G-equivariant homotopy between G-spaces

Question

I apologize for asking too many questions in a single post. I am not very conversant with equivariant homotopy theory. While discussing with some faculty I was told that certain fact is true. All spaces mentioned below are simply connected and finite type CW complex.

Statement: Let $G$ be a compact Lie group acting on topological spaces $X$ and $Y$. If:

1. The fixed point sets $X^H$ and $Y^H$ are homotopy equivalent (in the non-equivariant sense) for every subgroup $H$ of $G$
2. $H^G(X;\underline{\mathbb{Q}})\cong H^G(Y;\underline{\mathbb{Q}})$, where $H^G(X;\underline{\mathbb{Q}})(G/H):=H(X^H;\mathbb{Q})$ and $\underline{\mathbb{Q}}$ is the constant rational coefficient system i.e., $\underline{\mathbb{Q}}(G/H):=\mathbb{Q}$ for all subgroup $H$ of $G$

then there exists a $G$-homotopy equivalence $f: X \to Y$.

However, I could not find any proper reference of the same. Maybe I misunderstood the statement. Here are some questions:

1. Is this statement true?

2. If $G=C_p$ the cyclic group of order prime, does the similar hold?

3. For $G=C_p$, what additional conditions are required to obtain a $G$-homotopy equivalence $f$ between $X$ and $Y$?

I will be really grateful for any proper refernce/ proof/ valuable suggestions.

Answer 1

I believe that (3) has already been answered in the comments. Elmendorfs theorem concludes that there is a weak $G$-homotopy equivalence and under suitable cofibrancy conditions we get a $G$-homotopy equivalence. So the additional assumptions are the assumptions of Elmendorfs theorem.

I also have some trouble understanding the notation of (2). Let me sketch how I understood it. There is a covariant functor $\underline{\mathbb{Q}}$ from the orbit category that sends every object to $\mathbb{Q}$ and every morphism to the identity. Further any $G$-space can be viewed as a contravariant functor from the orbit category, sending $G/H$ to the fixed point set $X^H$. On morphisms this functor still remembers the action of the Weyl group. Using this we can define Bredon-homology: $H^G_*(X,\underline{\mathbb{Q}})=H^*(C^*(X;\mathbb{Q})\otimes_{Or(G)}\underline{\mathbb{Q}})$.

With this definition the tensor product eats up the $G/H$-argument, so I do not understand, what $H^G_*(X,\underline{\mathbb{Q}})(G/H)$ means.

Let us have a look at what $H^G_*(X,\underline{\mathbb{Q}})$ is for a free space and $G=\mathbb{Z}/p$. Then the orbit category has two elements $G/G$ and $G/1$.

Since the action is free, the contravariant functor $X$ sends $G/G$ to the empty set. Thus the tensor product over the orbit category above really takes into account only one object $G/1$. This object still has automorphisms, namely the entire group $G$. Thus the tensor product above simplifies to $H^G_*(X,\underline{\mathbb{Q}})=H^*(C^*(X(G/1);\mathbb{Q})\otimes_{G}\underline{\mathbb{Q}})=H^*(C^*(X/G))=H^*(X/G)$.

This means that if my interpretation of (2) is correct, then (2) means for free $\mathbb{Z}/p$-spaces, that their quotients have isomorphic rational homology. Then a counterexample is given by lens spaces. There is a list of free $G$-actions on $S^3$ depending on some numbers. The total space for all these actions is $S^3$, so the first assumption holds. Second the homology of their quotients is actually independent of the chosen numbers, so the second assumption holds. However, for some choices of these numbers their quotients are not homotopy equivalent, so the total spaces could not be $G$-homotopy equivalent.

This should answer the questions 1 and 2 negatively.

What are the diffenrences between the assumptions in this question and the assumptions of Elmendorffs theorem? Philosophically I think the main difference is that in this question the isomorphism might be an arbitrary isomorphism of the homology groups, while in Elmendorffs theorem they have to be induced by maps of spaces.