Question: Is there a group $G$ and a CW-complex $X$ such that
1) $X$ is homotopy equivalent to the circle $S^{1}$.
2) $G$ acts on $X$
3) the space of fixed points $X^{G}$ is weakly equivalent to $S^{2}$ ?
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2$\begingroup$ Is there any particular motivation? Do we know the answer for $S^0$ - $S^1$ variant of your question? $\endgroup$– user43326Commented Nov 14, 2014 at 15:53
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$\begingroup$ Oh, sorry I did not see you comment! Unfortunately I don't have a concrete and clear motivation. I found this question interesting in it self (at least for me). I would love to hear the answer to your second question too! $\endgroup$– MaxCommented Nov 14, 2014 at 16:50
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2 Answers
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A very general answer is given by Tony Elmendorf's paper Systems of Fixed Point Sets, http://www.ams.org/journals/tran/1983-277-01/S0002-9947-1983-0690052-0/. Very loosely speaking, it says that, if you can write down a reasonable system of fixed sets, it can be realized up to homotopy equivalence.
Take, for example, $G = \mathbb{Z}/2$. Define a contravariant functor $\mathscr{X}$ on the category of orbits of $G$ with $\mathscr{X}(G/e) = S^1$ and $\mathscr{X}(G/G) = S^2$; let the self-map of $G/e$ act on $S^1$ by reflection, with two fixed points, and let the map corresponding to the projection $G/e\to G/G$ map $S^2$ to one of the fixed points in $S^1$. Elmendorf's construction then gives a $G$-CW complex $X$ whose system of fixed sets maps to $\mathscr{X}$ with $X\to \mathscr{X}(G/e)$ and $X^G\to \mathscr{X}(G/G)$ both being homotopy equivalences.
answered Nov 14, 2014 at 18:09
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$\begingroup$ Very elegant! exactly was I was looking for, Thank you! Just one more question: is $X^{G}$ a CW complex in this case? $\endgroup$– MaxCommented Nov 14, 2014 at 18:27
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1$\begingroup$ Yes. A $G$-CW complex has cells of the form $G/H\times D^n$, which implies that its fixed sets are all CW complexes. $\endgroup$ Commented Nov 14, 2014 at 18:35
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How about the flow of
$$ X=(\cos(2\pi\theta)+2-|x|^2)\partial_\theta $$
on $B^3\times S^1$, with coordinates $x\in B$ and $\theta\in S^1$? The solutions of this system exist for all time, hence this defines an $\mathbb{R}$ action. The fix point set coincides with the zero set of the vector field, which is $S^2\times\{1/2 \}$. By removing the non-fixed points which are not on periodic orbits (i.e. $S^2\times (S^1\setminus\{\frac{1}{2}\})$) we get an $S^1$ action with the required properties.
answered Nov 14, 2014 at 16:03
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$\begingroup$ Unfortunately, my knowledge about dynamics is limited! I will be grateful if you can give more details! thank you nay way! $\endgroup$– MaxCommented Nov 14, 2014 at 16:25
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1$\begingroup$ Fixed points of the flow are exactly the zeros of the vector field. The vector field is only zero at points with $\theta=1/2$ and $|x|^2=1$. $\endgroup$ Commented Nov 14, 2014 at 16:31
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1$\begingroup$ I am on a mobile now, hence not introducing the variables properly. I'll edit this later $\endgroup$ Commented Nov 14, 2014 at 16:32
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$\begingroup$ Once @Thomas (bravo!) said $\ B^3\times S^1\ $ the rest is easy. One may give a generalization of this too. $\endgroup$ Commented Nov 14, 2014 at 19:15
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1$\begingroup$ @Max: yes, solutions of ODE's are $\mathbb{R}$ actions. $\endgroup$ Commented Nov 14, 2014 at 20:53