In the CMU DDG course, for an immersed surface defined by the map \(f\) which is a vector-valued 0-form, the relation between Riemannian metric \(g\), area form \(dA\) and complex structure \(\mathcal{J}\) is given as

\[g(X,Y) = dA(X, \mathcal{J}Y).\]

Here \(X\) and \(Y\) are vectors in the local coordinate chart assigned to the surface patch. \(g(\cdot,\cdot)\) is a covariant rank-2 tensor, which returns the inner product of the two pushed forward vectors related to its two operands, i.e.

\[g(X, Y) = \left\langle df(X), df(Y) \right\rangle.\]

\(dA(\cdot,\cdot)\) is the area form, which returns the absolute area of the parallelogram spanned by the two pushed forward vectors related to its two operands.

The complex structure \(\mathcal{J}\) is defined as rotating the pushed forward vector by \(\frac{\pi}{2}\) around the normal vector:

\[df(\mathcal{J}X) := N \times df(X).\]

According the rule for the wedge product of two vector-valued 1-forms, we have

\[\left( df \wedge df \right)(X, Y) = df(X)\times df(Y) - df(Y)\times df(X) = 2 df(X)\times df(Y) = 2 N dA(X, Y),\]

where \(N\) is the unit normal vector.

Take inner product with respect to \(N\) at both sides of the above equation, we have

\[\left\langle N, df(X) \times df(Y) \right\rangle = dA(X, Y).\]

Then the right hand side of the first equation becomes

\[\rhs = \left\langle N, df(X) \times df(\mathcal{J}Y) \right\rangle = \left\langle N, df(X) \times \left( N \times df(Y) \right) \right\rangle.\]

Using the mixed product identity \(a\cdot(b\times c) = b\cdot(c\times a) = -b\cdot(a\times c)\), we have

\[\rhs = -\left\langle df(X), N\times(N\times df(Y)) \right\rangle.\]

Note that applying \(N\times\) two times is equivalent to rotating the original pushed forward vector by \(\pi\), hence

\[\rhs = -\left\langle df(X), -df(Y) \right\rangle = \left\langle df(X), df(Y) \right\rangle,\]

which is equal to the left hand side \(g(X,Y)\) of the first equation.