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**Biharmonic maps on principal \(G\)-bundles over complete Riemannian manifolds of nonpositive Ricci curvature.**
*(English)*
Zbl 1472.58013

This article considers principal \(G\)-bundles, equipped with a Sasaki-type metric, over a Riemannian manifold and the canonical projection \(\pi\), which is then a Riemannian submersion.

The problem investigated is to find conditions such that \(\pi\) biharmonic implies \(\pi\) harmonic.

The first theorem proved is that if the principal \(G\)-bundle is compact and has non-positive Ricci curvature then \(\pi\) biharmonic implies \(\pi\) harmonic. The reader will notice in the proof that the one-form \(\alpha\) defined by Equation (3.7) is not quite well-defined on the base manifold unless the tension field of \(\pi\) is actually basic. This should then be compared with [C. Oniciuc, An. Ştiinţ. Univ. Al. I. Cuza Iaşi, Ser. Nouă, Mat. 48, No. 2, 237–248 (2002; Zbl 1061.58015)].

The second set of conditions is to assume the principal \(G\)-bundle is non-compact complete with non-positive Ricci curvature and the energy and bienergy of \(\pi\) are finite. Then \(\pi\) biharmonic implies \(\pi\) harmonic. The proof of this second theorem is really only a rehash of N. Nakauchi et al. [Geom. Dedicata 169, 263–272 (2014; Zbl 1316.58012)].

The problem investigated is to find conditions such that \(\pi\) biharmonic implies \(\pi\) harmonic.

The first theorem proved is that if the principal \(G\)-bundle is compact and has non-positive Ricci curvature then \(\pi\) biharmonic implies \(\pi\) harmonic. The reader will notice in the proof that the one-form \(\alpha\) defined by Equation (3.7) is not quite well-defined on the base manifold unless the tension field of \(\pi\) is actually basic. This should then be compared with [C. Oniciuc, An. Ştiinţ. Univ. Al. I. Cuza Iaşi, Ser. Nouă, Mat. 48, No. 2, 237–248 (2002; Zbl 1061.58015)].

The second set of conditions is to assume the principal \(G\)-bundle is non-compact complete with non-positive Ricci curvature and the energy and bienergy of \(\pi\) are finite. Then \(\pi\) biharmonic implies \(\pi\) harmonic. The proof of this second theorem is really only a rehash of N. Nakauchi et al. [Geom. Dedicata 169, 263–272 (2014; Zbl 1316.58012)].

Reviewer: Eric Loubeau (Brest)

Full Text:
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