I am currently reading "Restrictions on harmonic maps of surfaces" by J. Eells and J. C. Wood and don't understand how they get to the definition of the energy functional. The first paragraph of the paper states:
Let $X$ and $Y$ be closed orientable smooth surfaces, and $\phi: X \to Y$ a smooth map. Relative to Riemannian metrics $g,h$ on $X,Y$ we say that $\phi$ is harmonic if the divergence of its differential vanishes identically. In isothermal charts, writing $h = \sigma^2(w)dwd\overline{w}$ and representing $\phi$ in the form $z \to w(z)$, that condition is expressed by $$ w_{z\overline{z}} + (2\sigma_w/\sigma)w_z w_\overline{z} = 0, $$ being the Euler-Lagrange equation associated to the energy functional $$ E(\phi) = \frac{1}{2}\int \vert d\phi(x) \vert^2 dx = \frac{1}{2} \int_X \sigma^2(w(z))\left[ \vert w_z \vert^2+ \vert w_\overline{z}\vert^2 \right] dx dy. $$
where subscripts denote partial derivatives and $w_z := \frac{1}{2}(w_x - iw_y), w_\overline{z} := \frac{1}{2}(w_x + iw_y)$, etc.
What I don't understand is this:
- The general definition of the energy integral for harmonic maps is $E(\phi) = \frac{1}{2}\int \vert d\phi \vert^2 d\mu_g = \frac{1}{2}\int_M g^{ij}\frac{\partial \phi^\alpha}{\partial x^i}\frac{\partial \phi^\beta}{\partial x^j}h_{\alpha\beta}d\mu_g$ where $\mu_g$ represents the standard measure induced in $M$ from the Riemannian metric $g$. I tried plugging in the metrics in their isothermal representation but I don't seem to be able to get anywhere. How do they arrive at the last equation in the definition of the energy functional?
This is a standard calculation, but a bit hard to pin down in the literature. Maybe see p. 8 of Schoen and Yau's "Lectures on Harmonic Maps".
Here is my rewriting of it. Note I got the answer off by 2 from Eells and Wood.
As you naturally wanted, I will work everything in the real version. So let $$ w=u+iv,\ z=x+iy, $$ and the isothermal metrics on target and domain are \begin{gather*} \sigma^2(u, v)(du^2+dv^2)=\sigma^2(w)dwd\bar w,\\ \tau^2(x, y)(dx^2 + dy^2) = \tau^2(z)dzd\bar z. \end{gather*} (Since the energy is conformal invariant in dimension 2, the $\tau$ would not matter at the end, but for concreteness, we introduce it.)
Then a direct calculation (for just complex numbers) gives $$ |w_z|^2 + |w_{\bar z}|^2 = \frac{1}{2}(u_x^2 + u_y^2 + v_x^2 + v_y^2). $$
Then applying your general formula in this case, \begin{align*} |d\phi|^2 &= h_{\alpha\beta}\frac{\partial \phi^\alpha}{\partial x^i}\frac{\partial \phi^\beta}{\partial x^j}g^{ij}\\ &=\sigma^2 (u_x^2 + u_y^2 + v_x^2 + v_y^2) \tau^{-2}\\ &=2\sigma^2 (|w_z|^2 + |w_{\bar z}|^2) \tau^{-2}. \end{align*} Now $$ d\mu_g = \sqrt{\det g}\,dxdy = \tau^2 dxdy. $$
Therefore, the energy of the map is \begin{align*} E(\phi) &= \frac{1}{2}\int 2\sigma^2( |w_z|^2 + |w_{\bar z}|^2) \,dxdy\\ &=\int_X \sigma^2(w(z))( |w_z|^2 + |w_{\bar z}|^2) \,dxdy. \end{align*}