= Solution
Choose a <Riemannian metric> $h$ on $\Sigma$ compatible with its <complex structure> $j$, and use $g_J=\omega(\cdot,J\cdot)$ on the target. The <Dirichlet energy of a map> is
$$
E(u)=\frac12\int_\Sigma|du|_{h,g_J}^2\,d\operatorname{vol}_h.
$$
It is independent of the particular conformal representative $h$: rescaling $h$ multiplies the squared differential norm by the inverse factor and the area element by the same factor. For a <J-holomorphic curve>, the <Energy identity for a J-holomorphic curve> is
$$
\boxed{E(u)=\int_\Sigma u^*\omega.}
$$
To prove it, take an oriented $h$-orthonormal frame $e_1,e_2=je_1$ and put $a=du(e_1)$, $b=du(e_2)$. The <J-holomorphic curve> equation says $b=Ja$. The pointwise energy density is therefore $\frac12(|a|_{g_J}^2+|Ja|_{g_J}^2)=|a|_{g_J}^2$, while the pulled-back area density is $\omega(a,Ja)=|a|_{g_J}^2$. Integrating proves the identity.
More generally, the same frame gives
$$
\frac12(|a|^2+|b|^2)-\omega(a,b)=\frac12|b-Ja|^2.
$$
With the full tensor <norm> of $\bar\partial_Ju$, its two frame components are $(a+Jb)/2$ and $(b-Ja)/2$, so their squared norms sum to $\frac12|b-Ja|^2$. Hence the full identity is
$$
E(u)=\int_\Sigma u^*\omega+\int_\Sigma|\bar\partial_Ju|^2\,d\operatorname{vol}_h.
$$
This also fixes the normalization of the error term. \b[The energy is nonnegative and vanishes exactly when $du=0$.] The formulas apply whenever the relevant integrals are defined, in particular on compact source surfaces.
Back to article page
