Transformation for isothermal coordinates of energy of harmonic maps between Riemann surfaces

Question

I am currently reading section "3.6 Maps Between Surfaces. The Energy Integral. Definition and Simple Properties of Harmonic Maps" in "Compact Riemann surfaces" by Jürgen Jost and he defines the energy of a map as follows

Let $\Sigma_1$ and $\Sigma_2$ be Riemann surfaces; suppose $\Sigma_2$ carries a metric, given in local coordinates by $$ \rho^2(u)du d\overline{u}. $$ Let $z = x + iy$ be a local conformal parameter on $\Sigma_1$. Now let $$ u: \Sigma_1 \to \Sigma_2 $$ be a map, of class $C^1$ to start with. We define the energy integral of $u$ as $$ \begin{align} E(u) &:= \int_{\Sigma_1} \rho^2(u) (u_z \overline{u}_{\overline{z}} + \overline{u}_z u_{\overline{z}}) \frac{i}{2} dz d\overline{z} \\ &= \frac{1}{2} \int_{\Sigma_1} \rho^2(u(z))(u_x \overline{u}_x + u_y \overline{u}_y)dxdy \end{align} $$ ($u_z := \frac{1}{2} (u_x − iu_y), u_{\overline{z}} := \frac{1}{2} (u_x + iu_y)$, etc., subscripts denoting partial derivatives.)

What I don't understand basically boils down to one question:

• By plugging in $u_z := \frac{1}{2} (u_x − iu_y), u_{\overline{z}} := \frac{1}{2} (u_x + iu_y)$, etc., in the first line of the definition of the energy integral I managed to get the $(u_x \overline{u}_x + u_y \overline{u}_y)$ part of the second line but I don't understand how the differentials transform and how they seem to "suck up" the imaginary $i$.

Answer 1

With the little nudge from Ted Shifrin in the comments I can answer my own question:

By plugging in $u_z := \frac{1}{2} (u_x − iu_y), u_{\overline{z}} := \frac{1}{2} (u_x + iu_y)$, etc., in the first line of the definition of the energy integral I managed to get the $(u_x \overline{u}_x + u_y \overline{u}_y)$ part of the second line but I don't understand how the differentials transform and how they seem to "suck up" the imaginary $i$.

We plug in $u_z := \frac{1}{2} (u_x − iu_y), u_{\overline{z}} := \frac{1}{2} (u_x + iu_y)$, etc., and $z = x + iy$ to get $$ \begin{align} E(u) :&= \int_{\Sigma_1} \rho^2(u)(u_z\overline{u}_\overline{z} + \overline{u}_z u_\overline{z})\frac{i}{2}dzd\overline{z} \\ &= \int_{\Sigma_1} \rho^2(u)(u_z\overline{u}_\overline{z} + \overline{u}_z u_\overline{z})\frac{i}{2}dz \wedge d\overline{z} \\ &= \int_{\Sigma_1} \rho^2(u(z))\left(\frac{1}{2}(u_x -iu_y)\frac{1}{2}(\overline{u}_x + i\overline{u}_y) + \frac{1}{2}(\overline{u}_x - i\overline{u}_y)\frac{1}{2}(u_x + iu_y)\right)\frac{i}{2}d(x+iy) \wedge d(x - iy) \\ &= \int_{\Sigma_1} \rho^2(u(z))\left(\frac{1}{4}(u_x\overline{u}_x + iu_x\overline{u}_y -i\overline{u}_xu_y + u_y\overline{u}_y ) + \frac{1}{4}(u_x\overline{u}_x + i\overline{u}_xu_y - iu_x\overline{u}_y + u_y\overline{u}_y )\right)\frac{i}{2}d(x+iy) \wedge d(x - iy) \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)\frac{i}{2}d(x+iy) \wedge d(x - iy) \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)\left(\frac{i}{2}(dx+idy \wedge dx - idy)\right) \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)\left(\frac{i}{2}(dx\wedge dx -idx \wedge dy +idy\wedge dx + dy \wedge dy)\right) \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)\left(\frac{i}{2}(0 - idx \wedge dy - idx\wedge dy + 0)\right) \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)dx\wedge dy \\ &= \int_{\Sigma_1} \rho^2(u(z))\frac{1}{2}(u_x\overline{u}_x + u_y\overline{u}_y)dxdy \end{align} $$

which is exactly the second line.