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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](http://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem). (In particular, no nontrivial group acts freely on any closed ball.)2012-08-21
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    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......2012-08-21
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    Why would you need that? What does it have to do with freeness?2012-08-21
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    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”).2012-08-21
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    I'm looking to see if $S^1$ fixes a point $x_0\in D^n$, i.e. for all $g\in S^1$ we have $gx_0=x_0$.2012-08-21
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    Well, it seems like we have a definition issue then. The definition of free action I know is the same as the one on Wikipedia – no element but identity has any fixed points.2012-08-21
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    I'm sure everyone can edit their own questions. See also: [related thread](http://math.stackexchange.com/questions/166793/possibilities-of-an-action-of-s1-on-a-disk)2012-08-21
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    @MikeJolsteen: If you see „edit” button, you can edit. :) Everyone can edit their own posts.2012-08-21

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