**Question**: Suppose you have a **simply connected**, closed, orientable, smooth manifold $M$. What are some restrictions to the existence of a smooth (non-trivial) $S^{1}$-action on $M$?


**Note**: 1.There are some well known restrictions in the case that we require this action to have isolated fixed points (or preserve some additional structures), but I am not aware of any obstruction in this generality. 


2. I am aware of the classification of manifolds with $S^{1}$-actions in dimension $2$ and $3$. In higher dimensions (say $\geq 6$) it seems to become less clear.
3. I am particularly interested in dimension 6. I would like to get a picture of which manifolds in this dimension (satisying the above restrictions), have a smooth $S^{1}$-action.


**Edit**: Out of curiosity, using Puppe's methods, is it possible to construct a simply connected, smooth, complex projective 3-fold such that the underlying smooth manifold has no circle actions?


**Edit 2** It turns out that (the underlying smooth 6-manifold of) any smooth hypersurface in $\mathbb{C}\mathbb{P}^4$ of degree atleast 3 has no smooth (non-trivial) circle actions. It follows from the main result of <https://arxiv.org/pdf/1108.5327.pdf>.



Obstruction to a general S^1-action
V. Puppe, in [Simply connected manifolds without $S^1$-symmetry. Algebraic topology and transformation groups, Proc. Conf., Göttingen/FRG 1987, Lect. Notes Math. 1361, 261-268 (1988)](https://zbmath.org/?q=an:0666.57026) proved the following:



> 
> There exist a simply connected, closed, oriented smooth 6-dimensional manifold $M$ such that no closed, orientable manifold with the same rational cohomology algebra as $M$ admits a non-trivial circle action.
> 
> 
> 


