Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 / 6

An almost complex structure is a smooth bundle endomorphism $J:TM\to TM$ with $J^2=-I$. It is an $\omega$-compatible almost complex structure whenThese conditions make $g_J(v,w)=\omega (v,Jw)$ a Riemannian metric. In particular it is symmetric: invariance and $J^2=-I$ give $\omega (Jv,w)=-\omega (v,Jw)$, and skew-symmetry then gives $g_J(w,v)=g_J(v,w)$. Positivity is the second condition. Moreover $J$ is an isometry for $g_J$. Compatibility does not require that the almost complex structure be integrable.

For a Riemann surface $(\Sigma ,j)$ and an almost complex manifold $(M,J)$, a J-holomorphic curve is a smooth map $u:\Sigma \to M$ satisfyingThus its differential is complex-linear at every point. In oriented local coordinates $s,t$ with $j{\partial }_s={\partial }_t$, this is $u_t=J{u}_s$, equivalently $u_s+J{u}_t=0$. No integrability of the target almost complex structure is required. Constant maps satisfy the definition; some usages reserve the word curve for nonconstant maps, so that restriction should be stated separately when intended.

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 isIt 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 isTo prove it, take an oriented $h$-orthonormal frame $e_1,e_2=j{e}_1$ and put $a=du(e_1)$, $b=du(e_2)$. The J-holomorphic curve equation says $b=Ja$. The pointwise energy density is therefore $\frac{1}{2}(|a{|}_{g_J}^2+|Ja{|}_{g_J}^2)=|a{|}_{g_J}^2$, while the pulled-back area density is $\omega (a,Ja)=|a{|}_{g_J}^2$. Integrating proves the identity.

More generally, the same frame givesWith the full tensor norm of ${\bar{\partial }}_{J}u$, its two frame components are $(a+Jb)/2$ and $(b-Ja)/2$, so their squared norms sum to $\frac{1}{2}|b-Ja{|}^2$. Hence the full identity isThis also fixes the normalization of the error term. 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.

No nonconstant J-holomorphic curve with this closed connected source exists. The standard symplectic form on ${\mathbb{R}}^{2n}$ is exact; for instanceThe Energy identity for a J-holomorphic curve and the Generalized Stokes theorem giveThe target metric from the compatible almost complex structure is positive definite, so the nonnegative continuous energy density must vanish everywhere. Thus $du=0$. Connectedness of $\Sigma$ makes $u$ constant. This argument uses exactness and compatibility, and works even when $J$ is nonintegrable; it does not require the ordinary holomorphic maximum principle on the target.