Is there a contractible space with a free circle action?

Question

Question in title. Seems no to me (some vague intuition here about contracting orbits to a fixed point), but I can't prove it. I'd prefer to be wrong.

(I'm curious because I am thinking about group quotients by non-free actions, so was thinking about the action of the circle on the complex numbers ... And this is because I am learning about algebraic stacks.)

Answer 1

Consider the free action of $S^1$ on the odd sphere $S^{2n+1} = \{(z_0, \dots, z_n)\in \mathbb{C}^n:\ |z_0|^2 + \cdots + |z_n|^2 = 1\}$ by $$e^{i\theta}.(z_0, \dots, z_n) = (e^{i\theta} z_0, \dots, e^{i\theta} z_n).$$ These actions extend across the direct limit $S^\infty = \bigcup S^{2n+1}$, which is contractible.

Answer 2

Just take $X$ to be the total space of the universal bundle $ES^1\rightarrow BS^1$ where $BS^1$ is the infinite dimensional complex projective space which is the classifying space of $S^1$.