cup \{\infty\}$
The area two form of $S^2$ normalized to unity, expressed in stereographic co-ordinates is given by
$A = \frac{1}{2\pi i}\frac{d\phi^*\wedge d\phi}{(1+\phi^*\phi)^2}$
This is given in a paper about electromagnetic knots, I would like to know how to get to this area two form, I'm not sure where to start.
 A: $\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\dd}{\partial}$If $w = u + iv$ denotes a complex coordinate on the unit sphere $S^{2} \subset \Reals^{3}$, then stereographic projection from the north pole $(N = (0, 0, 1)$ is given by
$$
f(u, v) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1}
  = \frac{(2w, |w|^{2} - 1)}{|w|^{2} + 1}.
\tag{1}
$$
(By a small abuse of notation, $\Reals^{3}$ has been idenfitied with $\Cpx \times \Reals$.) The parametrization (1) identifies the unit sphere with the Riemann sphere $\Cpx \cup \{\infty\}$. A straightforward calculation (see below) shows that the induced area form on $\Cpx$ is
$$
\omega = \|f_{u} \times f_{v}\|\, du \wedge dv
  = \frac{2i\, dw \wedge d\bar{w}}{(|w|^{2} + 1)^{2}}.
$$
(Out of habit, I've used $\bar{w} = u - iv$ instead of  $w^{*}$ to denote complex conjugation.) Normalizing to unit area gives
$$
A = \frac{\omega}{4\pi}
  = \frac{1}{2\pi}\, \frac{i\, dw \wedge d\bar{w}}{(|w|^{2} + 1)^{2}}
  = \frac{1}{2\pi i}\, \frac{d\bar{w} \wedge dw}{(|w|^{2} + 1)^{2}}.
\tag{2}
$$
(Moving the $i$ from the numerator to the denominator introduces a sign, which is absorbed by swapping the order of the differentials. Also, $|w|^{2} = \bar{w}w$.) If $w = \phi(x_{1}, x_{2}, x_{3})$ is a complex scalar field, substitution in (2) gives your formula.

Edit: Here's a fairly detailed sketch, in real coordinates. By the quotient rule,
$$
\frac{\dd f}{\dd u}
  = \frac{2(-u^{2} + v^{2} + 1, -2uv, 2u)}{(u^{2} + v^{2} + 1)^{2}},\qquad
\frac{\dd f}{\dd v}
  = \frac{2(-2uv, u^{2} - v^{2} + 1, 2v)}{(u^{2} + v^{2} + 1)^{2}}.
$$
The cross product is
\begin{align*}
\frac{\dd f}{\dd u} \times \frac{\dd f}{\dd v}
  &= \frac{4}{(u^{2} + v^{2} + 1)^{4}}
  \left\lvert\begin{array}{@{}ccc@{}}
  i & j & k \\
  -u^{2} + v^{2} + 1 & -2uv & 2u \\
  -2uv & -u^{2} + v^{2} + 1 & 2v \\
  \end{array}\right\rvert \\
  &= \frac{4}{(u^{2} + v^{2} + 1)^{4}}
  \bigl(-2u(u^{2} + v^{2} + 1), -2v(u^{2} + v^{2} + 1), 1 - (u^{2} + v^{2})^{2}\bigr) \\
  &= -\frac{4}{(u^{2} + v^{2} + 1)^{3}}\, (2u, 2v, u^{2} + v^{2} - 1) \\
  &= -\frac{4}{(u^{2} + v^{2} + 1)^{2}}\, \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1}.
\tag{3}
\end{align*}
Since (1) has unit magnitude (as a parametrization of the unit sphere), (3) has magnitude $\dfrac{4}{(u^{2} + v^{2} + 1)^{2}}$.
To convert this to complex form, use $w = u + iv$ to write $dw = du + i\, dv$, $d\bar{w} = du - i\, dv$, and
$$
dw \wedge d\bar{w} = -2i\, du \wedge dv,\qquad
i\, dw \wedge d\bar{w} = 2\, du \wedge dv,
$$
so that
$$
\omega = \|f_{u} \times f_{v}\|\, du \wedge dv
  = \frac{4\, du \wedge dv}{(u^{2} + v^{2} + 1)^{2}}
  = \frac{2i\, dw \wedge d\bar{w}}{(|w|^{2} + 1)^{2}}.
$$
Incidentally, the round metric $g$ on $\Cpx \simeq \Reals^{2}$ induced by stereographic projection is conformally Euclidean:
$$
g = \frac{4\, (du^{2} + dv^{2})}{(u^{2} + v^{2} + 1)^{2}}.
$$
This "explains" why the area element is the indicated multiple of the Cartesian area form, and (in other contexts) is useful for calculating the Gaussian curvature of the round sphere.
