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Let $X$ be a compact Kahler manifold of complex dimension $n$. The Aubin--Calabi--Yau theorem says that if we fix a smooth form $\rho$ in the Chern class $c_1(X)$, then every Kahler class on $X$ contains a unique Kahler metric $\omega$ whose Ricci-form is $\rho$. Alternatively, one may fix a volume form $dV$ on $X$, then the theorem gives the existence of a unique metric $\omega$ in each Kahler class whose volume form is a constant multiple of $dV$, or $dV_\omega = c dV$ where $c > 0$ is a constant:

Indeed, if we have $\rho$, let $dV = dV_\omega$ for any Kahler metric $\omega$ whose Ricci-form is $\rho$. If we have $dV$, consider the smooth hermitian metric $h$ on the canonical bundle $K_X$ defined by the equality $i^{n^2} \alpha \wedge \overline \beta = h(\alpha,\overline \beta) dV$, and take $\rho$ to be its curvature form.

Since there are at least three ways to define the Ricci tensor of a hermitian metric, but the volume form of any hermitian metric $\omega$ is $dV_\omega = \omega^n/n!$, we'll fix a volume form $dV$ such that $Vol(X,dV) = 1$.

Question: The ACY theorem gives Kahler metrics $\omega$ with $dV_\omega = dV$. Can there be non Kahler metrics on $X$ whose volume form is $dV$?

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2 Answers 2

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A property of a hermitian metric to be conformally Kähler is a very restrictive property and a generic hermitian metric is not conformally Kähler. The property of the volume form to be equal to some fixed volume form is senseless if we allow to multiply the metric by a conformal coefficient. Combining this, we see that a generic hermitian metric with some fixed volume form it not Kähler, neither in the local nor in the global setup.

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I see how the second statement implies that every conformal class of hermitian metrics contains at most one representative with the given volume form, but is it clear that we have at least one such metric in each conformal class? –  Gunnar Þór Magnússon Apr 16 '12 at 11:25
Oh of course, I'm sorry, I was being thick. –  Gunnar Þór Magnússon Apr 16 '12 at 11:29

I really should think more about these things before asking. The answer is "yes".

K. Yang considers the flag manifold $F := F_{1,2,3} := SU(3)/S(U(1)^3)$ in Invariant Kahler metrics and projective embeddings of the flag manifolds. He shows that the space of hermitian metrics on $F$ is parametrized by $(\mathbb R_+)^3$, that is, any hermitian metric on $F$ is given by $$ h_{a,b,c} = a^2 \theta_1\otimes\overline\theta_1 + b^2 \theta_2\otimes\overline\theta_2 + b^2 \theta_3\otimes\overline\theta_3 $$ where $\theta_j$ are holomorphic 1-forms on $F$ and $a$, $b$ and $c$ are positive constants. He also shows that if $h$ is a Kahler metric on $F$, then $h$ is a constant multiple of $h_{1,\sqrt 2,1}$.

Note that the volume form of $h_{a,b,c}$ is $(abc)^2 dV$, where $dV$ is the volume form obtained by wedging the $(1,1)$-forms $(i/2)\theta_j \wedge \overline\theta_j$ together. A Kahler metric $h = \lambda h_{1,\sqrt 2,1}$ thus has the volume form $dV_h = 2\lambda^2 dV$ and we can find lots of hermitian metric $h_{a,b,c}$ with that same volume form.

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In the expression for $h_{a,b,c}$, should the coefficient of $\theta_3\otimes\bar{\theta}_3$ be $c^2$ instead of $b^2$? –  Michael Albanese Apr 19 '12 at 12:06
Yes, thank you. –  Gunnar Þór Magnússon Apr 19 '12 at 15:00
maybe this question help mathoverflow.net/questions/154093/… –  Hassan Jolany Jan 9 '14 at 21:15

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