Let $N_H^G$ be the norm functor from orthogonal $H$-spectra to orthogonal $G$-spectra. We know the category of orthogonal $G$-spectra $\mathcal{S}_G$ is enriched over the category of based $G$-spaces $\mathcal{T}_G$ (working with compactly generated weak Hausdorff spaces). We also know that $\mathcal{S}_G$ is tensored over $\mathcal{T}_G$ using the level-wise smash product $$(X\wedge E)(V)=X\wedge E(V),$$ where $X$ is a $G$-space and $E$ is a $G$-spectrum, and $\mathcal{S}_G$ is also cotensored over $\mathcal{T}_G$ using the internal Hom space $F(-,-)$ again applied level-wise $$(F(X,E))(V)=F(X,E(V)).$$ On page 140, remark A.55 of Hill, Hopkins and Ravenel (Kervaire Invariant One v4) it is stated that because $N_H^G$ is strong symmetric monoidal, then it extends to a functor of enriched categories, which implies that we have a map $$\phi:N_H^G(\mathcal{S}_H(E,E'))\longrightarrow \mathcal{S}_G(N_H^GE,N_H^GE').$$ By specialising to the case where the $E$ above is a suspension spectrum, using the shift desuspension-evaluation (or maybe generalised suspension-loop) adjunction $$\mathcal{S}_G(\Sigma^{\infty+V}X,E)\cong\mathcal{T}_G(X,E(V))$$ and the interaction of $N_H^G$ with free spectra $\Sigma^{\infty+V}X$, I would like to make a statement about this map $\phi$ is the case below. $$\Phi:N_H^G(F(X,E))\longrightarrow F(N_H^G X, N_H^G E)$$ Specifically I would love to know when this map is a weak (stable) equivalence of orthogonal $G$-spectra. I have a feeling it will be when $E$ is cofibrant and maybe even $X$ is also cofibrant. I cannot seem to tick the right boxes and piece it all together. I appreciate anyone's insight.
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$\begingroup$ I suspect that you might need X to be dualizable for this to hold. The right-hand side takes filtered colimits in X to limits, whereas on the left-hand side I don't see how a limit would pass outside the norm in general. $\endgroup$– Tyler LawsonSep 29, 2015 at 16:02
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$\begingroup$ Thank you for the comment @Tyler. In my case $X$ would be a representation sphere, so maybe there is hope yet! $\endgroup$– Jack DaviesSep 30, 2015 at 9:30
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