Symplectic 4-manifolds with Hermitian Weyl tensor

Abstract

Abstract. It is proved that any compact almost Kähler, Einstein 4-manifold whose fundamental form is a root of the Weyl tensor is necessarily Kähler.

Untitled

“…By the contracted Bianchi identity, C À is a section of V ð1Þ M, and so the last statement of Lemma 1 can be rephrased as follows. In particular it follows that on M 0 , the self-dual Weyl tensor of g g, with the orientation induced by I, is degenerate, and equal to 3 By the conformal covariance of the Weyl tensor, it follows that on M 0 , the anti-self-dual Weyl tensor of g (using the orientation induced by J) is given by…”

“…To any J-invariant 2-form j ¼ j 0 þ 3 2 s o, we may associate a normalized 2-form j j ¼ 1 2 j 0 þ 1 4 s o. For example, if j is the Ricci form r, thenj j is the 2-formr r associated to the normalized Ricci tensor:r rðÁ; ÁÞ ¼ hðJÁ; ÁÞ.…”

“…in particular, the Bach tensor vanishes whenever W þ , W À or Ric 0 vanishes. If ðM; g; J Þ is a Ka¨hler surface, the Bach tensor is easily computed by using the above identity and the fact that W þ ¼ 3 4 so 0 o. Indeed, if B þ and B À , denote the J-invariant and J-anti-invariant parts of B, we get…”

“…In this section we classify weakly self-dual Ka¨hler surfaces of nowhere constant scalar curvature s, but for which p and s are not independent. As in the previous section (when we assumed p and s were independent) we do this by finding an explicit formula for a Ka¨hler surface ðM; g; J; oÞ with a Hamiltonian 2-form j ¼ j 0 þ 3 2 so such that K 1 :¼ J grad g s has no zero, but K 2 :¼ J grad g p ¼ bK 1 , where p is the Pfaffian of j j and b is (necessarily) constant.…”

“…e.g. [5]) shows that an almost Ka¨hler 4-manifold has constant Lagrangian sectional curvature if and only if the Ricci tensor is J-invariant, the Weyl tensor is self-dual, and the Ka¨hler form is one of its roots (i.e., M has Hermitian Weyl tensor in the sense of [3]).…”