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We show that a para-Hermitian algebraic curvature model satisfies the para-Gray identity if and only if it is geometrically realizable by a para-Hermitian manifold. This requires extending the Tricerri-Vanhecke curvature decomposition to the para-Hermitian setting. Additionally, the geometric realization can be chosen to have constant scalar curvature and constant *-scalar curvature.
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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.
We show that every Kaehler algebraic curvature tensor is geometrically realizable by a Kaehler manifold of constant scalar curvature. We also show that every para-Kaehler algebraic curvature tensor is geometrically realizable by a para-Kaehler manifold of constant scalar curvature
We show that every Kaehler affine curvature model can be realized geometrically.
We study geometric realization questions of curvature in the affine, Riemannian, almost Hermitian, almost para Hermitian, almost hyper Hermitian, almost hyper para Hermitian, Hermitian, and para Hermitian settings. We also express questions in Ivanov-Petrova geometry, Osserman geometry, and curvature homogeneity in terms of geometric realizations.
We show any Riemannian curvature model can be geometrically realized by a manifold with constant scalar curvature. We also show that any pseudo-Hermitian curvature model, para-Hermitian curvature model, hyper-pseudo-Hermitian curvature model, or hyper-para-Hermitian curvature model can be realized by a manifold with constant scalar and *-scalar curvature.
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