# Symplectic form on a Kähler manifold can be not real analytic?

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.

• What about a small disk in $\mathbb{C}$ with $\omega = \overline{\partial}\partial \left(\|z\|^2 + \varepsilon e^{-1/|z|}\right)$? – Bertram Arnold Dec 1 '18 at 14:00
• @BertramArnold Yes, you're right but it should be $e^{-1/|z|^2}$ instead of $e^{-1/|z|}$. Just a misprint. – anna abasheva Dec 2 '18 at 13:49

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.

• Why doesn't the second argument work for a Riemann surface? – Deane Yang Dec 2 '18 at 2:53
• Dear Deane, I have not said that it doesn't work for surfaces, it does. – Dmitri Panov Dec 2 '18 at 11:11

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.