Some examples of Hermitian surfaces
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SOME EXAMPLES OF HERMITIAN SURFACES Jaeman Kim1 Department of Mathematics Education, Kangwon National University Kangwon Do, 200-701, Korea E-mail: [email protected] (Received March 9, 2010; Accepted March 29, 2011) [Communicated by J´ anos Szenthe]
Abstract The Riemannian version of the Goldberg–Sachs theorem says that a compact Einstein Hermitian surface is locally conformal K¨ ahler. In contrast to the compact case, we show that there exists an Einstein Hermitian surface which is not locally conformal K¨ ahler. On the other hand, it is known that on a compact Hermitian surface M 4 , the zero scalar curvature defect implies that M 4 is K¨ ahler. Contrary to the compact case, we show that there exists a non-K¨ ahler Hermitian surface with zero scalar curvature defect.
1. Introduction Let M 4 =(M 4 , J, g) be a Hermitian surface (i.e., a Hermitian manifold of real dimension four) with Riemannian metric g and g-orthogonal complex structure J. We denote by Ω the corresponding K¨ahler form, defined by Ω(X, Y ) = g(X, JY ). Then we have dΩ = θ ∧ Ω, where the 1-form θ is called the Lee form of M 4 . Note ahler if and only if θ = 0; M 4 is locally conformal K¨ ahler (lck) if and that M 4 is K¨ only if dθ = 0 and globally conformal K¨ahler (gck) if and only if θ = df for some smooth function f on M 4 (in this case e−f g is a K¨ahler metric). The Riemannian version of the Goldberg–Sachs theorem [1] says that the self-dual part of dθ on an Einstein Hermitian surface vanishes identically. If one assumes that M 4 is a compact Einstein Hermitian surface, then one has 2 ||dθ|| = dθ ∧ ∗(dθ) = dθ ∧ (−dθ) = − d(θ ∧ dθ) = 0. M4
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Hence one gets dθ = 0. Consequently, a compact Einstein Hermitian surface M 4 must be lck. Therefore it is natural to raise the question whether a non-compact Mathematics subject classification numbers: 53A30, 53B35, 53C25, 53C55, 53C56. Key words and phrases: Goldberg–Sachs theorem, Einstein Hermitian surface, locally conformal K¨ ahler, zero scalar curvature defect, K¨ ahler. 1 This study is supported by Kangwon National University. 0031-5303/2012/$20.00 c Akad´emiai Kiad´o, Budapest
Akad´ emiai Kiad´ o, Budapest Springer, Dordrecht
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J. KIM
Einstein Hermitian surface is lck or not. To the author’s knowledge, the above mentioned question still remains unanswered. In [2], T. Sato obtains an Einstein ¯ g) induced from a space of constant holomorphic Hermitian surface M¯ 4 = (D4 , J, sectional curvature. One of the purposes of this note is to show that M¯ 4 is in fact not lck and hence the answer to the above mentioned question is negative. On the other hand, the Riemannian curvature tensor R, the Ricci tensor Ric and the scalar curvature s are defined by R(X, Y )Z = [∇X , ∇Y ]Z − ∇[X,Y ] Z, Ric(X, Y ) = Trace{Z → R(Z, X)Y }, s = Traceg Ric for vector fields X, Y, Z. Furthermore, we define the ∗-Ricci tensor Ric∗ and ∗-scalar curvature s∗ of (J, g) by Ric∗ (X, Y ) = Trace{Z → −JR(Z, X)JY }, s∗ = Traceg Ric∗ . It is known [3] that on a Hermitian surface M 4 , the relation s − s∗ =
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