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Fact: For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.
I have thought of a possible argument that re-proves this, but am not sure how to complete it:

Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. Now here is where I want to say: Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$, and denseness will be sufficient by continuity of the action.)

1) Is it possible to fill this gap, i.e. can this 'proof' make sense? Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.

2) Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?

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up vote 6 down vote accepted

You might have a look at Chapter VI of Borel's "Seminar on transformation groups". This is the chapter on "Isotropy groups of toral actions" by E. E. Floyd.

In particular Theorem VI.1.2 seems to be saying that the fixed point sets eventually stabilize. (I can give more details if you don't have the reference to hand.)

Added later: If you are prepared to assume that your $S^1$-action is locally smooth, then the proof that the fixed-point sets eventually stabilize is a little easier, and contains most of the ideas of the general case. A reference is Section IV.1 of Bredon's book "Introduction to compact transformation groups".

To get around the boundary problem, I suggest the following approach. The pair $(D^n,\partial D^n)=(D,\partial D)$ is of course a mod $p$ cohomology $n$-disk, so Smith theory tells you that $(D^{U_p},\partial D^{U_p})$ is a mod $p$ cohomology $r$-disk for some $0\le r\le n$. This implies in particular that $D^{U_p} \neq\partial D^{U_p}$, and so there is a $U_p$-fixed-point in the interior of the disk.

Now note that any group action on a disk must preserve the boundary and the interior. So by restriction you have an $S^1$-action on $\operatorname{int} D \approx \mathbb{R}^n$, which by the above argument has $U_p$-fixed-points for all primes $p$. Now you can apply Theorem IV.1.4 in Bredon to conclude that there is an $S^1$-fixed-point.

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Ah yes, I have this neat book (as well as Bredon's one) which is where I picked up Smith theory 4 years ago. Two quick questions: Do we actually have a rigorously defined $\lim U_p =S^1$ ? (I'm not sure what the definition of limit is being used here). If that fits with the limit-hypothesis of the stated theorem, then this theorem is actually the desired proof for both of my questions, right? (I believe our $X$ is an n-cohomology manifold over $\mathbb{Z}$ since $H^{n-k}_c(X)=H_k(X)$, and we can choose the compact subset $C$ to be $X$ itself). –  Chris Gerig Jan 16 '13 at 21:33
    
Actually I'm not sure the Poincare duality between homology and compactly supported cohomology is true, since X has boundary. –  Chris Gerig Jan 17 '13 at 5:21
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The definition of limit being used is in subsection 2.2. It seems that $\lim U_p = S^1$ since every point of $S^1$ has a neighbourhood which intersects all but finitely many of the $U_p$. And the disk should certainly be a $n$-cohomology manifold with boundary, which just leaves the question of whether these results go over to the boundary case. Perhaps this is covered in the given references. –  Mark Grant Jan 17 '13 at 7:09
    
I hope so. As I cannot fully follow their proof, would you mind (in your answer) giving the gist of how their proof works? –  Chris Gerig Jan 17 '13 at 20:10
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Not quite your question, but I'll say it anyway.

The "fixed-points of actions on a $p$-acyclic space are $p$-acyclic" part of Smith theory easily extends to arbitrary $p$-groups. By induction on the order of the group: if $P$ acts on a $p$-acyclic $X$, choose a non-trivial proper normal subgroup $Z \leq P$ (these always exist; if $P$ is non-abelian take its centre), then $X^P = (X^Z)^{P/Z}$.

Let $U(1)$ act continuously on $D^n$, and $X(n)$ be the $\mathbb{Z}/p^n$ fixed points. This is a compact subset, and non-empty by Smith theory. Thus $X=\cap_{n=1}^\infty X(n)$ is also non-empty, by Cantor's intersection theorem. A point $x \in X$ is fixed by the subgroup of $U(1)$ of $p$-power-torsion points; this is dense, so $x$ is fixed by the whole of $U(1)$.

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Cool! I appreciate the similar flavor of the proof. –  Chris Gerig Jan 16 '13 at 22:15
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