The starting point of my question is the following fact: suppose $G$ is a finite group and let $H,K \leq G$ be arbitrary subgroups, then there exists an isomorphism of $G$-sets as follows
\begin{equation}
G/H \times G/K \cong \coprod_{H \backslash G/K}G/(H^g\cap K).
\end{equation}
The proof is not difficult: given an element $(xH,yK)$ we see it has stabilizer $xHx^{-1}\cap yKy^{-1}$ which is isomorphic to $H^{x^{-1}y}\cap K$. Then we map the orbit of $(xH,yK)$ to the coset $G/(H^{x^{-1}y}\cap K)$ indexed at the element $Hx^{-1}yK$ of the double coset. It is basic group calculus to verify that the function provided in this way is bijective and $G$-equivariant.
This fact has the following consequence in equivariant stable homotopy theory: it allows us to show that, for a fixed family $\mathcal{F}$ of subgroups of $G$, then $\text{Thick}(G/H_+ : H \in \mathcal{F})$, the thick subcategory of compact $G$-spectra generated by the cosets $G/H_+$ for all $H \in \mathcal{F}$, is a tensor ideal. In fact, knowing that the thick subcategory of compact $G$-spectra is generated by $G/K_+$ for the various $K \leq G$, using the above formula we see the smash product $G/H_+\wedge G/K_+$ decomposes as a wedge of cosets $G/L_+$ where $L \in \mathcal{F}$.
I wanted to prove the same claim in the case $G$ is a compact Lie group and the subgroups are assumed to be closed. In this situation we could define a map as above, going to the coproduct of the orbits, but we have no guarantee it is continuous. Indeed, I believe that the topology of the double coset $H\backslash G/K$ should be taken into account to produce a homeomorphic $G$-space.
It is know that the double coset admits a decomposition as follows: first we consider $G/K$ with the obvious left $H$-action, the stabilizer of $gK$ is given by $H_{gK}=H\cap gKg^{-1}$. We say that two orbits have the same type if the stabilizers of their points are conjugate in $H$. Also, we say that an orbit $H\backslash L_1$ has type lesser than or equal to $H \backslash L_2$ if $L_2$ is conjugate to a subgroup of $L_1$. Given that, it has been proved that the double coset $H\backslash G/K$ can be decomposed as a disjoint union of manifolds of possibly various dimensions. Each of these manifolds is given by double cosets in the same orbit type and they could be not connected. Moreover, the closure of an orbit type manifold is given by the manifolds with lesser or equal orbit type. Nevertheless, the number of types and the connected components of each manifold are finite.
This is described in the paper "The Transfer and Compact Lie Groups" of Feshbach, where he gives a reference to Bredon's "Introduction to compact transformation groups" for a proof of the finiteness of the connected components and the orbit types.
I tried to to use the manifold decomposition of the double coset to produce a suitable $G$-space and concoct an homeomorphism with the product to cosets, but without results. I am not able to present an explicit formula as in the finite case.
Then I fell back into just showing that the product $G/H\times G/K$ admits a structure of finite $G$-CW complex: since the stabilizers are again in the form $G/L_+$ with $L$ subconjugated to $H$ and $K$ the induced suspension spectrum must belong to the thick subcategory generated by the various $G/L_+$. However, I did not manage to find an appropriate result in the literature stating that the product of two $G$-CW complexes is still a $G$-CW complex in the compact case.
Indeed, when we deal with the product of cells and try to show that a generic product $(G/H \times D^{n_1}) \times (G/K \times D^{n_2})$ admits a decomposition in equivariant cells in the end we have to treat the product of cosets $G/H\times G/K$, so we go back to the previous problem.
To conclude, I have two questions:
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Can you provide a complete reference for a result showing that the product of two $G$-CW complexes is still a $G$-CW complex in the general case where $G$ is a compact Lie group?
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Does there exist some sort of known analogue of the double coset decomposition formula in the finite case?
Best Answer
Your question (1) may not be true as stated. One approach that I like is to think of the product of two $G$-CW complexes first as a $(G\times G)$-CW complex, which works because $G/H\times G/K \approx (G\times G)/(H\times K)$. Then you can consider it as a $G$-space by restricting along the diagonal $G\to G\times G$. However, in "Restricting the transformation group in equivariant CW complexes," Sören Illman showed that the restriction of a $G$-CW complex to an $H$-space does not generally give an $H$-CW complex. He then gave a construction of an $H$-homotopy equivalent $H$-CW complex with nice properties.
Related, and related to your question (2), is Illman's paper "Existence and uniqueness of equivariant triangulations of smooth proper $G$-manifolds with some applications to equivariant Whitehead torsion." One consequence of this paper is that $G/H\times G/K$ has a $G$-triangulation, so a $G$-CW structure. The problem that arises when you look at a product of two $G$-CW complexes is that the triangulations of the products of orbits don't have to interact well with the original CW structures.