pull back of a Kähler form by a smooth circle group action on a compact Kähler manifold 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 $t$.
 A: 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 $p$.
