Let $M$ be a Kähler manifold. The complex structure on it naturally gives rise to the real analytic structure. I wonder if there exist Kähler manifolds such that the associated symplectic $2$-form $\omega$ is $C^\infty$-smooth but not real analytic.
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1$\begingroup$ What about a small disk in $\mathbb{C}$ with $\omega = \overline{\partial}\partial \left(\|z\|^2 + \varepsilon e^{-1/|z|}\right)$? $\endgroup$– Bertram ArnoldCommented Dec 1, 2018 at 14:00
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$\begingroup$ @BertramArnold Yes, you're right but it should be $e^{-1/|z|^2}$ instead of $e^{-1/|z|}$. Just a misprint. $\endgroup$– cllCommented Dec 2, 2018 at 13:49
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2 Answers
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The answer is positive for any Kähler manifold.
Consider first surfaces. Take a compact Riemann surface $\Sigma$, then any symplectic form on it is associated to a Kähler form, so the answer is yes, since there are plenty of non-analytic $2$-forms.
More generally, for any Kähler manifold $M$ take any Kähler form $\omega$ and let $\omega_1$ be a closed $(1,1)$-form supported in a ball $B\subset M$. Then $\omega+\varepsilon \omega_1$ is Kähler for small $\varepsilon$, but not analytic.
answered Dec 1, 2018 at 22:12
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$\begingroup$ Why doesn't the second argument work for a Riemann surface? $\endgroup$ Commented Dec 2, 2018 at 2:53
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1$\begingroup$ Dear Deane, I have not said that it doesn't work for surfaces, it does. $\endgroup$ Commented Dec 2, 2018 at 11:11
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Not quite an answer, but something related to it.
There is this paper where the authors prove, that for every symplectic manifold $(M,\omega)$ there is an analytical manifold $M^a$ and an analytical symplectic form $\omega^a$ such that $(M,\omega)$ and $(M^a,\omega^a)$ are symplectomorphic.
Edit: The manifold $(M^a,\omega^a)$ is unique up to symplectomorphisms.
answered Dec 1, 2018 at 16:11