
> 
> I am looking for a smooth **closed** 4-manifold $M$ with a distinguished point $x\in M$, endowed with an $\mathbb{S}^1$ action such that the stabilizer of $p\in M\setminus\{x\}$ is trivial and $x$ is fixed.
> 
> 
> 


*A naive attempt:*


If we consider the action given by $\mathbb{S}^1$ on $\mathbb{S}^3\subset \mathbb{C}^2$ that gives the Hopf fibration we can extend this action to the 4-ball (it is an unitary action), and it will have the origin as single fixed point. Unfortunately this manifold is not closed.



Example of closed 4 manifold with $\mathbb{S}^1$ action with 1 fixed point and free away from it
Such a closed $4$-manifold does not exist, and this follows from:


Church, P., & Lamotke, K. (1974). Almost free actions on manifolds. Bulletin of the Australian Mathematical Society, 10(2), 177-196


Let me present the argument anyway. The answer breaks down into a local and global part. The local question is to understand what happens near the single fixed point of the $S^1$-action. The global question is about whether we can find, for each local model near the fixed point, a $4$-manifold which closes off the boundary of the model.


**Claim 1:** The only local model near the fixed point is the "naive attempt" from your question statement, i.e. the standard $S^1$-action on $\mathbb{C}^2$.


**Proof of Claim 1:** Up to coordinate change, we may assume the $S^1$-action acts by orthogonal transformations on $\mathbb{R}^4$. In particular, it preserves and is free on $S^3$, and the only free $S^1$-action on $S^3$ is the Hopf (or anti-Hopf if we invert one coordinate) fibration. So your naive attempt really is the only local model.


**Claim 2:** There is no closed $4$-manifold with the property you ask for.


**Proof of Claim 2:** Suppose there were such a closed $4$-manifold. Removing a ball around $p$ corresponding to the local model, $\widetilde{M} := M \setminus B\_p$. Then we obtain a fibration


$$S^1 \rightarrow \widetilde{M} \rightarrow X,$$


such that on the boundary the fibration is just the Hopf fibration $\partial\widetilde{M} = S^3$ over $\partial{X} = S^2$. The fibration over $X$ comes with its classifying map $X \rightarrow BS^1$, and the composition


$$S^2 = \partial X \hookrightarrow X \rightarrow BS^1$$


classifies the Hopf fibration. But the image of $S^2$ under the classifying map for the Hopf fibration $S^2 \rightarrow BS^1$ (which in more down-to-earth language is just $\mathbb{C}\mathbb{P}^1 \hookrightarrow \mathbb{C}{P}^{\infty}$) represents a nontrivial homology class, while $X \rightarrow BS^1$ is a nullhomology of this class. So we arrive at a contradiction.


If you prefer characteristic classes, it suffices to consider the first Chern class in this argument. In addition, we see that if we have a $4$-manifold with an "almost free" $S^1$-action, then the number of fixed points is even, since each fixed point either adds or subtracts $1$ from the first Chern class, depending upon orientations. Conversely, this argument can be boosted to prove you can find a closed $4$-manifold with any even number of fixed points.




---


**EDIT (responding to comment)**


You can ask this question for other group actions, and again, there’s the local and global parts to consider. Let me do this for the case of $\mathbb{Z}\_p$ (the cyclic group, not the $p$-adics for anybody who might be confused) which was asked in a comment. We claim that in this case, again, there is no $4$-manifold $M$ with a $\mathbb{Z}\_p$ action which is free except for a single fixed point. (Hopefully I haven't made a mistake, which is entirely possible, so feel free to be skeptical. I'm sure an algebraic topologist on this site has a better argument for the last part.)


Again, we start with the local models, which arise from the representation theory of $\mathbb{Z}\_p$. For any finite cyclic group, the irreducible real representations are either 1-dimensional (act by the +1 or -1, the latter if p is even) or 2-dimensional (a rotation of order $p$). Since you want every point to be free except the fixed point, you can't use the 1-dimensional representations, since they have order 1 and 2 respectively, so you have a direct sum of two rotations by some 2$\pi k\_1/p$ and $2\pi k\_2/p$ where $k\_1$ and $k\_2$ are coprime to $p$. Up to $\mathrm{Aut}(\mathbb{Z}\_p)$, we may assume $k\_1 = 1$, and we will simply write $k\_2 = k$. At the boundary $S^3$ of this local model, we obtain the fibration


$$\mathbb{Z}\_p \rightarrow S^3 \rightarrow L(p;k)$$


where $L(p;k)$ is a Lens space. Now for the global part. Assume there exists such a $4$-manifold $M$. Then we have the classifying map for the bundle $S^3 = \partial \widetilde{M} \rightarrow \partial X = L(p;k)$ factors through the classifying map for $X$ (which is now itself a $4$-manifold):


$$\phi \colon L(p;k) = \partial X \hookrightarrow X \rightarrow B\mathbb{Z}\_p = K(\mathbb{Z}\_p,1).$$


The cohomology ring $H^\*(K(\mathbb{Z}\_p,1);\mathbb{Z}\_p)$ can be completely understood in terms of Steenrod squares and the Bockstein homomorphism, and in particular, we find that for $u \in H^1(K(\mathbb{Z}\_p,1);\mathbb{Z}\_p)$ the fundamental class, the class


$$v:= u \smile \beta(u) \in H^3(K(\mathbb{Z}\_p;1)\mathbb{Z}\_p)$$


is a nontrivial element (in fact a generator). Then one can check (e.g. from the cohomology ring structure of $L(p;k)$) that $\phi^\*v \neq 0$ as well. Dually, the pushforward of the mod $p$ fundamental class represents a nontrivial element


$$\phi\_\*[L(p;k)] \neq 0 \in H\_3(K(\mathbb{Z}\_p,1);\mathbb{Z}\_p),$$


and so $\phi$ cannot factor through a $4$-manifold $X$. So no such $4$-manifold $M$ exists.


