2010
DOI: 10.2969/jmsj/06230851
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Geometric realizations of Hermitian curvature models

Abstract: We show that a Hermitian algebraic curvature model satisfies the Gray identity if and only if it is geometrically realizable by a Hermitian manifold. Furthermore, such a curvature model can in fact be realized by a Hermitian manifold of constant scalar curvature and constant ⋆-scalar curvature which satisfies the Kaehler condition at the point in question.

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Cited by 4 publications
(4 citation statements)
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“…if the para Nijenhuis tensor NJ Gray [41] showed that the curvature tensor of a Hermitian manifold has an additional symmetry given below in Equation (7.a); it is quite striking that a geometric integrability condition imposes an additional algebraic symmetry on the curvature tensor. We refer to [41] for the proof of the first Assertion and to [12] for the proof of the second Assertion in the following Theorem: We refer to [10] for the proof of the first Assertion and to [12] for the proof of the second Assertion in Theorem 7.2; this result provides a useful converse to Theorem 7.1. Again we shall focus on the scalar curvature and the ⋆-scalar curvature:…”
Section: Hermitian Geometrymentioning
confidence: 99%
“…if the para Nijenhuis tensor NJ Gray [41] showed that the curvature tensor of a Hermitian manifold has an additional symmetry given below in Equation (7.a); it is quite striking that a geometric integrability condition imposes an additional algebraic symmetry on the curvature tensor. We refer to [41] for the proof of the first Assertion and to [12] for the proof of the second Assertion in the following Theorem: We refer to [10] for the proof of the first Assertion and to [12] for the proof of the second Assertion in Theorem 7.2; this result provides a useful converse to Theorem 7.1. Again we shall focus on the scalar curvature and the ⋆-scalar curvature:…”
Section: Hermitian Geometrymentioning
confidence: 99%
“…These tensors are anti-symmetric in the last two indices so ρ 13 = −ρ 14 . We verify that 0 = A 1 − A 2 ∈ W 9 ∩ J + which establishes Assertion (2). Next, we take ̺ = ( 1 2 , − 1 2 , 1 2 ) to create a tensor such that: A 3 (e 1 , f 1 , e 1 , f 2 ) = A 3 (e 1 , f 1 , f 2 , e 1 ) = 1, A 3 (e 1 , f 1 , f 1 , e 2 ) = A 3 (e 1 , f 1 , e 2 , f 1 ) = A 3 (e 1 , e 2 , f 1 , f 1 ) = −1, A 3 (e 1 , e 2 , e 1 , e 1 ) = A 3 (f 1 , f 2 , e 1 , e 1 ) = A 3 (f 1 , f 2 , f 1 , f 1 ) = −1, ρ 14 (A 3 )(e 1 , e 2 ) = ρ 14 (A 3 )(f 1 , f 2 ) = 1, ρ 14 (A 3 )(e 2 , e 1 ) = ρ 14 (A 3 )(f 2 , f 1 ) = −1.…”
Section: The Proof Of Theorem 18mentioning
confidence: 59%
“…Gray [11] showed that the integrability of the (para)-complex structure gives rise to the additional curvature identity G(R g ) = 0. Although his result was originally stated only in the Hermitian setting, it extends easily to the para/pseudo-Hermitian setting [2,4]. In fact, this identity remains valid in the context of Weyl geometry:…”
Section: Introductionmentioning
confidence: 92%