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Is much known about $S^1$-actions on the following simple spaces?:
1) $D^2$ the disk
2) More generally $D^n$ the n-ball
3) $S^1$ the circle

In particular, does every $S^1$-action on the disk (or general n-ball) have a fixed point? I.e. there is an $x_0\in D^2$ (or $D^n$) such that for all $g\in S^1$ we have $gx_0=x_0$.

Are the general actions just rotations about the origin? I believe $S^1$ must at least send the boundary to itself, because the $\epsilon$-neighborhood of a boundary point is different than the $\epsilon$-neighborhood of an interior point. And is it just rotations for $S^1$ on the circle?

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You might want to get acquainted with Brouwer fixed-point theorem. (In particular, no nontrivial group acts freely on any closed ball.) –  tomasz Aug 21 '12 at 2:29
    
That's for a single map. It is not immediately obvious (nor do I see a proof) that a closed loop of maps $D^n\to D^n$ preserves the fixed point...... –  Matt Fahrad Aug 21 '12 at 2:31
    
Why would you need that? What does it have to do with freeness? –  tomasz Aug 21 '12 at 2:35
    
Also, what you say is not obvious, I think is intuitively false: it is not hard to imagine an action which drags a single point on the disk along an arbitrary closed curve (including one passing through all fixed points of one of the transformations corresponding to the „drag”). –  tomasz Aug 21 '12 at 2:40
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I'm sure everyone can edit their own questions. See also: related thread –  user31373 Aug 21 '12 at 3:08

1 Answer 1

I assume by $S^1$-action you mean a continuous map $S^1 \times D^n \to D^n$ satisfying the usual axioms.

In that case, every $S^1$-action on $D^n$ has a fixed point. Given such an action let $f : D^n \to D^n$ be a map coming from an irrational angle (by which I mean an irrational multiple of $\pi$) in $S^1$. By the Brouwer fixed point theorem, $f$ has a fixed point $x$, hence every power $f^k$ of $f$ has the same fixed point $x$. But since the multiples of an irrational angle are dense in $S^1$, it follows by continuity that $x$ is fixed by every element of $S^1$.

There are more general actions than rotations about the origin. I am most familiar with the case $n = 2$, although what I'm about to say should generalize: the interior of $D^2$ has a natural conformal structure and its group of conformal automorphisms acts transitively (and also acts on the boundary). Explicitly, thinking of $D^2 \subset \mathbb{C}$, the map

$$z \mapsto \frac{z - a}{\bar{a} z - 1}$$

switches $0$ and $a$, where $a$ is in the interior, so two such maps switch any two points in the interior. Consequently one can conjugate the group of rotations about the origin to get groups of "conformal rotations" about any point in the interior, so in particular any point of the interior may be a fixed point of an $S^1$-action.

More generally, any action can be conjugated by a homeomorphism. In particular an action of $S^1$ on $S^1$ can be conjugated by a homeomorphism $S^1 \to S^1$, and it is easy to write down weird examples, e.g. $e^{i \theta} \mapsto e^{i f(\theta)}$ where $f : [0, 2\pi] \to [0, 2\pi]$ is any increasing continuous function with $f(0) = 0, f(2\pi) = 2\pi$ (actually I think this exhausts all examples up to rotation).

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And it's also good to note (as you mentioned to me privately) that for smooth $S^1$-actions we have a different argument: The Lie-group action generates a vector field on $D^n$, and by the Poincare-Hopf index theorem it must have a fixed point (because the Euler characteristic of the n-disk is 1). –  Chris Gerig Aug 24 '12 at 4:59

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