$E^G_\ast(E)$ tensored with the rationals

Question

Lemma 17.19 of Switzer's "Algebraic topology - Homology and Homotopy" states that $E_\ast(F)\otimes\mathbb{Q}$ is isomorphic to $\pi_\ast(E)\otimes\pi_\ast(F)\otimes\mathbb{Q}$. I wanted to know whether the result holds for the equivariant case, where the underlying group $G$ is a finite abelian group over integer grading. Explicitly, for $G$-spectra $E_G$ and $F_G$, is the natural transformation between the homology functors

$\pi_\ast^G(E_G)\otimes\pi_\ast^G(-)\otimes\mathbb{Q}\longrightarrow\pi_\ast^G(E_G\wedge -)\otimes\mathbb{Q}=E^G_\ast(-)\otimes\mathbb{Q}$

an isomorphism over integer gradings? Equivalently, is

$\pi_\ast^G(E_G)\otimes\pi_\ast^G(F_G)\otimes\mathbb{Q}\longrightarrow\pi_\ast^G(E_G\wedge F_G)\otimes\mathbb{Q}=E^G_\ast(F_G)\otimes\mathbb{Q}$

an isomorphism?

Specifically, let $KU_G$ denote the $G$-spectrum representing equivariant complex $K$-theory. Looking at the homology functors, for $E_G=KU_G$ and $F_G=\mathbb{S}_G$, the sphere spectrum, we get

$\pi_\ast^G(KU_G)\otimes\pi_\ast^G(\mathbb{S}_G)\otimes\mathbb{Q}\cong R(G)[t,t^{-1}]\otimes A(G)\otimes\mathbb{Q}$

as $\pi_0^G(\mathbb{S}_G)\cong A(G)$, the Burnside ring and $\pi_\ast^G(\mathbb{S}_G)$ is torsion in non-zero gradings, and

$K^G_\ast(\mathbb{S}_G)\otimes\mathbb{Q}\cong R(G)[t,t^{-1}]\otimes\mathbb{Q}$. Does the result hold for equivariant $K$-theory?

Answer 1

For any finite abelian $G$ and $H\leq G$ we have a geometric fixed-point functor $\phi^H\colon\text{Sp}_G\to\text{Sp}$ which preserves smash products and sends the equivariant sphere $S^0_G$ to $S^0$. By combining these for all $H$, we get a functor $\Phi\colon\text{Sp}_G\to\prod_{H\leq G}\text{Sp}$. It is standard that $\Phi$ induces an equivalence of the rational subcategories. In particular, for $X\in\mathbb{Q}\text{Sp}_G$ we have $$ \pi_n^G(X) = \text{Sp}_G(S^n_G,X) = \prod_H\text{Sp}(S^0,\phi^H(X)) = \prod_H \pi_n(\phi^H(X)). $$ It follows that $\pi^G_*(X)\otimes\pi^G_*(Y)$ contains terms $\pi_*(\phi^H(X))\otimes\pi_*(\phi^K(Y))$ with $H\neq K$, which do not appear in $\pi^G_*(X\wedge Y)$, so we do not usually have $\pi^G_*(X\wedge Y)\simeq\pi^G_*(X)\otimes\pi^G_*(Y)$. However, in this context you should think of $\pi_*(\phi^H(X))$ as a more natural invariant that $\pi^G_*(X)$ anyway.

In the case of $K$-theory, when $H$ is cyclic of order $n$ the ring $\pi_*(\phi^H(KU))\otimes\mathbb{Q}$ is $\mathbb{Q}(\mu_n)[t,t^{-1}]$, where $\mathbb{Q}(\mu_n)$ is the cyclotomic field. If $H$ is not cyclic then $\phi^H(KU)$ is rationally trivial.

Analogous statements are also true for finite nonabelian groups of equivariance, although they take a little more work to formulate.