TITLE: Change of groups for naive G-spectra
QUESTION [6 upvotes]: 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:

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

REPLY [6 votes]: 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.

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.
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.
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.
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.
{"set_name": "stack_exchange", "score": 6, "question_id": 221462}