Does every continuous action of $S^1$ on $R^n$ have a fixed point?

Question

I certainly can't think of one that doesn't. I am aware that there are decompositions of $R^n$ as a union of embedded $S^1$'s, but none of these seem like they would support a continuous action.

($S^1$ is the Lie group.)

Answer 1

If we further assume that the action is differentiable, I have an idea involving Poincare-Hopf index theorem.

If we think of an action of an infinitesimal element of $S^1$, then we can visualize it as a vector field in $\mathbb{R}^n$. Suppose that the vector field is non-vanishing. We can also think of it as a vector field $X$ in $S^n$ (via one-point compactification). In this case we have a singular point (a point where the vector field vanishes) at $\infty$.

Because $S^1$ acts on $S^n$, every flowline of $X$ must come back to the original point after certain time. This implies that flowlines around $\infty$ must be circular flows going around $\infty$ (because no flow can go into $\infty$ nor coming out of it). This means that the index of $X$ at $\infty$ is $1$ or $-1$. (More precisely, if we think of a little sphere $s^{n-1}$ near $\infty$ the vector field around $\infty$ induces a map $s^{n-1} \rightarrow s^{n-1}$, and it has no fixed point. This means that the index at $\infty$ equals $(-1)^{n}$) Now, due to Poincare-Hopf index theorem, the sum of all the indices of $X$ must be equal to the Euler-Poincare characteristic of $S^n$, which is 0 for $n$ odd and 2 for $n$ even. It is a contradiction because $X$ has only one singular point and its index is odd as we have observed.

Therefore $X$ must have a singular point $x_0$ somewhere in $\mathbb{R}^n$, and this means that every differentiable action of $S^1$ on $\mathbb{R}^n$ must have a fixed point. (For action of $S^1$ is a one-parameter group generated by $X$, $x_0$ is a fixed point for any action of $S^1$.)