6
$\begingroup$

Let $G$ be a discrete group and $X$ be a Tychonoff $G$-space. Then there exists a $G$-action on Stone-Čech compactification $\beta X$. If the fixed point set $X^{G}\neq \emptyset $, then the Stone-Čech compactification of the fixed point set is the fixed point set of the Stone-Čech compactification, i.e., $\beta \left( X^{G}\right) =\left( \beta X\right) ^{G}$?

$\endgroup$
  • 1
    $\begingroup$ $\beta$ is a left adjoint functor, and taking $G$-fixed points is a limit. So I suspect this isn't true as in general left adjoint don't preserve limits. $\endgroup$ – Niall Oct 23 '18 at 10:12
  • $\begingroup$ To start with, one has a canonical continuous map $\beta(X^G)\to (\beta X)^G$. Also the assumption $X^G\neq\emptyset$ is irrelevant as it can be artificially enforced by adding an isolated fixed point to $X$. $\endgroup$ – YCor Oct 23 '18 at 22:51
5
$\begingroup$

Not without further assumptions. First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ say. Then make the product $X\times X$ and take out $\langle\Omega,\Omega\rangle$. You can rotate the resulting space $Y$ around that hole over $\pi$ by mapping $\bigl<\langle i,\alpha\rangle,\langle j,\beta\rangle\bigr>$ to $\bigl<\langle 1-i,\alpha\rangle,\langle 1-j,\beta\rangle\bigr>$ and similarly for points with $\Omega$ as a coordinate. This map $f:Y\to Y$ has no fixed points but $\beta Y=X$ and $\beta f$ does have a fixed point: $\langle\Omega,\Omega\rangle$.

Now take the sum $Y\cup\{0\}$ and extend $f$ by $f(0)=0$. Then $f$ has one fixed point and $\beta f$ has two. As $f$ is its own inverse this gives us an action of $\mathbb{Z}/2\mathbb{Z}$ that is a counterexample.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.