Totally umbilical semi-invariant submanifolds of a nearly trans-Sasakian manifold (Q2466282)
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| Language | Label | Description | Also known as |
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| English | Totally umbilical semi-invariant submanifolds of a nearly trans-Sasakian manifold |
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Totally umbilical semi-invariant submanifolds of a nearly trans-Sasakian manifold (English)
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14 January 2008
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Let \(\overline M\) be a manifold with an almost contact metric structure \((\varphi,\xi,\eta,g)\); that means: \(g\) is a Riemannian metric on \(\overline M\), \(\xi\) a unit vector field, \(\eta= g(\cdot,\xi)\) and \(\varphi\) a skew-symmetric tensor field of type \((1,1)\) with the following properties: \(\varphi(\xi)= 0\) and \(\varphi|_{\xi^\perp}\) is a complex structure on the subbundle \(\xi^\perp\subset T\overline M\). Moreover, let \(M\) be a semi-invariant submanifold of \(\overline M\); that means: \(\xi|_M\) is a vector field on \(M\) and its orthogonal complement in \(TM\) is the orthogonal sum of two distributions \({\mathcal D}\) and \({\mathcal D}^\perp\) such that \(\varphi({\mathcal D})={\mathcal D}\) and \(\varphi({\mathcal D}^\perp)\subset TM^\perp\). The authors show: If \(\overline M\) is even a nearly trans-Sasakian manifold, i.e. \(\nabla\varphi\) satisfies some additional relation, and if \(M\) is totally umbilical, then one of the following assertions is true: \(M\) is totally geodesic or \({\mathcal D}= \{0\}\) or (\(\dim{\mathcal D}^\perp= 1\) and \({\mathcal D}\oplus\langle\xi\rangle\) is not integrable). It should be mentioned that cosymplectic, \(\alpha\)-Sasakian and \(\beta\)-Kenmotsu manifolds are trans-Sasakian, and these are nearly trans-Sasakian.
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almost contact metric manifold
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semi-invariant submanifold
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totally umbilical submanifold
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