Can a metric conformal to a Kahler metric be Kahler? - MathOverflow

Can a metric conformal to a Kahler metric be Kahler?

2011-07-08T13:01:27Z

Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?

If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.

If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?

Answer by Francesco Polizzi for Can a metric conformal to a Kahler metric be Kahler?

2011-07-08T13:27:08Z

There are examples in real dimension $4$ of manifolds having two conformally equivalent Kahler metrics, inducing the same conformal structure but with opposite orientation.

See the paper Ambikahler geometry, ambitoric surfaces and Einstein 4-orbifolds by Apostolov, Calderbank and Gauduchon.

Answer by BS for Can a metric conformal to a Kahler metric be Kahler?

2011-07-08T13:49:59Z

You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$.

This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$.

The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).

This said, as in Francesco's answer, you can have non proportional conformal riemannian metrics that are Kähler with respect to different complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.

Answer by Spiro Karigiannis for Can a metric conformal to a Kahler metric be Kahler?

2011-07-08T14:08:44Z

The paper by Apostolov, Calderbank, and Gauduchon that Francesco mentions find different Kaehler structures whose associated Riemannian metrics are conformal to each other. But they correspond to different complex structures $J_+$ and $J_-$.

I believe what Gunnar is asking is whether or not one can have $f \omega$ be closed and thus Kaehler with respect to the same complex structure $J$ associated to $\omega$. The answer is no, and this has nothing at all to do with compactness or the hard Lefschetz theorem. On any almost Hermitian manifold $(M, J, \omega, g)$, it is a fact that the wedge product with the Kaehler form $\omega$ on the space of $1$-forms is injective, regardless of the compactness of $M$, the integrability of $J$, or the closedness of $\omega$. This follows, for example, from the identity

$$ \ast( \omega \wedge (\ast ( \omega \wedge \alpha) ) = - (m-1) \alpha $$

where $\alpha$ is any $1$-form on $M$, where $\ast$ is the Hodge star operator, and the real dimension of $M$ is $2m$. (One sees that the only requirement is that $m>1$.)