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Let $G$ be a discrete group and $X$ be a $G$-CW complex. For any $x\in X$ and open neighborhood $U$ of $x$, I am interested in the question that whether we can find a $G_x$-invariant open neighborhood $V$ of $x$ that is contained in $U$. This is not true in general, for example, if $X$ is the cone over $\mathbb R$ with the $G=\mathbb Z$ action by translation, then the cone point $x_0$ has a neighborhood which does not contain any $G_{x_0}$-invariant open neighborhood. But I am wondering whether the following is true:

Any $G$-CW complex $X$ is $G$-homotopy equivalent to a $G$-CW complex $Y$ so that any open neighborhood $U$ of $y\in Y$ contains a $G_y$-invariant open neighborhood $V$.

As a special case, I am wondering whether the following is true:

Let $\mathcal F$ be a family of subgroups of $G$ which is closed under conjugation and finite intersection. Can we always find a model $E_{\mathcal F}G$ for the universal $G$-CW complex relative to $\mathcal F$ so that any open neighborhood of $x\in E_{\mathcal F}G$ contains a $G_x$-invariant open neighborhood?

Thank you!

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I will call the property of a $G$-CW-complex that inside every neighborhood of a point one can find a $G$-invariant neighborhood property $A$.

As in your example, a graph where an edge stabilizer has infinite index in one of the adjacent vertex stabilizers does not have property $A$.

Further $G$-subcomplexes of $G$-CW-complexes with property $A$ have property $A$; given an open set in the subcomplex, extend it to an open set of the whole complex, choose that invariant neighborhood there and intersect back.

Now look at the free group in $a,b$ with the family of subgroups $\mathcal{F}$ containing the trivial group and all conjugates of $\langle a \rangle$ and $\langle b \rangle$. One model for $E_FG$ is given by the Bass-Serre tree, which does not have property $A$. But why can't there be a better model?

Then its one-skeleton would also have property $A$. But by the defining property of $E_FG$ we can find a point $p_a$ with $\langle a \rangle$ is contained in the stabilizer of $p_a$. Since $\langle a \rangle$ is maximal in $F$, it actually follows that $\langle a \rangle$ is the stabilizer of $p_a$. Choose $p_b$ analogously.

Since the one-skeleton is connected, we can choose a finite path from $p_a$ to $p_b$. we want to show that on that path there is some edge, whose stabilizer group has infinite index in the stabilizer of one adjacent vertex. If this was not the case, then $\langle a \rangle$ and $\langle b \rangle$ would be comensurable in the free group, which they are not.

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  • $\begingroup$ Thanks, Henrik. The reason that I asked this question is I was reading the proof of Proposition 7.4 of the paper arxiv.org/abs/math/0701434 (I guess you probably have read this paper). On page 28, it says "Let $\tilde{U}$ be the preimage of $U\subset E$ under the $G$-equivariant map $e\mapsto h_s(p(y), e)$". But I don't think this map is $G$-equivariant when $y$ is fixed (it is $G$-equivariant as a map $(e, y)\mapsto h_s(p(y), e)$). So $\tilde{U}$ won't be $G_{\tilde e}$-invariant. Do you have any thought about this or there is something missing in my understanding. $\endgroup$
    – Kun Wang
    Jul 19, 2020 at 12:55
  • $\begingroup$ Without having a detailed look I would guess it could one of the two: 1) What is the $G$-action used on a product of $G$-spaces. One would guess its the diagonal action, but maybe they choose just to act on $G$. 2)Maybe its just a typo and $\tilde{U}$ should just be $G_y$-invariant ? would this suffice? $\endgroup$ Jul 19, 2020 at 14:04
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    $\begingroup$ @KunWang Does your gmail Address still work? (We exchanged a couple of emails 7 years ago). I just found in my PDF folder the slides of a talk I gave in 2014 about this. Of course a talk should be less formal than the paper, so maybe these are exactly the opposite of what you are looking for, but I can send them to you anyway. $\endgroup$ Jul 19, 2020 at 15:03
  • $\begingroup$ Yes, Henrik. That would be very helpful. Thanks a lot! $\endgroup$
    – Kun Wang
    Jul 19, 2020 at 16:55

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