Change of groups for naive G-spectra

Question

Let $H$ be a subgroup of $G$ a compact Lie group and let $\text{spectra}[G]$ be the category of naive $G$-spectra (ie G-objects in the category of spectra). Then there is a forgetful functor $i^*$ from $\text{spectra}[G]$ to $\text{spectra}[H]$ with a left adjoint $G_+ \wedge_H -$.

What can be said about the adjunction $(G_+ \wedge_H - , i^*)$ on homotopy categories?

For example:

1. I believe the forgetful functor $i^*$ to be faithful, does anyone have a reference?
2. Is the left adjoint also faithful?
3. I've been told this adjunction (probably) satisfies some form of homotopy descent. What does this tell me? (Presumably that some spectral sequence converges.)
4. Is every $O(n)$-spectrum in the image of the left adjoint (on homotopy categories)?

(The specific case I'm interested in is $H$ is the symmetric group on $n$ letters and $G$ is the orthogonal group $O(n)$.)

Answer 1

The following will assume that you are using the underlying weak equivalence structure on $G$-spectra and $H$-spectra, so that an equivariant map $X \to Y$ is an equivalence if and only if it is so after forgetting the action. Some of what I will say below can be altered if you instead use maps which are weak equivalences on fixed points, but it makes things harder.

1. The functor $i^*$ is not faithful. Take the inclusion $\{e\} \to C_2$ from the symmetric group on one letter to the 1-by-1 orthogonal group, giving both the Eilenberg-Mac Lane spectrum $H\Bbb F_2$ and the sphere spectrum $\Bbb S$ the trivial $C_2$-action. We can calculate and find that $$ [\Bbb S, \Sigma H\Bbb F_2]_{C_2} \cong H^1(C_2, \Bbb F_2) \cong \Bbb F_2 $$ but $$ [\Bbb S, \Sigma H\Bbb F_2]_{\{e\}} \cong H^1(\{e\}, \Bbb F_2) = 0. $$ Therefore, the restriction cannot be faithful.

2. The left adjoint is also not faithful. For this we can take the inclusion $\Sigma_3 \to O(3)$ from the symmetric group on 3 letters to the 3-by-3 orthogonal group. Take the two $\Sigma_3$-spectra given by $\Bbb S[\Sigma_3] = \Sigma^\infty_+(\Sigma_3)$ and $M = H\Bbb Z^3$, where $\Bbb Z^3$ is given the permutation action of $\Sigma_3$. We find that the left adjoint sends these to $\Sigma^\infty_+(O(3))$ and $\Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} M$ respectively. We can calculate and find that $$ [\Bbb S[\Sigma_3], M]_{\Sigma_3} \cong [\Bbb S, H\Bbb Z^3] \cong \Bbb Z^3 $$ by the adjunction, but $$ [\Bbb S[O(3)], \Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} M]_{O(3)} \cong [\Bbb S, \Sigma^\infty_+(O(3)) \wedge_{\Sigma_3} H\Bbb Z] $$ is a proper quotient of $\Bbb Z^3$, roughly because smashing with $O(3)$ over $\Sigma_3$ explicitly adds a path between $(x,y,z)$ and $\sigma x = (y,z,x)$ for any $(x,y,z) \in \pi_0 M$ and $\sigma$ generating the alternating subgroup.

3. The adjunction does satisfy a form of descent. If $H$ is normal in $G$, then this takes the form of a homotopy fixed-point spectral sequence (analogous to the Lyndon-Hochschild-Serre spectral sequence) with $E_2$-term $$ H^s(G/H; [\Sigma^t i^* M,i^* N]_H) \Rightarrow [\Sigma^{t-s} M,N]_G. $$ However, you've listed groups that certainly don't satisfy normality. We do know that there always exists some descent-type spectral sequence for taking a group $G$ with a subgroup $H$ and calculating the group cohomology of $G$; however, the terms in this spectral sequence are built out of some kind of tangled web involving the group cohomologies of finite intersections of conjugates of $H$. In less obtuse terms, there's a classifying space $E{\cal F}$ for the family of subgroups of $G$ which are conjugate to a subgroup of $H$, and $E{\cal F}$ is weakly equivalent to a point. Therefore, we get an isomorphism $$ [X,Y]_G \xrightarrow{\sim} [E{\cal F}_+ \wedge X, Y]_G, $$ and the cellular filtration on $E{\cal F}$ gives us a spectral sequence of mixed practicality for calculating the left-hand side.

4. No, this is not true. For example, again when $n=2$ every $O(2)$-spectrum in the image of this map is obtained by extending a $C_2$-action freely to a little bit more than a circle action, and the image of such a spectrum always has Euler characteristic $0$ when defined. The spectrum $\Bbb S$ with the trivial $O(2)$-action cannot be in the image.