Suppose a compact Kähler manifold $(M,\omega)$ admits a smooth circle action $g_t,~t\in S^1$. So the pull back of the Kähler form $g_t^{\ast}(\omega)$ is a nondegenerate two-form. Since the circle action is only smooth and may not preserve the complex structure, $g_t^{\ast}(\omega)$ may not be $(1,1)$-from. My question is if it is possible for some special manifolds or under some special cases $g_t^{\ast}(\omega)$ is $(1,1)$ for any $t$.
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$\begingroup$ Certainly it's possible and happens all the time when the circle action does preserve the complex structure and the metric (which is what one normally means when one says that a Kähler manifold (as opposed to just the underlying manifold) admits a circle action. Do you mean to assume that the circle action does not preserve the complex structure but does preserve $\omega$? $\endgroup$– Robert BryantCommented Sep 13, 2013 at 1:43
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$\begingroup$ @RobertBryant Here I only assume that this circle action is a smooth action and may neither preserve the complex structure nor the Kahler form $\omega$. If it preserves $\omega$, then the circle action is Killing and its generating vector field is real holomorphic and thus necessarily preserves the complex structure. (Am I right??) $\endgroup$– KevinCommented Sep 13, 2013 at 2:02
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1$\begingroup$ Are you assuming that the circle action preserves the underlying metric of the Kähler structure? (In that case, it is Killing by definition, whether it preserves $\omega$ or not.) Or are you starting out assuming that the circle action is completely unrelated to any of the metric, the complex structure, or the $2$-form, and you just want to know whether it is possible that $g_t(\omega)$ remains always of type (1,1) with respect to the initial holomorphic structure? $\endgroup$– Robert BryantCommented Sep 13, 2013 at 5:16
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$\begingroup$ @RobertBryant Your last sentence is exactly what I want to know. I only assume that the cirlce action $\subset\text{Diff}^+(M)$ and is completely unrelated to any of the metric, complex structure or the Kahler form and I want to know if it is possible $g_t^{\ast}(\omega)$ remains always of type $(1,1)$ with respect to the original holomorphic structure. $\endgroup$– KevinCommented Sep 13, 2013 at 8:03
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1$\begingroup$ Take $\phi\in \mathrm{Diff}^+(S^2)$, and consider the pullbacks $\phi^*J$ and $\phi^*\omega$ of the standard complex structure $J$ and Kaehler form $\omega$. Let $S^1$ act by rotation around the $z$-axis. The action will usually not preserve $\phi^*J$, but $g_t^*(\phi^*\omega)$ is trivially a $(1,1)$-form. $\endgroup$– Tim PerutzCommented Sep 13, 2013 at 21:11
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Now that I fully understand your question, which (apparently) is whether every circle action on a compact Kähler manifold $(M,J,\omega)$ must preserve the $J$-type of $\omega$, I can answer it. The answer is 'no'.
Consider the $4$-dimensional case. Observe that a circle action that preserves all of the $(1,1)$-forms must be holomorphic since, for a circle action, preserving all of the $(1,1)$-forms is equivalent to preserving all of the $(2,0)$-forms, and doing the latter is exactly equivalent to preserving $J$.
Now, start with a circle action $g_t$ on $(M^4,J,\omega)$ that has an isolated fixed point $p$. (For example, choose one of the standard ones on $\mathbb{CP}^2$.) Let $X$ be the vector field that generates $g_t$, so that $X(p)=0$, and $X$ induces a nontrivial circle action on $T_pM$. If this circle action does preserve $J_p$, then conjugate $g_t$ by a nonholomorphic diffeomorphism that fixes $p$ to get a new circle action such that the flow of its $X$ does not preserve $J$ at $p$. Then the flow of this $X$ will not preserve all of the $(1,1)$-forms at $p$. If this $X$ does not preserve the $J$-type of $\omega_p$, then we are done. Otherwise, let $\alpha$ be a $(1,1)$-form at $p$ such that the flow of $X$ does not keep $\alpha$ of $J$-type $(1,1)$. Choose any closed $(1,1)$-form $\eta$ on $M$ such that $\eta_p = \alpha$. Then, for some small $\epsilon\not=0$, the form $\omega{+}\epsilon\eta$ will be a positive $(1,1)$-form on $(M,J)$. Then $(M, J, \omega{+}\epsilon\eta)$ is a Kähler manifold, but the circle action generated by $X$ will not keep $\omega{+}\epsilon\eta$ of $J$-type $(1,1)$ because this doesn't hold at $p$.
answered Sep 14, 2013 at 8:30