Rigidity of transversally biharmonic maps between foliated Riemannian manifolds (Q1797250)

A harmonic map \(\varphi : M \to N\) between Riemannian manifolds is a critical point of the energy functional \(E(\varphi)= \frac{1}{2}\int_M |d\varphi|^2 \, dv_g\), and a map is called biharmonic if it is a critical point of the bienergy functional \[ E_2(\varphi) = \frac{1}{2}\int_M |\tau(\varphi)|^2\, dv_g. \] Here \(\tau(\varphi)\) is the tension field of \(\varphi\). A harmonic map is biharmonic by definition, but the converse is not true in general. There is a conjecture in this direction called the generalized Chen's conjecture. This says that every biharmonic submanifold of a Riemannian manifold of nonpositive curvature must be harmonic. In the paper under review, the authors consider the generalized Chen's conjecture for smooth foliated maps between foliated Riemannian manifolds. Let \((M, g, {\mathcal F})\) be a complete, possibly noncompact, foliated Riemannian manifold, and let \((N, h, {\mathcal G})\) be another foliated Riemannian manifold. A smooth foliated map \(\varphi : (M, g, {\mathcal F}) \to (N, h, {\mathcal G})\) is transversally harmonic if it is a critical point of the transversal energy, and it derives the Euler-Lagrange equation \(\tau_b(\varphi) = 0\). Here \(\tau_b(\varphi)\) denotes the transversal tension field of \(\varphi\). The transversal bienergy can be also defined as \(1/2\int_M |\tau_b(\varphi)|^2\, dv_g\), and the transversally biharmonic maps are critical of the transversal bienergy. Recall that \(\varphi\) is a foliated map if, for every leaf \(\ell \in {\mathcal F}\), there eixsts a leaf \(\ell' \in {\mathcal G}\) such that \(\varphi(\ell) \subset \ell'\). In this paper, the authors prove that if a smooth foliated map \(\varphi : (M, g, {\mathcal F}) \to (N, h, {\mathcal G})\) is transversally biharmonic, and both its energy and bienergy are finite, then \(\varphi\) is transversally harmonic provided that the transversal sectional curvature of \((N, h, {\mathcal G})\) is nonpositive.