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**On surfaces of low genus whose twistor lifts are harmonic sections.**
*(English)*
Zbl 1198.53048

The author considers the twistor space \(Z\) of an oriented four-dimensional Riemannian manifold \(\tilde{M}\) with the canonical (“product”) Riemannian metric, and the canonical almost complex structure. Each immersed surface \(M\) on \(\tilde{M}\) has a natural twistor lift to \(Z\), and it is called a superminimal surface (twistor holomorphic surface, respectively) if its twistor lift is a horizontal map (holomorphic map, respectively) from \(M\) to \(Z\). The twistor lift is a harmonic section of \(Z\) if it is a stationary map for the energy functional among all sections of the twistor space. Supposing \(\tilde{M}\) is a hyper-Kähler manifold, and \(M\) is a compact surface of genus zero, if the twistor lift is a harmonic section of \(Z\), the author concludes that \(M\) is respectively a non-superminimal minimal surface, a superminimal surface, and a non-superminimal twistor holomorphic surface when \(\chi(T^{\bot}M)\geq 4\), \(\chi(T^{\bot}M)=2\), and \(\chi(T^{\bot}M)\leq 0\), respectively. This completely determines all genus zero surfaces with harmonic twistor lifts, since \(\chi(T^{\bot}M)\) is an even integer for \(\tilde{M}\) a hyper-Kähler manifold. This is applied to the case \(\tilde{M}=\mathbb{R}^4\) obtaining some conclusions by using the fact that no compact minimal submanifold exists, or to obtain an extension of Hopf’s theorem for a constant mean curvature surface of genus zero to surfaces with parallel mean curvature.

As another application in the case \(\tilde{M}=\mathbb{C}^2\), the author obtains the result of Castro and Urbano that, if \(M\) is a compact Lagrangian surface of genus zero and the Maslov form on \(M\) is conformal, then \(M\) is congruent to the Whitney immersion.

As another application in the case \(\tilde{M}=\mathbb{C}^2\), the author obtains the result of Castro and Urbano that, if \(M\) is a compact Lagrangian surface of genus zero and the Maslov form on \(M\) is conformal, then \(M\) is congruent to the Whitney immersion.

Reviewer: Isabel Salavessa (Lisboa)