Fixed point of $S^1$-action using roots of unity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:11:58Z http://mathoverflow.net/feeds/question/119050 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119050/fixed-point-of-s1-action-using-roots-of-unity Fixed point of $S^1$-action using roots of unity Chris Gerig 2013-01-16T10:36:06Z 2013-01-18T10:14:11Z <p><strong>Fact:</strong> For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.<br> I have thought of a possible argument that re-proves this, but am not sure how to complete it:</p> <p>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 $p$. <em>Now here is where I want to say:</em> 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.)</p> <p>1) <strong>Is it possible to fill this gap, i.e. can this 'proof' make sense?</strong> Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.</p> <p>2) <strong>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 $p>p_0$?</strong></p> http://mathoverflow.net/questions/119050/fixed-point-of-s1-action-using-roots-of-unity/119055#119055 Answer by Mark Grant for Fixed point of $S^1$-action using roots of unity Mark Grant 2013-01-16T11:50:00Z 2013-01-18T10:14:11Z <p>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.</p> <p>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.)</p> <p><strong>Added later</strong>: 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".</p> <p>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 <em>in the interior</em> of the disk.</p> <p>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. </p> http://mathoverflow.net/questions/119050/fixed-point-of-s1-action-using-roots-of-unity/119058#119058 Answer by Oscar Randal-Williams for Fixed point of $S^1$-action using roots of unity Oscar Randal-Williams 2013-01-16T12:31:44Z 2013-01-16T13:34:14Z <p>Not quite your question, but I'll say it anyway. </p> <p>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 $X^P = (X^Z)^{P/Z}$.</p> <p>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 $U(1)$.</p>