Locally free circle actions on Euclidean space

Question

I've been wandering whether or not there are locally free smooth circle actions on $\mathbb{R}^n$, or, more specifically, if an Euclidean space would admit a smooth nonvanishing vector field $X$ whose orbits are all periodic.

I know this cannot happen in $\mathbb{R}^2$, and I suspect it is true in every dimension. I know that, since $S^1$ is compact, we can construct an invariant metric on $\mathbb{R}^n$ using an averaging procedure, and therefore consider the $S^1$-action to be an action by isometries. If we could somehow show that these isometries must necessarily be some sort of rotation then they would have fixed points, and therefore could not be locally free. This seems to work in $\mathbb{R}^3$, but I'm not sure how to turn this idea into a rigorous argument, or if it would still hold in higher dimensions.

Any tips, better arguments or counterexamples would be welcome.

A related but more general question is whether or not $\mathbb{R}^n$ admit regular foliations by circles.

Answer 1

Here is a proof that there are no such actions (I am not even assuming smoothness of the action, only continuity). First, the definition (taken from here):

Definition. Let $G$ be a topological group, $X$ a topological space and $\mu: G\times X\to X$ a (continuous) action of $G$ on $X$. Such an action is called locally free if there exists a neighborhood $U$ of the identity $1\in G$ such that each $g\in U\setminus \{1\}$ acts freely on $X$, i.e. the homeomorphism $\mu(g, \cdot): X\to X$ has no fixed points in $X$.

You can also find the definition for instance in

Candel, Alberto; Conlon, Lawrence, Foliations I, Graduate Studies in Mathematics. 23. Providence, RI: American Mathematical Society (AMS). xiv, 402 p. (2000). ZBL0936.57001.

Suppose that $G=S^1$ (with the usual topological group structure) and $X= E^n$, the $n$-dimensional Euclidean space. Suppose that $\mu: G\times X\to X$ is a locally free action. Let $U$ be a neighborhood of $1\in S^1=G$ as in the definition of a locally free action. Then there exist a prime number $p$ such that $$ g=e^{2\pi i/p}\in U. $$ The existence of such $p$ is totally clear (from the infinitude of prime numbers). Then $g$ has no fixed points in $X$. Let $H$ denote the subgroup of $G$ generated by $g$. This group is isomorphic to ${\mathbb Z}_p$, the finite cyclic group of the order $p$ (I am using the topological notation here for finite cyclic groups). I claim that $H$ acts freely on $X$ (via the restriction of the action $\mu$). Indeed, suppose that $h\in H\setminus \{1\}$ is an element such that $h$ fixes a point $x\in X$. Since $H$ has prime order, every nontrivial element of $H$ is a generator. Hence, $g=h^k$ for some $k$. Since $\mu(h,x)=x$, it follows that $\mu(g,x)=\mu(h^k,x)=x$ as well, which is a contradiction to the fact that $g$ has no fixed points in $X$. Thus, $H$ acts freely on $X$. But finite nontrivial groups cannot act freely on Euclidean spaces, see my answer here: I give two proofs of this standard fact from the theory of transformation groups. We conclude that a locally free action of $S^1$ on $E^n$ does not exist.

Answer 2

I believe the answer is no. From your choice of words I assume that you were not looking for a topological proof but anyway: If $\mathbb{R}^n$ admits a free (smooth) circle action the quotient $X=\mathbb{R}^n/S^1$ is a $n-1$ dimensional (smooth) manifold. Since $\mathbb{R}^n$ is contractible the space $X$ is a classifying space for $S^1$-principal bundles (or complex line bundles). This 'space' is usually denoted by $BS^1 = BU(1)$ and it is known that it is unique up to homotopy equivalence. Its cohomology is known to be $H^\bullet(BU(1);\mathbb{Z}) = \mathbb{Z}[x]$ where $x \in H^2(BU(1);\mathbb{Z})$. This is usually done by representing $BU(1)$ by $\mathbb{CP}^\infty$.

Since $X$ is homotopy equivalent to $BU(1)$ it has the same cohomology ring, but at the same time $X$ is a finite dimensional manifold. A finite dimensional manifold cannot have an unbounded cohomology ring.