Can a metric conformal to a Kahler metric be Kahler? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:57:00Z http://mathoverflow.net/feeds/question/69792 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler Can a metric conformal to a Kahler metric be Kahler? Gunnar Magnusson 2011-07-08T13:01:27Z 2013-01-05T05:53:29Z <p>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?</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69796#69796 Answer by Francesco Polizzi for Can a metric conformal to a Kahler metric be Kahler? Francesco Polizzi 2011-07-08T13:27:08Z 2011-07-08T13:34:17Z <p>There are examples in real dimension $4$ of manifolds having two conformally equivalent Kahler metrics, inducing the same conformal structure but with opposite orientation.</p> <p>See the paper <a href="http://www.cirget.uqam.ca/~apostolo/papers/ambitoric.pdf" rel="nofollow">Ambikahler geometry, ambitoric surfaces and Einstein 4-orbifolds</a> by Apostolov, Calderbank and Gauduchon. </p> http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69798#69798 Answer by BS for Can a metric conformal to a Kahler metric be Kahler? BS 2011-07-08T13:49:59Z 2011-07-08T13:57:30Z <p>You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$. </p> <p>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 $\dim_R X \geq 4$. </p> <p>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).</p> <p>This said, as in Francesco's answer, you can have non proportional conformal <em>riemannian</em> metrics that are Kähler with respect to <em>different</em> complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.</p> http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler/69799#69799 Answer by Spiro Karigiannis for Can a metric conformal to a Kahler metric be Kahler? Spiro Karigiannis 2011-07-08T14:08:44Z 2011-07-08T14:08:44Z <p>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 $J_-$.</p> <p>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</p> <p>$$\ast( \omega \wedge (\ast ( \omega \wedge \alpha) ) = - (m-1) \alpha$$</p> <p>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 $m>1$.)</p>