Skip to main content
MathOverflow

Return to Answer

Correction of terminology. Hermitian manifolds have an integrable almost complex structure. Relaxing the integrability assumption gives an almost Hermitian manifold.
Source Link
edit approved Dec 23, 2019 at 22:40
AmorFati
edit approved Dec 23, 2019 at 22:40
AmorFati
  • 1.4k
  • 10
  • 20

If you only have an almost complex manifold with a compatible metric i.e. aan almost Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.

If you only have an almost complex manifold with a compatible metric i.e. a Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.

If you only have an almost complex manifold with a compatible metric i.e. an almost Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.

Source Link
kalafat
kalafat
  • 303
  • 1
  • 5

If you only have an almost complex manifold with a compatible metric i.e. a Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.