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September 10, 2025Contemporary Mathematics0 citations

Riemannian Manifolds Isometric to a Sphere

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SASuha B. Al-Shaikh

Key Points

  • The study shows that if a tangential vector field is an eigenvector, the manifold is isometric to a sphere.
  • It provides evidence that the condition on the Ricci curvature leads to the manifold being a sphere of constant curvature.
  • Using Nash's embedding theorem, any Riemannian manifold can be isometrically immersed in a higher Euclidean space.
  • The results include both direct implications and their converses in the context of Riemannian geometry.

Abstract

To investigate the geometry of a Riemannian manifold(Nm, g), sometimes it is convenient to isometrically immerse it into the Euclidean space Em+n, which is always possible through Nash's embedding Theorem provided that the codimension n is taken to be sufficiently high. The isometric immersion ψ : (Nm, g) → Rm+n can be treated as the position vector of points of Nm in Rm+n and therefore can be expressed as ψ = ξ + ψ⊥, where ξ is tangential to Nm, whereas ψ⊥ is normal to Nm. The Ricci tensor Ric of the Riemannian manifold (Nm, g) yields a symmetric operator Q, known as the Ricci operator, which satisfies the relation Ric (X, Y) = g(QX, Y). In this article, we consider the isometric immersion ψ : (Nm, g) → Rm+n of a compact Riemannian manifold (Nm, g) and show that if the tangential vector field ξ on (Nm, g) is an eigenvector of Q with constant eigenvalue , that is, Qξ = λξ , and the Ricci curvature Ric (ξ , ξ ) satisfies , then (Nm, g) is necessarily isometric to the Euclidean sphere Sm(c) of constant curvature. Moreover, the converse also holds.

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Cite This Study

Suha B. Al-Shaikh (2025) studied this question.

synapsesocial.com/papers/68c1a90c54b1d3bfb60e24bahttps://doi.org/10.37256/cm.6420257418
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